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+ : a + (1 )b, 0 1}.
which gives Shephards lemma.
Ex. 1.46 Since di is homogeneous of degree zero in p
and y, for any > 0 and for i = 1, . . . , n,
and suppose that for all x A, x a. Then % is convex
di (p, y) = 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 , we have
function exists. Moreover, since % (a) = % (b) is convex,
di (p, y)
there exists a supporting hyperplane H = {x Rn+ :
p di (p, y)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 = 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.
= 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
pR++
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
pj +
=0
y
= v(p)y,
e(p)
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
L = p
1 p2 (1 p1 x1 p2 x2 ),
with the first-order conditions
Ex. 1.66 (b) By definition y = E(p , u ), Therefore
p1
p2 + x1 = 0
1
E(p1 , u0 )
y1
>
y0
E(p0 , u0 )
and
1
+ x2 = 0.
p
1 p2
means that y 1 > E(p1 , u0 ). Since the indirect utility
function V is increasing in income y, it follows that
Eliminating from the first-order conditions gives
p2 =
u1 = V (p1 , y 1 ) > V (p1 , E(p1 , u0 )) = u0 .
x1
p1 .
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
1
,
+ x1
p2 =
1
.
+ 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
U (x) =
x
1 x2 ,
( + )+
u0 1/8
43
=
.
2u0 1
80
2
which is the CES function with = 1/2. You should
verify with Example 1.3 on p. 3941 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 Eulers 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 = p2 /(p1 + p2 ) so that 1 =
p1 /(p1 + p2 ) and 0 < < 1. The minimization problem
becomes
x1
x2
+
:0<<1 .
(8)
min
Ex. 4.5 Let w be the vector of factor prices and p
be the output price. Then the cost function of a typical 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 py c(w)y = max y[p c(w)].
Notice that for any x1 > 0 and x2 > 0,
x1
x2
lim
+
=
0
1
and
lim
x1
x2
+
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 < < 1. The
first-order condition for minimization is
(9)
x2
x1
+
= 0,
2
(1 )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
2 x2 = (1 )2 x1 .
with necessary condition
Taking the square root on both sides gives
1/2
x2
10 15q (J 1)
q 15q 2q = 0.
1/2
= (1 )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 = [40/(J +31)]2
1. In the long-run = 0 so that J = 9.
Rearranging gives
1/2
1/2
x1
1/2
x1
1/2
and
x2
1=
+ x2
1/2
x1
1/2
+ x2
It is clear that is indeed between 0 and 1. Putting Ex. 5.11 (a) The necessary condition for a Paretoand 1 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 2s 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
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 1s 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
py
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 , ij , 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 +
ij j (p),
x
jJ
with necessary condition
U i (x) = p.
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.
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