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ECON 5113 Advanced Microeconomics Winter 2018 Answers to Selected Exercises

The following questions are taken from Geoffrey A. Jehle and Philip J. Reny (2011) Advanced Microeconomic Theory, Third Edition, Harlow: Pearson Education Limited. The updated version is available at the course web page: http://flash.lakeheadu.ca/∼kyu/E5113/Main.html Ex. 1.11 Suppose that p ≥ 0 is a limit point of A. Then for every  > 0, there exists a point q 6= p in (p − , p + ) such that q ∈ A. This means that for every neighbourhood B (pe) ∈ Rn+ , there is a bundle qe in B (pe) ∩ % (x). Hence pe is a limit point of % (x). Since % is continuous so that % (x) is a closed set, pe ∈ % (x), which implies that p ∈ A. Therefore A is closed. The proof for set B is similar. Ex. 1.14 Let U be a continuous utility function that represents %. Then for all x, y ∈ Rn+ , x % y if and only if U (x) ≥ U (y). First, suppose x, y ∈ Rn+ . Then U (x) ≥ U (y) or U (y) ≥ U (x), which means that x % y or y % x. Therefore % is complete. Second, suppose x % y and y % z. Then U (x) ≥ U (y) and U (y) ≥ U (z). This implies that U (x) ≥ U (z) and so x % z, which shows that % is transitive. Finally, let x ∈ Rn+ and U (x) = u. Then U −1 ([u, ∞))

= {z ∈ Rn+ : U (z) ≥ u} = {z ∈ Rn+ : z % x} = % (x).

Since [u, ∞) is closed and U is continuous, % (x) is closed. Similarly (I suggest you to try this), - (x) is also closed. This shows that % is continuous.

Instructor: Kam Yu

% to be convex or strictly convex, therefore the utility function exists. Moreover, since % (a) = % (b) is convex, there exists a supporting hyperplane H = {x ∈ Rn+ : pT x = y} such that a, b ∈ H. Since H is an affine set, A ⊂ H. This means that every bundle in A is a solution to the utility maximization problem. Ex. 1.341 Suppose on the contrary that E is bounded above in u, that is, for some p  0, there exists M > 0 such that M ≥ E(p, u) for all u in the domain of E. Let u∗ = V (p, M ). Then E(p, u∗ ) = E(p, V (p, M )) = M = pT x∗ , where x∗ is the optimal bundle. Since U is continuous, there exists a bundle x0 in the neighbourhood of x∗ such that U (x0 ) = u0 > u∗ . Since U strictly increasing, E is strictly increasing in u, so that E(p, u0 ) > E(p, u∗ ) = M . This contradicts the assumption that M is an upper bound. Ex. 1.37 (a) Since x0 is the solution of the expenditure minimization problem when the price is p0 and utility level u0 , it must satisfy the constraint U (x0 ) ≥ u0 . Now by definition E(p, u0 ) is the minimized expenditure when price is p, it must be less than or equal to pT x0 since x0 is in the feasible set, and by definition equal when p = p0 . (b) Since f (p) ≤ 0 for all p  0 and f (p0 ) = 0, it must attain its maximum value at p = p0 . (c) ∇f (p0 ) = 0. (d) We have ∇f (p0 ) = ∇p E(p0 , u0 ) − x0 = 0, which gives Shephard’s lemma.

Ex. 1.17 Suppose that a and b are two distinct bundle such that a ∼ b. Let A = {x ∈ Rn+ : αa + (1 − α)b, 0 ≤ α ≤ 1}. and suppose that for all x ∈ A, x ∼ a. Then % is convex but not strictly convex. Theorem 1.1 does not require

Ex. 1.46 Since di is homogeneous of degree zero in p and y, for any α > 0 and for i = 1, . . . , n, di (αp, αy) = di (p, y). 1 It

may be helpful to review the proof of Theorem 1.8.

(a) The Lagrangian is

Differentiate both sides with respect to α, we have ∇p di (αp, αy)T p +

∂di (αp, αy) y = 0. ∂y

L=A

∂pj

j=1

pj +

∂di (p, y) y = 0. ∂y

The necessary conditions for maximization are the budget constraint and Qn xα i ∂L = αj A i=1 i − λpj = 0, ∂xj xj

(1)

for j = 1, . . . , n. Consider two goods i and j, the above necessary condition implies that

Dividing each term by di (p, y) yields the result.

pi αi /xi = , αj /xj pj

Ex. 1.47 Suppose that U (x) is a linearly homogeneous utility function. (a) Then E(p, u)

which can be rearranged to    αj pi xj = xi . αi pj

T

=

min{p x : U (x) ≥ u}

=

min{upT x/u : U (x/u) ≥ 1}

T i xα i − λ(p x − y).

i=1

Put α = 1 and rewrite the dot product in summation form, the above equation becomes n X ∂di (p, y)

n Y

x x

Substitute this relation for j = 1, . . . , n in the budget constraint, we have       pi pi αn α1 xi +· · ·+pi xi +· · ·+pn xi = y, p1 αi p1 αi pn

= u min{pT x/u : U (x/u) ≥ 1} x

= u min{pT x/u : U (x/u) ≥ 1}

(2)

x/u

= u min{pT z : U (z) ≥ 1} z

(3)

or

= uE(p, 1)



= ue(p)

pi αi

 X n

! αk

xi = y.

k=1

In (2) above it does not matter if we choose x or x/u Since Pn αk = 1, the above equation gives the Mark=1 directly as long as the objective function and the con- shallian demand function of good i as straint remain the same. We can do this because of the αi y objective function is linear in x. In (3) we simply rewrite , di (p, y) = xi = pi x/u as z. (b) Using the duality relation between V and E and for i = 1, . . . , n. the result from Part (a) we have Ex. 1.66 (b) By definition y 0 = E(p0 , u0 ), Therefore y = E(p, V (p, y)) = V (p, y)e(p) E(p1 , u0 ) y1 > so that y0 E(p0 , u0 ) y = v(p)y, V (p, y) = e(p) means that y 1 > E(p1 , u0 ). Since the indirect utility where we have let v(p) = 1/e(p). The marginal utility function V is increasing in income y, it follows that of income is

u1 = V (p1 , y 1 ) > V (p1 , E(p1 , u0 )) = u0 .

∂V (p, y) = v(p), ∂y

Ex. 1.67 It is straight forward to derive the expenditure function, which is

which depends on p but not on y. Ex. 1.54 The utility maximization problem is max x

A

n Y

E(p, u) = p2 u − i xα i

(4)

(a) For p0 = (1, 2) and y 0 = 10, we can use (4) to obtain u0 = 11/2. Therefore, with p1 = (2, 1),

i=1 T

subject to p x = y, where A > 0 and

p22 . 4p1

Pn

I=

i=1 αi = 1.

2

u0 − 1/8 43 = . 2u0 − 1 80

with the first-order conditions (b) It is clear from part (a) that I depends on u0 . (c) Using the technique similar to Exercise 1.47, it can −αpα−1 pβ2 + λx1 = 0 be shown that if U is homothetic, E(p, u) = e(p)g(u), 1 where g is an increasing function. Then and β−1 −βpα + λx2 = 0. e(p1 ) e(p1 )g(u0 ) 1 p2 = , I= 0 0 0 e(p )g(u ) e(p ) Eliminating λ from the first-order conditions gives which means that I is independent of the reference utility level.

p2 =

β x1 p1 . α x2

Ex. 2.1 Consider the case of one good. Let the demand Substitute this p2 into the constraint equation, we get function be √ y α 1 d(p, y) = √ . , p1 = p α + β x1 It is homogenous of degree zero but it does not satisfy budget balancedness. Conversely, consider the two-good case that the demand function is given by   y log p1 y log p2 , d(p, y) = . p1 log p2 + p2 log p1 p1 log p2 + p2 log p1

and p2 =

β 1 . α + β x2

The utility function is therefore   αα β β −β U (x) = x−α 1 x2 , (α + β)α+β

Then

which is a Cobb-Douglas function. yp1 log p2 yp2 log p1 p · d(p, y) = + p1 log p2 + p2 log p1 p1 log p2 + p2 log p1 Ex. 2.6 We want to maximize utility u subject to the = y constraint pT x ≥ E(p, u) for all p ∈ Rn++ . That is, and therefore satisfies budget balancedness. It is straight forward to verify that d(p, y) is not a homogenous function.

p1 x1 + p2 x2 ≥ Rearranging gives

Ex. 2.2 For i = 1, . . . , n, the i-th row of the matrix multiplication S(p, y)p is n  X ∂di (p, y)

∂pj

j=i

=

pj +

n X ∂di (p, y) j=i n X

∂pj

u≤

p1 + p2 p1 + p2 x1 + x2 p2 p1

for all p ∈ Rn++ . This implies that   p1 + p2 p1 + p2 u ≤ min x1 + x2 . p1 ,p2 p2 p1

 ∂di (p, y) pj dj (p, y) ∂y n

pj +

up1 p2 . p1 + p2

∂di (p, y) X pj dj (p, y) ∂y j=i

(7)

Therefore u attains its maximum value when equality holds in (7). To find the minimum value on the right= (5) hand side of (7), write α = p /(p + p ) so that 1 − α = 2 1 2 j=i p1 /(p1 + p2 ) and 0 < α < 1. The minimization problem =0 (6) becomes   x2 x1 where in (5) we have used the budget balancedness and + :0 0,   x1 x2 lim + =∞ α→0 α 1−α

Ex. 2.3 By (T.1’) on p. 82, U (x) = min {V (p, 1) : p · x = 1} . n p∈R++

The Lagrangian is

and

 lim

β L = −pα 1 p2 − λ(1 − p1 x1 − p2 x2 ),

α→1

3

x1 x2 + α 1−α

 =∞

P where y = i yi is aggregate income. It is clear that market demand depends on aggregate income y but not on income distribution. The market level income elasticity of demand for good j is

so that the minimum value exists when 0 < α < 1. The first-order condition for minimization is −

x2 x1 + = 0, α2 (1 − α)2

which can be written as

ηj =

∂(fj (p)y) y = 1. ∂y fj (p)y

α2 x2 = (1 − α)2 x1 . Ex. 4.2 If preferences are homothetic but not identical, the demand function of consumer i is

Taking the square root on both sides gives 1/2

αx2

1/2

= (1 − α)x1 .

di (p, yi ) = f i (p)yi .

Rearranging gives α=

1/2 x1 1/2 1/2 x1 + x2

and

1−α=

Market demand is therefore X X di (p, yi ) = f i (p)yi .

1/2 x2 . 1/2 1/2 x1 + x2

i

i

It is clear that α is indeed between 0 and 1. Putting α In this case market demand depends on income distribuand 1−α into the objective function in (8) give the direct tion. utility function Ex. 4.5 Let w be the vector of factor prices and p  2 1/2 1/2 be the output price. Then the cost function of a typU (x1 , x2 ) = x1 + x2 , ical firm with constant returns-to-scale technology is C(w, y) = c(w)y where c is the unit cost function. The which is the CES function with ρ = 1/2. You should profit maximization problem can be written as verify with Example 1.3 on p. 39–41 that the expenditure function is indeed as given. max py − c(w)y = max y[p − c(w)]. y

Ex. 3.2 Constant returns-to-scale means that f is linearly homogeneous. So by Euler’s theorem x1 ∂y/∂x1 + x2 ∂y/∂x2 = y.

y

For a competitive firm, as long as p > c(w), the firm will increase output level y indefinitely. If p < c(w), profit is negative at any level of output except when y = 0. If p = c(w), profit is zero at any level of output. In fact, market price, average cost, and marginal cost are all equal so that the inverse supply function is a constant function of y. Therefore the supply function of the firm does not exist and the number of firm is indeterminate.

(9)

Since average product y/x1 is rising, its derivative respect to x1 is positive, that is, (x1 ∂y/∂x1 − y)/x21 > 0. From (9) we have

Ex. 4.7 Suppose that all firms have the same technology and therefore the same cost function. Given market price x2 ∂y/∂x2 = −(x1 ∂y/∂x1 − y) < 0, p, profit of a typical firm j is pqj − c(qj ). (a) Suppose that a > 0, b < 0, and there are J firms in which means that the marginal product ∂y/∂x2 is negathe industry. The short-run profit maximization for firm tive. j is ! J X Ex. 4.1 If preferences are identical and homothetic, each max α − β qi qj − (aqj + bqj2 ). consumer’s preferences can be represented by a linearly qj i=1 homogeneous utility function. Using an argument similar to exercise 1.47, the ordinary demand function of The necessary condition for profit maximization is consumer i is separable in p and yi , that is, J X α − βq − β qi − a − 2bqj = 0. (10) i j d (p, yi ) = f (p)yi . i=1

Market demand is therefore X X di (p, yi ) = f (p) yi = f (p)y, i

By symmetry, q1 = q2 = · · · = qJ . Equation (10) becomes α − β(J + 1)qj − a − 2bqj = 0,

i

4

which gives

(b) Suppose that both firms have fixed costs F > 0. Then both firms stay in production in equilibrium and charge a price equal to the solution in equation (12). They share the market with q1 = q2 = (α−βp)/2. Notice that in equation (12), p = c is the solution if F = 0. (c) If firm 1 has a lower marginal cost so that c2 > c1 > 0, then firm 1 will charge a price p1 = c2 just to keep firm 2 out of the market. Therefore q2 = 0 and q1 = α − βc2 .

α−a . qj = β(J + 1) + 2b

The market output is q=

J(α − a) , β(J + 1) + 2b

and the market price is p=α−

βJ(α − a) , β(J + 1) + 2b

Ex. 4.14 The profit maximization problem for a typical firm is

(b) The average cost of production is

max [10 − 15q − (J − 1)¯ q ]q − (q 2 + 1), q

c(q) = a + bq, q

with necessary condition 10 − 15q − (J − 1)¯ q − 15q − 2q = 0.

which is a decreasing function if a > 0 and b < 0. Since there is no fixed cost, a firm can potentially increase output until the average (or total) cost of production is zero, which is at the output level q = −a/b. Notice that at zero market price, consumer demand is α/β. Therefore the long-run equilibrium market price and number of firms depend on the relative values of −a/b and α/β. (c) If a > 0 and b > 0, the minimum efficiency scale is at q = 0. Therefore the long-run equilibrium market price and number of firms are indeterminate.

(a) Since all firms are identical, by symmetry q = q¯. This gives the Cournot equilibrium of each firm q ∗ = 10/(J + 31), with market price p∗ = 170/(J + 31). (b) Short-run profit of each firm is π = [40/(J +31)]2 − 1. In the long-run π = 0 so that J = 9.

Ex. 4.19 The Marshallian demands are x∗ = 1/p and m∗ = y − 1. The indirect utility function is V (p, y) = y − log p − 1. If the price of x rises from p0 to p1 , the Ex. 4.12 In the Bertrand duopoly model of section 4.2.1, compensating variation is implicitly defined as the firms have no fixed costs and equal marginal cost c. V (p1 , y + CV) = V (p0 , y), In equilibrium both firms charge p = c and share the market equally. or (a) Now suppose that firm 1 has fixed costs F > 0. If y + CV − log p1 − 1 = y − log p0 − 1. it stays in production, in equilibrium the price it charges 1 0 must be the same as firm 2. This implies that p1 = p2 = This gives CV = log(p /p ), which is equal to the change in consumer surplus. p. But the zero profit condition for firm 1 is Ex. 5.11 (a) The necessary condition for a Paretoefficient allocation is that the consumers’ MRS are equal. Therefore

pq1 − cq1 − F = 0, or p=c+

F > c. q1

∂U 2 (x21 , x22 )/∂x21 ∂U 1 (x11 , x12 )/∂x11 = , ∂U 1 (x11 , x12 )/∂x12 ∂U 2 (x21 , x22 )/∂x22

Assume that the firm share the market equally, q1 = q2 = Q/2 so that 2F 2F p=c+ =c+ . Q α − βp

or

(11)

x12 x22 = . x11 2x21

(13)

The feasibility conditions for the two goods are

Since firm 2 has no fixed cost, it can charge a price x11 + x21 = e11 + e21 = 18 + 3 = 21, (14) slightly lower than this and capture the whole market. 1 2 1 2 In equilibrium firm 2 charges a price that satisfies equax2 + x2 = e2 + e2 = 4 + 6 = 10. (15) tion (11), which can be rearranged to the quadratic equaExpress x21 in (14) and x22 in (15) in terms of x11 and x12 tion 2 βp − (α + βc)p + (αc + 2F ) = 0. (12) respectively, (13) becomes x12 10 − x12 = , x11 2(21 − x11 )

Therefore p2 = p, the solution in equation (12), q1 = 0, and q2 = α − βp. 5

is given by  C(e) = (x11 , x12 , x21 , x22 ) : x12 =

10x11 , 42 − x11

14.16 ≤ x11 ≤ 15.21, x11 + x21 = 21, x12 + x22 = 10. (c) Normalize the price of good 2 to p2 = 1. The demand functions of the two consumers are: y1 2p1 y1 x12 = 2p2 y2 x21 = 3p1 2y 2 x22 = 3p2 x11 =

Figure 1: Contract Curve and the Core

or x12 =

10x11 . 42 − x11

p1 e11 + p2 e12 18p1 + 4 = 2p1 2p1 p1 e11 + p2 e12 18p1 + 4 = = 2p2 2 2 2 p1 e1 + p2 e2 3p1 + 6 = = 3p1 3p1 2 2 2(p1 e1 + p2 e2 ) 2(3p1 + 6) = = 3p2 3 =

In equilibrium, excess demand z1 (p) for good 1 is zero. Therefore

(16)

18p1 + 4 3p1 + 6 + − 18 − 3 = 0, Eq. (16) with domain 0 ≤ x11 ≤ 21, (14), and (15) com2p1 3p1 pletely characterize the set of Pareto-efficient allocations which gives p1 = 4/11 (check that market 2 also clears). A (contract curve). That is, The Walrasian equilibrium is p = (p1 , p2 ) = (4/11, 1).  From the demand functions above, the WEA is 10x11 1 1 2 2 1 1 A = (x1 , x2 , x1 , x2 ) : x2 = , 0 ≤ x1 ≤ 21, 42 − x11 x = (x11 , x12 , x21 , x22 ) = (14.5, 5.27, 5.6, 4.73). 1 2 1 2 x1 + x1 = 21, x2 + x2 = 10. (d) It is easy to verify that x ∈ C(e). (b) The core is the section of the curve in (16) between the points of intersections with the consumers’ indifference curves passing through the endowment point. For example, in Figure 1, if G is the endowment point, the core is the portion of the contract curve between points W and Z. Consumer 1’s indifference curve passing through the endowment is

Ex. 5.23 Let Y ⊆ Rn be a strongly convex production set. For any p ∈ Rn++ , let y1 ∈ Y and y2 ∈ Y be two distinct profit-maximizing production plans. Therefore p · y1 = p · y2 ≥ p · y for all y ∈ Y . Since Y is strongly ¯ ∈ Y such that for all t ∈ (0, 1), convex, there exists a y ¯ > ty1 + (1 − t)y2 . y

(x11 x12 )2 = (18 · 4)2 ,

Thus ¯ > tp · y1 + (1 − t)p · y2 p·y

or x12 = 72/x11 . Substituting this into (16) and rearranging give

= tp · y1 + (1 − t)p · y1 = p · y1 ,

5(x11 )2 + 36x11 − 1512 = 0.

which contradicts the assumption that y1 is profitmaximizing. Therefore y1 = y2 .

Solving the quadratic equation gives one positive value of 14.16. Consumer 2’s utility function can be written as x21 (x22 )2 . This can be expressed in terms of x11 and x12 using (14) and (15). The indifference curve passing through endowment becomes

Ex. 5.31 Let E = {(U i , ei , θij , Y j )|i ∈ I, j ∈ J } be the production economy and p ∈ Rn++ be the Walrasian equilibrium. (a) For any consumer i ∈ I, the utility maximization (21 − x11 )(10 − x12 )2 = (21 − 18)(10 − 4)2 = 108. problem is X Putting x12 in (16) into the above equation and solving max U i (x) s. t. p · x = p · ei + θij π j (p), x for x11 give x11 = 15.21. Therefore the core of the economy j∈J 6

with necessary condition ∇U i (x) = λp. The MRS between two goods l and m is therefore ∂U i (x)/∂xl pl = . i ∂U (x)/∂xm pm Since all consumers observe the same prices, the MRS is the same for each consumer. (b) Similar to part (a) by considering the profit maximization problem of any firm. (c) This shows that the Walrasian equilibrium prices play the key role in the functioning of a production economy. Exchanges are impersonal. Each consumer only need to know her preferences and each firm its production set. All agents in the economy observe the common price signal and make their own decisions. This minimal information requirement leads to the lowest possible transaction costs of the economy.

Other Exercises Theorem 2.1 Here is a suggested proof that the utility function generated by the expenditure function is unbounded: On the contrary suppose that U is bounded. That is, there exists a utility level u ¯ such that for all x ∈ Rn+ , 0 U (x) ≤ u ¯. Let u > u ¯ and p0  0. Then by the concavity of E in p, for any p  0, E(p, u0 ) ≤ E(p0 , u0 ) + ∇p E(p0 , u0 )T (p − p0 ) = E(p0 , u0 ) + ∇p E(p0 , u0 )T p −∇p E(p0 , u0 )T p0 = E(p0 , u0 ) + ∇p E(p0 , u0 )T p − E(p0 , u0 ) = ∇p E(p0 , u0 )T p, where in the second last equality we apply Euler’s theorem to E since it is linearly homogeneous in p. Define x0 = ∇p E(p0 , u0 ) so that we have E(p, u0 ) ≤ pT x0 for all p  0. In other words, u0 is feasibility set of the maximization problem U (x0 ) = max{u : pT x0 ≥ E(p, u) ∀ p  0}. Therefore U (x0 ) ≥ u0 , which contradicts the assumption that u ¯ is an upper bound of U . c

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