ECON 5113 Advanced Microeconomics
Winter 2015
Answers to Selected Exercises
Instructor:
Kam Yu
The following questions are taken from Geoffrey A. Jehle and Philip J. Reny (2011) Advanced Microeconomic The-ory, 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. 1.14 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. There-fore % 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.
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 % 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
+ :
pTx = 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
1It may be helpful to review the proof of Theorem 1.8.
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 = pTx∗,
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 pTx0 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) = ∇
pE(p0, u0) − x0= 0,
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, di(αp, αy) = di(p, y).
Differentiate both sides with respect to α, we have ∇pdi(αp, αy)Tp +
∂di(αp, αy)
∂y y = 0.
Put α = 1 and rewrite the dot product in summation form, the above equation becomes
n X j=1 ∂di(p, y) ∂pj pj+ ∂di(p, y) ∂y y = 0. (1)
Dividing each term by di(p, y) yields the result.
Ex. 1.47 Suppose that U (x) is a linearly homogeneous utility function. (a) Then E(p, u) = min x {p Tx : U (x) ≥ u} = min x {up T x/u : U (x/u) ≥ 1} = u min x {p Tx/u : U (x/u) ≥ 1} = u min x/u {pTx/u : U (x/u) ≥ 1} (2) = u min z {p Tz : U (z) ≥ 1} (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 con-straint 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
y = E(p, V (p, y)) = V (p, y)e(p) so that
V (p, y) = y
e(p) = v(p)y,
where we have let v(p) = 1/e(p). The marginal utility of income is
∂V (p, y)
∂y = v(p), which depends on p but not on y.
Ex. 1.66 (b) By definition y0= E(p0, u0), Therefore
y1
y0 >
E(p1, u0)
E(p0, u0)
means that y1 > E(p1, u0). Since the indirect utility function V is increasing in income y, it follows that
u1= V (p1, y1) > V (p1, E(p1, u0)) = u0.
Ex. 1.67 It is straight forward to derive the expenditure function, which is E(p, u) = p2u − p2 2 4p1 . (4)
(a) For p0 = (1, 2) and y0 = 10, we can use (4) to
obtain u0= 11/2. Therefore, with p1= (2, 1),
I =u
0− 1/8
2u0− 1 =
43 80.
(b) It is clear from part (a) that I depends on u0. (c) Using the technique similar to Exercise 1.47, it can be shown that if U is homothetic, E(p, u) = e(p)g(u), where g is an increasing function. Then
I =e(p
1)g(u0)
e(p0)g(u0)=
e(p1)
e(p0),
which means that I is independent of the reference utility level.
Ex. 2.2 For i = 1, . . . , n, the i-th row of the matrix multiplication S(p, y)p is n X j=i ∂di(p, y) ∂pj pj+ ∂di(p, y) ∂y pjdj(p, y) = n X j=i ∂di(p, y) ∂pj pj+ ∂di(p, y) ∂y n X j=i pjdj(p, y) = n X j=i ∂di(p, y) ∂pj pj+ ∂di(p, y) ∂y y (5) = 0 (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 p∈Rn ++ {V (p, 1) : p · x = 1} . The Lagrangian is L = −pα1p β 2− λ(1 − p1x1− p2x2),
with the first-order conditions −αpα−1 1 p β 2 + λx1= 0 and −βpα 1p β−1 2 + λx2= 0.
Eliminating λ from the first-order conditions gives p2= β α x1 x2 p1.
Substitute this p2 into the constraint equation, we get
p1= α α + β 1 x1 , and p2= β α + β 1 x2 . The utility function is therefore
U (x) =
ααββ
(α + β)α+β
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, p1x1+ p2x2≥ up1p2 p1+ p2 . Rearranging gives u ≤ p1+ p2 p2 x1+ p1+ p2 p1 x2 for all p ∈ Rn
++. This implies that
u ≤ min p1,p2 p1+ p2 p2 x1+ p1+ p2 p1 x2 . (7)
Therefore u attains its maximum value when equality holds in (7). To find the minimum value on the right-hand side of (7), write α = p2/(p1+ p2) so that 1 − α =
p1/(p1+ p2) and 0 < α < 1. The minimization problem
becomes min α x1 α + x2 1 − α : 0 < α < 1 . (8)
Notice that for any x1> 0 and x2> 0,
lim α→0 x1 α + x2 1 − α = ∞ and lim α→1 x1 α + x2 1 − α = ∞
so that the minimum value exists when 0 < α < 1. The first-order condition for minimization is
−x1 α2 +
x2
(1 − α)2 = 0,
which can be written as
α2x2= (1 − α)2x1.
Taking the square root on both sides gives αx1/22 = (1 − α)x1/21 . Rearranging gives α = x 1/2 1 x1/21 + x1/22 and 1 − α = x 1/2 2 x1/21 + x1/22 .
It is clear that α is indeed between 0 and 1. Putting α and 1−α into the objective function in (8) give the direct utility function U (x1, x2) = x1/21 + x1/22 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.
Ex. 3.2 Constant returns-to-scale means that f is lin-early homogeneous. So by Euler’s theorem
x1∂y/∂x1+ x2∂y/∂x2= y. (9)
Since average product y/x1is rising, its derivative respect
to x1 is positive, that is,
(x1∂y/∂x1− y)/x21> 0.
From (9) we have
x2∂y/∂x2= −(x1∂y/∂x1− y) < 0,
which means that the marginal product ∂y/∂x2 is
nega-tive.
Ex. 4.5 Let w be the vector of factor prices and p be the output price. Then the cost function of a typ-ical firm with constant returns-to-scale technology is C(w, y) = c(w)y where c is the unit cost function. The profit maximization problem can be written as
max
y py − c(w)y = maxy y[p − c(w)].
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. Ex. 4.14 The profit maximization problem for a typical firm is
max
q [10 − 15q − (J − 1)¯q]q − (q 2+ 1),
with necessary condition
10 − 15q − (J − 1)¯q − 15q − 2q = 0.
(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. 5.11 (a) The necessary condition for a Pareto-efficient allocation is that the consumers’ MRS are equal. Therefore ∂U1(x1 1, x12)/∂x11 ∂U1(x1 1, x 1 2)/∂x 1 2 = ∂U 2(x2 1, x22)/∂x21 ∂U2(x2 1, x 2 2)/∂x 2 2 ,
Figure 1: Contract Curve and the Core or x1 2 x1 1 = x 2 2 2x2 1 . (10)
The feasibility conditions for the two goods are
x11+ x21= e11+ e21= 18 + 3 = 21, (11) x12+ x22= e12+ e22= 4 + 6 = 10. (12) Express x2
1 in (11) and x22 in (12) in terms of x11 and x12
respectively, (10) becomes x12 x1 1 = 10 − x 1 2 2(21 − x1 1) , or x12= 10x 1 1 42 − x1 1 . (13)
Eq. (13) with domain 0 ≤ x11 ≤ 21, (11), and (12)
com-pletely characterize the set of Pareto-efficient allocations A (contract curve). That is,
A = (x11, x12, x12, x22) : x12= 10x 1 1 42 − x1 1 , 0 ≤ x11≤ 21, x11+ x21= 21, x12+ x22= 10.
(b) The core is the section of the curve in (13) be-tween the points of intersections with the consumers’ in-difference 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
(x11x12)2= (18 · 4)2, or x1
2= 72/x11. Substituting this into (13) and
rearrang-ing give
5(x11)2+ 36x11− 1512 = 0.
Solving the quadratic equation gives one positive value of 14.16. Consumer 2’s utility function can be written as x2
1(x22)2. This can be expressed in terms of x11 and
x1
2 using (11) and (12). The indifference curve passing
through endowment becomes
(21 − x11)(10 − x12)2= (21 − 18)(10 − 4)2= 108. Putting x1
2 in (13) into the above equation and solving
for x1
1give x11= 15.21. Therefore the core of the economy
is given by C(e) = (x11, x12, x12, x22) : x12= 10x 1 1 42 − x1 1 , 14.16 ≤ x11≤ 15.21, x1 1+ x 2 1= 21, x12+ x22= 10.
(c) Normalize the price of good 2 to p2 = 1. The
demand functions of the two consumers are: x11= y 1 2p1 = p1e 1 1+ p2e12 2p1 =18p1+ 4 2p1 x12= y 1 2p2 = p1e 1 1+ p2e12 2p2 =18p1+ 4 2 x21= y2 3p1 = p1e 2 1+ p2e22 3p1 =3p1+ 6 3p1 x22=2y 2 3p2 = 2(p1e 2 1+ p2e22) 3p2 = 2(3p1+ 6) 3
In equilibrium, excess demand z1(p) for good 1 is zero.
Therefore 18p1+ 4 2p1 +3p1+ 6 3p1 − 18 − 3 = 0,
which gives p1= 4/11 (check that market 2 also clears).
The Walrasian equilibrium is p = (p1, p2) = (4/11, 1).
From the demand functions above, the WEA is x = (x11, x12, x21, x22) = (14.5, 5.27, 5.6, 4.73).
(d) It is easy to verify that x ∈ C(e).
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
convex, there exists a ¯y ∈ Y such that for all t ∈ (0, 1), ¯ y > ty1+ (1 − t)y2. Thus p · ¯y > tp · y1+ (1 − t)p · y2 = tp · y1+ (1 − t)p · y1 = p · y1,
which contradicts the assumption that y1 is profit-maximizing. Therefore y1= y2.
Ex. 5.31 Let E = {(Ui, ei, θij, Yj)|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 max x U i(x) s. t. p · x = p · ei+X j∈J θijπj(p),
with necessary condition
∇Ui(x) = λp.
The MRS between two goods l and m is therefore ∂Ui(x)/∂xl ∂Ui(x)/∂x m = pl 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 maxi-mization problem of any firm.
(c) This shows that the Walrasian equilibrium prices play the key role in the functioning of a production econ-omy. Exchanges are impersonal. Each consumer only need to know her preferences and each firm its produc-tion set. All agents in the economy observe the common price signal and make their own decisions. This mini-mal information requirement leads to the lowest possible transaction costs of the economy.
c
