Jehle Solutions

ECON 5113 Advanced Microeconomics Winter 2015 Answers to Selected Exercises

Instructor: Kam Yu

The following questions are taken from Geoffrey A. Jehle above in u, that is, for some p  0, there exists M > 0 and Philip J. Reny (2011) Advanced Microeconomic The- such that M ≥ E(p, u) for all u in the domain of E. ory, Third Edition, Harlow: Pearson Education Limited. Let u∗ = V (p, M ). Then The updated version is available at the course web page: E(p, u∗ ) = E(p, V (p, M )) = M = pT x∗ , http://flash.lakeheadu.ca/∼kyu/E5113/Main.html where x∗ is the optimal bundle. Since U is continuous, there exists a bundle x0 in the neighbourhood of x∗ such Ex. 1.14 Let U be a continuous utility function that that U (x0 ) = u0 > u∗ . Since U strictly increasing, E is represents %. Then for all x, y ∈ Rn+ , x % y if and only strictly increasing in u, so that E(p, u0 ) > E(p, u∗ ) = if U (x) ≥ U (y). M . This contradicts the assumption that M is an upper First, suppose x, y ∈ Rn+ . Then U (x) ≥ U (y) or bound. U (y) ≥ U (x), which means that x % y or y % x. Therefore % is complete. Ex. 1.37 (a) Since x0 is the solution of the expenditure Second, suppose x % y and y % z. Then U (x) ≥ U (y) minimization problem when the price is p0 and utility and U (y) ≥ U (z). This implies that U (x) ≥ U (z) and level u0 , it must satisfy the constraint U (x0 ) ≥ u0 . Now so x % z, which shows that % is transitive. by definition E(p, u0 ) is the minimized expenditure when Finally, let x ∈ Rn+ and U (x) = u. Then 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 . = {z ∈ Rn+ : z % x} (b) Since f (p) ≤ 0 for all p  0 and f (p0 ) = 0, it = % (x). must attain its maximum value at p = p0 . (c) ∇f (p0 ) = 0. Since [u, ∞) is closed and U is continuous, % (x) is closed. (d) We have Similarly (I suggest you to try this), - (x) is also closed. This shows that % is continuous. ∇f (p0 ) = ∇p E(p0 , u0 ) − x0 = 0, U −1 ([u, ∞))

= {z ∈ Rn+ : U (z) ≥ u}

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}.

which gives Shephard’s lemma. Ex. 1.46 Since di is homogeneous of degree zero in p and y, for any α > 0 and for i = 1, . . . , n,

and suppose that for all x ∈ A, x ∼ a. Then % is convex di (αp, αy) = di (p, y). but not strictly convex. Theorem 1.1 does not require % to be convex or strictly convex, therefore the utility Differentiate both sides with respect to α, we have function exists. Moreover, since % (a) = % (b) is convex, ∂di (αp, αy) there exists a supporting hyperplane H = {x ∈ Rn+ : ∇p di (αp, αy)T p + y = 0. T ∂y p 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 Put α = 1 and rewrite the dot product in summation to the utility maximization problem. form, the above equation becomes Ex. 1.341 Suppose on the contrary that E is bounded 1 It

may be helpful to review the proof of Theorem 1.8.

n X ∂di (p, y) j=1

∂pj

pj +

∂di (p, y) y = 0. ∂y

(1)

(b) It is clear from part (a) that I depends on u0 . (c) Using the technique similar to Exercise 1.47, it can Ex. 1.47 Suppose that U (x) is a linearly homogeneous be shown that if U is homothetic, E(p, u) = e(p)g(u), utility function. where g is an increasing function. Then (a) Then e(p1 ) e(p1 )g(u0 ) = , I= T E(p, u) = min{p x : U (x) ≥ u} e(p0 )g(u0 ) e(p0 ) x Dividing each term by di (p, y) yields the result.

=

min{upT x/u : U (x/u) ≥ 1}

which means that I is independent of the reference utility level.

x

= u min{pT x/u : U (x/u) ≥ 1} x

= u min{pT x/u : U (x/u) ≥ 1}

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) ∂di (p, y) pj + pj dj (p, y) ∂pj ∂y j=i

(2)

x/u

= u min{pT z : U (z) ≥ 1} z

(3)

= uE(p, 1) = ue(p)

=

In (2) above it does not matter if we choose x or x/u directly as long as the objective function and the constraint remain the same. We can do this because of the objective function is linear in x. In (3) we simply rewrite x/u as z. (b) Using the duality relation between V and E and the result from Part (a) we have

=

V (p, y) =

∂di (p, y) X pj dj (p, y) ∂y j=i

∂di (p, y) ∂di (p, y) pj + y ∂pj ∂y

(5) (6)

where in (5) we have used the budget balancedness and (6) holds because of homogeneity and (1) in Ex. 1.46. Ex. 2.3 By (T.1’) on p. 82, U (x) = min {V (p, 1) : p · x = 1} . n p∈R++

where we have let v(p) = 1/e(p). The marginal utility of income is ∂V (p, y) = v(p), ∂y which depends on p but not on y. 0

∂pj

n

pj +

=0

y = v(p)y, e(p)

0

j=i n X j=i

y = E(p, V (p, y)) = V (p, y)e(p) so that

n X ∂di (p, y)

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

with the first-order conditions

0

Ex. 1.66 (b) By definition y = E(p , u ), Therefore

−αpα−1 pβ2 + λx1 = 0 1

E(p1 , u0 ) y1 > y0 E(p0 , u0 )

and β−1 + λx2 = 0. −βpα 1 p2

means that y 1 > E(p1 , u0 ). Since the indirect utility function V is increasing in income y, it follows that

Eliminating λ from the first-order conditions gives p2 =

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

β x1 p1 . α x2

Substitute this p2 into the constraint equation, we get Ex. 1.67 It is straight forward to derive the expenditure function, which is E(p, u) = p2 u −

p22 . 4p1

α 1 , α + β x1

p2 =

β 1 . α + β x2

and

(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=

p1 =

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

u0 − 1/8 43 = . 2u0 − 1 80 2

which is the CES function with ρ = 1/2. You should verify with Example 1.3 on p. 39–41 that the expenditure function is indeed as given.

which is a Cobb-Douglas function. Ex. 2.6 We want to maximize utility u subject to the constraint pT x ≥ E(p, u) for all p ∈ Rn++ . That is, p1 x1 + p2 x2 ≥

Ex. 3.2 Constant returns-to-scale means that f is linearly homogeneous. So by Euler’s theorem

up1 p2 . p1 + p2

x1 ∂y/∂x1 + x2 ∂y/∂x2 = y.

Rearranging gives u≤

Since average product y/x1 is rising, its derivative respect to x1 is positive, that is,

p1 + p2 p1 + p2 x1 + x2 p2 p1

(x1 ∂y/∂x1 − y)/x21 > 0.

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

From (9) we have (7)

x2 ∂y/∂x2 = −(x1 ∂y/∂x1 − y) < 0, which means that the marginal product ∂y/∂x2 is negative.

Therefore u attains its maximum value when equality holds in (7). To find the minimum value on the righthand side of (7), write α = p2 /(p1 + p2 ) so that 1 − α = p1 /(p1 + p2 ) and 0 < α < 1. The minimization problem becomes   x1 x2 + :0 0,   x1 x2 lim + =∞ α→0 α 1−α and

 lim

α→1

x1 x2 + α 1−α

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.

 =∞

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

(9)

x2 x1 + = 0, α2 (1 − α)2

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

which can be written as max [10 − 15q − (J − 1)¯ q ]q − (q 2 + 1), q

α2 x2 = (1 − α)2 x1 .

with necessary condition

Taking the square root on both sides gives 1/2

αx2

10 − 15q − (J − 1)¯ q − 15q − 2q = 0.

1/2

= (1 − α)x1 .

(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.

Rearranging gives 1/2

α=

1/2

x1 1/2

x1

1/2

and

x2

1−α=

+ x2

1/2

x1

1/2

.

+ x2

It is clear that α is indeed between 0 and 1. Putting α Ex. 5.11 (a) The necessary condition for a Paretoand 1−α into the objective function in (8) give the direct efficient allocation is that the consumers’ MRS are equal. Therefore utility function U (x1 , x2 ) =



1/2 x1

+

1/2 x2

2

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

, 3

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 (11) and (12). The indifference curve passing through endowment becomes (21 − x11 )(10 − x12 )2 = (21 − 18)(10 − 4)2 = 108. Putting x12 in (13) into the above equation and solving for x11 give x11 = 15.21. Therefore the core of the economy is given by  10x11 , C(e) = (x11 , x12 , x21 , x22 ) : x12 = 42 − x11 14.16 ≤ x11 ≤ 15.21, x11 + x21 = 21, x12 + x22 = 10.

Figure 1: Contract Curve and the Core or

x2 x12 = 22 . 1 x1 2x1

(c) Normalize the price of good 2 to p2 = 1. The demand functions of the two consumers are: (10)

y1 2p1 y1 x12 = 2p2 y2 x21 = 3p1 2y 2 x22 = 3p2 x11 =

The feasibility conditions for the two goods are x11 + x21 = e11 + e21 = 18 + 3 = 21,

(11)

x12

(12)

+

x22

=

e12

+

e22

= 4 + 6 = 10.

Express x21 in (11) and x22 in (12) in terms of x11 and x12 respectively, (10) becomes x12 10 − x12 = , 1 x1 2(21 − x11 ) or x12 =

10x11 . 42 − x11

p1 e11 + p2 e12 18p1 + 4 = 2p1 2p1 1 1 p1 e1 + p2 e2 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 18p1 + 4 3p1 + 6 + − 18 − 3 = 0, 2p1 3p1

(13)

Eq. (13) with domain 0 ≤ x11 ≤ 21, (11), and (12) com- which gives p1 = 4/11 (check that market 2 also clears). pletely characterize the set of Pareto-efficient allocations The Walrasian equilibrium is p = (p1 , p2 ) = (4/11, 1). From the demand functions above, the WEA is A (contract curve). That is,  x = (x11 , x12 , x21 , x22 ) = (14.5, 5.27, 5.6, 4.73). 10x11 1 A = (x11 , x12 , x21 , x22 ) : x12 = , 0 ≤ x ≤ 21, 1 42 − x11 (d) It is easy to verify that x ∈ C(e). x11 + x21 = 21, x12 + x22 = 10. Ex. 5.23 Let Y ⊆ Rn be a strongly convex production (b) The core is the section of the curve in (13) be- set. For any p ∈ Rn , let y1 ∈ Y and y2 ∈ Y be two ++ tween the points of intersections with the consumers’ in- distinct profit-maximizing production plans. Therefore difference curves passing through the endowment point. p · y1 = p · y2 ≥ p · y for all y ∈ Y . Since Y is strongly For example, in Figure 1, if G is the endowment point, convex, there exists a y ¯ ∈ Y such that for all t ∈ (0, 1), the core is the portion of the contract curve between points W and Z. Consumer 1’s indifference curve passing ¯ > ty1 + (1 − t)y2 . y through the endowment is Thus (x11 x12 )2 = (18 · 4)2 , ¯ > tp · y1 + (1 − t)p · y2 p·y 1 1 or x2 = 72/x1 . Substituting this into (13) and rearrang= tp · y1 + (1 − t)p · y1 ing give = p · y1 , 5(x11 )2 + 36x11 − 1512 = 0. 4

which contradicts the assumption that y1 is profitmaximizing. Therefore y1 = y2 . 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 problem is X max U i (x) s. t. p · x = p · ei + θij π j (p), x

j∈J

with necessary condition ∇U i (x) = λp. The MRS between two goods l and m is therefore pl ∂U i (x)/∂xl = . ∂U i (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.

c

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