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Deeper Understanding, Faster Calc: SOA MFE and CAS Exam 3F Yufeng Guo July 14, 2009

Contents Introduction

ix

9 Parity and other option relationships 9.1 Put-call parity . . . . . . . . . . . . . . . . . . . 9.1.1 Option on stocks . . . . . . . . . . . . . . 9.1.2 Options on currencies . . . . . . . . . . . 9.1.3 Options on bonds . . . . . . . . . . . . . . 9.1.4 Generalized parity and exchange options . 9.1.5 Comparing options with respect to style, strike . . . . . . . . . . . . . . . . . . . .

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10 Binomial option pricing: I 10.1 One-period binomial model: simple examples 10.2 General one-period binomial model . . . . . . 10.2.1 Two or more binomial trees . . . . . . 10.2.2 Options on stock index . . . . . . . . 10.2.3 Options on currency . . . . . . . . . . 10.2.4 Options on futures contracts . . . . .

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35 35 36 49 64 67 71

11 Binomial option pricing: II 11.1 Understanding early exercise . . . . . . . . . . . 11.2 Understanding risk-neutral probability . . . . . . 11.2.1 Pricing an option using real probabilities 11.2.2 Binomial tree and lognormality . . . . . . 11.2.3 Estimate stock volatility . . . . . . . . . . 11.3 Stocks paying discrete dividends . . . . . . . . . 11.3.1 Problems with discrete dividend tree . . . 11.3.2 Binomial tree using prepaid forward . . .

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79 79 80 81 88 91 95 96 98

12 Black-Scholes 12.1 Introduction to the Black-Scholes formula . . . . 12.1.1 Call and put option price . . . . . . . . . 12.1.2 When is the Black-Scholes formula valid? 12.2 Applying the formula to other assets . . . . . . .

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1 1 1 11 12 13 19

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CONTENTS

12.3

12.4

12.5

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12.2.1 Black-Scholes formula in terms of prepaid forward price 12.2.2 Options on stocks with discrete dividends . . . . . . . . 12.2.3 Options on currencies . . . . . . . . . . . . . . . . . . . 12.2.4 Options on futures . . . . . . . . . . . . . . . . . . . . . Option the Greeks . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.2 Gamma . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.3 Vega . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.4 Theta . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.5 Rho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.6 Psi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.7 Greek measures for a portfolio . . . . . . . . . . . . . . 12.3.8 Option elasticity and volatility . . . . . . . . . . . . . . 12.3.9 Option risk premium and Sharp ratio . . . . . . . . . . 12.3.10 Elasticity and risk premium of a portfolio . . . . . . . . Profit diagrams before maturity . . . . . . . . . . . . . . . . . . 12.4.1 Holding period profit . . . . . . . . . . . . . . . . . . . . 12.4.2 Calendar spread . . . . . . . . . . . . . . . . . . . . . . Implied volatility . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5.1 Calculate the implied volatility . . . . . . . . . . . . . . 12.5.2 Volatility skew . . . . . . . . . . . . . . . . . . . . . . . 12.5.3 Using implied volatility . . . . . . . . . . . . . . . . . . Perpetual American options . . . . . . . . . . . . . . . . . . . . 12.6.1 Perpetual calls and puts . . . . . . . . . . . . . . . . . . 12.6.2 Barrier present values . . . . . . . . . . . . . . . . . . .

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107 108 108 110 110 111 111 112 112 112 112 112 113 114 115 115 115 118 119 119 120 121 121 121 125

13 Market-making and delta-hedging 13.1 Delta hedging . . . . . . . . . . . . . . . . . . . . 13.2 Examples of Delta hedging . . . . . . . . . . . . 13.3 Textbook Table 13.2 . . . . . . . . . . . . . . . . 13.4 Textbook Table 13.3 . . . . . . . . . . . . . . . . 13.5 Mathematics of Delta hedging . . . . . . . . . . . 13.5.1 Delta-Gamma-Theta approximation . . . 13.5.2 Understanding the market maker’s profit

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129 129 129 138 140 141 141 142

14 Exotic options: I 14.1 Asian option (i.e. average options) 14.1.1 Characteristics . . . . . . . 14.1.2 Examples . . . . . . . . . . 14.1.3 Geometric average . . . . . 14.1.4 Payoff at maturity T . . . . 14.2 Barrier option . . . . . . . . . . . . 14.2.1 Knock-in option . . . . . . 14.2.2 Knock-out option . . . . . . 14.2.3 Rebate option . . . . . . . .

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145 145 145 146 146 147 147 147 147 148

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CONTENTS 14.2.4 Barrier parity . . 14.2.5 Examples . . . . 14.3 Compound option . . . 14.4 Gap option . . . . . . . 14.4.1 Definition . . . . 14.4.2 Pricing formula . 14.4.3 How to memorize 14.5 Exchange option . . . .

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148 148 149 151 151 151 151 152

18 Lognormal distribution 18.1 Normal distribution . . . . . . . . . . . . . . . . . . . 18.2 Lognormal distribution . . . . . . . . . . . . . . . . . . 18.3 Lognormal model of stock prices . . . . . . . . . . . . 18.4 Lognormal probability calculation . . . . . . . . . . . . 18.4.1 Lognormal confidence interval . . . . . . . . . . 18.4.2 Conditional expected prices . . . . . . . . . . . 18.4.3 Black-Scholes formula . . . . . . . . . . . . . . 18.5 Estimating the parameters of a lognormal distribution

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165 165 166 167

19 Monte Carlo valuation 19.1 Example 1 Estimate E (ez ) . . . . . . . . . . . . . . . . . . . 19.2 Example 2 Estimate π . . . . . . . . . . . . . . . . . . . . . . 19.3 Example 3 Estimate the price of European call or put options 19.4 Example 4 Arithmetic and geometric options . . . . . . . . . 19.5 Efficient Monte Carlo valuation . . . . . . . . . . . . . . . . . 19.5.1 Control variance method . . . . . . . . . . . . . . . . . 19.6 Antithetic variate method . . . . . . . . . . . . . . . . . . . . 19.7 Stratified sampling . . . . . . . . . . . . . . . . . . . . . . . . 19.7.1 Importance sampling . . . . . . . . . . . . . . . . . . . 19.8 Sample problems . . . . . . . . . . . . . . . . . . . . . . . . .

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173 173 177 180 184 193 193 197 199 199 199

18.6 How are asset prices distributed 18.6.1 Histogram . . . . . . . . 18.6.2 Normal probability plots 18.7 Sample problems . . . . . . . .

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20 Brownian motion and Ito’s Lemma 20.1 Introduction . . . . . . . . . . . . . . . . . 20.1.1 Big picture . . . . . . . . . . . . . 20.2 Brownian motion . . . . . . . . . . . . . . 20.2.1 Stochastic process . . . . . . . . . 20.2.2 Definition of Brownian motion . . 20.2.3 Martingale . . . . . . . . . . . . . 20.2.4 Properties of Brownian motion . . 20.2.5 Arithmetic Brownian motion and motion . . . . . . . . . . . . . . . .

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205 205 206 207 207 207 209 210 214

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CONTENTS 20.2.6 Ornstein-Uhlenbeck process . . . . . . . . . 20.3 Definition of the stochastic calculus . . . . . . . . . 20.4 Properties of the stochastic calculus . . . . . . . . 20.5 Ito’s lemma . . . . . . . . . . . . . . . . . . . . . . 20.5.1 Multiplication rules . . . . . . . . . . . . . 20.5.2 Ito’s lemma . . . . . . . . . . . . . . . . . . 20.6 Geometric Brownian motion revisited . . . . . . . 20.6.1 Relative importance of drift and noise term 20.6.2 Correlated Ito processes . . . . . . . . . . . 20.7 Sharpe ratio . . . . . . . . . . . . . . . . . . . . . . 20.8 Risk neutral process . . . . . . . . . . . . . . . . . 20.9 Valuing a claim on S a . . . . . . . . . . . . . . . . 20.9.1 Process followed by S a . £. . . . ¤. . . . . . . 20.9.2 Formula for S A (t) and E S A (t) . . . . . . 20.9.3 Expected return of a claim on S A (t) . . . . 20.9.4 Specific examples . . . . . . . . . . . . . . .

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215 216 222 223 223 223 225 225 226 230 233 233 233 234 235 235

21 Black-Scholes equation 21.1 Differential equations and valuation under certainty . 21.1.1 Valuation equation . . . . . . . . . . . . . . . . 21.1.2 Bonds . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Dividend paying stock . . . . . . . . . . . . . . 21.2 Black-Scholes equation . . . . . . . . . . . . . . . . . . 21.2.1 How to derive Black-Scholes equation . . . . . 21.2.2 Verifying the formula for a derivative . . . . . . 21.2.3 Black-Scholes equation and equilibrium returns 21.3 Risk-neutral pricing . . . . . . . . . . . . . . . . . . .

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245 245 245 246 246 246 246 247 250 252

22 Exotic options: II 253 22.1 All-or-nothing options . . . . . . . . . . . . . . . . . . . . . . . . 253 23 Volatility 24 Interest rate models 24.1 Market-making and bond pricing . . . . . . . . . . . . . 24.1.1 Review of duration and convexity . . . . . . . . . 24.1.2 Interest rate is not so simple . . . . . . . . . . . 24.1.3 Impossible bond pricing model . . . . . . . . . . 24.1.4 Equilibrium equation for bonds . . . . . . . . . . 24.1.5 Delta-Gamma approximation for bonds . . . . . 24.2 Equilibrium short-rate bond price models . . . . . . . . 24.2.1 Arithmetic Brownian motion (i.e. Merton model) 24.2.2 Rendleman-Bartter model . . . . . . . . . . . . . 24.2.3 Vasicek model . . . . . . . . . . . . . . . . . . . 24.2.4 CIR model . . . . . . . . . . . . . . . . . . . . . 24.3 Bond options, caps, and the Black model . . . . . . . .

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CONTENTS 24.3.1 Black formula . . . . . 24.3.2 Interest rate caplet . . 24.4 Binomial interest rate model 24.5 Black-Derman-Toy model . .

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Introduction This study guide is for SOA MFE and CAS Exam 3F. Before you start, make sure you have the following items: 1. Derivatives Markets, the 2nd edition. 2. Errata of Derivatives Markets. You can download the errata at http:// www.kellogg.northwestern.edu/faculty/mcdonald/htm/typos2e_01. html. 3. Download the syllabus from the SOA or CAS website. 4. Download the sample MFE problems and solutions from the SOA website. 5. Download the recent SOA MFE and CAS Exam 3 problems.

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Chapter 9

Parity and other option relationships 9.1

Put-call parity

9.1.1

Option on stocks

Notation t=0 T S0 ST K CEur (K, T ) CEur (K, 0) PEur (K, T ) PEur (K, 0) r δ F0,T

Current date (date when an option is sold or bought) Option expiration date (maturity date) Current price of the underlying asset The price of the underlying asset at the option expiration date Strike price or exercise price Premium of a European call option with strike price K and T years to expiration Premium of a European call option on the expiration date Premium of a European put option with strike price K and T years to expiration Premium of a European put option on the expiration date The continuously compounded annual risk-free interest rate The continuously compounded annual dividend rate Delivery price in a forward contract expiring in T

Put-call parity The textbook gives the following formula CEur (K, T ) − PEur (K, T ) = P V0,T (F0,T − K) = e−rT (F0,T − K)

(9.1)

The textbook explains the intuition behind Equation 9.1. If we set the forward price F0,T as the common strike price for both the call and the put 1

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

(i.e. K = F0,T ) , then CEur (K, T ) − PEur (K, T ) = P V0,T (F0,T − K) = 0. Buying a call and selling a put with K = F0,T synthetically creates a forward contract and the premium for a forward contract is zero. However, often I find that Equation 9.1 is not intuitive at all. In fact, it’s annoyingly complex and hard to memorize. So I like to rewrite 9.1 as follows: CEur (K, T ) + P V (K) = PEur (K, T ) + S0

(9.2)

Later I will explain the intuition behind 9.2. First, let’s prove 9.2. The proof is extremely important. Suppose at time zero we have two portfolios: • Portfolio #1 consists of a European call option on a stock and P V (K) = Ke−rT . P V (K) is the present value of the strike price K and r is the continuously compounded risk free interest rate per year. • Portfolio #2 consists of a European put option on the stock and one share of the stock with current price S0 . • Both the call and put have the same underlying stock, the same strike price K, and the same expiration date T . Notice that at time zero Portfolio #1 is worthy CEur (K, T ) + P V (K); Portfolio #2 is worth PEur (K, T ) + S0 . Since it’s difficult to compare the value of Portfolio #1 and the value of Portfolio #2 at time zero, let’s compare them at the expiration date T . We’ll soon see that the two portfolios have the same value at T . Payoff of Portfolio #1 at the expiration date T If ST < K If ST ≥ K Call payoff is max (0, ST − K) 0 ST − K Payoff of P V (K) K K Total K ST If you have P V (K) at t = 0, you’ll have K at T . Payoff of Portfolio #2 at the expiration date T If ST < K If ST ≥ K Put payoff max (0, K − ST ) K − ST 0 Payoff of S0 ST ST Total K ST If you have one stock worth S0 at t = 0, you’ll have one stock at T worth ST . You see that Portfolio #1 and Portfolio #2 have identical payoffs of max (K, ST ) at T . If ST < K, both portfolios are worth the strike price K; if ST ≥ K, both

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9.1. PUT-CALL PARITY

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portfolios are worth the stock price ST . Since Portfolio #1 and #2 are worth the same at T , to avoid arbitrage, they must be worth the same at any time prior to T . Otherwise, anyone can make free money by buying the lower priced portfolio and selling the higher priced one. So Portfolio #1 and #2 are worth the same at time zero. Equation 9.2 holds. I recommend that from this point now you throw Equation 9.1 away and use Equation 9.2 instead. How to memorize Equation 9.2: Tip 9.1.1. Many candidates have trouble memorizing Equation 9.2. For example, it’s very easy to write a wrong formula CEur (K, T ) + S0 = PEur (K, T ) + P V (K). To memorize Equation 9.2, notice that for a call to work, a call must go hand in hand with the strike price K. When exercising a call option, you give the call seller both the call certificate and the strike price K. In return, the call seller gives you one stock. Similarly, for a put to work, a put must go hand in hand with one stock. When exercising a put, you must give the put seller both the put certificate and one stock. In return, the put seller gives you the strike price K. Tip 9.1.2. Another technique that helps me memorize Equation 9.2 is the phrase “Check (CK) Please (PS).” At expiration T , CEur (K, 0)+K = PEur (K, 0)+ ST . Discounting this equation to time zero, we get: CEur (K, T ) + P V (K) = PEur (K, T ) + S0 .

Example 9.1.1. The price of a 6-month 30-strike European call option is 12.22. The stock price is 35. The continuously compounded risk-free interest rate is 8% per year. What”s the price of a 6-month 30-strike European put option on the same stock? Solution. K = 30 T = 6/12 = 0.5 S0 = 35 r = 0.08 CEur = 12.22 CEur + P V (K) = PEur + S0 → 12.22 + 30e−0.08(0.5) = PEur + 35, PEur = 6. 043 Parity if the stock pays discrete dividends If the stock pays discrete dividend, the parity formula is CEur (K, T ) + P V (K) = PEur (K, T ) + S0 − P V (Div)

(9.3)

This is why we need to subtract the term P V (Div). At expiration, CEur (K, T )+ K = PEur (K, T ) + ST . If we discount ST from T to time zero, we’ll get

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

S0 − P V (Div). If you have one stock worth S0 at time zero, then during [0, T ], you’ll receive dividend payments. Then at T , you not only have one share of stock, you also have the accumulated value of the dividend. To get exactly one stock at T , you need to have S0 − P V (Div) at time zero. Discounting this equation back to time zero, we get Equation 9.3 Please note that Equation 9.3 assumes that both the timing and the amount of each dividend are 100% known in advance.

Example 9.1.2. The price of a 9-month 95-strike European call option is 19.24. The stock price is 100. The stock pays dividend of $1 in 3 months and $2 in 6 months. The continuously compounded risk-free interest rate is 10% per year. What”s the price of a 9-month 95-strike European put option on the same stock? Solution.

T = 9/12 = 0.75 K = 95 S0 = 100 CEur = 19.24 P V (Div) = e−0.1(3/12) + 2e−0.1(6/12) = 2. 877 8

r = 0.1

CEur (K, T ) + P V (K) = PEur (K, T ) + S0 − P V (Div) 19.24 + 95e−0.1(0.75) = PEur (K, T ) + 100 − 2. 877 8, PEur (K, T ) = 10. 25

Example 9.1.3. The price of a 9-month 83-strike European put option is 13.78. The stock price is 75. The stock pays dividend of $1 in 3 months, $2 in 6 months, $3 in 9 months, and $4 in 12 months. The continuously compounded risk-free interest rate is 6% per year. What”s the price of a 9-month 95-strike European call option on the same stock? Solution.

T = 9/12 = 0.75 K = 83 S0 = 75 r = 0.06 PEur = 13.78 P V (Div) = e−0.06(3/12) + 2e−0.06(6/12) + 3e−0.06(9/12) = 5. 794 CEur (K, T ) + P V (K) = PEur (K, T ) + S0 − P V (Div) CEur (K, T ) + 83e−0.06(0.75) = 13.78 + 75 − 5. 794 CEur (K, T ) = 3. 64 Tip 9.1.3. When calculating P V (Div) in Equation 9.3, discard any dividend paid after the option expiration date. In this example, the $4 is paid in 12 months, which is after the expiration date of the option. This dividend is ignored when we use Equation 9.3.

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9.1. PUT-CALL PARITY

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Parity if the stock pays continuous dividends If the stock pays dividends at a continuously compounded rate of δ per year, the parity formula is: CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT

(9.4)

At expiration, CEur (K, T ) + K = PEur (K, T ) + ST . If we discount ST from T to time zero, we’ll get S0 e−δT . If you have e−δT share of a stock, by investing dividends and buying additional stock, you’ll have exactly one stock at T . This concept is called tailing. Refer to Derivatives Markets Section 5.2 about tailing. Example 9.1.4. The price of a 6-month 90-strike European put option is 5.54. The stock price is 110. The stock pays dividend at a continuously compounded rate of 2% per year. The continuously compounded risk-free interest rate is 6% per year. What”s the price of a 6-month 90-strike European call option on the same stock? Solution. T = 6/12 = 0.5 K = 90 S0 = 110 r = 0.06 δ = 0.02 PEur = 5.54 CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT CEur (K, T ) + 90e−0.06(0.5) = 5.54 + 110e−0.02(0.5) CEur (K, T ) = 27. 11 Example 9.1.5. The price of a 3-month 40-strike European call option is 6.57. The stock price is 44. The stock pays dividend at a continuously compounded rate of 5% per year. The continuously compounded risk-free interest rate is 8% per year. What”s the price of a 3-month 40-strike European put option on the same stock? Solution. T = 3/12 = 0.25 K = 40 S0 = 44 r = 0.08 δ = 0.08 CEur = 6.57 CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT 6.57 + 40e−0.08(0.25) = PEur (K, T ) + 44e−0.05(0.25) PEur (K, T ) = 2. 32

Synthetic stock Rearranging Equation 9.3, we get: S0 = CEur (K, T ) − PEur (K, T ) + P V (K) + P V (Div) To understand the meaning of Equation 9.5, notice

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(9.5)

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Symbol +CEur (K, T ) −PEur (K, T ) +P V (K) +P V (Div)

Meaning at t = 0 buy a K-strike call expiring in T sell a K-strike put expiring in T buy a zero-coupon bond that pays K at T buy a zero-coupon bond that pays Div at T

Next, notice CEur (K, T ) − PEur (K, T ) + P V (K) is worth ST . At T , you always receive K from the zero-coupon bond seller. However, the call and put values depend on whether ST ≥ K. • If ST ≥ K, the sold put expires worthless; you exercise the call, paying K and receiving one stock. • If on the other hand, ST ≤ K, the purchased call expires worthless and the sold put is exercised against you. You pay the put holder K and receive one stock from him. • Either way, if at time zero you buy a call, sell a put, and invest P V (K) in a zero-coupon bond, then at T you are guaranteed to have one stock. Once you understand CEur (K, T )−PEur (K, T )+P V (K) is worth one stock at T , the meaning of Equation 9.5 is obvious: if you buy one stock at time zero, then at time T , you’ll have one stock worth ST . In addition, you’ll receive the future value of the dividends due to owning a stock. Example 9.1.6. The price of a 9-month 52-strike European call option on a non dividend-paying stock is 33.4420. The price of a 9-month 52-strike European put option on the same stock is 15.1538. The continuously compounded risk-free interest rate is 6% per year. How can you synthetically create one stock at time zero? What’s the price of this synthetically created stock at time zero? Solution. The parity for a non-dividend paying stock is CEur (K, T ) + P V (K) = PEur (K, T ) + S0 Rearranging this equation, we get: S0 = CEur (K, T ) + P V (K) − PEur (K, T ) To synthetically create the ownership of one stock, you need to do the following at time zero: • Buy a 9-month 52-strike European call option • Sell a 9-month 52-strike European put option • Invest 52e−0.06(0.75) = 49. 711 9 in an account earning risk-free interest rate (i.e. buying a zero coupon bond that pays 52 in 9 months) The current price of this synthetically created stock is: S0 = 33.4420 + 49. 711 9 − 15.1538 = 68

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Synthetic T-bill Rearranging Equation 9.3, we get: P V (Div) + P V (K) = S0 + PEur (K, T ) − CEur (K, T )

(9.6)

According to Equation 9.6, buying one stock, buying a K-strike put, and selling a K-strike call synthetically creates a zero coupon bond with a present value equal to P V (Div) + P V (K). Creating synthetic T-bill by buying the stock, buying a put, and selling a call is called a conversion. If we short the stock, buy a call, and sell a put, we create a short position on T-bill. This is called a reverse conversion. Example 9.1.7. Your mother-in-law desperately wants to borrow $1000 from you for one year. She’s willing to pay you 50% interest rate for using your money for one year. You really want to take her offer and earn 50% interest. However, state anti-usury laws prohibits any lender from charging an interest rate equal to or greater than 50%. Since you happen to know the put-call parity, you decide to synthetically create a loan and circumvent the state anti-usury law. Explain how you can synthetically create a loan and earn 50% interest. You want to lend $1000 at time zero and receive $1000 (1.5) = 1500 at T = 1. To achieve this, at time zero you can • have your mother-in-law sell you an asset that’s worth $1000 • have your mother-in-law sell you a 1500-strike, 1-year to expiration put option on the asset • sell your mother-in-law a 1500-strike, 1-year to expiration call option on the asset. Let’s see what’s happens at T = 1. • If ST ≥ 1500, you exercise the put and sell the asset to your mother-in-law for 1500 • if ST ≤ 1500, your mother-in-law exercises the call and buys the asset from you for 1500. The net effect is that you give your mother-in-law $1000 at time zero and receive $1500 from her at T = 1. In this example, you have a long position on the synthetically created Tbill. This is an example of conversion. In contrast, your mother-in-law has a short position in the synthetically created T-bill. This is an example of reverse conversion.

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

Synthetic call option Rearranging Equation 9.3, we get: CEur (K, T ) = PEur (K, T ) + S0 − P V (Div) − P V (K) Example 9.1.8. The price of a 6-month 75-strike European put option on a dividend-paying stock is 8.06. The stock price is 80. The continuously compounded risk-free interest rate is 5% per year. The continuously compounded dividend rate is 3% per year. Explain how you can create a synthetic 6-month 75-strike European call option on the stock. Calculate the premium for such a synthetic call option. Solution.

The put-call parity for a stock paying continuous dividend is CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT Rearranging this equation, we get: CEur (K, T ) = PEur (K, T ) + S0 e−δT − P V (K) = 8.06 + 80e−0.03(0.5) − 75e−0.05(0.5) = 13. 72 This is how to synthetically create a 6-month 75-strike European call symbol at t = 0 +PEur (K, T ) buy a 6-month 75-strike European put +S0 e−δT buy e−0.03(0.5) = 0.985 share of stock −P V (K) sell a bond that pays 75 in 6 months Let’s see why a call is synthetically created. 0.985 share of stock at t = 0 will grow into one stock if you reinvest the dividend and buy additional share of stock. If at T = 0.5 the stock price is greater than 75 (i.e. ST > 75), then two things happen: your purchased put expires worthless; the bond matures and you need to pay the bond holder K = 75. So the net effect is that if ST > 75 then at T you pay K = 75 and own one stock. This is the same as if ST > 75 you exercise the call, paying K = 75 and receiving one stock. If ST ≤ 75, then two things will happen. You exercise the put, selling your stock for K = 75; the bond matures and you pay the bond holder K = 75. So the net effect is that if ST ≤ 75 you net cash flow is zero and you don’t own a stock. This is the same as if ST ≤ 75 you do nothing and let your call expire worthless. Example 9.1.9. The price of a 9-month 110-strike European put option on a dividend-paying stock is 42.81. The stock price is 100. The continuously compounded risk-free interest rate is 8% per year. The stock will pay $1 dividend in 3 months and $1 dividend in 6 months. Explain how you can create a synthetic 9-month 110-strike European call option on the stock. Calculate the premium for such a synthetic call option.

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CEur (K, T ) = PEur (K, T ) + S0 − P V (Div) − P V (K) = 42.81 + 100 − e−0.08(0.25) − e−0.08(0.5) − 110e−0.08(0.75) = 37. 27 How to synthetically create a 9-month 110-strike European call symbol at t = 0 +PEur (K, T ) buy a 9-month 110-strike European put +S0 buy one share of stock paying 100 −P V (Div) sell Bond #1 that pays $1 in 3 months and $1 in 6 months −P V (K) sell Bond #2 that pays $110 in 6 months Let’s see why a synthetic call is created. If at expiration date ST > 110 • You’ll receive $1 dividend at the end of Month 3 and $1 dividend at the end of Month 6. You use the dividends to pay off Bond #1 • Your purchased put expires worthless • You have one stock at T = 0.75 • Your pay K = 110 at T = 0.75 The net effect is that you’ll pay K = 110 at T = 0.75 and own one stock. This is the same as owning a call option and ST > 110. If at expiration date ST ≤ 110 • You’ll receive $1 dividend at the end of Month 3 and $1 dividend at the end of Month 6. You use the dividends to pay off Bond #1 • You have one stock at T = 0.75 • You exercise your put, surrendering one stock and receiving K = 110 • Your pay K = 110 at T = 0.75 The net effect is that you have zero cash left and don’t own one stock. This is the same as owning a call option and ST ≤ 110. Synthetic put option Rearranging Equation 9.3, we get: PEur (K, T ) = CEur (K, T ) − S0 + P V (Div) + P V (K) If the stock pays continuous dividend, then PEur (K, T ) = CEur (K, T ) + P V (K) − S0 e−δT

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

Example 9.1.10. The price of a 6-month 45-strike European call option on a dividend-paying stock is 18.62. The stock price is 50. The continuously compounded risk-free interest rate is 6% per year. The stock will pay $1 dividend in 3 months. Explain how you can create a 6-month 45-strike European put option on the stock. Calculate the premium for such a synthetic put option. Solution. PEur (K, T ) = CEur (K, T ) − S0 + P V (Div) + P V (K) = 18.62 − 50 + e−0.06(0.25) + 45e−0.06(0.5) = 13. 28 How to create a synthetic 6-month 45-strike European put option symbol +CEur (K, T ) −S0 +P V (Div) +P V (K)

at t = 0 buy a 6-month 45-strike European call short sell one share receiving 50 and invest in a savings account buy Bond #1 that pays $1 in 3 months sell Bond #2 that pays $45 in 6 months

This is why CEur (K, T ) − S0 + P V (Div) + P V (K) behaves like a put. If ST ≥ 45 • (1) You will receive K = 45 at T = 0.5 from Bond #2 • (2) You exercise the call at T = 0.5, paying K = 45 which you get from (1) and receiving one stock • (3) at T = 0.5 you give the stock you get from (2) to the broker from whom you borrowed the stock for short sale • (4) Bond #1 pays you $1 dividend at the end of Month 3. After receiving this dividend, you Immediately pay this dividend to the original owner of the stock you sold short (2)+(3) will close out your short position on the stock The net effect is that if ST ≥ 45 you keep the proceeds from the short sale, which you can use to buy a stock. This is "keeping your asset (i.e. proceeds from the short sale)." This is the same as if you own a 6-month 45-strike European put and ST ≥ 45. If you own a 6-month 45-strike European put and ST ≥ 45, you let the put option expire worthless and you still own a stock. If ST < 45 • (1) You will receive K = 45 at T = 0.5 from Bond #2 • (2) You let the call expire worthless • (3) At T = 0.5 you buy a stock from the open market using the proceeds from the short sale; you give the stock to the broker from whom you borrowed the stock for short sale

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9.1. PUT-CALL PARITY

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• (4) Bond #1 pays you $1 dividend at the end of Month 3. After receiving this dividend, you Immediately pay this dividend to the original owner of the stock you sold short (3)+(4) will close out your short position on the stock. The net effect is that if ST ≥ 45 you receive K = 45 and you spent the proceeds from the short sale. This is "giving up an asset (proceeds from the short sale) and getting the strike price K."

9.1.2

Options on currencies

The put-call parity when currencies are underlying is CEur (K, T ) + P V (K) = x0 e−r€ + PEur (K, T )

(9.7)

In Equation 9.7, the underlying asset is 1 euro. The call holder has the right, at T , to buy the underlying (i.e. 1 euro) by paying a fixed dollar amount K. The premium of this call option is CEur (K, T ) dollars. Similarly, the put holder has the right, at T , to sell the underlying (i.e. 1 euro) for a fixed dollar amount K. The premium of this put option is PEur (K, T ) dollars. x0 is the price, in dollars, of buying the underlying (i.e. 1 euro) at time zero. r€ is the continuously compounded euro interest rate per year earned by the underlying (i.e. 1 euro). To understand the term x0 e−r€ in 9.7, notice that to have 1€ at T , you need to have e−r€ € at time zero. Since the cost of buying 1€ at time zero is x0 dollars, the cost of buying e−r€ € at time zero is x0 e−r€ dollars. Equation 9.7 is very similar to Equation 9.4. If you set S0 = x0 and δ = r€ , Equation 9.4 becomes 9.7. This shouldn’t surprise us. Both S0 and x0 refer to the price of an underlying asset at time zero. S0 is the dollar price of a stock; x0 is the dollar price of 1 euro. Both δ and r€ measure the continuous rate of reward earned by an underlying asset. Tip 9.1.4. How to memorize Equation 9.7. Just memorize Equation 9.4. Next, set S0 = x0 and δ = r€ . Tip 9.1.5. When applying Equation 9.7, remember that K, CEur (K, T ), PEur (K, T ), and x0 are in U.S. dollars. To remember this, assume that you are living in the U.S. (i.e. US dollar is your home currency); that your goal is to either buy or sell the underlying asset (i.e.1€) with a fixed dollar amount. Also remember that r€ is the euro interest rate on euro money. Example 9.1.11. The current exchange rate is 1€= 1.33 US dollars. The dollar-denominated 6-month to expiration $1.2-strike European call option on one euro has a premium $0.1736. The continuously compounded risk-free interest rate on dollars is 6% per year. The continuously compounded risk-free

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

interest rate on euros is 4% per year. Calculate the premium for the dollardenominated 6-month to expiration $1.2-strike European put option on one euro. Solution. CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT K = 1.2 T = 0.5 CEur (K, T ) = $0.1736 r = 0.06 δ = 0.04 S0 = 1.33 0.1736 + 1.2e−0.06(0.5) = PEur (K, T ) + 1.33e−0.04(0.5) PEur (K, T ) = 0.03 447 Example 9.1.12. The current exchange rate is $1 = 0.78€. The dollar-denominated 9-month to expiration $1-strike European put option on one euro has a premium $0.0733. The continuously compounded risk-free interest rate on dollars is 7% per year. The continuously compounded risk-free interest rate on euros is 3% per year. Calculate the premium for the dollar-denominated 9-month to expiration $1-strike European call option on one euro. CEur (K, T ) + P V (K) = PEur (K, T ) + S0 e−δT K=1 T = 0.5 PEur (K, T ) = 0.0733 1 r = 0.07 δ = 0.03 S0 = = 1. 282 05 0.78 1 −0.03(0.75) e CEur (K, T )+e−0.07(0.75) = 0.0733+ 0.78

9.1.3

CEur (K, T ) = $0.3780

Options on bonds

This is an option contract where the underlying asset is a bond. Other than having different underlying assets, a stock option and a bond option have no major differences. You buy a call option on stock if you think the stock price will go up; you buy a put option if you think the stock price will drop. Similarly, you buy a call option on a bond if you think that the market interest rate will go down (hence the price of a bond will go up); you buy a put option on a bond if you think that the market interest rate will go up (hence the price of the bond will go down). The coupon payments in a bond are like discrete dividends in a stock. Using Equation 9.3, we get the CEur (K, T ) + P V (K) = PEur (K, T ) + B0 − P V (Coupon)

(9.8)

Please note that in Equation 9.8, in the term P V (Coupon), "coupon" refers to the coupon payments during [0, T ]. In other words, only the coupons made during the life of the options are used in Equation 9.8. Coupons made after T are ignored.

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Example 9.1.13. A 10-year $1,000 par bond pays 8% annual coupons. The yield of the bond is equal to the continuously compounded risk-free rate of 6% per year. A 15-month $1,000-strike European call option on the bond has a premium $180. Calculate the premium for a 15-month $1,000-strike European put option on the bond. Solution. The annual effective interest rate is i = e0.06 − 1 = 6. 184% ¢ ¡ 1 − 1.06184−10 B0 = 1000 (0.08) a10| +1000v 10 = 1000 (0.08)× +1000 1.06184−10 = 0.06184 1132. 50 We need to be careful about calculating P V (Coupon). The bond matures in 10 years. There are 10 annual coupons made at t = 1, 2, ..., 10. However, since the option expires at T = 15/12 = 1. 25, only the coupon paid at t = 1 is used in Equation 9.8. P V (Coupon) = e−0.06 (1000) (0.08) = 75. 34 CEur (K, T ) + P V (K) = PEur (K, T ) + B0 − P V (Coupon) 180 + 1000e−0.06(1.25) = PEur (K, T ) + 1132. 50 − 75. 34 PEur (K, T ) = 50. 59

9.1.4

Generalized parity and exchange options

General definition of call and put Most times, the strike price is a constant and we can easily tell whether an option is a call or a put. For example, an option that gives you the privilege of buying one Google stock in 1 year for $35 is a call option; an option that gives you the privilege of selling one Google stock in 1 year for $35 is a put option. However, occasionally, the strike price is not a constant and it’s hard to determine whether an option is a call or put. For example, you purchase an option that gives you the privilege of receiving one Google stock by surrendering one Microsoft stock in 1 year. If at T = 1, the price of a Google stock is higher than that of a Micros0ft stock (i.e. STGoogle >.STMicrosof t ), it’s advantageous for you to exercise the option. To exercise the call, you buy one Microsoft stock from the open market for STM icrosof t , give it to the option writer. In return, the option writer gives you one Google stock worth STGoogle . Your payoff is STGoogle −.STMicrosof t . If, on the other hand, STGoogle ≤.STMicrosof t , you let your option expires worthless and your payoff is zero. Is this option a call or a put? It turns out that this option can be labeled as either a call or a put. If you view the Google stock as the underlying asset and the Microsoft stock as the strike asset, then it’s a call option. This option gives you the privilege of buying, at T = 1, one Google stock by paying one Microsoft stock. If you view

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

the Microsoft stock as the underlying asset and the Google stock as the strike asset, then it’s a put option. This option gives you the privilege of selling, at T = 1, one Microsoft stock for the price of one Google stock. Definition 9.1.1. An option gives the option holder the privilege, at T , of surrendering an asset AT and receiving an asset BT (we denote the option as AT → BT ). This is a call option if we view B as the underlying asset and A as the strike asset (the option holder has the privilege of buying BT by paying AT ). This is a put option if we view A as the underlying asset and B as the strike asset (the option holder has the privilege of selling AT and receiving BT ). Please note that the payoff of Option AT → BT is max (0, BT − AT ). Example 9.1.14. An option gives the option holder the privilege, at T = 0.25 (i.e. 3 months later), of buying €1 with $1.25. Explain why this option can be viewed (perhaps annoyingly) as either a call or a put.

Solution. This option is $1.25 →€1. The option holder has the privilege, at T = 0.25, of surrendering $1.25 and receiving €1 (i.e. give $1.25 and get €1). This is a call option if we view €1 as the underlying asset. The option holder has the privilege of buying €1 by paying $1.25. This is also a put option if we view $1.25 as the underlying asset. The option 1 holder has the privilege of selling $1.25 for €1 (i.e. selling $1 for € =€ 1.25 0.8). Example 9.1.15. An option gives the option holder the privilege, at T = 0.25, of buying one Microsoft stock for $35. Explain why this option can be viewed (perhaps annoyingly) as either a call or a put. This option is $35 → 1 M icrosof t stock (give $35 and get 1 Microsoft stock). If we view the Microsoft stock as the underlying asset, this is a call option. The option holder has the privilege of buying one Microsoft stock by paying $35. This is also a put option if we view $35 as the underlying. The option holder has the privilege of selling $35 for the price of one Microsoft stock. Example 9.1.16. An option gives the option holder the privilege, at T = 0.25, of selling one Microsoft stock for $35. Explain why this option can be viewed as either a call or a put. This option is 1 M icrosof t stock → $35 (give 1 Microsoft stock and get $35). If we view the Microsoft stock as the underlying asset, this is a put option. The option holder has the privilege of selling one Microsoft stock for $35. This is also a call option if we view $35 as the underlying. The option holder has the privilege of buying $35 by paying one Microsoft stock.

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Generalized put and call parity (AT → BT )0 + P V (AT ) = (BT → AT )0 + P V (BT )

(9.9)

In Equation 9.9, (AT → BT )0 is the premium paid at t = 0 for the privilege of giving AT and getting BT at T . Similarly, (BT → AT )0 is the premium paid at t = 0 for the privilege of giving BT and getting AT at T . P V (AT ) is the present value of AT . If you have P V (AT ) at time zero and invest it from time zero to T , you’ll have exactly AT . Similarly, if you have P V (BT ) at time zero and invest it from time zero to T , you’ll have exactly BT . This is the proof of Equation 9.9. Suppose we have two portfolios. Portfolio #1 consists a European option AT → BT and the present value of the asset AT . Portfolio #2 consists a European option BT → AT and the present value of the asset BT . Payoff of Portfolio #1 at T If AT ≥ BT If AT < BT Option AT → BT 0 BT − AT P V (AT ) AT AT Total AT BT

Payoff of Portfolio #2 at T If AT ≥ BT Option BT → AT AT − BT P V (BT ) BT Total AT

If AT < BT 0 BT BT

Portfolio #1 and Portfolio #2 have the identical payoff at T . To avoid arbitrage, these two portfolios must cost us the same to set up at any time prior to T . The cost of setting up Portfolio #1 is (AT → BT )0 + P V (AT ). The cost of setting up Portfolio #2 is (BT → AT )0 +P V (BT ). Hence Equation 9.9 holds. Example 9.1.17. Stock A currently sells for $30 per share. It doesn’t pay any dividend. Stock B currently sells for $50 per share. It pays dividend at a continuously compounded rate of 5% per year. The continuously compounded risk-free interest rate is 6% per year. A European option gives the option holder the right to surrender one share of Stock B and receive one share of Stock A at the end of Year 1. This option currently sells for $8.54. Calculate the premium for another European option that gives the option holder the right to surrender one Stock A and receive one Stock B at the end of Year 1. (AT → BT )0 + P V (AT ) = (BT → AT )0 + P V (BT ) (AT → BT )0 + 30 = 8.54 + 50e−0.05 (AT → BT )0 = 26. 1 Please note that the risk free interest rate 6% is not needed for solving the problem. In addition, you don’t need to decide whether to call the option "give

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Stock B and receive Stock A" or "give Stock A and receive Stock B" as a call or put. Equation 9.9 holds no matter you call the option AT → BT or BT → AT a call or put. Example 9.1.18. Stock A currently sells for $55 per share. It pays dividend of $1.2 at the end of each quarter. Stock B currently sells for $72 per share. It pays dividend at a continuously compounded rate of 8% per year. The continuously compounded risk-free interest rate is 6% per year. A European option gives the option holder the right to surrender one share of Stock A and receive one share of Stock B at the end of Year 1. This option currently sells for $27.64. Calculate the premium for another European option that gives the option holder the right to surrender one Stock B and receive one Stock A at the end of Year 1. (AT → BT )0 + P V (AT ) = (BT → AT )0 + P V (BT ) P V (AT ) = A0 − 1.2a4|i = 55 − 1.2a4|i i is the effective interest rate per quarter. i = e0.25r − 1 = e0.25(0.06) − 1 = 1. 511 µ 3% ¶ 1 − 1.01 511 3−4 P V (AT ) = 55 − 1.2a4|i = 55 − 1.2 = 50. 38 0.01 511 3 P V (BT ) = B0 e−δB T = 72e−0.08 = 66. 46 27.64 + 50. 38 = (BT → AT )0 + 66. 46 (BT → AT )0 = 11. 56 Currency options Example 9.1.19. Let’s go through the textbook example. Suppose that a 1-year dollar-denominated call option on €1 with the strike price $0.92 is $0.00337. The current exchange rate is €1 = $0.9. What’s the premium for a 1-year 1 euro-denominated put option on $1 with strike price € = €1. 087? 0.92 First, let’s walk through the vocabulary. The phrase "dollar-denominated option" means that both the strike price and the option premium are expressed in U.S. dollars. Similarly, the phrase "euro-denominated option" means that both the strike price and the option premium are expressed in euros. Next, let’s summarize the information using symbols. "dollar-denominated call option on €1 with the strike price $0.92" is $0.92 →€1. The premium for this option is $0.00337. This is represented by ($0.92 → €1)0 =$0.0337. 1 = €1. 087" can "Euro-denominated put option on $1 with strike price € 0.92µ ¶ 1 1 be represented by $1 →€ . The premium for this option is $1 → € 0.92 0.92 0 Now the solution be simple. µ ¶ should µ ¶ 1 1 1 $1 → € = $1 × 0.92 → € × 0.92 0.92 0 0.92 0.92 0 1 1 = ($0.92 → €1)0 = ×$0.0337 0.92 0.92

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9.1. PUT-CALL PARITY 1 Since the exchange rate is €1 = $0.9 or $1 =€ , we have: 0.9 ¶ µ 1 1 1 1 = ×$0.0337 = × 0.0337 × € = €0.04 07 $1 → € 0.92 0 0.92 0.92 0.9

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18

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS General formula: 1 The current exchange rate is €1 = $x0 or $1=€ x0 A dollar-denominated $K strike call on €1 has a premium of $a m ( $K → €1)0 = $a or ( $K → €1)$0 = a 1 strike put on $1 has a premium of €b K m µ ¶ ¶€ µ 1 1 =€b or $1 → € =b $1 → € K 0 K 0

A euro-denominated €

It then follows: µ

$1 → €

1 K

¶€

µ

0

× x0 =

µ

$1 → €

¶€

1 K

¶$ 0

=

1 × ( $K → €1)$0 K

(9.10)

1 is the euro-cost of "give $1, get € ." Since K 0 µ ¶€ 1 × x0 is the dollar cost of "give the exchange rate is €1 = $x0 , $1 → € K 0 1 $1, get ." Similarly, ( $K → €1)$0 is the dollar-cost of "give $K, get €1." K 1 1 Equation 9.10 essentially says that the dollar cost of "give $1, get " is of K K the dollar cost "give $K, get €1." This should make intuitive sense. µ ¶ 1 1 Tip 9.1.6. The textbook gives you the complex formula C$ (x0 , K, T ) = x0 KPf , ,T . x0 K Do not memorize this formula or Equation 9.10. Memorizing complex formulas is often prone to errors. Just translate options into symbols. Then a simple solution should emerge. See the next example. In Equation 9.10,

1 $1 → € K

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Example 9.1.20. The current exchange rate is €0.9 per dollar. A European euro-denominated call on 1 dollar with a strike price €0.8 and 6 months to expiration has a premium €0.0892. Calculate the price of a European dollardenominated put option on 1 euro with a strike price $1.25. Just translate the options into symbols. Then you’ll see a solution. The current exchange rate is €0.9 per dollar. $1 =€0.9 or €1 = $

1 0.9

Euro-denominated call on 1 dollar with strike price €0.8 has a premium €0.0892 (€0.8 → $1)0 =€0.0892 Calculate the price of a dollar-denominated put on 1 euro with strike price $1.25 (€1 → $1.25)0 = $? ¶ µ 1 → $1 = 1.25 × (€0.8 → $1)0 (€1 → $1.25)0 = 1.25 × € 1.25 0 1 = 1.25 × €0.0892 = 1.25 × 0.08928 × $ = $0.124 0.9

9.1.5

Comparing options with respect to style, maturity, and strike

European vs American options American options can be exercised at any time up to (and including) the maturity. In contrast, European options can be exercised only at the maturity. Since we can always convert an American option into a European option by exercising the American option only at the maturity date, American options are at least as valuable as an otherwise identical European option. CAmer (K, T ) ≥ CEur (K, T )

(9.11)

PAmer (K, T ) ≥ PEur (K, T )

(9.12)

Equation 9.11 and Equation 9.12 are not earth-shaking observations. You shouldn’t have trouble memorizing them. Maximum and minimum price of a call 1. The price of a call option is always non-negative. CAmer (K, T ) ≥ CEur (K, T ) ≥ 0. Any option (American or European, call or put) is a privilege with non-negative payoff. The price of a privilege can never be negative. The worst thing you can do is to throw away the privilege. 2. The price of a call option can’t exceed the current stock price. S0 ≥ CAmer (K, T ) ≥ CEur (K, T ). The best you can do with a call option is to own a stock. So a call can’t be worth more than the current stock.

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS 3. The price of a European call option must obey the put call parity. For a non-dividend paying stock, the parity is CEur (K, T ) = PEur (K, T ) + S0 − P V (K) ≥ S0 − P V (K) Combining 1, 2, and 3, we have: For a non-dividend paying stock S0 ≥ CAmer (K, T ) ≥ CEur (K, T ) ≥ max [0, S0 − P V (K)]

(9.13)

For a discrete-dividend paying stock S0 ≥ CAmer (K, T ) ≥ CEur (K, T ) ≥ max [0, S0 − P V (Div) − P V (K)] (9.14) For a continuous-dividend paying stock £ ¤ S0 ≥ CAmer (K, T ) ≥ CEur (K, T ) ≥ max 0, S0 e−δT − P V (K)

(9.15)

Tip 9.1.7. You don’t need to memorize Equation 9.13, 9.14, or 9.15. Just memorize basic ideas behind these formulas and derive the formulas from scratch. Maximum and minimum price of a put 1. The price of a put option is always non-negative. PAmer (K, T ) ≥ PEur (K, T ) ≥ 0. 2. The price of a European put option can’t exceed the present value of the strike price. PEur (K, T ) ≤ P V (K). The best you can do with a European put option is to get the strike price K at T . So a European put can’t be worth more than the present value of the strike price. 3. The price of an American put option can’t exceed the strike price. K ≥ CAmer (K, T ). The best you can do with an American put option is to exercise it immediately after time zero and receive the strike price K. So an American put can’t be worth more than the strike price. 4. The price of a European put option must obey the put call parity. For a non-dividend paying stock, the parity is PEur (K, T ) = CEur (K, T ) + P V (K) − S0 ≥ P V (K) − S0 Combining 1, 2, and 3, we have: K ≥ PAmer (K, T ) ≥ PEur (K, T ) ≥ 0 For a non-dividend paying stock

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(9.16)

9.1. PUT-CALL PARITY

P V (K) ≥ PEur (K, T ) ≥ max [0, P V (K) − S0 ]

21

(9.17)

For a discrete-dividend paying stock P V (K) ≥ PEur (K, T ) ≥ max [0, P V (K) + P V (Div) − S0 ]

(9.18)

For a continuous-dividend paying stock ¤ £ P V (K) ≥ CEur (K, T ) ≥ max 0, P V (K) − S0 e−δT

(9.19)

Tip 9.1.8. You don’t need to memorize Equation 9.16, 9.17, 9.18, or 9.19. Just memorize the basic ideas behind these formulas and derive the formulas from scratch. Early exercise of American options Suppose an American option is written at time zero. The option expires in date T . Today’s date is t where 0 ≤ t < T . The stock price at the option expiration date is ST . Today’s stock price is St . The continuously compounded risk-free interest rate is r per year. Proposition 9.1.1. It’s never optimal to exercise an American call early on a non-dividend paying stock. If at t you exercise an American call option , your payoff is St − K. If at t you sell the remaining £ call option, you’ll ¤ get at least CEur (St , K, T ) ≥ max [0, St − P V (K)] = max 0, St − Ke−r(T −t) , the premium for a European call option written in t and expiring in T . £ ¤ Clearly, CEur (St , K, T ) ≥ max [0, St − P V (K)] = max 0, St − Ke−r(T −t) > St − K for r > 0 It’s not optimal to exercise an American call option early if you can sell the remaining call option. What if you can’t sell the remaining call option? If you can’t sell the call option written in t and expiring in T ., you can short sell one stock at t, receiving St and accumulating to St er(T −t) . Then at T , if ST > K, you exercise your call, paying K and receiving one stock. Next, you return the stock to the broker. Your total profit is St er(T −t) − K > St − K for a positive r; if at T , ST < K, you let your call option expire worthless, purchase a stock in the market, and return it to the broker. Your profit is St er(T −t) − ST > St er(T −t) − K > St − K for a positive r. Intuition behind not exercising American call option early. If you exercise the call option early at t, you pay the strike price K at t and gain physical possession of the stock at t. You lose the interest you could have earned during [t, T ] had you put K in a savings account, yet you gain nothing by physically owning a

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

stock during [t, T ] since the stock doesn’t pay any dividend. In addition, by exercising the call option t, you throw away the remaining call option during [t, T ]. Example 9.1.21. You purchase a 30-strike American call option expiring in 6 months on a non-dividend paying stock. 2 months later the stock reaches a high price of $90. You are 100% sure that the stock will drop to $10 in 4 months. You are attempted to exercise the call option right now (i.e. at t = 2/12 = 1/6 year) and receive 90 − 30 = 60 profit. The continuously compounded risk-free interest rate is 6% per year. Explain why it’s not optimal to exercise the call early. Solution. The problem illustrates the pitfall in common thinking "If you I know for sure that the stock price is going to fall, shouldn’t I exercise the call now and receive profit right away, rather than wait and let my option expire worthless?" Suppose indeed the stock price will be $10 at the call expiration date T = 6/12 = 0.5. If you exercise the call early at t = 2/12 = 1/6, you’ll gain St − K = 90 − 30 = 60, which will accumulate to 60e0.06(4/12) = 61. 212 at T = 0.5. Instead of exercising the call early, you can short-sell the stock at t = 2/12 = 1/6. Then you’ll receive 90, which will accumulate to 90e0.06(4/12) = 91. 818 at T = 0.5. Then at T = 0.5, you purchase a stock from the market for 10 and return it to the brokerage firm where you borrow the stock for short sale. Your profit is 91. 818 − 10 = 81. 818, which is the greater than 61. 212 by 81. 818 − 61. 212 = 20. 606. Proposition 9.1.2. It might be optimal to exercise an American call option early for a dividend paying stock. Suppose the stock pays dividend at tD . Time 0 ... ... tD ... ... T Pro and con for exercising the call early at tD . • +. If you exercise the call immediately before tD , you’ll receive dividend and earn interest during [tD , T ] • −. You’ll pay the strike price K at tD , losing interest you could have earned during [tD , T ] • −. You throw away the remaining call option during [tD , T ]. Had you waited, you would have the call option during [tD , T ] However, if the accumulated value of the dividend is big enough, then it can optimal to exercise the stock at tD . Proposition 9.1.3. If it’s optimal to exercise an American call early, then the best time to exercise the call is immediately before the dividend payment.

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Here is the proof. Suppose the dividend is paid at tD Time 0 t1 ... ... tD t2 ... ... T It’s never optimal to exercise an American call at t1 . If you exercise the call at t1 instead of tD , you’ll • lose interest that can be earned on K during [t1 , tD ] • lose a call option during [t1 , tD ] • gain nothing (there’s no dividend during [t1 , tD ]) It’s never optimal to exercise an American call at t2 . If you exercise the call at t2 instead of tD , you’ll • gain a tiny interest that can be earned during [tD , t2 ] but lose the dividend that can be earned if the call is exercised at tD So for a dividend paying stock, if it’s ever worthwhile to exercise an American call early, you should exercise the call immediately before the dividend payment, no sooner or later. Combining these two proposition, we have: Proposition 9.1.4. It’s only optimal to exercise an American call option either at maturity or immediately before a dividend payment date. Any other time is not optimal. Proposition 9.1.5. It might be optimal to exercise an American put early. Time 0 t1 ... ... t ... ... T The pros and cons of exercising an American put at t instead of T • +. You receive K at t and earn interest during [t, T ] • −. You lose the remaining put option during [t, T ]. If you wait and delay exercising the option, you’ll have a put option during [t, T ] and can decide whether to exercise it or discarding it. This is especially painful if ST > K. If ST > K and you exercise the put at t, you’ll get K at t, which accumulates to Ker(T −t) at T . If you wait and ST > K, you’ll let the put option expire worthless and have ST at T . If ST > Ker(T −t) , then you lose money by exercising the put at t. So one danger of exercising the put at t is that the stock might be worth more than K after t. However, if the interest earned on K during [t, T ] is big enough, it can be optimal to exercise the put at t instead of T .

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Appendix 9.A: Parity bound for American options Appendix 9.5A (and 9.5B) is not on the syllabus. However, I still recommend that you study it. Appendix 9.5A has two important ideas 1. Why the put-call parity doesn’t hold for American options 2. How to calculate the non-arbitrage boundary price for American options Why the put-call parity doesn’t hold for American options The putcall parity such as Equation 9.2 holds only for European options. It doesn’t hold for American options. To understand why, let’s start from European options. For Equation 9.2 to hold, among other things, the call and the put must be exercised at the same time. Recall our proof of the put-call parity. At time t we have two portfolios. Portfolio #1 consists a European call option on a stock and P V (K), the present value of the strike price K. Portfolio #2 consists a European put option on the stock and one share of the stock with current price St . Both the call and put have the same underlying stock, have the same strike price K, and the same expiration date T . We have found that Portfolio #1 and Portfolio #2 have an identical payoff of max (K, ST ) at the common exercise date T . To avoid arbitrage, the two portfolios must cost us the same at time zero. Hence we have Equation 9.2. Now suppose the call is exercised at T1 and the put is exercised at T2 where T1 6= T2 . Portfolio #1 consists of a European call option and P V (K) = Ke−rT1 ; Portfolio #2 consists an American put and one stock worth S0 . Then at T1 , Portfolio #1 has a payoff of max (K, ST1 ); Portfolio #2 has a payoff of max (K, ST2 ). Now the two payoffs differ in timing and the amount. As a result, we don’t know whether the two portfolios have the same set-up cost. Now you should understand why Equation 9.2 doesn’t hold for American options. American options can be exercised at any time up to (and including) the maturity. Even when an American call and an American put have the same maturity T , the American call can be exercised at T1 where 0 ≤ T1 ≤ T ; the American put can be exercised at T1 where 0 ≤ T2 ≤ T . Hence an American call and an American put can be exercised at different times, they don’t follow the put-call parity. Non-arbitrage boundary price for American options For a stock that pays discrete dividend, the key formula is S0 − P V (K) ≥ CAme (K, T ) − PAme (K, T ) ≥ S0 − P V (Div) − K

(9.20)

Many people find it hard to understand or memorize Equation 9.20. Here is an intuitive proof without using complex math.

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First, we establish the boundary of American options: CEur (K, T ) + P V (Div) ≥ CAme (K, T ) ≥ CEur (K, T ) PEur (K, T ) + K − P V (K) ≥ PAme (K, T ) ≥ PEur (K, T )

(9.21)

(9.22)

This is why Equation 9.21 holds. Clearly, CAme (K, T ) ≥ CEur (K, T ). This is because an American call option can always be converted to a European call option. To understand why CEur (K, T ) + P V (Div) ≥ CAme (K, T ), consider an American call option and an otherwise identical European call option. Both call options have the same underlying stock, the same strike price K, the same expiration date T . The European call option is currently selling for CEur (K, T ). How much more can the American call option sell for? The only advantage of an American option over an otherwise identical European call option is that the American call option can be exercised early. The only good reason for exercising an American call early is to get the dividend. Consequently, the value of an American call option can exceed the value of an otherwise identical European call option by no more than the present value of the dividend. So CEur (K, T ) + P V (Div) ≥ CAme (K, T ). A rational person will pay no more than CEur (K, T ) + P V (Div) to buy the American call option. So Equation 9.21 holds. Similarly, an American put is worth at least as much as an otherwise identical European put. In addition, the value of an American put exceeds the value of an otherwise identical European put by no more than K − P V (K). The only advantage of an American put over an otherwise identical European put is that the American put can be exercised early. The only good reason for exercising an American put early is to receive the strike price K immediately at time zero (as opposed to receiving K at T in a European put) and earn the interest on K from time zero to T . The maximum interest that can be earned on K during [0, T ] is K − P V (K) = K − Ke−rT . Consequently, PEur (K, T ) + K − P V (K) ≥ PAme (K, T ). Equation 9.22 holds. Next, we are ready to prove Equation 9.20. CAme (K, T ) − PAme (K, T ) reaches its minimum value when CEur (K, T ) reaches it minimum value CEur (K, T ) and PAme (K, T ) reaches its maximum value PEur (K, T ) + K − P V (K): CAme (K, T ) − PAme (K, T ) ≤ CEur (K, T ) − [PEur (K, T ) + K − P V (K)] From the put-call parity, we have: CEur (K, T ) + P V (K) = PEur (K, T ) + S0 − P V (Div) → CEur (K, T ) = PEur (K, T ) + S0 − P V (Div) − P V (K) → CEur (K, T ) − [PEur (K, T ) + K − P V (K)] = PEur (K, T ) + S0 − P V (Div) − P V (K)

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS − [PEur (K, T ) + K − P V (K)] = S0 − P V (Div) − K → CAme (K, T ) − PAme (K, T ) ≤ S0 − P V (Div) − K

Similarly, CAme (K, T )−PAme (K, T ) reaches it maximum value when CAme (K, T ) reaches its maximum value and PAme (K, T ) reaches it minimum value. CAme (K, T ) − PAme (K, T ) ≤ CEur (K, T ) + P V (Div) − PEur (K, T ) CEur (K, T )+P V (Div)−PEur (K, T ) = [PEur (K, T ) + S0 − P V (Div) − P V (K)] +P V (Div) − PEur (K, T ) = S0 − P V (Div) Hence S0 − P V (K) ≥ CAme (K, T ) − PAme (K, T ) ≥ S0 − P V (Div) − K. Equation 9.20 holds. Tip 9.1.9. If you can’t memorize Equation 9.20, just memorize Equation 9.21, Equation 9.22, and Equation 9.3. Then can derive Equation 9.20 on the spot. Example 9.1.22. If the interest rate is zero. Is it ever optimal to exercise an American put on a stock? Solution. According to Equation 9.22, if the risk-free interest rate is zero, then K = P V (K) and PEur (K, T ) ≥ PAme (K, T ) ≥ PEur (K, T ). This gives us PAme (K, T ) = PEur (K, T ). So it’s never optimal to exercise an American put early if the interest rate is zero. Example 9.1.23. Is it ever optimal to exercise an American call on a nondividend paying stock? Solution. According to Equation 9.21, if Div = 0, then CEur (K, T ) ≥ CAme (K, T ) ≥ CEur (K, T ) or CAme (K, T ) = CEur (K, T ). It’s never optimal to exercise the American call option. Example 9.1.24. An American call on a non-dividend paying stock with exercise price 20 and maturity in 5 months is worth 1.5. Suppose that the current stock price is 1.9 and the risk-free continuously compounded interest rate is 10% per year. Calculate the non-arbitrage boundary price of the American put option on the stock with strike price 20 and 5 months to maturity. Solution. If you can memorize Equation 9.20, then S0 − P V (K) ≥ CAme (K, T ) − PAme (K, T ) ≥ S0 − P V (Div) − K 19 − 20e−0.1(5/12) ≥ 1.5 − PAme (K, T ) ≥ 19 − 0 − 20 −0.184 ≥ 1.5 − PAme (K, T ) ≥ −1 1 ≥ PAme (K, T ) − 1.5 ≥ 0.184

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1 + 1.5 ≥ PAme (K, T ) ≥ 0.184 + 1.5 2. 5 ≥ PAme (K, T ) ≥ 1. 684 If you can’t memorize Equation 9.20, this is how to solve the problem using basic reasoning. The American put is worth at least the otherwise identical European put. PAme (K, T ) ≥ PEur (K, T ) Using the put-call parity: PEur (K, T ) = CEur (K, T ) + P V (K) − S0 Since the stock doesn’t pay any dividend, CAme (K, T ) = CEur (K, T ) = 1.5 PEur (K, T ) = CEur (K, T ) + P V (K) − S0 = 1.5 + 20e−0.1(5/12) − 19 = 1. 684 So PAme (K, T ) ≥ PEur (K, T ) = 1.684 The value of an American put can exceed the value of an otherwise identical European put by no more the early-exercise value. Since the only possible reason to exercise an American put early is to receive K and earn interest K − P V (K) during [0, T ], so the maximum early-exercise value is K − P V (K) = 20 − 20e−0.1(5/12) = 0.816 → PAme (K, T ) ≤ PEur (K, T ) + K − P V (K) = 1.684 + 0.816 = 2. 5 Time to expiration American option An American option (call or put) has more time to expiration is at least as valuable as an otherwise identical American option with less time to expiration. If options are on the same stock and T1 > T2 , we have: CAmer (K, T1 ) > CEur (K, T2 ) PAmer (K, T1 ) > PEur (K, T2 )

(9.23) (9.24)

European option A European call on a non-dividend paying stock will be at least as valuable as an otherwise identical European call option with a shorter time to expiration. This is because for a non-dividend paying stock, an European call option is worth the same as an otherwise identical American call option. And an American call option with a longer time to expiration is more valuable than an otherwise identical American option with a shorter time to expiration. If both options are on the same non-dividend paying stock and T1 > T2 , we have: CEur (K, T1 ) > CEur (K, T2 ) (9.25) If the stock pays dividend, then a longer-lived European option may be less valuable than an otherwise identical but shorter-lived European option. The textbook gives two good examples. In Example #1, a stock is valuable only because of its dividend. The stock pays a dividend at the end of Week 2. Once the dividend is paid, the stock dies

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

and is worth nothing. In this case, if T > 2 weeks, CEur (K, T ) = 0. If if T ≤ 2 weeks, CEur (K, T ) might be worth something depending on how high the strike price K is. In Example #2, a put is written on the asset of a bankrupt company. The asset of the bankrupt company is a constant c. The put is worth P V (c) = ce−rT . If the risk-free interest rate r > 0, ce−rT will decrease if time to expiration T increases.

European options when the strike price grows over time Typically, a call or put has a fixed strike price K. However, there’s nothing to prevent someone from inventing a European option whose strike price changes over time. Consider a European option whose strike price grows with the risk-free interest rate. That is, KT = KerT . What can we say about the price of such a European option? For a European option with strike price KT = KerT , a longer-lived European option is at least as valuable as an otherwise identical but shorter-lived European option. ¡ ¢ ¡ ¢ CEur KerT1 , T1 ≥ CEur KerT2 , T2 if T1 > T2

¡ ¢ ¡ ¢ PEur KerT1 , T1 ≥ PEur KerT2 , T2 if T1 > T2

(9.26) (9.27)

This is why Equation 9.26 holds. Suppose at time zero we buy two European calls on the same stock. The first call expires at T1 and has a strike price KerT1 . The second call expires at T2 and has a strike price KerT2 , where T1 > T2 . Let’s choose a common time T1 and compare¡ the payoffs of ¢these two calls at T1 . The payoff of the longer-lived call is max 0, ST1 − KerT1 . The payoff of the shorter-lived call is calculated as follows. First, we calculate its payoff at T2 . Next, we accumulate this payoff from ¡ T2 to T1 . rT ¢ The payoff of the shorter-lived call at T is max 0, ST2 − Ke 2 . Next, we 2¢ ¡ rT2 accumulate this payoff max 0, ST2 − Ke from T2 to T1 . As we’ll soon see, it’s much harder for a short-lived call to have a positive payoff at T1 . The longer-lived call will have a positive payoff at ST1 − KerT1 at T1 if ST1 > KerT1 ; all else, the payoff is zero. On the other hand, the shorter-lived call will have a positive payoff ST1 − KerT1 at T1 only if the following two conditions are met • ST2 > KerT2 • ST1 > KerT1

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If ST2 ≤ KerT2 , the shorter-lived call will expire worthless, leading to zero payoff at T2 , which accumulates to zero payoff at T1 . If ST2 > KerT2 , we’ll receive a positive payoff ST2 − KerT2 at T2 . If we accumulate ST2 − KerT2 from T2 to T1 , ST2 will accumulate to ST1 and KerT2 to KerT2 er(T1 −T2 ) = KerT1 , leading to a total amount ST1 − KerT1 at T1 . The total payoff amount ST1 − KerT1 is positive if ST1 > KerT1 . In summary, both calls can reach the common positive payoff ST1 −KerT1 at T1 . The longer-lived call will reach this payoff if ST1 > KerT1 . The shorter-lived call will reach this payoff if both ST2 > KerT2 and ST1 > KerT1 . Consequently, the long-lived call has a better payoff and should be at least as valuable as the shorter-lived call. Hence Equation 9.26 holds. If you still have trouble understanding why the longer-lived call has a richer payoff, you can draw the following payoff table: The accumulated payoff of the shorter-lived call at T1 If ST1 ≤ KerT1 If ST1 > KerT1 rT2 If ST2 ≤ Ke P ayof f = 0 P ayof f = 0 If ST2 > KerT2 P ayof f = ST1 − KerT1 ≤ 0 P ayof f = ST1 − KerT1 > 0 The payoff of the longer-lived call at T1 If ST1 ≤ KerT1 If ST1 > KerT1 rT2 If ST2 ≤ Ke P ayof f = 0 P ayof f = ST1 − KerT1 > 0 rT2 If ST2 > Ke P ayof f = 0 P ayof f = ST1 − KerT1 > 0 You can see that the longer-lived call has a slightly better payoff than the shorter-lived payoff. To avoid arbitrage, the longer-lived call can’t sell for less than the shorter-lived call. Hence Equation 9.26 holds. Similarly, you can prove Equation 9.27. Different strike price Proposition 9.1.6. A call (European or American) with a low strike price is at least as valuable as an otherwise identical call with a higher strike price. However, the excess premium shouldn’t exceed the excess strike price. 0 ≤ C (K1 , T ) − C (K2 , T ) ≤ K2 − K1 if K1 < K2

(9.28)

Equation 9.28 should make intuitive sense. The lower-strike call allows the call holder to buy the underlying asset by the guaranteed lower strike price. Clearly, the payoff of a lower-strike call can never be less than the payoff of an otherwise identical but higher-strike call. Consequently, 0 ≤ C (K1 , T ) − C (K2 , T ). And this is why C (K1 , T ) − C (K2 , T ) ≤ K2 − K1 . The only advantage of a K1 -strike call over the K2 -strike call is that the guaranteed purchase price of the underlying asset is K2 − K1 lower in the K1 -strike call; to buy the asset, the K1 -strike call holder can pay K1 yet the K2 -strike call holder will pay

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K2 . Consequently, no rational person will pay more than C (K2 , T ) + K2 − K1 to buy the K1 -strike call. Proposition 9.1.7. A European call with a low strike price is at least as valuable as an otherwise identical call with a higher strike price. However, the excess premium shouldn’t exceed the present value of the excess strike price.

0 ≤ CEur (K1 , T ) − CEur (K2 , T ) ≤ P V (K2 − K1 ) if K1 < K2

(9.29)

The only advantage of a K1 -strike European call over the K2 -strike European call is that the guaranteed purchase price of the underlying asset is K2 − K1 lower in the K1 -strike call at T . Consequently, no rational person will pay more than CEur (K2 , T ) + P V (K2 − K1 ) to buy the K1 -strike call. Please note that CEur (K1 , T )−CEur (K2 , T ) ≤ P V (K2 − K1 ) doesn’t apply to American call options because two American options can be exercised at different dates. Proposition 9.1.8. A put (European or American) with a higher strike price is at least as valuable as an otherwise identical put with a lower strike price. However, the excess premium shouldn’t exceed the excess strike price.

0 ≤ P (K2 , T ) − P (K1 , T ) ≤ K2 − K1 if K1 < K2

(9.30)

Clearly, a higher strike put is at least as valuable as an otherwise identical put with a lower strike price. Since the only advantage of a K2 -strike put over the K1 -strike put is that the guaranteed sales price of the underlying asset is K2 −K1 high in the K2 -strike put, no rational person will pay C (K1 , T ) + K2 − K1 to buy the K2 -strike put. Proposition 9.1.9. A European put with a higher strike price is at least as valuable as an otherwise identical put with a lower strike price. However, the excess premium shouldn’t exceed the present value of the excess strike price.

0 ≤ PEur (K2 , T ) − PEur (K1 , T ) ≤ P V (K2 − K1 ) if K1 < K2

(9.31)

Please note that PEur (K2 , T ) − PEur (K1 , T ) ≤ P V (K2 − K1 ) won’t apply to two American put options because they can be exercised at two different dates. Proposition 9.1.10. A diversified option portfolio is at least as valuable as one undiversified option. For K1 < K2 and 0 < λ < 1

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C [λK1 + (1 − λ) K2 ] ≤ λC (K1 ) + (1 − λ) C (K2 )

(9.32)

P [λK1 + (1 − λ) K2 ] ≤ λP (K1 ) + (1 − λ) P (K2 )

(9.33)

Please note that Equation 9.32 and Equation 9.33 apply to both European options and American options. A call portfolio consists of λ portion of K1 -strike call and (1 − λ) portion of K2 -strike call. The premium of this portfolio, λC (K1 ) + (1 − λ) C (K2 ), can be no less than the premium of a single call with a strike price λK1 + (1 − λ) K2 . Similarly, a call portfolio consists of λ portion of K1 -strike put and (1 − λ) portion of K2 -strike put. The premium of this portfolio, λP (K1 ) + (1 − λ) P (K2 ), can be no less than the premium of a single call with a strike price λK1 + (1 − λ) K2 . Before proving Equation 9.32, let’s look at an example. Example 9.1.25. (Textbook example 9.8 K2 and K3 switched) K1 = 50, K2 = 65, K3 = 0.4 (50) + 0.6 (65) = 59. C (K1 , T ) = 14, C (K2 , T ) = 5. Explain why C (59) ≤ 0.4C (50) + 0.6C (65). Payoff 59-strike call (a)

ST < 50 0

50 ≤ ST < 59 0

59 ≤ ST < 65 ST − 59

ST ≥ 65 ST − 59

50-strike call (b) 65-strike call (c) 0.4b + 0.6c

0 0 0

ST − 50 0 0.4 (ST − 50)

ST − 50 0 0.4 (ST − 50)

ST − 50 ST − 65 ST − 591

(0.4b + 0.6c) − a

0

0.4 (ST − 50) ≥ 0

0.6 (65 − ST ) > 02

0

The above table says the following: If we buy 0.4 unit of 50-strike call, buy 0.6 unit of 65-strike call, and sell 1 unit of 59-strike call, our initial cost is (0.4b + 0.6c) − a and our payoff at T is always non-negative. To avoid arbitrage, the position of always having a non-negative payoff at expiration T surely has a non-negative cost at t = 0. Imagine what happens otherwise. For example, you always have a non-negative payoff at T and it costs you −$10 (i.e. you receive $10) to set up this position at t = 0. Then you’ll make at least $10 free money. Hence we have (0.4b + 0.6c) − a ≥ 0 no matter what ST is. Clearly, the call portfolio consisting of 40% 50-strike call and 60% 65-strike call is at least as good as the 59-strike call. Consequently, the portfolio is at least as valuable as the 59-strike call. C (59) ≤ 0.4C (50) + 0.6C (65) . 1 0.4 (S T 2 0.4 (S T

− 50) + 0.6 (ST − 65) = ST − 59 − 50) − (ST − 59) = 0.6 (65 − ST )

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

Example 9.1.26. Prove Equation 9.32. Let K3 = λK1 + (1 − λ) K2 . Clearly, K1 < K3 < K2 Payoff ST < K1 K1 ≤ ST < K3 K3 ≤ ST < K2 K3 -strike call (a) 0 0 ST − K3

ST ≥ K2 ST − K3

K1 -strike call (b) K2 -strike call (c) λb + (1 − λ) c

0 0 0

ST − K1 0 λ (ST − K1 )

ST − K1 0 λ (ST − K1 )

ST − K1 ST − K2 ST − K3

λb + (1 − λ) c − a

0

λ (ST − K1 ) ≥ 0

(1 − λ) (K2 − ST ) > 0

0

Please note λ (ST − K1 ) + (1 − λ) (ST − K2 ) = ST − [λK1 + (1 − λ) K2 ] = ST − K3 In addition, λ (ST − K1 ) − (ST − K3 ) = K3 − λK1 − (1 − λ) ST = λK1 + (1 − λ) K2 − λK1 − (1 − λ) ST = (1 − λ) K2 − (1 − λ) ST = (1 − λ) (K2 − ST ) If we buy λ unit of K1 -strike call, buy (1 − λ) unit of K2 call, and sell 1 unit of K3 = λK1 + (1 − λ) K2 strike call, we’ll have a non-negative payoff at T . To avoid arbitrage, the initial cost must be non-negative. Hence λb+(1 − λ) c−a > 0. Anyway, the call portfolio consisting of λ portion of K1 -strike call and 1 − λ portion of K2 -strike call is at least as good as the call with the strike price K3 = λK1 + (1 − λ) K2 . To avoid arbitrage, C [λK1 + (1 − λ) K2 ] ≤ λC (K1 ) + (1 − λ) C (K2 ). Example 9.1.27. (Textbook example 9.9 K2 and K3 switched) K1 = 50, K2 = 70, K3 = 0.75 (50) + 0.25 (70) = 55. P (K1 , T ) = 4, P (K2 , T ) = 16. Explain why P (55) ≤ 0.75P (50) + 0.25P (70). Payoff 55-strike put (a)

ST < 50 55 − ST

50-strike put (b) 70-strike put (c) 0.75b + 0.25c

50 − ST 70 − ST 55 − ST

0 70 − ST 0.25 (70 − ST )

0 70 − ST 0.25 (70 − ST )

0 0 0

0

0.75 (ST − 50) > 0

0.25 (70 − ST ) > 03

0

(0.75b + 0.25c) − a

50 ≤ ST < 55 55 − ST

Please note 0.75 (50 − ST ) + 0.25 (70 − ST ) = 55 − ST 0.25 (70 − ST ) − (55 − ST ) = 0.75 (ST − 50) 3 0.4 (S T

− 50) − (ST − 59) = 0.6 (65 − ST )

Yufeng Guo, Fall 09 MFE, actuary88.com

55 ≤ ST < 70 0

ST ≥ 70 0

9.1. PUT-CALL PARITY

33

(0.75b + 0.25c) − a ≥ 0 no matter what ST is. Clearly, the put portfolio consisting of 75% 50-strike put and 25% 70-strike put is at least as good as the 55-strike put. Consequently, to avoid arbitrage, the portfolio is at least as valuable as the 55-strike put. P (55) ≤ 0.75P (50) + 0.25P (70). Example 9.1.28. Prove Equation 9.33. Let K3 = λK1 + (1 − λ) K2 . Clearly, K1 < K3 < K2 Payoff ST < K1 K1 ≤ ST < K3 K3 ≤ ST < K2 K3 -strike put (a) K3 − ST K3 − ST 0

ST ≥ K2 0

K1 -strike put (b) K2 -strike put (c) λb + (1 − λ) c

K1 − ST K2 − ST K3 − ST

0 K2 − ST (1 − λ) (K2 − ST )

0 K2 − ST λ (K2 − ST )

0 0 ST − K3

λb + (1 − λ) c − a

0

λ (ST − K1 ) ≥ 0

λ (K2 − ST ) > 0

0

Please note λ (K1 − ST ) + (1 − λ) (K2 − ST ) = K3 − ST In addition, (1 − λ) (K2 − ST ) − (K3 − ST ) = (1 − λ) (K2 − ST ) − λK1 − (1 − λ) K2 + ST = λ (ST − K1 ) λ (ST − K1 ) − (ST − K3 ) = K3 − λK1 − (1 − λ) ST = λK1 + (1 − λ) K2 − λK1 − (1 − λ) ST = (1 − λ) K2 − (1 − λ) ST = (1 − λ) (K2 − ST ) The payoff is always non-negative. Consequently, the call portfolio consisting of λ portion of K1 -strike put and 1 − λ portion of K2 -strike put is at least as good as the put with the strike price K3 = λK1 +(1 − λ) K2 . To avoid arbitrage, P [λK1 + (1 − λ) K2 ] ≤ λP (K1 ) + (1 − λ) P (K2 ). Exercise and moneyness Proposition 9.1.11. If it’s optimal to exercise an option, it’s also optimal to exercise an otherwise identical option that’s more in-the-money. This is just common sense. The textbook gives an example. Suppose it’s optimal to exercise a 50-strike American call on a dividend paying stock. The current stock price is 70. If it’s optimal to exercise the American a 50-strike American call, then it must also be optimal to exercise an otherwise identical call but with a strike price 40.

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

Yufeng Guo, Fall 09 MFE, actuary88.com

Chapter 10

Binomial option pricing: I This chapter is one of the easiest chapters in Derivatives Markets. The textbook did a good job explaining the mechanics of how to calculate the option price in a one-period binomial model. Besides learning the mechanics of option pricing, you should focus on understanding two basic ideas: the non-arbitrage pricing and the risk-neutral probabilities.

10.1

One-period binomial model: simple examples

Example 10.1.1. Suppose we want to find the price of a 12-month European call option on a stock with strike price $15. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. 30

15

20 Time 0

? 10 T =1

Time 0

Stock price binomial tree

0 T =1

Call payoff binomial tree

It’s hard to directly calculate the price of the call option. So let’s build something that behaves like a call, something that has the same payoff pattern as the call. Suppose at time zero we create a portfolio by buying X stocks and putting Y dollars in a savings account. We want this portfolio to have the exact payoff as the call. Y e0.1

30X 20X Time 0

+ 10X T =1

Y Time 0

15 =

Y e0.1 T =1 35

Yufeng Guo, Fall 09 MFE, actuary88.com

? Time 0

0 T =1

36

CHAPTER 10. BINOMIAL OPTION PRICING: I

In the above diagram, 2X stocks at time zero are worth either 30X or 10X at T = 1. Putting Y dollars in a savings account at time zero will produce Y e0.1 at T = 1. We want our portfolio to behave like a call. So the payoff of our portfolio should as the payoff of the call. We set up the following equation: ½ the same 0.1 30X + Y e = 15 X = 0.75 Y = −10 (0.75) e−0.1 = −6. 786 10X + Y e0.1 = 0 Y = −6. 786 means that we borrow 6. 786 at t = 0 and pays 6. 786e0.1 = 7. 5 at T = 1 If at t = 0 we buy 0.75 share of a stock and borrow $6. 786, then at T = 1, this portfolio will have the same payoff as the call. To avoid arbitrage, the portfolio and the call should have the same cost at t = 0. C = 20X + Y = 20 (0.75) − 6. 786 = 8. 214 Example 10.1.2. Find the price of a 12-month European put option on a stock with strike price $15. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. Suppose at time zero we create a replicating portfolio by buying X stocks and investing Y dollars in a savings account. We want this portfolio to have the exact payoff as the put. Y e0.1

30X 20X

+

Time 0 ½

10X T =1

30X + Y e0.1 = 0 10X + Y e0.1 = 5

Y

0 =

Time 0

Y e0.1 T =1

X = −0.25

? Time 0

5 T =1

Y = 6. 786

So the replicating portfolio is at t = 0 by short-selling 0.25 stock and investing 6. 786 in a savings account. The price of the put at t = 0 is: P = 20X + Y = 20 (−0.25) + 6. 786 = 1. 786 Let’s check whether the put-call parity holds. C + P V (K) = 8. 214 + 15e−0.1 = 21. 787 P + S0 = 1. 786 + 20 = 21. 786 Ignoring the rounding difference, we get: C + P V (K) = P + S0

10.2

General one-period binomial model

Suppose we have two points in time, t = 0 (today) and t = h (some point in the future). The continuously compounded risk-free interest rate per year is r ( a positive constant). We have two assets: a stock that pays zero dividend and

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10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

37

a savings account. The savings account is the same as a zero-coupon bond. At time h, the stock price is Sh ; the bond price is Bh . The bond price is deterministic: B0 = 1 Bh = erh The stock price is½stochastic: Su S0 = S Sh = Sd So at h the stock price either goes up to Su ("up state") or goes down to Sd ("down state"). Su S Time 0

Sd Time h

We assume there’s no tax, no transaction cost, or margin requirements; one is allowed to short sell any security and receive full proceeds. In addition, we assume that anyone can buy or sell any number of securities without affecting the market price. We assume that the market is an invisible cop automatically enforcing Sd < Serh < Su . For example, if Su > Sd > Serh , then the stock’s return is guaranteed to be higher than the risk-free interest rate. If this is the case, then the risk-free interest rate was set too low and stock’s return is too good. Everyone will jump on this opportunity, withdraw all his money from his savings account, and invest it in the stock. This will instantaneously bid up the price of the stock and the risk-free interest rate, forcing Sd < Serh < Su to hold. What’s the non-arbitrage condition is the stock pays dividend continuously at the yield δ? Suppose you buy e−δh share of a stock at t = 0 (thus you pay Se−δh ) , buy reinvesting dividend in the stock, you’ll have exactly one e−δh eδh = 1 stock at time h, which is worth either Su or Sd . Su Se−δh Time 0

Sd Time h

Suppose you invest Se−δh in a savings account, then your wealth at h is ¡ −δh ¢ rh simply Se e = Se(r−δ)h . Se(r−δ)h Se−δh Se(r−δ)h Time 0 Time h To avoid arbitrage, the following condition needs to be met:

Yufeng Guo, Fall 09 MFE, actuary88.com

38

CHAPTER 10. BINOMIAL OPTION PRICING: I • At good times (i.e. when the stock goes up), the return you earn from the stock should exceed the risk free interest rate. So Su > Se(r−δ)h • At bad times (i.e. when the stock goes down), the return you earn from the stock should be less than the risk free interest rate. So Sd < Se(r−δ)h

If the condition is not met, we’ll end up in a weird situation where the stock is always better off than the savings account or the savings account is always better off than the stock. Then everyone will invest his money in the better performing asset, instantly bidding up the price of the lower-performing asset and forcing the above condition to be met. To avoid arbitrage, Su , Sd , and r need to satisfy the following condition: Su > Se(r−δ)h > Sd

(10.1)

If Su = uS and Sd = dS, then Equation 10.1 becomes: u > e(r−δ)h > d

(10.2)

Equation 10.2 is the textbook Equation 10.4. Let’s continue. Let C represent the option price at time zero. Let Cu and Cd represent the payoff at time h of an option in the up state and down state respectively: Cu C Time 0

Cd Time h

Our task is to determine C by setting a portfolio that replicates the option payoff. We build the replicating portfolio by buying 4 shares of the stock and investing $B in a zero-coupon bond. So we set up the following equation: 4Su Berh Cu 4S + B = C 4Sd Berh Cd t=0 t=h t=0 t=h t=0 t=h ½

4Su + Berh = Cu 4Sd + Berh = Cd

Solving these equations, we get: 4=

Yufeng Guo, Fall 09 MFE, actuary88.com

Cu − Cd Su − Sd

(10.3)

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

B = e−rh

Su Cd − Sd Cu Su − Sd

39

(10.4)

Rearranging Equation 10.3, we get 4Su − Cu = 4Sd − Cd . This equation is critical as we’ll soon see. The value of the portfolio at the up state is 4Su + Berh = 4Su + e−rh (Cu − 4Su ) erh = Cu The value of the portfolio at the down state is 4Sd + Berh = 4Sd + e−rh (Cu − 4Sd ) erh = Cd Using Equation 10.3 and Equation 10.4, we get (verify this for yourself): µ rh ¶ Su − Serh Se − Sd −rh Cu + Cd (10.5) C = 4S + B = e Su − Sd Su − Sd

Define

Serh − Sd Su − Sd rh S u − Se πd = q∗ = Su − Sd

π u = p∗ =

(10.6) (10.7)

Because market automatically enforces Sd < Serh < Su , we have 0 < πu < 1 0 < πp < 1 πu + πd = 1

(10.8) (10.9) (10.10)

According to Equation 10.8 , we can pretend that π u and π d are probabilities. Then Equation 10.5 becomes C = e−rh (π u Cu + π d Cd )

(10.11)

Doing some algebra, we also get (verify this for yourself):

S=e

−rh

µ

Su − Serh Serh − Sd Su + Sd Su − Sd Su − Sd

¶

= e−rh (π u Su + π d Sd )

(10.12)

According to Equation 10.11 and 10.12, once we set up faked probabilities π u and π d , the call price at t = 0 is simply the expected present value of the

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40

CHAPTER 10. BINOMIAL OPTION PRICING: I

call payoffs discounted at the risk-free interest rate; the stock price at t = 0 is simply the expected present value of the future stock prices discounted at the risk-free interest rate. Since the discounting rate is risk-free interest rate, π u and πd are called risk neutral probabilities. Please note that π u and πd are not real probabilities. They are artificially created probabilities so that Equation 10.11 and 10.12 have simple and intuitive explanations. Example 10.2.1. Using the risk-neutral probabilities, find the price of a 12month European call option on a stock with strike price $15. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. Solution. Stock price tree Time 0

T Su = 30

Option terminal payoff Time 0

S = 20

T Cu = max (0, 30 − 15) = 15

C =? Sd = 10 Cd = max (0, 10 − 15) = 0 Serh − Sd 20e0.1(1) − 10 πu = = = 0.605 Su − Sd 30 − 10 πd = 1 − 0.605 = 0.395 Cu = 30 − 15 = 15 Cd = 0 C = e−rh (π u Cu + π d Cd ) = e−0.1(1) (0.605 × 15 + 0.395 × 0) = 8. 211 Example 10.2.2. Using the risk-neutral probabilities, find the price of a 12month European put option on a stock with strike price $15. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. Stock price tree Time 0

T Su = 30

S = 20

Option terminal payoff Time 0 C =?

Sd = 10

T Cu = max (0, 15 − 30) = 0 Cd = max (0, 15 − 10) = 5

Serh − Sd 20e0.1(1) − 10 = = 0.605 Su − Sd 30 − 10 πd = 1 − 0.605 = 0.395

πu =

Cu = 0 Cd = 5 C = e−rh (π u Cu + π d Cd ) = e−0.1(1) (0.605 × 0 + 0.395 × 5) = 1. 787

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10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

41

Suppose the stock pays dividends at a continuously compounded rate δ per year. At time zero, our replicating portfolio consists of 4 shares of stocks and $B in a bond (or a savings account). If we continuously reinvest dividends and buy additional stocks during [0, h], our 4 shares at time zero will grow into 4eδh shares at time h. Our 4eδh shares will be worth either 4eδh Su in the up state or 4eδh Sd in the down state. We want our replicating portfolio to have the same payoff as the option at time h. ½

4eδh Su + Berh = Cu 4eδh Sd + Berh = Cd Solving the equations, we get: Cu − Cd Su − Sd

(10.13)

Su Cd − Sd Cu Su − Sd

(10.14)

4 = e−δh B = e−rh

Notice whether the stock pays dividend or not, at time zero, we always need Su Cd − Sd Cu Su Cd − Sd Cu in a savings account, which grows into to have e−rh Su − Sd Su − Sd Cu − Cd dollars at t = h. If the stock doesn’t pay dividend, at t = 0 we hold Su − Sd Cu − Cd shares of stock, which is shares of stock at t = h ; if the stock pays Su − Sd Cu − Cd dividend at a continuously compounded rate δ, at t = 0 we hold e−δh , Su − Sd Cu − Cd which grows shares of stock at t = h. Su − Sd Cu − Cd Su Cd − Sd Cu So at time h we need to have units of stocks and Su − Sd Su − Sd dollars in a savings account (or a bond), regardless of whether the stock pays dividend or not. To see why, suppose our replicating portfolio at t = h (not t = 0) consists of U shares of stocks and V dollars in a savings account. Then regardless of whether the stock pays dividend or not, we need to have: ½

U Su + V = Cu U Sd + V = Cd

This gives us: Cu − Cd U= Su − Sd

V =

Su Cd − Sd Cu Su − Sd

To have V dollars in a savings account at t = h, we need to have V e−rh = Cu − Cd Su Cd − Sd Cu e−rh at t = 0. To have U = shares of stocks at t = h, Su − Sd Su − Sd

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CHAPTER 10. BINOMIAL OPTION PRICING: I

if the stock pays dividend at a continuously compounded rate of δ, we need to Cu − Cd have U e−δh = e−δh shares of stocks at t = 0. If we use the dividends Su − Sd received to buy additional stocks, then we’ll have U shares of stock at t = h. Now let’s find the cost of the option on a stock that pays dividends at a continuously compounded rate δ per year. The option cost at time zero is Cu − Cd Su Cd − Sd Cu 4S + B = e−δh × S + e−rh S − S Su − Sd d µ u (r−δ)h ¶ Se S − S − Se(r−δ)h d u −rh =e Cu + Cd Su − Sd Su − Sd C = 4S + B = e−rh (πu Cu + πd Cd )

(10.15)

where

πu =

Se(r−δ)h − Sd Su − Sd

(10.16)

πd =

Su − Se(r−δ)h Su − Sd

(10.17)

If Su = Su and Sd = Sd, then Equation 10.16 and 10.17 becomes: πu =

e(r−δ)h − d u−d

(10.18)

πd =

u − e(r−δ)h u−d

(10.19)

Tip 10.2.1. If you don’t want to memorize Equation 10.12, 10.15, 10.16, 10.17, just set up the replication portfolio and calculate 4 and B from scratch. Example 10.2.3. Find the price of a 12-month European call option on a stock with strike price $15. The stock pays dividends at a continuously compounded rate 6% per year. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. Stock price tree Time 0

T Su = 30

S = 20

Option terminal payoff Time 0 C =?

Sd = 10

Yufeng Guo, Fall 09 MFE, actuary88.com

T Cu = max (0, 30 − 15) = 15 Cd = max (0, 10 − 15) = 0

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Replicating portfolio Time 0 (4, B) = (0.706 32, −6. 786 28)

43

T (4u , Bu ) = (0.75, −7. 50)

(4d , Bd ) = (0.75, −7. 50) Our replicating portfolio at t = 0 consists of 4 shares of stocks and $B in a savings account. Cu − Cd 15 − 0 4 = e−δh = e−0.06(1) = 0.706 32 Su − Sd 30 − 10 Su Cd − Sd Cu 30 (0) − 10 (15) = e−0.1(1) B = e−rh = −6. 786 28 Su − Sd 30 − 10 C = 4S + B = 0.706 3 (20) − 6. 786 3 = 7. 34 The replicating portfolio at T is: 4u = 4d = 4eδh = 0.706 32e0.06(1) = 0.75 Bu = Bd = Berh = −6. 786 28e0.1(1) = −7. 5 Verify that the replicating portfolio and the option have the same value: The value of the replicating portfolio at the up state: 4u Su + Bu = 0.75 (30) − 7. 50 = 15 = Cu The value of the replicating portfolio at the down state: 4d Sd + Bd = 0.75 (10) − 7. 50 = 0 = Cd The value of the portfolio at t = 0 : C = 4S + B = 0.706 32 (20) − 6. 78628 = 7. 34 Alternatively, Se(r−δ)h − Sd 20e(0.1−0.06)1 − 10 πu = = = 0.540 81 Su − Sd 30 − 10 π d = 1 − πu = 1 − 0.540 81 = 0.459 19 C = e−rh (πu Cu + πd Cd ) = e−0.1(1) (0.540 81 × 15 + 0.459 19 × 0) = 7. 34 Example 10.2.4. Find the price of a 12-month European put option on a stock with strike price $15. The stock pays dividends at a continuously compounded rate 6% per year. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. Solution. Stock price tree Time 0

T Su = 30

S = 20

Option terminal payoff Time 0 C =?

Sd = 10

T Cu = max (0, 15 − 30) = 0 Cd = max (0, 15 − 10) = 5

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44

CHAPTER 10. BINOMIAL OPTION PRICING: I Replicating portfolio Time 0

T (4u , Bu ) = (−0.25, 7. 50)

(4, B) = (−0.235 4, 6. 786 3) (4d , Bd ) = (−0.25, 7. 50) Our replicating portfolio at t = 0 consists of 4 shares of stocks and $B in a savings account. Cu − Cd 0−5 4 = e−δh = e−0.06(1) = −0.25e−0.06(1) = −0.235 4 Su − Sd 30 − 10 Su Cd − Sd Cu 30 (5) − 10 (0) = e−0.1(1) = 6. 786 3 B = e−rh Su − Sd 30 − 10 C = 4S + B = −0.235 4 (20) + 6. 786 3 = 2. 078 The replicating portfolio at T is: 4u = 4d = 4eδh = −0.235 4e0.06(1) = −0.25 Bu = Bd = Berh = 6. 786 3e0.1(1) = 7. 5 Verify that the replicating portfolio and the option have the same value: The value of the replicating portfolio at the up state: 4u Su + Bu = −0.25 (30) + 7. 50 = 0 = Cu The value of the replicating portfolio at the down state: 4d Sd + Bd = −0.25 (10) + 7. 50 = 5 = Cd The value of the portfolio at t = 0 : C = 4S + B = −0.235 4 (20) + 6. 786 3 = 2. 078 Alternatively, Se(r−δ)h − Sd 20e(0.1−0.06)1 − 10 = = 0.540 8 πu = Su − Sd 30 − 10 πd = 1 − π u = 1 − 0.540 8 = 0.459 2 C = e−rh (π u Cu + π d Cd ) = e−0.1(1) (0.540 8 × 0 + 0.459 2 × 5) = 2. 078 Arbitrage a mispriced option If an option sells for more or less than the price indicated by Equation 10.15, we can make money by "buy low, sell high." Example 10.2.5. A 12-month European call option on a stock has strike price $15. The stock pays dividends at a continuously compounded rate 6% per year. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. This call currently sells for $8. Design an arbitrage strategy.

Solution.

Stock price tree

Option terminal payoff

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10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Time 0

T Su = 30

S = 20

Time 0 C =?

Sd = 10 Replicating portfolio Time 0 (4, B) = (0.706 32, −6. 786 28)

45

T Cu = max (0, 30 − 15) = 15 Cd = max (0, 10 − 15) = 0 T (4u , Bu ) = (0.75, −7. 50) (4d , Bd ) = (0.75, −7. 50)

There are two calls. One is in the market selling for $8 at t = 0. The other is a synthetic call, which consists, at t = 0, of holding 0.706 3 stock and borrowing $6. 786 3 at risk-free interest rate. The synthetic call sells for $7. 34 at t = 0. These two calls have identical payoffs at t = 1. To make a riskless profit, we buy low and sell high. At t = 0, we sell a call for $8 (sell high). Then at t = 1, if the stock price is $30, the call holder exercises the call and our payoff is 15 − 30 = −15; if the stock is $10, the call expires worthless and our payoff is zero.

8

−15

0 Time 0 Time t = 1 Payoff of a written call Simultaneously, at t = 0 we buy 0.706 3 stock and borrow 6. 786 3 dollars at risk-free interest rate (buy low). This costs us 0.706 3 (20) − 6. 786 3 = 7. 34 at t = 0. At t = 1, our initial 0.706 3 stock becomes 0.706 3e0.06(1) stock and our initial debt 6. 786 3 grows into 6. 786 3e0.1(1) . Our portfolio is worth 0.706 3e0.06(1) S1 − 6. 786 3e0.1(1) . If S1 = 30, our portfolio is worth 0.706 3e0.06(1) (30) − 6. 786 3e0.1(1) = 15 If S1 = 10, our portfolio is worth 0.706 3e0.06(1) (10) − 6. 786 3e0.1(1) = 0 15 −7.34 Time 0

0 Time t = 1

Payoff of a replicating portfolio So at t = 0, we gain 8 − 7. 34 = 0.66. At t = 1, the portfolio exactly offsets our payoff in the call. We earn 0.66 sure profit at t = 0.

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CHAPTER 10. BINOMIAL OPTION PRICING: I

Example 10.2.6. A 12-month European call option on a stock has strike price $15. The stock pays dividends at a continuously compounded rate 6% per year. The stock currently sells for $20. In 12 months, the stock can either go up to $30 or go down to $10. The continuously compounded risk-free interest rate per year is 10%. This call currently sells for $7. Design an arbitrage strategy. Stock price tree Time 0

Option terminal payoff

T Su = 30

S = 20

Time 0 C =?

Sd = 10 Replicating portfolio Time 0 (4, B) = (0.706 32, −6. 786 28)

T Cu = max (0, 30 − 15) = 15 Cd = max (0, 10 − 15) = 0 T (4u , Bu ) = (0.75, −7. 50) (4d , Bd ) = (0.75, −7. 50)

There are two calls. One is in the market selling for $7 at t = 0. The other is a synthetic call, which consists, at t = 0, of holding 0.706 3 stock and borrowing 6. 786 3 dollars at risk-free interest rate. The synthetic call sells for $7. 34 at t = 0. These two calls have identical payoffs at t = 1. To make a riskless profit, we buy low and sell high. At t = 0, we buy a call for $7 (buy low). Then at t = 1, if the stock price is $30, the call holder exercises the call and our payoff is 30 − 15 = 15; if the stock is $10, the call expires worthless and our payoff is zero. 15 −7

0 Time 0 Time t = 1 Payoff of a purchased call Simultaneously, at t = 0 we sell the replicating portfolio. We short sell 0.706 3 stock and lend 6. 786 3 dollars at risk-free interest rate (sell high). We gain 0.706 3 (20) − 6. 786 3 = 7. 34 at t = 0. At t = 1, our initially borrowed 0.706 3 stock becomes 0.706 3e0.06(1) stock and our lent principal 6. 786 3 grows into 6. 786 3e0.1(1) . Our portfolio is worth 6. 786 3e0.1(1) − 0.706 3e0.06(1) S1 . If S1 = 30, our portfolio is worth 6. 786 3e0.1(1) − 0.706 3e0.06(1) (30) = −15 If S1 = 10, our portfolio is worth

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

47

6. 786 3e0.1(1) − 0.706 3e0.06(1) (10) = 0 7.34

−15

Time 0

0 Time t = 1

Payoff of a replicating portfolio So at t = 0, we gain7.34 − 7 = 0.34. At t = 1, the call payoff exactly offsets our liabilities in the replicating portfolio We earn 0.34 sure profit at t = 0. Tip 10.2.2. An option and its replicating portfolio are exactly the same in terms of payoff and cost. If an option in the market sells for more than the fair price indicated in Equation 10.15, we can make a sure profit by buying the replicating portfolio and selling the option. If an option in the market sells for less than the fair price indicated in Equation 10.15, we can make a sure profit by selling the replicating portfolio and buying the option. Risk neutral probability and forward price If we use π u and π d to calculate the undiscounted stock price, we get: Se(r−δ)h − Sd Su − Se(r−δ)h Su + Sd Su − Sd Su − Sd (r−δ)h = Se = F0,h

π u Su + π d Sd =

πu Su + πd Sd = Se(r−δ)h = F0,h

(10.20)

F0,h is the delivery price at t = h of a forward contract signed at t = 0. The textbook uses the symbol Ft,t+h to indicate that the forward contract is signed at time t and the asset is to be delivered at t + h. If we treat the contract initiation date t as time zero, then the asset deliver date is time h. So the notation doesn’t matter. If you want to use the textbook notation, you’ll have πu Su + πd Sd = Se(r−δ)h = Ft,t+h

(10.21)

If you want to use my notation, you’ll get Equation 10.20. Suppose you enter into a forward contract as a seller agreeing to deliver one stock for a guaranteed price F0,h at t = h. To ensure you indeed can deliver one stock at time h, you’ll want to buy e−δh stock at time zero. If you reinvest dividend and buy additional stocks, your initial e−δh stock will grow into exactly one stock at time h. Your cost of buying e−δh stock at time zero is Se−δh . Since you’ll tie up your money Se−δh during [0, h], you’ll want the forward price to include the interest you could otherwise earn on Se−δh . So the forward price is just the future value of Se−δh : F0,h = Se−δh erh = Se(r−δ)h

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CHAPTER 10. BINOMIAL OPTION PRICING: I

According to Equation 10.20, the undiscounted stock price equals the forward price under the risk neutral probability. If a problem gives you a forward price, you can use Equation 10.20 to calculate the risk-neutral probability. Constructing a binomial tree Suppose we are standing at time t. If we can be 100% certain about the stock price at time t + h, then investing in stocks doesn’t have any risk. Then stocks must earn a risk-free interest rate. Since the stock already pays dividends at rate δ, to earn a risk free interest rate, the stock price just needs to grow at the rate of r − δ. Hence the stock price at t + h is Se(r−δ)h , which is just the forward price Ft,t+h because Ft,t+h is also equal to Se(r−δ)h . However, the stock price is a random variable and we generally can’t be 100% certain about a stock’s future price. To incorporate uncertainty, we use Formula 11.16 in Derivatives Markets: ln

√ St+h = (r − δ) h ± σ h St

(Textbook 11.16)

St+h is the continuously compounded rate of return St during [t, t√+ h]. This return consists of a known element (r − δ) h and a random element σ h, where σ is the annualized standard deviation of the continuously compounded stock return. The variance of the stock return in one year is σ 2 . The variance during the interval [t, t + h] (which is h year long) is σ 2 h and this is why. [t, t + h] can be broken down into h intervals, with each interval being one year long. Assume stock return during each year is independent identically distributed. The total return during [t, t + h] is the sum of the returns over h intervals. Then the total variance is just the sum of the variance over h intervals, σ 2 h. In the above equation, ln

√

√

√

e±σ h . In the binomial So St+h = St e(r−δ)h±σ h = Se(r−δ)h±σ h = Ft,t+h √ √ (r−δ)h+σ h σ h model, the stock price either goes √up to Se = Ft,t+h e or goes √ down to Se(r−δ)h−σ h = Ft,t+h e−σ h . So we have

uSt = Ft,t+h eσ

√ h

dSt = Ft,t+h e−σ

√ h

(10.22) (10.23)

If we set volatility σ to zero, then Equation 10.22 and 10.23 becomes uSt = dSt = Ft,t+h . This means that if the stock price is 100% certain, then the stock price is just the forward price. Apply Equation 10.21 to Equation 10.22 and 10.23, we get:

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

u = e(r−δ)h+σ d = e(r−δ)h−σ

10.2.1

√ h

√ h

49

(10.24) (10.25)

Two or more binomial trees

Example 10.2.7. Let’s reproduce Derivatives Markets Figure 10.4. Here is the recap of the information on a European call option. The current stock price is 41. The strike price K = 40. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded dividend rate per year is δ = 0%.The option expiration date is T = 2 years. Use a 2-period binomial tree to calculate the option premium. T Each period is h = = 1 year long. 2 Step 1 Draw a stock price tree. √

√

u = e(r−δ)h+σ√ h = e(0.08−0)1+0.3√1 = 1. 462 28 d = e(r−δ)h−σ h = e(0.08−0)1−0.3 1 = 0.802 52 Stock price Period 0 1 2 Period 0 1 Su2 Su Su = 59.9537 =⇒ S Sud S = 41 Sd Sd = 32.9033 2 Sd

2 Suu = 87.6693 Sud = 48.1139 Sdd = 26.4055

Step 2 Draw the terminal payoffs at T = 2. 47.6693 = max (0, 87.6693 − 40) 8.1139 = max (0, 48.1139 − 40) Option Payoff Period 0 1 Cuu

2 = 47.6693

Cu C =?

Cud = 8.1139 Cd Cdd = 0

Step 3 Work backward from right to left (called backwardization). Calculate the premium using the formula C = e−rh (πu Cu + πd Cd ) e(r−δ)h − d e(0.08−0)1 − 0.802 52 πu = = = 0.425 56 u−d 1. 462 28 − 0.802 52

Yufeng Guo, Fall 09 MFE, actuary88.com

50

CHAPTER 10. BINOMIAL OPTION PRICING: I πd = 1 − π u = 1 − 0.42556 = 0.574 44 Option premium Period 0

1

2 Cuu = 47.6693

Cu = 23.0290 C = 10.7369

Cud = 8.1139 Cd = 3.1875 Cdd = 0

Cu = e−rh (π u Cuu + πd Cud ) = e−0.08(1) (0.425 56 × 47.6693 + 0.574 44 × 8.1139) = 23. 029 0 Cd = e−rh (π u Cud + π d Cud ) = e−0.08(1) (0.425 56 × 8.1139 + 0.574 44 × 0) = 3. 187 5 The call premium is C = e−rh (π u Cu + π d Cd ) = e−0.08(1) (0.425 56 × 23.0290 + 0.574 44 × 3.1875) = 10.7369

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

51

Step 4

Calculate 4 and B using the formula − Cd Su Cd − Sd Cu 4=e B = e−rh Su − Sd Su − Sd To avoid errors, put the stock price tree and the premium tree side by side: −δh Cu

Period 0

1

2 Suu = 87.6693

Period 0

Sud = 48.1139

C = 10.7369

Su = 59.9537 S = 41

1

2 Cuu = 47.6693

Cu = 23.0290

Sd = 32.9033

Cud = 8.1139 Cd = 3.1875

Sdd = 26.4055 Period 0

(4, B) = (0.7335, −19.3367)

Cdd = 0 1

(4u , Bu ) = (1.0, −36.9247) (4d , Bd ) = (0.3738, −9.1107)

Cuu − Cud 47.6693 − 8.1139 = e−0(1) = 1.0 Suu − Sdd 87.6693 − 48.1139 Cud − Cdd 8.1139 − 0 4d = e−δh = e−0(1) = 0.373 8 Sud − Sdd 48.1139 − 26.4055 Cu − Cd 23.0290 − 3.1875 4 = e−δh = e−0(1) = 0.733 5 Su − Sd 59.9537 − 32.9033 4u = e−δh

Suu Cud − Sud Cuu Suu − Sud −0.08(1) 87.6693 (8.1139) − 48.1139 (47.6693) =e = −36.9247 87.6693 − 48.1139

Bu = e−rh

−36.9247 means that $36.9247 needs to be borrowed at a risk-free rate. Sdu Cdd − Sdd Cdu Sdu − Sdd −0.08(1) 48.1139 (0) − 26.4055 (8.1139) =e = −9. 110 7 48.1139 − 26.4055 Su Cd − Sd Cu B = e−rh Su − Sd −0.08(1) 59.9537 (3.1875) − 32.9033 (23.0290) =e = −19. 3367 59.9537 − 32.9033 Bd = e−rh

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52

CHAPTER 10. BINOMIAL OPTION PRICING: I

Step 5 Verify that the portfolio replicates the premium tree. Here is recap of the information: Period 0 1 2 0 1 2 Suu = 87.6693 Cuu = 47.6693 Su = 59.9537 Cu = 23.0290 S = 41 Sud = 48.1139 C = 10.7369 Cud = 8.1139 Sd = 32.9033 Cd = 3.1875 Sdd = 26.4055 Cdd = 0 Period 0

(4, B) = (0.7335, −19.3367)

1 (4u , Bu ) = (1.0, −36.9247) (4d , Bd ) = (0.3738, −9.1107)

The value of (4, B) at time zero: 4S + B = 0.7335 (41) − 19.3367 = 10. 736 8 = C The value of(4u , Bu ) at Node u (up state): 4u Su + Bu = 1.0 (59.9537) − 36.9247 = 23. 029 = Cu The value of(4d , Bd ) at Node d (down state): 4d Sd + Bd = 0.3738 (32.9033) − 9.1107 = 3. 188 = Cd Finally, let’s verify that the portfolio replicates the terminal payoff. First, we need to find the replicating portfolio at the expiration date. Period 1 2 (4uu , Buu ) = (1.0, −40) (4u , Bu ) = (1.0, −36.9247) (4ud , Bud ) = (1.0, −40) = (0.3738, −9. 869 5) (4d , Bd ) = (0.3738, −9.1107) (4dd , Bdd ) = (0.3738, −9. 869 5) If we reinvest dividends, 4u stocks grows into 4u eδh after h years; Bu grows into Bu erh after h¡ years. ¢ ¡ ¢ (4uu , Buu ) = 4u eδh , Bu erh = 1.0e0(1) , −36.9247e0.08(1) = (1.0, −40) 4uu Suu + Buu = 1.0 (87.6693) − 40 = 47. ¢ 669 ¡ 3 = Cuu ¢ ¡ Similarly, (4dd , Bdd ) = 4d eδh , Bd erh = 0.3738e0(1) , −9.1107e0.08(1) = (0.373 8, −9. 869 5) 4dd Sdd + Bdd = 0.3738 (26.4055) − 9. 869 5 = 0.00 = Cdd Finally, (4ud , Bud ) = (1.0, −40) = (0.3738, −9. 869 5) Portfolios (1.0, −40) and (0.3738, −9. 869 5) have equal values at Node ud. 1.0 (48.1139) − 40 = 8. 113 9 = Cud 0.3738 (48.1139) − 9. 869 5 = 8. 115 = Cud (ignore rounding difference)

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

53

Tip 10.2.3. For a multi-binomial tree, using the risk neutral probability to find the premium is faster than using the replicating portfolio. The risk neutral probabilities π u and π d are constant cross nodes. However, the replicating portfolio (4, B) varies by node. Tip 10.2.4. For European options, you can calculate the premium using the terminal payoffs. This is how to quickly find the premium for this problem. Node at T = 2 uu ud dd Total

Payoff at T Cuu = 47.6693 Cud = 8.1139 Cdd = 0

Risk neutral probability of reaching this node1 π 2u = 0.425 562 = 0.181 1 2π u πd = 2 (0.425 56) (0.574 44) = 0.488 9 2 π 2d = 0.574 442 = 0.3300 2 (πu + π d ) = 0.181 1 + 0.488 9 + 0.3300 = 1 3

The premium is the expected present value of the terminal payoff using the risk neutral probability. ¡ ¢ C = e−rT π2u Cuu + 2π u πd Cud + π 2d Cdd ¢ ¡ = e−0.08(2) 0.425 562 × 47.6693 + 0.488 9 × 8.1139 + 0.33 × 0 = 10. 736 9 Tip 10.2.5. If you purchased the textbook Derivatives Markets, you should see a CD attached to the back cover of the book. Install the CD in your computer. Run the spreadsheet titled "optall2’" or "optbasic2." These two spreadsheets can calculate European and American option prices. When you solve a practice problem, you can use either of these two spreadsheets to double check you answer. Please note that these two spreadsheets don’t calculate the replicating portfolio (4, B). So you can’t use them to verify your calculation of the replicating portfolio. Example 10.2.8. Let’s reproduce Derivatives Markets Figure 10.5. Here is the recap of the information on a European call. The current stock price is 41. The strike price K = 40. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded dividend rate per year is δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. Solution. Each period is h = u = e(r−δ)h+σ

1 year long. 3

√ h

√ = e(0.08−0)1/3+0.3 1/3 = 1. 221 246

probabilities in this column are the 3 terms of (π u + πd )2 = π 2d + 2π d π u + π 2u are two ways of reaching Node ud: up and dow or down and up. 3 The total probability is one. Otherwise, you made an error. 1 The

2 There

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54

CHAPTER 10. BINOMIAL OPTION PRICING: I √

d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 Stock price Period 0 1 2

√

1/3

= 0.863 693 3 Su3 = 74.6781

Su2 = 61.1491 Su2 d = 52.8140

Su = 50.0711 S = 41

Sud = 43.2460 Sd2 u = 37.3513

Sd = 35.4114 2

Sd = 30.5846 Sd3 = 26.4157 Calculate the premium by working backward from right to left. Period 0

1

2 Cu2 = 22. 201 6

Cu = 12. 889 5 C = 7. 073 9

Cud = 5. 699 5 Cd = 2. 535 1 Cd2 = 0

−rh

3 Cu3 = max (0, 74.6781 − 40) = 34. 678 1 Cu2 d = max (0, 52.8140 − 40) = 12. 814 Cd2 u = max (0, 37.3513 − 40) = 0 Cd3 = max (0, 26.4157 − 40) = 0

C=e (π u Cu + π d Cd ) e(r−δ)h − d e(0.08−0)1/3 − 0.863 693 πu = = = 0.456 806 u−d 1. 221 246 − 0.863 693 πd = 1 − π u = 1 − 0.456 806 = 0.543 194

Cu2 = e−rh (π u Cu3 + πd Cu2 d ) = e−0.08(1/3) (0.456 806 × 34. 678 1 + 0.543 194 × 12. 814) = 22. 201 7 Cud = e−rh (π u Cu2 d + π d Cd2 u ) = e−0.08(1/3) (0.456 806 × 12. 814 + 0.543 194 × 0) = 5. 699 5 Cd2 = e−rh (πCd2 u + πd Cd3 ) = 0 Cu = e−rh (π u Cu2 + π d Cud ) = e−0.08(1/3) (0.456 806 × 22. 201 6 + 0.543 194 × 5. 699 5) = 12. 889 5 Cd = e−rh (π u Cud + π d Cd2 ) = e−0.08(1/3) (0.456 806 × 5. 699 5 + 0.543 194 × 0) = 2. 535 1 C = e−rh (πu Cu + πd Cd ) = e−0.08(1/3) (0.456 806 × 12. 889 5 + 0.543 194 × 2. 535 1) = 7. 073 9

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

55

Now I’m going to tell you a calculator shortcut I used when I was preparing for the old Course 6. The above calculations are intense and prone to errors. To quickly and accurately calculate the premium at each node, use TI-30X IIS calculator because TI-30X IIS allows you to modify formulas easily. For example, to calculate Cu2 = e−rh (π u Cu3 + π d Cu2 d ), enter eˆ(−0.08/3)(0.456 806 × 34. 678 1 + 0.543 194 × 12. 814 Please note for TI-30 IIS, the above expression is the same as eˆ(−0.08/3)(0.456 806 × 34. 678 1 + 0.543 194 × 12. 814) In other words, you can omit the ending parenthesis ")". I tell you this because occasionally people emailed me saying they discovered a typo. This is not a typo. Now you have entered the formula eˆ(−0.08/3)(0.456 806 × 34. 678 1 + 0.543 194 × 12. 814. Press "=" and you should get: 22.20164368 Next, to calculate Cud = e−rh (πud Cu2 d + πd Cd2 u ), you don’t need to enter a brand new formula. Just reuse the formula eˆ(−0.08/3) (0.456 806 × 34. 678 1 + 0.543 194 × 12. 814) Change 34. 678 1 to 12. 8140 (so 0.456 806 ×34. 678 1 becomes 0.456 806 × 12. 8140). Change 12. 814 into 00.000 (so 0.543 194×12. 814 becomes 0.543 194×00.000). Now the modified formula is eˆ(−0.08/3) (0.456 806 × 12. 8140 + 0.543 194 × 00.000) Press "=" and you should get: 5.6994813 To calculate Cu = e−rh (πu Cu2 + π d Cud ), once again reuse a previous formula. Change the formula eˆ(−0.08/3) (0.456 806 × 12. 8140 + 0.543 194 × 00.000) into eˆ(−0.08/3) (0.456 806 × 22. 201 6 + 0.543 194 × 05. 699 5) Press "=" and you should get:12.8894166 Reusing formulas avoids the need to retype e−rh , π u , and π d (these three terms are constant across all nodes) and increases your speed and accuracy. By reusing formulas, you should quickly find C = 7. 073 9.

Yufeng Guo, Fall 09 MFE, actuary88.com

56

CHAPTER 10. BINOMIAL OPTION PRICING: I Next, we calculate the replicating portfolio. Cu − Cd Su Cd − Sd Cu 4 = e−δh B = e−rh Su − Sd Su − Sd To avoid errors, put the stock price tree and the premium tree side by side: Period 0

1

2

3 Su3 = 74.6781

Su2 = 61.1491 Su2 d = 52.8140

Su = 50.0711 S = 41

Sud = 43.2460 Sd2 u = 37.3513

Sd = 35.4114 2

Sd = 30.5846 Sd3 = 26.4157 Period 0

1

2

3 Cu3 = 34. 678 1

Cu2 = 22. 201 6 Cu = 12. 889 5 C = 7. 073 9

Cu2 d = 12. 814 Cud = 5. 699 5

Cd = 2. 535 1

Cd2 u = 0 Cd2 = 0 Cd3 = 0

Period 0

(4, B) = (0.706 3, −21. 885 2)

1

(4, B)u = (0.921 8, −33. 263 6) (4, B)d = (0.450 1, −13. 405 2)

2 (4, B)u2 = (1, −38. 947 4) (4, B)ud = (0.828 7, −30. 138 6) (4, B)d2 = (0, 0)

Cu3 − Cu2 d 34. 678 1 − 12. 814 = e−0(1/3) =1 Su3 − Su2 d 74.6781 − 52.8140 Su3 Cu2 d − Su2 dCu2 d 74.6781 (12. 814) − 52.8140 (34. 678 1) = e−0.08(1/3) = Bu2 = e−rh Su3 − Su2 d 74.6781 − 52.8140 −38. 947 4 4u2 = e−δh

Cu2 d − Cd2 u 12. 814 − 0 = e−0(1/3) = 0.828 7 2 2 Su d − Sd u 52.8140 − 37.3513 2 2 Su dCd2 u − Sd uCu2 d 52.8140 (0) − 37.3513 (12. 814) = e−0.08(1/3) = Bud = e−rh 2 2 Su d − Sd u 52.8140 − 37.3513 −30. 138 6 4ud = e−δh

4d2 = e−δh

Cd2 u − Cd3 =0 Su2 d − Sd3

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Bd2 = e−rh

57

Su2 dCd3 − Sd3 Cd2 u =0 Su2 d − Sd3

Cu2 − Cud 22. 201 6 − 5. 699 5 = e−0(1/3) = 0.921 8 Su2 − Sud 61.1491 − 43.2460 Su2 Cud − SudCu2 61.1491 (5. 699 5) − 43.2460 (22. 201 6) Bu = e−rh = e−0.08(1/3) = 2 Su − Sud 61.1491 − 43.2460 −33. 263 6 4u = e−δh

Cud − Cd2 5. 699 5 − 0 = e−0(1/3) = 0.450 1 Sud − Sd2 43.2460 − 30.5846 SudCd2 − Sd2 Cud 43.2460 (0) − 30.5846 (5. 699 5) = e−0.08(1/3) = Bd = e−δh 2 Sud − Sd 43.2460 − 30.5846 −13. 405 2 4d = e−δh

Cu − Cd 12. 889 5 − 2. 535 1 = e−0(1/3) = 0.706 3 Su − Sd 50.0711 − 35.4114 Cu − Cd 50.0711 (2. 535 1) − 35.4114 (12. 889 5) B = e−rh = e−0.08(1/3) = Su − Sd 50.0711 − 35.4114 −21. 885 2 4 = e−δh

If our goal is to calculate the premium without worrying about the replicating portfolio, then we just need to know the risk neutral probability and the terminal payoff. Node u3 u2 d ud2 d3

Payoff Cu3 = 34. 678 1 Cu2 d = 12. 814 Cd2 u = 0 Cd3 = 0

Risk neutral prob of reaching this node4 3 π 3u¡= 0.456 806 ¢ 2 2 3π u πd = 3 0.456 806 0.543 194 3π u π 2d = 3 (0.456 806) 0.543 1942 π 3d = 0.543 1943

The option premium is just the expected present value of the terminal payoffs using the risk¡neutral probability. ¢ C = e−rT Cu3 π 3u + Cu2 d × 3π 2u πd + Cd2 u × 3πu π 2d + Cd3 π 3d ¢ ¡ = e−0.08(1) 34. 678 1 × 0.456 8063 + 12. 814 × 3 × 0.456 8062 × 0.543 194 = 7. 073 9

Example 10.2.9. Let’s reproduce Derivatives Markets Figure 10.6. Here is the recap of the information on a European put. The current stock price is 41. The strike price K = 40. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded dividend rate per year is δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. 4 The probabilities of this column is just the four terms in (π + π )3 = π 3 + 3π π 2 + u d d u u 3π 2d π u + π 3d = 1.

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CHAPTER 10. BINOMIAL OPTION PRICING: I

Solution.

Each period is h =

1 year long. 3

√ √ u = e(r−δ)h+σ√ h = e(0.08−0)1/3+0.3√1/3 = 1. 221 246 d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 1/3 = 0.863 693 Stock price Period 0 1 2 3 Su3 = 74.6781 Su2 = 61.1491 Su = 50.0711 Su2 d = 52.8140 S = 41 Sud = 43.2460 Sd = 35.4114 Sd2 u = 37.3513 2 Sd = 30.5846 Sd3 = 26.4157 Calculate the premium by working backward from right to left. Period 0

1

2 Cu2 = 0

Cu = 0.740 9 C = 2. 998 5

Cud = 1. 400 9 Cd = 5. 046 2 Cd2 = 8. 362 9

3 Cu3 = max (0, 40 − 74.6781) = 0 Cu2 d = max (0, 40 − 52.8140) = 0 Cd2 u = max (0, 40 − 37.3513) = 2. 648 7 Cd3 = max (0, 40 − 26.4157) = 13. 584 3

C = e−rh (π u Cu + π d Cd ) e(r−δ)h − d e(0.08−0)1/3 − 0.863 693 πu = = = 0.456 806 u−d 1. 221 246 − 0.863 693 πd = 1 − π u = 1 − 0.456 806 = 0.543 194 0

Cu2 = e−rh (π u Cu3 + πd Cu2 d ) = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 0) =

Cud = e−rh (π u Cu2 d + π d Cd2 u ) = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 2. 648 7) = 1. 400 9 Cd2 = e−rh (πu Cd2 u + πd Cd3 ) = e−0.08(1/3) (0.456 806 × 2. 648 7 + 0.543 194 × 13. 584 3) = 8. 362 9 Cu = e−rh (π u Cu2 + π d Cud ) = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 1. 400 9) = 0.740 9 Cd = e−rh (π u Cud + π d Cd2 ) = e−0.08(1/3) (0.456 806 × 1. 400 9 + 0.543 194 × 8. 362 9) = 5. 046 2

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

59

C = e−rh (πu Cu + π d Cd ) = e−0.08(1/3) (0.456 806 × 0.740 9 + 0.543 194 × 5. 046 2) = 2. 998 5 Next, we calculate the replicating portfolio. Cu − Cd Su Cd − Sd Cu B = e−rh 4 = e−δh Su − Sd Su − Sd Period 0

1

2 (4, B)u2 = (0, 0)

(4, B)u = (−0.07 82, 4. 658 9) (4, B) = (−0.293 7, 15. 039 5)

(4, B)ud = (−0.171 3, 8. 808 8) (4, B)d = (−0.549 9, 24. 517 3) (4, B)d2 = (1, 38. 947 4)

Cu3 − Cu2 d 0−0 = e−0(1/3) =0 Su3 − Su2 d 74.6781 − 52.8140 Su3 Cu2 d − Su2 dCu2 d 74.6781 (0) − 52.8140 (0) = e−rh = e−0.08(1/3) =0 Su3 − Su2 d 74.6781 − 52.8140

4u2 = e−δh Bu2

Cu2 d − Cd2 u 0 − 2. 648 7 = e−0(1/3) = −0.171 3 2 2 Su d − Sd u 52.8140 − 37.3513 2 2 Su dCd2 u − Sd uCu2 d 52.8140 (2. 648 7) − 37.3513 (0) = e−rh = e−0.08(1/3) = 2 2 Su d − Sd u 52.8140 − 37.3513

4ud = e−δh Bud 8. 808 8

Cd2 u − Cd3 2. 648 7 − 13. 584 3 = e−0(1/3) = −1.0 Su2 d − Sd3 37.3513 − 26.4157 2 3 Su dCd3 − Sd Cd2 u 37.3513 (13. 584 3) − 26.4157 (2. 648 7) = e−0.08(1/3) = Bd2 = e−rh Su2 d − Sd3 37.3513 − 26.4157 38. 947 4 C 2 − Cud 0 − 1. 400 9 = e−0(1/3) = −0.07 82 4u = e−δh u2 Su − Sud 61.1491 − 43.2460 2 Su Cud − SudCu2 61.1491 (1. 400 9) − 43.2460 (0) = e−0.08(1/3) = Bu = e−rh 2 Su − Sud 61.1491 − 43.2460 4. 659 4d2 = e−δh

Cud − Cd2 1. 400 9 − 8. 362 9 = e−0(1/3) = −0.549 9 Sud − Sd2 43.2460 − 30.5846 SudCd2 − Sd2 Cud 43.2460 (8. 362 9) − 30.5846 (1. 400 9) = e−0.08(1/3) = Bd = e−δh Sud − Sd2 43.2460 − 30.5846 24. 517 3 4d = e−δh

Cu − Cd 0.740 9 − 5. 046 2 = e−0(1/3) = −0.293 7 Su − Sd 50.0711 − 35.4114 Cu − Cd 50.0711 (5. 046 2) − 35.4114 (0.740 9) = e−0.08(1/3) = 15. B = e−rh Su − Sd 50.0711 − 35.4114 039 5 4 = e−δh

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CHAPTER 10. BINOMIAL OPTION PRICING: I

If our goal is to calculate the premium without worrying about the replicating portfolio, then we just need to know the risk neutral probability and the terminal payoff. Node u3 u2 d ud2 d3

Payoff Cu3 = 0 Cu2 d = 0 Cd2 u = 2. 648 7 Cd3 = 13. 584 3

Risk neutral prob of reaching this node5 3 π3u¡= 0.456 806 ¢ 2 2 3πu π d = 3 0.456 806 0.543 194 3π u π2d = 3 (0.456 806) 0.543 1942 π3d = 0.543 1943

The option premium is just the expected present value of the terminal payoffs using the risk¡neutral probability. ¢ C = e−rT Cd2 u × 3π u π 2d + Cd3 π 3d ¡ ¢ = e−0.08(1) 2. 648 7 × 3 × 0.456 806 × 0.543 1942 + 13. 584 3 × 0.543 1943 = 2. 998 5

Example 10.2.10. Let’s reproduce Derivatives Markets Figure 10.7. Here is the recap of the information on an American put. The current stock price is 41. The strike price K = 40. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded dividend rate per year is δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. Solution. Each period is h =

1 year long. 3

√ √ u = e(r−δ)h+σ√ h = e(0.08−0)1/3+0.3√1/3 = 1. 221 246 d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 1/3 = 0.863 693 Step 1 Find the stock price tree Period 0 1 2 3 3 Su = 74.6781 Su2 = 61.1491 Su = 50.0711 Su2 d = 52.8140 S = 41 Sud = 43.2460 Sd = 35.4114 Sd2 u = 37.3513 2 Sd = 30.5846 Sd3 = 26.4157 5 The probabilities of this column is just the four terms in (π + π )3 = π 3 + 3π π 2 + u d d u u 3π 2d π u + π3d = 1.

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Step 2 Period 0

61

Calculate the payoff at expiration (terminal payoff) 1

2

3 Vu3 = max (0, 40 − 74.6781) = 0

Vu2 Vu V

Vu2 d = max (0, 40 − 52.8140) = 0

Vud Vd Vd2

Vd2 u = max (0, 40 − 37.3513) = 2. 648 7 Vd3 = max (0, 40 − 26.4157) = 13. 584 3

Step 3 Calculate the value of the American put one step left to the expiration. An American put can be exercised immediately. The value of an American option if exercised immediately is called the exercise value (EV ) or intrinsic value. At Period 2, we compare the value calculated using backwardization and the exercise value. We take the greater of the two as the value of the American put.

0

The backwardized values at Period 2 are: Cu2 = e−rh (πu Cu3 + π d Cu2 d ) = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 0) =

Cud = e−rh (πu Cu2 d + πd Cd2 u ) = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 2. 648 7) = 1. 400 9 Cd2 = e−rh (π u Cd2 u + πd Cd3 ) = e−0.08(1/3) (0.456 806 × 2. 648 7 + 0.543 194 × 13. 584 3) = 8. 362 9 The exercise values at Period 2 are: ¡ ¢ EVu2 = max 0, K − Su2 = max (0, 40 − 61.1491) = 0 EVud = max ¡(0, K − Sud) ¢ = max (0, 40 − 43.2460) = 0 EVd2 = max 0, K − Sd2 = max (0, 40 − 30.5846) = 9. 415 4 We take the greater of the two as the value of the American put. Vu2 = max (Cu2 , EVu2 ) = max (0, 0) = 0 Vud = max (Cud , EVud ) = max (1. 400 9, 0) = 1. 400 9 Vd2 = max (Cd2 , EVd2 ) = max (8. 362 9, 9. 415 4) = 9. 415 4 Now we have: Period 0 1

2

3 Vu3 = 0

Vu2 = 0 Vu V

Vu2 d = 0 Vud = 1. 400 9

Vd

Vd2 u = 2. 648 7 Vd2 = 9. 415 4 Vd3 = 13. 584 3

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CHAPTER 10. BINOMIAL OPTION PRICING: I

Step 3 Move one step to the left. Repeat Step 2. Compare the backwardized value and the exercise value. Choose the greater. The backwardized values are: Cu = e−0.08(1/3) (0.456 806 × 0 + 0.543 194 × 1. 400 9) = 0.740 9 Cu = e−0.08(1/3) (0.456 806 × 1. 400 9 + 0.543 194 × 9. 415 4) = 5. 602 9 The exercise value at Period 1 is: EVu = max (0, K − Su ) = max (0, 40 − 50.0711) = 0 EVd = max (0, K − Sd ) = max (0, 40 − 35.4114) = 4. 588 6 Vu = max (Cu , EVu ) = max (0.740 9, 0) = 0.740 9 Vud = max (Cd , EVd ) = max (5. 602 9, 4. 588 6) = 5. 602 9 Hence we have: Period 0

1

2

3 Vu3 = 0

Vu2 = 0 Vu = 0.740 9 V

Vu2 d = 0 Vud = 1. 400 9

Vd = 5. 602 9

Vd2 u = 2. 648 7 Vd2 = 9. 415 4

Vd3 = 13. 584 3 Step 4 Repeat Step 3. Move one step left. Compare the backwardized value and the exercise value. Choose the greater value. The backwardized value at t = 0 is: C = e−0.08(1/3) (0.456 806 × 0.740 9 + 0.543 194 × 5. 602 9) = 3. 292 9 The exercise value is: EV = max (0, K − S) = max (0, 40 − 41) = 0 Hence the premium for the American put is: V = max (C, EV )= max (3. 292 9, 0) = 3. 292 9

Now we have: Period 0

1

2

3 Vu3 = 0

Vu2 = 0 Vu = 0.740 9 V = 3. 292 9

Vu2 d = 0 Vud = 1. 400 9

Vd = 5. 602 9

Vd2 u = 2. 648 7 Vd2 = 9. 415 4 Vd3 = 13. 584 3

Step 5

Calculate the replicating portfolio at each node.

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL 4 = e−δh

Cu − Cd Su Cd − Sd Cu B = e−rh Su − Sd Su − Sd Period 0

63

1

2 (4, B)u2 = (0, 0)

(4, B)u = (−0.07 82, 4. 658 9) (4, B) = (−0.331 66, 16. 890 9)

(4, B)ud = (−0.171 3, 8. 808 8) (4, B)d = (−0.632 99, 28. 017 8) (4, B)d2 = (1, 38. 947 4)

Vu3 − Vu2 d 0−0 = e−0(1/3) =0 3 2 Su − Su d 74.6781 − 52.8140 Su3 Vu2 d − Su2 dVu2 d 74.6781 (0) − 52.8140 (0) = e−rh = e−0.08(1/3) =0 3 2 Su − Su d 74.6781 − 52.8140

4u2 = e−δh Bu2

Vu2 d − Vd2 u 0 − 2. 648 7 = e−0(1/3) = −0.171 3 2 2 Su d − Sd u 52.8140 − 37.3513 2 2 Su dVd2 u − Sd Vu2 d 52.8140 (2. 648 7) − 37.3513 (0) = e−rh = e−0.08(1/3) = 2 2 Su d − Sd u 52.8140 − 37.3513

4ud = e−δh Bud 8. 808 8

Vd2 u − Vd3 2. 648 7 − 13. 584 3 = e−0(1/3) = −1.0 Su2 d − Sd3 37.3513 − 26.4157 Su2 dVd3 − Sd3 Vd2 u 37.3513 (13. 584 3) − 26.4157 (2. 648 7) = e−0.08(1/3) Bd2 = e−rh = Su2 d − Sd3 37.3513 − 26.4157 38. 947 4 V 2 − Vud 0 − 1. 400 9 = e−0(1/3) = −0.07 82 4u = e−δh u2 Su − Sud 61.1491 − 43.2460 Su2 Vud − SudVu2 61.1491 (1. 400 9) − 43.2460 (0) Bu = e−rh = e−0.08(1/3) = Su2 − Sud 61.1491 − 43.2460 4. 659 4d2 = e−δh

Vud − Vd2 1. 400 9 − 9. 415 4 = e−0(1/3) = −0.632 99 2 Sud − Sd 43.2460 − 30.5846 2 SudVd2 − Sd Vud 43.2460 (9. 415 4) − 30.5846 (1. 400 9) = e−0.08(1/3) Bd = e−δh = 2 Sud − Sd 43.2460 − 30.5846 28. 017 8 Vu − Vd 0.740 9 − 5. 602 9 4 = e−δh = e−0(1/3) = −0.331 66 Su − Sd 50.0711 − 35.4114 Vu − Vd 50.0711 (5. 602 9) − 35.4114 (0.740 9) B = e−rh = e−0.08(1/3) = 16. Su − Sd 50.0711 − 35.4114 890 9 4d = e−δh

Please note that for an American option, you can’t use the following approach to find the option price:

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Node Payoff Risk neutral prob of reaching this node 3 u3 Vu3 = 0 π3u¡= 0.456 806 ¢ 2 2 2 u d Vu2 d = 0 3πu π d = 3 0.456 806 0.543 194 ud2 Vd2 u = 2. 648 7 3π u π2d = 3 (0.456 806) 0.543 1942 3 d Vd3 = 13. 584 3 π3d = 0.543 1943 ¡ ¢ V = e−rT Cd2 u × 3π u π2d + Cd3 π3d ¡ ¢ = e−0.08(1) 2. 648 7 × 3 × 0.456 806 × 0.543 1942 + 13. 584 3 × 0.543 1943 = 2. 998 5 This approach is wrong because it ignores the possibility that the American option can be exercised early.

10.2.2

Options on stock index

The price of an option on stock index can be calculated the same way as the price of an option on a stock is calculated. Example 10.2.11. Let’s reproduce Derivatives Markets Figure 10.8. Here is the recap of the information on an American call. The current stock index is 110. The strike price K = 100. The annualized standard deviation of the continuously compounded stock index return is σ = 30%. The continuously compounded risk-free rate per year is r = 5%. The continuously compounded dividend rate per year is δ = 3.5%.The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. Solution. Each period is h =

1 year long. 3

√ √ u = e(r−δ)h+σ√ h = e(0.05−0.035)1/3+0.3√1/3 = 1. 195 07 d = e(r−δ)h−σ h = e(0.05−0.035)1/3−0.3 1/3 = 0.845 18 e(r−δ)h − d e(0.05−0.035)1/3 − 0.845 18 πu = = = 0.456 807 u−d 1. 195 07 − 0.845 18 πd = 1 − π u = 1 − 0.456 807 = 0.543 193 Step 1 Find the stock price tree Period 0 1 2 3 Su3 = 187.7471 Su2 = 157.1013 Su = 131.4577 Su2 d = 132.7789 S = 110 Sud = 111.1055 Sd = 92.9699 Sd2 u = 93.9042 2 Sd = 78.5763 Sd3 = 66.4112 Step 2

Calculate the payoff at expiration (terminal payoff)

Yufeng Guo, Fall 09 MFE, actuary88.com

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Period 0

1

2 Vu2

Vu V

Vud

65

3 Vu3 = max (0, 187.7471 − 100) = 87. 747 1 Vu2 d = max (0, 132.7789 − 100) = 32. 778 9

Vd

Vd2 u = max (0, 93.9042 − 100) = 0

Vd2

Vd3 = max (0, 66.4112 − 100) = 0 Step 3 Calculate the value of the American call at Period 2 by taking the greater of the backwardized value and the exercise value at each node. The backwardized values at Period 2 are: Cu2 = e−rh (πu Vu3 + π d Vu2 d ) = e−0.05(1/3) (0.456 807 × 87. 747 1 + 0.543 193 × 32. 778 9) = 56. 931 9 Cud = e−rh (πu Vu2 d + πd Vd2 u ) = e−0.05(1/3) (0.456 807 × 32. 778 9 + 0.543 193 × 0) = 14. 726 1 Cd2 = e−rh (π ud Vd2 u + π d Vd3 ) = e−0.05(1/3) (0.456 807 × 0 + 0.543 193 × 0) = 0 The exercise values at Period 2 are: ¡ ¢ EVu2 = max 0, Su2 − K = max (0, 157.1013 − 100) = 57. 101 3 EVud = max ¡(0, Sud − K) ¢ = max (0, 111.1055 − 100) = 11. 105 5 EVd2 = max 0, Sd2 − K = max (0, 78.5763 − 100) = 0 We take the greater of the two as the values. Vu2 = max (Cu2 , EVu2 ) = max (56. 931 9, 57. 101 3) = 57. 101 3 Vud = max (Cud , EVud ) = max (14. 726 1, 11. 105 5) = 14. 726 1 Vd2 = max (Cd2 , EVd2 ) = max (0, 0) = 0 Now we have: Period 0 1

2

3 Vu3 = 87. 747 1

Vu2 = 57. 101 3 Vu V

Vu2 d = 32. 778 9 Vud = 14. 726 1

Vd

Vd2 u = 0 Vd2 = 0 Vd3 = 0

Step 4 Calculate the value of the American call at Period 1 by taking the greater of the backwardized value and the exercise value at each node. Cu = e−rh (πu Vu2 + πd Vud ) = e−0.05(1/3) (0.456 807 × 57. 101 3 + 0.543 193 × 14. 726 1) = 33. 520 02

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Cd = e−rh (π u Vud + π d Vd2 ) = e−0.05(1/3) (0.456 807 × 14. 726 1 + 0.543 193 × 0) = 6. 615 8 EVu = max (0, Su − K) = max (0, 131.4577 − 100) = 31. 457 7 EVd = max (0, Sd − K) = max (0, 92.9699 − 100) = 0 Vu = max (Cu , EVu ) = max (33. 520 02, 31. 457 7) = 33. 520 02 Vd = max (Cd , EVd ) = max (6. 615 8, 0) = 6. 615 8 Now we have: Period 0

1

2

3 Vu3 = 87. 747 1

Vu2 = 57. 101 3 Vu = 33. 520 02 V

Vu2 d = 32. 778 9 Vud = 14. 726 1

Vd = 6. 615 8

Vd2 u = 0 Vd2 = 0 Vd3 = 0

Step 5 Calculate the value of the American call at Period 0 by taking the greater of the backwardized value and the exercise value at each node. C = e−rh (π u Vu + π d Vd ) = e−0.05(1/3) (0.456 807 × 33. 520 02 + 0.543 193 × 6. 615 8) = 18. 593 35 EV = max (0, S − K) = max (0, 110 − 100) = 10 V = max (C, EV ) = max (18. 593 35, 10) = 18. 593 35 Step 6 Calculate the replicating portfolio Our goal is to replicate the following values: Period 0 1 2

3 Vu3 = 87. 747 1

Vu2 = 57. 101 3 Vu = 33. 520 02 V = 18. 593 35

Vu2 d = 32. 778 9 Vud = 14. 726 1

Vd = 6. 615 8

Vd2 u = 0 Vd2 = 0 Vd3 = 0

Cu − Cd Su − Sd Period 0

4 = e−δh

4 = 0.690 923

B = e−rh 1

Su Cd − Sd Cu Su − Sd 2

4u = 0.910 598 4d = 0.447 45

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4u2 = 0.988 401 4ud = 0.988 401 4d2 = 0

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Period 0

B = −57. 408 1

1

Bu = −86. 185 1 Bd = −34. 983 9

67

2 Bu2 = −98. 347 1 Bud = −77. 870 8 Bd2 = 0

Vu3 − Vu2 d 87. 747 1 − 32. 778 9 = e−0.035(1/3) = 0.988 401 3 2 Su − Su d 187.7471 − 132.7789 3 2 Su Vu2 d − Su dVu3 187.7471 (32. 778 9) − 132.7789 (87. 747 1) = e−0.05(1/3) = Bu2 = e−rh 3 2 Su − Su d 187.7471 − 132.7789 −98. 347 1 4u2 = e−δh

Vu2 d − Vud2 32. 778 9 − 0 = e−0.035(1/3) = 0.988 401 Su2 d − Sud2 187.7471 − 132.7789 3 2 Su dVu2 d − Su dVu3 187.7471 (0) − 132.7789 (32. 778 9) = e−0.05(1/3) = Bud = e−rh Su3 − Su2 d 187.7471 − 132.7789 −77. 870 8 4ud = e−δh

Vd2 u − Vd3 =0 Sd2 u − Sd3 Sd2 uVd3 − Sd3 Vd2 u = e−rh =0 Sd2 u − Sd3

4d2 = e−δh Bd2

Vu2 − Vud 57. 101 3 − 14. 726 1 = e−0.035(1/3) = 0.910 598 2 Su − Sud 157.1013 − 111.1055 2 Su Vud − SudVu2 157.1013 (14. 726 1) − 111.1055 (57. 101 3) = e−0.05(1/3) = Bu = e−rh 2 Su − Sud 157.1013 − 111.1055 −86. 185 1 4u = e−δh

Vud − Vd2 14. 726 1 − 0 = e−0.035(1/3) = 0.447 45 Sud − Sd2 111.1055 − 78.5763 2 Su Vud − SudVu2 111.1055 (0) − 78.5763 (14. 726 1) = e−0.05(1/3) = Bd = e−rh Su2 − Sud 111.1055 − 78.5763 −34. 983 9 4d = e−δh

Vu − Vd 33. 520 02 − 6. 615 8 = e−0.035(1/3) = 0.690 923 Su − Sd 131.4577 − 92.9699 SuVd − SdVu 131.4577 (6. 615 8) − 92.9699 (33. 520 02) B = e−rh = e−0.05(1/3) = Su − Sd 131.4577 − 92.9699 −57. 408 1 4 = e−δh

10.2.3

Options on currency

Now let’s find the price of a European call option on €1. The underlying asset is €1. The option expires in h years. The current dollar value of the underlying is

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S (so €1 = $S at t = 0). In h years the dollar value of the underlying asset €1 can go up to $Su or go down to $Sd . The strike price is $K. The continuously compounded risk-free interest rate on dollars is r per year (so dollars earn a return r). The continuously compounded interest rate on the underlying asset of €1 is δ per year (so euros earn a return δ). Let’s calculate the call price using one-period binomial tree. The asset price tree and the option payoff tree are as follows: Asset price tree (in dollars) Option payoff (in dollars) time 0

h Su

time 0

S

h Cu = max (0, Su − K)

C Sd Cd = max (0, Sd − K) To replicate the payoff, at t = 0 we’ll buy ∆ units of the underlying (i.e. buy ∆ euros) and simultaneously invest B dollars in a savings account. Since the underlying asset €∆ earns interest at a continuous interest δ, it will grow into €∆eδh at T , which is worth $∆eδh Su in the up state and $∆eδh Sd in the down state. We want our replicating portfolio to match the option payoff. So we have: ∆eδh Su Berh Cu 4S + B = C ∆eδh Sd Berh Cd t=0 t=h t=0 t=h t=0 t=h ½

∆eδh Su + Berh = Cu ∆eδh Sd + Berh = Cd Solving these equations, we get: Cu − Cd Su − Sd

(10.26)

Su Cd − Sd Cu Su − Sd

(10.27)

4 = e−δh B = e−rh

Equation 10.26 and 10.27 are exactly the same as Equation 10.13 and 10.14. This tells us that if we treat the currency as a stock and treat the euro return δ as the stock’s dividend rate, we can find the currency option’s price and the replicating portfolios using all the formulas available for a stock option. Example 10.2.12. Reproduce Derivatives Markets Figure 10.9. Here is the recap of the information on an American put option on €1. The current exchange rate is S = $1.05/€. The strike price is K = $1.10. The annualized standard deviation of the continuously compounded return on dollars is σ = 10%. The continuously compounded risk-free rate on dollars is r = 5% per year. The continuously compounded return on euros is δ = 3.1% per year. The option

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69

expiration date is T = 0.5 year. Use a 3-period binomial tree to calculate the option premium. Solution. The length period is h =

T 1 = 3 6

√ √ u = e(r−δ)h+σ√ h = e(0.055−0.031)1/6+0.1√1/6 = 1. 045 845 d = e(r−δ)h−σ h = e(0.055−0.031)1/6−0.1 1/6 = 0.963 845 e(r−δ)h − d e(0.055−0.031)1/6 − 0.963 845 πu = = = 0.489 795 u−d 1. 045 845 − 0.963 845 π d = 1 − πu = 1 − 0.489 795 = 0.510 205 Step 1 Find the underlying asset price tree Period 0 1 2 3 Su3 = 1.2011 Su2 = 1.1485 Su = 1.0981 Su2 d = 1.1070 S = 1.05 Sud = 1.0584 Sd = 1.0120 Sd2 u = 1.0202 2 Sd = 0.9754 Sd3 = 0.9402 Step 2 Period 0

Calculate the payoff at expiration (terminal payoff) 1

2 Vu2

Vu V

Vud Vd Vd2

3 Vu3 = max (0, 1.10 − 1.2011) = 0 Vu2 d = max (0, 1.10 − 1.1070) = 0 Vd2 u = max (0, 1.10 − 1.0202) = 0.079 8 Vd3 = max (0, 1.10 − 0.9402 ) = 0.159 8

Step 3 Calculate the value of the American put at Period 2 by taking the greater of the backwardized value and the exercise value at each node.

0

The backwardized values at Period 2 are: Cu2 = e−rh (πu Vu3 + π d Vu2 d ) = e−0.055(1/6) (0.489 795 × 0 + 0.510 205 × 0) =

Cud = e−rh (πu Vu2 d + πd Vd2 u ) = e−0.055(1/6) (0.489 795 × 0 + 0.510 205 × 0.079 8) = 0.04 034 Cd2 = e−rh (π ud Vd2 u + π d Vd3 ) = e−0.055(1/6) (0.489 795 × 0.079 8 + 0.510 205 × 0.159 8) = 0.119 5

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70

CHAPTER 10. BINOMIAL OPTION PRICING: I The exercise values at Period 2 are: ¡ ¢ EVu2 = max 0, K − Su2 = max (0, 1.10 − 1.1485) = 0 EVud = max ¡(0, K − Sud) ¢ = max (0, 1.10 − 1.0584) = 0.041 6 EVd2 = max 0, K − Sd2 = max (0, 1.10 − 0.9754) = 0.124 6

We take the greater of the two as the values. Vu2 = max (Cu2 , EVu2 ) = max (0, 0) = 0 Vud = max (Cud , EVud ) = max (0.04 034 , 0.041 6) = 0.041 6 Vd2 = max (Cd2 , EVd2 ) = max (0.119 5, 0.124 6) = 0.124 6 Now we have: Period 0 1

2

3 Vu3 = 0

Vu2 = 0 Vu V

Vu2 d = 0 Vud = 0.041 6

Vd

Vd2 u = 0.079 8 Vd2 = 0.124 6 Vd3 = 0.159 8

Step 4 Calculate the value of the American put at Period 1 by taking the greater of the backwardized value and the exercise value at each node. Cu = e−rh (π u Vu2 + π d Vud ) = e−0.055(1/6) (0.489 795 × 0 + 0.510 205 × 0.041 6) = 0.02 10 Cd = e−rh (π u Vud + π d Vd2 ) = e−0.055(1/6) (0.489 795 × 0.041 6 + 0.510 205 × 0.124 6) = 0.08 31 EVu = max (0, K − Su ) = max (0, 1.10 − 1.0981) = 0.001 9 EVd = max (0, K − Sd ) = max (0, 1.10 − 1.0120 ) = 0.088 Vu = max (Cu , EVu ) = max (0.02 10 , 0.001 9) = 0.021 Vd = max (Cd , EVd ) = max (0.088 , 0.021) = 0.088 Now we have: Period 0

1

2

3 Vu3 = 0

Vu2 = 0 Vu = 0.021 V

Vu2 d = 0 Vud = 0.041 6

Vd = 0.088

Vd2 u = 0.079 8 Vd2 = 0.124 6 Vd3 = 0.159 8

Step 5 Calculate the value of the American put at Period 0 by taking the greater of the backwardized value and the exercise value at each node.

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C = e−rh (πu Vu + πd Vd ) = e−0.055(1/6) (0.489 795 × 0.021 + 0.510 205 × 0.088) = 0.05 47 EV = max (0, K − S ) = max (0, 1.10 − 1.05) = 0.05 V = max (C, EV ) = max (0.05 47, 0.05) = 0.054 7 Now we have: Period 0

1

2

3 Vu3 = 0

Vu2 = 0 Vu = 0.021 V = 0.054 7

Vu2 d = 0 Vud = 0.041 6

Vd = 0.088

Vd2 u = 0.079 8 Vd2 = 0.124 6 Vd3 = 0.159 8

You can verify that the replicating portfolios are as follows: Period 0 1 2

4 = −0.7736

Period 0

4u = −0.4592 4d = −0.9948

1

4u2 = 0.0000 4ud = −0.9151 4d2 = −0.9948 2 Bu2 = $0.0000

Bu = $0.5253 B = $0.8669

Bud = $1.0089 Bd = $1.0900 Bd2 = $1.0900

10.2.4

Options on futures contracts

Suppose we want to find the price of a European call option on a stock futures contract. The underlying asset is futures. The option expires in h years. The current price of the underlying asset is F0,h = F , where F0,h is the price of a futures contract signed at t = 0 and expiring on date h.In h years the futures price can go up to Fh,h = Fu = F u or go down to Fh,h = Fd = F d, where Fh,h is the price of a futures contract signed at t = h and expiring on date h years (i.e. expiring immediately). Fh,h is equal to the spot price Sh . The fact that Fh,h can be either Fu or Fd is the same as fact that the stock price at t = h is

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CHAPTER 10. BINOMIAL OPTION PRICING: I

either Fu or Fd . The continuously compounded risk-free interest rate is r per year. Stocks pay dividend at a continuously compounded rate of δ per year. The strike price is K. The find the European call price on the futures, we draw the price tree of the underlying asset and the payoff tree. Asset price tree Option payoff time 0

h Fu = F u

time 0

F

h Cu = max (0, Fu − K)

C Fd = F d Cd = max (0, Fd − K) We form a replicating portfolio at t = 0 by entering ∆ futures contracts as a buyer and simultaneously putting $B in the savings account. Assume that no margin account is needed before one enters a futures contract. Then the cost of entering a futures contracts is zero. At the contract expiration date h, the ∆ futures contracts are settled in cash. If the futures price at expiration is Fu > F , then the seller in the futures pays ∆ (Fu − F ) to us, the buyer. If on the other hand, the futures price at h is Fd < F , then we pay the seller ∆ (F − Fd ). Paying ∆ (F − Fd ) is the same as receiving ∆ (Fd − F ). We assume that Fd < F < Fu holds so there’s no arbitrage. So the cash flow of the underlying asset (futures) is: time 0 h $∆ (Fu − F ) $0 $∆ (Fd − F ) We want the replicating portfolio ∆ (Fu − F ) 0 + B ∆ (Fd − F ) t=0 t=h t=0 ½

and the option have Berh = C Berh t=h t=0

the same payoff. Cu Cd t=h

∆ (Fu − F ) + Berh = Cu ∆ (Fd − F ) + Berh = Cd

Solving these equations, we get: Cu − Cd Cu − Cd = Fu − Fd F (u − d) µ ¶ 1−d u−1 −rh Cu + Cd B=e u−d u−d µ ¶ 1−d u−1 + Cd V = 4 × 0 + B = B = e−rh Cu u−d u−d 4=

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(10.28) (10.29) (10.30)

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL

73

Define

πu =

F − Fd 1−d = Fu − Fd u−d

(10.31)

πd =

Fu − 1 u−1 = Fu − Fd u−d

(10.32)

Then V = e−rh (Cu πu + Cd π d )

(10.33)

Equation 10.30 is the same as Equation 10.16 if we set r − δ = 0 and S = F . Consequently, we can find the price of a futures option by using Equation 10.16 Cu − Cd if we set δ = r. We just need to remember that 4 = instead of Fu − Fd¶ µ Cu − Cd 1−d u−1 and that B = V = e−rh Cu + Cd instead of 4 = e−rδ Fu − Fd u−d u−d Su Cd − Sd Cu . B = e−rh Su − Sd How we can specify u or d? u=

up price of Fh,h F0,h

u=

We know that Fh,h = Sh =

(

down price of Fh,h F0,h √

S0 e(r−δ)h+σ√h . In addition, F0,h = S0 e(r−δ)h . S0 e(r−δ)h−σ h √

√ up price of Fh,h S0 e(r−δ)h+σ h σ h u= = = e F0,h S0 e(r−δ)h √ √ down price of Fh,h S0 e(r−δ)h−σ h −σ h d= = = e F0,h S0 e(r−δ)h

u = eσ

√ h

d = e−σ

√ h

(10.34) (10.35)

Equation 10.34 and 10.35 are the same as Equation 10.24 and 10.25 if we set δ = r. We can use the stock option’s formula on u and d for futures options. Tip 10.2.6. To find the price of a futures option, just use the price formula for a stock option and set δ = r. However, remember that for aµ futures option, 4 = ¶ Cu − Cd Cu − Cd 1−d u−1 instead of 4 = e−rh and B = V = e−rh Cu + Cd Fu − Fd Fu − Fd u−d u−d Su Cd − Sd Cu . instead of B = e−rh Su − Sd

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CHAPTER 10. BINOMIAL OPTION PRICING: I

Example 10.2.13. Let’s reproduce Derivatives Markets Figure 10.10. Here is the recap of the information on an American call on a futures contract. The current futures price is S = 300. The strike price K = 300. The annualized standard deviation of the continuously compounded stock index return is σ = 10%. The continuously compounded risk-free rate per year is r = 5%.The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. Solution. We’ll reuse the stock option formula and set δ = r. 1 year long. 3 √ √ √ h = e0.1 1/3 u = e(r−δ)h+σ√ h = eσ √ √ = 1. 059 434 d = e(r−δ)h−σ h = e−σ h = e−0.1 1/3 = 0.943 900 e(r−δ)h − d 1−d 1 − 0.943 900 πu = = = = 0.485 57 u−d u−d 1. 059 434 − 0.943 900 πd = 1 − π u = 1 − 0.485 57 = 0.514 43 Each period is h =

Step 1 Find the underlying asset price tree Period 0 1 2

3 Su3 = 356.7330

Su2 = 336.7203 Su2 d = 317.8303

Su = 317.8303 F = 300

Sud = 300.0000 Sd2 u = 283.1700

Sd = 283.1700 2

Sd = 267.2842 Sd3 = 252.2895 Step 2 Period 0

Calculate the payoff at expiration (terminal payoff) 1

2 Vu2

Vu V

Vud Vd Vd2

3 Vu3 = max (0, 356.7330 − 300) = 56. 733 Vu2 d = max (0, 317.8303 − 300) = 17. 830 3 Vd2 u = max (0, 283.1700 − 300) = 0 Vd3 = max (0, 252.2895 − 300) = 0

Step 3 Calculate the value of the American call at Period 2 by taking the greater of the backwardized value and the exercise value at each node. The backwardized values at Period 2 are:

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75

Cu2 = e−rh (πu Vu3 + π d Vu2 d ) = e−0.05(1/3) (0.485 57 × 56. 733 + 0.514 43 × 17. 830 3) = 36. 113 3 Cud = e−rh (πu Vu2 d + πd Vd2 u ) = e−0.05(1/3) (0.485 57 × 17. 830 3 + 0.514 43 × 0) = 8. 514 7 Cd2 = e−rh (π ud Vd2 u + π d Vd3 ) = e−0.05(1/3) (0.485 57 × 0 + 0.514 43 × 0) = 0 The exercise values at Period 2 are: ¡ ¢ EVu2 = max 0, Su2 − K = max (0, 336.7203 − 300) = 36. 720 3 EVud = max (0, Sud − K) = max (0, 300.0000 − 300) = 0 ¡ ¢ EVd2 = max 0, Sd2 − K = max (0, 267.2842 − 300) = 0 We take the greater of the two as the values. Vu2 = max (Cu2 , EVu2 ) = max (36. 113 3, 36. 720 3) = 36. 720 3 Vud = max (Cud , EVud ) = max (8. 514 7, 0) = 8. 514 7 Vd2 = max (Cd2 , EVd2 ) = max (0, 0) = 0 Now we have: Period 0 1

2

3 Vu3 = 56. 733

Vu2 = 36. 720 3 Vu V

Vu2 d = 17. 830 3 Vud = 8. 514 7

Vd

Vd2 u = 0 Vd2 = 0 Vd3 = 0

Step 4 Calculate the value of the American call at Period 1 by taking the greater of the backwardized value and the exercise value at each node. Cu = e−rh (πu Vu2 + πd Vud ) = e−0.05(1/3) (0.485 57 × 36. 720 3 + 0.514 43 × 8. 514 7) = 21. 843 4 Cd = e−rh (π u Vud + π d Vd2 ) = e−0.05(1/3) (0.485 57 × 8. 514 7 + 0.514 43 × 0) = 4. 066 2 EVu = max (0, Su − K) = max (0, 317.8303 − 300) = 17. 830 3 EVd = max (0, Sd − K) = max (0, 283.1700 − 300) = 0 Vu = max (Cu , EVu ) = max (21. 843 4, 17. 830 3) = 21. 843 4 Vd = max (Cd , EVd ) = max (4. 066 2, 0) = 4. 066 2 Now we have:

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CHAPTER 10. BINOMIAL OPTION PRICING: I Period 0

1

2

3 Vu3 = 56. 733

Vu2 = 36. 720 3 Vu = 21. 843 4 V

Vu2 d = 17. 830 3 Vud = 8. 514 7

Vd = 4. 066 2

Vd2 u = 0 Vd2 = 0 Vd3 = 0

Step 5 Calculate the value of the American call at Period 0 by taking the greater of the backwardized value and the exercise value at each node. C = e−rh (π u Vu + π d Vd ) = e−0.05(1/3) (0.485 57 × 21. 843 4 + 0.514 43 × 4. 066 2) = 12. 488 4 EV = max (0, S − K) = max (0, 300 − 300) = 0 V = max (C, EV ) = max (12. 488 4, 0) = 12. 488 4 Now we have: Period 0 1 2 3 Vu3 = 56. 733 Vu2 = 36. 720 3 Vu = 21. 843 4 Vu2 d = 17. 830 3 V = 12. 488 4 Vud = 8. 514 7 Vd = 4. 066 2 Vd2 u = 0 Vd2 = 0 Vd3 = 0 Next, we need to find the replicating portfolio. Our goal is to replicate the following values: Period 0 1 2 3 Vu3 = 56. 733 Vu2 = 36. 720 3 Vu = 21. 843 4 Vu2 d = 17. 830 3 V = 12. 488 4 Vud = 8. 514 7 Vd = 4. 066 2 Vd2 u = 0 Vd2 = 0 Vd3 = 0 Using Equation 10.29 and 10.29, you should get: Period 0 1 2

4 = 0.5129

4u = 0.7681 4d = 0.2603

4u2 = 1 4ud = 0.5144

Yufeng Guo, Fall 09 MFE, actuary88.com

4d2 = 0

10.2. GENERAL ONE-PERIOD BINOMIAL MODEL Period 0

1

77

2 Bu2 = $36. 720 3

Bu = $21. 843 4 B = $12. 488 4

Bud = $8. 514 7 Bd = $4. 066 2 Bd2 = $0

For example, V 3 − Vu2 d 56. 733 − 17. 830 3 4u2 = u = = 1.0 Su3 − Su2 d 356.7330 − 317.8303 Bu2 = Vu2 = $36. 720 3 21. 843 4 − 4. 066 2 Vu − Vd = = 0.512 9 Su − Sd 317.8303 − 283.1700 B = V = $12. 488 4 I recommend that you reproduce my replicating portfolio in each node. 4=

Options on commodities The textbook is brief on this topic. So you don’t need to spend lot of time on it. This is the main idea: we can price commodity options using the same framework for pricing stock options if we can build a replicating portfolio using commodities and bonds with zero transaction cost. In reality, it’s hard to build a replicating portfolio using commodities. Unlike stocks, commodities such as corn may incur storage cost or other cost. It may be impossible to short sell commodities. As such, our ability to build a replicating portfolio is limited. However, if we can build any replicating portfolios using commodities instantaneously and effortlessly, commodity options and stock options are conceptually the same. We can use the same framework to calculate the price of a commodity option and the price of a stock option. Example 10.2.14. Here is the information on an American call on a commodity. The current commodity price is 110. The strike price is K = 100. The annualized standard deviation of the continuously compounded return on the commodity is σ = 30%. The continuously compounded risk-free rate per year is r = 5%. The continuously compounded lease rate of the commodity is δ = 3.5% per year. The option expiration date is T = 1 year. Use a 3-period binomial tree to calculate the option premium. Assume that you can effortlessly and instantaneously build any replicating portfolio using the commodity and the bond. We just treat the commodity as a stock. The lease rate δ = 3.5% is the same as the stock dividend rate. We can use the framework for pricing stock options to price this commodity option. The solution to this problem is in the textbook Figure 10.8.

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Options on bonds The textbook points out two major differences between bonds and stocks: 1. A bond’s volatility decreases over time as the bond approaches its maturity. A stock’s volatility doesn’t have this pattern. 2. When pricing a stock option, we assume that the interest rate is constant over time. The random variable is the stock price under a fixed interest. However, when pricing a bond, we can’t assume that the interest rate is constant any more. If the interest rate is constant, then the bond’s price is known with 100% certainty. If the bond price doesn’t change randomly, an option on the bond has zero value. Who wants to buy a call or put on an asset whose price is known with 100% certainty? Because of these differences, options on bonds should be treated differently from options on stocks. This is all you need to know about bond options right now. Chapter 24 will cover more on options on bonds.

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Chapter 11

Binomial option pricing: II 11.1

Understanding early exercise

Pros and cons of exercising an American call option early Pro: • Receive the stock and future dividends Cons • Pay the strike price early and lost interest that could have earned on the strike price. • Lose the insurance implicit in the call. If you hold the option, the stock price might be below the strike price at expiration, in which case you would not exercise the option. However, if you exercise the American call early, you lose the privilege of not exercising it. To understand this point, suppose you go to a garage sale and find a book you like that sells for only $1. You think "How cheap the book is. I must buy it." You pay $1 and buy the book. You think you get a good deal. However, if you resist the temptation to buy the book immediately and wait till the end of the garage sale, the book’s price may drop to $0.25. Better yet, you may even get the book for free. That same thing may happen when you exercise an American call early. At the moment, the stock price is high and you might be attempted to exercise the call. However, if we wait for a while, the stock price may drop below the strike price. Next, the textbook gives us a rule to determine when it’s optimal to exercise a perpetual 1 American call early. For a perpetual American call with zero volatility, it’s optimal to exercise a perpetual American call early if the dividend to be received exceeds the interest savings on the strike price: 1 The

textbook errata say the formulas work for an infinitely-lived American call option.

79

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80

CHAPTER 11. BINOMIAL OPTION PRICING: II δST > rK

¢ ¡ δ The annual dividend you ¶ gain if you exercise the call early is ST e − 1 = µ 1 ST 1 + δ + δ 2 + ... − 1 ≈ δST for a small δ. The annual interest earned on 2 µ ¶ 1 r K is K (e − 1) = K 1 + r + r2 + ... − 1 ≈ rK for a small r. If you early 2 exercise, you’ll pay K and receive ST ; during a year you’ll receive δST dividend but you will lose rK interest. Hence early exercise is optimal if the annual dividend exceeds the annual interest, δST > rK. And it’s optimal to defer exercising a perpetual American call if the interest savings on the strike price exceeds dividends lost: δST < rK Example 11.1.1. A perpetual American call option on a stock has a strike price $50. The stock pays dividend at a continuously compounded rate of 8% per year. The continuously compounded risk-free interest rate is 6% per year. The volatility of the stock price is zero. When is it optimal to exercise this American call option early? When is it optimal to defer exercise? Sδ > rK rK 0.06 (50) S> = = 37. 5 δ 0.08 Once the stock price becomes greater than 37. 5, then it’s optimal to exercise this perpetual American call option early. If the stock price is less than 37. 5, then it’s optimal to defer exercising this perpetual American call option early. Please note that zero volatility doesn’t mean that the stock price is a constant. It means that the stock price is known in advance with 100% certainty. Next, the textbook says that the decision to exercise a perpetual American call option early is complex if the volatility of the stock price is greater than zero. In this case, the insurance in the call option is greater than zero. For each non-zero volatility, there’s a lowest stock price at which the early exercise is optimal.

11.2

Understanding risk-neutral probability

Risk neutral probability is explained in the previous chapter. There’s not much new information about the risk neutral probability in this chapter. The key point to remember is that risk neutral probability is a shortcut or a math trick that enables us to quickly find the price of an option. We can use risk neutral probability to find the correct price of an option whether consumers are really risk neutral or not. The risk neutral probability is similar to the moment generating function in Exam P. The moment generating function (M GF ) is merely a math trick that

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81

allows us to quickly find the mean and variance of a random variable (hence the name "moment generating"). If we don’t use GM F , we can still find the mean and variance, but we have to work a lot harder. With the help of GM F , we can quickly find the mean and variance. Similarly, if we don’t use the risk neutral probability, we can still find the option price, but we have to work a lot harder. Once we use risk-neutral probability, we can quickly find an option’s price. Risk neutral probability is merely a math risk. By the way, one investment consultant told me that risk neutral probability is often hard to non-technical clients to understand. If you tell a non-technical client that an option price is calculated using risk-neutral probability and that the risk neutral probability not real, the client often immediately ask "So the price you calculated is wrong then?" It may take the consultant a while to explain why the risk neutral probability is not real yet the price is still correct.

11.2.1

Pricing an option using real probabilities

Next, the textbook answers a frequently asked question: Can we calculate the option price as the expected payoff using real probabilities of the stock price? The answer is "Yes if you know the discount rate." Let p and q = 1−p represent the real probability of stock going up and down respectively. Let γ represent the real discount rate (instead of the risk-free rate, which is used to discount payoff in the risk neutral world). uS with real probability p S t=0

Cu with real probability p C =?

dS with real probability q t=h

t=0

Cd with real probability q t=h

Then the price of a European option expiring in h years is: C = e−γh (pCu + qCd ) How can we find the real probability p and the real discount rate γ? Suppose we know that the expected return on the stock during [0, h] is α. Assume that the continuously compounded dividend rate is δ per year. If we have one stock at t = 0, then at t = h we’ll have eδh stocks. The value at t = 0 is the expected value at t = h discounted at rate α. uSeδh with real probability p S dSeδh with real probability q Value t = 0 Value t = h ¡ ¢ e(α−δ)h − d →p= S = e−αh puSeδh + qdSeδh u−d p=

e(α−δ)h − d u−d

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(11.1)

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CHAPTER 11. BINOMIAL OPTION PRICING: II

q=

u − e(α−δ)h u−d

(11.2)

Tip 11.2.1. If we set r = α Equation 10.16 and 10.17 become Equation 11.1 and 11.2. So you just need to memorize Equation 10.16 and 10.17. To get the formulas for the real probability, just set r = α. We can use the replicating portfolio to find γ. Suppose the replicating portfolio at t = 0 consists of ∆ shares of stock and putting $B in a savings account. We already know that we can calculate ∆ and B using Equation 10.13 and 10.13. At t = 0, the replicating portfolio is worth ∆S + B. At t = h, the replicating portfolio consists of ∆eδh shares of stock and $Berh in a savings account, which is worth ∆eδh dS + Berh in the up state and ∆eδh dS − Berh in the down state. uS∆eδh + Berh with real probability p ∆S + B Value t = 0

dS∆eδh + Berh with real probability q Value t = h

The value at value at t = h discounted at rate γ: ¢ ¡ ¢¤ £ t¡ = 0 is the expected ∆S + B = p uS∆eδh + Berh + q dS∆eδh + Berh e−γh £ ¡ ¤ ¢ = ∆ puSeδh + qdSeδh + Berh (p + q) e−γh ¡ ¢ = ∆Seαh + Berh e−γh e−γh =

∆S + B ∆Seαh + Berh

(11.3)

∆S + B (pCu + qCd ) ∆Seαh + Berh Since C = ∆S + B, we just need to prove that pCu + qCd = ∆Seαh + Berh .

C = e−γh (pCu + qCd ) =

pCu + qCd e(α−δ)h − d u − e(α−δ)h = Cu + C u−d u−d ∙ (r−δ)h ¸ ∙ ¸ (α−δ)h e e(r−δ)h − e(α−δ)h e u − e(r−δ)h −d − e(r−δ)h = Cu + Cu + Cd + Cd u−d u−d u−d u−d ¸ ∙ (r−δ)h e −d u − e(r−δ)h e(α−δ)h − e(r−δ)h = Cu + Cd + (Cu − Cd ) u−d u−d u−d According to Equation 10.15: u − e(r−δ)h e(r−δ)h − d Cu + Cd = erh (∆S + B) u−d u−d

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11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY

83

Cu − Cd = ∆eδh S u−d pCu + qCd £ ¤ = erh (∆S + B) + ∆eδh¡ S e(α−δ)h¢ − e(r−δ)h = erh (∆S + B) + ∆S eαh − erh = eαh ∆S + Berh According to Equation 10.13,

∆S + B (pCu + qCd ) = ∆S + B ∆Seαh + Berh The above derivation tell us that →C=

• Real probabilities lead to the same answer as the risk neutral probability • Any consistent pair of (α, γ) will produce the correct answer. The above derivation doesn’t require that α has to be reasonable or precise. Any α will generate a corresponding γ. Together, α and γ will produce the option price correctly. • The simplest calculation is to set α = γ = r. Setting α = γ = r means using risk neutral probabilities.

Example 11.2.1. Reproduce the textbook Figure 11.3 (which is the same as the textbook Figure 10.5). A European call option has strike price K = 40. The current price is S = 41. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded dividend rate per year is δ = 0%. The continuously compounded expected return on the stock per year is α = 15%. The option expiration date is T = 1 year. Use a 1-period binomial tree and real probabilities to calculate the option premium. Solution. √

√

u = e(r−δ)h+σ √h = e(0.08−0)1+0.3√1 = 1. 462 3 u = e(r−δ)h−σ h = e(0.08−0)1−0.3 1 = 0.802 5 1. 462 3 (41) = 59. 954 with real probability p 41 0.802 5 (41) = 32. 9023 with real probability q t=0 t=h=1

C =?

Cu = max (0, 59. 954 − 40) = 19. 954 with real probability p

t=0

Cd = max (0, 32. 9023 − 40) = 0 with real probability q t=h=1

Calculate the real probabilities: 41e0.15(1) = 59. 954p + 32. 9023 (1 − p)

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84

CHAPTER 11. BINOMIAL OPTION PRICING: II p = 0.544 6

q = 1 − p = 1 − 0.544 6 = 0.455 4

Calculate the replicating portfolio: Cu − Cd 19. 954 − 0 4 = e−δh = e−0(1) = 0.737 6 Su − Sd 59. 954 − 32. 9023 Su Cd − Sd Cu 59. 954 (0) − 32. 9023 (19. 954) = e−0.08(1) B = e−rh = −22. Su − Sd 59. 954 − 32. 9023 403 6 Calculate the ¢ ¡ discounting rate: ∆S + B = ∆Seαh + Berh e−γh (∆S + B) eγh = ∆Seαh + Berh (0.737 6 × 41 − 22. 403 6) eγ(1) = 0.737 6 × 41e0.15(1) − 22. 403 6e0.08(1) eγ(1) = 1. 386 γ = 0.326 4 Calculate the option price: C = e−γh (pCu + qCd ) = e−0.326 4(1) (0.544 6 × 19. 954 + 0.455 4 × 0) = 7. 84 This option price is the same as the price in the textbook Figure 10.3.

Example 11.2.2. Reproduce the textbook Figure 11.4 (the risk neutral solution to an otherwise identical European call is in the textbook Figure 10.5). Here is the recap of the information on an American call. The current stock price is 41. The strike price K = 40. The annualized standard deviation of the continuously compounded stock return is σ = 30%. The continuously compounded risk-free rate per year is r = 8%. The continuously compounded expected return on the stock per year is α = 15%.The continuously compounded dividend rate per year is δ = 0%.The option expiration date is T = 1 year. Use a 3-period binomial tree and real probabilities to calculate the option premium. Solution. Each period is h =

1 year long. 3

√ √ u = e(r−δ)h+σ√ h = e(0.08−0)1/3+0.3√1/3 = 1. 221 246 d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 1/3 = 0.863 693 Stock price Period 0 1 2 3 Su3 = 74.6781 Su2 = 61.1491 Su = 50.0711 Su2 d = 52.8140 S = 41 Sud = 43.2460 Sd = 35.4114 Sd2 u = 37.3513 2 Sd = 30.5846 Sd3 = 26.4157

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11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY

85

The replicating portfolios are copied over from the textbook Figure 10.5. Period 0

1

(4, B) = (0.706 3, −21. 885 2)

(4, B)u = (0.921 8, −33. 263 6) (4, B)d = (0.450 1, −13. 405 2)

2 (4, B)u2 = (1, −38. 947 4) (4, B)ud = (0.828 7, −30. 138 6) (4, B)d2 = (0, 0)

In addition, we need to calculate the discount rate for each node. We put the stock price table and the replicating portfolio table side by side: Stock price (4, B) Period 0 1 2 3 Period 0 1 74.6781 61.1491 50.0711 52.8140 (0.921 8, −33. 263 6) 41 43.2460 (0.706 3, −21. 885 2) 35.4114 37.3513 (0.450 1, −13. 405 2) 30.5846 26.4157

0:

Calculate the common discounting factor Node u →Node 0 and Node d →Node e−γ(1/3) =

∆S + B 0.706 3 (41) − 21. 885 2 = = ∆Seαh + Berh 0.706 3 (41) e0.15(1/3) − 21. 885 2e0.08(1/3)

0.887 87 e−γ(1/3) = 0.887 87

γ = 0.356 8

Calculate the common discounting factor Node u2 →Node u and Node ud →Node u : e−γ(1/3) =

∆S + B 0.921 8 1 (50.0711) − 33. 263 6 = = αh rh ∆Se + Be 0.921 8 (50.0711) e0.15(1/3) − 33. 263 6 2e0.08(1/3)

0.897 84 e−γ(1/3) = 0.897 84

γ = 0.323 3

Calculate the common discounting factor Node ud →Node d and Node d2 →Node d : e−γ(1/3) =

∆S + B 0.450 1 (35.4114) − 13. 405 2 = = ∆Seαh + Berh 0.450 1 (35.4114) e0.15(1/3) − 13. 405 2 64e0.08(1/3)

0.847 79 e−γ(1/3) = 0.847 79

γ = 0.495 4

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2 (1, −38. 947 4) (0.828 7, −30. 138 6) (0, 0)

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CHAPTER 11. BINOMIAL OPTION PRICING: II

Calculate the common discounting factor Node u3 →Node u2 and Node u d →Node u2 : 2

∆S + B 1 (61.1491) − 38. 947 4 = = ∆Seαh + Berh 1 (61.1491) e0.15(1/3) − 38. 947 4e0.08(1/3)

e−γ(1/3) =

0.914 24 e−γ(1/3) = 0.914 24

γ = 0.2690

Calculate the common discounting factor Node u2 d →Node ud and Node ud →Node ud : 2

e−γ(1/3) =

∆S + B 0.828 7 (43.2460) − 30. 138 6 = = αh rh ∆Se + Be 0.828 7 (43.2460) e0.15(1/3) − 30. 138 6e0.08(1/3)

0.847 83 e−γ(1/3) = 0.847 83

γ = 0.495 2

Calculate the common discounting factor Node ud2 →Node d2 and Node d →Node d2 : 3

∆S + B 0 (30.5846) − 0 = = N/A ∆Seαh + Berh 0 (30.5846) e0.15(1/3) − 0 6e0.08(1/3) = N/A γ = N/A

e−γ(1/3) = e−γ(1/3)

We calculate the premium by working backward from right to left. At each node, we take the greater of the backwardized value and the exercise value. Calculate the common real probability of stock going up and down at each node. 41e0.15(1/3) = 50.0711p + 35.4114 (1 − p) p = 0.524 6 q = 1 − 0.524 6 = 0.475 4 Period 0

1

2 γ = 0.2690

3 Cu3 = max (0, 74.6781 − 40) = 34. 678 1

γ = 0.495 2

Cu2 d = max (0, 52.8140 − 40) = 12. 814

γ = N/A

Cd2 u = max (0, 37.3513 − 40) = 0

γ = 0.323 3 γ = 0.356 8 γ = 0.495 4

Cd3 = max (0, 26.4157 − 40) = 0 Period 0

1

2

3 Su3 = 74.6781

Su2 = 61.1491 Su2 d = 52.8140

Su = 50.0711 S = 41

Sud = 43.2460 Sd2 u = 37.3513

Sd = 35.4114 2

Sd = 30.5846 Sd3 = 26.4157

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11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY Cu2 = (34. 678 1 × 0.524 6 + 12. 814 × 0.475 4) e−0.2690(1/3) = 22. 201 2 EVu2 = max (0, 61.1491 − 40) = 21. 149 1 Vu2 = max (22. 201 2, 21. 149 1) = 22. 201 2 Cud = (12. 814 × 0.524 6 + 0 × 0.475 4) e−0.495 2(1/3) = 5. 699 4 EVud = max (0, 43.2460 − 40) = 3. 246 Vud = max (5. 699 4, 3. 246) = 5. 699 4 Cd2 = 0 EVd2 = max (0, 30.5846 − 40) = 0 Vd2 = max (0, 0) = 0 Now we have: Period 0 1

2

3 Vu3 = 34. 678 1

Vu2 = 22. 201 2 γ = 0.323 3 γ = 0.356 8

Vu2 d = 12. 814 Vud = 5. 699 4

γ = 0.495 4

Vud2 = 0 Vd2 = 0 Vd3 = 0

Similarly, Cu = (22. 201 2 × 0.524 6 + 5. 699 4 × 0.475 4) e−0.323 3(1/3) = 12. 889 6 EVu = max (0, 50.0711 − 40) = 10. 071 1 Vu = max (12. 889 6, 10. 071 1) = 12. 889 6 Cd = (5. 699 4 × 0.524 6 + 0 × 0.475 4) e−0.495 4(1/3) = 2. 534 8 EVd = max (0, 35.4114 − 40) = 0 Vd = max (2. 534 8, 0) = 2. 534 8 Now we have: Period 0 1

2

3 Vu3 = 34. 678 1

Vu2 = 22. 201 2 Vu = 12. 889 6 γ = 0.356 8

Vu2 d = 12. 814 Vud = 5. 699 4

Vd = 2. 534 8

Vud2 = 0 Vd2 = 0

Vd3 = 0 Finally, C = (12. 889 6 × 0.524 6 + 2. 534 8 × 0.475 4) e−0.356 8(1/3) = 7. 0734 EV = max (0, 41 − 40) = 1

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87

88

CHAPTER 11. BINOMIAL OPTION PRICING: II V = max (7. 0734, 1) = 7. 073 4 So we have: Period 0 1

2

3 Vu3 = 34. 678 1

Vu2 = 22. 201 2 Vu = 12. 889 6 V = 7. 073 4

Vu2 d = 12. 814 Vud = 5. 699 4

Vd = 2. 534 8

Vud2 = 0 Vd2 = 0 Vd3 = 0

Tip 11.2.2. Real probability pricing requires intensive calculation. Not only do we need to find the real probability of up and down, we also need to find the replicating portfolio at each node. In contrast, in risk neutral pricing, we either use risk neutral probabilities or use replicating portfolio but not both. In comparison, risk neutral pricing is more efficient than real probability pricing.

11.2.2

Binomial tree and lognormality

Random Walk model Here’s a brief review of the random walk model. There are 3 schools of thoughts: the chartist approach (or technical analysis), fundamental analysis, and the random walk model. Those who use the chartist approach draw charts to predict stock future prices. They believe that history repeats itself and that past stock prices help predict future stock prices. Fundamental analysis believes that at any point the stock has an intrinsic value that depends on the earning potential of the stock. Random Walk model, on the other hand, believes that the price of a stock is purely random and that past price can’t help predict the stock price in the future. Is the random walk model true? Is stock price purely random? Some scholars challenged the random walk theory. If interested, you can look into the book A Non-Random Walk Down Wall Street at Amazon.com http://www.amazon. com/ Now let’s look at the random walk math model. Imagine that an object zero starts from point zero and travels along a straight line (such as Y axis). At each step, the object either move up by 1 unit with probability p or move down by 1 unit with probability 1 − p. An equivalent description of the movement is this: at each move a coin is tossed. If we get a head, the object moves up by 1 unit; if we get a tail, the object moves down by 1 unit. Let Y½ i represent the movement in the i-th step. Then 1 with probability p Yi = −1 with probability 1 − p Let Zn represent the position of the article after n steps. Then Zn = Y1 + Y2 + ... + Yn Y1 , Y2 , ...Yn are independent identically distributed.

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11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY

89

To apply the random walk model to stock prices, we can use Yi to represent the price movement during an interval of time. We can use Zn to represent the ending price of a stock after n equal intervals. To get a good feel of the random variable Zn , check out the simulation of the random walk model at http://math.furman.edu/~dcs/java/rw.htmll Just type in the url in your web browser. Modeling stock as a random walk There are 3 problems if we use Zn = Y1 + Y2 + ... + Yn to model the price of a stock: 1. Zn can be negative yet the price of a stock can’t be negative. 2. The incremental change of any stock price Yi is either 1 or −1. Though $1 change might be OK for modeling the change of a low priced stock, it may be inappropriate to model the change of a high priced stock. 3. The expected return on stock whose price following a random walk is zero, that is E (Zn ) = nE (Y ) = n × 0 = 0. However, stock on average should have a positive return. Continuously compounded returns Let rt,t+h represent the continuously compounded return earned during the time interval [t, t + h]. Let St and St+h represent the stock price at time t and t + h respectively. Then St+h St+h St ert,t+h = St+h ert,t+h = rt,t+h = ln St St Continuously compounded returns are additive. Consider the time interval [t, t + nh]. We have: St+nh St+h St+2h St+3h St+nh ert,t+nh = = × × × ... × St St St+h St+2h St+(n−1)h = ert,t+h ert+h,t+2h ert+2h,t+3h ...ert+(n−1)h,t+nh ⇒ rt,t+nh = rt,t+h +rt+h,t+2h +rt+2h,t+3h + ... + rt+(n−1)h,t+nh Continuously compounded returns can be negative. Even if r < 0, we still have er > 0. Thus if ln S follows a random walk, S can’t be negative. Standard deviation of returns The annual return is the sum of the returns in each of the 12 months: rannual = rJan + rF eb + ... + rDec Assume each monthly return is independent identical distributed with common variance σ 2Monthly . Let σ 2 represent the variance of the return over 1 year period.

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CHAPTER 11. BINOMIAL OPTION PRICING: II

V ar (rannual ) = V ar (rJan + rF eb + ... + rDec ) = 12σ 2Monthly σ σ 2 = 12σ2M onthly σMonthly = √ 12 Suppose we split one year into n intervals with each interval being h long (so nh = 1). Let σ h represent the standard deviation of the return over an interval of h long, then rannual = r0,h + rh,2h + r2h,3h £ + ... + r(n−1)h,nh ¤ 2 σ = V ar (rannual ) = V ar r0,h + rh,2h + r2h,3h + ... + r(n−1)h,nh = n σ 2h σ σh = √ n √ σ However, nh = 1. So we have: σ h = √ = σ h n Binomial model √

√

Previously, we set u = e(r−δ)h+σ h and d = e(r−δ)h−σ h . Now let’s see why do√ √ (r−δ)h+σ h (r−δ)h−σ h and d = e is equivalent ing so is reasonable. Setting u = e to setting √ St+h = St e(r−δ)h±σ h , which is equivalent to √ rt,t+h = (r − δ) h ± σ h (11.4) Let’s see why Equation 11.4 solves the three problems in the random walk model: 1. Even when rt,t+h is negative, the stock price St+h is always positive. 2. The change priceiis proportional to the stock price. ∆S = St+h − h in stock √ (r−δ)h±σ h −1 . St = St e

3. The expected return during [t, t + h] is largely driven by the constant term (r − δ) h. Hence the expected return is no long always zero. Alternative binomial tree The Cox-Ross-Rubinstein binomial tree u = eσ

√ h

d = e−σ

√

(11.5)

h

(11.6)

The lognormal tree (also called the Jarrow-Rudd binomial model) 2 u = e(r−δ−0.5σ )h+σ

2 d = e(r−δ−0.5σ )h−σ

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√ h

√

h

(11.7) (11.8)

11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY

11.2.3

91

Estimate stock volatility

Formulas n+1 number of stock prices observed: S0 , S1 ,...,Sn−1 ,and Sn n number of stock returns observed. Continuously compounded return per period is estimated as S1 S2 S3 Sn r1 = ln r2 = ln r3 = ln ...rn = ln S0 S1 S2 Sn−1 Since our focus is stock returns not stock price, the number of observations is n (i.e. the number of stock prices observed minus one). Remember this point. The expected return is: r1 + r2 + ... + rn ∧ r= n The v estimated standard deviation is: ´2 ³ ´ ³ ´ u³ ∧ 2 ∧ 2 u r −∧ r + r − r + ... + r − r 1 2 n t ∧ σ= n−1 Example 11.2.3. Reproduce the textbook Figure 11.1 and estimate the standard deviation of the continuously compounded return per year earned by S&P 50 index. ´ ³ St ∧ 2 Week S&P 500 price rt =ln rt − r St−1 0 829.85 1 804.19 −0.0314 0.001846 2 874.02 0.0833 0.005143 3 869.95 −0.0047 0.000263 4 880.9 0.0125 0.000001 5 865.99 −0.0171 0.000819 6 879.91 0.0159 0.000019 7 919.02 0.0435 0.001020 8 916.92 −0.0023 0.000191 9 929.62 0.0138 0.000005 10 939.28 0.0103 0.000001 11 923.42 −0.0170 0.000817 12 953.22 0.0318 0.000409 Total 0.1386 0.010534 The expected continuously compounded return per week is estimated as: r1 + r2 + ... + r12 0.1386 ∧ r= = = 0.011 55 n 12 The standard deviation of the continuously compounded return per week is estimated as:

Yufeng Guo, Fall 09 MFE, actuary88.com

92

CHAPTER 11. BINOMIAL OPTION PRICING: II r

0.010534 = 0.03 095 12 − 1 1 Year = 52 Weeks ∧

σ=

Let Y represent the continuously compounded return per year and Xi represent the continuously compounded return earned in the i-th week. Then Y = X1 + X2 + ... + X52 Where X1 , X2 , ..., X52 are assumed to be independent identically distributed. V ar (Y√) = V ar (X √ 1 + X2 + ... + X52 ) = 52V ar (X) σ Y = 52σ Y = 52 (0.03 095) = 0.223 18 Please note my calculation was done using Excel. If you can’t perfectly reproduce my numbers, that’s fine. By the way, in Excel, the formula for the mean is AVERAGE; the formula for the sample variance is VAR This is how to use Excel to calculate the continuously compounded return per year earned by S&P 500 index. Suppose the stock prices are entered in Column C (from C3 to C15) and the weekly returns are calculated in Column D (from D4 to D15). 1

Column B

Column C

2

Week

S&P 500 price

Column D St rt =ln St−1

3 0 829.85 4 1 804.19 −0.0314 5 2 874.02 0.0833 6 3 869.95 −0.0047 7 4 880.9 0.0125 8 5 865.99 −0.0171 9 6 879.91 0.0159 10 7 919.02 0.0435 11 8 916.92 −0.0023 12 9 929.62 0.0138 13 10 939.28 0.0103 14 11 923.42 −0.0170 15 12 953.22 0.0318 The expected continuously compounded return per week is: ∧ r =AVERAGE(D4:D15)= 0.01155006 The sample variance of the continuously compounded return per week is estimated as: ∧2 σ =VAR(D4:D15)= 0.000957682 √ ∧ σ = 0.000957682 = 0.030946434 Make sure you don’t use the population variance formula: VARP(D4:D15)= 0.000877875

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11.2. UNDERSTANDING RISK-NEUTRAL PROBABILITY

93

In Excel, VARP is: ´ ³ ´ ³ ´ ³ ∧ 2 ∧ 2 ∧ 2 r1 − r + r2 − r + ... + rn − r V ARP = n While VAR is³calculated ´ as ³ ´ ³ ´ ∧ 2 ∧ 2 ∧ 2 r1 − r + r2 − r + ... + rn − r V AR = n−1 So V AR = V ARP ×

n n−1

Finally, I’ll give you a BA II Plus (or BA II Plus Professional) calculator shortcut for quickly finding the sample mean and the sample standard deviation. BA II Plus and BA II Plus Professional have a 1-V Statistics Worksheet. This worksheet can calculate the sample mean, the sample standard deviation, the population mean (which is the same as the sample mean), and the population standard deviation. In 1-V Statistics Worksheet, enter: 1 Column B Column C Column D St 2 Week S&P 500 price rt =ln St−1 3 0 829.85 4 1 804.19 −0.0314 5 2 874.02 0.0833 6 3 869.95 −0.0047 7 4 880.9 0.0125 8 5 865.99 −0.0171 9 6 879.91 0.0159 10 7 919.02 0.0435 11 8 916.92 −0.0023 12 9 929.62 0.0138 13 10 939.28 0.0103 14 11 923.42 −0.0170 15 12 953.22 0.0318 804.19 X01 = ln = −0.03 140 940 829.85 Y 01 = 1 (we observed X01 only once) 874.02 X02 = ln = 0.08 326 770 804.19 Y 02 = 1 (we observed X02 only once)

So on and so forth. The final entry is: 953.22 X12 = ln = 0.03 176 156 923.42 Y 12 = 1 (we observed X12 only once)

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CHAPTER 11. BINOMIAL OPTION PRICING: II

Press 2nd STAT of the calculator key, you should get: n = 12 X = 0.01155006 This is the sample mean (or population mean) of the continuously compounded return per week. SX = 0.03094643 This is the estimated standard deviation (or sample standard deviation) of the continuously compounded return per week. σ X = 0.02962895 This is the estimated standard deviation (or population standard deviation) of the continuously compounded return per week. You should use SX and discard σ X when estimating the stock volatility. Example 11.2.4. Reproduce the textbook Figure 11.1 and estimate the standard deviation of the continuously compounded return per year earned by IBM. Week

S&P 500 price

0 1 2 3 4 5 6 7 8 9 10 11 12

77.73 75.18 82 81.55 81.46 78.71 82.88 85.75 84.9 86.68 88.7 86.18 87.57

rt =ln

St St−1

−0.0334 0.0868 −0.0055 −0.0011 −0.0343 0.0516 0.0340 −0.0100 0.0207 0.0230 −0.0288 0.0160

In BA II Plus (or BA II Plus Professional) 1-V Statistics Worksheet, enter: X01 = −0.0334 Y 01 = 1 X02 = 0.0868 Y 02 = 1 X03 = −0.0055 Y 03 = 1 ...... X13 = 0.0160 Y 12 = 1 You should get: SX = 0.0365 So the estimated standard deviation (or sample standard deviation) of the continuously compounded return per week is: σ X = 0.0365 The standard deviation of the continuously compounded return per year is:

Yufeng Guo, Fall 09 MFE, actuary88.com

11.3. STOCKS PAYING DISCRETE DIVIDENDS

95

√ 52 (0.0365) = 0.263 2

11.3

Stocks paying discrete dividends

Previously, we assume that dividends are paid continuously at rate δ. However, in reality, dividends are paid discretely (such as quarterly or annually). Now let’s build a binomial tree to calculate the price of a stock that pays discrete dividends. Suppose we have a European option on a stock that pays a discrete dividend. The option is written at t (today) and expires in t+h. The stock pays a dividend or several dividends during [t, t + h]. The future value of the dividends at t + h is D. The continuously compounded risk-free interest rate per year is r ( a positive constant). The stock price today is St . At t + h, the stock price either goes up to Stu = uSt or goes down to Std = dSt . The standard deviation of the continuously compounded return earned by the stock per year is σ. We want to calculate the option price.

Using Equation 10.22√ and 10.23, we have: √ Stu = uSt = Ft,t+h eσ h Std = dSt = Ft,t+h e−σ h However, Ft,t+h = St erh − D This gives us: ¡ ¢ √ Stu = St erh − D eσ h

√ ¡ ¢ Std = St erh − D e−σ h

(11.9) (11.10)

To find the price of the European option, we calculate the cost of the replicating portfolio. We have two assets: the stock and a savings account. The savings account is the same as a zero-coupon bond. At time t + h, the stock price is Sh ; the bond price is Bt+h . The bond price is deterministic: Bt+h = erh Bt = 1 The stock ( price at¡ t + h is stochastic: ¢ √ Stu = St erh − D eσ √h ¡ ¢ St+h = Std = St erh − D e−σ h So at t + h the stock price either goes up to Stu ("up state") or goes down to Std ("down state").

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CHAPTER 11. BINOMIAL OPTION PRICING: II Stu St Time t

Std Time t + h

Our task is to determine C by setting a portfolio that replicates the option payoff of Cu in the up state and Cd in the down state. We build the replicating portfolio by buying 4 stocks and investing $B in a zero-coupon bond.

If we own one stock at t, then at t + h our total wealth is St+h + D. We not only can sell the stock in the market for St+h , we’ll also have D, the future value of the dividend earned during [t, t + h]. So we need to set up the following equation: 4St t ½

4 (Stu + D)

¡ ¢ 4 Std + D t+h

Berh +

B

Cu =

Berh t+h

t

C t

Cd t+h

4 ¡(Stu + D)¢ + Berh = Cu 4 Std + D + Berh = Cd Solving these equations, we get:

4=

B = e−rh

11.3.1

µ

Cu − Cd Stu − Std

Stu Cd − Std Cu − 4D Stu − Std

(11.11)

¶

(11.12)

Problems with discrete dividend tree

One major problem with the stock price tree using Equation 11.9 and 11.10 is that the tree doesn’t complete recombine after the discrete dividend. Example 11.3.1. Reproduce the textbook Figure 11.1 but just for the 2 periods. Show that the stock price tree doesn’t recombine at Period 2. This is the recap of the information. The current stock price is 41. The stock pays a dividend during Period 1 and Period 2. The future value of the dividend accumulated at the risk-free interest rate r from Period 1 to Period 2 is 5. Other data are: r = 0.08, σ = 0.3, t = 1,and h = 1/3.

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11.3. STOCKS PAYING DISCRETE DIVIDENDS Period 0

1

Stu

2 Stuu = 55. 203 = 50. 071 Stud

97

3 Stuuu = 67. 417 15 Stuud = Studu = 47. 678 91

= 39. 041 Studd = 33. 719 59

St = 41

Std

Stdu = 37. 300 = 35. 411 Stdd

Stduu = 45. 553 05 Stdud = Stddu = 32. 216 14

= 26. 380 Stddd = 22. 783 97

No dividend is paid during Period 0 and Period 1. √ ¡ ¢ √ ¡ ¢ Stu = St erh − D eσ h = 41e0.08×1/3 − 0 e0.3 1/3 = 50. 071 09 √ √ ¡ ¢ ¡ ¢ Std = St erh − D e−σ h = 41e0.08×1/3 − 0 e−0.3 1/3 = 35. 411 39

Dividend is paid during Period 1 to Period 2, whose value √ at Period 2 is 5. ¡ ¢ √ ¡ ¢ Stuu = Stu erh − D eσ h = 50. 071 09e0.08×1/3 − 5 e0.3 1/3 = 55. 203 57 √ √ ¡ ¢ ¡ ¢ Stud = Stu erh − D e−σ h = 50. 071 09e0.08×1/3 − 5 e−0.3 1/3 = 39. 041 20 √ ¡ ¢ √ ¡ ¢ Stdu = Std erh − D eσ h = 35. 411 39e0.08×1/3 − 5 e0.3 1/3 = 37. 300 47 √ √ ¡ ¢ ¡ ¢ Stdd = Std erh − D e−σ h = 35. 411 39e0.08×1/3 − 5 e−0.3 1/3 = 26. 379 73

Now you see that Stud 6= Stdu , . The tree doesn’t recombine. No dividend is paid during Period 2 and Period 3. √ ¡ ¢ √ ¡ ¢ Stuuu = Stuu erh − D eσ h = 55. 203 57e0.08×1/3 − 0 e0.3 1/3 = 67. 417 15 √ √ ¡ ¢ ¡ ¢ Stuud = Stuu erh − D e−σ h = 55. 203 57e0.08×1/3 − 0 e−0.3 1/3 = 47. 678 91 √ ¡ ¢ √ ¡ ¢ Studu = Stud erh − D eσ h = 39. 041 20e0.08×1/3 − 0 e0.3 1/3 = 47. 678 91 √ √ ¡ ¢ ¡ ¢ Studd = Stud erh − D e−σ h = 39. 041 20e0.08×1/3 − 0 e−0.3 1/3 = 33. 719 59 √ ¡ ¢ √ ¡ ¢ Stduu = Stdu erh − D eσ h = 37. 300 47e0.08×1/3 − 0 e0.3 1/3 = 45. 553 05 √ √ ¡ ¢ ¡ ¢ Stdud = Stdu erh − D e−σ h = 37. 300 47e0.08×1/3 − 0 e−0.3 1/3 = 32. 216 14 √ ¡ ¢ √ ¡ ¢ Stddu = Stdd erh − D eσ h = 26. 379 73e0.08×1/3 − 0 e0.3 1/3 = 32. 216 14 √ √ ¡ ¢ ¡ ¢ Stddd = Stdd erh − D e−σ h = 26. 379 73e0.08×1/3 − 0 e−0.3 1/3 = 22. 783 97 Please note that in this example Stuud = Studu and Stdud = Stddu . This is not a coincidence.

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98

CHAPTER 11. BINOMIAL OPTION PRICING: II Let’s verify that Stuud = Studu . √ ¡ ¢ √ Stu = St erh − 0 eσ h = St³erh eσ h ´ √ √ ¡ ¢ √ Stuu = Stu erh − 5 eσ h = St erh eσ h − 5 eσ h ³³ ´ √ ´ √ √ √ ¡ ¢ Stuud = Stuu erh − 0 e−σ h = St erh eσ h − 5 eσ h erh e−σ h ³ ´ √ = St erh eσ h − 5 erh

³ ´ √ √ √ ¡ ¢ Stud = Stu erh − 5 e−σ h = St erh eσ h − 5 e−σ h

³³ ´ ´ √ √ √ ¡ ¢ √ Studu = Stud erh − 0 eσ h = St erh eσ h − 5 e−σ h erh − 0 eσ h ³ ´ √ = St erh eσ h − 5 erh

Clearly, Stuud = Studu . Similarly, you can verify for yourself that Stdud =

Stddu .

In this problem, Period 2 had 4 prices. If the stock pays continuous dividend, Period 2 will have only 3 prices. Similarly, in this problem, Period 3 has 5 distinct prices. In contrast, if the stock pays continuous dividend, Period 3 will have only 4 prices. In addition to non-combining stock prices, the above method may produce negative stock prices.

11.3.2

Binomial tree using prepaid forward

Schroder presents a method that overcomes the two shortcomings of the above method. This is the idea behind Schroder’s method. Instead of directly building a stock price tree (which proves to be non-combining after the discrete dividend is paid), we’ll build a tree of a series of prepaid forward prices on the same stock. Hopefully, the prepaid forward price ³ ´ tree is recombining and looks like this: P Ft,T

³ ´u P Ft+h,T ³ ´d P Ft+h,T

In the above table,

P Ft+2h,T

uu

³ ´ud P Ft+2h,T ³ ´dd P Ft+2h,T

P • Ft,T is the prepaid price of a forward contract signed at t expiring on date T

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11.3. STOCKS PAYING DISCRETE DIVIDENDS

99

P • Ft+h,T is the prepaid price of a forward contract signed at t + h expiring on date T P • Ft+2h,T is the prepaid price of a forward contract signed at t + 2h expiring on date T

Since there’s a one-to-one relationship between a prepaid forward price and the stock price, we can change the (recombining) prepaid forward price tree into a (recombining) stock price tree. Once we have a recombining stock price tree, we’ll can easily find the price of the option using either risk-neutral probability or the replicating portfolio. First, let’s find the relationship between the stock price and the prepaid forward price on the stock. Suppose today is time t . At t we enter into a forward contract agreeing to buy a stock on date T . The stock will pay dividend D on date TD where TD < T . Suppose we want to prepay the seller at t. The price of this prepaid forward contract is the current stock price minus the present value of the dividend: ½ St −r(T D −t) if TD ≥ t De P = St − P Vt (D) = St − Ft,T 0 if TD < t ½ De−r(TD −t) if TD ≥ t P → St = Ft,T + 0 if TD < t Similarly,

½

De−r(TD −t−h) = St+h − P Vt+h (D) = St+h − 0 ½ −r(TD −t−h) if TD ≥ t + h De P + → St+h = Ft+h,T 0 if TD < t + h P Ft+h,T

if TD ≥ t + h if TD < t + h

So there’s a one-to-one mapping between the prepaid forward price and the stock price. Next, let’s find out how to build a prepaid forward price tree. We need to know how the prepaid forward price changes over time. Suppose today is time t . At t + h we enter into a forward contract agreeing to buy a stock at date T where T > t + h. The stock will pay dividend D in date TD where t < TD < T . If the stock volatility is zero (meaning that the future stock price is known today with 100% certainty), then the price of the prepaid forward contact at t + h is P P rh Ft+h,T = Ft,T e P Ft+h,T P Ft,T

= erh

(11.13)

(11.14)

P at t, we’ll receive one stock This is why Equation 11.13 holds. If we pay Ft,T P P at T . This gives us Ft,T = ST e−r(T −t) . Similarly, if we pay Ft+h,T at t + h, we’ll

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100

CHAPTER 11. BINOMIAL OPTION PRICING: II

P also receive a stock at T . This gives us Ft+h,T = ST e−r(T −t−h) . Then Equation 11.13 holds. Now suppose that the forward price has a volatility of σ F per year. Then it’s reasonable ( to assume that √ P rh σ h P F e e = Ft,T u in the up state t,T P √ Ft+h,T = P rh −σ h P Ft,T e e = Ft,T d in the down state

where u = erh+σF P Similarly, Ft+2h,T

√ h

√

and d = erh−σF h ½ P Ft+h,T u in the up state = P d in the down state Ft+h,T

Typically, we know the volatility of the stock, σ S . We need to calculate σ F using the following approximation: St P Ft,T Now the prepaid forward price³tree is: ´ uu P P = Ft,T u2 Ft+2h,T ³ ´u P P = Ft,T u Ft+h,T ³ ´ud ³ ´du P P P P Ft,T = Ft+2h,T = Ft,T ud Ft+2h,T ³ ´d P P = Ft,T d Ft+h,T ³ ´dd P P 2 = Ft,T d Ft+2h,T

σF = σS ×

Once we have the prepaid forward price tree, we’ll transform it into the stock price tree:

P St = Ft,T + P Vt (D)

³ ´u u P St+h = Ft+h,T + P Vt+h (D) ³ ´d d P St+h = Ft+h,T + P Vt+h (D)

³ ´uu uu P St+2h = Ft+2h,T + P Vt+2h (D) ³ ´ud ud P St+2h = Ft+2h,T + P Vt+2h (D) ³ ´dd dd P St+2h = Ft+2h,T + P Vt+2h (D)

Example 11.3.2. Let’ s reproduce the textbook Figure 11.11. Here is the recap of the information on an American call option. The stock pays a dividend of $5 in 8 months. Current stock price is $41. The strike price K = $40. The stock volatility is σS = 0.3. The continuously compounded risk-free rate is r = 0.08. The option expires in T = 1 year. Use a 3-period binomial tree to calculate the option price.

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11.3. STOCKS PAYING DISCRETE DIVIDENDS First, we’ll build a prepaid forward price tree.

σF = σS ×

St P Ft,T

P Ft,T = St − P Vt (D) = 41 − 5e−0.08(8/12) = 36. 26 St 41 = 0.339 2 σ F = σ S × P = 0.3 × 36. 26 Ft,T √ √ u = erh+σF √ h = e0.08(1/3)+0.3392 √1/3 = 1. 249 20 d = erh−σF h = e0.08(1/3)−0.3392 1/3 = 0.844 36

Prepaid forward price tree:

P Ft,T

³ ´u P Ft+h,T ³ ´d P Ft+h,T

³ ´uu P Ft+2h,T ³ ´ud P Ft+2h,T ³ ´dd P Ft+2h,T

³ ´uuu P Ft+3h,T ³ ´uud P Ft+3h,T ³ ´udd P Ft+3h,T ³ ´ddd P Ft+3h,T

³ ´u P P = Ft,T u = 36. 26 × 1. 249 20 = 45. 296 Ft+h,T ³ ´d P P Ft+h,T = Ft,T d = 36. 26 × 0.844 36 = 30. 616 ³ ´uu ´u ³ P P P = Ft,T u2 = Ft+h,T u = 45. 296 × 1. 249 20 = 56. 584 Ft+2h,T ³ ´ud ³ ´u P P Ft+2h,T = Ft+h,T d = 45. 296 × 0.844 36 = 38. 246 ³ ³ ´dd ´d P P 2 P Ft+3h,T = Ft,T d = Ft+h,T d = 30. 616 × 0.844 36 = 25. 851 ´uuu ³ ´uu ³ P P = Ft+2h,T u = 56. 584 × 1. 249 20 = 70. 685 Ft+3h,T ³ ´uud ³ ´uu P P Ft+3h,T = Ft+2h,T d = 56. 584 × 0.844 36 = 47. 777 ³ ´udd ³ ´ud P P Ft+3h,T = Ft+2h,T d = 38. 246 × 0.844 36 = 32. 293 ³ ´ddd ³ ´dd P P Ft+3h,T = Ft+2h,T d = 25. 851 × 0.844 36 = 21. 828 Prepaid forward price tree:

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101

102

CHAPTER 11. BINOMIAL OPTION PRICING: II 70. 685 56. 584 45. 296 36. 26

47. 777 38. 246

30. 616

32. 293 25. 851

21. 828 Next, we change the prepaid forward price tree into a stock price tree. The one-to-one mapping between the prepaid forward price and the stock price is ½ −r(TD −t−∆t) if TD ≥ t + ∆t De P + St+∆t = Ft+∆t,T 0 if TD < t + ∆t where 0 ≤ ∆t ≤ T − t ¡ ¢uuu St+3/3 ¡ ¢uu St+2/3 ¡ ¢u ¡ ¢uud St+1/3 St+3/3 ¡ ¢ud St St+2/3 ¡ ¢d ¡ ¢udd St+1/3 St+3/3 ¡ ¢dd St+2/3 ¡ ¢ddd St+3/3 8 + t, D = 5,T = t + 1 12 −r(TD −t) + De = 36. 26 + 5e−0.08(8/12) = 41. 000

In this problem, TD = P St = Ft,T

´u ¢u ³ P ¡ + De−r(TD −t−1/3) = 45. 296 + 5e−0.08(8/12−1/3) St+1/3 = Ft+1/3,T 50. 164 ´d ¡ ¢d ³ P St+1/3 = Ft+1/3,T + De−r(TD −t−1/3) = 30. 616 + 5e−0.08(8/12−1/3) 35. 484 ´uu ¡ ¢uu ³ P = Ft+2/3,T +De−r(TD −t−2/3) = 56. 584+5e−0.08(8/12−2/3) St+2/3 61. 584 ´ud ¡ ¢ud ³ P St+2/3 = Ft+2/3,T +De−r(TD −t−2/3) = 38. 246+5e−0.08(8/12−2/3) 43. 246 ´dd ¡ ¢dd ³ P St+2/3 = Ft+2/3,T + De−r(TD −t−2/3) = 25. 851 + 5e−0.08(8/12−2/3) 30. 851 ´uuu ¡ ¢uuu ³ P = Ft+3/3,T = 70. 685 (because TD < t + 3/3) St+3/3 Similarly, ´uud ¡ ¢uud ³ P St+3/3 = Ft+3/3,T = 47. 777 ³ ´ udd ¡ ¢udd P St+3/3 = Ft+3/3,T = 32. 293 ³ ´ddd ¡ ¢ddd P St+3/3 = Ft+3/3,T = 21. 828

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= = = = =

11.3. STOCKS PAYING DISCRETE DIVIDENDS

103

So the stock price tree is: 70. 685 61. 584 50. 164 41. 000

47. 777 43. 246

35. 484

32. 293 30. 851 21. 828

After getting the stock price tree, we calculate the price of the American call option as usual. We work backward from right to left. At each node, we compare the backwardized value with the exercise value, taking the maximum of the two. The risk neutral probabilities are: erh − d e0.08(1/3) − 0.844 36 πu = = = 0.451 2 u−d 1. 249 20 − 0.844 36 π d = 1 − 0.451 2 = 0.548 8 Payoff tree: Vuuu = 30. 685 Vuu = 21. 584 Vu = 11. 308 V = 5.770

Vuud = 7. 777 Vud = 3. 417

Vd = 1. 501

Vudd = 0 Vdd = 0 Vddd = 0

Vuuu = Cuuu = max (0, 70. 685 − 40) = 30. 685 Vuud = Cuud = max (0, 47. 777 − 40) = 7. 777 Vudd = Cudd = max (0, 32. 293 − 40 ) = 0 Vddd = Cddd = max (0, 21. 828 − 40) = 0 Cuu = (30. 685 × 0.451 2 + 7. 777 × 0.548 8) e−0.08(1/3) = 17. 636 EVuu = max (0, 61. 584 − 40) = 21. 584 Vuu = max (17. 636 , 21. 584) = 21. 584 Cud = (7. 777 × 0.451 2 + 0 × 0.548 8) e−0.08(1/3) = 3. 417 EVud = max (0, 43. 246 − 40) = 3. 246 Vud = max (3. 417 , 3. 246) = 3. 417 Cdd = 0 EVdd = max (0, 30. 851 − 40) = 0 Vud = 0 Cu = (21. 584 × 0.451 2 + 3. 417 × 0.548 8) e−0.08(1/3) = 11. 308 EVu = max (0, 50. 164 − 40) = 10. 164

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CHAPTER 11. BINOMIAL OPTION PRICING: II

Vu = max (11. 308, 10. 164) = 11. 308 Cd = (3. 417 × 0.451 2 + 0 × 0.548 8) e−0.08(1/3) = 1. 501 EVd = max (0, 35. 484 − 40) = 0 Vd = max (1. 501 , 0) = 1. 501 C = (11. 308 × 0.451 2 + 1. 501 × 0.548 8) e−0.08(1/3) = 5. 770 EV = max (0, 41. 000 − 40) = 1.0 V = max (5. 770, 1.0) = 5. 77

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Chapter 12

Black-Scholes Except the option Greeks and the barrier option price formula, this chapter is an easy read.

12.1

Introduction to the Black-Scholes formula

12.1.1

Call and put option price

The price of a European call option is: C (S, K, σ, r, T, δ) = Se−δT N (d1 ) − Ke−rT N (d2 )

(12.1)

The price of a European put option is: P (S, K, σ, r, T, δ) = −Se−δT N (−d1 ) + Ke−rT N (−d2 ) ¶ µ S 1 2 ln + r−δ+ σ T K 2 √ d1 = σ T √ d2 = d1 − σ T

(12.2)

(12.3) (12.4)

Notations used in Equation 12.1, 12.3, and 12.4: • S, the current stock price (i.e. the stock price when the call option is written) • K, the strike price • r, the continuously compounded risk-free interest rate per year • δ, the continuously compounded dividend rate per year 105

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106

CHAPTER 12. BLACK-SCHOLES

• σ, the annualized standard deviation of the continuously compounded stock return (i.e. stock volatility) • T , option expiration time • N (d) = P (z ≤ d) where z is a standard normal random variable • C (S, K, σ, r, T, δ), the price of a European call option with parameters (S, K, σ, r, T, δ) • P (S, K, σ, r, T, δ), the price of a European put option with parameters (S, K, σ, r, T, δ) Tip 12.1.1. To help memorize Equation 12.2, we can rewrite Equation 12.2 similar to Equation 12.1 as P (S, K, σ, r, T, δ) = (−S) e−δT N (−d1 )+(−K) e−rT N (−d2 ). In other words, change S,K,d1 ,and d2 in Equation 12.1 and you’ll get Equation 12.2.

Example 12.1.1. Reproduce the textbook example 12.1. This is the recap of the information. S = 41, K = 40,r = 0.08, σ = 0.3, T = 0.25 (i.e. 3 months), and δ = 0. Calculate the price of the price of a European call option. ¶ µ S 1 2 ln + r−δ+ σ T K 2 √ d1 = ¶ µ σ T 1 41 + 0.08 − 0 + × 0.32 0.25 ln 40 2 √ = 0.3730 = 0.3 0.25 √ √ d2 = d1 − σ T = 0.3730 − 0.3 0.25 = 0.2230 N (d2 ) = 0.588 2 N (d1 ) = 0.645 4 −0(0.25) 0.645 4 − 40e−0.08(0.25) 0.588 2 = 3. 399 C = 41e

Example 12.1.2. Reproduce the textbook example 12.2. This is the recap of the information. S = 41, K = 40,r = 0.08, σ = 0.3, T = 0.25 (i.e. 3 months), and δ = 0. Calculate the price of the price of a European put option. N (−d1 ) = 1 − N (d1 ) = 1 − 0.645 4 = 0.354 6 N (−d2 ) = 1 − N (d2 ) = 1 − 0.588 2 = 0.411 8 P = −41e−0(0.25) 0.354 6 + 40e−0.08(0.25) 0.411 8 = 1. 607

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12.2. APPLYING THE FORMULA TO OTHER ASSETS

12.1.2

107

When is the Black-Scholes formula valid?

Assumptions under the Black-Scholes formula: Assumptions about the distribution of stock price: 1. Continuously compounded returns on the stock are normally distributed (i.e. stock price is lognormally distributed) and independent over time 2. The volatility of the continuously compounded returns is known and constant 3. Future dividends are known, either as a dollar amount (i.e. D and TD are known in advance) or as a fixed dividend yield (i.e. δ is a known constant) Assumptions about the economic environment 1. The risk-free rate is known and fixed (i.e. r is a known constant) 2. There are no transaction costs or taxes 3. It’s possible to short-sell costlessly and to borrow at the risk-free rate

12.2

Applying the formula to other assets

12.2.1

Black-Scholes formula in terms of prepaid forward price

P The prepaid forward price for the stock is: F0,T (S) = Se−δT P The prepaid forward price for the strike asset is: F0,T (K) = P V (K) = −rT Ke √ Define V (T ) = σ T The price of a European call option in terms of repaid forward is:

¢ ¡ P P P P (S) , F0,T (K) , V (T ) = F0,T (S) N (d1 ) − F0,T (K) N (d2 ) C F0,T

(12.5)

The price of a European put option in terms of prepaid forward is: ¡ P ¢ P P P P F0,T (S) , F0,T (K) , V (T ) = −F0,T (S) N (−d1 ) + F0,T (K) N (−d2 ) (12.6) ln d1 =

P F0,T (S)

P F0,T

1 + V 2 (T ) (K) 2 V (T )

d2 = d1 − V (T )

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(12.7) (12.8)

108

12.2.2

CHAPTER 12. BLACK-SCHOLES

Options on stocks with discrete dividends

When the stock pays discrete dividends, the prepaid forward price is: P (S) = S − P V0,T (Div) F0,T P (S) in Equation 12.5 and 12.6, you should get the price of the Apply F0,T European call and put where the stock pays discrete dividends. Example 12.2.1. Reproduce the textbook example 12.3. Here is the recap of the information. S = 41, K = 40, σ = 0.3, r = 0.08, and T = 0.25 (i.e. 3 months). The stock pays dividend of 3 in 1 month, but makes no other payouts over the life of the option (so δ = 0). Calculate the price of the European call and put. P (S) = S − P V0,T (Div) = 41 − 3e−(0.08)1/12 = 38. 020 F0,T P (K) = P V (K) = Ke−rT = 40e−0.08(0.25) = 39. 208 F0,T √ √ V (T ) = σ T = 0.3 0.25 = 0.15 P (S) F0,T 1 38. 020 1 + V 2 (T ) ln P ln + 0.152 F0,T (K) 2 39. 208 2 d1 = = = −0.130 1 V (T ) 0.15 d2 = d1 − V (T ) = −0.130 1 − 0.15 = −0.280 1 N (d2 ) = 0.389 7 N (d1 ) = 0.448 2 N (−d1 ) = 1 − 0.448 2 = 0.551 8 N (−d2 ) = 1 − 0.389 7 = 0.610 3 P P (S) N (d1 ) − F0,T (K) N (d2 ) C = F0,T = 38. 020 (0.448 2) − 39. 208 (0.389 7) = 1. 76 P P P = −F0,T (S) N (−d1 ) + F0,T (K) N (−d2 ) = −38. 020 (0.551 8) + 39. 208 (0.610 3) = 2. 95

12.2.3

Options on currencies

Notation • x, the current dollar value of €1 • K, the strike price in dollars of €1 • r, the continuously compounded risk-free rate earned by $1 • rf , the continuously compounded risk-free rate earned by €1 • σ, the annualized standard deviation of the continuously compounded return on dollars • T , the option expiration date • C (x0 , K, σ, r, T, rf ), the price of a European call option with parameters (x0 , K, σ, r, T, rf )

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12.2. APPLYING THE FORMULA TO OTHER ASSETS

109

• P (x0 , K, σ, r, T, rf ), the price of a European put option with parameters (x0 , K, σ, r, T, rf ) The price of a European call option is: C (x, K, σ, r, T, rf ) = xe−rf T N (d1 ) − Ke−rT N (d2 )

(12.9)

The price of a European put option is: P (x, K, σ, r, T, rf ) = −xe−rf T N (−d1 ) + Ke−rT N (−d2 ) ln d1 =

¶ µ 1 x + r − rf + σ2 T K 2 √ σ T √ d2 = d1 − σ T

(12.10)

(12.11) (12.12)

Tip 12.2.1. For currency options, just set S = x and δ = rf and apply the Black-Scholes formulas on European call and put. The same thing happened in Equation 10.26 and 10.27. Example 12.2.2. Reproduce the textbook example 12.4. Here is the recap of the information. The current dollar price of €1 is $0.92. The strike dollar price of €1 is $0.9. The annualized standard deviation of the continuously compounded return on dollars is σ = 0.1. The continuously compounded risk-free rate earned by dollars is r = 6%. The the continuously compounded risk-free rate earned by €1 is rf = 3.2%. The option expires in 1 year. Calculate the price of the European call and put on €1. ¶ µ 1 2 x + r − rf + σ T ln K 2 √ d1 = σ T ¶ µ 0.92 1 ln + 0.06 − 0.032 + × 0.12 1 0.9 2 √ = = 0.549 8 0.1 1 √ √ d2 = d1 − σ T = 0.549 8 − 0.1 1 = 0.449 8 N (d1 ) = 0.708 8 N (−d1 ) = 1 − N (d1 ) = 1 − 0.708 8 = 0.291 2 N (d2 ) = 0.673 6 N (−d2 ) = 1 − N (d2 ) = 1 − 0.673 6 = 0.326 4 C = xe−rf T N (d1 ) − Ke−rT N (d2 ) = 0.92e−0.032(1) 0.708 8 − 0.9e−0.06(1) 0.673 6 = 0.06 06 P = −xe−rf T N (−d1 ) + Ke−rT N (−d2 ) = −0.92e−0.032(1) 0.291 2 + 0.9e−0.06(1) 0.326 4 = 0.017 2

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12.2.4

Options on futures

For a futures contract, the prepaid price is just the present value of the futures P P price. Set F0,T (F ) = F e−rT and F0,T (K) = Ke−rT , we get: ln d1 =

P F0,T (F ) P F0,T

1 + σ2T (K) 2 √ σ T

(12.13)

√ d2 = d1 − σ T

(12.14)

C (F, K, σ, r, T ) = F e−rT N (d1 ) − Ke−rT N (d2 )

(12.15)

P (F, K, σ, r, T ) = −F e−rT N (−d1 ) + Ke−rT N (−d2 )

(12.16)

Example 12.2.3. Reproduce the textbook example 12.5. Here is the recap of the information about the European option on a 1-year futures contract. The current futures price for natural gas is $2.10. The strike price is K = 2.10. The volatility is σ = 0.25. r = 0.055, T = 1. Calculate the price of the European call and put. 2.10 1 F 1 ln + σ2T + × 0.252 (1) K 2 2.10 2¡√ ¢ √ = d1 = = 0.125 0.25 1 σ T √ ¡√ ¢ d2 = d1 − σ T = 0.125 − 0.25 1 = −0.125 N (d2 ) = 0.450 3 N (d1 ) = 0.549 7 N (−d1 ) = 1 − N (d1 ) = 1 − 0.549 7 = 0.450 3 N (−d2 ) = 1 − N (d2 ) = 1 − 0.450 3 = 0.549 7 C = F e−rT N (d1 ) − Ke−rT N (d2 ) = 2.10e−0.055(1) 0.549 7 − 2.10e−0.055(1) 0.450 3 = 0.197 6 P = −F e−rT N (−d1 ) + Ke−rT N (dd2 ) = −2.10e−0.055(1) 0.450 3 + 2.10e−0.055(1) 0.549 7 = 0.197 6 ln

12.3

Option the Greeks

The learning objectives in the SOA’s syllabus is to Interpret the option Greeks. The learning objective in CAS Exam 3 Financial Economics is to Interpret the option Greeks and elasticity measures. How to derive the option Greeks is explained in the textbook Appendix 12.B, but Appendix 21.B is excluded from the SOA MFE and CAS FE. So you might want to focus on the learning objective of the exam. Of all the Greeks, delta and gamma are the most important.

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12.3. OPTION THE GREEKS

12.3.1

111

Delta

Cu − Cd when we try to find the replicating Su − Sd portfolio of a European call or put. It’s the number of stocks you need to own at time zero to replicate the discrete payoff of a European call or put at ∂C ∂C . Here ∆ = expiration date T . If the payoff is continuous, then ∆ = ∂S ∂S is the number of stocks you need to have now to replicate the payoff of the next instant (i.e. the payoff one moment later). The European call price is ∂C C = Se−δT N (d1 ) − Ke−rT N (d2 ) and the delta for a call is ∆call = = ∂S e−δT N (d1 ). One less visible thing to know is that d1 is also a function of S. So it’ll tame ∂C some work to derive ∆call = = e−δT N (d1 ). One naive approach is treat ∂S ∂C N (d1 ) as a constant and get ∆call = = e−δT N (d1 ). Interestingly, this gives ∂S the correct answer! Since deriving delta is not on the syllabus, you don’t need to go through the ∂C = e−δT N (d1 ). Just memorize that ∆call = messy math and prove ∆call = ∂S e−δT N (d1 ) for a European call and ∆put = −e−δT N (−d1 ) = −e−δT [1 − N (d1 )] = ∆call − e−δT for a European put. Other results you might want to memorize: 0 ≤ ∆call ≤ 1 −1 ≤ ∆put ≤ 0 Delta ∆. You already see ∆ =

Example 12.3.1. Calculate the delta of the following European call and put. The information is: S = 25,K = 20, σ = 0.15, r = 6%, δ = 2%, and T = 1 year. ¶ ¶ µ µ 25 S 1 2 1 2 ln ln + r−δ+ σ T + 0.06 − 0.02 + × 0.15 1 K 2 20 2 √ √ = = d1 = 0.15 1 σ T 1. 829 3 N (d1 ) = 0.966 3 N (−d1 ) = 1 − 0.966 3 = 0.033 7 ∆call = e−δT N (d1 ) = e−0.02(1) 0.966 3 = 0.947 2 ∆put = −e−δT N (−d1 ) = −e−0.02(1) 0.033 7 = −0.03 30

12.3.2

Gamma

Gamma is a measure of the change in delta regarding change in the underlying stock price. ∂∆ ∂2C Γ= = ∂S ∂S 2 If gamma is too large a small change in stock price will cause a big change in ∆. The bigger Γ, the more often you need to adjust your holding of the underlying stocks.

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Since ∆put = ∆call − e−δT , then

12.3.3

∂∆put ∂∆call = or Γcall = Γput . ∂S ∂S

Vega

Vega is the change of option price for 1% change of stock volatility (you can think that the letter V stands for volatility). ∂C V ega = 100∂σ

12.3.4

Theta

Theta is the change of option price regarding change in time when the option is written (you can think that the letter T represents time). Let t represent the time when the option is written and T the expiration date. Then ∂C (T − t) θ= ∂t

12.3.5

Rho

Rho is a measure of the change in option value regarding a 1% change in the risk free interest rate (you can think the letter R represent r) ∂C ρ= 100∂r

12.3.6

Psi

Psi is a measure of the change in option value regarding a 1% change in the dividend yield. ∂C Ψ= 100∂δ

12.3.7

Greek measures for a portfolio

The portfolio’s Greek is just the sum of the individual Greek. Example 12.3.2. A portfolio consists of 10 European calls and 50 otherwise identical puts. The information about the European call and put is as follows: S = 60,K = 65, σ = 0.25, r = 6%, δ = 4%, and T = 0.75 year. Calculate the portfolio delta. ¶ ¶ µ µ 60 S 1 2 1 2 ln ln + r−δ+ σ T + 0.06 − 0.04 + × 0.25 0.75 K 2 65 2 √ √ = d1 = = 0.25 0.75 σ T −0.192 2 N (d1 ) = N (−0.192 2) = 1 − N (0.192 2) N (0.192 2) = 0.576 2 N (d1 ) = 1 − 0.576 2 = 0.423 8 N (−d1 ) = N (0.192 2) = 0.576 2

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113

∆call = e−δT N (d1 ) = e−0.04(0.75) 0.423 8 = 0.411 3 ∆put = −e−δT N (−d1 ) = −e−0.04(0.75) 0.576 2 = −0.559 2 ∆portf olio = 10 (0.411 3) + 50 (−0.559 2) = −23. 847 Example 12.3.3. You buy 20 European calls and simultaneously write 35 European puts on the same stock. The call expires in 3 months. The put both expires in 9 months. The current stock price is 40. The call strike price is 35. The put strike price is 45. The volatility is 20%. r = 8%, δ = 3%. Calculate the delta of your portfolio. −δT Calculate ∆call N (d¶ 1) ¶ µ=e µ 40 S 1 2 1 2 ln ln + r−δ+ σ T + 0.08 − 0.03 + × 0.2 0.25 K 2 35 2 √ √ d1 = = = 0.2 0.25 σ T 1. 510 3 N (d1 ) = 0.934 5 ∆call = e−δT N (d1 ) = e−0.03(0.25) 0.934 5 = 0.927 5 −δT Calculate ∆put N (−d ¶ 1) ¶ µ = −e µ 40 1 2 1 S ln + r−δ+ σ T + 0.08 − 0.03 + × 0.22 0.75 ln K 2 45 2 √ √ = d1 = = 0.2 0.75 σ T −0.376 9 N (−d1 ) = 0.646 9 ∆put = −e−0.03(0.75) 0.646 9 = −0.632 5 ∆portf olio = 20 (0.927 5) − 35 (−0.632 5) = 40. 687 5 Since your write 35 puts, the delta of 35 puts is −35 (−0.632 5).

12.3.8

Option elasticity and volatility

The elasticity is % change in option price Ω= % change in stock price Suppose the stock price increase by where Then the option price will change by ∆. % change in stock price= S ∆ % change in option price= C ∆ ∆S C = Ω= C S

can be positive or negative.

(12.17)

The option volatility is σ option = σ stock × |Ω|

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(12.18)

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Example 12.3.4. Calculate the elasticity and the volatility of a European call option and an otherwise identical European put option. The information is: S = 80,K = 70, σ = 0.35, r = 5%, δ = 3%, and T = 0.5. µ µ ¶ ¶ 1 1 80 S + r − δ + σ2 T + 0.05 − 0.03 + × 0.352 0.5 ln ln K 2 70 2 √ √ d1 = = = 0.35 0.5 σ T 0.703 7 N (−d1 ) = √ 1 − 0.759 2 = 0.240 8 N (d1 ) = 0.759 √ 2 d2 = d1 − σ T = 0.703 7 − 0.35 0.5 = 0.456 2 N (−d2 ) = 1 − 0.675 9 = 0.324 1 N (d2 ) = 0.675 9 C = Se−δT N (d1 )−Ke−rT N (d2 ) = 80e−0.03(0.5) 0.759 2−70e−0.05(0.5) 0.675 9 = 13. 687 P = −Se−δT N (−d1 )+Ke−rT N (−d2 ) = −80e−0.03(0.5) 0.240 8+70e−0.05(0.5) 0.324 1 = 3. 150 ∆call = e−δT N (d1 ) = e−0.03(0.5) 0.759 2 = 0.747 9 ∆put = −e−δT N (−d1 ) = −e−0.03(0.5) 0.240 8 = −0.237 2 ∆call S 0.747 9 × 80 Ωcall = = = 4. 371 C 13. 687 ∆put S −0.237 2 × 80 = = −6. 024 Ωput = P 3. 150 σ call = σ stock × |Ωcall | = 0.35 (4. 371) = 1. 530 σ put = σstock × |Ωput | = 0.35 (6. 024) = 2. 108

12.3.9

Option risk premium and Sharp ratio

At t = 0 the option is worth C = ∆S + B. Suppose the option expires in h years. Let α represent the expected annual return on the stock, r the continuously compounded risk-free rate per year, and γ the expected continuously compounded return earned on the option per year. According to Equation 11.3, we have ∆S + B e−γh = ∆Seαh + Berh However, ∆S + B = C µ ¶ ∆Seαh + Berh ∆S ∆S αh eγh = = e + 1− erh C C C eγh = Ωeαh + (1 − Ω) erh

(12.19)

Equation 12.19 holds for any h. Using the Taylor’s expansion, we have: 1 1 + γh + (γh)2 + ... 2 ∙ ¸ ∙ ¸ 1 1 = Ω 1 + ah + (ah)2 + ... + (1 − Ω) 1 + rh + (rh)2 + ... 2 2

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12.4. PROFIT DIAGRAMS BEFORE MATURITY

115

For the above equation to hold for any h, it seems reasonable to assume that 1 = Ω + (1 − Ω) γh = Ωah + (1 − Ω) rh

∙ ¸ ∙ ¸ 1 1 1 2 2 2 (γh) + ... = Ω (ah) + ... + (1 − Ω) (rh) + ... 2 2 2 So we have γh = Ωah + (1 − Ω) rh or γ − r = Ω (α − r)

(12.20)

The Sharp ratio of an asset is the asset’s risk premium divided by the asset’s volatility: Sharp Ratio =

α−r σ

(12.21)

The Sharp ratio of an option is

Sharp Ratiooption =

Ω (α − r) α−r = = Sharp Ratiostock Ωσ stock σ stock

(12.22)

So the Sharp ratio of an option equals the Sharp ratio of the underlying stock.

12.3.10

Elasticity and risk premium of a portfolio

The elasticity of a portfolio is the weighted average of the elasticities of the portfolio components. In contrast, the Greek of a portfolio is just the sum of the Greeks of the portfolio components. The risk premium of a stock portfolio is just the portfolio’s elasticity times the stock’s risk premium: (γ − r)portf olio = Ωportf olio (α − r)

12.4

Profit diagrams before maturity

12.4.1

Holding period profit

(12.23)

Example 12.4.1. You buy a European call option that expires in 1 year and hold it for one day. Calculate your holding profit. Information is: • The stock price is 40 when you buy the option.

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• The stock price is still 40 one day later. • K = 40 • r = 0.08 • δ=0 • σ = 30% Solution. At time zero, you buy a 1-year European option. Your purchase price is the call price. ¶ µ S 1 2 ln + r−δ+ σ T K 2 √ d1 = ¶ µ σ T 40 1 ln + 0.08 − 0 + × 0.32 1 40 2 √ = = 0.416 7 0.3 1 √ √ d2 = d1 − σ T = 0.416 7 − 0.3 1 = 0.116 7 N (d1 ) = 0.661 6 N (d2 ) = 0.546 5 C = 40e−0(1) 0.661 6 − 40e−0.08(1) 0.546 5 = 6. 285 One day later,µyour option is worth: ¶ 364 40 1 ln + 0.08 − 0 + × 0.32 40 2 365 r = 0.416 1 d1 = 364 0.3 365 r √ 364 d2 = d1 − σ T = 0.416 1 − 0.3 = 0.116 5 365 N (d1 ) = 0.661 33 N (d2 ) = 0.546 37 C = 40e−0(364/365) 0.661 33 − 40e−0.08(364/365) 0.546 37 = 6. 274 Suppose you buy the option at t = 0 by paying 6. 285 and sell the option one day later for 6. 274 . Your holding period profit is: 6. 274 − 6. 285e0.08(1/365) = −0.01 2 You buy a European call option that expires in 1 year and hold it for 6 months. Calculate your holding profit. Information is: • The stock price is 40 when you buy the option. • The stock price is 40 after 6 months. • K = 40 • r = 0.08

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117

• δ=0 • σ = 30% At time zero, you buy a 1-year European option. Your purchase price is the call price. As calculated before, the call price is 6. 285 6 months later, µ the call is worth: ¶ 40 1 ln + 0.08 − 0 + × 0.32 0.5 40 2 √ d1 = = 0.294 6 0.3 0.5 √ √ d2 = d1 − σ T = 0.294 6 − 0.3 0.5 = 0.082 5 N (d1 ) = 0.615 9 N (d2 ) = 0.532 9 C = 40e−0(0.5) 0.615 9 − 40e−0.08(0.5) 0.532 9 = 4. 156 Your holding profit is: 4. 156 − 6. 285e0.08(0.5) = −2. 385 You buy a European put option that expires in 1 year and hold it for 6 months. Calculate your holding profit. Information is: • The stock price is 40 when you buy the option. • The stock price is 42 after 6 months. • K = 40 • r = 0.08 • δ = 0.02 • σ = 30% At time zero, you buy a 1-year European option. Your purchase price is the call price. ¶ µ 1 S + r − δ + σ2 T ln K 2 √ d1 = ¶ µ σ T 40 1 2 ln + 0.08 − 0.02 + × 0.3 1 40 2 √ = = 0.35 0.3 1 √ √ d2 = d1 − σ T = 0.35 − 0.3 1 = 0.05 N (−d1 ) = 0.363 2 N (−d2 ) = 0.480 1 P = −40e−0.02(1) 0.363 2 + 40e−0.08(1) 0.480 1 = 3. 487 One day later, your ¶ µ option is worth: 42 1 ln + 0.08 − 0.02 + × 0.32 0.5 40 2 √ = 0.477 5 d1 = = 0.3 0.5√ √ d2 = d1 − σ T = 0.477 5 − 0.3 0.5 = 0.265 4

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N (−d1 ) = 0.316 5 N (−d2 ) = 0.395 4 P = −42e−0.02(0.5) 0.316 5 + 40e−0.08(0.5) 0.395 4 = 2. 035 Suppose you buy the option at t = 0 by paying 3. 487 and sell the option 6 months later for 2. 035. Your holding period profit is: 2. 035 − 3. 487e0.08(0.5) = −1. 59

12.4.2

Calendar spread

The textbook has an intimidating diagram (Figure 12.14). Don’t worry about this diagram. Just focus on understanding what a calendar spread is. A calendar spread (also called time spread or horizontal spread) is an option strategy that takes advantage of the deteriorating time value of options. A calendar spread involves selling one option that has a shorter expiring date and simultaneously buying another option that has a longer expiration date, with both options on the same stock and having the same strike price. Suppose that Microsoft is trading for $40 per share. To have a calendar spread, you can sell a $40-strike call on a Microsoft stock with option expiring in 2 month. Simultaneously, you buy a $40-strike call on a Microsoft stock with option expiring in 3 months. Suppose the price of a $40-strike 2-month to expiration call is $2; the price of a $40-strike 3-month to expiration call is $5. So your net cost of having a calendar spread at time zero is $3. Then as time goes by, suppose the stock price doesn’t move much and is still around $40, then the value of your sold call and purchased call both deteriorate but at a different deteriorating speed. The value of the $40-strike 2-month to expiration call deteriorates much faster. With each day passing, this option has less and less value left. If there are only several days left before expiration, the value of the sold call will be close to zero. With each day passing, the value of the $40-strike 3-month to expiration call also decreases but at a slower speed. For example, one month later, the sold call has 1 month to expiration and is worth only $1. The purchased call has 2-month to expiration and is worth $4.5. Now the calendar spread is worth $3.5. You can close out your position by buying a $40-strike 1-month to expiration call (price: $1) and sell a $40-strike 2-month to expiration call (price: $4.5). If you close out your position, you’ll get $3.5. At time zero, you invest $3 to set up a calendar spread. One month later, you close out your position and get $3.5. Your profit (assuming no transaction cost) is $0.5. Time zero: your cost is $3 $40-strike call $40-strike call Value $2 $5 Time to expiration 2 months 3 months

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12.5. IMPLIED VOLATILITY

119

One month later: your total value is $3.5 $40-strike call $40-strike call Value $1 $4.5 Time to expiration 1 month 2 months A calendar spread can create value because as time passes the sold option (which is your liability) can quickly become worthless yet the purchased option (your asset) is still worth something. For more examples, please refer to • http://www.optionsxpress.com/educate/strategies/calendarspread. aspx • http://www.highyieldstrategy.com/artclndrsprds.htm.

12.5

Implied volatility

12.5.1

Calculate the implied volatility

Volatility cannot be observed. One approach to estimating volatility is use past returns to calculate the historical volatility. However, past volatility may be a poor estimate of future volatility because the market condition may change. Another approach to estimating volatility is to calculate the implied volatility. The call and put values depend on (S, K, T, r, δ, σ). Given (S, K, T, r, δ) and the option price, we can calculate σ. This is called the implied volatility. Calculate the implied volatility given the following information about a European call. • CEuropean = 7.25 • S = 60 • K = 55 • T = 0.75 (i.e. 9 months) • r = 0.06 • δ = 0.02 Solution. This is a difficult problem to solve manually. However, the calculation procedure is conceptually simple. Implied volatility is solved by trial and error. You use a trial σ and see whether the computed option price under the trial σ reproduces the actual option price. If the computed option price is lower than the observed option price, use a higher trial σ and try again; if the computed option price is higher

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than the observed option price, use a lower trial σ and try again. Keep doing this until you find a σ such the computed option price equals the observed option price. First, let’s tryµσ = 10% ¶ 60 1 2 ln + 0.06 − 0.02 + 0.1 0.75 55 2 √ d1 = = 1. 3944 0.1 0.75 √ √ d2 = d1 − σ T = 1. 3944 − 0.1 0.75 = 1. 307 8 N (d1 ) = 0.918 4 N (d2 ) = 0.904 5 −0.02(0.75) C = 60e 0.918 4 − 55e−0.06(0.75) 0.904 5 = 6. 725 6. 725 < CEuropean = 7.25. So increase σ and try again. Try σ = 20% µ ¶ 1 2 60 + 0.06 − 0.02 + 0.2 0.75 ln 55 2 √ d1 = = 0.762 2 0.2 0.75 √ √ d2 = d1 − σ T = 0.762 2 − 0.2 0.75 = 0.5890 N (d1 ) = 0.777 0 N (d2 ) = 0.722 1 C = 60e−0.02(0.75) 0.777 0 − 55e−0.06(0.75) 0.722 1 = 7. 958 7. 958 > CEuropean = 7.25. So decrease σ and try again Try σ = 15% µ ¶ 1 60 + 0.06 − 0.02 + 0.152 0.75 ln 55 2 √ d1 = = 0.965 7 0.15 0.75 √ √ d2 = d1 − σ T = 0.965 7 − 0.15 0.75 = 0.835 8 N (d1 ) = 0.832 9 N (d2 ) = 0.798 4 C = 60e−0.02(0.75) 0.832 9 − 55e−0.06(0.75) 0.798 4 = 7. 25 7. 25 = CEuropean = 7.25. So the implied σ is 15%.

12.5.2

Volatility skew

Volatility skew refers to • Options on the same stock with different strike price and expiration date should have the same implied volatility. However, in reality, options on the same stock with different strike price and expiration date don’t have the same implied volatility. • In addition, the different implied volatilities often form a pattern such as "smiles," "frowns," and "smirks." Volatility is explained more in Derivatives Markets Chapter 23.

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12.6. PERPETUAL AMERICAN OPTIONS

12.5.3

121

Using implied volatility

Implied volatility is important because it helps us • We can generate option price that’s consistent with the price of other similar options • We can quote the option in terms of volatility rather than a dollar price • Volatility skew helps us see how well an option pricing formula works. Volatility skew shows that the Black-Scholes formula and assumptions are not perfect. This is all you need to know about how to use the implied volatility.

12.6

Perpetual American options

Perpetual American options are excluded from the exam syllabus. Please ignore this chapter. I included this section for completeness, but you don’t need to read it.

12.6.1

Perpetual calls and puts

Our task here is to derive the price formula for a perpetual American call with strike price K. A perpetual American option never expires and the option holder can exercise the option at any time. It’s a typical American option where the expiration date T = +∞. The theoretical framework behind the perpetual option formula is the BlackScholes partial differential equation (called the Black-Scholes PDE). The BlackScholes PDE is Derivatives Market Equation 21.11 (page 682): 1 Vt + σ 2 S 2 VSS + (r − δ) SVS − rV = 0 2

(Textbook 20.11)

The above equation is also presented in Derivatives Market Equation 13.10 (page 430): 1 θ + σ 2 St2 Γt + rSt ∆t − rV (St ) = 0 2

(Textbook 13.10)

Please note that Equation 13.10 assumes δ = 0. In addition,it uses C (instead of V ) to represent the option price. The Black-Scholes PDE is commonly written as: ∂V (t, St ) 1 2 2 ∂ 2 V (t, St ) ∂V (t, St ) + (r − δ) St − rV (t, St ) = 0 (12.24) + σ St ∂t 2 ∂St2 ∂St

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In the above equation, V (t, St ) is the option price at time t where the stock price is St . If you are interested in learning how to derive the Black-Scholes PDE, refer to the textbook. For now let’s accept Equation 12.24. ∂V For a perpetual option, its value doesn’t depends on time. Hence = 0. ∂t The Black-Scholes PDE becomes an ordinary differential equation: 1 2 2 d2 V (t, St ) dV (t, St ) + (r − δ) St − rV (t, St ) = 0 σ St 2 dSt2 dSt

(12.25)

To find the solution to Equation 12.25, let’s simplify the equation as d2 V (t, St ) dV (t, St ) + St − V (t, St ) = 0 St2 dSt2 dSt We can guess the solution is in the form of V (t, St ) = Sth . Then d2 V (t, St ) dV (t, St ) = hSth−1 = h (h − 1) Sth−2 dSt dSt2 d2 V (t, St ) dV (t, St ) St2 + St − V (t, St ) dSt2 dSt = St2 h (h − 1) Sth−2 + St hSth−1¡ − Sth ¢ = Sth [h (h − 1) + h − 1] = Sth h2 − 1 ¡ ¢ d2 V (t, St ) dV (t, St ) So as long as h2 − 1 = 0, or h = ±1, Equation St2 +St − 2 dSt dSt V (t, St ) = 0 has a solution. Of course, if Sth is a solution, aSth must also be a solution. Similarly, we can guess that the solution to Equation 12.25 is in the form of V (t, St ) = Sth . Some brilliant thinker guessed the following solution: µ ¶h St H∗ − K h ∗ V (t, St ) = (H − K) = St = aSth h H∗ (H ∗ ) Here H ∗ the stock price where exercise is optimal (H ∗ is a constant). H ∗ −K µ ¶h St is the terminal payoff at exercise time. is an indicator telling us how H∗ ∗ close the stock price approaches H . d ¡ h¢ d2 V (t, St ) dV (t, St ) = = ah (h − 1) Sth−2 aSt = ahSth−1 dSt dS dSt2 Equation 12.25 becomes: 1 2 2 σ St ah (h − 1) Sth−2 + (r − δ) St ahSth−1 − raSth = 0 2 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 ¶ µ 1 2 2 1 2 σ h + r−δ− σ h−r =0 2 2 sµ ¶2 1 1 2 r − δ − σ 2 + 2σ 2 r σ − (r − δ) ± 2 2 h= 2 σ

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12.6. PERPETUAL AMERICAN OPTIONS 1 r−δ = − 2 ± 2 σ 1 r−δ = − 2 ± 2 σ

sµ sµ

r−δ 1 − σ2 2 1 r−δ − 2 2 σ

¶2

¶2

+

2r σ2

+

2r σ2

123

But how can we find H ∗ ? Since the perpetual American option can be exercised at any time, the option holder will choose H ∗ such that V (t, St ) = µ ¶h St dV (t, St ) reaches its maximum value. This requires setting = (H ∗ − K) ∗ H dH ∗ 0. " µ ¶h # dV (t, St ) St d ∗ (H − K) = dH ∗ dH ∗ H∗ h i d ∗ 1−h ∗ −h ) − K (H ) (H = Sth hdH ∗ i −h −h−1 h = St (1 − h) (H ∗ ) − K (−h) (H ∗ ) =0 (1 − h) (H ∗ )−h − K (−h) (H ∗ )−h−1 = 0 (1 − h) H ∗ − K (−h) = 0 K (−h) h H∗ = = K 1−h h−1 h 1 H∗ − K = K −K = K h−1 h−1 ∗

So V (t, St ) = (H − K)

µ

St H∗

¶h

K = h−1

µ

h − 1 St h K

¶h

Set t = 0. Let S represent the stock price at time zero (i.e. S = S0 ). Then the option value at time zero is µ ¶h µ ¶h S K h−1 S V (0, S) = (H ∗ − K) = H∗ h−1 h K But how do we choose h since there are two possible value of h? H ∗ − K = 1 K. To avoid h − 1 becoming negative, we choose the bigger h: h−1 sµ ¶2 1 r−δ 1 r−δ 2r + + 2 hcall = − − 2 2 σ2 2 σ σ A similar logic can be applied to a perpetual American put. We can guess the put value is: ¶h µ St K − H∗ h ∗ V (t, St ) = (K − H ) = St = bSth h H∗ (H ∗ ) Once again, we get the following equation:

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1 2 σ h (h − 1) + (r − δ) h − r = 0 2 ¶ µ 1 1 2 2 σ h + r − δ − σ2 h − r = 0 2 s2µ ¶2 2r 1 r−δ 1 r−δ + 2 h= − 2 ± − 2 σ 2 σ2 σ dV (t, St ) = 0: dH ∗ " ¶h # µ h i d dV (t, St ) S t h d ∗ ∗ ∗ −h (K − H = (S = ) ) ) (H ) (K − H t dH ∗ dH ∗ H∗ dH ∗ h i d K (H ∗ )−h − (H ∗ )1−h = (St )h hdH ∗ i h −h−1 −h = (St ) K (−h) (H ∗ ) − (1 − h) (H ∗ ) =0

Set

K (−h) (H ∗ )

−h−1

−h

− (1 − h) (H ∗ )

=0

h → H∗ = K K (−h) − (1 − h) H ∗ = 0 h − 1 µ ¶ h 1 K= K → K − H∗ = 1 − h−1 1−h To avoid 1 − h becoming we choose the smaller h: sµ negative, ¶ 2 1 r−δ 2r 1 r−δ − + 2 − hput = − 2 σ2 2 σ2 σ Summary of the formulas for perpetual American calls and perpetual American puts:

Cperpetual =

∗ (HCall

− K)

µ

S ∗ HCall

¶hcall

K = hcall − 1

µ

hcall − 1 S hcall K

¶hcall

(12.26)

hcall K hcall − 1 µ ¶ sµ ¶2 1 r−δ 2r 1 r−δ = + 2 + − − 2 2 2 σ 2 σ σ ∗ = HCall

hcall

¢ ¡ ∗ Pperpetual = K − Hput

µ

S ∗ Hput

¶hput

∗ Hput =

Yufeng Guo, Fall 09 MFE, actuary88.com

=

K 1 − hput

hput K hput − 1

µ

hput − 1 S hput K

(12.27)

(12.28) ¶hput

(12.29)

(12.30)

12.6. PERPETUAL AMERICAN OPTIONS

hput =

µ

1 r−δ − 2 2 σ

hcall and hput satisfies:

¶

−

sµ

1 r−δ − 2 2 σ

125 ¶2

+

2r σ2

1 2 σ h (h − 1) + (r − δ) h − r = 0 2

(12.31) (12.32)

Example 12.6.1. Calculate the price of a perpetual American call and the price of an otherwise identical perpetual American put. The information is as follows. The current stock price is S = 50. The strike price is K = 45. The continuously compounded risk-free rate is r = 6%. The continuously compounded dividend yield is 2%. The stock volatility is σ = 25%. Solution. Solve for h. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.252 h (h − 1) + (0.06 − 0.02) h − 0.06 = 0 2 h2 = −1. 532 7 h1 = 1. 252 7 Use the bigger h for call and the smaller h for put. Next, calculate the stock price where exercising the option is optimal. hcall 1. 252 7 ∗ HCall = K= × 45 = 223. 08 hcall − 1 1. 252 7 − 1 hput −1. 532 7 ∗ = Hput K= × 45 = 27. 23 hput − 1 −1. 532 7 − 1 µ µ ¶hcall ¶1. 252 7 S 50 ∗ = (223. 08 − 45) = Cperpetual = (HCall − K) ∗ HCall 223. 08 27. 35 ¶hput µ µ ¶−1. 532 7 ¢ ¡ S 50 ∗ Pperpetual = K − Hput = (45 − 27. 23 ) = 7. ∗ Hput 27. 23 00 Tip 12.6.1. The CD attached to the textbook Derivatives Markets has a spreadsheet that calculates the price of a perpetual American call and a perpetual American put. The spreadsheet is titled "optbasic2." You can use this spreadsheet to double check your solution.

12.6.2

Barrier present values

Consider the following barrier option. If the stock price first reaches a preset price H from below, then the payoff of $1 is received. This is a special case of a ∗ perpetual American call option by setting the terminal payoff HCall − K as $1 ∗ and by setting HCall = H.

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The value at time zero of $1 received when the stock price first reaches H from below (i.e. the stock first rises to H) is µ h1 =

µ

1 r−δ − 2 2 σ

¶

S H +

¶h1

sµ

(12.33) 1 r−δ − 2 2 σ

¶2

+

2r σ2

(12.34)

Similarly, the value at time zero of $1 received when the stock price first reaches H from above (i.e. the stock first falls to H) is µ h2 =

µ

1 r−δ − 2 2 σ

h1 and h2 satisfies:

¶

S H −

¶h2

sµ

(12.35) 1 r−δ − 2 2 σ

¶2

+

2r σ2

1 2 σ h (h − 1) + (r − δ) h − r = 0 2

(12.36) (12.37)

Example 12.6.2. Calculate the value of a $1 paid if the stock price first reaches $100 and $60 respectively. The information is: • S = 80 • r = 6% • δ = 2% • σ = 30% Solution. • calculate the value of a $1 paid if the stock price first reaches $100 H = 100 > S. So we need to calculate the price of $1 payoff when the stock price first rises to H from below 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 2 0.3 h (h − 1) + (0.06 − 0.02) h − 0.06 = 0 2 h2 = −1. 100 5 h1 = 1. 211 6 Use the bigger h µ ¶h1 µ ¶1. 211 6 80 S = = 0.763 H 100

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127

• calculate the value of a $1 paid if the stock price first reaches $90 H = 60 < S.So we need to calculate the price of $1 payoff when the stock price first falls to H from above. h1 = 1. 211 6 h2 = −1. 100 5 Use the smaller h µ ¶h2 µ ¶−1. 100 5 S 80 = = 0.729 H 60 Tip 12.6.2. The CD attached to the textbook Derivatives Markets has a spreadsheet that calculates the price of a barrier option. The spreadsheet is titled "optbasic2." You can use this spreadsheet to double check your solution.

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Chapter 13

Market-making and delta-hedging 13.1

Delta hedging

∂V Here is the main idea behind delta hedging. Delta of an option is ∆ = . ∂S Hence for a small change in stock price, the change of the option value is approximately V1 − V0 ≈ ∆ (S1 − S0 ). Suppose you sell one European call option. If you hold ∆ shares of stock, you are immunized against a small change of the stock price. If the stock price goes up from S0 to S1 , then the call will be more valuable to the buyer and your are exposed to more risk. Suppose the value of the call goes up from V0 to V1 as the stock price goes up from S0 to S1 , then your liability will increase by V1 − V0 . At the same time, the value of your ∆ shares of stock will go up by ∆ (S1 − S0 ). Because V1 − V0 ≈ ∆ (S1 − S0 ), the increase of your liability will be roughly offset by the increase of your asset. However, under delta hedging, you are immunized against only a small change of the stock price (just like immunization by duration matching assets and liabilities is only good for a small change of interest rate). If a big change of stock price knocks off your hedging, you’ll need to rebalance your hedging. However, in the real world, continuously rebalancing the hedging portfolio is impossible. Traders can only do discrete rebalancing.

13.2

Examples of Delta hedging

Make sure you can reproduce the textbook calculation of delta-hedging for 2 days. In addition, make sure you can reproduce the textbook table 13.2 and 13.3 Exam problems may ask you to outline hedging transactions or calculate the hedging profit. 129

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The major difficulty many candidates face is not knowing how to hedge. They wonder "Should the market-maker buy stocks? Should he sell stocks?" To determine how to hedge a risk, use the following ideas: • The goal of hedging is to break even. If a trader makes money on option, he must lose money on stock; if he loses money on option, he must make money on stock. • To determine whether a trader should buy stocks or sell stocks, ask "If the stock price go up (or down), will the trader make money or lose money on the option?" • If a trader loses money on the option as the stock price goes up, then the trader needs to initially own (i.e. buy) stocks. This way, the value of the trader’s stocks will go up and the trader will make money on his stocks. He can use this profit to offset his loss in the option. • If a trader makes money on the option as the stock price goes up, then the trader needs to initially short sell stocks. This way, as the stock price goes up, the trader will lose money on the short sale (because he needs to buy back the stocks at a higher price). His loss in short sale can offset his profit in option. Example 13.2.1. The trader sells a call. How can he hedge his risk,? If the stock price goes up, the call payoff is higher and the trader will lose money. To hedge this risk, the trader should buy stocks. This way, if the stock price goes up, the trader makes money in the stocks. This profit can be used to offset the trader’s loss in the written call. You can also ask the question "If the stock goes down, will the trader make money or lose money on the option?" If the stock goes down, the call payoff is lower and the trader will make money. To eat up his profit in the option, the trader needs to buy stocks. This way, as the stock price goes down, the value of the trader’s stocks will go down too and the trader will lose money in his stocks. This loss will offset the trader’s profit in the written call. Example 13.2.2. The trader sells a put. How can he hedge his risk? If the stock price goes down, the put payoff is higher and the trader will make money. To hedge his risk, the trader should short sell stocks. This way, if the stock price goes down, the trader can buy back stocks at lower price, making a profit on stocks. This profit can be used to offset the trader’s loss in the written put. You can also ask the question "If the stock goes up, will the trader make money or lose money on the option?" If the stock goes up, the put payoff is lower and the trader will make money on the written put. To eat up his profit in the option, the trader needs to short sell stocks. This way, as the stock price

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131

goes up, the trader needs to buy back stocks at a higher price. The trader will lose money in his stocks. This loss will offset the trader’s profit in the written put. Example 13.2.3. The trader buys a call. How can he hedge his risk? If a trader buys a call, the most he can lose is his premium and there’s no need to hedge. This is different from selling a call, where the call seller has unlimited loss potential. However, if a trader really wants to hedge his limited risk, he can short sell stocks. Example 13.2.4. The trader buys a put. How can he hedge his risk? If a trader buys a put, the most he can lose is his premium and there’s no need to hedge. This is different from selling a put, where the call seller has a big loss potential. However, if a trader really wants to hedge his limited risk, he can buy stocks.

Example 13.2.5. Reproduce the textbook example of delta-hedging for 2 days. Here is the recap of the information. At time zero the market maker sells 100 European call options on a stock. The option expires in 91 days. • K = $40 • r = 0.08 • δ=0 • σ = 0.3 • The stock price at t = 0 is $40 • The stock price at t = 1/365 (one day later) is $40.50 • The stock price at t = 2/365 (two days later) is $39.25 The market-maker delta hedges its position daily. Calculate the marketmaker’s daily mark-to-market profit

Solution. First, let’s calculate the call premium and delta at Day 0, Day 1, and Day 2. I used my Excel spreadsheet to do the following calculation. If you can’t fully match my numbers, it’s OK.

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CHAPTER 13. MARKET-MAKING AND DELTA-HEDGING Time t Expiry T (Yrs) ¶ St µ 1 St + 0.08 + × 0.32 T ln 40 2 √ d1 = 0.3 T N (d √1 ) d2 = d1 − 0.3 T N (d2 ) C = St N (d1 ) − 40e−0.08T N (d2 ) ∆ = e−δ(T −t) N (d1 )

Day 0 (t = 0) 91/365 40

Day 1 90/365 40.50

Day 2 89/365 39.25

0.208048

0.290291

0.077977

0.582404 0.058253 0.523227 2.7804 0.58240

0.614203 0.141322 0.556192 3.0621 0.61420

0.531077 −0.070162 0.472032 2.3282 0.53108

The above table can be simplified as follows: Time t Day 0 t = 0 Day 1 t = 1/365 Expiry T T0 = 91/365 T1 = 90/365 St S0 = 40 S1 = 40.50 Ct C0 = 100 × 2.7804 = 278. 04 C1 = 306.21 ∆t ∆0 = 100 × 0.58240 = 58. 24 ∆1 = 61.420

Day 2 t = 2/365 T2 = 89/365 S2 = 39.25 C2 = 232.82 ∆2 = 53.108

Beginning of Day 0 Trader #0 goes to work Day 0 t = 0 T0 = 91/365 S0 = 40 C0 = 278.04 ∆0 = 58.240 Trader #0 goes to work. The brokerage firm (i.e. the employer of Trader #0) gives Trader #0 C0 = $278. 04. This is what the call is worth today. Trader #0 needs to hedge the risk of the written call throughout Day 0. To hedge the risk, Trader #0 buys ∆0 stocks, costing ∆0 S0 = 58. 24×40 = $ 2329. 6. Since Trader #0 gets $278. 04 from the brokerage firm, he needs to borrow: ∆0 S0 − C0 = 2329. 6 − 278. 04 = $2051. 56. The trader can borrow $2051. 56 from a bank or use this own money. Either way, this amount is borrowed. The borrowed amount earns a risk free interest rate. Now Trader #0’s portfolio is: Component Value ∆0 = 58.24 stocks 2487. 51 call liability −278. 04 borrowed amount 2051. 56 Net position 0 End of Day 0 (or Beginning of Day 1)

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133

Mark to market without rebalancing the portfolio Day 1 t = 1/365 T1 = 90/365 S1 = 40.50 C1 = 306.21 Method 1 To cancel out his position, Trader #0 can at t = 0 • buy a call (we call this the 2nd call) from the market paying C1 = $306. 21. At expiration, the payoff of this 2nd call will exactly offset the payoff of the 1st call. The 1st call is the call sold by the brokerage firm at t = 0 to the customer who bought the call. For example, if at expiration the stock price is ST = 100, then both calls are exercised. The trader gets ST − K = 100 − 40 = 60 from the 2nd call. The liability of the first call is also $60. These two calls cancel each other out. • sell out ∆0 = 58. 24 stocks for ∆0 S1 = 58. 24 × 40.5 = $2358. 72 • pay off the loan. The payment is (∆0 S0 − C0 ) erh = 2051. 56e0.08×1/365 = 2052. 01 At the end of Day 0, the trader’s profit is −C1 + ∆0 S1 − (∆0 S0 − C0 ) erh = −306. 21 + 2358. 72 − 2052. 01 = 0.5 Trader #0 hands in $0.5 profit to his employer and goes home. Method 2

We consider the change between Day 0 and Day 1.

• In the beginning of Day 0, the trader’s stock is worth ∆0 S0 = 58. 24 (40) = 2329. 6. In the end of Day 0 (or the beginning of Day 1), the trader’s stock is worth ∆0 S1 = 58. 24 (40.5) = 2358. 72. The value of the trader’s 58. 24 stocks goes up by 58. 24 (40.5 − 40) = 29. 12. This is good for the trader. • In the beginning of Day 0, the call is worth C0 = $278. 04. In the end of Day 0 (or the beginning of Day 1), the call is worth C1 = 306.21. The call value is the trader’s liability. Now the trader’s liability increases by 306. 21 − 278. 04 = 28. 17. So the trader has a loss 28. 17 (or a gain of −28.75). Recall under Method 1, the trader has to buy a call for 306. 21 to cancel out the call he sold. So call value increase is bad for the trader. • In the beginning of Day 0, the borrowed amount is ∆0 S0 − C0 = 2051. 56. In the end of Day 0 (or the beginning of Day 1), this borrowed amount grows to (∆0¡S0 − C0 ¢) erh = 2051. 56e0.08×1/365 = 2052. 01. The increase (∆0 S0 − C0 ) erh − 1 = 2051. 56e0.08×1/365 − 2051. 56 = 0.45. This is the interest paid on the amount borrowed. No matter the trader borrows money from a bank or uses his own money, the borrowed money needs to earn a risk free interest rate.

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0.5

CHAPTER 13. MARKET-MAKING AND DELTA-HEDGING

At the end of Day 0, the trader’s profit is: ¡ ¢ ∆0 S1 − ∆0 S0 − (C1 − C0 ) − (∆0 S0 − C0 ) erh − 1 = 29. 12 − 28. 17 − 0.45 =

You can verify that ¡ ¢ ∆0 S1 −∆0 S0 −(C1 − C0 )−(∆0 S0 − C0 ) erh − 1 = −C1 +∆0 S1 −(∆0 S0 − C0 ) erh Method 3

On Day 0, the trader owns ∆0 = 58.240 stocks to hedge the call liability C0 = $278. 04. The trader’s net asset is M V (0) = ∆0 S0 − C0 = 58. 24 × 40 − 278. 04 = $2051. 56 In the end of Day 0 (or the beginning of Day 1) before the trader rebalances his portfolio, the trader’s asset is: M V BR (1) = ∆0 S1 − C1 = 58. 24 × 40.5 − 100 × 3.0621 = 2052. 51 BR stands for before rebalancing. The trader’s profit at the end of Day 0 is: M V BR (1) − M V (0) e0.08×1/365 = (∆0 S1 − C1 ) − (∆0 S0 − C0 ) erh = 2052. 51 − 2051. 56e0.08×1/365 = 0.50 Please note that Trader #0 doesn’t need to rebalance the portfolio. The portfolio is rebalanced by the next trader. Beginning of Day 1 Trader #1 goes to work Day 1 t = 1/365 T1 = 90/365 S1 = 40.50 C1 = 306.21 ∆1 = 61.420 Trader #1 goes to work. He starts from a clean slate. The portfolio is automatically rebalanced since Trader #1 starts from scratch. The brokerage firm gives Trader #1 C1 = $306.21. Trader #1 needs to hedge the risk of the written call throughout Day 1. To hedge the risk, Trader #1 buys ∆1 = 61.420 stocks, costing ∆1 S1 = 61.42 × 40.5 = $2487. 51. Since Trader #1 gets $306.21 from the brokerage firm, he needs to borrow: ∆1 S1 − C1 = 2487. 51 − 306.21 = 2181. 3 The trader can borrow $2181. 3 from a bank or use this own money. Either way, this amount is borrowed. The borrowed amount earns a risk free interest rate. Now Trader #1’s portfolio is:

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13.2. EXAMPLES OF DELTA HEDGING component ∆1 = 61.42 stocks call liability borrowed amount Net position

135

value 2487. 51 −306.21 2181. 3 0

One question arises, "What if Trader #1 doesn’t start from a clean slate?" Next, we’ll answer this question. Instead of starting from scratch, Trader #1 can start off with Trader #0’s portfolio. At the end of Day 0, Trader #0 has • ∆0 = 58.24 stocks • a borrowed amount (∆0 S0 − C0 ) erh = 2051. 56e0.08×1/365 = 2052. 01 • 0.5 profit At the end of Day 0 or the beginning of Day 1, ∆1 = 61.420. So Trader #1 buys additional shares: ∆1 − ∆0 = 61.42 − 58.24 = 3. 18 The cost of these additional shares is (∆1 − ∆0 ) S1 = 3. 18 × 40.5 = 128. 79 Since these additional shares are bought at the current market price, Trader #1 can sell these shares at the same price he bought them. This doesn’t affect the mark-to-market profit. Trader #1 can borrow 128. 79 to pay for the purchase of 3. 18 stocks. Now Trader #1 has a total of ∆1 = 61.42 shares worth ∆1 S1 = 61.42×40.5 = 2487. 51 His liability is now C1 = 306.21 The borrowed amount is now 2052. 01 + 128. 79 + 0.5 = 2181. 3. The 0.5 (the profit made by Trader #0) is the amount of money Trader #1 borrows from Trader #0. Now Trader #1’s portfolio is: component value ∆1 = 61.42 stocks $2487. 51 call liability −306.21 borrowed amount 2181. 3 Net position 0 The portfolio is the same as the portfolio if Trader #1 starts from scratch. Calculation is simpler and cleaner if we start from scratch. End of Day 1 (or Beginning of Day 2) Mark to market without rebalancing the portfolio Day 2 t = 2/365 T2 = 89/365 S2 = 39.25 C2 = 232.82

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Method 1 To cancel out his position, the trader can • buy a call from the market paying C2 = 232.82. The payoff of this call will exactly offset the payoff of the call sold by the brokerage firm to the customer. These two call have the common expiration date T1 = 89/365 and the same payoff. They will cancel each other out. • sell out ∆1 = 61.420 stocks for ∆1 S2 = 61.420 × 39.25 = 2410. 735 • pay off the loan. The payment is (∆1 S1 − C1 ) erh = 2181. 3e0.08×1/365 = 2181. 778 The trader’s profit at the end of Day 1 is −C2 + ∆1 S2 − (∆1 S1 − C1 ) erh = 232.82 + 2410. 735 − 2181. 778 = −3. 863 Trader #1 hands in −3. 863 profit to his employer and goes home. Method 2 We consider the change between the beginning of Day 1 and the end of Day 1 (or the beginning of Day 2). • In the beginning of Day 1, the trader’s stock is worth ∆1 S1 ; In the end of Day 1 (or the beginning of Day 2), the trader’s stock is worth ∆1 S2 . The value of the trader’s stocks goes up by ∆1 S2 − ∆1 S1 • In the beginning of Day 1, the call is worth C1 ; In the end of Day 1 (or the beginning of Day 2), the call is worth to C2 . The call value is the trader’s liability. Now the trader’s liability increases by C2 − C1 • In the beginning of Day 1, the borrowed amount is ∆1 S1 − C1 . In the end of Day 1 (or the beginning of Day 2), this borrowed ¡ ¢amount grows to (∆1 S1 − C1 ) erh . The increase is (∆1 S1 − C1 ) erh − 1 . This is the interest paid on the amount borrowed. No matter the trader borrows money from a bank or uses his own money, the borrowed money needs to earn a risk free interest rate. The trader’s profit at the end of Day 1: ¡ ¢ (∆1 S2 − ∆1 S1 ) − (C2 − C1 ) − (∆1 S1 − C1 ) erh − 1 = −C2 + ∆1 S2 − (∆1 S1 − C1 ) erh = −3. 863 Method 3 In the beginning of Day 1, the trader’s net asset is M V (1) = ∆1 S1 − C1 = 61.420 × 40.5 − 306.21 = 2181. 3 In the end of Day 1 before the trader rebalances his portfolio, the trader’s asset is: M V BR (2) = ∆1 S2 − C2 = 61.420 × 39.25 − 232.82 = 2177. 915 The trader’s profit at the end of Day 1 is: M V BR (2) − M V (1) erh = 2177. 915 − 2181. 3e0.08×1/365 = −3. 863

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137

You can verify that M V BR (2) − M V (1) erh ¡ ¢ = (∆1 S2 − ∆1 S1 ) − (C2 − C1 ) − (∆1 S1 − C1 ) erh − 1 = −C2 + ∆1 S2 − (∆1 S1 − C1 ) erh Method 3 is often faster. Under this method, Profit during a day=Asset at the end of Day before rebalancing the portfolio - Future value of the asset at the beginning of the day Asset at the end of Day before rebalancing the portfolio = delta at the beginning of the day × stock price at the end of the day Profit at the end of Day t = M V BR (t + 1)−M V (t) erh = (∆t St+1 − Ct+1 )− (∆t St − Ct ) erh Now this completes the textbook example on page 417 to 418. Next, let’s do additional calculations and calculate the profit at the end of Day 2, 3, and 4 (or the profit at the beginning of Day 3,4,5) Day Time t Expiry T St Ct ∆t Profit end of Day

Day 2 t = 2/365 T2 = 89/365 S2 = 39.25 C2 = 232.82 ∆2 = 53.108 0.40

Day 3 t = 3/365 T3 = 88/365 S3 = 38.75 C3 = 205. 46 ∆3 = 49. 564 −4. 0

Day 4 t = 4/365 T4 = 87/365 S4 = 40 C4 = 271.04 ∆4 = 58.06 1. 32

Day 5 t = 5/365 T5 = 86/365 S5 = 40 C5 = 269.27 ∆5 = 58.01

Profit at the end of Day 2 (or beginning of Day 3). • Asset at the beginning of Day 2: M V (2) = ∆2 S2 − C2 = 53.108 × 39.25 − 232.82 = 1851. 669 • Asset at the end of Day 2 before rebalancing: M V BR (3) = ∆2 S3 − C3 = 53.108 × 38.75 − 205. 46 = 1852. 475 • Profit at the end of Day 2: M V BR (3) − M V (2) erh = 1852. 475 − 1851. 669e0.08/365 = 0.400 1

Profit at the end of Day 3 (or beginning of Day 4). • Asset at the beginning of Day 3: M V (3) = ∆3 S3 − C3 = 49. 564 × 38.75 − 205. 46 = 1715. 145 • Asset at the end of Day 3 before rebalancing: M V BR (4) = ∆3 S4 − C4 = 49. 564 × 40 − 271.04 = 1711. 52 • Profit at the end of Day 3: M V BR (4) − M V (2) erh = 1711. 52 − 1715. 145e0.08/365 = −4. 000 96

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Profit at the end of Day 4 (or beginning of Day 5). • Asset at the beginning of Day 4: M V (4) = ∆4 S4 − C4 = 58.06 × 40 − 271.04 = 2051. 36 • Asset at the end of Day 4 before rebalancing: M V BR (3) = ∆4 S5 − C5 = 58.06 × 40 − 269.27 = 2053. 13 • Profit at the end of Day 4: M V BR (5) − M V (4) erh = 2053. 13 − 2051. 36e0.08/365 = 1. 320 3 You can verify that the profit at the end of Day 2, 3, 4 calculated above matches Derivatives Markets Table 13.2. However, in Table 13.2, the profit at the end of Day 0 is posted in Day 1 column. Similarly, the profit at the end of Day 1 is posted in Day 1 column. So on and so forth.

13.3

Textbook Table 13.2

Next, we want to reproduce the textbook Table 13.2. Profit at the end of Day t (or the beginning of Day t + 1) can be broken down into two parts: = M V BR (t + 1) − M V (t) erh = M V BR (t + 1) − M V (t) − M V (t) erh + M V (t) ¡ ¢ = M V BR (t + 1) − M V (t) − M V (t) erh − 1 ¡ ¢ −M V (t) erh − 1 = M V BR (t + 1) − M V (t) + {z } | {z } | capital gain at end of Day t

Interest earned at the end of Day t

Define M V BR (t + 1) − M V (t) = CapitalGain earned at the end of Day t CapitalGain (t) is also equal to: M V BR (t + 1)−M V (t) = (∆t St+1 − Ct+1 )−(∆t St − Ct ) = ∆t (St+1 − St )− (Ct+1 − Ct )

¡ ¢ Define −M V (t) erh − 1 as the interest earned at the end of Day t. If the trader invests money at the beginning of Day t, then M V (t) is positive and ¡ ¢ −M V (t) erh − 1 is negative. The negative interest earned is just the interest expense incurred by the trader. Define M V (t) as the investment made at the beginning of Day t. If M V (t) is negative, then it means that the trader receives money. Example 13.3.1. Reproduce the textbook Table 13.2 We already reproduced the daily profit in Table 13.2. We just need to reproduce the investment, interest, and the capital gain.

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Day Time t Expiry T St Ct ∆t Investment (beginning of the day) interest earned during the day Capital gain (end of the day) Profit (end of the day)

0 t=0 T0 = 91/365 S0 = 40 C0 = 278.04 ∆0 = 58.2404 2051. 56 −0.45 0.95 0.50

1 t = 1/365 T1 = 90/365 S1 = 40.5 C1 = 306.21 ∆1 = 61.42 2051. 56 −3. 385 −0.478 −3. 863

2 t = 2/365 T2 = 89/365 S2 = 39.25 C2 = 232.82 ∆2 = 53.108 1851. 669 −0.41 0.81 0.40

Day 0 We already know: M V (0) = ∆0 S0 − C0 = 58. 24 × 40 − 278. 04 = $2051. 56 M V BR (1) = ∆0 S1 − C1 = 58. 24 × 40.5 − 100 × 3.0621 = 2052. 51 The trader’s profit at the end of Day 0 is: M V BR (1) − M V (0) erh = 2052. 51 − 2051. 56e0.08×1/365 = 0.50 To find the capital gain and the interest earned at the end of Day 0, we just need to break down the profit M V BR (1) − M V (0) erh into two parts: M V BR (1) − M V (0) erh = M V BR (1) − M V (0) − M V (0) erh + M V (0) ¡ rh ¢ BR = MV (1) − M V (0) + −M V (0) e − 1 {z } | {z } | capital gain

interest earned

The capital gain at the end of Day 0: M V BR (1) − M V (0) = 2052. 51 − 2051. 56 = 0.95

The interest ¡ credited ¢ at the end¡of Day 0: ¢ −M V (0) erh − 1 = −2051. 56 e0.08×1/365 − 1 = −0.449 7 = −0.45 Investment at the beginning of Day 0: M V (0) = 2051. 56 Please note the textbook shows the interest credited at the end of Day 0, capital gain earned at the end of Day 0, and daily profit at the end of Day 0 in Day 1 column. Day 1 M V (1) = ∆1 S1 − C1 = 61.420 × 40.5 − 306.21 = 2181. 3 M V BR (2) = ∆1 S2 − C2 = 61.420 × 39.25 − 232.82 = 2177. 915 The trader’s profit at the end of Day 1 is: M V BR (2) − M V (1) erh = 2177. 915 − 2181. 3e0.08×1/365 = −3. 863

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To find the capital gain and the interest earned at the end of Day 0, we just need to break down the profit M V BR (1) − M V (0) erh into two parts: ¡ ¢ M V BR (2) − M V (1) erh = M V BR (2) − M V (1) + −M V (1) erh − 1 {z } | {z } | capital gain

interest earned

Capital gain at the end of Day 1: M V BR (2) − M V (1) = 2177. 915 − 2181. 3 = −3. 385

¡ ¢ ¡ ¢ Interest earned at the end of Day 1: −M V (1) erh − 1 = −2181. 3 e0.08×1/365 − 1 = −0.478 Investment at the beginning of Day 1: M V (1) = ∆1 S1 − C1 = 61.420 × 40.5 − 306.21 = 2181. 3 You should be able to reproduce Table 13.2 for the other days. Day 3 4 Time t t = 3/365 t = 4/365 Expiry T T3 = 88/365 T4 = 87/365 St S3 = 38.75 S4 = 40 Ct C3 = 205. 46 C4 = 271.04 ∆t ∆3 = 49. 564 ∆4 = 58.06 Investment (beginning of the day) 1715. 15 2051. 36 interest earned during the day −0.38 −0.45 Capital gain (end of the day) −3. 63 1. 77 Profit (end of the day) −4. 01 1. 32

13.4

5 t = 5/365 T5 = 86/365 S5 = 40 C5 = 269.27 ∆5 = 58.01

Textbook Table 13.3

The stock price of Table 13.3 follows the binomial tree with σ = 0.3. On Day 0, the stock price is S0 = 40 On Day 1 the√stock moves up 1 σ√ S1 = S0 erh+σ h = 40e0.08/365+0.3 1/365 = 40.642 Day 2 the stock moves down 1 σ √ √ rh−σ h S2 = S1 e = 40.642e0.08/365−0.3 1/365 = 40. 018 Day 3 the stock moves down 1 σ √ √ S3 = S2 erh−σ h = 40. 018e0.08/365−0.3 1/365 = 39. 403 Day 4 the stock moves down 1 σ √ √ rh−σ h S4 = S3 e = 39. 403e0.08/365−0.3 1/365 = 38. 797 Day 5 the stock moves up 1 σ √ √ S5 = S4 erh+σ h = 38. 797e0.08/365+0.3 1/365 = 39. 420 If you use the same method for reproducing Table 13.2, you should be able to reproduce Table 13.3. When reading Table 13.3, remember the interest, the capital gain, and the daily profit on Day 1 is the interest, the capital gain, and the daily profit at the end of Day 0 (or the beginning of Day 1). Similarly, the

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interest, the capital gain, and the daily profit on any other day is the interest, the capital gain, and the daily profit at the end of the previous day or the beginning of that day. The author of the textbook uses Table 13.3 to show us that if the stock price moves up or down 1 σ daily, then the trader’s profit is zero.

13.5

Mathematics of Delta hedging

13.5.1

Delta-Gamma-Theta approximation

First, let’s understand the textbook Equation 13.6: 1 V (St+h , T − t − h) ≈ V (S0 , T − t) + ∆t + θh + Γt 2

2

(Textbook 13.6)

Let’s consider an option written at time t expiring on date T (i.e. expiring in T − t years). The value of this option is V (St , T − t), where St is the stock price at t. Suppose a tiny time interval h (such as 0.00001 second) has passed and we are now standing at t + h. Now the option has a remaining life T − t − h years and is worth V (St+h , T − t − h), where St+h is the stock price at t + h. Suppose St+h = St + . Time t t+h T Stock price St St+h = St + Option value V (St , T − t) V (St+h , T − t − h) We want to estimate V (St+h , T − t − h). By Taylor series, we have: ∂f (x0 , y0 ) ∂f (x0 , y0 ) f (x0 + x , y0 + y ) ≈ f (x0 , y0 ) + x+ y ∂x ∂y 2 2 1 ∂ f (x0 , y0 ) 2 1 ∂ f (x0 , y0 ) 2 + x+ y 2 ∂x2 2 ∂y 2 Similarly, V (St+h , T − t − h) = V [St+h , − (t + h) + T ] ∂V (St , T − t) ∂V (St , T − t) ≈ V (St , T − t) + + h ∂S ∂t 2 2 1 ∂ ∂ V (St , T − t) 2 1 V (St , T − t) 2 + h + 2 ∂S 2 2 ∂t2 However, ∂V (St , T − t) = ∆t ∂S

∂V (St , T − t) =θ ∂t

1 ∂ 2 V (St , T − t) 2 h since it’s close to zero. However, 2 ∂t2 1 ∂ 2 V (St , T − t) 2 is not close to zero. The reason that h 2 ∂t2

We decide to ignore 1 ∂ 2 V (St , T − t) 2 ∂S 2

2

∂ 2 V (St , T − t) = Γt ∂S 2

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1 ∂ 2 V (St , T − t) 2 is close to zero but is not close to zero will be explained 2 ∂S 2 in Derivatives Markets Chapter 20 when we derive Ito’s Lemma. For now just accept it. Now we have: 1 V (St+h , T − t − h) ≈ V (St , T − t) + ∆t + θh + Γt 2

13.5.2

2

Understanding the market maker’s profit

Suppose a trader sets up a hedging portfolio at time t. The trader’s profit after a short interval h (i.e. at time t + h) is: P rof it (t + h) = M V BR (t + h) − M V (t) erh = (∆t St+h − Ct+h ) − (∆t St − Ct ) e¡rh ¢ = (∆t St+h − Ct+h ) − (∆t St − Ct ) erh − 1 + (∆ Ct ) t St − ¢ ¡ rh = ∆t (St+h − St ) − (Ct+h − Ct ) − (∆t St − Ct ) e − 1 For a small h, using Taylor series, we get erh ≈ 1 + rh P rof it (t + h) = ∆t (St+h − St ) − (Ct+h − Ct ) − rh (∆t St − Ct ) According to the textbook Equation 13.6, 1 Ct+h − Ct = ∆t + θh + Γt 2 2 µ

1 P rof it (t + h) = ∆t (St+h − St ) − ∆t + θh + Γt 2 2 Since St+h = St + , we have: P rof it (tµ+ h) ¶ 1 2 = ∆t − ∆t + θh + Γt − rh (∆t St − Ct ) 2 ¶ µ 1 2 − rh (∆t St − Ct ) = − θh + Γt 2 ¶ µ 1 2 Γt + θh − rh [∆t St − Ct ] P rof it (t + h) ≈ − 2 However, St+h = St erh+σ

¶

− rh (∆t St − Ct )

(Textbook 13.7)

√ h

Using the Taylor series, we have: erh+σ ...

√ h

³ √ ´ 1³ √ ´2 = 1+ rh + σ h + rh + σ h + 2

√ ´2 1³ rh + σ h and higher order terms all approach zero. 2 √ √ In addition, as h approaches 0, h is much larger than h. For example, 0.0001 = 0.01 is much larger than 0.0001.Hence, we can discard rh but keep √ σ h. As h → 0,

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¶ µ ³ ³ √ ´ 1³ √ ´ √ ´2 = St 1 + rh + σ h + rh + σ h + ≈ St 1 + σ h 2 √ → = St+h − St ≈ St σ h → 2 ≈ St2 σ2 h Plug the above equation in Textbook Equation 13.7, we get: → St+h = St erh+σ

√ h

P rof it (t + h) ≈ −

µ

¶ 1 2 2 St σ hΓt + θh − rh [∆t St − Ct ] 2

(Textbook 13.9)

From the textbook Table 13.3, we know that if the stock price moves up or down µ by 1 σ, the trader’s profit is zero. ¶So 1 2 2 − S σ hΓt + θh − rh [∆t St − Ct ] = 0 2 t This gives us the Black-Scholes PDE: 1 2 2 S σ Γt + θ − r [∆t St − Ct ] = 0 2 t

(Textbook 13.10)

Next, the textbook has the following topics: • Delta-hedging of American options • The advantage to frequent rehedging • Delta-hedging in practice • Market making as insurance These topics are minor ideas. I recommend that you skip them.

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Chapter 14

Exotic options: I Any option that is not a plain vanilla call or put is called an exotic option. There are usually no markets in these options and they are purely bought OTC (over-the-counter). They are much less liquid than standard options. They often have discontinuous payoffs and can have huge deltas near expiration which make them difficult to hedge. Before studying this chapter, make sure you understand the learning objective. SOA’s learning outcome for this chapter: • Explain the cash flow characteristics of the following exotic options: Asian, barrier, compound, gap and exchange

If you want to cut corners, you can skip the pricing formula for exotic options because calculating the exotic option price is out of the scope of the learning outcome or learning objective.

14.1

Asian option (i.e. average options)

14.1.1

Characteristics

Asian options (also called average options) have payoffs that are based on the average price of the underlying asset over the life of the option. The average price can be the average stock price or the average strike price. The average can be arithmetic or geometric. • The unique characteristic of an average price option is that the underlying asset prices are averaged over some predefined time interval. • Averaging dampens the volatility and therefore average price options are less expensive than standard options 145

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• Average price options are path-dependent, meaning that the value of the option at expiration depends on the path by which the stock arrives at its final price. The price path followed by the underlying asset is crucial to the pricing of the option. • Average price options are useful in situations where the trader/hedger is concerned only about the average price of a commodity which they regularly purchase.

14.1.2

Examples

Examples are based on

1

• A 9-month European average price contract calls for a payoff equal to the difference between the average price of a barrel of crude oil and a fixed exercise price of USD18. The averaging period is the last two months of the contract. The impact of this contract relative to a standard option contract is that the volatility is dampened by the averaging of the crude oil price, and therefore the option price is lower. The holder gains protection from potential price manipulation or sudden price spikes. • A Canadian exporting firm doing business in the U.S. is exposed to Can$/US$ foreign exchange risk every week. For budgeting purposes the treasurer must pick some average exchange rate in which to quote Can$ cash flows (derived from US$ revenue) for the current quarter. Suppose the treasurer chooses an average FX rate of Can$1.29/US$1.00. If the US$ strengthens, the cash flows will be greater than estimated, but if it weakens, the company’s Can$ cash flows are decreased. Arithmetic average We record the stock price every h periods from time 0 to time T . There are N = T /h periods. The arithmetic average is: n 1 X A (T ) = Sih N i=1

14.1.3

Geometric average

G (T ) = (Sh S2h ...SN h )1/N 1 Based on http://www.fintools.com/doc/exotics/exoticsAbout_Average_Options.html

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14.2. BARRIER OPTION

14.1.4

147

Payoff at maturity T

Let AV G (T ) represent the average stock price. For arithmetic average, AV G (T ) = n 1 X A (T ) = Sih . For geometric average, AV G (T ) = (Sh S2h ...SN h )1/N . N i=1 Option average average average average

14.2

price call price put strike call strike put

Payoff max [0, AV G (T ) − K] max [0, K − AV G (T )] max [0, ST − AV G (T )] max [0, AV G (T ) − ST ]

Barrier option

Barrier options are similar to standard options except that they are extinguished or activated when the underlying asset price reaches a predetermined barrier or boundary price. As with average options, a monitoring frequency is defined as part of the option which specifies how often the price is checked for breach of the barrier. The frequency is normally continuous but could be hourly, daily, etc.

14.2.1

Knock-in option

• "In" option starts its life worthless and becomes active only if the barrier price is reached • Technically, this type of contract is not an option until the barrier price is reached. So if the barrier price is never reached it is as if the contract never existed. • down-and-in (spot price starts above the barrier level and has to move down for the option to become activated.) • up-and-in (spot price starts below the barrier level and has to move up for the option to become activated.)

14.2.2

Knock-out option

• "Out" option starts its life active and becomes null and void if the barrier price is reached • The option will expire worthless if the asset price exceeds the barrier price • down-and-out (spot price starts above the barrier level and has to move down for the option to become null and void) • up-and-out (spot price starts below the barrier level and has to move up for the option to be knocked out)

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CHAPTER 14. EXOTIC OPTIONS: I

Rebate option

• Barrier options are sometimes accompanied by a rebate, which is a payoff to the option holder in case of a barrier event. • Rebates can either be paid at the time of the event or at expiration.

14.2.4

Barrier parity

”Knock − in” option + ”Knock − out” option = Ordinary Option

(14.1)

Example 14.2.1. Explain why up-and-in option + up-and-out option = Ordinary option. T =option expiration time tH =time when the stock price first reaches the barrier, where the barrier price is greater than the current stock price During the time interval [0, tH ) • the up-and-out option is alive • the up-and-in option is dead During the time interval [tH , T ] • the up-and-out option is dead (because the barrier is reached) • the up-and-in option is alive (because the barrier is reached) If we have a down-and-in option and a down-and-out option with two options on the same stock, having the same barrier price and the same strike price, then at any moment in the interval [0, T ], we’ll always have exactly one ordinary option alive. Hence down-and-in option + down-and-out option = Ordinary option. Similarly, down-and-in option + down-and-out option = Ordinary option.

14.2.5

Examples

1. A European call option is written on an underlying with spot price $100, and a knockout barrier of $120. This option behaves in every way like a vanilla European call, except if the spot price ever moves above $120, the option "knocks out" and the contract is null and void. Note that the option does not reactivate if the spot price falls below $120 again. Once it is out, it’s out for good.

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2. A bank purchases an at-the-money 9-month Nikkei call option struck at 17,000 with a down-and-out barrier price of 16,000. If the price of the Nikkei falls to 16,000 or below, during the 9-month period, the bank will no longer have the benefit of Nikkei price appreciation since the call option will have been knocked out. 3. An airline is concerned that events in the Middle East might drive up the price of fuel. An up-and-in call would allow the airline to buy crude oil futures at a fixed price if some knock-in boundary price is reached. The price of the U&I call would be less than a standard call with the same expiration and exercise price so it might be viewed as a cost effective hedging instrument.

14.3

Compound option

Definition • With a compound option one has the right to buy an ordinary option at a later date • Compound options are more expensive to purchase than the underlying option, as the purchaser has received a price guarantee and effectively extended the life of the option Make sure you understand the textbook Figure 14.2. For a compound call (i.e. call on call) to be valuable, the following two conditions need to be met: • St1 > S ∗ • St1 > K Compound option parity The key formula is DM 14.12: CallOnCall − P utOnCall + xe−rt1 = BSCall

(DM 14.12)

This formula looks scary but is actually easy to remember. We know the put-call parity C + Ke−rT = P + S0 . If we treat the standard BSCall as the underlying asset, then applying the standard put-call parity, we get: CallOnCall + xe−rt1 = P utOnCall + BSCall This is DM 14.12. Similarly, we have: CallOnP ut + xe−rt1 = P utOnP ut + BSP ut

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Options on dividend-paying stocks This section is a minor part of Exam MFE. I recommend that you skip it. However, if you don’t want to skip, here is the main idea. At t1 , the value of an American call option CA (St1 , T − t1 ) is (see the textbook’s explanation for this formula):

CA (St1 , T − t1 ) = max [CE (St1 , T − t1 ) , St1 + D − K]

(DM 14.13)

Here CE (St1 , T − t1 ) is a European call option. Under the put-call parity, we get: CE (St1 , T − t1 ) = PE (St1 , T − t1 ) + St1 − Ke−r(T −t) So £ ¤ CA (St1 , T − t1 ) = max PE (St1 , T − t1 ) + St1 − Ke−r(T −t) , St1 + D − K £ ¤ = St1 +D−K+max PE (St1 , T − t1 ) + St1 − Ke−r(T −t) − (St1 + D − K) , 0 £ ¡ ¡ ¢¢ ¤ = St1 + D − K + max PE (St1 , T − t1 ) − D − K 1 − e−r(T −t) , 0

h ³ ³ ´´ i CA (St1 , T − t1 ) = St1 +D−K+max PE (St1 , T − t1 ) − D − K 1 − e−r(T −t) , 0 (DM 14.14) £ ¡ ¡ ¢¢ ¤ max PE (St1 , T − t1 ) − D − K 1 − e−r(T −t) , 0 is payoff of a call on put ¡ ¢ with strike price equal to D − K 1 − e−r(T −t) . Next, the author made two points: Exercising call on put at t1 is the same as not exercising the American call at t ¡ 1 . If you ¡ exercise the ¢¢ call on put, then it must be true that PE (St1 , T − t1 )− D − K 1 − e−r(T −t) ≥ 0 (otherwise you won’t exercise it). Then DM 14.14 becomes ¡ ¡ ¢¢ CA (St1 , T − t1 ) = St1 + D − K + PE (St1 , T − t1 ) − D − K 1 − e−r(T −t) = St1 + PE (St1 , T − t1 ) − Ke−r(T −t) = CE (St1 , T − t1 ) So exercising the call on put at t1 means that not early exercising the American call option at t1 (so the American call option becomes a European call option). The second point. If you don’t the call on ¡ exercise ¡ ¢¢ put at t1 , then it must be true that PE (St1 , T − t1 ) − D − K 1 − e−r(T −t) < 0. This gives us CA (St1 , T − t1 ) = St1 + D − K. This equation means that we receive dividend D at t1 and immediately exercise the call at t1 (so the American call option at t1 is equal to St1 + D − K). So not exercising the call on put at t1 is the same as early exercising the American call option at t1 . This is all you need to know about this section.

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DM Example 14.2 shows you how to use DM 14.14 to calculate the price of an American call option with a single dividend. When reading this example, please note two things: (1) DM Example 14.2 has errors. Make sure you download the errata. The DM textbook errata can be found at the SOA website. (2) The call on put price is calculated using the Excel spreadsheet (so don’t worry about how to manually calculate the call on put).

14.4

Gap option

14.4.1

Definition

• An option in which one strike price K1 determines the size of the payoff and another strike price K2 determines whether or not the payoff is made. • Example. A call option pays ST − 5 if ST > 8 and pays zero otherwise. The strike price K1 = 5 determines the size of the payoff but a different strike price K2 = 8 determines whether or not the payoff is made. K2 is called the payment trigger. • Example. A put option pays 8 − ST if ST > 5 and pays zero otherwise. The strike price K1 = 8 determines the size of the payoff but a different strike price K2 = 8 determines whether or not the payoff is made. • As the footnote in Page 457 of the textbook indicates, a gap call or put option is really not a option; once the payment trigger is satisfied, the owner of a gap option MUST exercise the option. Hence the premium of a gap option can be negative.

14.4.2

Pricing formula

To find the price of a gap option, you can modify a standard option’s formula by changing d1 : C (K1 , K2 ) = Se−δT N (d1 ) − K1 e−rT N (d2 ) −rT −δT N (−d¶ N (−d1 ) P (K1 , K2 ) = K 2 ) − Se µ1 e 1 2 S + r−δ+ σ T ln K2 2 √ d1 = √ σ T d2 = d1 − σ T

14.4.3

How to memorize the pricing formula

• The call payoff is 0 if ST ≤ K2 and ST − K1 if ST > K2 . The call option price is just the risk-neutral expected discounted value of the payoff. So the call price is Se−δT N (d1 ) − K1 e−rT N (d2 ). Similarly, the put price is K1 e−rT N (−d2 ) − Se−δT N (−d1 ).

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You are given the following information on a gap option on a stock • The gap call expires in 6 months • The stock is currently selling for $40 • The payment trigger is $50 • The strike price is $60 • The continuously compounded risk-free interest rate is 6% per year • The continuously compounded dividend yield is 2% per year • The stock volatility σ = 30% Calculate • the price of the gap call option. • the price of the gap put option K2 = 50 r = 0.06 K1 = 60 T = 0.5 µ ¶ ¶ µ 1 2 40 S 1 2 + r−δ+ σ T ln ln + 0.06 − 0.02 + × 0.3 0.5 K2 2 50 2 √ √ = d1 = = 0.3 0.5 σ T −0.851 6 √ √ d2 = d1 − σ T = −0.851 6 − 0.3 0.5 = −1. 063 7 N (d1 ) = 0.197 2 N (d2 ) = 0.143 7 N (−d1 ) = 0.802 8 N (−d2 ) = 0.856 3 C = Se−δT N (d1 ) − K1 e−rT N (d2 ) = 40e−0.02×0.5 0.197 2 − 60e−0.06×0.5 0.143 7 = −0.56 S = 40 δ = 0.02

P = K1 e−rT N (−d2 ) − Se−δT N (−d1 ) = 60e−0.06×0.5 0.856 3 − 40e−0.02×0.5 0.802 8 = 18. 07

14.5

Exchange option

• Allows the holder of the option to exchange one asset for another • Pricing p formula. Use the standard Black-Scholes formula except changing σ to σ 2S + σ 2K − 2ρσ S σ K . Also, make sure you know which is the stock asset and which is the strike asset. If you give up Asset 2 and receive Asset 1, Asset 1 is the stock and Asset 2 is the strike asset.

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153

Let’s walk through DM Example 14.3. S = 95.92 K = 3.5020 × 27.39 = 95. 92 = S (i.e. at-the-money call) δ S = 0.75% δ K = 1.17% σ = 0.1694 T =1 95.92e−0.0075×1 1 + × 0.16942 × 1 −0.0117×1 95.92e √2 d1 = = 0.109 5 0.1694 1 √ √ d2 = d1 − σ T = 0.109 5 − 0.1694 1 = −0.059 9 N (d1 ) = 0.543 6 N (d2 ) = 0.476 1 C = Se−δS T N (d1 ) − K1 e−δK T N (d2 ) = 95.92e−0.0075×1 × 0.543 6 − 95.92e−0.0117×1 × 0.476 1 = 6. 616 ln

This is slightly different from the textbook call price 6.6133 due to rounding. You can verify the call price using the Excel worksheet "Exchange." Inputs: Underlying Asset Price 95.92 Volatility 20.300% Dividend Yield 0.750% Strike Asset Price 95.92 Volatility 22.270% Dividend Yield 1.170% Other Correlation 0.6869 Time to Expiration (years)

1

Output: Exchange Option Black-Scholes Call Put Price 6.6144 6.2154 So the exchange call price is 6.6144; the exchange put price is 6.2154.

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154

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CHAPTER 14. EXOTIC OPTIONS: I

Chapter 18

Lognormal distribution This chapter is an easy read. I’m going to highlight the major points.

18.1

Normal distribution

The normal distribution has the following pdf (probability density function) (DM 18.1): " µ ¶2 # 1 1 x−μ φ (x; μ, σ) = √ exp − 2 σ σ 2π In the standard normal distribution, μ = 0 and σ = 1. Its pdf is: ¶ µ 1 1 2 φ (x; 0, 1) = √ exp − x 2 2π Its cdf (cumulative probability density function) is DMµ18.2: ¶ Ra Ra 1 1 P (z 6 a) = N (a) = −∞ φ (x; 0, 1) dx = −∞ √ exp − x2 dx 2 2π A frequently used formula is DM 18.3: N (−a) = 1 − N (a) How to convert a¡ normal ¢ random variable to the standard normal random variable. If x ∼ N μ, σ2 , then we can transform x into a standard normal random variable using DM 18.4: x−μ z= σ 1 1 x − μ is normal σ σ (linear combination of¶normal random variables are also normal). Next, µ x−μ 1 E (z) = E = [E (x) − μ] = 0 σ σ We can verify that z ∼ N (0, 1). First, notice that z =

155

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CHAPTER 18. LOGNORMAL DISTRIBUTION

V ar (z) = V ar So z ∼ N (0, 1)

µ

x−μ σ

¶

=

1 1 1 V ar (x − μ) = 2 V ar (x) = 2 × σ 2 = 1 2 σ σ σ

DM 18.4 can be rewritten as DM 18.7: x = μz + σ variables is still normal. For example, if x1 ∼ ¡Sum of ¡ ¢ normal random ¢ 2 N μ1 , σ 21 and x2 ∼ N μ , σ , then 2 2 ¡ ¢ ax1 + bx2 ∼ N aμ1 + bμ2 , aσ 21 + bσ 22 + 2abρσ1 σ 2

Next, the textbook briefly explains the central limit theorem, a concept you most likely already know.

18.2

Lognormal distribution

If x is normal, then y = ex is lognormal. If you say y is lognormal, it means that ln y is normal. Let R (0, t) represent the not-annualized continuously compounded return over the interval [0, t], the stock price at time t is (DM 18.12): St = S0 eR(0,t) St This leads to DM 18.11: R (0, t) = ln S0 Next comes an important formula: ´ ³ 2 ¡ ¢ 2 2 If x ∼ N m, v 2 , then E (ex ) = em+0.5v and V ar (ex ) = e2m+v ev − 1 . You don’t need to know how to derive these formulas. Just memorize them.

18.3

Lognormal model of stock prices

Define: • α, the expected (annualized) continuously compounded return earned by the stock • δ, the (annualized) continuously compounded dividend yield • σ, the stock’s volatility • z, the standard normal random variable Then the (not annualized) continuously compounded return earned during ¡ ¢ [0, t] is normally distributed with mean α − δ − 0.5σ 2 t and standard deviation √ σ t (DM 18.18): ¡¡ ¢ ¢ St ∼ N α − δ − 0.5σ 2 t, σ 2 t ln S0 random variable with mean ¡ Using DM2 ¢18.7, we can rewrite the normal √ α − δ − 0.5σ t and standard deviation σ t as:

Yufeng Guo, Fall 09 MFE, actuary88.com

18.4. LOGNORMAL PROBABILITY CALCULATION

157

¡ ¢ √ x = α − δ − 0.5σ2 t + σ tz

Then we have DM 18.19: ¡ ¢ √ St ln = x = α − δ − 0.5σ 2 t + σ tz S0

The stock’s price at time t is (DM 18.20): √ ¤ £¡ ¢ St = S0 ex = S0 exp α − δ − 0.5σ 2 t + σ tz

Using DM 18.13, ¡ we £¡ get DM 18.22:¢ ¡¡ ¢ √ ¤¢ ¢ E (St ) = S0 E exp α − δ − 0.5σ 2 t + σ tz = S0 exp α − δ − 0.5σ2 t + 0.5σ 2 t = S0 e(α−δ)t

18.4

Lognormal probability calculation

Let’s go through some formulas. First, let’s look at DM 18.23: " ¡ ¢ # ¶ µ ∧ ln K − ln S0 − α − δ − 0.5σ 2 t √ P (St < K) = N = N −d 2 σ t ¢ S0 ¡ ¡ ¢ ln + α − δ − 0.5σ 2 t ln K − ln S0 − α − δ − 0.5σ 2 t K √ √ =− where = σ t σ t £ ¡ ¤ ¢ Here’s how to derive it. ln St ∼ N ln S0 + α − δ − 0.5σ 2 t, σ 2 t ∧ d2

The real world probability of St < K " is: ¡ ¢ ¢# ¡ ln K − ln S0 + α − δ − 0.5σ2 t √ P (St < K) = P (ln St < ln K) = N σ t

In the above formula, if we set α = r, we’ll get the risk neutral probability of St < K: " ¡ ¢ ¢# ¡ ln K − ln S0 + r − δ − 0.5σ 2 t ∗ √ P (St < K) = N = N (−d2 ) σ t Next, let’s calculate the real world probability of P (St > K). Please note that P (St = K) = 0. This is because St is a continuous random variable; the probability for a continuous random variable to take on a specific value is zero. So P (St > K) = 1 − P (St < K) − P (St = K) = 1 − P (St < K). The real world probability of St > K is: ⎛ ¢ ⎞ S0 ¡ 2 µ ¶ ∙ µ ¶¸ µ ¶ ln t + α − δ − 0.5σ ∧ ∧ ∧ ⎜ ⎟ √ = N d2 = N ⎝ K P (St > K) = 1−N −d2 = 1− 1 − N d2 ⎠ σ t

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CHAPTER 18. LOGNORMAL DISTRIBUTION

If we set α = r, we’ll get the risk neutral probability of P (St > K): ⎛ ¢ ⎞ S0 ¡ ln + r − δ − 0.5σ2 t ⎟ ⎜ √ P ∗ (St > K) = N (d2 ) = N ⎝ K ⎠ σ t

18.4.1

Lognormal confidence interval

The textbook’s explanation is a bit confusing. Here is how to build a confidence interval of the stock price. First, let’s consider building a 95% confidence interval for the standard normal random variable z. We need to find a value a positive value b such that P (−b < z < b) = 0.95. P (−b < z < b) = N (b) − N (−b) = N (b) − [1 − N (b)] = 2N (b) − 1 = 0.95 1 + 0.95 =⇒ N (b) = = 0.975 2 b = N −1 (0.975) = 1.96. Then the 95% confidence interval is −1.96 < z < 1.96. Alternatively, we can find −b. P (−b < z < b) = N (b) − N (−b) = [1 − N (−b)] − N (−b) = 1 − 2N (−b) = 0.95 1 − 0.95 Then N (−b) = = 0.025 2 −1 −b = N (0.025) = −1.96 So the 95% confidence interval is −1.96 < z < 1.96. Let’s generalize the result. The interval µ p confidence µ of ¶the standard normal random variable z is ¶ 1−p 1+p −1 −1 N K)

S0 e

Black-Scholes formula

This section derives the Black-Scholes formula with simple math. The BlackScholes formula was originally derived using stochastic calculus and partial differential equation. Many years after the Black-Scholes formula was published, someone came up with this simple proof (hindsight is always 20-20).

Yufeng Guo, Fall 09 MFE, actuary88.com

18.5. ESTIMATING THE PARAMETERS OF A LOGNORMAL DISTRIBUTION163 The European R ∞ call option price in the risk-neutral world, is: C = e−rt K (St − K)×g ∗ (St ; S0 ) dSt = e−rt E ∗ (St − K|St > K) P ∗ (St > K) Here any term with the ∗ sign indicates the term is valuated in the riskneutral world (i.e. by setting α = r). E ∗ (St − K|St > K) P ∗ (St > K) = E ∗ (St |St > K) P ∗ (St > K)−E ∗ (K|St > K) P ∗ (St > K) E ∗ (St |St > K) = =⇒

S0 e(r−δ)t N (d1 ) N (d2 )

P ∗ (St > K) = N (d2 )

E ∗ (St |St > K) P ∗ (St > K) = S0 e(r−δ)t N (d1 )

E ∗ (K|St > K) = K =⇒ E ∗ (K|St > K) P ∗ (St > K) = KN (d2 ) =⇒

¤ £ C = e−rt S0 e(r−δ)t N (d1 ) − KN (d2 ) = S0 e−δt N (d1 )−Ke−rt N (d2 )

Similarly, the European put price is: P = Ke−rt N (−d2 ) − S0 e−δt N (−d1 )

18.5

Estimating the parameters of a lognormal distribution

Consider the stock price over the time interval [t − h, t]. 2 St = St−h e(α−δ−0.5σ )h+σ

√ hz

where z is the standard normal random variable with E (z) = 0 and V ar (Z) = 1 =⇒

ln

√ ¡ ¢ St = α − δ − 0.5σ 2 h + σ hz St−h

µ ¶ h¡ √ i ¡ ¢ ¢ St =⇒ E ln = E α − δ − 0.5σ2 h + σ hz = α − δ − 0.5σ 2 h+ St−h √ ¡ ¢ 2 h σ hE (z) = α −µδ − 0.5σ¶ h¡ √ i ¢ St =⇒ V ar ln = V ar α − δ − 0.5σ 2 h + σ hz St−h ¡ ¢ Since α − δ − 0.5σ 2 h is a constant, we have: h¡ √ i ³ √ ´2 ¢ V ar α − δ − 0.5σ 2 h + σ hz = σ h V ar (z) = σ 2 h Now let’s go through DM Table 18.2.

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CHAPTER 18. LOGNORMAL DISTRIBUTION

week

S

1 2 3 4 5 6 7 sum

100 105.04 105.76 108.93 102.5 104.8 104.13 0

St ln St−1 0.049171 0.006831 0.029533 −0.060843 0.022191 −0.006414 0.040470

µ

St ln St−1

¶2

0.0024178 0.0000467 0.0008722 0.0037018 0.0004924 0.0000411 0.0075721

Weµhave: ¶ ¡ ¢ St 0.040470 E ln = α − δ − 0.5σ 2 h = = 0.006 745 St−h 6 0.006 745 =⇒ α = δ + 0.5σ 2 + h In the above equations, α is the stock’s expected (annual) continuously compounded return δ is the stock’s (annual) continuously compounded dividend yield (δ = 0 in this problem) σ is the stock’s (annual) volatility h = 1/52 (52 weeks in one year) ⎛ ⎛ ⎞2 ⎞ P St µ ¶ ¶2 µ ln ⎟ ⎜ St n ⎜ St−1 ⎟ ⎜ 1 × P ln St ⎟ ⎟= V ar ln −⎜ = σ2h = ⎝ ⎠ ⎠ St−h n − 1 ⎝n St−1 n 6 5

µ

1 × 0.0075721 − 0.006 7452 6

=⇒ =⇒ 0.388 7

¶

= 0.001 459 83

0.001 459 83 h 0.006 745 0.001 459 83 0.006 745 α = δ + 0.5σ 2 + = 0 + 0.5 × + = h 1/52 1/52

σ2 =

The textbook talks about making two adjustments before calculating α. I found this approach that you use my approach. r confusing. I recommend r 0.001 459 83 0.001 459 83 =⇒ σ= = = 0.275 5 h 1/52 So the stock’s expected return per year is α = 38.8 7% and its annual volatility is σ = 0.275 5. ¶ µ 1P 1P St St = = Please note that E ln ln (ln St − ln St−h ) = St−h n St−h n ¢ 1 Sn ¡ 1 = α − δ − 0.5σ 2 h (ln Sn − ln S0 ) = ln n n S0

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18.6. HOW ARE ASSET PRICES DISTRIBUTED

165

In µthis problem, we could have estimated ¶ ¡ ¢ St 1 104.13 1 Sn E ln = ln = α − δ − 0.5σ2 h = ln = 0.006745 St−h n S0 6 100 In addition, we have: 1 1 Sn Sn = ln α − δ − 0.5σ 2 = ln nh S0 T S0 where T = nh is the length of the observation. We see that when we estimate α − δ − 0.5σ 2 , only the following 3 factors matter: • the length of the observation T • the stock’s starting price S0 • the stock’s ending price Sn The stock prices between S0 and Sn are irrelevant. And for a given T , increasing the number of observations n doesn’t affect our estimate of α − δ − 0.5σ 2 (because as n goes up, h goes down, and T = nh is a constant). Frequent observations don’t improve our estimate of α − δ − 0.5σ 2 . To improve our estimate of α − δ − 0.5σ 2 , we need to increase T . When we estimate σ, the in-between stock prices do matter and more frequent observations do improve our estimate.

18.6

How are asset prices distributed

The lognormal stock price model assumes that the stock returns are normally distributed. Are stock returns really normally distributed?

18.6.1

Histogram

One method to assess whether stock returns are normally distributed is to plot the continuously compounded returns as a histogram. Look at DM Figure 18.4. The histograms in Figure 18.4 don’t appear normal. The textbook offers two explanations: • Stock prices can jump discretely from time to time • Stock returns are normally distributed, but the variance of the return changes over time. That’s all your need to know about this section.

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166

18.6.2

CHAPTER 18. LOGNORMAL DISTRIBUTION

Normal probability plots

Another method to check whether stock returns are normally distributed is to draw a normal probability plot. To see what a normal probability plot is, let’s go through an example. We know that the 99% confidence interval for a standard normal random variable is [−2.58, 2.58]. Suppose we take the following samples from the range [−2.58, 2.58] with the step equal to 0.01: −2.58, −2.57, −2.56, −2.55, ..., 0, 0.01, 0.02, ..., 2.57, 2.58 The for each number xi sampled, we calculate yi = P (z < xi ) = N (xi ). If we plot the data pairs (xi , yi ), what do we get? Answer: we’ll get roughly a straight line. You can experiment this using Excel. Here is a screen shot: A B 1 xi yi = P (Z < xi ) 2 −2.58 0.00494 3 −2.57 0.005085 4 −2.56 0.005234 ... ... ... 260 0 0.5 261 0.01 0.503989 262 0.02 0.507978 ... ... ... 517 2.57 0.994915 518 2.58 0.99506 Sample formulas. Cell A2 = −2.58 A3 = A2 + 0.01 ... A518 = A517 + 0.01 B2 = N ORM DIST (A2, 0, 1, 1) B3 = N ORM DIST (A3, 0, 1, 1) N ORM DIST (A518, 0, 1, 1)

...

Next, plot the range "A2:B518" as xy diagram in Excel. The plot you got should look like the 2nd diagram in DM Figure 18.2 (see DM page 589). You can see that from diagram from z = −2.58 to z = 2.58 is roughly a straight line. The point of this Excel experiment: if you plot samples from a normal distribution in a normal probability plot, you’ll get roughly a straight-line. If you plot samples from an unknown distribution in a normal probability plot and get roughly a straight line, then the samples are from approximately normal distribution. Steps on how to construct a normal probability plot: 1. Sort the samples from smallest to biggest. Number the samples as i = 1, 2, 3, ..., n. So you samples are x1 , x2 , ..., xn .

Yufeng Guo, Fall 09 MFE, actuary88.com

B518 =

18.7. SAMPLE PROBLEMS

167

i − 0.5 .Here n −1 0.5 is the continuity adjustment. Calculate the corresponding zi = N (yi ).

2. Assign the cumulative probability to xi as yi = P (x < xi ) =

3. Graph the data pairs (xi , yi ) or (xi , zi ) If the graph is roughly a straight line, then x1 , x2 , ..., xn are approximately normal. Here’s a table to summarize the 3 steps for building a normal probability plot: index

xi

1

x1

2

x2

3

x3

...

...

n

xn

yi = P (x < xi ) = 0.5 y1 = n 1.5 y2 = n 2.5 y3 = n ... n − 0.5 yn = n

i − 0.5 n

zi = N −1 (yi ) µ ¶ 0.5 z1 = N −1 µ n ¶ 1.5 z2 = N −1 µ n ¶ 2.5 z3 = N −1 n ... µ ¶ n − 0.5 −1 zn = N n

Let’s walk through DM Example 18.9. i − 0.5 yi = P (x < xi ) = zi = N −1 (yi ) n 0.5 1 3 N −1 (0.1) = −1.2816 = 0.1 5 1.5 2 4 N −1 (0.3) = −0.5244 = 0.3 5 2.5 3 5 N −1 (0.5) = 0 = 0.5 5 3.5 4 7 N −1 (0.7) = 0.5244 = 0.7 5 4.5 5 11 N −1 (0.9) = 1.2816 = 0.9 5 Next, plot the data pairs (xi , yi ) or (xi , zi ). The result is the top two diagrams in DM Figure 18.6. i

18.7

xi

Sample problems

Problem 1

• Stock prices follow the lognormal distribution • S0 = 100

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CHAPTER 18. LOGNORMAL DISTRIBUTION

• α = 0.12 • δ = 0.06 • σ = 0.25 • T =2 Construct the 95% confidence interval of ST . Solution First we find the 95% confidence interval for the standard normal random variable z. We need to b such that P (−b < z < b) = 0.95. P (−b < z < b) = N (b) − N (−b) = N (b) − [1 − N (b)] = 2N (b) − 1 = 0.95 1 + 0.95 N (b) = = 0.975 b = N −1 (0.975) = 1. 96 2 So the 95% confidence interval for z is (−1.96, 1.96). ln

£¡ ¤ ¢ ST ∼ N a − δ − 0.5σ 2 T, σ 2 T S0

¡ ¢ ¡ ¢ a − δ − 0.5σ 2 T = 0.12 − 0.06 − 0.5 × 0.252 2 = 0.057 5 σ 2 T = 0.252 × 2 ST The 95% confidence interval for ln is: S0 √ 0.057 5 − 1.96√0.252 × 2 = −0.635 46 0.057 5 + 1.96 0.252 × 2 = 0.750 46 The 95% confidence interval for ST is: 100e−0.635 46 = 52. 97 100e0.750 46 = 211. 80 The 95% confidence interval for ST is (52. 97, 211. 80) Problem 2 (Spring 2007 Exam C #4) You are given the following information for a stock with current price 0.25: • The price of the stock is lognormally distributed with continuously compounded expected annual rate of return α = 0.15. • The dividend yield of the stock is zero. • The volatility of the stock is σ = 0.35 .

Yufeng Guo, Fall 09 MFE, actuary88.com

18.7. SAMPLE PROBLEMS

169

Using the procedure described in the McDonald text, determine the upper bound of the 90% confidence interval for the price of the stock in 6 months. Solution 90% confidence interval for a standard normal random variable is (−1.645, 1.645). £¡ ¤ ¢ ST ln ∼ N a − δ − 0.5σ 2 T, σ 2 T S0 ¢ ¡ ¢ ¡ a − δ − 0.5σ 2 T = 0.15 − 0 − 0.5 × 0.352 0.5 = 0.044 375 σ 2 T = 0.352 × 0.5

√ √ ¢ ST ¡ is 0.044 375 − 1.645 × 0.352 × 0.5, 0.044 375 + 1.645 × 0.352 × 0.5 . S0 So the upper bound√for ST is: 2 0.25e0.044 375+1.645× 0.35 ×0.5 = 0.392 7

90% confidence interval for ln

Problem 3 A European call option and a European put option are written on the same stock. You are given: • S0 = 100 • K = 105 • α = 0.08 • δ=0 • σ = 0.3 • T = 0.5 Calculate the probability that the call option will be exercised. Calculate the probability that the put option will be exercised. Solution just need to use The call will be exercised is S ⎛T > K. We ⎞ DM 18.24. ¢ ¡ S 0 µ ¶ ln + α − δ − 0.5σ 2 t ∧ ⎜ ⎟ √ P (ST > K) = N d2 = N ⎝ K ⎠ σ t ⎛ ⎞ ¢ 100 ¡ ln + 0.08 − 0 − 0.5 × 0.32 0.5 ⎜ ⎟ √ = N ⎝ 105 ⎠ 0.3 0.5

= N (−0.147 5 ) = 1 − N (0.147 5 ) = 1 − NormalDist (0.147 5 ) = 0.441 4

The put option will exercise is S µ be ¶ µT < ¶K ∧ ∧ P (ST < K) = N −d2 = 1 − N d2 = 1 − 0.441 4 = 0.558 6

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CHAPTER 18. LOGNORMAL DISTRIBUTION

Alternatively, P (ST < K) = 1 − P (ST > K) = 0.558 6 Problem 4 • S0 = 50 • α = 0.1 • δ=0 • σ = 0.4 Calculate the conditional expected value of the stock at T = 3 given ST > 75. Calculate the conditional expected value of the stock at T = 3 given ST < 75. Solution Need to use DM 18.28 and 18.29. µ ¶ ∧ S0 e(α−δ)t N −d1 µ ¶ E (St |St < K) = ∧ N −d 2 ∧ d1

ln =

µ ¶ ∧ S0 e(α−δ)t N d1 µ ¶ E (St |St > K) = ∧ N d2

¢ ¢ S0 ¡ 50 ¡ ln + α − δ + 0.5σ 2 t + 0.1 − 0 + 0.5 × 0.42 3 K 75 √ √ = 0.194 2 = σ t 0.4 3

¢ ¢ S0 ¡ 50 ¡ ln + α − δ − 0.5σ 2 t + 0.1 − 0 − 0.5 × 0.42 3 K 75 √ √ = = −0.498 6 = σ t 0.4 3 50e(0.1−0)3 N (−0.194 2) 50e(0.1−0)3 0.423 0 = = 41. 32 E (St |St < K) = N (0.498 6) 0.6910 ∧ d2

03

ln

E (St |St > K) =

50e(0.1−0)3 N (0.194 2) 50e(0.1−0)3 (1 − 0.423 0) == = 126. N (−0.498 6) 1 − 0.6910

Problem 5 Stock prices follow lognormal distribution. You are given: • S0 = 100 • α = 0.08 • δ=0 • σ = 0.3 • T =5

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18.7. SAMPLE PROBLEMS

171

Calculate the 75 percentile of ST . Calculate the median of ST . Solution 75 percentile of a standard normal random variable z is N −1 (0.75) = 0.674 5. £¡ ¤ ¢ ST ln ∼ N a − δ − 0.5σ 2 T, σ 2 T S0 ¡ ¢ ¡ ¢ a − δ − 0.5σ 2 T = 0.08 − 0 − 0.5 × 0.32 5 = 0.175 σ 2 T = 0.32 × 5 ST is S0 √ x = 0.175 + 0.674 5 × 0.32 × 5 = 0.627 47 75 percentile of ln

75 percentile of ST is: 100ex = 100e0.627 47 = 187. 29 Median is 50 percentile, which corresponds to z = 0 and x = 0.175. Median of ST is: 100e0.175 = 119. 12 Problem 6 (Spring 2007 C #34) The price of a stock in seven consecutive months is: Month Price 1 54 2 56 3 48 4 55 5 60 6 58 7 62 Based on the procedure described in the McDonald text, calculate the annualized expected return of the stock. (A) Less than 0.28 (B) At least 0.28, but less than 0.29 (C) At least 0.29, but less than 0.30 (D) At least 0.30, but less than 0.31 (E) At least 0.31 Solution

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CHAPTER 18. LOGNORMAL DISTRIBUTION month

price

1 2 3 4 5 6 7 sum

54 56 48 55 60 58 62

St St−1

x2

0.036368 −0.154151 0.136132 0.087011 −0.033902 0.066691 0.138150

0.001323 0.023762 0.018532 0.007571 0.001149 0.004448 0.056785

x = ln

¶ µ ¡ ¢ 0.138150 St = α − δ − 0.5σ 2 h = E ln = 0.023 025 St−h 6 0.023 025 =⇒ α = δ + 0.5σ 2 + h In the above equations, α is the stock’s expected (annual) continuously compounded return δ is the stock’s (annual) continuously compounded dividend yield (δ = 0 in this problem) σ is the stock’s (annual) volatility h = 1/12 (12 months in one year) ⎛ ⎛ ⎞2 ⎞ P St µ ¶ ¶2 µ ln ⎟ ⎜ St n ⎜ St−1 ⎟ ⎜ 1 × P ln St ⎟ ⎟= V ar ln −⎜ = σ2h = ⎠ ⎝ ⎝ ⎠ St−h n−1 n St−1 n 6 5

µ

1 × 0.056785 − 0.023 0252 6

¶

= 0.010 720 8

0.010 720 8 h 0.023 025 0.010 720 8 0.023 025 =⇒ α = δ + 0.5σ 2 + = 0 + 0.5 × + = h 1/12 1/12 0.340 624 8 σ2 =

The answer is E.

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Chapter 19

Monte Carlo valuation In this chapter, I’m going to walk you through some examples. If you understand my examples, you’ll know enough for the exam.

19.1

Example 1 Estimate E (ez )

Let’s forget about option pricing for now and focus on a random variable X. Suppose we want to find out E (X), the mean of X, but we don’t have an easy formula to do so. For example, we have a random variable X = ez where z is a standard normal random variable with mean 0 and variance 1. ¢ ¡ R∞ R∞ 1 E (X) = −∞ x (z) f (z) dz = −∞ exp (z) √ exp −0.5z 2 dz 2π ¡ ¢ Suppose we haven’t heard of DM 18.13 E (ex ) = exp m + 0.5v 2 . ¢ ¡ R∞ 1 We don’t know how to do the integration −∞ exp (z) √ exp −0.5z 2 dz. 2π How can we calculate E (X)? One approach is to create n instances of z , µ calculate ¶ the corresponding n 1 P z Xi . Based on the cenvalue of X = e , and find the sample mean μ = n i=1 ¶ µn 1 P Xi is normally distributed with mean E (μ) = tral limit theorem, μ = n i=1 µn ¶ ¶ µn P P 1 1 1 Xi = (nE (X)) = E (X) and variance V ar (μ) = 2 V ar Xi = E n n n i=1 i=1 1 V ar (X) × nV ar (X) = . Since E (μ) = E (X), we can calculate the mean of 2 n n the sample mean E (μ) and use it to approximate the population mean E (X). Let’s come back to the example of estimating E (ez ). I’m going to produce 10 sample means and the use the average of these 10 sample means as an estimate to E (ez ). To produce each sample mean, I’m going to randomly draw 10, 000 z 0 s (so I have 10, 000 values of ez to produce each sample mean). Another way 173

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to describe this process is to say that I’m going to have 10 trials with each trial having 10, 000 samples. I’m going to do this in Microsoft Excel. The following table shows how to calculate the sample mean for one trial.

1 2 3 4 5 6 7 8 ...... 10001 10002 10003

A simulation i 1 2 3 4 5 6 7 ... 10000 Total sample mean

B zi −0.042457 0.253383 0.017632 −1.167590 0.273333 −0.206665 0.305234 ... −1.506862

C Xi = ezi 0.958432 1.288376 1.017788 0.311116 1.314338 0.813292 1.356942 ... 0.221604 16, 479.279610 1.647928

Sample formulas in Excel: B2=NORMINV(RAND(),0,1) = −0.042457

C2 = e−0.042457 = 0.958 432

Let’s walk through B2. Rand() is Excel’s formula for a random draw in a uniform distribution over (0, 1). Norminv(p, μ, σ) is Excel’s formula for the inverse normal distribution. For example, for a standard normal random variable z ∼ N (0, 1), P (z ≤ 1.96) = N (1.96) = 0.975. Then Norminv(0.975, 0, 1) = N −1 ¡ (0.975) ¢ = 1.96. Another example. For a normal random variable Z ∼ N 2, 0.32 , P (Z ≤ 2) = 0.5. Then Norminv(0.5, 2, 0.3) = 2. NORMINV(RAND(),0,1) works this way. First, Rand() generates a number from a uniform distribution over (0, 1). Say this random number is 0.4831. Then Excel finds that P (z ≤ −0.042457) = 0.4831. Hence a =NORMINV(0.4831,0,1) = −0.042457. Please note that Excel has a similar formula normsinv for calculating the inverse normal random variable. The "s" in normsinv stands for standard. So normsinv produces the inverse standard normal distribution, while norminv produces the inverse normal distribution with mean μ and standard deviation σ. In other words, normsinv(p) =norminv(p, μ = 0, σ = 1). So normsinv(0.975) =norminv(0.975, 0, 1) = 1.96. Similarly, the formula for B3 and C3 are: B3=NORMINV(RAND(),0,1) = 0.253383 C3 = e0.253383 = 1. 288 376 B10001=NORMINV(RAND(),0,1) = −1.506862 C10001 = e−1.506862 = 0.221 604 C10002=sum(C2:C10001)=16, 479.279610

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¡ ¢ 19.1. EXAMPLE 1 ESTIMATE E E Z

175

The sampleµmean ¶ is n P 1 16, 479.279610 μ= Xi = = 1.647928 10000 i=1 10000

Next, I’m going to produce 9 more trials. Here is the snapshot of Excel:

i 1 2 3 ... 10000 P X

Trial 1 zi −0.042457 0.253383 0.017632 −1.506862

μ

Now I have 10 Trial μ 1 1.6479 2 1.6512 3 1.6163 4 1.6325 5 1.6821 6 1.6626 7 1.6656 8 1.5994 9 1.6459 10 1.6554 total 16.4589

Xi = ezi 0.958432 1.288376 1.017788

Trial 2 zi −0.985719 −1.324269 −0.157924

Xi = ezi 0.373171 0.265997 0.853915

0.221604 16, 479

−1.673002

0.187683 16, 512

1.6479

...

Trial 10 zi −0.783098 0.186120 0.504305 1.177804

1.6512

Xi = ezi 0.456988 1.204567 1.655835 3.247236 16, 554 1.6554

sample means (k = 10): μ2 2.715574 2.726461 2.612426 2.665056 2.82946 2.764239 2.774223 2.55808 2.708987 2.740349 27.09486

E (X) = E (μ) '

1.6479 + 1.6512 + ... + 1.6554 1 P μi = = 1.64589 10 10

So E (X) ' 1.64589. The correct value of E (X) according to DM 18.13 is E (X) = e0.5 = 1. 648 721. We can also construct the confidence interval of E (μ). The estimated variance of the sample mean μ Ã ¢ is P P¡ P µ ¶2 ! 2 μi − μ 1P μ2i − k ( μi )2 k 1P 2 V ar (μ) ' = = = μi − μi k−1 k−1 k−1 k k ¢ 10 ¡ 2.709486 − 1.645892 = 0.000 591 2 9 p √ V ar (μ) ' 0.000 591 2 = 0.02 431 The 95% confidence interval is [1. 598 24, 1. 693 54]. This is calculated as follows:

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First, we calculate the critical value z ∗ where P (z ≤ z ∗ ) = 0.025 . This gives us z ∗ = −1.96. Then lower bound of the 95% confidence interval is 1.64589 − 1.96 (0.02 431) = 1. 598 242 4

1 − 0.95 = 2

Then upper bound of the 95% confidence interval is 1.64589 + 1.96 (0.02 431) = 1. 693 537 6 So we are 95% certain that the sample mean E (μ) is in the range [1. 598 24, 1. 693 54]. Since E (X) = E (μ), we are 95% certain that E (X) is in the range [1. 598 24, 1. 693 54]. The 99% confidence interval is calculated as follows: 1 − 0.99 First, we calculate the critical value z ∗ where P (z ≤ z ∗ ) = = 2 ∗ 0.005 . This gives us z = −2.58. Then lower bound of the 99% confidence interval is 1.64589 − 2.58 (0.02 431) = 1. 583 170 2 Then upper bound of the 99% confidence interval is 1.64589 + 2.58 (0.02 431) = 1. 708 609 8 So we are 99% certain that E (X) is in the range [1. 583 2, 1. 708 6] Some other key points in this example. This method to generate a normal random variable is called the inversion method (DM textbook page 622 and 623). This method works as follows: • Step 1. Generate a random number p from the uniform distribution over (0, 1). Say p = 0.4831. • Step 2. Ask "For a normal random variable Z with mean μ and standard deviation σ, what’s the number a such that P (Z ≤ a) = p?" In this example, we find P (z ≤ a) = 0.4831. Then a = −0.042457 is the simulated normal random variable. We can also generate the normal random variable as the sum of 12 uniformly distributed random variables minus 6 (DM textbook 622): ∼

Z = u1 + u2 + ... + u12 − 6 where ui is an independent identically distributed uniform random variable over (0, 1). ¡ ¢ R1 R1 R1 R1 E (ui ) = 0 uf (t) du = 0 udu = 1/2 E u2i = 0 u2 f (t) du = 0 u2 du = 1/3 µ ¶2 1 1 1 V ar (ui ) = − = 3 2 12 Based the central limit theorem, u1 + u2 + ... + u12 is approximately normal. The normal random variable with a constant −6 added is still normal. Hence ∼ Z is normal.

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19.2. EXAMPLE 2 ESTIMATE π

177

µ ¶ ∼ 1 E Z = E (u1 + u2 + ... + u12 ) − 6 = 12 × − 6 = 0 2 µ ¶ ∼ 1 V ar Z = V ar (u1 + u2 + ... + u12 ) = 12 × =1 12 ∼

So Z roughly a standard normal random variable. The following snapshot shows how to create 5 standard normal random variables in Excel: ui trial 1 trial 2 trial 3 trial 4 trial 5 1 0.6316 0.3308 0.0062 0.3634 0.3207 2 0.0247 0.6775 0.0296 0.8678 0.7361 3 0.4223 0.3489 0.8446 0.4283 0.8308 4 0.1326 0.5315 0.3436 0.3937 0.5658 5 0.1944 0.6660 0.2083 0.4484 0.6079 6 0.5996 0.3129 0.5807 0.3446 0.2882 7 0.7959 0.7749 0.9899 0.5553 0.9242 8 0.5468 0.2398 0.7499 0.5410 0.7900 9 0.1719 0.0645 0.0679 0.8638 0.0394 10 0.0964 0.4656 0.1779 0.8405 0.5718 11 0.1059 0.9432 0.3354 0.6266 0.4902 12 0.6106 0.0310 0.6455 0.8713 0.7856 sum 4.3328 5.3868 4.9794 7.1447 6.9508 sum-6 -1.6672 -0.6132 -1.0206 1.1447 0.9508 Each uniform random variable is created using Excel’ formula Rand(). The last row show the 5 standard normal random variable created. For example, the first standard normal random variable is created as follows: (0.6316 + 0.0247 + 0.4223 + ... + 0.6106) − 6 = −1.6672 This method is more cumbersome than the inversion method.

19.2

Example 2 Estimate π

We decide to estimate π using the Monte Carlo simulation (we can treat π as a random variable that happens to be a constant). We’ll estimate π using a classic "throwing the dart" method. Imagine a square whose size is two units. The center of this square is the origin (0, 0). Inside this square sits a unit circle x2 + y 2 = 1. The center of this circle is also (0, 0). If we randomly throw a dart at the square, what’s the probability that the dart falls within the unit circle?

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y

1.0 0.8 0.6 0.4 0.2

-1.0 -0.8 -0.6 -0.4 -0.2 -0.2

0.2

0.4

0.6

0.8

1.0

x

-0.4 -0.6 -0.8 -1.0

Of all the darts falling into the square (we ignore the darts falling out of the square), the probability that the dart falls within the circle ¡ ¢ is the area of the π 12 π circle divided by the area of the square, which is P = = . 22 4 We can also focus on the first quadrant. Of all the darts falling in square in 1 ¡ 2¢ π 1 the first quadrant, the probability that they fall in the circle is P = 4 2 = 1 π ∧ . Then π = 4P . 4 Next, we need to simulate darts falling in the first quadrant. Once again, we’ll do the simulation in Excel. We plan to produce 20 estimates of π. For each estimate, we’ll use 10, 000 simulations. Here’s how to produce one trial (one trial=10, 000 simulations). • Create the first point (x1 , y1 ), where 0 < x1 < 1 and 0 < y1 < 1. We can create x1 and y1 by randomly drawing two numbers from U ∼ (0, 1), a uniform distribution over [0, 1]. • If x21 + y12 ≤ 1, then (x1 , y1 ) falls in the circle in the first quadrant. • Similarly, create the 2nd point (x2 , y2 ) by randomly drawing two numbers from U ∼ (0, 1). Determine whether (x2 , y2 ) falls in the circle in the first quadrant. • ......

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19.2. EXAMPLE 2 ESTIMATE π

179

• Create n-th point (xn , yn ) by randomly drawing two numbers from U ∼ [0, 1]. Determine whether (xn , yn ) falls in the circle in the first quadrant. Set n = 65, 000. • Count m, the total number of points that fall in the circle. • P ≈

m ∧ . Then π = 4P . n

Here’s the snapshot A 1 throw i 2 1 3 2 4 3 5 4 6 5 7 6 8 7 ... ... 10, 001 10, 000

of Excel. B C xi yi 0.2854 0.1694 0.5147 0.2787 0.1915 0.2699 0.6501 0.8420 0.0007 0.6580 0.1520 0.9772 0.3179 0.5713 ... ... 0.5560 0.5679

D x2i + yi2 0.1102 0.3426 0.1095 1.1317 0.4330 0.9780 0.4274 ... 0.6317

E Fall in the circle? (1=Yes, 0=No) 1 1 1 0 1 1 1 ... 1

The number of simulations: n = 10, 000. We find that the number of darts falling in the circle in the first quadrant: m = 7, 894 m 7, 894 P ≈ = = 0.7894 n 10, 000 ∧ π = 4 (0.7894) = 3. 157 6 Sample formulas in the above Excel spreadsheet are: B2=rand()=0.2854 C2=rand()=0.1694 D2=B2^2+C2^2=0.1102 E2=if(D2>=1,1,0)=1 B3=rand()=0.5147 C3=rand()=0.2787 D3=B3^2+C3^2=0.3426 E3=if(D3>=1,1,0)=1 m=SUM(E2:E10001)=7, 894 We produce 19 more trials. The total number of trials is k = 20. Here is the snapshot of Excel: ³ ∧ ´2 ∧ trial mi pi πi πi 1 7,894 0.7894 3.1576 9.97043776 2 7,818 0.7818 3.1272 9.77937984 3 7,830 0.783 3.1320 9.80942400 4 7,752 0.7752 3.1008 9.61496064 5 7,879 0.7879 3.1516 9.93258256 6 7,879 0.7879 3.1516 9.93258256 7 7,778 0.7778 3.1112 9.67956544

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180 8 9 10 11 12 13 14 15 16 17 18 19 20 total

CHAPTER 19. MONTE CARLO VALUATION 7,878 7,891 7,896 7,844 7,880 7,920 7,821 7,825 7,870 7,886 7,822 7,839 7,762

0.7878 0.7891 0.7896 0.7844 0.788 0.792 0.7821 0.7825 0.787 0.7886 0.7822 0.7839 0.7762

3.1512 9.93006144 3.1564 9.96286096 3.1584 9.97549056 3.1376 9.84453376 3.1520 9.93510400 3.1680 10.03622400 3.1284 9.78688656 3.1300 9.79690000 3.1480 9.90990400 3.1544 9.95023936 3.1288 9.78938944 3.1356 9.83198736 3.1048 9.63978304 62.7856 197.10829728

The estimated mean of the sample mean is ∧ ³∧´ P π 62.7856 i π≈E π = = = 3. 139 28 k 20

The estimated variance of the sample mean is: Ã µ ¶2 ! ¶ µ ³∧´ P P k 1 20 197.10829728 1 = μ2i − μi − 3. 139 282 = V ar π ' k−1 k k 19 20 0.0003536 27 p √ V ar (μ) ' 0.0003536 27 = 0.01880 5 The 95% confidence interval is [3. 102 4, 1. 3. 176 1]. This is calculated as follows: 3. 139 28 − 1.96 (0.01880 5) = 3. 102 422 2 3. 139 28 + 1.96 (0.01880 5) = 3. 176 137 8 The 99% confidence interval is [3. 090 8, 3. 187 8]. This is calculated as follows: 3. 139 28 − 2.58 (0.01880 5) = 3. 090 763 1 Then upper bound of the 99% confidence interval is 3. 139 28 + 2.58 (0.01880 5) = 3. 187 796 9 Now you have some ideas about the Monte Carlo simulation. Let’s look at the textbook.

19.3

Example 3 Estimate the price of European call or put options

Monte Carlo simulation can be used to price European options, especially when there’s no simple formula for the option price such as the arithmetic Asian option.

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19.3. EXAMPLE 3 ESTIMATE THE PRICE OF EUROPEAN CALL OR PUT OPTIONS181 First, let’s use the Monte Carlo method to calculate the price of a European call option and put option. The inputs are (See DM page 377 Example 12.1): • S0 = 41 • K = 40 • σ = 0.3 • r = 8% • δ=0 • T = 0.25 (3 months) Solution We’ll use DM Equation 19.6 to calculate the option price: n n 1 ¡ ¢ ¢ ¡ P 1 −rT P V STi , T = e−rT e V STi , T n t=1 t=1 n n 1 ¡ i ¢ P V ST , T is the average terminal payoff. So the price of the European t=1 n option is just the discounted value of the terminal payoff.

V (S0 , 0) =

We’re are going to produce 5 sample means. For each sample mean, we’ll use 10 simulations. So we have 5 trials with 20 simulations per trial. For each trial, we need to generate the stock’s terminal price at T = 0.25. We generate hthe terminal stock price using DM 19.3: √ i ¡ ¢ ST = S0 exp α − δ − 0.5σ 2 T + σ T Z

Please note that the above formula produces the real world stock price at T . If we use the real world terminal stock price to calculate the option price, you have to use the real world discount rate, which is path-dependent (see DM Table 19.1). Path dependent discount rates are difficult to calculate. To avoid the difficulty of finding the path-dependent discount rates, we’ll want to live in the risk neutral world where everything earns the risk-free rate. To move into the risk neutral hworld, we just set α = r andi change DM 19.3 into: √ ¡ ¢ ST = S0 exp r − δ − 0.5σ2 T + σ T Z √ ¡¡ ¢ ¢ S0.25 = 41 exp 0.08 − 0.5 × 0.32 0.25 + 0.3 0.25Z = 41 exp (0.15Z + 0.008 75)

Next, we’ll produce 10 simulations of Z and S0.25 . The following is the snapshot of Excel simulation: index Z 1 0.67830 2 0.65327 3 1.21158

S0.25 Call payoff 45.7901 5.7901 45.6185 5.6185 49.6034 9.6034

put payoff 0.0000 0.0000 0.0000

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4 1.50347 51.8234 11.8234 0.0000 5 0.62495 45.4251 5.4251 0.0000 6 -1.89286 31.1369 0.0000 8.8631 7 0.57282 45.0712 5.0712 0.0000 8 -0.84887 36.4154 0.0000 3.5846 9 -0.34767 39.2586 0.0000 0.7414 10 -0.34540 39.2720 0.0000 0.7280 Total 429.4146 43.3317 13.9171 Sample calculations. If Z = 0.67830, S0.25 = 41 exp (0.15 × 0.67830 + 0.008 75) = 45. 790 1 The call payoff is 45. 790 1 − 40 = 5. 790 1 The put payoff is zero. 43.3317 = 4. 333 17 10 −rT So the call price is 4. 333 17e = 4. 333 17e−0.08×0.25 = 4. 247 4

The average call payoff at T is:

13.9171 = 1. 391 71 10 −0.08×0.25 = 1. 364 2 So the put price is 1. 391 71e The average put payoff at T is:

Now I’m going to repeat this process 4 more times. This is my final result: trials call price put price 1 4. 247 4 1. 364 2 2 3.9132 1.2391 3 0.5436 2.4875 4 1.6850 1.8342 5 2.4911 1.9250 The mean of the sample mean for the call price is: 4. 247 4 + 3.9132 + 0.5436 + 1.6850 + 2.4911 = 2. 58 5 Compare this with the correct price using the Black-Scholes formula (see DM page 377): C = 3.40 The mean of the sample mean for the put price is: 1. 364 2 + 1.2391 + 2.4875 + 1.8342 + 1.9250 = 1. 77 5 Compare this with the correct price using the Black-Scholes formula (see DM page 377): P = 1.61 The prices from the Monte Carlo simulations are off because the number of simulations per trial is small. After running 10 trials (5,000 simulations per trial), I got the following: trials call call^2 put put^2 1 3.4132 11.6502 1.5976 2.5523

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19.3. EXAMPLE 3 ESTIMATE THE PRICE OF EUROPEAN CALL OR PUT OPTIONS183 2 3 4 5 6 7 8 9 10

3.3931 3.4085 3.4002 3.4019 3.4224 3.4089 3.4140 3.3998 3.4018

11.5134 11.6182 11.5614 11.5729 11.7129 11.6206 11.6551 11.5587 11.5725

1.6003 1.6007 1.6249 1.5815 1.6056 1.6158 1.6039 1.5879 1.6216

2.5611 2.5622 2.6402 2.5012 2.5778 2.6108 2.5725 2.5214 2.6297

Sum 34.0639 116.03585 16.0398 25.7292 So the mean of the sample mean for call price is: 34.0639 = 3. 40639 10 Theµvariance of the sample ¶ mean for the call price is: 10 116.03585 − 3. 406 392 = 1. 024 1 × 10−4 9 10 √ 1. 024 1 × 10−4 = 0.01 So the 95% confidence interval for the call price is [3. 39, 1. 3. 43]. This is calculated as follows: 3. 40639 − 1.96 (0.01) = 3. 39 3. 40639 + 1.96 (0.01) = 3. 43 So the mean of the sample mean for put price is: 16.0398 = 1. 603 98 10 Theµvariance of the sample ¶ mean for the put price is: 10 25.7292 − 1. 603 982 = 0.000 186 844 9 10 √ 0.000 186 844 = 0.014 So the 95% confidence interval for the call price is [1.58, 1.63]. This is calculated as follows: 1. 603 98 − 1.96 (0.014) = 1. 58 1. 603 98 + 1.96 (0.014) = 1. 63 By the way, there’s no need for us to use Monte Carlo summations to calculate a European call or put option. The Black-Scholes formula can produce the price of a European call or put option. Here we calculate European option prices to illustrate how to use the Monte Carlo simulation. Please note that using the Monte Carlo simulation to calculate the American option price is covered in DM 1936, which is out of the scope of exam MFE.

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19.4

CHAPTER 19. MONTE CARLO VALUATION

Example 4 Arithmetic and geometric options

Let’s use the Monte Carlo method to calculate the price of an arithmetic call option and an arithmetic put option. The inputs are (See DM page 629 Table 19.3): • S0 = 40 • K = 40 • σ = 0.3 • r = 8% • δ=0 • T = 0.25 (3 months) • The average stock price is the average stock prices at the end of Month 1, Month2, and Month 3 Solution The monthly interval is h = T /3 = 0.25/3 The stock prices (the real world price) at the end of Month 1,´ 2, and 3 are: ³¡ √ ¢ end of Month 1: Sh = S0 exp α − δ − 0.5σ 2 h + σ hz1 ³¡ √ ´ ¢ end of Month 2: S2h = Sh exp α − δ − 0.5σ 2 h + σ hz2 ³¡ √ ´ ¢ end of Month 3: S3h = ST = S2h exp α − δ − 0.5σ 2 h + σ hz3 where z1 , z2 , and z3 are three separate random draws of the standard normal distribution. Since our goal is to calculate the option price, we don’t need the real world stock price. We just need the risk-neutral stock price. We change the stock’s expected return α into the risk free rate r. The risk neutral stock prices at the end of Month 1,³2, and 3 are: √ ´ ¡ ¢ Sh = S0 exp r − δ − 0.5σ 2 h + σ hz1 ³¡ √ ´ ¢ S2h = Sh exp r − δ − 0.5σ 2 h + σ hz2 ³¡ √ ´ ¢ S3h = S2h exp r − δ − 0.5σ 2 h + σ hz3 The average stock price is S = The arithmetic call price is: The arithmetic call price is:

Sh + S2h + S3h 3 ¡ ¢ C = e−rT max S − K, 0

¡ ¢ P = e−rT max K − S, 0

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19.4. EXAMPLE 4 ARITHMETIC AND GEOMETRIC OPTIONS

185

The following snapshot of Excel shows one trial (5000 simulations per trial) of the call/put price: i z1 Sh z2 S2h z3 S3h S 1 0.8424 43.1529 0.258 44.2568 −2.4041 36.0434 41.1510 2 0.066 40.3468 −1.4415 35.7157 −2.0925 29.8830 35.3152 3 0.5357 42.0218 0.1255 42.6051 −0.8295 39.7677 41.4649 4 0.0014 40.1217 1.0027 43.8893 −0.8446 40.9128 41.6413 5 −1.598 34.9321 −1.5906 30.5258 1.2832 34.2134 33.2238 6 1.2843 44.8364 −0.6896 42.3605 −0.3961 41.0516 42.7495 7 −0.0712 39.8702 0.2363 40.8134 0.0147 40.9848 40.5561 8 −1.0075 36.7649 −0.5005 35.3082 0.6539 37.4745 36.5159 ... ... ... ... ... ... ... ... 5000 −0.0168 40.0585 −1.2203 36.1464 0.6609 38.3874 38.1974

C pay 1.1510 0.0000 1.4649 1.6413 0.0000 2.7495 0.5561 0.0000 ... 0.0000

P pay 0.0000 4.6848 0.0000 0.0000 6.7762 0.0000 0.0000 3.4841 ... 1.8026

Sample calculations for the first row. First, we draw z1 = 0.8424, z2 = 0.258,and z3 = −2.4041 from the standard normal distribution. Ã ! r ¡ ¢ 0.25 0.25 Sh = 40 exp 0.08 − 0 − 0.5 × 0.32 + 0.3 × 0.8424 = 40 exp (0.075 87) = 3 3 43. 152 9 Ã ! r ¡ ¢ 0.25 2 0.25 + 0.3 × 0.258 = S2h = 43. 152 9 exp 0.08 − 0 − 0.5 × 0.3 3 3 ¢ ¡ −2 = 44. 256 8 43. 152 9 exp 2. 526 01 × Ã 10 ! r ¡ ¢ 0.25 2 0.25 S3h = 44. 256 8 exp 0.08 − 0 − 0.5 × 0.3 + 0.3 × (−2.4041) = 3 3 44. 256 8 exp (−0.205 28) = 36. 0434 43. 152 9 + 44. 256 8 + 36. 0434 S= = 41. 151 0 3 index 1 2 3 4 5 6 7 8 ... 5000 sum

Call payoff 1.1510 0.0000 1.4649 1.6413 0.0000 2.7495 0.5561 0.0000 ... 0.0000 10, 111.9410

put payoff 0.0000 4.6848 0.0000 0.0000 6.7762 0.0000 0.0000 3.4841 ... 1.8026 7, 390.6006

C 1.1282 0.0000 1.4359 1.6088 0.0000 2.6951 0.5451 0.0000 ... 0.0000 9, 911.7119

Sample calculations for the first row.

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P 0.0000 4.5920 0.0000 0.0000 6.6420 0.0000 0.0000 3.4151 ... 1.7669 7, 244.2548

C2 1.2728 0.0000 2.0618 2.5882 0.0000 7.2636 0.2971 0.0000 ... 0.0000 61, 373.4417

P2 0.0000 21.0865 0.0000 0.0000 44.1162 0.0000 0.0000 11.6629 ... 3.1219 32, 916.8904

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CHAPTER 19. MONTE CARLO VALUATION

Call price: e−rT ×call payoff= e−0.08(0.25) × 1.1510 = 1. 128 2 Put payoff: e−rT ×put payoff= e−0.08(0.25) × 0.0000 = 0 C 2 = 1. 128 22 = 1. 272 8 P 2 = 02 = 0 The estimated option prices are calculated as follows.

index Call payoff put payoff C sum 10, 111.9410 7, 390.6006 9, 911.7119 The average call price of these 5000 simulations:

e−rT ×

or

P 7, 244.2548

C2 61, 373.4417

P2 32, 916.8904

1 P 10111.9410 call payoff= e−0.08(0.25) × = 1. 982 34 5000 5000

9911.7119 1 P C= = 1. 982 34 5000 5000

The average put price of these 5000 simulations:

e−rT ×

or

1 P 7390.6006 put payoff= e−0.08(0.25) × = 1. 448 9 5000 5000

1 P 7244.2548 P = = 1. 448 9 5000 5000

The estimated variance of the call price per simulation: Ã µ P ¶2 ! ¶ µ C 5000 1 P 2 1 n = × C − × 61373.4417 − 1. 982 342 = n−1 n n 5000 − 1 5000 8. 346 7

The estimated variance of the put price per simulation: Ã Ã µ P ¶2 ! µ ¶2 ! n 5000 P 1 P 2 1 7244.2548 = × P − × 32916.8904 − n−1 n n 5000 − 1 5000 5000 = 4. 485 1

The following is the snapshot of Excel for 30 trials (5000 simulations per trial) for the arithmetic European call and put prices:

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19.4. EXAMPLE 4 ARITHMETIC AND GEOMETRIC OPTIONS i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 sum

C 1.9823 1.9409 1.9467 2.0098 1.9894 1.9515 1.9825 2.0462 2.1270 1.9583 1.9878 1.9664 2.0147 1.9664 1.9915 1.9504 1.9414 1.9804 1.9666 2.0276 1.9940 1.9307 1.9221 1.9710 1.9185 1.9529 1.9791 1.9776 1.9713 1.9689 59.3139

P 1.4489 1.4486 1.4920 1.4211 1.4447 1.4886 1.4813 1.4408 1.3565 1.4675 1.4691 1.4496 1.3785 1.4327 1.4253 1.4592 1.4800 1.4186 1.4724 1.3779 1.4217 1.4815 1.4903 1.4491 1.4526 1.4638 1.4538 1.4676 1.5092 1.4640 43.5069

C2 3.929513 3.767093 3.789641 4.039296 3.957712 3.808352 3.930306 4.186934 4.524129 3.834939 3.951349 3.866729 4.059016 3.866729 3.966072 3.804060 3.769034 3.921984 3.867516 4.111162 3.976036 3.727602 3.694468 3.884841 3.680642 3.813818 3.916837 3.910902 3.886024 3.876567 117.3193

187

P2 2.099311 2.098442 2.226064 2.019525 2.087158 2.215930 2.194250 2.075905 1.840092 2.153556 2.158255 2.101340 1.900262 2.052629 2.031480 2.129265 2.190400 2.012426 2.167962 1.898608 2.021231 2.194842 2.220994 2.099891 2.110047 2.142710 2.113534 2.153850 2.277685 2.143296 63.1309

For the remaining part of the calculation, all you need to know is the following: Trial C P C2 P2 sum 59.3139 43.5069 117.3193 63.1309

The estimated arithmetic call price is: ¡ ¢ 59.3139 = 1. 977 13 E C = 30 The estimated variance of the call price (per trial) is:

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188

CHAPTER 19. MONTE CARLO VALUATION ⎛

⎞2 ⎞ ¶ µ Ci ⎟ ⎟ ⎟ ¡ ¢ P 2 ⎜ n ⎜ 1 ⎜ 1 n=30 i=1 2 ⎟ ⎟ = 30 Ci − ⎜ V ar C = = × 117.3193 − 1. 977 1 ⎜ × ⎝ n ⎠ ⎠ 30 − 1 30 n − 1 ⎝n i=1 ⎛ n=30 P

0.001 778 √ The standard deviation is: 0.001 778 = 0.04 217

We can also estimate the variance of the call price as follows: For each trial, the average call price of the 5, 000 simulations is used to P 1 5,000 estimate the call price. C = Ci . 5, 000 i=1 So the variance of the price¶per trial is: µ call 5,000 ¡ ¢ P 1 1 1 V ar C = Ci = ×5, 000V ar (C) = V ar V ar (C) = 2 2 5, 000 5, 000 5000 i=1 8. 346 7 = 0.001 669 5000 √ The standard deviation is: 0.001 669 = 0.0406 Please note that the two methods of estimating the variance of the call price per trial often produce close, but not identical, results. The results are not identical because the estimated variance per trial changes depending how many trial you have and how many simulations per trial. The estimated arithmetic put price is: ¡ ¢ 43.5069 E P = = 1. 450 23 30 The estimated variance of the put price (per trial) is: ⎛ ⎛ n=30 ⎞2 ⎞ P ¶ µ P i ⎟ ⎟ ¡ ¢ P 2 ⎜ 1 30 n ⎜ ⎜ 1 n=30 ⎟ i=1 2 ⎜ ⎟ Pi − ⎝ × 63.1309 − 1. 450 23 = V ar P = ⎜ × ⎟= n − 1 ⎝n n ⎠ ⎠ 30 − 1 30 i=1

0.001 237 5 √ The standard deviation is: 0.001 237 5 = 0.03 518

We can also estimate the variance of the put price as follows: For each trial, the average call price of the 5, 000 simulations is used to P 1 5,000 estimate the call price. P = Pi . 5, 000 i=1 µ ¶ 5,000 ¡ ¢ P 1 1 1 V ar P = V ar Pi = ×5, 000V ar (P ) = V ar (P ) = 2 2 5, 000 5, 000 5000 i=1 4. 4668 = 0.000 89 5000 √ The standard deviation is: 0.000 89 = 0.0298 Now let’s use Monte Carlo simulation to estimate the geometric call and put prices. I’m going to use the same inputs:

Yufeng Guo, Fall 09 MFE, actuary88.com

19.4. EXAMPLE 4 ARITHMETIC AND GEOMETRIC OPTIONS

189

• S0 = 40 • K = 40 • σ = 0.3 • r = 8% • δ=0 • T = 0.25 (3 months) • The average stock price is the average stock prices at the end of Month 1, Month2, and Month 3

In addition, I’m going to use the same random draws of standard normal random variables z1 , z2 , and z3 as used for simulating the arithmetic call/put prices. Though I don’t have to use the z1 , z2 , and z3 used for estimating the arithmetic call/put prices, reuse saves me time. Here is a snapshot of Excel doing one trial: i z1 Sh z2 S2h 1 0.8424 43.1529 0.258 44.2568 2 0.066 40.3468 −1.4415 35.7157 3 0.5357 42.0218 0.1255 42.6051 4 0.0014 40.1217 1.0027 43.8893 5 −1.598 34.9321 −1.5906 30.5258 6 1.2843 44.8364 −0.6896 42.3605 7 −0.0712 39.8702 0.2363 40.8134 8 −1.0075 36.7649 −0.5005 35.3082 ... ... ... ... ... 5000 −0.0168 40.0585 −1.2203 36.1464 sum Sample calculations. The 1st simulation. S = (43.1529 × 44.2568 × 36.0434)1/3 = 40. 983 1 Call payoff: 40. 983 1 − 40 = 0.983 1 Put payoff: 0 The 5000-th simulation: i z1 Sh 5000 −0.0168 40.0585

z2 −1.2203

S2h 36.1464

S = (40.0585 × 36.1464 × 38.3874)1/3 = 38. 163 6 Call payoff: 0 Put payoff: 40 − 38. 163 6 = 1. 836 4

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z3 −2.4041 −2.0925 −0.8295 −0.8446 1.2832 −0.3961 0.0147 0.6539 ... 0.6609

z3 0.6609

S3h 36.0434 29.8830 39.7677 40.9128 34.2134 41.0516 40.9848 37.4745 ... 38.3874

S3h 38.3874

S 40.9831 35.0508 41.4466 41.6101 33.1662 42.7209 40.5532 36.5047 ... 38. 163 6

S 38. 163 6

C pay 0.9831 0 1.4466 1.6101 0 2.7209 0.5532 0 .... 0 9, 911.3146

C pay 0

P pay 0 4.9492 0 0 6.8338 0 0 3.4953 ... 1. 836 4 7, 528.5134

P pay 1. 836 4

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CHAPTER 19. MONTE CARLO VALUATION

The next snapshot: i call payoff 1 0.9831 2 0 3 1.4466 4 1.6101 5 0 6 2.7209 7 0.5532 8 0 ... ... 5000 0 sum 9, 911.3146

put payoff 0 4.9492 0 0 6.8338 0 0 3.4953 ... 1. 836 4 7, 528.5134

C 0.9636 0 1.418 1.5782 0 2.667 0.5422 0 ... 0 9, 715.0581

P 0 4.8512 0 0 6.6985 0 0 3.4261 ... 1. 800 0 7, 379.4385

C2 0.9285 0 2.0107 2.4907 0 7.1129 0.2940 0

P2 0 23.5341 0 0 44.8699 0 0 11.7382

0 59, 337.3153

3.24 33, 852.0593

Sample calculation. The 1st simulation. Call price: 0.9831e−0.08×0.25 = 0.963 6 Put price: 0 × e−0.08×0.25 = 0 The 5000-th simulation. Call price: 0 Put price: 1. 836 4e−0.08×0.25 = 1. 800 0 The average call price of these 5000 simulations: 1 P 10111.9410 e−rT × call payoff= e−0.08(0.25) × = 1. 982 34 5000 5000 or

1 P 9911.7119 C= = 1. 982 34 5000 5000

The following calculation uses this part of the table: i call payoff put payoff C P sum 9, 911.3146 7, 528.5134 9, 715.0581 7, 379.4385

C2 59, 337.3153

P2 33, 852.0593

The average call price of these 5000 simulations: 1 P 9911.3146 e−rT × call payoff= e−0.08(0.25) × = 1. 943 0 5000 5000 or

1 P 9715.0581 C= = 1. 943 0 5000 5000

The estimated variance of the call price per simulation is: Ã µ P ¶2 ! ¶ µ C 5000 1 P 2 1 n = × C − × 59337.3153 − 1. 943 02 = n−1 n n 5000 − 1 5000 8. 093 8

Yufeng Guo, Fall 09 MFE, actuary88.com

19.4. EXAMPLE 4 ARITHMETIC AND GEOMETRIC OPTIONS The average put price of these 5000 simulations: e−rT × or

1 P 7528.5134 put payoff= e−0.08(0.25) × = 1. 475 9 5000 5000

1 P 7379.4385 P = = 1. 475 9 5000 5000

The estimated variance of the put price per ¶ simulation is: µ 1 5000 × 33852.0593 − 1. 475 92 = 4. 593 0 5000 − 1 5000 Next is the snapshot of the 30 trials (1 trial=5,000 simulations): i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 sum

C 1. 943 0 1.9001 1.9080 1.9698 1.9506 1.9112 1.9436 2.0058 2.0854 1.9191 1.9487 1.9281 1.9748 1.9272 1.9519 1.9116 1.9039 1.9424 1.9274 1.9867 1.9535 1.8924 1.8845 1.9328 1.8804 1.9133 1.9399 1.9398 1.9318 1.9279 58.1356

P 1.4759 1.4754 1.5200 1.4476 1.4727 1.5171 1.5091 1.4675 1.3830 1.4958 1.4959 1.4775 1.4058 1.4608 1.4523 1.4863 1.5072 1.4453 1.5007 1.4049 1.4483 1.5100 1.5182 1.4771 1.4809 1.4916 1.4814 1.4947 1.5376 1.4921 44.3327

C2 3.775249 3.610380 3.640464 3.880112 3.804840 3.652685 3.777581 4.023234 4.348893 3.682945 3.797432 3.717570 3.899835 3.714100 3.809914 3.654215 3.624835 3.772918 3.714871 3.946977 3.816162 3.581178 3.551340 3.735716 3.535904 3.660717 3.763212 3.762824 3.731851 3.716798 112.7048

P2 2.178281 2.176805 2.310400 2.095546 2.168845 2.301592 2.277383 2.153556 1.912689 2.237418 2.237717 2.183006 1.976274 2.133937 2.109175 2.209088 2.271652 2.088892 2.252100 1.973744 2.097573 2.280100 2.304931 2.181824 2.193065 2.224871 2.194546 2.234128 2.364214 2.226362 65.5497

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191

192

CHAPTER 19. MONTE CARLO VALUATION

C2 P2 112.7048 65.5497 58.1356 The estimated call price is: = 1. 937 85 30 44.3327 = 1. 477 76 The estimated put price is: 30 trial sum

C 58.1356

P 44.3327

The estimated variance of the call price per trial: ¶ µ 30 1 2 × 112.7048 − 1. 937 85 = 0.001618 30 − 1 30 √ The standard deviation is: 0.001618 = 0.040 22 The estimated variance of the call per trial can also be calculated as follows: For each trial, the average call price of the 5, 000 simulations is used to P 1 5,000 Ci . estimate the call price. C = 5, 000 i=1 So the variance of the price¶per trial is: µ call 5,000 ¡ ¢ P 1 1 8. 093 8 V ar C = Ci = V ar V ar (C) = = 0.001 619 2 5, 000 5000 5000 i=1 The standard deviation is:

√ 0.001 619 = 0.04 02

The estimated variance of the put¶price per trial: µ 30 1 × 65.5497 − 1. 477 762 = 0.001 257 3 30 − 1 30 The standard deviation is:

√ 0.001 257 3 = 0.03545 8

The estimated variance of the call per trial can also be calculated as follows: For each trial, the average put price of the 5, 000 simulations is used to P 1 5,000 Pi . estimate the put price. P = 5, 000 i=1 So the variance of the price¶per trial is: µ put 5,000 ¡ ¢ P 1 1 4. 593 0 V ar P = Pi = V ar V ar (P ) = = 0.000 918 6 2 5, 000 5000 5000 i=1 The standard deviation is:

√ 0.000 918 6 = 0.0303

Here’s a question. We know that the estimated standard √ variance of the put price per simulation is 4. 593 0 (the standard deviation is 4. 593 0 = 2. 143 1). If we want the standard deviation of the put price per trial is 0.02, how many simulations should be performed in one trial? Here is how to find it. ¡ ¢ 1 V ar C = 2 n

µ

V ar

n P

Ci

i=1

¶

=

1 V ar (C) × nV ar (C) = n2 n

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19.5. EFFICIENT MONTE CARLO VALUATION

Set 5

r

V ar (C) = 0.02. We have: n

r

193

4.5930 4. 593 0 = 11482. = 0.02 or n = n 0.022

We need to have about 11,500 simulations per trial.

19.5

Efficient Monte Carlo valuation

19.5.1

Control variance method

Suppose we have n simulations per trial. These n simulations produce n option prices V1 , V2 , ..., Vn . We can use the sample mean of these option prices to estimate the true option price. So the true option is estimated as: V1 + V2 + ... + Vn V = n The variance of V is: ¡ ¢ V ar V = V ar

µ

¶

V ar (V ) nV ar (V ) = 2 n n σV The standard deviation of the sample mean is σ V = √ , where n is the numn ber of the simulations and σ V is the standard deviation of the option price per simulation. To decrease σ V , we need to increase n by doing more simulations. Doing more simulations costs time. However, there are techniques out there to reduce σ V without increasing n. One method is called the control variate method. It goes like this. Suppose we have two random similar variables X and Y . We need to calculate E (X) and E (Y ). We can calculate E (X) easily because there’s a formula for E (X). E (Y ), on the other hand, doesn’t have a formula and needs to be calculated through the Monte Carlo simulation. Since X and Y are similar, we expect that our errors in estimating E (X) are similar to our errors in estimating E (Y ): ∧

V1 + V2 + ... + Vn n

=

∧

E (X) − μX ≈ E (Y ) − μY In the above formula, E (X) and E (Y ) are the true means of X and Y ; ∧

∧

μX and μY are the estimated means of X and Y based on the Monte Carlo simulation. Since our goal his to find E (Y i ), we ³ arrange´the above formula into: ∧

∧

∧

∧

E (Y ) ≈ μY + E (X) − μX = μY − μX + E (X)

³∧ ´ ∧ So we can use μY − μX + E (X) to estimate E (Y ). We define this new ³∧ ´ ∧∗ ∧ estimate as μY = μY − μX + E (X).

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194

CHAPTER 19. MONTE CARLO VALUATION

³∧ ´ ∧∗ ∧ Why is the new estimate E (Y ) ≈ μY = μY − μX + E (X) better than ∧

the old estimate E (Y ) ≈ μY ? It turns out the new estimate often has lower ∧ ∧ variance if μX and μY are positively correlated.

The ³ variance estimate is: i ´ of the h³new ´ ³∧ ´ ³∧ ´ ∧∗ ∧ ∧ ∧ V ar μY = V ar μY − μX + E (X) = V ar μY − μX = V ar μX + ³∧ ´ ³∧ ∧ ´ V ar μY − 2Cov μX ,μY The above formula holds because E (X) is a constant. Now let’s go through an example. There’s no formula for the arithmetic European call option price; there’s a formula for the geometric European call option price. So we’ll use the Monte Carlo simulation to estimate the arithmetic European call option. We’ll use the geometric European price as the control variate (i.e. a dummy variable). Let AC=arithmetic European call price price. ∧∗ ∧ ∧ μAC = μAC − μGC + E (GC) ∧

GC=geometric European call

∧

We’ll find μAC and μGC through Monte Carlo simulation. We’ll calculate E (GC) using the equation DM 14.18 and DM 14.19 (see Derivatives Markets Appendix 14.A). Once again, the inputs are: • S0 = 40 • K = 40 • σ = 0.3 • r = 8% • δ=0 • T = 0.25 (3 months) • The average stock price is the average stock prices at the end of Month 1, Month2, and Month 3 (so N = 3) First, we’ll calculate the geometric European call option price. By the way, the appendix 14.5 is not on the syllabus of the exam MFE. You can ignore my calculation and just accept that the geometric call option price is 1. 938 5. Using DM 14.18, we get: µ ¶ ¢ 4 0.32 4 × 7 1 2 ¡ δ∗ = 0.08 × + 0 + 0.5 × 0.32 − 2 × = 0.03 333 2 3 3 3 6 Using DM 14.19, we have:

Yufeng Guo, Fall 09 MFE, actuary88.com

19.5. EFFICIENT MONTE CARLO VALUATION σ∗ =

0.3 3

r

195

4×7 = 0.216 025 6 ∗

C (S, K, σ ∗ , r, T, δ ∗ ) = Se−δ T N (d1 ) − Ke−rT N (d2 ) ¶ ¶ µ µ 1 40 1 S ln + r − δ ∗ + (σ ∗ )2 T + 0.08 − 0.03 333 + × 0.216 0252 0.25 ln K 2 40 2 √ √ d1 = = = 0.216 025 0.25 σ T 0.162 026

N (d1 ) = NormalDist (0.162 026) = 0.564 357

√ √ d2 = d1 − σ ∗ T = 0.162 026 − 0.216 025 0.25 = 0.054 013 5

N (d2 ) = NormalDist (0.054 013 5) = 0.521 538

C = 40e−0.03333×0.25 × 0.564 357 − 40e−0.08×0.25 × 0.521 538 = 1. 938 5

So the geometric call option price is 1. 938 5.

Now we have: ∧∗

∧

∧

μAC = μAC − μGC + 1. 938 5

Next, we perform 30 trials (1 trial =5,000 simulations).

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196

CHAPTER 19. MONTE CARLO VALUATION AC − GC 0.0393 0.0408 0.0387 0.0400 0.0388 0.0403 0.0389 0.0404 0.0416 0.0392 0.0391 0.0383 0.0399 0.0392 0.0396 0.0388 0.0375 0.0380 0.0392 0.0409 0.0405 0.0383 0.0376 0.0382 0.0381 0.0396 0.0392 0.0378 0.0395 0.0410 1.1783

(AC − GC)2 0.001544 0.001665 0.001498 0.001600 0.001505 0.001624 0.001513 0.001632 0.001731 0.001537 0.001529 0.001467 0.001592 0.001537 0.001568 0.001505 0.001406 0.001444 0.001537 0.001673 0.001640 0.001467 0.001414 0.001459 0.001452 0.001568 0.001537 0.001429 0.001560 0.001681 0.046313

AC GC AC − GC 59.3139 58.1356 1.1783 59.3139 = = 1. 977 13 30

(AC − GC)2 0.046313

i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 sum

AC 1.9823 1.9409 1.9467 2.0098 1.9894 1.9515 1.9825 2.0462 2.1270 1.9583 1.9878 1.9664 2.0147 1.9664 1.9915 1.9504 1.9414 1.9804 1.9666 2.0276 1.9940 1.9307 1.9221 1.9710 1.9185 1.9529 1.9791 1.9776 1.9713 1.9689 59.3139

GC 1.9430 1.9001 1.9080 1.9698 1.9506 1.9112 1.9436 2.0058 2.0854 1.9191 1.9487 1.9281 1.9748 1.9272 1.9519 1.9116 1.9039 1.9424 1.9274 1.9867 1.9535 1.8924 1.8845 1.9328 1.8804 1.9133 1.9399 1.9398 1.9318 1.9279 58.1356

So i sum ∧

μAC

58.1356 ∧ μGC = = 1. 937 85 30 Hence the updated estimated price of the geometric European call option is: ∧∗ ∧ ∧ μAC = μAC − μGC + 1. 938 5 = 1. 977 13 − 1. 937 85 + 1. 938 5 = 1. 977 78 Alternatively,

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19.6. ANTITHETIC VARIATE METHOD ∧

197

1.1783 = 0.03927 67 30 = 0.03927 67 + 1. 938 5 = 1. 977 78 ∧

μAC − μGC = ∧∗

μAC

∧∗

The variance of μAC is: ³∧∗ ´ ´ ³∧ ³∧ ´ ∧ ∧ V ar μAC = V ar μAC − μGC + 1. 938 5 = V ar μAC − μGC =

1. 145 7 × 10−6

30 30 − 1

µ

In contrast, the estimated variance of the geometric European call price without ³using´the control variate method is, as calculated before, ∧ V ar μGC = 0.001618 ³∧ ´ ³ ∧∗ ´ We can see that V ar μAC is much smaller than V ar μAC .

³∧ ´ ∧∗ ∧ Boyle points out that μY = μY − μX + E (X) doesn’t always produce ∧

lower variance than the variance of the original estimate μY . Boyle recommends the following new³estimate: ´ ∧∗ ∧ ∧ μY = μY + β E (X) − μX If we use the¡ textbook ¢ notation, the above new estimate is DM 19.10: A∗ = A + β G − G

Since G is constant (here G is equivalent to E (X)), we have DM 19.11: £ ¡ ¢¤ ¡ ¢ ¡ ¢ ¡ ¢ V ar (A∗ ) = V ar A + β G − G = V ar A − βG = V ar A +β 2 V ar G − ¢ ¡ 2βCov A, G

¢ ¡ Cov A, G ¡ ¢ . The textbook says that V ar (A ) is minimized if we set β = V ar G How do we find β? Typically, we perform a small number of Monte Carlo simulations, run regression of DM 19.10, and estimate β. Then we apply the estimated β to DM 19.10, run more simulations, and calculate A∗ . This is all you need to know about the Boyle’s improved control variate method. ∗

19.6

Antithetic variate method

Antithetic variate method is simple. Suppose we want to estimate E (X) and have generated random numbers from a symmetric distribution such as a standard normal random variables. We have n random draws z1 , z2 ,...,zn from the

Yufeng Guo, Fall 09 MFE, actuary88.com

1 × 0.046313 − 0.03927 672 30

¶

=

198

CHAPTER 19. MONTE CARLO VALUATION

standard normal distributions. Since z1 , z2 ,...,zn are random draws from the standard normal distribution, −z1 , -z2 ,...,−zn are also random draws from the standard normal distribution. Next, we calculate two samples means, X 1 (using z1 , z2 ,...,zn ) and X 2 (using −z1 , -z2 ,...,−zn ). Then we can estimated E (X) as the average of X 1 and X 2 : X1 + X2 E (X) ≈ 2 This method is called the antithetic variate method. Here’s an example. We want to estimate E (ez ) where z is the standard normal random variable. We have generated the following 10 standard normal random variables: i zi 1 −0.0183 2 2.0478 3 −0.4849 4 −0.6583 5 −0.4666 6 −0.3757 7 1.1392 8 2.1566 9 0.1096 10 −1.5389 This is i 1 2 3 4 5 6 7 8 9 10 sum

how to estimate E (ez ) using the antithetic variate method: zi Xi = exp (zi ) −zi Yi = exp (−zi ) −0.0183 0.981866 0.0183 1.018468 2.0478 7.750831 −2.0478 0.129018 −0.4849 0.615759 0.4849 1.624013 −0.6583 0.517731 0.6583 1.931506 −0.4666 0.627131 0.4666 1.594563 −0.3757 0.686808 0.3757 1.456010 1.1392 3.124268 −1.1392 0.320075 2.1566 8.641706 −2.1566 0.115718 0.1096 1.115832 −0.1096 0.896193 −1.5389 0.214617 1.5389 4.659462 24.276548 13.745027

X=

24.276548 = 2. 427 654 8 10

Y =

13.745027 = 1. 374 502 7 10

E (ez ) ≈

2. 427 654 8 + 1. 374 502 7 = 1. 901 078 75 2

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19.7. STRATIFIED SAMPLING

19.7

199

Stratified sampling

In stratified sampling, we divide the population into several non-overlapping group. Each group is called a strata. Then we take a sample from each group. Example. We divide the range [0, 1] into 100 groups: [0, 0.01), [0.01, 0.02), [0.02, 0.03),...,[0.99, 1). Next, we draw 100 numbers u1 , u2 , ..., u100 from the uniform distribution [0, 1]. All these 100 numbers are divided by 100. So we u1 u2 u3 u100 i−1 ui have , , , ..., .Next, we add to the i-th number . Now 100 100 100 100 100 100 we have: u3 u100 u1 u2 , + 0.01, + 0.02, ..., + 0.99 100 100 100 100 ui < 0.01. Since 0 ≤ ui < 1, 0 ≤ 100 u1 So falls in the group [0, 0.01). 100 u2 u2 +0.01 < 0.02. So +0.01 falls into group [0.01, 0.02). Similarly, 0.01 ≤ 100 100 u3 u3 Similarly, 0.02 ≤ +0.02 < 0.03. So +0.01 falls into group [0.02, 0.03). 100 100 So on and so forth. The end result is that we take one sample from [0, 0.01), one sample from [0.01, 0.02), ..., and one sample from [0.99, 1). This is all you need to know about the stratified sample for the purpose of passing Exam MFE.

19.7.1

Importance sampling

In the importance sampling, we perform simulations from a conditional distribution, not from the original distribution. For example, we want to estimate the price of an option that is deep out of the money. If we perform simulations from the original distribution, then most of the simulated payoffs will be zero. This is a waste of our time. To make the simulations more efficient, we’ll draw random numbers from the conditional distribution where the payoff is not zero. This is all you need to know about the importance sampling. The textbook also mentioned Latin hypercube sampling and low discrepancy sequences. Since the textbook merely mentioned these terms without providing much explanation, I don’t think SOA expects you to know much about term. Skip these terms and move on.

19.8

Sample problems

Problem 1

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You are simulating the standard normal random variable z by taking random draws from a uniform distribution over (0, 1). Let a represent the simulated value of z. Calculate P (z ≤ a). You are given: i 1 2 3 4 5 6 7 8 9 10 11 12

ui 0.3763 0.1349 0.414 0.0405 0.5225 0.0423 0.2041 0.9282 0.6792 0.3368 0.1535 0.157

Solution P ui = 0.3763+ 0.1349 + 0.414 + 0.0405+ 0.5225 + 0.0423 + 0.2041+ 0.9282 + 0.6792 + 0.3368 + 0.1535 + 0.157 = 3. 989 3 P The simulated value is a = ui − 6 = 3.9893 − 6 = −2. 010 7 P (z ≤ a) = N (−2. 010 7) = 1 − N (2. 010 7) = 0.022

Problem 2 You are simulating E (ez ) by taking random 10 draws from a uniform distribution over (0, 1). Calculate the simulated value of E (ez ). You are given: i 1 2 3 4 5 6 7 8 9 10

ui 0.878 0.762 0.069 0.8 0.048 0.22 0.178 0.661 0.191 0.258

Solution

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201

zi = N −1 (ui ) 1.17 0.71 −1.48 0.84 −1.66 −0.77 −0.92 0.42 −0.87 −0.65

xi = ezi 3.221993 2.033991 0.227638 2.316367 0.190139 0.463013 0.398519 1.521962 0.418952 0.522046 11.314619 11.314619 The estimated value is E (ez ) = = 1. 131 461 9 10 Sample calculation. u1 = 0.878. Look at the normal table. You see that roughly P (z < 1.17) = 0.878 so z1 = 1.17. i 1 2 3 4 5 6 7 8 9 10 sum

ui 0.878 0.762 0.069 0.8 0.048 0.22 0.178 0.661 0.191 0.258

By the way, I’m using Excel to find zi = N −1 (ui ) for me so I don’t have to look at the normal table. So if you can’t match my z, that’s OK. For example, for u1 = 0.878, you might get z1 = 1.16 or z1 = 1.165. However, your final estimated E (ez ) should be close to mine. x1 = e1.17 = 3. 221 993 u3 = 0.069. Since 0.069 < 0.5, you know that z3 = N −1 (u3 ) < 0. You can’t directly look up z3 from the normal table (the normal table lists only the positive z values). Use the formula N (−z3 ) = 1 − N (z3 ) = 1 − 0.069 = 0.931 . From the normal table you see that P (z < 1.48) = 0.931. So −z3 = 1.48 or z3 = −1.48. Problem 3 You are simulating the real-world price ST . Inputs are: • Stock prices are lognormally distributed • S0 = 100 • α = 0.06 • r = 0.08 • δ = 0.02 • T = 0.5 • σ = 0.3

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•

CHAPTER 19. MONTE CARLO VALUATION i 1 2 3 4 5

ui ∼ U (0, 1) 0.6515 0.8839 0.7621 0.3922 0.1748

Calculate the average of the simulated real-world stock prices at T . Solution The real world ³¡ price at T is: ¢ √ √ ´ 2 ST = S0 exp α − δ − 0.5σ 2 T + σ T z = 100e(0.06−0.02−0.5×0.3 )0.5+0.3 0.5z = √ 0.5z

100e−0.002 5+0.3 i 1 2 3 4 5 Total

u1 ∼ U (0, 1) 0.6515 0.8839 0.7621 0.3922 0.1748

zi = N −1 (u1 ) 0.39 1.19 0.71 −0.27 −0.94

√

−0.002 5+0.3 0.5z ST = 100e √ −0.002 5+0.3 0.50.39 100e = 108. 353 8 128. 394 5 115.9645 94.1976 81.7173 528.6276

The average of the simulated stock prices is:

528.6276 = 105. 73 5

Please note that the risk free rate r is not needed for solving this problem. Problem 4 You are simulating the price of an arithmetic average stock price European call option and put option. You are given: • Stock prices are lognormally distributed • S0 = 40 • K = 40 • α = 0.06 • r = 0.08 • δ=0 • T =1 • σ = 0.3

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• Stock prices at the end of Month 4, 8, and 12 are averaged • The 3 random draws from a uniform distribution (0, 1) are 0.4828, 0.6177, and 0.9345, which are used to simulate the stock price at the end of Month 4, 8, and 12 respectively. Calculate the simulated values of this arithmetic Asian call option and put option. Solution The option is: µ price ¶ ∗ ∗ + S2/3 + S1∗ S1/3 −rT e max − K, 0 3 stock price at t. St∗ is the risk-neutral ³¡ √ √ ´ ¢ 2 St∗ = St−h exp r − δ − 0.5σ 2 h + σ hz = St−h e(0.08−0−0.5×0.3 )h+0.3 hz u1 0.4828

z1 −0.04

∗ S1/3 40.1900

u2 0.6177

z2 0.3

∗ S2/3 42.8303

u3 0.9345

z3 1.51

S1∗ 56.2864

Sample calculation. √ 2 ∗ = 40e(0.08−0−0.5×0.3 )1/3+0.3 1/3(−0.04) = 40. 1900 S1/3 √ 2 ∗ = 40.1900e(0.08−0−0.5×0.3 )1/3+0.3 1/3(0.3) = 42. 830 3 S2/3 √ 2 S ∗ = 42.8303e(0.08−0−0.5×0.3 )1/3+0.3 1/3(1.51) = 56. 286 4 1

40. 1900 + 42.8303 + 56.2864 = 46. 435 6 3 The call payoff: max (46. 435 6 − 40, 0) = 6. 435 6 The put payoff: max (40 − 46. 435 6, 0) = 0 Average stock price:

The simulated value of the call price is: e−0.08×1 6. 435 6 = 5. 940 8 The simulated value of the put price is: e−0.08×1 0 = 0 Problem 5 (spring 2007 Exam C #19) The price of a non dividend-paying stock is to be estimated using simulation. It is known that: • The price St follows the lognormal distribution • S0 = 50, α = 0.15, and σ = 0.30.

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Using the following uniform (0, 1) random numbers and the inversion method, three prices for two years from the current date are simulated 0.9830, 0.0384, 0.7794. Calculate the mean of the three simulated prices. (A) Less than 75 (B) At least 75, but less than 85 (C) At least 85, but less than 95 (D) At least 95, but less than 115 (E) At least 115 Solution ST = S0 exp √ 2z

50e0.21+0.3

³¡ √ √ ´ ¢ 2 α − δ − 0.5σ 2 T + σ T z = 50e(0.15−0−0.5×0.3 )2+0.3 2z = zi = N −1 (u1 ) 2. 12 −1. 77 0.77

√

ST √= 50e0.21+0.3 2z 2×2.12 50e0.21+0.3 = 151. 63 √ 0.21+0.3 2×(−1. 77) 50e = 29. 11 √ 0.21+0.3 2×(0. 77) 50e = 85. 52 151. 63 + 29. 11 + 85. 52 = 266. 26 266. 26 The average of the simulated prices is = 88. 75. The answer is C. 3 i 1 2 3 Total

u1 ∼ U (0, 1) 0.9830 0.0384 0.7794

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Chapter 20

Brownian motion and Ito’s Lemma 20.1

Introduction

According to Wikipedia, Brownian motion (named in honor of the botanist Robert Brown) is either the random movement of particles suspended in a fluid or the mathematical model used to describe such random movements, often called a Wiener process. In 1827, while examining pollen grains suspended in water under a microscope, Brown observed minute particles in the pollen grains executing a continuous jittery motion. He observed the same motion in particles of dust, enabling him to rule out the hypothesis that the motion was due to pollen being alive. Although he did not provide a theory to explain the motion, the phenomenon is now known as Brownian motion in his honor. Brownian motion is a useful tool for modeling the stock price. The price of a stock is constantly hit by random events, just as a particle in the water is constantly hit by water molecules. In fact, the Brownian motion is to stochastic processes as the standard normal distribution to random variables. Brownian motion is an abstract concept. The first step toward learning the Brownian motion is to see it and experiment it. You can easily find Brownian motion simulations in the internet. Here are some simulations in the internet: • http://www.phy.ntnu.edu.tw/java/gas2d/gas2d.html • http://www.aip.org/history/einstein/brownian.htm • http://www.stat.umn.edu/~charlie/Stoch/brown.html • http://www.matter.org.uk/Schools/Content/BrownianMotion There’s a large body of knowledge on the internet about the Brownian motion. For example, you can check out Wikipedia’s explanation at: 205

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http://en.wikipedia.org/wiki/Brownian_motion To help you experiment with the Brownian motion, I designed a spreadsheet titled "Simulate Brownian Motion." Download this spreadsheet. My spreadsheet uses the following Excel functions: • Rand() returns a random number equal to or greater than 0 but less than 1. In other words, Rand() simulates a random variable uniformly distributed over [0, 1). • NORMINV(probability,mean,standard_dev) returns the inverse of the normal cumulative distribution for the specified mean and standard deviation.

20.1.1

Big picture

When studying the Brownian motion and Ito’s lemma, remember the following big picture. The Black-Scholes option pricing formula Equation 12.1 relies on the Black-Scholes PDE (Equation 12.24). Equation 12.24 assumes the following price model for a risk-free asset and a risky asset (i.e. stock): The price of a risk free asset (i.e. the savings account or a bond) is B (t) = ert

(20.1)

The price of a risky asset (i.e. stock) at time t is: #

α−

S (t) = S (0) e

1 2 σ 2

$

√ t+σ tY (t)

(20.2)

In the above equation: • S (0) is the stock price at time zero • S (t) is the stock price at time t • α is the expected (annualized continuously compounded) ∙ return ¸ of the 1 S (t) stock. In other words, E [S (t)] = S (0) eαt or α = ln E t S (0) • r is the (annualized continuously compounded) risk-free interest rate. • σ is the stock’s volatility. • Y (t) is a random draw of a standard normal random variable. Y (t1 ) and Y (t2 ) are independent for t1 6= t2 Equation 20.2 is another form of Equation 20.1 in the textbook: dS (t) = αdt + σdZ (t) S (t)

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(Textbook 20.1)

20.2. BROWNIAN MOTION

207

20.2

Brownian motion

20.2.1

Stochastic process

A stochastic process is a family of random variables indexed by time. For example, the temperature out side your house X (t) is a stochastic process. Let t = 0 represent now and t = 1 represent the next time (such as next hour, next day, next week). Then for a series of time points t = 1, 2, 3, ... there is a family of random temperatures X (0), X (1), X (2), X (3) .... This is the major difference between a stochastic process and a deterministic process. In a stochastic process you’ll see a series of random variables; for each time t there’s a corresponding random variable X (t). In contrast, in a deterministic process, you’ll see only one random variable.

20.2.2

Definition of Brownian motion

Consider a particle that jumps, at discrete times, up or down along the vertical line. At t = 0 the particle is at position zero. After each h-long time period, the particle jumps up or down by a constant distance of k and with equal probability of 0.5. That is, at t = h, 2h, 3h, ..., nh, the particle either moves up by k or moves down by k, with up and down movements having an equal probability of 0.5. Let Z (t) represent the height of the article from the position zero at time t. Clearly Z (0) = 0. We like to find Z (T ), the height of the article at time T = nh. The particle’s height 4k 3k 2k k 0

0 −k

time 0

2k k

h

−2k 2h

0 −k −3k 3h

−k −4k 4h

Let’s walk through the above table. At t = 0 the particle is at the position zero. At t = h, the particle’s height is either k or −k. At t = 2h, the k node either goes up to 2k or goes down to 0. Similarly, the −k node either goes up to 0 or goes down to −2k. So on and so forth. The jump at t = h is: Z (h) − Z (0) = Y (h) k Here Y (h) is a direction indicator. If Y (h) = 1, then the particle moves up by k; if Y (h) ½ = −1, then the particle moves down by k: 1 Probability 0.5 Y (h) = −1 Probability 0.5

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The jump½at t = 2h is: Z (2h) − Z (h) = Y (2h) k 1 Probability 0.5 Y (2h) = −1 Probability 0.5 The jump at t = nh = T is: Z (nh) − Z [(n − 1) h] = Y (nh) k Here Y (nh) ½ is a direction indicator: 1 Probability 0.5 Y (nh) = −1 Probability 0.5 The direction indicators Y (h), Y (2h), Y (3h), ..., Y (nh) are independent identically distributed binomial random variables. For i = 1 to n, E h(ih) =i(1) 0.5 + (−1) 0.5 = 0

E (ih)2 = (1)2 0.5 + (−1)2 0.5 = 1 i h 2 V ar (ih) = E (ih) − E 2 (ih) = 1 The particle’s height at time T is: Z (T ) = [ Z (h) − Z (0)]+[ Z (2h) − Z (h)]+[ Z (3h) − Z (2h)]+...+[ Z (nh = T ) − Z [(n − 1) h]] = Y (h) k + Y (2h) k + Y (3h) k + ... + Y (nh) k = [Y (h) + Y (2h) + Y (3h) + ... + Y (nh)] k

According to the central limit theorem, Z (T ) is approximately normal with mean and variance as: E [Z (T )] = kE [Y (h) + Y (2h) + Y (3h) + ... + Y (nh)] = 0 V ar [Z (T )] = k 2 V ar [Y (h) + Y (2h) + Y (3h) + ... + Y (nh)] k2 T = k2 n = k2 = T h h 2 k We want V ar [Z (T )] = T to exist (i.e. not to become infinite) as n → ∞. h 2 k k2 To achieve this, we set to a positive constant: = c. To make our h h model simple, we set c = 1. Hence √ k2 =1 k= h h Now we have: √ Z (ih) − Z [(i − 1) h] = Y (ih) h Now Z (T ) is approximately normal with mean zero and variance Z (T ) ∼ N (0, T )

(20.3) k2 T = T: h

Please note that another way to specify the model is treat Y (ih) as a random draw of a standard normal random variable (instead of a binomial random variable):

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20.2. BROWNIAN MOTION

√ Z (ih) − Z [(i − 1) h] = Y (ih) h and Y (ih) ∼ N (0, 1)

209

(20.4)

My spreadsheet for simulating the Brownian motion uses both Equation 20.3 and Equation 20.4. The random process Z (t) as n → ∞ is called the Brownian motion or the Wiener process. Next, let’s formally define the Brownian motion. Definition 20.2.1. A stochastic process Z (t) is a Brownian motion or a Wiener process if 1. Z (0) = 0 . Brownian motion starts at zero (this is merely for our convenience). 2. Z (t + h) − Z (t) is normally distributed with mean 0 and variance h. This means that the increments over a time interval h is normally distributed with mean 0 and variance h. This stands true no matter how small or big h is. 3. Z (t + s1 ) − Z (t) is independent of Z (t) − Z (t − s2 ) where s1 , s2 > 0. 4. Z (t) is continuous.

20.2.3

Martingale

The following part is based on Wikipedia. A martingale is a stochastic process (i.e., a sequence of random variables) such that the conditional expected value of an observation at some time t, given all the observations up to some earlier time s, is equal to the observation at that earlier time s. Originally, martingale referred to a betting strategy popular in 18th century France. The rule of the game is that the gambler wins his stake if a coin comes up heads and loses it if the coin comes up tails. The martingale strategy had the gambler double his bet after every loss, so that the first win would recover all previous losses plus win a profit equal to the original stake. Since eventually a gambler will win at least once, the martingale betting strategy was thought to be sure way of winning. In reality, however, the exponential growth of the bets would eventually bankrupt the gambler. A stochastic process (i.e., a sequence of random variables) X (t) is a martingale if the following holds: E [X (t) |X (s)] = X (s) for t > s. It can be shown that the Brownian motion Z (t) is martingale.

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For t > s, we have: E [Z (t) |Z (s)] = E [Z (t) − Z (s) + Z (s) |Z (s)] = E [Z (t) − Z (s) |Z (s)] + E [Z (s) |Z (s)] E [Z (s) |Z (s)] = Z (s) Z (t) − Z (s) is independent of Z (s). → E [Z (t) − Z (s) |Z (s)] = E [Z (t) − Z (s)] = 0 → E [Z (t) |Z (s)] = E [Z (t) − Z (s) |Z (s)] + E [Z (s) |Z (s)] = Z (s) Hence Z (t) is martingale. E [Z (t) |Z (s)] = Z (s) means that the best estimate of the future value of a Brownian motion is its current value.

20.2.4

Properties of Brownian motion

1. The Brownian motion is continuous everywhere yet differentiable nowhere. 2. The first-order variation is infinite: lim |Z (h)−Z (0) |+|Z (2h)−Z (h) |+ n→∞

... + |Z (nh) − Z [(n − 1) h] | → ∞

3. The second-order variation (called quadratic variation) is equal to the 2 2 length of the time interval: lim {[Z (h) − Z (0)] + [Z (2h) − Z (h)] + n→∞

... + (Z (nh) − Z [(n − 1) h])2 } = T

4. Cov [Z (s) , Z (t)] = min (s, t). The covariance of two Brownian motions is the shorter time interval. ½ 0 if n is odd 5. The higher moments of Z (t) is: E [Z n (t)] = tn/2 (n − 1) (n − 3) ...1 if n is even Let’s look at the first property. If you look at the simulation of Brownian motion over the internet, you’ll find that the Brownian motion is always continuous yet it’s not differentiable anywhere. Can you imagine that a function is continuously anywhere yet differentiable nowhere? If I hadn’t studied the Brownian motion, I would have never thought that such a function exists. We can explain the non-differentiality using the following equation (it’s textbook Equation 20.4): √ dZ (t) = Y (t) dt

(20.5)

The above equation is another form of √ Equation 20.3 or 20.4: dZ (t) = Z (t + dt) − Z (t) = Y (t) dt The textbook explains in the footnote that you can treat Y (t) as a binomial

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211

random variable or a standard normal random variable. Either way, dZ (t) is a normal random variable. From Equation 20.5, we have: dZ (t) Y (t) = √ → ∞ as dt → 0 dt dt Let’s look at the 2nd property. The following is an intuitive but not rigorous proof. According to the definition of the Brownian motion, Z (h) − Z (0), Z (2h) − Z (h), ..., Z (nh) − Z [(n − 1) h] are independent identically distributed normal random variable with mean 0 and variance h. On average, |Z [(i + 1) h]−Z (ih) | approaches E|X| where X ∼ N (0, h). Let f (x) represent the probability density functionR of X, then ∞ E|X| = −∞ |x|f (x) dx > 0 (since |x| ≥ 0, it’s mean must be positive) Then lim (|Z (h) − Z (0) | + |Z (2h) − Z (h) | + ... + |Z (nh) − Z [(n − 1) h] |) n→∞

approaches nE|X|, which approaches ∞ (since n → ∞, a positive constant E|X| times n must also approach infinity.) √ Property 2 should be easy to understand. Since ∆Z (t)√= Y (t) ∆t, for a √ 0.000001 tiny interval ∆t, ∆t is much large than ∆t. For example, = 1, 000. 0.000001 Hence during any short interval, the Brownian motion can move up or down by an infinitely large amount. In contrast, for a continuously differentiable function y = f (t), we have: ∆y = f 0 (t) ∆t → 0 as ∆t → 0 Let’s get an intuitive ¡ feel ¢ of the 3rd property. On average, {Z [(i + 1) h] − 2 2 approaches E X . Z (ih)} ¡ ¢ E X 2 = E 2 (X) + V ar (X) = 02 + h = h 2 2 2 lim [Z (h) − Z (0)] + [Z (2h) − Z (h)] + ... + (Z (nh) − Z [(n − 1) h]) apn→∞ proaches h + h + ... + h = nh = T In contrast, the 2nd order variation of a differentiable function is zero. For example, we can find the 2nd order variation of the function y = x is zero. Divide the interval [0, T ] into [0, h],[h, 2h],...,[(n − 1) h, nh = T ] µ ¶2 T 2 2 2 2 lim {(h − 0) + (2h − h) +... +[nh − (n − 1) h]} = nh = lim n = n→∞ n→∞ n 0 It can be proven that the quadratic variation of any continuously differentiable function is zero. The quadratic variation of any continuously differentiable function is zero because such a function is roughly linear at any point. For a continuously differentiable function y = f (t), we have: 2 ∆y = f 0 (t) ∆t → (∆y)2 = [f 0 (t)] (∆t)2 2 2 For a tiny interval ∆t, (∆t) → 0 must faster than ∆t → 0. Hence (∆y) → 0

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√ 2 In contrast, for a Brownian motion we have ∆Z (t) = Y (t) ∆t and [∆Z (t)] = P P 2 2 [Y (t)] ∆t → ∆t. Hence [∆Z (t)] → ∆t = T Brownian motion Property 3 reconfirms the idea that Brownian motion is not differentiable anywhere. If it’s differentiable, then its second order variation would be zero. √ Tip 20.2.1. Just memorize Equation 20.5 dZ (t) = Y (t) dt. This equation tells you that the Brownian motion Z (t) is not differentiable anywhere, its second order variation is t, and its first order variation is infinite. √ Tip 20.2.2. To get an intuitive feel of the equation dZ (t) = Y (t) dt, imagine you are looking at the Brownian motion under a magnifying class. If you zoom in on the Brownian motion by shrinking the time interval dt, no matter how much you reduce dt, you’ll see a jigsaw. In comparison, if you zoom in on a continuously differentiable function such as y = t2 , you’ll see a straight line. Please note that Derivatives Markets explains Property 2 and 3 using the following formula: √ Z [(i + 1) h] − Z (ih) = Y [(i + 1) h] h Since Y [(i + 1) h] is a√binomial random variable having a value of ±1, then |Z [(i + 1) h] − Z (ih) | = h and {Z [(i + 1) h] − Z (ih)} = h. Hence √ lim (|Z (h) − Z (0) | + |Z (2h) − Z (h) | + ... + |Z (nh) − Z [(n − 1) h] |) = n h → n→∞ ∞ 2 2 2 lim {[Z (h) − Z (0)] + [Z (2h) − Z (h)] + ... + (Z (nh) − Z [(n − 1) h]) } = n→∞ nh = T The problem with this explanation is that it works if we treat Y [(i + 1) h] as binomial random variable whose value is ±1. Such explanation won’t work if we treat Y [(i + 1) h] as a random draw of a standard normal random variable. The explanation I provided here works no matter if you treat Y [(i + 1) h] as a binomial random variable or a standard normal random variable. Let’s look at Property 4. Suppose s ≤ t Cov [Z (s) , Z (t)] = Cov{Z (s) , Z (s) + [Z (t) − Z (s)]} Using the formula Cov (a, b + c) = Cov (a, b) + Cov (a, c), we get: Cov{Z (s) , Z (s)+[Z (t) − Z (s)]} = Cov{Z (s) , Z (s)}+Cov{Z (s) , [Z (t) − Z (s)]} Cov{Z (s) , Z (s)} = V ar [Z (s)] = s Since Z (s) and [Z (t) − Z (s)] are independent (Brownian motion definition Point #3), Cov{Z (s) , [Z (t) − Z (s)]} = 0 → Cov [Z (s) , Z (t)] = s = min (s, t) Property 5 is based on the moment formula for a standard normal random variable φ ½ 0 if n is odd E (φn ) = (20.6) (n − 1) (n − 3) ...1 if n is even We can find the n-th moment of a random variable X using the moment generating function (MGF):

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20.2. BROWNIAN MOTION

213 ∙

dn E (X ) = MX (t) dtn n

¸

(20.7)

t=0

The MGF of a normal random variable X with mean μ and the standard deviation σ is: 1 ¡ tX ¢ μt+ σ 2 t2 2 MX (t) = E e =e

(20.8)

1 2 t → Mφ (t) = e 2 Using Equation 20.7, you can verify that Equation 20.6 holds. Since Z (t) is a normal random variable with mean 0 and variance t, then Z (t) √ is a standard normal random variable. Hence t ½ 0 if n is odd (20.9) E [Z n (t)] = n/2 (n − 1) (n − 3) ...1 if n is even t Example 20.2.1. Calculate P [Z (3) > 1] Z (3) is a normal random variable with mean µ 0 and¶variance 3. 1−0 P [Z (3) > 1] = 1 − P [Z (3) ≤ 1] = 1 − Φ √ = 1 − Φ (0.577 35) = 3 0.281 9 Example 20.2.2. Calculate P [Z (1) ≤ 0 ∩ Z (2) ≤ 0] Z (1) is a normal random variable with mean 0 and variance 1. Let X = Z (1) Z (2) = Z (1) + [Z (2) − Z (1)] Z (2) − Z (1) is a normal random variable with mean 0 and variance 1. Z (2) = Z (1) + [Z (2) − Z (1)] ≤ 0 → [Z (2) − Z (1)] ≤ −Z (1) Let Y = Z (2) − Z (1). X and Y are independent. P [Z (1) ≤ 0 ∩ Z (2) ≤ 0] = P (X ≤ 0 ∩ Y ≤ −X) To have X ≤ 0 ∩ Y ≤ −X, we first fix X at a tiny interval (x, x + dx) where −∞ < x < 0. Next, we set Y < −x. Then we are guaranteed to have X ≤ 0∩Y ≤ −X. Let f (x) and Φ (x) represent the probability density function (pdf) and the cumulative density function (cdf) of a standard normal random variable. R0 R0 P (X ≤ 0 ∩ Y ≤ −X) = −∞ P (x < X < x + dx) P (Y < −x) = −∞ [f (x) dx] P (Y < −x) However, f (x) dx = dΦ (x) and P (Y < −x) = Φ (−x) = 1 − Φ (x) R0 R0 R0 [f (x) dx] P (−x) = −∞ [dΦ (x)] [1 − Φ (x)] = −∞ [1 − Φ (x)] dΦ (x) = −∞ R0 [1 − Φ (x)] dΦ (x) −∞ ∙ ¸0 1 2 = Φ (x) − Φ (x) 2 −∞

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¤ 1£ 2 = [Φ (0) − Φ (−∞)] − Φ (0) − Φ2 (−∞) 2 # ∙ ¸ " µ ¶2 1 1 1 2 = −0 −0 − 2 2 2 µ ¶2 1 1 1 3 = − = 2 2 2 8 Example 20.2.3. Calculate Cov [Z (5) , Z (2)] Cov [Z (5) , Z (2)] = min (5, 2) = 2 £ ¤ Example 20.2.4. Calculate E Z 4 (t) . £ ¤ E Z 4 (t) = t4/2 (4 − 1) = 3t2

20.2.5

Arithmetic Brownian motion and Geometric Brownian motion

A standard Brownian motion is normally distributed with mean 0 and variance 1. Now we want to extend the Brownian motion to allow for non-zero mean and an arbitrary variance. Define a stochastic √ process: X (t + h) − X (t) = αh + σY (t + h) h Breaking down [0, T ] into small intervals [0, h],[h, 2h],...,[(n − 1) h, nh = T ], we have: √ i √ i Pn h Pn h X (T ) − X (0) = i=1 αh + σY (ih) h = αT + σ i=1 Y (ih) h √ i P h As n → ∞, ni=1 Y (ih) h → Z (T ) X (T ) − X (0) = αT + σZ (T )

(20.10)

dX (t) = αdt + σdZ (t)

(20.11)

Equation 20.10 and 20.11 are called arithmetic Brownian motion. α is the instantaneous mean per unit of time; σ is the instantaneous standard deviation per unit of time. Equation 20.10 and 20.11 indicate that X (T )−X (0) is normally distributed. Its mean and variance are: E [X (T ) − X (0)] = E [αT + σZ (T )] = αT + σE [Z (T )] = αT + σ × 0 = αT V ar [X (T ) − X (0)] = V ar [αT + σZ (T )] = V ar [σZ (T )] = σ2 V ar [Z (T )] = σ2T The textbook lists the major properties and weaknesses of Equation 20.11. Major properties: 1. X (t) is normally distributed.

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20.2. BROWNIAN MOTION

215

2. We can change the variance by changing the parameter σ. 3. We can change the mean by changing the parameter α. Now the mean is no longer zero if α 6= 0. And we have E [X (T )] − E [X (0)] = αT . This means that after time T , the stock price drifts away from the price at time zero. Major weaknesses: 1. The stock price X (t) can be negative. Since X (t) is normally distributed,−∞ < X (t) < ∞. Equation 20.11 allows a negative stock price. Of course, the stock price can’t become negative. 2. The expected change of the stock price does not depend on the stock price. In reality, the expected change of the stock price should be proportional to the stock price. The higher the stock price, the higher the expected change. So we like to have E [dX (t)] = αX (t). We need to modify Equation 20.11 to allow E [dX (t)] = αX (t). 3. The variance of the stock price does not depend on the stock price. In reality, the variance should be proportional to the stock price. So we need to modify Equation 20.11 to allow σ [X (t) , t] = σX (t). Major Weakness #2 and #3 can also be stated this way. Equation 20.11 can dX (t) α σ dX (t) be rewritten as = dt + dZ (t). This indicates that , X (t) X (t) X (t) X (t) the percentage return on the stock depends on the stock price X (t). However, in reality, we think that the stock return on average shouldn’t depend on the stock price. In other words, instead of Equation 20.11, we like to see dX (t) = αdt + σdZ (t) X (t)

(20.12)

dX (t) = X (t) αdt + σX (t) dZ (t)

(20.13)

or

Equation 20.12 and 20.13 are called the geometric Brownian motion.

20.2.6

Ornstein-Uhlenbeck process

We can modify Equation 20.11 to allow for mean reversion. It’s reasonable for us to assume that the stock price or the interest rate will revert to the mean. For example, if the stock price is too high, then it might go down; if the stock price is too low, it might go up. We modify the drift term in Equation 20.11: dX (t) = λ [α − X (t)] dt + σdZ (T )

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(20.14)

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If α = 0, 20.14 is called the Ornstein-Uhlenbeck process: dX (t) = −λX (t) dt + σdZ (T )

(20.15)

The Ornstein-Uhlenbeck process is the most widely used mean reverting stochastic process in financial modeling.

20.3

Definition of the stochastic calculus

We can rewrite Equation 20.12 as: RT RT RT dX (t) = 0 X (t) αdt + 0 σX (t) dZ (t) 0 RT RT RT RT or X (T )−X (0) = 0 X (t) αdt+ 0 σX (t) dZt = α 0 X (t) dt+σ 0 X (t) dZt RT But what’s the meaning of 0 X (t) dZ (t)? Or generally, what’s the meaning Rb of a g (t) dZ (t)? To answer this question, let’s R 1take a step back and find out the meaning of a simple deterministic calculus 0 x2 dx. R1 2 x dx is the area of the function x2 bounded by x = 0 and x = 1. To find 0 this area, we divide the interval [0, 1] into n intervals [0, h], [h, 2h], [2h, 3h], ..., [(n − 1) h, nh = 1]. Then we approximate the area with the sum of n rectangles. The area of function x2 over the interval [(i − 1) h, ih] is roughly the area of the rectangular with height [(i − 1) h]2 and width ih − (i − 1) h = h. R ih

x2 dx ≈ [(i − 1) h]2 h = (i − 1)2 h3 Pn R ih Pn 2 3 2 2 = lim i=1 (i−1)h x dx = lim i=1 (i − 1) h = lim {0 h +

(i−1)h R1 2 x dx 0 2 2

n→∞

n→∞

n→∞

2

h h + (2h) h + ... + [(n − 1) h] h} 02 h +hh2 h + (2h)2 h + ... + [(n − 1)i h]2 h = h3 02 + 12 + 22 + ... + (n − 1)2 Using the famous formula: n (n + 1) (2n + 1) 12 + 22 + 32 + ... + n2 = h i6 (n − 1) (n) (2n − 1) 2 3 2 2 2 h 0 + 1 + 2 + ... + (n − 1) = h3 6 1 Since h = , we have: n R1

1 (n − 1) (n) (2n − 1) x2 dx = lim 3 n¶ ∙ n→∞ µ µ 6 ¶¸ 1 1 1 1 = lim 1− (1) 2 − = n→∞ 6 n n 6 R ih There are other ways to approximate (i−1)h x2 dx. For example, the area of function x2 over the interval [(i − 1) h, ih] is roughly the area of the rectangular with height (ih)2 and width ih − (i − 1) h = h. R ih 2 x2 dx ≈ (ih) h = i2 h3 (i−1)h 0

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20.3. DEFINITION OF THE STOCHASTIC CALCULUS R1 0

x2 dx = lim

n→∞

R ih i=1 (i−1)h

Pn

x2 dx = lim

n→∞

Pn

i=1

217

n (n + 1) (2n + 1) 3 h n→∞ 6

i2 h3 = lim

1 n (n + 1) (2n + 1) 1 = lim = n→∞ 6 n3 R 6 ih The 3rd way to approximate (i−1)h x2 dx is to take the average height of the rectangular. So the area of function x2 over the interval [(i − 1) h, ih] is 2 2 (i − 1) h2 + (ih) and width roughly the area of the rectangular with height 2 ih − (i − 1) h = h. 2 2 2 (i − 1) h2 + (ih) (i − 1) + i2 2 = h 2 2 # " 2 R ih (i − 1) + i2 2 2 x dx ≈ h h (i−1)h 2 " # 2 2 R1 2 Pn R ih P (i − 1) + i n 2 x dx = lim lim h2 h i=1 (i−1)h x dx = n→∞ i=1 0 n→∞ 2 i Pn h Pn £ 2 2 ¤ 1 1 2 2 h (i − 1) h + lim i h h = lim i=1 n→∞ 2µ 2 n→∞ i=1 ¶ µ ¶ 1 1 1 1 1 = + = 2 6 2 6 6 It seems natural that we extend this logic of deterministic integration to a define a stochastic integration. Suppose we partition [a, b] into a = t0 < t1 < t2 < ... < tn = b. We can make the partition intervals [t0, t1 ], [t1, t2 ], ..., [tn−1, tn ] have the same length b−a tk+1 − tk = . We can also have the partition intervals [t0, t1 ], [t1, t2 ], ..., n [tn−1, tn ] have different lengths. Rb Pn−1 g (t) dZ (t) = lim k=0 g (tk ) [Z (tk+1 ) − Z (tk )] a n→∞

Definition 20.3.1. Suppose g (t) is a simple process, meaning that g (t) is piecewise-constant ¤ Rb £ but may have jumps at a = t0 < t1 < t2 < ... < tn = b. If a E g 2 (t) dt < ∞, Rb then the stochastic integral a g (t) dZ (t) is defined as Z

b

g (t) dZ (t) =

a

n−1 X k=0

g (tk ) [Z (tk+1 ) − Z (tk )]

(20.16)

¤ Rb £ In the above definition, a E g 2 (t) dt < ∞ is the sufficient condition for Pn−1 k=0 g (tk ) [Z (tk+1 ) − Z (tk )] to exist. Example 20.3.1. Calculate

RT 0

dZ (t)

Solution.

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Here g (t) = 1. RT R T ¡ 2¢ E 1 dt = T < ∞. So 0 dZ (t) exists. 0 RT Pn−1 dZ (t) = k=0 [Z (tk+1 ) − Z (tk )] = Z (T ) 0

Example 20.3.2. ⎧ ⎨ 1 if 0 ≤ t ≤ 1 2 if 1 < t ≤ 2 X (t) = ⎩ 3 if 2 < t ≤ 3 R3 Calculate 0 X (t) dZ (t)

Solution.

Dividend [0, 3] into (0, 1), (1, 2), and (2, 3). Then X (t) is constant during the Rinterval and jumps at t = 1 and t = 2. Hence 3 X (t) dZ (t) 0 = X (0) [Z (1) − Z (0)] + X (1) [Z (2) − Z (1)] + X (2) [Z (3) − Z (2)] = 1 [Z (1) − Z (0)] + 2 [Z (2) − Z (1)] + 3 [Z (3) − Z (2)] = 1Z (1) + 2 [Z (2) − Z (1)] + 3 [Z (3) − Z (2)] = 3Z (3) − Z (2) − Z (1) If g (tk ) is not a simple process, then we define the stochastic integral as follows. If E lim (X − a)2 = 0, we say that X approach a in mean square. n→∞ Explain why the sample mean approaches the population mean in mean square. Suppose we take n random samples X1 , X2 ,...,Xn from the population X. 1 Pn Xk . It Let μ represent the population mean. Then the sample mean is n k=1 can be shown that ¶2 µ 1 Pn =0 E lim k=1 Xk − μ n→∞ n Toµsee why, notice ¶ Pn 1 Pn 1 1 E Xk = E ( k=1 Xk ) = nμ = μ n k=1 n n ¶2 1 Pn Xk − μ E n µk=1 ¶ P 1 Pn 1 = V ar = 2 V ar ( nk=1 Xk ) k=1 Xk n n V ar (X) 1 = 2 nV ar (X) = n n µ ¶2 1 Pn → E lim X − μ =0 k k=1 n→∞ n So the sample mean approaches the population mean in mean square. µ

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20.3. DEFINITION OF THE STOCHASTIC CALCULUS

219

Definition 20.3.2. Let a = t0 < t1 < t2 < ... < tn = b represent a partition of [a, b]. Define P random variable In = n−1 k=0 g (tk ) [Z (tk+1 ) − Z (tk )], where g (tk ) is a simple or a complex process and g (tk ) and Z (tk+1 ) − Z (tk ) are independent. If Rb £ 2 ¤ E g (t) dt < ∞ and the mean squared difference between In and U is zero a as n → ∞: 2

E lim (In − U ) = 0 n→∞

(20.17)

Then we say •

Rb a

g (t) dZ (t) = U

• In converges to

Rb a

g (t) dZ (t) in mean square

The above definition holds whether g (t) is a simple or complex process. Please note that in the term g (tk ) [Z (tk+1 ) − Z (tk )], g is evaluated at the left of the interval [tk , tk+1 ]: Rb Pn−1 g (t) dZ (t) = lim k=0 g (tk ) [Z (tk+1 ) − Z (tk )] a n→∞ Rb Pn−1 g (t) dZ (t) 6= lim k=0 g (tk+1 ) [Z (tk+1 ) − Z (tk )] a n→∞ Rb Pn−1 g (tk ) + g (tk+1 ) g (t) dZ (t) 6= lim [Z (tk+1 ) − Z (tk )] k=0 a n→∞ 2 R ih This is different from the deterministic calculus (i−1)h x2 dx, which can be approximated using 3 heights: • The left height [(i − 1) h]2 • The right height (ih)2 • The average height

(i − 1)2 h2 + (ih)2 2

In addition, we require that g (tk ) and Z (tk+1 ) − Z (tk ) are independent. In other words, we require that g (tk ) not depend on the future Brownian increment Z (tk+1 ) − Z (tk ). The requirements that we evaluate g at the left of the interval [tk , tk+1 ] and that g (tk ) and Z (tk+1 ) − Z (tk ) are independent agree with our intuition. We calculating g (tk ) [Z (tk+1 ) − Z (tk )], g (tk ) is based on the information available to us during [0, tk ] and is independent of the future Brownian increment Z (tk+1 ) − Z (tk ). RT 2 Example 20.3.3. Calculate 0 [dZ (t)] .

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RT

RT 2 2 [dZ (t)] = 0 g (t) [dZ (t)] where g (t) = 1 Partition [0, T ] into [0, h], [h, 2h],...,[(n − 1) h, nh = T ]. P Pn 2 2 Let In = n−1 k=0 {Z [(k + 1) h] − [Z (kh)]} = k=1 {Z (kh) − [Z (k − 1) h]} Z (kh) − [Z (k − 1) h] is a normal random variable with mean 0 and variance h. Using Equation 20.9, we have 2 E{Z (kh) Pn − [Z (k − 1) h]} = h In = k=1 {Z (kh) − [Z (k − 1) h]}2 RT 2 Since E (In ) = nh = T , we guess that 0 [dZ (t)] = T The difficult part is to verify that the mean square error is zero as n → ∞: hP i2 2 lim E ( nk=1 {Z (kh) − [Z (k − 1) h]}) − T = 0 0

n→∞

Let ∆k = Z (kh) − [Z (k − 1) h] Pn 2 → ( k=1 {Z (kh) − [Z (k − 1) h]}) = ∆21 + ∆22 + ... + ∆2n hP i2 2 n → ( k=1 {Z (kh) − [Z (k − 1) h]}) − T £¡ ¡ ¢ ¤2 ¡ ¢2 ¢ = ∆21 + ∆22 + ... + ∆2n − T = ∆21 + ∆22 + ... + ∆2n +T 2 −2T ∆21 + ∆22 + ... + ∆2n

¢2 ¡ 2 ∆1 + ∆22 + ... + ∆2n = ∆41 + ∆42 + ...∆4n + 2∆21 ∆22 + 2∆21 ∆23 + ... ¡ ¢ ¡ ¢2 ¢ ¡ → E ∆21 + ∆22 + ... + ∆2n = E ∆41 + ∆42 + ...∆4n +2E ∆21 ∆22 + ∆21 ∆23 + ... ∆k is¡normal with mean 0 and variance h ¢ → E ¡∆4k = 3h2 ¢ → E ∆41 + ∆42 + ...∆4n = 3nh2 ∆i and ∆j where i 6= j are two independent normal random variables (Point 3 of the Brownian ¡ ¢ motion ¡ ¢definition) ¡ ¢ → E ∆2i ∆2j = E ∆2i E ∆2j = h × h = h2 1 There are n (n − 1) pairs of ∆i and ∆j where i 6= j 2 ¢ ¡ 1 → 2E ∆21 ∆22 + ∆21 ∆23 + ... = 2 × n (n − 1) h2 = n (n − 1) h2 2 ¡ ¢2 E ∆21 + ∆22 + ... + ∆2n = 3nh2 + n (n − 1) h2 £ ¡ ¢¤ ¡ ¢ E 2T ∆21 + ∆22 + ... + ∆2n = 2T E ∆21 + ∆22 + ... + ∆2n = 2T (nh) = 2T 2 hP i2 2 n → E ( k=1 {Z (kh) − [Z (k − 1) h]}) − T

= 3nh2 + n (n − 1) h2 + T 2 − 2T = [3n + n (n − 1)] h2 − T 2 However, h =

T n

µ ¶2 T 3n + n (n − 1) 2 → [3n + n (n − 1)] h −T = [3n + n (n − 1)] −T 2 = T − n n2 2

T2

As n → ∞,

2

3n + n (n − 1) →1 n2

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3n + n (n − 1) 2 T − T2 → 0 n2

20.3. DEFINITION OF THE STOCHASTIC CALCULUS

221

hP i2 2 lim E ( nk=1 {Z (kh) − [Z (k − 1) h]}) − T = 0 n→∞ RT Hence 0 [dZ (t)]2 = T . RT Example 20.3.4. Calculate 0 Z (t) dZ (t)

Partition [0, T ] into [0, h], [h, 2h],...,[(n − 1) h, nh = T ]. Pn−1 Pn Let In = k=0 Z (kh) {Z [(k + 1) h]−Z (kh)} = k=1 Z [(k − 1) h] {Z (kh)− Z [(k − 1) h]} ¡ ¢ 2 (a + b) − a2 + b2 Use the formula: ab = 2 Let a = Z [(k − 1) h] b = Z (kh) − Z [(k − 1) h] a P+n b = Z (kh) k=1 Z [(k − 1) h] {Z (kh) − Z [(k − 1) h]} 2 2 P [Z (kh)] − Z [(k − 1) h] − {Z (kh) − Z [(k − 1) h]}2 = nk=1 2 1 Pn 1 Pn 2 2 {[Z (kh)] − Z [(k − 1) h] }− {Z (kh) − Z [(k − 1) h]}2 = k=1 2 2 k=1 1 1 Pn 2 = [Z (nh)] − {Z (kh) − Z [(k − 1) h]}2 2 2 k=1 1 1 Pn 2 = [Z (T )] − {Z (kh) − Z [(k − 1) h]}2 2 2 k=1 → In =

1 1 Pn 2 {Z (kh) − Z [(k − 1) h]}2 [Z (T )] − 2 2 k=1

Pn lim lim In k=1 Z [(k − 1) h] {Z (kh) − Z [(k − 1) h]} = n→∞ P 1 1 2 n = [Z (T )] − lim {Z (kh) − Z [(k − 1) h]}2 2 2 n→∞ k=1 Pn From the previous example, we know that k=1 {Z (kh)−Z [(k − 1) h]}2 apRT RT proaches 0 [dZ (t)]2 = T in the mean square. So we guess that 0 Z (t) dZ (t) = 1 1 [Z (T )]2 − T 2 2 µ ∙ ¶¸2 1 1 2 Next, we need to prove that E lim In − =0 [Z (T )] − T n→∞ 2 2 µ ¶ 1 1 1 1 Pn 2 In − {Z (kh) − Z [(k − 1) h]}2 [Z (T )] − T = T − 2 2 2 2 k=1 In − E (In ) µ ¶ 1 1 Pn 1 1 2 2 = [Z (T )]2 − {Z (kh) − Z [(k − 1) h]} − − [Z (T )] T 2 2 k=1 2 2 ¢ Pn 1¡ 2 T − k=1 {Z (kh) − Z [(k − 1) h]} = 2 From the previous example, we found that ¡ ¢2 Pn lim E T − k=1 {Z (kh) − Z [(k − 1) h]}2 = 0 n→∞ µ ∙ ¶¸2 1 1 → E lim In − =0 [Z (T )]2 − T n→∞ 2 2 n→∞

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So we have

Z

T

Z (t) dZ (t) =

0

1 1 2 [Z (T )] − T 2 2

(20.18)

Equation 20.18 is surprising. In the deterministic calculus, we have 1 2 T . 2

RT 0

xdx =

1 Equation 20.18 has an extra term − T . This is why this extra term is 2 needed. Taking expectation of Equation 20.18: i 1 hR 1 T 2 E 0 Z (t) dZ (t) = E [Z (T )] − T 2 2

i hR T As to be explained later, E 0 Z (t) dZ (t) = 0. We already know that hR i 1 1 T 2 2 E [Z (T )] = T . Hence E 0 Z (t) dZ (t) = E [Z (T )] − T = 0. The extra 2 2 1 term − T is needed so the expectations of both sides of Equation 20.18 are 2 equal.

20.4

Properties of the stochastic calculus

1. Lineality.

RT 0

[c1 g (t) + c2 h (t)] dZ (t) = c1

2. Zero mean property. If 0

RT 0

RT 0

g (t) dZ (t)+c2

RT 0

h (t) dZ (t)

³R ´ £ ¤ T E X 2 (t) dt < ∞, then E 0 X (t) dZ (t) =

The proof is complex. However, for a simple process g (t) and h (t), Property #1 holds due to the definition of the stochastic integral. For a simple process g (t), Property #2 can be easily established. RT P g (t) dZ (t) = n−1 k=0 g (tk ) [Z (tk+1 ) − Z (tk )] 0 g (tk ) is constant during each partition interval and independent of Z (tk+1 )− Z (tk ). Then E{g (tk ) [Z (tk+1 ) − Z (tk )]} = E [g (tk )] E [Z (tk+1 ) − Z (tk )] = E [g (tk )] 0 = 0 RT £ ¤ RT RT Let’s apply Property 2 to 0 Z (t) dZ (t). Since 0 E Z 2 (t) dt = 0 tdt = h i RT 1 2 T < ∞, we have E 0 Z (t) dZ (t) = 0 2

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20.5. ITO’S LEMMA

223

20.5

Ito’s lemma

20.5.1

Multiplication rules

dZ dt 0

dZ dt

dt 0 0

The above table means: [dZ (t)]2 = dt dZ (t) dt = 0

(dt)2 = 0

Example 20.5.1. Explain why [dZ (t)]2 = dt dZ (t) is a normal random variable with mean 0 and variance dt (see the footnote of Derivatives Markets Page 652). Hence E [dZ (t)]2 = dt 2 E [dZ (t) − dt] = V ar [dZ (t)] = dt → 0 2 2 Hence dt approach [dZ (t)] in mean square. So [dZ (t)] = dt Example 20.5.2. Explain why dZ (t) dt = 0 E [dZ (t) dt] = E [dZ (t)] dt = 0 (here dt is treated as a constant) 2 2 2 3 E [dZ (t) dt − 0] = E [dZ (t)] (dt) = (dt) → 0 Hence 0 approaches dZ (t) dt in mean square. dZ (t) dt = 0. 2

Example 20.5.3. Explain why (dt) = 0 2

(dt) doesn’t contain any Brownian motion term dZ. So we need to calculate (dt)2 according to the deterministic calculus. In the deterministic calculus, 2 2 (dt) → 0 as dt → 0. Hence (dt) = 0. The textbook Derivatives Markets also gives the following formula: 0

dZ × dZ = ρdt

(20.19)

The above formula will be explained later.

20.5.2

Ito’s lemma

In essence, Ito’s lemma is a Taylor series applied to Brownian motion. ∧ Suppose that a stock has an expected instant return α [S (t) , t], dividend ∧

∧

yield δS (t) , t, and instant volatility σ [S (t) , t] follows geometric Brownian motion: µ ¶ ∧ ∧ ∧ dS (t) = α − δ dt + σdZ (t) (20.20) ∧ ∧

∧

Here α, δ, and σ are function of the stock price S (t) and time t.

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA

Let C [S (t) , t] represent the value of a call or put option. We want to find out the change of the option value given there’s a small change of the stock price and a small change of time. Using Taylor series, we have: ∂C 1 ∂2C ∂2C ∂C 1 ∂2C 2 2 (dS) + (dt) + dS + dt+ dSdt (20.21) ∂S ∂t 2 ∂S 2 2 ∂t2 ∂S∂t ∙µ ¶ ¸2 µ ¶ µ ¶ ∧ ∧ 2 ∧ ∧ ∧ ∧ ∧ ∧ 2 2 [dS (t)] = α − δ dt + σdZ (t) = α − δ (dt) +2 α − δ σdZ (t) dt+

dC [S (t) , t] =

∧2

2

σ [dZ (t)] µ ¶ ∧ ∧ ∧ 2 dSdt = α − δ (dt) + σdZ (t) dt

2

Using the multiplication rules: (dt) = 0 2

∧2

[dS (t)] = σ dt Now we have:

2

[dZ (t)] = dt

dZ (t) dt = 0

dSdt = 0

∂C ∂C ∂C 1 ∂2C ∂C 1 ∂ 2 C ∧2 2 (dS) = σ dt dS + dt + dS + dt + ∂S ∂t 2 ∂S 2 ∂S ∂t 2 ∂S 2 (20.22) Next, apply Equation∙µ 20.20 to¶Equation ??:¸ ∧ ∂C ∂C 1 ∂ 2 C ∧2 ∧ ∧ → dC [S (t) , t] = α − δ dt + σdZ (t) + σ dt dt + ∂S ∂t 2 ∂S 2 ∙µ ¶ ¸ ∧ ∂C ∂C 1 ∧2 ∂ 2 C ∂C ∧ ∧ = α−δ + σdZ (t) + σ dt + ∂S 2 ∂S 2 ∂t ∂S ¶ ¸ ∙µ ∧ ∂C ∂C 1 ∧2 ∂ 2 C ∂C ∧ ∧ α−δ + σdZ (t) (20.23) + σ dt + dC [S (t) , t] = 2 ∂S 2 ∂S ∂t ∂S

dC [S (t) , t] =

Equation 20.23 is called the Ito’s lemma. Tip 20.5.1. Don’t bother memorizing Equation 20.23. Just derive Equation 20.23 on the spot. First, write down the Taylor series Equation 20.22. Next, apply the multiplication rules. Then you’ll get Equation 20.23. If S (t) follows a geometric Brownian motion, we have: ∧

• α [S (t) , t] = αS (t) ∧

• δ [S (t) , t] = δS (t) ∧

• σ [S (t) , t] = σS (t) Equation 20.23 becomes: ∙ ¸ ∂C ∂C 1 ∂C ∂2C dC [S (t) , t] = (α − δ) S + σ2S 2 2 + dt + σSdZ (t) (20.24) ∂S 2 ∂S ∂t ∂S

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Tip 20.5.2. Don’t bother memorizing Equation 20.24. Just derive Equation 20.24 on the spot. Please note that ∂C ∂C ∂2C =Γ =∆ =θ ∂S ∂S 2 ∂t Then Equation 20.22 becomes:

(option Greeks)

1 2 dC [S (t) , t] = ∆dS + θdt + Γ (dS) 2

(20.25)

Consider a tiny interval dt ≈ h. We have: dC [S (t) , t] ≈ C [S (t + h) , t + h] − C [S (t) , t] dS ≈ S (t + h) − S (t) Equation 20.25 becomes: 1 2 C [S (t + h) , t + h]−C [S (t) , t] ≈ ∆ [S (t + h) − S (t)]+θh+ Γ [S (t + h) − S (t)] 2 (20.26) Equation 20.26 is just the textbook Equation 13.6 (Derivatives Markets page 426). If S (t) were deterministic (i.e. if Equation 20.20 didn’t have dZ (t) term), ∂2C then → 0 and Equation 20.22 would be: ∂S 2 ∂C ∂C dC [S (t) , t] = dS + dt ∂S ∂t

20.6

Geometric Brownian motion revisited

There are two minor concepts under geometric Brownian motion. SOA can easily write a question on these concepts. So let’s study them.

20.6.1

Relative importance of drift and noise term

Consider a discrete geometric Brownian motion: X (t + h) − X (t) =

αX (t) h | {z }

deterministic component

√ + σX (t) Y (t) h {z } | random comp onent

One phrase you need to know is called "the ratio of the standard deviation to the drift" (the drift is actually the deterministic component). The ratio of the standard deviation to the drift is defined as: √ σX (t) h σ (20.27) = √ αX (t) h α h

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20.6.2

CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA

Correlated Ito processes

Let W1 (t) and W2 (t) represent two independent Brownian motions. Suppose Z (t) = W1 (t) 0

Z (t) = ρW1 (t) +

p 1 − ρ2 W2 (t)

(20.28) (20.29)

Please note that Z (t) is normally distributed with mean 0 and variance t because W1 (t) is normally distributed with mean 0 and variance p t. You might wonder why Equation 20.29 has constants ρ and 1 − ρ2 . These 0

two constants are needed to make Z (t) normally distributed with variance t. Because W1 (t) and W2 (t) areptwo independent normal random variables, the 2 linearhcombination i hρW1 (t) + p1 − ρ W2 (t) iis also normally distributed. 0 E Z (t) = E ρW1 (t) + 1 − ρ2 W2 (t) hp i = E [ρW1 (t)] + E 1 − ρ2 W2 (t) p = ρE [W1 (t)] + 1 − ρ2 E [W2 (t)] p = ρ0 + 1 − ρ2 0 = 0 h 0 i h i p V ar Z (t) = V ar ρW1 (t) + 1 − ρ2 W2 (t) hp i = V ar [ρW1 (t)] + V ar 1 − ρ2 W2 (t) ¡ ¢ = ρ2 V ar [W1 (t)] + 1 − ρ2 V ar [W2 (t)] ¡ ¢ = ρ2 t + 1 − ρ2 t = t

h i 0 0 The covariance between Z (t) and Z (t) is: Cov Z (t) , Z (t)

Using the formula: Yi) = E (XY ) −hE (X)iE (Y ) h standard i h Cov (X, 0 0 0 → Cov Z (t) , Z (t) = E Z (t) Z (t) − E [Z (t)] E Z (t) h i h i 0 0 = E Z (t) Z (t) − 0 × 0 = E Z (t) Z (t)

h i p p 0 Z (t) Z (t) = W1 (t) ρW1 (t) + 1 − ρ2 W2 (t) = ρ [W1 (t)]2 + 1 − ρ2 W1 (t) W2 (t) h i ³ i ´ hp 0 E Z (t) Z (t) = E ρ [W1 (t)]2 + E 1 − ρ2 W1 (t) W2 (t) p 2 = ρE [W1 (t)] + 1 − ρ2 E [W1 (t) W2 (t)] E [W1 (t)]2 = t E [W1 (t) W2 (t)] = E [W1 (t)] E [W2 (t)] = 0 × 0 = 0 i h 0 (20.30) E Z (t) Z (t) = ρt

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h i h i 0 0 The textbook says calls Cov Z (t) , Z (t) = E Z (t) Z (t) = ρt the corre0

lation between Z (t) and Z (t). However, the term correlation between X and Y typically means the following: Cov (X, Y ) ρX,Y = σX σY 0 If wehuse the typical i definition, the correlation between Z (t) and Z (t) is: 0

Cov Z (t) , Z (t)

ρt = √ √ =ρ σ Z(t) σ Z 0 (t) t t Since Derivatives Markets is the textbook, we have to adopt its definition 0 that the correlation between Z (t) and Z (t) is ρt. 0

dZ (t) and dZ (t) are both normal random variables with mean 0 and variance dt. Applying Equation 20.30 and replacing t with dt, we have: i h 0 (20.31) E dZ (t) dZ (t) = ρdt 0

Finally, let’s explain why Equation 20.19 dZ × dZ = ρdt holds. i h 0 E dZ (t) dZ (t) = ρdt i2 h i h i2 i h h 0 0 0 0 E dZ (t) dZ (t) − ρdt = V ar dZ (t) dZ (t) = E dZ (t) dZ (t) −E 2 dZ (t) dZ (t) 0

dZ (t) dZ (t) h i p = dW1 (t) × d ρW1 (t) + 1 − ρ2 W2 (t) h i p = dW1 (t) × ρdW1 (t) + 1 − ρ2 dW2 (t) p = ρ [dW1 (t)]2 + 1 − ρ2 dW1 (t) dW2 (t) h i2 0 dZ (t) dZ (t) p ¢ 4 ¡ 2 2 3 = ρ2 [dW1 (t)] + 1 − ρ2 [dW1 (t)] [dW2 (t)] +2ρ 1 − ρ2 [dW1 (t)] dW2 (t) h i2 0 E dZ (t) dZ (t)

³ ´ ´ p ¡ ¢ ³ = ρ2 E [dW1 (t)]4 + 1 − ρ2 E [dW1 (t)]2 [dW2 (t)]2 +2ρ 1 − ρ2 E [dW1 (t)]3 dW2 (t) p ¡ ¢ = ρ2 E [dW1 (t)]4 + 1 − ρ2 E [dW1 (t)]2 E [dW2 (t)]2 +2ρ 1 − ρ2 E [dW1 (t)]3 E [dW2 (t)] p ¡ ¢ = ρ2 3 (dt)2 + 1 − ρ2 dt × dt + 2ρ 1 − ρ2 0 × 0 ¡ ¢ 2 = ρ2 3 (dt) + 1 − ρ2 dt × dt ¢¤ ¡ ¡ ¢ £ = 3ρ2 + 1 − ρ2 (dt)2 = 2ρ2 + 1 (dt)2 h i 0 E dZ (t) dZ (t) = ρdt i2 i h h 0 0 E dZ (t) dZ (t) − E 2 dZ (t) dZ (t) ¢ ¡ ¢ ¡ 2 2 2 = 2ρ2 + 1 (dt) − ρ2 (dt) = ρ2 + 1 (dt)

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA 2

(dt) = 0 h i2 i h 0 0 → E dZ (t) dZ (t) − E 2 dZ (t) dZ (t) = 0 h i2 h i 0 0 → E dZ (t) dZ (t) − ρdt = V ar dZ (t) dZ (t) = 0 0

Hence dZ (t) dZ (t) approach ρdt in mean square. Hence 0

dZ × dZ = ρdt

(20.32)

Suppose the stock price S (t) follows a geometric Brownian motion: dS (t) = αdt + σdZ (t) S (t) 1. Apply Ito’s lemma to ln S (t) £¡ and derive that ¢ ¤ ln S (t) is a normal random variable with mean ln S (0) + α − 0.5σ 2 t and variance σ 2 t. 2. Derive that the mean of S (t) is E [S (t)] = S (0) eαt

As explained in my book and my solution manual, you don’t need to mem2 orize Ito’s lemma. Just use the Taylor expansion but keep the (dZ) term. First, use Taylor expansion: ∂ ln S ∂ ln S 1 ∂ 2 ln S 2 d ln S = (dS) dS + dt + ∂S ∂t 2 ∂S 2 ∂ ln S ∂ ln S In the deterministic calculus, you just write d ln S = dS + dt. ∂S ∂t 2 2 However, in the stochastic calculus, we can’t ignore (dZ) since (dZ) = dt. In this problem, d ln S is a linear function of dZ, so: h i 2 2 2 2 (dS) = [Sαdt + σdZ] = S 2 α2 (dt) + 2ασdtdZ + σ 2 (dZ) Using the multiplication rule (DM 20.17 a, ib, c), we get: h ¤ £ 2 2 2 2 2 (dS) = S α (dt) + 2αdtσdZ + σ 2 (dZ) = S 2 α2 × 0 + 2ασ × 0 + σ 2 dt =

S 2 σ 2 dt

So we have to keep the term

1 ∂ 2 ln S 2 (dS) . 2 ∂S 2

Anyway, Taylor expansion gives us: ∂ ln S 1 ∂ 2 ln S 2 2 ∂ ln S S σ dt dS + dt + d ln S = ∂S ∂t 2 ∂S 2 2 1 ∂ ln S 1 ∂ ln S ∂ ln S = =− 2 =0 Next, ∂S S ∂S S ∂t ¶ µ 1 2 1 2 dS 1 2 → d ln S = − σ dt = αdt + σdZ − σ dt = α − σ dt + σdZ S 2 2 2 During a tiny time interval [t, t + dt], the change of ln S is d ln S. d ln S is a normal random variable. This is why. √ First, dZ ∼ N (0, dt) .According to DM 20.4, dZ = Y dt. So dZ is a normal random variable with mean 0 and variance dt.

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229

During the fixed interval [t, t + dt], dt is a constant. Since a constant plus a random variable is also a random variable, µ ¶ µ ¶ 1 1 d ln S = α − σ 2 dt+σdZ is a normal random variable with mean α − σ 2 dt 2 2 and variance σ2 dt Now consider the time interval [0, t]. µ ¶ Rt Rt Rt 1 2 ln S (t) = ln S (0) + 0 d ln S = ln S (0) + 0 α − σ ds + 0 σdZ 2 µ ¶ ¡ ¢ Rt 1 2 Rt = ln S (0) + α − σ ds + σ 0 dZ = ln S (0) + α − 0.5σ 2 t + σZ 0 2 Z is a standard ¢ normal random variable with mean 0 and variance 1, that ¡ is, Z ∼ N 0, σ 2 t . Hence σZ is a normal random variable ¡with mean¢ 0 and variance σ 2 t. And ln S (t) is normal with mean ln S (0) + α − 0.5σ 2 t and variance σ 2 t: £ ¡ ¤ ¢ ln S (t) ∼ N ln S (0) + α − 0.5σ 2 t, σ 2 t ¡ ¢ Next, from ln S (t) = ln S (0) + α − 0.5σ 2 t + σZ, we get: 2 eln S(t) = eln S(0)+(α−0.5σ )t+σZ h i 2 2 → S (t) = S (0) e(α−0.5σ )t+σZ = S (0) e(α−0.5σ )t eσZ h i ¡ ¢ 2 → E [S (t)] = S (0) e(α−0.5σ )t E eσZ

¢ ¡ We can use DM Equation 18.13 to calculate E eσZ . DM Equation ¡ 18.13 ¢ says that if x is normal with mean m and variance v 2 , that is x ∼ N m, v 2 , then E (ex ) = em+0.5v

2

(DM 18.13)

¡ ¢ 2 → E eσZ = e0+0.5σ t h i 2 2 → E [S (t)] = S (0) e(α−0.5σ )t e0+0.5σ = S (0) eαt £ ¡ ¤ ¢ Alternative method to calculate E [S (t)]. Since ln S (t) ∼ N S (0) + α − 0.5σ 2 t, σ 2 t , using DM 18.13, we have: 2 2 E [S (t)] = eS(0)+(α−0.5σ )t+0.5σ t = S (0) eαt

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I want you to memorize the following results (these results are used over and over in Exam MFE): dZ ∼ N (0, dt)

(20.33)

Z ∼ N (0, t)

(20.34)

¡ ¢ x ∼ N m, v 2

→ E (ex ) = em+0.5v

S (t) is geometric Brownian motion

(20.35)

(20.36)

(20.37)

¤ ¢ £ ¡ → ln S (t) ∼ N ln S (0) + α − 0.5σ 2 t, σ 2 t (20.38)

dS (t) = αdt + σdZ (t) S (t)

2 → S (t) = S (0) e(α−0.5σ )t+σZ

dS (t) = αdt + σdZ (t) S (t)

20.7

dS (t) = αdt + σdZ (t) S (t)

¢ ¡ → d ln S (t) = α − 0.5σ 2 dt + σdZ

dS (t) = αdt + σdZ (t) S (t) dS (t) = αdt+σdZ (t) S (t)

→

2

→ E [S (t)] = S (0) eαt

(20.39)

(20.40)

Sharpe ratio

An asset’s Sharpe ratio is equal to the asset’s risk premium α − r divided by the asset’s volatility σ: α−r (20.41) σ If two non-dividend paying assets are perfectly corrected (i.e. they are driven by the same Brownian motion Z (t)), then their Sharpe ratios are equal. Let’s derive this. The price processes of Asset 1 and Asset 2 are: SR =

dS1 = α1 S1 dt + σ1 S1 dZ

(20.42)

dS2 = α2 S2 dt + σ2 S2 dZ

(20.43)

We can form a riskless portfolio by removing the random Brownian motion dZ. Rewrite Equation 20.42 and 20.43 as:

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20.7. SHARPE RATIO

231 µ

µ

1 σ 1 S1 1 σ 2 S2

¶ ¶

dS1 =

α1 dt + dZ σ1

(20.44)

dS2 =

α2 dt + dZ σ2

(20.45)

Suppose at time zero we buy N1 =

1 units of Asset 1 and short sell σ 1 S1

1 units of Asset 2. If we hold one unit of Asset 1 at time zero, then σ 2 S2 after a tiny interval dt, the value of Asset 1 increases by the amount dS1 . If 1 units of Asset 1 at time zero, then after dt the value of we hold N1 = σ 1 S1 µ ¶ 1 1 α1 N1 = units of Asset 1 will increase by dt + dZ. Notice dS1 = σ 1 S1 σ 1 S1 σ1 that the increase of the value of Asset 1 has a random component dZ, where dZ is a normal random variable with mean 0 and variance dt. 1 Similarly, if we short sell N2 = units of Asset 2 at time zero, after dt, σµ 2 S2 ¶ 1 α2 the value of Asset 2 will increase by dt + dZ. The increase of dS2 = σ 2 S2 σ2 the value of Asset 2 has a random component dZ, where dZ is a normal random variable with mean 0 and variance dt. Suppose at time zero we simultaneously buy N1 units of Asset 1 and short sell N2 units of Asset 2. Then at dt, we close our position by selling N1 units of Asset 1 in the open market and buying N2 units of Asset 2 from the open market. The cash flow at time zero: µ ¶ 1 1 dollar to buy N1 units of Asset 1. S1 = • We pay N1 S1 = σ 1 S1 σ1 N2 =

• We receive N2 S2 =

1 dollars for short selling N2 units of Asset 2 σ2

1 1 1 1 − dollars. If − < 0, then we receive • The net cost is σ2 σ1 σ2 ¶ σ1 µ 1 1 − net cash. To avoid tying up our capital, we go to a bank − σ1 σ2 µ ¶ 1 1 1 1 1 1 and borrow − dollars. If − < 0, then we lend − − . σ1 σ2 σ1 σ2 σ1 σ2 • Our net cash outgo is zero. Our payoff at time dt • At time dt, we sell off N1 units of Asset 1 in the open market for the price 1 α1 of S1 + dS1 , receiving N1 (S1 + dS1 ) = N1 S1 + N1 dS1 = + dt + dZ σ1 σ1

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• At time dt we buy N2 units of Asset 2 in the open market at the price 1 α2 + dt + dZ S2 + dS2 . We pay N2 (S2 + dS2 ) = N2 S2 + N2 dS2 = σ2 σ2 • At time dt, we pay back the bank both the principal µ and the¶accrued 1 1 interest. The sum of the principal and the interest is − erdt σ1 σ2 Our profit at dt is: P rof it =

µ

¶ µ ¶ µ ¶ 1 α1 1 α2 1 1 + dt + dZ − + dt + dZ − − erdt (20.46) σ1 σ1 σ2 σ2 σ1 σ2

µ ¶ 1 1 1 1 − < 0, we’ll lend − − at t = 0. Then at dt, we’ll receive If σ1 σ2 µ σ 1 σ¶2 1 1 − − erdt from the borrower. σ1 σ2 The Taylor expansion: r2 r3 2 3 erdt = 1 + rdt + (dt) + (dt) + ... = 1 + rdt 2! 3! Equation 20.46 ¶ can µ be rewritten ¶ as: µ ¶ µ α1 1 α2 1 1 1 + dt − + dt − − erdt σµ1 σ 1 ¶ µσ 2 σ2 ¶ σ σ 1 2 µ ¶ 1 1 1 α1 α2 1 = − − − + dt − erdt σ2 ¶ µ σ1 σ2 ¶ µ σ1 µ σ1 σ2 ¶ µ ¶ 1 1 1 1 α1 α2 1 1 = − − − − + dt − − rdt σµ1 σ2 ¶ σ1 σ2 σ1 σ2 µ σ1 σ2 ¶ α1 α2 1 1 = − − dt − rdt σ2 µ σ1 ¶σ 1 σ 2 α1 − r α2 − r = − dt σ1 σ2 = (SR1 − SR2 ) dt µ ¶ α1 − r α2 − r P rof it = dt = (SR1 − SR2 ) dt − σ1 σ2

(20.47)

Equation 20.46 and 20.47 don’t have the random term dZ (t), indicating that the profit is surely made. Our cash outgo is zero at t = 0, but we’ll have the profit indicated by 20.47 at dt. µ ¶ α1 − r α2 − r Wealth 0 −→ − dt = (SR1 − SR2 ) dt σ1 σ2 Time 0 −→ dt To avoid arbitrage, the profit needs to be zero: α1 − r α2 − r → = SR1 = SR2 σ1 σ2

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20.8. RISK NEUTRAL PROCESS

20.8

233

Risk neutral process

This section is difficult because the author tired to put too many complex concepts in this small section. It seems that the author was in a big hurry to finish this chapter. The author mentioned many concepts (such as martingale, Girsanov’s theorem) but he didn’t really explain them, leaving us hanging in the air asking why. The only thing worth studying is how to transform a standard geometric Brownian motion into a risk neutral process. The standard geometric Brownian motion is: dS (t) = (α − δ) dt + σdZ (t) S (t) This is how to transform: dS (t) = (α − δ) dt + σdZ (t) S (t) = (r − δ) dt + σdZ ∙ (t) + (α − δ) dt¸ α−δ dt = (r − δ) dt + σ dZ (t) + σ ∼ α−δ dt Define dZ (t) = dZ (t) + σ ∼ dS (t) → = (α − δ) dt + σdZ (t) = (r − δ) dt + σdZ (t) S (t) Just learn this transformation and move on.

20.9

Valuing a claim on S a

In a call or put, the payoff at T is a linear function of the stock price S (T ). What if the payoff is linear? For example, what if the payoff at T is S (T ) raised to some power a?

20.9.1

Process followed by S a

Though the textbook uses S a , I’m going to use S A . The reason is explained later. Suppose the stock price follows geometric Brownian motion: dS = (α − δ) dt + σdZ S This is why I use S A . If you use S a , the letter a in S a is very similar to dS the α (alpha) in the equation = (α − δ) dt + σdZ. This can easily lead to S confusion and mistake. As a matter of fact, when I was deriving the formula for E (S a ), I couldn’t match the textbook’s formula. It took me a while to figure out that I accidentally switched a and α. We need to determine the process followed by payoff at T is S A .

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∂S A 1 1 ∂2S A 2 2 (dS) = AS A−1 dS + A (A − 1) S A−2 (dS) dS + ∂S 2 ∂S 2 2 ¡ ¢ 2 2 However, (dS) = [(α − δ) dt + σdZ] S 2 = σ 2 dt S 2 = σ 2 S 2 dt

dS A =

1 → dS A = AS A−1 dS + A (A − 1) S A−2 σ 2 S 2 dt 2 1 = AS A−1 dS + A (A − 1) S A σ 2 dt 2 →

dS A dS 1 =A + A (A − 1) σ 2 dt SA S 2 2 =A £ (α − δ) dt + AσdZ + 0.5A2(A ¤ − 1) σ dt = A (α − δ) + 0.5A (A − 1) σ dt + AσdZ

We see that S A follows a geometric Brownian motion with drift A (α − δ) + 0.5A (A − 1) σ 2 and risk component AσdZ.

20.9.2

£ ¤ Formula for S A (t) and E S A (t)

Next, we apply Equation 20.37. Replace α with A (α − δ) + 0.5A (A − 1) σ 2 and σ with Aσ. £ ¤ d ln S A (t) =£ A (α − δ) + 0.5A (A ¤− 1) σ 2 − 0.5A2 σ2 dt + AσdZ = A (α − δ) dt − 0.5Aσ2 dt + AσdZ Using Equation £ 20.38, we £get: ¤ ¤ ln S A (t) ∼ N ln S A (0) + A (α − δ) − 0.5Aσ 2 t, A2 σ 2 t Using Equation 20.39: 2 S A (t) = S A (0) e[A(α−δ)−0.5Aσ ]t+AσZ

Using Equation 20.40 just replacing α with A (α − δ) + 0.5A (A − 1) σ 2 £ ¤ 2 E S A (t) = S A (0) e[A(α−δ)+0.5A(A−1)σ ]t £ ¤ 2 We can also derive E S A (t) using S A (t) = S A (0) e[A(α−δ)−0.5Aσ ]t+AσZ ³ ´ £ ¤ 2 E S A (t) = E S A (0) e[A(α−δ)−0.5Aσ ]t+AσZ ´ ³ 2 = S A (0) E e[A(α−δ)−0.5Aσ ]t+AσZ ³ ´ 2 = S A (0) E e[A(α−δ)−0.5Aσ ]t eAσZ ¡ ¢ 2 = S A (0) e[A(α−δ)−0.5Aσ ]t E eAσZ

AσZ is a normal random variable with mean 0 and variance V ar [AσZ (t)] = A2 σ 2 V ar [Z (t)] = A2 σ2 t. Using Equation 20.35, we get: ¡ ¢ 2 2 E eAσZ = e0.5A σ t £ A ¤ 2 2 2 2 → E S (t) = S A (0) e[A(α−δ)−0.5Aσ ]t e0.5A σ t = S A (0) e[A(α−δ)+0.5A(A−1)σ ]t

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20.9. VALUING A CLAIM ON S A

235

£ ¤ Alternative method to calculate E S A (t) . Using Equation 20.39, we get: 2 dS = (α − δ) dt + σdZ → S (t) = S (0) e(α−δ−0.5σ )t+σZ S 2 → S A (t) = S A (0)³eA(α−δ−0.5σ )t+AσZ ´ £ A ¤ 2 → E S (t) = E S A (0) eA(α−δ−0.5σ )t+AσZ ¡ ¢ 2 = S A (0) eA(α−δ−0.5σ )t E eAσZ 2 2 2 = S A (0) e[A(α−δ)−0.5Aσ ]t e0.5A σ t 2 = S A (0) e[A(α−δ)+0.5A(A−1)σ ]t

20.9.3

Expected return of a claim on S A (t)

Consider two geometric Brownian motions S (t) and S A (t). dS = (α − δ) dt + σdZ S where α is the expected return on a claim on S (t) and δ is the continuous dividend yield earned by S (t) ¤ dS A £ = A (α − δ) + 0.5A (A − 1) σ 2 dt + AσdZ A S = (γ − δ ∗ ) dt + AσdZ where γ is the expected return on a claim on S A (t) and δ ∗ is the continuous dividend yield earned by S A (t). The textbook calls δ ∗ the lease rate. These two processes have the same risk dZ. Consequently, they have the same Sharpe ratio: α−r γ−r = → γ = r + A (α − r) σ Aσ where r is the continuously compounded risk-free interest rate.

20.9.4

Specific examples

The textbook keeps mentioning Jensen’s inequality. So let’s first talk about Jensen’s inequality. Jensen’s inequality is in the appendix C of the textbook. You can also find information at http://en.wikipedia.org/wiki/Convex_ function Jensen’s inequality says 1. if f (x) is convex, then for any probability distribution, we have E [f (x)] ≥ f [E (x)] 2. if f (x) is concave, then for any probability distribution, we have E [f (x)] ≤ f [E (x)]

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What’s a convex function? What’s a concave function?

If at any point you draw a tangent line, the function f (x) stays above the tangent line, then f (x) is a convex function.

If at any point you draw a tangent line, the function f (x) stays below the tangent line, then f (x) is a concave function.

A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative there.

A twice differentiable function of one variable is concave on an interval if and only if its second derivative is negative there.

Here’s a simple explanation of Jensen’s inequality.

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d2 y = 2 > 0, y = x2 is a convex function. Suppose dx2 weµtake two points A (1, 1) and ¶ B (4, 16). The mid point of the line AB is 1+4 1 + 16 C = 2.5, = 8.5 . The fact that y = x2 is a convex function means 2 2 that y = x2 curves up. Then it follows that the point C must be above the point ¡ ¢ 12 + 42 D 2.5, 2.52 = 6. 25 . From the graph below, we clearly sees that = 8.5 2 ¶2 µ 1+4 = 6.25 (which (which is the height of the point C) is greater than 2 is the height of the point D). This is an example where the mean of a convex function is greater than the function of the average. Consider y = x2 . Sine

y

25

20

B 15

10

C D

5

A 0 0

1

12 + 42 > 2

µ

2

1+4 2

¶2

3

4

, an example of E [f (x)] ≥ f [E (x)].

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5

x

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√ Next, let’s consider a concave function two points!A (1, 1) Ãy = x. Consider √ √ 1+4 1+ 4 and B (4, 2). The mid point of AB is C = 2.5, = 1.5 . Since 2 2 µ ¶ √ √ 1+4 y = x curves down, then it follows that C must be lower than D = 2.5, 2.5 = 1. 58 . 2 √ √ 1+ 4 From the graph below, we clearly sees that = 1.5 (which is the height 2 r 1+4 = 1. 58 (which is the height of the point of the point C) is less than 2 D). This is an example where the mean of a concave function is less than the function of the average.

y

B

2.0

D C

1.5

A

1.0

0.5

0.0 0

1

2

3

4

r √ √ 1+ 4 1+4 < , an example of E [f (x)] ≤ f [E (x)] 2 2

5

x

By now you should have intuitive feel of Jensen’s inequality. Let’s move on.

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The textbook considers the following examples: A = −1, 0, 1. If A = 1, the time 0 value of the claim S (T ) at T is: P [S (T )] = e−rT S (0) e[(r−δ)]T = S (0) e−δT V (0) = F0,T This is just DM Equation 5.4. If A = 0, then the claim is just one dollar: S 0 (T ) = 1 P [1] = e−rT → V (0) = F0,T So the time 0 value of getting $1 at T is e−rT . V (0) is equal $1 discounted back to time 0 at the risk-free rate. If A = 2, the time 0 value of the claim S 2 (T ) at T is: ¤ £ 2 2 P S (T ) = e−rT S 2 (0) e[2(r−δ)+σ ]T V (0) = F0,T As a general rule, the forward price of any asset at T is just the prepaid forward price accumulating at the risk-rate from time 0 to T : P erT F0,T = F0,T We use the risk-free rate r in the above equation. To get an asset at T , we P as a buyer can either pay the seller F0,T at time 0 or pay the seller F0,T at time T . To avoid the arbitrage, the two payments should differ only in timing. So P F0,T = F0,T erT . ¤ ¤ £ 2 £ 2 2 P S (T ) erT = S 2 (0) e[2(r−δ)+σ ]T → F0,T S (T ) = F0,T P F0,T [S (T )] = F0,T [S (T )] erT = S (0) e−δT erT = S (0) e(r−δ)T ¤ £ 2 2 2 → F0,T S (T ) = S 2 (0) e[2(r−δ)+σ ]T = (F0,T [S (T )])2 eσ T ¤ £ 2 Since eσ T ≥ 1, we have F0,T S 2 (T ) ≥ (F0,T [S (T )])2 . This agrees with Jensen’s inequality. Roughly speaking 1 , F0,T = E (ST ). d2 2 S = 2 > 0. Hence S 2 is convex. dS £ ¤ 2 According to Jensen’s inequality, we have E S 2 (T ) ≥ (E [S (T )]) . This ¤ £ 2 ¤ £ 2 2 2 leads to F0,T S (T ) = E S (T ) ≥ (F0,T [S (T )]) = (E [S (T )]) . S 2 is twice differentiable and

1 at T is: If A = −1, the time 0 value of the claim S (T ) ∙ ¸ 1 1 [−(r−δ)+σ2 ]T 1 2 P = e−rT eσ T V (0) = F0,T = e−rT e (r−δ)T S (T ) S (0) S (0) e ∙ ∙ ¸ ¸ 1 1 1 2 P → F0,T eσ T = F0,T erT = S (T ) S (T ) S (0) e(r−δ)T (r−δ)T However, ∙ F0,T [S ¸ (T )] = S (0) e 1 1 1 2 → F0,T = eσ T ≥ S (T ) F0,T [S (T )] F0,T [S (T )]

1 Experts don’t agree whether F 0,T = E (ST ). Some say the forward price is the unbiased estimate of the expected future spot price, that is, F0,T = E (ST ). Others disagree. However, it’s safe to see that F0,T is very close to E (ST ).

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1 Since is concave, according to Jensen’s inequality, we have: S (T ) ∙ ∙ ¸ ∙ ¸ µ ¸¶2 µ ∙ ¸¶2 1 1 1 1 = E . F0,T =E ≥ F0,T S (T ) S (T ) S (T ) S (T ) d2 −1 2 S = 3 > 0. Hence S 2 is convex. 2 dS S∙ ¸ µ ∙ ¸¶2 1 1 According to Jensen’s inequality, we have E . This ≥ E S (T ) S (T ) ∙ ∙ ¸ ∙ ¸ µ ¸¶2 µ ∙ ¸¶2 1 1 1 1 leads to F0,T = E . =E ≥ F0,T S (T ) S (T ) S (T ) pS (T ) If A = 0.5, the the claim S (T ) at T is: hptime 0ivalue of p 2 P −rT S (T ) = e S (0)e[0.5(r−δ)+0.5×0.5(0.5−1)σ ]T V (0) = F0,T p 2 = he−rT S i(0)e[0.5(r−δ)−0.125 σ ]T p p p 2 → F0,T S (T ) = S (0)e[0.5(r−δ)−0.125 σ ]T = S (0)e0.5(r−δ)T p p 2 σ2 T = S (0) e(r−δ)T e−0.125 = F0,T [S (T )]e−0.125 σ T i h p p 2 Since e−0.125 σ T ≤ 1, we see that F0,T S (T ) ≤ F0,T [S (T )] This agrees with Jensen’s inequality. p p d2 √ S (T ) is twice differentiable. S = −4S −1.5 < 0. Hence S (T ) is dS 2 concave. hp i hp i p hp i S (T ) = E S (T ) ≤ F0,T [S (T )] = E (S (T )) → F0,T S −1 is twice differentiable and

Example 20.9.1. The price of a stock follows a geometric Brownian motion: dS (t) = (0.1 − 0.04) dt + 0.3dZ S (t) The current price of the stock is 10. The continuously compounded dividend yield is 0.04 per year. The continuously compounded risk-free rate is 0.06 per year. The seller and the buyer enter a forward contract. The contract requires the seller to pay the buyer S 2 (5) five years from now. Calculate • the prepaid forward price • the forward price • γ • δ∗

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• The probability that S 2 (5) ≤ 160. Solution. dS (t) = (0.1 − 0.04) dt + 0.3dZ S (t) ¤ £ A 2 P S (T ) = e−rT S A (0) e[A(r−δ)+0.5A(A−1)σ ]T F0,T £ 2 ¤ 2 P F0,5 S (5) = e−0.06×5 102 e(2(0.06−0.04)+0.5×2(2−1)×0.3 )5 = 141. 906 8 ¤ £ 2 ¤ £ P F0,5 S (5) = F0,5 S 2 (5) e0.06×5 = 141. 906 8e0.06×5 = 191. 554 1 γ = r + A£ (α − r) = 0.06 + 2 (0.1 − 0.06) ¤ = 0.14 δ ∗ = γ − A ¡(α − δ) + 0.5A (A − 1) σ 2 ¢ = 0.14 − 2 (0.1 − 0.04) + 0.5 × 2 (2 − 1) × 0.32 = −0.07 It’s OK for δ ∗£to be negative. £ ¤ ¤ ln S A (t) ∼ N ln S A (0) + A (α − δ) − 0.5Aσ 2 t, A2 σ 2

£ £ ¤ ¤ 2 2 2 2 2 ln ¡ S (5) ∼ N ln 10 + 2 (0.1 2−¢ 0.04) − 0.5 × 2 × 0.3 5, 2 × 0.3 × 5 2 (0.1 −¡0.04) − 0.5 × 2 × 0.3 5 = 0.15 ¢ ln 102 + 2 (0.1 − 0.04) − 0.5 × 2 × 0.32 5 = ln 100 + 0.15 = 4. 755 17

¤ £ ¤ £ P S 2 (5) ≤ 160 = P ln S 2 (5) ≤ ln 160 = Φ (z) ln 160 − 4. 755 17 = 0.238 5 z= √ 22 × 0.32 × 5 £ ¤ Φ (z) = 0.594 3 → P S 2 (5) ≤ 160 = 0.594 3 Example 20.9.2.

The price of a stock follows a geometric Brownian motion: dS (t) = (0.12 − 0.05) dt + 0.25dZ S (t) The current price of the stock is 20. The continuously compounded dividend yield is 0.05 per year. The continuously compounded risk-free rate is 0.07 per year. The seller and the buyer enter a forward contract. The contract requires the 1 four years from now. seller to pay the buyer S (4) Calculate • the prepaid forward price • the forward price • γ

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• δ∗ • The probability that

1 ≤ 0.03. S (4)

Solution. dS (t) = (0.12 − 0.05) dt + 0.25dZ S (t) ¤ £ 2 P F0,T S A (T ) = e−rT S A (0) e[A(r−δ)+0.5A(A−1)σ ]T ∙ ¸ 2 1 1 P F0,4 = e−0.07×4 e(−1(0.07−0.05)+0.5×(−1)(−1−1)×0.25 )4 = 0.04479 ∙ S (4) ¸ ∙ 20 ¸ 1 1 P F0,4 = F0,4 e0.07×4 = 0.04479e0.07×4 = 0.05 926 S (4) S (4) γ = r + A£ (α − r) = 0.07 − 1 (0.12 − 0.07) = 0.02 ¤ δ ∗ = γ − A ¡(α − δ) + 0.5A (A − 1) σ 2 ¢ = 0.02 − −1 (0.12 − 0.05) + 0.5 × (−1) (−1 − 1) × 0.252 = 0.027 5

£ £ ¤ ¤ ln S A (t) ∼ N ln S A (0) + A (α − δ) − 0.5Aσ 2 t, A2 σ 2 ¸ ∙ ¤ £ 1 1 2 ln ∼ N ln + −1 (0.12 − 0.05) − 0.5 × (−1) × 0.252 4, (−1) × 0.252 × 4 20 ¡ S (4) ¢ −1 (0.12 − 0.05) − 0.5 × (−1) × 0.252 4 = −0.155 ¤ 1 1 £ ln + −1 (0.12 − 0.05) − 0.5 × (−1) × 0.252 4 = ln −0.155 = −3. 150 7 20 20 ¸ ∙ ¸ ∙ 1 1 ≤ 0.03 = P ln ≤ ln 0.03 = Φ (z) P S (4) S (4) ln 0.03 − (−3. 150 7) = −0.711 7 z=q 2 (−1) × 0.252 × 4 ∙ ¸ 1 Φ (z) = 0.238 3 →P ≤ 0.03 = 0.238 3 S (4)

Example 20.9.3. The price of a stock follows a geometric Brownian motion: dS (t) = (0.15 − 0.04) dt + 0.35dZ S (t) The current price of the stock is 10. The continuously compounded dividend yield is 0.04 per year. The continuously compounded risk-free rate is 0.08 per year. The seller and the buyer p enter a forward contract. The contract requires the seller to pay the buyer S (6) six years from now. Calculate

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• the prepaid forward price • the forward price • γ • δ∗ • The probability that

p S (6) ≤ 4.

Solution. dS (t) = (0.15 − 0.04) dt + 0.35dZ S (t) ¤ £ 2 P F0,T S A (T ) = e−rT S A (0) e[A(r−δ)+0.5A(A−1)σ ]T i hp √ 2 P F0,6 S (6) = e−0.08×6 10e(0.5(0.08−0.04)+0.5×0.5(0.5−1)×0.35 )6 = 2. 012 6 i i hp hp P F0,6 S (6) = F0,6 S (6) e0.08×6 = 2. 012 6e0.08×6 = 3. 252 5 γ = r + A£ (α − r) = 0.08 + 0.5 (0.15 −¤0.08) = 0.115 δ ∗ = γ − A (α¡ − δ) + 0.5A (A − 1) σ 2 ¢ = 0.115 − 0.5 (0.15 − 0.04) + 0.5 × 0.5 (0.5 − 1) × 0.352 = 0.075 3 £ £ ¤ ¤ ln S A (t) ∼ N ln S A (0) + A (α − δ) − 0.5Aσ 2 t, A2 σ 2

p £ √ £ ¤ ¤ ln S (6) ∼ N ln 10 + 0.5 (0.15 −¢0.04) − 0.5 × 0.5 × 0.352 5, 0.52 × 0.352 × 6 ¡ 0.5 (0.15 − 0.04) − 0.5 × 0.5 × 0.352 6 = 0.146 25 √ √ £ ¤ ln 10 + 0.5 (0.15 − 0.04) − 0.5 × 0.5 × 0.352 5 = ln 10 + 0.146 25 = 1. 297 5 hp i h p i P S (6) ≤ 4 = P ln S (6) ≤ ln 4 = Φ (z) ln 4 − 1. 297 5 = 0.207 1 z=√ 0.52 × 0.352 × 6 h i p S (6) ≤ 4 = 0.582 0 Φ (z) = 0.582 0 →P

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Chapter 21

Black-Scholes equation 21.1

Differential equations and valuation under certainty

21.1.1

Valuation equation

Let’s consider a risk-free world where all assets just earn the risk free interest rate. Support at time t we spend S (t) to buy one share of a stock. Then at t+h, we receive D (t + h) h amount of the dividend, where D (t + h) represents the dividend accumulated per unit of time during [t, t + h]. At t + h after receiving D (t + h) h dividend, we sell the stock and receive S (t + h). The total amount of money we have at t + h is D (t + h) h + S (t + h). Had we put S (t) amount of money in a savings account, we would have S (t) (1 + rh ) amount of money at t+h, where rh represent the (not-annualized) risk-free interest rate per h period. To avoid arbitrage, we need to have S (t) (1 + rh ) = D (t + h) h + S (t + h). Rearranging this equation, we get: D (t + h) h + S (t + h) 1 + rh

(DM 21.1)

S (t + h) − S (t) + D (t + h) h = rh S (t)

(DM 21.2)

S (t) =

hr i S (t + h) − S (t) dS (t) h = lim = lim S (t) − D (t + h) = rS (t) − D (t) h→0 h→0 h dt h (DM 21.3) 1 In Equation 21.3, r = × rh represents the annualized continuously comh pounded interest rate per year. rh is the interest rate per h period; the total 1 1 number of h lengths in a year is . Hence r = × rh represents the annualized h h continuously compounded interest rate per year. 245

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21.1.2

CHAPTER 21. BLACK-SCHOLES EQUATION

Bonds

For a zero-coupon bond, D (t) = 0. DM Equation 21.3 becomes This gives us S (t) = S (T ) e−r(T −t)

dS (t) = rS (t). dt (DM 21.4)

Since the boundary condition is S (T ) = 1, we have S (t) = e−r(T −t) . This equation says that $1 at T is worth e−r(T −t) at time t.

21.1.3

Dividend paying stock

At time t, you spend S (t) and buy one share of a stock. This gives you two things: • at time T you can sell the stock and get S (T ), which is worth S (T ) e−r(T −t) at t • you accumulate dividend at a continuous rate of D (s), where D (s) is the instant dividend earned per unit of time at time s. The total value of the continuous dividend earned during the interval [s, s + ds] is D (s) ds, which is worth [D (s) ds] e−r(s−t) = D (s) e−r(s−t) ds at time t. The present value at time t of the total continuous dividend earned during the interval RT [t, T ] is t D (s) e−r(s−t) ds The PV of this continuous flow of dividend is 1 − e−r(T −t) DaT −t|r = D r RT To avoid arbitrage, we have S (t) = S (T ) e−r(T −t) + t D (s) e−r(s−t) ds. RT Please note that if D (s) = D is a constant, then t D (s) e−r(s−t) ds = RT 1 − e−r(T −t) D t e−r(s−t) ds = DaT −t|r = D . r RT Anyway, t D (s) e−r(s−t) ds is a continuous annuity. If you have trouble RT understanding t D (s) e−r(s−t) ds, refer to your FM book.

21.2

Black-Scholes equation

21.2.1

How to derive Black-Scholes equation

This section derives DM Equation 21.11. Vt + 0.5σ 2 S 2 VSS + (r − δ) SVS − rV = 0 This is the outline of how to derive this formula. At time t • We buy one option on the stock. We pay V

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(DM 21.11)

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247

• We buy N shares of the stock (a negative N means short selling stocks). We pay N S . • We deposit W means into a savings account. We pay W . To have zero-financing, we set our total initial cost to zero I = V + NS + W = 0

(DM 21.7)

Next, let’s consider the change of I during [t, t + dt]: dI = dV + N (dS + δSdt) + dW

(DM 21.8)

dI is the interest earned on the option during [t, t + dt]. Since W is invested in a savings account, we have dW = rW dt. This says that the interest earned on W during [t, t + dt] is rW dt. Notice that the change of S is dS + δSdt (the sum of the change of the stock price dS and the dividend received δSdt). Apply It’o lemma: dV = Vt dt + VS dS + 0.5σ 2 S 2 VSS dt → dI = Vt dt + VS dS + 0.5σ2 S 2 VSS dt + N (dS + δSdt) + dW = Vt dt + (VS + N ) dS + 0.5σ 2 S 2 VSS dt + N δSdt + dW Set N = −VS . Then dS term becomes zero and W = − (V − VS S). Because our initial cost is zero, the interest we earned dI should be zero. dI = Vt dt + 0.5σ 2 S 2 VSS dt − VS δSdt − r (V − VS S) dt = 0 This leads to DM 21.11. Vt + 0.5σ2 S 2 VSS + (r − δ) SVS − rV = 0

21.2.2

Verifying the formula for a derivative

Simple PV calculation Verification that the price of a zero-coupon bond satisfies the Black-Scholes equation DM 21.11. $1 at time T is worth V (t) = e−r(T −t) at t. =⇒ VS = VSS = 0 Vt = re−r(T −t) = rV 2 2 =⇒ Vt + 0.5σ S VSS + (r − δ) SVS − rV = 0 Verification that the price of a prepaid forward contract satisfies the BlackScholes equation DM 21.11. V (S, t) = S (t) e−δ(T −t) =⇒ VS = e−δ(T −t) VSS = 0 Vt = δS (t) e−δ(T −t) =⇒ Vt +0.5σ 2 S 2 VSS +(r − δ) SVS −rV = δSe−δ(T −t) +(r − δ) Se−δ(T −t) − rSe−δ(T −t) = 0

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Call option The textbook explains that the price of a European call option satisfies (1) the boundary condition and, (2) DM 21.11. V = S (t) e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ) Verification that the call price formula meets the boundary condition. The boundary condition is that at the call expiration date T the call is worth S (T ) − K (T ) if S (T ) > K (T ) and zero otherwise.

¶ µ 1 S (t) + r − δ + σ 2 (T − t) K 2 √ d1 = σ T −t √ d2 = d1 − σ T − t ¶ µ S (T ) 1 2 ln + r − δ + σ (T − t) K 2 √ , d2 = d1 , If t approaches T , then d1 = σ T −t and µ ¶ 1 2 √ S (T ) r−δ+ σ T −t ln S (T ) S (T ) 2 • If S (T ) > K, then > 1, ln > 0, d1 = √ K + = K K σ σ T −t +∞ , N (d1 ) = N (d2 ) = 1, V = S − T. ln

S (T ) • If S (T ) < K, then ln < 0, d1 = −∞ , N (d1 ) = N (d2 ) = 0 and K V =0 By the way, we don’t need to worry about S (T ) = K because the probability of S (T ) = K is zero. The probability that a continuous random variable takes on a fixed value is zero. When we talk about the probability regarding a continuous random variable X, we talk about the probability that X falls in a range [a, b], not the probability that X takes on a single value. If the probability that X takes a single value is not zero, then the total probability that a < X < b will be infinite because there are infinite number of single values in the range [a, b]. We have proved that the call price satisfies the boundary condition. For the verification that the call price satisfies DM 21.11, see my solution to DM Problem 21.5, 21.6, and 21.7. We can also verify that the European put price satisfies DM 21.11. The put price is V = Ke−r(T −t) N (−d2 ) − S (t) e−δ(T −t) N (−d1 ) Using the formula N (−x) = 1 − N (x), we can rewrite the put price as V = Ke−r(T −t) [1 − N (d2 )] − S (t) e−δ(T −t) [1 − N (d1 )] = Ke−r(T −t) − S (t) e−δ(T −t) + S (t) e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 )

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249

You can also derive the above equation using the put-call parity. Notice that S (t) e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ) is the call price. Each of the three terms, Ke−r(T −t) , S (t) e−δ(T −t) , and S (t) e−δ(T −t) N (d1 )− −r(T −t) Ke N (d2 ), satisfies the BS PDE. Hence the put price satisfies the BS PDE. I won’t prove that the put price satisfies the boundary condition V (t = T ) = K − S (T ) if K > S (T ) and zero otherwise. You can easily prove this yourself. Key formula to remember: If t → T , then d1 = d2 and • If S (T ) > K, N (d1 ) = N (d2 ) = 1 • If S (T ) < K, N (d1 ) = N (d2 ) = 0 All or nothing option The textbook points out that S (t) e−δ(T −t) N (d1 ) and e−r(T −t) N (d2 ) each satisfy the BS PDE (see my solution to DM Problem 21.5, 21.6). So S (t) e−δ(T −t) N (d1 ) can be a price of a derivative; e−r(T −t) N (d2 ) can be a price of another derivative. What derivatives are priced as S (t) e−δ(T −t) N (d1 ) and e−r(T −t) N (d2 ) respectively? Let’s check the boundary condition. As t → T , • S (t) e−δ(T −t) N (d1 ) → S (T ) if S (T ) > K • S (t) e−δ(T −t) N (d1 ) → 0 if S (T ) < K S (t) e−δ(T −t) N (d1 ) must be the price of an option that pays S (T ) if S (T ) > K and zero otherwise. Such an option is an asset-or-nothing option. Similarly, as t → T • e−r(T −t) N (d2 ) → 1 if S (T ) > K • e−r(T −t) N (d2 ) → 0 if S (T ) < K e−r(T −t) N (d2 ) must be the price of an option that pays 1 if S (T ) > K and zero otherwise. Such an option is an cash-or-nothing option. We can break down a European call option into one asset-or-nothing option and several cash-or-nothing options. Buying a European call option is equivalent to buying one asset-or-nothing option and selling K units of cash-or-nothing option (you can verify that they have the same payoff). Hence the price of the European call option is S (t) e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ). Similarly, we can break down a gap option (gap option is explained in Chapter 14). In a gap call, the payoff is S (T ) − K1 if S (T ) > K2 . This is equivalent

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to buying one asset-or-nothing call and selling K1 units of cash-or-nothing option. So the price of this gap call is S¢ (t) e−δ(T −t) N (d1 ) − K1 e−r(T −t) N (d2 ), ¡ −δ(T −t) √ /K2 e−r(T −t) + 0.5σ 2 (T − t) ln St e √ and d2 = d1 −σ T − t. where d1 = σ T −t • P ∗ (ST > K) = N (d2 ). This is the risk-neutral probability that the call will be exercised (i.e. call will finish in the money) • P ∗ (ST < K) = 1 − N (d2 ) = N (−d2 ). This is the risk-neutral probability that the call will NOT be exercised. • Ke−r(T −t) N (d2 ) is the strike price multiplied by the risk-neutral probability that the strike price will ever be paid. This is the expected value of the strike price *if* the call will be exercised • S (t) e−δ(T −t) N (d1 ) is the expected value of the stock price *if* the call will be exercised (i.e. if ST > K) • N (d1 ) is the expected fractional share of the stock *if* the call will be exercised • If an option pays one stock when ST > K and zero otherwise, then this option is worth S (t) e−δ(T −t) N (d1 ) • If an option pays $1 when ST > K and zero otherwise, then this option is worth PV of $1 (which is e−r(T −t) ) multiplied by P ∗ (ST > K)

21.2.3

Black-Scholes equation and equilibrium returns

This section derives the BS PDE using the idea that the actual expected return on an option should be equal to the equilibrium expected return. The expected continuously compounded return on an option is αoption = E (dV ) . Here dV is the increase of the option value (i.e. the interest earned V dt on the option) during [t, t + dt]. At time t, if you spend V dollars and buy an option, then during the tiny interval [t, t + dt], your option value will go up by dV dV . The (not annualized) return earned on the option per dt period is . V 1 Since the number of dt periods in one year is , the annualized (continuously dt dV 1 dV compounded) rate of return per $1 invested in the option is × = . V dt V dt ¶ µ E (dV ) dV = . Here V and dt are The expected return on the option is E V dt V dt treated as constants. dV is a random variable. To calculate E (dV ), we use Ito’s lemma:

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21.2. BLACK-SCHOLES EQUATION

251

£ ¤ dV = 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt dt + SVS σdZ ¤ ¢ ¡£ → E [dV ] = E 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt dt + E (SVS σdZ) £ ¤ = 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt dt + SVS σE (dZ) We know that E (dZ) = 0. Recall dZ (t) is a normal random variable with mean 0 and variance dt (see the footnote of Derivatives Markets Page 652). E (dV ) 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt = V dt V SVS σdZ The (not annualized) unexpected return on the option is . Here we V didn’t divide SVS σdZ by dt because the unexpected return is not annualized. SVS σ Define = σ option . V Hence the (not annualized) unexpected return on the option is σoption dZ. The Ito’s lemma can be written as: dV = αoption dt + σoption dZ V Hence αoption =

By the way, it seems that the textbook has a typo in DM 21.18: E (dV ) dV SVS σdZ − = V V V

(DM 21.18)

E (dV ) SVS σdZ dV − = V V V Consider two assets, an stock and an option on the stock. dS The process of the stock price: = αdt + σdZ (DM 20.1) S dV The process of the option price: = αoption dt + σ option dZ V These two processes are driven by the same dZ. According to Chapter 20, the stock and the option on the stock must have the same Sharpe ratio. Hence α−r αoption − r = σ σ option The correct formula should be:

=⇒

=⇒

=⇒ =⇒

V α−r αoption − r α−r = (αoption − r) = SV σ σ SVS S V µ ¶ V 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt α−r = −r SVS V (α − r) SVS = 0.5σ 2 S 2 VSS + (α − δ) SVS + Vt − rV Vt + 0.5σ 2 S 2 VSS + (r − δ) SVS − rV = 0 (BS PDE)

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SVS By the way, according to DM 12.8, the option elasticity is Ω = . Hence V SVS σ σ option = = σΩ (DM 21.20). DM 21.20 is slightly different from DM V 12.9: (DM 12.9) σ option = σ|Ω| σ option = σΩ (DM 21.20) Obviously, the author of Derivatives Markets changed his definition of σoption . In Chapter 12, σ option is non negative; in Chapter 21, σ option can be positive, zero, or negative.

21.3

Risk-neutral pricing

This section repeats the old idea of risk neutral pricing. Under risk neutral pricing, we can set the expected return on the stock α to r and get the correct price of a derivative. This section contains many formulas. Make sure you understand the meaning of each formula. The textbook lists the following equations: ˜ dS = (r − δ) dt + σdZ S

d ∗ E (dV ) = Vt + 0.5σ2 S 2 VSS + (r − δ) SVS dt d ∗ E (dV ) = rV dt

(DM 21.28) (DM 21.30) (DM 21.31)

These equations are risk-neutral version of similar formulas. Next, the textbook introduced a new symbol f (ST : St ). f (ST : St ) is the conditional probability of the stock’s terminal price ST given that the price today is St . You’ll want to memorize the following formulas: R∞ P (ST > K) = K f (ST : St ) dST (this holds whether f (ST : St ) is riskneutral probability or the true probability) The price of a derivative is Vt = e−r(T −t) E ∗ (VT ) The price of a European call is RK R∞ Vt = e−r(T −t) E ∗ (VT ) = e−r(T −t) 0 0× f ∗ (ST : St ) dST +e−r(T −t) K [S (T ) − K] R ∞ f ∗ (ST : St ) dST = e−r(T −t) K [S (T ) − K] f ∗ (ST : St ) dST Please note that the price of the call at expiration is VT = 0 if S (T ) < K and S (T ) − K if S (T ) > K. Similarly, the price of a European put is RK Vt = e−r(T −t) E ∗ (VT ) = e−r(T −t) 0 [K − S (T )] × f ∗ (ST : St ) dST

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Chapter 22

Exotic options: II This is an easy chapter.

22.1

All-or-nothing options

Cash-or-nothing option. • A cash call pays $1 at T is ST > K and $0 if ST ≤ K • A cash put pays $1 at T is ST < K and $0 if ST > K Asset-or-nothing option. • An asset-or-nothing call pays ST at T is ST > K and $0 if ST ≤ K • An asset-or-nothing put pays ST at T is ST < K and $0 if ST > K Price of cash-or-nothing option. • The price of a cash call option is e−r(T −t) N (d2 ), where N (d2 ) = P ∗ (ST > K) (i.e. the risk neutral probability of ST > K) • The price of a cash put option is e−r(T −t) [1 − N (d2 )] = e−r(T −t) N (−d2 ), where N (−d2 ) = P ∗ (ST < K) (i.e. the risk neutral probability of ST < K) Price of asset-or-nothing option. • The price of an asset call option is St e−δ(T −t) N (d1 ) • The price of an asset put option is St e−δ(T −t) [1 − N (d1 )] = St e−δ(T −t) N (−d1 ) 253

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Price of a European call and put • Buying an ordinary European call option is equivalent to buying one asset-or-nothing call and selling K cash-or-nothing calls. The call price is St e−r(T −t) N (d1 ) − Ke−r(T −t) N (d2 ) • Buying an ordinary European put option is equivalent to buying K cashor-nothing puts and selling one asset-or-nothing put. The put price is Ke−r(T −t) N (−d2 ) − St e−r(T −t) N (−d1 ) Price of a gap call and put • Buying a gap call option is equivalent to buying one asset-or-nothing call with strike price K2 and selling K1 cash-or-nothing calls with strike price K2 . The gap call price is St e−r(T −t) N (d1 ) − K1 e−r(T −t) N (d2 ) . • Buying a gap put option is equivalent to buying K1 asset-or-nothing puts with strike price K2 and selling one asset-or-nothing put with strike price K2 . The gap put price is K1 e−r(T −t) N (−d2 ) − St e−r(T −t) N (−d1 ) . ¢ ¡ ln St e−δ(T −t) /K2 e−r(T −t) + 0.5σ 2 (T − t) √ • For both a gap call and put, d1 = σ T −t √ and d2 = d1 − σ T − t. Delta-hedging all-or-nothing options It’s difficult to hedge an all-or-nothing option because the payoff is not continuous. The textbook explains this well. Refer to the textbook.

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Chapter 23

Volatility The concept of implied volatility is explained in DM section 12.5. The implied volatility of an option is the volatility implied by the market price of the option based on the Black-Scholes formula. If you plug the implied volatility into the Black-Scholes formula, the formula should produce the option price that is equal to the current market price of the option. The Black-Scholes formula assumes that a given stock has a constant volatility. Under the BS formula, the stock’s volatility is a inherent characteristic of the stock; it doesn’t depend on other factors (such as the stick price K and the option’s expiry T ). In reality, however, the implied volatility of a stock depends on K and T . The volatility smile is a long-observed pattern where at-the-money options tend to have lower implied volatilities than in- or out-of-the-money options. • The implied volatility is lowest when the option is at the money • The implied volatility gets higher and higher as the option gets more and more in-the-money • The implied volatility gets higher and higher as the option gets more and more out-of-the money. If we plot the implied volatility in a 2-D plane (setting the implied volatility σ as Y and the strike price K as X), the diagram looks like a letter U (the letter U looks like a smile). To see a diagram of the volatility smile, refer to • http://www.optiontradingpedia.com/volatility_smile.htm • http://en.wikipedia.org/wiki/Volatility_smile Volatility surface is a 3-D diagram of the implied volatility σ as a function of the strike price K and time to maturity T . 255

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To account for the factor that the implied volatility depends on K and T , dS (t) dS (t) we can rewrite DM 20.25 = (α − δ) dt + σdZ (t) as DM 23.1 = S (t) S (t) (α − δ) dt+σ (St , Xt, t) dZ (t). In DM 23.1, the instant volatility σ (St , Xt, t) dZ (t) is a function of the stock price St , another factor Xt, , and the ∙ timent. ¸ 1 1 P 2 ˆ2 Historical volatility. Please note that DM 23.2 σ = ε is not h n − 1 i=1 i new. This formula is already used in DM Chapter 11 "Estimating Volatility." In n 1 P ε2 is the estimated variance per h period. Since the number DM 23.2, n − 1 i=1 i ¸ ∙ n 1 1 1 P ε2i . of h periods in one year is , the variance per year is h h n − 1 i=1 This is all you need to know about Chapter 23.

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Chapter 24

Interest rate models 24.1

Market-making and bond pricing

24.1.1

Review of duration and convexity

Duration and convexity are explained in Derivatives Markets Section 7.3. Section 7.3 is excluded from the MFE syllabus. However, Derivatives Markets page 781 mentions the duration hedging: ...hedging a bond portfolio based on duration does not result in a perfect hedge. Recall that the duration of a zero-coupon bond is the bond’s time to maturity... Because Chapter 24 mentions duration and Chapter 24 is on the syllabus, SOA may test your knowledge about duration hedging. So read DM Section 7.3 and take a quick review of duration and convexity. Next, let’s solve a few problems. Example 24.1.1. The yield to maturity of a 5-year zero-coupon bond is 6%. The face amount of the bond is 100. Calculate • the value of the bond • the Macaulay duration • the modified duration • the convexity 257

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In addition, calculate the bond value if the yield to maturity • moves up to 7% • moves down to 5%. Solution. The bond value is P0 = 100 × 1.06−5 = 74. 725 8 The (Macaulay) duration of a zero-coupon bond is its maturity T . So the (Macaulay) duration is DMac = T = 5 years. The modified duration is DMac 5 Dmod = = = 4. 716 98 years 1 + yield 1.06 The convexity of a zero-coupon bond is: 5×6 T (T + 1) C= 2 = 1.062 = 26. 699 893 (1 + y) If the yield to maturity is now 7%, the bond value is: P1 = 100 × 1.07−5 = 71. 298 6 We can also use duration and convexity to approximate the bond value under the new yield. The new bond value µis ¶ ∆P P1 = P0 + ∆P = P0 1 + P0 1 1 ∆P = −Dmod ∆y+ C (∆y)2 = −4. 716 98×0.01+ ×26. 699 893×(0.01)2 = P0 2 2 −0.045 8 → P1 = 74. 725 8 (1 − 0.045 8) = 71. 303 3 This is very close to the correct amount of 71. 298 6. If the yield to maturity is now 5%, the bond value is: P1 = 100 × 1.05−5 = 78. 352 6 We can also use duration and convexity to approximate the bond value under the new yield. ∆P 1 1 2 = −Dmod ∆y + C (∆y) = −4. 716 98 × (−0.01) + × 26. 699 893 × P0 2 2 (−0.01)2 = 0.048 5 → P1 = 74. 725 8 (1 + 0.048 5) = 78. 350 0 This is very close to the correct amount of 78. 352 6. Example 24.1.2.

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259

The yield to maturity of a 4-year zero-coupon bond is 8%. The face amount of the bond is 100. Calculate • the value of the bond • the Macaulay duration • the modified duration • the convexity In addition, calculate the bond value if the yield to maturity • moves up to 9% • moves down to 7%. Solution. The bond value is P0 = 100 × 1.08−4 = 73. 5030 DMac = T = 4 years. DMac 4 Dmod = = = 3. 703 7 years 1 + yield 1.08 4×5 T (T + 1) C= 2 = 1.072 = 17. 468 8 (1 + y) If the yield to maturity is now 9%, the bond value is: P1 = 100 × 1.09−4 = 70. 842 5 We can also use duration and convexity to approximate the bond value under the new yield. The new bond value µis ¶ ∆P P1 = P0 + ∆P = P0 1 + P0 1 1 ∆P 2 2 = −Dmod ∆y + C (∆y) = −3. 703 7 × 0.01 + × 17. 468 8 × (0.01) = P0 2 2 −0.036 2 → P1 = 73. 5030 (1 − 0.036 2) = 70. 842 2 This is very close to the correct amount of 70. 842 5. If the yield to maturity is now 7%, the bond value is: P1 = 100 × 1.07−4 = 76. 289 5 We can also use duration and convexity to approximate the bond value under the new yield. ∆P 1 1 = −Dmod ∆y + C (∆y)2 = −3. 703 7 × (−0.01) + × 17. 468 8 × P0 2 2 2 (−0.01) = 0.037 9 → P1 = 73. 5030 (1 + 0.037 9) = 76. 288 8 This is very close to the correct amount of 76. 289 5.

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Example 24.1.3. Portfolio A consists of one unit of 4-year zero coupon bond. Portfolio B consists of x units of 1-year zero coupon bond and y units of 8-year zero coupon bond. Portfolio B is used to duration hedge Portfolio A. Each of the three zero-coupon bonds above has 100 face amount. The current annual effective interest rate is 5% for all three bonds. Assume that the yield curve changes by a uniform amount if there’s a change. Demonstrate why duration-hedging leads to arbitrage under two scenarios: • the annual effective interest rate moves up to 6% immediately after the three bonds are issued. • the annual effective interest rate instantly moves down to 4% immediately after the three bonds are issued.

Solution. To duration hedge Portfolio A, we need to satisfy two conditions: • Portfolio A and B have the same present value • Portfolio A and B have the same duration So we set up the following equations: 100 100 100 x+ y= 1.051 1.058 1.054 100 100 x y 8 1.051 1.05 ×1+ ×8=4 100 100 100 100 x + y x + y 1.051 1.058 1.051 1.058 This is the meaning of the second equation. Portfolio B consists of two bonds. If a portfolio consists of multiple bonds, then the portfolio’s duration is just the weighted average of the each bond’s duration, with weight equal to the present value of each bond. The 2nd equation can be simplified as

µ

¶ µ ¶ 100 100 100 x 1+ y 8 = 4× 1 8 1.05 1.05 1.054

100 100 100 x+ y= 1 8 1.05 1.05¶4 ¶1.05 µ µ 100 100 100 x 1+ y 8=4× 1.051 1.058 1.054

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→ x = 0.493 62 , y = 0.520 93

24.1. MARKET-MAKING AND BOND PRICING

261

So Portfolio B needs to consist of 0.493 62 unit of 1-year zero coupon bond and 0.520 93 units of 8-year zero coupon bond. If the interest rate moves up to 6%: 100 Portfolio A is worth: = 79. 2094 1.064 100 100 × 0.493 62 + × 0.520 93 = 79. 251 7 1.061 1.068 To arbitrage, at t = 0 (when the interest rate is 5%), we buy low and sell high: Portfolio B is worth:

• buy Portfolio B. We pay

100 100 100 x+ y= = 82. 270 2 1.051 1.058 1.054

• sell Portfolio A. We receive

100 = 82. 270 2 1.054

So our net cost is zero. Then instantly later at t = 0+ , the interest rate moves up to 6%. Then Portfolio B (our asset) is worth 79. 251 7; Portfolio A (our liability) is worth 79. 2094. Our net profit is 79. 251 7 − 79. 2094 = 0.042 3 If the interest rate moves down to 4%: 100 Portfolio A is worth: = 85. 480 4 1.044 100 100 × 0.493 62 + × 0.520 93 = 85. 527 3 1 1.04 1.048 To arbitrage, at t = 0 (when the interest rate is 5%), we buy low and sell high: Portfolio B is worth:

• buy Portfolio B. We pay

100 100 100 x+ y= = 82. 270 2 1 8 1.05 1.05 1.054

• sell Portfolio A. We receive

100 = 82. 270 2 1.054

So our net cost is zero. Then instantly later at t = 0+ , the interest rate moves down to 4%. Then Portfolio B (our asset) is worth 85. 527 3; Portfolio A (our liability) is worth 85. 480 4. Our net profit is 85. 527 3 − 85. 480 4 = 0.046 9 So no matter the interest rate moves up to 6% or moves down to 4%, we can always make free money by buying Portfolio B and selling Portfolio A.

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Why is Portfolio B better than Portfolio A? It turns out B has high convexity. Convexity of A: TA (TA + 1) 4×5 = = 18. 1406 CA = 2 2 1.05 (1 + y) Convexity of B is just the weighted average convexities of the two bonds, with weights beingµthe present value¶of the bond.µ ¶ 100 8×9 100 1×2 × × 0.493 62 + × × 0.520 93 1.052 1.051 1.052 1.058 CB = 100 100 × 0.493 62 + × 0.520 93 1.051 1.058 µ ¶ µ ¶ 1×2 100 8×9 100 × × 0.493 62 + × × 0.520 93 1.052 1.051 1.052 1.058 = = 29. 024 9 100 1.054 CB > CA If the interest rate moves up by µ ¶ ∆y, then the present value of a portfolio is: ∆P P1 = P0 + ∆P = P0 1 + P0 1 A 1 ∆P B ∆P A 2 A B = −D ∆y + (∆y) = −Dmod ∆y + C B (∆y)2 C mod A B 2 2 P0 P0 4 A B B A However, Dmod = Dmod = = 3. 809 5 C >C 1.05 ∆P B ∆P A > B P0 P0A A Since P0 = P0B → ∆P B > ∆P A B A → P1 > P1 So Portfolio B always worth more than Portfolio A under a flat yield curve. For example, if the interest rate moves up to 6%, then ∆P A 1 = −3. 809 5 × 0.01 + × 18. 1406 × 0.012 = −0.037 2 2 P0A 1 ∆P B = −3. 809 5 × 0.01 + × 29. 024 9 × 0.012 = −0.036 6 B 2 P0 →

P1A =

100 (1 − 0.037 2) = 79. 2098 1.054

P1B =

100 (1 − 0.036 6) = 79. 259 2 1.054

P1B − P1A = 79. 259 2 − 79. 2098 = 0.049 4 Key point to remember:

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263

1. Two portfolios can have the same present value, the same duration, but different convexities. 2. Under the assumption of the parallel shift of a flat yield curve, we can always make free money by buying the high-convexity portfolio and sell the low-convexity portfolio. 3. The parallel shift of a flat yield curve assumption leads to arbitrage. Information about three zero-coupon bonds: Maturity (Yrs) Face 2 100 4 100 7 100 The current interest rate is 7% for all three bonds. Assume a parallel shift of a flat yield curve. Design an arbitrage strategy. We need to form 2 portfolios. These two portfolios have the same PV, the same duration, but different convexity. We can make free money by buying the high convexity portfolio and selling the low convexity portfolio. The low convexity portfolio is the 4-year bond (Portfolio A). The high convexity portfolio consists of x unit of 2-year bond and y unit of 7-year bond (Portfolio B). This is called a barbell. A barbell bond portfolio combines short maturities (low duration) with long maturities (high duration) for a blended, moderate maturity (moderate duration) Portfolio

Maturity (Yrs)

A

4

B

2

B

7

PV 100 = 76. 29 1.074 100 = 87. 34 1.072 100 = 62. 27 1.077

100 100 100 x+ y= 1.072 ¶1.077µ 1.07¶4 µ 100 100 100 x 2+ y 7=4× 2 7 1.07 1.07 1.074

Duration 4 2 7

Convexity 4×5 = 17. 468 8 1.072 2×3 = 5. 240 6 1.072 7×8 = 48. 912 6 1.072

x = 0.524 1, y = 0.490 0

At time 0, we buy Portfolio B (which consists of 0.524 1 unit of 2-year bond and 0.490 0 unit of 7-year bond). Simultaneously, we sell Portfolio A. The convexity of Portfolio A is: C A = 17. 468 8 The convexity of Portfolio B is:

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Units 1 x y

264

CB

CHAPTER 24. INTEREST RATE MODELS 100 7×8 100 2×3 × × 0.524 1 + × 0.490 0 × 2 2 2 7 1.07 1.07 1.07 1.07 = = 22. 708 9 100 1.074

For example, if the new interest rate is 7.25% for all bonds with different maturities, then 100 100 P1B = × 0.524 1 + 0.490 0 × = 75. 584 1 1.07252 1.07257 100 = 75. 580 9 1.07254 B A P1 > P1 Our profit is 75. 584 1 − 75. 580 9 = 0.003 2 P1A =

If the new interest rate is 6.5% for all bonds with different maturities, then 100 100 P1B = × 0.524 1 + 0.490 0 × = 77. 739 6 2 1.065 1.0657 100 = 77. 732 3 1.0654 P1B > P1A Our profit is 77. 739 6 − 77. 732 3 = 0.007 3 P1A =

Either way, we make money. By now you should see that the parallel shift of a flat yield curve is a bad model.

24.1.2

Interest rate is not so simple

We all know what an interest rate is, yet a derivative on interest rate is surprisingly difficult. To get a sense of the difficulty, suppose we want to calculate the price of a European call on a 1-year zero-coupon bond that pays $100 one year from now. Here are the inputs: • The call expires in one year • The strike price is $100 • The continuous risks-free rate is r = 6% per year • The current price of the bond is 100e−0.06 = 94. 18. • The volatility of the bond return is σ = 30% Using the Black-Scholes formula, we find: d2 = −0.15 N (d1 ) = 0.5596 N (d2 ) = 0.4404 d1 = 0.15 → C = 94. 18 × 0.5596 − 94. 18 × 0.4404 = 11. 23

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265

Everything looks fine. However, after further thinking, you realize that the call price C should be zero. At T = 1, the bond pays $100. Hence the 100-strike call value is zero. Why should anyone buy a 100-strike call on an asset worth $100 at call expiration? What went wrong? It turns out that we can’t use the Black-Scholes formula to calculate the price of an interest rate derivative: • The Black-Scholes option formula assume that the term structure of the interest rate is flat and deterministic (see DM page 379). However, If the interest rate is known and constant, there won’t be any need for interest rate derivatives. This is similar to the idea that if the stock price is known and constant, there won’t be any need for call or put option. • The standard deviation of the return σ is a constant. However, the standard deviation of the return σ is not constant. Unlike a stock, a bond has a finite maturity. At the maturity date, the bond value is its face amount. Hence the standard deviation of a bond’s return decreases as the bond approaches its maturity. The key point. It’s much harder to calculate the price of an option interest rate because interest rate is not a tradeable underlying asset. For call and put on stocks, we can buy or short sell stocks to set up the hedge portfolio. However, we can’t go out and buy a 5% interest rate to hedge an option on interest rate. Now let’s go to the textbook.

24.1.3

Impossible bond pricing model

A bond is a derivative on interest rate. We normally don’t think this way, but a bond derives its value from interest rate. If the market interest rate goes up, the bond value goes down; if the market interest rate goes down, the bond value goes up. So our starting point is to set up a stochastic random variable called the short interest rate. The textbook keeps using the phrase "short interest rate" or "short rate" without giving a clear definition. Here is the definition: Definition 24.1.1. A short interest rate or short rate is just the instantaneous interest rate r (t) over a short (hence the name "short rate") interval [t, t + dt]. First, we assume that the short interest rate r (t) follows the Ito’s process: dr = α (r) dt + σ (r) dZ

(24.1)

Next, the text book explains that we can’t assume r (t) follows a flat yield curve ("impossible bond pricing model"). A flat yield curve means that the interest rate is independent of time. To have a flat yield curve, the following two conditions are met:

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• The initial interest rate r is a constant regarding time, that is, r (t) = t • If the interest rate changes, the change is also independent of time (i.e. parallel shift of the yield curve) If the continuously compounded interest rate r is constant, then the present value at time t of $1 to be received at T is: P (t, T ) = e−r(T −t)

(24.2)

For example, if a zero coupon bond pays us $1 at T = 2, the PV of this bond at time zero is P (0, 2) = e−2r . The PV of this bond at t = 1 is P (1, 2) = e−r . Next, let’s review the textbook’s proof and find out why Equation 24.1 and 24.2 won’t work together. Suppose we want to delta hedge a bond. Whereas delta hedging a call means buying ∆ shares of a stock, delta hedging a bond means buying ∆ units of a bond with a different maturity date. Suppose at time t we buy a bond maturing at T2 . "A bond maturing at T2 " just means that we, the bond holder, will receive $1 at T2 . So the price of this bond at time t is P (t, T2 ) = e−r(T2 −t) . To delta hedge this bond, at t, we buy ∆ bonds maturing at T1 (please note the textbook uses N instead of ∆). The cost is ∆P (t, T1 ) = ∆e−r(T1 −t) The total cost of buying two bonds is ∆P (t, T1 ) + P (t, T2 ) = ∆e−r(T1 −t) + −r(T2 −t) e . To avoid tying up our capital, at time t we borrow ∆e−r(T1 −t) + −r(T2 −t) e from a bank to finance the purchase of two bonds. So at time t, our portfolio is: • buy 1 bond maturing at T2 • buy ∆ bonds maturing at T1 • borrow e−r(T2 −t) + ∆e−r(T1 −t) from a bank Our net position is zero: £ I = ∆P (t, T1 ) + P (t, T ¤ 2 ) + W = 0 where W = − [∆P (t, T1 ) + P (t, T2 )] = − ∆e−r(T1 −t) + e−r(T2 −t) . As seen before, if we borrow money, we use a negative number. If we lend money, we use the a positive number. This is why W is negative. In the next instant dt, the interest rate moves up to r + dr. Now change of our portfolio value is: dI = ∆ × dP (t, T1 ) + dP (t, T2 ) + dW Using Ito’s lemma, we have: 2 2 1) 1) 1) dt + ∂P (t,T dr + 12 ∂ P∂r(t,T (dr) dP (t, T1 ) = ∂P (t,T 2 ∂t ∂r ∂P (t,T1 ) ∂ −r(T1 −t) = ∂t e = rP (t, T1 ) ∂t ∂P (t,T1 ) ∂ −r(T1 −t) = e = −P (t, T1 ) (T1 − t) ∂r ∂r

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24.1. MARKET-MAKING AND BOND PRICING ∂ 2 P (t,T1 ) ∂r2 2

2

267

2

2

∂ −r(T1 −t) = ∂r = er(t−T1 ) (T1 − t) = P (t, T1 ) (T1 − t) 2e 2 2 (dr) = (αdt + σdZ) = σ2 (dZ) = σ 2 dt → dP (t, T1 ) = rP (t, T1 ) dt − P (t, T1 ) (T1 − t) dr + 12 P (t, T1 ) (T1 − t)2 σ 2 dt

Similarly, 2 dP (t, T2 ) = rP (t, T2 ) dt − P (t, T2 ) (T2 − t) dr + 12 P (t, T2 ) (T2 − t) σ 2 dt However, dW = rW dt. To see why, notice that dW is the interest earned on W during the interval [t, t + dt]. During [t, t + dt], the continuous interest rate r can be treated as a discrete interest rate. The interest earned is just the initial capital × interest rate × length of the interval = rW dt Now wehhave DM Equation 24.6: i 2 dI = ∆ rP (t, T1 ) dt − P (t, T1 ) (T1 − t) dr + 12 P (t, T1 ) (T1 − t) σ2 dt h i + rP (t, T2 ) dt − P (t, T2 ) (T2 − t) dr + 12 P (t, T2 ) (T2 − t)2 σ 2 dt +rW dt We’ll want to have dI = 0. If dI = 0, then the value of our portfolio won’t change during the same during [t, t + dt]. To make dI = 0, we first choose ∆ such that the dr term is zero. → ∆ [−P (t, T1 ) (T1 − t)] + [−P (t, T2 ) (T2 − t)] = 0 2 −t)P (t,T2 ) → ∆ = − (T (T1 −t)P (t,T1 )

This negative delta means that we need to sell maturing at T1 .

(T2 −t)P (t,T2 ) (T1 −t)P (t,T1 )

units of bond

2 −t)P (t,T2 ) By the way, ∆ = − (T (T1 −t)P (t,T1 ) is similar to DM Equation 7.13 (Derivatives Markets page 227):

N =−

D1 B1 (y1 ) / (1 + y1 ) D2 B2 (y2 ) / (1 + y2 )

(DM 7.13)

So buying ∆ bonds is really duration-hedging. Next, h we want to set the dt term to zero: i h i 2 2 ∆ rP (t, T1 ) + 12 P (t, T1 ) (T1 − t) σ 2 + rP (t, T2 ) + 12 P (t, T2 ) (T2 − t) σ 2 + rW = 0 2

0

2

→ r [∆P (t, T1 ) + P (t, T2 ) + W ]+ 12 ∆P (t, T1 ) (T1 − t) σ 2 + 12 P (t, T2 ) (T2 − t) σ 2 = However,∆P (t, T1 ) + P (t, T2 ) + W = 0 2 2 → 12 ∆P (t, T1 ) (T1 − t) σ 2 + 12 P (t, T2 ) (T2 − t) σ 2 = 0 2 (T −t)P (t,T ) → − 12 (T21 −t)P (t,T21 ) P (t, T1 ) (T1 − t) σ2 + 12 P (t, T2 ) (T2 − t)2 σ 2 = 0 2

→ − 12 (T2 − t) P (t, T2 ) (T1 − t) σ 2 + 12 P (t, T2 ) (T2 − t) σ2 = 0 → − 12 (T2 − t) P (t, T2 ) [(T1 − t) − (T2 − t)] σ 2 = 0

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→ − 12 (T2 − t) P (t, T2 ) (T1 − T2 ) σ 2 = 0 → T1 = T2 ∆ = −1 ∆ = −1 means that sell one bond. If we buy a bond maturing in T2 years, the only way to hedge the risk is to sell this bond! So one moment you buy a bond. The next moment you sell it. Your net position is zero. Of course you are hedged against all risks, but this hedging is really doing nothing type of hedging. This tells us that the bond pricing model based on Equation 24.1 and 24.2 are impossible. In our words, Equation 24.1 is OK. But Equation 24.2 is bad. This confirms that a parallel shift of a flat yield curve is a bad assumption for pricing a bond.

24.1.4

Equilibrium equation for bonds

Since it’s bad to assume that r is a flat yield curve, we switch our gears and assume that r isn’t a flat yield curve. Now we just assume that Equation 24.1 holds. dP (r, t, T ) =

∂P ∂P 1 ∂2P (dr)2 dr + dt + ∂r ∂t 2 ∂t2

2

2

2

(dr) = [α (r) dt + σ (r) dZ] = σ 2 (r) (dZ) = σ 2 (r) dt ∂P ∂P ∂P 1 ∂2P 2 σ (r) dt = → dP (r, t, T ) = dr + dt + [α (r) dt + σ (r) dZ] + ∂r ∂t 2 ∂t2 ∂r 2 1∂ P 2 ∂P σ (r) dt dt + ∂t 2 ∂t2 ∙ ¸ ∂P ∂P 1 ∂2P 2 ∂P σ (r) + dP (r, t, T ) = α (r) + dt + σ (r) dZ ∂r 2 ∂t2 ∂t ∂r

(24.3)

Define α (r, t, T ) =

∙ ¸ ∂P ∂P 1 ∂2P 2 1 σ (r) + α (r) + P (r, t, T ) ∂r 2 ∂t2 ∂t

(24.4)

∂P 1 σ (r) P (r, t, T ) ∂r

(24.5)

q (r, t, T ) =

Now Equation 24.3 becomes: dP (r, t, T ) = α (r, t, T ) dt + q (r, t, T ) dZ P (r, t, T )

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(24.6)

24.1. MARKET-MAKING AND BOND PRICING

269

dP (r, t, T ) is the bond’s return. Equation 24.6 says that the bond’s return P (r, t, T ) is the sum of a drift term α (r, t, T ) dt and a random component q (r, t, T ) dZ. ∂P is Please also note that generally q (r, t, T ) is negative. This is because ∂r negative. If r goes up, the bond price goes down; if r goes down, P goes up. Next, we are going to derive DM Equation 24.16: α2 (r, t, T2 ) − r α1 (r, t, T1 ) − r = q1 (r, t, T1 ) q2 (r, t, T2 )

(24.7)

Suppose we have two bonds, Bond #1 and Bond #2 maturing at T1 and T2 respectively. Each bond follows DM Equation 24.6: dP (r, t, T1 ) = α1 (r, t, T1 ) dt + q1 (r, t, T1 ) dZ P (r, t, T1 ) dP (r, t, T2 ) = α2 (r, t, T2 ) dt + q2 (r, t, T2 ) dZ P (r, t, T2 ) At t = 0, we form a self-financing portfolio consisting of • buying ∆ units of Bond #1 • buying 1 Bond #2 • borrowing ∆P (r, t, T1 ) + P (r, t, T2 ) from a bank at the riskless short rate r The value of this portfolio is I dP (r, t, T1 ) dP (r, t, T2 ) dI = ∆P (r, t, T1 ) + P (r, t, T2 ) P (r, t, T1 ) P (r, t, T2 ) − [∆P (r, t, T1 ) + P (r, t, T2 )] rdt = ∆P (r, t, T1 ) [α1 (r, t, T1 ) dt + q1 (r, t, T1 ) dZ] +P (r, t, T2 ) [α2 (r, t, T2 ) dt + q2 (r, t, T2 ) dZ]−[∆P (r, t, T1 ) + P (r, t, T2 )] rdt = ∆P (r, t, T1 ) [α1 (r, t, T1 ) − r] dt + P (r, t, T2 ) [α2 (r, t, T2 ) − r] dt + [∆P (r, t, T1 ) q1 (r, t, T1 ) + P (r, t, T2 ) q2 (r, t, T2 )] dZ Choose ∆ such that ∆P (r, t, T1 ) q1 (r, t, T1 )+P (r, t, T2 ) q2 (r, t, T2 ) = 0. This removes the stochastic random term dZ. Now dI is deterministic. Since the portfolio is self-financing and riskless, it earns zero interest rate. Hence ∆P (r, t, T1 ) [α1 (r, t, T1 ) − r] dt + P (r, t, T2 ) [α2 (r, t, T2 ) − r] dt = 0 P (r, t, T2 ) q2 (r, t, T2 ) →− P (r, t, T1 ) [α1 (r, t, T1 ) − r] dt+P (r, t, T2 ) [α2 (r, t, T2 ) − r] dt = P (r, t, T1 ) q1 (r, t, T1 ) 0 q2 (r, t, T2 ) →− [α1 (r, t, T1 ) − r] dt + [α2 (r, t, T2 ) − r] dt = 0 q1 (r, t, T1 ) α2 (r, t, T2 ) − r α1 (r, t, T1 ) − r = → q1 (r, t, T1 ) q2 (r, t, T2 )

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Define the following term as Sharp ratio: α (r, t, T ) − r = φ (r, t) q (r, t, T )

(24.8)

Apply Equation 24.4 and 24.6 to 24.8: ∙ ¸ ∂P ∂P 1 ∂2P 2 1 σ (r) + α (r) + −r P (r, t, T ) ∂r 2 ∂t2 ∂t = φ (r, t) ∂P 1 σ (r) P (r, t, T ) ∂r ¸ ∙ ∂P 1 ∂2P 2 ∂P ∂P σ (r) + + − rP = φ (r, t) σ (r) → α (r) ∂r 2 ∂t2 ∂t ∂r ∂2P ∂P ∂P 1 2 σ (r) 2 + [α (r) − σ (r) φ (r, t)] + − rP = 0 (24.9) 2 ∂r ∂r ∂t Any interest-dependent securities (not just zero coupon bonds) must satisfy Equation 24.9. To solve the bond price P (r, t, T ), we need to use Equation 24.9 together with the following boundary condition: P (r, T, T ) = 1

(24.10)

The solution to Equation 24.9 and 24.10 is:

P (r, t, T ) = Et∗

Ã

" Z exp −

#!

T

r (s) ds

t

= Et∗ (exp [−R (t, T )])

(24.11)

where E ∗ means that the expectation is based on the risk neutral probability and Z T

R (t, T ) =

r (s) ds

(24.12)

t

Equation 24.11 is the general formula for a bond price.

24.1.5

Delta-Gamma approximation for bonds

The key formula is: 1 ∗ (24.13) E (dP ) = rP dt Don’t worry about how to prove Equation 24.13. Just memorize its meaning. Equation 24.13 says that under the risk neutral distribution, the bond is priced to earn a risk-free rate.

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24.2. EQUILIBRIUM SHORT-RATE BOND PRICE MODELS

271

24.2

Equilibrium short-rate bond price models

24.2.1

Arithmetic Brownian motion (i.e. Merton model)

A simple model is to assume that the short rate r (t) follows arithmetic Brownian motion: dr (t) = αdt + σdZ

(24.14)

In addition, we assume φ (r, t) = φ Advantage: • The model is simple. Disadvantages: • r (t) can go negative. Since r (t) is normally distributed with mean r0 + αt and variance σ 2 t, r (t) can be negative. A negative interest rate won’t make any sense. • r (t) is not mean reverting. • V ar [r (t)] = σ 2 t. This is undesirable because the volatility of the interest should depend on the interest rate. If the interest rate is high, then the volatility is high. We can derive the bond price under the Merton model. Under this model, Equation 24.9 now becomes: 1 2 ∂2P ∂P ∂P 2 σ ∂r2 + [α − σφ] ∂r + ∂t − rP = 0 We guess the solution is P (r, t, T ) = A (T − t) e−B(T −t)r(t) where A (T − t) and B (T − t) are two functions of T − t. ∂2P ∂P ∂P 0 = AB 2 e−Br = −ABe−Br = rAB e−Br − A0 e−Br 2 ∂r ∂r ∂t 0 → 12 σ 2 AB 2 e−Br − [α − σφ] ABe−Br + rAB e−Br − A0 e−Br − rAe−Br = 0 ³ ´ 0 1 2 σ AB 2 e−Br − [α − σ] ABe−Br − A0 e−Br = rA e−Br − B e−Br 2

(24.15)

Equation 24.15 should hold for any r. The only way to make it work for any r is: 1 2 2 0 2 σ³ AB −´[α − σφ] AB − A = 0 A 1−B

0

=0

Using the boundary condition is P (r, T, T ) = 1, we get A (0) e−B(0)r = 1. For this equation ³ to hold ´ for any r, we need to have A (0) = 1 and B (0) = 0. 0 Hence from A 1 − B = 0, we get:

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B =1 →B =T −t From 12 σ 2 AB 2 − [α − σφ] AB − A0 = 0. we get: A0 2 = 12 σ 2 B 2 − [α − σφ] B = 12 σ 2 (T − t) − [α − σφ] (T − t) A 2 → d ln A (T − t) = 12 σ 2 (T − t) − (α − σφ) (T − t) 1 3 2 → ln A (T − t) = 16 σ 2 (T − t) − (α − σφ) (T − t) + C where C is a con2 stant. ∙ ¸ 1 3 2 1 2 → A = A (0) exp 6 σ (T − t) − (α − σφ) (T − t) 2 ∙ ¸ 1 3 2 1 2 = exp 6 σ (T − t) − (α − σφ) (T − t) 2 Finally, we have:∙ ¸ 1 3 2 1 2 P (r, t, T ) = exp 6 σ (T − t) − (α − σφ) (T − t) e−(T −t)r(t) 2 You don’t need to memorize the above formula. Just understand how to derive the formula in case SOA or CAS gives you a tough problem.

24.2.2

Rendleman-Bartter model

This model assume that r (t) follows a geometric Brownian motion: dr (t) = αdt + σdZ r (t)

(24.16)

Advantage: • r (t) can’t be negative • V ar [r (t)] = r2 (t) σ2 t. The variance increases if r (t) increases. This is desirable because the volatility of the interest is high if the interest rate is high. Disadvantage: • not mean-reverting

24.2.3

Vasicek model

According to Wikipedia, Vasicek’s model was the first one to capture mean reversion, an essential characteristic of the interest rate that sets it apart from other financial prices. Stock prices can rise indefinitely. However, interest rates cannot. Excessively high interest rate would hamper economic activity, prompting a decrease in interest rates. Similarly, interest rates can not decrease indefinitely. If interest rates are too low, few people are willing to lend their money. This tends to push up the interest rate. As a result, interest rates move in a limited range, showing a tendency to revert to a long run value.

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24.2. EQUILIBRIUM SHORT-RATE BOND PRICE MODELS

273

The model is: dr (t) = α (b − r) dt + σdZ

(24.17)

2

The variance of r (t) is σ t Advantage: • mean reverting Disadvantage: • can produce a negative interest rate • V ar [r (t)] = σ 2 t. So the variance doesn’t increase if r (t) increases. We can also derive the bond price P (t, T ) under the Vasicek model. However, the derivation is far more complex than in the Merton model. Here is the outline of the derivation. Under the Vasicek model, Equation 24.9 becomes: 1 2 ∂2P ∂P ∂P 2 σ ∂r2 + [α (b − r) − σφ] ∂r + ∂t − rP = 0 subject to the boundary condition P (r, T, T ) = 1 Once again, we guess the solution is P (r, t, T ) = A (T − t) e−B(T −t)r(t) ∂P ∂2P ∂P 0 → = AB 2 e−Br = −ABe−Br = rAB e−Br − A0 e−Br 2 ∂r ∂r ∂t0 → 12 σ 2 AB 2 e−Br −[α (b − r) − σφ] ABe−Br +rAB e−Br −A0 e−Br −rAe−Br = 0 0

→ 12 σ 2 AB 2 − [α (b − r) − σφ] AB + rAB − A³0 − rA = 0 ´ 0 → 12 σ 2 AB 2 − (αb − σφ) AB + σφAB − A0 = A − AB − aAB r For the above equation to hold for any r, we need to have: 0 A − AB − αAB = 0 1 2 2 0 2 σ AB − (αb − σφ) AB + σφAB − A = 0 0

0

→ 1 − B − αB = 0 A − AB − αAB = 0 The boundary condition is B (0) = 0 1 − e−α(T −t) This gives us: B = α Equation 12 σ2 AB 2 − (αb − σφ) AB + σφAB − A0 = 0 is hard to solve. Someone solved this equation for us. The result is DM Equation 24.26. You don’t need to memorize the solution for A or B. Just memorize P (r, t, T ) = A (T − t) e−B(T −t)r(t) . Example 24.2.1. May 2007 SOA MFE #13 Let P (r, t, T ) denote the price at time t of $1 to be paid with certainty at time T , t ≤ T , if the short rate at time t is equal to r. For a Vasicek model you are given: P (0.04, 0, 2) = 0.9445

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P (0.05, 1, 3) = 0.9321 P (r∗ , 2, 4) = 0.8960 Calculate r∗ . The price of any bond must satisfy Equation 24.9. We guess that the solution to Equation 24.9 is P (r, t, T ) = A (T − t) e−B(T −t)r(t) P (0.04, 0, 2) = 0.9445 → A (2) e−B(2)0.04 = 0.9445 P (0.05, 1, 3) = 0.9321 → A (2) e−B(2)0.05 = 0.9321 −B(2)0.04 0.9445 0.9445 A (2) e = eB(2)0.01 = → −B(2)0.05 0.9321 0.9321 A (2) e µ µ ¶ 1 ¶100 0.9445 0.01 0.9445 → eB(2) = = 3. 749 26 = 0.9321 0.9321 £ ¡ ¤0.04 ¢ → A (2) = 0.9445eB(2)0.04 = 0.9445 eB(2) = 0.9445 3. 749 260.04 = 0.995 77 P (r∗ , 2, 4) = 0.8960 £ ¤−r∗ ∗ → A (2) e−B(2)r = A (2) eB(2) = 0.8960 ¢ ¡ −r∗ = 0.8960 → 0.995 77 3. 749 26 ¡ ¢ 0.8960 −r∗ → 3. 749 26 = = 0.899 81 0.995 77 ∗ → −r ln 3. 749 26 = ln 0.899 81 ln 0.899 81 → r∗ = − = 0.07988 4 = 0.08 ln 3. 749 26 Please note r∗ can be solved without using any specifics of the Vasicek model. As a matter of fact, this problem can be rewritten as: Let P (r, t, T ) denote the price at time t of $1 to be paid with certainty at time T , t ≤ T , if the short rate at time t is equal to r. You are given: P (0.04, 0, 2) = 0.9445 P (0.05, 1, 3) = 0.9321 P (r∗ , 2, 4) = 0.8960 Calculate r∗ .

24.2.4

CIR model

The model assumes that the short interest rate r (t) follows the stochastic differential equation: p dr (t) = α (b − r) dt + σ r (t)dZ (24.18) In CIR model, V ar [r (t)] = σ 2 r (t) t. The drift factor α (b − r) in the CIR model is the same as the drift factor in the Vasicek model. It ensures mean reversion of the interest rate towards the long run value b, with speed of adjustment governed by the strictly positive parameter a. p The standard deviation factor, σ r (t), corrects the main drawback of Vasicek’s model, ensuring that the interest rate cannot become negative. Thus, at low values of the interest rate, the standard deviation becomes close to zero,

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24.3. BOND OPTIONS, CAPS, AND THE BLACK MODEL

275

canceling the effect of the random shock on the interest rate. Consequently, when the interest rate gets close to zero, its evolution becomes dominated by the drift factor, which pushes the rate upwards. Advantage: • Mean-reverting • Not allow a negative interest rate • Variance of V ar [r (t)] = σ2 r (t) t. The higher the r (t), the higher the V ar [r (t)]. This is a desirable feature. To solve the bond price under CIR model, again we guess the solution is P (r, t, T ) = A (T − t) e−B(T −t)r(t) . The solution is listed in DM page 788. You don’t need to memorize the solution. Just memorize P (r, t, T ) = A (T − t) e−B(T −t)r(t) .

24.3

Bond options, caps, and the Black model

24.3.1

Black formula

Notations: • Pt (T, T + s). This is the price agreed upon at time t, which will be paid at T in order to receive $1 at T + s. Simply put, Pt (T, T + s) is the present value of $1 discounted from T + s to T using the interest rate available at t. For example, at time zero we know that the annual interest rate from t = 1 to t = 3 is 10% compounded continuously. Then P0 (1, 1 + 2) is just PV of $1 discounted from t = 3 to t = 1 using 10% continuously compounded interest rate. So P0 (1, 1 + 2) = e−0.1(2) = 0.818 73 • PT (T, T + s). This is PV $1 discounted from T + s to T . Generally, we simplify the symbol PT (T, T + s) as P (T, T + s) Consider a call option with strike price K, expiring at time T , on a zerocoupon bond paying $1 at time T + s. If you buy this call option, then at T you have the right to buy a bond maturing at time T + s for the guaranteed price K. Since time T cost of a bond maturing in T + s is PV of $1 discounted from T + s to T , buying a bond maturing at time T + s for the guaranteed price K really means "give up K and receive PV of $1 discounted from T + s to T ." The PV of $1 discounted from T + s to T is PT (T, T + s). So the call payoff is Call P ayof f = max [0, PT (T, T + s) − K]

(DM 24.30)

Next, the textbook has a difficult formula: Ft,T [P (T, T + s)] =

P (t, T + s) P (t, T )

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(DM 24.31)

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To understand the meaning of DM 24.31, use an example. Suppose t = 0, T = 1, and s = 2. In addition, assume that the interest rate is always 10% compounded continuously. • Ft,T [P (T, T + s)] = F0,1 [P (1, 3)] is the price agreed upon at t = 0 to be paid at T = 1 in order to receive $1 at T + s = 3. In other words, if you pay F0,1 [P (1, 3)] at T = 1, you should receive PV of $1 discounted from T + s = 3 to T = 1. To avoid arbitrage, F0,1 [P (1, 3)] = e−0.1(2) = e−0.02 • P (t, T + s) = P (0, 3) is PV of $1 discounted from T + s = 3 to time zero. P (0, 3) = e−0.1(3) = e−0.03 • P (t, T ) = P (0, 1) is PV of $1 discounted from T = 1 to time zero. P (0, 1) = e−0.1 −0.03

Clearly we see that e−0.02 = ee−0.1 . Next, the textbook gives us the price of the call on a bond. Consider a call option is written at time zero with strike price K, expiring at time T , on a zero-coupon bond paying $1 at time T + s. If you buy this call option at time zero, then at T you have the right to pay K and receive the PV of $1 discounted from T + s to T . To find the call price, we’ll use DM Equation 12.5: P P C = F0,T (S) N (d1 ) − F0,T (K) N (d2 ) ln

d1 =

P (S) F0,T +0.5σ 2 T F P (K) 0,T

√ σ T

√ d2 = d1 − σ T

This is how I memorize DM Equation 12.5: C =Time zero cost of what you get at T ×N (d1 ) −Time zero cost of what you give³at T ×N (d2 ) ´ √ Tim e zero cost of what you get at T 2 d1 = ln Time zero cost of what you give at T + 0.5σ T /σ T √ d2 = d1 − σ T

Let’s apply DM Equation 12.5 to call on bond. • What we get at T is P (T, T + s), PV of $1 discounted from T + s to T • Time zero cost of P (T, T + s) is P (0, T + s), PV of $1 discounted from T + s to time 0 • What we give at T is K • Time zero cost of what we give at T is KP (0, T ), PV of K discounted from T to time 0

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24.3. BOND OPTIONS, CAPS, AND THE BLACK MODEL

277

C =Time zero cost of what you get at T ×N (d1 ) −Time zero cost of what you give³at T ×N (d2 ) ´ √ Time zero cost of what you get at T 2 d1 = ln Time + 0.5σ T /σ T zero cost of what you give at T √ d2 = d1 − σ T

This gives us the price of the call on a bond: C = P (0, T + s) N (d1 ) − P (0, T ) KN (d2 ) d1 =

ln

P (0,T +s) 2 P (0,T ) +0.5σ T

√ σ T

√ d2 = d1 − σ T

We can also calculate the price of the put on the bond: P = P (0, T ) KN (−d2 ) − P (0, T + s) N (−d1 ) Please note that DM Equation 24.32 uses the following formula: C = P (0, T ) [F N (d1 ) − KN (d2 )] d1 =

ln

2 F T K +0.5σ √

σ T

This is how to derive these formulas. Notice, F = F0,T [P (T, T + s)] = P (0,T +s) P (0,T )

→ C = P (0, T + s) N (d1 ) − P (0, T ) KN (d2 ) = F P (0, T ) N (d1 ) − P (0, T ) KN (d2 ) = P (0, T ) [F N (d1 ) − KN (d2 )] Example 24.3.1. SOA May 2007 MFE #7 You are given the following information: Bond maturity (years) 1 2 Zero-coupon bond price 0.9434 0.8817 A European call option, that expires in 1 year, gives you the right to purchase a 1-year bond for 0.9259. The bond forward price is lognormally distributed with volatility σ = 0.05. Using the Black formula, calculate the price of the call option. Solution. If you buy this option, then at T = 1, you can pay K = 0.9259 and buy a 1-year bond. This 1-year bond will give you $1 at time T + s = 2. .

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• The value of this bond at T = 1 is PV of $1 discounted from T + s = 2 to T = 1.Time zero cost of PV of $1 discounted from T + s = 2 to T = 1 is just PV of $1 discounted from T + s = 2 to t = 0. The cost is 0.8817 • Time zero cost of the strike price K at T = 1 is just PV of K discounted from T = 1 to time zero. So P V (K) = K × 0.9434 = 0.9259 × 0.9434 C =Time zero cost of what you get at T ×N (d1 ) −Time zero cost of what you give³at T ×N (d2 ) ´ √ Tim e zero cost of what you get at T 2 d1 = ln Time zero cost of what you give at T + 0.5σ T /σ T → d1 =

ln

2 0.8817 0.9259×0.9434 +0.5×0.05 ×1

√ = 0.212 0 √ 0.05 1 √ → d2 = d1 − σ T = 0.212 0 − 0.05 1 = 0.162 0

N (d2 ) = 0.564 3 N (d1 ) = 0.583 9 C = 0.8817 × 0.583 9 − 0.9259 × 0.9434 × 0.564 3 = 0.021 9 We can also calculate the put price. P =Time zero cost of what you give at T ×N (−d2 ) −Time zero cost of what you get at T ×N (−d1 ) P = 0.9259 × 0.9434 × (1 − 0.564 3) − 0.8817 × (1 − 0.583 9) = 0.013 7 If you want to use the formula C = P (0, T ) [F N (d1 ) − KN (d2 )], this is how: • P (0, T ) = P (0, 1) is PV of $1 discounted from T = 1 to time zero. So P (0, 1) = 0.9434 • F = F0,T [P (T, T + s)] = F0,1 [P (1, 2)] is the forward price of 1-year bond. This is the price agreed up at time zero, paid at T = 1 in order to receive $1 at T + s = 2. Using Equation DM 24/31, we have F0,1 [P (1, 2)] = P (0,2) 0.8817 P (0,1) = 0.9434 d1 =

ln

F K

+0.5σ 2 T √ σ T

=

ln

0.8817 0.9434 0.9259

+0.5×0.052 ×1 √ 0.05 1

= 0.212 0

√ d2 = 0.212 0 − 0.05 1 = 0.162 0 N (d1 ) = 0.583 9 N (d2 ) = 0.564 3 C = P (0,¡T ) [F N (d1 ) − KN (d2 )] ¢ = 0.9434 0.8817 0.9434 × 0.583 9 − 0.9259 × 0.564 3 = 0.0220

P = P (0,¡T ) [KN (d2 ) − F N (d1 )] = 0.9434 0.9259 × (1 − 0.564 3) − = 0.013 7

Yufeng Guo, Fall 09 MFE, actuary88.com

0.8817 0.9434

¢ × (1 − 0.583 9)

24.4. BINOMIAL INTEREST RATE MODEL

24.3.2

279

Interest rate caplet

Notation • Rt (T, T + s). The (not annualized) interest rate pre-agreed upon at time t where t ≤ T that applies to the future time interval [T, T + s]. • RT (T, T + s). The (not annualized) interest rate agreed upon at time T that applies to the time interval [T, T + s]. • Caplet. A caplet gives the buyer the right to buy the time-T market interest rate RT (T, T + s) by paying a fixed strike interest rate KR . If KR ≥ RT (T, T + s), the caplet expires worthless. The payoff of the caplet at T + s is max [0, RT (T, T + s) − KR ]. The payoff of the caplet at T is max [0, RT (T, T + s) − KR ] 1 + RT (T, T + s) To calculate the price of the caplet, we first modify the payoff: ∙ ¸ max [0, RT (T, T + s) − KR ] RT (T, T + s) − KR = (1 + KR ) max 0, = 1 + R∙T (T, T + s) (1 + RT (T, T∙ + s)) (1 + KR ) ¸ ¸ 1 1 1 − − PT (T, T + s) (1 + KR ) max 0, = (1 + KR ) max 0, 1 + KR 1 + RT¸(T, T + s) 1 + KR ∙ 1 max 0, − PT (T, T + s) is the payoff of a put on a bond. This put 1 + KR gives the buyer the right, at T , to sell a bond that matures at T + s for a 1 guaranteed price 1+K . Let P represent the price of this put. Hence the price R of the caplet is (1 + KR ) P .

24.4

Binomial interest rate model

Binomial interest rate tree is really simple. Unfortunately, the author of the textbook uses too many math symbols and formulas, making this section hard to read. What I will do here is to walk you through a few examples. If you understand these examples, you are fine. Example 24.4.1. (DM Example 24.3) The following is the 3-period interest rate tree (DM Figure 24.3) t=0 t=1 t=2 0.18 0.14 0.10 0.10 0.10 0.06 0.02

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Make sure you understand the above table. The 10% interest rate at t = 0 applies to the interval [t = 0, t = 1]. The 14% and 6% interest rates apply to the interval [t = 1, t = 2]. The interest rates 0.18, 0.10, and 0.02 at t = 2 apply to the interval [t = 2, t = 3]. The price of the 1-year bond is just the PV of $1 discounted from t = 1 to t = 0. Hence P (0, 1) = e−0.1 The price of the 2-year bond is just the PV of $1 discounted from t = 2 to t = 0. What’s tricky is that we have two interest rates during [t = 1, t = 2]. • If the path is u, then $1 at t = 2 travels back to t = 0 through 2 interest rates: 0.14, and 0.1. The PV of $1 discounted from t = 2 to t = 1 is e−0.14 . The PV of e−0.14 discounted from t = 1 to t = 0 is e−0.14 e−0.11 = e−(0.14+0.1) . So the 2-year bond is worth e−(0.14+0.1) at t = 0 • If the path is d, then $1 at t = 2 travels back to t = 0 through 2 interest rates: 0.06 and 0.1. The 2-year bond is worth e−(0.06+0.1) at t = 0 Since the risk-neutral probability of up or down is 50%, then the 2-year bond is worth the following at t = 0: P (0, 2) = 0.5e−(0.14+0.1) + 0.5e−(0.06+0.1) = 0.819 4 The term 0.5e−(0.14+0.1) + 0.5e−(0.06+0.1) can be irewritten as: h S −(0.14+0.1) −(0.06+0.1) ∗ − 2i=0 ri h 0.5e + 0.5e =E e with h = 1 The price of the 3-year bond is just the PV of $1 discounted from t = 3 to t = 0. • If the path is uu, then $1 at t = 3 travels back to t = 0 through 3 interest rates: 0.18, 0.14, and 0.1. The PV of $1 discounted from t = 3 to t = 0 is e−(0.18+0.14+0.1) . So the 3-year bond is worth e−(0.18+0.14+0.1) at t = 0 • If the path is ud, then $1 at t = 3 travels back to t = 0 through 3 interest rates: 0.1, 0.14, and 0.1. The 3-year bond is worth e−(0.1+0.14+0.1) at t = 0 • If the path is du, then $1 at t = 3 travels back to t = 0 through 3 interest rates: 0.1, 0.06, and 0.1. The 3-year bond is worth e−(0.1+0.06+0.1) at t = 0 • If the path is dd, then $1 at t = 3 travels back to t = 0 through 3 interest rates: 0.02, 0.06, and 0.1. The 3-year bond is worth e−(0.02+0.06+0.1) at t=0 Since the risk-neutral probability for each is 0.52 = 0.25, the price of a 3-year bond is the following ¡ at t = 0: ¢ P (0, 3) = 0.25 e−(0.18+0.14+0.1) + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1) = 0.743 8 ¡ ¢ The term 0.25 e−(0.18+0.14+0.1) + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1) can be rewritten as:

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24.4. BINOMIAL INTEREST RATE MODEL

281

¡ −(0.18+0.14+0.1) ¢ + e−(0.1+0.14+0.1) + e−(0.1+0.06+0.1) + e−(0.02+0.06+0.1) = h0.25Se3 i E ∗ e− i=0 ri h The general formula is DM 24.44:

h Sn i P (0, n) = E ∗ e− i=0 ri h

(DM 24.31)

Option pricing example

Example 24.4.2. (SOA May 2007 #9) You use a binomial interest rate model to evaluate a 7.5% interest rate cap on a $100 three-year loan. You are given: (i) The interest rates for the binomial tree are as follows: • r0 = 6% • ru = 7.704% • rd = 4.673% • ruu = 9.892% • rud = rdu = 6.000% • rdd = 3.639% (ii) All interest rates are annual effective rates. (iii) The risk-neutral probability that the annual effective interest rate moves up or down is 0.5. (iv) The loan interest payments are made annually. Using the binomial interest rate model, calculate the value of this interest rate cap. t=0 t=1 t=2 9.892% 7.704% 6.000% 6% 6.000% 4.673% 3.639% Once again, the interest rate at t in the above table applies to the period [t, t + 1]. For example, the 6% rate applies to [0, 1] (i.e. Year 1). First, let’s understand what’s an interest rate cap. Imagine you borrowed $100 from a bank. Your interest accrued on the loan each year is not based on a fixed interest rate (such as 8%), but is based on the then market interest rate. For example, if the market interest rate is 6% in Year 1, then your interest payment at the end of Year 1 is 100 × 0.06 = 6.0.

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CHAPTER 24. INTEREST RATE MODELS

As a borrower, you are worried that the market interest may go up. For example, if the Year 1 interest rate is 20%, then your interest payment at the end of Year 1 is 100 × 0.2 = 20. How can you reduce your risk? One thing you can do is to buy an interest cap. Suppose you buy an interest rate cap of 7.5%. This is what happens: • If the market interest rate is at or below 7.5%, then the cap doesn’t kick in. So you get nothing from the cap • if the market interest rate is above 7.5%, then the party who sold you the cap will pay you the excess of the market interest rate over the cap rate. For example, if the market interest rate in Year 1 is 10%, then the seller of the cap will pay you 100 (0.1 − 0.075) = 2. 5 at the end of Year 1. You still own the bank 100 × 0.1 = 10 at the end of Year 1, but you pay 100 × 0.075 = 7. 5 out of your own pocket. The cap seller pays you 100 (0.1 − 0.075) = 2. 5. So together you get 7. 5 + 2.5 = 10. You mail a $10 check to the bank. In summary, if you buy an interest capped of 7.5%, then the interest on your loan is capped at 7.5% regardless of the market interest rate (because any excess of the market rate over the cap 7.5% is paid by the cap seller). Now you know what an interest cap is, let’s solve this problem.

The first year market rate 6% is below the cap rate. At the end of Year 1, you get nothing from the cap

If the 2nd year market rate is 7.704%, then at the end of year 2 the cap seller pays you 100 (0.07704 − 0.075) = 0.204 . Notice that 0.204 occurs at t = 2. To help keep track of payments, we’ll discount this payment to t = 1. 0.204 The discounted value is at t = 1. 1 + 0.07704 If the 2nd year market rate is 4.673%, then the payment at the end of year 2 is zero.

Let’s calculate the cap payoff in Year 3. Of the 4 interest rates during [t = 2, t = 3] , only when the market interest rate is 9.892% do we get a payoff of 100 (0.09892 − 0.075) = 2. 392. This payment occurs at t = 3. The PV of 2. 392 . this payment at t = 2 is 1 + 0.09892 Now we can draw a payoff table:

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24.5. BLACK-DERMAN-TOY MODEL t=0

t=1

283 t=2 9.892% Payoff:

7.704% Payoff:

100 (0.07704 − 0.075) 1 + 0.07704

100 (0.09892 − 0.075) 1 + 0.09892

6.000% 6% 6.000% 4.673% 3.639% 100 (0.07704 − 0.075) = 0.189 4 1 at t = 1 1 + 0.07704 100 (0.07704 − 0.075) t = 0 is . The risk neutral probability (1 + 0.09892) (1 + 0.06) node is 0.5. 100 (0.09892 − 0.075) = 2. 176 68 at t = 2 The PV of payoff 1 + 0.09892 100 (0.09892 − 0.075) t = 0 is . The risk neutral (1 + 0.09892) (1 + 0.07704) (1 + 0.06) reaching uu node is 0.52 = 0.25 The PV of payoff

discounted to of reaching u discounted to probability of

Hence risk neutral based expected present value of the cap payoff is: 100 (0.07704 − 0.075) 100 (0.09892 − 0.075) ×0.5+ ×0.52 = (1 + 0.07704) (1 + 0.06) (1 + 0.09892) (1 + 0.07704) (1 + 0.06) 0.565 99 = 0.57 So the price of the cap is $0.57

24.5

Black-Derman-Toy model

The BDT model is a procedure to produce a discrete binomial interest rate tree that matches the observed term structure of interest rates. Once again, to quickly explain the essence of the BDT model, I’m walk you through an example. If you understand this example, you are ready for the exam. Suppose we gathered the following data from the market (DM Table 24.2): Maturity n (Yrs) YTM Bond Price Volatility in Yr 1 on (n − 1)-Yr bond 1 10% 0.9091 2 11% 0.8116 10% 3 12% 0.7118 15% 4 12.5% 0.6243 14% We want to produce an interest rate tree that will match the above table. What should we do? First, let’s go through some terms.

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YTM (yield to maturity) is the discrete annual effective interest rate earned by a bond. The formula is P (0, T ) = (1 + Y T M )−T For example, a 2-year bond in the table is worth 0.8116. This means that PV of $1 discounted from t = 2 to t = 0 is 0.8116. To find YTM, we solve the following equation: −2 0.8116 = (1 + i) → i = 0.11 So the YTM for this 2-year bond is 11% The volatility in Yr 1 on (n − 1) bond is the standard deviation of the natural log of the YTM on an (n − 1)-year bond issued at t = 1 and maturity at time n. For example, 10% volatility in the table means that standard deviation of the natural log of the YTM on a 1-year bond issued at t = 1 maturing in t = 2 is 10%. This concept will be clear to you later. Next, next use the BDT model to build an interest rate tree. We’ll first find the interest rate ru and rd , the two possible interest rates for Year 2 (i.e. for the time interval [t = 1, t = 2]). t=0 t=h=1 ru r0 = 10% rd

Once again, rt is the interest for the interval [t, t + 1]. For example, r0 = 10% is the interest rate for Year 1. ru and rd are the two interest rates for Year 2 (i.e. from t = 1 to t = 2). We assume the risk neutral probability of up and down is 0.5. We want to find ru and rd to satisfy the condition that standard deviation of the natural log of the YTM on a 1-year bond issued at t = 1 maturing in t = 2 is 10%. The price of a 1-year bond issued at t = 1 maturing in t = 2 is 1 1 P (1, 2) = or 1 + ru 1 + rd The YTM is: 1 −1 → Y T Mu = ru (1 + Y T Mu ) = 1 + ru 1 −1 or (1 + Y T Md ) = → Y T Md = rd 1 + rd We know that the standard deviation of ln Y T Mu = ln ru and ln Y T Md = ln rd is 10%. The mean of the log of the YTM on a 1-year bond issued at t = 1 maturing in t = 2 rate is: E (ln r1 ) = 0.5 ln ru + 0.5 ln rd = 0.5 (ln ru + ln rd )

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24.5. BLACK-DERMAN-TOY MODEL

285 2

We use the variance formula V ar (X) = E [X − E (X)] to calculate the variance of ln Y T Mu = ln ru and ln Y T Md = ln rd : 0.5 [ln ru − E (ln r1 )]2 + 0.5 [ln rd − E (ln r1 )]2 2 2 = 0.5 (ln ru − 0.5 ln ru − 0.5 ln rd ) + 0.5 (ln rd − 0.5 ln ru − 0.5 ln rd ) 2 2 = 0.5 (0.5 ln ru − 0.5 ln rd ) + 0.5 (0.5 ln rd − 0.5 ln ru ) = 0.5 (0.5 ln ru − 0.5 ln rd )2 + 0.5 (0.5 ln ru − 0.5 ln rd )2 µ ¶2 ru 2 = (0.5 ln ru − 0.5 ln rd ) = 0.5 ln rd The standard deviation of ln Y T Mu = ln ru and ln Y T Md = ln rd is σ 1 = 10%. ru ru = σ1 ln = 2σ 1 ru = rd e2σ1 = rd e0.2 → 0.5 ln rd rd Now we see that matching volatility requires ru = rd e2σ . This relationship holds for every node. Next, we want to reproduce the 2-year bond price of 0.8116. If the Year 2 interest rate is ru , then the PV of $1 discounted from t = 2 to 1 t = 0 is (1 + ru ) (1 + r0 ) If the Year 2 interest rate is rd , then the PV of $1 discounted from t = 2 to 1 t = 0 is (1 + rd ) (1 + r0 ) The risk-neutral probability of up and down is 0.5. The expected 2-year bond price is: 1 1 0.5 × + 0.5 × (1 + ru ) (1 + r0 ) (1 + rd ) (1 + r0 ) To match the observed price P (0, 2) = 0.8116, we have: 1 1 0.5 × + 0.5 × = P (0, 2) (1 + ru ) (1 + r0 ) (1 + rd ) (1 + r0 ) To ⎧ sum up, we have two equations: ⎨ ru = rd e2σ1 1 1 + 0.5 × = P (0, 2) ⎩ 0.5 × (1 + ru ) (1 + r0 ) (1 + rd ) (1 + r0 ) Or ⎧ ⎨ ru = rd e0.2 ⎩ 0.5 ×

1 1 + 0.5 × = 0.8116 (1 + ru ) (1 + 0.1) (1 + rd ) (1 + 0.1)

1 1 + = 2 × 0.8116 (1 + 0.1) 1 + rd 1 + rd e0.2 → rd = 0.108 265 = 10.83%

→

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→ ru = 0.108 265e0.2 = 0.132 2 = 13.22% We can use the same procedure to calculate ruu , rud , rdu , rdd . To simplify our model, we arbitrary set rud = rdu (i.e. we assume the tree is recombining). t=0 t=h=1 t = 2h = 2 ruu = rdd e4σ2 ru = 13.22% r0 = 10% rud = rdu = rdd e2σ2 rd = 10.83% rdd We want to choose rdd such that the standard deviation of the natural log of the YTM on a 2-year bond issued at t = 1 and maturing at t = 3 is 15%. First, we calculate P (1, 3), the price of a 2-year bond issued at t = 1 and maturing at t = 3. If the 2nd year interest rate is ru = 13.22%, then the 3rd year rate is either ruu or rud with equal risk-neutral probability of 0.5. Then the expected PV of $1 discounted from t = 3 to t = 1 is 1 1 + 0.5 × P (1, 3, ru ) = 0.5 × (1 + r ) (1 + r ) (1 + r ) u uu ¶ u (1 + rud ) µ 1 0.5 1 + = 1.1322 1 + rdd e4σ2 1 + rdd e2σ2 The YTM can solved as follows: P (1, 3, ru ) = (1 + i)−2 → i = P (1, 3, ru )−0.5 − 1 If the 2nd year interest rate is rd = 10.83%, then the 3rd year rate is either rdu or rdd with equal risk-neutral probability of 0.5. Then the expected PV of $1 discounted from t = 3 to t = 1 is 1 1 P (1, 3, rd ) = 0.5 × + 0.5 × (1 + rd ) (1 + r¶ (1 + rd ) (1 + rdd ) du ) µ 1 0.5 1 + = 1.1083 1 + rdd e2σ2 1 + rdd The YTM can solved as follows: −2

P (1, 3, rd ) = (1 + i)

→ i = P (1, 3, rd ) −0.5

−0.5

−1

P (1, 3, ru ) −1 → 0.15 = 0.5 ln −0.5 P (1, 3, rd ) −1 ¶¸−0.5 ∙ µ 1 0.5 1 + −1 1.1322 1 + rdd e4σ2 1 + rdd e2σ2 = 0.5 ln ∙ ¶¸−0.5 µ 0.5 1 1 + −1 2σ 2 1.1083 1 + rdd e 1 + rdd By the way, please note that σ 2 is not 10%.

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24.5. BLACK-DERMAN-TOY MODEL

287

We should also match the 3-year bond price P (0, 3) = 0.7118 The PV of $1 at t = 3 discounted to t = 0 is •

1 if the path is 0 → u → uu (1 + r0 ) (1 + ru ) (1 + ruu )

•

1 if the path is 0 → u → ud (1 + r0 ) (1 + ru ) (1 + rud )

•

1 if the path is 0 → d → du (1 + r0 ) (1 + rd ) (1 + rdu )

•

1 if the path is 0 → d → dd (1 + r0 ) (1 + rd ) (1 + rdd )

Each path has a risk neutral probability of 0.25. Now we have: 0.25 0.25 + (1 + r0 ) (1 + ru ) (1 + ruu ) (1 + r0 ) (1 + ru ) (1 + rud ) 0.25 0.25 + + = P (0, 3) (1 + r0 ) (1 + rd ) (1 + rdu ) (1 + r0 ) (1 + rd ) (1 + rdd ) 0.25 0.25 + (1.1) (1.1322) (1 + rdd e4σ ) (1.1) (1.1322) (1 + rdd e2σ ) 0.25 0.25 + + = 0.7118 (1.1) (1.1083) (1 + rdd e2σ ) (1.1) (1.1083) (1 + rdd ) Now our equations are: ¶¸−0.5 µ ∙ 1 1 0.5 + −1 1.1322 1 + rdd e4σ2 1 + rdd e2σ2 0.15 = 0.5 ln ∙ ¶¸−0.5 µ 1 0.5 1 + −1 1.1083 1 + rdd e2σ2 1 + rdd 0.25 0.25 + 4σ 2 (1.1) (1.1322) (1 + rdd e ) (1.1) (1.1322) (1 + rdd e2σ2 ) 0.25 0.25 + + = 0.7118 2σ 2 (1.1) (1.1083) (1 + rdd e ) (1.1) (1.1083) (1 + rdd ) These equations are hard to solve manually. Special software is needed to solve them. σ 2 = 0.195 0 You can verify that the solutions are: rdd = 0.0925 ¶¶−0.5 µ µ 1 1 0.5 + −1 1.1322 1 + 0.0925e4×0.1950 1 + 0.0925e2×0.1950 = 0.149 97 = 0.5 ln µ µ ¶¶−0.5 0.5 1 1 + −1 1.1083 1 + 0.0925e2×0.1950 1 + 0.0925 0.15

Yufeng Guo, Fall 09 MFE, actuary88.com

288

CHAPTER 24. INTEREST RATE MODELS

0.25 0.25 + (1.1) (1.1322) (1 + 0.0925e4×0.1950 ) (1.1) (1.1322) (1 + 0.0925e2×0.1950 ) 0.25 0.25 + + (1.1) (1.1083) (1 + 0.0925e2×0.1950 ) (1.1) (1.1083) (1 + 0.0925) = 0.711 755 = 0.7118

Then the year 3 interest rates are: rdd = 0.0925 rdu = rud = 0.0925e2×0.1950 = 0.136 6 ruu = 0.0925e4×0.1950 = 0.201 8 We can use the same logic and calculate the Year 4 interest rates. However, I’m not going to do the calculation because the calculation is overly intensive. Make sure you can reproduce the Yr 2 and Yr 3 rates. Yr 2 rates can be easily reproduced. SOA or CAS can ask you to calculate the Year 2 rates using the BDT model. Yr 3 rates are harder. A full calculation of Year 3 rates by hand is difficult. However, SOA or CAS can give you some partial information on Year 3 rates and ask you to calculate the rest.

Yufeng Guo, Fall 09 MFE, actuary88.com

Solution to Derivatives Markets: SOA Exam MFE and CAS Exam 3 FE Yufeng Guo July 14, 2009

Contents Introduction

vii

9 Parity and other option relationships

1

10 Binomial option pricing I

23

11 Binomial option pricing II

91

12 Black-Scholes formula

107

13 Market making and delta hedging

125

14 Exotic options: I

149

18 Lognormal distribution

161

19 Monte Carlo simulation

177

20 Brownian motion and Ito’s lemma

187

21 The Black-Scholes equation

193

22 Exotic options: II

203

23 Volatility

205

24 Interest rate models

209

iii

Preface This is Guo’s solution to Derivatives Markets (2nd edition ISBN 0-321-28030-X) for SOA MFE or CAS Exam 3 FE. Unlike the official solution manual published by Addison-Wesley, this solution manual provides solutions to both the evennumbered and odd-numbered problems for the chapters that are on the SOA Exam MFE and CAS Exam 3 FE syllabus. Problems that are out of the scope of the SOA Exam MFE and CAS Exam 3 FE syllabus are excluded. Please report any errors to [email protected]. This book is the exclusive property of Yufeng Guo. Redistribution of this book in any form is prohibited.

v

Introduction Recommendations on using this solution manual: 1. Obviously, you’ll need to buy Derivatives Markets (2nd edition) to see the problems. 2. Make sure you download the textbook errata from http://www.kellogg. northwestern.edu/faculty/mcdonald/htm/typos2e_01.html

vii

Chapter 9

Parity and other option relationships Problem 9.1. S0 = 32 C = 2.27

T = 6/12 = 0.5 K = 35 r = 0.04 δ = 0.06

C + P V (K) = P + S0 e−δT 2.27 + 35e−0.04(0.5) = P + 32e−0.06(0.5)

P = 5. 522 7

Problem 9.2. S0 = 32 T = 6/12 = 0.5 K = 30 C = 4.29 P = 2.64 r = 0.04 C + P V (K) = P + S0 − P V (Div) 4.29 + 30e−0.04(0.5) = 2.64 + 32 − P V (Div) P V (Div) = 0.944 Problem 9.3. S0 = 800 r = 0.05 δ=0 T =1 K = 815 C = 75 P = 45 a. Buy stock+ sell call+buy put=buy P V (K) C + P V (K) = P + S0 → P V (K = 815) = S0 + −C + |{z} P = 800 + (−75) + 45 = 770 |{z} |{z} buy stock

sell call

buy put

So the position is equivalent to depositing 770 in a savings account (or buying a bond with present value equal to 770) and receiving 815 one year later. 770eR = 815 R = 0.056 8 1

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS So we earn 5.68%. b. Buying a stock, selling a call, and buying a put is the same as depositing P V (K) in the savings account. As a result, we should just earn the risk free interest rate r = 0.05. However, we actually earn R = 0.056 8 > r. To arbitrage, we "borrow low and earn high." We borrow 770 from a bank at 0.05%. We use the borrowed 770 to finance buying a stock, selling a call, and buying a put. Notice that the net cost of buying a stock, selling a call, and buying a put is 770. One year later, we receive 770eR = 815. We pay the bank 770e0.05 = 809. 48. Our profit is 815 − 809. 48 = 5. 52 per transaction. If we do n such transactions, we’ll earn 5. 52n profit. Alternative answer: we can burrow at 5% (continuously compounding) and lend at 5.6 8% (continuously compounding), earning a risk free 0.68%. So if we borrow $1 at time zero, our risk free profit at time one is e0.0568 − e0.05 = 0.00717 3; if we borrow $770 at time zero, our risk free profit at time one is 0.00717 3×770 = 5. 52. If we borrow n dollars at time zero, we’ll earn 0.00717 3n dollars at time one. c. To avoid arbitrage, we need to have: P V (K = 815) = S0 + |{z} −C + |{z} P = 815e−0.05 = 775. 25 |{z} buy stock

sell call

buy put

→ C − P = S0 − P V (K) = 800 − 775. 25 = 24. 75

d. C − P = S0 − P V (K) = 800 − Ke−rT = 800 − Ke−0.05 If K = 780 C − P = 800 − 780e−0.05 = 58. 041 If K = 800 C − P = 800 − 800e−0.05 = 39. 016 If K = 820 C − P = 800 − 820e−0.05 = 19. 992 If K = 840 C − P = 800 − 840e−0.05 = 0.967 Problem 9.4. To solve this type of problems, just use the standard put-call parity. To avoid calculation errors, clearly identify the underlying asset. The underlying asset is €1. We want to find the dollar cost of a put option on this underlying. The typical put-call parity: C + P V (K) = P + S0 e−δT C, K, P , and S0 should all be expressed in dollars. S0 is the current (dollar price) of the underlying. So S0 = $0.95. C = $0.0571 K = $0.93 δ is the internal growth rate of the underlying asset (i.e. €1). Hence δ = 0.04 Since K is expressed in dollars, P V (K) needs to be calculated using the dollar risk free interest r = 0.06. 0.0571 + 0.93e−0.06(1) = P + 0.95e−0.04(1) P = $0.02 02 www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Problem 9.5. As I explained in my study guide, don’t bother memorizing the following complex formula: µ ¶ 1 1 C$ (x0 , K, T ) = x0 KPf , ,T x0 K Just use my approach to solve this type of problems. Convert information to symbols: 1 The exchange rate is 95 yen per euro. Y 95 =€1 or Y 1 =€ 95 Yen-denominated put on 1 euro with strike price Y100 has a premium Y8.763 → (€1 → Y 100)0 =Y8.763 What’s the strike price of a euro-denominated call on 1 yen? €K → 1Y Calculate the price of a euro-denominated call on 1 yen with strike price €K (€K → 1Y )0 = €? 1 →Y1 100 1 The strike price of the corresponding euro-denominated yen call is K =€ =€0.01 100 ¶ µ 1 1 1 →Y1 = × (€1 → Y 100)0 = (Y 8.763) € 100 100 100 0

€1 → Y 100

→

€

1 , we have: 95 µ ¶ 1 1 1 (Y 8.763) = (8.763) € =€9. 224 2 × 10−4 100µ 100 95 ¶ 1 → € → Y 1 =€9. 224 2 × 10−4 100 0

Since Y 1 =€

So the price of a euro-denominated call on 1 yen with strike price K =€ is €9. 224 2 × 10−4

www.actuary88.com

c °Yufeng Guo

1 100

3

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Problem 9.6. The underlying asset is €1. The standard put-call parity is: C + P V (K) = P + S0 e−δT C, K, P , and S0 should all be expressed in dollars. S0 is the current (dollar price) of the underlying. δ is the internal growth rate of the underlying asset (i.e. €1). We’ll solve Part b first. b. 0.0404 + 0.9e−0.05(0.5) = 0.0141 + S0 e−0.035(0.5) S0 = $0.920 04 So the current price of the underlying (i.e. €1) is S0 = $0.920 04. In other words, the currency exchange rate is $0.920 04 =€1 a. According to the textbook Equation 5.7, the forward price is: F0,T = S0 e−δT erT = 0.920 04e−0.035(0.5) e0.05(0.5) = $0.926 97 Problem 9.7. The underlying asset is one yen. a. C + Ke−rT = P + S0 e−δT 0.0006 + 0.009e−0.05(1) = P + 0.009e−0.01(1) 0.0006 + 0.008561 = P + 0.008 91 P = $0.00025

b. There are two puts out there. One is the synthetically created put using the formula: P = C + Ke−rT − S0 e−δT The other is the put in the market selling for the price for $0.0004. To arbitrage, build a put a low cost and sell it at a high price. At t = 0, we: • Sell the expensive put for $0.0004 • Build a cheap put for $0.00025. To build a put, we buy a call, deposit Ke−rT in a savings account, and sell e−δT unit of Yen.

Sell expensive put Buy call Deposit Ke−rT in savings Short sell e−δT unit of Yen Total

t=0 0.0004 −0.0006 −0.009e−0.05(1) 0.009e−0.01(1) $0.00015

T =1 ST < 0.009 ST − 0.009 0 0.009 ST 0

T =1 ST ≥ 0.009 0 ST − 0.009 0.009 ST 0

0.0004 − 0.0006 − 0.009e−0.05(1) + 0.009e−0.01(1) = $0.00015 www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS At t = 0, we receive $0.00015 yet we don’t incur any liabilities at T = 1 (so we receive $0.00015 free money at t = 0). c. At-the-money means K = S0 (i.e. the strike price is equal to the current exchange rate). Dollar-denominated at-the-money yen call sells for $0.0006. To translate this into symbols, notice that under the call option, the call holder can give $0.009 and get Y 1. "Give $0.009 and get Y 1" is represented by ($0.009 → Y 1). This option’s premium at time zero is $0.0006. Hence we have: ($0.009 → Y 1)0 = $0.0006 We are asked to find the yen denominated at the money call for $1. Here the call holder can give c yen and get $1. "Give c yen and get $1" is represented by (Y c → $1). This option’s premium at time zero is (Y c → $1)0 . First, we need to calculate c, the strike price of the yen denominated dollar 1 call. Since at time zero $0.009 = Y 1, we have $1 = Y . So the at-the0.009 1 money yen denominated call on $1 is c = . Our task is to find this option’s 0.009 ¶ µ 1 → $1 =? premium: Y 0.009 0 We’ll find the premium for Y 1 →$0.009, the option of "give 1 yen and get $0.009." Once we find this premium, we’ll scale it and find the premium of "give 1 yen and get $1." 0.009 We’ll use the general put-call parity: (AT → BT )0 + P V (AT ) = (BT → AT )0 + P V (BT ) ($0.009 → Y 1)0 + P V ($0.009) = (Y 1 → $0.009)0 + P V (Y 1) P V ($0.009) = $0.009e−0.05(1) Since we are discounting $0.009 at T = 1 to time zero, we use the dollar interest rate 5%. P V (Y 1) = $0.009e−0.01(1) If we discount Y1 from T = 1 to time zero, we get e−0.01(1) yen, which is equal to $0.009e−0.01(1) . So we have: $0.0006+$0.009e−0.05 = (Y 1 → $0.009)0 + $0.009e−0.01(1) (Y = $2. 506 16 × 10−4 µ 1 → $0.009)0 ¶ 1 1 2. 506 16 × 10−4 = Y 1 → $1 (Y 1 → $0.009)0 = $ = $2. 0.009 0.009 0.009 0 −2 2. 784 62 × 10 = Y 3. 094 784 62 × 10−2 = Y 0.009 www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS So the yen denominated at the money call for $1 is worth $2. 784 62 × 10−2 or Y 3. 094. We are also asked to identify the relationship between the yen denominated at the money call for $1 and the dollar-denominated yen put. The relationship is that we use the premium of the latter option to calculate the premium of the former option. Next, we calculate the premium for the yen denominated at-the-money put for $1: µ $→Y

1 0.009

¶

=

0

1 ($0.009 → Y 1)0 0.009

1 = × $0.0006 = $ 0.0 666 7 0.009 1 = Y 7. 407 8 = Y 0.0 666 7 × 0.009 So the yen denominated at-the-money put for $1 is worth $ 0.0 666 7 or Y 7. 407 8. I recommend that you use my solution approach, which is less prone to errors than using complex notations and formulas in the textbook. Problem 9.8. The textbook Equations 9.13 and 9.14 are violated. This is how to arbitrage on the calls. We have two otherwise identical calls, one with $50 strike price and the other $55. The $50 strike call is more valuable than the $55 strike call, but the former is selling less than the latter. To arbitrage, buy low and sell high. We use T to represent the common exercise date. This definition works whether the two options are American or European. If the two options are American, we’ll find arbitrage opportunities if two American options are exercised simultaneously. If the two options are European, T is the common expiration date. The payoff is: Transaction Buy 50 strike call Sell 55 strike call Total

t=0 −9 10 1

T ST < 50 0 0 0

T 50 ≤ ST < 55 ST − 50 0 ST − 50 ≥ 0

T ST ≥ 55 ST − 50 − (ST − 55) 5

At t = 0, we receive $1 free money. At T , we get non negative cash flows (so we may get some free money, but we certainly don’t owe anybody anything at T ). This is clearly an arbitrage. www.actuary88.com

c °Yufeng Guo

6

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS This is how to arbitrage on the two puts. We have two otherwise identical puts, one with $50 strike price and the other $55. The $55 strike put is more valuable than the $50 strike put, but the former is selling less than the latter. To arbitrage, buy low and sell high. The payoff is: T T T Transaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55 Buy 55 strike put −6 55 − ST 55 − ST 0 Sell 50 strike put 7 − (50 − ST ) 0 0 Total 1 5 55 − ST > 0 0 At t = 0, we receive $1 free money. At T , we get non negative cash flows (so we may get some free money, but we certainly don’t owe anybody anything at T ). This is clearly an arbitrage. Problem 9.9. The textbook Equation 9.15 and 9.16 are violated. We use T to represent the common exercise date. This definition works whether the two options are American or European. If the two options are American, we’ll find arbitrage opportunities if two American options are exercised simultaneously. If the two options are European, T is the common expiration date. This is how to arbitrage on the calls. We have two otherwise identical calls, one with $50 strike price and the other $55. The premium difference between these two options should not exceed the strike difference 15 − 10 = 5. In other words, the 50-strike call should sell no more than 10 + 5. However, the 50-strike call is currently selling for 16 in the market. To arbitrage, buy low (the 55-strike call) and sell high (the 50-strike call). The $50 strike call is more valuable than the $55 strike call, is selling less than the latter. The payoff is: T T Transaction t = 0 ST < 50 50 ≤ ST < 55 Buy 55 strike call −10 0 0 Sell 50 strike call 16 0 − (ST − 50) Total 6 0 − (ST − 50) ≥ −5

but the former

T ST ≥ 55 ST − 55 − (ST − 50) −5

So we receive $6 at t = 0. Then at T , our maximum liability is $5. So make at least $1 free money. This is how to arbitrage on the puts. We have two otherwise identical calls, one with $50 strike price and the other $55. The premium difference between these two options should not exceed the strike difference 15 − 10 = 5. In other www.actuary88.com

c °Yufeng Guo

7

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS words, the 55-strike put should sell no more than 7 + 5 = 12. However, the 55-strike put is currently selling for 14 in the market. To arbitrage, buy low (the 50-strike put) and sell high (the 55-strike put). The payoff is: T T T Transaction t = 0 ST < 50 50 ≤ ST < 55 ST ≥ 55 Buy 50 strike put −14 50 − ST 0 0 Sell 55 strike put 7 − (55 − ST ) − (55 − ST ) 0 Total 7 −5 − (55 − ST ) < −5 0 So we receive $7 at t = 0. Then at T , our maximum liability is $5. So make at least $2 free money. Problem 9.10. Suppose there are 3 options otherwise identical but with different strike price K1 < K2 < K3 where K2 = λK1 + (1 − λ) K2 and 0 < λ < 1. Then the price of the middle strike price K2 must not exceed the price of a diversified portfolio consisting of λ units of K1 -strike option and (1 − λ) units of K2 -strike option: C [λK1 + (1 − λ) K3 ] ≤ λC (K1 ) + (1 − λ) C (K3 ) P [λK1 + (1 − λ) K3 ] ≤ λP (K1 ) + (1 − λ) P (K3 ) The above conditions are called the convexity of the option price with respect to the strike price. They are equivalent to the textbook Equation 9.17 and 9.18. If the above conditions are violated, arbitrage opportunities exist. We are given the following 3 calls: Strike K1 = 50 K2 = 55 Call premium 18 14

K3 = 60 9.50

λ50 + (1 − λ) 60 = 55 → λ = 0.5 0.5 (50) + 0.5 (60) = 55 Let’s check: C [0.5 (50) + 0.5 (60)] = C (55) = 14 0.5C (50) + 0.5C (60) = 0.5 (18) + 0.5 (9.50) = 13. 75 C [0.5 (50) + 0.5 (60)] > 0.5C (50) + 0.5C (60) So arbitrage opportunities exist. To arbitrage, we buy low and sell high. The cheap asset is the diversified portfolio consisting of λ units of K1 -strike option and (1 − λ) units of K3 -strike option. In this problem, the diversified portfolio consists of half a 50-strike call and half a 60-strike call. The expensive asset is the 55-strike call. www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Since we can’t buy half a call option, we’ll buy 2 units of the portfolio (i.e. buy one 50-strike call and one 60-strike call). Simultaneously,we sell two 55strike call options. We use T to represent the common exercise date. This definition works whether the options are American or European. If the options are American, we’ll find arbitrage opportunities if the American options are exercised simultaneously. If the options are European, T is the common expiration date. The payoff is: T ST < 50

T 50 ≤ ST < 55

T 55 ≤ ST < 60

T ST ≥ 60

−18 −9.5 −27. 5

0 0 0

ST − 50 0 ST − 50

ST − 50 0 ST − 50

ST − 50 ST − 60 2ST − 110

2 (14) = 28 0.5

0 0

0 ST − 50 ≥ 0

−2 (ST − 55) 60 − ST > 0

−2 (ST − 55) 0

Transaction buy two portfolios buy a 50-strike call buy a 60-strike call Portfolio total

t=0

Sell two 55-strike calls Total

−27. 5 + 28 = 0.5 ST − 50 − 2 (ST − 55) = 60 − ST 2ST − 110 − 2 (ST − 55) = 0 So we get $0.5 at t = 0, yet we have non negative cash flows at the expiration date T . This is arbitrage. The above strategy of buying λ units of K1 -strike call, buying (1 − λ) units of K3 -strike call, and selling one unit of K2 -strike call is called the butterfly spread. We are given the following 3 puts: Strike K1 = 50 K2 = 55 Put premium 7 10.75

K3 = 60 14.45

λ50 + (1 − λ) 60 = 55 → λ = 0.5 0.5 (50) + 0.5 (60) = 55 Let’s check: P [0.5 (50) + 0.5 (60)] = P (55) = 10.75 0.5P (50) + 0.5P (60) = 0.5 (7) + 0.5 (14.45) = 10. 725 P [0.5 (50) + 0.5 (60)] > .5P (50) + 0.5P (60) So arbitrage opportunities exist. To arbitrage, we buy low and sell high. The cheap asset is the diversified portfolio consisting of λ units of K1 -strike put and (1 − λ) units of K3 -strike put. In this problem, the diversified portfolio consists of half a 50-strike put and half a 60-strike put. The expensive asset is the 55-strike put. www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Since we can’t buy half a option, we’ll buy 2 units of the portfolio (i.e. buy one 50-strike put and one 60-strike put). Simultaneously,we sell two 55-strike put options. The payoff is:

Transaction buy two portfolios buy a 50-strike put buy a 60-strike put Portfolio total

t=0

T ST < 50

T 50 ≤ ST < 55

T 55 ≤ ST < 60

T ST ≥ 60

−7 −14.45 −21. 45

50 − ST 60 − ST 110 − 2ST

0 60 − ST 60 − ST

0 60 − ST 60 − ST

0 0 0

−2 (55 − ST ) 0

−2 (55 − ST ) ST − 50 ≥ 0

0 60 − ST > 0

0 0

Sell two 55-strike puts 2 (10.75) Total 0.05 −21. 45 + 2 (10.75) = 0.05 50 − ST + 60 − ST = 110 − 2ST

−21. 45 + 2 (10.75) = 0.05 110 − 2ST − 2 (55 − ST ) = 0 60 − ST − 2 (55 − ST ) = ST − 50 So we get $0.05 at t = 0, yet we have non negative cash flows at the expiration date T . This is arbitrage. The above strategy of buying λ units of K1 -strike put, buying (1 − λ) units of K3 -strike put, and selling one unit of K2 -strike put is also called the butterfly spread.

Problem 9.11. This is similar to Problem 9.10. We are given the following 3 calls: Strike K1 = 80 K2 = 100 Call premium 22 9

K3 = 105 5

80λ + 105 (1 − λ) = 100 → λ = 0.2 0.2 (80) + 0.8 (105) = 100 C [0.2 (80) + 0.8 (105)] = C (100) = 9 0.2C (80) + 0.8C (105) = 0.2 (22) + 0.8 (5) = 8. 4 C [0.2 (80) + 0.8 (105)] > 0.2C (80) + 0.8C (105) So arbitrage opportunities exist. To arbitrage, we buy low and sell high. The cheap asset is the diversified portfolio consisting of λ units of K1 -strike option and (1 − λ) units of K3 -strike option. In this problem, the diversified portfolio consists of 0.2 unit of 80-strike call and 0.8 unit of 105-strike call. The expensive asset is the 100-strike call. www.actuary88.com

c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS Since we can’t buy a fraction of a call option, we’ll buy 10 units of the portfolio (i.e. buy two 80-strike calls and eight 105-strike calls). Simultaneously,we sell ten 100-strike call options. We use T to represent the common exercise date. This definition works whether the options are American or European. If the options are American, we’ll find arbitrage opportunities if the American options are exercised simultaneously. If the options are European, T is the common expiration date. The payoff is: T ST < 80

T 80 ≤ ST < 100

Transaction buy ten portfolios buy two 80-strike calls buy eight 105-strike calls Portfolio total

t=0 −2 (22) −8 (5) −84

0 0 0

2 (ST − 80) 0 2 (ST − 80)

Sell ten 100-strike calls Total

10 (9) 6

0 0

0 2 (ST − 80) ≥ 0

Transaction buy ten portfolios buy two 80-strike calls buy eight 105-strike calls Portfolio total

t=0

T 100 ≤ ST < 105

T ST ≥ 105

−2 (22) −8 (5) −84

2 (ST − 80) 0 2 (ST − 80)

2 (ST − 80) 8 (ST − 105) 10ST − 1000

−10 (ST − 100) 8 (105 − ST ) > 0

−10 (ST − 100) 0

Sell ten 100-strike calls 10 (9) Total 6 −2 (22) − 8 (5) = −44 − 40 = −84

−84 + 10 (9) = −84 + 90 = 6 2 (ST − 80) + 8 (ST − 105) = 10ST − 1000 2 (ST − 80) − 10 (ST − 100) = 840 − 8ST = 8 (105 − ST ) 10ST − 1000 − 10 (ST − 100) = 0 So we receive $6 at t = 0, yet we don’t incur any negative cash flows at expiration T . So we make at least $6 free money. We are given the following 3 put: Strike K1 = 80 K2 = 100 Put premium 4 21

K3 = 105 24.8

80λ + 105 (1 − λ) = 100 → λ = 0.2 0.2 (80) + 0.8 (105) = 100 P [0.2 (80) + 0.8 (105)] = P (100) = 21 0.2P (80) + 0.8P (105) = 0.2 (4) + 0.8 (24.8) = 20. 64 www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS P [0.2 (80) + 0.8 (105)] > 0.2P (80) + 0.8P (105) So arbitrage opportunities exist. To arbitrage, we buy low and sell high. The cheap asset is the diversified portfolio consisting of λ units of K1 -strike option and (1 − λ) units of K3 -strike option. In this problem, the diversified portfolio consists of 0.2 unit of 80-strike put and 0.8 unit of 105-strike put. The expensive asset is the 100-strike put. Since we can’t buy half a fraction of an option, we’ll buy 10 units of the portfolio (i.e. buy two 80-strike puts and eight 105-strike puts). Simultaneously,we sell ten 100-strike put options. The payoff is:

Transaction buy ten portfolios buy two 80-strike puts buy eight 105-strike puts Portfolio total

t=0

T ST < 80

T 80 ≤ ST < 100

−2 (4) −8 (24.8) −84

2 (80 − ST ) 8 (105 − ST ) 1000 − 10ST

0 8 (105 − ST ) 8 (105 − ST )

Sell ten 100-strike puts Total

10 (21) 3. 6

−10 (100 − ST ) 0

−10 (100 − ST ) 2 (ST − 80) ≥ 0

Transaction buy ten portfolios buy two 80-strike puts buy eight 105-strike puts Portfolio total

t=0

T 100 ≤ ST < 105

T ST ≥ 105

−2 (4) −8 (24.8) −84

0 8 (105 − ST ) 8 (105 − ST )

0 0 0

Sell ten 100-strike puts Total

10 (21) 3. 6

0 8 (105 − ST ) > 0

0 0

−2 (4) − 8 (24.8) = −206. 4 2 (80 − ST ) + 8 (105 − ST ) = 1000 − 10ST −206. 4 + 10 (21) = 3. 6 1000 − 10ST − 10 (100 − ST ) = 0 8 (105 − ST ) − 10 (100 − ST ) = 2 (ST − 80) We receive $3. 6 at t = 0, but we don’t incur any negative cash flows at T . So we make at least $3. 6 free money. Problem 9.12. For two European options differing only in strike price, the following conditions must be met to avoid arbitrage (see my study guide for explanation): 0 ≤ CEur (K1 , T ) − CEur (K2 , T ) ≤ P V (K2 − K1 ) if K1 < K2 www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS 0 ≤ PEur (K2 , T ) − PEur (K1 , T ) ≤ P V (K2 − K1 ) if K1 < K2 a. Strike Call premium

K1 = 90 10

K2 = 95 4

C (K1 ) − C (K2 ) = 10 − 4 = 6 K2 − K1 = 95 − 90 = 5 C (K1 ) − C (K2 ) > K2 − K1 ≥ P V (K2 − K1 ) Arbitrage opportunities exist. To arbitrage, we buy low and sell high. The cheap call is the 95-strike call; the expensive call is the 90-strike call. We use T to represent the common exercise date. This definition works whether the two options are American or European. If the two options are American, we’ll find arbitrage opportunities if two American options are exercised simultaneously. If the two options are European, T is the common expiration date. The payoff is: T T T Transaction t = 0 ST < 90 90 ≤ ST < 95 ST ≥ 95 Buy 95 strike call −4 0 0 ST − 95 Sell 90 strike call 10 0 − (ST − 90) − (ST − 90) Total 6 0 − (ST − 90) ≥ −5 −5 We receive $6 at t = 0, yet we our max liability at T is −5. So we’ll make at least $1 free money. b. T =2 r = 0.1 Strike K1 = 90 Call premium 10

K2 = 95 5.25

C (K1 ) − C (K2 ) = 10 − 5.25 = 4. 75 K2 − K1 = 95 − 90 = 5 P V (K2 − K1 ) = 5e−0.1(2) = 4. 094 C (K1 ) − C (K2 ) > P V (K2 − K1 ) Arbitrage opportunities exist. Once again, we buy low and sell high. The cheap call is the 95-strike call; the expensive call is the 90-strike call. The payoff is: T T T Transaction t=0 ST < 90 90 ≤ ST < 95 ST ≥ 95 Buy 95 strike call −5.25 0 0 ST − 95 Sell 90 strike call 10 0 − (ST − 90) − (ST − 90) 0.1(2) 0.1(2) Deposit 4. 75 in savings −4. 75 4. 75e 4. 75e 4. 75e0.1(2) Total 0 5. 80 95. 80 − ST > 0 0.80 www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS 4. 75e0.1(2) = 5. 80 − (ST − 90) + 4. 75e0.1(2) = 95. 80 − ST ST − 95 − (ST − 90) + 4. 75e0.1(2) = 0.80 Our initial cost is zero. However, our payoff is always non-negative. So we never lose money. This is clearly an arbitrage. It’s important that the two calls are European options. If they are American, they can be exercised at different dates. Hence the following non-arbitrage conditions work only for European options: 0 ≤ CEur (K1 , T ) − CEur (K2 , T ) ≤ P V (K2 − K1 ) if K1 < K2 0 ≤ PEur (K2 , T ) − PEur (K1 , T ) ≤ P V (K2 − K1 ) if K1 < K2 c. We are given the following 3 calls: Strike K1 = 90 K2 = 100 Call premium 15 10

K3 = 105 6

1 λ90 + (1 − λ) 105 = 100 λ= 3 1 2 → (90) + (105) = 100 3 ¸ ∙3 2 1 (90) + (105) = C (100) = 10 C 3 3 2 1 2 1 C (90) + C (105) = (15) + (6) = 9 3∙ 3 3 ¸ 3 2 1 2 1 (90) + (105) > C (90) + C (105) C 3 3 3 3 Hence arbitrage opportunities exist. To arbitrage, we buy low and sell high. 1 The cheap asset is the diversified portfolio consisting of unit of 90-strike 3 2 call and unit of 105-strike call. 3 The expensive asset is the 100-strike call. Since we can’t buy a partial option, we’ll buy 3 units of the portfolio (i.e. buy one 90-strike call and two 105-strike calls). Simultaneously,we sell three 100-strike calls. The payoff at expiration T : T ST < 90

T 90 ≤ ST < 100

T 100 ≤ ST < 105

T ST ≥ 105

−15 2 (−6) −27

0 0 0

ST − 90 0 ST − 90

ST − 90 0 ST − 90

ST − 90 2 (ST − 105) 3ST − 300

3 (10) 3

0 0

0 ST − 90 ≥ 0

−3 (ST − 100) 2 (105 − ST ) > 0

−3 (ST − 100) 0

Transaction buy 3 portfolios buy one 90-strike call buy two 105-strike calls Portfolios total

t=0

Sell three 100-strike calls Total www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS −15 + 2 (−6) = −27 ST − 90 + 2 (ST − 105) = 3ST − 300 −27 + 3 (10) = 3 ST − 90 − 3 (ST − 100) = 210 − 2ST = 2 (105 − ST ) 3ST − 300 − 3 (ST − 100) = 0 So we receive $3 at t = 0, but we incur no negative payoff at T . So we’ll make at least $3 free money. Problem 9.13. a. If the stock pays dividend, then early exercise of an American call option may be optimal. Suppose the stock pays dividend at tD . Time 0 ... ... tD ... ... T Pro and con for exercising the call early at tD . • +. If you exercise the call immediately before tD , you’ll receive dividend and earn interest during [tD , T ] • −. You’ll pay the strike price K at tD , losing interest you could have earned during [tD , T ]. If the interest rate, however, is zero, you won’t lose any interest. • −. You throw away the remaining call option during [tD , T ]. Had you waited, you would have the call option during [tD , T ] If the accumulated value of the dividend exceeds the value of the remaining call option, then it’s optimal to exercise the stock at tD . As explained in my study guide, it’s never optimal to exercise an American put early if the interest rate is zero. Problem 9.14. a. The only reason that early exercise might be optimal is that the underlying asset pays a dividend. If the underlying asset doesn’t pay dividend, then it’s never optimal to exercise an American call early. Since Apple doesn’t pay dividend, it’s never optimal to exercise early. b. The only reason to exercise an American put early is to earn interest on the strike price. The strike price in this example is one share of AOL stock. Since AOL stocks won’t pay any dividends, there’s no benefit for owning an AOL stock early. Thus it’s never optimal to exercise the put. If the Apple stock price goes to zero and will always stay zero, then there’s no benefit for delaying exercising the put; there’s no benefit for exercising the www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS put early either (since AOL stocks won’t pay dividend). Exercising the put early and exercising the put at maturity have the same value. If, however, the Apple stock price goes to zero now but may go up in the future, then it’s never optimal to exercise the put early. If you don’t exercise early, you leave the door open that in the future the Apple stock price may exceed the AOL stock price, in which case you just let your put expire worthless. If the Apple stock price won’t exceed the AOL stock price, you can always exercise the put and exchange one Apple stock for one AOL stock. There’s no hurry to exercise the put early. c. If Apple is expected to pay dividend, then it might be optimal to exercise the American call early and exchange one AOL stock for one Apple stock. However, as long as the AOL stock won’t pay any dividend, it’s never optimal to exercise the American put early to exchange one Apple stock for one AOL stock. Problem 9.15. This is an example where the strike price grows over time. If the strike price grows over time, the longer-lived valuable as the shorter lived option. Refer to Derivatives We have two European calls: Call #1 K1 = 100e0.05(1.5) = 107. 788 T1 = 1.5 Call #2 K2 = 100e0.05 = 105. 127 T2 = 1

option is at least as Markets Page 298. C1 = 11.50 C2 = 11.924

The longer-lived call is cheaper than the shorter-lived call, leading to arbitrage opportunities. To arbitrage, we buy low (Call #1) and sell high (Call #2). The payoff at expiration T1 = 1.5 if ST2 < 100e0.05 = 105. 127 Transaction Sell Call #2 buy Call #1 Total

t=0 11.924 −11.50 0.424

T2 0

T1 ST1 < 100e0.05(1.5) 0 0 0

T1 ST1 ≥ 100e0.05(1.5) 0 ST1 − 100e0.05(1.5) ST1 − 100e0.05(1.5) ≥ 0

We receive $0.424 at t = 0, yet our payoff at T1 is always non-negative. This is clearly an arbitrage. The payoff at expiration T1 = 1.5 if ST2 ≥ 100e0.05 = 105. 127 Transaction Sell Call #2 buy Call #1 Total www.actuary88.com

t=0 11.924 −11.50 0.424

T2 100e0.05 − ST2

c °Yufeng Guo

T1 ST1 < 100e0.05(1.5) 100e0.05(1.5) − ST1 0 100e0.05(1.5) − ST1 < 0

T1 ST1 ≥ 100e0.05(1.5) 100e0.05(1.5) − ST1 ST1 − 100e0.05(1.5) 0 16

CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS If ST2 ≥ 100e0.05 , then payoff of the sold Call #2 at T2 is 100e0.05 − ST2 . From T2 to T1 , ¡ ¡ ¢ ¢ • 100e0.05 grows into 100e0.05 e0.05(T1 −T2 ) = 100e0.05 e0.05(0.5) = 100e0.05(1.5) • ST2 becomes ST1 (i.e. the stock price changes from ST2 to ST1 )

We receive $0.424 at t = 0, yet our payoff at T1 can be negative. This is not an arbitrage. So as long as ST2 < 100e0.05 = 105. 127 , there’ll be arbitrage opportunities. Problem 9.16. Suppose we do the following at t = 0: 1. Pay C a to buy a call 2. Lend P V (K) = Ke−rL at rL 3. Sell a put, receiving P b 4. Short sell one stock, receiving S0b The net cost is P b + S0b − (C a + Ke−rL ). The payoff at T is: Transactions Buy a call Lend Ke−rL at rL Sell a put Short sell one stock Total

t=0 −C a −KerL Pb S0b P b + S0b − (C a + Ke−rL )

If ST < K

If ST ≥ K

0 K ST − K −ST 0

ST − K K 0 −ST 0

The payoff is always zero. To avoid arbitrage, we need to have P b + S0b − (C a + Ke−rL ) ≤ 0 Similarly, we can do the following at t = 0: 1. Sell a call, receiving C b 2. Borrow P V (K) = Ke−rB at rB 3. Buy a put, paying P a 4. Buy one stock, paying S0a www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS ¡ ¢ ¡ ¢ The net cost is C b + Ke−rB − P b + S0b . The payoff at T is: Transactions Sell a call Borrow Ke−rB at rB Buy a put Buy one stock Total

t=0 Cb Ke−rB −P a a −S ¡ 0b ¢ ¡ ¢ C + Ke−rB − P b + S0b

If ST < K

If ST ≥ K

0 −K K− ST ST 0

K − ST −K 0 ST 0

The ¡ b payoff−ris ¢always ¡ zero. ¢To avoid arbitrage, we need to have C + Ke B − P b + S0b ≤ 0 Problem 9.17. a. According to the put-call parity, the payoff of the following position is always zero: 1. Buy the call 2. Sell the put 3. Short the stock 4. Lend the present value of the strike price plus dividend The existence of the bid-ask spread and the borrowing-lending rate difference doesn’t change the zero payoff of the above position. The above position always has a zero payoff whether there’s a bid-ask spread or a difference between the borrowing rate and the lending rate. If there is no transaction cost such as a bid-ask spread, the initial gain of the above position is zero. However, if there is a bid-ask spread, then to avoid arbitrage, the initial gain of the above position should be zero or negative. The ¡ b initial ¢ gain of the position is: P + S0b − [C a + P VrL (K) + P VrL (Div)] There’s no ¡ b ¢ arbitrage if P + S0b − [C a + P VrL (K) + P VrL (Div)] ≤ 0 In this problem, we are given • rL = 0.019 • rB = 0.02 • S0b = 84.85. We are told to ignore the transaction cost. In addition, we are given that the current stock price is 84.85. So S0b = 84.85. • The dividend is 0.18 on November 8, 2004. www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS To find the expiration date, you need to know this detail. Puts and calls are called equity options at the Chicago Board of Exchange (CBOE). In CBOE, the expiration date of an equity option is the Saturday immediately following the third Friday of the expiration month. (To verify this, go to www.cboe.com. Click on "Products" and read "Production Specifications.") If the expiration month is November, 2004, the third Friday is November 19. Then the expiration date is November 20. 11/20/2004 − 10/15/2004 36 T = = = 0.09863 365 365 If the expiration month is January, 2005, the third Friday is January 21. Then the expiration date is January 22, 2005. 1/22/2005 − 10/15/2004 T = 365 Calculate the days between 1/22/2005 and10/15/2004 isn’t easy. Fortunately,we can use a calculator. BA II Plus and BA II Plus Professional have "Date" Worksheet. When using Date Worksheet, use the ACT mode. ACT mode calculates the actual days between two dates. If you use the 360 day mode, you are assuming that there are 360 days between two dates. When using the date worksheet, set DT1 (i.e. Date 1) as October 10, 2004 by entering 10.1504; set DT2 (i.e. Date 2) as January 22, 2004 by entering 1.2204. The calculator should tell you that DBD=99 (i.e. the days between two days is 99 days). 1/22/2005 − 10/15/2004 99 So T = = = 0.271 23 365 365 If you have trouble using the date worksheet, refer to the guidebook of BA II Plus or BA II Plus Professional. 11/8/2004 − 10/15/2004 24 = = 0.06 575 The dividend day is tD = 365 365 P VrL (Div) = 0.18e−0.06575(0.019) = 0.18 −0.019T P ¡ VbrL (K) ¢ = Kea b P + S0 − [C ¡+ P VrL (K) + P VrL (Div)] ¢ = P b + 84.85 − C a + Ke−0.019T + 0.18 ¡ b ¢ a K T Ca Pb + P VrL (Div)] ¢ P + S0b − [C ¡ + P VrL (K) 75 0.0986 10.3 0.2 0.2 + 84.85 − ¡10.3 + 75e−0.019×0.0986 + 0.18 ¢ = −0.29 −0.019×0.0986 80 0.0986 5.6 0.6 0.6 + 84.85 − ¡5.6 + 80e + 0.18¢ = −0.18 85 0.0986 2.1 2.1 2.1 + 84.85 − ¡2.1 + 85e−0.019×0.0986 + 0.18 ¢= −0.17 90 0.0986 0.35 5.5 5.5 + 84.85 − ¡0.35 + 90e−0.019×0.0986 + 0.18 ¢ = −1. 15 −0.019×0.271 2 75 0.271 2 10.9 0.7 0.7 + 84.85 − 10.9 ¡ + 75e−0.019×0.271 2 + 0.18¢ = −0.14 80 0.271 2 6.7 1.45 1.45 + 84.85 −¡ 6.7 + 80e + 0.18 ¢ = −0.17 85 0.271 2 3.4 3.1 3.1 + 84.85 − ¡3.4 + 85e−0.019×0.271 2 + 0.18 ¢= −0.19 90 0.271 2 1.35 6.1 6.1 + 84.85 − 1.35 + 90e−0.019×0.271 2 + 0.18 = −0.12

b. According to the put-call parity, the payoff of the following position is always zero: www.actuary88.com

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS 1. Sell the call 2. Borrow the present value of the strike price plus dividend 3. Buy the put 4. Buy one stock If there is transaction cost such as the bid-ask spread, then to avoid arbitrage, the initial gain of the above position is zero. However, if there is a bid-ask spread, the initial gain of the above position can be zero or negative. The initial gain of the position is: C b + P VrB (K) + P VrB (Div) − (P a + S0a ) There’s no arbitrage if C b + P VrB (K) + P VrB (Div) − (P a + S0a ) ≤ 0 P VrB (Div) = 0.18e−0.06575(0.02) = 0.18 P VrL (K) = Ke−0.02T C b +P VrB (K)+P VrB (Div)−(P a + S0a ) = C b +Ke−0.02T +0.18−(P a + 84.85) K 75 80 85 90 75 80 85 90

T 0.0986 0.0986 0.0986 0.0986 0.271 2 0.271 2 0.271 2 0.271 2

Cb 9.9 5.3 1.9 0.35 10.5 6.5 3.2 1.2

Pa 0.25 0.7 2.3 5.8 0.8 1.6 3.3 6.3

C b + P VrB (K) + P VrB (Div) − (P a + S0a ) 9.9 + 75e−0.02×0.0986 + 0.18 − (0.25 + 84.85) = −0.17 5.3 + 80e−0.02×0.0986 + 0.18 − (0.7 + 84.85) = −0.23 1.9 + 85e−0.02×0.0986 + 0.18 − (2.3 + 84.85) = −0.24 0.35 + 90e−0.02×0.0986 + 0.18 − (5.8 + 84.85) = −0.30 10.5 + 75e−0.02×0.271 2 + 0.18 − (0.8 + 84.85) = −0.38 6.5 + 80e−0.02×0.271 2 + 0.18 − (1.6 + 84.85) = −0.20 3.2 + 85e−0.02×0.271 2 + 0.18 − (3.3 + 84.85) = −0.23 1.2 + 90e−0.02×0.271 2 + 0.18 − (6.3 + 84.85) = −0.26

Problem 9.18. Suppose there are 3 options otherwise identical but with different strike price K1 < K2 < K3 where K2 = λK1 + (1 − λ) K2 and 0 < λ < 1. Then the price of the middle strike price K2 must not exceed the price of a diversified portfolio consisting of λ units of K1 -strike option and (1 − λ) units of K2 -strike option: C [λK1 + (1 − λ) K3 ] ≤ λC (K1 ) + (1 − λ) C (K3 ) P [λK1 + (1 − λ) K3 ] ≤ λP (K1 ) + (1 − λ) P (K3 ) The above conditions are called the convexity of the option price with respect to the strike price. They are equivalent to the textbook Equation 9.17 and 9.18. If the above conditions are violated, arbitrage opportunities exist. K 80 85 90

T 0.271 2 0.271 2 0.271 2

Cb 6.5 3.2 1.2

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Ca 6.7 3.4 1.35 c °Yufeng Guo

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS 85 = λ (80) + (1 − λ) (90) → λ = 0.5 a. If we buy a 80-strike call, buy a 90-strike call, sell two 85-strike calls • A 80-strike call and a 90-strike call form a diversified portfolio of calls, which is always as good as two 85-strike calls • So the cost of buying a 80-strike call and a 90-strike call can never be less than the revenue of selling two 85-strike calls What we pay if we buy a 80-strike call and a 90-strike call: 6.7 + 1.35 = 8. 05 What we get if we sell two 85-strike calls: 3.2 × 2 = 6. 4 8. 05 > 6. 4 So the convexity condition is met. I recommend that you don’t bother memorizing textbook Equation 9.17 and 9.18. b. If we sell a 80-strike call, sell a 90-strike call, buy two 85-strike calls • A 80-strike call and a 90-strike call form a diversified portfolio of calls, which is always as good as two 85-strike calls • So the revenue of selling a 80-strike call and a 90-strike call should never be less than the cost of buying two 85-strike calls. What we get if we sell a 80-strike call and a 90-strike call: 6.5 + 1.2 = 7. 7 What we pay if we buy two 85-strike calls: 3.4 × 2 = 6. 8 7. 7 > 6.8 So the convexity condition is met. c. To avoid arbitrage, the following two conditions must be met: C [λK1 + (1 − λ) K3 ] ≤ λC (K1 ) + (1 − λ) C (K3 ) P [λK1 + (1 − λ) K3 ] ≤ λP (K1 ) + (1 − λ) P (K3 ) These conditions must be met no matter you are a market-maker or anyone else buying or selling options, no matter you pay a bid-ask spread or not.

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CHAPTER 9. PARITY AND OTHER OPTION RELATIONSHIPS

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Chapter 10

Binomial option pricing I Problem 10.1. The stock price today is S = 100. The stock at T is either • Su = uS = 1.3 × 100 = 130 • Sd = dS = 0.8 × 100 = 80 a. For a call, the payoff at T is • Vu = max (0, Su − K) = max (0, 130 − 105) = 25 • Vd = max (0, Sd − K) = max (0, 80 − 105) = 0 We hold a replicating portfolio (4, B) at t = 0. This portfolio will have value Vu if the stock goes up to Su or Vd if the stock goes down to Sd . We set up the following equations: ½ 4130 + Be0.08(0.5) = 25 4Su + BerT = Vu → rT 4Sd + Be = Vd 480 + Be0.08(0.5) = 0 → B = −38. 431 6 4 = 0.5 So the option premium is: V = 4S + B = 0.5 × 100 + (−38. 431 6) = 11. 568 4 ½

b. For a put, the payoff at T is either • Vu = max (0, K − Su ) = max (0, 105 − 130) = 0 • Vd = max (0, Sd − K) = max (0, 105 − 80) = 25 23

CHAPTER 10. BINOMIAL OPTION PRICING I We hold a replicating portfolio (4, B) at t = 0. This portfolio will have value Vu if the stock goes up to Su or Vd if the stock goes down to Sd . We set up the following equations: ½ 4130 + Be0.08(0.5) = 0 4Su + BerT = Vu rT 4Sd + Be = Vd 480 + Be0.08(0.5) = 25 B = 62. 451 3 4 = −0.5 So the option premium is: V = 4S + B = −0.5 × 100 + 62. 451 3 = 12. 451 3 ½

Problem 10.2. The stock price today is S = 100.The stock at T • Su = uS = 1.3 × 100 = 130 • Sd = dS = 0.8 × 100 = 80 a. Payoff at T is either • Vu = max (0, Su − K) = max (0, 130 − 95) = 35 • Vd = max (0, Sd − K) = max (0, 80 − 95) = 0 We hold a replicating portfolio (4, B) at t = 0. This portfolio will have value Vu if the stock goes up to Su or Vd if the stock goes down to Sd . We set up the following equations: ½ 4130 + Be0.08(0.5) = 35 4Su + BerT = Vu rT 4Sd + Be = Vd 480 + Be0.08(0.5) = 0 B = −53. 804 2 4 = 0.7

½

So the option premium is: V = 4S + B = 0.7 × 100 + (−53. 804 2) = 16. 195 8 b. There are two calls out there. One can be synthetically built by buying 0.7 share of a stock and borrowing $53. 804 2 from a bank. The other is in the market selling for $17. To arbitrage, we buy low and sell high. • Buy low. Buy 0.7 share of a stock and borrow $53. 804 2 from a bank. Our initial cash outgo is 0.7 × 100 + (−53. 804 2) = 16. 195 8. This grows to either (0.7) 130−53. 804 2e0.08(0.5) = 35 if the stock goes up or (0.7) 80−53. 804 2e0.08(0.5) = 0 if the stock goes down at T . • Sell high. We sell a call for 17, receiving $17 at time zero. We pay either max (0, 130 − 95) = 35 if the stock goes up or max (0, 80 − 95) = 0 if the stock goes down at T . www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I The net cash flow at T is zero. The net cash inflow at time zero is 17 − 16. 195 8 = 0.804 2. So we receive $0.804 2 at time zero without incurring any liability at T ; we have made $0.804 2 free money. Millions of investors will copy this arbitraging strategy, which will bid down the call price from 17 to the fair price of 16. 195 8. c. There are two calls out there. One can be synthetically built by buying 0.7 share of a stock and borrowing $16. 195 8 from a bank. The other is in the market selling for $15.5. To arbitrage, we buy low and sell high. • Sell high. Short sell 0.7 share of a stock and deposit $53. 804 2 in a bank. Our initial cash inflow is 0.7 × 100 + (−53. 804 2) = 16. 195 8. At T , we need to buy back 0.7 share of a stock from the market to close our short position. In addition, our bank account grows into 53. 804 2e0.08(0.5) . So our cash inflow at T is either (−0.7) 130 + 53. 804 2e0.08(0.5) = −35 if the stock goes up or (−0.7) 80 + 53. 804 2e0.08(0.5) = 0 if the stock goes down. • Buy low. We buy a call for 15.5 at time zero, paying $15.5. We get either max (0, 130 − 95) = 35 if the stock goes up or max (0, 80 − 95) = 0 if the stock goes down at T . The net cash flow at T is zero. The initial cash inflow is 16. 195 8 − 15.5 = 0.695 8. So we receive $0.695 8 at time zero without incurring any liability at T ; we have made $0.695 8 free money. Millions of investors will copy this arbitraging strategy, which will bid up the call price from 15.5 to the fair price of 16. 195 8. Problem 10.3. The stock price today is S = 100. The stock at T is • Su = uS = 1.3 × 100 = 130 • Sd = dS = 0.8 × 100 = 80 a. Payoff at T is • Vu = max (0, K − Su ) = max (0, 95 − 130) = 0 • Vd = max (0, K − Sd ) = max (0, 95 − 80) = 15 We hold a replicating portfolio (4, B) at t = 0. This portfolio will have value Vu if the stock goes up to Su or Vd if the stock goes down to Sd . We set up the equations: ½ following rT ½ 4Su + Be = Vu 4130 + Be0.08(0.5) = 0 4Sd + BerT = Vd 480 + Be0.08(0.5) = 15 B = 37. 470 9 4 = −0.3 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I So the option premium is: V = 4S + B = −0.3 × 100 + 37. 470 9 = 7. 470 9 b. There are two puts out there. One can be synthetically built by short selling 0.3 share of a stock and depositing $37. 470 9 in a bank. The other is in the market selling for $8. To arbitrage, we buy low and sell high. • Buy low. Short sell 0.3 share of a stock and deposit $37. 470 9 in a bank. Our initial cash inflow is 0.3 × 100 − 37. 470 9 = −7. 470 9. At T , we need to buy back 0.3 share of a stock to close our short position. In addition, our initial deposit grows to 37. 470 9e0.08(0.5) = 39. If the stock goes up, our cash outgo at T is (−0.3) 130 + 37. 470 9e0.08(0.5) = 0; if the stock goes down to 80 at T , our cash outgo is (−0.3) 80 + 37. 470 9e0.08(0.5) = 15. • Sell high. We sell a put for 8, receiving $8 at time zero. At T , we pay either max (0, 95 − 130) = 0 if the stock goes up or max (0, 95 − 80) = 15 if the stock goes down. The net cash flow of "buy low, sell high" is zero at T . The net cash inflow at time zero is 8 − 7. 470 9 = 0.529 1. So we receive $0.529 1 at time zero without incurring any liability at T ; we have made $0.529 1 free money. Millions of investors will copy this arbitraging strategy, which will bid down the put price from 8 to the fair price of 7. 470 9. c. There are two puts out there. One can be synthetically built by short selling 0.3 share of a stock and depositing $37. 470 9 in a bank. The other is in the market selling for $6. To arbitrage, we buy low and sell high. • Buy low. We buy a put for $6 at time zero. At T we receive either max (0, 95 − 130) = 0 if the stock goes up or max (0, 95 − 80) = 15 if the stock goes down. • Sell high. At time zero we buy 0.3 share of a stock and borrow $37. 470 9 in a bank. Our initial cash inflow is −0.3 × 100 + 37. 470 9 = 7. 470 9. At T , we sell 0.3 share of a stock, receiving 0.3 × 130 = 39 if the stock goes up or receiving 0.3 × 80 = 24 if the stock goes down. In addition, we need to pay the bank 37. 470 9e0.08(0.5) = 39. So our net cash inflow at T is 39 − 39 = 0 if the stock goes up or 24 − 39 = −15 if the stock goes down. The net cash flow of "buy low, sell high" is zero at T . The net cash inflow at time zero is 7. 470 9 − 6 = 1. 470 9. So we receive $1. 470 9 at time zero without incurring any liability at T ; we have made $1. 470 9 free money. Millions of investors will copy this arbitraging strategy, which will bid up the put price from 6 to the fair price of 7. 470 9. Problem 10.4. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I The problem doesn’t say the option is European or American. Let’s assume the option is European. T =1 n=2 →period length h = T /2 = 0.5 Period 0 1 2 ¡ ¢ Suu = 100 1.32 = 169 Su = 100 (1.3) = 130 Vuu = 169 − 95 = 74 Vu =? (∆u , Bu ) Sud = 100 (1.3) (0.8) = 104 S = 100 Vud = 104 − 95 = 9 V =? (∆, B) =? Sd = 100 (0.8) = 80 ¡ ¢ Vd =? Sdd = 100 0.82 = 64 (∆d , Bd ) Vdd = 0 We start from right to left. Step 1 Calculate (∆u , Bu ), the replicating portfolio at Node u which will produce the payoff Vud and Vdd . ½ ½ 4u Suu + Bu erh = Vuu 4u 169 + Bu e0.08(0.5) = 74 → rh 4u Sud + Bu e = Vud 4u 104 + Bu e0.08(0.5) = 9 Bu = −91. 2750 4u = 1 The premium at Node u is: Vu = 4u Su + Bu = 1 (130) − 91. 2750 = 38. 725 Step 2 Calculate(∆d , Bd ), the replicating portfolio at Node d which will produce the payoff Vud and Vdd . ½ ½ 4d Sud + Bd erh = Vud 4d 104 + Bd e0.08(0.5) = 9 → rh 4d Sdd + Bd e = Vdd 4d 64 + Bd e0.08(0.5) = 0 4d = 0.225 , Bd = −13. 835 4 The premium at Node d is: Vd = 4d Sd + Bd = 0.225 (80) − 13. 835 4 = 4. 164 6 Step 3 Calculate(∆, B), the replicating portfolio at time zero, which will produce the payoff Vu and Vd . ½ ½ 4Su + BerT = Vu 4130 + Be0.08(0.5) = 38. 725 → rT 4Sd + Be = Vd 480 + Be0.08(0.5) = 4. 164 6 4 = 0.691 2 B = −49. 127 1 The premium at time zero is: V = 4S + B = 0.691 2 (100) − 49. 127 1 = 19. 992 9 The final diagram is: www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 0

1

S = 100 V = 19. 992 9 ∆ = 0.691 2 B = −$49. 127 1

Su = 100 (1.3) = 130 Vu = 38. 725 ∆u = 1 Bu = −$91. 2750

Sd = 100 (0.8) = 80 Vd = 4. 164 6 ∆d = 0.225 Bd = −$13. 835 4

2 ¢ ¡ Suu = 100 1.32 = 169 Vuu = 169 − 95 = 74

Sud = 100 (1.3) (0.8) = 104 Vud = 104 − 95 = 9 ¡ ¢ Sdd = 100 0.82 = 64 Vdd = 0

Problem 10.5. This question asks us to redo the previous problem by setting the initial stock price to 80, 90, 110, 120, and 130. • S = 80. Period 0

1 Su = 80 (1.3) = 104 Vu =? (∆u , Bu ) =?

2 ¡ ¢ Suu = 80 1.32 = 135. 2 Vuu = 135. 2 − 95 = 40. 2 Sud = 80 (1.3) (0.8) = 83. 2 Vud = 0

S = 80 V =? (∆, B) =? Sd = 80 (0.8) = 64 Vd =? (∆d , Bd ) =?

¡ ¢ Sdd = 80 0.82 = 51. 2 Vdd = 0

½

4u 135. 2 + Bu e0.08(0.5) = 40. 2 Bu = −61. 7980 4u = 0.773 08 4u 83. 2 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0.773 08 (104) − 61. 7980 = 18. 602 32

½

4d 83. 2 + Bd e0.08(0.5) = 0 4d = 0.0 4d 51. 2 + Bd e0.08(0.5) = 0 Vd = 4d Sd + Bd = 0 (64) − 0 = 0 ½

4104 + Be0.08(0.5) = 18. 602 32 464 + Be0.08(0.5) = 0

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Bd = 0

4 = 0.465 058

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B = −28. 596 7 28

CHAPTER 10. BINOMIAL OPTION PRICING I V = 4S + B = 0.465 058 (80) − 28. 596 7 = 8. 607 94 The final diagram is: Period 0 1 Su = 80 (1.3) = 104 Vu = 18. 602 32 ∆u = 0.773 08 Bu = −61. 798 S = 80 V = 8. 607 94 ∆ = 0.465 058 B = −28. 596 7

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Sd = 80 (0.8) = 64 Vd = 0 ∆d = 0 Bd = 0

2 ¡ ¢ Suu = 80 1.32 = 135. 2 Vuu = 135. 2 − 95 = 40. 2

Sud = 80 (1.3) (0.8) = 83. 2 Vud = 0

¡ ¢ Sdd = 80 0.82 = 51. 2 Vdd = 0

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 90. Period 0

1 Su = 90 (1.3) = 117 Vu =? 4u =? Bu =?

S = 90 V =? ∆ =? B =?

2 ¡ ¢ Suu = 90 1.32 = 152. 1 Vuu = 152. 1 − 95 = 57. 1

Sud = 90 (1.3) (0.8) = 93. 6 Vud = 0

Sd = 90 (0.8) = 72 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 90 0.82 = 57. 6 Vdd = 0

½

4u 152. 1 + Bu e0.08(0.5) = 57. 1 Bu = −87. 777 7 4u = 0.976 068 4u 93. 6 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0.976 068 (117) − 87. 777 7 = 26. 422 26

½

4d 93. 6 + Bd e0.08(0.5) = 0 4d = 0.0 4d 57. 6 + Bd e0.08(0.5) = 0 Vd = 4d Sd + Bd = 0 (72) − 0 = 0 ½

Bd = 0

4117 + Be0.08(0.5) = 26. 422 26 4 = 0.587 161 472 + Be0.08(0.5) = 0 V = 4S + B = 0.587 161 (90) − 40. 617 97 = 12. 226 52 The final diagram is: Period 0 1

S = 90 V = 12. 226 52 4 = 0.587 161 B = −40. 617 97

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Su = 90 (1.3) = 117 Vu = 26. 422 26 4u = 0.976 068 Bu = −87. 777 7

Sd = 90 (0.8) = 72 Vd = 0 ∆d = 0 Bd = 0

B = −40. 617 97

2 ¡ ¢ Suu = 90 1.32 = 152. 1 Vuu = 152. 1 − 95 = 57. 1

Sud = 90 (1.3) (0.8) = 93. 6 Vud = 0

¢ ¡ Sdd = 90 0.82 = 57. 6 Vdd = 0 0

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 110. Period 0

1 Su = 110 (1.3) = 143 Vu =? 4u =? Bu =?

S = 110 V =? ∆ =? B =?

¡2 ¢ Suu = 110 1.32 = 185. 9 Vuu = 185. 9 − 95 = 90. 9

Sud = 110 (1.3) (0.8) = 114. 4 Vud = 114. 4 − 95 = 19. 4 Sd = 110 (0.8) = 88 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 110 0.82 = 70. 4 Vdd = 0

½

4u 185. 9 + Bu e0.08(0.5) = 90. 9 4u = 1 4u 114. 4 + Bu e0.08(0.5) = 19. 4 Vu = 4u Su + Bu = 1 (143) − 91. 2750 = 51. 725

Bu = −91. 2750

½

4d 114. 4 + Bd e0.08(0.5) = 19. 4 4d = 0.440 909 4d 70. 4 + Bd e0.08(0.5) = 0 Vd = 4d Sd + Bd = 0.440 909 (88) − 29. 822 9 = 8. 977 092

½

4143 + Be0.08(0.5) = 51. 725 4 = 0.777 235 488 + Be0.08(0.5) = 8. 977 092 V = 4S + B = 0.777 235 (110) − 57. 089 7 = 28. 406 15

The final diagram is: Period 0 1

S = 110 28. 406 15 4 = 0.777 235 B = −57. 089 7

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Su = 110 (1.3) = 143 Vu = 51. 725 4u = 1 Bu = −91. 2750

Sd = 110 (0.8) = 88 Vd = 8. 977 092 ∆d = 0.440 909 Bd = −29. 822 9

Bd = −29. 822 9

B = −57. 089 7

2 ¡ ¢ Suu = 110 1.32 = 185. 9 Vuu = 185. 9 − 95 = 90. 9

Sud = 110 (1.3) (0.8) = 114. 4 Vud = 114. 4 − 95 = 19. 4 ¡ ¢ Sdd = 110 0.82 = 70. 4 Vdd = 0

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 120. Period 0

1 Su = 120 (1.3) = 156 Vu =? 4u =? Bu =?

S = 120 V =? ∆ =? B =?

2 ¡ ¢ Suu = 120 1.32 = 202. 8 Vuu = 202. 8 − 95 = 107. 8

Sud = 120 (1.3) (0.8) = 124. 8 Vud = 124. 8 − 95 = 29. 8 Sd = 120 (0.8) = 96 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 120 0.82 = 76. 8 Vdd = 0

½

4u 202. 8 + Bu e0.08(0.5) = 107. 8 4u = 1 4u 124. 8 + Bu e0.08(0.5) = 29. 8 Vu = 4u Su + Bu = 1 (156) − 91. 275 = 64. 725

Bu = −91. 275

½

4d 124. 8 + Bd e0.08(0.5) = 29. 8 Bd = −45. 810 44 4d = 0.620 833 4d 76. 8 + Bd e0.08(0.5) = 0 Vd = 4d Sd + Bd = 0.620 833 (96) − 45. 810 44 = 13. 789 528

½

4156 + Be0.08(0.5) = 64. 725 4 = 0.848 925 496 + Be0.08(0.5) = 13. 789 528 V = 4S + B = 0.848 925 (120) − 65. 052 4 = 36. 818 6

The final diagram is: Period 0

S = 120 V = 36. 818 6 4 = 0.848 925 B = −65. 052 4

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1

Su = 120 (1.3) = 156 Vu = 64. 725 4u = 1 Bu = −91. 275

B = −65. 052 4

¡2 ¢ Suu = 120 1.32 = 202. 8 Vuu = 202. 8 − 95 = 107. 8

Sud = 120 (1.3) (0.8) = 124. 8 Vud = 124. 8 − 95 = 29. 8

Sd = 120 (0.8) = 96 Vd = 13. 789 528 4d = 0.620 833 Bd = −45. 810 44 c °Yufeng Guo

¡ ¢ Sdd = 120 0.82 = 76. 8 Vdd = 0

32

CHAPTER 10. BINOMIAL OPTION PRICING I • S = 130. Period 0

1 Su = 130 (1.3) = 169 Vu =? 4u =? Bu =?

S = 130 V =? ∆ =? B =?

2 ¡ ¢ Suu = 130 1.32 = 219. 7 Vuu = 219. 7 − 95 = 124. 7

Sud = 130 (1.3) (0.8) = 135. 2 Vud = 135. 2 − 95 = 40. 2 Sd = 130 (0.8) = 104 Vd =? ∆d =? Bd =?

¢ ¡ Sdd = 130 0.82 = 83. 2 Vdd = 0

½

4u 219. 7 + Bu e0.08(0.5) = 124. 7 Bu = −91. 275 4u = 1 4u 135. 2 + Bu e0.08(0.5) = 40. 2 Vu = 4u Su + Bu = 1 (169) − 91. 275 = 77. 725 ½ 4d 135. 2 + Bd e0.08(0.5) = 40. 2 Bd = −61. 7980 4d = 0.773 077 4d 83. 2 + Bd e0.08(0.5) = 0 Vd = 4d Sd + Bd = 0.773 077 (104) − 61. 7980 = 18. 602 ½ 4169 + Be0.08(0.5) = 77. 725 4 = 0.909 585 B = −73. 015 4104 + Be0.08(0.5) = 18. 602 V = 4S + B = 0.909 585 (130) − 73. 015 = 45. 231 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 130 1.32 = 219. 7 Su = 130 (1.3) = 169 Vuu = 219. 7 − 95 = 124. 7 Vu = 77. 725 4u = 1 Bu = −91. 275 S = 130 Sud = 130 (1.3) (0.8) = 135. 2 V = 45. 231 Vud = 135. 2 − 95 = 40. 2 4 = 0.909 585 B = −73. 015 Sd = 130 (0.8) = 104 ¡ ¢ Vd = 18. 602 Sdd = 130 0.82 = 83. 2 4d = 0.773 077 Vdd = 0 Bd = −61. 7980 Notice that the delta for a call is always positive. A positive delta means buying stocks. If you sell a call, the risk you face is that the stock price may go www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I up to infinity, at which case the call holder will pay only the strike price to buy a priceless stock from you. To hedge your risk, you need to already own some stocks at t = 0. If indeed the future stock price goes up at T , your stock will also go up in value, offsetting your loss in the call. Intuitively, the higher the initial stock price, everything else equal, the higher the stock price in the future, the higher the payoff of the call. Hence to hedge the risk, the market maker needs to buy more shares of stocks at t = 0. So the higher the initial stock price, the higher the initial delta, the more stocks the market maker needs to buy at t = 0.

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CHAPTER 10. BINOMIAL OPTION PRICING I Problem 10.6. Period 0

1 Su = 100 (1.3) = 130 Vu =? 4u =? Bu =?

S = 100 V =? ∆ =? B =?

2 ¡ ¢ Suu = 100 1.32 = 169 Vuu = 0

Sud = 100 (1.3) (0.8) = 104 Vud = 0

Sd = 100 (0.8) = 80 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 100 0.82 = 64 Vdd = 95 − 64 = 31

½

4u 169 + Bu e0.08(0.5) = 0 4u = 0 4u 104 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0 (130) + 0 = 0

Bu = 0

½

4d 104 + Bd e0.08(0.5) = 0 Bd = 77. 439 6 4d = −0.775 4d 64 + Bd e0.08(0.5) = 31 Vd = 4d Sd + Bd = −0.775 (80) + 77. 439 6 = 15. 439 6 ½

4130 + Be0.08(0.5) = 0 4 = −0.308 792 B = 38. 568 932 480 + Be0.08(0.5) = 15. 439 6 V = 4S + B = −0.308 792 (100) + 38. 568 932 = 7. 689 732 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 100 1.32 = 169 Su = 100 (1.3) = 130 Vuu = 0 Vu = 0 4u = 0 Bu = 0 S = 100 Sud = 100 (1.3) (0.8) = 104 V = 7. 689 732 Vud = 0 4 = −0.308 792 B = 38. 568 932 Sd = 100 (0.8) = 80 ¢ ¡ Vd = 15. 439 6 Sdd = 100 0.82 = 64 4d = −0.775 Vdd = 95 − 64 = 31 Bd = 77. 439 6

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CHAPTER 10. BINOMIAL OPTION PRICING I Problem 10.7. • S = 80 Period 0

1 Su = 80 (1.3) = 104 Vu =? 4u =? Bu =?

S = 80 V =? ∆ =? B =?

2 ¡ ¢ Suu = 80 1.32 = 135. 2 Vuu = 0

Sud = 80 (1.3) (0.8) = 83. 2 Vud = 95 − 83. 2 = 11. 8 Sd = 80 (0.8) = 64 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 80 0.82 = 51. 2 Vdd = 95 − 51. 2 = 43. 8

½

4u 135. 2 + Bu e0.08(0.5) = 0 4u = −0.226 923 4u 83. 2 + Bu e0.08(0.5) = 11. 8 Vu = 4u Su + Bu = −0.226 923 (104) + 29. 477 02 = 5. 877

½

4d 83. 2 + Bd e0.08(0.5) = 11. 8 4d = −1 4d 51. 2 + Bd e0.08(0.5) = 43. 8 Vd = 4d Sd + Bd = −1 (64) + 91. 275 = 27. 275

Bu = 29. 477 02

Bd = 91. 275

½

4104 + Be0.08(0.5) = 5. 877 4 = −0.534 95 B = 59. 1 464 + Be0.08(0.5) = 27. 275 V = 4S + B = −0.534 95 (80) + 59. 1 = 16. 304 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 80 1.32 = 135. 2 Su = 80 (1.3) = 104 Vuu = 0 Vu = 5. 877 4u = −0.226 923 Bu = 29. 477 02 S = 80 Sud = 80 (1.3) (0.8) = 83. 2 V = 16. 304 Vud = 95 − 83. 2 = 11. 8 4 = −0.534 95 B = 59. 1 Sd = 80 (0.8) = 64 ¡ ¢ Vd = 27. 275 Sdd = 80 0.82 = 51. 2 4d = −1 Vdd = 95 − 51. 2 = 43. 8 Bd = 91. 275 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 90 Period 0

1 Su = 90 (1.3) = 117 Vu =? 4u =? Bu =?

S = 90 V =? ∆ =? B =?

2 ¡ ¢ Suu = 90 1.32 = 152. 1 Vuu = 0

Sud = 90 (1.3) (0.8) = 93. 6 Vud = 95 − 93. 6 = 1. 4 Sd = 90 (0.8) = 72 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 90 0.82 = 57. 6 Vdd = 95 − 57. 6 = 37. 4

½

4u 152. 1 + Bu e0.08(0.5) = 0 Bu = 3. 497 27 4u = −0.02 393 4u 93. 6 + Bu e0.08(0.5) = 1. 4 Vu = 4u Su + Bu = −0.02 393 (117) + 3. 497 27 = 0.697 5 ½

4d 93. 6 + Bd e0.08(0.5) = 1. 4 4d = −1 4d 57. 6 + Bd e0.08(0.5) = 37. 4 Vd = 4d Sd + Bd = −1 (72) + 91. 275 = 19. 275

Bd = 91. 275

½

4117 + Be0.08(0.5) = 0.697 5 4 = −0.412 833 B = 47. 077 722 472 + Be0.08(0.5) = 19. 275 V = 4S + B = −0.412 833 (90) + 47. 077 722 = 9. 923 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 90 1.32 = 152. 1 Su = 90 (1.3) = 117 Vuu = 0 Vu = 0.697 5 4u = −0.02 393 Bu = 3. 497 27 S = 90 Sud = 90 (1.3) (0.8) = 93. 6 V = 9. 923 Vud = 95 − 93. 6 = 1. 4 4 = −0.412 833 B = 47. 077 722 Sd = 90 (0.8) = 72 ¡ ¢ Vd = 19. 275 Sdd = 90 0.82 = 57. 6 4d = −1 Vdd = 95 − 57. 6 = 37. 4 Bd = 91. 275

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 110 Period 0

1 Su = 110 (1.3) = 143 Vu =? 4u =? Bu =?

S = 110 V =? ∆ =? B =?

2 ¡ ¢ Suu = 110 1.32 = 185. 9 Vuu = 0

Sud = 110 (1.3) (0.8) = 114. 4 Vud = 0

Sd = 110 (0.8) = 88 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 110 0.82 = 70. 4 Vdd = 95 − 70. 4 = 24. 6

½

4u 185. 9 + Bu e0.08(0.5) = 0 4u = 0 4u 114. 4 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0 (143) + 0 = 0

½

Bu = 0

4d 114. 4 + Bd e0.08(0.5) = 0 4d = −0.559 091 4d 70. 4 + Bd e0.08(0.5) = 24. 6 Vd = 4d Sd + Bd = −0.559 091 (88) + 61. 452 1 = 12. 252

Bd = 61. 452 1

½

4143 + Be0.08(0.5) = 0 4 = −0.222 76 B = 30. 606 14 488 + Be0.08(0.5) = 12. 252 V = 4S + B = −0.222 76 (110) + 30. 606 14 = 6. 103 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 110 1.32 = 185. 9 Su = 110 (1.3) = 143 Vuu = 0 Vu = 0 4u = 0 Bu = 0 S = 110 Sud = 110 (1.3) (0.8) = 114. 4 V = 6. 103 Vud = 0 4 = −0.222 76 B = 30. 606 14 Sd = 110 (0.8) = 88 ¡ ¢ Vd = 12. 252 Sdd = 110 0.82 = 70. 4 4d = −0.559 091 Vdd = 95 − 70. 4 = 24. 6 Bd = 61. 452 1

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CHAPTER 10. BINOMIAL OPTION PRICING I • S = 120 Period 0

1 Su = 120 (1.3) = 156 Vu =? 4u =? Bu =?

S = 120 V =? ∆ =? B =?

2 ¡ ¢ Suu = 120 1.32 = 202. 8 Vuu = 0

Sud = 120 (1.3) (0.8) = 124. 8 Vud = 0

Sd = 120 (0.8) = 96 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 120 0.82 = 76. 8 Vdd = 95 − 76. 8 = 18. 2

½

4u 202. 8 + Bu e0.08(0.5) = 0 4u = 0 4u 124. 8 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0 (156) + 0 = 0

½

Bu = 0

4d 124. 8 + Bd e0.08(0.5) = 0 4d = −0.379 167 4d 76. 8 + Bd e0.08(0.5) = 18. 2 Vd = 4d Sd + Bd = −0.379 167 (96) + 45. 464 56 = 9. 065

Bd = 45. 464 56

½

4156 + Be0.08(0.5) = 0 4 = −0.151 083 B = 22. 644 85 496 + Be0.08(0.5) = 9. 065 V = 4S + B = −0.151 083 (120) + 22. 644 85 = 4. 514 89 The final diagram is: Period 0 1 Su = 120 (1.3) = 156 Vu = 0 4u = 0 Bu = 0 S = 120 V = 4. 514 89 4 = −0.151 083 B = 22. 644 85

Sud = 120 (1.3) (0.8) = 124. 8 Vud = 0

Sd = 120 (0.8) = 96 Vd = 9. 065 4d = −0.379 167 Bd = 45. 464 56

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2 ¡ ¢ Suu = 120 1.32 = 202. 8 Vuu = 0

c °Yufeng Guo

¡ ¢ Sdd = 120 0.82 = 76. 8 Vdd = 95 − 76. 8 = 18. 2

39

CHAPTER 10. BINOMIAL OPTION PRICING I • S = 130 Period 0

1 Su = 130 (1.3) = 169 Vu =? 4u =? Bu =?

S = 130 V =? ∆ =? B =?

2 ¡ ¢ Suu = 130 1.32 = 219. 7 Vuu = 0

Sud = 130 (1.3) (0.8) = 135. 2 Vud = 0

Sd = 130 (0.8) = 104 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 130 0.82 = 83. 2 Vdd = 95 − 83. 2 = 11. 8

½

4u 219. 7 + Bu e0.08(0.5) = 0 4u = 0 4u 135. 2 + Bu e0.08(0.5) = 0 Vu = 4u Su + Bu = 0 (169) + 0 = 0

½

Bu = 0

4d 135. 2 + Bd e0.08(0.5) = 0 4d = −0.226 923 4d 83. 2 + Bd e0.08(0.5) = 11. 8 Vd = 4d Sd + Bd = −0.226 923 (104) + 29. 477 0 = 5. 877

Bd = 29. 477 0

½

4169 + Be0.08(0.5) = 0 4 = −0.09 041 5 B = 14. 681 05 4104 + Be0.08(0.5) = 5. 877 V = 4S + B = −0.09 041 5 (130) + 14. 681 05 = 2. 927 1 The final diagram is: Period 0 1 2 ¡ ¢ Suu = 130 1.32 = 219. 7 Su = 130 (1.3) = 169 Vuu = 0 Vu = 0 4u = 0 Bu = 0 S = 130 Sud = 130 (1.3) (0.8) = 135. 2 V = 2. 927 1 Vud = 0 4 = −0.09 041 5 B = 14. 681 05 Sd = 130 (0.8) = 104 ¡ ¢ Vd = 5. 877 Sdd = 130 0.82 = 83. 2 4d = −0.226 923 Vdd = 95 − 83. 2 = 11. 8 Bd = 29. 477 0

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CHAPTER 10. BINOMIAL OPTION PRICING I Notice that the delta for a put is always negative. A negative delta means shorting sell stocks. If you sell a put, the risk you face is that the stock price may go down to zero, at which case the put holder sells you a worthless stock for the strike price. To hedge your risk, you need to short sell some stocks at t = 0. If indeed the future stock price goes down at T , you can buy back stocks in the market at low price to close your short position on stocks, earning a profit. Your profit from the short sale can offset your loss in the put. Intuitively, the higher the initial stock price, everything else equal, the higher the stock price in the future, the lower the payoff of the put. Hence to hedge the risk, the market maker needs to short sell fewer shares of stocks at t = 0. So the higher the initial stock price, the lower the absolute value of the delta for a put, the fewer of the stocks that the market maker needs to short sell initially. Problem 10.8. Period 0

1 Su = 100 (1.3) = 130 EVu = 0 Vu =? 4u =? Bu =?

S = 100 EV = 0 V =? ∆ =? B =?

2 ¡ ¢ Suu = 100 1.32 = 169 Vuu = 0

Sud = 100 (1.3) (0.8) = 104 Vud = 0

Sd = 100 (0.8) = 80 EVd = 95 − 80 = 15 Vd =? ∆d =? Bd =?

¡ ¢ Sdd = 100 0.82 = 64 Vdd = 95 − 64 = 31

½

4u 169 + Bu e0.08(0.5) = 0 Bu = 0 4u = 0 4u 104 + Bu e0.08(0.5) = 0 Vu = max (4u Su + Bu , EVu ) = max [0 (130) + 0, 0] = 0 This is the logic behind the calculation of Vu . The American put can be exercised at the end of Period 1 and end of Period 2. • If exercised at the end of Period 1, the put is worth EVu = max (0, 95 − 130) = 0 at the end of Period 1. • If exercised at Period 2, the put is worth 4u Su + Bu = 0 (130) + 0 = 0 at the end of Period 1. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I We compare the two values and take the greater. ½

4d 104 + Bd e0.08(0.5) = 0 Bd = 77. 439 6 4d = −0.775 4d 64 + Bd e0.08(0.5) = 31 Vd = max (4d Sd + Bd , EVd ) = max [−0.775 (80) + 77. 439 6, 15] = 15. 439 6 ½

4130 + Be0.08(0.5) = 0 4 = −0.308 792 B = 38. 568 932 480 + Be0.08(0.5) = 15. 439 6 V = max (4S + B, EV ) = max [−0.308 792 (100) + 38. 568 932 , 0] = 7. 689 732

The final diagram is: Period 0 1 Su = 100 (1.3) = 130 EVu = 0 Vu = 0 4u = 0 Bu = 0 S = 100 EV = 0 V = 7. 689 7 ∆ = −0.308 792 B = 38. 568 932

2 ¡ ¢ Suu = 100 1.32 = 169 Vuu = 0

Sud = 100 (1.3) (0.8) = 104 Vud = 0

Sd = 100 (0.8) = 80 EVd = 95 − 80 = 15 Vd = 15. 439 6 4d = −0.775 Bd = 77. 439 6

¡ ¢ Sdd = 100 0.82 = 64 Vdd = 95 − 64 = 31

Problem 10.9. a. time 0

T Su = 100 (1.2) = 120 Vu = 120 − 50 = 70

S = 100 V =? ∆ =?, B =?

Sd = 100 (1.05) = 105 Vd = 105 − 50 = 55 Here d = 1.05 > 1. Is there anything wrong with this? No necessarily. The non-arbitrage requirement is textbook Equation 10.4: u > e(r−δ)h > d Let’s check. e(r−δ)h = e(0.07696−0)1 = 1. 08 u = 1.2 > e(r−δ)h > d = 1.05 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I So it’s OK to have d > 1 as long as u > e(r−δ)h > d is met. ½ 4120 + Be0.07696(1) = 70 4=1 B = −46. 296 3 4105 + Be0.07696(1) = 55 V = 4S + B = 1 (100) − 46. 296 3 = 53. 703 7 The final diagram is: time 0

S = 100 V = 53. 703 7 4 = 1, B = −46. 296 3

T Su = 100 (1.2) = 120 Vu = 120 − 50 = 70

Sd = 100 (1.05) = 105 Vd = 105 − 50 = 55

b. Now the stock price has a bigger increase and a bigger decrease. We are not clear whether the call premium will increase or decrease. We have to calculate the premium. time 0 T Su = 100 (1.4) = 140 Vu = 140 − 50 = 90 S = 100 V =? ∆ =?, B =? Sd = 100 (0.6) = 60 Vd = 60 − 50 = 10 ½

4140 + Be0.07696(1) = 90 4=1 B = −46. 296 3 460 + Be0.07696(1) = 10 V = 4S + B = 1 (100) − 46. 296 3 = 53. 703 7 The call premium is the same. What’s going on? It turns out that as long Su > Sd > K, the call premium is fixed regardless of how you choose u and d. time 0 T Su = uS > K Vu = uS − K S V =? ∆ =?, B =? Sd = dS > K Vd = dS − K ½

4uS + Berh = uS − K 4dS + Berh = dS − K

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CHAPTER 10. BINOMIAL OPTION PRICING I → 4 = 1 B = −Ke−rh 50e = 53. 703 7 −0.07696(1)

C = 4S + B = S − Ke−rh = 100 −

We can explain this using the put-call parity. If Su > Sd > K, the put is never exercised. C + P V (K) = P + S Since P = 0, we have C = S −P V (K) = S −Ke−rh = 100−50e−0.07696(1) = 53. 703 7 c. I think the question wants us to compare b and c, not a and c. Compared with b, now u is the same but d gets smaller. In fact, now at Node d, the call expires worthless. time 0

S = 100 V =? ∆ =?, B =?

T Su = 100 (1.4) = 140 Vu = 140 − 50 = 90

Sd = 100 (0.4) = 40 Vd = 0 We might think the call option should be worth less since it expires worthless at Node d. ½ 4140 + Be0.07696(1) = 90 4 = 0.9 B = −33. 333 440 + Be0.07696(1) = 0 V = 4S + B = 0.9 (100) − 33. 333 = 56. 667 > 53. 703 7 Why did the call premium goes up to 56. 667? Does this lead to arbitrage? First, I’m going to show you this is not an arbitrage. Now we have two call options: Call b Call c time 0 T time 0 T Su = 100 (1.4) = 140 Su = 100 (1.4) = 140 Vu = 140 − 50 = 90 Vu = 140 − 50 = 90 S = 100 S = 100 V = 53. 703 7 V = 56. 667 4 = 1, B = −46. 296 3 4 = 0.9, B = −33. 333 Sd = 100 (0.6) = 60 Sd = 100 (0.4) = 40 Vd = 60 − 50 = 10 Vd = 0 What if we buy Call b and simultaneously sell Call c? Does this lead to arbitrage? The answer is No because these two calls are on two different stocks. If these two options have the same underlying asset, then we’ll make free money www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I by "buy low and sell high." However, these two calls are on two different stocks; if the stock is the same, there won’t be two different values of d0 s. Call b is on a stock whose up price is 140 and down price is 60. Call c is on the stock whose up price is 140 and down price is 40. Clearly, the stock under Call c is more volatile than the stock under Call b. Hence Call c is more valuable than Call b. We can also explain why Call b is more valuable using the put-call parity. C + P V (K) = P + S For b, the put is never exercised. Hence P = 0 and C = S − P V (K). For c, the put is exercised Node d. Hence P > 0 and C = P + S − P V (K). So Call c exceeds Call b.

Problem 10.10. h = T /n = 1/3 u = e(r−δ)h+σ d = e(r−δ)h−σ πu =

√ h

√ h

√ = e(0.08−0)1/3+0.3 1/3 = 1. 221 246 √ = e(0.08−0)1/3−0.3 1/3 = 0.863 693

e(r−δ)h − d e(0.08−0)1/3 − 0.863 693 = = 0.456 806 u−d 1. 221 246 − 0.863 693

π d = 1 − πu = 1 − 0.456 806 = 0.543 194 a. Calculate the call premium Period 2 ¡ ¢ Suu = 100 1. 221 2462 = 149. 144 2 EVuu = 149. 144 2 − 95 = 54. 144 2 Vuu =? Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2 EVud = 105. 478 2 − 95 = 10. 478 2 Vud =? ¡ ¢ Sdd = 100 0.863 6932 = 74. 596 6 EVdd = 0 Vdd =?

3 ¡ ¢ Suuu = 100 1. 221 2463 = 182. 141 7 Vuuu = 182. 141 7 − 95 = 87. 141 7

¡ ¢ Suud = 100 1. 221 2462 (0.863 693) = 128. 814 8 Vuud = 128. 814 8 − 95 = 33. 814 8 ¡ ¢ Sudd = 100 (1. 221 246) 0.863 6932 = 91. 100 8 Vudd = 0 ¡ ¢ Sddd = 100 0.863 6933 = 64. 428 5 Vddd = 0

Let R represent the roll-back value. R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 806 × 87. 141 7 + 0.543 194 × 33. 814 8) = 56. 644 02 ¡ R ¢ Vuu = max Vuu , EVuu = max (56. 644 02, 54. 144 2) = 56. 644 02 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I R Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.08)1/3 (0.456 806 × 33. 814 8 + 0.543 194 × 0) = 15. 040 33 ¢ ¡ R , EVud = max (15. 040 33, 10. 478 2) = 15. 040 33 Vud = max Vud

0

R = e−rh (π u Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 0) = Vdd

¡ R ¢ Vdd = max Vdd , EVdd = max (0, 0) = 0

Now we have: Period 1

Su = 100 (1. 221 246) = 122. 124 6 EVu = 122. 124 6 − 95 = 27. 124 6 Vu =?

Sd = 100 (0.863 693) = 86. 369 3 EVd = 0 Vd =?

2 ¡ ¢ Suu = 100 1. 221 2462 = 149. 144 2 EVuu = 149. 144 2 − 95 = 54. 144 2 Vuu = 56. 644 02 Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2 EVud = 105. 478 2 − 95 = 10. 478 2 Vud = 15. 040 33 ¡ ¢ Sdd = 100 0.863 6932 = 74. 596 6 EVdd = 0 Vdd = 0

VuR = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 806 × 56. 644 02 + 0.543 194 × 15. 040 33) = 33. 149 27 ¡ ¢ Vu = max VuR , EVu = max (33. 149 27, 27. 124 6) = 33. 149 27

VdR = e−rh (πu Vud + πd Vdd ) = e−(0.08)1/3 (0.456 806 × 15. 040 33 + 0.543 194 × 0) = 6. 689 72 ¡ ¢ Vd = max VdR , EVd = max (6. 689 72, 0) = 6. 689 72 Now we have: Period 0

S = 100 EV = 100 − 95 = 5 V =?

1 Su = 100 (1. 221 246) = 122. 124 6 EVu = 122. 124 6 − 95 = 27. 124 6 Vu = 33. 149 27

Sd = 100 (0.863 693) = 86. 369 3 EVd = 0 Vd = 6. 689 72

V R = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 806 × 33. 149 27 + 0.543 194 × 6. 689 72) = 18. 282 51 ¡ ¢ V = max V R , EV = max (18. 282 51, 5) = 18. 282 51 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I From the above calculation you can see that the exercise value is never greater than the roll back value. Hence the American call option is never exercised early; the American call and European call have the same value of $18. 282 51. Generally, if a stock doesn’t pay dividend, then an American call and an otherwise identical European call on this stock are worth the same.

c. Calculate the American put premium.

Period 2 ¢ ¡ Suu = 100 1. 221 2462 = 149. 144 2 EVuu = 0 Vuu =? Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2 EVud = 0 Vud =? ¡ ¢ Sdd = 100 0.863 6932 = 74. 596 6 EVdd = 95 − 74. 596 6 = 20. 403 4 Vdd =?

0

3 ¡ ¢ Suuu = 100 1. 221 2463 = 182. 141 7 Vuuu = 0

¡ ¢ Suud = 100 1. 221 2462 (0.863 693) = 128. 814 8 Vuud = 0 ¡ ¢ Sudd = 100 (1. 221 246) 0.863 6932 = 91. 100 8 Vudd = 95 − 91. 100 8 = 3. 899 2 ¡ ¢ Sddd = 100 0.863 6933 = 64. 428 5 Vddd = 95 − 64. 428 5 = 30. 571 5

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 0) =

¡ R ¢ , EVuu = max (0, 0) = 0 Vuu = max Vuu

R = e−rh (π u Vuud + πd Vudd ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 3. 899 2) = Vud 2. 062 29 ¡ R ¢ , EVud = max (2. 062 29, 0) = 2. 062 29 Vud = max Vud R = e−rh (πu Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 806 × 3. 899 2 + 0.543 194 × 30. 571 5) = Vdd 17. 903 58 ¢ ¡ R , EVdd = max (17. 903 58, 20. 403 4) = 20. 403 4 Vdd = max Vdd

At the dd node, the exercise value is greater than the roll-back value. The put is exercised at dd. Now we have: www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

Su = 100 (1. 221 246) = 122. 124 6 EVu = 0 Vu =?

Sd = 100 (0.863 693) = 86. 369 3 EVd = 95 − 86. 369 3 = 8. 630 7 Vd =?

2 ¢ ¡ Suu = 100 1. 221 2462 = 149. 144 2 EVuu = 0 Vuu = 0 Sud = 100 (1. 221 246) (0.863 693) = 105. 478 2 EVud = 0 Vud = 2. 062 29 ¡ ¢ Sdd = 100 0.863 6932 = 74. 596 6 EVdd = 95 − 74. 596 6 = 20. 403 4 Vdd = 20. 403 4

VuR = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 2. 062 29) = 1. 090 75 ¡ ¢ Vu = max VuR , EVu = max (1. 090 75, 0) = 1. 090 75 VdR = e−rh (πu Vud + πd Vdd ) = e−(0.08)1/3 (0.456 806 × 2. 062 29 + 0.543 194 × 20. 403 4) = 11. 708 64 ¡ ¢ Vd = max VdR , EVd = max (11. 708 64, 8. 630 7) = 11. 708 64 Now we have: Period 0 1 Su = 100 (1. 221 246) = 122. 124 6 EVu = 0 Vu = 1. 090 75 S = 100 EV = 0 V =? Sd = 100 (0.863 693) = 86. 369 3 EVd = 95 − 86. 369 3 = 8. 630 7 Vd = 11. 708 64 V R = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 806 × 1. 090 75 + 0.543 194 × 11. 708 64) = 6. 677 85 ¡ ¢ V = max V R , EV = max (6. 677 85, 0) = 6. 677 85 b. Calculate the European put premium. Verify that the put-call parity holds. Since the option is European, we just calculate the roll-back value. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 2

3 Vuuu = 0

Vuu =? Vuud = 0 Vud =? Vudd = 3. 899 2 Vdd =? −rh

Vuu = e 0

Vddd = 30. 571 5 (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 0) =

Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 3. 899 2) = 2. 062 29 Vdd = e−rh (πu Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 806 × 3. 899 2 + 0.543 194 × 30. 571 5) = 17. 903 58 Now we have: Period 1 2 Vuu = 0 Vu =? Vud = 2. 062 29 Vd =? Vdd = 17. 903 58 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 806 × 0 + 0.543 194 × 2. 062 29) = 1. 090 75 Vd = e−rh (π u Vud + πd Vdd ) = e−(0.08)1/3 (0.456 806 × 2. 062 29 + 0.543 194 × 17. 903 58) = 10. 386 48 Now we have: Period 0 1 Vu = 1. 090 75 V =? Vd = 10. 386 48 −rh V =e (π u Vu + π d Vd ) = e−(0.08)1/3 (0.456 806 × 1. 090 75 + 0.543 194 × 10. 386 48) = 5. 978 6 Check whether the put-call parity-holds: C + Ke−rT = 18. 282 51 + 95e−(0.08)1 = 105. 978 6 P + S = 5. 978 6 + 100 = 105. 978 6 → C + Ke−rT = P + S Problem 10.11. h = T /n = 1/3 u = e(r−δ)h+σ

√ h

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= e(0.08−0.08)1/3+0.3

√

1/3

= 1. 189 11

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CHAPTER 10. BINOMIAL OPTION PRICING I d = e(r−δ)h−σ

πu =

√ h

√ = e(0.08−0.08)1/3−0.3 1/3 = 0.840 965

e(r−δ)h − d e(0.08−0.08)1/3 − 0.840 965 = = 0.456 807 u−d 1. 189 11 − 0.840 965

πd = 1 − π u = 1 − 0.456 807 = 0.543 193 a. • Calculate the price of the American call option Period 2 ¡ ¢ Suu = 100 1. 189 112 = 141. 398 3 EVuu = 141. 398 3 − 95 = 46. 398 3 Vuu =? Sud = 100 (1. 189 11) (0.840 965) = 100 EVud = 100 − 95 = 5 Vud =? ¡ ¢ Sdd = 100 0.840 9652 = 70. 722 2 EVdd = 0 Vdd =?

3 ¡ ¢ Suuu = 100 1. 189 113 = 168. 138 1 Vuuu = 168. 138 1 − 95 = 73. 138 1

¡ ¢ Suud = 100 1. 189 112 (0.840 965) = 118. 9110 Vuud = 118. 9110 − 95 = 23. 911 ¡ ¢ Sudd = 100 (1. 189 11) 0.840 9652 = 84. 096 5 Vudd = 0 ¡ ¢ Sddd = 100 0.840 9653 = 59. 474 9 Vddd = 0

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 807 × 73. 138 1 + 0.543 193 × 23. 911) = 45. 177 3 ¡ R ¢ , EVuu = max (45. 177 3, 46. 398 3) = 46. 398 3 Vuu = max Vuu The call is early exercised at the uu node.

R = e−rh (πu Vuud + π d Vudd ) = e−(0.08)1/3 (0.456 807 × 23. 911 + 0.543 193 × 0) = Vud 10. 635 3 ¡ R ¢ , EVud = max (10. 635 3, 5) = 10. 635 3 Vud = max Vud

0

R Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 0) =

¡ R ¢ , EVdd = max (0, 0) = 0 Vdd = max Vdd

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 46. 398 3

Su = 100 (1. 189 11) = 118. 911 EVu = 118. 911 − 95 = 23. 911 Vu =? Vud = 10. 635 3 Sd = 100 (0.840 965) = 84. 096 5 EVd = 0 Vd =? Vdd = 0 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.08)1/3 (0.456 807 × 46. 398 3 + 0.543 193 × 10. 635 3) = 26. 262 3 ¡ ¢ Vu = max VuR , EVu = max (26. 262 3, 23. 911) = 26. 262 3 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 10. 635 3 + 0.543 193 × 0) = 4. 730 4 ¡ ¢ Vd = max VdR , EVd = max (4. 730 4, 0) = 4. 730 4 Period 0

1 Vu = 26. 262 3

S = 100 EV = 100 − 95 = 5 Vu =? Vd = 4. 730 4 V R = e−rh (π u Vu + π d Vd ) = e−(0.08)1/3 (0.456 807 × 26. 262 3 + 0.543 193 × 4. 730 4) = 14. 183 0 ¡ ¢ V = max V R , EV = max (14. 183 0, 5) = 14. 183 0 • Calculate the price of the European call option Period 0

1

2

Vuu Vu V

Vud Vd Vdd

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3 ¡ ¢ Suuu = 100 1. 189 113 = 168. 138 1 Vuuu = 168. 138 1 − 95 = 73. 138 1

¢ ¡ Suud = 100 1. 189 112 (0.840 965) = 118. 9110 Vuud = 118. 9110 − 95 = 23. 911 ¢ ¡ Sudd = 100 (1. 189 11) 0.840 9652 = 84. 096 5 Vudd = 0 ¡ ¢ Sddd = 100 0.840 9653 = 59. 474 9 Vddd = 0 c °Yufeng Guo

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CHAPTER 10. BINOMIAL OPTION PRICING I Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 807 × 73. 138 1 + 0.543 193 × 23. 911) = 45. 177 3 Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.08)1/3 (0.456 807 × 23. 911 + 0.543 193 × 0) = 10. 635 3 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 0) = 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 807 × 45. 177 3 + 0.543 193 × 10. 635 3) = 25. 719 3 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 10. 635 3 + 0.543 193 × 0) = 4. 730 4 V = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 807 × 25. 719 3 + 0.543 193 × 4. 730 4) = 13. 941 5 Since the American call is early exercised at Node uu, the price of the American call 14. 183 0 is greater than the price of the European call 13. 941 5. c. Calculate the American put premium. Period 2 ¡ ¢ Suu = 100 1. 189 112 = 141. 398 3 EVuu = 0 Vuu =? Sud = 100 (1. 189 11) (0.840 965) = 100 EVud = 0 Vud =? ¡ ¢ Sdd = 100 0.840 9652 = 70. 722 2 EVdd = 95 − 70. 722 2 = 24. 277 8 Vdd =? R Vuu

0

3 ¡ ¢ Suuu = 100 1. 189 113 = 168. 138 1 Vuuu = 0

¡ ¢ Suud = 100 1. 189 112 (0.840 965) = 118. 9110 Vuud = 0 ¡ ¢ Sudd = 100 (1. 189 11) 0.840 9652 = 84. 096 5 Vudd = 95 − 84. 096 5 = 10. 903 5

¡ ¢ Sddd = 100 0.840 9653 = 59. 474 9 Vddd = 95 − 59. 474 9 = 35. 525 1 −rh −(0.08)1/3 =e (π u Vuuu + π d Vuud ) = e (0.456 807 × 0 + 0.543 193 × 0) =

¡ R ¢ Vuu = max Vuu , EVuu = max (0, 0) = 0

R = e−rh (πu Vuud + π d Vudd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 10. 903 5) = Vud 5. 766 9 ¡ R ¢ Vud = max Vud , EVud = max (5. 766 9, 0) = 5. 766 9

R = e−rh (π u Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 807 × 10. 903 5 + 0.543 193 × 35. 525 1) = Vdd 23. 638 9 ¡ R ¢ Vdd = max Vdd , EVdd = max (23. 638 9, 24. 277 8) = 24. 277 8 The American put is early exercised at Node dd.

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 0

Su = 100 (1. 189 11) = 118. 911 EVu = 0 Vu =? Vud = 5. 766 9 Sd = 100 (0.840 965) = 84. 096 5 EVd = 95 − 84. 096 5 = 10. 903 5 Vd =? Vdd = 24. 277 8 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 5. 766 9) = 3. 050 1 ¡ ¢ Vu = max VuR , EVu = max (3. 050 1, 0) = 3. 050 1

VdR = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 5. 766 9 + 0.543 193 × 24. 277 8) = 15. 405 6 ¡ ¢ Vd = max VdR , EVd = max (15. 405 6, 10. 903 5) = 15. 405 6 Period 0

1 Vu = 3. 050 1

S = 100 EV = 0 Vu =? Vd = 15. 405 6 V R = e−rh (π u Vu + π d Vd ) = e−(0.08)1/3 (0.456 807 × 3. 050 1 + 0.543 193 × 15. 405 6) = 9. 504 7 ¡ ¢ V = max V R , EV = max (9. 504 7, 0) = 9. 504 7 b. Calculate the European put premium. Verify that the put-call parity holds. Period 0 1 2 3 ¡ ¢ Suuu = 100 1. 189 113 = 168. 138 1 Vuuu = 0 Vuu ¢ ¡ Vu Suud = 100 1. 189 112 (0.840 965) = 118. 9110 Vuud = 0 V Vud ¢ ¡ Sudd = 100 (1. 189 11) 0.840 9652 = 84. 096 5 Vd Vudd = 95 − 84. 096 5 = 10. 903 5 Vdd ¡ ¢ Sddd = 100 0.840 9653 = 59. 474 9 Vddd = 95 − 59. 474 9 = 35. 525 1 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I

0

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 0) =

Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 10. 903 5) = 5. 766 9 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.08)1/3 (0.456 807 × 10. 903 5 + 0.543 193 × 35. 525 1) = 23. 638 9 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 5. 766 9) = 3. 050 1 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 5. 766 9 + 0.543 193 × 23. 638 9) = 15. 067 6 V = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 807 × 3. 050 1 + 0.543 193 × 15. 067 6) = 9. 325 9 So the European put is worth 9. 325 9. Verify the put-call parity: C + Ke−rT = 13. 941 5 + 95e−(0.08)1 = 101. 637 55 P + Se−δT = 9. 325 9 + 100e−(0.08)1 = 101. 637 53 Ignoring rounding difference, we get C + Ke−rT = P + Se−δT The put-call parity holds. Problem 10.12. a. h = T /n =√0.5/2 = 0.25 √ u = e(r−δ)h+σ h = e(0.08)0.25+0.3 0.25 = 1. 185 305 d = e(r−δ)h−σ πu =

√ h

√ 0.25

= e(0.08)0.25−0.3

= 0.878 095

e(r−δ)h − d e(0.08)0.25 − 0.878 095 = = 0.462 57 u−d 1. 185 305 − 0.878 095

πd = 1 − π u = 1 − 0.462 57 = 0.537 43 b. Since the stock doesn’t pay dividend, the American call and an otherwise identical European call have the same value. Period 0 1 2 ¡ ¢ Suu = 40 1. 185 3052 = 56. 197 92 Vuu = 56. 197 92 − 40 = 16. 197 92 Vu =? Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4 V =? Vud = 41. 632 4 − 40 = 1. 632 4 Vd =?

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¡ ¢ Sdd = 40 0.878 0952 = 30. 842 03 Vdd = 0 c °Yufeng Guo

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CHAPTER 10. BINOMIAL OPTION PRICING I Vu = e−rh (πu Vuu + π d Vud ) = e−(0.08)0.25 (0.462 57 × 16. 197 92 + 0.537 43 × 1. 632 4) = 8. 204 2 Vd = e−rh (π u Vud + πd Vdd ) = e−(0.08)0.25 (0.462 57 × 1. 632 4 + 0.537 43 × 0) = 0.740 1 V = e−rh (π u Vu + π d Vd ) = e−(0.08)0.25 (0.462 57 × 8. 204 2 + 0.537 43 × 0.740 1) = 4. 109 7 So both the American call and the European call are worth 4. 109 7. We also Node uu ud dd

calculate the European call price using the following shortcut: Payoff Risk Neutral Prob 16. 197 92 π 2u = 0.462 572 1. 632 4 2π u πd = 2 × 0.462 57 × 0.537 43 0 π 2d = 0.537 432

P V = e−rT ¡payof f × RiskN eutralP rob ¢ = e−0.08(0.5) 16. 197 92 × 0.462 572 + 1. 632 4 × 2 × 0.462 57 × 0.537 43 + 0 × 0.537 432 = 4. 109 8 Please note that the above shortcut works only for European options. c. Even if the stock doesn’t pay dividend, it may be still optimal to exercise an American put early. • Calculate the premium of the American put Period 0

1 Su = 40 (1. 185 305) = 47. 412 2 EVu = 0 Vu =?

S = 40 EV = 0 V =?

2 ¡ ¢ Suu = 40 1. 185 3052 = 56. 197 92 Vuu = 0 Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4 Vud = 0

Sd = 40 (0.878 095) = 35. 123 8 EVd = 40 − 35. 123 8 = 4. 876 2 Vd =?

¡ ¢ Sdd = 40 0.878 0952 = 30. 842 03 Vdd = 40 − 30. 842 03 = 9. 157 97

VuR = e−rh (π u Vuu + πd Vud ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 0) = 0

¢ ¡ Vu = max VuR , EVu = max (0, 0) = 0

VdR = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 9. 157 97) = 4. 843 6 ¡ ¢ Vd = max VdR , EVd = max (4. 843 6, 4. 876 2) = 4. 876 2 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I So the American put is early exercised at Node d. V R = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 4. 876 2) = 2. 579 0 ¡ ¢ V = max V R , EV = max (2. 579 0, 0) = 2. 579 0 The American put is worth 2. 579 0.

• Calculate the premium of the European put Period 0

1

2 ¢ ¡ Suu = 40 1. 185 3052 = 56. 197 92 Vuu = 0

Vu

Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4 Vud = 0

V

¡ ¢ Sdd = 40 0.878 0952 = 30. 842 03 Vdd = 40 − 30. 842 03 = 9. 157 97

Vd

Vu = e−rh (πu Vuu + π d Vud ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 0) = 0 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 9. 157 97) = 4. 843 6 V = e−rh (πu Vu + πd Vd ) = e−(0.08)1/3 (0.456 807 × 0 + 0.543 193 × 4. 843 6) = 2. 561 8 The European put is worth 2. 561 8. Problem 10.13. a. From the previous problem, we know: Period 0 1

S = 40 V = 4. 109 7 (∆, B)

Su = 40 (1. 185 305) = 47. 412 2 Vu = 8. 204 2 (∆u , Bu )

Sd = 40 (0.878 095) = 35. 123 8 Vd = 0.740 1 (∆d , Bd )

2 ¢ ¡ Suu = 40 1. 185 3052 = 56. 197 92 Vuu = 56. 197 92 − 40 = 16. 197 92 Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4 Vud = 41. 632 4 − 40 = 1. 632 4 ¢ ¡ Sdd = 40 0.878 0952 = 30. 842 03 Vdd = 0

First, let find the replicating portfolio. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I • The replicating portfolio at t = 0: ½ 4Su + Berh = Vu 4Sd + Berh = Vd ½

447. 412 2 + Be0.08(0.25) = 8. 204 2 435. 123 8 + Be0.08(0.25) = 0.740 1

4 = 0.607 41

B = −20. 186 7

• The replicating portfolio Node u: ½ 4u Suu + Bu erh = Vuu 4u Sud + Berh = Vud ½

207 9

4u 56. 197 92 + Bu e0.08(0.25) = 16. 197 92 4u 41. 632 4 + Bu e0.08(0.25) = 1. 632 4

4u = 1

Bu = −39.

• The replicating portfolio Node d: ½ 4d Sud + Bd erh = Vud 4d Sdd + Berh = Vdd ½

4d 41. 632 4 + Bd e0.08(0.25) = 1. 632 4 4d 30. 842 03 + Bd e0.08(0.25) = 0

4d = 0.151 28

Bd = −4.

573 5 At t = 0, we

• sell the over-priced call for $5 • build a synthetic call by buying 4 = 0.607 41 share of the stock and borrowing $20. 186 7 from a bank. Cost: 4S + B = 0.607 41 (40) − 20. 186 7 = 4. 109 7. So we receive 5 − 4. 109 7 = 0.890 3 at t = 0. b. • Suppose the call in the market at time h is fairly priced To liquidate our position at t = h (i.e. the end of Period 1) yet hedge our sold call, we’ll do the following: 1. Sell our 4 = 0.607 41 share of the stock, receiving 4Sh 2. Pay the bank Berh = 20. 186 7e0.08(0.25) = 20. 594 5 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I 3. Buy the call from the open market for the fair price of Vh , which is either Vu = 8. 204 2 if the stock price goes up or Vd = 0.740 1 if the stock price goes down.

Suppose the stock price goes up to Su and we liquidate our position at Node u. We’ll 1. Sell our 4 = 0.607 41 share of the stock, receiving 4Su = 0.607 41 (47. 412 2) = $28. 798 6 2. Pay the bank Berh = 20. 186 7e0.08(0.25) = $20. 594 5 3. Buy the call from the open market for the fair price of Vu = $8. 204 2 The net receipt is: 28. 798 6 − 20. 594 5 − 8. 204 2 = −0.000 1 = $0 (if we ignore rounding errors) So we receive $0.890 3 free money at t = 0 without incurring any liability at time h. Suppose the stock price goes up to Sd and we liquidate our position at Node u. We’ll 1. Sell our 4 = 0.607 41 share of the stock, receiving 4Sd = 0.607 41 (35. 123 8) = $21. 334 5 2. Pay the bank Berh = 20. 186 7e0.08(0.25) = $20. 594 5 3. Buy the call from the open market for the fair price of Vd = $0.740 1 The net receipt is: 21. 334 5 − 20. 594 5 − 0.740 1 = −0.000 1 = 0 (if we ignore rounding errors) So we receive $0.890 3 free money at t = 0 without incurring any liability at time h.

• Suppose the call in the market at h is over priced We’ll not liquidate our position at h. Instead, we rebalance our replicating portfolio at t = h and liquidate our position at T = 2h. We’ll consider two situation. Situation #1 — the stock price goes up to Su = 40 (1. 185 305) = 47. 412 2 at time h We’ll change our replicating portfolio from (∆, B) = (0.607 41, B = −$20. 186 7) to (∆u , Bu ) = (1, −$39. 207 9) at h. We need to buy 1 − 0.607 41 = 0.392 59 share of the stock. The cost is 0.392 59 (47. 412 2) = 18. 613 6. We’ll borrow 18. 613 6 from the bank to pay for this. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I So at time h, our total debt to the bank is: 18. 613 4 + 20. 186 7e0.08(0.25) = 39. 207 9. The number of stocks we have is 1. Now at time h, our replicating portfolio is exactly (∆u , Bu ) = (1, −$39. 207 9). Then at T = ¡ 2h ¢ If Suu = 40 1. 185 3052 = 56. 197 92

1. Our written call is exercised against us. We need to pay the call holder Vuu = 56. 197 92 − 40 = 16. 197 92. 2. We sell ∆u stocks in the market and pay off our loan from the bank. Our net cash receipt is ∆u Suu + Bu erh = 1 × 56. 197 92 − 39. 207 9e0.08(0.25) = 16. 197 97. 3. Our net cash flow is zero (ignoring rounding). If Sud = 40 (1. 185 305) (0.878 095) = 41. 632 4 1. Our written call is exercised against us. We need to pay the call holder Vud = 41. 632 4 − 40 = 1. 632 4

2. We sell ∆u stocks in the market and pay off our loan from the bank. Our net cash receipt is ∆u Sud + Bu erh = 1 × 41. 632 4 − 39. 207 9e0.08(0.25) = 1. 632 4 3. Our net cash flow is zero. So we receive $0.890 3 free money at t = 0 without incurring any liability at T. Situation #2 — the stock price goes down to Sd = 40 (0.878 095) = 35. 123 8 at time h We’ll change our replicating portfolio from (∆, B) = (0.607 41, B = −$20. 186 7) to (∆d , Bd ) = (0.151 28, −$4. 573 5) at h. We need to sell 0.607 4 − 0.151 28 = 0.456 12 share of the stock, receiving 0.456 12 (35. 123 8) = $16. 020 7. We immediately send a check of $16. 020 7 to the bank to partially pay our debt. Now our remaining debt to the bank is 20. 186 7e0.08(0.25) − 16. 020 7 = 4. 573 8 = 4. 573 5 (ignore rounding) Now at time h, our replicating portfolio is exactly (∆d , Bd ) = (0.151 28, −$4. 573 5) Then at T = 2h If Sdu = 40 (1. 185 305) (0.878 095) = 41. 632 4 1. Our written call is exercised against us. We need to pay the call holder Vdu = 41. 632 4 − 40 = 1. 632 4 2. We sell ∆d stocks in the market and pay off our loan from the bank. Our net cash receipt is ∆d Sdu +Bd erh = 0.151 28×41. 632 4−4. 573 5e0.08(0.25) = 1. 632 3 3. Our net cash flow is zero (ignoring rounding). www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I ¡ ¢ If Sdd = 40 0.878 0952 = 30. 842 03 1. Our written call expires worthless. 2. We sell ∆d stocks in the market and pay off our loan from the bank. Our net cash receipt is ∆d Sdd +Bd erh = 0.151 28×30. 842 03−4. 573 5e0.08(0.25) ≈ 0 3. Our net cash flow is zero. So we receive $0.890 3 free money at t = 0 without incurring any liability at T.

c. If the call option is under priced at h, we just sell off our replicating portfolio (i.e. sell ∆ share of the stock and pay back our loan to the bank) and purchase a call from the market. Because the value of our replicating portfolio at h represents the fair price of a call and the call in the market is lower than the fair price, the money we get from selling our replicating portfolio exceeds the call price in the market. So we’ll make some profit. Now we are standing at time h. The call we buy at time h has life from time h to T = 2h. The call we initially sold at time zero has life from h to T = 2h. These two calls exactly offset each other so our liability at T is zero.

Problem 10.14. We can use the general binomial tree formula by setting S = 0.92 and δ = r€ = 0.03. h = T /n = 0.75/3 = 0.25

u = 1.2 d = 0.9 e(r−δ)h − d e(0.04−0.03)0.25 − 0.9 πu = = = 0.341 677 u−d 1.2 − 0.9 πd = 1 − π u = 1 − 0.341 677 = 0.658 323 We’ll calculate part b first. b. Calculate the American call premium. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 2 ¡ ¢ Suu = 0.92 1. 22 = 1. 324 8 EVuu = 1. 324 8 − 0.85 = 0.474 8 Vuu =? Sud = 0.92 (1. 2) (0.9) = 0.993 6 EVud = 0.993 6 − 0.85 = 0.143 6 Vud =? ¡

¢ Sdd = 0.92 0.92 = 0.745 2 EVdd = 0 Vdd =?

3 ¢ ¡ Suuu = 0.92 1. 23 = 1. 589 76 Vuuu = 1. 589 76 − 0.85 = 0.739 76

¡ ¢ Suud = 0.92 1. 22 (0.9) = 1. 192 32 Vuud = 1. 192 32 − 0.85 = 0.342 32 ¡ ¢ Sudd = 0.92 (1. 2) 0.92 = 0.894 24 Vudd = 0.894 24 − 0.85 = 0.044 24 ¡ ¢ Sddd = 0.92 0.93 = 0.670 68 Vddd = 0

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.04)0.25 (0.341 677 × 0.739 76 + 0.658 323 × 0.342 32) = 0.473 359 ¡ R ¢ Vuu = max Vuu , EVuu = max (0.473 359, 0.474 8) = 0.474 8 So the American call is early exercised at the uu node. R Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.04)0.25 (0.341 677 × 0.342 32 + 0.658 323 × 0.044 24) = 0.144 633

¡ R ¢ Vud = max Vud , EVud = max (0.144 633, 0.143 6) = 0.144 633 R Vdd = e−rh (πu Vudd + πd Vddd ) = e−(0.04)0.25 (0.341 677 × 0.044 24 + 0.658 323 × 0) = 0.01 496 5 ¡ R ¢ Vdd = max Vdd , EVdd = max (0.01 496 5, 0) = 0.014 965 Period 1

2 Vuu = 0.474 8

Su = 0.92 (1. 2) = 1. 104 EVu = 1. 104 − 0.85 = 0.254 Vu =? Vud = 0.144 633 Sd = 0.92 (0.9) = 0.828 EVd = 0 Vd =? Vdd = 0.01 496 5 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.04)0.25 (0.341 677 × 0.474 8 + 0.658 323 × 0.144 633) = 0.254 882 ¡ ¢ Vu = max VuR , EVu = max (0.254 882 , 0.254) = 0.254 882 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.04)0.25 (0.341 677 × 0.144 633 + 0.658 323 × 0.01 496 5) = 0.05 8680 ¡ ¢ Vd = max VdR , EVd = max (0.05 8680 , 0) = 0.058 68 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 0

1 Vu = 0.254 882

S = 0.92 EV = 0.92 − 0.85 = 0.07 V =? Vd = 0.058 68 V R = e−rh (πu Vu + πd Vu ) = e−(0.04)0.25 (0.341 677 × 0.254 882 + 0.658 323 × 0.058 68) = 0.124 467 ¢ ¡ V = max V R , EV = max (0.124 467 , 0.07) = 0.124 467 So the American call premium is 0.124 467 .

a. Calculate the European call premium. Period 0 1 2 3 ¡ ¢ Suuu = 0.92 1. 23 = 1. 589 76 Vuuu = 1. 589 76 − 0.85 = 0.739 76 Vuu ¡ ¢ Suud = 0.92 1. 22 (0.9) = 1. 192 32 Vu Vuud = 1. 192 32 − 0.85 = 0.342 32 V

Vud Vd Vdd

¡ ¢ Sudd = 0.92 (1. 2) 0.92 = 0.894 24 Vudd = 0.894 24 − 0.85 = 0.044 24 ¡ ¢ Sddd = 0.92 0.93 = 0.670 68 Vddd = 0

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.04)0.25 (0.341 677 × 0.739 76 + 0.658 323 × 0.342 32) = 0.473 359 Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.04)0.25 (0.341 677 × 0.342 32 + 0.658 323 × 0.044 24) = 0.144 633 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.04)0.25 (0.341 677 × 0.044 24 + 0.658 323 × 0) = 0.01 496 5 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.04)0.25 (0.341 677 × 0.473 359 + 0.658 323 × 0.144 633) = 0.254 394 VdR = e−rh (πu Vud + πd Vdd ) = e−(0.04)0.25 (0.341 677 × 0.144 633 + 0.658 323 × 0.01 496 5) = 0.05 8680 V = e−rh (πu Vu + πd Vu ) = e−(0.04)0.25 (0.341 677 × 0.254 394 + 0.658 323 × 0.058 68) = 0.124 302 So the European call premium is 0.124 302 . www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Problem 10.15. We can use the general binomial tree formula by setting S = 0.92 and δ = r€ = 0.03. h = T /n = 0.75/3 = 0.25

u = 1.2 d = 0.9 e(r−δ)h − d e(0.04−0.03)0.25 − 0.9 πu = = = 0.341 677 u−d 1.2 − 0.9 π d = 1 − πu = 1 − 0.341 677 = 0.658 323 b. Calculate the American put premium. Period 2 3 ¡ ¢ Suuu = 0.92 1. 23 = 1. 589 76 ¡ 2¢ Suu = 0.92 1. 2 = 1. 324 8 Vuuu = 0 EVuu = 0 ¡ ¢ Vuu =? Suud = 0.92 1. 22 (0.9) = 1. 192 32 Vuud = 0 Sud = 0.92 (1. 2) (0.9) = 0.993 6 EVud = 1 − 0.993 6 = 0.006 4 Vud =? ¡ 2¢ S = 0.92 (1. 2) 0.9 = 0.894 24 udd ¡ ¢ Sdd = 0.92 0.92 = 0.745 2 Vudd = 1 − 0.894 24 = 0.105 76 EVdd = 1 − 0.745 2 = 0.254 8 ¡ ¢ Vdd =? Sddd = 0.92 0.93 = 0.670 68 Vddd = 1 − 0.670 68 = 0.329 32

0

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0) =

¡ R ¢ , EVuu = max (0, 0) = 0 Vuu = max Vuu

R = e−rh (π u Vuud + πd Vudd ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0.105 76) = Vud 0.06893 1

¡ R ¢ , EVud = max (0.06893 1, 0.006 4) = 0.068 931 Vud = max Vud

R = e−rh (πu Vudd + πd Vddd ) = e−(0.04)0.25 (0.341 677 × 0.105 76 + 0.658 323 × 0.329 32) = Vdd 0.250 418 ¡ R ¢ , EVdd = max (0.250 418, 0.254 8) = 0.254 8 Vdd = max Vdd The American put is early exercised at the dd node.

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 0

Su = 0.92 (1. 2) = 1. 104 EVu = 0 Vu =? Vud = 0.068 931 Sd = 0.92 (0.9) = 0.828 EVd = 1 − 0.828 = 0.172 Vd =? Vdd = 0.254 8 VuR = e−rh (πu Vuu + π d Vud ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0.068 931 ) = 0.04 492 7 ¡ ¢ Vu = max VuR , EVu = max (0.04 492 7 , 0) = 0.04 492 7 VdR = e−rh (πu Vud + πd Vdd ) = e−(0.04)0.25 (0.341 677 × 0.068 931 + 0.658 323 × 0.254 8) = 0.189 389 ¡ ¢ Vd = max VdR , EVd = max (0.189 389 , 0.172) = 0.189 389 Period 0

1 Vu = 0.04 492 7

S = 0.92 EV = 1 − 0.92 = 0.08 V =? Vd = 0.189 389 V R = e−rh (πu Vu + πd Vu ) = e−(0.04)0.25 (0.341 677 × 0.04 492 7 + 0.658 323 × 0.189 389) = 0.138 636 ¡ ¢ V = max V R , EV = max (0.138 636 , 0.08) = 0.138 636 So the American put premium is 0.138 636 .

a. Calculate the European put premium. Period 0 1 2 3 ¡ ¢ Suuu = 0.92 1. 23 = 1. 589 76 Vuuu = 0 Vuu ¡ ¢ Vu Suud = 0.92 1. 22 (0.9) = 1. 192 32 Vuud = 0 V Vud ¢ ¡ Sudd = 0.92 (1. 2) 0.92 = 0.894 24 Vd Vudd = 1 − 0.894 24 = 0.105 76 Vdd ¡ ¢ Sddd = 0.92 0.93 = 0.670 68 Vddd = 1 − 0.670 68 = 0.329 32 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0) = 0 Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0.105 76) = 0.06893 1 Vdd = e−rh (πu Vudd + πd Vddd ) = e−(0.04)0.25 (0.341 677 × 0.105 76 + 0.658 323 × 0.329 32) = 0.250 418 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.04)0.25 (0.341 677 × 0 + 0.658 323 × 0.068 931 ) = 0.04 492 7 Vd = e−rh (π u Vud + πd Vdd ) = e−(0.04)0.25 (0.341 677 × 0.068 931 + 0.658 323 × 0.250 418) = 0.186 533 V = e−rh (π u Vu + π d Vu ) = e−(0.04)0.25 (0.341 677 × 0.04 492 7 + 0.658 323 × 0.186 533) = 0.136 775 So the European put premium is 0.136 775 . Problem 10.16. This problem looks scary, but it’s actually simple if you know what’s the underlying asset. The underlying asset is $1. The call option gives the call holder the right to purchase $1 with a guaranteed price (i.e. strike price) of 120 Yen. So the strike price and the option price are expressed in Yen. To simplify the problem, we can treat the underlying asset $1 as a stock and translate the original problem into the following: • Now we are living in Japan and are interested in options on a stock. • The current stock price is S = 120 Yen. • The continuously compounded risk free interest rate is rf = 0.01 (remember we are living in Japan). • The stock’s volatility is σ = 0.1. • The stock’s continuous dividend yield is δ = 0.05; the dollar interest rate is like the dividend yield. • T =1 • h = T /3 = 1/3 Now we can use the standard binomial formula. √ √ u = e(r−δ)h+σ h = e(0.01−0.05)1/3+0.1 1/3 = 1. 045 402 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I d = e(r−δ)h−σ πu =

√ h

√ = e(0.01−0.05)1/3−0.1 1/3 = 0.931 398

e(r−δ)h − d e(0.01−0.05)1/3 − 0.931 398 = = 0.485 57 u−d 1. 045 402 − 0.931 398

πd = 1 − π u = 1 − 0.485 57 = 0.514 43

a. Calculate the price of the American call option Period 2 3 ¡ ¢ Suuu = 120 1. 045 4023 = 137. 098 0 ¡ ¢ Suu = 120 1. 045 4022 = 131. 143 8 Vuuu = 137. 098 0 − 120 = 17. 098 EVuu = 131. 143 8 − 120 = 11. 143 8 ¡ ¢ Vuu =? Suud = 120 1. 045 4022 (0.931 398) = 122. 147 1 Vuud = 122. 147 1 − 120 = 2. 147 1 Sud = 120 (1. 045 402) (0.931 398) = 116. 842 2 EVud = 0 Vud =? ¡ ¢ Sudd = 120 (1. 045 402) 0.931 3982 = 108. 826 6 ¡ ¢ Sdd = 120 0.931 3982 = 104. 100 3 Vudd = 0 EVdd = 0 ¡ ¢ Vdd =? Sddd = 120 0.931 3983 = 96. 958 8 Vddd = 0 R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.01)1/3 (0.485 57 × 17. 098 + 0.514 43 × 2. 147 1) = 9. 375 5 ¡ R ¢ Vuu = max Vuu , EVuu = max (9. 375 5, 11. 143 8) = 11. 143 8 So the American call is early exercised at the uu node. R = e−rh (πu Vuud + π d Vudd ) = e−(0.01)1/3 (0.485 57 × 2. 147 1 + 0.514 43 × 0) = Vud 1. 039 1 ¡ R ¢ , EVud = max (1. 039 1, 0) = 1. 039 1 Vud = max Vud

0

R = e−rh (πu Vudd + π d Vddd ) = e−(0.01)1/3 (0.485 57 × 0 + 0.514 43 × 0) = Vdd

¡ R ¢ Vdd = max Vdd , EVdd = max (0, 0) = 0 Period 1

2 Vuu = 11. 143 8

Su = 120 (1. 045 402) = 125. 448 2 EVu = 125. 448 2 − 120 = 5. 448 2 Vu =? Vud = 1. 039 1 Sd = 120 (0.931 398) = 111. 767 8 EVd = 0 Vd =? Vdd = 0 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I VuR = e−rh (π u Vuu + π d Vud ) = e−(0.01)1/3 (0.485 57 × 11. 143 8 + 0.514 43 × 1. 039 1) = 5. 925 9 ¡ ¢ Vu = max VuR , EVu = max (5. 925 9 , 5. 448 2) = 5. 925 9 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.01)1/3 (0.485 57 × 1. 039 1 + 0.514 43 × 0) = 0.502 9 ¡ ¢ Vd = max VdR , EVd = max (0.502 9 , 0) = 0.502 9 Period 0

1 Vu = 5. 925 9

S = 120 EV = 0 V =? Vd = 0.502 9 V R = e−rh (π u Vu + π d Vu ) = e−(0.01)1/3 (0.485 57 × 5. 925 9 + 0.514 43 × 0.502 9) = 3. 125 7 ¡ ¢ V = max V R , EV = max (3. 125 7 , 0) = 3. 125 7 So the American call premium is 3. 125 7 Yen.

Calculate the price of the European call Period 0 1 2 3 ¡ ¢ Suuu = 120 1. 045 4023 = 137. 098 0 Vuuu = 137. 098 0 − 120 = 17. 098 Vuu ¡ ¢ Vu Suud = 120 1. 045 4022 (0.931 398) = 122. 147 1 Vuud = 122. 147 1 − 120 = 2. 147 1 V Vud ¡ ¢ Sudd = 120 (1. 045 402) 0.931 3982 = 108. 826 6 Vd Vudd = 0 Vdd ¡ ¢ Sddd = 120 0.931 3983 = 96. 958 8 Vddd = 0 Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.01)1/3 (0.485 57 × 17. 098 + 0.514 43 × 2. 147 1) = 9. 375 5 Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.01)1/3 (0.485 57 × 2. 147 1 + 0.514 43 × 0) = 1. 039 1 Vdd = e−rh (π u Vudd + π d Vddd ) = e−(0.01)1/3 (0.485 57 × 0 + 0.514 43 × 0) = 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.01)1/3 (0.485 57 × 9. 375 5 + 0.514 43 × 1. 039 1) = 5. 070 1 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Vd = e−rh (π u Vud + π d Vdd ) = e−(0.01)1/3 (0.485 57 × 1. 039 1 + 0.514 43 × 0) = 0.502 9 V = e−rh (πu Vu + πd Vu ) = e−(0.01)1/3 (0.485 57 × 5. 070 1 + 0.514 43 × 0.502 9) = 2. 711 5 So the European call premium is 2. 711 5 Yen. b. Calculate the price of the American put Period 2 ¡ ¢ Suu = 120 1. 045 4022 = 131. 143 8 EVuu = 0 Vuu =? Sud = 120 (1. 045 402) (0.931 398) = 116. 842 2 EVud = 120 − 116. 842 2 = 3. 157 8 Vud =? ¡ ¢ Sdd = 120 0.931 3982 = 104. 100 3 EVdd = 120 − 104. 100 3 = 15. 899 7 Vdd =?

0

3 ¡ ¢ Suuu = 120 1. 045 4023 = 137. 098 0 Vuuu = 0

¡ ¢ Suud = 120 1. 045 4022 (0.931 398) = 122. 147 1 Vuud = 0 ¡ ¢ Sudd = 120 (1. 045 402) 0.931 3982 = 108. 826 6 Vudd = 120 − 108. 826 6 = 11. 173 4 ¡ ¢ Sddd = 120 0.931 3983 = 96. 958 8 Vddd = 120 − 96. 958 8 = 23. 041 2

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.01)1/3 (0.485 57 × 0 + 0.514 43 × 0) =

¡ R ¢ Vuu = max Vuu , EVuu = max (0, 0) = 0

R = e−rh (πu Vuud + π d Vudd ) = e−(0.01)1/3 (0.485 57 × 0 + 0.514 43 × 11. 173 4) = Vud 5. 728 8 ¡ R ¢ , EVud = max (3. 157 8, 5. 728 8) = 5. 728 8 Vud = max Vud

R = e−rh (π u Vudd + πd Vddd ) = e−(0.01)1/3 (0.485 57 × 11. 173 4 + 0.514 43 × 23. 041 2) = Vdd 17. 221 1 ¡ R ¢ Vdd = max Vdd , EVdd = max (17. 221 1, 15. 899 7) = 17. 221 1

Period 1

2 Vuu = 0

Su = 120 (1. 045 402) = 125. 448 2 EVu = 0 Vu =? Vud = 5. 728 8 Sd = 120 (0.931 398) = 111. 767 8 EVd = 120 − 111. 767 8 = 8. 232 2 Vd =? Vdd = 17. 221 1 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I VuR = e−rh (π u Vuu + π d Vud ) = e−(0.01)1/3 (0.485 57 × 0 + 0.514 43 × 5. 728 8) = 2. 937 3 ¡ ¢ Vu = max VuR , EVu = max (2. 937 3 , 0) = 2. 937 3

VdR = e−rh (π u Vud + π d Vdd ) = e−(0.01)1/3 (0.485 57 × 5. 728 8 + 0.514 43 × 17. 221 1) = 11. 602 0 ¡ ¢ Vd = max VdR , EVd = max (11. 602 0 , 8. 232 2) = 11. 602 Period 0

1 Vu = 2. 937 3

S = 120 EV = 0 V =? Vd = 11. 602 V R = e−rh (π u Vu + π d Vu ) = e−(0.01)1/3 (0.485 57 × 2. 937 3 + 0.514 43 × 11. 602) = 7. 370 1 ¢ ¡ V = max V R , EV = max (7. 370 1 , 0) = 7. 370 1

Since the exercise value is never greater than the roll-back value, the American put is not early exercised. Hence the American put and the European put have the same value. Both the American put and European put are worth 7. 370 1 Yen.

c. Since the underlying asset generates dividend (the Yen interest rate is like the dividend rate), it may be optimal to early exercise an American call. However, even if the asset generates dividend, it may not be optimal to early exercise an American put.

Problem 10.17. Method 1 Apply the stock binomial formula to futures As explained in my study guide, we can apply the stock binomial formula to options on futures if we do the following three things: • Set the dividend yield equal to the risk free rate (i.e. δ = r) Cu − Cd Cu − Cd instead of 4 = e−rh Su − Sd Su − Sd µ ¶ 1−d u−1 Su Cd − Sd Cu • B = V = e−rh Cu . + Cd instead of B = e−rh u−d u−d Su − Sd

• 4=

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CHAPTER 10. BINOMIAL OPTION PRICING I h = 1 year long. √

√

√

u = e(r−δ)h+σ√ h = eσ h√= e0.1 1 = 1. 105 170 9 d = e(r−δ)h−σ h = e−0.1 1 = 0.904 837 4 e(r−δ)h − d 1−d 1 − 0.904 837 4 πu = = = = 0.475 021 u−d u−d 1. 105 170 9 − 0.904 837 4 πd = 1 − π u = 1 − 0.475 021 = 0.524 979 Period 0

S = 300 EV = 300 − 290 = 10 V =? (∆, B) =?

1 Su = 300 (1. 105 170 9) = 331. 551 27 Vu = 331. 551 27 − 290 = 41. 551 27

Sd = 300 (0.904 837 4) = 271. 451 22 Vd = 0 V R = e−rh (πu Vu + πd Vu ) = e−(0.06)1 (0.475 021 × 41. 551 27 + 0.524 979 × 0) = 18. 588 29 ¡ ¢ V = max V R , EV = max (18. 588 29 , 10) = 18. 588 29 So the call premium is $18. 588 29 at t = 0. 41. 551 27 − 0 Cu − Cd = = 0.691 368 Su − Sd 331. 551 27 − 271. 451 22 B = V = 18. 588 29

4=

So to form a replicating portfolio at time zero, we buy 0.691 368 unit of futures contract (i.e. enter 0.691 368 unit of futures contract as a buyer) and invest $18. 588 29 in a savings account. Method 2 Just as√in Method 1, we calculate √ h = e0.1 1 √= 1. 105 170 9 u = eσ √ d = e−σ h = e−0.1 1 = 0.904 837 4 Next, we build a futures price tree. Period 0 1 u F1,1 = 300 (1. 105 170 9) = 331. 551 27 Vu = 331. 551 27 − 290 = 41. 551 27 F0,1 = 300 EV = 300 − 290 = 10 V =? (∆, B) =? d F1,1 = 300 (0.904 837 4) = 271. 451 22 Vd = 0 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I If we buy ∆ futures contract at t = 0, our gain in these futures contracts at T is: ¢ ¡ u − F0,1 = ∆F0,1 (u − 1) in the u node • ∆ F1,1 ¡ d ¢ • ∆ F1,1 − F0,1 = ∆F0,1 (d − 1) in the d node

If we put $B in a savings account, we’ll get BerT at T . We want our replicating portfolio and the call option have the same payoff at T½: ¡ ¢ ½ u − F0,1 ¢ + BerT = Vu ∆ ¡F1,1 ∆F0,1 (u − 1) + BerT = Vu → d rT ∆F0,1 (d − 1) + BerT = Vd ∆ F1,1 − F0,1 + Be = Vd ½

∆300 (1. 105 170 9 − 1) + Be0.06(1) = 41. 551 27 ∆300 (0.904 837 4 − 1) + Be0.06(1) = 0 ∆ = 0.691 368 B = 18. 588 29

Since it costs nothing to enter a future contract, the call premium is equal to B: V = B = 18. 588 29

The statement "Replicating a call option always entails borrowing to buy the underlying asset" is true for calls on stocks. However, it’s not true for calls on futures; it costs nothing to enter a futures contract (so you don’t need to borrow money to finance your transaction on futures). If you write a call on futures, you need to enter ∆ futures contracts and simultaneously deposit the call premium into a savings account. This way, at call expiration date T , the combined payoff of your ∆ futures contracts and your deposit in the savings account will replicate the payoff of a call on futures.

Problem 10.18. We’ll apply the stock binomial formula to futures. We need to do the following three things: • Set the dividend yield equal to the risk free rate (i.e. δ = r) Cu − Cd Cu − Cd instead of 4 = e−rh Su − Sd Su − Sd µ ¶ 1−d u−1 Su Cd − Sd Cu • B = V = e−rh Cu . + Cd instead of B = e−rh u−d u−d Su − Sd

• 4=

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CHAPTER 10. BINOMIAL OPTION PRICING I h = 1/3 √ √ √ = e0.3 1/3 = 1. 189 110 u = e(r−δ)h+σ√ h = eσ h√ d = e(r−δ)h−σ h = e−0.3 1/3 = 0.840 965 Notice that ud = 1 e(r−δ)h − d 1−d 1 − 0.840 965 = = = 0.456 807 πu = u−d u−d 1. 189 110 − 0.840 965 πd = 1 − π u = 1 − 0.456 807 = 0.543 193

• calculate the American call premium Period 2 Suu = 1000u2 = 1413. 982 6 EVuu = 1413. 982 6 − 1000 = 413. 982 6 Vuu =? Sud = 1000ud = 1000 EVud = 0 Vud =?

3 Suuu = 1000u3 = 1681. 380 8 Vuuu = 1681. 380 8 − 1000 = 681. 380 8 Suud = 1000u2 d = 1000u = 1189. 1100 Vuud = 1189. 11 − 1000 = 189. 11

Sudd = 1000ud2 = 1000d = 840. 965 Vudd = 0

2

Sdd = 1000d = 707. 222 1 EVdd = 0 Vdd =?

¡ ¢ Sddd = 1000d3 = 1000 0.840 965 3 = 594. 749 Vddd = 0

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1/3 (0.456 807 × 681. 380 8 + 0.543 193 × 189. 11) = 407. 140 2 ¡ R ¢ Vuu = max Vuu , EVuu = max (407. 140 2, 413. 982 6) = 413. 982 6

The American call is early exercised at Node uu. R Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1/3 (0.456 807 × 189. 11 + 0.543 193 × 0) = 84. 958 9 ¡ R ¢ , EVud = max (84. 958 9, 0) = 84. 958 9 Vud = max Vud

0

R = e−rh (π u Vudd + πd Vddd ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 0) = Vdd

¡ R ¢ Vdd = max Vdd , EVdd = max (0, 0) = 0

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 413. 982 6

Su = 1000 (1. 189 11) = 1189. 11 EVu = 1189. 11 − 1000 = 189. 11 Vu =? Vud = 84. 958 9 Sd = 1000 (0.840 965 ) = 840. 965 EVd = 0 Vd =? Vdd = 0 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.05)1/3 (0.456 807 × 413. 982 6 + 0.543 193 × 84. 958 9) = 231. 370 7 ¡ ¢ Vu = max VuR , EVu = max (231. 370 7 , 189. 11) = 231. 370 7 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.05)1/3 (0.456 807 × 84. 958 9 + 0.543 193 × 0) = 38. 168 4 ¡ ¢ Vd = max VdR , EVd = max (38. 168 4 , 0) = 38. 168 4 Period 0

1 Vu = 231. 370 7

S = 1000 EV = 0 V =? Vd = 38. 168 4 V R = e−rh (π u Vu + π d Vu ) = e−(0.05)1/3 (0.456 807 × 231. 370 7 + 0.543 193 × 38. 168 4) = 124. 334 9 ¡ ¢ V = max V R , EV = max (124. 334 9 , 0) = 124. 334 9 So the American call premium is 124. 334 9. • Calculate the European call premium Period 0

1

2

Vuu Vu V

Vud Vd Vdd

www.actuary88.com

3 Suuu = 1000u3 = 1681. 380 8 Vuuu = 1681. 380 8 − 1000 = 681. 380 8 Suud = 1000u2 d = 1000u = 1189. 1100 Vuud = 1189. 11 − 1000 = 189. 11 Sudd = 1000ud2 = 1000d = 840. 965 Vudd = 0 ¡ ¢ Sddd = 1000d3 = 1000 0.840 965 3 = 594. 749 Vddd = 0 c °Yufeng Guo

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CHAPTER 10. BINOMIAL OPTION PRICING I Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1/3 (0.456 807 × 681. 380 8 + 0.543 193 × 189. 11) = 407. 140 2 Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1/3 (0.456 807 × 189. 11 + 0.543 193 × 0) = 84. 958 9 Vdd = e−rh (π u Vudd + πd Vddd ) = 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.05)1/3 (0.456 807 × 407. 140 2 + 0.543 193 × 84. 958 9) = 228. 296 7 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.05)1/3 (0.456 807 × 84. 958 9 + 0.543 193 × 0) = 38. 168 4 V = e−rh (πu Vu + πd Vu ) = e−(0.05)1/3 (0.456 807 × 228. 296 7 + 0.543 193 × 38. 168 4) = 122. 953 9 So the European call premium is $122. 953 9. • Time zero replicating portfolio for the European call

B = V = $122. 953 9 Vu − Vd 228. 296 7 − 38. 168 4 4= = = 0.546 118 Su − Sd 1000 (1. 189 11) − 1000 (0.840 965 ) • calculate the American put premium Period 2

3 Suuu = 1000u3 = 1681. 380 8 Vuuu = 0

Suu = 1000u2 = 1413. 982 6 EVuu = 0 Vuu =?

Suud = 1000u2 d = 1000u = 1189. 1100 Vuud = 0

Sud = 1000ud = 1000 EVud = 0 Vud =? 2

Sdd = 1000d = 707. 222 1 EVdd = 1000 − 707. 222 1 = 292. 777 9 Vdd =?

0

Sudd = 1000ud2 = 1000d = 840. 965 Vudd = 1000 − 840. 965 = 159. 035

¡ ¢ Sddd = 1000d3 = 1000 0.840 965 3 = 594. 749 Vddd = 1000 − 594. 749 = 405. 251

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 0) =

¡ R ¢ Vuu = max Vuu , EVuu = max (0, 0) = 0 R Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 159. 035) = 84. 958 9 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I ¡ R ¢ Vud = max Vud , EVud = max (84. 958 9, 0) = 84. 958 9 R = e−rh (πu Vudd + πd Vddd ) = e−(0.05)1/3 (0.456 807 × 159. 035 + 0.543 193 × 405. 251) = Vdd 287. 938 62 ¡ R ¢ Vdd = max Vdd , EVdd = max (287. 938 62, 292. 777 9) = 292. 777 9 The American put is early exercised at the node dd.

Period 1

2 Vuu = 0

Su = 1000 (1. 189 11) = 1189. 11 EVu = 0 Vu =? Vud = 84. 958 9 Sd = 1000 (0.840 965 ) = 840. 965 EVd = 1000 − 840. 965 = 159. 035 Vd =? Vdd = 292. 777 9 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 84. 958 9) = 45. 386 3 ¡ ¢ Vu = max VuR , EVu = max (45. 386 3 , 0) = 45. 386 3 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.05)1/3 (0.456 807 × 84. 958 9 + 0.543 193 × 292. 777 9) = 194. 574 6 ¡ ¢ Vd = max VdR , EVd = max (194. 574 6 , 159. 035) = 194. 574 6 Period 0

1 Vu = 45. 386 3

S = 1000 EV = 0 V =? Vd = 194. 574 6 V R = e−rh (π u Vu + π d Vu ) = e−(0.05)1/3 (0.456 807 × 45. 386 3 + 0.543 193 × 194. 574 6) = 124. 334 7 ¡ ¢ V = max V R , EV = max (124. 334 7 , 0) = 124. 334 7 So the American put premium is 124. 334 7.

• Calculate the European put premium. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 0

1

2

3 Suuu = 1000u3 = 1681. 380 8 Vuuu = 0

Vuu Suud = 1000u2 d = 1000u = 1189. 1100 Vuud = 0

Vu V

Vud Vd Vdd

0

Sudd = 1000ud2 = 1000d = 840. 965 Vudd = 1000 − 840. 965 = 159. 035

¡ ¢ Sddd = 1000d3 = 1000 0.840 965 3 = 594. 749 Vddd = 1000 − 594. 749 = 405. 251

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 0) =

Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 159. 035) = 84. 958 9 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.05)1/3 (0.456 807 × 159. 035 + 0.543 193 × 405. 251) = 287. 938 62 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.05)1/3 (0.456 807 × 0 + 0.543 193 × 84. 958 9) = 45. 386 3 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.05)1/3 (0.456 807 × 84. 958 9 + 0.543 193 × 287. 938 62) = 191. 989 4 V = e−rh (πu Vu + πd Vu ) = e−(0.05)1/3 (0.456 807 × 45. 386 3 + 0.543 193 × 191. 989 4) = 122. 9537 So the European call premium is $122. 9537. • Time zero replicating portfolio for the European call B = V = $122. 9537 45. 386 3 − 191. 989 4 Vu − Vd = = −0.421 1 4= Su − Sd 1000 (1. 189 11) − 1000 (0.840 965 ) So at t = 0 we need to enter 0.421 1 futures contract as a seller and deposit 122. 9537 in a savings account. • Why the European call and a European put have the same premium The standard put-call parity is: C + P V (K) = P + S S is the price of the underlying asset at time zero. For the futures contract, S is the present value of the forward price: S = P V (F0,T ) In this problem, the forward price and the strike price are equal (i.e. K = F0,T ). Hence P V (K) = P V (F0,T ) = S. This gives us C = P . www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Problem 10.19. For options on√ a stock index, you can use the standard binomial formula. √ (r−δ)h+σ h (0.05−0.03)1+0.3 1 u=e = 1. 377 128 √ =e √ d = e(r−δ)h−σ h = e(0.05−0.03)1−0.3 1 = 0.755 784 πu =

e(r−δ)h − d e(0.05−0.03)1 − 0.755 784 = = 0.425 557 u−d 1. 377 128 − 0.755 784

π d = 1 − πu = 1 − 0.425 557 = 0.574 443 a. Calculate the price of the European call option Period 0 1 2 3 ¡ ¢ Suuu = 100 1. 377 1283 = 261. 169 8 Vuuu = 261. 169 8 − 95 = 166. 169 8 Vuu ¡ ¢ Vu Suud = 100 1. 377 1282 (0.755 784) = 143. 333 0 Vuud = 143. 333 0 − 95 = 48. 3330 V Vud ¡ ¢ Sudd = 100 (1. 377 128) 0.755 7842 = 78. 662 9 Vd Vudd = 0 Vdd ¡ ¢ Sddd = 100 0.755 7843 = 43. 171 1 Vddd = 0 Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1 (0.425 557 × 166. 169 8 + 0.574 443 × 48. 333) = 93. 676 4 Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.05)1 (0.425 557 × 48. 3330 + 0.574 443 × 0) = 19. 565 3 Vdd = e−rh (πu Vudd + πd Vddd ) = 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.05)1 (0.425 557 × 93. 676 4 + 0.574 443 × 19. 565 3) = 48. 611 4 Vd = e−rh (π u Vud + πd Vdd ) = e−(0.05)1 (0.425 557 × 19. 565 3 + 0.574 443 × 0) = 7. 920 1 V = e−rh (π u Vu + π d Vu ) = e−(0.05)1 (0.425 557 × 48. 611 4 + 0.574 443 × 7. 920 1) = 24. 005 8 So the European call premium is $24. 005 8.

b. Calculate the European put premium www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 0

1

2

Vuu Vu V

Vud Vd Vdd

0

3 ¢ ¡ Suuu = 100 1. 377 1283 = 261. 169 8 Vuuu = 0

¡ ¢ Suud = 100 1. 377 1282 (0.755 784) = 143. 333 0 Vuud = 0 ¡ ¢ Sudd = 100 (1. 377 128) 0.755 7842 = 78. 662 9 Vudd = 95 − 78. 662 9 = 16. 337 1 ¡ ¢ Sddd = 100 0.755 7843 = 43. 171 1 Vddd = 95 − 43. 171 1 = 51. 828 9

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1 (0.425 557 × 166. 169 8 + 0.574 443 × 48. 333) =

Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1 (0.425 557 × 0 + 0.574 443 × 16. 337 1) = 8. 927 0 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.05)1 (0.425 557 × 16. 337 1 + 0.574 443 × 51. 828 9) = 34. 934 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.05)1 (0.425 557 × 0 + 0.574 443 × 8. 927 0) = 4. 8780 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.05)1 (0.425 557 × 8. 927 0 + 0.574 443 × 34. 934 0) = 22. 702 6 V = e−rh (πu Vu + πd Vu ) = e−(0.05)1 (0.425 557 × 4. 8780 + 0.574 443 × 22. 702 6) = 14. 379 9 So the European put premium is $14. 379 9. c. If we switch S and K and switch r and δ and recalculate the option price, what happens? After the switch, we have: • S = 95 • K = 100 • r = 3% • δ = 5% √

√

u = e(r−δ)h+σ√ h = e(0.03−0.05)1+0.3√1 = 1. 323 13 d = e(r−δ)h−σ h = e(0.03−0.05)1−0.3 1 = 0.726 149 πu =

e(r−δ)h − d e(0.03−0.05)1 − 0.726 149 = = 0.425 557 u−d 1. 323 13 − 0.726 149

πd = 1 − π u = 1 − 0.425 557 = 0.574 443 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I • Calculate the European call premium Period 0

1

2

Vuu Vu V

Vud Vd Vdd

3 ¡ ¢ Suuu = 95 1. 323 133 = 220. 0550 Vuuu = 220. 0550 − 100 = 120. 055

¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Vuud = 120. 768 7 − 100 = 20. 768 7 ¡ ¢ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 Vudd = 0 ¢ ¡ Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 0

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.03)1 (0.425 557 × 120. 055 + 0.574 443 × 20. 768 7) = 61. 158 1 Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.03)1 (0.425 557 × 20. 768 7 + 0.574 443 × 0) = 8. 577 1 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 0) = 0 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.03)1 (0.425 557 × 61. 158 1 + 0.574 443 × 8. 577 1) = 30. 038 5 Vd = e−rh (π u Vud + πd Vdd ) = e−(0.03)1 (0.425 557 × 8. 577 1 + 0.574 443 × 0) = 3. 542 2 V = e−rh (π u Vu + π d Vd ) = e−(0.03)1 (0.425 557 × 30. 038 5 + 0.574 443 × 3. 542 2) = 14. 3799 The call premium after the switch is equal to the put premium before the switch. • Calculate the European put premium Period 0

1

2

Vuu Vu V

Vud Vd Vdd

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3 ¡ ¢ Suuu = 95 1. 323 133 = 220. 0550 Vuuu = 0

¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Vuud = 0 ¢ ¡ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 Vudd = 100 − 66. 279 3 = 33. 720 7 ¡ ¢ Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 100 − 36. 374 8 = 63. 625 2 c °Yufeng Guo

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CHAPTER 10. BINOMIAL OPTION PRICING I

0

Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 0) =

Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 33. 720 7) = 18. 798 1 Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.03)1 (0.425 557 × 33. 720 7 + 0.574 443 × 63. 625 2) = 49. 394 8 Vu = e−rh (πu Vuu + π d Vud ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 18. 798 1) = 10. 479 3 Vd = e−rh (π u Vud + π d Vdd ) = e−(0.03)1 (0.425 557 × 18. 798 1 + 0.574 443 × 49. 394 8) = 35. 299 1 V = e−rh (πu Vu + πd Vd ) = e−(0.03)1 (0.425 557 × 10. 479 3 + 0.574 443 × 35. 299 1) = 24. 005 8 The put premium after the switch is equal to the call premium before the switch. By the way, you can also use the textbook’s spreadsheet "optbasics2" to calculate the European option premium and verify • the European call price is $14. 379 9 (which is the European put price before the switch) • the European put price is $24. 005 8 (which is the European call price before the switch) What a coincidence, you might wonder. Why? This can be explained using the Black-Scholes option formulas: The price of a European call option is:δ C = Se−δT N (d1 ) − Ke−rT N (d2 )

(Textbook 12.1)

The price of a European put option is: P = Ke−rT N (−d2 ) − Se−δT N (−d1 )

(Textbook 12.3)

¶ µ S 1 2 Se−δT Se−δT 1 ln + r−δ+ σ T ln ln + σ2T −rT −rT 1 √ K 2 Ke √ 2 Ke √ √ = = + σ T d1 = 2 σ T σ T σ T (Textbook 12.2a) √ d2 = d1 − σ T

(Textbook 12.2b)

Before the switch: www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I C = Se−δT N (d1 ) − Ke−rT N (d2 ) P = Ke−rT N (−d2 ) − Se−δT N (−d1 ) Se−δT −rT 1 √ Ke √ d1 = + σ T 2 σ T Se−δT Se−δT ln ln √ √ √ −rT −rT 1 1 √ Ke Ke √ √ + σ T −σ T = − σ T d2 = d1 − σ T = 2 2 σ T σ T ⎛ ⎞ ⎛ ⎞ Se−δT Se−δT ⎜ ln Ke−rT ⎜ ln Ke−rT 1 √ ⎟ 1 √ ⎟ √ + σ T⎟ − σ T⎟ −Ke−rT N ⎜ → C = Se−δT N ⎜ ⎝ σ√T ⎠ ⎝ ⎠ 2 2 σ T ⎛ ⎞ ⎛ ⎞ Se−δT Se−δT ⎜ ln Ke−rT ⎜ ln 1 √ ⎟ 1 √ ⎟ −rT ⎟−Se−δT N ⎜− Ke √ √ → P = Ke−rT N ⎜ + − σ T⎟ − T σ ⎝ ⎠ ⎝ ⎠ 2 2 σ T σ T ln

After£ we switch S and K and switch and δ: ¤r←r→δ C = Se−δT N (d1 ) − Ke−rT N (d2 ) K←→S = Ke−rT N (d1 ) − Se−δT N (d2 ) £ ¤r←→δ P = Ke−rT N (−d2 ) − Se−δT N (−d1 ) K←→S = Se−δT N (−d2 )−Ke−rT N (−d1 ) ⎡

⎤r←→δ Se−δT Ke−rT ln ⎢ ln Ke−rT √ ⎥ −δT 1 1 √ Se √ + σ T⎥ + σ T = d1 = ⎢ ⎣ σ √T ⎦ 2 2 σ T K←→S

Ke−rT Se−δT ln ln √ −δT −rT 1 1 √ Ke √ √ − σ T = − σ T → −d1 = − Se 2 2 σ T σ T Ke−rT Ke−rT ln ln √ √ √ −δT −δT 1 1 √ Se Se √ √ + σ T −σ T = − σ T d2 = d1 − σ T = 2 2 σ T σ T Ke−rT Ke−rT Se−δT ln − ln ln √ √ −δT −rT 1 Se−δT + 1 σ T = Ke √ √ √ + σ T = + → −d2 = − Se 2 2 σ T σ T σ T 1 √ σ T 2 ⎛ ⎞ ⎛ ⎞ Ke−rT Ke−rT ln ln ⎜ Se−δT ⎜ Se−δT 1 √ ⎟ 1 √ ⎟ √ + σ T⎟ − σ T⎟ −Se−δT N ⎜ → C = Ke−rT N ⎜ ⎝ σ √T ⎠ ⎝ ⎠ 2 2 σ T ⎛

⎞ ⎛ ⎞ Se−δT Se−δT ⎜ ln Ke−rT ⎜ ln Ke−rT 1 √ ⎟ 1 √ ⎟ √ → P = Se−δT N ⎜ + σ T⎟ − σ T⎟ −Ke−rT N ⎜ ⎝ σ √T ⎠ ⎝ ⎠ 2 2 σ T

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CHAPTER 10. BINOMIAL OPTION PRICING I You can see that the call premium after the switch is equal the put premium before the switch; the put premium after the switch is equal the call premium before the switch. This conclusion applies to both European options and American options. It’s more complex to prove this is true for American option. However, we are not going to worry about the proof. Intuitively, you can think that the by switching S and K and switching r and δ, we are switching the strike asset and the underlying asset. So after-the-switch put is like a before-the-switch call; after-the-switch call is like a before-the-switch put. By the way, the payoff of the before-switch call is not equal to the payoff of the after-switch put (however the premiums are the same): Period 3 European call ¢payoff before switch ¡ Suuu = 100 1. 377 1283 = 261. 169 8 Vuuu = 261. 169 8 − 95 = 166. 169 8

Period 3 European call ¡ ¢ payoff after switch Suuu = 95 1. 323 133 = 220. 0550 Vuuu = 0

¡ ¢ Sudd = 100 (1. 377 128) 0.755 7842 = 78. 662 9 Vudd = 0

¡ ¢ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 Vudd = 100 − 66. 279 3 = 33. 720 7

¡ ¢ Suud = 100 1. 377 1282 (0.755 784) = 143. 333 0 Vuud = 143. 333 0 − 95 = 48. 3330

¡ ¢ Sddd = 100 0.755 7843 = 43. 171 1 Vddd = 0

the payoff of the before-switch put is not equal switch call (however the premiums are the same): Period 3 European put payoff before switch Suuu = 1000u3 = 1681. 380 8 Vuuu = 0 Suud = 1000u2 d = 1000u = 1189. 1100 Vuud = 0 Sudd = 1000ud2 = 1000d = 840. 965 Vudd = 1000 − 840. 965 = 159. 035

¡ ¢ Sddd = 1000d3 = 1000 0.840 965 3 = 594. 749 Vddd = 1000 − 594. 749 = 405. 251

¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Vuud = 0

¡ ¢ Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 100 − 36. 374 8 = 63. 625 2 to the payoff of the after-

Period 3 European call ¡ ¢ payoff after switch Suuu = 95 1. 323 133 = 220. 0550 Vuuu = 220. 0550 − 100 = 120. 055

¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Vuud = 120. 768 7 − 100 = 20. 768 7 ¡ ¢ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 Vudd = 0 ¡ ¢ Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 0

Problem 10.20. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I For options on√ a stock index, you can use the standard binomial formula. √ u = e(r−δ)h+σ√ h = e(0.05−0.03)1+0.3√ 1 = 1. 377 128 d = e(r−δ)h−σ h = e(0.05−0.03)1−0.3 1 = 0.755 784 πu =

e(r−δ)h − d e(0.05−0.03)1 − 0.755 784 = = 0.425 557 u−d 1. 377 128 − 0.755 784

π d = 1 − πu = 1 − 0.425 557 = 0.574 443 a. Calculate the price of the American call option Period 2 3 ¢ ¡ Suuu = 100 1. 377 1283 = 261. 169 8 ¡ ¢ Suu = 100 1. 377 1282 = 189. 648 2 Vuuu = 261. 169 8 − 95 = 166. 169 8 EVuu = 189. 648 2 − 95 = 94. 648 2 Vuu =? ¡ ¢ Suud = 100 1. 377 1282 (0.755 784) = 143. 333 0 Sud = 100 (1. 377 128) (0.755 784) = 104. 081 1 Vuud = 143. 333 0 − 95 = 48. 3330 EVud = 104. 081 1 − 95 = 9. 081 1 Vud =? ¡ ¢ Sudd = 100 (1. 377 128) 0.755 7842 = 78. 662 9 ¡ ¢ Sudd = 100 0.755 7842 = 57. 120 9 Vudd = 0 EVdd = 0 ¡ ¢ Vdd =? Sddd = 100 0.755 7843 = 43. 171 1 Vddd = 0 R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1 (0.425 557 × 166. 169 8 + 0.574 443 × 48. 333) = 93. 676 4 ¡ R ¢ Vuu = max Vuu , EVuu = max (93. 676 4, 94. 648 2) = 94. 648 2 R = e−rh (π u Vuud + πd Vudd ) = e−(0.05)1 (0.425 557 × 48. 3330 + 0.574 443 × 0) = Vud 19. 565 3 ¡ R ¢ Vud = max Vud , EVud = max (19. 565 3, 9. 081 1) = 19. 565 3 R Vdd = e−rh ¡(πu Vudd + ¢πd Vddd ) = 0 R Vdd = max Vdd , EVdd = 0

Period 1

2 Vuu = 94. 648 2

Su = 100 (1. 377 128) = 137. 712 8 EVu = 137. 712 8 − 95 = 42. 712 8 Vu =? Vud = 19. 565 3 Sd = 100 (0.755 784) = 75. 578 4 EVd = 0 Vd =? Vdd = 0 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I VuR = e−rh (πu Vuu + π d Vud ) = e−(0.05)1 (0.425 557 × 94. 648 2 + 0.574 443 × 19. 565 3) = 49. 004 8 ¡ ¢ Vu = max VuR , EVu = 49. 004 8 VdR = e−rh (πu Vud + πd Vdd ) = e−(0.05)1 (0.425 557 × 19. 565 3 + 0.574 443 × 0) = 7. 920 1 ¡ ¢ Vd = max VdR , EVd = 7. 920 1 Period 0

1 Vu = 49. 004 8

S = 100 EV = 100 − 95 = 5 V =? Vd = 7. 920 1 V R = e−rh (πu Vu + πd Vu ) = e−(0.05)1 (0.425 557 × 49. 004 8 + 0.574 443 × 7. 920 1) = 24. 165 0 ¡ ¢ V = max V R , EV = 24. 165 0 So the American call premium is $24. 165 0. b. Calculate the American put premium Period 2 ¡ ¢ Suu = 100 1. 377 1282 = 189. 648 2 EVuu = 0 Vuu =? Sud = 100 (1. 377 128) (0.755 784) = 104. 081 1 EVud = 0 Vud =? ¡

¢ Sudd = 100 0.755 7842 = 57. 120 9 EVdd = 95 − 57. 120 9 = 37. 879 1 Vdd =?

0

3 ¡ ¢ Suuu = 100 1. 377 1283 = 261. 169 8 Vuuu = 0 ¡ ¢ Suud = 100 1. 377 1282 (0.755 784) = 143. 333 0 Vuud = 0 ¡ ¢ Sudd = 100 (1. 377 128) 0.755 7842 = 78. 662 9 Vudd = 95 − 78. 662 9 = 16. 337 1 ¡ ¢ Sddd = 100 0.755 7843 = 43. 171 1 Vddd = 95 − 43. 171 1 = 51. 828 9

R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.05)1 (0.425 557 × 0 + 0.574 443 × 0) =

¡ R ¢ Vuu = max Vuu , EVuu = 0 R Vud = e−rh (πu Vuud + π d Vudd ) = e−(0.05)1 (0.425 557 × 0 + 0.574 443 × 16. 337 1) = 8. 927 0 ¢ ¡ R Vud = max Vud , EVud = 8. 927 0 R Vdd = e−rh (π u Vudd + πd Vddd ) = e−(0.05)1 (0.425 557 × 16. 337 1 + 0.574 443 × 51. 828 9) = 34. 934 0 ¡ R ¢ Vud = max Vdd , EVdd = 37. 879 1

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 0

Su = 100 (1. 377 128) = 137. 712 8 EVu = 0 Vu =? Vud = 8. 927 0 Sd = 100 (0.755 784) = 75. 578 4 EVd = 85 − 75. 578 4 = 9. 421 6 Vd =? Vdd = 37. 879 1 VuR = e−rh (π u Vuu + π d Vud ) = e−(0.05)1 (0.425 557 × 0 + 0.574 443 × 8. 927 0) = 4. 8780 ¡ ¢ Vu = max VuR , EVu = 4. 8780 VdR = e−rh (π u Vud + π d Vdd ) = e−(0.05)1 (0.425 557 × 8. 927 0 + 0.574 443 × 37. 879 1) = 24. 311 8 ¡ ¢ Vd = max VdR , EVd = 24. 311 8 Period 0

1 Vu = 4. 8780

S = 100 EV = 0 V =? Vd = 24. 311 8 V R = e−rh (π u Vu + π d Vu ) = e−(0.05)1 (0.425 557 × 4. 8780 + 0.574 443 × 24. 311 8) = 15. 259 3 ¡ ¢ V = max V R , EV = 15. 259 3 So the American put premium is $15. 259 3

c. If we switch S and K and switch r and δ and recalculate the option price, what happens? After the switch, we have: • S = 95 • K = 100 • r = 3% • δ = 5% √

√

u = e(r−δ)h+σ√ h = e(0.03−0.05)1+0.3√ 1 = 1. 323 13 d = e(r−δ)h−σ h = e(0.03−0.05)1−0.3 1 = 0.726 149 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I

πu =

e(r−δ)h − d e(0.03−0.05)1 − 0.726 149 = = 0.425 557 u−d 1. 323 13 − 0.726 149

πd = 1 − π u = 1 − 0.425 557 = 0.574 443 Calculate the price of the American call option Period 2 3 ¡ ¢ Suuu = 95 1. 323 133 = 220. 0550 ¡ ¢ Suu = 95 1. 323 132 = 166. 313 9 Vuuu = 220. 0550 − 100 = 120. 055 EVuu = 166. 313 9 − 100 = 66. 313 9 Vuu =? ¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Sud = 95 (1. 323 13) (0.726 149) = 91. 275 0 Vuud = 120. 768 7 − 100 = 20. 768 7 EVud = 0 Vud =? ¡ ¢ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 ¡ ¢ Sudd = 95 0.726 1492 = 50. 092 8 Vudd = 0 EVdd = 0 ¡ ¢ Vdd =? Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 0 R Vuu = e−rh (π u Vuuu + π d Vuud ) = e−(0.03)1 (0.425 557 × 120. 055 + 0.574 443 × 20. 768 7) = 61. 158 1 ¡ R ¢ Vuu = max Vuu , EVuu = 66. 313 9 R = e−rh (πu Vuud + π d Vudd ) = e−(0.03)1 (0.425 557 × 20. 768 7 + 0.574 443 × 0) = Vud 8. 577 1 ¡ R ¢ Vud = max Vud , EVud = 8. 577 1 R Vdd = e−rh ¡(π u Vudd + ¢πd Vddd ) = 0 R Vdd = max Vdd , EVdd = 0

Period 1

2 Vuu = 66. 313 9

Su = 95 (1. 323 13) = 125. 697 4 EVu = 125. 697 4 − 100 = 25. 697 4 Vu =? Vud = 8. 577 1 Sd = 95 (0.726 149) = 68. 984 2 EVd = 0 Vd =? Vdd = 0 VuR = e−rh (πu Vuu + π d Vud ) = e−(0.03)1 (0.425 557 × 66. 313 9 + 0.574 443 × 8. 577 1) = 32. 167 7 ¡ ¢ Vu = max VuR , EVu = 32. 167 7 www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I VdR = e−rh (π u Vud + π d Vdd ) = e−(0.03)1 (0.425 557 × 8. 577 1 + 0.574 443 × 0) = 3. 542 2 ¡ ¢ Vd = max VdR , EVd = 3. 542 2 Period 0

1 Vu = 32. 167 7

S = 95 EV = 0 V =? Vd = 3. 542 2 V R = e−rh (π u Vu + π d Vu ) = e−(0.03)1 (0.425 557 × 32. 167 7 + 0.574 443 × 3. 542 2) = 15. 2593 ¡ ¢ V = max V R , EV = 15. 2593 So the American call premium is $15. 2593, which is equal to the American put premium before the switch. Calculate the American put premium Period 2 ¡ ¢ Suu = 95 1. 323 132 = 166. 313 9 EVuu = 0 Vuu =? Sud = 95 (1. 323 13) (0.726 149) = 91. 275 0 EVud = 100 − 91. 275 0 = 8. 725 Vud =? ¡ ¢ Sudd = 95 0.726 1492 = 50. 092 8 EVdd = 100 − 50. 092 8 = 49. 907 2 Vdd =?

0

3 ¡ ¢ Suuu = 95 1. 323 133 = 220. 0550 Vuuu = 0 ¡ ¢ Suud = 95 1. 323 132 (0.726 149) = 120. 768 7 Vuud = 0 ¡ ¢ Sudd = 95 (1. 323 13) 0.726 1492 = 66. 279 3 Vudd = 100 − 66. 279 3 = 33. 720 7 ¡ ¢ Sddd = 95 0.726 1493 = 36. 374 8 Vddd = 100 − 36. 374 8 = 63. 625 2

R Vuu = e−rh (πu Vuuu + πd Vuud ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 0) =

¢ ¡ R Vuu = max Vuu , EVuu = 0 R Vud = e−rh (π u Vuud + πd Vudd ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 33. 720 7) = 18. 798 1 ¡ R ¢ Vud = max Vud , EVud = 18. 798 1 R Vdd = e−rh (πu Vudd + πd Vddd ) = e−(0.03)1 (0.425 557 × 33. 720 7 + 0.574 443 × 63. 625 2) = 49. 394 8 ¡ R ¢ , EVdd = 49. 907 2 Vud = max Vdd www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Period 1

2 Vuu = 0

Su = 95 (1. 323 13) = 125. 697 4 EVu = 0 Vu =? Vud = 18. 798 1 Sd = 95 (0.726 149) = 68. 984 2 EVd = 100 − 68. 984 2 = 31. 015 8 Vd =? Vdd = 49. 907 2 VuR = e−rh (πu Vuu + π d Vud ) = e−(0.03)1 (0.425 557 × 0 + 0.574 443 × 18. 798 1) = 10. 479 3 ¡ ¢ Vu = max VuR , EVu = 10. 479 3 VdR = e−rh (πu Vud + πd Vdd ) = e−(0.03)1 (0.425 557 × 18. 798 1 + 0.574 443 × 49. 907 2) = 35. 584 8 ¡ ¢ Vd = max VdR , EVd = 35. 584 8 Period 0

1 Vu = 10. 479 3

S = 95 EV = 100 − 95 = 5 V =? Vd = 35. 584 8 V R = e−rh (πu Vu + πd Vu ) = e−(0.03)1 (0.425 557 × 10. 479 3 + 0.574 443 × 35. 584 8) = 24. 165 04 ¡ ¢ V = max V R , EV = 24. 165 0

So the American put premium is $24. 165 0, which is equal to the American call premium before the switch. Problem 10.21.

Suppose u < e(r−δ)h . Since d < u, we have d < u < e(r−δ)h . This means that the savings account is always better than the stock. So at t = 0, we short sell e−δh share of stock and investment the short sale proceeds Se−δh into the savings account. Then at time h, we close our short position by buying one stock from the market. The stock price at time h is either uS or dS. t=0 t = h, u mode t = h, d mode short e−δh stock Se−δh −uS −dS deposit Se−δh in savings −Se−δh Se¡(r−δ)h ¢ Se¡−δh ¢ −δh Total 0 S e − u > 0 S e−δh − d > 0

So initial cost is zero yet we have positive payoff at time h. This is an arbitrage opportunity. www.actuary88.com

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CHAPTER 10. BINOMIAL OPTION PRICING I Suppose d > e(r−δ)h . Since d < u, we have u > d > e(r−δ)h . This means that the investing in stocks is always better off than investing in a savings account. So at t = 0, we buy e−δh share of stock and borrow money from a bank to finance the purchase. t=0 t = h, u mode t = h, d mode buy e−δh stock −Se−δh uS dS (r−δ)h −δh borrow Se−δh in savings Se−δh −Se −Se ¡ ¡ ¢ ¢ −δh Total 0 S u−e > 0 S d − e−δh > 0 So initial cost is zero yet we have positive payoff at time h. This is an arbitrage opportunity.

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Chapter 11

Binomial option pricing II Problem 11.1. √

√

u = e(r−δ)h+σ√ h = e(0−0.08)1+0.3√1 = 1. 246 077 d = e(r−δ)h−σ h = e(0−0.08)1−0.3 1 = 0.683 861 e(r−δ)h − d e(0−0.08)1 − 0.683 861 πu = = = 0.425 56 u−d 1. 246 077 − 0.683 861 π d = 1 − πu = 1 − 0.425 56 = 0.574 44 • K = 70 t=0

S = 100 EV = 100 − 70 = 30 V

t=1 Su = 100 (1. 246 077) = 124. 607 7 Vu = 124. 607 7 − 70 = 54. 607 7 Sd = 100 (0.683 861 ) = 68. 386 1 Vd = 0

V R = e−rh (π u Vu + π d Vu ) = e−0(1) (0.425 56 × 54. 607 7 + 68. 386 1 × 0) = 23. 24 ¡ ¢ V = max V R , EV = max (23. 24, 30) = 30 • K = 80 t=0

S = 100 EV = 100 − 80 = 20 V

t=1 Su = 100 (1. 246 077) = 124. 607 7 Vu = 124. 607 7 − 80 = 44. 607 7 Sd = 100 (0.683 861 ) = 68. 386 1 Vd = 0 91

CHAPTER 11. BINOMIAL OPTION PRICING II V R = e−rh (π u Vu + π d Vu ) = e−0(1) (0.425 56 × 44. 607 7 + 68. 386 1 × 0) = 18. 98 ¡ ¢ V = max V R , EV = max (18. 98, 20) = 20 • K = 90 t=0

S = 100 EV = 100 − 90 = 10 V

t=1 Su = 100 (1. 246 077) = 124. 607 7 Vu = 124. 607 7 − 90 = 34. 607 7 Sd = 100 (0.683 861 ) = 68. 386 1 Vd = 0

V R = e−rh (π u Vu + π d Vu ) = e−0(1) (0.425 56 × 34. 607 7 + 68. 386 1 × 0) = 14. 73 ¡ ¢ V = max V R , EV = max (14. 73, 10) = 14. 73 • K = 100 t=0

S = 100 EV = 100 − 100 = 0 V

t=1 Su = 100 (1. 246 077) = 124. 607 7 Vu = 124. 607 7 − 100 = 24. 607 7 Sd = 100 (0.683 861 ) = 68. 386 1 Vd = 0

V R = e−rh (π u Vu + π d Vu ) = e−0(1) (0.425 56 × 24. 607 7 + 68. 386 1 × 0) = 10. 47 ¡ ¢ V = max V R , EV = max (10. 47, 0) = 10. 47 a. Early exercise occurs at t = 0 with K = 70, 80

b. The European options satisfy the put-call parity: CEur + Ke−rT = PEur + Se−δT CEur = PEur + Se−δT − Ke−rT = PEur + 100e−0.08 − K = PEur − K + 92. 311 6 Since K = 70, 80, 90, 100 ≤ 100, we have EV = max (0, S − K) = max (0, 100 − K) = 100 − K To have EV > CEur , we need to have: 100 − K > PEur − K + 92. 311 6 → PEur < 100 − 100e−0.08 = 7. 69 To early exercise the American call at t = 0, the European put premium needs to be less than 7. 69. www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Clearly, the smaller the European put premium is, the more likely the early exercise of the American call is optimal at t = 0. Everything else equal, the higher the strike price, the higher the price of a European put. As a result, it can be optimal to early exercise the American call with K = 70, 80, while it’s not optimal to early exercise the American with K = 90, 100. Use the European put premium using the Black-Scholes option formula (the textbook Equation 12.3): P = Ke−rT N (−d2 ) − Se−δT N (−d1 ) We find the following put price (S = 100, T = 1, r = 0, δ = 0.08, σ = 0.3): K PEur 70 $2.3087 80 $5.2537 90 $9.7517 100 $15.7113 Clearly, the condition PEur < 7. 69 is met when K = 70, 80 and violated when K = 90, 100. By the way, the price calculated using the Black-Scholes option formula won’t match the price calculated under the binomial option formula under h = 1. This is because the Black-Scholes option formula is the binomial option pricing method where n → ∞ and h = T /n → 0 . c.To early exercise the American call, we need to have PEur < 7. 69. This condition is met if K = 70, 80 and violated when K = 90, 100. Problem 11.2. Now r = 0.08 instead of r = 0. This increases the cost of early exercising an American call. By early exercising an American call, you lost interest on the strike asset K. We expect that it’s still not optimal to early exercise the American call at K = 90, 100 (it’s not optimal to early exercise the American call at these strike prices even when r = 0, let alone when r = 0.08). However, we are not clear whether it’s optimal to early exercise the American call when K = 70, 80. We have to check. √

√

u = e(r−δ)h+σ√ h = e(0.08−0.08)1+0.3√ 1 = 1. 349 859 d = e(r−δ)h−σ h = e(0.08−0.08)1−0.3 1 = 0.740 818 e(r−δ)h − d e(0.08−0.08)1 − 0.740 818 πu = = = 0.425 558 u−d 1. 349 859 − 0.740 818 π d = 1 − πu = 1 − 0.425 558 = 0.574 442 www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II • K = 70 t=0

S = 100 EV = 100 − 70 = 30 V

t=1 Su = 100 (1. 349 859) = 134. 985 9 Vu = 134. 985 9 − 70 = 64. 985 9 Sd = 100 (0.740 818 ) = 74. 081 8 Vd = 74. 081 8 − 70 = 4. 081 8

V R = e−rh (πu Vu + πd Vu ) = e−0.08(1) (0.425 558 × 64. 985 9 + 0.574 442 × 4. 081 8) = 27. 69 ¢ ¡ V = max V R , EV = max (27. 69, 25) = 27. 69 • K = 80 t=0

S = 100 EV = 100 − 80 = 20 V

t=1 Su = 100 (1. 349 859) = 134. 985 9 Vu = 134. 985 9 − 80 = 54. 985 9 Sd = 100 (0.740 818 ) = 74. 081 8 Vd = 0

V R = e−rh (πu Vu + πd Vu ) = e−0.08(1) (0.425 558 × 54. 985 9 + 0.574 442 × 0) = 21. 60 ¡ ¢ V = max V R , EV = max (21. 60, 20) = 21. 60 • K = 90 t=0

S = 100 EV = 100 − 90 = 10 V

t=1 Su = 100 (1. 349 859) = 134. 985 9 Vu = 134. 985 9 − 90 = 44. 985 9 Sd = 100 (0.740 818 ) = 74. 081 8 Vd = 0

V R = e−rh (πu Vu + πd Vu ) = e−0.08(1) (0.425 558 × 44. 985 9 + 0.574 442 × 0) = 17. 67 ¡ ¢ V = max V R , EV = max (17. 67, 10) = 17. 67 • K = 100 www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II t=0

S = 100 EV = 0 V

t=1 Su = 100 (1. 349 859) = 134. 985 9 Vu = 134. 985 9 − 100 = 34. 985 9 Sd = 100 (0.740 818 ) = 74. 081 8 Vd = 0

V R = e−rh (π u Vu + π d Vu ) = e−0.08(1) (0.425 558 × 34. 985 9 + 0.574 442 × 0) = 13. 74 ¡ ¢ V = max V R , EV = max (13. 74, 0) = 13. 74 a. Early exercise occurs at t = 0 with K = 70. Now the risk free rate is no longer zero and we’ll lose interest on the strike price by exercising the American call early. The higher the strike price, the more interest we lose. In contrast, the previous problem has r = 0 and we don’t lose any interest by early exercise. It makes sense that if r is not zero (i.e. positive) then fewer strike prices will lead to optimal exercise than if r = 0. b, c. The European options satisfy the put-call parity: CEur + Ke−rT = PEur + Se−δT CEur = PEur + Se−δT − Ke−rT = PEur + 100e−0.08 − Ke−0.08 Since K = 70, 80, 90, 100 ≤ 100, we have EV = max (0, S − K) = max (0, 100 − K) = 100 − K To have EV > CEur , we need to have: −0.08 100 − K > PEur + 100e−0.08 − Ke ¡ ¢ −0.08 → PEur < 100 − 100e − K 1 − e−0.08 = 7. 69 − 0.0769K

Using the Black-Scholes option pricing formula, we find the following put price (S = 100, T = 1, r = 0.08, δ = 0, σ = 0.3): K 70 80 90 100

PEur 0.7752 2.0904 4.4524 8.0229

7. 69 − 0.0769K 7. 69 − 0.0769 (70) = 2. 307 7. 69 − 0.0769 (80) = 1. 538 7. 69 − 0.0769 (90) = 0.769 7. 69 − 0.0769 (100) = 0

Among the 4 strike prices, only K = 70 satisfies the condition PEur < 7. 69 − 0.0769K. Hence of the 4 strike prices given, only K = 70 leads to optimal early exercise. Problem 11.3. If δ = 0, then the stock doesn’t pay any dividend. It’s never optimal to early exercise an American call on a non-dividend paying stock. So early exercise will never occur. www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Problem 11.4. √

√

u = e(r−δ)h+σ√ h = e(0.08−0)1+0.3√1 = 1. 462 285 d = e(r−δ)h−σ h = e(0.08−0)1−0.3 1 = 0.802 519 e(r−δ)h − d e(0.08−0)1 − 0.802 519 πu = = = 0.425 557 u−d 1. 462 285 − 0.802 519 πd = 1 − π u = 1 − 0.425 557 = 0.574 443 • K = 100 t=0

S = 100 EV = 0 V

t=1 Su = 100 (1. 462 285) = 146. 228 5 Vu = 0

Sd = 100 (0.802 519 ) = 80. 251 9 Vd = 100 − 80. 251 9 = 19. 748 1

V R = e−rh (πu Vu + πd Vu ) = e−0.08(1) (0.425 557 × 0 + 0.574 443 × 19. 748 1) = 10. 4720 ¡ ¢ V = max V R , EV = max (10. 4720, 0) = 10. 4720 • K = 110 t=0

S = 100 EV = 110 − 100 = 10 V

t=1 Su = 100 (1. 462 285) = 146. 228 5 Vu = 0

Sd = 100 (0.802 519 ) = 80. 251 9 Vd = 110 − 80. 251 9 = 29. 748 1

V R = e−rh (πu Vu + πd Vu ) = e−0.08(1) (0.425 557 × 0 + 0.574 443 × 29. 748 1) = 15. 774 8 ¡ ¢ V = max V R , EV = max (15. 774 8, 10) = 15. 774 8 • K = 120 t=0

S = 100 EV = 120 − 100 = 20 V

www.actuary88.com

t=1 Su = 100 (1. 462 285) = 146. 228 5 Vu = 0

Sd = 100 (0.802 519 ) = 80. 251 9 Vd = 120 − 80. 251 9 = 39. 748 1 c °Yufeng Guo

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CHAPTER 11. BINOMIAL OPTION PRICING II V R = e−rh (π u Vu + π d Vu ) = e−0.08(1) (0.425 557 × 0 + 0.574 443 × 39. 748 1) = 21. 077 5 ¡ ¢ V = max V R , EV = max (21. 077 5, 20) = 21. 077 5 • K = 130 t=0

t=1 Su = 100 (1. 462 285) = 146. 228 5 Vu = 0

S = 100 EV = 130 − 100 = 30 V

Sd = 100 (0.802 519 ) = 80. 251 9 Vd = 130 − 80. 251 9 = 49. 748 1

V R = e−rh (π u Vu + π d Vu ) = e−0.08(1) (0.425 557 × 0 + 0.574 443 × 49. 748 1) = 26. 380 3 ¡ ¢ V = max V R , EV = max (26. 380 3, 30) = 30 a. Early exercise occurs at t = 0 with K = 130.

b, c. The European options satisfy the put-call parity: CEur + Ke−rT = PEur + Se−δT PEur = CEur + Ke−rT − Se−δT = CEur + Ke−0.08(1) − 100e−0(1) = CEur + −0.08 Ke − 100 The early exercise value is EV = max (0, K − S) Since K = 100, 100, 120, 130 ≥ 100, we have EV = max (0, K − S) = K − S = K − 100 To have EV > CEur , we need to have: −0.08 K − 100 > CEur ¡ + Ke ¢ − 100 −0.08 → CEur < K 1 − e = 0.07 69K

Using the Black-Scholes option pricing formula, we find the following put price (S = 100, T = 1, r = 0.08, δ = 0, σ = 0.3): K 100 110 120 130

CEur 15.7113 11.2596 7.8966 5.4394

0.07 69K 0.07 69 (100) = 7. 69 0.07 69 (110) = 8. 459 0.07 69 (120) = 9. 228 0.07 69 (130) = 9. 997

Among the 4 strike prices, only K = 130 satisfies the condition CEur < 0.07 69K. Hence of the 4 strike prices given, only K = 130 leads to optimal early exercise. www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II The main reason for early exercising an American put is to earn interest on the strike price. Hence higher the strike price, everything else equal, the more likely an American put may be early exercised. Problem 11.5. Now δ = 0.08 as opposed to δ = 0 in the previous problem. This increases the cost of early exercising an American put. If you early exercise an American put, you can earn interest on the strike asset, but you also lose the opportunity of earning dividend on the stock. With δ = 0.08, we expect that it’s still not optimal to early exercise the American put when K = 100, 110, 120. Because it’s not optimal to early exercise the American put with these strike prices even when δ = 0, let alone δ = 0.08. However, we are not clear whether it’s optimal to early exercise the American put when K = 130. We have to check. The put-call parity is: CEur + Ke−rT = PEur + Se−δT CEur + Ke−0.08 = PEur + 100e−0.08 → PEur = CEur + Ke−0.08 − 100e−0.08 EV = max (0, K − S) = K − S = K − 100 Early exercise if: K − 100 > CEur + Ke−0.08 − 100e−0.08 → CEur < K − Ke−0.08 − 100 + 100e−0.08 = 0.07 69K − 7. 69 Using the Black-Scholes option pricing formula, we find the following put price (S = 100, T = 1, r = 0.08, δ = 0.08, σ = 0.3): K CEur 0.07 69K 0.07 69K − 7. 69 100 15.6584 0.07 69 (100) − 7. 69 = 0 110 11.2165 0.07 69 (110) − 7. 69 = 0.769 120 7.8627 0.07 69 (120) − 7. 69 = 1. 538 130 5.4136 0.07 69 (130) − 7. 69 = 2. 307 The condition CEur < 0.07 69K − 7. 69 is met only when K = 130. So it’s optimal to early exercise the American put only when K = 130. Problem 11.6. The only reason to early exercise an American put is to earn interest on the strike asset. If r = 0 we’ll never earn any interest on the strike price. So it’s never optimal to early exercise an American put if r = 0. We can also use the put-call parity to verify that it’s never optimal to early exercise the American put when r = 0. CEur + Ke−rT = PEur + Se−δT CEur + K = PEur + 100e−0.08 www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II → PEur = CEur + K − 100e−0.08 EV = max (0, K − S) = K − S = K − 100 Early exercise if: K − 100 > CEur + K − 100e−0.08 → CEur < −100 + 100e−0.08 = −7. 688 Since CEur ≥ 0, it’s never optimal to early exercise the American put. Problem 11.7. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.8. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.9. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.10. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.11. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.12. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 10 period binomial model in the exam. Problem 11.13. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 50 period binomial model in the exam. Problem 11.14. www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II The textbook Section 10 Equation 10.7 says that the undiscounted riskneutral stock price is the forward price: time t time t + h Su S Sd πu Su + πd Sd = Se(r−δ)h = Ft,t+h

(Textbook Equation 10.7)

We are reusing the data in the textbook Figure 11.4 except that we set S = 100. We have: S = 100 r = 0.08

σ = 0.3 δ=0

T =1

h = T /n = 1/3

Then: √ √ u = e(r−δ)h+σ√ h = e(0.08−0)1/3+0.3√1/3 = 1. 221 246 d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 1/3 = 0.863 693 e(r−δ)h − d e(0.08−0)1/3 − 0.863 693 = = 0.456 806 πu = u−d 1. 221 246 − 0.863 693 πd = 1 − π u = 1 − 0.456 806 = 0.543 194 The stock price tree is: Period 0 1 S u = 100 (1. 221 246) = 122. 124 6 S = 100 S d = 100 (0.863 693) = 86. 369 3 Period 2 ¡ ¢ S uu = 100 1. 221 2462 = 149. 144 2

S ud = 100 (1. 221 246) (0.863 693) = 105. 478 2 S dd

¡ ¢ = 100 0.863 6932 = 74. 596 6

3 ¡ ¢ S uuu = 100 1. 221 2462 = 182. 141 7

¡ ¢ S uud = 100 1. 221 2462 (0.863 693) = 128. 814 8 ¡ ¢ S udd = 100 (1. 221 246) 0.863 6932 = 91. 100 8 ¡ ¢ S ddd = 100 0.863 6933 = 64. 428 5

a.The forward price is calculated using the textbook Equation 5.7: F0,T = S0 e(r−δ)T The 4-month forward price is: F0,1/3 = 100e(0.08−0)1/3 = 102. 702 5 The 8-month forward price is: F0,2/3 = 100e(0.08−0)2/3 = 105. 478 1 www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II The 1-year forward price is: F0,1 = 100e(0.08−0)1 = 108. 328 7 b. √ √ u = e(r−δ)h+σ√ h = e(0.08−0)1/3+0.3√1/3 = 1. 221 246 d = e(r−δ)h−σ h = e(0.08−0)1/3−0.3 1/3 = 0.863 693 e(r−δ)h − d e(0.08−0)1/3 − 0.863 693 πu = = = 0.456 806 u−d 1. 221 246 − 0.863 693 π d = 1 − πu = 1 − 0.456 806 = 0.543 194 t = 1/3 Stock Price Risk neutral prob S u = 100 (1. 221 246) = 122. 124 6 πu S d = 100 (0.863 693) = 86. 369 3 πd Total 1 The undiscounted risk-neutral expected stock price at t = 1/3 is: π u Su + π d Sd = 0.456 806 (122. 124 6) + 0.543 194 (86. 369 3) = 102. 702 53 = F0,1/3 t = 2/3 Stock Risk neutral prob ¡ Price ¢ S uu = 100 1. 221 2462 = 149. 144 2 π 2u ud S = 100 ¡(1. 221 246)¢(0.863 693) = 105. 478 2 2π u πd S dd = 100 0.863 6932 = 74. 596 6 π 2d Total 1 The undiscounted risk-neutral expected stock price at t = 2/3 is: π 2u Suu +π 2d Sdd +2π u πd S ud = 0.456 8062 (149. 144 2)+2 (0.456 806) (0.543 194) (105. 478 2)+ 0.543 1942 (74. 596 6) = 105. 478 1 = F0,2/3 t = 1 Stock ¡Price ¢ S uuu = 100 ¡ 1. 221 2462¢ = 182. 141 7 S uud = 100 1. 221 2462¡ (0.863 693) ¢ = 128. 814 8 2 S udd = 100 (1. 221 246) 0.863 693 = 91. 100 8 ¡ ¢ S ddd = 100 0.863 6933 = 64. 428 5 Total

Risk neutral prob π 3u 3π 2u πd 3π u π2d π 3d 1

The undiscounted risk-neutral expected stock price at t = 2/3 is: π 3u Suuu + 3π 2u πd S uud + 3π u¡π2d S udd + π¢3d S ddd ¡ ¢ 3 = 0.456 141 7)+3 0.456 8062 (0.543 194) 128. 814 8+3 (0.456 806) 0.543 1942 91. ¡ 806 (182. ¢ 100 8 + 0.543 1943 64. 428 5 = 108. 328 7 = F0,1 Problem 11.15. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 50 period binomial model in the exam. Problem 11.16. www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 8 period binomial model in the exam. Problem 11.17. Skip. This is a spreadsheet problem. Most likely SOA and CAS won’t ask you to build a 8 period binomial model in the exam. Problem 11.18. Skip. If you understand the volatility calculation examples in my study guide, you are fine. Problem 11.19. Skip. If you understand the volatility calculation examples in my study guide, you are fine. Problem 11.20. This is a labor intensive problem. However, it’s a good practice problem for using the Schroder method. The solution is similar to the textbook Figure 11.11. First, we’ll build a prepaid forward price tree. St P Ft,T P Ft,T = St − P Vt (D) = 50 − 4e−0.08(0.25) = 46. 079 21 St 50 σ F = σ S × P = 0.3 × = 0.325 5265 46. 079 2 Ft,T σF = σS ×

√

√

u = erh+σF √ h = e0.08(0.25)+0.325 5265 √0.25 = 1. 200 53 d = erh−σF h = e0.08(0.25)−0.325 5265 0.25 = 0.866 959 Prepaid forward price tree: Time 0

0.25

0.5

0.75

1 95.71879

79.73044 66.41269 55.31947 46.07921

69.12305 57.57710

47.95973 39.94880

49.91701 41.57914

34.63398

36.04742 30.02625 26.03154

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CHAPTER 11. BINOMIAL OPTION PRICING II For ³ example, ´u the prepaid forward price at t = 0.25 is calculated as follows: P P = Ft,T u = 46. 079 21 × 1. 200 53 = 55. 319 47 Ft+h,T ³ ´d P P Ft+h,T = Ft,T d = 46. 079 21 × 0.866 959 = 39. 948 8 The prepaid ³ ´uu forward price at t = 0.5 is calculated as follows: P P Ft+2h,T = Ft,T u2 = 46. 079 21 × 1. 200 532 = 66. 412 7 ³ ´ud P P Ft+2h,T = Ft,T ud = 46. 079 21 × 1. 200 53 × 0.866 959 = 47. 959 73 ³ ´dd P P 2 Ft+2h,T = Ft,T d = 46. 079 21 × 0.866 9592 = 34. 633 98

(My numbers are calculated using Excel so you might not be able to fully match mine.) Next, convert the prepaid forward price tree into a stock price tree. The one-to-one mapping between the prepaid forward price and the stock price is ½ De−r(TD −t−∆t) if TD ≥ t + ∆t P P St+∆t = Ft+∆t,T +P V (Div) = Ft+∆t,T + 0 if TD < t + ∆t The PV of the dividend at each interval is: Time 0 0.25 0.5 0.75 Dividend time 0.25 Dividend amount $4 P V (Div) 3. 920 79 4 0 0

1

0

P Vt=0 (Div) = 4e−0.08(0.25−0) = 3. 920 79 We discount $4 from t = 0.25 to time zero P Vt=0.25 (Div) = 4e−0.08(0.25−0.25) = 4 We discount $4 from t = 0.25 to t = 0.25 For t > 0.25, we have P Vt (Div) = 0 Next, we add the P V (Div) to the prepaid forward price: Time 0 0.25 0.5 0.75

1 95.71879

79.73044 66.41269 55.31947 + 4 46.07921 + 3. 920 79

69.12305 57.57710

47.95973 39.94880 + 4

49.91701 41.57914

34.63398

36.04742 30.02625 26.03154

Now the stock price tree is: www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Time 0

0.25

0.5

0.75

1 95.718794

79.730440 66.412694 59.319474 50

69.123048 57.577104

47.959733 43.948797

49.917007 41.579138

34.633981

36.047421 30.026253 26.031540

The risk neutral probabilities are: erh − d e0.08(0.25) − 0.866 959 πu = = = 0.459 399 u−d 1. 200 53 − 0.866 959 πd = 1 − 0.459 399 = 0.540 601 From this point on, we can just use the standard binomial tree formula. First, we calculate the European call premium. We start from right to left, calculating the roll-back value. Time 0 0.25 0.5 0.75 1 50.71879 35.62150 23.17717 24.12305 14.27212 13.46816 8.43381 7.23801 4.91701 3.78761 2.21414 0.99703 0 0 0 In the above table, the final column is the call payoff. For example, 50.71879 = max (0, 95.718794 − 45) = 50. 718 794 The first 4 columns are the roll-back values. For example, 35.62150 = e−0.08(0.25) (50.71879 × 0.459 399 + 24.12305 × 0.540 601) = 35. 621 504 8.43381 = e−0.08(0.25) (14.27212 × 0.459 399 + 3.78761 × 0.540 601) = 8. 433 809 The European call premium is 8.43381. To calculate the American call premium, we still work from right to left. However, we’ll need to compare the roll back value and the exercise value and take the greater of the two. The American call premium is calculated as follows: www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Time 0

0.25

0.5

0.75

1 50.71879

35.62150 23.17717 max (14.27212, 14.31947) 8.45513

24.12305 13.46816

7.23801 3.78761

4.91701 2.21414

0.99703

0 0 0

It’s optimal to exercise the American call at the upper node at t = 0.25. The roll back value is 14.27212. The exercise value is 59.319474 − 45 = 14. 319 474, which is greater than the roll back value. The premium of 8.45513 is calculated as follows: The roll back value is: e−0.08(0.25) (14. 319 474 × 0.459 399 + 3.78761 × 0.540 601) = 8. 455 132 8 The early exercise value at t = 0 is 50 − 45 = 5. We take the greater of the two. So the American call premium is 8.45513. By the way, if you bother to calculate the European put and the American put premium with strike price K = 45, here are the results: The prepaid forward price tree and the stock price tree won’t change whether the option is a call or put. The European put premium: Time 0 0.25 0.5

0.75

1 0

0 0

0

1.33205 3.89484

0 2.51380

6.21822

0 4.74394

9.59857

8.95258 14.08269 18.96846

So the European put premium is 3.89484. Check the put-call parity: C + P V (K) = P + S − P V (Div) C + P V (K) = 8.43381 + 45e−0.08(1) = 49. 974 05 P + S − P V (Div) = 3.89484 + 50 − 3. 920 79 = 49. 974 05 So C + P V (K) = P + S − P V (Div) holds. The American put premium: www.actuary88.com

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CHAPTER 11. BINOMIAL OPTION PRICING II Time 0

0.25

0.5

0.75

1 0

0 0

0

1.33205 4.11033

0 2.51380

6.62489

0 4.74394 8.95258

10.36602 14.97375

18.96846 The two bold numbers indicate that early exercise is optimal. The American put premium is 4.11033.

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Chapter 12

Black-Scholes formula Problem 12.1. Skip this spreadsheet problem. Problem 12.2. Skip this problem but remember the following key point. As n gets bigger, the price calculated using the discrete binomial tree method approach the price calculated using the Black-Scholes formula. Problem 12.3. a. r = 8% δ=0 T European call price 1 7.8966 10 56.2377 100 99.9631 1, 000 100 1, 0000 100 1, 0000 100 As T → ∞, the European call premium approaches the current stock price. b. r = 8% T 1 10 100 1, 000 1, 0000 10, 0000

δ = 0.1% European call price 7.8542 55.3733 90.4471 36.7879 0.0045 0 107

CHAPTER 12. BLACK-SCHOLES FORMULA Dividend reduces the value of the stock. Over a long period of time, the value of the underlying stock is reduced to zero; the value of the call is reduced to zero. Problem 12.4. a. r = 0% δ = 8% T European call price 1 18.6705 10 10.1571 100 0.0034 1, 000 0.0000 Dividend reduces the value of the stock. Over a long period of time, the value of the underlying stock is reduced to zero; the value of the call is reduced to zero. b. r = 0.1% δ = 8% T European call price 1 18.7281 10 10.2878 100 0.0036 1, 000 0.0000 Dividend reduces the value of the stock, but you can earn interest on the strike asset with r = 0.1% (vs. r = 0 in the previous problem). Consequently, the call is more valuable when r = 0.1% than when r = 0. However, even with r = 0.1%, over a long period of time, the value of the underlying stock is reduced to zero due to the dividend paid out; the call value approaches zero. Problem 12.5. a. We need to find the 90-strike yen-denominated euro put. • The underlying asset is 1 euro. • The strike asset is expressed in yen. • The current price of the underlying is 95 yen. S = 95 • The strike asset is 90 yen. K = 90 • The strike asset 90 yen earns 1.5%. So the risk free rate is r = 1.5% (always remember that r is the earning rate of the strike asset) www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA • The underlying (i.e. 1 euro) earns 3.5%. So the dividend rate is δ = 3.5% (always remember that δ is the earning rate of the underlying asset) • T = 0.5 ln d1 = 0.658 560

¶ ¶ µ µ 95 1 1 S ln + r − δ + σ2 T + 0.015 − 0.035 + × 0.12 0.5 K 2 90 2 √ √ = = 0.1 0.5 σ T

N (d1 ) = NormalDist (0.658 560) = 0.744 910 8 N (−d1 ) = 1√− N (d1 ) = 1 − 0.744√910 8 = 0.255 089 d2 = d1 − σ T = 0.658 560 − 0.1 0.5 = 0.587 849 N (d2 ) = 0.721 683 N (−d2 ) = 1 − N (d2 ) = 1 − 0.721 683 = 0.278 317 P = −Se−δT N (−d1 ) + Ke−rT N (−d2 ) = −95e−0.035(0.5) 0.255 089 + 90e−0.015(0.5) 0.278 317 = 1. 048 3 (yen) 1 b. Now we need to find the price of a strike euro-denominated yen call 90 with 6 months to expiration. • The underlying asset is 1 yen. The earning rate of the underlying asset is 1.5%. So δ = 1.5% 1 1 euro. So S = • The current price of the underlying asset is 95 95 1 1 • The strike asset is euro. So K = . The earning rate of the strike 90 90 asset is 3.5%. So r = 3.5% • Volatility is still 10% C = Se−δT N (d1 ) − Ke−rT N (d2 ) ¶ ¶ µ µ 1/95 S 1 2 1 2 ln ln + r−δ+ σ T + 0.035 − 0.015 + × 0.1 0.5 K 2 1/90 2 √ √ = d1 = = 0.1 0.5 σ T −0.587 849 N (d1 ) = 0.278 316 8 √ √ d2 = d1 − σ T = −0.587 849 − 0.1 0.5 = −0.658 559 7 N (d2 ) = 0.255 089 C = Se−δT N (d1 )−Ke−rT N (d2 ) = 0.255 089 = 0.0001 226 1 (euro)

1 −0.015(0.5) 1 0.278 316 8− e−0.035(0.5) e 95 90

c. A 90-strike yen-denominated euro put is worth 1. 048 3 (yen). This put is "give 1 euro and get 90 yen." www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA • The privilege of "give 1 euro and get 90 yen" at T = 0.5 is worth 1. 048 3 (yen) at t = 0 (statement a) • This can be expressed at (1€ → 90Y ) = 1. 048 3Y 1 strike euro-denominated yen call is worth 0.0001 226 1 (euro). This 90 1 call is "give euro and get 1 yen." 90 A

1 euro and get 1 yen" at T = 0.5 is worth 0.0001 90 226 1 (euro) at t = 0 (statement b) µ ¶ 1 • This can be expressed at € → 1Y = 0.0001 226 1€ 90 • The privilege of "give

Statement a and b are essentially the same statement. Using Statement a, we can derive Statement b; using Statement b, we can derive Statement a. First, we derive Statement b from Statement a. (1€ → 90Y (statement a) µ ) = 1. 048¶3Y 1 1. 048 3 1 € → 1Y = (1€ → 90Y ) = Y ⇒ 90 90 90 The above equation means this: If you can sell 1€ for a guaranteed price 1 € for a guaranteed price of 1Y . This 90Y , then you must be able to sell 90 should make intuitive sense. 1 1. 048 3 Y into euros. Since at t = 0, 1€=95Y or 1Y = €, We convert 90 95 then 1. 048 3 1 1. 048 3 Y = × €= 0.0001226 1€ 90 µ 90 ¶ 95 1 1. 048 3 1 € → 1Y = (1€ → 90Y ) = Y = 0.0001226 1€ ⇒ 90 90 90 This is exactly Statement b Next, we derive ¶ Statement b from Statement a. µ 1 € → 1Y = 0.0001 226 1€ (statement b) 90 ¶ µ 1 € → 1Y = 90 (0.0001 226 1€) = 90 (0.0001 226 1) 95Y = ⇒ (1€ → 90Y ) = 90 90 1. 048 3Y This is exactly Statement a.

Problem 12.6. www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA a. C = Se−δT N (d1 ) − Ke−rT N (d2 ) ¶ ¶ µ µ 100 1 1 S ln + r − δ + σ2 T + 0.06 − 0 + × 0.42 1 ln K 2 105 2 √ √ = = 0.228 024 589 ≈ d1 = 0.4 1 σ T 0.228 025 N (d1 ) = 0.590 186 6 √ √ d2 = d1 − σ T = 0.228 025 − 0.4 1 = −0.171 975 N (d2 ) = 0.431 729 C = Se−δT N (d1 ) − Ke−rT N (d2 ) = 100e−0(1) 0.590 186 6 − 105e−0.06(1) 0.431 729 = 16. 327 019 b. F0,T = Se(r−δ)T = 100e(0.06−0)1 = 106. 183 65 c According to the textbook Equation 12.7, if the underlying asset is futures instead of stocks, we can use the general Black-Scholes formula except • set F = S (replace the stock price with futures price) • set δ = r

¶ µ 1 S F 100e(0.06−0)1 1 1 2 + r − δ + σ2 T ln ln T + σ + × 0.42 (1) K 2 K √2 105 √ √ 2 d1 = = = = 0.4 1 σ T σ T ¶ µ 1 100 + 0.06 − 0 + × 0.42 1 ln 105 2 √ = 0.228 024 589 ≈ 0.228 025 0.4 1 N (d1 ) = 0.590 186 6 √ √ d2 = d1 − σ T = 0.228 025 − 0.4 1 = −0.171 975 N (d2 ) = 0.431 729 C = Se−rT N (d1 ) − Ke−rT N (d2 ) = 100e−0(1) 0.590 186 6 − 105e−0.06(1) 0.431 729 = 16. 327 019 ln

The call premium in c is equal to the call premium in a. Why? We can prove this mathematically. Replace S with F = Se(r−δ)T and set δ = r , we get: ¶ ¶ µ µ S 1 1 Se(r−δ)T ln + r − r + σ2 T + r − δ + σ2 T ln K 2 K 2 √ √ d1 = = σ T σ T √ d2 = d1 − σ T £ ¤ C = Se(r−δ)T e−rT N (d1 ) − Ke−rT N (d2 ) = Se−δT N (d1 ) − Ke−rT N (d2 )

www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA This is the asset is a stock. ¤ formula when ¡the underlying ¢ £ same call P = − Se(r−δ)T e−rT N (−d1 ) + Ke−rT N (−d2 ) = −Se−δT N (−d1 ) + Ke−rT N (−d2 ) This is the same put formula when the underlying asset is a stock. We can also prove that the futures option premium is equal to the underlying stock option premium intuitively. On the maturity date T , the futures price is equal to the stock price. So if we stand at T , the payoff of a futures option and the payoff of an otherwise identical stock option are identical. Consequently, the premium of a futures option is equal to the premium of an otherwise identical stock option. Problem 12.7. a.

¶ ¶ µ µ 100 1 1 S ln + r − δ + σ2 T + 0.08 − 0.03 + × 0.32 0.75 K 2 95 2 √ √ d1 = = = 0.3 0.75 σ T 0.471 669 √ √ d2 = d1 − σ T = 0.471 669 − 0.3 0.75 = 0.211 861 N (d1 ) = 0.681 418 N (d2 ) = 0.583 892 C = Se−δT N (d1 ) − Ke−rT N (d2 ) = 100e−0.03(0.75) 0.681 418 − 95e−0.08(0.75) 0.583 892 = 14. 386 295 ln

b.

µ ¶ ¶ µ 1 100e−0.03(0.75) 1 S 2 + 0 − 0 + 0.75 ln + r − δ + σ2 T × 0.3 K 2 2 95e−0.08(0.75) √ √ = d1 = = 0.3 0.75 σ T 0.471 669100 √ √ d2 = d1 − σ T = 0.471 669 − 0.3 0.75 = 0.211 861 ln

N (d1 ) = 0.681 418 N (d2 ) = 0.583 892 C = Se−δT N (d1 ) − Ke−rT N (d2 ) = 100e−0.03(0.75) 0.681 418 − 95e−0.08(0.75) 0.583 892 = 14. 386 295 The call premium is b is the same as the call premium in a. This is because the call and ¡ put¢premium ¡formula ¢can be rewritten as: C = Se−δT N (d1 ) − Ke−rT N (d2 ) ¡ ¢ ¢ ¡ P = − Se−δT N (−d1 ) + Ke−rT N (−d2 ) µ ¶ ¶ µ 1 2 Se−δT S 1 + 0 − 0 + ln T ln + r − δ + σ2 T σ K 2 Ke−rT 2 √ √ = d1 = σ T √ σ T d2 = d1 − σ T www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA From the above equations, you see that instead of using r and δ, we can replace S with Se−δT , replace K with Ke−rT , and set r = δ = 0. This will also give us the correct option premium. Problem 12.8. a. F0,T =0.75 = Se(r−δ)T = 100e(0.08−0.03)0.75 = 103. 821 20 b. To find the futures option premium, we replace S with F0,T and replace δ with r in the standard Black-Scholes formula: ¶ ¶ µ µ 103. 821 20 S 1 1 ln ln + r − δ + σ2 T + 0.08 − 0.08 + × 0.32 0.75 K 2 95 2 √ √ = = d1 = 0.3 0.75 σ T 0.471 669 √ √ d2 = d1 − σ T = 0.471 669 − 0.3 0.75 = 0.211 861 N (d1 ) = 0.681 418 N (d2 ) = 0.583 892 C = Se−δT N (d1 )−Ke−rT N (d2 ) = 103. 821 20e−0.08(0.75) 0.681 418−95e−0.08(0.75) 0.583 892 = 14. 386 295 Here is another method. Since future option premium is equal to the stock option premium, we can just calculate the stock option premium. Actually, the stock call option premium is calculated in 12.7 a. c. 12.7 a and 12.8 b have the same premium. Problem 12.9. a.When a stock pays discrete dividend, we’ll use the textbook Equation 12.5. P N (d1 ) − P V (K) N (d2 ) C = F0,T P F0,T = S0 − P V (Div) = 50 − P V (Div) P V (Div) = 2e−0.08(1/360) = 1. 999 555 6 = 2 P F0,T (S) = 50 − 2 = 48 P F0,T 1 1 48 + σ2 T + × 0.32 × 0.5 ln −0.08(0.5) P V (K) 2 40e √ 2 √ d1 = = = 1. 154 1 0.3 0.5 σ T √ √ d2 = d1 − σ T = 1. 154 1 − 0.3 0.5 = 0.9420

ln

N (d1 ) = 0.875 77 N (d2 ) = 0.826 90 C = 48 (0.875 77) − 40e−0.08(0.5) (0.826 90) = 10. 258 So the European call premium is 10. 258 www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA If it’s ever optimal to early exercise an American call, the best time to early exercise is immediately before the dividend payment. Since the dividend is paid tomorrow, the best time to early exercise is today. The exercise value is EV = 50 − 40 = 10, which is less than the European call premium 10. 258. So it’s not optimal to early exercise the American call. Some of you might think that it’s optimal to early exercise the American call. If we early exercise today, we get a stock, which will pay us a dividend 2. So the total exercise value is 12. This reasoning is flawed. If you exercise the call today and take ownership of a stock, you’ll get $2 dividend tomorrow. However, after the dividend is paid, the price of your stock is reduced by the amount of the dividend $2 (so you also lose $2). So your exercise value is 10, not 12. And it’s not optimal to early exercise the American call. P b. C = F0,T N (d1 ) − P V (K) N (d2 ) P F0,T = S0 − P V (Div) = 50 − P V (Div) P V (Div) = 2e−0.08(1/360) = 1. 999 555 6 = 2 P F0,T (S) = 50 − 2 = 48 P F0,T 1 1 48 + σ2T + × 0.32 × 0.5 ln −0.08(0.5) P V (K) 2 2 40e √ √ d1 = = = 1. 154 1 0.3 0.5 √σ T √ d2 = d1 − σ T = 1. 154 1 − 0.3 0.5 = 0.9420

ln

N (d1 ) = 0.875 77 N (d2 ) = 0.826 90 C = 48 (0.875 77) − 40e−0.08(0.5) (0.826 90) = 10. 258 The exercise value is EV = 60 − 40 = 20 > C Hence it’s optimal to early exercise the American put at t = 0. c. It’s optimal to early exercise the American call at t = 0 if the exercise value is greater than the European call premium. It’s not optimal to exercise the American call at t = 0 if the exercise value is equal to or less than the he European call premium. However, keep in mind that if the stock doesn’t pay dividend, then it’s never optimal to early exercise an American call. Problem 12.10. The statement means that the absolute value |θ (t) | = | ∂V ∂t | reaches its maximum value when t → T . In other words, the closer to the expiration date, the higher the |θ (t) |. This statement is not correct. I don’t know of any intuitive way to explain why this statement is not correct. That’s why the textbook asks you to test this statement using a spreadsheet. www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA If you want to test the statement using a spreadsheet, you can use the spreadsheet titled "optbasic2." Then you can make up a case and validate the statement. However, for the purpose of passing the exam, you can ignore this problem. Problem 12.11. a. If you look at the Appendix 12.A (which is excluded from both CAS and SOA exam), you can see that Vega is the derivative of an option price regarding volatility: V ega = ∂V ∂σ (σ− ) The formula V ega = V (σ+ )−V is an approximation. For this approx2 imation to work, needs to be small. Since Appendix 12.A is excluded from both CAS and SOA exam, you can ignore this part. b. If you want to solve this problem, you can set up some test cases and compare the approximated Vega with the actual Vega (using the spreadsheet). For the purpose of passing the exam, you can ignore this part. Problem 12.12. Since Appendix 12.A is excluded from both CAS and SOA exam, you can ignore this problem. Problem 12.13. Let’s not worry about drawing a diagram and focus on how to calculate the profit. I’ll do some sample calculations. Let’s calculate the profit after 6 months (i.e. at expiration) assuming the stock price after 6 months is $60. Using the Black-Scholes formula, we can find: • The 40-strike call premium is 4.1553. This call premium is calculated using the Black-Scholes formula by setting S = 40, K = 40, T = 0.5, r = 0.08, σ = 0.3, δ = 0 • The 45-strike call premium is 2.1304. This call premium is calculated using the Black-Scholes formula by setting S = 40, K = 45, T = 0.5, r = 0.08, σ = 0.3, δ = 0 • The net cost of the bull spread at t = 0 is 4.1553 − 2.1304 = 2. 024 9 • Its future value at expiration is 2. 024 9e0.08(0.5) = 2. 107 5 The stock price at expiration is 60. www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA • The 40-strike call payoff is 60 − 40 = 20 • The 45-strike call payoff is − (60 − 45) = −15 • The total payoff is 20 − 15 = 5 So our profit is: −2. 107 5 + 5 = 2. 892 5 Let’s calculate the profit after 3 months assuming the stock price after 3 months is $60. Using the Black-Scholes formula, we can find: • The 40-strike call premium is 4.1553 • The 45-strike call premium is 2.1304 • The net cost of the bull spread at t = 0 is 4.1553 − 2.1304 = 2. 024 9 • Its future value after 3 months is 2. 024 9e0.08(0.25) = 2. 065 8 3 months later (i.e. at t = 0.25), we close our position. Right now, our purchased 40-strike call and sold 45-strike call both have 3 months to expiration. To close our position (i.e. to cancel out our position), at t = 0.25, we sell a 40strike call and buy a 45-strike call. After this, our net position is zero. At t = 0.25, the stock price is 60. • At t = 0.25, the 40-strike call with 3 months to expiration is worth 20.7969 (so we receive 20.7969). This call premium is calculated using the BlackScholes formula by setting S = 60, K = 40 T = 0.25, r = 0.08, σ = 0.3, δ = 0. • At t = 0.25, the 45-strike call with 3 months to expiration is worth 15.9480 (so we pay 15.9480). This call premium is calculated using the BlackScholes formula by setting S = 60, K = 45 T = 0.25, r = 0.08, σ = 0.3, δ = 0 • The net receipt is 20.7969 − 15.9480 = 4. 848 9 So our profit is: −2. 065 8 + 4. 848 9 = 2. 783 1 Let’s calculate the profit after 1 day assuming the stock price after 1day is $60. Using the Black-Scholes formula, we can find: • The 40-strike call premium is 4.1553 • The 45-strike call premium is 2.1304 • The net cost of the bull spread at t = 0 is 4.1553 − 2.1304 = 2. 024 9 • Its future value after 1 day is 2. 024 9e0.08(1/365) = 2. 025 34 www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA 1 day later (i.e. at t = 364/365), we close our position. Right now, our purchased 40-strike call and sold 45-strike call both have 364 days to expiration. To close our position (i.e. to cancel out our position), at t = 364/365 we sell a 40-strike call and buy a 45-strike call. After this, our net position is zero. At t = 364/365, the stock price is 60. • At t = 364/365, the 40-strike call with 364 days to expiration is worth 23.3775 (so we receive 23.3775). This call premium is calculated using the Black-Scholes formula by setting S = 60, K = 40 T = 364/365, r = 0.08, σ = 0.3, δ = 0. • At t = 364/365, the 45-strike call with 364 days to expiration is worth 19.2391 (so we pay 19.2391). This call premium is calculated using the Black-Scholes formula by setting S = 60, K = 45 T = 364/365, r = 0.08, σ = 0.3, δ = 0 • The net receipt is 23.3775 − 19.2391 = 4. 138 4 So our profit is: −2. 038 44 + 4. 138 4 = 2. 099 96 Problem 12.14. Appendix 12.A is excluded from both CAS and SOA exam. So SOA and CAS can’t ask you to calculate option Greeks using formulas in Appendix 12.A. (SOA and CAS can ask you to calculate delta ∆ using the textbook formula 10.1). However, SOA and CAS can ask you to calculate the Greeks for a portfolio using the formula presented in Page 389. Page 389 is on the syllabus. So you need to learn how to calculate portfolio’s Greeks using the following formula (in the textbook Page 389): Greekoption = a.

S = 40 ω1

Price Delta Gamma Vega Theta Rho

Pn

1 1 1 1 1 1

i=1

ω i Greeki

Greek1 bought 40K call 4.1553 0.6159 0.0450 0.1080 −0.0134 0.1024

ω2 1 1 1 1 1 1

Greek2 sold 45K call −2.1304 −0.3972 −0.0454 −0.1091 0.0120 −0.0688

Pn Greekoption = i=1 ω i Greeki Portfolio Price and Greeks 1 (4.1553) + 1 (−2.1304) = 2. 024 9 1 (0.6159) + 1 (−0.3972) = 0.218 7 1 (0.0450) + 1 (−0.0454) = −0.000 4 1 (0.1080) + 1 (−0.1091) = −0.001 1 1 (−0.0134) + 1 (0.0120) = −0.001 4 1 (0.1024) + 1 (−0.0688) = 0.033 6

Column 3 and 5 are calculated using the "optbasic2" spreadsheet. You don’t need to worry about these two columns. YouP just need to focus on the n last column and practice the formula Greekoption = i=1 ω i Greeki . www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA b.

S = 45 ω1

Price Delta Gamma Vega Theta Rho

1 1 1 1 1 1

Greek1 bought 40K call 7.7342 0.8023 0.0291 0.0885 −0.0135 0.1418

ω2 1 1 1 1 1 1

Greek2 sold 45K call −4.6747 −0.6159 −0.0400 −0.1216 0.0150 −0.1152

P Greekoption = ni=1 ω i Greeki Portfolio Price and Greeks 1 (7.7342) + 1 (−4.6747) = 3. 059 5 1 (0.8023) + 1 (−0.6159) = 0.186 4 1 (0.0291) + 1 (−0.0400) = −0.010 9 1 (0.0885) + 1 (−0.1216) = −0.033 1 1 (−0.0135) + 1 (0.0150) = 0.001 5 1 (0.1418) + 1 (−0.1152) = 0.026 6

c. Ignore this part. Not sure what this problem wants to accomplish. Problem 12.15. a.

S = 40 ω1

Price Delta Gamma Vega Theta Rho b.

1 1 1 1 1 1

S = 45 ω1

Price Delta Gamma Vega Theta Rho

1 1 1 1 1 1

Greek1 bought 40K call 2.5868 −0.3841 0.0450 0.1080 −0.0049 −0.0898

ω2

Greek1 bought 40K call 1.1658 −0.1977 0.0291 0.0885 −0.0051 −0.0503

ω2

1 1 1 1 1 1

1 1 1 1 1 1

Greek2 sold 45K call −5.3659 0.6028 −0.0454 −0.1091 0.0025 0.1474

P Greekoption = ni=1 ω i Greeki Portfolio Price and Greeks 1 (2.5868) + 1 (−5.3659) = −2. 779 1 1 (−0.3841) + 1 (0.6028) = 0.218 7 1 (0.0450) + 1 (−0.0454) = −0.000 4 1 (0.1080) + 1 (−0.1091) = −0.001 1 1 (−0.0049) + 1 (0.0025) = −0.002 4 1 (−0.0898) + 1 (0.1474) = 0.057 6

Greek2 sold 45K call −2.9102 0.3841 −0.0400 −0.1216 0.0056 0.1010

Pn Greekoption = i=1 ω i Greeki Portfolio Price and Greeks 1 (1.1658) + 1 (−2.9102) = −1. 744 4 1 (−0.1977) + 1 (0.3841) = 0.186 4 1 (0.0291) + 1 (−0.0400) = −0.010 9 1 (0.0885) + 1 (−0.1216) = −0.033 1 1 (−0.0051) + 1 (0.0056) = 0.000 5 1 (−0.0503) + 1 (0.1010) = 0.050 7

c. Ignore Problem 12.16. There’s no easy way to solve this problem manually. This type of problems shouldn’t show up in the exam. Problem 12.17. There’s no easy way to solve this problem manually. This type of problems shouldn’t show up in the exam. www.actuary88.com

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CHAPTER 12. BLACK-SCHOLES FORMULA Problem 12.18. S = 50

K = 60

r = 0.06

σ = 0.4

δ = 0.03

a. Solve for h. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.42 h (h − 1) + (0.06 − 0.03) h − 0.06 = 0 2 h = −0.608 182 5 h = 1. 233 182 5 Use the bigger h for call and the smaller h for put. hcall 1. 233 182 5 ∗ HCall = K= × 60 = 317. 309 19 hcall − 1 1. 233 182 5 − 1 ∗ − K) Cperpetual = (HCall

26. 351 83

µ

S ∗ HCall

¶hcall

= (317. 309 19 − 60)

µ

50 317. 309 19

¶1. 233 182 5

=

The call should be exercised when the stock price reaches 317. 309 19. This price is called the barrier. The call premium is 26. 351 83. b. Now δ = 0.04 (instead of δ = 0.03). Everything else is the same as in a. Higher dividend yield means that the stock price will decrease more quickly. Recall that the stock price drops by the dividend amount immediately after the dividend payment time. We expect that under δ = 0.04 the optimal stock price (i.e. barrier) is lower than the barrier when δ = 0.03. In addition, we expect that the option price under δ = 0.04 is lower than the option price under δ = 0.03. Solve for h. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.42 h (h − 1) + (0.06 − 0.04) h − 0.06 = 0 2 h = −0.568 729 3 h = 1. 318 729 3 Use the bigger h for call and the smaller h for put. hcall 1. 318 729 3 ∗ HCall = K= × 60 = 248. 247 52 hcall − 1 1. 318 729 3 − 1 Cperpetual =

∗ (HCall

22. 751 3

− K)

µ

S ∗ HCall

¶hcall

= (248. 247 52 − 60)

µ

50 248. 247 52

¶1. 318 729 3

The call should be exercised when the stock price reaches 248. 247 52. The call premium is 22. 751 28 c.Now r = 0.07 (instead of r = 0.06). Everything else is the same as in a. www.actuary88.com

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=

CHAPTER 12. BLACK-SCHOLES FORMULA When the risk free interest rate goes up, stocks are expected to generate high returns. Consequently, the value of a stock goes up; the call option premium goes up (while the put option premium goes down). We expect that both the barrier and the call premium go up when r goes up from 0.06 to 0.07. Solve for h. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.42 h (h − 1) + (0.07 − 0.03) h − 0.07 = 0 2 h = 1. 218 245 8 h = −0.718 245 8 Use the bigger h for call and the smaller h for put. hcall 1. 218 245 8 ∗ = HCall K= × 60 = 334. 919 4 hcall − 1 1. 218 245 8 − 1 Cperpetual =

∗ (HCall

− K)

27. 100 1

µ

S ∗ HCall

¶hcall

= (334. 919 4 − 60)

µ

50 334. 919 4

¶1. 218 245 8

=

The call should be exercised when the stock price reaches 334. 919 4 The call premium is 27. 1001 d. The higher the volatility, the more valuable an option is. As σ goes up from 0.4 to 0.5, we expect both the barrier and the call option premium will go up. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.52 h (h − 1) + (0.06 − 0.03) h − 0.06 = 0 2 h = 1. 170 189 9 h = −0.410 189 9 Use the bigger h for call and the smaller h for put. hcall 1. 170 189 9 ∗ = K= × 60 = 412. 547 36 HCall hcall − 1 1. 170 189 9 − 1 ∗ − K) Cperpetual = (HCall

29. 835 5

µ

S ∗ HCall

¶hcall

= (412. 547 36 − 60)

µ

50 412. 547 36

¶1. 170 189 9

The call should be exercised when the stock price reaches 412. 547 36 The call premium is 29. 835 5 Problem 12.19. S = 50

K = 60

a.Solve for h. www.actuary88.com

r = 0.06

σ = 0.4

δ = 0.03

1 2 σ h (h − 1) + (r − δ) h − r = 0 2 c °Yufeng Guo

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=

CHAPTER 12. BLACK-SCHOLES FORMULA 1 × 0.42 h (h − 1) + (0.06 − 0.03) h − 0.06 = 0 2 h = −0.608 182 5 h = 1. 233 182 5 Use the bigger h for call and the smaller h for put. hput −0.608 182 5 K= × 60 = 22. 690 80 HP∗ ut = hcall − 1 −0.608 182 5 − 1 Pperpetual = (K − HP∗ ut ) 23. 074 7

µ

S HP∗ ut

¶hput

= (60 − 22. 690 80)

µ

50 22. 690 80

¶−0.608 182 5

The put should be exercised when the stock price reaches 22. 690 80. The put premium is 23. 074 7 b. Now δ = 0.04 (instead of δ = 0.03). Everything else is the same as in a. Higher dividend yield means that the stock price will decrease more quickly. Since the value of a put goes up if the stock price goes down, we expect that the barrier price will go down (i.e. exercise occurs later because we want to exercise when the stock price is low) and the put premium will go up. 1 2 Solve for h. σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.42 h (h − 1) + (0.06 − 0.04) h − 0.06 = 0 2 h = −0.568 729 3 h = 1. 318 729 3 Use the bigger h for call and the smaller h for put. hput −0.568 729 3 K= × 60 = 21. 752 5 HP∗ ut = hcall − 1 −0.568 729 3 − 1 Pperpetual = (K − HP∗ ut ) 23. 824 8

µ

S HP∗ ut

¶hput

= (60 − 21. 752 5)

µ

50 21. 752 5

¶−0.568 729 3

The put should be exercised when the stock price reaches 21. 752 5 The put premium is 23. 824 8 c. As r goes up, people expect to get increased return from the stock. The stock price goes up and the put option premium goes down. We expect the barrier will go up (compared with a, meaning that exercise occurs sooner). 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.42 h (h − 1) + (0.07 − 0.03) h − 0.07 = 0 2 h = 1. 218 245 8 h = −0.718 245 8 Use the bigger h for call and the smaller h for put. hput −0.718 245 8 K= × 60 = 25. 080 67 HP∗ ut = hcall − 1 −0.718 245 8 − 1 www.actuary88.com

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=

=

CHAPTER 12. BLACK-SCHOLES FORMULA

Pperpetual = (K −

HP∗ ut )

23. 824 8

µ

S HP∗ ut

¶hput

= (60 − 25. 080 67)

µ

50 25. 080 67

¶−0.568 729 3

The put should be exercised when the stock price reaches 25. 080 67 The put premium is 23. 824 8 d. The higher the volatility, the more valuable an option is. As σ goes up from 0.4 to 0.5, we expect both the barrier will go down (i.e. exercise occurs later) and the put option premium will go up. 1 2 σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.52 h (h − 1) + (0.06 − 0.03) h − 0.06 = 0 2 h = 1. 170 189 9 h = −0.410 189 9 Use the bigger h for call and the smaller h for put. hput −0.410 189 9 HP∗ ut = K= × 60 = 17. 452 5 hcall − 1 −0.410 189 9 − 1 Pperpetual = (K − HP∗ ut ) 27. 629 4

µ

S HP∗ ut

¶hput

= (60 − 17. 452 5)

µ

50 17. 452 5

¶−0.410 189 9

The put should be exercised when the stock price reaches 17. 452 5 The put premium is 27. 629 4

Problem 12.20. For a and b, if you use the Black-Scholes formula, you’ll find that call and put are both worth 17.6988. For part c. After we switch S and K and switch r and δ, the put after the switch and the call before the switch have the same value. This is not a coincidence. It’s explained in my solution to Problem 10.19.

Problem 12.21. 1 2 a. σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.32 h (h − 1) + (0.08 − 0.05) h − 0.08 = 0 2 h = −1. 177 043 h = 1. 510 376 Use the bigger h for call and the smaller h for put. hcall 1. 510 376 ∗ = HCall K= × 90 = 266. 340 6 hcall − 1 1. 510 376 − 1 www.actuary88.com

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=

=

CHAPTER 12. BLACK-SCHOLES FORMULA

Cperpetual =

∗ (HCall

− K)

µ

S ∗ HCall

40. 158 9 The call premium is 40. 158 9

¶hcall

= (266. 340 6 − 90)

µ

100 266. 340 6

¶1. 510 376

1 2 b. σ h (h − 1) + (r − δ) h − r = 0 2 1 × 0.32 h (h − 1) + (0.05 − 0.08) h − 0.05 = 0 2 h = −0.510 38 h = 2. 177 04 Use the bigger h for call and the smaller h for put. hput −0.510 38 HP∗ ut = K= × 100 = 33. 791 5 hcall − 1 −0.510 38 − 1 Pperpetual = (K −

HP∗ ut )

µ

St HP∗ ut

40. 158 9 The put premium is 40. 158 9

¶hput

= (100 − 33. 791 5)

µ

90 33. 791 5

¶−0.510 38

c. After we switch S and K and switch r and δ, the perpetual American put after the switch and the perpetual American call before the switch have the same value. This is a minor fact probably not worth memorizing. We are not going to worry about the proof.

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=

=

CHAPTER 12. BLACK-SCHOLES FORMULA

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Chapter 13

Market making and delta hedging Problem 13.1. ¶ ¶ µ µ 91 40 S 1 1 ln + r − δ + σ2 T + 0.08 − 0 + × 0.32 K 2 45 2 365 r √ = = d1 = σ T 91 0.3 365 −0.578 25 r √ 91 = −0.728 04 d2 = d1 − σ T = −0.578 25 − 0.3 365 N (d1 ) = 0.281 55 N (d2 ) = 0.233 29 C = Se−δT N (d1 ) − Ke−rT N (d2 ) = 40e−0(91/365) 0.281 55 − 45e−0.08(91/365) 0.233 29 = 0.971 3 ln

A call on 100 stocks is worth 100 (0.971 3) = 97. 13 Use the formula in Appendix 12.8 to find ∆ ∆ = e−δT N (d1 ) = e−0×91/365 N (d1 ) = 0.281 55 (positive delta means buying stocks) Suppose a trader sells a call option on 100 stocks. To hedge his risk, the trader should at t = 0 • sell the call and receive 100 (0.971 3) = 97. 13 • buy 0.281 55 (100) = 28. 155 stocks costing 40 (28. 155) = 1126. 2 • borrow 1126. 2 − 97. 13 = 1029. 07 from the bank 125

CHAPTER 13. MARKET MAKING AND DELTA HEDGING The trader’s net position is zero at t = 0 If the stock price is 39 on Day 1. To close his short call position, the trader can buy, from the open market, a 45-strike call expiring in 90 days. This new (purchased) call and the original (sold) call have the same underlying, same expiration date, same strike price. They will cancel out each other and the trader doesn’t have any liabilities associated either call. The cost of theµnew call is 70.55. It’s calculated ¶ ¶ µ as follows: 90 39 1 2 1 S 2 ln + r−δ+ σ T + 0.08 − 0 + × 0.3 ln K 2 45 2 365 r √ d1 = = = σ T 90 0.3 365 −0.753 706 r √ 90 d2 = d1 − σ T = −0.753 706 346 − 0.3 = −0.902 675 365 N (d1 ) = 0.774 49 N (d2 ) = 0.183 349 C = Se−δT N (d1 ) −Ke−rT N (d2 ) = 39e−0(90/365) 0.225 513−45e−0.08(90/365) 0.183 349 = 0.705 5 A call on 100 stocks is worth 100 (0.705 5) = 70. 55 So the trader needs to pay 70. 55 to buy a new call to cancel out the original call he sold. At t = 0, the trader owns 28. 155 stocks. Now one day later, the stocks are worth 28. 155 (39) = 1098. 045. So the trader sells out his stocks, receiving 1098. 05. In addition, the trader needs to pay back the loan borrowed from the bank. The future value of the loan one day later is: 1029. 07e0.08(1/365) = 1029. 30 So the trader’s net wealth at t = 1/365 is: 1098. 05 − (1029. 30 + 70. 55) = −1. 8 So the trader lost $1.8. If you want to following the textbook calculation on Page 417 "Day 1: Marking-to-market," here it is: Day 1 Gain on 28. 155 shares 28. 155 (39 − 40) = −28. 155 Gain on the written call 97. 13 − ¡70. 55 = 26. 58 ¢ Interest 1029. 07 e0.08(1/365) − 1 = 0.226 Overnight profit −28. 155 + 26. 58 − 0.226 = −1. 8 If the stock price is 40.5 on Day 1. To close his short call position, the trader can buy, from the open market, a 45-strike call expiring in 90 days. This new (purchased) call and the original (sold) call have the same underlying, same expiration date, same strike price. www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING They will cancel out each other and the trader doesn’t have any liabilities associated either call. The cost of theµ new call is 110.46. It’s calculated ¶ ¶ µ as follows: 40.5 90 S 1 2 1 2 ln ln + r−δ+ σ T + 0.08 − 0 + × 0.3 K 2 45 2 365 r √ = = d1 = 90 σ T 0.3 365 −0.500 363 r √ 90 d2 = d1 − σ T = −0.500 363 − 0.3 = −0.649 332 365 N (d1 ) = 0.308 41 N (d2 ) = 0.258 062 C = Se−δT N (d1 )−Ke−rT N (d2 ) = 40.5e−0(90/365) 0.308 41−45e−0.08(90/365) 0.258 062 = 1. 104 65 A call on 100 stocks is worth 100 (1. 104 65) = 110. 465 One Day 1, the trader • pays 110. 46 and buy a new call to cancel out the original call he sold • sells out his 28. 155 stocks, receiving 28. 155 (40.5) = 1140. 28 • pays 1029. 07e0.08(1/365) = 1029. 30 to the bank to payoff the loan So the trader’s net wealth on Day 1 is: 1140. 28 − (1029. 30 + 110. 46) = 0.52 If you want to use the textbook notation, here you go: Day 1 Gain on 28. 155 shares 28. 155 (40.5 − 40) = 14. 077 5 Gain on the written call 97. 13 − ¡110. 465 = −13.¢335 Interest 1029. 07 e0.08(1/365) − 1 = 0.226 Overnight profit 14. 077 5 − 13. 335 − 0.226 = 0.52 So the trader gained $0.52. Problem 13.2.

¶ ¶ µ µ 91 40 1 1 S ln + r − δ + σ2 T + 0.08 − 0 + × 0.32 K 2 40 2 365 r √ = = d1 = σ T 91 0.3 365 0.208 05 r √ 91 d2 = d1 − σ T = 0.208 05 − 0.3 = 0.05 825 365 N (−d2 ) = 0.476 77 N (−d1 ) = 0.417 59 P = 40e−0.08(91/365) 0.476 77 − 40e−0(91/365) 0.417 59 = 1. 990 6 ln

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING A put on 100 stocks is worth 100 (1. 990 6) = 199. 06 Use the formula in Appendix 12.8 to find ∆ ∆ = −e−δT N (−d1 ) = −e−0×91/365 0.417 59 = −0.417 6 (negative delta means short selling stocks) Suppose a trader sells a put option on 100 stocks. To hedge his risk, the trader should at t = 0 • sell the put and receive 100 (1. 990 6) = 199. 06 • buy −0.417 6 (100) = −41. 76 stocks (i.e. short sell 41. 76 stocks) receiving 41. 76 (40) = 1670. 4 • lend 199. 06 + 1670. 4 = 1869. 46 to a bank The trader’s net position is zero at t = 0 If the stock price is 39.5 on Day 1. To close his short put position, the trader can buy, from the open market, a 40-strike put expiring in 90 days. This new (purchased) put and the original (sold) put have the same underlying, same expiration date, same strike price. They will cancel out each other and the trader doesn’t have any liabilities associated either put. The cost of the as follows: ¶ µ new put is ¶ µ 90 39 S 1 2 1 ln ln + r−δ+ σ T + 0.08 − 0 + × 0.32 K 2 40 2 365 r √ d1 = = 0.0 = σ T 90 0.3 365 3695 r √ 90 d2 = d1 − σ T = 0.03695 − 0.3 = −0.112 02 365 N (−d2 ) = 0.544 60 N (−d1 ) = 0.485 26 P = Ke−rT N (−d2 )−Se−δT N (−d1 ) = 40e−0.08(90/365) 0.544 60−39e−0(90/365) 0.485 26 = 2. 433 4 On Day 1 a put on 100 stocks is worth 100 (2. 433 4) = 243. 34 One Day 1, the trader • pays 243. 34, buying a new put to cancel out the original put he sold • buys back 41. 76 stocks to close the short sale position, paying 41. 76 (39) = 1628. 64 • receives 1869. 46e0.08(1/365) = 1869. 87 from the bank www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING So the trader’s net wealth on Day 1 is: −243. 34 − 1628. 64 + 1869. 87 = −2. 11 So the trader lost 2. 11. If you want to use the textbook notation, here you go: Day 1 Gain on 41. 76 shares 41. 76 (40 − 39) = 41. 76 Gain on the written put 199. 06 −¡ 243. 34 = −44.¢28 Interest income 1869. 46 e0.08(1/365) − 1 = 0.41 Overnight profit 41. 76 − 44. 28 + 0.41 = −2. 11 If the stock price is 40.5 on Day 1. To close his short call position, the trader can buy, from the open market, a 40-strike call expiring in 90 days. This new (purchased) call and the original (sold) call have the same underlying, same expiration date, same strike price. They will cancel out each other and the trader doesn’t have any liabilities associated either call. The cost of theµ new call is calculated as follows: ¶ ¶ µ 90 40.5 S 1 2 1 2 ln ln + r−δ+ σ T + 0.08 − 0 + × 0.3 K 2 40 2 365 r √ d1 = = = σ T 90 0.3 365 0.290 29 r √ 90 d2 = d1 − σ T = 0.290 29 − 0.3 = 0.141 32 365 N (−d1 ) = 0.385 80 N (−d2 ) = 0.443 81 P = Ke−rT N (−d2 )−Se−δT N (−d1 ) = 40e−0.08(90/365) 0.443 81−40.5e−0(90/365) 0.385 80 = 1. 780 8 On Day 1 a put on 100 stocks is worth 100 (1. 780 8) = 178. 08 One Day 1, the trader • pays 178. 08, buying a new put to cancel out the original put he sold • buys back 41. 76 stocks to close the short sale position, paying 41. 76 (40.5) = 1691. 28 • receives 1869. 46e0.08(1/365) = 1869. 87 from the bank So the trader’s net wealth on Day 1 is: −178. 08 − 1691. 28 + 1869. 87 = 0.51 So the trader gains 0.51. If you want to use the textbook notation, here you go: Day 1 Gain on 41. 76 shares 41. 76 (40 − 40.5) = −20. 88 Gain on the written put 199. 06 −¡ 178. 08 = 20. 98 ¢ Interest income 1869. 46 e0.08(1/365) − 1 = 0.41 Overnight profit −20. 88 + 20. 98 + 0.41 = 0.51 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Next, I’m going to redo Problem 13.2 using the strike price K = 45. This way, Problem 13.2 is similar to Problem 13.1 except that in Problem 13.1 there’s a call and in Problem ¶ a put. ¶ µ 13.2 there’s µ 91 40 S 1 2 1 ln ln + r−δ+ σ T + 0.08 − 0 + × 0.32 K 2 45 2 365 r √ = = d1 = σ T 91 0.3 365 −0.578 25 r √ 91 d2 = d1 − σ T = −0.578 25 − 0.3 = −0.728 04 365 N (−d1 ) = 0.718 45 N (−d2 ) = 0.766 71 P = Ke−rT N (−d2 )−Se−δT N (−d1 ) = 45e−0.08(91/365) 0.766 71−40e−0(91/365) 0.718 45 = 5. 082 6 A put on 100 stocks is worth 100 (5. 082 6) = 508. 26 Use the formula in Appendix 12.8 to find ∆ ∆ = −Se−δT N (−d1 ) = −e−0×91/365 0.718 45 = −0.718 45 (negative delta means short selling stocks) Suppose a trader sells a put option on 100 stocks. To hedge his risk, the trader should at t = 0 • sell the call and receive 100 (5. 082 6) = 508. 26 • buy −0.718 45 (100) = −71. 845 stocks (i.e. short sell 71. 845 stocks) receiving 71. 845 (40) = 2873. 8 • lend 2873. 8 + 508. 26 = 3382. 06 to a bank The trader’s net position is zero at t = 0 If the stock price is 39.5 on Day 1. To close his short put position, the trader can buy, from the open market, a 45-strike put expiring in 90 days. This new (purchased) put and the original (sold) put have the same underlying, same expiration date, same strike price. They will cancel out each other and the trader doesn’t have any liabilities associated either put. The cost of theµnew call is 110.46. It’s calculated ¶ ¶ µ as follows: 90 39 S 1 2 1 2 ln ln + r−δ+ σ T + 0.08 − 0 + × 0.3 K 2 45 2 365 r √ d1 = = = σ T 90 0.3 365 −0.753 706 r √ 90 d2 = d1 − σ T = −0.753 706 346 − 0.3 = −0.902 675 365 N (−d1 ) = 0.774 49 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING N (−d2 ) = 0.816 65 P = Ke−rT N (−d2 )−Se−δT N (−d1 ) = 45e−0.08(90/365) 0.816 65−39e−0(90/365) 0.774 49 = 5. 826 3 On Day 1 a put on 100 stocks is worth 100 (5. 826 3) = 582. 63 One Day 1, the trader • pays 582. 63, buying a new put to cancel out the original put he sold • buys back 71. 845 stocks to close the short sale position, paying 71. 845 (39) = 2801. 955 • receives 3382. 06e0.08(1/365) = 3382. 801 from the bank So the trader’s net wealth on Day 1 is: −582. 63 − 2801. 955 + 3382. 801 = −1. 784 So the trader lost 1.78. If you want to use the textbook notation, here you go: Day 1 Gain on 71. 845 shares 71. 845 (40 − 39) = 71. 845 Gain on the written put 508. 26 −¡ 582. 63 = −74.¢37 Interest income 3382. 06 e0.08(1/365) − 1 = 0.741 Overnight profit 71. 845 − 74. 37 + 0.741 = −1. 784 If the stock price is 40.5 on Day 1. To close his short call position, the trader can buy, from the open market, a 45-strike call expiring in 90 days. This new (purchased) call and the original (sold) call have the same underlying, same expiration date, same strike price. They will cancel out each other and the trader doesn’t have any liabilities associated either call. The cost of theµ new call is 110.46. It’s calculated ¶ ¶ µ as follows: 90 40.5 1 2 1 S 2 ln + r−δ+ σ T + 0.08 − 0 + × 0.3 ln K 2 45 2 365 r √ d1 = = = σ T 90 0.3 365 −0.500 363 r √ 90 = −0.649 332 d2 = d1 − σ T = −0.500 363 − 0.3 365 N (−d1 ) = 0.691 59 N (−d2 ) = 0.741 94 P = Ke−rT N (−d2 )−Se−δT N (−d1 ) = 45e−0.08(90/365) 0.741 94−40.5e−0(90/365) 0.691 59 = 4. 725 76 On Day 1 a put on 100 stocks is worth 100 (4. 725 76) = 472. 576 One Day 1, the trader www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING • pays 472. 576, buying a new put to cancel out the original put he sold • buys back 71. 845 stocks to close the short sale position, paying 71. 845 (40.5) = 2909. 72 • receives 3382. 06e0.08(1/365) = 3382. 801 from the bank So the trader’s net wealth on Day 1 is: −472. 576 − 2909. 72 + 3382. 801 = 0.505 So the trader gains 0.505. If you want to use the textbook notation, here you go: Day 1 Gain on 71. 845 shares 71. 845 (40 − 40.5) = −35. 92 Gain on the written put 508. 26 −¡ 472. 576 = 35. ¢684 Interest income 3382. 06 e0.08(1/365) − 1 = 0.741 Overnight profit −35. 92 + 35. 684 + 0.741 = 0.505 Problem 13.3. buy a 40-45 bull spread=buy a 40-strike call and sell a 45-strike put I used a spreadsheet to calculate the following so you might not be able to fully match my result Day 0: Expiration T = 91/365 Stock price 40 Position Strike Call premium buy 40 −2.7804 sell 45 0.9710 Net (one stock) −2.7804 + 0.9710 = −1. 809 4 Net (100 stocks) 100 (−1. 809 4) = −180. 94

Delta −0.58240 0.28155 −0.28155 + 0.58240 = −0.300 85 100 (−0.300 85) = −30. 085

In the above table • positive premium means cash inflow (i.e. receiving premium) • negative premium means cash outgo (i.e. paying premium) • positive delta means buying stocks • negative delta means short selling stocks On Day 0, the trader: • Buy a 40-strike call and sell a 45-strike call, paying 180. 94 • Short sell 30. 085 stocks, paying 30. 085 (40) = 1203. 4. The negative delta means short selling. www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING • Deposit 1203. 4 − 180. 94 = 1022. 46 to a savings bank to earn a risk free rate The trader’s net position is zero. Day 1: Expiration T = 90/365 Stock price 39 Position Strike price Call premium Sell 40 2.2144 Buy 45 −0.7054 Net (one stock) 2.2144 − 0.7054 = 1. 509 Net (100 stocks) 100 (1. 509) = 150. 9 Day 1, the trader closes out his position • Sell a 40-strike call to cancel out the original 40-strike call he bought; buy a 45-strike call to cancel out the original 45-strike call he sold. The trader receives 150. 9 • Buy 30. 085 stocks from the market to close the short sale, paying 30. 085 (39) = 1173. 315 • Receive 1022. 46e0.08(1/365) = 1022. 684 from the savings account The trader’s net profit: 150. 9 − 1173. 315 + 1022. 684 = 0.269 If the stock price is 39 on Day 1, the trader gets 0.269 profit. Day 1: Expiration T = 90/365 Stock price 40.5 Position Strike price Call premium Sell 40 3.0621 Buy 45 −1.1046 Net (one stock) 3.0621 − 1.1046 = 1. 957 5 Net (100 stocks) 100 (1. 957 5) = 195. 75 Day 1, the trader closes out his position • Sell a 40-strike call to cancel out the original 40-strike call he bought; buy a 45-strike call to cancel out the original 45-strike call he sold. The trader receives 195. 75 • Buy 30. 085 stocks from the market to close the short sale, paying 30. 085 (40.5) = 1218. 442 5 • Receive 1022. 46e0.08(1/365) = 1022. 684 from the savings account The trader’s net profit: 195. 75 − 1218. 442 5 + 1022. 684 = −0.008 5 = −0.01 If the stock price is 39 on Day 1, the trader losses 0.01. Problem 13.4. www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Day 0: Expiration T = 91/365 Stock price 40 Position Strike Put premium buy 45 −5.0824 sell 40 1.9905 Net −5.0824 + 1.9905 × 2 = −1. 101 4 Net 100 (−1. 101 4) = −110. 14

Delta − (−0.71845) = 0.718 451 −0.41760 0.71845 − 0.41760 × 2 = −0.116 75 100 (−0.116 75) = −11. 675

On Day 0, the trader: • Buy a 45-strike put and sell two 45-strike put, paying 110. 14 • Short sell 11. 675 stocks, receiving 11. 675 (40) = 467 • Deposit 467 − 110. 14 = 356. 86 to a savings bank to earn a risk free rate The trader’s net position is zero. Day 1: Expiration T = 90/365 Stock price 39 Position Strike price Call premium Sell 45 5.8265 Buy 40 −2.4331 Net 5.8265 − 2.4331 × 2 = 0.960 3 Net 100 (0.960 3) = 96. 03 Day 1, the trader closes out his position • Sell a 45-strike put to cancel out the original 40-strike put he bought; buy two 40-strike puts to cancel out the original two 40-strike puts he sold. The trader receives 96. 03 • Buy 11. 675 stocks from the market to close the short sale, paying 11. 675 (39) = 455. 325 • Receive 356. 86e0.08(1/365) = 356. 938 2 from the savings account The trader’s net profit: 96. 03 − 455. 325 + 356. 938 2 = −2. 36 If the stock price is 39 on Day 1, the trader loses 2.36. Day 1: Expiration T = 90/365 Stock price 40.5 Position Strike price Call premium Sell 45 4.7257 Buy 40 −1.7808 Net 4.7257 − 1.7808 × 2 = 1. 164 1 Net 100 (1. 164 1) = 116. 41 Day 1, the trader closes out his position 1 The delta formula in Appendix 12.B is the delta for the market maker if he sells an option. If the market maker buys an option, add a negative sign to the formula. In this case, the delta is -0.71845 if the trader sells the 45-strike put. However, since the trader buys a 45-strike put, the delta is − (−0.91845) = 0.918 45

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING • Sell a 45-strike put to cancel out the original 40-strike put he bought; buy two 40-strike puts to cancel out the original two 40-strike puts he sold. The trader receives 116. 41 • Buy 11. 675 stocks from the market to close the short sale, paying 11. 675 (40.5) = 472. 84 • Receive 356. 86e0.08(1/365) = 356. 938 2 from the savings account The trader’s net profit: 116. 41 − 472. 84 + 356. 938 2 = 0.51 If the stock price is 39 on Day 1, the trader gains 0.51. Problem 13.5. T = 91/365 r = 8% σ = 30% K = 40

δ=0

Let’s walk through Day 0 and Day 1 calculations. Day 0 1 stock $40.00 $40.50 put $199.05 $178.08 delta −0.417596 −0.385797 Investment −1, 869.43 −$1, 740.56 Interest credited (end of the day) $0.41 $0.381 5 Capital gain (end of the day) $0.09 −$4. 245 4 Daily profit (end of the day) $0.50 −$3. 864 Day 0: Using the Black-Scholes formula, you can verify the put premium is C0 = $199.05 and delta per stock is −0.417596. Negative delta means buying negative number of stocks (i.e. the trader needs to short sell stocks). The trader buys ∆0 = 100 (−0.417596) = −41. 759 6 stocks (i.e. short-sell 41. 759 6 stocks), receiving 41. 759 6×40 = 1670. 384. In addition, the trader receives put premium C0 =$199.05. So at t = 0 the trader’s asset (i.e. investment) M V (0) = ∆0 S0 − C0 = (−41. 759 6) 40 − 199.05 = −1869. 434 The negative amount means that the trader receives 1869. 434 The trader lends out $1869. 434 (i.e. depositing $1869. 434) in a savings account. Now his net position is zero. At the end of Day 0, the stock price goes up from S0 = 40 to S1 = 40.5. The put liability goes down from C0 = $199.05 to C1 = $178.08. The trader’s asset before he rebalances his portfolio is (BR stands for before rebalance): M V BR (1) = ∆0 S1 − C1 = (−41. 759 6) 40.5 − 178.08 = −1869. 343 8 The trader’s profit at the end of Day 0 is: M V BR (1)−M V (0) erh = −1869. 343 8−(−1869. 434) e0.08(1/365) = 0.499 98 = 0.5 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING To find the capital gain and the interest earned at the end of Day 0, we just need to break down the profit at the end of Day 0 into two parts: ¡ ¢ M V BR (1) − M V (0) erh = M V BR (1) − M V (0) + −M V (0) erh − 1 {z } | | {z } capital gain

interest earned

Capital gain at the end of Day 0: M V BR (1) − M V (0) = −1869. 343 8 − (−1869. 434) = 0.090 2 ¢ ¡ ¢ ¡ Interest earned at the end of Day 0: −M V (0) erh − 1 = − (−1869. 434) e0.08/365 − 1 = 0.409 78 = 0.41 The investment at the beginning of Day 0 is: M V (0) = 1869. 434 Day 1: In the beginning of Day 1, the trader starts from a clean slate. He buys sells ∆1 = 100 × (−0.385797) = −38. 579 7 stocks (i.e. short sell stocks) and receives the put premium C1 = $178.08. His asset (or investment) is M V (1) = ∆1 S1 − C1 = (−38. 579 7) 40.5 − 178.08 = −1740. 557 85 The trader’s asset at the end of Day 1 before he rebalances the portfolio (i.e. before he starts over from a clean slate the next day) is: M V BR (2) = ∆1 S2 − C2 = (−38. 579 7) 39.25 − 230.55 = −1744. 803 225 The trader’s profit at the end of Day 1 is: M V BR (2) − M V (1) erh = −1744. 803 225 − (−1740. 557 85) e0.08(1/365) = −3. 864 To find the capital gain and the interest earned at the end of Day 1, we just need to break down the profit at the end of Day 1 into two parts: ¡ ¢ M V BR (2) − M V (1) erh = M V BR (2) − M V (1) + −M V (1) erh − 1 {z } | | {z } capital gain

interest earned

The capital gain credited at the end of Day 1 is:

M V BR (2) − M V (1) = −1744. 803 225 − (−1740. 557 85) = −4. 245 4 The interest ¡ earned ¢ at the end of Day 1¡is: ¢ −M V (1) erh − 1 = − (−1740. 557 85) e0.08(1/365) − 1 = 0.381 5

I’ll omit the calculations for the other days. Here is the result for all days: Day 0 1 2 stock $40.00 $40.50 $39.25 put $199.05 $178.08 $230.55 delta −0.417596 −0.385797 −0.46892 Investment −1, 869.43 −$1, 740.56 −$2, 071.07 Interest credited (end of the day) $0.41 $0.381 5 $0.45 Capital gain (end of the day) $0.09 −$4. 245 4 −$0.05 Daily profit (end of the day) $0.50 −$3. 864 $0.40 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Day stock put delta Investment Interest credited (end of the day) Capital gain (end of the day) Daily profit (end of the day)

3 $38.75 $254.05 −0.50436 −$2, 208.46 $0.48 −$4.48 −$4.00

4 $40.00 $195.49 −0.41940 −$1, 873.10 $0.41 $0.91 $1.32

5 $40.00 $194.58 −0.41986 −$1, 874.02

Day stock put delta Investment Interest credited (end of the day) Capital gain (end of the day) Daily profit (end of the day)

0 $40.00 $199.05 −0.41760 −$1, 869.43 $0.41 −$0.42 −$0.01

1 $40.642 172.6644 −0.37684 −$1, 704.22 $0.37 −$0.35 $0.02

2 40.018 196.5319 −0.41731 −$1, 866.51 $0.41 −$0.40 $0.01

Day stock put delta Investment Interest credited (end of the day) Capital gain (end of the day) Daily profit (end of the day)

3 39.403 222.5962 −0.45918 −$2, 031.89 $0.45 −$0.45 $0.00

4 $38.80 250.8701 −0.50202 −$2, 198.56 $0.48 −$0.48 $0.00

5 39.420 220.0727 −0.45940 −$2, 031.02

Problem 13.6.

Problem 13.7. If SOA tests this type of problems in the exam, they’ll need to give you at least Γ and θ. You can calculate the option premium V and delta ∆. Once you have Greeks, just use the formula: 1 V (St+h , T − t − h) ≈ V (St , T − t) + ∆t + θh + Γt 2

2

(Textbook 13.6)

I’m not going to do all the parts in the problem. I’m just going to show you some examples. First, you’ll need to get Greeks. You can use the Black-Scholes formula and find the option premium V and delta ∆. To find Γ and θ, you’ll need to use the spreadsheet attached to the textbook. σ = 0.3 r = 8% δ=0 K = 40 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Day 0 1 time t 0 h = 1/365 Expiration T − t 180/365 = 0.493 150 684 9 179/365 = 0.490 410 958 9 stock price St 40 44 Call price Vt 4.1217 6.8991 Delta ∆t 0.6151 0.7720 Gamma Γt 0.0454 0.0327 Theta θt −0.0134 −0.0137 V (St , T − t) = V (40, 180/365 − 0) = 4.1217 ¡ ¢ V (St+h , T − t − h) = V S0+1/365 , 180/365 − 0 − 1/365 = 6.8991 = St+h − St = 44 − 40 = 4 ∆t = 0.6151 θt = −0.0134 × 365 (The spreadsheet gives the per-day theta; we need to annualize it.) Γt = 0.0454 1 V (St , T − t) + ∆t + θt h + Γt 2 2 1 1 = 4.1217 + 0.6151 × 4 − 0.0134 × 365 × + × 0.0454 × 42 365 2 = 6. 931 9 The true value is V (St+h , T − t − h) = 6.8991 The error percentage is: 6. 931 9 − 6.8991 = 0.4754 % 6.8991 σ = 0.3

r = 8%

Day time t Expiration T − t stock price St Call price Vt Delta ∆t Gamma Γt Theta θt

δ=0

K = 40

0 0 180/365 = 0.493 150 684 9 40 4.1217 0.6151 0.0454 −0.0134

5 h = 5/365 175/365 = 0.479 452 054 8 44 6.8440 0.7726 0.0330 −0.0138

V (St , T − t) = V (40, 180/365 − 0) = 4.1217 ¡ ¢ V (St+h , T − t − h) = V S0+1/365 , 180/365 − 0 − 5/365 = 6.8440 = St+h − St = 44 − 40 = 4 θt = −0.0134 × 365 Γt = 0.0454 ∆t = 0.6151 1 V (St , T − t) + ∆t + θt h + Γt 2 2 5 1 = 4.1217 + 0.6151 × 4 − 0.0134 × 365 × + × 0.0454 × 42 365 2 = 6. 878 3 The true value is V (St+h , T − t − h) = 6.8440 The error percentage is: www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING 6. 878 3 − 6.8440 = 0.5 01% 6.8440 Problem 13.8. I’m not going to do some examples. σ = 0.3 r = 8% Day time t Expiration T − t Stock price St Put price Vt Delta ∆t Gamma Γt Theta θt

all the parts in the problem. I’m just going to show you δ=0 K = 40 0 0 180/365 = 0.493 150 684 9 40 2.5744 −0.3849 0.0454 −0.0050

1 h = 1/365 179/365 = 0.490 410 958 9 44 1.3602 −0.2280 0.0327 −0.0053

V (St , T − t) = V (40, 180/365 − 0) = 2.5744 ¡ ¢ V (St+h , T − t − h) = V S0+1/365 , 180/365 − 0 − 1/365 = 1.3602 = St+h − St = 44 − 40 = 4 θt = −0.0050 × 365 Γt = 0.0454 ∆t = −0.3849 1 2 V (St , T − t) + ∆t + θt h + Γt 2 1 1 = 2.5744 − 0.3849 × 4 − 0.0050 × 365 × + × 0.0454 × 42 365 2 = 1. 393 The true value is V (St+h , T − t − h) = 1.3602 The error percentage is: 1. 393 − 1.3602 = 0.2 41% 1.3602 σ = 0.3

r = 8%

δ=0

K = 40

Day 0 5 time t 0 h = 5/365 Expiration T − t 180/365 = 0.493 150 684 9 175/365 = 0.479 452 054 8 Stock price St 40 44 Put price Vt 2.5744 1.3388 Delta ∆t −0.3849 −0.2274 Gamma Γt 0.0454 0.0330 Theta θt −0.0050 −0.0054 V (St , T − t) = V (40, 180/365 − 0) = 2.5744 ¡ ¢ V (St+h , T − t − h) = V S0+1/365 , 180/365 − 0 − 1/365 = 1.3602 = St+h − St = 44 − 40 = 4 θt = −0.0050 × 365 Γt = 0.0454 ∆t = −0.3849 1 2 V (St , T − t) + ∆t + θt h + Γt 2 www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING 5 1 = 2.5744 − 0.3849 × 4 − 0.0050 × 365 × + × 0.0454 × 42 365 2 = 1. 373 The true value is V (St+h , T − t − h) = 1.3602 The error percentage is: 1.3388 1. 373 − 1.3388 = 2. 55% 1.3388 Problem 13.9. I’m going to do one set of calculation assuming the stock price one day later is $30. However, I’m not going to produce a graph. σ = 0.3

S = 40

r = 8%

δ=0

K = 40

Day 0 1 time t 0 h = 1/365 Expiration T − t 91/365 90/365 Stock price St 40 30 Call price Vt 2.7804 0.0730 Delta ∆t 0.5824 0.0423 Gamma Γt 0.0652 0.0202 Theta θt −0.0173 −0.9134 a. The price of a 40-strike, 90 day to expiration call is worth 0.0730 b. Use delta approximation: = 30 − 40 = −10 0 V = V0 + ∆t = 2.7804 + 0.5824 (−10) = −3. 043 6 We get a nonsense value of −3. 043 6. This is because the delta approximation is good when is small. Here we have a large change of = −10. c. Use delta-gamma approximation: 1 0 V = V0 + ∆t + Γt 2 2 1 = 2.7804 + 0.5824 (−10) + × 0.0652 × (−10)2 = 0.216 4 2 d. Use delta-gamma-theta approximation: 1 0 V = V0 + ∆t + Γt 2 + θt h 2 1 1 2 = 2.7804 + 0.5824 (−10) + × 0.0652 × (−10) − 0.0173 × 365 × 2 365 = 0.199 1 We see in a, b, c, and d, the approximation is not good. This is because the approximation is good when is small. Here we have a large change of = −10. Problem 13.10. www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING I’m not going to produce a graph. I’ll do one set of calculation assuming the stock price one day later is $41. σ = 0.3

S = 40

r = 8%

δ=0

K = 40

Day 0 1 time t 0 h = 1/365 Expiration T − t 1 364/365 Stock price St 40 30 Call price Vt 6.2845 6.9504 Delta ∆t 0.6615 0.6909 Gamma Γt 0.0305 0.0287 Theta θt −0.0104 −0.0106 a. The price of a 40-strike, 364 day to expiration call is worth 6.9504 b. Use delta approximation: = 41 − 40 = 1 0 V = V0 + ∆t = 6.2845 + 0.6615 (1) = 6. 946 c. Use delta-gamma approximation: 1 0 V = V0 + ∆t + Γt 2 2 1 = 6.2845 + 0.6615 (1) + × 0.0305 × (1)2 = 6. 961 3 2 d. Use delta-gamma-theta approximation: 1 0 V = V0 + ∆t + Γt 2 + θt h 2 1 1 = 6.2845 + 0.6615 (1) + × 0.0305 × (1)2 − 0.0104 × 365 × = 6. 950 9 2 365 We see in a, b, c, and d, the approximation is good. This is because is small. Problem 13.11. I’m going to do one set of calculation assuming the stock price one day later is $30. σ = 0.3

S = 40

r = 8%

Day 0 time t 0 Expiration T − t 91/365 Stock price St 40 Put price Vt 1.9905 Delta ∆t −0.4176 Gamma Γt 0.0652 Theta θt −0.0088 a. The price of a 40-strike, 90 www.actuary88.com

δ=0

K = 40

1 h = 1/365 90/365 30 9.2917 −0.9577 0.0202 0.0061 day to expiration put is worth 9.2917

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING b. Use delta approximation: = 30 − 40 = −10 0 V = V0 + ∆t = 1.9905 − 0.4176 (−10) = 6. 166 5 c. Use delta-gamma approximation: 1 0 V = V0 + ∆t + Γt 2 2 1 = 1.9905 − 0.4176 (−10) + × 0.0652 × (−10)2 = 9. 426 5 2 d. Use delta-gamma-theta approximation: 1 0 V = V0 + ∆t + Γt 2 + θt h 2 1 1 = 1.9905 − 0.4176 (−10) + × 0.0652 × (−10)2 − 0.0088 × 365 × 2 365 = 9. 417 7 Problem 13.12. I’m going to do one set of calculation assuming the stock price one day later is $41. σ = 0.3

S = 40

r = 8%

δ=0

K = 40

Day 0 1 time t 0 h = 1/365 Expiration T − t 1 364/365 Stock price St 40 41 Put price Vt 3.2092 2.8831 Delta ∆t −0.3385 −0.3091 Gamma Γt 0.0305 0.0287 Theta θt −0.0023 −0.0025 a. The price of a 40-strike, 364 day to expiration put is worth 2.8831 b. Use delta approximation: = 41 − 40 = 1 0 V = V0 + ∆t = 3.2092 − 0.3385 (1) = 2. 870 7 c. Use delta-gamma approximation: 1 0 V = V0 + ∆t + Γt 2 2 1 2 = 3.2092 − 0.3385 (1) + × 0.0305 × (1) = 2. 885 95 2 d. Use delta-gamma-theta approximation: 1 0 V = V0 + ∆t + Γt 2 + θt h 2 1 1 = 3.2092 − 0.3385 (1) + × 0.0305 × (1)2 − 0.0023 × 365 × 2 365 = 2. 883 65 Problem 13.13. www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Please note that in DM 13.9, θ is annualized. However, in Table 13.1, θ is a per-day theta. (DM page 411 says that if the expiry is measured in years, then theta is the annualized change in option value; to get the per-day theta, divide by 365. The θ in Table 13.1 and in the author’s Excel spreadsheet is daily µ theta.) So we need to annualize θ before using DM 13.9. ¶ 1 2 2 − × 0.3 × 40 × 0.0652 − 0.0173 × 365 + 0.08 (0.5824 × 40 − 2.7804) × 2 1 = −5. 79 4 × 10−5 ≈ 0 365 Problem 13.14. We’ll use the Excel spreadsheet attached in the DM textbook and calculate the following: Inputs: Stock Price 40 Exercise Price 45 Volatility 30% Risk-free interest rate 8% Time to Expiration (years) 0.5 Dividend Yield 0% Outputs: Black-Scholes (European) Call Put Price 2.1304 5.3659 Delta 0.3972 −0.6028 Gamma 0.0454 0.0454 Vega 0.1091 0.1091 Theta −0.0120 −0.0025 Rho 0.0688 −0.1474 Psi −0.0794 0.1206 Elasticity 7.4578 −4.4936 DM ¶ µ 13.9 is: 1 − × 0.32 × 402 × 0.0454 − 0.0025 × 365 + 0.08 (−0.6028 × 40 − 5.3659) × 2

1 = 5. 29 × 10−6 ≈ 0 365 Problem 13.15.

Use DM’s spreadsheet. The inputs are:

www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING Stock Price Exercise Price Volatility Risk-free interest rate Time to Expiration (years) Dividend Yield

40 45 30% 8% = 180/365 0%

We’ll get the following outputs: Black-Scholes (European) 40-strike call Price 4.1217 Delta 0.6151 Gamma 0.0454

45-strike call 2.1004 0.3949 0.0457

We bought one 45-strike call. Suppose we need to sell (i.e. write) X unit of 40-strike call. The Gamma of the bought 45-strike call is −0.0457 (negative because we bought a call). The Gamma of the sold X unit of 40-strike call is 0.0454X. The total Gamma of our portfolio is 0.0454X − 0.0457. To Gamma hedge, set 0.0454X − 0.0457 = 0. This gives us X = 1. 006 6. The total Delta of our portfolio is 0.6151 × 1. 006 6 − 0.3949 = 0.224 3. To delta hedge, we need to buy 0.224 3 share of the underlying stock. This is our final portfolio at time zero: Transactions at time zero Cost Buy a 45-strike call 2.1004 Sell 1. 006 6 unit of 40-strike call (to Gamma hedge) −1. 006 6 × 4.1217 = −4. 148 9 Buy 0.224 3 share of the stock (to Delta hedge) 0.224 3 × 40 = 8. 972 Borrow 2.1004 − 4. 148 9 + 8. 972 = 6. 923 5 −6. 923 5 Total 2.1004 − 4. 148 9 + 8. 972 − 6. 923 5 = 0 One day later, you close out your position: Transactions (one day later) Revenue 0 Sell a 45-strike call C1 0 Buy 1. 006 6 unit of 40-strike call −1. 006 6C2 0 Sell 0.224 3 share of the stock 0.224 3S Repay the borrowed amount 6. 923 5e0.08/365 0 0 0 Total C1 − 1. 006 6C2 + 0.224 3S − 6. 923 5e0.08/365 0

C1 is the price of a 45-strike call with 179 day to expiration. 0 C2 is the price of a 40-strike call with 179 day to expiration. 0 S is the price of the stock one day later. 0 0 If you want to draw the diagram of the overnight profit C1 − 1. 006 6C2 + 0 0 0.224 3S − 6. 923 5e0.08/365 , you can get different profit by changing S . For example, www.actuary88.com

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING 0

assume S = 50. Also assume that the stock volatility and risk free interest rate are not changed. Using the Black-Scholes formula (use T = 179/365), we get: 0 0 C1 = 8.1511 C2 = 12.0043 The overnight profit is: 8.1511 − 1. 006 6 × 12.0043 + 0.224 3 × 50 − 6. 923 5e0.08/365 = 0.36 0

If S = 50, then the overnight profit is 0 − 1. 006 6 × 0 + 0.224 3 × 0 − 6. 923 5e0.08/365 = −6. 93. 0

By changing S , you’ll get different profits. Then you draw a graph on how 0 the overnight profit varies by S . Since it’s time-consuming to calculate the 0 overnight profit by changing S , I’m not going to do it. Problem 13.16. Use DM’s spreadsheet. The inputs are: Stock Price Exercise Price Volatility Risk-free interest rate Time to Expiration (years) Dividend Yield

40 45 30% 8% = 180/365 0%

We’ll get the following outputs: Black-Scholes (European) 40-strike call 45-strike put Price 4.1217 5.3596 Delta 0.6151 −0.6051 Gamma 0.0454 0.0457 We sell one 45-strike put and buy X unit of 40-strike call. The total Gamma is 0.0457 − 0.0454X. To Gamma hedge, set 0.0457 − 0.0454X = 0. This gives us X = 1. 006 6. The delta of 1. 006 6 unit of 40-strike call is −0.6151 × 1. 006 6. The total Delta is −0.6151 × 1. 006 6 − 0.6051 = −1. 224 3, meaning that we need to short sell 1. 224 3 unit of the underlying stock. Final portfolio at time zero: Transactions at time zero Sell a 45-strike put Buy 1. 006 6 unit of 40-strike call (to Gamma hedge) Short sell 1. 224 3 share of the stock (to Delta hedge) Lend 5.3596 − 4. 148 9 + 48. 972 = 50. 182 7 Total www.actuary88.com

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Revenue 5.3596 −1. 006 6 × 4.1217 = −4. 148 9 1. 224 3 × 40 = 48. 972 −50. 182 7 5.3596 − 4. 148 9 + 48. 972 − 50. 182 7 = 0 145

CHAPTER 13. MARKET MAKING AND DELTA HEDGING One day later, you close out your position: Transactions (one day later) Revenue 0 Buy a 45-strike put −P1 0 Sell 1. 006 6 unit of 40-strike call 1. 006 6C2 0 Buy 1. 224 3 share of the stock 1. 224 3S Collect loan 50. 182 7e0.08/365 0 0 0 Total −P1 + 1. 006 6C2 − 1. 224 3S + 50. 182 7e0.08/365 0

P1 is the price of a 45-strike put with 179 day to expiration. 0 C2 is the price of a 40-strike call with 179 day to expiration. 0 S is the price of the stock one day later. Problem 13.17. Assume the butterfly spread is asymmetric. K3 − K2 45 − 40 λ= = = 0.5 K3 − K1 45 − 35 Written butterfly spread consists of selling one 35-strike call, buying two 40-strike calls, and selling one 45-strike call. Inputs: σ = 0.3 S = 40 r = 8% δ=0 T = 91/365 strike K 35 40 45 butterfly spread call price 6.1315 2.7804 0.9710 6.1315 − 2 (2.7804) + 0.9710 = 1. 541 7 Delta 0.8642 0.5824 0.2815 0.8642 − 2 (0.5824) + 0.2815 = −0.019 1 Gamma 0.0364 0.0652 0.0563 0.0364 − 2 (0.0652) + 0.0563 = −0.037 7 Theta −0.0134 −0.0173 −0.0134 −0.0134 − 2 (−0.0173) − 0.0134 = 0.007 8 Inputs: σ = 0.3 S = 40 strike K 49 call price 4.1217 Delta 0.6151 Gamma 0.0454 Theta −0.0134

r = 8%

δ=0

T = 180/365

Suppose we sell X units of 180 day to expiration call. The total Gamma of the written butterfly spread and written X units of 180 day to expiration call is −0.037 7 + 0.0454X. To Gamma hedge, set −0.037 7 + 0.0454X = 0, X = 0.830 4. The total Delta of the written butterfly and the written X units of 180 day to expiration call is −0.019 1 + 0.830 4 × 0.6151 = 0.491 7. So we need to buy 0.491 7 share of the underlying stock. Transactions at time zero Sell the spread and buy 0.491 7 stock Borrow 18. 126 3 from a bank Total www.actuary88.com

Cost (negative cost means receiving money) −1. 541 7 + 0.491 7 × 40 = 18. 126 3 −18. 126 3 0

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CHAPTER 13. MARKET MAKING AND DELTA HEDGING One day later, we close out our position: Transactions (one day later) Buy a 35-strike 90 day to expiry call Sell two 40-strike 90 day to expiry calls 1. 006 6 Buy a 45-strike 90 day to expiry call Sell 0.491 7 stock Repay the borrowed amount Total 0

0

0

Revenue 0 −C1 0 2C2 0 −C3 0 0.491 7S −18. 126 3e0.08/365 0 0 0 0 −C1 + 2C2 − C3 + 0.491 7S − 18. 126 3e0.08/365 0

So the total overnight profit is −C1 +2C2 −C3 +0.491 7S −18. 126 3e0.08/365 . Problem 13.18. inputs: σ = 0.3 S = 40 r = 8% δ=0 T = 91/365 strike K 40 45 ratio spread put price 1.9905 5.0824 2 × 1.9905 − 5.0824 = −1. 101 4 Delta −0.4176 −0.7185 2 × (−0.4176) − (−0.7185) = −0.116 7 Gamma 0.0652 0.0563 2 × 0.0652 − 0.0563 = 0.074 1 Theta −0.0088 −0.0037 2 × (−0.0088) − (−0.0037) = −0.013 9 in puts: σ = 0.3 S = 40 strike K 40 call price 4.1217 Delta 0.6151 Gamma 0.0454 Theta −0.0134

r = 8%

δ=0

T = 180/365

We buy X unit of 40-strike 180-day to expiration call. The total Gamma of the ratio spread and bought X unit of 40-strike 180-day to expiration call is 0.074 1 + 0.0454X. To Gamma hedge, set 0.074 1 + 0.0454X = 0, X = −1. 632 2. So we need to short sell 1. 632 2 shares of the underlying stock. Transactions at time zero Enter the ratio spread and short 1. 632 2 stock Deposit 66. 389 4 in a savings account Total One day later, we close out our position: Transactions (one day later) Buy two 40-strike 90 day to expiry puts Sell a 45-strike 90 day to expiry put Buy 1. 632 2 stock Receive money from the bank Total www.actuary88.com

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Revenue 1. 101 4 + 1. 632 2 × 40 = 66. 389 4 −66. 389 4 0

Revenue 0 −2P1 0 P2 0 −1. 632 2S 66. 389 4e0.08/365 0 0 0 −2P1 + P2 − 1. 632 2S + 66. 389 4e0.08/365 147

CHAPTER 13. MARKET MAKING AND DELTA HEDGING 0

0

0

So the total overnight profit is −2P1 + P2 − 1. 632 2S + 66. 389 4e0.08/365 . Skip Problem 13.19 and 13.20. Vega hedge and rho hedge are far outside the scope of the Exam MFE syllabus.

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Chapter 14

Exotic options: I Problem 14.1. For n non-negative numbers (such as stock prices) S1 , S2 , ..., Sn , the arithmetic mean can never be less than the geometric mean: S1 + S2 + ... + Sn 1/n ≥ (S1 S2 ...Sn ) n If and only if S1 = S2 = ... = Sn , we have: S1 + S2 + ... + Sn = (S1 S2 ...Sn )1/n n The proof can be found at Wikipedia: http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_ means For example, this is the proof for n = 2. (S1 − S2 )2 ≥ 0 → S12 + S22 ≥ 2S1 S2 2 (S1 + S2 ) = S12 + S22 + 2S1 S2 ≥ 4S1 S2 µ ¶2 S1 + S2 S1 + S2 √ ≥ S1 S2 ≥ S1 S2 2 2 Problem 14.2. Arithmetic average: 5+4+5+6+5 = 5.0 5 Geometric average: (5 × 4 × 5 × 6 × 5)1/5 = 4. 959 Ignore the question "What happens to the difference between the two measures of the averages as the standard deviation of the observations increase?" This question is vague. I’m not sure what the author is after. 149

CHAPTER 14. EXOTIC OPTIONS: I Problem 14.3. √

√

u = e(r−δ)h+σ √h = e(0.08−0)0.5+0.3√0.5 = 1. 286 765 9 u = e(r−δ)h−σ h = e(0.08−0)0.5−0.3 0.5 = 0.841 868 0 e(r−δ)h − d e(0.08−0)0.5 − 0.841 868 πu = = = 0.447 165 u−d 1. 286 766 − 0.841 868 πd = 1 − π u = 1 − 0.447 165 = 0.552 835 t=0

t = 0.5

t=1 Suu = 165.57665

Su = 128.67659 S = 100

Sud = 108.32871 Sd = 84.18680 Sdd = 70.87417

The four arithmetic averages are: 128.67659 + 165.57665 = 147. 126 62 2

(The path is uu or up up)

128.67659 + 108.32871 = 118. 502 65 2

(ud)

84.18680 + 108.32871 = 96. 257 8 2 84.18680 + 70.87417 = 77. 530 485 2 Path uu ud du dd

Average asset 147. 126 62 118. 502 65 96. 257 8 77. 530 485

Call Payoff 47. 126 62 18. 502 65 0 0

(du)

(dd) Risk Neutral Prob π 2u πu πd πu πd π 2d

The price of the Asian arithmetic average price call: e−0.08 (47. 126 62 × 0.447 1652 + 18. 502 65 × 0.447 165 × 0.552 835 +0 × 0.447 165 × 0.552 835 + 0 × 0.552 8352 ) = 12. 921 If we were to calculate the price of the Asian arithmetic average asset put, then Path Average asset Put Payoff Risk Neutral Prob uu 147. 126 62 0 π2u ud 118. 502 65 0 πu πd du 96. 257 8 3. 742 2 πu πd dd 77. 530 5 22. 469 5 π2d The price of the Asian arithmetic average price put: www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I e−0.08 (0 × 0.447 1652 + 0 × 0.447 165 × 0.552 835 +3. 742 2 × 0.447 165 × 0.552 835 + 22. 469 5 × 0.552 8352 = 7. 193 The √ four geometric averages are: 128.67659 × 165.57665 = 145. 965 2

(uu)

√ 128.67659 × 108.32871 = 118. 065 1

(ud)

√ 84.18680 × 108.32871 = 95. 497 9

(du)

√ 84.18680 × 70.87417 = 77. 244 2 Path uu ud du dd

Average asset 145. 965 2 118. 065 1 95. 497 9 77. 244 2

(dd)

Call Payoff 45. 965 2 18. 065 1 0 0

Risk Neutral Prob π2u πu πd πu πd π2d

The price of the Asian geometric average price call: e−0.08 (45. 965 2 × 0.447 1652 + 18. 065 1 × 0.447 165 × 0.552 835 +0 × 0.447 165 × 0.552 835 + 0 × 0.552 8352 ) = 12. 607 If we were asked to calculate the price Path Average asset Call Payoff uu 145. 965 2 0 ud 118. 065 1 0 du 95. 497 9 4. 502 1 dd 77. 244 2 22. 755 8

of the Asian average price put, then: Risk Neutral Prob π2u πu πd πu πd π2d

The price of the Asian geometric average price put: e−0.08 (0 × 0.447 1652 + 0 × 0.447 165 × 0.552 835 +4. 502 1 × 0.447 165 × 0.552 835 + 22. 755 8 × 0.552 8352 ) = 7. 447 Problem 14.4. a. Asian arithmetic Path ST uu 165.57665 ud 108.32871 du 108.32871 dd 70.87417

average strike K 147. 126 62 118. 502 65 96. 257 8 77. 530 485

call: Call Payoff 18. 450 03 0 12. 070 91 0

Risk Neutral Prob π 2u πuπd πuπd π 2d

The price of the Asian arithmetic average strike call: e−0.08 (18. 450 03 × 0.447 1652 + 0 × 0.447 165 × 0.552 835 +12. 070 91 × 0.447 165 × 0.552 835 + 0 × 0.552 8352 ) = 6. 160 www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I If we were to calculate the price of the Asian arithmetic average strike put, then Path ST K Put Payoff Risk Neutral Prob uu 165.57665 147. 126 62 0 π 2u ud 108.32871 118. 502 65 10. 173 94 πu πd du 108.32871 96. 257 8 0 πu πd dd 70.87417 77. 530 485 76. 656 315 π 2d The price of the Asian arithmetic average strike put: e−0.08 (0 × 0.447 1652 + 10. 173 94 × 0.447 165 × 0.552 835 +0 × 0.447 165 × 0.552 835 + 6. 656 315 × 0.552 8352 ) = 4. 20 b. The price of the Asian geometric average strike call: Path ST K Call Payoff Risk Neutral Prob uu 165.57665 145. 965 2 19. 611 45 π2u ud 108.32871 118. 065 1 0 πu πd du 108.32871 95. 497 9 12. 830 81 πu πd dd 70.87417 77. 244 2 0 π2d The price of the Asian geometric average strike call: e−0.08 (19. 611 45 × 0.447 1652 + 0 × 0.447 165 × 0.552 835 +12. 830 81 × 0.447 165 × 0.552 835 + 0 × 0.552 8352 ) = 6. 548 If we were to calculate the price then Path ST K uu 165.57665 145. 965 2 ud 108.32871 118. 065 1 du 108.32871 95. 497 9 dd 70.87417 77. 244 2

of the Asian geometric average strike put, Put Payoff 0 9. 736 39 0 6. 370 03

Risk Neutral Prob π 2u πu πd πu πd π 2d

The price of the Asian geometric average strike put: e−0.08 (0 × 0.447 1652 + 9. 736 39 × 0.447 165 × 0.552 835 +0 × 0.447 165 × 0.552 835 + 6. 370 03 × 0.552 8352 ) = 4. 019 Problem 14.5. Stock price tree: 182.14179 165.57665 128.67659 100

128.814749 108.32871

84.18680

91.10067 70.87417 64.42843

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CHAPTER 14. EXOTIC OPTIONS: I Calculate the price of the arithmetic average asset call. Path Arithmetic Average Payoff Risk Neutral Prob uuu 151.1369 51.1369 π 3u = 0.0953 uud 133.3612 33.3612 π 2u π d = 0.1133 udu 118.8058 18.8058 π 2u π d = 0.1133 udd 106.2345 6.2345 π u π 2d = 0.1348 duu 106.8874 6.8874 π 2u π d = 0.1133 dud 94.3160 0 π u π 2d = 0.1348 ddu 84.0221 0 π u π 2d = 0.1348 ddd 75.1314 0 π 3d = 0.1603 Total 1 The price of the arithmetic average asset call is: e−0.08 (51.1369 × 0.0953 + 33.3612 × 0.1133 +18.8058 × 0.1133 + 6.2345 × 0.1348 + 6.8874 × 0.1133) = 11. 45 Calculate the price of the geometric average asset call. Path Arithmetic Average Payoff Risk Neutral Prob uuu 149.1442 49.1442 π 3u = 0.0953 uud 132.8796 32.8796 π 2u π d = 0.1133 udu 118.3887 18.3887 π 2u π d = 0.1133 udd 105.4781 5.4781 π u π 2d = 0.1348 duu 105.4781 5.4781 π 2u π d = 0.1133 dud 93.9754 0 π u π 2d = 0.1348 ddu 83.7272 0 π u π 2d = 0.1348 ddd 74.5965 0 π 3d = 0.1603 Total 1 The price of the geometric average asset call is: e−0.08 (49.1442 × 0.0953 + 32.8796 × 0.1133 +18.3887 × 0.1133 + 5.4781 × 0.1348 + 5.4781 × 0.1133) = 10. 94 Problem 14.6. a. Using the Black-Scholes formula, we find that the call price is $4.1293. b and c. knock-in call + knock-out call = ordinary call However, the knock-out call is worthless because the barrier 44 is lower than the strike price 45. For the call to be worth something, the stock price must exceed the strike price 45. However, as soon as the stock reaches the barrier 44, the call is knocked out dead (i.e. the call doesn’t exist any more). Hence this knock-out call will never have a positive payoff. Hence the knock-in call becomes the ordinary call. The premium of the knock-in call is also $4.1293. Key point to remember: www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I 1. A knock-out call is worthless if the barrier is less than or equal to the strike price. 2. A knock-in call is just an ordinary call if the barrier is less than or equal to the strike price.

Problem 14.7. DM Chapter 14 doesn’t have any formula on the price of a barrier option. To calculate the barrier option price, you have to use the textbook’s spreadsheet. So this problem is out of the scope of Exam MFE. I used the textbook’s Excel spreadsheet and calculated the following: T 0.25 0.5 1 2 3 4 5 100

BS 0.9744 2.1304 4.1293 7.4398 10.2365 12.6969 14.9010 39.9861

DO 0.7323 1.2482 1.8217 2.4505 2.8529 3.1559 3.4003 5.3112

BS/DO 1.3306 1.7067 2.2667 3.0360 3.5881 4.0232 4.3823 7.5286

BS: the price of a standard option using the Black-Scholes formula. DO: Down-and-out barrier option. Since the current stock price 40 exceeds the barrier 38, the knock-out call with 38 barrier is really a down-and-out option (if the stock price drops to 38, the call is dead). The greater the expiration, the greater the ratio of BS/DO. As the expiration increases, both the standard option and the down-and-out option become more valuable. However, as T increases, the value of a standard option goes up much faster than a down-and-out option. The value of a down-and-out option has a moderate increase. As T increase, the stock price may be more volatile, increasing the value of the barrier option. However, there’s also more chance that the stock price may hit the barrier.

Problem 14.8. Out of the scope of MFE. However, I used the textbook spreadsheet and found the following: www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I T BS UO BS/U O 0.25 5.0833 3.8661 1.3149 0.5 5.3659 3.4062 1.5753 1 5.6696 2.8626 1.9806 2 5.7862 2.2233 2.6026 3 5.6347 1.8109 3.1115 4 5.3736 1.5094 3.5601 5 5.0654 1.2761 3.9695 100 0.0012 0.0001 24.4960 As T increases, the BS/U O increases too (just like in Problem 14.7). However, if T is too big, the value of a standard put and the value of an up-and-out put both approach zero. If T is too big, we lose the time value of money (the money we spent in buying a put could have been invested elsewhere and earned lot of interest).

Problem 14.9. I used the spreadsheet and calculated the following (T is expressed in months): T BS UO BS/UO 1 0.1727 0.1727 1.0003 2 0.5641 0.5479 1.0296 3 0.9744 0.8546 1.1401 4 1.3741 1.0384 1.3234 5 1.7593 1.1243 1.5649 6 2.1304 1.1468 1.8577 7 2.4886 1.1316 2.1991 8 2.8353 1.0954 2.5885 9 3.1718 1.0482 3.0260 10 3.4991 0.9962 3.5124 11 3.8180 0.9430 4.0488 12 4.1293 0.8906 4.6365 As T goes up, the BS/UO ratio goes up too.

Problem 14.10. With K = 0.9, the standard put is worth 0.0188. With barrier 1 or 1.05, the up-and-out barrier is also worth 0.0188. Why? DM page 452 and 453 have an explanation. Here is the main point. If the exchange rate never hits 1, then the standard put and the up-and-out put with 1 or 1.5 barrier have the same value. The only way that the standard put is more valuable than the up-and-out put is when the exchange rate goes up from 0.9 to 1 or to 1.05 and then falls below 0.9, in which case the up-and-out is dead yet the standard put has a positive payoff of K − ST = 0.9 − ST . However, such a scenario is rare and 0.9 − ST is www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I small. Hence the standard put and the up-and-out put have roughly the same value. In contrast, when the strike price K = 1, the payoff of the standard put at the above mentioned scenario is 1 − ST , which is greater than 0.9 − ST . Hence the standard put is slightly more valuable than the up-and-out put. Problem 14.11. a.Using the Black-Scholes formula, we find the call premium is $9.6099. b. If we buy the compound option, then we can, at t1 = 1, buy, for a guaranteed price of x = 2, a call that expires at T = 2. We’ll exercise the compound option at t1 = 1 only if the stock price at t = 1 is such that the call we are entitled to buy is equal to or greater than the premium 2. We’ll use the textbook Equation 14.11 to find S ∗ , the minimum price such that exercising the compound option is worthwhile. C (S ∗ , K, T − t1 ) = x C (S ∗ , 40, 2 − 1) = 2 C (S ∗ , 40, 1) = 2 To solve for S ∗ , we have use the trial-and-error approach. I found that ∗ S = 31.723. In other words, if the stock price at t1 = 1 is 31.723, then a call option written at t1 = 1 and expires at T = 2 is exactly worth $2.In this case, we’ll not exercise the compound option. We’ll let the compound option expire worthless. If the stock price at t1 = 1 is less than 31.723, then the call written at t1 = 1 and expires at T = 2 is worth less than $2 and we’ll not exercise the compound option. We’ll let the compound option expire worthless. Only if the stock price at t1 = 1 is greater than 31.723 should we exercise the compound option. c. To find the price of the compound call (i.e. call on call), we’ll use worksheet called "Compound" provided by the DM textbook. The inputs are: Stock Price 40 Exercise Price to buy asset 40 Exercise Price to buy option 2 Volatility 30% Risk-free interest rate 8% Expiration for Option on Option (years) 1 Expiration for Underlying Option (years) 2 Dividend Yield 0%

Compound Option Prices Call on Call 7.9482 Put on Call 0.1845 Call on Put 2.2978 www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I Put on Put 0.4484 Critical S for compound call 31.723 Critical S for compound put 44.3494 So the compound call premium is 7.9482. (From the output, we see that the critical stock price for the compound call is 31.723, which matches our answer in part b). d. From Part c, we see that the price of the put on call is 0.1845. Alternatively, we can use DM 14.12 to calculate the put on the call, given we know the price of the call on call. CallOnCall − P utOnCall + xe−rt1 = BSCall 7.9482 − P utOnCall + 2e−0.08×1 = 9.6099 P utOnCall = 0.184 5 Problem 14.12. a. Use the Black-Scholes formula, we find the put price is 3.6956. b. Use the "Compound" worksheet. Inputs Stock Price 40 Exercise Price to buy asset 40 Exercise Price to buy option 2 Volatility 30% Risk-free interest rate 8% Expiration for Option on Option (years) 1 Expiration for Underlying Option (years) 2 Dividend Yield 0% Outputs: Call on Call 7.9482 Put on Call 0.1845 Call on Put 2.2978 Put on Put 0.4484 Critical S for compound call Critical S for compound put

31.723 44.3494

So if S1 < 44.3494, we’ll exercise the put at t = 1. c. The put on put price is 0.4484. Problem 14.13. a. DM 14.15 gives you the price formula for a gap call. A gap put formula is P (K1 , K2 ) = K1 e−rT N (−d2 ) − Se−δT N (−d1 ) www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I µ ¶ 1 S + r − δ + σ2 T K2 2 √ d1 = σ T √ d2 = d1 − σ T For foreign currency, δ = r€ , r = r$ , and S = x0 . ln

b. The gap put payoff is 0.8 − x if x < 1. If σ = 0, then x = x0 = 0.9. The gap out payoff is −0.1. The gap put premium is 0.8e−0.06×0.5 − 0.9e−0.03×0.5 = −0.110 2. As the volatility increases, the value of the gap put increases. Problem 14.14. Skip. Problem 14.15. Skip. Problem 14.16. Price of a standard 40-strike call on S. The inputs to the Black Scholes formulas are: S = K = 40 σ = 0.3 r = 0.08 T =1 δ=0 The call price is 6.2845. The price of an exchange option with S as underlying and 0.667Q as the strike asset is 7. 577. S =√ 40 K = 0.667 × 60 = 40 σ = 0.32 + 0.52 − 2 × 0.5 × 0.3 × 0.5 = 0.435 89 40e−0×1 1 ln + × 0.435 892 × 1 −0.04×1 40e 2 √ d1 = = 0.309 7 0.435 89 1 √ d2 = 0.309 7 − 0.435 89 1 = −0.126 2 N (d1 ) = 0.621 6 N (d2 ) = 0.449 8 C = 40e−0×1 × 0.621 6 − 40e−0.04×1 × 0.449 8 = 7. 577 Problem 14.17. a. Inputs to the exchange call formula: σ S = 0.3 S = 40 δS = 0 σ K = 0.5 K = 60 δK = 0 ρ = 0.5 T =1 www.actuary88.com

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CHAPTER 14. EXOTIC OPTIONS: I You should get: C = 2. If you change δ S = 0.1, then the call price is C = 1.2178 In the replicating portfolio, the number of underlying stocks to hold at time zero is e−δS T . If we increase δ S , then we hold fewer shares of the underlying asset and the call price drops. b. inputs to the exchange call formula: S = 40 δS = 0 σ S = 0.3 K = 60 δK = 0 σ K = 0.5 ρ = 0.5 T =1 Once again, C = 2. If δ K = 0.1, then C = 2.86. If δ K goes up, then the present value of the strike price goes down, making the call more valuable. c. inputs to the exchange call formula: S = 40 δS = 0 σ S = 0.3 K = 60 δK = 0 σ K = 0.5 ρ = 0.5 T =1 Once again, C = 2. If ρ = −0.5, then C = 5.79. p If ρ is negative, then σ = σ 2S + σ 2K − 2ρσ S σ K is higher then if ρ is positive. So a negative ρ increases σ, making an option more valuable. Problem 14.18. ρ = 1p √ σ = σ 2S + σ 2K − 2ρσ S σ K = 0.32 + 0.32 − 2 × 1 × 0.3 × 0.3 = 0 a. The call price is zero. Because V ar [ln (S/K)] = σ 2S + σ 2K − 2ρσ S σ K = 0, S/K is a constant during [0, T ]. Since at time zero S = K, then S = K during [0, T ]. So the payoff of the call is zero and the call is worthless. p √ b. Now σ = σ 2S + σ 2K − 2ρσS σK = 0.42 + 0.32 − 2 × 1 × 0.4 × 0.3 = 0.1. Using DM 14.16, we find that the exchange call premium is 1.60. Skip the remaining problems (Problem 14.19 through 14.22).

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CHAPTER 14. EXOTIC OPTIONS: I

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Chapter 18

Lognormal distribution Problem 18.1. x−μ Using DM 18.4 z = , we find the equivalent draws from the a standard σ normal distribution are: −11 − (−8) −3 − (−8) −7 − (−8) √ √ √ = 0.258 2 = −0.774 6 = 1. 2910 15 15 15 2 − (−8) √ = 2. 5820 15

−15 − (−8) √ = −1. 807 4 15

Problem 18.2. x−μ , we get x = μ + zσ. The equivalent draws are: Using DM 18.4 z = σ √ √ 0.8 + (−1.7) √25 = −7. 7 0.8 + (0.55) √ 25 = 3. 55 0.8 + (−0.3)√ 25 = −0.7 0.8 + (−0.02) 25 = 0.7 0.8 + (0.85) 25 = 5. 05 Problem 18.3. Linear combination of normal random variables is also normal. x1 + x2 is normal. Its mean is E (x1 + x2 ) = E (x1 ) + E (x2 ) = 1 − 2 = −1. Its variance is V ar (x1 + x2 ) = V ar (x1 ) + V ar (x2 ) + 2Cov (x1 , x2 ) = 5 + 2 + 2 (1.3) = 9. 6 x1 − x2 is normal. Its mean is E (x1 − x2 ) = E (x1 ) − E (x2 ) = 1 − (−2) = 3. Its variance is V ar (x1 − x2 ) = V ar (x1 ) + V ar (x2 ) − 2Cov (x1 , x2 ) = 5 + 2 − 2 (1.3) = 4. 4 Problem 18.4. 161

CHAPTER 18. LOGNORMAL DISTRIBUTION x1 + x2 is normal with mean E (x1 + x2 ) = E (x1 ) + E (x2 ) = 2 + 8 = 10 and variance V ar (x1 + x2 ) = V ar (x1 ) + V ar (x2 ) + 2Cov (x1 , x2 ) Cov (x1 , x2 ) = ρσ x1 σ x2 √ √ ⇒ V ar (x1 + x2 ) = 0.5 + 14 + 2 (−0.3) 0.5 14 = 12. 91 x1 − x2 is normal with mean E (x1 − x2 ) = 2 − 8 = −6 and variance V ar (x1 − x2 ) = V ar (x1 ) + V ar (x2 ¡√ ) − 2Cov ¢ ¡√(x1 ,¢x2 ) V ar (x1 − x2 ) = 0.5 + 14 − 2 (−0.3) 0.5 14 = 16. 09 Problem 18.5. Use DM 18.10. x1 + x2 + x3 is normal with mean E (x1 + x2 + x3 ) = E (x1 ) + E (x2 ) + E (x3 ) = 1 + 2 + 2.5 = 5. 5 variance V ar (x1 + x2 + x3 ) = V ar (x1 )+V ar (x2 )+V ar (x3 )+2Cov (x1 , x2 )+2Cov (x1 , x3 )+ 2Cov (x2 , x3 ) ¡√ ¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ = 5 + 3 + 7 + 2 (0.3) 5 3 + 2 (0.1) 5 7 + 2 (0.4) 3 7 = 22. 17 x1 + 3x2 + x3 is normal with mean E (x1 + 3x2 + x3 ) = E (x1 ) + 3E (x2 ) + E (x3 ) = 1 + 3 (2) + 2.5 = 9. 5 variance V ar (x1 + 3x2 + x3 ) = V ar (x1 )+32 V ar (x2 )+V ar (x3 )+2×3Cov (x1 , x2 )+ 2Cov (x1 , x3 ) + 2 × 3Cov (x2¡√ , x3 )¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ ¡√ ¢ 2 3 +2 (0.1) 5 7 +2×3 (0.4) 3 7 = = 5+3 ×3+7+2×3 (0.3) 5 58. 15 x1 + x2 + 0.5x3 is normal with mean E (x1 + x2 + 0.5x3 ) = E (x1 ) + E (x2 ) + 0.5E (x3 ) = 1 + 2 + 0.5 (2.5) = 4. 25 variance V ar (x1 + x2 + 0.5x3 ) = V ar (x1 )+V ar (x2 )+0.52 V ar (x3 )+2Cov (x1 , x2 )+ 2 × 0.5Cov (x1 , x3 ) + 2 × 0.5Cov (x¡√ 2 , x¢3 )¡√ ¢ ¡√ ¢ ¡√ ¢ = 5 + 3 + 0.52 × 7 + 2 (0.3) 5 3 + 2 × 0.5 (0.1) 5 7 +2× ¡√ ¢ ¡√ ¢ 0.5 (0.4) 3 7 = 14. 50 Problem 18.6. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION Use DM 18.13. E (ex ) = e2+0.5(5) = 90. 017 1 Let a represent the median of y = ex . Then P (y ≤ a) = 0.5. y = ex is an increasing function. Hence of the median of ex corresponds to the median of x. Let b represent the median of x. P (x ≤ b) = 0.5. Hence b = 2 (the median of a normal random variable is its mean). =⇒ a = eb = e2 = 7. 389 1

Problem 18.7. There’s a typo in the problem. The table should say "Month" instead of "Day" (i.e. should be Month 0, Month 1, ...,Day 4, not Day 0, Day 1, ...,Day 4). This is similar to DM Table 11.1. a. Stock A: ´ ³ St ∧ 2 Month Stock price rt =ln rt − r St−1 0 100 105 1 105 ln = 0.04879 016 0.00238048 100 102 2 102 ln = −0.02898 75 0.00084028 105 97 3 97 ln = −0.05026 183 0.00252625 102 100 4 100 ln = 0.03 045 921 0.00092776 97 Total 0.00000000 0.00667477 4 P

rt

0 mean monthly continuously compounded return: r = = =0 4 4 v uP ´ ∧ 2 u 4 ³ r u rt − r t t=1 0.00667477 = = 4. 716 909 3× monthly standard deviation: 4−1 3 ∧

t=1

10−2 √ ¡ ¢ annual standard deviation: 12 4. 716 909 3 × 10−2 = 16.34% b. Stock B www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION

Month

Stock price

0 1 2 3 4 Total

100 105 102 97 100

rt =ln

St St−1

0.04879016 0.35667494 −0.43592432 0.00000000 0

³

´ ∧ 2

rt − r

0.00238048 0.12721702 0.19003001 0.00092776 0.32055527 4 P

rt 0 ∧ mean monthly continuously compounded return: r = t=1 = = 0 4 4 v uP ´ ∧ 2 u 4 ³ r u rt − r t t=1 0.32055527 monthly standard deviation: = = 0.326 881 87 4−1 3 √ annual standard deviation: 12 (0.326 881 87) = 1. 132 4 = 113.24% c. The statement is correct. For example, the mean monthly continuously compounded return for Stock A is

∧

4 P

rt

r = t=1 4 100 ln 100 = 0 4

=

4 P

t=1

4 P St S4 (ln St − ln St−1 ) ln ln S4 − ln S0 St−1 S0 t=1 = = = = 4 4 4 4

ln

So the mean monthly continuously compounded return for Stock A depends only on S4 and S0 ; S1 , S2 , and S3 are irrelevant. However, S1 , S2 , and S3 matters whenvwe are calculating the monthly stanuP ´ ∧ 2 u 4 ³ u rt − r t t=1 dard deviation in the following formula: . 4−1 Problem 18.8. ¡ ¢ ln St is normally distributed with mean ln S0 + α − δ − 0.5σ 2 t and variance σ 2 t. £ ¡ ¤ ¢ ln St ∼ N ln S0 + α − δ − 0.5σ 2 t, σ 2 t Then for a given stock price c

P (St ≤ c) = P (ln St ≤ ln c) = Φ www.actuary88.com

Ã

¡ ¢ ! ln c − ln S0 − α − δ − 0.5σ 2 t √ σ t

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CHAPTER 18. LOGNORMAL DISTRIBUTION One subtle point. Since St is continuous, the probability is zero that St takes on a single value. Hence P (St = c) = 0, P (St ≤ c) = P (St < c), and P (St > c) = P (St > c). In this problem, S0 = 100, α = 0.08, δ = 0, σ = 0.3 £ ¡ ¤ ¢ ln St ∼ N £ln 100 + 0.08 − 0 − 0.5 ¤× 0.32 t, 0.32 t ln St ∼ N 4. 605 17 + 0.035 t, 0.32 t P (St ≤ c) = Φ

µ

ln c − 4. 605 17 − 0.035 t √ 0.3 t

For t = 1, £we have: ¤ ln S1 ∼ N 4. 640 17, 0.32 P (S1 ≤ c) = P (ln S1 ≤ ln c) = Φ P (S1 ≤ 105) = Φ 0.518 3

µ

µ

¶

ln c − 4. 640 17 0.3

ln 105 − 4. 640 17 0.3

¶

¶

= Φ (0.0460) = NormalDist (0.0460) =

P (S1 > 105) = 1 − 0.518 3 = 0.481 7 Next, we consider how P (St >Ãc) changes if we change t or σ. ! ¡ ¢ ln c − ln S0 − α − δ − 0.5σ 2 t √ We know that P (St ≤ c) = Φ σ t ¡ ¢ ln c − ln S0 − α − δ − 0.5σ 2 t √ , where z is a normal random variSet z = σ t able. ¡ ¢ α − δ − 0.5σ 2 α − δ − 0.5σ 2 d √ d ln c − ln S0 − α − δ − 0.5σ 2 t d √ √ t=− =− z= dt dt σ dt σ t 2σ t For this problem, α = 0.08, δ = 0, σ = 0.3 d 0.005 833 0.08 − 0 − 0.5 × 0.32 √ √ =− z=− dt 2 × 0.3 t t So z is a decreasing function of t. Remember that the accumulative normal distribution Φ (z) is an increasing function of z, we conclude: The higher the t, the lower the z, the lower the P (St ≤ c) = Φ (z), and the higher the P (St > c) = 1 − Φ (z). Similarly, www.actuary88.com

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¡ ¡ ¢ ¢ √ d α − δ − 0.5σ 2 d − α − δ − 0.5σ 2 t d d ln c − ln S0 − α − δ − 0.5σ 2 t √ √ = =− t z= dσ dσ µ dσ¶ dσ σ σ¶ t σ t µ √ d α−δ √ α−δ =− t − 0.5σ = t 0.5 + dσ σ σ2 For this problem, α = 0.08, δ = 0 so

µ ¶ √ 0.08 d > 0. z = t 0.5 + 2 dσ σ

So z is an increasing function of σ. The higher the σ, the higher the z, the higher the P (St ≤ c) = Φ (z), and the lower the P (St > c) = 1 − Φ (z). Problem 18.9. Use DM 18.30.

µ ¶ ∧ N d1 E (St |St > K) = Se(α−δ)t µ∧ ¶ N d2 ∧ d1

ln =

¢ S0 ¡ + α − δ + 0.5σ 2 t K √ σ t

∧ d2

∧ √ = d1 − σ T

In this problem, ¢ 100 ¡ + 0.08 − 0 + 0.5 × 0.32 1 105 √ = 0.254 0 = 0.3 1 µ ¶ ∧ N d1 = NormalDist (0.2540) = 0.600 3

∧ d1

ln

∧ √ d2 µ = 0.2540 − 0.3 1 = −0.046 ¶ ∧ N d2 = NormalDist (−0.046 )

= 0.481 7

E (S1 |S1 > 105) = 100e(0.08−0)1 ×

0.600 3 = 135 0.481 7

Next, we wan to analyze how E (St |St > 105) changes if we change t (while keeping other parameters unchanged), σ (while keeping other parameters unchanged), and α (while keeping other parameters unchanged). We see that E (St |St > 105) increases if we increase t, σ, or α. We consider t = 0.25, 0.5, 0.75, 1, ..., 6. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION t 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6

E (S1 |S1 > 105) 117.19 123.98 129.74 135.00 139.97 144.76 149.42 154.01 158.54 163.05 167.54 172.04 176.54 181.06 185.61 190.19 194.81 199.48 204.19 208.95 213.76 218.64 223.58 228.58

σ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4

E (S1 |S1 > 105) 115.15 124.65 135.00 146.25 158.48 171.78 186.26 202.04 219.26 238.06 258.61 281.11 305.77 332.82 362.55 395.25 431.27 471.02 514.93 563.51 617.34 677.07 743.46 817.37

α 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24

E (S1 |S1 > 105) 131.63 132.08 132.54 133.01 133.49 133.98 134.49 135.00 135.53 136.07 136.63 137.19 137.78 138.37 138.98 139.60 140.24 140.90 141.57 142.25 142.95 143.67 144.41 145.16

Problem 18.10. Use DM 18.23. P (St < K) = N (−d2 )

∧ d1

0.484 0

ln =

¢ ¢ S0 ¡ 100 ¡ ln + α − δ + 0.5 × σ 2 t + 0.08 − 0 + 0.5 × 0.32 1 K 98 √ √ = = σ t 0.3 1

∧ ∧ √ √ d2 µ = d1 − ¶ σ t = 0.484 0 − 0.3 1 ∧ N −d2 = NormalDist (−0.184

= 0.184 ) = 0.427 0

P (St < 98) = 0.427 0 Next, we analyze how P (St < 98) changes if we change t. We find that P (St < 98) increases with t initially and then decreases with t. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION t 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6

d2 0.1930 0.1777 0.1788 0.1840 0.1907 0.1979 0.2052 0.2126 0.2199 0.2271 0.2341 0.2410 0.2477 0.2543 0.2607 0.2670 0.2732 0.2792 0.2852 0.2910 0.2967 0.3023 0.3078 0.3133

P (St < 98) = N (−d2 ) 0.4235 0.4295 0.4290 0.4270 0.4244 0.4216 0.4187 0.4158 0.4130 0.4102 0.4075 0.4048 0.4022 0.3996 0.3972 0.3947 0.3924 0.3900 0.3878 0.3855 0.3833 0.3812 0.3791 0.3770

Problem 18.11. Use DM 18.28.

µ ¶ ∧ N −d1 E (St |St < K) = Se(α−δ)t µ ∧ ¶ N −d2 If K = 98: ¢ ¢ S0 ¡ 100 ¡ ln ln + α − δ + 0.5 × σ 2 t + 0.08 − 0 + 0.5 × 0.32 1 ∧ K 98 √ √ d1 = = = σ t 0.3 1 0.484 0 ∧ √ √ = d1 − σ t = 0.484 0 − 0.3 1 = 0.184 ¶ µ ∧ N −d1 = NormalDist (−0.484 0 ) = 0.3142 µ ¶ ∧ N −d2 = NormalDist (−0.184 ) = 0.427 0 ∧ d2

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CHAPTER 18. LOGNORMAL DISTRIBUTION E (S1 |S1 < 98) = 100e(0.08−0)1 ×

0.3142 = 79. 71 0.427 0

If K = 120: ¢ 100 ¡ ln + 0.08 − 0 + 0.5 × 0.32 1 ∧ 120 √ d1 = = −0.191 1 0.3 1 ∧ ∧ √ √ d2 = d1 − σ t = −0.191 1 − 0.3 1 = −0.491 1 N N

µ

µ

∧

−d 1 ∧

−d 2

¶ ¶

= NormalDist (0.191 1 ) = 0.575 8 = NormalDist (0.491 1 ) = 0.688 3

E (S1 |S1 < 120) = 100e(0.08−0)1 ×

0.575 8 = 90. 62 0.688 3

If K = 98. We find that E (S1 |S1 < K) decreases with t. t d1 d2 E (S1 |S1 < K) 0.25 0.4235 0.1930 88.12 0.5 0.4295 0.1777 84.41 0.75 0.4290 0.1788 81.79 1 0.4270 0.1840 79.71 1.25 0.4244 0.1907 77.97 1.5 0.4216 0.1979 76.47 1.75 0.4187 0.2052 75.15 2 0.4158 0.2126 73.97 2.25 0.4130 0.2199 72.90 2.5 0.4102 0.2271 71.92 2.75 0.4075 0.2341 71.01 3 0.4048 0.2410 70.18 3.25 0.4022 0.2477 69.39 3.5 0.3996 0.2543 68.66 3.75 0.3972 0.2607 67.97 4 0.3947 0.2670 67.32 4.25 0.3924 0.2732 66.71 4.5 0.3900 0.2792 66.12 4.75 0.3878 0.2852 65.57 5 0.3855 0.2910 65.04 5.25 0.3833 0.2967 64.53 5.5 0.3812 0.3023 64.04 5.75 0.3791 0.3078 63.58 6 0.3770 0.3133 63.13 If K = 120. We find that E (S1 |S1 < K) decreases with t. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION t 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6

d1 0.8764 0.7814 0.7260 0.6883 0.6602 0.6381 0.6198 0.6044 0.5910 0.5792 0.5687 0.5591 0.5504 0.5424 0.5350 0.5281 0.5216 0.5156 0.5098 0.5044 0.4992 0.4942 0.4895 0.4850

d2 −1.1571 −0.7770 −0.6007 −0.4911 −0.4131 −0.3533 −0.3051 −0.2647 −0.2302 −0.1999 −0.1730 −0.1488 −0.1268 −0.1066 −0.0879 −0.0705 −0.0543 −0.0390 −0.0246 −0.0109 0.0021 0.0145 0.0263 0.0377

E (S1 |S1 < K) 98.14 95.09 92.64 90.62 88.88 87.36 86.00 84.78 83.66 82.63 81.68 80.80 79.97 79.20 78.46 77.77 77.11 76.49 75.89 75.32 74.78 74.26 73.76 73.28

The following table lists the value of E (St |St < K) www.actuary88.com

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t 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6

K = 98 σ = 0.3 88.12 84.41 81.79 79.71 77.97 76.47 75.15 73.97 72.90 71.92 71.01 70.18 69.39 68.66 67.97 67.32 66.71 66.12 65.57 65.04 64.53 64.04 63.58 63.13

K = 120 σ = 0.3 98.14 95.09 92.64 90.62 88.88 87.36 86.00 84.78 83.66 82.63 81.68 80.80 79.97 79.20 78.46 77.77 77.11 76.49 75.89 75.32 74.78 74.26 73.76 73.28

K = 98 σ = 0.1 95.26 94.22 93.52 92.99 92.56 92.21 91.90 91.64 91.40 91.19 91.00 90.83 90.68 90.53 90.40 90.28 90.16 90.06 89.96 89.86 89.78 89.69 89.61 89.54

K = 120 σ = 0.1 102.01 103.69 104.72 105.35 105.74 105.99 106.16 106.28 106.37 106.43 106.48 106.51 106.53 106.55 106.56 106.57 106.57 106.57 106.57 106.57 106.57 106.57 106.56 106.56

The diagram:

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Problem 18.12. This problem states that KT = S0 erT , but it doesn’t specify r .To solve the problem, we set r = α.

As T increases, P (ST < KT ) increases, P (ST > KT ) decreases, and the call/put price increases. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION t 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6

d2 −0.0750 −0.1061 −0.1299 −0.1500 −0.1677 −0.1837 −0.1984 −0.2121 −0.2250 −0.2372 −0.2487 −0.2598 −0.2704 −0.2806 −0.2905 −0.3000 −0.3092 −0.3182 −0.3269 −0.3354 −0.3437 −0.3518 −0.3597 −0.3674

P (ST < KT ) 0.5299 0.5422 0.5517 0.5596 0.5666 0.5729 0.5786 0.5840 0.5890 0.5937 0.5982 0.6025 0.6066 0.6105 0.6143 0.6179 0.6214 0.6248 0.6281 0.6313 0.6345 0.6375 0.6405 0.6433

P (ST > KT ) 0.4701 0.4578 0.4483 0.4404 0.4334 0.4271 0.4214 0.4160 0.4110 0.4063 0.4018 0.3975 0.3934 0.3895 0.3857 0.3821 0.3786 0.3752 0.3719 0.3687 0.3655 0.3625 0.3595 0.3567

Call price $5.98 $8.45 $10.34 $11.92 $13.32 $14.58 $15.73 $16.80 $17.80 $18.75 $19.64 $20.50 $21.32 $22.10 $22.85 $23.58 $24.29 $24.97 $25.63 $26.27 $26.89 $27.50 $28.09 $28.67

Put price $5.98 $8.45 $10.34 $11.92 $13.32 $14.58 $15.73 $16.80 $17.80 $18.75 $19.64 $20.50 $21.32 $22.10 $22.85 $23.58 $24.29 $24.97 $25.63 $26.27 $26.89 $27.50 $28.09 $28.67

Why do the European call and the European put have the same price? Use the call-put parity: Ke−rT C = S0 e−δT N (d1 ) − Ke−rT N (d2 ) Since K = S0 erT and δ = 0, we have: C = S0 N (d1 ) − S0 N (d2 ) µ ¶ ¶ µ 1 2 1 2 S0 + r − δ + T T −rT + r + σ σ ln √ S0 erT 2 2 √ √ d1 = = = 0.5σ T σ T √ σ T √ d2 = d2 − σ T = −0.5σ T ³ ³ ³ ³ ³ √ ´ √ ´ √ ´ √ ´´ = =⇒ C = S0 N 0.5σ T −S0 N −0.5σ T = S0 N 0.5σ T −S0 1 − N 0.5σ T h ³ i √ ´ S0 2N 0.5σ T − 1 ³ ³ √ ´ √ ´ P = −S0 e−δT N (−d1 )+Ke−rT N (−d2 ) = −S0 N −0.5σ T +S0 N −0.5σ T = ³ i ³ ³ h ³ √ ´ √ ´´ √ ´ −S0 1 − N 0.5σ T + S0 N −0.5σ T = S0 2N 0.5σ T − 1

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CHAPTER 18. LOGNORMAL DISTRIBUTION For example, for T = 0.25

C = P = 100 (2N (0.075) − 1) = 100 (2 × 0.529 9 − 1) = 5. 98 By the way, we can use the put-call parity to see why C = P . C + Ke−rT = P + S0 e−δT Since K = S0 erT and δ = 0, we have: C + S0 = P + S0 and C = P .

Typically, the strike price K is a fixed amount (as opposed to the increasing amount K = S0 erT ); the call price and put price move in opposite directions. In this problem, K = S0 erT increases with T ; the call and the put options both become more valuable as T increases. This is a pure mathematical coincidence. Intuitively, how to reconcile the fact that as T increases, P (ST < KT ) increases, P (ST > KT ) decreases, and the call and the put prices both increase? Please note that the call/put price not only depends on the probability of the option being in the money, not also depends on the payoff. For example, even though P (ST > KT ) decreases with T , if ST − KT increases at a faster speed, the call price will go up.

Problem 18.13.

Parameters: S = 100

K = 90

σ = 0.3

α = 4.5%

δ=0

As T increases, P (St < K) increases and then decreases. So the impact of T on P (St < K) is ambiguous. However, E(St |St < K) decreases over time. www.actuary88.com

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CHAPTER 18. LOGNORMAL DISTRIBUTION T 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12

d2 0.5438 0.4179 0.3684 0.3426 0.3275 0.3182 0.3124 0.3089 0.3070 0.3061 0.3061 0.3067 0.3077 0.3091 0.3108 0.3127 0.3148 0.3171 0.3194 0.3219 0.3244 0.3270 0.3296 0.3323

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P (St < K) = N (−d2) 0.2933 0.3380 0.3563 0.3659 0.3716 0.3752 0.3774 0.3787 0.3794 0.3798 0.3798 0.3795 0.3791 0.3786 0.3780 0.3772 0.3764 0.3756 0.3747 0.3738 0.3728 0.3718 0.3708 0.3698

E(St |St < K) $79.20 $74.64 $71.45 $68.96 $66.89 $65.13 $63.59 $62.23 $61.00 $59.88 $58.86 $57.91 $57.04 $56.22 $55.46 $54.74 $54.06 $53.42 $52.82 $52.24 $51.70 $51.17 $50.68 $50.20

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Chapter 19

Monte Carlo simulation Please note that due to rounding, you may not be able to fully reproduce my result. For example, when you see my z = 0.1234, the actual z used in my calculation could be 0.123436. If you plug in z = 0.1234, you may not be able to reproduce my result.

Problem 19.1. Here is the snapshot of the simulation done in Excel:

1 2 3 4 5 6 ... 1000 1001 1002

A i 1 2 3 4 5 ... 999 1000 Total

B u 0.104689 0.491579 0.085629 0.878402 0.199163 ... 0.132422 0.869963 498.294032

C u2 0.010960 0.241650 0.007332 0.771590 0.039666 ... 0.017536 0.756835 334.277115

Sample formulas Cell B2 = rand() B3 = rand() C2 = B2ˆ2 C3 = B3ˆ2 B1002 = sum(B2 : B1001) C1002 = sum(C2 : C1001) E (u) =

P

498.294032 ui = = 0.498 294 n 1000 177

CHAPTER 19. MONTE CARLO SIMULATION n V ar (u) = n−1 0.08 606 6

Ã

1P 2 ui − n

µP

ui n

¶2 !

1000 = 999

µ

334.277115 − 0.498 2942 1000

¶

The correct mean of u ∼ (0, 1) is: E (u) = 0.5 1 The correction variance: V ar (u) = = 0.08333 3 12 Next, we graph the histogram. D E F 1 bin range frequency 2 0.1 0 < u < 0.1 104 3 0.2 0.1 ≤ u < 0.2 106 4 0.3 0.2 ≤ u < 0.3 94 5 0.4 0.3 ≤ u < 0.4 100 6 0.5 0.4 ≤ u < 0.5 90 7 0.6 0.5 ≤ u < 0.6 108 8 0.7 0.6 ≤ u < 0.7 91 9 0.8 0.7 ≤ u < 0.8 93 10 0.9 0.8 ≤ u < 0.9 114 11 1 0.9 ≤ u < 1 100 12 Total 1000 Sample formulas for Column F. F 2 = countif (B2 : B1001, ” < ”&D2) F 3 = countif (B2 : B1001, ” < ”&D3) − F 2 F 4 = countif (B2 : B1001, ” < ”&D4) − (F 2 + F 3) F 5 = countif (B2 : B1001, ” < ”&D5) − (F 2 + F 3 + F 4) ... F 11 = countif (B2 : B1001, ” < ”&D11) − (F 2 + F 3 + F 4 + ... + F 10) Problem 19.2. 1 2 3 4 5 6 7 ... 1000 Total

u1 0.6922 0.0553 0.6137 0.2610 0.7009 0.1991 0.8552 ... 0.4822

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u2 0.1621 0.1343 0.5859 0.5720 0.5851 0.8738 0.3089 ... 0.9639

u3 0.0013 0.5132 0.1285 0.5668 0.4098 0.1330 0.2594 ... 0.9245

... ... ... ... ... ... ... ... ... ...

u12 0.1888 0.4550 0.6732 0.3890 0.1757 0.0924 0.6550 ... 0.7599

c °Yufeng Guo

x=

P

ui − 6 −0.8461 −0.2583 0.5420 −1.1239 −0.6006 0.2029 0.4612 ... 0.9899 45.381885

x2 0.715842 0.066742 0.293817 1.263203 0.360681 0.041173 0.212701 ... 0.979991 949.311003 178

=

CHAPTER 19. MONTE CARLO SIMULATION

Figure 19.1:

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CHAPTER 19. MONTE CARLO SIMULATION

Figure 19.2: 45.381885 = 0.045 382 (close to zero) 1000 µ ¶ 1 1000 2 × 949.311003 − 0.045 382 = 0.948 200 (close to 1) V ar (x) = 999 1000 The histogram looks like a bell curve. E (x) =

Problem 19.3. Snapshot of Excel for the simulation: A B C 1 i x1 x2 2 1 −0.88028 0.535024 3 2 −0.71431 0.209528 4 3 0.114864 −0.1811 5 4 −0.8346 0.156575 6 5 0.646959 −1.65712 7 6 0.628469 0.48616 ... ... ... 2001 2000 0.779131 0.509077 2002 sum

D ex1 0.414668 0.489528 1.121721 0.434047 1.909724 1.874738 ... 2.179578 3, 210.5069

E ex2 1.70749 1.233096 0.834356 1.169498 0.190687 1.626059 ... 1.663754 17, 452.8493

G (ex1 )2 0.171949 0.239637 1.258259 0.188397 3.647045 3.514643 ... 4.750559 13, 501.1984

Sample formulas. www.actuary88.com

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H (ex2 )2 2.915522 1.520526 0.696150 1.367726 0.036361 2.644069 ... 2.768078 1, 753, 120.4685

CHAPTER 19. MONTE CARLO SIMULATION Cell B2 = N ormInv (Rand(), 0, 1) D2 = exp (B2) E2 = exp (C2)

C2 = N ormInv (Rand(), 0.7, 0.3ˆ0.5)

3210.5069 = 1. 605 3 2000 The correct mean (using DM 18.13): E (ex1 ) = e0.5 = 1. 648 7

E (ex1 ) =

The estimated variance: ¶ µ 2000 1 x1 2 V ar (e ) = = 4. 1757 × 13501.1984 − 1. 605 3 1999 2000 The correct variance (DM18.14) : V ar (ex1 ) = e (e − 1) = 4. 670 8 17452.8493 = 8. 726 4 2000 The correct mean: E (ex2 ) = e0.7+0.5×3 = 9. 025 0

E (ex2 ) =

The estimated variance: ¶ µ 2000 1 V ar (ex2 ) = × 1753120.4685 − 9. 025 0 2 = 795. 507 4 1999 2000 The correct variance ¡is: ¢ V ar (ex2 ) = e2×0.7+3 e3 − 1 = 1554. 533 6

Our estimate of E (ex2 ) and V ar (ex2 ) are way off. To improve our estimate, we need to increase the number of simulations.

Problem 19.4. I performed 5,000 i z 1 −0.2273 2 −0.0408 3 1.7371 4 −0.8544 5 −0.5836 ... ... 5000 0.4792 sum

simulations. ST put payoff 38.9990 1.0010 40.1055 0.0000 52.3623 0.0000 35.4978 4.5022 36.9693 3.0307 ... ... 43.3588 0.0000 10, 197.3969

P 0.9812 0.0000 0.0000 4.4131 2.9707 ... 0.0000 9, 995.4749

P2 0.9627 0.0000 0.0000 19.4754 8.8248 ... 0.0000 61, 151.4337

The z column in the above table is generated using Excel’s formula N ormInv (Rand () , 0, 1) Sample calculations for Row 1. Excel’s formula N ormInv√(Rand () , 0, 1) happens to return √z = −0.2273. 2 2 S = S e(r−δ−0.5σ )T +σ T z = 40e(0.08−0−0.5×0.3 )0.25+0.3 0.25(−0.2273) = T

0

38. 9990 www.actuary88.com

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CHAPTER 19. MONTE CARLO SIMULATION

Figure 19.3:

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CHAPTER 19. MONTE CARLO SIMULATION

Figure 19.4: The put payoff: 40 − 38. 9990 = 1. 001 The put price: P = e−0.08×0.25 × 1. 001 = 0.9812 P 2 = 0.98122 = 0.962 7 The estimate put price is:

9995.4749 = 1. 999 1 5000

You can verify that the put price based on the Black-Scholes formula is 1.9927. The estimated variance of the put price ¶ per simulation is: µ 1 5000 × 61151.4337 − 1. 999 12 = 8. 235 5 4999 5000 Suppose we want to perform n simulations and take the average put price of these n simulations as an estimate of the put price. P1 + P2 + ... + Pm Then P = n here Pi is the put price calculated from the i-th simulation =⇒ 0.012 =

¡ ¢ nV ar (P per simulation) V ar (P per simulation) = V ar P = n2 n

8. 235 5 n

n=

8. 235 5 = 82355 0.012

So we need to perform roughly 82, 400 simulations. www.actuary88.com

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CHAPTER 19. MONTE CARLO SIMULATION Problem 19.5. i 1 2 3 4 5 6 7 8 ... 5000 sum

z −0.03768 0.531987 2.062351 −2.46722 2.579892 −1.23246 0.477051 −1.37135 ... 1.421245

S1 40.9592 48.5928 76.9061 19.7610 89.8235 28.6210 47.7985 27.4530 ... 63.4500

1/S1 0.0244 0.0206 0.0130 0.0506 0.0111 0.0349 0.0209 0.0364 ... 0.0158

F = e−rT (1/S1 ) 0.0225 0.0190 0.0120 0.0467 0.0103 0.0323 0.0193 0.0336 ... 0.0145 116.9386

F2 0.000508 0.000361 0.000144 0.002182 0.000106 0.001040 0.000373 0.001131 ... 0.000212 2.9935

For example, if z = −0.03768, then √ √ 2 2 S1 = S0 e(r−δ−0.5σ )T +σ T z = 40e(0.08−0−0.5×0.3 )1+0.3 1(−0.03768) = 40. 959 2 The forward price is: 1 F = e−rT (1/S1 ) = e−0.08×1 = 0.0225 40.9592 116.9386 = 0.023 39 5000 The estimated variance of the forward price per simulation is: ¶ µ 5000 1 × 2.9935 − 0.023 392 = 0.00005162 4999 5000

The estimated forward price is:

We can calculate the true forward price using DM 20.30:

2 P F0,T [S a (T )] = e−rT S a (0) e[a(r−δ)+0.5a(a−1)σ ]T Set a = −1. The true forward price is: ¤ £ −1 ¡ ¢ 2 P S (T ) = e−0.08 40−1 e(−1(0.08−0)+0.5×(−1)(−1−1)0.3 )1 = 0.02331 F0,T

Problem 19.6. a. www.actuary88.com

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CHAPTER 19. MONTE CARLO SIMULATION i 1 2 3 4 5 6 7 8 ... 5000 sum

z 0.5992 −1.3882 −1.2717 0.3870 −0.7793 0.4702 1.1824 0.4278 ... −0.1042

S1 49.5825 27.3147 28.2864 46.5239 32.7893 47.7004 59.0623 47.0974 ... 40.1502

S12 2, 458.4195 746.0939 800.1177 2, 164.4688 1, 075.1384 2, 275.3243 3, 488.3595 2, 218.1649 ... 1, 612.0360

¡ ¢ F = e−rT S12 2, 269.4072 688.7315 738.6017 1, 998.0565 992.4778 2, 100.3891 3, 220.1616 2, 047.6243 ... 1, 488.0968 9, 517, 170.0248

F2 5, 150, 209.1192 474, 351.0158 545, 532.5311 3, 992, 229.8443 985, 012.2592 4, 411, 634.2656 10, 369, 440.9962 4, 192, 765.3642 ... 2, 214, 432.0185 26, 459, 271, 681.8572

The estimate forward price at time zero for a claim paying S12 at T = 1 is: 9517170.0248 = 1903. 43 5000 You can verify that the true forward price (using DM 20.30) is 1, 896.49 The estimated variance of the forward price per ¶ simulation is: µ 1 5000 × 26, 459, 271, 681.8572 − 1903. 432 = 166, 9142. 40 4999 5000 b. i 1 2 3 4 5 6 7 ... 5000 sum

z 0.112 −0.8019 0.0557 −0.9828 2.0755 −1.0444 1.0135 ... −1.2727

S1 42.8403 32.5673 42.1228 30.8470 77.2100 30.2822 56.1445 ... 28.2776

S10.5 6.5453 5.7068 6.4902 5.5540 8.7869 5.5029 7.4930 ... 5.3177

¡ ¢ F = e−rT S10.5 6.0420 5.2680 5.9912 5.1270 8.1114 5.0798 6.9169 ... 4.9088 30, 059.267012

F2 36.506096 27.752022 35.894682 26.286079 65.794022 25.804789 47.843187 ... 24.096581 184, 880.552079

The estimate forward price at time zero for a claim paying S10.5 at T = 1 is: 30059.267012 = 6. 01185 5000 You can verify that true forward price (using DM 20.30) is 6.0086 The estimated variance of the forward price per simulation is: ¶ µ 1 5000 × 184880.552079 − 6.011852 = 0.833 9 4999 5000 www.actuary88.com

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CHAPTER 19. MONTE CARLO SIMULATION c. i 1 2 3 4 5 6 7

z 0.0111 −0.1893 −0.6352 0.4972 −1.2449 0.1526 −0.5234

S1 41.5630 39.1378 34.2374 48.0883 28.5144 43.3653 35.4052

S1−2 0.0006 0.0007 0.0009 0.0004 0.0012 0.0005 0.0008

... 8 sum

... 0.2497

... 44.6471

... 0.0005

The estimated forward price is:

¡ ¢ F = e−rT S1−2 0.0005 0.0006 0.0008 0.0004 0.0011 0.0005 0.0007

F2 0.00000029 0.00000036 0.00000062 0.00000016 0.00000129 0.00000024 0.00000054

... 0.0005 3.216862

... 0.00000021 0.002918

3.216862 = 0.000 643 5000

You can verify that the true forward price is 0.000644 The estimated variance of the forward ¶ price per simulation is: µ 1 5000 2 × 0.002918 − 0.000644 = 1. 69 × 10−7 4999 5000 Skip the remaining problems.

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Chapter 20

Brownian motion and Ito’s lemma Problem 20.1. I don’t want you to memorize Ito’s lemma. For problems related to It’s lemma, all you need to know is two things: 2

1. Unlike a deterministic random variable X where (dX) = 0, for a stochastic random variable Z, the term (dZ)2 is not zero and hence can’t be ignored. In fact, the textbook and my study guide have explained that 2 3 (dZ) = dt. However, you can ignore higher order such as (dZ) . 2. Ito’s lemma is just the stochastic counterpart of the Taylor series. To derive Ito’s lemma, first write the Taylor series. For a stochastic random variable X and a function y = f (t, X), first write the Taylor expansion: ∂f 1 ∂2f ∂f 1 ∂2f dy = d f (t, X) = (dt)2 + (dX)2 +....Here we dt+ dX + 2 ∂t ∂X 2 ∂t 2 ∂X 2 3 3 ignored the higher order terms (dt) , (dX) , and above. Next, throw away the term (dt)2 since t is a deterministic random variable and (dt)2 → 0. ∂f ∂f 1 ∂2f Now we have dy = d f (t, X) = (dX)2 . This is dt + dX + ∂t ∂X 2 ∂X 2 Ito’s lemma. With this point in mind, let’s solve the problem. a. If the stock price S follows the textbook Equation 20.8, then: dS (t) = αdt + σdZ (t)

(DM 20.8)

dS (t) is a linear function of dZ (t). Since [dZ (t)]2 = dt, we should keep 2 (dS) in the Taylor expansion but throw away all other higher order terms: 187

CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA ∂ ln S ∂ ln S 1 ∂ 2 ln S (dS)2 dt + dS + ∂t ∂S 2 ∂S 2 ∂ ln S ∂ ln S 1 Since ln S doesn’t contain t, = 0. In addition, = and ∂t ∂S S 2 ∂ 1 1 ∂ ln S = =− 2 ∂S 2 ∂S S S ∂ ln S 1 1 ∂ 2 ln S 1 1 2 2 → d ln S = (dS) = dS − (dS) dS + 2 ∂S 2 ∂S S 2 S2 → d ln S =

(dS)2 = (αdt + σdZ)2 = α2 (dt)2 + 2ασ (dZ) (dt) + σ 2 (dZ)2 Next, use the multiplication rule: dZ × dt = 0

(DM 20.17a)

(dt) = 0

2

(DM 20.17b)

2

(DM 20.17c)

(dZ) = dt 2

2

2

→ (dS) = (αdt + σdZ) = σ 2 (dZ) = σ 2 dt 1 1 1 2 → d ln S = (αdt + σdZ) − σ dt S 2 S2 ¶ µ σ 1 1 1 2 σ dt + dZ = α− S 2 S2 S b.If the stock price S follows the textbook Equation 20.9, then: dS (t) = λ (α − S) dt + σdZ → d ln S =

(DM 20.9)

1 1 ∂ 2 ln S 1 1 ∂ ln S 2 2 (dS) = dS − (dS) dS + ∂S 2 ∂S 2 S 2 S2

(dS)2 = [λ (α − S) dt + σdZ]2 = λ2 (α − S)2 (dt)2 + 2λ (α − S) dt × σdZ + σ2 (dZ)2 = σ 2 dt 1 1 1 2 → d ln S = dS − (dS) S 2 S2 λ (α − S) dt + σdZ 1 σ 2 dt = − S ∙ ¸ 2 S2 λ (α − S) 1 σ 2 σ = dt + dZ − S 2 S2 S c.If the stock price S follows the textbook Equation 20.27, then: ¸ ∙ ∧ ∧ ∧ (DM 20.8) dS (t) = α (S, t) − δ (S, t) dt + σ (S, t) dZ (t) When S (t) follows a geometric Brownian motion, www.actuary88.com

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA ∧

∧

α (S, t) = αS (t) δ (S, t) = δS (t) Hence DM Equation 20.8 becomes:

∧

σ (S, t) = σS (t)

dS (t) = S (α − δ) dt + σSdZ (t) d ln S =

(20.1)

∂ ln S 1 1 ∂ 2 ln S 1 1 2 2 (dS) = dS − (dS) dS + ∂S 2 ∂S 2 S 2 S2

(dS)2 = [S (α − δ) dt + σSdZ (t)]2 = σ 2 S 2 (dZ)2 = σ 2 S 2 dt 1 1 1 → d ln S = dS − (dS)2 S 2 S2 S (α − δ) dt + σSdZ (t) 1 σ2 S 2 dt = − S ¶ 2 S2 µ 1 2 = α − δ − σ dt + σdZ 2 Problem 20.2. ∂S 2 ∂S 2 1 ∂2S2 2 (dS) dt + dS + ∂t ∂S 2 ∂S 2 ∂S 2 1 ∂2S2 = (dS)2 = 2SdS + (dS)2 dS + ∂S 2 ∂S 2 a. Under DM Equation 20.8: dS (t) = αdt + σdZ (t) (dS)2 = σ 2 dt → dS 2 = 2SdS + (dS)2 ¡ ¢ = 2S (αdt + σdZ) + σ 2 dt = 2αS + σ 2 dt + 2σSdZ dS 2 =

b. Under DM Equation 20.9: dS (t) = λ (α − S) dt + σdZ (dS)2 = σ 2 dt 2 → dS 2 = 2SdS + (dS) = £2S [λ (α − S) dt + ¤σdZ] + σ 2 dt = 2Sλ (α − S) + σ 2 dt + 2SσdZ

c. Under Geometric Brownian motion dS (t) = S (α − δ) dt + σSdZ (t) (dS)2 = σ 2 S 2 dt → dS 2 = 2SdS + (dS)2 = 2S [S (α − δ) dt + σSdZ] + σ 2 S 2 dt + σ2 S 2 dt = 2S 2£ [(α − δ) dt + σdZ] ¤ 2 2 = S 2 (α − δ) + σ dt + 2σS 2 dZ Problem 20.3. www.actuary88.com

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA ∂S −1 ∂S −1 1 ∂ 2 S −1 (dS)2 dt + dS + ∂t ∂S 2 ∂S 2 ∂S −1 ∂ 2 S −1 ∂S −1 = 2S −3 =0 = −S −2 ∂t ∂S ∂S 2 → dS −1 = −S −2 dS + S −3 (dS)2 a. Under DM Equation 20.8: dS −1 =

dS (t) = αdt + σdZ (t) (dS)2 = σ 2 dt 2 → dS −1 = −S −2 dS + S −3 (dS) −2 −3 2 = −S (αdt + σdZ)¢ + S σ dt ¡ = −S −2 α + S −3 σ 2 dt − S −2 σdZ

b. Under DM Equation 20.9: dS (t) = λ (α − S) dt + σdZ (dS)2 = σ 2 dt 2 → dS −1 = −S −2 dS + S −3 (dS) −3 2 = ¡−S −2 (λ (α − S) dt + σdZ) σ dt ¢ + S −2 −2 −3 2 = −S λ (α − S) + S σ dt − S σdZ

c. Under Geometric Brownian motion dS (t) = S (α − δ) dt + σSdZ (t) (dS)2 = σ 2 S 2 dt 2 → dS −1 = −S −2 dS + S −3 (dS) −2 = −S [S (α − δ) dt + σSdZ (t)] + S −3 σ 2 S 2 dt S −1 σ 2 dt = −S −1 £ [(α − δ) dt + 2σdZ ¤ (t)] + −1 −1 =S − (α − δ) + σ dt − S σdZ (t) Problem 20.4. ∂S 0.5 ∂S 0.5 1 ∂ 2 S 0.5 2 (dS) dt + dS + ∂t ∂S 2 ∂S 2 ∂S 0.5 ∂ 2 S 0.5 ∂S 0.5 = −0.25S −1.5 =0 = 0.5S −0.5 ∂t ∂S ∂S 2 → dS 0.5 = 0.5S −0.5 dS − 0.125S −1.5 (dS)2 a. Under DM Equation 20.8:

dS 0.5 =

dS (t) = αdt + σdZ (t) (dS)2 = σ 2 dt → dS 0.5 = 0.5S −0.5 [αdt + σdZ (t)] − 0.125S −1.5 σ 2 dt b. Under DM Equation 20.9: dS (t) = λ (α − S) dt + σdZ 2 (dS) = σ 2 dt → dS 0.5 = 0.5S −0.5 [λ (α − S) dt + σdZ] − 0.125S −1.5 σ 2 dt www.actuary88.com

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA c. Under Geometric Brownian motion dS (t) = S (α − δ) dt + σSdZ (t) (dS)2 = σ 2 S 2 dt → dS 0.5 = 0.5S −0.5 [S (α − δ) dt + σSdZ (t)] − 0.125S −1.5 σ 2 S 2 dt = 0.5S 0.5 [(α − δ) dt + σdZ (t)] − 0.125S 0.5 σ 2 dt Problem 20.5. dS (t) = S (αs − δ s ) dt + σ s SdZS

(DM 20.37)

dQ (t) = Q (αQ − δ Q ) dt + σ Q QdZQ

(DM 20.38)

dSdQ = ρdt (DM 20.17d) ¡ 2 0.5 ¢ ¡ ¢ ¡ 2 0.5 ¢ ¡ 2 0.5 ¢ 2 ¡ 2 0.5 ¢ ∂ S Q ∂ S Q ∂ S Q 1 ∂ 2 S 2 Q0.5 d S Q (dS)2 + = dS+ dQ+ dSdQ+ ∂S ∂Q ∂S∂Q 2 ∂S 2 ¡ ¢ 1 ∂ 2 S 2 Q0.5 (∂Q)2 2 ∂Q2 ¡ ¡ ¢ ¢ ¡ ¢ ∂ S 2 Q0.5 ∂ S 2 Q0.5 ∂ 2 S 2 Q0.5 = 2SQ0.5 = 0.5Q−0.5 S 2 = ∂S ∂Q ∂S∂Q SQ−0.5 ¡ ¢ ¡ ¢ ∂ 2 S 2 Q0.5 ∂ 2 S 2 Q0.5 0.5 = 2Q = −0.25S 2 Q−1. 5 ∂S 2 ∂Q2 ¡ ¢ → d S 2 Q0.5 = 2SQ0.5 [S (αs − δ s ) dt + σ s SdZS ] +0.5Q−0.5 S 2 [Q (αQ − δ Q ) dt + σ Q QdZQ ] +SQ−0.5 ρdt +Q0.5 (σ Q Q)2 dt 2 −0.125S 2 Q−1. 5 [Q (αQ − δ Q ) dt + σ Q QdZQ ] Problem 20.6. From Problem 20.1.c, we ¶ have: µ 1 2 d ln S = αS − δ S − σ S dt + σ S dZS 2 µ ¶ 1 d ln Q = αQ − δ Q − σ2Q dt + σQ dZQ 2 ¶ ¶ µ µ 1 2 1 2 → d ln (SQ) = d ln S+d ln Q+ αS − δ S − σ S dt+σ S dZS + αQ − δ Q − σ Q dt+ 2 2 σ Q dZQ

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CHAPTER 20. BROWNIAN MOTION AND ITO’S LEMMA Problem 20.7. dS (t) = (α − δ) dt + σdZ S (t) ¤ £ 2 P F0,T S A (T ) = e−rT S A (0) e[A(r−δ)+0.5A(A−1)σ ]T Assume T = 1 If A £= 2 ¤ 2 P F0,1 S 2 (1) = e−0.06×1 1002 e(2(0.06−0)+0.5×2(2−1)0.4 )1 = 12460. 77

If A £= 0.5 ¤ 2 P F0,1 S 0.5 (1) = e−0.06×1 1000.5 e(0.5(0.06−0)+0.5×0.5(0.5−1)0.4 )1 = 9. 51

If A £= −2 ¤ 2 P F0,1 S −2 (1) = e−0.06×1 100−2 e(−2(0.06−0)+0.5(−2)(−2−1)0.4 )1 = 1. 35 × 10−4

The textbook asks you to compare your solution to the solution to Problem 19.7. However, Problem 19.7 is out of the scope of Exam MFE. So you don’t to do such a comparison. Skip the remaining problems (Problem 20.8 and beyond); they are out of the scope of Exam MFE.

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Chapter 21

The Black-Scholes equation Problem 21.1. Equation 21.12 is V (t, T ) = e−r(T −t) 1 BS PDE is Equation 21.11: Vt + σ2 S 2 VSS + (r − δ) SVS − rV = 0 2 We want to prove that V (t, T ) = e−r(T −t) satisfies Equation 21.11. V (t, T ) = e−r(T −t) ∂V ∂2V ∂V =⇒ VS = =0 Vt = =0 VSS = = re−r(T −t) = rV 2 ∂S ∂S ∂t 1 =⇒ Vt + σ 2 S 2 VSS + (r − δ) SVS − rV = rV − rV = 0 2 The boundary condition is that V (T, T ) = 1. Clearly, V (t, T ) = e−r(T −t) satisfies this boundary condition. Problem 21.2. V (t, T ) = AS a eγt ¡ ¢ aV VS = a AS a−1 eγt = S Vt = γ (AS a eγt ) = γV

0

¡ ¢ a (a − 1) V VSS = a (a − 1) AS a−2 eγt = S2

1 1 Vt + σ 2 S 2 VSS + (r − δ) SVS − rV = γV + σ 2 a (a − 1) V + (r − δ) aV − rV 2 2 To satisfy BS PDE, we just need to choose the a such that 1 1 γV + σ 2 a (a − 1) V +(r − δ) aV −rV = 0 or γ+ σ 2 a (a − 1)+(r − δ) a−r = 2 2 µ ¶ 1 2 2 1 σ a + r − δ − σ 2 a + (γ − r) = 0 2 2 Solving this equation, we get: 193

CHAPTER 21. THE BLACK-SCHOLES EQUATION ¶ sµ ¶2 µ 1 1 2 r − δ − σ 2 − 2σ 2 (γ − r) µ − r−δ− σ ± ¶ 2 2 1 r−δ a= = − 2 ± σ2 2 σ sµ ¶2 2 (γ − r) r−δ 1 − − σ 2 σ2 If we set a above, then V (t, T ) = AS a eγt satisfies the BS PDE. Problem 21.3. According to Proposition 20.30, the prepaid forward price at time zero value of a claim paying S a (T ) is 2 F0,T [S a (T )] = e−rT S a (0) e[a(r−δ)+0.5a(a−1)σ ]T In the above formula, if we replace T with (T − t) AS a eγt and S (0) with S (t), we’ll get the prepaid forward price at time t value of a claim paying S a (T ): 2 V (t, T ) = Ft,T [S a (T )] = e−r(T −t) S a (t) e[a(r−δ)+0.5a(a−1)σ ](T −t) We need to prove that Ft,T [S a (T )] satisfies the BS PDE. Notice that V (t, T ) is in the form of AS a eγt where γ = r − a (r − δ) − 0.5a (a − 1) σ According to Problem 21.2, V (t, T ) satisfies the BS PDE if we set µ ¶ sµ ¶2 1 r−δ 2 (γ − r) r−δ 1 a= − ± − 2 − 2 σ σ 2 σ2 Problem 21.4. First, let’s prove that if V 1 (S, t, T ) and V 2 (S, t, T ) each satisfy the BS PDE 1 Vt + σ2 S 2 VSS + (r − δ) SVS − rV = 0, then for any constants k1 and k2 the 2 linear combination V (S, t, T ) = k1 V 1 (S, t, T ) + k2 V 2 (S, t, T ) also satisfies the BS PDE. Proof. 1 2 Vt = k1 Vt1 + k2 Vt2 VS = k1 VS1 + k2 VS2 VSS = k1 VSS + k2 VSS 1 Vt + σ 2 S 2 VSS + (r − δ) SVS − rV ∙2 ¸ ∙ ¸ 1 2 2 1 1 2 2 2 1 1 1 2 2 2 = k1 Vt + σ S VSS + (r − δ) SVS − rV +k2 Vt + σ S VSS + (r − δ) SVS − rV 2 2 = k1 × 0 + k2 × 0 = 0 In this problem, Ke−r(T −t) and S (t) e−δ(T −t) are each in the form of AS a eγt . According to Problem 21.2, Ke−r(T −t) and S (t) e−δ(T −t) each satisfy the BS PDE. Hence the linear combination Ke−r(T −t) + S (t) e−δ(T −t) satisfy the BS PDE. The boundary condition is that V (S, T, T ) = K + S. www.actuary88.com

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CHAPTER 21. THE BLACK-SCHOLES EQUATION Problem 21.5. Some basics. 1 R x −s2 /2 e ds N (x) = √ 2π −∞ 1 d Rx 0 2 N (x) = √ e−x /2 Generally f (s) ds = f (x) dx a 2π 1 00 − x2 1 d −x2 /2 1 0 N (x) = √ = − √ xe 2 = −xN (x) e 2π dx 2π Now let’s prove. V = S (t) e−δ(T −t) N (d1 )

¶ µ 1 2 S (t) + r − δ + σ (T − t) ln K 2 √ d1 = σ T −t

∂ 1 ∂d1 1 S (t) 1 × × = √ ln = √ ∂S (t) K σ T − t ∂S (t) σ T − t S (t) 1 ∂ ∂d1 1 0 0 × N (d1 ) = N (d1 ) = N (d1 ) × √ ∂S ∂S S (t) σ T −t 00 00 1 ∂d1 1 ∂ 0 × N (d1 ) = N (d1 ) = N (d1 ) × √ ∂S ∂S S (t) σ T −t # " ∙ ¸ 0 ∂N (d N ) (d ) 1 1 → VS = e−δ(T −t) N (d1 ) + S (t) = e−δ(T −t) N (d1 ) + √ ∂S σ T −t # " 0 N (d1 ) =⇒ (r − δ) SVS = (r − δ) Se−δ(T −t) N (d1 ) + √ σ T −t (r − δ) 0 Se−δ(T −t) N (d1 ) = (r − δ) V + √ σ T −t ¸ ∙ ∂ ∂ 0 1 → VSS = e−δ(T −t) N (d1 ) + √ N (d1 ) ∂S σ T − t ∂S ∙ ¸ 00 1 1 1 1 1 0 = e−δ(T −t) N (d1 ) × √ × × N (d1 ) × √ × + √ σ T − t S (t) σ T − t σ T − t S (t) ¸ ∙ 1 1 1 1 1 0 0 × × d1 N (d1 ) × √ × = e−δ(T −t) N (d1 ) × √ − √ S (t) S (t) σµ T − t σ T −t ¶ σ T −t −δ(T −t) d1 e 0 √ N (d1 ) 1 − √ = S (t) σ T − t σ T −t

¶ µ 1 2 2 1 2 2 e−δ(T −t) 0 d1 √ =⇒ N (d1 ) 1 − √ σ S VSS = σ S 2 2 Sσ T − t σ T −t ¶ µ e−δ(T −t) 0 d1 = σS √ N (d1 ) 1 − √ 2 T −t σ T −t www.actuary88.com

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CHAPTER 21. THE BLACK-SCHOLES EQUATION ∙ ¸ ∂ ∂ ∂ → Vt = S (t) e−δ(T −t) N (d1 ) = S (t) N (d1 ) e−δ(T −t) + e−δ(T −t) N (d1 ) ∂t ∂t¸ ∂t ∙ ∂ −δ(T −t) −δ(T −t) 0 = S (t) N (d1 ) δe +e N (d1 ) d1 ∂t Please note that S (t) is a fixed constant for a given time t. ¶ µ S (t) 1 ln + r − δ + σ 2 (T − t) ∂ ∂ K 2 √ d1 = ∂t ∂t µ σ T −t ¶ 1 S (t) r − δ + σ 2 (T − t) ln ∂ ∂ 2 √ K + √ = ∂t σ T − t ∂t σ µ T −t ¶ √ ∂ T −t 1 S (t) 1 = ln + r − δ + σ2 3 K 2 ∂t σ 2σ (T − t) 2

µ ¶ 1 2 S (t) 1 = ln − √ σ +r−δ 3 K 2σ T − t 2 2σ (T − t) 2 ⎤ ⎡ S (t) √ ¶ µ ⎥ ⎢ ln K T −t 1 2 1 ⎢ − = σ +r−δ ⎥ ⎦ ⎣ 1 2 (T − t) σ 2 σ (T − t) 2µ ¶ ⎤ ⎡ S (t) 1 2 ln (T − t) + r − δ + σ ⎥ ⎢ 1 K 2 ⎥ ⎢ = ⎦ 1 2 (T − t) ⎣ 2 σ (T − t) ¶ µ ⎤ ⎡ 1 2 √ ¶ µ (T − t) σ r − δ + ⎥ ⎢ T −t 1 2 1 2 ⎥ ⎢ − + r − δ + σ ⎦ 1 2 (T − t) ⎣ σ 2 2 σ (T − t) ¶ µ ⎤ ⎡ 1 2 r − δ + σ2 ⎥ ⎢ √ 1 2 ⎢d1 − T − t⎥ = ⎦ ⎣ 2 (T − t) σ 1

∙ ¸ ∂ −δ(T −t) −δ(T −t) 0 =⇒ Vt = S (t) N (d1 ) δe +e N (d1 ) d1 ¶ µ ∂t ⎤ ⎡ 1 2 2 r − δ + σ ⎥ ⎢ √ 1 0 2 ⎢d1 − T − t⎥ = δV + S (t) e−δ(T −t) N (d1 ) ⎦ ⎣ 2 (T − t) σ

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CHAPTER 21. THE BLACK-SCHOLES EQUATION 1 Vt + σ 2 S 2 VSS + (r − δ) SVS 2

¶ µ ⎤ 1 2 2 r − δ + σ ⎢ ⎥ √ 1 0 2 ⎢d1 − = δV + S (t) e−δ(T −t) N (d1 ) T − t⎥ ⎣ ⎦ 2 (T − t) σ ⎡

¶ µ d1 e−δ(T −t) 0 N (d1 ) 1 − √ +σS √ 2 T −t σ T −t

(r − δ) 0 + (r − δ) V + √ Se−δ(T −t) N (d1 ) σ T −t = rV End of the proof.

Problem 21.6. V = e−r(T −t) N (d2 ) √ d2 = d1 − σ T − t ¢ ∂ ∂ ∂ ¡ √ ∂ 1 σ d2 = d1 − σ T − t = d1 + √ ∂t ∂t ∂t ∂t 2 T −t 1 ∂ 1 1 ∂ √ × d2 = d1 = √ = ∂S ∂S σ T − t S (t) Sσ T − t From the previous problem, we know that ¶ µ ⎤ 1 2 σ 2 r − δ + ⎥ ⎢ √ 1 ∂ 2 ⎢d1 − T − t⎥ d1 = ⎦ ⎣ ∂t 2 (T − t) σ ⎡

¶ µ ⎤ 1 2 2 r − δ + σ ⎥ ⎢ √ √ 1 2 ⎢d2 + σ T − t − = T − t⎥ ⎦ ⎣ 2 (T − t) σ ⎡

=

¸ ∙ 1 2 (r − δ) √ T −t d2 − 2 (T − t) σ

→

¸ ∙ 1 2 (r − δ) √ 1 σ ∂ T −t + √ d2 = d2 − ∂t 2 (T − t) σ 2 T −t

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CHAPTER 21. THE BLACK-SCHOLES EQUATION ¶ µ ⎤ 1 2 2 r − δ + σ ⎥ ⎢ √ 1 2 ⎢d1 − = T − t⎥ ⎦ ⎣ 2 (T − t) σ ⎡

¶ µ ⎤ 1 2 2 r − δ + σ ⎥ ⎢ √ √ 1 2 ⎢d2 + σ T − t − = T − t⎥ ⎦ ⎣ 2 (T − t) σ ⎡

¶ µ ⎤ 1 2 2 r−δ+ σ ⎢ ⎥ √ √ 1 2 ⎢d2 + σ T − t − = T − t⎥ ⎣ ⎦ 2 (T − t) σ ⎡

¶ µ ⎤ 1 2 2 r − δ − σ ⎢ ⎥ √ 1 2 ⎢d2 − ⎥ = T − t ⎦ 2 (T − t) ⎣ σ ⎡

¸ ∙ ∂ ∂ 0 rN (d2 ) + N (d2 ) = rV + e−r(T −t) N (d2 ) d2 ⇒ Vt = e ∂t ∂t ¶ µ ⎤ ⎡ 1 2 σ 2 r − δ − ⎥ ⎢ √ 1 0 2 ⎢d2 − = rV + e−r(T −t) N (d2 ) T − t⎥ ⎦ ⎣ 2 (T − t) σ −r(T −t)

0

VS = e−r(T −t) N (d2 )

0

N (d2 ) ∂ d2 = e−r(T −t) √ ∂S Sσ T − t

1 1 0 × =⇒ (r − δ) SVS = (r − δ) Se−r(T −t) N (d2 ) √ σ T −t S 1 (r − δ) e−r(T −t) 0 0 √ = N (d2 ) = (r − δ) e−r(T −t) N (d2 ) √ σ T −t σ T −t # " 0 # 0 e−r(T −t) ∂ N (d2 ) N (d2 ) √ = √ VSS = e S Sσ T − t σ T − t ∂S " 00 # 0 N (d2 ) e−r(T −t) N (d2 ) ∂ d2 − = √ S ∂S S2 σ T −t " # 0 0 N (d2 ) e−r(T −t) −d2 N (d2 ) 1 √ − = √ S S2 σ T −t Sσ T − t −r(T −t)

0

= −N (d2 )

∂ ∂S

"

e−r(T −t) √ S 2σ T − t

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µ

¶ ¢ √ d e−r(T −t) ¡ 0 √ 2 + 1 = −N (d2 ) 2 2 d2 + σ T − t S σ (T − t) σ T −t c °Yufeng Guo

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CHAPTER 21. THE BLACK-SCHOLES EQUATION 1 Vt + σ 2 S 2 VSS + (r − δ) SVS 2

¶ µ ⎤ 1 2 σ 2 r − δ − ⎢ ⎥ √ 1 0 2 ⎢d2 − = rV + e−r(T −t) N (d2 ) T − t⎥ ⎦ 2 (T − t) ⎣ σ ⎡

¢ √ 1 e−r(T −t) ¡ 0 − σ 2 S 2 N (d2 ) 2 2 d2 + σ T − t 2 S σ (T − t) 0 N (d2 ) = rV + (r − δ) Se−r(T −t) √ Sσ T − t Problem 21.7.

S (t) e−δ(T −t) N (d1 ) and e−δ(T −t) N (d2 ) each satisfy BS PDE. Hence their linear combination S (t) e−δ(T −t) N (d1 )−Ke−δ(T −t) N (d2 ) also satisfies BS PDE. You can verify on your own that S (t) e−δ(T −t) N (d1 )−Ke−δ(T −t) N (d2 ) satisfies the boundary condition when t = T . Problem 21.8. (r−δ)T The forward price is Se(r−δ)T ,µ then ¶ . If K = Se ¶ µ 1 2 S S 1 2 ln (r−δ)T + r − δ + σ T ln + r−δ+ σ T K 2 2 Se √ √ = d1 = σµ T σ T ¶ 1 2 − (r − δ) T + r − δ + σ T 1√ 2 √ = Tσ = 2 σ T √ √ 1√ 1√ Tσ − σ T = − Tσ d2 = d1 − σ T = 2 2 (r−δ)T a. Bet #1: Getting $1 if ST > K = Se and zero otherwise. Bet #2: Getting $1 if ST < K = Se(r−δ)T and zero otherwise.

Why Bet #1 is always worth less than Bet #2? Bet #1 is worth e−r(T −t) N (d2 ). Bet #2 is worth e−r(T −t) N (−d2 ). Since d2 < 0, we have −d2 > 0 > d2 . Since the normal cdf is increasing function, N (−d2 ) > N (d2 ). Hence Bet #2 is greater than Bet #1. b. To make the bet fair, we need to have P ∗ (ST > K) = N (d2 ) = 0.5 so there’ll be equal risk-neutral chance of ST >√ K and ST < K. To have N (d2 ) = 0.5, we need to have ¶ d1 = σ T . µ d2 = 0 and 1 2 S ¶ µ + r−δ+ σ T ln √ S 1 2 K 2 √ =σ T ln + r − δ + σ T = =⇒ d1 = K 2 σ T σ2T www.actuary88.com

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CHAPTER 21. THE BLACK-SCHOLES EQUATION ¶ µ S 1 ln = − r − δ − σ2 T K 2 #

r−δ−

So x = K = Se

1 2 σ 2

¶ µ K 1 ln = r − δ − σ2 T S 2

#

r−δ−

K = Se

1 2 σ 2

$

T

$

T

c. price of an asset call option is Se−r(T −t) N (d1 ). N (d1 ) is the expected fractional share of the stock *if* the call is exercised. To make the bet µfair, set N (d1¶) = 0.5. This gives us d1 = 0 S 1 ¶ µ ln + r − δ + σ2 T S 1 K 2 √ d1 = =0 ln + r − δ + σ2 T K 2 σ T ¶ ¶ µ µ S K 1 1 ln ln = − r − δ + σ2 T = r − δ + σ2 T K 2 S 2 # $ 1 2 r−δ+ σ T 2 =⇒ K = Se #

r−δ+

So we need to set x = K = Se

1 2 σ 2

$

T

Problem 21.9. ln Since K = Se(r−δ)T is the forward price of the stock, d1 = √ √ 1√ 1√ 1√ T σ and d2 = d1 − σ T = Tσ − σ T = − T σ. 2 2 2

S Se(r−δ)T

µ ¶ 1 + r − δ + σ2 T 2 √ = σ T

value´is V1 = Se−rT N (d1 )−Ke−rT N (d2 ) = Bet #1 ST´−K if ST > K. ³ pays ³ Its √ √ Se−δT N 0.5 T σ − Ke−rT N −0.5 T σ

−rT Bet #2 pays K − ST ³ if K > S . Its value N (−d2 ) − 2 = ´Ke ³ is V√ √ ´T −rT −δT Se N (−d1 ) = Ke N 0.5 T σ − Se N −0.5 T σ h³ √ ´ ³ ³ √ ´i h ³ √ ´i √ ´ V1 −V2 = Se−δT N 0.5 T σ + N −0.5 T σ −Ke−rT N −0.5 T σ + N 0.5 T σ −δT

Using the formula N (−x) = 1 − N (x), we get: V1 − V2 = Se−δT − Ke−rT = Se − e−rT Se(r−δ)T = 0 The two bets have the same value. −δT

Problem 21.10. Please note that the continuous payment rate Γ in this problem is not the option Gamma. To avoid confusion, we’ll use β to represent the option’s continuous payment rate. At time t www.actuary88.com

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CHAPTER 21. THE BLACK-SCHOLES EQUATION • We buy one option on the stock. We pay V • We buy N shares of the stock (a negative N means short selling stocks). We pay N S . • We deposit W means into a savings account. We pay W . To have zero-financing, we set our total initial cost to zero I = V + NS + W = 0

(DM 21.7)

Next, let’s consider the change of I during [t, t + dt]: dI = βdt + dV + N (dS + δSdt) + dW

(DM 21.8)

dI is the interest we earned during [t, t + dt]. Since W is invested in a savings account, we have dW = rW dt. This says that the interest earned on W during [t, t + dt] is rW dt. βdt is the payment we receive from the derivative during [t, t + dt]. Notice that the change of S is dS + δSdt (the sum of the change of the stock price dS and the dividend received δSdt). Apply Ito’s lemma: dV = Vt dt + VS dS + 0.5σ 2 S 2 VSS dt → dI = βdt + Vt dt + VS dS + 0.5σ 2 S 2 VSS dt + N (dS + δSdt) + dW = βdt + Vt dt + (VS + N ) dS + 0.5σ 2 S 2 VSS dt + N δSdt + rW dt Set N = −VS . Then dS term becomes zero and W = − (V − VS S). Because our initial cost is zero, the interest we earned dI should be zero. βdt + Vt dt + 0.5σ 2 S 2 VSS dt − VS δSdt − r (V − VS S) dt = 0 This gives us: β + Vt + 0.5σ 2 S 2 VSS − VS δS − r (V − VS S) = 0 ⇒ Vt + 0.5σ 2 S 2 VSS + (r − δ) VS S + β − rV = 0 Skip the remaining problems.

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CHAPTER 21. THE BLACK-SCHOLES EQUATION

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Chapter 22

Exotic options: II Skip all the problems.

203

CHAPTER 22. EXOTIC OPTIONS: II

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Chapter 23

Volatility Problem 23.1. This problem can be solved using the approach used in DM Table 11.1 (DM page 361). In other words, if you can reproduce DM Table 11.1, you should be able to solve this problem. Problem 23.2. This problem can be solved using the approach used in DM Table 11.1 (DM page 361). Problem 23.3. This problem is out of the scope of the exam MFE. However, if you want to solve it, you can use DM 23.6 to find the answer. Problem 23.4. Out of the scope of the exam MFE. See DM Example 23.2 if you want to know how to solve it. Problem 23.5. Out of the scope of the exam MFE. See DM Example 23.2 if you want to know how to solve it. Problem 23.6. 205

CHAPTER 23. VOLATILITY a. You can do this using the option price formula. Alternatively, you can find the call price using the put-call parity C + P V (K) = P + S. Since the maturity is very short, P is close to zero; there’s little chance that the stock price will drop from $100 to below $50 during T = 0.01. Hence C = S − P V (K) = 100 − 50e−0.06×0.01 = 50. 03. b. Using the spreadsheet in the CD that comes with the textbook, you should find vega is almost zero. Vega measures the sensitivity of the option price to volatility (see DM section 12.3). Vega is almost zero because the option maturity T = 0.01 is very short. It’s hard for the stock price to change significantly during this short amount of time (unless the volatility is very very big). c. Under 5% or 100% volatility, the option price is still about 50.03. The volatility isn’t huge enough to move the stock price during T = 0.01. However, under 500% volatility, the option price is 51.3. The volatility is huge; the stock price can change significantly during T = 0.01 d. It’s difficult to calculate the implied volatility for a call option that’s deep in the money and that has a short time to maturity. Problem 23.7. a. Using the put-call parity, we have C = P + S − P V (K) > S − P V (K) = 100 − 50e−0.06×0.01 = 50. 03. So the call is worth at east 50.03; it can never be worth less than 50. 03. Hence the bid price 50 is never possible. We can’t find the implied volatility under this bid price. b. Once again, we use the put-call parity. The ask price 50.1 is greater than the minimum value of the call 50.03. In order for the call to be worth more than 50. 03, the put value must be greater than zero. To make this happen, there must be some change that the stock price will drop from 100 to below 50 during T = 0.01. The only way this can happen is that the stock volatility is very high. c. Skip. d. It’s difficult to calculate the implied volatility for a deep in-the-money call with a short maturity. Problem 23.8. a. For the call to have any value, the stock price must move up from $50 and be greater than K = 100 at maturity. There’s little chance that the stock price will move up by that much during the short maturity T = 0.01 and under a small volatility of 30%. Hence the call price is virtually zero. www.actuary88.com

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CHAPTER 23. VOLATILITY b. The maturity is short. The option price is not very sensitive to the volatility. Hence vega is close to zero. c. The implied volatility is about 30% if the bid price is zero; the implied volatility must be huge if the ask price is 0.05 (there must be some chance that the stock price can move up by more than $50 at maturity in order for the call to be worth 0.05). d. The market maker wants to buy the option for $0 price and sell the option for $0.05. The market-maker thinks that the option isn’t worth much. At the same time, the market maker still wants to make a little profit. e. It’s difficult to calculate and interpret the implied volatility for a deep out-of-the-money option that has a short maturity. Skip the remaining problems.

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CHAPTER 23. VOLATILITY

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Chapter 24

Interest rate models Problem 24.1. a. We need to find the price of a 1-year bond issued at t = 1. The price of this bond is just the PV of $1 discounted from t = 2 to t = 1. P (0, 2) P (1, 2) = P (0, 1) To understand this formula, notice that P (0, 2) = P (1, 2) P (0, 1). This equation means that to calculate the PV of $1 discounted from t = 2 to t = 0, we first discount $1 from t = 2 to t = 1 and next discount it from t = 1 to t = 0. P (0, 2) 0.8495 P (1, 2) = = = 0.917 485 7 P (0, 1) 0.9259 b. C =Time zero cost of what you get at T ×N (d1 ) −Time zero cost of what you ³ give at T ×N (d2 ) ´ √ Time zero cost of what you get at T 2 d1 = ln Time + 0.5σ T /σ T zero cost of what you give at T

√ d2 = d1 − σ T Make sure you know the formula for σ. DM page 790 shows that p σ = V ar (ln (Ft,T (P [T, T + s]))) If you buy this option, then at T = 1, you can pay K = 0.9009 and buy a 1-year bond. This 1-year bond will give you $1 at time T + s = 2. • The value of this bond at T = 0 is P (0, 2) = 0.8495, the PV of $1 discounted from T + s = 2 to T = 1 • Time zero cost of the strike price K at T = 1 is just PV of K discounted from T = 1 to time zero. So P V (K) = 0.9009 × 0.9259 C =Time zero cost of what you get at T ×N (d1 ) −Time zero cost of what you give at T ×N (d2 ) 209

CHAPTER 24. INTEREST RATE MODELS → d1 =

ln

2 0.8495 0.9009×0.9259 +0.5×0.1 ×1

√ = 0.232 43 √ 0.1 1 √ → d2 = d1 − σ T = 0.232 43 − 0.1 1 = 0.132 43

N (d2 ) = 0.552 68 N (d1 ) = 0.591 90 C = 0.8495 × 0.591 90 − 0.9009 × 0.9259 × 0.552 68 = 0.0418 c. P =Time zero cost of what you give at T ×N (−d2 ) −Time zero cost of what you get at T ×N (−d1 ) P = 0.9009 × 0.9259 × (1 − 0.552 68) − 0.8495 × (1 − 0.591 90) = 0.0264 Alternative calculation using the put-call parity. C + P (0, T ) K = P + P (0, T + s) 0.0418 + 0.9009 × 0.9259 = P + 0.8495 P = 0.0264 d. Let’s walk through the notations and formula. Notation • RT =1 (T = 1, T + s = 2). The (not annualized) interest rate agreed upon at time T = 1 that applies to the time interval [T = 1, T + s = 2]. • Caplet. A Caplet gives the buyer the right to buy the time-T = 1 market interest rate RT (T, T + s) = RT =1 (T = 1, T + s = 2) by paying a fixed strike interest rate KR = 11%. If KR ≥ RT =1 (T = 1, T + s = 2), the caplet expires worthless. The payoff of the caplet at T + s = 2 is max [0, RT =1 (T = 1, T + s = 2) − 0.11]. The payoff of the caplet at T is max [0, RT =1 (T = 1, T + s = 2) − 0.11] RT =1 (T = 1, T + s = 2)

Please note that the actual interest rate during [T = 1, T + s = 2] is a random variable. One might be tempted to think that we already know the Year 2 interest rate as: 1 − 1 = 8. 993 5% P (1, 2) However, this thinking is flawed. 8. 993 5% is the implied Yr 2 interest rate based on the information available to us at t = 0. This rate can be different from the Year 2 spot rate. To calculate the price of the caplet, we first modify ∙the payoff: ¸ RT =1 (T = 1, T + s = 2) − 1.11 max [0, RT =1 (T = 1, T + s = 2) − 0.11] = 1.11 max 0, = RT =1 (T = 1, T + s = 2) 1.11 ∙ RT =1 (T = 1, T + s = 2) ¸ 1 1 1.11 max 0, − 1.11 RT =1 (T = 1, T + s = 2) www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS ∙ ¸ ∙ ¸ 1 1 1 max 0, − = max 0, 0.900 9 − 1.11 RT =1 (T = 1, T + s = 2) RT =1 (T = 1, T + s = 2) is the payoff of a put on a bond. This put gives the buyer the right, at T = 1, 1 to sell a bond that matures at T + s = 2 for a guaranteed price = 0.900 9. 1.11 From Part c, we already know that this put price is 0.0264. Hence the caplet price is: 1.11 × 0.0264 = 0.02 93 Problem 24.2. a. P (2, 3) =

P (0, 3) 0.7722 = = 0.909 01 P (0, 2) 0.8495

b. C = P (0, T + s) N (d1 ) − P (0, T ) KN (d2 ) P (0, T + s) = P (0, 2 + 1) = P (0, 3) = 0.7722 P (0, T ) = P (0, 2) = 0.8495 → C = 0.7722N (d1 ) − 0.8495 × 0.9N (d2 ) ln d1 =

P (0, T + s) 0.7722 + 0.5σ2 T ln + 0.5 × 0.1052 × 2 P (0, T ) K 0.8495 × 0.9 √ √ = = 0.141 29 0.105 × 2 σ T

√ √ d2 = d1 − σ T = 0.141 29 − 0.105 × 2 = −0.007 2 N (d1 ) = 0.556 18 N (d2 ) = 0.497 13

→ C = 0.7722 × 0.556 18 − 0.8495 × 0.9 × 0.497 13 = 0.04 94 c. P = P (0, T ) KN (−d2 ) − P (0, T + s) N (−d1 ) = 0.8495 × 0.9 × (1 − 0.497 13) − 0.7722 × (1 − 0.556 18) = 0.04175 1 Alternative calculation using the put-call parity. C + P (0, T ) K = P + P (0, T + s) 0.04 94 + 0.8495 × 0.9 = P + 0.7722 P = 0.041 75 d. To calculate the price of the caplet, we first modify the payoff: ∙ ¸ 1 1 max [0, RT =2 (T = 2, T + s = 3) − 0.11] = 1.11 max 0, − 1.11 RT =2 (T = 2, T + s = 3)¸ ¸ ∙ ∙ RT =2 (T = 2, T + s = 3) 1 1 1 − = max 0, 0.900 9 − max 0, 1.11 RT =2 (T = 2, T + s = 3) RT =2 (T = 2, T + s = 3) is the payoff of a put on a bond. This put gives the buyer the right, at T = 2, 1 to sell a bond that matures at T + s = 3 for a guaranteed price = 0.900 9. 1.11 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS 1 We can estimate the put price. Since = 0.900 9 is close to 0.9, from Part 1.11 c we know that P = 0.041 75. Hence the price of the caplet is 1.1×0.041 75 = 0.0 459 Or we can calculate the put price. P = P (0, T ) KN (−d2 ) − P (0, T + s) N (−d1 ) = 0.8495 × 0.900 9N (−d2 ) − 0.7722N (−d1 ) P (0, T + s) 0.7722 ln + 0.5σ 2 T ln + 0.5 × 0.1052 × 2 P (0, T ) K 0.8495 × 0.900 9 √ √ d1 = = = 0.105 × 2 σ T 0.134 56 √ √ d2 = d1 − σ T = 0.134 56 − 0.105 × 2 = −1. 393 2 × 10−2 N (−d1 ) = 0.446 48 N (−d2 ) = 0.505 56 P = 0.8495 × 0.900 9 × 0.505 56 − 0.7722 × 0.446 48 = 0.04214 Hence the price of the caplet is 1.1 × 0.04214 = 0.0464 Problem 24.3. Make sure you understand that this problem is different from SOA May 2007 #9. In SOA May 2007 #9, the first cash flow occurs at t = 1.The first year market rate 6% is below the cap rate 7.5%. At the end of Year 1 (at t = 1), we get nothing from the cap . In this problem, the author wants us to repeatedly use using DM Equation 24.36 to calculate the cap price. So the author wants the first cash flow to occur at t = 2. The cap contract is signed at t = 0. During Yr 2, the cap rate 11.5% is compared with the actual Yr 2 interest. The payoff at t = 2 is the max (0, 11.5% − Yr 2 rate). The PV of this payoff at t = 1 is: µ ¶ max (0, 11.5% − Yr 2 rate) 1 1 = 1.115 max 0, − 1 + Yr 2 rate 1.115 1 + Yr 2 rate µ ¶ 1 1 max 0, − is the payoff a put. This put gives the buyer 1.115 1 + Yr 2 rate the right at T = 1 to buy a one-year bond maturing at T + s = 2. You can verify that the time zero cost of this put is 0.0248. P = P (0, 1) KN (−d2 ) − P (0, 2) N (−d1 ) 0.8495 ln + 0.5 × 0.12 × 1 P (0, 2) 1 2 ln + 0.5σ T 0.9259 × P (0, 1) K 1.115 √ √ d1 = = = 0.277 36 0.1 × 1 σ T √ √ d2 = d1 − σ T = 0.277 36 − 0.1 × 1 = 0.177 36 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS 1 N (−0.177 36) − 0.8495N (−0.277 36) 1.115 1 = 0.9259 × × 0.429 61 − 0.8495 × 0.390 75 = 0.0248 1.115 So the time zero cost of the payoff at t = 2 is 1.115 × 0.0248. P = 0.9259 ×

Similarly, the time zero cost of the payoff at t = 3 is 1.115 × 0.0404 P = P (0, 2) KN (−d2 ) − P (0, 3) N (−d1 ) The put gives the buyer the right at T = 2 to buy a bond maturing at T +s=3 0.7722 + 0.5 × 0.1052 × 2 ln P (0, 3) 1 2 ln + 0.5σ T 0.8495 × P (0, 2) K 1.115 √ √ = d1 = = 0.164 82 0.105 × 2 σ T √ √ d2 = d1 − σ T = 0.164 82 − 0.105 × 2 = 0.01633 1 P = 0.8495 × N (−0.01633) − 0.7722N (−0.164 82) 1.115 1 = 0.8495 × × 0.493 49 − 0.7722 × 0.434 54 = 0.040 4 1.115 The time zero cost of the payoff at t = 4 is 1.115 × 0.0483 P = P (0, 2) KN (−d2 ) − P (0, 3) N (−d1 ) The put gives the buyer the right at T = 3 to buy a bond maturing at T +s=4 0.7020 ln + 0.5 × 0.112 × 3 P (0, 4) 1 ln + 0.5σ 2 T 0.7722 × P (0, 3) K 1.115 √ √ = d1 = = 0.166 35 0.11 × 3 σ T √ √ d2 = d1 − σ T = 0.166 35 − 0.11 × 3 = −0.02417 1 P = 0.7722 × N (0.02417) − 0.7020N (−0.166 35) 1.115 1 = 0.7722 × × 0.509 64 − 0.7020 × 0.433 94 = 0.04832 8 1.115 The time zero cost of the cap is: 1.115 × (0.0248 + 0.0404 + 0.0483) = 0.126 6 Problem 24.4. At time zero, we • But a 3-year bond. Cost: P (t, T2 ) = e−0.08×3 = 0.786 63 (T2 − t) P (t, T2 ) 1 × e−0.08×3 = −0.635 62 =− (T1 − t) P (t, T1 ) 2 × e−0.08×6 unit of 6-year bond (i.e. sell 0.635 62 unit of 6-year bond), receiving 0.635 62e−0.08×6 = 0.393 31

• To hedge, we buy N = −

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CHAPTER 24. INTEREST RATE MODELS • Borrow 0.786 63 − 0.393 31 = 0.393 32 from a bank at 8% interest rate. • Our net position is 0.786 63 − (0.393 31 + 0.393 32) = 0 If at the end of the day the risk-free rate is 8.25%, we’ll close off our position. • Sell the (3 − 1/365)-year bond, receiving e−0.0825×(3−1/365) = 0.780 93 • Buy back the (6 − 1/365)-year bond, paying 0.635 62e−0.0825×(6−1/365) = 0.387 54 • Pay off the loan, paying 0.393 32e0.08×1/365 = 0.393 41 (we borrow the loan for only one day). • Our net position: 0.780 93 − 0.387 54 − 0.393 41 = −0.000 02. So we lose 0.000 02

Why do we lose money? Because of the convexity mismatch. We buy the 3-year bond. The convexity is (see DM 7.14 and DM Example 7.9): 3×4 = 10. 288 07 1.082 We sell 0.635 62 unit of the 6-year bond. The convexity is: 6×7 = 22. 887 55 0.635 62 × 1.082 Even though we hedged the duration, we didn’t hedge convexity. The two portfolios have different convexities. As explained in my study guide, the highconvexity bond has a better value. To arbitrage, we should have bought the high-convexity bond and sold the low-convexity bond. So to arbitrage, we need to reverse our position (i.e. at t = 0, sell one unit of 3-year bond and buy 0.635 62 unit of 6-year bond). Then we’ll earn 0.000 02 free money during Day 1. How to reverse our position. At time zero, we • Sell a 3-year bond, receiving P (t, T2 ) = e−0.08×3 = 0.786 63 • To hedge, we buy 0.635 62 unit of 6-year bond, paying 0.635 62e−0.08×6 = 0.393 31 • Lend 0.786 63 − 0.393 31 = 0.393 32 at 8% interest rate. • Our net position is 0.786 63 − (0.393 31 + 0.393 32) = 0 At the end of the day, the risk-free rate is 8.25%. We close off our position. • Buy the (3 − 1/365)-year bond, paying e−0.0825×(3−1/365) = 0.780 93 • Sell the (6 − 1/365)-year bond, receiving 0.635 62e−0.0825×(6−1/365) = 0.387 54 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS • Close the loan, receiving 0.393 32e0.08×1/365 = 0.393 41. • Our net position: 0.387 54 + 0.393 41 − 0.780 93 = 0.000 02. So we gain 0.000 02 At time zero, our cost is zero. At the end of the day, we receive 0.000 02 profit. This is arbitrage. If at the end of the day the risk-free rate is 7.75%, we’ll close off our position. • Sell the (3 − 1/365)-year bond, receiving e−0.0775×(3−1/365) = 0.792 72 • Buy back the (6 − 1/365)-year bond, paying 0.635 62e−0.0775×(6−1/365) = 0.399 34 • Pay off the loan, paying 0.393 32e0.08×1/365 = 0.393 41 (we borrow the loan for only one day). • Our net position: 0.792 72 − 0.399 34 − 0.393 41 = −0.000 03. So we lose 0.000 03

To arbitrage, we reverse out position and gain 0.000 03 free money. This example shows that the assumption of the parallel shift of a flat yield curve leads to arbitrage. To build a good model for bond price, we need to throw away this bad assumption. Problem 24.5. a.4-year 5% annual ¡ coupon bond with yield 6% (Bond 1) ¢ Price: P1 = 0.05 e−0.06×1 + e−0.06×2 + e−0.06×3 + e−0.06×4 + e−0.06×4 = 0.959 16 Macaulay duration: ¢ ¡ 0.05 e−0.06×1 + 2e−0.06×2 + 3e−0.06×3 + 4e−0.06×4 + 4e−0.06×4 D1 = = 3. 0.959 16 716 7 8-year 7% annual coupon bond with yield 6% (Bond 2) Price: ¡ ¢ P2 = 0.07 e−0.06×1 + e−0.06×2 + e−0.06×3 + e−0.06×4 + ... + e−0.06×8 +e−0.06×8 = 1. 050 3 Macaulay duration: ¢ ¡ 0.07 e−0.06×1 + 2e−0.06×2 + 3e−0.06×3 + ... + 8e−0.06×8 + 8e−0.06×8 D2 = = 1. 050 3 6. 433 2 b. Use DM 7.13: D1 B1 (y1 ) (1 + y1 ) N =− D2 B2 (y2 ) (1 + y2 ) www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS Please also note that DM 24.7 is the special case of DM 7.13. If the two bonds each have one cash flow and two bonds have the same yield (i.e. y1 = y2 ), then DM 7.13 becomes DM 24.7. In this problem, the yield is the same for the two bonds. So y1 = y2 . D1 P1 (y) 3. 716 7 × 0.959 16 D1 B1 (y1 ) (1 + y1 ) =− =− = −0.527 6 N =− D2 B2 (y2 ) (1 + y2 ) D2 P2 (y) 6. 433 2 × 1. 050 3 we

So we need to buy −0.527 6 (i.e. sell 0.527 6) unit of Bond 2. At time zero, • sell 0.527 6 unit of Bond 2, receiving 0.527 6 × 1. 050 3 = 0.554 14 • buy one Bond 1, paying 0.959 16 • borrow the difference 0.959 16 − 0.554 14 = 0.405 02 from a bank 0.959 16 − 0.527 6 × 1. 050 3 = 0.405 021 72 Our net position is zero.

If at the end of the day, the yield is 6.25%: Back back 0.527 6 unit of Bond 2, which has (8 − 1/365) year to maturity. Now we are at t = 1/365. Bond 2’s price now is: ¡ standing ¡ ¢ ¢ P2 = 0.07 e−0.0625×1 + e−0.0575×2 + e−0.0625×3 + ... + e−0.0625×8 + e−0.0625×8 e0.0625/365 = 1. 034 39 Sell Bond 1, which has (4 − 1/365) year to maturity The price is¡ ¡ ¢ ¢ P1 = 0.05 e−0.0.0625×1 + e−0.0625×2 + e−0.0625×3 + e−0.0625×4 + e−0.0625×4 e0.0625/365 = 0.953 48 Repay the bank: 0.405 02e0.06/365 = 0.405 09 The net profit at the end of Day 1 is: 0.953 48−0.527 6×1. 034 39−0.405 09 = 0.002 65 If at the end of the day, the yield is 5.75%: Back back 0.527 6 unit of Bond 2, which has (8 − 1/365) year to maturity. Now we are standing at t = 1/365. Bond 2’s price now is: £ ¤ P2 = 0.07 e−0.0575×(1−1/365) + e−0.0575×(2−1/365) + ... + e−0.0575×(8−1/365) + e−0.0575×(8−1/365) ¡ ¡ ¢ ¢ = 0.07 e−0.0575×1 + e−0.0575×2 + e−0.0575×3 + ... + e−0.0575×8 + e−0.0575×8 e0.0575/365 = 1. 067 54 Sell Bond 1, which has (4 − 1/365) year to maturity The price is¡ ¡ ¢ ¢ P1 = 0.05 e−0.0575×1 + e−0.0575×2 + e−0.0575×3 + e−0.0575×4 + e−0.0575×4 e0.0575/365 = 0.968 27 Repay the bank: 0.405 02e0.06/365 = 0.405 09 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS The net profit at the end of Day 1 is: 0.968 27−0.527 6×1. 067 54−0.405 09 = −0.000 05 If we reserve our position, we can have 0.000 05 at the end of Day 1. Refer to Problem 24.4 to see how to reverse our position. Summary: If the yield moves up to 6.25% in Day 1, we can make 0.002 65 profit; if the yield moves down to 5.75% in Day 1, we can make 0.000 05 profit. Once again, the assumption of the flat yield curve leads to arbitrage. Problem 24.6. Vasicek model: a. Zero risk premium means that α (r, t, T ) = r. Hence φ (r, t) = α (r, t, T ) − r =0 q (r, t, T ) σ2 0.12 r = b − 0.5 2 = 0.1 − 0.5 × = −0.025 DM 24.26: α 0.22 2-year bond: 1 − e−α(T −t) 1 − e−0.2(2) B (T − t = 2) = aT −t|α = = = 1. 648 4 α 0.2 aT −t|α is a T − t year continuous annuity with the force of interest α. 2

2

2

2

A (T − t = 2) = er[B(2)−2]−B (2)σ /4α = e−0.025(1. 648 4−2)−1. 648 4 ×0.1 /(4×0.2) = 0.975 14 The 2-year bond is worth: P (0, 2) = A (2) e−B(2)r = 0.975 14e−1. 648 4×0.05 = 0.897 99 The delta is: d d P (0, 2) = A (2) e−B(2)r = −B (2) A (2) e−B(2)r = −B (2) P (0, 2) = dr dr −1. 648 4 × 0.897 99 = −1. 480 2 The gamma is: d2 d P (0, 2) = −B (2) P (0, 2) = B 2 (2) P (0, 2) = 1. 648 42 × 0.897 99 = 2. 2 dr dr 44 10-year bond: 1 − e−0.2×10 = 4. 323 3 0.2 2 2 r[B(10)−10]−B 2 (10)σ 2 /4α = e−0.025(4. 323 3−10)−4. 323 3 ×0.1 /(4×0.2) = A (10) = e 0.912 36 The 10-year bond is worth: P (0, 10) = A (10) e−B(10)r = 0.912 36e−4. 323 3×0.05 = 0.735 The delta is: d P (0, 10) = −B (10) P (0, 10) = −4. 323 3 × 0.735 = −3. 177 6 dr The gamma is: d2 P (0, 10) = B 2 (10) P (0, 10) = 4. 323 32 × 0.735 = 13. 738 dr2 B (10) = a10|0.2 =

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CHAPTER 24. INTEREST RATE MODELS The delta is: d d P (0, 2) = A (2) e−B(2)r = −B (2) A (2) e−B(2)r = −B (2) P (0, 2) = dr dr −1. 648 4 × 0.897 99 = −1. 480 2 The gamma is: d d2 P (0, 2) = −B (2) P (0, 2) = B 2 (2) P (0, 2) = 1. 648 42 × 0.897 99 = 2. dr2 dr 44 b. If we buy a 2-year bond, to duration hedge our risk, we need to buy 2P (0, 2) 2 × 0.897 99 =− = −0.244 35 10P (0, 10) 10 × 0.735 So we need to sell 0.244 35 unit of 10-year bond.

N =−

At t = 0, we • buy a 2-year bond, paying P (0, 2) = 0.897 99 • sell 0.244 35 unit of 10-year bond, receiving 0.244 35 × 0.735 = 0.179 60 • borrow 0.897 99 − 0.179 60 = 0.718 39 at 5% • Our net position is zero. p The one pstandard deviation of the interest rate under Vasicek is r±σ 1/365 = 0.05 ± 0.1 1/365 p ru = 0.05 + 0.1p 1/365 = 0.055 rd = 0.05 − 0.1 1/365 = 0.045 p Under ru = 0.05 + 0.1 1/365 = 0.055 2 − 1/365 year bond: 1 − e−α(T −t) 1 − e−0.2(2−1/365) B (2 − 1/365) = = = 1. 647 α 0.2 −0.025(1. 646 7−2+1/365)−1. 6472 ×0.12 /(4×0.2) = 0.975 2 A (2 − 1/365) = e The 2 − 1/365-year bond is worth: P (0, 2 − 1/365) = 0.975 2e−1. 6467×0.055 = 0.890 8 10 − 1/365-year bond: 1 − e−0.2×(10−1/365) B (10 − 1/365) = = 4. 323 0 0.2 2 2 r[B(10−1/365)]−B 2 (10−1/365)σ 2 /4α = e−0.025(4. 323 −10+1/365)−4. 323 ×0.1 /(4×0.2) = A (10 − 1/365) = e 0.912 34 The 10 − 1/365-year bond is worth: P (0, 10 − 1/365) = A (10 − 1/365) e−B(10−1/365)r = 0.912 34e−4. 323 ×0.055 = 0.719 3 At the end of the day, we close our position: www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS • sell a 2 − 1/365-year bond, receiving 0.890 8 • buy 0.244 35 unit of 10−1/365-year bond, paying 0.244 35 × 0.719 3 = 0.175 8 • pay back the loan, paying 0.718 39e0.05/365 = 0.718 5 • our net position is: 0.890 8 − 0.175 8 − 0.718 5 = −0.003 5 • So we lose 0.003 5 p Under rd = 0.05 − 0.1 1/365 = 0.045 2 − 1/365 year bond: 1 − e−α(T −t) 1 − e−0.2(2−1/365) B (2 − 1/365) = = = 1. 647 α 0.2 −0.025(1. 646 7−2+1/365)−1. 6472 ×0.12 /(4×0.2) = 0.975 2 A (2 − 1/365) = e The 2 − 1/365-year bond is worth: P (0, 2 − 1/365) = 0.975 2e−1. 6467×0.045 = 0.905 6 10 − 1/365-year bond: 1 − e−0.2×(10−1/365) B (10 − 1/365) = = 4. 323 0 0.2 2 2 r[B(10−1/365)]−B 2 (10−1/365)σ 2 /4α = e−0.025(4. 323 −10+1/365)−4. 323 ×0.1 /(4×0.2) = A (10 − 1/365) = e 0.912 34 The 10 − 1/365-year bond is worth: P (0, 10 − 1/365) = A (10 − 1/365) e−B(10−1/365)r = 0.912 34e−4. 323 ×0.045 = 0.751 1 At the end of the day, we close our position: • sell a 2 − 1/365 year bond, receiving 0.905 6 • buy 0.244 35 unit of 10 − 1/365 year bond, paying 0.244 35 × 0.751 1 = 0.183 5 • pay back the loan, paying 0.718 39e0.05/365 = 0.718 5 • our net position is: 0.905 6 − 0.183 5 − 0.718 5 = 0.003 6 • So we gain 0.003 6 Delta hedge. At t = 0, we buy a 2-year bond and buy N units of 10-year bond. We already calculated the following: • The delta of the 2-year bond is −1. 480 2 • The delta of the 10-year bond is −3. 177 6 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS The total delta of the portfolio is: −1. 480 2 + N (−3. 177 6). To delta hedge, set −1. 480 2 + N (−3. 177 6) = 0 → N = −0.465 8 So at t = 0, we • buy a 2-year bond, paying P (0, 2) = 0.897 99 • sell 0.465 8 unit of 10-year bond, receiving 0.465 8 × 0.735 = 0.342 36 • borrow 0.897 99 − 0.342 36 = 0.555 63 at 5% • Our net position is zero.

p Under ru = 0.05 + 0.1 1/365 = 0.055 The 2 − 1/365-year bond is worth: P (0, 2 − 1/365) = 0.890 8 The 10 − 1/365-year bond is worth: P (0, 10 − 1/365) = 0.719 3 At the end of the day, we close our position: • sell a 2 − 1/365-year bond, receiving 0.890 8 • buy 0.465 8 unit of 10−1/365-year bond, paying 0.465 8×0.719 3 = 0.335 05 • pay back the loan, paying 0.555 63e0.05/365 = 0.555 71 • our net position is: 0.890 8 − 0.335 05 − 0.555 71 = 0.000 04 • So we gain 0.000 04 p Under rd = 0.05 − 0.1 1/365 = 0.045 The 2 − 1/365-year bond is worth: P (0, 2 − 1/365) = 0.905 6 The 10 − 1/365-year bond is worth: P (0, 10 − 1/365) = 0.751 1 At the end of the day, we close our position: • sell a 2 − 1/365 year bond, receiving 0.905 6 • buy 0.465 8 unit of 10−1/365 year bond, paying 0.465 8×0.751 1 = 0.349 86 • pay back the loan, paying 0.555 63e0.05/365 = 0.555 71 • our net position is: 0.905 6 − 0.349 86 − 0.555 71 = 0.000 03 • So we gain 0.000 03 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS CIR: a. Zero risk premium means that α (r, t, T ) = r. Hence φ (r, t) = 0

α (r, t, T ) − r = q (r, t, T )

√ DM 24.28: φ (r, t) = φ r/σ. Hence φ = 0 DM q 24.29: q 2 2 γ = (a + φ) + 2σ 2 = (0.2 + 0) + 2 × 0.447212 = 0.663 32

a + φ + γ = 0.2 + 0 + 0.663 32 = 0.863 32 2-year bond: "

2γe(a+φ+γ )(T −t)/2 ¢¡ ¢ A (T − t = 2) = ¡ a + φ + γ eγ(T −t) − 1 + 2γ ∙

2 × 0.663 32e0.863 32×2/2 = 0.863 32 (e0.663 32×2 − 1) + 2 × 0.663 32

# 2ab σ2

¸

2 × 0.2 × 0.1 0.447212

= 0.967 18

¢ ¡ 2 eγ(T −t) − 1 ¢¡ ¢ B (T − t = 2) = ¡ a + φ + γ eγ(T −t) − 1 + 2γ

¢ ¡ 2 e0.663 32×2 − 1 = = 1. 489 72 0.863 32 (e0.663 32×2 − 1) + 2 × 0.663 32

P (0, 2) = A (2) e−B(2)r = 0.967 18e−1. 489 72×0.05 = 0.897 76 The delta is: d d P (0, 2) = A (2) e−B(2)r = −B (2) P (0, 2) = −1. 489 72 × 0.897 76 = dr dr −1. 337 41 The gamma is: d d2 P (0, 2) = −B (2) P (0, 2) = B 2 (2) P (0, 2) = 1. 489 722 × 0.897 76 = dr2 dr 1. 992 37 10-year bond: 2 × 0.2 × 0.1 ¸ 2 × 0.663 32e0.863 32×10/2 0.447212 = A (T − t = 10) = 0.663 32×10 0.863 32 (e − 1) + 2 × 0.663 32 0.685 54 ¢ ¡ 2 e0.663 32×10 − 1 B (T − t = 10) = = 2. 311 96 0.863 32 (e0.663 32×10 − 1) + 2 × 0.663 32 ∙

P (0, 10) = A (10) e−B(10)r = 0.685 54e−2. 311 96×0.05 = 0.610 70 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS The delta is: d P (0, 10) = −B (10) P (0, 10) = −2. 311 96 × 0.610 70 = −1. 411 91 dr The gamma is: d2 P (0, 10) = B 2 (10) P (0, 10) = 2. 311 962 × 0.610 70 = 3. 264 29 dr2 b. If we buy a 2-year bond, to duration hedge our risk, we need to buy N units of 10 year bond. 2P (0, 2) 2 × 0.897 76 =− = −0.294 01 10P (0, 10) 10 × 0.610 70 So we need to sell 0.294 01 unit of 10-year bond.

N =−

At t = 0, we • buy a 2-year bond, paying P (0, 2) = 0.897 76 • sell 0.294 01 unit of 10-year bond, receiving 0.294 01 × 0.610 70 = 0.179 55 • borrow 0.897 76 − 0.179 55 = 0.718 21 at 5% • Our net position is zero. √ Notice that under CIR, dr = a (b − r) dt√+p σ rdz. The one standard √ deviation of the interest rate under CIR is r ± σ r 1/365 = 0.05 ± 0.44721 0.05 × p 1/365 p √ 0.44721 0.05 × 1/365 = 0.005 234 2 ru = 0.05 + 0.005 234 2 = 0.055 23 rd = 0.05 − 0.005 234 2 = 0.044 77 Under ru = 0.055 23 The 2 − 1/365-year bond is worth: "

0.863 32×(2−1/365)/2

# 2 × 0.2 × 0.1 0.447212

2 × 0.663 32e ¡ ¢ 0.863 32 e0.663 32×(2−1/365) − 1 + 2 × 0.663 32 0.967 26 ¡ ¢ 2 e0.663 32×(2−1/365) − 1 ¡ ¢ B (T − t = 2 − 1/365) = = 0.863 32 e0.663 32×(2−1/365) − 1 + 2 × 0.663 32 1. 488 40 A (T − t = 2 − 1/365) =

→ P (0, 2 − 1/365) = A (2 − 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.055 23 = 0.890 93 The 10 − 1/365-year bond is worth: www.actuary88.com

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=

CHAPTER 24. INTEREST RATE MODELS

A (T − t = 10 − 1/365) = 0.685 63

"

0.863 32×(10−1/365)/2

2 × 0.663 32e ¡ ¢ 0.863 32 e0.663 32×(10−1/365) − 1 + 2 × 0.663 32

# 2 × 0.2 × 0.1 0.447212

¡ ¢ 2 e0.663 32×(10−1/365) − 1 ¡ ¢ B (T − t = 10 − 1/365) = = 0.863 32 e0.663 32×(10−1/365) − 1 + 2 × 0.663 32 2. 311 95 → P (0, 10 − 1/365) = 0.685 63e−2. 311 95×0.055 23 = 0.603 44 At the end of the day, we close our position: • sell a 2 − 1/365-year bond, receiving 0.890 93 • buy 0.294 01 unit of 10 − 1/365-year bond, paying 0.294 01 × 0.603 44 = 0.177 42 • pay back the loan, paying 0.718 21e0.05/365 = 0.718 31 • our net position is: 0.890 93 − 0.177 42 − 0.718 31 = −0.004 8 • So we lose 0.004 8 Under rd = 0.044 77 P (0, 2 − 1/365) = A (2 − 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.044 77 = 0.904 91 P (0, 10 − 1/365) = 0.685 63e−2. 311 95×0.044 77 = 0.618 21 At the end of the day, we close our position: • sell a 2 − 1/365-year bond, receiving 0.904 91 • buy 0.294 01 unit of 10 − 1/365-year bond, paying 0.294 01 × 0.618 21 = 0.181 76 • pay back the loan, paying 0.718 21e0.05/365 = 0.718 31 • our net position is: 0.904 91 − 0.181 76 − 0.718 31 = 0.004 84 • So we gain 0.004 84 Delta hedge: −1. 337 41 + N (−1. 411 91) = 0 N = −0.947 23 So we need to sell 0.947 23 unit of the 10-year bond. At t = 0, we www.actuary88.com

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=

CHAPTER 24. INTEREST RATE MODELS • buy a 2-year bond, paying P (0, 2) = 0.897 76 • sell 0.947 23 unit of 10-year bond, receiving 0.947 23 × 0.610 70 = 0.578 47 • borrow 0.897 76 − 0.578 47 = 0.319 29 at 5% • Our net position is zero. One day later, under ru = 0.055 23 P (0, 2 − 1/365) = A (2 − 1/365) e−B(2−1/365)r = 0.967 26e−1. 488 40×0.055 23 = 0.890 93 P (0, 10 − 1/365) = 0.685 63e−2. 311 95×0.055 23 = 0.603 44 At the end of the day, we close our position: • sell a 2 − 1/365-year bond, receiving 0.890 93 • buy 0.947 23 unit of 10 − 1/365-year bond, paying 0.947 23 × 0.603 44 = 0.571 60 • pay back the loan, paying 0.319 29e0.05/365 = 0.319 33 • our net position is: 0.890 93 − 0.571 60 − 0.319 33 = 0 • So we gain 0 Under rd = 0.044 77 P (0, 2 − 1/365) = 0.967 26e−1. 488 40×0.044 77 = 0.904 91 P (0, 10 − 1/365) = 0.685 63e−2. 311 95×0.044 77 = 0.618 21 At the end of the day, we close our position: • sell a 2 − 1/365-year bond, receiving 0.904 91 • buy 0.947 23 unit of 10 − 1/365-year bond, paying 0.947 23 × 0.618 21 = 0.585 59 • pay back the loan, paying 0.319 29e0.05/365 = 0.319 33 • our net position is: 0.904 91 − 0.585 59 − 0.319 33 ≈ 0 • So we gain 0 Problem 24.7. www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS t=0

t=1

t=2

t=3 0.16

0.04 0.12 0.12 0.12 0.10 0.08 0.10 0.12 0.10 0.08 0.08 0.08 0.06 0.04 Price of 1-Yr bond: P (0, 1) = e−0.1 = 0.904 84 The yield for 1-Yr bond is 0.1. Price of Path 0→u 0→d

2-Yr bond: Prob Price 0.5 e−0.12 e−0.1 0.5 e−0.08 e−0.1

P (0, 2) = 0.5e−0.12 e−0.1 + 0.5e−0.08 e−0.1 = 0.818 89 The yield for 2-Yr bond is y ln 0.818 89 y=− = 9.990 3% e−2y = 0.818 89 2 Price of 3-Yr bond: Path Prob 0 → u → uu 0.25 0 → u → ud 0.25 0 → d → du 0.25 0 → d → dd 0.25

Price e−0.14 e−0.12 e−0.1 e−0.1 e−0.12 e−0.1 e−0.1 e−0.08 e−0.1 e−0.06 e−0.08 e−0.1

¡ ¢ P (0, 5) = 0.25 e−0.14 e−0.12 e−0.1 + e−0.1 e−0.12 e−0.1 + e−0.1 e−0.08 e−0.1 + e−0.06 e−0.08 e−0.1 = 0.741 56 1 Yield for the 3-Yr bond: − ln 0.741 56 = 9. 966 6% 3 Price of 4-Yr bond: www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS Path 0 → u → uu → uuu 0 → u → uu → uud 0 → u → ud → udu 0 → u → ud → udd 0 → d → du → duu 0 → d → du → dud 0 → d → dd → ddu 0 → d → dd → ddd

Prob 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

Price e−0.16 e−0.14 e−0.12 e−0.1 e−0.12 e−0.14 e−0.12 e−0.1 e−0.12 e−0.10 e−0.12 e−0.1 e−0.08 e−0.10 e−0.12 e−0.1 e−0.12 e−0.10 e−0.08 e−0.1 e−0.08 e−0.10 e−0.08 e−0.1 e−0.08 e−0.06 e−0.08 e−0.1 e−0.04 e−0.06 e−0.08 e−0.1

= 0.594 52 = 0.618 78 = 0.644 04 = 0.670 32 = 0.670 32 = 0.697 68 = 0.726 15 = 0.755 78

0.594 52 + 0.618 78 + 0.644 04 + 0.670 32 + 0.670 32 + 0.697 68 + 0.726 15 + 0.755 78 P (0, 4) = = 8 0.672 20 1 Yield for the 4-Yr bond: − ln 0.672 20 = 9. 930 0% 4 Yields decline with maturity. This is explained in DM page 796. The average of the exponential interest rates is less than the exponentiated average. Problem 24.8. Instead of working path by path, we can work backwards. At t = 3, the bond price is: Vuuu = 0.8331 Vuud = 0.8644 Vddu = 0.8906 At t = 2, the bond price is: 0.8331 + 0.8644 Vuu = 0.8321 × = 0.706 24 2 Vud = Vdu = 0.8798 × Vdd = 0.9153 ×

Vddd = 0.9123

0.8644 + 0.8906 = 0.772 02 2

0.8906 + 0.9123 = 0.825 10 2

At t = 1, the bond price is: 0.706 24 + 0.772 02 Vu = 0.8832 × = 0.652 80 2 Vd = 0.9023 ×

0.772 02 + 0.825 10 = 0.720 54 2

At t = 0, the bond price is: 0.652 80 + 0.720 54 Vu = 0.9091 × = 0.624 25 = 0.6243 2 Problem 24.9. www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS Next year, the bond price is either Vu = 0.652 80 with a yield of 0.652 80−1/3 − 1 = 0.152 76 or Vd = 0.720 54 with a yield of 0.720 54−1/3 − 1 = 0.115 44. 0.152 76 Using DM 24.48, the yield volatility is 0.5 ln = 0.140 06 = 0.14 0.115 44 Problem 24.10. See DM page 805. Problem 24.11. P (0, 3) 0.7118 −1 = −1 = 0.140 16. P (0, 4) 0.6243 Since the interest rate cap applies during the interval [t = 3, t = 4], the fair cap rate must be the implied forward rate rA = 0.140 16 during [t = 3, t = 4] . We can verify that rA = 0.140 16 using the binomial tree. Let r (3, 4) represent the actual interest rate during the interval [t = 3, t = 4]. The reference rate rA is the 1-year forward rate 3 years hence. So rA is also the rate during [t = 3, t = 4]. This is the difference between r (3, 4) and rA . r (3, 4) is the actual interest rate observed in the market during Year 3; rA is rate agreed upon at t = 0 that applies to Year 3. Set the notional principal to $100. At t = 4, the payoff is 100 [r (3, 4) − rA ] = 100r (3, 4) − 100rA . We need to find rA such that the PV of the payoff is zero. First, we calculate PV of 100r (3, 4). At t = 4, 100r (3, 4) has 4 possible values: 20.03,15.68,12.28, and 9.62. Discounting these 4 values to t = 3, we get the 4 values: 100 × 0.2003 100 × 0.1568 = 16. 687 = 13. 555 1 + 0.2003 1 + 0.1568 DM Page 806 and 807 explain that rA =

100 × 0.1228 = 10. 937 1 + 0.1228

100 × 0.0962 = 8. 775 8 1 + 0.0962

We have 3 values at t = 2 (using 0.5 as the risk-neutral probability of up or down) µ ¶ 16. 687 13. 555 Vuu = 0.5 + = 12. 583 1 + 0.2017 1 + 0.2017 Vud = 0.5

Vdd = 0.5

µ

µ

13. 555 10. 937 + 1 + 0.1366 1 + 0.1366 10. 937 8. 775 8 + 1 + 0.0925 1 + 0.0925

¶

¶

= 10. 774

= 9. 021 9

The valueµat t = 1: ¶ 12. 583 10. 774 + = 10. 315 Vu = 0.5 1 + 0.1322 1 + 0.1322 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS

Vd = 0.5

µ

10. 774 9. 021 9 + 1 + 0.1082 1 + 0.1082

¶

= 8. 931 6

The value µ at t = 0: ¶ 10. 315 8. 931 6 V = 0.5 + = 8. 748 5 1 + 0.1 1 + 0.1 So PV of 100r (3, 4) is 8. 748 5. PV of 100rA is 100rA × P (0, 4) = 100rA × 0.6243 P (0, 4) = 0.6243 is from DM Table 24.2. 100rA × 0.6243 = 8. 748 5

rA = 0.140 13

Problem 24.12. I just solve for Tree #1. Once you understand the logic, you can do Tree #2. 1-Yr bond price: P (0, 1) =

1 = 0.925 93 1.08

2-Yr bond price: 1 1 Vd = Vu = 1 + 0.07676 1 + 0.10363 µ ¶ Vu + Vd 1 1 1 V = 0.5 × = 0.5 × + = 0.849 45 1.08 1.08 1 + 0.07676 1 + 0.10363 3-Yr bond price: 1 Vdd = 1 + 0.08170

1 1 + 0.13843 µ ¶ 1 1 1 Vd = 0.5 × + = 0.849 1 + 0.07676 µ 1 + 0.08170 1 + 0.10635 ¶ 1 1 1 Vu = 0.5 × + = 0.807 46 1 + 0.10363 1 + 0.10635 1 + 0.13843 1 (0.849 + 0.807 46) = 0.766 88 V = 0.5 × 1.08 Vud =

1 1 + 0.10635

Vuu =

4-Yr bond price: Value at t = 3 1 1 + 0.07943

1 1 + 0.09953

1 1 + 0.12473

1 1 + 0.15630

Value at t = 2 Vdd = 0.5 ×

1 1 + 0.08170

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µ

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¶

= 0.848 62 228

CHAPTER 24. INTEREST RATE MODELS µ ¶ 1 1 1 + = 0.812 84 1 + 0.10635 µ 1 + 0.09953 1 + 0.12473 ¶ 1 1 1 = 0.5 × + = 0.770 33 1 + 0.13843 1 + 0.12473 1 + 0.15630

Vud = 0.5 × Vuu

Value at t = 1 1 (0.848 62 + 0.812 84) = 0.771 51 Vd = 0.5 × 1 + 0.07676 1 (0.812 84 + 0.770 33) = 0.717 26 Vu = 0.5 × 1 + 0.10363 Value at t = 0 V = 0.5 ×

1 (0.771 51 + 0.717 26) = 0.689 25 1.08

5-Yr bond price: value at t = 4 1 1 1 + 0.07552 1 + 0.09084 1 1 1 + 013143 1 + 0.15809 value at t = 3 µ 1 1 0.5 × 1 + 0.07943 µ 1 + 0.07552 1 1 0.5 × 1 + 0.09953 µ 1 + 0.09084 1 1 0.5 × 1 + 0.12473 µ 1 + 0.10927 1 1 0.5 × 1 + 0.15630 1 + 0.13143

1 1 + 0.10927 ¶ 1 = 0.855 32 1 + 0.09084 ¶ 1 + = 0.826 82 1 + 0.10927 ¶ 1 + = 0.793 67 1 + 0.13143 ¶ 1 + = 0.755 57 1 + 0.15809 +

value at t = 2 1 (0.855 32 + 0.826 82) = 0.777 54 Vdd = 0.5 × 1 + 0.08170 1 Vud = 0.5 × (0.826 82 + 0.793 67) = 0.732 36 1 + 0.10635 1 (0.793 67 + 0.755 57) = 0.680 43 Vuu = 0.5 × 1 + 0.13843 value at t = 1 1 Vd = 0.5 × (0.777 54 + 0.732 36) = 0.701 13 1 + 0.07676 1 (0.732 36 + 0.680 43) = 0.640 07 Vu = 0.5 × 1 + 0.10363 Value at t = 0 1 V = 0.5 × (0.701 13 + 0.640 07) = 0.620 93 1.08 www.actuary88.com

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CHAPTER 24. INTEREST RATE MODELS Problem 24.13. I’m going to solve for only Tree #1. DM Page 799 states that the volatility in Year 1 is the standard deviation of the natural log of the yield for that bond 1 year hence. So the volatility in Yr 1 for an n-year bond is the standard deviation of the natural log of the yield of an (n − 1)-year bond issued at t = 1. Yr-1 yield for 1-yr bond is unknown. Yr-1 yield for 2-yr bond . The up yield of an 2 − 1 = 1 year bond issued at t = 1 is ru = 0.10362; the down yield is 0.07676.Using DM 24.48, we get: 0.10362 ru = 0.5 ln = 0.150 02 0.5 ln rd 0.07676 Yr-1 yield for a 3-yr bond. We first calculate the price of a 2-yr bond issued at t = 1. From the previous problem µ regarding the price of the¶ 3-year bond, we know 1 1 1 + = 0.849 Vd = 0.5 × 1 + 0.07676 µ 1 + 0.08170 1 + 0.10635 ¶ 1 1 1 + = 0.807 46 Vu = 0.5 × 1 + 0.10363 1 + 0.10635 1 + 0.13843 ru = 0.807 46−1/2 − 1 = 0.112 86 rd = 0.849−1/2 − 1 = 0.08529 1 Volatility: 0.5 ln

ru 0.112 85 = 0.5 ln = 0.140 00 rd 0.08529 1

Yr-1 yield for a 4-yr bond. We first calculate the price of a 3-yr bond issued at t = 1. From the previous problem about the price of a 4-yr bond, 1 (0.848 62 + 0.812 84) = 0.771 51 Vd = 0.5 × 1 + 0.07676 −1/3 − 1 = 9. 031 7 × 10−2 rd = 0.771 51 1 Vu = 0.5 × (0.812 84 + 0.770 33) = 0.717 26 1 + 0.10363 rd = 0.717 26−1/3 − 1 = 0.117 14 ru 0.117 14 = 0.5 ln = 0.130 02 rd 9. 031 7 × 10−2 Yr-1 yield for a 5-yr bond. We first calculate the price of a 4-yr bond issued at t = 1. From the previous problem about the price of a 5-yr bond, 1 Vd = 0.5 × (0.777 54 + 0.732 36) = 0.701 13 1 + 0.07676 rd = 0.701 13−1/4 − 1 = 9. 282 4 × 10−2 Volatility: 0.5 ln

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CHAPTER 24. INTEREST RATE MODELS 1 (0.732 36 + 0.680 43) = 0.640 07 1 + 0.10363 −1/4 − 1 = 0.118 ru = 0.640 07 Vu = 0.5 ×

ru 0.118 = 0.5 ln = 0.119 99 = 0.12 rd 9. 282 4 × 10−2 Don’t worry about the question "Can you unambiguously say that rates in one tree are more volatile than the other?" Volatility: 0.5 ln

Problem 24.14. Skip. This problem is not worth your time.

Problem 24.15. I’ll solve for only Tree #1. time 0 1 2

3

4 0.15809

0.15630 0.13843

0.13143

0.10362 0.08

0.12473 0.10635

0.10927

0.07676

0.09953 0.08170

0.09084 0.07943 0.07552

Set the notional amount to $1. The payoff at each node is Payoff: time 0

1

2

3

0.13843 − 0.105 1 + 0.13843 0 0.10635 − 0.105 1 + 0.10635

0 0

0.15630 − 0.105 1 + 0.15630 0.12473 − 0.105 1 + 0.12473

1 max (0, r − 0.105) 1+r

4 0.15809 − 0.105 1 + 0.15809 0.13143 − 0.105 1 + 0.13143 0.10927 − 0.105 1 + 0.10927

0 0

0 0 0

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CHAPTER 24. INTEREST RATE MODELS 0.15630 − 0.105 1 + 1 + 0.15630 1 + 0.15630 7. 428 977 9 × 10−2 Vuuu =

µ ¶ 0.15809 − 0.105 0.13143 − 0.105 0.5 × + 0.5 × = 1 + 0.15809 1 + 0.13143

µ ¶ 0.12473 − 0.105 1 0.13143 − 0.105 0.10927 − 0.105 Vuud = + 0.5 × + 0.5 × = 1 + 0.12473 1 + 0.12473 1 + 0.13143 1 + 0.10927 2. 963 786 7 × 10−2 µ ¶ 1 0.10927 − 0.105 Vdud = 0 + 0.5 × + 0.5 × 0 = 1. 750 465 4 × 1 + 0.09953 1 + 0.10927 10−3

¢ ¡ 0.13843 − 0.105 1 + 0.5 × 7. 428 977 9 × 10−2 + 0.5 × 2. 963 786 7 × 10−2 = 1 + 0.13843 1 + 0.13843 7. 501 016 6 × 10−2 Vuu =

¢ ¡ 0.10635 − 0.105 1 + 0.5 × 2. 963 786 7 × 10−2 + 0.5 × 1. 750 465 4 × 10−3 = 1 + 0.10635 1 + 0.10635 1. 540 576 3 × 10−2 Vud =

¢ ¡ 1 0.5 × 1. 750 465 4 × 10−3 + 0.5 × 0 = 8. 091 270 2×10−4 1 + 0.08170 ¢ ¡ 1 Vu = 0.5 × 7. 501 016 6 × 10−2 + 0.5 × 1. 540 576 3 × 10−2 = 1 + 0.10362 4. 096 334 3 × 10−2 Vdd =

¢ ¡ 1 0.5 × 1. 540 576 3 × 10−2 + 0.5 × 8. 091 270 2 × 10−4 = 1 + 0.07676 7. 529 482 0 × 10−3 Vd =

¢ ¡ 1 0.5 × 4. 096 334 3 × 10−2 + 0.5 × 7. 529 482 0 × 10−3 = 2. 1 + 0.08 245 038 2 × 10−2 V =

If the notional amount is $1, the interest cap is worth 2. 245 038 2 × 10−2 at t = 0. Since the notional amount is 250 million, the interest cap is worth at t=0 250 × 2. 245 038 2 × 10−2 = 5. 612 595 5 (million)

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