Solutions Manual for Microeconomic Theory - A. Mas-Colell, M. D. Whinston, J. R. Green

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• 
o utions anua 
for 
• 
~ as-Colell, inston, and Green 
Prepared by: 
Chiaki Hara 
Cambridge University 
Ilya Segal 
University of California, Berkeley 
Steve Tadelis 
Han-·ard University 
New York Oxford OXFORD UNIVERSITY PRESS 1997 
We could never overestimate the of work which to be done to 
complete this solution t:..:>Ok, but the satisfaction in seeing the finished 
product more than makes up for the many hours of work. We have tried to be 
both co~x:ise and exhaustive, and we hope that these two objectives do not 
conflict too often ln this solution book. On rare occasions. we refer the 
reader to a book or uticle containing a well-presented solution. 
Chiaki Hara has done a lion•s share of the work, preparing solutions for 
parts I and IV. Ilya Segal has provided the solutions for chapters 10 and 
22. Steve Tadelis has the solutions for chapters 11, 13, 14, 21 and 
23, and has completed the work on part II. Finally, llya and Steve. have 
together prepared solutions for chapter 12. Some of the early work on the 
solutions had been done by Marc and we thank him for laying down 
foundations for solutions for II and chapter 12. 
We would like to thanJc: Andreu Mas-Colell and Mike Whinston for many hours 
of We are also to the many teaching fellows who taught 
Ec:2010a/b, the Harvard graduate sequence in microeconomic theory, for their 
input ove:- the years when earlier versions of the textt:..XJk had been used 
for instNction. The list of these teaching fellows is too long to inc:lude 
here. While some errors surely remain, we hope that both students and teachers 
will benefit from our solutions, which make an excellent textbook even more 
useful. 
Finally, some personal thanks: Chiaki Hara thanks his parents and 
colleagues at various places where he has spent the last four years. Ilya 
Segal thanks Olga. for her unwavering support during all times. Steve Tadelis 
thanks lrit, for always being there to encourage and support. 
Chiaki Hara 
Ilya Segal 
Steve Tade!is 
June 1996 
CHAPTER 1 
l.B.l Since y >- z implies y >- z, the transitivity implies that x >- z. 
- -
Suppose that z >- x. 
-
Since y ;t- z, the transitivity then implies that y >- x. 
-
But this contradicts x >- y. Thus we cannot have z >- x. 
-
Hence x >- z. 
l.B.2 By the completeness, x >- x for every x e X. Hence there is no x e X 
-
such that x >- x. Suppose that x >- y and y >- 2, then x >- y >- z. By (iii) of 
-
Proposition l.B.l, which was proved in Exercise I.B.l, we have x >- z. Hence >-
is transitive. Property (i) is now proved. 
As for (ii), since x >- x for every x e X, x - x for every x e X as welL 
-
Thus - is refiexive. Suppose that x - y and y - z. Tnen x >- y, y >- z. 
- -
V >- X ~ - . 
and z ;t- y. By the transitivity, this implies that x ~ 2 and 2 ~ x. Thus x -
z. Hence - is transitive. Suppose x that - y. Then x >- y and y >- x. 
- -
Thus y 
>- x and x >- y. Hence y - x. Thus - is symmetric Property (ii) is now 
- -
proved. 
l.B.3 Le~ x e X and y e X. Since u( ·) represents ~· x ?: y ;:· ,:;.:~d only if 
u(x} :!:: u(y). Since f( ·) is strictly increasing, u(x) i!: u(y) if and only if 
• 
\I{X) :!:: Y{y). Hence x >- y if and only if v(x) :!:: v(y). Therefor-e v( · l 
-
represents >-. 
-
l.B.4 Suppose first that x >- y. If, furthermore, y >- x, then x - y and hence 
- -
u(x) = u(y). If, on the contrary, we do not have y >- x, then x >- y. 
-
Hence 
u(x) > u(y). Thus, if x >- y, then u(x) <: u(y). 
-
Suppose conversely that u(x) :!:: u(y). If, furthermore, u{xl = u(yl, then 
1-1 
x - y and hence x ::; y. If, on the contrary, u(x) > u(y), then x >- y, and 
hence x >- y. Tnus, if u(x) l! u(y), then x >- y. So u{ ·) represents >-. 
- - -
l.B.S First, we shall prove by induction on the number N of the elements of X 
that, if there is no indifference between any two different elements of X, 
then there exists a utility function. If N = 1, there is nothing to prove: 
Just assign any number to the unique element. So let N > 1 and suppose that 
the above assertion is true for N - 1. We will show that it is still true for 
N. By the induction hypothesis, >- can be 
-
represented by a utility function u( ·) on the subset {xl' ... ,xN-l} . Without 
• 
loss of generality we can assume that u(x1) > uCx2 ) > ... > u(~-l ). 
Consider the following three cases: 
Case 1: For every i < N, xN >- xi" 
Case 2: For every i < N, x 1 >- ~· 
Case 3: There exist i < N and j < N such that x. >- XN >- x .. 
1 • . J 
Since there is no indifierence between two different elements, these _three 
cases are are exhaustive and mutually exclusive. We shall now show how the 
value of u(xN) should be 
reoresent >- on the whole X. 
. -
in each of the three cases, for u(·) to 
If Case 1 applies, then take u(xN) to be larger than u(x1). If Case 2 
applies, take u(~) to be smaller than u(xN_1). Suppose now that Case 3 
applies. Let I = {i e {1, ... , N - 1}: xi >- xN+l} and J = {j e {1, ... , N - 1}: 
XN t >- x .}. Completeness and the assumption that there is no indifference ) .... J 
• • • 
implies that I v J = {1, ... , N - 1}. The transitivity implies that both I and 
J are "inte;vals," in the sense that if i e I and i' < i. then i' e I; and if 
j e J and j' > j, then j' e J. Let i• = max I, then i• + 1 = min J. Take 
1-2 
u{xN) to lie in the open interval (u(x .• 1J,u(x .• )). • 1 + 1 Then it is ~- -·· -c:.::.y • tc see 
that u( · ) reoreser:ts >- on the whole X. 
. -
Suppose next that there may be indifference between some two elements of 
For each n = 1, ... ,N, define X = {x E X: x - x }. 
n m rr: r. 
Then, 
= X. Also, bv the 
• 
transitivity of - (Proposition l.B.Hiill, if X ~ X , then X " X = e. 
· n rn n m 
So 
let M be a subset of {!, ... ,N} such that X = u MX and X = X for-
me m m n 
any m e M 
and any n e M with m ~ n. Define an relation >-• on {X : m E M} by letting X 
- m m 
>-• X if and only if x >- x . In fact, by the definition of M, there is no 
- n m- n 
indifference between two different elements of {X : m e M}. Thus, by the 
m 
• 
preceding result, there exists a utility function u•( ·) that represents 
Then define u: X -7 lR by u(x ) = u•(x ) if m e M and x e X . 
n m n m 
It is easv to 
• 
show that, by the transitivity, u( ·) represents >-. 
-
l.C.l If y e C({x,y,z}), then the \VA would imply that y e C({x,y}). B:.1t 
contradicts the equality C({x,y}) = {x}. Hence y ~ C((x,y,z}). Tl'rJs 
C({x,y,z}) e {{x},{z},{x,z}}. 
I.C.2 The property in the question are equivalent to the followir.g property: 
If B e :E, B' e !13, X E B. y e B. X e B', y e 8', x e C(B), and y e C{B'), then 
X e C(B') and v e C(BL 
-
We shall thus prove the equivalence between this 
propeny and the Weak Axiom. 
Suppose first that the Weak Axiom is satisfied. Assume tha: B e 23, B' e 
13, x e 8, y e 8, x e 8', y e B', x e C(B), and y e C(B' ). If we apply the 
Weak Axiom twice, we obtain x e C(B') and y e C(Bl. Hence the above property 
is also satisfied. 
1-3 
Suppose conversely that the above property is satisfied. Let B e 13. x e 
8, y e B. x e B', and x e C(B). Furthermore, let 8' e :B, x e B', y e 8', and 
y e C(B'). Tnen the above condition implies that x e C(B') (and y e C(Bl). 
Thus the Weak Axiom is satisfied. 
l.C.3 (a} Suppose that x >-• y, then there is some B e :B such that x e 8, y e 
B, x e C(B), and y fl! C(B). Thus x >-• y. 
-
Suppose that y >-• x. then there 
-
exists B e 13 such that x e B, y e B and x e C(Bl. But the Weak Axiom implies 
that y e C(B), which is a contradiction. Hence if x >-• y, then we cannot have 
y >-• x. Hence x >-•• y. 
-
Converseiy, suppose that x >-•• y, then x >-• y but not y >-• x. Hence 
- -
there
is some B e :8 such that x e B, y e B, x e C(B) and if x e B' and y e B' 
for any B' e :B, then y E C(B'}. In particular, x e C(B) and y E C(B). Thus 
The equa!ity of the two relation is not guaranteed without the WA. As 
car. be seen from the above proof, the WA is not necessary to guarantee tha~ if 
x >-•• y, the:-1 x >-6 y. But the converse need not be true, as sho·wn by the 
following example. Define X = {X".y,z}, :B = {{x,y},{x,y,z}}, C({x,y}) = {x}. 
and C({x,y,z}) = {y}. Then x >-• y and y >-• x. But neither x >-• y nor y >-• x. 
(b) The relation >-• need not be transitive, as shown by the following example. 
Define X = {x,y,z}, :B = {{x,y},(y,z}}, C({x,y}) = {x} and C({y,z}} = (y}. 
But we do not have x >-• z (because neither of the two 
-
sets in 13 in::ludes {x,z}} and hence we do not have x >-• z either. 
(c) According to the proof of Proposition 1.0.2, if :B includes all three-
eler:1ent subset of X. then >-• is transitive. By Proposition l.B.Hil, >-•• is 
-
1-4 
transitive. Since >-• is equal to >-••, >-• is also transitive. 
An alternative proof is as follows: Let x e X, y e X, z e X, x >-• y, and 
y >-• z. Then (x,y,z} 4: ~ and, by (a), x >-•• y, and y >-•• z. Hence we have 
neither- y >-• x nor z >-• y. Since >-• rationalizes (!B,C( • )), this implies that 
- - -
y fl! C({x,y,z}) and z - C((x.y,z}). Since C({x,y,z}) :;e "· C{{x,y,z}) = {x). 
Thus x >-• z. 
1.0.1 The simplest example is X = {x,y}. !B = {{x},{y}}, C({x}) = {x}, C{{y}) 
= {y}. Then UJ rational relation of X rationalizes C( · ). 
1.0.2 By Exercise l.B.S, let u( ·) be a utility representation of t· Since X 
is finite, for any B c X with B ~ 0, there exists x e B such that u(x) l!: u(y) 
for a!l y e B. Then x e c-(B,_t} and hence c•(B.tl :;e "· (A direct proof with 
no use of utility representation is possible, but it is essentially the same 
as the proof m Exercise 1.8.5.) 
1.0.3 Su:~cose that the Weak Axiom holds.· If x e C(X), then x e C( {x,z} }, 
•• 
which cor:tradicts the equality C({x,z}) = {z}. If y e C(X), then y E 
C({x.y}), whid contradicts C({x,y}) = {x}. If z e C(X), ther.. z e C{ {v, z}), 
• 
which cont:radkts C( {x,z}) = {y}. Thus (~,C( ·)) must violate the Weak Axiom. 
1.0.4 Let t ntionali~ C{ ·) relative to !B. Let x e C(B1 v B2 l and y e C\B1) 
v CCB
2
), the~ x ~ y becango 8
1 
u B
2 
;::l C(B1l v CCB2 ). Tnus x e C(C(B1l v 
CCB2 )). 
Le: x e ctCCB1l v CCB2 )) and y e B1 v B2• then there a:-e four cases: 
Case 1. x e C(B1 ), y e Bl" 
• 
1-5 
Case 2. x·e C(B1), y E B2. 
Case 3. X e CCB2J. y e Br 
Case 4. x e C(B2l. y e 82. 
If either Case 1 or 4 is true, then x >- y follows directly fro:-::. 
-
rationalizability. If Case 2 is true, then pick any z e cm2 >. Then z >- y. -
• 
>- z. 
-
Hence, .by the transitivity, X >- y. 
-
If 
• 
Case 3 is true, then pick any z e C(B1l and do the same argurne:tt as for Case 
2. 
l.D.S (a) Assign probability 1/6 to each of the six possible preferences, 
which are x >- y >- z, x >- z >- y, y >- x >- z, y >- z >- x, z >- x >- y, and z >- y >-
x. 
(b) If the given stochastic choice function were rationalizable, then the 
probability that at least one of x >- y, y >- z, and z >- x holds would be at 
most 3 x {1/4) = 3/4. But, in fact, at least one of the three relations 
always holds, because, if the first two do not hold, then y >- x and z >- y. 
Hence the transitivity implies the third. Thus. the given stochastic choice 
function is not rationalizable . 
• 
{c) T!le same argument as in (b) can be used to show that a: 2: 1/3. Since 
C({x,y}) = C({y,z}} = C({z.x}) = (a:, 1 - a:) is equivalent to C({y,x}l = 
C( {z,y}) = C( {x,z}) = (1 - et, et), if we apply the same argurner.t as in (b) to y 
>- x, z >- y, a:1.d x >- z. then we can establish 1 - « ~ 1/3, that is. a: =: 2/3. 
Thus. in order for the given stochastic choice function is rationalizable, it 
is necessary that a: e [1/3,2/3]. Moreover, this condition is actually 
suffic!e:1.t: Fo:- any a: e [1/3,2/3), assign probability a: - 1/3 to each of x >-
1-6 
y >- z, y >- z >- x, and z >- x >- y; assign probability 2/3 - a to each of x >- z >-
y, y >- >- z, and z >- y >- x. Then we obtain the given stochastic choice 
function. 
1-7 
CHAPTER 2 
2.0.1 Let p2 be the price of the consumption good in period 2. measured in 
units of the consumption good in period 1. Let x1, x2 be the consum:~tion • 
levels in periods 1 and 2. respectively. Then his lifetime Walrasian budget 
• set 1s 
2.D.2 
2 
equal to {x e IR+: x 1 + PzXz :s w} . 
2 {(x,h) e IR : h :s 24, px + h :::!: 24}. 
+ 
2.0.3 (a) No. In fact, the budget set consists of the two points, each of 
which is the intersection of the budget line and an axis. 
(b) Let x e B , x' e B , and ~ e {0,1). Write x" = AX + (1 - ~)x'. Since p,w p,w 
X is convex, x" e X. Moreover, p·x" = ~(p·x) + (1 - ~)(p·x'J :::!: i\.w + (l - i\.)w 
= W. "T"\. .. !HUS X E B . p,w 
2.0.4 It follows from a direct calculation that consumption level M can be 
attained by (8 + (M - Ss)/s') hours of labor. It follows from the definition 
that (24,0} and {16 - (M - Ss)/s', M) are in the budget set. But their convex 
combination of these two consumption 
M - 8s 
• • • 
vectors Wltn rat1o 
s' 
-----;-M~-~8::::-s- ' 
8 + 
s' 
8 + 
8 
M- 8s 
s' 
is net in the budget set: the amount of leisure of this combination equals to 
16 (so the labor is eight hours), but the amount of the consumption good is 
8 M--~-~­M - 8s 8 + 
s 
8 
> M --.......,M...,...---~8::-s-
8 + 
8 
= M M/s = 8s. 
s 
2-1 
2.£.1 The homogeneity can be checked as follows: 
ctpl 
x (ctp ctw) = 
3 • ctp 1 + ctp2 + a.p3 
tl.W 
tl.W 
a.w 
-
-
-
-
-
-
w 
p 1 
w 
w 
To see if the demand function satisfies Walras' law, note that 
= x,(jJ,wl, 
l 
Hence p • x(p, w) = w if and only if (3 = 1. Therefore the demand function 
satisfies Walras' law if and only if J3 = 1. 
2.E.2 Multiply by plc/w both sides of (2.E.4), then we obtain 
l=l(Pr£'P·w)/wH8xl(p,w)/81\)(pk/xt'p,w)} + pkxk(p.w)/w = o. 
Hence l=lbl(p,wklk(p,w} + bk(p,w} = 0. 
By (2.E.6), l=l(plxl(p,w)/wHaxt(p,w)law)(w/xip,w)} = l. Hence 
l=l0!(p,w)clw(p, w) = 1. 
2.E.3. Tnere are two ways to verify that p · D x(p, wlp = - w. p 
One way is to post-multiply (2.E.5) by p, then p·Dpx(p,w} p + w = 0 by 
Walt as' law. 
Tne othe:- way is to pre-multiply (2.E.l) by pT, then p·D x{p,w)p + p 
p·DwJdp.w)w = 0. By Proposition 2.E.3, this is equal to p·Dpx(p,wjp + w = 0. 
An inte:-pretation is that, when all prices are doubled, in order for the 
to stay a: the same consumption, it is necessary to increase his 
wealth bv w . 
• 
2-2 
2.E.4 By differentiating the equation x(p,cxw} = cxx(p,w) with respect to a and 
evaluating at a = l, we obtain wD x(p, w} = x(p, w). Hence D x(o,w) = w \\' . 
(1/w)x(p,w). Hence ctw = (8xip,w)/8w)(w/xip,w)) = 1. This means that an 
• 
one-percent increase in wealth will increase the consumption level for- all 
goods by one per-cent. 
Since (1/w)x(p, w) = x(p,l} by the homogeneity assumption, D x(p,w) is a 
w 
function of p only. The assumption also implies that the wealth expansion 
path, E = {x(p, w): w > 0}, is a ray going through x(p,l). p 
2.E.S Since x(p, w) is homogeneous of degree one with respect to w, x(p,o:w) = 
o:x(p, w) for every o: > 0. Thus xt'p,w) = xip,l)w. Since 8xefp.l)/cpk = 
8cpip)/8pk = 0 whenever k :;e t, xt(p,l) is actually a function of pl alone. So 
we can write xt(p,w) = xt(pl Since x(p,w} is homogeneous of degree zero, 
xl(pl) must be homogeneous of degree - 1 (in pl). Hence there exists o:l > 0 
such that xt(pl} = «lPt· By Walras' law, Lt.Pt(alpt)w = w[lc::i = w. We must 
2.E.6 When a
= l, Walras' law and homogeneity hold. Hence the conclusions of 
• 
Propositions 2.E.l - 2.E.3 hold. 
2.E.7 Bv Walras' law, 
• 
Tnis demand function is thus homogeneous of degree zero. 
2-3 
2.E.8 For the fil"st part, note that 
Thus, by the chain rule, 
~ 
oxt 
a (p, w) • exp On pk) 
axl 
a {p,w)·pk d(ln xl(p,w)) pk pk 
~ lk(p, w) . • - - -- - -d(ln p,) xt(p,w) xl(p,w} 
iC 
Simiiarlv, 
-
cHIn x l( p, wl l 
d(ln\r,:} --
8xl 
aw (p, w) · exp(ln w) 
axl 
aw (p, w)w 
xl(p,w) = 
• 
Since o:1 = dOn xip,w))ld(ln p1), o:2 = d(ln xip,w))/d(ln p2 l, and a:3 = 
d(ln x iP· w) )/d(ln w), the assertion is established. 
Z.F.l We proved in Exercise l.C.Z that Definition l.C.l and the prope:-ty -in 
the exercise is equivalent. It is easy to see that the latter is equivalent 
to the foUowing property: For every B E :B and B' e ~. if C(B) " 8' :;c 0 and B 
" C\8') .: e, t!ien C(Bi "B' c C(B') and B" CCB') c CCB). If C{ · l is 
· , , · ~ · · · · 1 h f 11 . F smg.:.e-va.uec, t;.en trus property 1s equ1va ent to t e o owmg one: o:o-
ever-•: 8 E 73 anci B' e B, if C(8) c B' and B c C(B' ), then C(B) = C(B' ). In the 
- . 
. ' 
contex: of Walrasiar: demand functions, this can be resta-;ed as follows: For 
any (p,v:) and (p',w'), if p·x(p',w') :s wand p'·x(p,w) :s w', the:t x(p,w) = 
x(p', \II'). But this is the contraposition of the property stated in Definition 
2 - 1 . t .... He:1ce Definitions l.C.l and Z.F.l are equivalent. 
2.F.2 It is straightfor-ward to check that the Weak Axiom holds. In fact, if 
• • I I 
0 • • X"' 8 . . :!: anc l 
• 
• 
- J • P-e·~~:~~ ... -e., ·~ x ' & .....,, I - ~'W • 
• • 
= j, then pJ·x 1 = 9. S . 2 1 mce p ·x 
S . .. I lr:lllar" y r . I 3 8 I . Stnce p ·X = , X 15 
2-4 
= 8, x2 is revealed 
revealed prefe~:-ed to 3 x. 
But, since 3 2 3 2 p · x = 8, x i's revealed preferred to x . 
2.F.3 [First prin~ing errata: Add the sentence "Assume that the weak axiom is 
satisfied." in (b) and (c).] Denote the demand for good 2 in year 2 by y. 
(a) His behavior violates the weak axiom if 
100 · 120 + lOOy ::5 100 ·100 + 100 · 100 
and 
100·100 + 80·100 ::5 100·120 + SOy. 
That is, the Weak Axiom is violated if y e [75,80). 
(b) The bundle in year 1 is revealed preferred if 
100 ·120 + 100y ::5 100 ·100 + 100 ·100 
and 
100 ·100 + 80 ·100 > 100 ·120 + SOy, 
that is, y < 75. 
(c) The bundle in year 2 is revealed preferred if 
100·100 + 80 ·100 ::5 100 ·120 + SOy 
and 
100 ·120 + lOOy > 100 ·100 + 100 ·100, 
that is, v > 80 . 
• 
(d) For any value of y, we have sufficient information to justify exactly one 
of (a), (b). and (c). 
(e) We shall prove that if y < 75, then good 1 is an inferior good. Sc 
suppose that y < 75. Then 
2-5 
100·"120 + 100y :s 100·100 + 100·100 
and 
100 ·100 + 80 •100 > 100 ·120 + 80y. 
Hence the real wealth decreases from year 1 to 2. Also the rela:ive price of 
good 1 increases. But the demand for good 2, y, decreases because y < 75 < 
100. This means that the wealth effect on good 1 must be negative. Hence it 
is an inferior good. 
(f) We shall prove that if 80 < y < 100, then good 2 is an inferior good. So 
suppose that 80 < y < 100. Then 
100·100 + 80·100 :s 100 ·120 + 80v 
• 
and 
100 ·120 + 100y > 100 · 100 + 100 · 100. 
Hence the real wealth increases from year 1 to 2. Also the relative price of 
good 2 dec:-eases. But the demand for good 2, y, dec::-eases because y < 100. 
This mea .. ls tha-t the wealth effect on good 2 must be negative. Hence it is an 
. ~ . . m: er1or gooc. 
the co:1sume::- has a revealed preference for x 0 over x 1. 
(b) If PQ >!,then (p1·x1)/(p1·x0 ) > 1 and hence p1·x1 > p 1 ·x0 . Thus the 
consume:- has a revealed preference for x1 over x0 . 
= i\. Hence, by taking i\ 
larger or smalle; than one, we can make EQ larger or smaller than 
this obviously does not have any revealed preference relationship. 
2-6 
one. But 
2.F.5 We shall first prove the discrete version. By the homogeneity of 
degree one with respect to wealth, it is enough to show that 
(p' - p) · (x(p' ,1) - x(p,1}) ~ 0 for every p and p'. 
Since 
x(p' ,1) - x(p.ll = p . X p, . 
1 
+ (x(p, p' ·x(p,1) ) - x(p,l)), 
it is sufficient to show that 
(p' - p) • (x(p' ,p' · x(p,l)) - x(p,l)) ~ 0, 
and 
1 (p'- p}·(x(p,-p-;-·-.x......,-(p-,-:-1 ~1 )- x(p,l)) :s 0 . 
• 
For the first inequality, note that 
(p'- p)·(.~c(p',p'·x(p,1))- x(p,1)) =- p·x(p',p'·x(p,ll) + 1. 
If x(p',p' ·x{p,1)} = x(p,l), then the value is equal to zero. If 
x(p',p'·x(p,l)l ;= x(p,l), then the weak axiom implies that p·x(p',p'·x(p,lll > 
1. Hence the above value is negative. 
As for the second inequality, 
---.-___;,.----,..,.... ) - x( p ,1 } ) p ·x p, 
-
-
= 2 - (p' ·x(p,l) + 1 p' ·x(p,l) ) 
:s 2 - 2 (p' ·x(p,l))( 1 p' ·x(p,l} ) 
= 2 - 2 = 0. 
1 
+ 1 • p' · x(p,ll 
The infinitesimal version goes as follows. By differentiating x(p,aw) = 
ax(p, w) with respect to a and evaluating at a = 1, we obtain Dwx(p, w)w = 
x(p, w). Hence 
2-7 
Thus 
S(p, w) =· D x{p. w) p 
T 
+ D x(p, w)x(p, w) = D x(p, w) 
w p + (1/w)x{p,w)x(p,w)T. 
D x(p,w) = S(p,w) - (1/w)x(p,w}x(p,w)T. 
p 
By Proposition 2.F.2, S(p,w) is negative semidefinite. Moreover, since 
v·(x(p,w)x(p,w}T)v =- (v·x(p,w))2 , the matrix- (1/w)x(p,w)x(p,w)T is also 
negative semidefinite. Thus D x(p,w) p is negative semidefinite. • 
2.F.6 Clearly the weak axiom implies that there exists w > 0 such that for 
every p, p', and w', if p·x(p',w') ~wand x(p',w') :;e x(p,w), then p'·x(p,w) > 
w. 
• 
Conversely, suppose that such a w > 0 exists and that p · x{p', w'} ~ w and 
x(p',w') :;e x(p,w). Let a: = w' /w. -1 Then x(p' ,w') = x(p' ,a:w) = X(IX p', w} by 
the homogeneity assumption, -1 -1 and p·x(a: p',w} ~ w and x{a: p',w) :;e x{p,wl. But 
this implies that (IX-lp')·x(p,w} > w, or, equivalently, p'·x!p,w) > a.w = w'. 
Tnus the weak axiom holds. 
2.F.7 By Propositions 2.E.2 and 2.E.3, 
• 
- . 
T T p·.S(p,w) = p·Dpx(p,w) + p·Dwx(p,w)x(p,w) = p·Dpx(p,w) + x(p,\'11") = 0 
By Proposition 2.E.l and Walras' law, 
S(c w)c = 
• • • 
2.F.8 
D x(p,w}o 0 • 
T 
... Dwx(p,w)x(p,w) p = 
• 
-
-
pk 
D. OXi. 
• lC 
--.--..... -::::--- ( p, w) 
xl ( p, w) 8pk 
w 
= c //' ( 0. w) + 
'"K" xi.(p,w) 
= clk(p,w) + ci.w(p,w)bk(p,w). 
2-8 
D x(p,w)p + D x(o, w)w = 0 . p w . 
2 F 9 ( ) S . TAT ( TA )T TA . A . . . r" . . ~ . • a mce x x = x x = x x, a matr1x 1s negative ae. m1te 1. 
and only if x TAx + x TAT x < 0 for every x n e IR \{0). S . TA TAT 1nce x .. x + x x = 
T T T 
x (A + A }x, this is equivalent to the negative definiteness of A + A . Thus 
A is negative definite if and only if so is A + AT_ The case of negative 
definiteness can be proved similarly. 
The following examples shows that the determinant condition is not 
sufficient for the nonsynunetric case. Let A = - 1 0 3 _ 1 , then A11 = - 1 
and A22 = 1. 
semidefinite. 
But U, 1) - 1 3 
0 
- 1 
• 
1 
1 • 1. Hence A is not negative 
(b) Let S(p, w) be a substitution matrix. By Proposition 2.F.3, S(p, w)p = 0 
and hence s 12Cp,w) = (- P/P2 >s11(p,w). Also p·S(p,w) = 0 and hence s 21 Cp.w) 
2 2 
= (- p11p2 ls11Cp,w). Thus s22(p,w) = (p1/p2 >s11(p,w). Thus, for every v = 
vl 
v 2 
• 
v·S(p,w)v 
Now, su:ppose that S(p, w) is negative semidefinite and of rank one. Ac::ording 
to (•J, the nega~ive semidefiniteness implies that s 11(p,w} ~ 0. Being o:· 
rank one implies that s 11 (p,w} :;!: 0. Hence s 11(p,w) < 0. Tnus s 22Cp,w) < 0. 
Conversely, let s11Cp.w) < 0, then, by (•), v·S(p,w)v ~ 0 for every v. 
2.F.10 (a) If p = (1,1,1) and w = 1, then, by a straightforward caicu!ation, 
• • we ootam 
- 1 1
0 
S(p,w) = (1/3) 0 - 1 1 • 
1 0 - 1 
2-9 
Hence S(p, w) is not symmetric. Note that 
- 1 
0 
1 
- 1 
2 
=-v +vv-
1 1 2 
2 
v = - (v -2 1 
• 
2 3v2/4. 
Hence - 1 I 0 - 1 is negative definite. Thus, by Propositi"on 2.F.3 and 
Theorem M.D.4(iii), S(p,w) is negative semidefinite . 
... 
(b) Let p = (l,l,d and w = 1. Let S(p,w} be the 2 x 2 submatrix of S(p,wl 
obtained by deleting the last row and column. By a straightforward 
calculation, we obtain 
S(p, w) 
Thus, 
(1, 4, O)S(p,w) 
1 
4 
0 
- 2 - c 
0 
1 + 2c 
- 3c • 
A 
= (1, 4}S(p, w) 1 4 = (2 + cl-
2£2 - 4lc) > 0, 
if £ > 0 is sufficiently small. Then S(p, w} is not negative semidefinite and 
hence the demand function in Exercise 2.E.l does not satisfy the Weak Axiom. 
2.F.ll By Pro;>osition 2.F.3, S(p, w}p = 0 and hence s 12(p, w) = 
(- P/P2 )s11(p,w). Also p:S(p,w} = 0 and hence s 21 (p,w) = (- p1/p2 >s11£p,w). 
(We sa\'1 this in the answer for Exercise 2.F. 9 as well.) Thus s 12(p, w) = 
s21 (p, w). 
2.F.l2 By a;:~plying Proposition 1.0.1 to the Walrasian choice s:ructure, we 
know that x(p, w} satisfies the weak axiom in the sense of Defir.itior. 1. C.l. By 
Exercise 2.F.l. this implies that x(p,w} satisfies the weak axiom in the sense 
of Definition 2. F .1. 
2-10 
2.F.l3 [First printing errata: In the last part of condition (•) of (b), the 
inequality p · x > w should be p' · x > w'. Also, in the las7. part of (c), the 
relation x' e x(p,wl should be x' E x(p,w).} 
(a) We say that a Walrasian demand correspondence satisfies the weak axiom if 
the following condition is satisfied: For any (p, w} and (p', w' ), if x e 
.x(p,w), x' e x(p',w'J, p'·x::!: w', and p·x'::!: w, then x' e x(p,w). Or 
equivalently, for any (p,w} and (p',w'), if x E x(p,w), x' e x(p',w'), p·x' ::!: 
w, and x' E x(p, w), then p' · x > w'. 
(b) If x e x(p,w), x' e x(p',w'), and p·x' < w, then x' E x(p,w) by Walras' 
law. Thus p'·x > w'. 
(c) If x e x(p,w), x' e x(p',w'), and p'·x = w', then (p'- p)·(x'- xl = w-
p·x'. If, furthermore, x' e x(p,w), then Walras' law implies that p·x' = w. 
Hence (p' - p) · (x' - x) = 0. If, on the contrary, x' E x(p, w), then the 
generalized weak axiom implies that p·x' > w. Hence (p' - p) · (x' - x) < 0. 
(d) It can be shown in the same way as in the small-type discussion of the 
p::-oof of Proposition 2.F.l-that, in.order to verify the asse::-•.:~·: .. it is 
sufficient tc show that the generalized weak axiom holds fer aH compensated 
price changes. So suppose that x e x(p,w), x' e x(p',w'l, p' ·x = w', a:1d p·x' 
::!: w. Then (p' - p) · (x' - x) = w - p · x' 2:: 0. Hence, by the gene.-alized 
co::npensated law of Demand, we must have (p' - p) · (x' - x} = 0 and x' e x(p, w}. 
2.F.l4 Let p » 0, w 2!: o, and a> 0. Since p·x(p,w)::!: w an:i (o:pl·x(o:p,o:w) ~ 
o:w, we have o:p·x{p,wl ::!: o:w and p·x(a.p,o:w) ::!: w. .. . . . . The wea:< ax1o:r: r:ow 1mpnes 
2-11 
that x(p, w) = x(a:p,o:w). 
2.F.15 Since oxip, w)/8w = 0 for both l = 1,2, we have slk(p, w) = 
-
8x£'p,w)/8pk for both l = 1,2 and k = 1,2. Hence, let S(p,w) be the 2 x 2 
submatrix of S(p,w) obtained by deleting the last row and column, then S(p,w) 
-
-
- 1 
0 
1 
- 1 • Tnis matrix is negative definite because 
- l 
0 
1 
- 1 
vl 
v2 
2 
v -2-
(We saw this in the answer to Exercise 2.F.lO(a).) Hence, by Theorem 
-
M.D.4(iii), v·S(p,w)v < 0 for all v not proportional to p. Since S(p,w) is 
net S)-m.me:ric, S(p,w) is not symmetric either. 
2.F.l6 (a) The homogeneity can be checked as follows: 
xl (a:p,a:w) = o:p2/a:p3 = p2/p3 = xl (p, w), 
x 2£o:p,a.w) = - o:p/a:p3 = - P/P3 = x 2Cp,w), 
x3 (c:;::,o:w) = IXW/o:p3 = w/p3 = x 3 (p,w). 
As for Wal:-as' law, 
• 
~ v ,~ - •••. ::'-. • ./ y.-_ - 'IY 
.. ; ..) 
(b) p = (1,2,1), w = l, p' = (1,1,1), and w' = 2, then x(p,w) = 
(2, - l, 1} and x(p', w') = (1, - 1, 2). Thus p' · x(p, w) = 2 • = \V • anc 
p · x(p', w') = 1 = w. Hence the Weak Axiom is violated. 
-(c) Denote by Dx{p, wl the 2 x 2 submatrix of the Jacobian matrix Dx(p, w) 
obtained by de~eting the last row and column, then 
-
- 1 
0 
1 
- 1 • 
Let S(p, w} be the 2 x 2 submatrix of S(p, w) obtained by deleting the last row 
2-12 
-
and column, then S(p, w) - - 1 = Dx(p, w) = 0/p
3
) 0 
1 
- 1 1 because ax1(p,w)/8w = 
Note that 
" ., A - 2 
v·S(p,w)v = 0 for every v e ~ . Now let v 
Note that v = (v - Cv3/p3 )p) + Cv3/p3 )p and the third coordinate of v -
So denote its first two coordinates 
- ... ... 
Then, by Proposition 2.F.3, v·S(p,w)v = v·S(p,w)v = 0. 
- 2 by v e IR • 
(c) Suppose that p' ·x(p,w) ~ w' and p·x(p' ,w') ~ w. The first inequality 
3 
E !R • 
implies that C[tP£lw/C[lpt) ~ w', that is, w/([tpt) :s w' /([tPe>· Tne second 
inequality implies similarly that C[tpt)w' /([tpt) :s w, that is, w' /([tpt) :s 
w/([lpl). Therefore w/([lpl) = w'/([tpt). Hence x{p,w) = x(p',w'). Thus the 
weak axiom holds. 
(d) Bv 
• 
calculation, we obtain 
1 • • • 1 
2 • • D x(p,w) - (- w/([tpt) ) • • - • p • • 
1 • • • 1 
1 
• 
= x(p, w). • 
• 
1 
Hence S(p, w) = 0. It is symmetric, negative semidefinite, but not negative 
definite. 
2-13 
• CHAPTER 3 
3.B.l (a) Assume that t is strongly monotone and x » y. Then x ~ v and x ;: 
-
y. Hence x >- y. Thus >- is monotone. 
-
(b) Assume that >- is monotone, x e X, and £ > 0. 
-
L Let e = (1, ... ,1) e IR and 
• 
y = x + (c/v'L)e. Then U y - x II $ c and y >- x. rnus >- is locally nonsatiated. 
- . 
3.B.2 Suppose that x » y. Define £ = Min {x1 - y1, ... , ~ - yL} > 0, then, 
for every z e X, if hy - zH < £, then x » z. By the local nonsatiation, 
• 
there exists z• e X such that lly - z•u < £ and z• >- y. By x » z• and the 
weak monotonicity, x >- z•. By Proposition l.B.Uiiil (which is implied by the 
-
transitivity). x >- y. Thus >- is monotone. 
-
3.8.3 Foliowing is an example of a convex, locally nonsatiated preference 
relation that is not monotone in IR2. 
+ 
- • 
X 
• y 
0 
i 
0 
• 
• 
For example, x » y but y >- x. 
• 
XI 
Figure 3.B.3 
3-1 
• 
3.C.l Let ~ be a lexicographic ordering. To prove the completeness, suppose 
Thus y >- x. 
-
To prove the transitivity, suppose that x ?: y and y ?: z. Then x
1 
~ y
1 
• 
and z1• then x ?: z. 
Hence x2 !::: z2. 
z1, then X = 1 
Thus x >- z. 
-
To show that the strong monotonicity, suppose that x i!: y and y :;c x. This 
• 
either case x >- y. 
To show the strict convexity, suppose that y >- x, z >- x, y :;c z, and o: e 
- -
(0,1). Without loss of generality, assume that x :;e y. By the definition of 
the lexicographic: ordering, we have either "y1 > x1" or "y1 = x1 a.'ld y2 >. x2". 
On the other hand, since z z: x, we have either "z1 > x1" or "z1 = x1 and x 2 i!: 
V II 
J 2' Hence, we have either "a:y 1 + (1 - o:)z1 > X/' or "o:y 1 + (1 - cdz 1 = x 1 
and a:y2 + (1 - cdz2 > x 2. II Thus a.y + (1 - o:}z >- x. 
3.C.2 Take a sequence of pairs n n such that x >- y for all n, n X 
-
x. and yn .. y. Then u(xnl "!::: u'(yn) for all n, and the continuity of u( ·) 
impiies that u(xl !::: u(y). Hence x >- y. Thus >- is continuous. 
- -
• 
• 
3.C.3 One way to prove the assertion is to assume that >- is rnor.otone and 
-
notice that the p:-oof actually make use only of the closedness of uppe:- and 
lower contour sets. Tnen the proposition is applicable to >-, implying that it 
-
has a continuous utility function. Thus, by Exercise 3.C.2, >- is continuous. 
-
A mere direct proof [without assuming monotonicity or using a utility 
function) goes as follows. Suppose that there exist two sequences {xn} and 
3-2 
{yn} in X such that xn ?:' yn for
every n, xn -? x e X, yn -+ y e X. a:1d y >- x. 
Since {z: y >- z} is open, there exists a positive integer N
1 
such that y n >- X 
for every n > Nl" Since {z: z >- x} is open there exists a positive integer 1':2 
such that n y >- x for every n > N2. Conceivably, there are two cases on the 
n 
.sequence {y }: 
• 
Case 1: There exists a positive integer N3 such that yn >- y for every n > N . . - 3 
. k(n) k(n) Case 2: There ex1sts a subsequence (y ) such that y >- y for every n. 
If Case 1 applies, then, by Proposition l.B.Hiii), n n we have y >- x for every 
• 
n > Max {N1,N3 }. This is a contradiction. If Case 2 applles, then there 
n k(m) 
open, the:-e exists a positive integer N 4 such that y >- y ) fo:- every n > 
n n n k(m) . By x ?:' y and Proposition l.B.l(iii), x >- y for every n > N4. · 
. { k(m)} Smce z: z >- v 
- -
.. d' . con .. ra 1ct1on. 
. k(m) 1s closed, x >- y . 
-
But, since k(m) > N2, this is a 
3.C.4 We p:-cvid.e two examples. The first one is simpler, but the second o:1.e 
satisfies mono~:)nicity, which the first does not . 
• 
- . 
Exar:1-:>le 1. 
-
Let X = IR and define u( · ): IR -+ IR by letting u(xl = 0 fo:- x < ;,, 
+ + 
u(x) = 1 for x > 1, and u(l) be any number in (0,1}. Denote bv >- the 
- -
preference relation represented by u(- ). We shall now prove that >- is not 
-
continuous. Ir:. fact, if u(l) > 0, then consider a sequence {xn} with xn = 1 -
1/n fer everv r... 
-
Although n n x - 0 for every n and x -+ l, we have 1 >- 0. If 
u( l) < l, the:l consider a a sequence {xn} with xn = 1 + 1/n for every n. 
Although xn - 2 fer every n and xn ~ 1, we have 2 >- 1. Note that if u(x) = 0, 
then all lower contour sets are closed. If u(l) = 1, then all uppe:- contou:-
. ~ sets are Close ..... 
3-3 
• 
2. 
-
following rule: 
Case 1. If x 1 + x2 :s 2 and x :;e 0,1), then u(x) = x1 + x2 .. 
Case 2. If min{x1 ,x2> 3!::: 1 and x :;e (1,1), then u(x) = min{x1 ,x2} + 2. 
Case 3. If x1 + x2 > 2., min{xl'x2} < 1, and x1 > x2, then 
• 
• 
Case 5. u(l,l) e [2,31. 
The indifference curves of the prefezoence relation >- represented by u( · ) are 
-
described in the following picture: 
2 
I ... ·----····· . 
• 
• 
• 
0 
• 
0 
• 
• 
• 
• 0 
0 
) 2 
Figure 3.C.4 
• 
• 
XI 
It follows from this construction that u( ·) is continuous at every x :;e (1,1). 
Tne preference >- is convex and monotone. But, whateve: the choice of the 
-
value of u(l,l) is, it cannot be continuous at (1,1). ln fact, 
(1 - 1/n, 1 - l/n) -7 (l,ll and (1 + 1/n, 1 + 1/n) -7 (1,1) as n -7 cc, and 
u(l - 1/n, 1 - lin) = 2 - 2/n -+ 2; 
3-4 
u(l + 1/n, 1 + 1/n) = 1 + 1/n + 2 -+ 3. 
Hence, if 2 < u(l,l), then (2,0) >- (1 - 1/n, 1 - 1/n) but (l,ll >- (2,0}; if 
-
u(l,l) < 3, then (l + 1/n, 1 + 1/n) >- (2,1) but (2,1) >- (1,1). If u(l,l) = 3, 
-• 
then all upper contour sets of >- are closed; if u(l,l) = 2, then all lower 
... 
contour sets of >- are closed. 
-
• 
• 
3.C.S (a) Suppose first that u( ·) is homogeneous of degree one and let o: l!:: 0, 
L 
xe!R,y 
+ 
L 
e IR +, and X - y. Then u(x) = u(y) and hence cw(x) = o:u(y). By the 
homogeneity, u(o:x) = u(o:y). Thus o:x - a.y. 
Suppose conversely that >- is homothetic. We shall prove that the utility 
-
function constructed in the proof of Proposition 3.C.I is homogeneous of 
• 
degree one. Let x e IRL 
+ 
and o: > 0, then u(x)e - x and u(ax)e - ax. Since >- is 
-
homothetic, o:u{x)e - o:x. By the transitivity of - (Proposition l.B.l(ii)), 
• 
u(ax)e - au(x)e. T.'lus u(ax) = o:u(x). 
(b) Suppose fir-st that t is represented by a utility function of the form u(x) 
Let o: e IR, L L X E IR , y e IR , and X - y. 
+ + 
Then u(x) = u(v) 
• 
and hence u(x} + a = u{y} + a. ~ the functional form, 
u(xj + o: = (a + x1} + ~(x2, .... xL) = u(x + a.e11, 
u(yl + a = (a + y1) + ~(y2, ... ,yL) = u(y + ae1). 
where e. 
L 
= (l,O, ... ,0} e ae1 - y + 
Suppose conver-sely that ::;- is quasilinear with respect to the first 
c~mmodity. T."!e idea of the proof of this direction is the same as in (a) or 
Propositi or.. 3. C.l, in that we reduce comparison of commodity bundles on a line 
by fining ot:t i.-:.different bundles and then assigning utility levels along the 
line. But this proof turns out to exhibit more intricacies, partly because it 
3-5 
depends crucially on the connectedness of IRL-1, which appears in 
+ 
X = {- a:, m) 
• 
L-1 
x lR + . (Connectedness was mentioned in the first small-type discussion in 
the proof of Proposition 3.C.l.) The proof will be done in a series of steps. 
First, we show that comparison of bundles can be reduced to a line parallel to 
Then we show that the quasilinearity of >- implies the given functional 
-
form. 
• 
-Let >- be a quasilinear preference and a utility function u{ ·) represent 
-
->-. The existence of such a u( ·) is guaranteed by Proposition 3.C.l, but, of 
-
course, it need not be of the quasilinear form. 
• 
A L-1 
For each x e G< + , define 
,. - ... 
I(x) = {u(x1.x> e IR: x 1 e IR}, then J(x} is a nonempty open interval, by the 
• 
continuity and the strong monotonicity of ~ along e1 . 
• 
Step 1: For every x e 
= 0. 
L-1 IR 
+ 
... 
and y e 
- -
L-1 
IR • + i.f J(x) :;e l(y), then Hxl 
,. 
,... l(y) 
Proof: Suppose that l(x) :;e l(y). Without loss of generality, we can assume 
... 
that there exists u e J(x) such that u IC l(y). Then either u 2: s•;p I(y} or-
- ... • 
u ::5 infl{yl. Suppose that u 2: supl(y). (The other case can be treated 
-
simila.:-ly.) Then let xi e ~ sa~isfy. u = u(xi,x), then, for every y 1 e IR, 
- - .. 
(y1 - x 1 + xi, y). By the quasilinearity, this implies that (x1,x) >-
- - -
- -(y 1,y). Thus uCx 1.x) > u(y l'y). Hence l(x) n l(y) = 0. 
.. L-1 
For each x e ~. , 
... 
- - L-1 - -define E(x) = {y e IR : l(x) = l(y)}. 
+ 
Step 2: For every - L-1 "' x e IR , E(x} is 
+ 
· . L-1 
open tn IR 
+ 
• 
Proof: - L-1 Let x e IR , 
+ 
-x1 e IR, and u = u(xl'x) e Hx). 
- - Let c > 0 satisfy 
-(u - c, u + d c !(x). Since u( ·) is continuous, there exists o > 0 such that 
L -1 
- . u v e ~"~ 
- + 
-
- ... 
- -
. - • an c. llx - vii 
• 
- -< ~. then I u(xl'xJ - u(x1,yl I < £. Hence 
• 
• 
3-6 
• 
- - -
I(x) " l(y) :~ (u - c, u + c) r. Hyl = 0. 
-
... 
Thus, by Step 1, l(x) = l(y), or y E E(x). Hence - L-1 {y e IR+ : fix - vii < o} 
-
- -
c E(xl. Thus E(x) is open. 
Step 3: For every 
Proof: It is sufficient to show that for - L-1 - L-1 every x e IR anc! y e IR , 
+ + 
we 
S h th . . ... IRL-1 d uppose not, t en ere ex1st x e + an 
... L-1 
y e ~ such 
+ 
have E(x) = £(y). 
- - L-1 "' 
that E(x) = E(y), then the complement IR 'E(x) is nonempty. By Step 1, 
+ 
• 
L-1 "' " ... L 1 -lR+ ~(x) is equal to the union of those E(y) for which y e IR +- ~(x). By 
Step 2. this implies that IRL-l~(;) is open. Hence we have obtained a 
+ 
... L-1 ... L-1 
partition {E(x},IR + ~(x)} of IR + , both of whose e~ements are nonempty and 
open. This contradicts the connectedness of IRL-I. Hence E(~) = E(;) for 
+ 
• 
... L-1 
every x e IR 
+ 
- L-1 IR . 
+ 
and v e 
• 
-
- L-1 Thus for every x e IR , there 
+ 
By Ste? 3, I(x) I(O) for - L-1 every x e IR . 
+ 
... L-1 ... 
Define f/J: IR+ -+ IR by ¢(x)e1 -exists a unique a e IR such that o:e1 - (O,x}. 
-
every x e X. 
-L-1 e ;:'1 
... 
• Define u: X -+ IR by u(x) = x1 
• 
• 
Step 4: The function u.( ·l represents >-. 
-
• 
fc:-
Proof: Suppose that x e X, y E X, and x >- y. By the quasilinearity, this is 
-
equivalent to (x1 - y1, x2 , ... , xL) ~ (O,y2 , ... ,yL}. By the definition of 
¢( ·}. this
is equivalen-:: to Cx1 - y1, x2 , ... , xL} ~ ¢(y2 , ... ,yL)el" Again by 
the quasilinearity, this is equivalent to 
Again by the definition of ¢( · ), this is equivalent to 
¢(x2, ... , XL)el ?= (¢(y2, ... ,yL) + yl - xl)e1. 
Hence ¢(x2, ... , xL) 2: q)(y2, ... ,yL) + y1 - x1, that is, u(x) 2: u(yl. 
• 
• 
3-7 
These properties of u( · ) are cardinal, because they are not preserved 
under some monotone transformation, such as f(u(x)) = u(x?. 
3.C.6 (a} For p = 1, we have u(x) = o:1x 1 + o:2x 2. Thus the indifference 
• 
curves are linear. 
• 
(b) Since every monotonic ·transformations of a utility function represents the 
same preference, we shall consider 
By L'Hopital's rule, 
-lim u(x) 
p .. O 
- 1 im (o: 1x~lnx 1 - p ... O 
Since expUet1 + we have obtained a Cobb-Douglas utility 
function. 
T..1ere is an alternative proof to this proposition: Since both the CES and 
• 
the Cobb-Dougias utility functions are continuously differentiabie and 
• 
• 
• 
• 
homothetic, i: is sufficient to check the convergence of the marginal rate of 
substitu:ion at eve:y point. The marginal rate of substitution at (xl'x2 l 
• 
p-1 p-1 
with respect tc the CES utility function is equal to o:1 x 1 lc::2x 2 . The 
mar gina! rate of substitution at (x1,x2 ) with respect to the Cobb-Dougias 
p-1 p-1 
as p ~ 1. (In fact, a: 1 x 1 let2x 2 is well defined for eve:-y p a!1d is equal to 
o: 
1 
x
1
11X2x 2 when p = 1.) The proof is thus completed. 
St:-ictly speaking. there is a missing point in both proofs: We proved the 
conver-ge:.1ce o:~ preferences on the strictly positive 2 orthant {x e IR : x » 0}, 
3-8 
• 
but we did not prove the convergence on the horizontal and vertical axes. In 
fact, the convergence on the axes are obtained in such a way that all vectors 
there tend to be indifferent. To be more specific, compare, for example, x = 
Cx1,o> and y = (yl'O} with x1 > y1 > 0. According to the CES utility 
• 
function, x is preferred to y, regardless of the values of p. But, according 
to the Cobb-Douglas utility function, x and y are indifferent.. Futhermore, 
• 
the following is true: If x is in the strictly positive orthant and y is on an 
axis, then x is preferred to y for every p sufficiently close to 0. To see 
. 
this, simply note that if x = (x1,o) with x 1 > 0 and y » 0, then 
and a1~ + a2~ -+ a 1 + a2. The implication of this fact is that, as p -+ 0, 
every vector in the strictly positive orthant becomes preferred to all vectors 
on the axes. That is, unconditional preference towards strictly positive . 
vectors tends to hold, as it is true for the Cobb-Douglas utility function. 
(c) Suppose that x1 :s x2 . We want to show that 
. p p l/p 
x1 = 11m £o:1x 1 + o:2x 2 l -p-t -co 
Let p < 0. Since x1 ~ 0 and x2 ~ 0, 
p p 1/p . ' 
?! (o: 1 x 1 + a:2x 2 ) . On the other hand, 
• smce x1 :s x2 . 
Hence 
p (o:l + o:2)xl. 
Therefore, 
l/p > ("' xp p)l/p ( )1/p 
al xi - Yo! 1 + a2x2 i!: al + a:2 xl. 
p p)l/p Letting p -+ - CXl, we obtain lim Co:1x 1 + a 2x 2 = xl' because p-+ -co 
. . ' 
1lffi \C::l X •. L p-+- a: 
3.0.1 To check condition (i), 
3-9 
1 im 
P .... -co 
Thus 
x 2 (i\p,i\w} = (1 ...: o:.HAw)/(Ap2 ) = (1 - a:)w/p2 = x 2 (p,w). 
To check condition (ii), 
• 
plxl(p,w) + p2x2(p,w) = pla:w/pl + p2(1 - cx)w/p2 = w. 
Condition (iii) is obvious. 
3.D.2. To check condition (i), • 
• 
v(i\p,ll.w) 
= a:lna: + Cl - a:)ln(l - a:) + IM.w - cxlllAp -1 
= a:lna: + (1 - o:.)ln(l - et} + InA + lnw 
- a:lni\ - cdnp1 - (1 - «)lnA - (1 - cxHnp2 
• 
= edna: + (1 - a:HnCl - a:) + lnw - cdnp
1 
- (1 - a)lnp2 
c v(p,w). 
To check condition (ii), 
8v(p, w)/8w = 1/w > 0, 
cv(p, w)/8pl = - alp! < 0, 
ov(p, w)/8p2 = - (1 - a:)/p2 < 0. 
Condition (iv) follows the functional fox·m of v( • ). 
• 
In order- to ver-ify i iii)', by 'Property (i}, it is sufficient -;,~ !)!"'ove 
• 
tha:., for anv v e ~ and w > 0, the set {p e IRL : v(p,w) ~ v} is convex. Since 
. - ++ 
the logari:hrni:: function is concave, the set 
is convex for eve':"y v e IR. Since the other terms, 
a:lna: + (1 - a.)ln(l - IX) + lnw, 
de not depend on p, this implies that the set L {p e IR : 
++ 
v(p, w) ~ v} is convex. 
3.0.3 (a) We shall prove that for every p e w i!: 0, a i!: 0, and x ~L 1"f e "' ' + 
• 
3-10 
x = x(p,w), then ax = x(p;aw). Note first that p· (ax) :s aw, that is, ax is 
affordable at (p,aw). Let y L e IR and p · y :s a:w. 
+ 
• 
-! Then p· (a yl :s w. Hence 
-1 
u(IX yl :s u(x). Tnus, by the homogeneity, u(y) :s u(crx). Hence a:x = x(p,a:w) . 
• 
By this result, 
v(p,o:w) = u(x(p,a.w)} = u(cxx(p,w)) = «X"(x(p,w)) = av(p,w). 
Thus the indirect utility function is homogeneous of degree one in w. 
• 
-
• 
Given the above results, we can write x(p,w) = wx(p,l) = wx(p) and v(p,w) 
• 
-
= wv(p,l) = wv(p). Exercise 2.E.4 showed that the wealth expansion path 
-{x(p,w): w > 0} is a ray going through x(p). The wealth elasticity of demand 
• 
C JJ lS C.T/v• eoual to l. • 
(b) \Ve firs: prove that for every p e IRL , w :.?:. 0, and a. :: 0, we· have X(p,o:w) = 
++ 
ax(p,w}. In fact, since v(·,·) is homogeneous of degree one in w, Vpv(p,cxw) = 
aVpv(p,w) and Vwv(p,IXW) = Vwv(p,w). Thus, by Roy's identity, x(p,aw) = 
ax{p, w}. 
NOYl 1,.. -~ 
L 
X E IR • 
+ 
x' e IRL, u(x) = u(x'), and o: :: 0. 
+ 
Since u( · l is strictly 
quasiconcave, by the supporting hyperplane theorem (Theorem M.G.3), the::-e 
1 
exist p e r;.-io •• , 
++ P ' e !RL , w :: '0, and w' i!:: 0 such that x = x(c,w) and x' = ++ . 
x(p',w'}. Then u(x) = v(p,w) and u(x') = v(p',w'). Hence vlp,wi = v(p',w'} . 
• 
Tnu·s, by the homogeneity, v(p,a:w) = v(p' ,aw' ). But as we saw above. x(p,o:w) = 
lXX and x(p' ,aw') = ax'. Hence v(p,o:w) = u(a:x) and v(p',aw') = u(cxx' ). Thus 
u(IXX) = u.(ax'). Ther-efore u(x) is homogeneous of degree one. 
3.D.4 L (a) Le: e1 = (1,0, ... ,0) e IR . We shall prove that for every p e 
w e R, a e iR, and x e (- co, co) x IR:-1• if x = .x(p,w}, then x + ae1 = 
x(p, w + c::l. Note fir-st that, 
is affordable a! (p, w + a.e1L 
by p1 = 1, p· (o:x + e 1l :: o:w, that is, x + ae1 
L Let y e IR + and p · y :s w + a. Then p · (y - IXe 1 l 
3-11 
-
< w 
• Hence·x ~ y- o:.er Thus, by the quasilinearity, x + ae
1 >- y . - Hence 
x + o:e1 = x(p, w + o:). 
Therefore, for every t e {2, ... ,L}, w e IR, and w' e rR, xl(p,wl = 
XiP· w' ). That is, the Walrasian demand functions for goods 2, ... ,L are 
independent of wealth. As for good 1, we have 8x(p, w)/Bw = 1 for every (p, w). 
That is, any additional amount of money is spent on good 1. 
• 
(b) Define t/J(p) = u(x(p,O)). Since x(p,w) • x(p,O) + we
1 
and the preference 
relation can be represented by a utility function of the quasilinear form u(x) 
-
= x1 + u(x2 , .... ;_l (Exercise 3.C.S}, we have 
v(p,w) = u(x(p,w)) • 
-u(x2(p, w), ... ,xL (p, w)). 
-
= w + x 1(p,O) + u(x2 (p,O), ... ,xL(p,O)) 
= w + u(x(p,O}) = w + t/J(p). 
(c) The non-negativity constraint is binding if and only if p2x 2Cp,Ol > w. 
-1 Note that x 2 (p,O) = (1)') (p2 l, because p1 = 1. Hence the const:-aint is • 
-1 binding if and only if p2(1J') Cp2 ) > w. If so, the Walrasian demand is given • 
• 
by x(p,w) = (O,wlp2 l. Tlrus,, 'as w changes, the consumption level of the first 
good is unchanged and the consumption of the second good changes at rc.te llp2 
with w until the non-negativity constraint no longer binds. 
3.0.5 (a) Si!lce any monotone transformation of a utility function represents 
• 
the same prefer-ence relation, we may as well choose 
-By the first-or-der condition of the UMP with u( · ), 
x(p,w) = 
• 
• 
3-12 
where = p/(p - 1) e
(- co, 1). Plug this into u( · ), then we obtain 
• 
v(p,w) 
(b) To check the homogeneity of the function, 
~ ~ ~-1 tS-1 
x(IXp,o:w) = (aw/[(o:p
1
) + (o:p2 } )((o:p1} , (ap2l ) 
~-1 ~ tS ~ tS-1 ~-1 
= (a.·a: /(X }(w/(pl + PzlHpl ·P2 > 
• 
= x(p,w}. 
To check Walras' law, 
p·x(p,w) = w. 
The uniqueness is obvious. 
To check the homogeneity of the indirect utility function, 
v(o:p,aw) 
~ 
= a:w/( (o:pl) + 
= v(p,w) 
To check the monotonicity, 
~ ~ 1/o 8v(p,w)/8w = 1/(pl + Pz) > 0, 
- ~ ~-1 ~ 0 1/~+1 
ov(p,w)/opt = - wpl /(pl + p2) < 0 . 
• 
- . 
~)1/~ 
+ Pz 
• 
The continuity follows immediately from the derived functional form . 
• 
• 
In orde:- to prove the quasiconvexity, by property the homogeneity, it is 
sufficient to prove that, for any v e IR and w > 0, the set {p e IR2: v(p, w) :s 
\ . v, 1s convex. If o = 0, then the utility function is a Cobb-Douglas one, and 
the quasiconcavity was already established in Exercise 3.0.2. So we consider 
f( ) (f( .~>1/o . . f p = pJ ~ vt !S convex or every v. Since v(p,wl = w/f(p}, this 
implies that {p e IR2: v(p,w) :s v} is convex for every v and w. 
• 3-13 
~ a ~ If o < o, then f(p) = p 1 + Pz is a convex function. 
• 
Hence {p 2 E IR : 
o 1/C-a> . 1/f(p) = (f(p} } :s v} 1s convex for every v. Sir.ce v(p,w) = w/f(p), 
this implies that {p e Ill: v(p,w) :s v} is convex for every v and w. 
(c) For the linear indifference curves. we have 
(w/p
1
, 0} if pl < p2. 
x(p, w} - (0. w/p2) •r pl > P2' - h 
{(w/p1)(A.. 1 -A.): A e 10,1)} . .,. p1 - p2; 1. -
v(p,w) = max{w/p1,wlp2} . 
• 
For the Leontief preference. 
• 
As for the limit argument with respect to p. First consider the case 
with p < 1 and p -+ 1. Then a = p/(p - 1) -+ - co as p -+ 1. 
S. ' ... ,.. .n __ 
1 • 
... l IH 
a 1. we have (p2;p1) -+ 0. 
Cl-1 ~ 0 
Thus 
w/pl 
p 1 w/Cp 1 + p 2 l = 1 im ~...;. - = o-+- co 1 + 
• 
have ~ 1, we (p{p2 l -+ co . Thus • 
w/p2 
1 im ~-1 ~ a 1 im P2 w/(pl + Pz> -- 0 o-+ -co o-+-co (p1/p2) + 1 
- 0. 
-
Thus the CES Walrasian demands converge to the Walrasian demand of the linear 
indiffe::"e!"lce curves. As for the indirect utility functions, we showed in the 
. ~ 0 1/~ 
answer to Exercise 3.C.6(c) that (p1 + p2 ) -+ p1 for p1 :s p2 . Hence the 
CES indirect u"tilities converge to the indirect utility of the linea:-
indifference curves. 
Case 2. p1 > p2. 
De the sa:ne argument as in the Case 1. 
• 3-14 
Case 3. • 
ln this case, 
[w/2.p1Hl,l}. This consumption bundle belongs to the set of the Walrasian 
demal'lds cf the linear indifference curves when p1 = p2. As for the indirect 
utility functions, we showed in the answer to Exercise 3.C.6(c) that 
~ ~ 1/~ (pl + Pz> -+ Pl for PI :s Pz· 
• 
Let's next consider the case p -+ - co. Note that ~ = p/( p - 1) -+ 1 as p -+ 
1. So just plug ~ = 1 into the CES Walrasian demand functions and the 
. 
indirect utility functions. We then get the Walrasian demand function and the 
indirect utility function of the Leontief preference. 
(d) From the calculation of the Walrasian demand functions in (a) we get 
~-1 
x 1 (p, w)/x2 (p, w) = (p1/p2) , 
• 
~-2 £x1 (p,w)/x2(p,w)}/(p1/p2> = (P/Pzl , 
o-2 d(x 1 (p, w)/x2(p, w)]/d[p1 /p2J = (~ - 1Hp1 /p2) . 
Thus ~lZ(p,w} = - (~ - 1) = 1/(1 - p). Hence, ~ 12£p,w) = cz: for the linear, 
~ 12(p,w) ::: 0 for the Leontief, and ~12Cp,w) = I for the Cobb-Douglas u!ility 
f ... unc-.1cns. 
-3.0.6 (a) Define u(x) 
• 
-
• 
• 
• 
a' 
bl) (x2 
with a' = a/(a + {3 + a-l. {3' = {3/(a. + f3 + r>. 'l' = 7/(a + f3 + a-). Tnen c::' + W 
-
+ 7' = 1 a:1.d u( · l represents the same preferences as u( · ), because the 
&' 1/( a+f3+1') . f . 
! unction u -+ u 1s a monotone trans ormation. Tnus we can assume 
wi:hout loss of generality that a. + f3 + a- = 1. 
(b) Use anothe:- monotone transformation of the given utility function, 
• 3-15 
The first-order condition of the UMP yields the demand function 
x(p,w) = (b1,b2,b3 ) + (w - p·b)(ct!p1,,a/p2 ,7/p3 J, 
where p · b -= p1 b1 + p2b2 + p3b3• Plug this demand function to u( · ), then we 
obtain the indirect utility function 
v(p,w) = (w -
• 
(c) To check the homogeneity of the demand function, 
• 
x(A.p,hw) = Cb1,b2,b3 ) + CA.w - i\p· b)(a/i\pl',S/A.p2.r/i\.p3 ) 
= (b1,b2,b3 ) + [w- p·b}(a/p1,,S/p2,,./p3) = x(p,w). 
To check Walras law, 
= p·b + (w - p·b)(o: + f3 + 7} = w. 
The uniqueness is obvious. 
To check the homogeneity of the indirect utility function, 
v(i\p,hW) 
-
-
= (w -
To check the m:>notonicity, 
Civ(p,w)/op1 = v(p,w}·(- a/p1) < 0, 
Civ(p,w)/c3p2 = v(p,w)·(- {3/p2 ) < 0, 
8v(p,w)/8p3 = v(p,w}• (- 0 /p3 ) < 0. 
= v(p, w). 
> 0, 
• 
The continui:y follows directly from the given functional form. In order to 
prove the quasiccnvexity, it is sufficient to prove that, for any v e !R and w 
> 0, the set (p e IR3: v(p. w) ::= v} is convex. Consider 
lnv(p, wl = IX!niX + {3in(3 + 7ln'1 + ln[w - p· b) - cdnp1 - ,Slnp2 - a-lnp3. 
Since the logarithmic function is concave, the set 
• 
3-16 
3 {p e IR : lnlw - p·b) - cdnp1 - (3lnp2 - 'llnp3 ~ v} 
is convex for every v e IR. Since the other terms, edna: + (3lr:/3 + 'lin)', do not 
3 . 
depend on p, this implies that the set {p e ~ : lnv(p,w) :s v} is convex. 
Hence so is {p e IR3: v(p,w) :s v) 
3.0.7 . 1 0 1 1 0 0 1 o. (a) Smce p · x < w and x :;e x , the weak axiom implies p · x > w . 
Thus x 1 has to be on the bold line in the following figure. 
8 
6 
4 ··-········--·· • 
' • 
' • 
• 
• 
• 
• • 
• • 
' ' 
' 
• 
• • 
• • 
• • 
• 
0 2 4 8 Xl 
Figure 3.D.7(a) 
• 
• 
• • 
In the following four question; we assume the given preferer.::..: ···"'.., be a ... _ ... 
differentiable utility function u( • ). 
• 
(b) If the pre!"erence is quasilinear with respect to the first good, then we 
can take a utility function u( ·) so that 8u(x)/8x1 = 1 for every x CExe:-cise 
3.C.5(b)). Hence the first-order condition implies 8u(xt)/8x~ = p~/p~ for 
each t = 0,!. I ... .. S . ... :t .. .. 0 0 p /p < 2 1 
to be on the bold line in the following figure . 
• 
3-17 
0 1 
concave, x 2 > x2. T
. 1 
nus x has 
8 
4 
• 
• 
• • 
• • 
• • 
• 
' • ' • ' • • • 
• • • 
• • • 
' 
• • 
• 
' 
0 2 4 8 10 XI 
3.D.7(b) 
(c} If the preference is qnasilinear with respect to the second good, then 
then we can take a utility function u( ·) so that 8u(x)/8x2 = 1 for every x 
(Exercise 3. C.S(b)). Hence the first-order condition implies c3u(xt)/8x~ = 
0,1. Since 0 0 1 1 . P11P2 > p 11p2 and u( · l 1s concave, we must 
Thus x 1 has to be on the bold line in the following figu:-e. 
• 
• 
•• 
• 
8 
5.5 ..... ::~. :;:-:: 
1 0 (d) Since p · x 
• 
• 
• 
• 
• .................... 
• 
• 
• 
• 
• 
• • 
• • 
• • 
• • 
• • 
• • 
• • 
• • 
• • 
• 
__ . __ . ___ ....;..._ __________ _ 
0 
1 
<w 
2 4 8 X! 
Figure 3.D.7(c) 
and the relative price of good 1 decreased, x~ has to 
3-18 
• 
have 
increase if good 1 is normal. If good 2 is normal, then the wealth effect 
(positive) and th~ substitution effect (negative) go in opposite direction 
which gives us no additional infonnation about x 2. 
bold line in the following figure. 
• 
8 
................ 
• 
• 
• • 
• • 
• • 
• • 
• 
• • 
• • 
' 
• 
• • 
• • 
• • 
• • 
• • 
• • 
0 2 4 8 
Figure 3.D.7(d) 
1 Thus x has to be on the 
Xl 
(e) If the preference is homothetic, the the marginal rates of substitution at 
• 
• 
all vectors on a ray are the same, and they becomes less steep as the ray 
becomes flatte:-. 1 1 1 > P/P2. x has 
• 
0 
to be on the right side of the ray
that goes through x . Thus x 1 has to be on 
the bold line in the following figure . 
• 
• 
• 
• 
3-19 
• 
8 
• 
5.2 • • 
• 
• 
• 
····--·· • 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
0 2 4 5.2 
• • 
•• 
' 
• 
8 
•• 
• • 
• 
•• 
•• 
• 
... 
• • 
•• 
••• 
• 
• 
• 
• 
3.D.7(e) 
X! 
3.0.8 By Proposition 3.D.3(i}, v(ttp,tXW} = v(p,w} for all IX > 0. Bv 
-
differentiating t.'lis equality with respect to « and evaluating at et • 1, we 
obtain 'V v(p,wl·p + wc3v(p,w}/8w = 0. Thus w8v(p,w)/8w = - 'V v(p,wl·p. p p 
3.E.l The EMP is equivalent to the following maximization problem: 
Max - p·x s.t. u(x) ~ u and x ~ 0. 
Tne Kul:m-Tucker condition (Theorem M.K.2) implies that the fi!'st-order 
conditions ar-e that ther-e exists ~ L > 0 and IJ. e IR such that p = A'ii'u(x•) + 1-1 
+ • 
and IJ. • x• = 0. That is, for some 
This is the same as that of the UMP. 
3.E.2 To check the homogeneity of the expenditure function, 
e(Ao,u) 
• 
= e(p,ul. 
To check the monotonicity, 
3-20 
-« cx-1 « 1-a. Be(p, u)/au = a (1 - «} p 1 p2 > O, 
~ 1-IX et-1 a-1 1-a 
oe(p,u)/8p1 = et (1 - a.) p1 p2 > O, 
~ (X « « -(X 
oe(p,u)/8p2 = « (1 - o:J p 1 p2 > o. 
2 To check the concavity, it is easy to actually calculate Dpe(p,u) and then 
apply the condition in Exercise 2.F.9 to show that o2e(p,u) is negative 
. p 
semidefinite. An alternative way is to only calculate 
• 
_2 2 1-« « o:-2 1-a. 
o e(p,u)l8p 1 = - a. (1 - a.) p1 p2 < o. 
Then note that the homogeneity implies that D2e(p,u)p = 0. Hence we can apply p 
Theorem M.D.4(iii) to conclude that D2e(p,u) is negative semidefinite. The p 
continuity follows from the functional form. 
• 
To check the homogeneity of the Hicksian demand function, 
h
1 
(ll.p, u) = 
• 
h (ll.o u} = 2 . t 
(1 - « )ll.p1 
(1 - a.)ll.p1 
a.ll.p2 
To check the no excess utility, 
u(h(p,u)) = (1 - cx)p 
1 
• 
1-o: 
u 
1-a. 
u = (1 - tt)p1" 
(1 - a.)pl 
a.p2 
u = 
(1 - a)p1 
.cxp2 
u 
1-o: 
1-a. 
- {1 ~(X )a.-(X(l-(X) 
-
- ( 1 - o:)p 
1 
The unioueness is obvious . 
• 
-
et+(l-a.) 
u = u. 
• 
• 
-
• 
3.E.3 - L Let x e IR and 
+ 
u(xl i!:: u. Define A = {x e IRL: p·x s 
+ 
p·x and u(x) i!:: u}. 
-Then A :;e 0 by x e A. Furthermore, A is compact: The closedness follows from 
that of {x 
inclusion 
L 
e ~ : u!x) i!:: u} and L of IR • +' the boundedness follows from the 
L -A c {x e IR : 0 s xl s p·xlpt for every l = l, ... ,L}. 
3-21 
Now consider the truncated EMP: 
Min p·x s.t. x e A. 
Since p · x is a continuous function and A is a compact set, this problem has a 
solution, denoted by x• e A. We shall show that this is also a solution to 
the original EMP. L Let x e IR + and u(x) o= u. If x e A, then p · x :=: p· x• because 
-
x• is a solu:ion to the truncated EMP. If x t! A, then p · x > p · x and hence p ·-x 
• 
> p · x•. Thus x• is a solution of the original EMP. 
3 • .E.4 Suppose first that >- is convex and that x e h(p,u) and x' e h(p,u}. 
-
Then p·x = p·x' and u(x) 2: u, u(x'} 0!:: u. Let a. e [0,1] and define x" = ax + 
(1 - a}x'. Tnen p · x" = ap · x + (1 - a:)p · x' = p · x = p · x' and, by the convexity 
of >-, u(x"l ~ u. Thus x" e h(p,u). 
-
Suppose next that >- is strictly convex and that x e h(p,ul, x' e h(p,u}, 
-
= x', and u(x) o= u(x' l 0!:: u. By the argument above, x" = ax + (1 - o:)x' with 
a e (0,1) satis:ies p·x" = p·x = p·x' and, by the strict convexity of >-, we 
-
have x" >- x'. Since >- is continuous, (3x" >- x' for any (3 e ( 0 ,ll close enough 
-
to L But this implies that p · ((3x") < p · x and u((3x" l > u(x') 2: u, which 
cont:;:-adicts the fact that x. is a sol\ltion of the EMP. Hence .... : -- ~ 1 l mu-· 
···.--- ;l!L. be a 
-3.E.S [Fi:-st pfir..tinsz errata: The equality h(p,u) = h(p)u shoulc be h(o,u} = 
• 
- -
uh{p), because u is a scalar and h(p) is a vector.] We shall first prove 
tha:, for every p » 0, u 2: 0, a l!:. 0, and x 2: 0, if x = h(p,c), ther. ax = 
h(p,o:ul. In fact, note that u(ax) = a.u(x) o= au, that is, ax satisfies the 
constrain~ of t~e EMP for em. Let L y e IR and u(y) l!:. o:u. 
+ 
• 
-. 
u(o: ·y) :=: u. 
-1 Hence p· (a y) l!:. p·x. Thus p·y ~ p·(ax). Hence ax = h(p,au]. The:-e:!'ore 
3-22 
h(p,u) is homogeneous of degree one Jn u. 
By this result, 
e(p,o:u) = p•h(p,o:u) = p·(ah(p,u)} = o:(p·h(p,u)) = o:e(p,ul. 
Thus the expenditure function is homogeneous of degree one in u. 
- - -Now define h(p) = h(p,l) and e(p) = e(p,l), then h(p.u} = uh{p) and 
• 
-e(p,u) = ue(p) . 
• 
3.£.6 Define ~ = p/(p - 1}, then the expenditure function and the Hicksian 
demand function are derived from the first-order conditions of the EMP and 
they are as follows: 
~ ~ c 1-a)la ~-1 CJ-1 h(p,u) = uCp1 + P2 ) (p 1 .p2 ), 
1 o ~ ua e,p,u} = u(p1 + p2 ) . 
To check the homogeneity of the expenditure function, 
a ·I/o c ~ 
a. u pl CJ J11a _ + Pz - o:e(p, u). 
To check the monotonicity, 
~ o ~ I/a 
oe(p,ul/au = (p 1 + p2 l > o, 
oe(p,u}/opl = > 0 . 
• ~ . 
To check the concavity. it is a bit lengthy but easy to actuai lv ::..:1.lculate 
-
2 D e(o,u) p • 
is negative semidefinite. An alternative way is to only calculate 
bv cS < 1. 
• 
a>I/~-1 a-1£ a a,I/~-2(1/~ 
+ p2 + up! PI + Pz 
..,_ 1 
- l)Clpa • 
1 
2 The:-t note that the homogeneity implies that D e(p,u)p = 0. p 
semidefinite. Tne continuity follows from the functional form. 
3-23 
Hence 
To cheek the homogeneity of the Hicksian demand function, 
~ ~ (1-~}/~ ~-1 c3-l h(ap.u) = u((ap1) + (o:p2 ) ) ((ap1l ,(exp2 l ) 
cao-~>t~l+(o-u c a 
= ex up 1 
= h(p,u). 
To check no excess utility, 
(h( )) ( a a >(1-ol/oc ca-op ca-Ilp>l/p u p, u = u p 1 + Pz . P 1 + P2 
Since (c3 - l)/c3 = - 1/p, we obtain u(h(p,u)) = u. The uniqueness is obvious . 
• 
3.E.7 In Exercise 3.C.S(b), we showed that every quasilinear preference with 
respect to good 1 can be represented by a utility function of the u(x) = 
• 
-
x 1 + u(x2 , ... ,XL). We shall prove that for every 
• 
p » 0 with p1 = 
L-1 l, u e IX, ex e IR, and x e (- .,, .,) x IR , 
+ 
if x ~ h(p,u), then 
x + ae1 = h(p, u + a). Note first that u(x + cte1) :2:: u + o:, that is, x + ae1 
L 
satisfies the constraint of the EMP for (p, u + «). Let y e !R and u(y) ~ u + 
+ 
c::. Then u(y - ae.) :: u . 
• 
Hence x + ae
1 
= h(p, u + a). 
Therefore, for e.,.·e:-y l e {2, ... ,L}, u e IR, and u' e IR, hl(p, u) = 
That is, the Hick!>ian' demand functions for goods 2., ... ,L 
-
a
-· . ~ 
independent of utiiity levels. Thus, if we define h(p) = h(p,O), then h(p,u) 
-
= h(p) + uel' 
Since h(p, u + a) = h(p,u} + ae1, we have e(p, u + a) = e(p,u) + a. 
- -Thus, if we define e{p) = e(p,O), then e(p,u) = e(p) + u. 
3.E.8 We use the utility function u(x} = To prove (3.E.1l, 
• 
-a o:-1 a 1-o: « 1-a -a a-1 
e(p, v(p, w)) = a (1 - o:) p
1 
p
2 
(o: (1 - ctl p 1 p2 w) c w, 
a {p,e(p,u)) = o: (1 - 1"« -a o:-1 -a «} Pl Pz (a (1 
3-24 
a-1 a 1-a 
- «) P1P2 u) = u . 
To prove (3.t.3), 
x(p,e(p,u)) = (a:-c::(l 
-
-
h(p,v{p,w)) 
1-IX 
u, 
1-IX -a: a:-1 a> PI Pz w 
• 
u = h(p,u), 
1-a: 
• 
(1 - o:)p 
1 
0:. 
3.E.9 First, we shall prove that Proposition 3.0.3 implies Proposition 3.E.2 
via (3.E.Il. Let p » 0, p' » 0, u e IR, u' e IR, and a ~ 0 . 
• 
(i) Homogeneity of degree one in p: Let a: > 0. Define w = e(p,u), then u = 
v(p,w) by the second relation of (3.E.l). Hence 
e(o:p,u} = e(a:p,v(p,w)) = e(a:p,v(a:p,a:w)) = a.w = o:e(p,u}, 
whe:-e the second equality follows from the homogeneity of v( • , • } and the third 
from the first relation of (3. E.l).
(ii) ~fonotonicity: Let u' > u. Define w = e(p,u) and w' = e{p,u'}, t!\en L: = 
v(p, w) a.'1d u' = v(p,w'). By the monotonicity of v( ·, ·) in w, we must have w' 
• 
• • 
> w, that is, e{p' ,u) > e(p,u). 
Next let p' =: p. Define w = e(p,u) and w' = e(p' ,u), then, by the second 
• 
relation of (3.E.l), u = v(p,w) = v(p' ,w'). By the monotonicity of v( ·, · ), we 
must have w' =:. w, that is, e(p' ,u) :=: e(p,u). 
(iii} Concavity: Let o: e [0,1]. Define w = e(p,u) and w' = e(p',u). then u = 
• 
v(p,w) = v{p',w). Define p" = a.p + (1 - o:.)p" and w" = a:w + (1 - IX)w'. Then, 
by the quasiconvexity of v( ·, • ), v(p", w") ~ u. Hence, by the monotonicity of 
v( • , · l in w and the second relation of (3.E.l), w" ~ e(p•, ul. that is, 
e(ap + (1 - o:)p', u) :=: o:e(p,u) + (1 - a.)e(p' ,ul. 
3-25 
• (iv) Continuity: It is sufficient to prove the following statement: For any 
n n)}co . ( n n . n n 
sequence {(p ,u n=l wtth p .u ) ~ (p,u) and any w, 1f e(p ,u ) :s w for 
every n, then e(p,u) ~ w; if e(pn.un} =::: w for every n, then e(p,u) == w. 
n n Suppose that e(p , u ) :s w for every n. Then, by the monotonicity of v( ·,.) in 
w, and the second relation of (:l.E.ll, we have. un ~ v(pn,w) for every n. By 
the continuity of v( ·, • ), u :> v(p,w). By the second relation ~f (3.E.ll and 
• 
the monotonicity of v( ·, ·) in w, we must have e(p,u) :s w. The same argument 
can be applied for the case with e(pn,un) =::: w for every n. 
Let's next prove that Proposition 3.E.2 implies Proposition 3.0.3 via 
(3.E.ll. Let p » 0, p' » 0, w e R, w' e IR, and a. 2: 0. 
(i) Homogeneity: Let :> 0. Define u = v(p, w). Then, by the first relation 
of (3.E.l}, e(p,u} = w. Hence 
v(o:p,o:w) = v(«p,cte(p,w)} = v(o:p,e(o:p,u)} = u = v(p,w}, 
where the se~ond equality follows from the homogeneity of e{ ·, · l and the thi::-d 
from the second relation of (3.E.l). 
(ii) Monotonicity: Let w' > w. Define u = v(p,w) and u' = v(p,w'), then 
e(p,u) = w and e(p,u'} = w'. By the monotonicity of e( ·, ·) and w' > w, we 
must have u' > u, that is, v(p, w') > v(p, w). 
Nex:, assume that p' 2: p. Define u = v(p, w) and u' = v(p', wl, then 
e(p,u} = e(p',u'} = w. By the monotonicity of e(·,·l and p' 2:: p, we must have 
u' :s u, that is, v(p,wl == v(p',w). 
(iii} Quasiconvexity: Let o: e [0,1}. Define u = v(p,w) and u' = v(p',w'). 
Then e(p,u) = w and e(p,u') = w'. Without loss of generality, assume that u' 
2: u. Define p'' = o:p + ( 1 - a:)p' and w = o:w + ( 1 - ct)w'. Then 
e{p" ,u') 
== ae(p,u'} + (1 - o:)e(p' ,u') 
3-26 
i!:: ae(p, u) + (1 - ct)e(p', u') 
= o:w + (1 - ct)w' = w", 
where the first inequality follows from the concavity of e( · ,u), the second 
from the monotonicity of e( ·, ·) in u and u' i!:: u. We must thus have v(p", w") ~ 
• 
u'. 
(iv} Continuity: It is sufficient to prove the following statement. For anv 
• 
• 
nnca. nn nn 
sequence {(p ,w )}n=l Wlth (p ,w ) -+ (p,w) and any u, if v(p ,w ) ::5 u for 
every n, then v(p,w} ~ u; if v(pn,wn) i!:: u for every n, then v(p,w) i!:: u. 
n n · Suppose that v(p , w ) ~ u for every n. Then, by the monotonicity of e( ·, ·) in 
u and the first relation of (3.E.l), we have wn ~ e(pn,u) for eve:-y n. By the 
continuity of e( ·, · ), w ~ e(p, u). We must thus have v(p, w} ~ u. The same 
argument can be applied for the case with v(pn,wn) :=: u for every n. 
An alternative, simpler way to show the equivalence on the concavity/ 
quasiconvexity and the continuity uses what is sometimes called the 
epig:-aph. 
For the concavity/quasiconvexity, the concavity of e( · ,u) is equivalent 
to the convex!:y of the s~.t {(p,w):. e(p,u) i!:: w} and the quasi-convexity of 
v( ·:. is equivalent to the convexity of the set {(p,w): v(p,w) :s u} for eve!"y 
u. B•Jt (3.E.l} and the monotonicity imply that v(p,w) ::5 u if and only if · 
e(p,\.i) i!:: w. Hence the two sets coincide and the quasiconvexity of v( ·) is 
equi\•alent to the concavity of e( · ,u). 
As for the continuity, the function e( ·) is continuous if and only if 
both {(p,w,u): e(p,ul ~ w} and {(p,w,u): e(p,u) :=: w} are 'closed sets. The 
function v( ·} is continuous if and only if both {(p, w, u): v(p, w) i!:: u} and 
Hp. v:,u}: v(p, w) ::5. u} are closed sets. But, again by (3.£.1) and the 
monot~nici ty, 
3-27 
{(p,w,u):· e(p,u} ~ w) = {(p,w,u): v(p,w) 2: u}; 
{(p,w,u): e(p,u) 2: w} = {(p,w,u): v(p,w) s ti}. 
Hence the continuity of e( ·) is equivalent to that of v( •). 
3.E.10 [Fi:-st printing errata: Proposition 3.E.4 should be Proposition 
3.E.3.] 
• 
Let's first prove that Proposition 3.0.2 implies Proposition 3.E.3 via the 
relations of (3.E.l) and (3.E.4). Let L p e R and u e R. 
++ 
(i) Homogeneity: Let a. > 0. Define w = e(p.u), then u = v(p, w) by the second 
relation of (3.E.l). Hence 
. 
h(a.p,u) = x(a.p,e(a.p,u)) = x(a.p,a.e(p,u)) = x(p,e(p,u)) = h(p,u), 
• 
where the first equality follows from by the first relation of (3.E.4), the 
second from the homogeneity of e( · ,u), the third from the homogeneity of 
x( ·, · }, and the last from by the first relation of (3.E.4l. 
(ii) No excess utility: Let (p,u) be given and x e h(p,u}. Then x e 
x(p,e(p,ull by the fir-st relation of (3.E.4). Thus u(xl = v(p,e(p,u)) = u bv 
• 
the second relation of (3.£.1). 
(iii} Convexity/Uniqueness:. Obvious. 
Let's first prove that Proposition 3.E.3 implies Proposition 3.0.2. via 
the relations of (3.E.l) and (3.E.4). Let L p e IR and w e IR. 
++ 
(il Homogeneity: Let a. > 0 and define w = e(p,u), then v(p,w) = u. Hence 
x(a.p,a.w) = h(a.p,v(a.p,a:w)) = h(a:p,v(p,w)) = h(p,v(p,w)) = x(p,w}, 
where the fi!"st equality follows from the second relation of (3.£.4), the 
second from the homogeneity of v( · ), the third from the homogeneity of h( · l in 
p, and the last from the first relation of (3.E.4). 
( ii) Walras' law: Let (p, w) be given and x e x(p, w}. Then x e h(p, v(p, w) l by 
3-28 
the second relation of C3.E4}. Thus p·x = e(p,v(p,w)} == w by the definition 
of the Hicksian demand and the first relation of (3.E.ll. 
(iii) Convexity/Uniqueness: Obvious. 
3.F.l Denote by A the intersection of the half spaces that includes K. then 
-
clearly A ::l K. To show the inverse inclusion, let x e K, then, since K is a 
• 
closed convex set, the separating hyperplane theorem (Theorem M.G.2) implies 
-that there exists a p :;e 0 and c, such that p • x < c < p · x for every x e K. 
L Then {z e IR : p·z i!!: c} is a half space that includes K but does not contain x. 
-Hence x E A. Thus K ::l A. 
3.F.2 If K is not a convex set, then there exists x e K and y e K such that 
0/2)x + {l/2)y E K, as depicted in the figure below. The intersection of all 
the half -spaces containing K (which also means containing x and y) will 
contain the point (l/2)x + (l/2)y, since half-spaces are convex and the 
intersection of convex sets is convex. Therefore, the point (112lx + {1/2)y 
cannot be separ-ated from K. 
X 
(II2)x + (112}y K 
y 
Figme 3.F.2 
• 
3-29 
As for the second statement, if K is not convex, then there exist x e K, 
y e K, and o: e [0,11 such that o:x + (1 - o:)y e K. Since every half space that 
includes K also contains o:x + (1 - et)y, it cannot be separated from K. 
3.G.l Since the identity v(p,e(p,u)) = u holds for all p, differentiation 
with respect to p yields 
By Roy's identity, 
V v(p,e(p,u)) + (Bv(p,e(p,u))/8w)V e(p,u) = 0. p p 
(8v(p,e(p, u))/Bw)(- x(p,e(p, u)) + V e(p,u)) • 0. p 
By ov(p,e(p,u))/Bw > 0 and h(p,u} = x(p,e(p,u}}, we obtain h(p,u) = V e(p,u}. p 
• 
• 
3.G.2 from Examples 3.D.1 and 3.E.l, for the utility function u(x) 
we obtain 
D x(p,w) = p 
Ve(p,ul = 
a/pl 
[I - o:)/p2 • 
2 
- o:w/pl 
0 - (1 
D e(p,u)
= D h(p,u) p p 
0 
2 
- a:)w/p2 
• 
• 
- cx(l 
• 
o:(l - o:)/plp2 
0: 1-o: . The indirect utility function for u(x) = x 1x2 1s 
« 1-o: 
v(p,w) = Cp 1/et) (p2/0 - a.)) w. 
o:(l - (X)/pr P2 
2 
- o:(l - IX)/p2 
• 
(Note he:-e that the indirect utility function obtained in Example 3.0.2 is for 
the utility function u(x) = o:lnx1 + (1 - o:)lnx2.) Thus 
vpv(p,w) = v(p,w)(- a/pl. - (1 - o:)/p2)' 
3-30 
Vwv(p,w) = v(p,w)/w. 
Hence: 
h[p,u) = V e[p,u}, p 
o2e(p,u) = D h(p,u), which is negative semidefinite and symmetric, p p 
D h[p, u)p = 0, p 
D h(p,u) = D x(p,w) + D x(p,w)x(p,w)T, p p w • 
x l(p,w) = - (8v(p,u)/Bpt)/(8v(p,u)/8w). 
3.G.3 (a) Suppose that o: + f3 + '¥ = 1. Note that 
lnu(x) = cdn(x1 - b1) + 13lnCx2 - b2) + "¥ln{x3 - b3 l . 
• 
By the first-order condition of the EMP, 
Plug this into p·h(p,u), then we obtain the expenditure function 
• 
To check the homogeneity of the expenditure function, 
(-,. I e,,.p,u, 
To check the monotonicity, assume b1 ~ 0, b2 i!!: 0, and b3 i!!: 0. 
To c!leck the concavit.y, we can 2 show that D e(p,u) p 
3-31 
> 0, 
> 0, 
> 0. 
is eaual to 
• 
Then 
-
and then apply the condition in 
af3/plp2 
2 
- 13 c 1 - a ltp 2 
ise 2.F. 9 to show 
• 
negative semidefinite. An alternative way is to only calculate the 2 x 2 
• 
submatrix obtained from o2e(p,u) by deleting the last row and the last column p 
and apply the condition in Exercise 2.F.9 to show that it is negative 
definite. Then note that the homogeneity implies that o2e(p,u)p = 0. Hence p 
semidefinite. The continuity follows from the functional form. 
To che:::k the homogeneity of the Hicksian demand function. 
h( i\.p, u} 
tt+a+;-1 tt 13 ; 
= (bl,b2,b3) + ui\. (pl/tt) (p2//3) (p3/ir) (tt/pl,/3/p2'71p3) 
= h(p,u). 
To che:::k no excess utility, 
= u. 
The uniqueness is obvious. 
[b) We cal:::ulated the derivatives 8e(p,u)/8pl in (a). If we compare them with 
hip,u). then we can immediately see 8e(p,u)/8pl = hip,u). 
(c) Bv (b), D h(v,ul 
• p . 
3.0.6, we showed 
2 
= D e(p,u). p In (a), we calculated 
3-32 
2 D e(p,ul. p In Exercise 
2 0 0 a:/pl o:/pl 
2 D x(p, w) -(w-p·b) 0 0 f3/p2 - f3/p2 --p 2 0 7/p3 0 1'1P3 
Using these results, we can verify the Slutsky equation . 
• (d) Use D h(p,u) p and the explicit expression of 
• 
(bl,b2,b3). 
(a). 
(e) This follows 
that D2eCp, u) is p 
2 from S(p,u) = D h(p,u) = D e(p,u) and (a). in which we showed p p 
• 
negative semidefinite and has rank 2. 
3.G.4 (a) Let a > 0 and b e IR. Define ~: IRL ~ 1R bv u(x) = au(x} + b and, for + • 
- -
each l, u{ IR+ ~ IR by ut(xt} = aut(xt) + b/L. Then 
u(x) = artut(xt} + b = I:t'aul(xt) +biLl = r_i~l(xe). 
Thus any linear (to be exact, affine} transformation of a separable utility 
function is again separable. 
l\ext, we prove that if a monotone transformation of a sepa:-able utility 
function is again separable, then the monotone tra."l.sformatior. must be linea• 
(affine). To do this, let's assume that each u£' ·) is continuous Lid strongly 
monotone. Tnen, for each l, the range ut(IR) is a half -open inter-val. So let 
ul(~+} = [al' bel, where bl may be + co. Define cl = bl - at > 0, a = Ltal' b = 
L:eb t' and c = Etc 2. or some be is equal to + co, then b and c a!"e + (I) as 
well.) Then uCKL) = [a, b). Suppose that /: [a, b) ... IR is strongly monotone 
+ 
.. .. 
and the utility function u( ·) defined by u(x) = f(u(x}) is sepa:-abie. To 
simplify the proof, let's assume that f( ·) is differentiable. Define g: [O,c) 
-+ IR bv 
• 
g(v) = f(v + a) - f(a), 
3-33 
then gCO) = 0, g( ·) is diffel"'entiable, and 
g(u(x} - u(O)) = f(u(x)) - f(u(Q)) 
for L every x e IR .• 
,. 
Thus, in OI"'der to piOVe that f( ·) is linear (affine), it is 
sufficient to prove that g( ·) is lineal"'. For' this, it is sufficient tc show 
that the first-order derivatives g'(v) do not depend on the choice of 
v e [O,c). 
To this end, we shall fil"'st pi"'OVe that if vl e [O,ct) for' each t, then 
g([tvt) = Ltg(vl). For' this, it is sufficient to prove that 
g(u(x) - u(O)) == 
for' L every x e IR • 
+ 
In fact, by the separability assumption, for each i., there 
- - -
exists a monotone utility function ut,< ·} such that u(x) = Ltut'xt) for every x 
• 
L 
e IR . 
+ 
Fix an x e IRL and, for each t, 
+ 
. I. L I. l 
defme y e IR+ by Yt = xi and yk = 0 for 
any k :;c t. - l I. Since u(y ) = f(u(y )), 
' - -Subtracting Lk=luk(O) = u(O) = f(u(O)) from both sides and noticing that 
ul(xl) + Lk:;cfuk(Q} = uixtl - u,_!O) + k=luk(O), we obtain 
Summing over- l, we obtain 
Since 
we have 
g(u(x) - u(O)) == Et~Cut(xt) - u(O)). 
We have thus proved that g([lvl) == [,_gCv,_L 
To prove that the g'(v) do not depend on the choice of v e [O,c), note 
first that if vl e [O,cl) for each l and v == Etvl e [O,c), then g'(vl = g'(vll 
3-34 
for each t. This can be established by differentiating both sides of g([lv ll 
= Ltg(vl} with respect to vt" 
So let v e (O,c) and v' e [O,c}, then, for each l, there exist vf. e 
Then g'(v) = g'(v ) 
1 
- -
and g'(v') = g'(vil· Now, for some v2 e [O,c2 ), consider v1 + v2 e [O,c) and 
-vi + v2 e [O,cl. Then 
= g'(v + 1 
= g'(v' + 
1 
Thus g'(v ) = g'(v') and hence g'(v) = g'(v'). 1 1 
• 
-
= g'(v2}' 
-
= g'(v2). 
Note that the above proof by means of derivatives is u!:'lderlain by the 
• 
cardinal proper.:y of additively separable utility function, which is that, 
when moving from one commodity vector to another, if the loss in utility from 
some commodity is exactly compensated by the gain in utility from another, 
then this must be the case for any of its monotone transformations resulting 
in anothe~ additively separable utility function. (This fact is often much 
mo:-e shortly put into as: utility differences matter.) For example, consider-
L 
X e IR anc X' e 
• 
-
• 
• 
implies that u(x) = u(x'). By the equality g([lvl} = [lg(vtl• 
g(u1(x1l - u.1(0)} + g(u2Cx2>- u 2CO)) = g(u1(xi> - u1Wll + g~u2<xzl - u2(0)). 
Hence 
We have st:.own tha:, under the differentiability assumption, if this holds for 
(b) Define S = {l, ... ,L} and let T be a subset of S. The commodity vectors 
3-35 
for those in·s are represented by z
1 
= the like, and the 
commodity vectors for those outside S are represented by z2 = {zf)lET 
L-#T 
E IR 
+ 
and the like. We shall prove that for ~#T E ~'I. I 
.,. 
L-#T 
z2 e IR+ , 
L-#T 
and z2 e IR + , (z1,z2 l ?: (zi,z21 if and only if Cz1.z;21 ?: (zi,zil. In fact, 
since u( ·) represents ?:• (zl"z2 l ?: (zj.z2 l if and only if 
LteTut(zl) + Lt!CTut(zt) ·::!!:: LteTut(zt) + LtETul(zt). 
Likewise, Cz1,zz.) ?: (zi,z2_) if and only if 
LteTut(zl} + Lt!C~lCz£> l!: LteTu£'zt) + LteTulCz£l-
But both of these two inequalities are equivalent to 
Hence they are equivalent to each other. 
{c) Suppose that the wealth level w increases and all prices remain unchanged. 
Tne::-1 the demand for at least one good (say, good l) has to increase by the 
Wali"as' law. From (3.0.4) we know that uk_(xk(p,w)) = (pk/pelu£(xl(p,w)) for 
eve:-y k = l, ... ,L. Since xl(p,w) increased and ul( ·) is strictly concave, the 
right hand side will decrease. Hence, again since uk( ·) is strictly concave, 
• 
-
xk(p,w) will have to increase. Thus all goods are normal. 
(d) The first-o::-der condition of the UMP can be written as 
ll.(p,w)pt = u'(X£'P· w}), 
wher-e the Lagrange multiplier ll.(p, w) is a differentiable function of (p, wl: 
This can be easily seen in the proof of ·the differentiability of Wah· asian 
demand functions. which is contained in the Appendix. 
By diffe:--entia:ing the above first-order condition