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. both
co~x:ise
We have tried to be
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 foundations for solutions for
and we thank him for laying down 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
Since y >- z implies y >- z, the transitivity implies that x >- z.
l.B.l
-
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.
-
By the completeness, x >- x for every x e X.
l.B.2
Hence there is no x e X
-
such that x >- x.
Hence x >- z.
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. is transitive.
Property
(i)
Hence >-
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. and z ;t- y. z.
Suppose that
x -
y and y -
z. Tnen x
By the transitivity, this implies that x
Hence - is transitive.
>- x and x >- y.
-
-
Suppose x that - y.
Hence y - x.
~ 2
and
>- y, y >- z.
-
-
x.
2 ~
-
Property
-
(ii)
->- X .
Thus x -
Then x >- y and y >- x.
Thus - is symmetric
V
~
Thus y
is now
proved.
Le~
l.B.3
u(x}
:!::
x e X and y e X.
u(y).
Since
f( ·)
Since u( ·) represents
is strictly increasing, u(x)
~·
i!:
x
?: y ;:·
,:;.:~d
only if
u(y) if and only if
•
Hence x >- y
\I{X) :!:: Y{y).
-
if and only if
v(x) :!:: v(y).
Therefor-e
v( ·
l
represents >-.
-
Suppose first that x >- y.
l.B.4
-
u(x) = u(y). u(x)
>
u(y).
If, furthermore, y >- x, then x - y and hence
If, on the contrary, we do not have y
>- x, -
then x >- y.
Hence
Thus, if x >- y, then u(x) - y.
-
l.B.S
If, on the contrary, u(x) > u(y), then x >- y, and
Tnus, if u(x)
l!
u(y), then x >- y.
-
So u{ ·) represents >-.
-
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, If N = 1, there is nothing to prove:
then there exists a utility function.
So let N > 1 and suppose that
Just assign any number to the unique element. the above assertion is true for N - 1.
N.
We will show that it is still true for
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(x ) > uCx ) > ... > 1 2
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
in each of the three cases, for u(·) to
value of u(xN) should be
-
reoresent >on the whole X. . If Case 1 applies, then take u(xN) to be larger than u(x ). 1 applies, take applies.
u(~)
to be smaller than u(xN_ ). 1
If Case 2
Suppose now that Case 3
Let I = {i e {1, ... , N - 1}: xi >- xN+l} and J = {j e {1, ... , N - 1}:
XN t >x .}. ) .... • J •
Completeness and the assumption that there is no indifference
•
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.
1-2
Take
u{xN) to lie in the open interval (u(x .•
J,u(x .• )).
1 +1
•
Then it is
1
-··•
tc see
~-c:.::.y
-
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
n
m
E
X: x
= X. transitivity of - (Proposition l.B.Hiill, if X
·
n
let M be a subset of {!, ... ,N} such that X = u ~
and any n e M with m
~
>-• X if and only if x >- x . n m- n
r.
rr:
X , then X " X rn n m
me
MX
m
Then,
Also, bv• the
and X
m
Define an relation >-• on {X : m m
n.
- x }.
= e.
= Xn E
So
for- any m e M
M} by letting X
m
In fact, by the definition of M, there is no
indifference between two different elements of {X : m e M}. m
Thus,
by the
•
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
n
m
n
e X .
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}).
contradicts the equality C({x,y}) = {x}.
Hence y
~
C((x,y,z}).
B:.1t
Tl'rJs
C({x,y,z}) e {{x},{z},{x,z}}.
I.C.2 If B X
The property in the question are equivalent to the followir.g property:
e :E, B' e !13,
X E
B. y
e C(B') and -v e C(BL
e B.
X
e B', y e 8', x e C(B), and y e C{B'), then
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' ).
Weak Axiom twice, we obtain x e C(B') and y e C(Bl. is also satisfied.
1-3
If we apply the
Hence the above property
Suppose conversely that the above property is satisfied.
Furthermore, let 8' e :B, x e B', y e 8', and
8, y e B. x e B', and x e C(B). y e C(B').
Let B e 13. x e
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 Suppose that y
Thus x >-• y.
B, x e C(B), and y fl! C(B).
-
exists B e 13 such that x e B, y e B and x e C(Bl.
-
>-••
Hence x
-
x. then there
But the Weak Axiom implies
Hence if x >-• y, then we cannot have
that y e C(B), which is a contradiction. y >-• 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. car. be seen from the above proof, the WA is not necessary to guarantee 6
x >-•• y, the:-1 x >-
following example.
y.
tha~
if
But the converse need not be true, as sho·wn by the
Define X
and C({x,y,z}) = {y}.
As
= {X".y,z},
:B
= {{x,y},{x,y,z}},
Then x >-• y and y >-• x.
C({x,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 threeeler:1ent subset of X. then >-• is transitive.
-
1-4
By Proposition l.B.Hil, >-•• is
Since >-• is equal to >-••, >-• is also transitive.
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} neither- y >-•
-
x nor z
4: ~
and, by (a), x >-•• y, and y >-•• z.
>-• y.
-
Hence we have
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.
The simplest example is X = {x,y}. !B = {{x},{y}}, C({x}) = {x}, C{{y})
1.0.1
relation of X rationalizes C( · ).
= {y}. Then UJ rational
1.0.2
By Exercise l.B.S, let u( ·) be a utility representation of
is finite, for any B c X with B
~ 0,
t·
Since X
there exists x e B such that u(x)
for a!l y e B. Then x e c-(B,_t} and hence c•(B.tl
:;e "·
l!:
u(y)
(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.)
Su:~cose ••
1.0.3
If x e C(X), then x e C( {x,z} },
that the Weak Axiom holds.·
which cor:tradicts the equality C({x,z}) = {z}. C({x.y}), whid contradicts C({x,y}) = {x}.
which cont:radkts C( {x,z}) = {y}.
Thus
v CCB ), the~ x ~ y becango 8 u B 1 2 2
;::l
If z e C(X), ther.. z e C{ {v, z}), •
(~,C( ·))
Let t ntionali~ C{ ·) relative to !B.
1.0.4
If y e C(X), then y E
must violate the Weak Axiom.
Let x e C(B v B l and y e C\B ) 2 1 1
C(B l v CCB ). 1 2
Tnus x e C(C(B l v 1
CCB )). 2 Le: x e ctCCB l v CCB )) and y e B v B • then there a:-e four cases: 2 1 2 1 Case 1.
x e C(B ), y e Bl" 1 •
1-5
Case 2.
x·e C(B ), y
Case 3.
X
Case 4.
x e C(B l. y e 82. 2
1
E
B . 2
e CCB J. y e Br 2
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(B l and do the same argurne:tt as for Case 1 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 p
be the price of the consumption good in period 2. measured in
2
units of the consumption good in period 1. levels in periods 1 and 2. respectively. 2
•
set 1s equal to {x e IR+: x
1
Let x , x 1
be the
2
consum:~tion •
Then his lifetime Walrasian budget
+ PzXz :s w} .
2
2.D.2
{(x,h) e IR : h :s 24, px + h
2.0.3
(a) No.
+
:::!:
24}.
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
p,w
, x' e B
p,w
, and ~ e {0,1).
X is convex, x" e X.
Moreover, p·x" =
=
.
W.
2.0.4
"T"\.
!HUS X
..
E B
p,w
Write x" = AX + (1 -
~(p·x) + (1 -
~)(p·x'J :::!:
~)x'.
Since
i\.w + (l - i\.)w
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 vectors Wltn rat1o M - 8s
8
s' -----;-M~-~8::::-s-
8
+
' 8 +
s'
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--~-~
8 +
M - 8s s
8 > M --.......,M...,...---~8::-s- = M M/s 8 + s 8
2-1
= 8s.
2.£.1 The homogeneity can be checked as follows: tl.W
tl.W
x (ctp ctw) = 3
•
ctpl
a.w
ctp 1 + ctp2 + a.p3
w
-
p
= x,(jJ,wl, l
1
w
-
w
-
To see if the demand function satisfies Walras' law, note that
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)} Hence
+
pkxk(p.w)/w =
o.
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·DwJdp.w)w = 0.
p·D x{p,w)p p
+
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
By differentiating the equation x(p,cxw} = cxx(p,w) with respect to a and
2.E.4
evaluating at a = l, we obtain wD wx(p, w} = x(p, w).
Hence D\\' x(o,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. path, E
p
2.E.S
The assumption also implies that the wealth expansion
= {x(p, w): w > 0}, is a ray going through x(p,l).
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. 8cpip)/8pk = 0 whenever k we can write xt(p,w) =
Thus xt'p,w) :;e
= xip,l)w.
t, xt(p,l) is actually a function of pl alone.
xt(pl
2.E.6
So
Since x(p,w} is homogeneous of degree zero,
xl(pl) must be homogeneous of degree - 1 (in pl). such that xt(pl} =
Since 8xefp.l)/cpk =
Hence there exists o:l > 0
«lPt· By Walras' law, Lt.Pt(alpt)w = w[lc::i =
When a = l, Walras' law and homogeneity hold.
w.
We
must
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
d(ln xl(p,w))
pk
-
d(ln p,)
axl
a
(p, w) • exp On pk)
pk xl(p,w}
-
•
xt(p,w)
iC
{p,w)·pk
-
~ lk(p, w) .
Simiiarlv, -
cHIn x l( p, wl l d(ln\r,:}
8xl
-
aw
axl
aw (p, w)w
(p, w) · exp(ln w)
=
xl(p,w) •
Since o:
= dOn
1
= d(ln
xip,w))ld(ln p ), o: 1 2
=
xip,w))/d(ln p l, and a: 2 3
d(ln x
iP· w) )/d(ln
Z.F.l
We proved in Exercise l.C.Z that Definition l.C.l and the prope:-ty -in
w), the assertion is established.
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\8') .: e, t!ien C(Bi "B' c C(B') and B" CCB') c CCB).
:;c 0
and B
If C{ · l is
· , , · t;.en ~ · · property 1s · equ1va · 1ent to t h e f o 11 owmg . smg.:.e-va.uec, trus one: F 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'). 2 . -t
He:1ce Definitions l.C.l and Z.F.l are equivalent.
1 ....
2.F.2 •
But this is the contraposition of the property stated in Definition
It is straightfor-ward to check that the Weak Axiom holds. •
I
. .
I
8 anc
0 • • X"' :!: •
P
-
... -e.,
'-e·~~:~~ & .....,, I
J
-
·~ ~'W
x
l
•
•
•
•
•
=
j, then pJ·x
. .. Iy r S lr:lllar"
.
1
Stnce
= 9. p
I
·X
In fact, if
2 1 . 2 S mce p ·x = 8, x is revealed
3
= 8,
2-4
X
I .
15
revealed
prefe~:-ed
3
to x.
3
But, since p · x
2.F.3
2
= 8, x
3
2
i's revealed preferred to x .
[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. suppose that y < 75.
Then
2-5
Sc
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. good 1 increases. 100.
Also the rela:ive price of
But the demand for good 2, y, decreases because y < 75
100 · 100 + 100 · 100. Hence the real wealth increases from year 1 to 2. good 2 dec:-eases.
Also the relative price of
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.
.m: er1or . ~
Hence it is an
. gooc.
the co:1sume::- has a revealed preference for x
over x . 1 0
(b) If PQ >!,then (p ·x )/(p ·x ) > 1 and hence p ·x > p ·x . 1 0 1 1 1 0 1 1
consume:- has a revealed preference for x
Thus the
over x . 0 1 = i\.
Hence, by taking i\
larger or smalle; than one, we can make EQ larger or smaller than one. this obviously does not have any revealed preference relationship.
2-6
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-,-:-~ )- x(p,l)) :s 0 .
11
•
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. x(p',p'·x(p,l)l 1.
;=
If
x(p,l), then the weak axiom implies that p·x(p',p'·x(p,lll >
Hence the above value is negative. As for the second inequality, ---.-___;,.----,..,.... ) - x( p ,1 })
p ·x p,
1
-
p' · x(p,ll
= 2 - (p' ·x(p,l) :s 2 - 2
+
+
1•
1
p' ·x(p,l) )
(p' ·x(p,l))(
1
p' ·x(p,l} )
= 2 - 2 = 0.
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
S(p, w) =· D x{p. w) + D x(p, w)x(p, w)
T
w
p
= D x(p, w) + (1/w)x{p,w)x(p,w)T. p
Thus 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
2
v·(x(p,w)x(p,w}T)v =- (v·x(p,w)) , the matrix- (1/w)x(p,w)x(p,w)T is also negative semidefinite.
2.F.6
Thus D x(p,w) is negative semidefinite. • p
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'} x(p',w')
:;e
Let a: = w' /w.
x(p,w).
~
w and
-1
Then x(p' ,w') = x(p' ,a:w) = X(IX p', w} by
the homogeneity assumption, and p·x(a:
-1
p',w}
~
w and x{a:
-1
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,
-
p·.S(p,w)
= p·Dpx(p,w)
+
•
.
T p·Dwx(p,w)x(p,w) = p·Dpx(p,w)
+
T x(p,\'11")
=0
By Proposition 2.E.l and Walras' law,
T
D x(p,w}o D x(p,w)p + D x(o, w)w = 0 . S(c w)c = ... Dwx(p,w)x(p,w) p = • • • 0 • p
•
pk
2.F.8
-
OXi. --.--..... -::::--- ( p, w) xl ( p, w) 8pk •D.lC
= c //' (0. w) + '"K"
w
xi.(p,w)
= clk(p,w) + ci.w(p,w)bk(p,w).
2-8
w
.
. TAT ( TA )T TA . A . . . r" . .~ ( a ) S mce x x = x x = x x, a matr1x 1s negative ae. m1te 1.
2 . F •9
and only if x TAx T
+
n
x TATx < 0 for every x e IR \{0).
. TA S 1nce x .. x
+
T
TAT x x = T
x (A + A }x, this is equivalent to the negative definiteness of A + A . A is negative definite if and only if so is A + AT_
Thus
The case of negative
definiteness can be proved similarly. The following examples shows that the determinant condition is not - 1
Let A =
sufficient for the nonsynunetric case.
_
3
0 1
, then A
11
= - 1
•
and A
22
= 1.
- 1 3
But U, 1)
0
1 1
- 1
Hence A is not negative
• 1.
semidefinite. By Proposition 2.F.3, S(p, w)p = 0
(b) Let S(p, w) be a substitution matrix.
and hence s
= (-
Cp,w) = (- P/P >s (p,w). 12 2 11
p 1p ls Cp,w). 1
2
11
Thus s
22
(p,w)
=
Also p·S(p,w) = 0 and hence s
21
2 2 (p /p >s (p,w). Thus, for every v 1
2
11
Cp.w)
=
vl v
2
•
v·S(p,w)v
Now, su:ppose that S(p, w) is negative semidefinite and of rank one. to (•J, the
nega~ive
semidefiniteness implies that s (p,w} 11
rank one implies that s
11
(p,w}
:;!:
0.
Hence s (p,w) < 0. 11
Conversely, let s Cp.w) < 0, then, by (•), v·S(p,w)v 11
2.F.10 •
~
~
0.
Tnus s
Ac::ording
Being o:· 22
Cp,w) < 0.
0 for every v.
(a) If p = (1,1,1) and w = 1, then, by a straightforward caicu!ation, •
we ootam 1
0
0
- 1
1
1
0
- 1
S(p,w) = (1/3)
1
2-9
•
•
Hence S(p, w) is not symmetric. -
- 1 0
Hence
1
1
0
I - 1
Note that =-v
- 1
2
2
2
+ v v - v = - (v 2 1 1 1 2
is negative definite.
3v /4. 2
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
- 2 - c
S(p, w)
0
1 + 2c •
- 3c
Thus, (1, 4, O)S(p,w)
if
£
1 4 0
A
= (1, 4}S(p, w)
> 0 is sufficiently small.
1
4
2
= (2 + cl- £2 - 4lc) > 0,
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
(- P/P )s (p,w). 2
11
Also p:S(p,w} = 0 and hence s
12
(p, w) =
(p,w) = (- p /p >s £p,w). 21 1 2 11
(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). equivalently, for any (p,w} and (p',w'), if x
E
Or
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 p·x'.
= w',
then (p'- p)·(x'- xl
= w-
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. ::!: w.
So suppose that x e x(p,w), x' e x(p',w'l, p' ·x = w', a:1d p·x'
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.
2-11
.
.
The wea:< ax1o:r: r:ow
.1mpnes ..
that x(p, w) = x(a:p,o:w).
Since oxip, w)/8w = 0 for both l = 1,2, we have slk(p, w) =
2.F.15
8x£'p,w)/8pk for both l
= 1,2
and k = 1,2.
-S(p,w)
Hence, let
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
vl
- 1
v2
2 v 2Hence, by Theorem
(We saw this in the answer to Exercise 2.F.lO(a).)
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) = x
3
o:p/a:p
3
(c:;::,o:w) = IXW/o:p
3
= - P/P
= w/p
3
= x Cp,w), 3 2
= x (p,w). 3
As for Wal:-as' law, ~ v • ./,~ ::'-. y.-_
.;
(b)
p = (1,2,1), w = l, p'
= (1,1,1),
(2, - l, 1} and x(p', w') = (1, - 1, 2).
p · x(p', w') = 1
(c) Denote by obtained by
= w.
•••. 'IY
..)
and w' = 2, then x(p,w) = • • Thus p' · x(p, w) = 2 = \V anc
Hence the Weak Axiom is violated.
Dx{p, wl
de~eting
--
the 2 x 2 submatrix of the Jacobian matrix Dx(p, w)
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)
-
-
= Dx(p, w) = 0/p )
1
1 - 1
because ax (p,w)/8w = 0 3 1 2 Note that v·S(p,w)v = 0 for every v e ~ . Now let v E !R 3 • "
Note that v = (v - Cv /p )p) 3 3
+
.,
1
A
-
Cv /p )p and the third coordinate of v 3 3
-
2
So denote its first two coordinates by v e IR •
-
...
...
Then, by Proposition 2.F.3, v·S(p,w)v = v·S(p,w)v = 0.
(c) Suppose that p' ·x(p,w)
~
w' and p·x(p' ,w')
~
w.
The first inequality
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
D x(p,w)
p
-
(-
2
w/([tpt) )
• • •
1 • •
• •
•
•
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
y.
(a) Assume that t is strongly monotone and x » y.
-
Then Uy - x II
»
y.
£,
-
~
= Min {x - y , ... , 1 1
£
z•u
0, then,
and z• >- y.
By x » z• and the
By Proposition l.B.Uiiil (which is implied by the
-
transitivity). x >- y.
rnus >- is locally nonsatiated. .
then x » z.
e X such that lly -
weak monotonicity, x >- z•.
3.8.3
0.
c and y >- x.
$
Define
for every z e X, if hy - zH
L Let e = (1, ... ,1) e IR and
•
Suppose that x
there exists
-
-
(b) Assume that >- is monotone, x e X, and £
3.B.2
v and x ;:
~
Thus >- is monotone.
Hence x >- y.
y = x + (c/v'L)e.
Then x
Thus >- is monotone.
-
Foliowing is an example of a convex, locally nonsatiated preference 2
relation that is not monotone in IR . +
-
For example, x » y but y >- x. •
•
X
•
y 0
i
0
XI
Figure 3.B.3
•
•
3-1
•
Let ~ be a lexicographic ordering.
3.C.l
To prove the completeness, suppose
Thus y >- x.
-
To prove the transitivity, suppose that x ?: y and y ?: z.
Then x
1
~
y
1
•
z1• then x ?: z.
and
Hence x
!:::
2
z . 2
z , then X = 1 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
-
(0,1). Without loss of generality, assume that x
:;e
y.
-
:;c
z, and o: e
By the definition of
the lexicographic: ordering, we have either "y > x " or "y = x a.'ld y >. x ". 1 1 1 1 2 2
z:
On the other hand, since z V
J
2'
II
Hence, we have either "a:y
and a:y
3.C.2
2
+
(1 - cdz
> x . 2 2
1
+ (1 - o:)z
Thus a.y
II
+ (1 -
Take a sequence of pairs
x. and yn .. y. impiies that u(xl
3.C.3
x, we have either "z > x " or "z = x and x 1 1 1 1 2
Then u(xnl !:::
u(y).
"!:::
1
>
X/'
or "o:y
+ (1 - cdz
1
1
= x
1
o:}z >- x.
such that x
n
n >- y for all n,
-
X
n
u'(yn) for all n, and the continuity of u( ·)
Hence x >- y.
-
Thus >- is continuous.
-
•
•
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
i!:
{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 for every n > Nl"
1
such that y >-
X
n
Since {z: z >- x} is open there exists a positive integer 1':
n such that y >- x for every n > N .
2
Conceivably, there are two cases on the
2
•
n .sequence {y }: Case 1: There exists a positive integer N .
such that yn >- y for every n > N .
3
-
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), we have y
n
>- x
n
for every
•
n > Max {N ,N }. 1
3
This is a contradiction.
open, the:-e exists a positive integer N By x .
Smce
{
n
?:' y
n
--
such that y
and Proposition l.B.l(iii), x
k(m)} .
z: z >- v
4
If Case 2 applles, then there
1s closed, x >- y
-
k(m)
n
>- y
n
k(m) >- y ) fo:- every n
k(m)
>
. for every n > N . · 4
But, since k(m) > N , this is a
.
2
. con ....ra d'1ct1on.
3.C.4
We p:-cvid.e two examples.
satisfies
mono~:)nicity,
Exar:1-:>le -1.
Let X = IR
The first one is simpler, but the second o:1.e
which the first does not .
-
+
•
.
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}. preference relation represented by u(- ). continuous.
Denote bv >- the
--
We shall now prove that >- is not
-
Ir:. fact, if u(l) > 0, then consider a sequence {xn} with xn = 1 -
1/n fer everv r...
-
Although x
n
- 0 for every n and x
n
-+ 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. then all lower contour sets are closed. sets are
Note that if u(x) = 0,
If u(l) = 1, then all uppe:- contou:-
. Close ..... ~
3-3
•
2.
-
following rule: Case 1.
If x + x :s 2 and x 2 1
Case 2.
If min{x ,x > 3!::: 1 and x 1 2
Case 3.
If x
1
+
x
:;e
0,1), then u(x) = x + x . . 1 2 :;e (1,1),
then u(x) = min{x ,x } 1 2
2.
+
> 2., min{xl'x } < 1, and x > x , then 2
2
2
1
•
•
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
Tne preference >- is convex and monotone.
-
1/n, 1 - l/n)
-7
(l,ll and (1 + 1/n, 1 + 1/n)
u(l -
ln fact,
-7 (1,1)
1/n, 1 - lin) = 2 - 2/n -+ 2;
3-4
(1,1).
But, whateve: the choice of the
value of u(l,l) is, it cannot be continuous at (1,1). (1 -
:;e
as n
-7 cc,
and
+ 2 -+ 3.
u(l + 1/n, 1 + 1/n) = 1 + 1/n
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).
-
•
= 3,
If u(l,l)
...
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 L xe!R,y e IR +, and
X -
+
y.
homogeneity, u(o:x) = u(o:y).
Then u(x) = u(y) and hence cw(x) = o:u(y).
0,
l!::
By the
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,
X E
~
and hence u(x} + a = u{y} + a.
L L IR , y e IR , and +
+
X -
y.
Then u(x) = u(v) •
the functional form,
u(xj + o:
= (a
+
~(x , .... xL)
= u(x
+ a.e 1,
u(yl
= (a
+ y ) + ~(y , ... ,yL)
= u(y
+
+ a
x } + 1
2 2
1
where e. = (l,O, ... ,0} e
1
ae ). 1
ae
L
1
- 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
1 depends crucially on the connectedness of IRL- , which appears in X = {+
•
L-1 x lR + .
a:,
m)
(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.
•
-u{ ·)
Let >- be a quasilinear preference and a utility function
-
-u( ·)
The existence of such a
>-.
-
is guaranteed by Proposition 3.C.l, but, of A
For each x e
course, it need not be of the quasilinear form. •
-
,.
represent
L-1 G< + , define
...
I(x) = {u(x .x> e IR: x e IR}, then J(x} is a nonempty open interval, by the 1 1 •
continuity and the strong monotonicity of •
L-1
along e . 1
,.
L-1 e IR • i.f J(x) :;e l(y), then Hxl ,... l(y) +
and y
x e IR +
Step 1: For every
=
...
~
0.
Proof: Suppose that
l(x)
l(y).
:;e
Without loss of generality, we can assume
... that there exists u e J(x) such that u IC l(y). u
::5
infl{yl.
simila.:-ly.)
Suppose that u Then let
(y (y
1 1
-
- x
,y).
1
-uCx
..
u =
-u(xi,x),
then, for every y
~.
...
e IR, 1
.x)
1
> u(y l'y).
Hence
l(x)
n
l(y)
L-1 : l(x) , define E(x) = {y e IR +
= 0.
=
l(y)}.
L-1 "' · . L-1 , E(x} is open tn IR For every x e IR •
Step 2:
+
-
-
-
+
-
L-1 Let x e IR , x e IR, and u = u(xl'x) e Hx). 1 +
Proof:
(u - c, u + d c !(x).
.u- -v e - L . -1
-
(The other case can be treated
By the quasilinearity, this implies that (x ,x) >1
L-1
For each x e
~ sa~isfy.
s•;p I(y} or•
supl(y).
2:
2:
..
+ xi, y).
Thus
xi e
...
Then either u
~"~
+
•
-
Since
...
an c. llx - •vii
0 satisfy
is continuous, there exists
I u(xl'xJ
•
•
3-6
-
-u(x ,yl- I 1
0 Hence
such that
•
I(x)
l(y)
"
Thus, by Step 1, l(x) = l(y), c E(xl. Thus E(x) is open. Step 3:
:~
(u - c, u + c) r.
... or y
-
L-1 Hence {y e IR+ : fix - vii < o}
E(x).
E
Hyl = 0. -
For every -
L-1
L-1 anc! y e IR , we
It is sufficient to show that for every x e IR +
Proof:
.
have E(x) = £(y). -
.
...
Suppose not, t h en th ere ex1st x e
+
IRL-1
an
+
d ...
y e
~
L-1 +
such
-
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
=
•
IRL-l~(;)
Step 2. this implies that ...
L-1
partition {E(x},IR + open.
is open.
+
...
~(x)}
L-1
of IR +
,
Hence we have obtained a
both of whose
This contradicts the connectedness of IRL-I. •
+
-
e~ements
are nonempty and
E(~) =
Hence
E(;) for
... L-1 L-1 every x e IR . and •v e IR +
+
I(x)
By Ste? 3,
-
-
L-1 I(O) for every x e IR . +
Thus for every x e IR
...
exists a unique a e IR such that o:e - (O,x}. 1
-
-L-1 e ;:'1 ... •
L-1
Define f/J: IR+
Define u: X -+ IR by u(x) = x
L-1 +
, there ...
-+ IR by ¢(x)e
1
-
fc:-
1
every x e X. •
Step 4: The function
u.(
•
·l represents >-. •
Proof: Suppose that x e X, y equivalent to (x
1
E
-
X, and x >- y.
-
By the quasilinearity, this is
- y , x , ... , xL) ~ (O,y , ... ,yL}.
1
2
2
¢( ·}. this is equivalen-:: to Cx
By the definition of
- y , x , ... , xL} ~ ¢(y , ... ,yL)el" 2 1 1 2
the quasilinearity, this is equivalent to
Again by the definition of ¢( · ), this is equivalent to ¢(x2, ... , XL)el ?= (¢(y2, ... ,yL) + yl - xl)e1.
Hence ¢(x , ... , xL) 2
2:
q)(y
2
, ... ,yL) + y
- x , that is, u(x) 1 1
•
•
3-7
2:
u(yl.
Again by
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: x + o: x . 1 1 2 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: x~lnx p ... O
Since expUet
1
1
1
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'x l 2
p-1 p-1 with respect tc the CES utility function is equal to o: x lc:: x . 1 1 2 2
•
The
mar gina! rate of substitution at (x ,x ) with respect to the Cobb-Dougias 1 2
as p
~ 1.
(In fact, a: x
p-1
p-1
let x 2 2 1 1
o: x 11X x when p = 1.) 1 1 2 2
is well defined for eve:-y p a!1d is equal to
The proof is thus completed.
St:-ictly speaking. there is a missing point in both proofs: We proved the conver-ge:.1ce
o:~
2
preferences on the strictly positive 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 To be more specific, compare, for example, x =
there tend to be indifferent.
= (yl'O}
Cx ,o> and y 1
with x
1
> y > 0. 1
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 = (x ,o) with x > 0 and y 1 1 and a ~ + a ~ -+ a
1
2
1
a .
+
0, then
The implication of this fact is that, as p -+ 0,
2
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 x
1
:s x .
We want to show that
2
x
Let p < 0.
Since x
?! (o:
x
p
1 1
+
1
~
.
p l/p
= 11m £o: x + o: x l 1 2 2 1 1 p-t -co
0 and x p 1/p .
a: x ) 2 2
p
.
2
~
-
0,
Thus
'
•
On the other hand, smce x :s x . 1 2 p
(o:l + o:2)xl.
Therefore,
Hence al Letting p
.1lffi . p-+- a:
3.0.1
'
\C::l
-+ -
l/p
>
xi -
("'
xp
Yo! 1
p)l/p
+ a2x2 p
CXl,
i!:
(
p)l/p
we obtain lim Co: x + a x 2 1 1 2 p-+ -co X •. L
To check condition (i),
3-9
al + a:2
)1/p
xl.
= xl' because 1 im P.... -co
x (i\p,i\w} 2
= (1
...: o:.HAw)/(Ap )
2
= (1
- a:)w/p
= x 2 (p,w).
2
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\ - cdnp - (1 - «)lnA - (1 - cxHnp 1 2
= edna: + (1 - a:HnCl - a:) + lnw - cdnp - (1 - a)lnp 1
c
2
•
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 tha:., for anv v e .
-
-;,~
!)!"'ove
~ 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 {p e IR
3.0.3
(a) We shall prove that for every p e •
3-10
L
w
++
i!:
: v(p, w)
~
v} is convex.
~L 1"f 0, a i!: 0, and x e "' +'
x
= x(p,w),
= x(p;aw).
then ax
affordable at (p,aw). u(IX
-1
yl :s u(x).
Note first that p· (ax) :s aw, that is, ax is
Let y e IR
L +
and p · y :s a:w.
Then p· (a
Tnus, by the homogeneity, u(y) :s u(crx).
•
-!
yl :s w.
Hence
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.
wx(p)
•
Given the above results, we can write x(p,w) = wx(p,l) =
wv(p).
= wv(p,l) =
C JJ
C.T/v•
lS
and v(p,w)
•
Exercise 2.E.4 showed that the wealth expansion path
-x(p).
{x(p,w): w > 0} is a ray going through •
•
The wealth elasticity of demand
eoual to l. •
(b) \Ve firs: prove that for every p e IRL , w ++
ax(p,w}.
:.?:.
0, and a. :: 0, we· have X(p,o: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
X E
-~
L
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 exist p e x(p',w'}.
1
r;.io ••
, w :: '0, and w' ' e !RL P ++ ++ ,
i!::
0 such that x = x(c,w) and x' = .
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' ). lXX
and x(p' ,aw') = ax'.
u(IXX) = u.(ax').
3.D.4
But as we saw above. x(p,o:w) =
Hence v(p,o:w) = u(a:x) and v(p',aw') = u(cxx' ).
Ther-efore u(x) is homogeneous of degree one.
L
(a) Le: e = (1,0, ... ,0) e IR . 1
We shall prove that for every p e
1 w e R, a e iR, and x e (- co, co) x IR:- • if x = .x(p,w}, then x x(p, w + c::l.
Thus
Note fir-st that, by p
is affordable a! (p, w + a.e L 1
1
= 1, p· (o:x L
+
ae
1
=
e l :: o:w, that is, x 1
Let y e IR + and p · y :s w + a.
3-11
+
+
ae
1
Then p · (y - IXe l 1
< -
w•
Hence·x
x + o:e
1
~
y- o:.er
= x(p, w +
1
>- y .
-
Hence
o:).
Therefore, for every
XiP· w' ).
Thus, by the quasilinearity, x + ae
t e
w e IR, and w' e rR, xl(p,wl =
{2, ... ,L},
That is, the Walrasian demand functions for goods 2, ... ,L are As for good 1, we have 8x(p, w)/Bw = 1 for every (p, w).
independent of wealth.
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 and the preference 1
relation can be represented by a utility function of the quasilinear form u(x)
= x
1
+
-u(x
2
, .... ;_l (Exercise 3.C.S}, we have
v(p,w) = u(x(p,w))
•
-u(x = w + x (p,O) + 1
2
(p, w), ... ,xL (p, w)).
-u(x
2
(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 p x Cp,Ol > w. 2 2 Note that x (p,O) = (1)') 2
-1
(p l, because p = 1. 2 1
binding if and only if p (1J') 2 by x(p,w) = (O,wlp l. 2
-1
Cp ) > w. 2
Hence the const:-aint • is
If so, the Walrasian demand is given •
Tlrus,, 'as w changes, the consumption level of the first
good is unchanged and the consumption of the second good changes at rc.te llp 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 x(p,w) = •
•
3-12
-u( · ),
2
where
= p/(p - 1) e (- co, 1).
Plug this into u( · ), then we obtain
•
v(p,w)
(b) To check the homogeneity of the
x(IXp,o:w) = (aw/[(o:p ) 1
= (a.·a:
~-1
~
function, ~
~-1
+ (o:p } )((o:p } 2 1
, (ap l 2
tS-1
)
~
tS ~ tS-1 ~-1 /(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)
~
+
+
~)1/~
Pz
= v(p,w)
To check the monotonicity, ~
~
8v(p,w)/8w = 1/(pl + Pz)
1/o
> 0, •
~
-
~-1
ov(p,w)/opt = - wpl
~
0 1/~+1
/(pl + p2)
. immediately from
< 0.
•
The continuity follows •
the derived functional form .
•
In orde:- to prove the quasiconvexity, by property the homogeneity, it is 2
sufficient to prove that, for any v e IR and w > 0, the set {p e IR : 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.
f( p ) = (f( pJ.~>1/o
. . vt !S
~
f convex or every v.
2
Since v(p,wl = w/f(p}, this
implies that {p e IR : v(p,w) :s v} is convex for every v and w.
•
3-13
So we consider
If 1/f(p) =
o
(f(p} }
=
:s v}
Ill:
this implies that {p e
(c)
~
a p
.1s
+
1
~
•
Pz is a convex function.
convex for every v.
Hence {p
E
2
IR :
Sir.ce v(p,w) = w/f(p),
v(p,w) :s v} is convex for every v and w.
For the linear indifference curves. we have
x(p, w}
(w/p , 0}
if pl < p2.
(0. w/p2)
•r
1
-
h
{(w/p1)(A.. 1 -A.): A e 10,1)}
.1.
.,.
pl > P2' p ; p1 2
v(p,w) = max{w/p ,wlp }.
1
•
2
For the Leontief preference.
•
As for the limit argument with respect to p. with p < 1 and p -+ 1.
Then
a
First consider the case
= p/(p - 1) -+ - co as p -+ 1.
a )
..n ... ,.. S' __
1. we have (p ;p -+ 0. Thus 2 1 w/pl Cl-1 ~ 0 1 • ... l IH p w/Cp + p l = 1 im 1 2 1 ~...;. - = o-+- co 1 + •
1, we have (p{p l 2 ~-1
1 im P2 o-+ -co
~
~
w/(pl +
-+ co .
Thus
•
a Pz>
w/p2 1 im 0 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 (p + p ) -+ p for p :s p . 2 1 1 1 2
Hence the
CES indirect u"tilities converge to the indirect utility of the linea:indifference curves. Case 2.
p
1
> p . 2
De the sa:ne argument as in the Case 1.
•
3-14
•
Case 3. ln this case, [w/2.p Hl,l}.
This consumption bundle belongs to the set of the Walrasian
1
demal'lds cf the linear indifference curves when p = p . 1 2
As for the indirect
utility functions, we showed in the answer to Exercise 3.C.6(c) that ~
(pl
~ 1/~
+ Pz>
-+ Pl for PI :s
Pz·
•
Let's next consider the case p -+ - co. So just plug
1.
~
Note that
~
= p/( p - 1) -+ 1 as p -+
= 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
x (p, w)/x (p, w) = (p /p ) 1
2
1
~-1
2
•
,
£x (p,w)/x (p,w)}/(p /p > = (P/Pzl 1 2 1 2
~-2
,
d(x (p, w)/x (p, w)]/d[p /p J = (~ - 1Hp /p 1 2 1 2 1 2
Thus ~
12
~lZ(p,w}
(p,w) :::
Hence,
- p).
0 for the Leontief, and
~
12
-
•
(a) Define
&' !
7' = 1 a:1.d
12
£p,w)
= cz:
for the linear,
= I for the Cobb-Douglas u!ility
Cp,w)
•
•
-u(x)
bl)
with a' = a/(a + {3 + a-l. {3' = {3/(a. + f3 + +
~
•
... f unc-.1cns.
3.0.6
= 1/(1
= - (~ - 1)
o-2 ) .
-u( · l
unction u -+ u
r>.
a'
(x2
'l' = 7/(a + f3 + a-).
Tnen c::'
represents the same preferences as u( · ), because the f
1/( a+f3+1') .
.
1s a monotone trans ormation.
wi:hout loss of generality that a.
+
f3
+
Tnus we can assume
a- = 1.
(b) Use anothe:- monotone transformation of the given utility function,
•
+
3-15
W
The first-order condition of the UMP yields the demand function x(p,w) = (b ,b ,b ) + (w - p·b)(ct!p ,,a/p ,7/p J, 1 2 3 1 2 3
-=
where p · b
p b + p b + p b • 1 1 2 2 3 3
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) = Cb ,b ,b ) + CA.w - i\p· b)(a/i\pl',S/A.p .r/i\.p ) 1 2 3 2 3
= (b ,b ,b ) + [w- p·b}(a/p ,,S/p ,,./p ) = x(p,w). 1 2
3
1
2
3
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 -
= v(p, w).
To check the m:>notonicity,
> 0, Civ(p,w)/op
1
= v(p,w}·(- a/p ) < 0, 1
Civ(p,w)/c3p
= v(p,w)·(- {3/p ) < 0, 2 2
8v(p,w)/8p
= v(p,w}• (- 0 /p ) < 0. 3 3
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 3
> 0, the set (p e IR : v(p. w) ::= v} is convex.
Consider
lnv(p, wl = IX!niX + {3in(3 + 7ln'1 + ln[w - p· b) - cdnp
Since the logarithmic function is concave, the set
•
3-16
1
- ,Slnp
- a-lnp . 2 3
3
{p e IR : lnlw - p·b) - cdnp - (3lnp 1
2
- 'llnp
3
~ 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. 3 Hence so is {p e IR : v(p,w) :s v)
.
1
(a) Smce p · x
3.0.7
Thus x
1
0
< w
1
and x
1
:;e x
0
,
o. the weak axiom implies p · x > w . 0
1
has to be on the bold line in the following figure.
8 6
4 ··-········--·· • •' •' •• ••
0
• '•' ••
• • '•• • ••
2
4
8 Figure 3.D.7(a)
Xl
•
•
•
•
... _... be a In the following four question; we assume the given preferer.::..: ···"'..,
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)/8x 3.C.5(b)).
each
t
1
= 1 for every x CExe:-cise
Hence the first-order condition implies 8u(xt)/8x~ = p~/p~ for
= 0,!.
0 0 S ...:t .. .. p /p < 2 1
.
I
concave, x
... . .
to be on the bold line in the following figure .
•
3-17
0
2
1
> x .
2
.
T nus x
1
has
8
4
•• •
0
•• •
•• •• •• • • '• '
•• •
2
4
'' ••' •
•
•
•
8
10 3.D.7(b)
XI
(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)/8x = 1 for every x 2 Hence the first-order condition implies c3u(xt)/8x~ =
(Exercise 3. C.S(b)).
0,1.
Thus x
1
0 0 Since P 1P 1 2
>
1 1 . p 1p and u( · l 1s concave, we must have 1 2
has to be on the bold line in the following figu:-e. •
•
••
•
8 5.5 ..... ::~.:;:-:: • •• • • •• ••
.................... • ••• •• •• •
••• • •• •••
2
4
•
__.__.___ ___________ 0
1
(d) Since p · x
0
....;..._
8 Figure 3.D.7(c)
1
X!
<w and the relative price of good 1 decreased,
•
3-18
x~
has to
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
Thus x
1
has to be on the
bold line in the following figure.
• •
8
................ • •• ••• •• ••'
•
•
••
0
•
2
••• • •• •• •• ••
4
•
8 Figure 3.D.7(d)
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 1
becomes flatte:-.
> P/P 2 . x 0
to be on the right side of the ray that goes through x . the bold line in the following figure .
•
•
1
•
•
3-19
Thus x
1
1
has
has to be on
8 •'
••
••
•
••
••
•• •
...
•••
•••
••
• • •••
• •
5.2
•
• •• • •• •••
····--··
•
••
•• ••
••
0
2
4 5.2
8
X!
3.D.7(e)
3.0.8
By Proposition 3.D.3(i}, v(ttp,tXW} = v(p,w} for all
IX
> 0.
differentiating t.'lis equality with respect to « and evaluating at obtain 'V v(p,wl·p + wc3v(p,w}/8w = 0. p
Bv
-
et •
1, we
Thus w8v(p,w)/8w = - 'V v(p,wl·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 and IJ. • x• = 0.
~
> 0 and IJ. e IR
L +
such that p = A'ii'u(x•) + 1-1
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 p > O, 1 2 ~ 1-IX et-1 a-1 1-a oe(p,u)/8p = et (1 - a.) p p > O, 1 1 2 ~
oe(p,u)/8p
2
= «
(X
(1 - o:J
« «
p p
-(X
> o.
1 2
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 . semidefinite.
2
op e(p,u)
is negative
An alternative way is to only calculate •
_2 2 1-« « o:-2 1-a. o e(p,u)l8p = - a. (1 - a.) p p < 1 1 2 2 Then note that the homogeneity implies that D e(p,u)p = 0. p
o. Hence we can apply
2 Theorem M.D.4(iii) to conclude that D e(p,u) is negative semidefinite. p
The
continuity follows from the functional form. •
To check the homogeneity of the Hicksian demand function, 1-a. 1-o: h 1 (ll.p, u) =
(1 -
•
u
« )ll.p1
=
•
(1 -
tt)p1" •
(1 -
h 2 (ll.o u} = . t
a.)ll.p1
(1 -
u
a.ll.p2
=
a.)pl
a.p2
To check the no excess utility, 1-o:
u(h(p,u)) =
(1 - cx)p
-
u
.cxp2
1 •
( 1 - o:)p
(1 - a)p1
u
-
1-a.
{1 ~(X )a.-(X(l-(X)
u
et+(l-a.)
= u.
1
•
The unioueness is obvious . •
3.E.3
Then A
-
Let x e IR :;e
-x
0 by
that of {x e
~
L
:
L +
u(xl
and
e A. u!x)
i!::
u.
Define A = {x e IRL: p·x s +
p·x
and u(x)
i!:: u}.
Furthermore, A is compact: The closedness follows from i!::
u}
L
and of IR +'• the boundedness follows from the
inclusion L
-
A c {x e IR : 0 s xl s p·xlpt for every l = l, ... ,L}.
3-21
Now consider the truncated EMP: p·x
Min
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. the original EMP.
We shall show that this is also a solution to L
If x e A, then p · x :=: p· x• because
Let x e IR + and u(x) o= u.
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.
Suppose first that >- is convex and that x e h(p,u) and x' e h(p,u}.
3 •.E.4
-
Then p·x = p·x' and u(x) (1 -
2:
u, u(x'}
0!::
Let a. e [0,1] and define x" = ax +
u.
Tnen p · x" = ap · x + (1 - a:)p · x' = p · x = p · x' and, by the convexity
a}x'.
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'. to L
Since >- is continuous, (3x" >- x' for any (3 e (0 ,ll close enough
-
But this implies that p · ((3x") < p · x and u((3x" l > u(x')
cont:;:-adicts the fact that x. is a sol\ltion of the EMP.
3.E.S
uh{p),
[Fi:-st pfir..tinsz errata: The equality h(p,u) = because u is a scalar and
tha:, for every p » 0, u h(p,o:ul.
2:
0, a
-h(p) l!:.
is a vector.]
0, and x
2:
Hence
-h(p)u
2: u,
which
1 ~ ···.---
l mu-· ;l!L. be a
.... : --
shoulc be h(o,u} = •
We shall first prove
0, if x = h(p,c), ther. ax =
In fact, note that u(ax) = a.u(x) o= au, that is, ax satisfies the
constrain~
Hence p· (a
of -1
t~e
y) l!:.
EMP for em. p·x.
Let y e IR
Thus p·y
L +
~ p·(ax).
3-22
and u(y)
l!:.
o:u.
Hence ax = h(p,au].
-.•
u(o: ·y) :=: u.
The:-e:!'ore
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 e(p,u) =
-h(p)
= h(p,l) and
-e(p)
ue(p) .
= e(p,l), then h(p.u} =
uh{p)
and
•
•
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: h(p,u) = uCp 1 e,p,u} =
~
1
o u(p 1
+ +
P
p
~
2 ~
2
)
c1-a)la
ua ) .
(p
~-1
1
.p
CJ-1
2
),
To check the homogeneity of the expenditure function,
a ·I/o c ~ a. u pl
CJ J11a _ o:e(p, u). + Pz -
To check the monotonicity,
o (p +
~
oe(p,ul/au =
1
~
p l
I/a
2
> o,
> 0.
oe(p,u}/opl = ~
•
To check the concavity. it is a
. bit lengthy
::..:1.lculate but easy to actuai lv 2 D e(o,u) p •
is negative semidefinite.
An alternative way is to only calculate
a>I/~-1
+ p2
bv cS < 1. •
+
a-1£ a up! PI
..,_ 1
a,I/~-2(1/~ - l)Clpa • + Pz 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, h(ap.u) = u((ap )
~
1
= ex
+
~ (1-~}/~
(o:p ) ) 2
cao-~>t~l+(o-u
((ap l 1
~-1
,(exp l
c3-l
2
)
c a up
1
= h(p,u).
To check no excess utility, u (h( p, u )) =
a u (p 1
a + Pz
>(1-ol/oc ca-op . P
ca-Ilp>l/p + P 2
1
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(x
2
We shall prove that for every
, ... ,XL). •
p » 0 with p x + ae
1
1
= l, u e IX, ex e IR, and x e (- .,, .,) x IR
= h(p, u +
Note first that u(x + cte )
a).
1
:2::
satisfies the constraint of the EMP for (p, u + «). c::.
L-1 +
, if x
~
h(p,u), then
u + o:, that is, x + ae Let y e !R
L +
and u(y)
~
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. =
-h(p)
+
Thus, if we define
-h(p)
~
= h(p,O), then h(p,u)
uel'
Since h(p, u + a) = h(p,u} + ae , we have e(p, u + a) = e(p,u) + a. 1
Thus, if we define
3.E.8
-e{p)
= e(p,O), then e(p,u) =
We use the utility function u(x} =
-e(p)
+
u.
To prove (3.E.1l,
o:-1 a 1-o: « 1-a -a a-1 e(p, v(p, w)) = a (1 - o:) p p (o: (1 - ctl p p w) c w, 1 2 1 2 a-1 a 1-a a 1"« -a o:-1 -a {p,e(p,u)) = o: (1 - «} Pl Pz (a (1 - «) P1P2 u) = u .
-a
•
3-24
1 u +
To prove (3.t.3), x(p,e(p,u)) = (a:-c::(l 1-IX
-
u,
= h(p,u),
u
1-a:
a>
h(p,v{p,w))
1-IX -a: a:-1
PI Pz w
(1 - o:)p
•
0:.
1
•
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.
•
> 0.
(i) Homogeneity of degree one in p: Let a:
v(p,w) by the second relation of (3.E.l).
Define w = e(p,u), then u =
Hence
e(o:p,u} = e(a:p,v(p,w)) = e(a:p,v(a:p,a:w))
= a.w
whe:-e the second equality follows from the homogeneity of
= o:e(p,u}, v( • , • }
and the third
from the first relation of (3. E.l). (ii) ~fonotonicity:
Let u' > u.
v(p, w) a.'1d u' = v(p,w').
Define w = e(p,u) and w' = e{p,u'}, t!\en
By the monotonicity of •
•
v( ·, ·)
L:
=
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: v(p,w) = v{p',w).
e [0,1].
Define w = e(p,u) and w' = e(p',u). then u = •
Define p" = a.p + (1 - o:.)p" and w" = a:w + (1 - IX)w'.
by the quasiconvexity of v( ·, • ), v(p", w") v( • , ·
~
u.
Hence, by the monotonicity of
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
Then,
•
(iv) Continuity: It is sufficient to prove the following statement: For any
n n)}co . ( n n sequence {(p ,u n=l wtth p .u )
~
. n n (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. the continuity of
v( ·, • ),
u :> v(p,w).
By the second relation
~f
By
(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'
Homogeneity: Let
(i)
»
0, w e R, w' e IR, and a.
Define u = v(p, w).
:> 0.
of (3.E.l}, e(p,u} = w.
2:
0.
Then, by the first relation
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.
e(p,u)
=w
and e(p,u'}
must have u'
= w'.
Define u = v(p,w) and u' = v(p,w'), then By the monotonicity of e( ·, ·) and w' > w, we
> u, that is, v(p, w') > v(p, w).
Nex:, assume that p' e(p,u} = e(p',u'} = w.
2:
p.
Define u = v(p, w) and u' = v(p', wl, then
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}.
Then e(p,u) = w and e(p,u') = w'. 2:
u.
Define p''
= o:p
+ (1 -
Define u = v(p,w) and u' = v(p',w'). Without loss of generality, assume that u'
a:)p' and w = o:w
+ (1 -
e{p" ,u') == ae(p,u'} + (1 - o:)e(p' ,u')
3-26
ct)w'.
Then
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.
n
n
Suppose that v(p , w )
i!::
u for every n, then 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. continuity of e( ·, · ), w
~
e(p, u).
We must thus have v(p, w}
~
u.
By the
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 v( ·:.
u.
s~.t
{(p,w):. e(p,u)
i!::
w}
and the quasi-convexity of
is equivalent to the convexity of the set {(p,w): v(p,w) :s u} for eve!"y B•Jt (3.E.l} and the monotonicity imply that v(p,w)
e(p,\.i)
i!::
w.
::5
u if and only if ·
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. function v( ·} is continuous if and only if both {(p, w, u): v(p, w) Hp. v:,u}: v(p, w)
::5.
u} are closed sets.
i!::
u} and
But, again by (3.£.1) and the
monot~nici ty,
3-27
The
{(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( •).
[Fi:-st printing errata: Proposition 3.E.4 should be Proposition
3.E.10 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). (i) Homogeneity: Let a.
relation of (3.E.l).
> 0.
L
Let p e R
++
Define w
and u e R.
= e(p.u),
then u
= v(p, w)
by the second
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}.
x(p,e(p,ull by the fir-st relation of (3.E.4).
Then x e
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 p e IR
L ++
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}.
3-28
Then x e h(p, v(p, w) l by
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 L
Then {z e IR : p·z
i!!:
:;e
0 and c, such that
p •x
< c < p · x for every x e K.
c} is a half space that includes K but does not contain x.
Hence
-x
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
E
A.
Thus K
::l
A.
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
cannot be separ-ated from K.
X
(II2)x + (112}y
y Figme 3.F.2
•
3-29
K
+
{1/2)y
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 V v(p,e(p,u)) + (Bv(p,e(p,u))/8w)V e(p,u) = 0.
p
p
By Roy's identity,
(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 a/pl [I - o:)/p2
•
2 - o:w/pl D x(p,w) = p
0
0 - (1 - a:)w/p2
Ve(p,ul = D e(p,u) p
•
2
•
•
= Dp h(p,u) - cx(l
o:(l - (X)/pr P2
o:(l - o:)/plp2
2
•
- o:(l - IX)/p2
1-o: . The indirect utility function for u(x) = x x 1s 1 2 0:
« v(p,w) = Cp /et) (p /0 1
2
a.))
1-o:
w.
(Note he:-e that the indirect utility function obtained in Example 3.0.2 is for the utility function u(x) = o:lnx
1
+ (1 -
o:)lnx .) 2
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
2 o e(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 - b ) + "¥ln{x - b l . 2 3 3 2
lnu(x) = cdn(x - b ) + 13lnCx 1
1
•
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 b
1
~
0, b
2
i!!:
0, and b
3
i!!:
>
0.
0,
> 0, > 0.
2 to To c!leck the concavit.y, we can show that D e(p,u) is eaual • p
3-31
Then
-
af3/plp2
- 13 c1
and then apply the condition in
-
2
a ltp2
•
ise 2.F. 9 to show •
negative semidefinite.
An alternative way is to only calculate the 2 x 2
2 submatrix obtained from o e(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
semidefinite.
2
op e(p,u)p
= 0.
Hence
The continuity follows from the functional form.
To che:::k the homogeneity of the Hicksian demand function. h( i\.p, u}
= (bl,b2,b3) + ui\.
tt+a+;-1
tt
13
;
(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 .
2
= D e(p,u). p
2 In (a), we calculated D e(p,ul. p
3.0.6, we showed
3-32
In Exercise
2 o:/pl D x(p, w) p
-
-(w-p·b)
0
0 2
0
-
0
f3/p2
0
a:/pl
2
0
1'1P3
f3/p2
(bl,b2,b3).
7/p3
Using these results, we can verify the Slutsky equation . •
(d) Use D h(p,u) p
and the explicit expression of
(a).
•
2 (e) This follows from S(p,u) = D h(p,u) = D e(p,u) and (a). in which we showed p
p
2
•
that D eCp, u) is negative semidefinite and has rank 2. p
3.G.4
Let a > 0 and b e IR.
(a)
-u{
each l,
IR+ ~ IR by
-ut(xt}
~:
Define
= aut(xt)
u(x) = artut(xt} + b
+
IRL +
~
b/L.
= I:t'aul(xt)
1R bv u(x) = au(x} + b and, for •
Then +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). monotone. ul(~+}
To do this, let's assume that each u£' ·) is continuous
strongly
Tnen, for each l, the range ut(IR) is a half -open inter-val.
or
some be is equal to + co, then b and c a!"e +
Then uCKL) = [a, b). +
and the utility function
..u( ·)
So let
Define cl = bl - at > 0, a = Ltal' b =
= [al' bel, where bl may be + co.
L:ebt' and c = Etc . 2 well.)
Lid
(I)
as
Suppose that /: [a, b) ... IR is strongly monotone defined by
..u(x)
= f(u(x}) is sepa:-abie.
simplify the proof, let's assume that f( ·) is differentiable. -+ IR bv•
g(v) = f(v + a) - f(a),
3-33
To
Define g: [O,c)
then gCO) = 0, g( ·) is diffel"'entiable, and g(u(x} - u(O)) = f(u(x)) - f(u(Q)) L
for every x e IR ,..•
Thus, in OI"'der to piOVe that
sufficient to prove that g( ·) is lineal"'.
f( ·) is linear (affine), it is
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)) ==
L
for' 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 •
. I. L I. l e IR . Fix an x e IRL and, for each t, defme y e IR+ by Yt = xi and yk = 0 for + + I. l any k :;c t. Since u(y ) = f(u(y )), L
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([lvll
= 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 ) and g'(v') = g'(vil·
vi
+
-v e [O,cl. 2
Now, for some
-v
e [O,c ), consider v + 2 1 2
-v
1
2
e [O,c) and
•
Then
= g'(v + 1
= g'(v' + 1
=
g'(v2}'
=
g'(v2).
Thus g'(v ) = g'(v') and hence g'(v) = g'(v'). 1 1 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.) X
e IR
L
• •
anc
X'
For example, consider-
e
•
implies that u(x) = u(x'). g(u (x l - u. (0)} 1 1 1
+
By the equality g([lvl} = [lg(vtl•
g(u Cx >- u CO)) = g(u (xi> - u Wll + g~u <xzl - u (0)). 2 2 2 2 1 1 2
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.
3-35
The commodity vectors
for those in·s are represented by z
1
=
the like, and the
commodity vectors for those outside S are represented by z and the like. and z2 e IR +
~#T
We shall prove that for
L-#T
,
2
= {zf)lET
E ~'I.
.,.
I
1
2
IR
+
L-#T z 2 e IR+ ,
(z ,z l ?: (zi,z 1 if and only if Cz .z;21 ?: (zi,zil. 1 2
E
L-#T
In fact,
since u( ·) represents ?:• (zl"z l ?: (zj.z l if and only if 2 2 LteTut(zl) + Lt!CTut(zt)
·::!!::
LteTut(zt) + LtETul(zt).
Likewise, Cz ,zz.) ?: (zi,z2_) if and only if 1
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. xk(p,w) will have to
increase.
Hence, again since uk( ·) is strictly concave, •
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 with respect to Pf we obtain
3-36
... (Ci~(p,w)/Bpt)pl + ~(p,w)
= u"(xip,w))(ch::ip,w)/cpl
By differentiating the above first-order condition with respect to pk (k
:;o:
ll,
we obtain
...
(8~(p, w)/8pk)pt
= u"(xip,w))(Bxip,w)/8pk).
Thus d[p · x(p, w} }/dp,
iC
= xk(p,w) + pk(Bxk(p,w)/8pk) + Lt~kpiBxip,w)/pk}.
2
{8~(p,w}/Bpk)pk
+ ~(p,w)pk
...
+ Lt:;ek
u" (xk(p, w))
u"(xl(p,w))
2
>.(p,w)pk
Pt
+ (BA.(p, w)/Bpk)[l -=-~---.
-
u"(xl(p,wll
..
= u'(xk(p, w)) and hence this equals 2 u'(xk(p,w)) xk(p,w)u'(xk(p,w)} pl '"""::_:------- ( ... + 1) + (o>.(p, w)/8pk}Lt -::...: - - - - - . u"(xk(p,w)) u"(xk(p,w)) u''(xl(p,w))
By the first-order condition,
~[p, w)pk
A
By Walras' law, this equals zero.
By the strong monotonicity and the strict
...
2
u'[x. (p,w)) •
-
concavity,
-u ( . ) •
K
..
< 0 .and
-
Pt
Lt -,..~--- . u"(xl(p,w))
< 0.
xk(p, w)u' (xk ( p, w)) +
-u'(xk(p,wll
1 > 0.
Hence
... xk(p,w)u'(xk(p,w))
-.=-----. ( ---,. ,..------. u"(xk(p,w)} We must thus have 8:\(p,w)/Bpk < 0.
3.G.5
Bv on • the assumction •
+ 1)
< 0.
u"(xk(p,w)) Hence, by (•), Bxip,w)/cpk > 0.
(a) We shall show the following two statements: First, i: (x•,z•J is a
solution to
•
3-37
Max(
x,z
)
-u(x,z)
s.t. p·x
a:z ::= w,
+
then there exists y• such that q · y• ::= z• and (x• ,y• J is a solution to Max(x,yl u(x,y}
s. t. p· x
+
(aq l · y :: w;
+
(a:q l·y:: w,
0
second, if (x•,y•) is a solution to •
Max(
x,y 1
u(x,y)
s.t. p·x
0
then (x• ,q · y•) is a solution to 0
Max(x,z)
-u(x,zl
s.t. p·x
a:z :: w.
+
Suppose first that (x•,z•) is a solution to a:z :: w.
Then, by the definition of
z• and u(x•,y•) =
-u(x•,z•).
-u( • ),
X,Z )
u(x,y) ::
0
+
(aq )·y :: w. 0
inequality follows from p·x + a(q ·y) = p·x 0
Then
+
-u( · )
.
anc the second
(a:q )·y:: w and the definition 0
The first statement is thus established. se~ond
As for the u(x,yl
p·x +
·y) :: u(x•,z•) = u(x•,y•),
where the first inequality follows from the definition of
of (x•, z•).
s.t.
there exists y• such that q0 · y• -
0,
+ ~Ap /Ap
3
+ ~AWIApJ
= 100 - Sp 1p
1 3
+ ~p 1p
2 3
+ oAP 1Ap + ~~w/~p = a + Sp 1p + oP 1p 2 3 3 1 3 2 3
+ ~w/p .
3
+ ~w1p
3
.
•
(c) By Proposition 3.G.2 and 3.G.3, the Slutsky substitution matrix is symmetric.
Thus •
f3/p3 + (~/p3)(a. + flp1/p3 +
= (3/p3
+ (~/pJ)(lOO
7Pz1 P3
+ ~w/p3)
- Spl/pJ + ~p2/p3 + ~w/p3) . •
Hence by putting p
3
= 1 and rearranging terms, we obtain +
Since this equality must hold for all pl' p
f3 +
o:o
= f3 +
2
100~. 13~
Hence a == 100, f3 = - 5, and 71 = - 5.
2 ~ w.
and w, we have &
-
sa. ,.-a
= {3a.
Therefore,
•
Recall also that all diagonal entries of the Slutsky matrix must be nonoositive. We shall now derive from this fact that • the first diagonal element is equal to • .. • - 5 + ~(100 - Sp
I r~ o~
ilt
o, . . ·n_....._. o~
2
>
o
a!;,ove value is positive.
(d) Si::1.ce x
(x
1
= x
2
~
= 0.
Let p_ = l, then
•
1
an d h ence we can a 1ways We must thus have
o
r·m d
= 0.
(p ,p . w l su:::n · ·h • a~• t'ne 1 2 In conclusion,
for a!l prices, the consumer's indifference curves in the
,x )-plane must be L-shaped ones, with kinks on the diagonal.
1
.J
2
restricted preferen.ce is the Leontief one.) following fig•.:.re.
3-39
(So the
They are depicted in the
••
...
••
••
••
••
•• • ••
••
••
• • •
••
•
••
,•
. · -------• •• •
:..·
••
•• •• •
••
••
• .~---------------
••
••
•• ••
•• • •
••
0
XI
Figure 3.G.6(d)
(e) By (d}, for a fixed x , the preference for goods 1· and 2 can be 3 Moreover, there is no wealth effect on the demands •
for- goods 1 and 2.
We must thus have
or a monotone transformation of this.
3.G.7
By the first-order condition of the UMP, there exists A > 0 such that
hg(x) = 'i7u(x).
P::-emultiply both sides by x, then .Ax·g(x) = x·'i7u(x).
Walras' law, x·g{x) = 1 and he'nce A = x·Vu(x). -1
By
Thus 1
g(x) = A 'Qu(x) = ·x·Vulx) Vu(x) By Exercise 3.0.8, we have 8v(p,l)/8w = - p·V v(p,l). By Proposition p 1 1 3.G.4, x(p) = - av(p,l)/dw Vpv(p,l) = p·ti v{p,"l) Vpv(p,l}. p
•
3.G.8
Differentiate the equality v(p,a:w) = v(p, w) + lna with respect to a and
evaluate at a: = l, then we obtain (Bv(p,w)/8w)w = 1. By Proposition 3.G.4, x(p,ll = - 'i7 v(p,l). p
3-40
Hence 8v(p,ll/8w = 1.
3.G.9
Let p
•
» 0 and w > 0.
All the functions and derivatives below are
evaluated at (p, w).
By differ-entiating Roy's identity with respect to pk, we obtain
2
2
[8 v/8ptapk)(av/8w) -
(8vl8pl)(8 vlcplaw)
(a v/8w)
•
2
Or, in the matrix notation,
1
2
- - - - :::::- (V vD v - V vD V Dx=p 2
w
v)
(V
w
p
p w
p
v)
e IR
LxL
.
L
(Recall that V v is a colmnn vector of IR , and D v and D V v are row vectors p
p
of IRL (Section M.A).}
p w
By differentiating Roy's identity with respect to w, we
obtain
(8 ------------------2~----------.
(Bv/8w)
Or, in matrix notation, 1
- - -...:.~::::- (V Dw X y)2
('17
L
e R .
vV V v w p w
w
Hence
s
=-
(v
w
- •
2
1
1
•
(v
w
2 \/}
cv w vD pv
2 (v vD v w p 2 v}
- V vD V
p
-
•
-
p w •
y
-
2
1
vw v
(V vV V v)( D v) + p 'ii v w p w w
vw2 ·. vw v
(V vD 'i7 v + V V vD v) + p p w p w p
~
p
V v( p
1
v w \!
••iJ v). •D
It is noteworthy that we can know directly from Roy's identity and this eq•Jality that the Slutsky matrix S has all the properties stated in Proposition 3. G. 2.
To see this, note first that both V v(p, w) and V v(p, w}
c•
are homogeneous of degree - 1 (in (p,w)) by Theorem M.B.l.
w
Hence x(p,w) is
homogeneous of degree 0 (where we are regarding Roy's identity as defining x(o,w) from 'i7 v(o,w} and V v(p,w)). . p . w Proposition 2.E.l.
Thus (2.E.2) follows, as proved in
On the other hand, by Exercise 3.0.8, p·x(p,w) = w.
3-41
D v p
Hence, as proved in Propositions 2.£.2 and 2.£.3, this implies (2.E.5) and (2.E.7).
Now, as proved in Exercise 2.F.7. (2.E.2), (2.E.Sl, and (2.E.7l
together imply that S(p,w)p
=0
and p·S(p,w)
= 0.
2 Tne mati"ix S(p, w) is symmetric because D v. 'V vD 'V v + 1/ 'V vD v, and p
'V vD v are symmetric. p p
following way.
p
p w
p w
p
The negative semidefiniteness can be shown in the
Since v( ·) is quasiconvex, for every price-wealth pair (q,b), •
2
if D vq + 'V vb = 0, then (q,b)·D v (q,b) p w
= 0 if and onlv• if b = - Dp vq/'V wv.
2
2
q • D vq + 2b(D 'V vq) + V v b p pw w
2:
0 (Theorem M.C.4).
But D vq + 'V · vb p
w
2 Plug this into (q,b) · D v(q,b). then
2 2
2 = q· D vq p
-
1
'V
w
\1
20 'V vq (D vq) --:P::--_w__ D vq + V2 v -..:.P__·-::;-1/wv p w ('V v)2 w
2 (('V v)(q · D vq) - 2(D V vq)(D vq) + w p p w p
2 (D vq) p 2 ). ( 'V v) p
Hence the quasiconvexity of v( • ) implies that the above expression is nonnegative for every q.
On the other hand,
CD vql
1 2 q·S(p,w}q = - ----::::-({1/ v)(q·D vq)- 2£0 V vq)(D vq) ('V v) 2 w p p w p w
Thus q · S{p, w)q ::= 0.
3.G.l0
2 2 ).
.p
( v'IN \1)
There1=ore· S(p, w) is negative semidefinite.
We shall prove that a(p) is a constant function and b(p) is
homogeneous of degree - l, quasiconvex, and satisfies b(p)
:!::
0 a.11.d Vb(p)
~
0
for every p >> 0. We shall fi:-st pro\"e that a(p) must be homogeneous of degree zero. v(p, w) is homogeneity of degree zero, we must have a(ll.p) b(p)w fo:- all p » ~
0, w
2:
0, and ll. > 0.
+
Since
b(ll.p)hw = a(p) +
If there are p and i\ for which a(hp}
a(p), then this constitutes an immediate contradiction with w = 0.
3-42
Thus
a(p) is homogeneous of degree zero. Next, we show by contradiction that 'i7a(p)
0 for every p » 0.
~
Sir.ce
v(p, w) is nonincreasing in p, we must have 'i7a(p) + 'i7b(p)w ~ 0 for every p
and w
~
0.
If there are p
» 0 and t for which Ba(p)/8pl > 0, then this
constitutes an immediate contradiction with w = 0.
»
» 0
Thus a(pl == 0 for every p
0.
We shall now prove that the only function a(p) that is homogeneous of degree zero and satisfies Va(p) :s 0 for every p
»
0 is a constant function.
In fact, by differentiating both sides of a(i\p) = a(p) with respect to i\ and evaluating the:n at h = 1, we obtain Va(p)p = 0.
Since p » 0 and va(p) :s 0,
•
we must have Va(p) = 0.
Hence a(p)
be constant.
Given this result, the homogeneity of v(p, w) implies that b(p) is homogeneous of degree - 1. function of p.
Since v(p,w) is quasiconveX, so is a(p) as a
Finally, since 'i7pv(p,w)
= 'i7b(p)w
must have b(p} :: 0 and 'i7b(p) ::= 0 for every p
:s 0 and 'i7wv(p,w) = b(p), we
» 0.
This result implies that, up to a constant, v(p, w) = b(p)w and hence, if • the unde:-lying ctility function is quasiconcave, then it must be homogeneous
-
of degree one.
.
On the other hand, according to Exercise 3.D.4(b), if the
underlying utility function is quasilinear with respect to good l. then, for all w and p »
0 with p = 1, v(p,w) can be written in the form 41Cp •... ,pL) + 1 2
w.
You will thus wonder why we have ended up excluding this quasilinear case.
Th~
reason is that, when we derived v(p,w) = (/l(pl + w, we assumed that the
. . ( consumouon set 1s .
then Bv(p, w)/Bp
1
Ill,
) IRL-1 Ill x + .
Thus x (p,w) can be negative and, if so, 1
is positive, which we excluded at the beginning of our
analysis, following Proposition 3.0.3.
(Note that, when we established
Bv(p, w)/Bpe ::= 0 in Proposition 3.0.3, we assumed that the ccnsu:::ption set is
3-43
L
As x 1(p, w) is positive for sufficiently large w > 0, the c;uasil!nea:-
1R +.)
case could be accommodated in our analysis if we assume that the ineaualitv . ~
8v(p, w)/Bpl ~ 0 applies only for sufficiently large w > 0.
(lr:
this case, we
can show that a(p) is homogeneous of degree zero, and b(p) is homogeneous of degree - 1, quasiconvex, and satisfies b(p) ~ 0. and V'b(p) :s 0 for every p » 0.)
3.G.ll
~y
Suppose that v(p,w) = a(p) + b(p)w.
Roy's identity,
x(p, w) = - (1/b(p))V' a(p} - (w/b(p))V' b(p). p
p
Thus the wealth expansion path is linear in the direction of 'iJ b!p) and p intercept (- 1/b(p))V a(p). p
•
3.G.l2
Note first that, according to Exercise 3.G.ll, the wealth expansion
path is linear in the direction of V' b(p) and intercept (- 1/b(p))V' a(p). p
p
If the uncedying preference is homothetic, then (- 1/b(p))V'pa(p) = 0. Hence a(p) must be a constant function.
If the underlying utility function is
homogeneous of degree one in w, then v(p,w) must be homogeneous of degree one ~
in w by Exe:-cise 3.0.3(a). -Hence atpl = 0 for every p »
0.
If the pre!'e:-ence is quasilinear in good l, then please first go back to . the proviso given at the end of the answer to Exercise 3.G.IO.
After doing
so, note that, since the demand for goods 2, ... ,L do not depend on w,
c-
Ilb(pllV'Pb(p) =
c-
l/p ,o, ... ,o>. 1
or (8b(p)/op, l/b(pl = 1/p and 8b(p)/Bpl = 0 for eve:-y l > 1. 1 {3p~ +
r
for some 13
:;e
0, p :~: 0, and 7 e IR.
But, by Exercise 3.G.l0, b(p) must
be homogeneous of degree - 1, positive, and nonincreasing. 0, and (3 > 0.
Tr.a: is, b(p) = f3/p
1
with {3 > 0.
3-44
Hence b(p} =
Hence p
=-
1, 1· =
If the underlying utility
function is in the qnasilinear fonn x
+
1
-ucx
.
2
, ... ,xL), then, by Exercise
J.D. 4(b), v(p, w) must be written in the form (j)Cp , ... ,pL) + w for all p 2
with p
.3.G.l3
1
= 1.
»
0
= 1•
Thus (3
For each i e {0,1, ... ,n}, let a. be a differentiable function defined 1
on the strictly positi've ort~ant {p be an indirect utility flDlction.
» 0}. Let v(p, w)
IR : p
E
= ~ =Oai (p)w
. 1
Denoting the corresponding Walrasian demand
By Roy's identity,
function by x(p.w}. 1
V
) V v(p,w) wv p,w p
x(p,w) = -
(
J
=
L
ril
.
.
i
•
Li=liai{pJw
1-
1
I::t=ow~a.(p) 1
=
r:;=O ___ w_·--:. l . J-
...,....1
va.(p). 1
Hence, for any fixed p, the wealth expansion path is contained the linear
As for the interpretation, recall from Exercise 3.G.ll that, an indirect uti!ity function in the Gorman form exhibits linear wealth expansion curves. But the Gorman form is a polynomial of degree one on w and a linear wealth expansion curve is contained' in a· linear subspace of dimension two.
Hence
the above result implies that the indirect utility functions that are polyr:omials on w is a natural extension of the Gorman form.
3.G.14
Define
a.
b,
c. d. e, and
f so that
a
b
c
- 4
d
3
e
f
- 10
•
Since the substitution matrix is symmetric, we know b = 3, a = c, and e == d. Thus c
By Propositions 2.F.3, p·S(p, w) = 0.
= - 4 and a = c = - 4.
= 0.
For the
3-45
Hence e
s:::
= 0. 3
p f
2 and d = e = 2: Thus f = - 7/6.
Finally, for the third column, we have p 3 + p 2 + 1 2 Hence we have - 10
- 4
3
-
4
- 4
2
3
2
•
- 7/6
The matrix has all the properties of a substitution matrix, which are •
synuuetry, negative semidefiniteness, S[p, w)p
= 0,
•
and p • S(p, w)
= 0.
(For
•
negative semidefiniteness, apply the determinant test of Exercise 2.F.l0 and Theorem M.D.4Uii).}
•
2
2
•
•
To verify Roy's identit:'. 8v(p ,p ,w)/op = (w/p 1 2 1 1 A
•
+
-112 2 4w/p ) (- w/p ), 2 1
ov(pl,p2,w)/8w = (w/pl + 4w/p2)
3.G.l6
,.,.....e ... ,_
-1/2
(1/pl + 4/p2).
(a) I! is easv to check that •
Since e(p,u) is nondecreasing in p, this must be nonnegative for all (p,u). But, if ak < 0 and llpll is sufficiently small, then this becomes negative. Also, if
~k
< 0 and llpll is sufficiently big, then this becomes negative. 3-46
Tnerefore (I)
ak
~
0 and ak
0 for all k.
:!::
It is a little bit manipulation to show that
•
•
Since e(p,u) is homogeneous of degree one with respect to p, they must be equal for every (p,u) and A > 0.
Take, for example, p = (1, ..• ,1) and u = 1.
Then
1]3,_ log e(p,u) = CL(llllogi\ + A. log A.e(p, u) = logi\
1.
+
They must be equal for every i\ > 0.
(2)
Lf'Xt
,
Therefore
= 1,
Lf3t
=
o.
T:"'US
Hen~e
(3)
the expenditure function now takes the simplified form:
.
at
e(p,u) =·(expu)(Uipt
);
Lt_«t
= I,
o..t
~
0.
This is ir:.c:-easing with respect to u and concave in p.
(b) By equatior. (3.£.1), w =
Hence
(4)
(c) By differentiating e(p,u) with respect to p, we obtain
Since x(p,w) = h(p,v(p,w)} and e(p,v(p,w)) = w, (5)
x(p,w) = w(«/Pl·····~/pL).
Use equations (3), (4}, and (5) and follows the same method as in the answer
3-47
to Exercise 3.G.2 to verify Roy's identity and the Slutskf equation . •
hand side of the indirect utility function should be deleted.
That is, it
shoulo be v(p,w) = Also, in (b), the minus sign in front of the first term d the right-hand side of the expenditure function should be deleted.
That is, it should be
e(p,u) = p uexp(bp/p ) - (1/bHap + ap{b + p cl. 2 2 1 2 Finally. in (c), the minus sign in front of the first tel m af the right-hand •
side of the Hicksian function should be deleted.
That is. it should be
(a) Use
anc apply Roy's formula. •
•
•
(b) According to (3.E.l}, we· can obtain the expenditure fmctio:1 by solving •
(c)
A~ly
Pr-oposition 3.G.l to obtain the given Hicksian dm:land function for
the fit st good. •
3.G.l8 We p::--cve the assertion by contradiction.
Suppo2 that the!'e exist l e
{!, ... ,L} and k e {1, ... ,L} such that there is no chain of Slbstitutes
connecting t and k.
Define J = (f} u {j e {1, ... ,U: there is a chain of
substitutes connecting l and j}.
Since t e J and k IC J, mth J and its
•
•
3-48
Moreover, for any j e J ar:c any j'
complemst {l, ... ,U\J are nonempty. 8h j( p, u )lip
r
u,
2::
X
e Vt}
ii!::
u.
Define u• = Sup {t: x e Vt}.
u.
such that x e Vt.
u
e(p, u ) for all p. n
e(p,u) for all p.
3.H.2
Let n ..
If u• = u, then,. for
e (u - 1/n, u) such that x e V
u
then u
a~,
n
..
11
If u•
Since e( ·) is
increasing in utility levels, Vu .::> Vt and hence x e V . n
vu if
u.
2::
e Vt then Sup {t:
every n e IN, there exists u
That is, x e
, that is, p·x =:::
n
and, by continuity of e(p,u}, p • x
2::
Thus x e V . u
We show the contrapositive of the assertion.
If a preference is not
convex, then there exists at least one nooconvex upper contour set. be its corres;:>ollliing utility level.
Let u e
We can choose a price vector p so that
h(p,u) consists of more than one elements, as the following figure shows. According to Proposition 3.F.l, e( ·) is not differential at (p,u}.
•
3-50
[R
0
XI
Figure 3.H.2
3.H.3
By (3.E.l), for each p, take the inverse of e(p,u) with respect to .u.
3.H.4
The following method is analogous to that of "Recovering the
Expenditure Function from Demand" for L = 2. 0 0 b' . d . . . P lCK an a: ltrar-y consumpuon vector x an ass1gn a utilitv• value u to X
0
.
0
x .
. h . We will now recover the ,indifference -curve {x: u(x) = u 0} gomg t• :rougn
-
•
Assuming s-.:rong mono tonicity, this is equivalent to finding a function for
eve~y
x1 > 0.
•
Differ-entiate
8u(x
1
.~Cx
1
))/ax
1
+
(8u(x
1
,~£x
1
)}/Bx
2
l~'(x
1
l
= 0.
Hence
= -
eu(xl.~( xl) )/ox 1 Since
-
Bu(x .~£x ) }/8x 2 1 1 gl(xl.~(xl)) _
_
, we have
gz(xl.~(xl}}
3-51
•
gl(xl.~(xl})
-
~'(x l
1
= -
gzcxl.~(xl>l
•
or, by replacing
-x
by x , ·we obtain 1 1 ~·ex
gl(xl.~(xl))
) = 1
-
g2(xl.~(xl))
.
By solving this differential equation, we obtain the indifference curve going 0
through x .
By (3.E.l), we can recover the expenditure function by simply inverting
3.H.S
the indirect utility function. To recover the direct utility function, define u: IRL +
~
IR by u(x)
=
Min{v(p, w): p · x :s w}.
We shall prove that u(x) is the direct utility function
that generates v(p, w).
So let x•(p, w) be the demand function and v•(p, w) be It is sufficient to show
the indirect utility function generated by u(x). that v•(p, w) = v(p, w) for all p Let p » v(p, w).
0 and
w
:2::
»
:2::
It thus remains to show that v•(p, w)
Then x e
IR~ ~
0.
0, then p·x•(p,w) = w and hence v•(p,w) = u(x•(p,w)) ::
x = -
V
1 (
wv p,w
:2::
v(p, w).
Define
} V v(p, w).
p
the
by the monotonici ty.
homogeneity, p· x = w. let p'
0 and w
» 0 and
w~
It • is thus sufficient to show that u(x1 ::::: v(p,w). •
:: 0 satisfy p' ·x = w'.
Then (p' - p) · x = w' - w, or, by
the· definition of x. V v(p,w)·{p' - p} + V v(p,w)(w' - w) = 0. p
w
quasiconvexity of v(p,w}, v(p',w') :: v(p,w).
3.H.6
So
Hence, by the
Thus u(x) :: v(p,w).
Let's first prove that the symmetry condition on S(p,w) is satisfied.
By equation (3.H.2}, c3e(p,u)/8pl =
o:.~(p,u)/pt
for every l.
By
differ-entiating both sides with respect to pk' we obtain
On the ether- hand, by differentiating b~th sides of Be(p, u}/c3pk = '\e(p, u)/pk 3-52
•
with respect to pl' we obtain
Hence the symmetry condition is satisfied. We can thus apply the iterative method explained in the small-type discussion at the end of Section 3.H to derive the expenditure function. First, we shall prove by induction that, for every 1., there exists a function •
f t(pl+l' ... ,pL' u) such that lne(p,u) = Suppose first that l = 1.
Since (8e(p,u)/8p )/e(p,u) = «{p . 1 1
Hence, by
integrating both sides with respect to p , we obtain 1
•
Suppose next that l > 1 and the
Thus the equality is verified for I. = 1. equality holds for t - 1.
with
respec~
By differentiating both sides of
tc pl. we obtain ae(p,u)/Bpl = Bfl.-l(Pt····•PL'u)/8pt.
Since ce(p,u)/Bpt =
alpt1 thjs is _equivalent to
Hence, by integrating both sides with respect to Pt.• we know that there exists fl(pi+l' ... ,pL,ul such that o:llnpl = ft-l(pl'" .. ,pL,u) - ft.'Pt+!' ... ,pL,ul By plugging this into
we obtain lne(p,ul = Lk~t'1clnpk + ft(pl+I'"'"'pL,u). If t = L, ther1 this equality becomes lne(p,u) = Lk:sLa:klnpk + fL(ul. •
3-53
Or,
equivalently,· e(p,u) = In what follows, for every increasing function /L(u), we shall find the utility function that generates the expenditure function e(p, u) = To sta.-t, consider the utility function u•(xl = which appeared in Example 3.E.1 for the case of L = 2.
Denote its expenditure
function by e•(p,u), then
(We considered a similar expenditure function in Exercise 3. G.16.
Note that,
these similarities incidentally show that, for every increasing function
proper-ties of expenditure functions in Proposition 3.E.2, because it
a.t
corresponds one of the monotone transfonuations of u•(x) = ntxt
.)
Let g(u•)
be an monotone transformation and denote by e (p,u) the expenditure function
g
of the utility function (gou•)(x).
Then
-a.,_
-1
e g{p,u) :: e•(p,g (u)) = mflt
)(IIlpl
cxl
-1 )g
(ul.
a. =
By comparing u(x) if
This equality is equivalent to g
-1
. u • -- g-l(ul L ettmg
(u)
and solving this with respect to u•, we obtain
Thus -1
-1
u(x) = g(u•(x)l = /L (lnu•(x) - Lfltlna.l) = fL ([lcxl(lnxl- lna:l)).
a. to
Of = u •
3-54
3.H.7
(a) Let
-p = (1, ... ,1).
Since
x(p,L)
=
(1, ... ,1) and u(l, ... ,l)
according to Propositions 3.E.l and 3.G.l, D
p
e(p,l)
=
h(p,ll
= 1,
= (1, ... ,1).
Hence e(p,l) = L.
2
On the other hand, S(q, w) = D e(q,w) = 0 for every q p
2.F.l7(d).
»
0 by Exercise
•
Hence e(p,u) - e(q,u) = D e(q,u)(p - ql. p
for every q
»
0 and p
»
0.
Now take q =
-p,
then e(p,l) - L =
-p· (p
-
-p).
Thus e(p,ll = •
(b) The uppe:- contour set is equal to
•
•
L {x e IR : p·x :=: +
X ~ (1, ... ,1)}.
By the same method as deriving equation (3.1.3), we obtain
3.1.1
.01 01 11 £1i(p ,p ,w) = e[p ,u) - e(p ,u) 01
= e(p ,u l -
100
01
p1 0 0 0" 1 . . = J 0 h(pl'p2,p3, ... ,pL,u ) dpl +
p.!
•
100
I
01
11
e(p ,u l
p2 1 0 0 1 0 h(pl,p2,p3, ... ,pL,u l dp2; p2
.01 00 10 Cv(p ,p ,w) = e!p ,u ) - e(p •. u )
= e(p =
J
00
1 pl 0
Pr
100 00 100 00 10 ,u ) - e(p ,p ,p , ... ,pL,u ) + e(pl'p ,p , ... ,pL,u } - e{p ,u l 1 2 3 2 3 l 0 0 0 0 p2 1 0 0 0 h(pl'p2,p3, ... ,pL,u ) dpl + I 0 h(pl'p2,p3, ... ,pL,u ) dp2. P2
If there is no wealth effect for either good, then, by the first reiatior. of
(3.E.4l,
3-55
3.1.2
Denote the deadweight loss given in equation (3.1.5) by DW ( t) and that 1
in equation (3.1.6) by DW (t}. Then 0 1 0 1 0 DWi(t) = h[p + t, p_ , u ) - (h(p + t, p_l' u } + 1 1 1 0 I = - t·ahCp + t, p_ • u >!apr 1 1
+ t, •
•
> 0 for every p > 0,
Thus
1
>
0 for ever-y t
0.
It can be similarly shown that DW (0) = 0 and, if
0
0 Sh(pl'p _ ,u )/ap > 0 for every pi > 0, then DW £t) > 0 for ever-y t > 0. 1
0
1
A possible interpretation of this result is that the first-order •
derivatives of the deadweight loss at t = 0 may be a bit misleading approximation.
In fact, their being zero means that, approximately, there is
no deadweight loss.
Henc~
3.1.3 p
0
:;c
On the other hand, since those derivatives are positive
the deadweight losses are in fact positive.
Wr-ite u 1 p .
0
0
= v(p , w) and u
1
1
= v(p ,w), then u
0
< u
1
because p
0
1
c: p and
Thus
for every Pt. > 0.
Since good
t
is inferior.
By the first relation of (3.£.4). this is equivalent to 0
0
0
1
hiPt•P -l'u ) > ht'Pt•P -l'u ). 0 1 0 1 0 1 Hence, by (3.1.3), (3.1.4), and Pt < pl. we have CV(p ,p ,w) > EV(p ,p ,w).
3.1.4
We
shai~
give two examples, both of which have two commodities. •
3-56
The
first one is simpler, while ·the second one is more illustrative. In the first example, we consider a preference with "L-shaped"
indifference curves such that the vectors of indifference curves.
= 1.
Let u(l,l)
(1,1), (4,2), and (5,3) are kinks
Note that if one of the two prices
is equal to zero, then the demand is not a singleton.
We thus need to
•
consider a dema."ld correspondence x( p, w}. ·But this does not essentially change· •
•
our argument because we are working on expenditure functions, which is single-valued by its definition. Let p
0
1
= (1,1), p = U/2,0), p
1 (1,1), x(p ,w)
i!)
2 (4,2), x(p ,w)
2
112 and e(p ,ll = 2/3.
2
= (0,2/3), and w = 2.
2 (5,3), and v(p ,w)
>
0
Then x[p ,w)
1 v(p ,w).
i!)
1
But e(p ,1) =
•
Thus
•
0
1
0
2
CV(p ,p ,w) = 2 - 1/2 = 3/2, CV(p ,p ,w) = 2 - 2/3 = 4/3.
0
1
0
2
Hence CV(p ,p , w) > CV[p ,p , w). It is wo::-thwhi!e to remark that, although the given preference is neither
smooth, strongly monotone, nor strictly convex, it can be approximated by such one. In the se:ond example, we consider a utility function c{x' ·.vnich is •
quasilinear with respect to the first conunodity. corresponding indirect utility function.
Let v(p, w) b!! the
Starting from p
0
= (l,l) and w > 0,
. 1 1 2 2 we consider two other pnce vectors p = (pl"l) and p = (1,p ) such that 0 < 2 1 2 . 0 0 1 1 2 pl < 1. 0 < P2 < 1. and v(p ,w) = v(p ,w). Wr1te u = v(p ,w) and u = 1
2
v{p , w) = v(p ,w).
0
1
0
2
Then EV(p ,p ,w) = EV(p ,p ,w). •
0
1
0
2
We shall now show .. hat CV(p ,p , w) < CV(p ,p , w).
By p
1
1 < 1, CV(p ,p ,w)
1 0 2 0 2 Also, by the quasilinearity, CV(p ,p ,w} = EV(p ,p ,w)
(Exercise 3.!.5).
0
1
0
2
Hence CV(p ,p ,w) < CV(p ,p ,w).
3-57
0
0
1 0 2 It is worthwhile to remark that, although £V(p ,p ,w} = £rtp ,p ,w}, we 0
1
0 2 can obtain the strict reverse inequality EV(p ,p ,w) > £V(p ,p ,w). while
. 01 02 . 2 preservmg CV(p ,p ,w) < CV(p ,p ,wl, by decreasmg p only slightly.
2
3.1.5
According to Exercise 3.E. 7, we can write the expenditure function
-eCp
e(p,u) =
2
0
, ... ,pL) + u for p
1
1
= L
Hence
0
1 0 0 EV(p ,p ,w) = e(p ,u ) - e(p ,u )
- 0 0 1 - 0 0 0 = (e(p ..... pL) + u l - (eCp , ...• pL) + u ) 2 2 = u 0
1
0
- u .
1
1 1 1 0 CV(p ,p ,w) = e(p ,u } - e(p ,u )
- 1 1 1 = (e(p2, ... ,pL) + u ) -
= u
3.1.6
Let
0
1.!.
1
=
I
0
- u .
0
v.(p ,w.). 1
1
If
1 0 So define w: = e.(p ,u. ), then 1
3. I. 7
1
1
.CV. p ,p ,w.) - 0, then
11
[.w~ 11
1
.w. -
ll
1
.e.(p ,u. ). 1
0
1
0
s 'l.w. and v.(p ,w:) = u. = v.(p ,w.). ~1
1
1
1
1
1
(a) By appiying Walras' law and the homogeneity of degr-ee ze:"'o, we can
obtain the demand functions for all three good defined over the whole domain {(p, w)
e !R
.-.,
x IR: p
»
0}.
Thus we can obtain the whole 3 x 3 Slutskv matrix
as well from the demand function.
-
The 2 x 2 submatrix of the Slutsky matr-ix
that is obtained by deleting the last row and the last column is equal to b
e
c g
•
•
By the homogeneity and Walras' law, the 3 x 3 Slutsky r.1atrix
is symmetric if and only if this 2 x 2 matrix is symmetric.
Moreover, just as
in the oroof of Theorem M.D.4(iii), we can show that the 3 x 3 Siutsky matr-ix • 3
is negative semidefinite (on T , and hence on the whole IR ) if and cnlv- if the p
3-58
2 x 2 matrix is negative semidefinite.
2 that c = e, b =" 0, g s 0, and bg - c
Hence, utility maximization implies 0.
j!:
(b) First, we verify that the corresponding Hicksian demand functions for the first two commodities are independent of utility levels and, as functions of the pr-ices of the first two commodities alone, they are equal to the given Let p be any price vector and u, u' be any two
Walrasian demand functions. utility levels.
By (3.E.4), hip,u) = xip,e(p,u)) and hl(p,u') =
xip,e(p,u'}) for l
= 1,2.
Since the
xe 0 and p > 2 2 1 1 1 2 2
Thus c = b , d = d = 0. 1 2 2 1
o.
Then the negative semidefiniteness implies that =:;
0.
•
(c) Since the Walrasian demand functions and Hicksian demand functions are the •
same as we saw m Exercise 3. I. 7(b), we can define p' p'
cv P1
Pz
or, equivalently,
cv
•
{d) By the same calculation as in Exercise 3.1. 7(c), we obtain
EV In this case. EV (b)
~
EV
1
= 1/2, EV
+ EV . 1 2
= 1/2, EV3 2
= 3/2.
In the general case in which the conditions in
hold,
EV EV
= 2 3
az
= a
1
+ b2. + (3/2)c2, + (3/2)b
1
+ c
1
+ a
2
+ 2b
Her.ce EV + EV = EV if and only if b = c = 0. 3 1 2 2 1
+(3/2lc . 2 2
This condition is
equivalent tc saying that any change in the price of one good does not have any (cross) effect on the demand for the other.
the other components are zero. after-re~ate
For each t, define p(t) = p + tet"
Then the
income w(t) with tax t satisfies w(t) = w + xt(p(t),w(t)lt.
3-62
Hence p·x(p(t),w(t))
= (p(t')
- tet)·x(p(t),w(t))
= w(t)
Therefore x(p(t),w(t)) is at most as good as x(p,w).
- xl(p(t),w(t))t = w. In order to prove that
x(p(t),w(tll is strictly less preferred to x(p,w), it is sufficient to prove that these two are different, because the demand function is assumed to be single-valued. •
Now SU;:Jpose that there exists at> 0 such that x(p(t),w(t)) = x(p,w) . •
Let u = v(p,w). hip,u).
Tnen we have h(p(t),u)
= h(p,u).
Since the Hicksian demand function s
~
In particular, ht(p(t),u) ht(p(s),u) is nonincreasing,
this equality implies that hip(s),u} = hl(p,u) for every s e [0, t]. d[hl(p(s),u)J/ds = 8hl(p(s),u)/8pl. = sU(p, wl "" 0.
h(p(t),w(tll
3.1.10
~
=
But
Evaluating at s = 0, we have Sht(p,u)/8pt
This violates the assumption that sU(p, wl < 0.
Hence
h(p,w) for every t > 0.
We consider an example of a consumer who face the choices over two
goods and whose preference t and demand function x(pl'p ,wl satisfy the 2 following condition: Fo:-- every p
e [1,2), x(p ,1,2) 1 1 •
for e·1rery p
e fl,2}, x(l;p ,2} 2 2
= {(1 - c)/p1• 1 + c); = (1 - c, (1 + cVr:,::i;
• •
((1 - cl/2, 1 + c) >- (1 - c, (1 + c}/2} .
By using the figure below, you can convince yourself, perhaps with some
application of the weak axiom, that there actually exists such a preference.
3-63
2
J+E
I--
1 •
•• •• •
1-
0
Define p
0
1
E
1
= (2,1) and p = 0,2).
2
XI
Then
0
x£p .21 = ((1 - c)/2, 1 + c), 1
x£p ,21 = (1 - c, (1 + c}/2).
0
1 Thus x(p ,2) >- x(p ,2).
price-change path p 0
0
However, the area variation measure following the
-+ (1,1) -+ p
1
is
1
AV(p ,p ,2) =
•
•
. = - f(l - clln
p
2 1
1 p1=1
+ [(1 + din p
2
2 J • pl=~
= 2£ln2 > 0.
0 . . nk 1 Hence the area var1at1on measure ra s p over p .
3.1.11
If (p
1
0
- p )·x
1
0
1
> 0, then w > p · x .
The local non-satiation implies
•
that x
0
1 is oref erred to x . -
1 Hence the consumer must be worse off at (-c , w}.
As for the interpretation in term of the first-order approximation, since e(p,u) is concave in p, , 0
elp ,u
1)
1 1 1 1 0 ::s e(p ,u) + Ve(p ,u )·(p •
. 3-64
1 1 0 0 1 1 1 1 Since Ve(p ,u ) · Cp - p ) < 0, e(p ,u > < e(p ,u ) = w.
0 0 Thus u = v(p ,wl >
1
u.
This test is depicted in the picture below:
p
X
0
XI
Figure 3.1.11
3.1.12
Let u
0
0
0
= v(p ,w ) and u
0011 EV(p ,w ;p ,w )
0
0
1 "1
1
1
1
= v(p ,w ). 01
Then we define 00
0
01
= e(p ,u ) - e(p ,u ) = e(p ,u ) - w • .
1 1 1 0 1 1 0 CV(p ,w ;p ,w ) = .e(p ,u ) - e(p ,u ) = w - e(p ,u }.
1
The "partial information" test can be extended as follows: If p · x I
1
the consumer is better off at (p • w ).
0
1
<w •
This can be proved in three ways.
The first one is the same revealed-preference argument as in the proof of Proposition 3.I.l. •
The second way is to use the indirect utility function.
Since v(p, w} is
. . quas1convex, li (p
1
0
- p ). v
p
00 1 0 00 v(p • w } + (w - w )8v(p •w )/ aw > 0,
1 1 0 0 then we can conclude that v(p ,w ) > v(p ,w ).
3-65
But, by Roy's identity, this
sufficient condition is equal to - (p
1
00 1 0 00 - p )·(av(p ,w )/8w)x(p ,w) + (w - w Hov(p ,w )/8w) 0-00
0
0
0
0
= (Bv(p .w )/8w)(- p = (8v(p ,w )/Bw)(w
1
1
·x
0 1
+ w
0
- p ·x )
0
+ w
1
0
- w )
> 0.
. 1 o 1 1 1 o o· Hence, 1f p ·x < w , then v(p ,w ) > v(p ,w ).
The third way is to use the expenditure function.
1 1 0 0 v(p , w ) > v(p , w ) if
•
. . 111 I 00 ana only 1f e(p ,v(p ,w )) > e(p ,v(p ,w )). 100 v(p ,w ).
[First printing errata: The difficulty level should probably be B.]
It
follows immediately from the definition that if x(p, w) satisfies the strong axiom, then it satisfied the weak axiom.
Conversely, if x( p, w ) satisfies
the weak axiom (in addition to the homogeneity of degree zer-o and Wa1ras' law), ther. the Slutsky matrix is negative semidefinite and, by Exercise 2. F .11, symmetric.
Hence x(p, w) is integrable, implying that there exists a
preference relation that generates x( p, w ).
Thus x{p, w) satisfies the strong
axiom as wel:.
3.AA.l
If (p,w)
= (1,1,1),
then x(p,w)
= (0,1).
The locally cheaper
condition is not satisfied since B p,w
1}
y such that p · y < w, as depicted in the following figure.
3-66
and there is no
'
'
0
XI
Figure 3.AA.l
To check that the demand function is not continuous at (1,1,1). consider the n
n sequence (p ,w ') = (1 - 1/n, 1, 1 - 1/n).
n
n
n
Then (p ,w ) -. (1,1,1) and x(p· ,w )
= (1,0), but x(l,l,l) = (0,1).
This discontinuous change in demands arises
because the budget set B
consists of (1,0) for every n, but B(l,l),l = n
n p ,w
+ x a~
suddenly
3.AA.2
2
= 1}, so that the commodity bundle (0,1) becomes available
p = {l,l}.
!Firs: printing errata: The upper hemicontinuity of h(p,u) cannot be
guaranteed a: p
l!:
0, because the local boundedness condition in t.he definition
of uppe:- hemicontinuity need not be satisfied.
Hence the clause in the
bracket "eve::-: i: we replace minimum by infimum and allow p unde:-stood as concerning only with e(p,u).] .lS
' .:.UDD-. • •
h em.-~., . . . .Jl•. n-vUS. ,.-nr.~,
"i!:
e B
S'...i.C!":
that
-u
i!:
u for eve:-y (p, u l e B.
0} x IRl.
u for every (p,u) e B. For each l, define
3-67
i!:
0" should be
We shall first pr-ove that h(p, u)
Let 8 be a compact subset of the
(which is, in tu:-n, a subset of {p e IRL: p »
(p, u)
n
Let
XE
do:r.~ain
of h(p, u)
Ther. the:-e exists a h(p,u), then u(x)
i!:
-u
-Yt and
-Y = (yl, - ... ,yL) -
h(p, u), we ha·;e
-y
e IRL.
x.
=
e IR+: (p,u) e B}
We now show that, for every (p, u) e B and x e
+
l!::
Max{p·x/pl
u(x)
If fact, since
p·x
u,
l!::
Since p » 0 and
p · x.
i!:
•
•
Divide both side by pi, then we ootam
p · x/pl
-y l
xt and hence
i!:
i!:
xi..
We have therefore established the lo:a.l n
n
Next, let {(p ,u )}
boundedness condition of upper hemicontinuity.
be a
n
sequence of pairs of price vectors and utility levels, converging to (p,u). Let {xn}
n n n • IRL n be a sequence 1n , x e h(p ,u ) for every n, and x ... x.
n
+
Since u(xn)
sufficient to prove that x e h(p, u}. continuity, we obtain u(x) at (p,u}.
i!:
u.
u(y} > u, then u(y) > u
0
E
+
n
Hence p ·y l!::
p · x.
u.
i!:
p
i!:
If
n
n
·x for
S•Jppose then that
By the local nonsatiation, there exists a sequence {yn} n in IRL such +
n
-+
x.
Hence there exists a subseouence •
n n)} nsuch tatuy h ( n {( ? k(n) ,u k(n)),.tnof {'\p,u 1 k(n~
•
IRL and u(y l
for any sufficiently large n.
that u(y l > u(x) for every n and y
·x
un and u(x l -+ u(x) by the
Hence x satisfies the constraint of the EMP
n
o•
n
By taking the limit as n ~ co, we obtain p · y
u(y) = u.
k!n)
i!:
To show that it is cost-minimizing, let y
such n.
It is
B".,' taking the limit as n •
•
•
•
~ co,
l!::
u
k(n) •
Hence p
we obtain p · y
l!::
k(nl
p· x .
·y
n
l!::
Hen:e x is
•
cost-mmmuzmg. We now turn to the continuity of e(p,u). every p »
In fact. its continuity at
0 can be derived immediately the continuity of h(p,u), as the
latter is weH defined at every p »
0.
Thus the essential part of the
following proof is the case of nonnegative, but not strictly vectors.
posi~ive,
price
We sl':all establish the continuity with respect to p and that with
respect to u separately. Let u e IR be a utility level, p e IRL be a price vector, and {pr:} be a +
seque~ce
L . t of p;ice vectors in IR convergmg o p. +
3-68
We need to prove that
n
e(p ,u) -+ e(p,u).
As a preliminary result, let's first prove that if the
sequence {e(pn,u)} in IR+ converges, then it must do so to e(p,u). n
Let x e IR
the limit of {e(p ,u)}. every n.
L +
and u(x)
~
n
u.
Then p · x
Taking the limit as n -+ co, we obtain p·x
every x e IRL with u(x) +
~
•
subsequence {p
:5
k(n)
.
L
w, we use the concavity of e(p,u) in p e IR+.
k(n)
n
} of {p } such that (Pt
k(m)
)(
~
- Pt Pt
at most 2L sign patterns. - pl
Pt
~
Take a 0 for all
t e
That is, for each t e {l, ... ,L}, we •
Such a subsequence does actually exist because each p
k(n)
e(p ,ul for
~ u, we have e(p,u) ~ w. To prove the reverse
{1, ... ,L} and positive integers n and m.
zero).
n
Since this holds for
w.
•
inequality e(p,u)
l!:'
Let w be
n
- p has one of
Now, for each t e {l, ... ,U, let vl = 1 if •
k(n)
0 for every n; and vt = - 1 1f Pt
- Pt => 0 for every n.
Then,
•
k(n}
Pt So let
;!:
k(n)
= Pt
+
Jpt
- Ptlvt' l = vl and v lc = 0
0
for every k ;:: t, then k(n)
P
Now, define
. k(n)
S mce p
+
Ltzlv.
n
1 -
l=lzt, then p
.
= p
n l
k(n)
.
=
n ... p, zl -+ 0 for every
t e
{l, ...• U.
n Thus z 0
1.
Hence, for
. · 1arge n, z n > 0 an d p k(n) 1's a convex com b"matton · 1 every su ff 1c1ent.y oi- z n , 0 0
z7, ... ,
z~.
The:-efore, by the concavity, k(n) ) e (p ,u ~ z~e(p.u)
...... w. ..- ce e(pn,U'J ........ ...., .... , e(pk(nl,u) ...., 5.;••
Th e r1g · ht - h an d s1'de converges t o e(p . u) .
Therefore w s e(p,u). We have thus proved our preliminary fact that if the sequence in !R
+
converges. then it must do so to e(p,u).
3-69
Let's now prove by
n {e(p ·,u)}
0
contradiction that this implies that e(p ,u) -+ e(p,u).
So suppose not, then
there are a ~ > 0 and a subsequence {pk(n)} of {pn} such that Je(pk(n),u) - e(p,u)J ~ Cl for all n.
Since the subsequence {e(p
convergent subsequence. because I e(p
k[n)
k(n)
. , u)} 1s bounded, it has a further,
On the one hand, the limit can never be e(p,u),
,u) - e(p,u) I
~
o for
all• n.
On the other hand, our
preliminary result implies that the limit must be e(p,u). contradiction.
This is a
n
We must thus have e(p ,u) -+ e(p,u).
Let's now turn to the continuity of e(p,u) with respect to u.
Let p
iR
E
L +
be a price vector, u e IR be a utility level, and {un} be a sequence of utility •
levels in IR converging to u.
.
n
We need to prove that e[p,u ) -+ e(p,u).
before, it is sufficient to prove that if the sequence {e(p, un)} in IR converges, then it must. do so to e(p.u). L
Let c > 0, x e IR , u(x) +
~
By the local nonsatiation,
we can make u(x) > u while preserving p·x < e(p,u)
pos!~ive intege:- N such that u(x) > u i!::
n
e(p,'J ).
Take the limit as n-+
-
•
Ill,
.
+
Let w be the limit of {e(p,un)}.
u, and p·x < e(p,u) + c.
0
Just as
c.
+
for every n > N.
Then there exists a Thus, for such n, p · x
Thus e(p,u) + c > w .
then p·x :: w.
Since this holds for every c > 0, we must have e{p,u) reverse inequality e(p,u) :s w, we can assume that u
n
~
w.
To show the
:s u for every n.
reason is as follows: If there is a subsequence such that u
0
(The
:s u for every n
in the subsequence, then we can apply this case to subsequen=e.
If there is
no such subsequence, then there is a subsequence such that un
u for e·;e::-v n •
in the subsequence.
. n Hence e(p,u} :: e(p,u } for such n. -
L IR and u(x) i!!: u. +
i!!:
Taking the limit, we
obtain e(p,ul :: w.)
Now let x e
then B is compa·:::t.
This and un :s u implies that the truncated EMP Min p·x
s.t u(x) :: u
3-70
Define B = {x e
n
-X
->
' X),
has a solution, denoted by xn e B.
Then xn e h(p,un), that is, xn a solution
to the original, untruncated EMP, because p is a conver-gent subsequence {xk(n)} of {x
0
e[p,u
) and p·x
ldn)
.. p·x.
IR + • Since B is compact, there
E
Denote its limit bv x. -
}.
u( ·) is continuous, u[x) a: u and hence p·x k[n)
L
Moreover, p·xk(n) =
e[p,u).
i2::
Since
Thus w = p·x and hence w
2::
e[p,u).
Suppose that u(x) is strictly quasiconcave, twice continuously differentiable and that Vu(x)
*
0 for all x.
Then we know that h(p,u) is a
function and the Lagrange multiplier i\ of the EMP must be positive.
The
first-order condition for the EMP can be considered as a system of L + 1 eouations and L •
+ 1
unknowns: p - i\Vu(x) = 0 u(x) - u
0
.o::: •
By the implicit function theorem [Theorem M.E.ll, the solution h(p,u) as a function of the parameters (p,u) of the system is differentiable if the Jacobian of this system has a nonzero determinant 2 - D u(x) Vu(x)
- p T
~
0
0
•
But, then, p = i\.'i7u(x} and hence
at (p,x) satisfying the above· two eqllations. this condition is equivalent to
2
- D u(x)
'i7u(x)
- Vu(x) :;1!:
T
0.
0
that is,
o
2
uCx)
Vu(x)T
Vu(x)
:;1!:
0.
0 2
By Theorem M.D.3Ci), this inequality holds if o u(x} is negative definite on I
{y e ~-= V'u(x) · y = 0}.
Tnis sufficient condition is a stronger differential
version of quasic-oncavity, as the latter is equivalent to the condition that
3-71
2 D uCx) is negative semidefinite on {y e IRL: Vu(x) · y = 0}.
3-72
CHAPTER 4
By Roy's identity (Proposition 3.G.4) and vi(p,wi) = ai(p) + b(p)wi"
4.B.l
x (p,w) = i
1
I
V
w.
( ) 'il v.(p,w.) v.p,w. p1 1 1
l
Thus IJ
= -
1
I
x.(p w.) = - -:--:-(-p~) 'il p b (p) for all i. 'IN, 1 1 0
Since the right-hand side is
1
1
identical fo:r every i, the set of consumers exhibit parallel, straight expansion paths. As for the second part, by (3.E.I), e.(p,u.) = (u. - a.(p))/b(p). 1
1
1
1
Hence, by letting c(p) = I/b(p) and d.(p) = - a.(p)/b(p), we obtain e.(n, u. l = 1 • 1 1
1
c(p)u. + d.(p). 1
4.B.2
1
(a) Let p e IR
L
be a price vector and w
Consider two consumers, i and j.
!::
Consider two wealth distributions
Cw , ... ,w } a."':d (wi, ... ,wjl such that wi = wj = w 1 1 and wk = 0 for any k for eve:-y k.
:;c
j.
0 be an aggregate wealth.
2:
..... 0, wk = 0 f or any k ~ 1,
Since the preferences are homothetic, x(p,O,s. ) = 0 K
Thus the aggregate demand with (wl' ... ,w ) is x(p,w,si) and the 1
agg:-egate demand with (wi, ... ,wjl is x(p,w,sj).
Since aggregate demand
depends only on prices and aggregate wealth, we have x(p,w,s.) = x(p,w,s .). I J Since p and w were arbitrarily chosen. this means that i and j have the same demand fun::tion.
Hence they have the same preference.
Since i and j were
arbit::-a!"'ilv c:.:csen, we conclude that all consumers have the same preference.
-
(b) By analogy to the Gorman form, consider the following form of indirect utilitv funct!ons:
-
4-1
v.(p-,w.,s.} = a.(p) + b(p)w. + c(p)s .. 1
1
1
1
1
Note that b(p) and c(p) do not depend on i.
1
By this and Roy's ider.tity
(Proposition 3.G.4), x(p,w.,s.) 1
1
=-
1
V
w.
1
-( ) V v.(p,w.) v.p,w. pl 1 1
1
w.
1 b(p) Vai(p} -
--
s. I
1
'"";"b-;(-p~) 'i7 c (p )
b( pl Vb(p} -
Thus
[.w. [.x(p,w.,s.) 1
I
1
t::
1
b( p) Vb(p} -
-
1
L·1s.1
b(p) Vc(p).
Thus the aggregate demand depends only on r.w. and r.s. (and pl. 1 1
1 1
By the definition of a directional partial derivative,
4.C.I
Dpx(p,w)dp =
limE;.~(I/c)(x[p +
cdp, w) - x(p,w)).
Hence cp·Dpx(p,w}dp = dp·(limc.-+O(l/E;)(x(p + cdp, w)- x(p,w)))
=
lim£~(1/c)dp·(x(p +
cdp, w) - x{p,w))
But the ULD property implies that dp· (x(p + cdp, w} - x(p, v:)) 0.
Hence, by taking the limit c .-+ 0, we obtain dp · D x(p, w)dp p
i!::
:S
0 for all c >
0.
Tnus
D x(p,wl is negative semidefinite. p We shall p:-ove the converse by contradiction.
Suppose that the Jacobian L
D x(p, w) is negative definite for all (p, w) and that there exist p e IR , p' e p
iRL, and w. e !R suc:h that x.(p,w.) -:t: x.(p',w.) and 1
1
1
1
1
(p' - p) · (x.(p' ,w.) - x.(p,w.)} 1
1
1
i!::
1
0.
Let ~ > 1 be sufficiently close to 1 that for every A e [O.~J. demand is well defined at (1 - A)p + Ap'. price vectors, [O.~].)
-A is
(If demand is well defined at strictly positive
determined so that (1 - i\.)p + ll.p'
Define _o(i\.) == (1 - A)p + i\.p' and
4-2
»
0 for every A e
w .(h) = (p' - p)·(x.(p(~),w.)- x.(p,w.)). 1
1
1
1
1
~
Then the function w( ·} is differentiable, w.(O) = 0, w.(l) 1
= [p'
w~(~) 1
- p)·D
1
x.(p(~).w.)(p' p 1 1
0, and
- p).
We consider two cases: Case 1: w.(ll.) ::s: 0 for every 1
~
e
[O,ll.].
Then w.(l) = 0 and it is a maximum. 1
Thus w:(l) = 0, that is, 1
(p' - p) · D x(p', w)(p' - p)
•
p
= 0.
This is a contradiction to the negative definiteness. Case 2: w.(h) > 0 for some ll. e 1
(O,ll.].
Then, by the mean-value theorem, there exists 11.• e w.(O) = w:(ll.•)(h - O). 1
!
(0,~)
such that w.(h) 1
By w.(O) = 0 and w.(ll.) > 0, w:(ll.•) > 0. 1
1
1
(p' - p) · D x(p(h •), w)(p' - p) p
This is a contradiction to the negative definiteness.
That is,
> 0. Our proof is thus
completed.
4.C.2
It D x.(p,a..w) is negative definite on the whole IR p 1 1
L
for every i, then
the sum [.D x.(p,cr..w) is negative definite on the whole IRL. 1 p 1 1 ~.D x.(p,o:.w), L.1 p l 1
Since D x(p, w) = p I
D p x(o,w) is negative definite on the whole R-. i:-:.:). lyin.e: that . . ~
x(p,wl satisfies the ULD property.
To establish the WA, one way is simply to
notice that the ULD property implies the WA, as the latter conside:--s only compensated pr-ice changes. Another- \..,·ay is to pr-ove the given differential sufficien! condition. Let's ass;,1me that w > 0.
Define H = {v e IRL: v·x(p,w) = 0}, that is, H is the
hyperplar.e with normal x(p, w} that goes through the origin. because p · x(p, w) = w > 0.
Then P E H
Thus, if v e IRL and v is not proportional to p, IR
L
such that v
4-3
1
e H, v
1
~
0, v
2
is
?roportional to p, and v = v
1
+
v . 2
Since S(p,w)v
= 0 and v ·S(p,w) = 0 by 2 2
Proposition 2..F.3, we have v·S(p,w}v = Cv + v J·S(p,w)(v + v l 2 1 2 1
= v ·S(p,w)v + v ·S(p,wlv + v •S(p,w)v 1
1
2
1
2
1
+
v ·S(p,w)v 2 2
= v ·S(p,wlv . 1
1
But here, by v e H, 1
T
S(p,wlv = (Dpx(p,w) + Dwx(p,w)x(p,w) lv = Dpx(p,w)vl" 1
1
Hence v ·S(p,w)v = v ·Dpx(p,w)v < 0 because v 1
definite.
4.C.3
1
1
1
1
~
•
0 and Dpx(p,w) is negat1ve
Thus the WA holds.
A Giffe;1 good will be a most familiar example.
In the figure below,
good 1 is a Giffen good.
•• ••• •• •• ••
•
••
'•
0
XI
This example shows that the ULD property is actually not derived from the utility maximization.
4.C.4
It is a restriction on preferences.
The L-shaped indifference curves imply that for every strictiy positive
4-4
price vector, the consumer's demand is always be at the corner of a upper contour se!.
Hence no compensated price change will change the demand.
= D
S.(p,w.) = 0 for every (p,w.) and D x.(p,w.) 1 1 1 p 1 I
that there exists (p,w.) such that D
w.
1
p·D
w.
x.(p,w.) 1
1
1
:r::
w.
1
1
1
l
1
1
1
1
p·£1/w.)x(p,w.) = 1, this implies that D 1
w.
Suppose
1
1
Since
x.(p,w.) and
1
Hence there exists a v e IR
1
v·D
1
x.(p,w.) ;e (1/w.)x.(p,w.l.
1
x.(p,w.) are not proportional. 1
W,
T x.(p,w.)x.(p,w.) .
Thus
2
1
1
such that
x.(p,w.) < 0 and v·x.(p,w.) > 0, as illustrated in the figure below:
1
1
1
1
1
DwOO(p, Wi)
p
0
XI
Figure 4.C.4
v Thus v·D x.(p,w.)v = - v·D p
1
1
T x.(p,w.)x.(p,w.) v = - (v·D
w. 1 1
1
1
1
w.
1
which implies that the ULD property is not satisfied. satisfied, the:l we must have D
w.
x.(p,w.))(v·x.(o,w.)) > 0, 1
1
1.
!
Thus, if it is in fact
x.(p,w.) = (1/w.)x.(p,w.) for every (c,w.).
1
1
1
1
1
1
.
1
Thus the unique \':ealth expansion path, which is the set of the corners of the upper contour set.s, is a ray going through the origin. is homothetic.
4-5
Hence the prefe:-ence
Following the hint, we fix w = 1 and write x.(p) = x.(p.1).
4.C.S
I
1
1
the indirect demand function g.(x) =
V ( ) Vu.(x).
x· u. x
1
1
1
Consider
Since x.(p) = x if and 1
only if g.(x) = p, the tJLD property of x.(p) is equivalent to the following 1
1
property: if x property,
i~
:;e
.
y, then (g.(x) - g.(y)) · (x - y) < 0. 1
1
For this latter
2 is sufficient to show that D g.(x) is negative definite. 1
We
shall now establish this. By the chain rule (Appendix M.A), •
2 2 2 Dg.(x) = (x·Vu.(x))- (Cx·Vu.(x))D u.(x) - Vu.(x)V'u.(x)T - V'u.(x}xTo u.(x)) 1
1
1
1
1
1
l
1
2
Let q = V'u.(x) and C = D u.(x}, then this can be rewritten as 1
1
2 Dg.(x) = (x·ql- ((x·q)C - qqT- qxTC). 1
We need to show that v · Dg.(x}v < 0 for every v 1
1 v·Dg.(x)v = (x·q( v·Cv < 0. 1
-:;e
0.
If v · q = 0, then ·
(This property is equivalent to the negative •
definiteness of the bordered Hessian of u.( ·) and used to guarantee the 1
differentiability of the demand function, as explained in the Appendix to Chapter 3}.
Sc suppose that v · q
*
0.
By multiplying a scalar to v if
necessary, we can assume _that v · Q. = x · q.
•
Then
v·Dg.(x)v = (x·q)-l(v·Cv- v·q- x·Cvl. 1
> 0, we need to show that v·Cv- v·q - x·Cv < 0.
By x·q
Sb::e C is svmmetric, •
v·Cv - x·Cv = (v - (l/2)x)·C(v - (l/2)x) - (l/4lx·Cx. Since u. ( · l is co;,.cave, C is negative semidefinite and the first term in the l
above expression is non-positive.
Thus,
v·Cv - x·Cv
~
- (l/4)x·Cx.
v·Cv - v·q - x·Cv
~
x . - (1/4)x·Cx - c. •
Hence
Since -
x·Cx q·x
< 4, the right-hand side is negative.
4-6
Hence so is the left-
hand side.
4.C.6
By differentiating both sides of u(>.x) = i\.u(x) with respect to i\. and
taking i\. = 1, we obtain 'i7u(x) · x = u(x).
Then by differentiating both sides of
2
o
this equality with respect to x, we obtain
•
u(x)x + 'i7u(x) = 'i7u(x). Tnus
2
D u£x)x = 0 and hence cr(x) = 0 . •
4.C.7
Suppose that the distribution of wealth has a differentiable, •
nonincreasing density function /( ·) over the interval ;~t
[0, w ).
Let v e IR
L
and v
0, then, just as in the proof of Proposition 4.C.4, we have
v·Dx(p)v = I
-w
-w
o(v·S(p,w)v)f(w)dw - I o(v·Dwx(p,w))(v·x(p,w))f(w)dw.
Here, the first term is negative, unless v is proportional to p.
(This
property is equivalent to the negative definiteness of the bordered Hessian of w.( ·) l
and used to guarantee the differentiability of the demand function, as
explained ir. the Appendix to Chapter 3).
As for the second term. just as in
the proof of P;-oposition 4.C.4,
-
-
-
2 f(w}dw .
.
By integration by parts and
-
-x(p,O) .= •
-w
2 (l/2)(v·x(p,w)) f(w) - (l/2)J o(v·x(p,w)} f'(w)cw The
firs~
2
part of this is always nonnegative, and it is positive when v is
proportional to p. f'(w) :::
-
0, this is equai to
The integral of the second part is nonpositive because
0.
0.
Hence v·Dx(p)v < 0.
To see that there are unimodal density functions for which the conclusions o: this propositions do not hold. recall that
-
v · Dx(o)v = iw(v · S(p w)v)f(w)dw 0 •
w
1
-w
- .Ir 0 (v·D w x(o,w))(v·x(o,wj)j(w)dw. . .
IRL t then To be scecifi::, let v = (l, 0, ... , 0) e • •
4-7
-
-
-+ x
Suppose a!so that the graph of the function w
(p, w) is as shown below . 1
•• •• •• •
•• •• •• •• •• •• •• • •
w
w•
0
Figure 4.C.7
Dwx Cp,w)
Then
1
(w•,w].
c:: 0 for every w e [Q,w•] and
Dwx (p,w) 1
-w
:s 0 for every w e
w• Tnus I Dwx (p,w}x (p,w)f(w)dw:: 0 and Jw.Dwx (p,w)x Cp,w)f(w)dw s 0. 0
1
1
1
1
If the density function f( • ) is nonincreasing, the weight placed on the interval [O,w•J is sufficient to dominate the negative effect of the interval
-
[w•.~·), implying that J~Dw;l(p,w); Cp,w)f(w)dw c:: 0. •
•
1
If the dist::-ibution
function is not noninc:reasing, then the weight on the interva: i ·;:",~·I may
-
dominate that on the interval It could even dominate the substitution effect, in which case we have v·Dx(p}v =
4.C.8
J
-\'4/ 0
(v·S(p,w)v)f(w)dw- J
-w 0
(v·Dwx(p,w)){v·x(p,w))f(w}dw > 0.
By substitu-.ing (4.C.6) into (4.C.S), we obtain
S(p,w} = }:.S.(p,o:.wl - t.D l
1
1
x.(p,o:.w)x.(p,o:.w) + Ct.o:.D
1 W. 1
1
1
1
1
T
x.(p,o:.wl)x(p,wl .
1 1 W. 1 1
1
On the other- hand, the right-hand side of (4.C. 7) equals to r.s.(p,c:.w) 1 1
1
[,D . X;(p,o:.w)x.(p,o:.w) •
w. 1
1
1
. 1
4-8
+
([.o:.D 1 1
w.
x.(p,a..w))x(p,w)
1
1
1
T
· T Dwx(p,w)rf'.x.(p,a.w})
+
'L11
1
- r.o x.Cp,cx..w)x.(p,a:.w) l w. 1 1 1 1
= f.S.(p,a.w) "'i 1 1
T xfp,w))x(p,w)
- ([.a:.)D
11W
(~.o:.D x.fp,a..w)lx(p,w)T L.l 1 w. 1 1 .
+
1
1
We h• thas proved (4.C.8).
4.C.9 The homotheticity implies that D hence\\ r{p.w) = I:.o:.D ..
1 1
w.
x.(p,o:.w) = (1/a.w).x.(p,a.w) and
W. 1 1
1
. I
x.(p,a:.wl = (1/w}x(p,w). 1
1
1
1
Thus
I
ap.w) = [.o:.((l/a.w)x.(p.o:.w) - (1/w)x(p,w))((l/o:.)x.(p,o:.w) - x(p, 11
1
1
I
11
1
w)?
T = [.(a./w)((l/a.)x.(p,o:.w) - x(p,w})((l/o:.)x.(p,«.w) - x(p,w)) . 11
111
111
L
Tnus, Car every v e IR ,
•
v·C(p,w)v = l:.Co:./w)((Vo:.)x.(p,a.w) - x(p,wll·v) 1
Thereftre.
4.C.l0
ctp. w)
~t's
1
1
1
1
2
2:
0
•
is positive semidefinite.
start by formulating our continuum-of-consttue!"S situation.
take t2 iutu·val [0,21 as the set of (the names of) the Clll51tll1ers. popul
density is equal to 1/2 uniformly on [0,2).
pro
We
Tne
We assume that the
. af each consumer's wealth to the average weallh (whic!-1 is, in this
situatilll, a counterpart '"of .the aggregate wealth) is const.-:. r--::.;::c.rdiess of the
af the aggregate wealth.
wealths
to
-w
d7) =
Since
We further assume 1hat. when the "average" •
> 0, the wealth of consumer 11 e [0.21 is equa" to
-w,
the term "average" is justified.
demand is then defined by x(p,w) =
Slutskymatri." 16, then it is positive semidefinite, and if 8 < w < 16, then it is negative semidefinite.
(This can be shown by applying Theorem M.D.2(iil,
or by noticing that C(p,w)p = 0 and p·C(p,w) = 0 and then applying the
4-11
For example, if w = 12 and v =
argument in the proof of Exercise 2. F. 9(b}.} (1,0), then v·C(p,w)v =- 1.
Thus C(p,w) is not positive semidefinite.
saw in (b) that the aggregate demand function satisfies the WA.
We
We can thus
conclude that in order for an aggregate demand function to satisfy the WA, it •
. is not necessary that the matrix C(p, w) is positive semidefinite . •
(d) Here is a figure depicting the wealth expansions paths of the two consumers for p = (1,1).
= x (1,1,8l = (4,4). 2
They intersect each other at (4,4) because x (1,1,8) 1
Note that if 8 < w < 16, the Engel curves resemble those
of figure 4.C.2(b); if w > 16, the Engel curves resemble those of figure 4.C.2(a).
consumer 2 •
4
0
consumer 1
4
x:
Figure 4.C.ll(d)
4.C.l2
As suggested in the hint, our example is a two-commodity, two-consumer-
economy in whlc.h the two consumers have the same preference and the wealth dist;:-ibutior:. rule is such that when the agg1·egate wealth is equal to 4, the wealth of consumer 1 is equal to 1 and the wealth of consumer 2 is equal to 3. The example here is essentially the same as Example 4.C.l (which is
4-12
illustrated in Figure 4.C.l).
Example 4.C.l is not directly applicable to the
present context, because the two consumers have the same wealth but the different demands.
However, if wealth levels are different between the two
consumers, then it is possible that they have the same preference and yet c;iemands similar to those in Figure 4.C.l, yielding a violation of the WA in the aggregate.
This is illustrated in the figure below: X2
6
4
3
2 1 0
1
4
3
2
6
8
X:
Figure 4.C.12
4.C.l3
We consider a two-commodity, two-consumer- economy with the given
wealth distribution rule w (p,w} = wp/(p + p l and w (p,w) = wp /(p + P l. 2 1 2 2 1 1 2 The preferences of the consumers are represented by the following utility
functions: u £x ) = Min <x ,2x
1
11
1
u £x ) = Min {2x
2
2
12
21
,x
}.
22
}.
So the prefer-ences are homothetic and have L-shaped indifference curves. unique wealth expansion paths are depicted in the figu:-e below:
4-13
The
consumer 2 4 .................... . •
••• •• •
consumer 1
•
••
•• .,••......
...... 2 ·----· ••
•• •
•
••
•
••• ••
•••
2
4
••
•
0
XI
Figure 4.C.13 Just as we saw in the answer to Exercise 3.D.S(c), the individual demand functions are xl(p,wl) = (2.wl/(2.pl + p2)' wl/(2.pl + p2)), x2(p,w2) = (w2/(pl + 2.p2)' 2.w2/(pl + 2.p2)). Tne aggregate demand function is given by
We claim that the aggregate demand function does not satist"y the WA. . . d ~t~·1ne see tn1s. 6 and w = 1.
A
.~..:
= { p e IR2 : p ++
Tnen w.(p,wl I
1
= p.1
+
Pz
To
= 1} and our restrict attention to p e
fJJr both i and hence
limp-+(l,O)x(p,l) = limp-+[l,O)x Cp,w Cp,l)) = (U/:.::;, 1 1 limp-+(O,l)x(p,l}
Thus, if p e 1!, q e 1!, and p and p · x(q,ll < l.
2
= limp-+(O,l)')C:2 (p,w 2 (p,lll = (1/2,1}. and q
1
are sufficiently small, then q · x(p,l) < 1
Hence the aggregate demand function does not satisfy the
WA. Proposition 4.C.l does not apply to this example because the wealth distribution rule of this example depends on prices, while the proposition cioes r.ot aliow this.
4-14
It is easy to check from the budget constraints that th~ distribution
4.0.1
(x (p, w (p, w}), ... ,x (p, w (p, w))) of commodity bundles satisfies the
1
1
1
1
constraints of the maximization problem in this exercise, and, from the definition of v[p, w).
th~
indirect utility functions, that it attains the value
It thus remains to show that if (x , ... ,xi) satisfies the constraints 1 •
of the maximization problem in this exercise, then WCu (x ), ... ,u (xi}) :s 1
v(p,w).
Fo:-- each i, define
W. 1
= p•X.,1
then r.w.
11
= [.p·X. 1 1
1
1
= p•([.X.) 11
:S W,
that
is, Cw , ... ,w l satisfies the constraint of the maximization problem of 1 1
Hence, by the definition of v(p,w).
(4.0.1).
u.(x.) On the other ha.-ld, by the definition of the indirect utility functions, . 1 1
:s v.(p,w.) for ever-y 1
1
•
1.
This and the increasingness of W( • ) imply that
W(u (x ), ... ,u £x )) :s W(v £p,w ), .... v (p,w )J. 1 1 1 1 1 1 1 1
cx
Hence W(u £x ), ... ,u ll :s v(p,w). 1 1 1 1
4.0.2
This completes our proof.
To check that v(p, w) is increasing in w, let p be a price vector, w and
w' be two aggregate wealth levels with w :s w', and (w , ... ,\\'t} be a solution 1
to the social welfare maximization problem of (p,w).
\ < d • \' < AI so, L..w.- wan nence ~....w. - w. I
!
1
1
1
Then
Thus, by the definition of v( p, w'),
Hence v(p, wl :s v(p, w'). To check that v(p,w) is nonincreasing in p, let w be an aggregate wealth level, p and p' be two price vectors with and p'
;!:
p, and let (w , ... , wi) be a 1
solution to the social welfa•e maximization problem of (p', w).
•
4-15
Then
Also, v.(p' ,w".> ~ v.(p,w.) for every i because p' 1
1
1
i!::
1
p.
Since W( ·) is
increasing, this implies that W(v Cp',w ), ... , v (p',w }) 1 1 1 1
W{v Cp,w ), ... ,v (p,w )J. 1 1 1 1
~
By the definition of v(p,w),
Hence v(p',w)
~
v(p,w).
To verify the homogeneity of degree zero and the quasiconvexity, we apply the equivalence of the two maximization problems established in Exercise 4.0.1.
For any (p,w} and A > 0, the two price-wealth pairs (p,w) and (i\p,Aw)
give the same constraints to the maximization problem of Exercise 4. D.l. •
Hence v(p, w) = v(i\p,i\w), implying the homogeneity. let
11
e lR.
As for the quasiconvexity,
Let (p,w) and (p',w') be two price-wealth pairs such that v(p,w)
v and v(p' ,w') :> v.
Let i\ e [0,1} and define p"
= Ap
+ (1 - A)p' and w"
~
= AW
and hence AD· C[.x. l + (1 - i\)p' · (E.x.l ~ etw + ( 1 - o:}w'. •
1 1
We must thus have either w •
either WCu Cx ), ... ,u Cx ll 1 1 1 1
~
1 1
i!::
.
p· Ct.x.) or w' 1 1 •
p' · £l:.x.). 1 1
Hence we
v(p,w) or W(u £x ), ... ,u Cx )) :: v(p',w'). 1 1 1 1
eithe:-- case, we have W(u £x ), ... _,u cx ll 1 1 1 1
4.0.3
i!!:
:>
v.
mus~
have
In
Hence v(p,w) is quasiconvex.
The welfar-e maximization problem is now rewritten with nannegativity
constraints: •
W£v (p, w ), ... ,v Cp, w )} 1 1 1 1
s. t.
[.w. I
w
I
1
:> i!::
w, 0 for all i.
We assume tha: xi(p,O) = 0 for every i and every p
4-16
» 0. This assumption is
satisfied
• the consumption
0 for
»
p
sees are
all IRL. +
Then 9 v .(p.O) =
p
1
The Kuhn-1-.:ter conditions for the
0.
{Theorem M.K.2) ~that there exist A > 0 and IJ.
social
1
0 such t
~
.) - A + IJ. = 0 for~ i = 1, ... ,I,
. 1
1
= 0
~t.w. 1 1
where all
a.e
for every i =l, .... I.
evaluated at the
(w ..... w )
= {w (p,w)...••wlp.w)). 1 1
1
The En
em M. L 1) implies
•
/Bu. )(Bv ./ap,_L 1
Now
w)
(8W /8u.H
> 0}.
Since p.. • 0 far every i e J, 8v/8w = 1
Since BvlfPL = 0 for any i I! J, ovlapl =
i e J.
1
caw / IE 1
1
[.
. Thus. •
(BW/Bu. )(Bv./8pDJ 1
-
1
v.
l
Hence the
w.
.1
1
~
I·
=- LieJ (c3W/Ciu."iUlv./8w.)
Bv/Bw ilv/BP l a /8
(8W/8u.)(8v./8pl.) 1
1
1
= l . .x.(p.w.{p,w)) = [.x.(p,w.(p,w)). L..j.eJ
1
1
1 1
1
function derived from v(p,wl equals [ixi(p,wi(p,w))
and the
4.D.4
(a)
-+ W(u £x ), 1 1
Maximum If the
and W( ·) are monotone. lben so is u( ·).
-1lso continuous.
Thus.
a,
the Theorem of the
• , u( ·) is also con tin ....... • •
and
we . ) are
mc:w#one and concave. then
u{.)
is
•
4-17
concat This can be proved as follows. Definer"
= ll.x
Let x e IRL, x' e IRL, and A e [0,1].
+ (1
= each i Then
[.x~ ::5 1 1
Alx:
1
for
1
Ax + (1 - ll.)x' and hence u(~x + (1 - ~)x•) ~
By the
~x. + (1 -
WCu Cxi), ... ,u Cxl)) .. 1 1
vity of the u.( ·} and the monotonicity of W( • ), 1
(xi), ... ,u Cxjll 1
!:ltA.u Cx 1 + (1 - i\)u Cxil• ...• ;w (x > + 1 1 1 1 1
{1 -
A}u fxjll. 1
By theccavity of W( · ),
I{Au cx > + (1 - ~)u Cxi>·
1
1 1
... ,
~ cx > + (1 - ~lu Cxjl)
1 1
1
!:lu.(x) + (1 - A)u(x').
Hence
~ + (1 -
ll.)x')
ll.u(x) + (1 - i\)u[x').
Hence u( ·) is concave.
Its worthwhile to point out that the quasiconcavity of W( ·) and the ui ( · ) dll5 not imply that of u( · ).
2
2
As an example, let L
x 1 :· uJ&zl = x 22 • and WCu1,u2 ) = u1
+
= 2,
I
= 2,
u Cx ) = 1 1
u2 . Then
W(2,0) = W(u C2,0),u CO,O)) = W£4,0) = 4, 2 1 W(0,2)
= WCu1CO,O),u2 (0,2)) = W(0,4)
= 4,
W(l,l) = W(u U,O),u CO,l)) = W(l,l) = 2. 2 1
_LI ::-;
\..:-:-·1es tu..sat.~.::~.
~ L..x.
1 1
::5
· a so1ut1on · . . . x an d 1s to t h e maxmuzauon pro bl em of
Exer-cisr4.D.l, then x is a solution to the maximization problem of this part. •
1s a soiutioall the ma..ximization problem of Exercise 4.0.1.
4-18
Let x' e i:1
L
and p · x'
solu1ir 1D the maximization problem of this part.
Thus, by the above result, [.x.(p,w.(p,w)J is a
Pi"Oblem.
1 1
1
•
til
the maximization problem of this part.
Hence the Wal!·asian demand
prated from it is equal to the aggregate demand· function.
4.0.5 1Jere is no positive
consumer if the WA is violated.
4.Cl thus serves as an e-vaul'le for this exercise.
E
welfare maximization problem is now v.rritten as
4.0.6.
[.w.
s.t.
1
~
1
w.
conditions are that there exists A > 0 such that
The
.} = A for every i, where all derivatives are evaluated at a
1
solut•tw •... , w 1
>. 1
By the definitiCil of WC • ), 8W lou. = a./v.(p, w.). 1
1
1
Bv ~
l
•
l.D.3(b), v.(p,w.) is
Ex
1
c3·dp,~. 1 t 1
of degree one in w.
1
1
= v.(p,w.)/w.. 1
1
~•d
hence
Hence the left-hand sides of the abo...-e
1
mrsditions equal (o:./v.(p,w.))(v.(p,w.)/w.) = a.lw. for eve:-v i.
firs~
1111
Th '.J s .. = a..ll. 1
I
Since [.o:.
1 1
= 1 and li_Wi = W,
11
W
= 1/i\..
II
•
Hence w = a • w . l
Thus
I
w. (p, ..Pc.w. 1
J.
welfare maximizatim problem is now written as
4.0.7
r.a.(p) + b(p)~.w.) •... , w I ) 1 1 'Li. 1
s..t.
[.w. 1
1
~
w.
with [.w. = w is a solution to this problem and v(p,w) = 1
1
4--19
+ b(p)w.
U
Suppose that (p I w"lJIB the potential 1
I
e fi such that '"· w: L.l 1
Then there exists
fiR
definition of v(p
1 I
over
W
1
:S
w and v.(p' w:) > 1
1
•
•
l
1
WCv Cp ,wi), ... ,v Cp',-I)}· 1 1
) :!:
•
1
Hence
•
f,r} > v(p w). 1
Define A. = {x.:u.(x..llllill. then A = r:.A.. 1
1
1
1· 1
I
Ths, for every p
1 l
»
\x E -/- is a solution to ·
p·y anly if there exists
A x . . . x A such tiat 1 1
E
r,.xi
= x
C fer every i, x. is a soblidll. 1
. eA •
1 ~
Win{p·y: y e A} =
1
1
y. e A.}. 1
y.
1
i.
p·x .. Bv the d ~
1
E A.} = e.(p 1 u.(x.)) 1
1
1
1
Hence
[e.(p,u.(x.ll.
e A} =
1 1
l
1
ather hand, by the J e B} =
k A c B. Min{p • y: y e A)
·y: y e B}. •
g(p) p.
-
1 1
.-rmite.
ii'lG.JI s - r:.s:( 1 1
~
-
1
1
::5
0
Hence, by the
amy condition for a ~(Section
2 Since D g(p)
Hence
r:.e.(p,u.(x.ll
= 0.
Bv the •
e(p,u(x)).
2 M.K), D g(pl is aqative
r.s.(p,u.(x.ll 1 1
~negative
4-20
1
1
by
semidefinite.
3.G.3
4.0.10
As a saw in the answer to 2
is symmetric. and if dp e IR , dp dp · S( P, w }dp c 0. representati~
:;e
4.C.ll(b), for every lp.w), S(p, w) 0, and d.p is not proportional tap, then
Thus, according to
3.H, there exists a psitiw
But accordill( tD 'Exercise 4.C.ll(c), if I< w < 16 •
consumer.
•
then the matrix C(p, w} is not positive
e.
Hence, accordiag to the
small-type djsr:nssion in Section 4.0, tll!r-e- is no normative represmtatrn consumer.
4.0.11
We sbil give an example in wbkh L = 3 and
c12 Cp,w)
:;e
the definitioa.
Here,
by parts,
2
-
I C7JI2lax (p,,.,wl/Bw)dl'l
0
1
2 - = I 0("'1)/iW)(dxl(p,TIW)/dl))dT)
•
=
cuwJif.zw> -
Clll;>x (p.w>. 1
H..-n,..·· •
--
-
-(1/w)x (p,2wlx
-
(llw)x
1
2
(p, w) + (l/2w}x (p, w)x 1
2
(j), w).
Similarly,
Again by in
-(p,w)x
1
2
(p,2w) + OJZ.wlx Cp,wlx
by parts,
(p. nw )( ax
2
(p,1Jw)/8w)dll
1
2
Cp,w}.
c (p,w). 21
By
2 = J (1/2)X l (p, l)W)(l/w)(dx (p, l)W)/dl))dl) 0 2 - - - 'J=2 2 = ((1/2W)Xl (p, 'J)W)x £p,1JW}l7)=Q - JQ(l/2w)(dx (p,l)W)/dl))X (p, "')W)dl) 2 1 2
2 = (l/2w)x £p,2wlx [p,2wl - J' {1/2)(dx (p,l)w)/8w)x (p,l)W)dlJ 2
1
0
2
1
an example in which
It is thus sufficient to
(p, wlx
-
(l/w)x
+
-U/w)x (p,2wlx
1
2
(p,2w),
or, equivalently,
;t.
Sc
--(112w)x (p,2w)x 1
2
(p,2w)
1
•
2
(p,w)
-
-(l/w)x (p,w)x 1
ider a preference, a price vector p, and the average wealth
-w
2
(p,2w} .
such
that:
-xip,2w} x£'p,wl
= 2 for both l = 1,2, 2 = S0 £1/2lxt(p;qwld'J)
= 1 for
(Another resti"'iction will be given shortly.
both t = 1,2. The demand for good 3 is
detetmined by Walras' law.) Then the right-hand side of the above inequality is equal to
2/w.
It is then sufficient to show that
But if the graphs of the functions 1'1 -+
-x l(p,l)w} -
are as in the figure below,
then the first term can be wade as close to zero as needed. inequality can thus be established.
4-22
The above
JU{p,T)W)
2 --··------·-······· ••
'•'
•• •• •• •• •• • •• ••• ••• •• •• ••• •• •
w
2w
••
xr{p,TJw) • ••
•• •• •
••'
0
Figure 4.0.11
4-23
w
CHAPTER 5
S.B.l
The first example violates irreversibility and the secane! one satisfies
this property.
0
0
Figm e S.B.l
5.8.2
L-1 Let z e ~+
Suooose first that Y exhibits constant returns to scale. ••
and a: > 0.
The~
(- o:z, af(z}} e Y. place of z and f(o:z) :s c::f{z).
(- z, f(z)) e Y.
By the constant retUI"ns to scale,
Hence etf(z) :s f(o:z).
By applying this inequality to o:z in
1 i::J place of o:, we obtain et- /(o:z} Hence f(o:z) = o:f(z).
:'5
1 f(a- (o:z)) = f(z), or
The homogeneity of degree one is thus
obtained. Suppose con•;ersely that /( ·) is homogeneous of degree one. e Y and a
i!:
0, then q :s f(zl and hence o:q ::= a:f(z) = f(o:z}.
(- a:z, f(o:z)) e Y, we obtain (- o:z, o:q) e Y.
Let (- z. q)
Since
The constant returns to scale is
thus obta:nec.
S.B.3
Sunoose firs: that Y is convex. ••
Let
5-1
Z
•
Z
•
e lRL-1 and a e lO,l]. then +
(- z, f(zl) e Y and (- z', f(z'))
E
Y.
By the convexity,
(- (a:z + U - «lz}. a.flzl + U - a.)f(z)) e Y.
Thus, cxf(z) + U - o:)f(z) == f(o:z + (1 - «)z).
Hence f(z) is concave. Let (q, - z) e Y, (q'. - z') e
Y, and
11 E
[0,11, then q
:£
f(z) and q'
a.q + (1 - cr)q' :s
~
/(z'J.
Hence
..rrz> + U
- o:)f{z' }.
By the concavity, a.f(z) + (1 - «)f(z'l s ttuz + (1 - a:)z').
Thus
aq + (1 - a.)q' :s f(a.z + (1 - a}z').
Hence
= «(- z, q) +
(- (a:z + ( l - «)z'), o:q + [1- «)q')
(1 -a)(- z', q') e Y.
Therefore Y is convex.
S.B.4
Note first tha-: if Y itself is additive, then Y+ = Y by the definition
of the additive closure. 5. B. 7, and S.B.8.
This applies to
5.8.4, S.B.6(a), S.B.6(b),
So we shall not depict
Let's now take up the cases in
the production set is not additive.
Note first that Y+ is equal to the set of
of IKL that can be •
represented a.s the sum of finitely
of Y.
If the production set
is convex, as in Figures S.B.l. S.B.2fa), 5.B.2(b), 5.8.3(a), S.B.3(b), and
5. B.S(a) (which is the same as S.B-2(b}), then we have the following stronger production plans.
of all , far
positive integer n, define nY c IRL by nY = {n:y e
We then claim that Y+ = vr:o nY. n=1
Y.
~L:
y e Y}.
In fact, if 1 e nY for some n, then (1/n)y e
By the definition, (1/n)y e Y+ and h
vr:sJ nY. n= 1
To be more
y = n((l/n)y) e Y+.
Thus Y+ ::;:,
To m-nve that Y+ c vm_ nY, it is sufficient to show that um nY is ~n-1 · n= 1
5-2
additive, by the def'inition of Y+. ·
So let y e vm
n=l
nY and
v' J
the:--e exist positive integers n and n' such that y e nY and y' E n'Y. 0/n)y e Y and (Vn')y' e Y.
Thus
Since Y is convex and n/(n + n') + n'/(n + n') =
1, (nl(n + n')){(lJn)y) +
/{n + n'))((l/n'}y') e
Thas y + y' e (n + n')Y.
That is, (1/(n + n'))(y + y') e Y.
as in Figure S.B.S(b),
If the pa oduction set is not m
+
longer have Y = un=InY. +
have Y :> v
As we can see in the above proof, +
~
m
= nY, we need not have Y c un=lnY. 0 1
some production plans in Y+ can be
('i·'i} be
. . d. max1mtzc
we no
we still
That is, it may be true that
only by allocating
inputs to di!fe-eot p&oduction units. follows. Let
Y.
point can be foz·mulated as •
•
plan at which the average return is
the
is, Y211Yi1 > y 1ly 1 for any other y = (y ,y l e Y. 2
that the function that associates each
quasiconcave.
1 2
1
Assume
r 1 < 0 to the average retmn at y1 is It is e;uivalent to
(This appears to be true from the figure.
Let y < 0 be an 1
saying that the average return functiCil is single-peaked.}
(aggr-egate) input level and n be the positive integer such that n.Ji < y
1
~
allocation •f a;gregate inpu:
Yr
(The number J of productim onits to be used is also being
•
optimized here.)
We shall prove that aoe of the following three li'ses must
then apply: (i)
J = n - 1 and ylj
(ii l J (iii)
=n
and ylj
c:
y /(n - I) far eruy j. 1
= Y{n for every j.
J = n and there exists k
y j = Yj for any j 1
E
{l, ... ,J} sucb tha.t ylk = y
~ k.
To prove this, note first that we y : ::= Yi for everJ j. 1...
1
- (n • IlJi md
have either y lj ?!
Yi
for- ere.ry j, or
In fact, if neither of these applies, then a small
5-3
•
input reassignment from a production unit with y . < y• to one with v . . 1 1J . ~ 1J increases the (aggregate) output level, because the average return increases. If ylj
~
Yi for every j, then the average return is decreasing at every ylj
and hence the y,. must be all the same. ~J
maximum integer that satisfies ( i)
applies.
at every ylj'
If y lj
~ y~
Yi
By the quasiconcavity, J must be the
= Lllj (and ylj :s.
Yil.
Hence J = n -
1 and
for every j, then the average return is increasing
If the y j are all the same, then, by the quasiconvexity, J 1
must be the minimum integer that satisfies
= n and (ii) applies.
Yi = Lllj
(and ylj
!:
Yi).
So suppose that some of the ylj are d"ifferent.
there exist two production units for which y lj >
Yi.
Hence J If
then a small input
reassignment from one to the other would increase the output level, because the average return is increasing. which ylk > Yi· applies.
Hence there mnst exist at most orie k for
By the quasiconcavity, y j = 1
Yi
for any j
~~~'
k.
Thus
(iii)
This la5t case happens when the average return decreases very fast
after the input level goes beyond
Yj·
For Figure S.B.l
For Figure 5.B.2(a)
5-4
For Figme 5.B.2(b)
••
••
••
••
••
••
........
•• •
•'
• •• •
• • •• • • •• •• ••• •
•• •••
•••
•• •• •
••
••
For Figure S.B.3(a)
5-S
·.....• •
•• •
••
•
........... •
...... . •
••
••
•• • • • • - · ...... • •• • - - - - · • • • · •..••••• •r •• •• •• •• •• •• •
•• • ••
·-·····-~·-•
••
••'
••
•'
For Figure 5.B.3(b)
•• •
•• ••
• • • • • ••
•• •
'
•
For Figure S.B.S(b)
S-6
•
Let y E Y and v e - IRL.
S.B.S
e Y bv Y c - iR + •
Since Y is convex,
~
(1 -
1/n)y + (1/n)(nv) = (1 - 1/n)y + v e Y.
Since Y is closed, y + v = lim
S.B.6
+
+
L
•
Then, for every n e IN, r.v e - c:tL and hence nv
n..-
((1 - 1/n)y + v) e Y.
(a) From the given functions tf>.( ·) (i = 1,2), the production set is 1
defined as Y = {(yl'y ,q): there exist q 1 2 and - y.
1
2:.
t/). (q.) for both i}. 1
•
0 and q
2:
2
0 such that q
2:
1
+ q
2
2:
q
A three-dimensional production set is depicted
l
in the following picture, assuming that. the
~.( 1
·) are convex.
output input 2 input 1
Figure 5.B.6(a)
•
(b)
We ciaim that the condition that ¢.(q. + q: l 11
1
!S
t/>. (q.l 11
is su:'ficient fer additivity.
+ t/).(q:) 11
for all q.
1
0, q:
1
2:
0, and i = 1,2,
In fact, let (y ,y ,ql e Y and (Yi.Y2_.q'l e Y. 1 2
Then there exist (ql'qZ) such that q + ~ 1
2:
sue h ~· ..na t q • 1
Then
+
2:.
q . -> q • ana. - Y•i
2
"" ( qi') · 2: .,..i
q and - yi
2:
¢i(qi). and (qi· q2_l
and -
(V'. + y:) 2: "".(q.) + .,._(q:) 2:. 1/>.(q. + q:) . • 1
1
'f'l
1
'f'1
5-7
l
1
1
1
Thus (y 1
Yj·
+
Yz
+
q + q') e Y, establishing the additivity.
(c) Let the output price be p.
The first-order necessary conditions for
profit maximization are that, for both i,
0.
w.t/J~(q.} ~ 1 1 1
p, with equalitv if ~
qi >
The interpretation is that marginal cost (in monetary term) due to a unit
increase of output level must be smaller than or equal to the output price, and the former must be equal to the latter if the output level is positive . •
If both ¢ ( ·) and ¢ ( ·) are convex (so that the corresponding production 1 2 M.K.J, these first-order necessary
set is convex}, then, according to conditions are also sufficient.
(d) Let q
>
By renumbering i = 1.2 if necessary, we can assume that
0.
w ¢ (q) ::s: w ¢ £q}. 2 2 1 1
> wl¢1(q). ql/q
+
In order to prove the statement, it is sufficient to show •
In fact, since the ¢.( ·} are strictly concave (and ¢.(0) = 0) and 1
1
q2/q = l, +
w2¢2(q2) > wl(ql/q)¢l(q) + w2(q2/q)¢2(q} = (ql/q)wlt/Jl(q) + ~q2/q)w2¢2(q)
The statement is th'.ls proved.
~ w,¢,(q). •
•
The strict concavity of the ¢. ( · } is 1
inte.r:;:::retec as the increasing returns to scale, which makes the statement qui:e plausib!e: tJnde:- the increasing returns to scale, it is bette:- to concentrate on one technique.
The
isoq~ants
of the input use is drawn in the following figu:-e.
Note
that the additive se;:>arability imposes the same restriction on the isoquants as that alluded to in Exercise 3.G.4(b).
5-8
input2
•
input 1
0 Figme 5.B.6(d)
If the:-e is a production plan y e Y with p· y > 0, then, by us"ing a.y e Y
S.C.l
with a lar-ge o: > l, it is possible to attain any sufficiently large profit Hence rr(p) = co.
level.
:s 0.
If, on the contrary, p·y :s 0 for all y e Y, then rr{p)
Thus we ha•te either n(p} = +
S.C.2
Let p >> 0, p' »
rr{p) and p' • y :s n{p').
e~~
or ll(p) :s 0.
0, o: e [0,1}, and y e y(o:p + (1 - o:)p' ), then p • y :s
Thus,
(a.p + Cl - o:)p') · y = o:p · y + (1 - o:)p' · y :s o:rr(p) + (1 - o:m(p' l.
Since (o:p
(1 - o:}p')·y
+
= n(a.p
+ (1 - a.)p'),
n(o:p + (1 - cx)p') :s «ll(p) + (I - o:)n{p'). Hence n( • } is convex.
S.C.3
The hcmoge:1eity of c( ·} in q is implied by that of z( ·).
prove this latte:- homogeneity only. Let w z(w,q).
»
0, q
~
0, and o: > 0.
Since f( ·} is homogeneous of degree one, f(o:z) = o:f(z) ,L-1
every z' e =-
+
,
·~
1J
f(z'}
~
-1
-1
c:q, then f(o: z') = o: f(z')
S-9
\v'e shall thus
~
q.
?!
Let z e o:q.
Fer
Thus, by z e
-1
z(w,q), w·(o: z') a:: w·z.
Hence w•z' a:: w·(a:z).
Thus a:z e z{w,a:q).
So a:z(w,q)
•
c z(w,o:q).
By applying this inclusion to
-1
a.
-1
in place of a: and o:q in place of
-1
q, we obtain o: z(w,a:q) c z(w,a: (a:q)), or z(w,a.q) c o:z(w,q) and thus conclude that zlw,a:q) = az{w,q). We next prove property {viii). [0,11.
Let w e IR
Let z e z(w,q) and z' e z(w,q').
w·z.• and c(w,q'l = w·z'.
L-1 ++
, q a:: 0, q' a:: 0, and o: e
Then f(z)
q, /(z' l a:: q', c(w,q) =
k
Hence
cxc(w,q) + (I - a:)c(w,q') = a:(w·z) + (1 - o:Hw·z') = w·(a:z
+ (1 -
a:)z').
Since f( ·) is concave, f(a.z + (1 - o:)z') a:: o:f(z) + (1 - a:)f(z') a:: o:q + (1 - o:)q'.
Thus w · (o:z +
(1 -
o:lz') a:: c(w, o:q + (1 - «)q' ).
ac(w, q)
S.C.4
•
+ (1 -
o:}c(w, q')
ii!::
Tnat is,
c(w, o:.q
+ (1 -
a.)q').
[First printing errata: When there are multiple outputs, the function
f(z) need not be well defined because it is conceivably possible to produce
different combinations of outputs from a single combination of inputs. Assuming that Li.e first L - M commodities are inputs and the last M corr.modities are outputs, we should thus understand the set {z as
' {Z
z, q) Yfa) • •
E
Y}. 1 For each q
= {z e IR
L-M +
: (- z, q)
i!:
0: f(z) a:: q}
0, define
L-M e Y} = {z e IR : f(z} +
Then c( · ,q) is the support function of Y(q) for every q. foHows from the discussion of Section 3.F.
~
i!: G}.
Hence prope:--ty (ii l
Moreover, according to Exercise
3.F.l, if Y(q) is closed and convex. then
L-M Y(q) = {z e IR : w·z a:: c(w,ql for all w » 0}. + Since Y = {(- z, q}: q a:: 0 and z e Y(q)}, this implies property (iii}. To prove property (iv), let w » 0, q Y(q).
Then w·z ::s w·z'.
i!:
0, a > 0, z e z(w,q), and z'
Hence (a:w)·z ::s (o:w)·z'.
5-10
Thus z e z(a:w,q).
E
Therefore
z(w,q) c z(a:w,q).
By applying this inclusion to o:w in place of w and 1
place of o:, we obtain z(o:w,q) c: z(o:- Co:w),q) = z(w,q).
0:
-1 .
1n
Property (iv) thus
follows. Property (iv) implies the homogeneity of degree one of c{ ·) in w, which As for its second part, let w » 0, q
is the first part of property (i). q'
i!!:
0, and q'
2:
Then Y(q') c Y(q).
q.
~
0,
Since c( ·,q) and c( · ,q'} a:-e the
support functions, this inclusion implies that c(w,q')
2:
c(w,q).
Hence c( ·)
is nondecreasing in q. As for property (v}, note that z(w,q) = Y(q)
every w >> 0 and q
~
0.
n {z e IR : w ·z = c(w,q)} for
Since both of the two sets on the right-hand side is
convex, so is the intersection, and hence so is z(w,q). single-valuedness, let q
2:
L
0., w
As for the
» 0, z e z(w,q) , z' e z(w,q), and z
Also suppose that Y(q) is strictly convex.
:;t
z'.
By the convexity of z(w,q},
0/2)z + (1/Z)z' e z(w,q).
By the strict convexity of Y(q), there exists a z" e Y(q) such that
(l/2)z + (1/2)z' » z". Hence w·((l/2)z + (l/2)z') > w·z", which contradicts (l/2lz + (112}z' e z(w,q}.
Tnus z{w,q} must be single-valued.
Property (vi) foliows from the fact that c( · ,q) is the support function of Y(q) and the c•Jality theorem (Proposition 3.F.l). Property (vi) implies that if z( · ,q) is differentiable at -
-
2
-
D z(w,q) = D ('i7 c(w,q)) = D c(w,q). w w w w
By property (ivl,
Dwz(w,q}w
=
Property (vii) is thus established.
S.C.S
e lR
then
As a Hessian matrix, this is svmmetri::. -
By property (ii}, i
Theory" by Hukukane Nikaido.
Here, instead of simplemindedly quoting the
Hawkins-Simon condition, we shall provide a direct proof tha:
reli~s
on the
2 symmetry of o t£zfp,w)) (which is not assumed in the Hawkins-Simon condition).
5-13
2 Write H = D tCz(p,w)).
To show that all entries of H-l are negative, it
is sufficient ta prove that, for every v e RL-l. if Hv
«
i!:
In fact. then, for each l. we can choose v e IRL-
0.
vector whose l:th
txM dinate
0 and Hv 1
;t
O, then v
so that Hv is the
is equal to one and the other coordinates are
-1
-1
equal to zero. then H (Hv) is equal to the lth. column of H .
Of course, it
•
is also equal to v, which is claimed to be strictly negative. I column of H- is strictly negative.
Tnus every •
-1 • Hence all entries of H are negat1ve.
For this prupez ty, in turn, it is sufficient to prove that for every v e IRL-l. if Hv ~ 0. then v ~ 0. if there cYists a v e IR some l.
Then '
;t.
inequalities suffice.}
such that Hv
::!::
0, Hv * 0, v
0, and "l = 0 for
:5
~
0.
We shall now prove by contradiction that for every v e IR then v
l.
~
0.
In fact,
0 and hence
Hv
which
L-1
(That is.
Then there exists v e IR
L-1
such that Hv
i!:
L-1
, if Hv
i!:
0,
0, anc vl > 0 for some
By re-ordering the inputs if necessary, we can assume that the fir-st M
entries of v au: positive and the last L - 1 - M entries are nonpositive Write write v =
write H =
HI
whose entries
0, x
2
e IR
L-1-M
,
0. Also,
, where H is an M x M matrix, H is M x (L - 1 - M} matrix 2 1 all positive, H is an (L - 1 - M) x M matrix whose entries 3
are all positi"R. and H is an (L - 1 - Ml x (L - 1 - M) mat:-ix. 4 0, y
2
e IR
L-1-M
=
,
H1x1 + H2x2
H x + H x 4 2 3 1
5-14
Then
-
y
1 •
Let Hv =
positive. liDs. by x » 0, • 1
x s 0, which is a contradiction
1 1
•
definiteness of H. aud heDce that of H, .
to the
~C.B
~ ·H
•
mat
12'G:St
in .uat1l 95' is 2·55 + 2·40 = 190, but it
AI
could
output
with a lower cost by using the input
-
•.
comb
"' '
is that.
due to
aodlor some restrictions that AI faced
the
plans are not observed to
oatside 12
~
8llf ;md it is
ba.ve
Thus the problem we wili encounter
•
recover i1s
those observations in order to 5.C.2(iii) or S.C.l(iii) .
based oa
S.C.9. To lid 1d ·} and y( ·) for (a}. the fil:st-order condition (5.C.2) is •
DOt
rmd
very •u1.
constraint binds.
a.use one d the
l
Tt(.) . . y(.)
•
for (b), it is not even applicable because f(. )
1s
Also, to not
In both cases, bowe91r, btcanse of the nature of the
diff
productia.la::tions, it is quia easJ to salve their CMP (which is similar to those in
S.C.lO. ), and the c:ast fmv:tions c( · ) turn out t::> be
c!ff er-entiliR witb respect to
levels q.
a-der c01llli'ilrll5.C.6} (which
' aoly the differentiabilit::
.~:· ~he
cost
· to output levels} to f"lod profit maximizing p;:-oduction
function levels,
We can thus apply the first-
= the
and sapply correspondences.
profit
the answer, til! output priee is fixed to be equal to one. (;.} rr( w)
V4'11 if w s 1 1 = V4-w if w > 2 1
2
wi wz
l-l/4wl, 0, 112w
J(w}
-
~--~·
» 1
- z2' l/2W l): ,
i!.
0, ~
5-IS
2:
2 0, zl + z = l/4w } 2 1
·r
1·
w
<w· 2' 1
•r l!
w
1
w2;
(b) n(w) = 1/4(wl +
= (- l/4(wl
y(w)
wz>2
+ wz) •
(c) Note first that this production
• x:bibits constant r
Moreover, if p < 1, then the
scale.
CUilSti
If p = l, then this production
to
aint does not bind.
rise to the satne ·
that of (a}, and hence one of the DODDegativity cousu·aints binds.
as It is thus
easy to apply (5.C.6).
If p < 1, then
Tr(W)
ID
·r 1
0
if
o/(p-1)
w-1
+
=
o/(p-1) < 1·
W"z
•
) :!:: 1.
p-1) +
1
'f
0
=
y(w)
1
V(p-1) ( w { a, 1
p/(p-1) +
w1
·r p/(p-1) 1 w 1
+
o/(p-1)
.w-2
< 1
o/(p-I} _
w2
- 1;
{0)
If p = l, then 1l(w) =
0 if Min{wl'w } 2 (I) if Min{w ,w }
2
1
=
1
w2.
p-1) (1-I/p)
1
2
.
o/(p-1))(-Vp)( 1/(p-1)
91'2
Wl
•
ll{p-1))
,W2
•
.ue that c( ·) is t1rice continuously diff'a Ultiable.
By Proposition
5. C.2(vi). %{ ·} is continuously differentiable and
= (Btaq)(8c(w,q)/Bwl) = (a/awl)(8c(w,q)/Bq).
Bzl(w,ql/Bq
Hence Bzl(w,q)/Bq > 0 if and only if (8/Bwll(BC(w,q)/Bq) > 0, that is, marginal rost is increasing in w t:
S.C.l2 Suppose first that y
= 0.
E
E
"J
at p, then {p,n(p))· (y, - 1)
y' e Y, (p,n(p))·(fdy', -Ill = a(p·y' - tr{p))
Also, for every a. c: 0
Converseiy, if y
Y n1aximizes profit
and {y, .- 1) maximizes profit in Y' at (p,pL+l ), •
then (p.Il.+l)·(y,- 1) = p·y -1\.+l = 0 by the coostant returns to scale. Also, for every y' e Y, (p, pL·t"" [y:, - 1) = p · y' - 1\.+l
!S
0.
Hence y
•
m
profit in Y at
p and Jr(p)
= p· y = 'l.+r
S.C.l3 Denote the production fum. lion of the firm by f( · ), then its
optimization problem is
Tnis is analogous to the utility maximization problem i."l Section 3.0 and the function R( ·) corresponds to de indirect utility fmw:tion.
5-17
Hence,
analogously to lo}'s identity (Proposition l.G.4), the input demands are obtained as 1
S.D.l
We shall \Ee the differentiabilitJ of Cl..·l aniy at
differentiabilitJ is not necessary.
AC" (q)
The everywhert
By the M"Ulition,
1:
at q = q,
Thus, if the awnge cost is
C' (q) = C(q)/q a
q.
then
AC'(q) =
0 and hence
ICtq).
5.0.2
.•
·"
....
•• ••
••••
·-··
___ .. ....
•••••
•
••
• ••'
• • • -----------~ •• •• •• • •• •• • • • • • • • • • • •• •• •• • •• • • •• • •• ••
q
C'(q)
A
A
Figure 5.D.2
•
•
5.0.3
Let q(w,pl bt the profit-maximizing
are w and the initial output
price is p.
pri£z. z be •
wbe tie initial
input prices,
the initial long-run ilp"' demand at
output level at
be the initial p for every p (
Let
level when the input pric(
(w,p).
that q("w,p) > 0).
(w,p).
-p
be the
and
-q
By (5.C.6), Bc(w,q(w,p})/8q =
Thm. 'by differentiating both si• •
5-18
wilh
p,
:: to P 3lll tlaaaluating at p =
we obtain
2
(l•.q(w,iin;aq Haqcw.p)lap> -1
) On the other
t.t.
= 1.
.
the short-run cost function function
qs(w,q!z
aut the .
s ._did -a.bo"'e, we
1
in the hint.
) as 1
- - -
2 - - -
aqs{w,p I z1 )/8p ={aciw,q I z 1)/8
•
c s { w, q Iz ) 2
Just
q)
-1
•
cfw,ql
Now, by the
--c {w,qlz ).
Hence te flldan g(q) =
1
5
-q 3llld,
by the
5
c(w,q)
1
-
cs(w ~)
'a .
=2
cost.
GltJ. J
J
t
and c:idilm
g"(q)
:s 0,
costs (and C(O} . IC(q .).
J=
J
multiple firms aDI
=Yiliin{q ,q }, ·ther: C(·) exhibits dec:-easing ray 1
~
But le!-, =Ul,
2
= (8,1}, and q = q + q = (9,9). 2 1
Hwce C(q} > C(q l + C('lz)· 1
Hence
C{ ·
Then
l is
1 additive.
(cl We shall first prae • • q
r,ctqJ. In fact, J
M.Kl,
over j, we obtain C(q) :s
C(<Jt = C(Ciz) = 1 andf:tqJ=l. not
is maximized at q =
By the decreasing
q .•
is no w;r ta hit up the production of q
(b) Let M
=
1
S:D.4 (a) Suppose tilt (q ./q)C(q) :s J
c(w,q)
) for all q and
condition (see
2 that is. acl<w.q)la ,
-
:s c
(w,q I z
l
=
.q . and c . » 0 far everv j, then C( q l ~ L.J J 'J • 't"
f«.ry j, there exists 1. > 0 such that 7.q. » J J J
th~
> ctq).
By lie •
ctal. •J
•
eXIstS Cl ...
J
theu E.lltk. = I and :J J ~
C(q).
By lie:
Tnus, by the
and C(q) >
SEh that C(ajqj) = C(q}. 'Define f3 =
5-19
~
L}laj,
Thus, by the quasi::onvexity.
.} = {1/fl)q. J.
ss, f3
q.
1.
Hence, by the decreasing ray
average cost. I:.C(q.) ~ I:.U/o:JC(o:.q.) = l:.U/a.)C{q} = SC(q) ~ C(c). J J :.J J JJ J J .
For the general
in which some q. J
-applJ the above result to the q. J
and then
+ £e,
where
~
0
?!
not be strictly positive,
> 0 and e = 0,!, ... ,1) e IRM,
£
The continuity of C( ·} then implies that
the limit as c -+ 0.
•
:j=lC(qj)
S.D.S
i!!
C(q).
(a) The prcdTCtim function f( ·) exhibits
only if f(Az}
i!!
Af(z) for all z and all 1 i!! 1.
returns if and
Helice, lf z'
?!
z > 0, then
(1/z')f(z') = (1/z')/((z'/z)z) 2: U/z'Kz'/z)f(z) = (1/z)/(z).
Thus the average product is nondecreasing.
The ma!'ginal product may however
be decreasing on some region of output levels, as the following example shows:
q
.r··
••
• • ••
.•
•••••
• • ••
•
••
••
•••
••
••••
..... ··· •
•
0
(b)
SD.S(a)
Mathctaatically, the t:nmmmer's
ion p-ablem is
ma.xz!:O u(f(z1J -
The first-order
rewritten as u'(f(z)) 1
z
z.
condition is u'(f(z))f'(zl = 1, which ca::1 be
1 = rczl- . Since the
cost
1
is given by z =
f- (q), the marginal cost is equal to f'(z}- and the equality of marginal
5-20
cost and marginal utility is thus a necessary condition for a maximurr... Economically, the consumer will choose the output
l~vc!
at which the
marginal utility of an extra unit of the output is exactly equal to the disutility incurred by giving up the necessary amounts of input to produce it. But •
IS
~
lcnter is nothing but the marginal cost.
Hence the marginal utility
to the marginal cost.
(c) "Ibis assertion is wrong.
As the following figure shows. even if marginal
cost ami ma.-ginal utility are equal at an input level.
may be another
input level at which the consumer attains higher utility:
l..z)
A. q
•
Curves
0
The
z
SD.S(c)
for suboptimality is that the first-order necessary conditions
are DOt S".Uficie:-:.t when the production function exhibits increasing ret
(which gives rise to nonconvexity of the feasible set).
S.E.l By applying Hotelling's lemma (Proposition S.C.l(vi)) twice, we get y•(p} = tTn•(p}
= V([ .11" .(p)) J J
=
E.tTfl. .(p} = LY .(p). J
J
I J
S.E.2 "This is just a matter of going through the proof of Proposition S.E.l
5-21
and checking that convexity was never used.
Its interpretation was given
before the stat~ment of the proposition (p. 148).
It is a consequence of the
very definition of the aggregate production set, that is, it is the sum of the
J firms' production sets.
It is thus independent of convexity o:- any other
properties of the firms' production sets.
[First printing errata: We should assume that there is a p• »
S.E ..3
l!:
r.v .. J J
Otherwise, denoting the
L
LJ.Y.J
profit function of
p•. y for every y e
by n•( · ), the set {y e IR : p·y
~
n•(p) fo:r all p »
may be empty, and all we can obtain is the equality between p · y :s n•(p) for all p
l!:
0).
This assumption is also
validity of Proposition S.C.l(iii).
L/
necess~ry
0}
L
j and {y e IR :
for the
In fact, its proof should· go as follows:
It is sufficient to prove that, for every z e that p · z > n(p).
0 and y•
IR~Y.
there exist a p
»
0 such
Since Y is closed and convex, the separating hyperplane
theorem implies the existence of such a nonzero vector p. property implies that p must actually be nonnegative.
The free disposal
If it is not strictly
positive, then t";.ce the convex combination (1 - c}p + cp• with a sufficiently small
£
> 0.
1 ne:-:
it is strictly positive, and satisfies ( (1 - c l:=
+ ~p· J • z
>
nUl - c )p + £p*) by the upper semi continuity of n( • }, which is implied by its
conveX:itv• .l
Since each Y. is convex and satisfies the free disposal prope!'ty, J
[.Y. is also convex and satisfies the free disposal property. J J
Since it is
also assumed to be closed, Proposition S.C.l(iii) implies that L [jYj = {y e IR: p·y :s n•(p) for all p »
But here, by
Pr~p~sition
r J.Y.J
0}.
S.E.l, n•(p) = Ljn j(p) and hence
= {v e IRL: p· y :s -
LJ:nJ.(p)
for all p » 0}.
of the technoiogy with
S.E.4
5-22
istics z = (~;z_Ja
{(- 'r-~ y (w) 2
l)}
((-., -itr
=
ct}: ct
e (O,I])
if
•fi + wiz.2
if Wft +
{0}
rcr which
area of the
w-r2
= 1,
>
1.
the output .., be one is depicted
in the following pict'IJI'e.
- - - - - - - (10,
IIW1
• 0
)/WI
5.E.4
"More gener ally• 5ladd. be deleted. J
(h) [First printing
with
the profit of the by -.: (1(118
z
n (w)= z
nws.
•n --rz.
1 -
if w z + 1 1
0
the aggregate (or,
the integral of 1 - w \ 1
s
1}, which is depicted rr(w)
=
l/w 0
if w lzl + w2z2 ~ 1,
wr-2 >L
average) profit is
-n_•* area
Cll ~
rlgUre.
1
{z
E
(O.lOh
Thus the
by taking
wlzl + w2z2
profit is
=l/600w1w 2 .
I
5-23
(c) The agg1 egate input demand can also be obtained by integr-ating the input demands of the individual firms, but the following point is noteworthy: If a firm has characteristic z
= £z 1,z2 l
with w z + w z 11 2 2
input demands at input prices w = (w .w ). 1 2
= 1,
then it has multiple
But those firms constitute only a
negligible pmtion in the whole populaU.. Hence it is harmless to assume that such a firm has input demand z =
(~·~)
(in absolute values).
Hence the
aggregate demands are 11w 1
0
1
0
/w
1
2"
600w w 1 2
0
It is easy to check that these aggregate input demand functions can also be obtained by applying Hotelling's lemma to the aggregate profit function, which was obtained in (b).
(d) We need to find an aggregate production function whose input demand
function is the same as the aggregate input demand function in (c).
Denote
such a production function by f( · ), then the first order-conditions for profit maximization a1 e of(zrz )1azr = w. and Bf(zl'z l1Bz = w . 2 2 2 1 2
These
expressions, when evaluated at the inputs demanded, must hold for all w. 1
--
L "".
2
z = l/600w w and 1 2 1
Zz •
then
Thus
•
5-24
2 = l/600w w , 1 2
I
Thus
Therefore, f(z .z l 1 2
a:
3Cz z /600) 1 2
1/3
.
The aggregate production function is a
Cobb-Douglas one exhibiting decreasing returns to scale.
(a) Plant j"s marginal cost is MC .{q.) = « + ~ .q .. J J J J
S.E.S
Since f3. > 0 for J
every j, the first-order necessary and sufficient. conditions for cost minimization are that
L .q. J J
q and MC .{q.) = MC ., (q ., ) for all j and j". J J J J
a::
From
these, we obtain qj = (q/{3 j}/(l:hl/,Sh).
(b) (c) In both cases, it is cost-mininlizing to concentrate on plants with the •
smallest
.S.
J
< 0, because the average cost is decreasing at the highest rate at
such plants.
S.F.l
The production plan y in Figure S.F.Hbl is not efficient but it
maximizes profit for p = (0,1).
Throughout the answer, we fix the price of the input at one and denote
S.G.l
Suppose that there are I consumer
the price of the output by p.
Denote their shares by 9. > 0. .. , • 1
indexed by i = 1, ... ,I.
Of course,
r.e. = 1. 1 1
Since they have quasilinear utility functions, by Exercise 3.D.4(b), their indirect utilitv functions can be written as v.(p,w.) = w. + ¢.{pl. -
1
1
1
1
Note that
the demand fun:tior:. x.( ·) for the output of consumer i does not depend on the 1
weal~h
and satisfies xi(p) = - ¢i(p) by Roy's identity.
(a) Wher. the input is z, the utility level of consumer i is e.(p(z)f(z) - zJ + f/>.(p(z)). 1
1
(Her-e we are ass-rCzl - 11 -
But, since f(z) = lri(p(z)), this implies
first-order condition the summation of the
cf>~(p(z)). 1
1
o.
we obtain
~I(p(z))p'(zl
p(z)r~
- 1 = 0.
= o. Plugging this
into the first-order condition, we obtain e.p'(z)f(z) - x.(p(z)Jp"lz) = 0. l
1
Thus B. = x.(p(z))/f(z). 1
1
(b) We know from (a) that, if ownership share are identical, then, in order
for consumer-owners to unanimously agree on a production plan, it is necc ssary tha! they all cor.smne the same amount of the l:mput.
But if their tastes are
different for the output, then their consumption levels will be different. Hence they will instruct managers to carry out different output levels. (c) If p:-efer-ences and ownership
r~>hares
are
, then the first-order
conditions are also identical and hence the cons.aer-owners unanimously agree
unanimous agreement is that p(z) = Vf'(z).
The right-hand side is the
inverse of the marginal return, and hence equal to the marginaL nothing p( zl
bu~
profit maximization with respect to iaput., when the output
is taken as giveo.
S.AA.l
This is
Fr-::~m
the unit isoquant,
5-26
pri~
w . < w2, 1 ·r - w2, I• w
(2,1) •
-
z(w,ll
'#"
1&
{A(2,1) + U - A)(l,2)
IR: i\
E
E
[O,ll}
1
( 1,2)
if w
1
> ....... 2.
Thus 2w + w 2 1 w + 2w 1 2
c(w,l) =
if w
:.o;
w ,
1 2 if w > w . 1 2
This is differentiable at w = lwrw ) if and only if w ':f:. w . 2 1 2
Moreover-,
•
(a) We shall first prove that if t3 e IRL-l, (I - Alf3
S.AA.2 ':f:.
0, then b•f3 > 0, and that if {J
RL-l and (I -
E
A)~ =
0, and (I - A}{3
i!::
0, then b·t3 = 0.
In
fact, in the proof of Proposition S.AA.l, we showed that since A is
1
the inverse matrix U - Af exists and ail its entries are
productive, •
Thus, if {3 e
nonnegative.
(I - A)-l({l - A)(3} i!:: 0
e IR
L-1
1
~- • U -
and f3 • 0.
A)f3
l!:
Since b
0, and
»
(I -
A)f3
i!::
(I - A)o:' and (I - A)o:
AHa - o:') :;: 0.
A}o:' • tnen (I - A)(o: - a'} = 0.
anv• case, it is i:::1:>ossible that • •
I - A -
I - A -
•
0
':f:.
L-1 +
a •.
and a:'
E
If (I - Ala:=
Hence b • (o: - cr.') = 0, or r· - e:· = b·a:'. •
I - A - b
0: i!::
ln
I - A , d _ b o: an
Efficiency is thus established.
(b) By (a) and Proposition S.F.2, any production plan with o:
»
0 is profit-
L-1 be To establish its uniqueness, let p E IR
maximizing at some price vector.
+
•
a suppor-ting p&ice vector.
L-1 IR . +
(l - A)o:', then (I- A}(o: - o:') i!:: 0 and
Hence b·(c- o:') > 0, or b·o: > b·«'.
•
•0
If
and (I - Alt3 = 0, then f3 = 0 and hence b·f3 = 0.
If (I - A)o:
(I -
t3 =
0, then
0, this implies b·(3 > 0.
To derive efficiency from the above result, let a E IR
(I -
;t
By a:
»
0 and the zero-profit condition for
activities beir.g actcally used, we must have p· (I - Al = b, that is, p = ((I - Al-llTb (where b is now a column vector).
This implies the uniqueness
•
and the strict positivity of p,
all entries of (I - A)
5-27
-1
are
nonnegative, all its diagonal entries of are positive, and b » 0 .
•
•
(c) For each l, denote by et the vector in IR
the other components are zero.
L-1
whose lth camoonent ts one and • •
As we saw in the rema!'k following Proposition
5.AA.1, the total amounts of the producible goods necessary to realize a net output vector el e IR
L-1 +
is equal to (
Hence the total •
amount of labor embodied in these necessary amounts of the producible goods n
•
equals b· b ·(I
- A)
A )el = b ·(I - A) -1
-1
et = Pt·
=
Thus the price (row) vector p
can be interpreted as the amounts of primary factor directly or
indirectly embodied in the production of one unit of each producible good.
I - A' - b'
(d) Let
productive. (! - A')
-1
Lx[L-1) be 1 . h ' f .. ' e IR any a ternat1ve c otce o actiVities that is
, -. By the productivity, the inverse matrices (I - A) • and So denote them by C = [c
exis-:. and are nonnegative.
1
... cL_
1 and 1
C' = [c~ ... cL-l ), where the cl and ci (l = 1, ... , L - 1) are (L - ll-
dimensional column vectors.
Then
(I -
A)cl
= el
and (I - A' lc£
for each c > 0, define dl(c) = cl + c([k~lck) and dl(c) Then die) -+ c< and dt(c) -+
cl
as
£
= cl
+
= et
Now,
dl:k;!:lck).
Moreover, dl(c) >> 0, dt(£) »
-+ 0.
0,
and
Bv (a''• ~
1- A
-
By this strict positivity and the
- b
assumptior. tha: the activities
I - A - b
have been singled out by the
nonsubstitution theo&em, we must have b · dt'c) :s b' · dt(c).
Taking the limit as
assertion now fellows from (c).
5.AA.3
(a) Denote the activity levels by o: and o: . 1 2
for labor is a
1
+
2a:
= 10, or a 110 1 2
+
a. 15 = L 2
5-28
The resource constraint
Since the production level
10 •
is given by
-s
- 5
• the production possibility frontier is as
5
follows:
Y1 5 •
10 •• i y1 •• ••• ••
Figure S.AA.3(a)
(b) By Exercise S.AA.2(c), the equilibrium price vector is b ·(I - A)
-1
-
(4,6).
(c) The amount of labor embodied in each commodity equals its price, as shown in Exercise S.AA.2(c).. (d} The locus of amounts of good 1 aM labor necessary to produce one uni't of good 2 is equal to 2
{A(1,2) + (1 - i\)(l/2,fJ): i\ e {0,1]} - IR , +
assuming free disposal.
It is represented in the following figure:
5-29
-----·· ~
•••
------·!·-···· • ••• ••
•• •• •• •• • ••
---····----·--· •• •• •• •• •• •• •• •• •• ••
• ••
•• •• •
0
••• •• •• •• •• •• •• •• •• ••
••• ••• •• •
••• ••
1/2
1
••
0
13>2
s
(e) In the context of (d), the nonsubstitution theorem says that it is possible to choose one of the two techniques to produce good 2 (or a combination of the two with a fixed proportion) in such a way that any efficient production plan with positive net outputs of the two producible goods can be attained by using the technique chosen for good 2. We could determine which of the two techniques (or their mixtures) is efficient by actually plotting the frontier of the feasible output combinations from one unit of labor. 'In the following, however, we shall identify an effi::ie:1t technique based on Exercise S.AA.2(d). 0
For each i\ e {0,1], define A{i\) = (1 - i\)A + i\A,
112 b(i\) =
So let A' =
(1 -
1 A.)b + i\b', and p(i\) = b(i\) · (l - A(i\))- (where p(i\) is a row •
vector).
According tc Exercise S.AA.2(d), if i\• e [0,1) a.-'ld the convex
combination of the fir-st and the second technique with weight 1 - i\ • and i\ • is efficient, then p(i\)
~
p(i\ •) for every i\ e [0,1).
Hence the switch of
efficient techniques occurs precisely when the value of i\ • switches as varies.
We shall now find a value of
~
at which the value of i\• switches.
Just as ir.. (b), we can calculate
5-30
2
p(i\} =
2
+
((3 - 2)i\ + 4 (2(j - S)i\ + 6
213 - 8
and Dp(i\) =
1•
2
(i\ + 2)2
•
Hence: if (3 < 4, then A• = 0; if 13 > 4, then .A• = I: and if f3 = 4, then .A• can Thus the switching occurs at (3 = 4.
be any value in [0,1}.
More precisely:
if (3 < 4, then it is efficient to continue using the first technique: if (3 > 4, then it is efficient to switch to the second technique; and if 13 = 4, every •
mixtu•e of the two technique is efficient .
•
But y and y are not. 5 1
these three vectors are in the production set. this, suppose that y By a
2
1
::!'
L.o: .a.
Suppose next that Ys :s By a
(b) If p = (1,3,3,2), then p · a .
J
profit at p.
3
= 0.
LJ.o:J.a.J
But there is no cx
with a. := 0. J
= 0, according to ·good 4, a
4
According to good 2, o: = a. = 0. 1 2
with ex. := 0. J J J J
= 0, according to good 3, a
:5
To see
= 0.
3
4
!:
0 for which y
1
:5
According to good 1, a · • 1 But there is no
0 for all j and p · y = 0.
ct
2
:= 0 for
Hence y maximizes
By Proposition S.F.l, y is efficient.
(c) Since y = a , y is feasible.
But, since a
feasible, y ca..·mot be efficient.
(Note that a
1
•
2
+ a
2
3
+ a
3
+ a +
4
a
4
= (2, - 1, 0, 0} is represents a
round-about production of good 1 out of good 2.)
S.AA.S
[First p:-!nting
=·· The last elementary activity as ::
(- 2, - 4, 5, 2) should be a 3 = (- 2 , - 4, 5, - 2).]
(a) The set Y is defined as
y =
La. a e ... m m
o: a
m m
4 e IR : «
m
Y, and y' = }"1n o:'ma m e Y. Ay + (l - i\)y' =
!:
0 for each m}.
Then
Lm(i\Clrn +
5-31
(1 - i\)o:' )a
.
m m
Let A e [O,ll,
Since ~cr. m + (1 - ~)o:m' ~ 0 for every m, ~Y + (1 - A.)y' e Y.
Thus y is convex.
(b) Since all the activities use commodities 1 and 2 as an input, in order- to
produce any commodity in positive quantity, it is necessary to use commodities 1 and 2 as an inputs.
The no-free-lunch property thus follows .
•
(c) Note that it is impossible to dispose of one of commodities 3 and 4 without increasing the output· of the other, and that it is impossible to dispose of any one of commodities 1 and 2 without disposing the other.
Hence
Y does not satisfy the free-disposal property and it is necessary to add the four disposal activities to the given elementary activities in orde!"' for the free-disposal property to be satisfied.
(d) Note that 3a 2a , and a 6
7
:;e
1
~
2a . 6
a , 3a 5 1
:;e
Hence a , a , a 5
(e) We can che:k that (4/3)a + 3
a
3
(f)
+ (1/2)a
1
~
a , and a 2
3
~
a , a 5 2 4
8
a , a 4 2
and a
(5/24)~ ~
+ (1/2)~
* a . 2
6
:;e
a , 2a 3 4
~
a , 2a 3 8
*
a , 8
~ ~
are not efficient.
a , (4/3la + (5/24la_, :;: a , 1 3 1 Hence a
1
and a
2
are not efficient.
We shall pr-o':le that the set of the efficient production vectors is equal
production vector- belongs to this set.
Conversely, we can check that the
production vectcr-s in this set cannot be dominated by each othe:-.
Hence they
are all efficient and the set of the efficient production vectors is equal to cr._ ~ 0, 0:7 == 0} . .j
{g}
Since cr. a 3 3
+ cx._a..., I I
= (- cr.
3
problem of max:wizing the net production of the third commodity is as follows:
5-32
s.t.
::
~so.
5~ ~
300,
10~ ::
o.
a~
0,
a..,
~
0.
(h) The feasible set is shaded in the
CX7
(o:
3
.a_,l-space below.
2w + Sa., = 300
60
CXl
12 0
120 ISO
+ 8a.7 = 480 480
Figure S.AA.S(h)
The solut.ior. to this problem satisfies 2« + 3
5-33
SU.,
c=
300, o:
3
=
10~.
Thus
CHAPTER 6
6.8.1
Suppose fir-st that L >- L'.
A first application of the independence
axiom (in the "only-if" dil'ection in Definition 6.8.4) yields cd. + (1 - a)L" ..., >- aL' + (1 - o:)L" .
rr these two compound lotteries were indifferent, then a second application of the independence
(in the "if" direction) would
yield L' >- L, which contradicts L >- L'.
We must thus have
-
cd. + (1 - a)L" >- cd." + (1 - a)L".
>- aL' +
Suppose conversely that aL +
(1 - «)L"
independence axiom, L >- L'.
If these two simple lotteries were indifferent,
-
(1 -
a)L", then, by the
then the independence axiom would imply o:L' a contradiction.
+ (1 -
a)L" >- aL
-
+ (1 -
alL",
We must thus have L >- L'.
Suppose next that L - L', then L >- L' and L' >- L.
-
-
Hence by applying the
independence axiom twice (in the "only if" direction), we obtain aL +
(1 -
cdL" - aL' + (1 - o:)L" . •
Conversely, we can show that·if o:l.. + •
(1 - a)L" - cxL' + (1 - Gt!L"",
then L - L' .
For the last part of the exercise, suppose that L >- L' and L" >- L'", then, by the independence axiom and the first assertion of this exercise, aL
+ (1 -
o:}L" >- «L' + (1 - a)L"
and «L' + (1 - a)L" >- cxL' + (1 - o:)L"'.
Thus, by the transitivity of >- (Proposition l.B.l(i)), aL + (1 - o:)L" >- «L' +
6-1
(1 -
o:)L "'.
Assume that the preference relation
6.B.2
•
~
-
is represented by an v.N-M
expected utility function U(L) = Lounpn for every L = (p , ... ,pN) E !f.. 1
e !f., and Tnen L ~ L' if and only if \' u p ~ \' u p'. "nnn '1lnn cl() u p ) + (l - ex)() u p")
'Tlnn
'Tlnn
~
e
(0,1}.
This inequality is equivalent to
o:() u p') + "nnn
Hence L ~ L' if and only if cxL + (1 - a:)L'' ~ o:L' +
-
a.) (\' u p"}.
(1 -
(1 -
~nn
~
This latter inequality holds if and only if cd. + (1 - a)L"
-
et
Let L
-
o:L' + C1 - a.)L".
o:)L".
Thus the
independence axiom holds.
6.B.J
Since the set C of outcomes is finite, there are best and worst
outcomes in C.
Let
-L
be the lottery that yields a particular best outcome
with probability one and L be the lottery that yields a particular worst
-
We shall now prove that
outcome with probability one.
-L
~
-
L
~
L for every L e
- -
!f. by applying the following lemma:
Lem..:na: Let L ,L , ... , 0
. b'l' . prooa 1 1t1es
1
~
be (1 + K) lotteries and (o:l' ... ,~)
. h ~l( 1 Wit~ L..k=l~ = .
If
~
~ 0
be
t L0 for every k, then •
•
Pr-oof of Lemma: We shall 'prove th1s lemma by induction on K.
If K = 1, there
•
is nothing to prove. l.
So let K > 1 and -suppose that the lemma is true for K -
Assume tha: Lk ?: L
0
for every k.
By the definition of a compound lottery.
-1 ~ et. L = (1 - o:.,) k=l KK .IIi. k=l 1 - IX
By the induction hypothesis,
-1
a:k
K
+ a L k K K'
Hence, as our first •
application of the independence axiom, we obtain o:k a.K) k=l-=----1
(! -
L
Applying the axiom once again, we obtain
6-2
k=la.kLk .t L . 0
Hence, by the transitivity, verified.
The case of L
The first statement is thus
.t Lk can similarly be verified.
0
Now, for each n, let L n be the lottery that yields outcome n with Then L
probability one. outcomes.
Let
{First
-
L
because both of
can be identified with sure n L=}oL.
L
the above lemma.
6.B.4
~
n
"'n·n
Thus,
The same argument can be used to prove that L
:prin~~pg
~
-L
~
-
L
L.
--
errata: On the the 11th and the 12th line of the
exercise, the phrase "the lottery of B with probability q and D with •
•
probability 1 - q" should be "the lottery of A with "probability q and D with probability 1 - q".
Also, in the description of Criterion 2, the phrase "an
unnecessary evacuation in 57." should be "an unnecessary evacuation in 15'7.".1
(a} We can choose an assign utility levels (uA''1ruc·~) so that uA = 1 and u 0 = 0 as a normalization (Proposition 6.8.2).
uC = q·l +
(1-
Then u
8
= p ·1
+ (1 - p} · 0
= p and
q)·O = q.
{b) The probability distribution under Criterion 1 is
The orobabilitv distribution under Criterion 2 is • •
The expected utility under Criterion 1 is thus 0.891 + 0.099p + 0.009q. expected utility under Criterion 2 is thus 0. 8415 + 0.1485p + 0. 0095q. the age:1cy
wo~ld
The Hence
prefer Criterion 1 if and only if 99 > 99p + q, and it would
prefer Criterion 2 if and only if 99 < 99p + q.
6-3
6.B.S
(a) This follows from Exercise 6.8.1.
(b) The equivalence of the betweenness axiom and straight indifference curves can be established in the same way as in the part of Section 6. B on pp. 175-176 that explains how the independence axiom implies straight indifference •
curves .
(Note that the argument there does not use the fully fledged
independence axiom; as it is concerned with two indifferent lotter-ies, the betweenness axiom suffices.)
'I hose straight lines need not be parallel,
because the betweenness axiom imposes restrictions only on straight indifferent curves and nothing on the relative positions of different indifference lines.
In fact, the argument for Figure 6.B.S(c) is not
applicable to the betweenness axiom. (c) Any preference represented by straight. but not parallel indifference curves satisfies the betweenness axiom but does not satisfy the independence axiom.
Hence the former is weaker than the latter.
(d) Here is an example of a preference relation and its indiffer-ence map that satisfies the betweenness axiom and yields the choice of the Allais paradox.
6-4
(2 500 000 dollars) •
•• ••
•• ••
•
•••• •• •• •• •
••
·-....,
•••
••
•• ••
••
••
•••
•••
••
L'2 ..•
•
••
•• •
•
•• •
L1 (0 dollars)
(500 000 dollars) Figure 6.B.5(d)
6.B.6 U(p) = Max{p·c e IR: c e C} = - Min{p·c e IR: c e - C}. Hence U( ·) is equal to - J.l_c( • ), the support function (Section 3.F) of - C multiplied by - 1, where the domain of the support function is restricted to N
the simplex {p e IR : } p +
is convex.
(A
"'n n
= 1}.
Since any support function is concave, U( ·)
more direct proof is possible, which is essentially the same as
the proof of concavity of support functions in Section 3.F.) As ar. example of a nonlinear Bernoulli utility function, consider A = 8 = {1,2} and d~fine
u U) = u (2) = 1 and u 1
2
then U(L) = Max {p ,p }. 1 2
6.8.7
2
(1) =
u (2) = 0. 1
Let L = (p ,p l,
(This is essentially the same as Example 6.8.4. l
Sin::e the individual prefers L to L' and is indifferent between L and
xL and bet·.vee!:: L' and XL' by Proposition l.B.Hiii), he orefers • the monotcnicity, this this is equivalent to XL > xL,.
6.C.l
1 2
If « :::: D > 0 (complete insurance), then
6-5
Bv •
- q(l - n)u'(w - o:q) + n(l - q)u'(w - D + o:(l - q)} •
= - q(l - n)u'[w - Dq)
+ n(l -
q)u'(w - Dql
= u'(w - DqHnU - q) - q(l - n}) < 0 = u'(w - Dq)(n - q) < 0
by q > n.
Thus the first-order condition is not satisfied at o: = D.
Hence
the individual will not insure completely . •
6.C.2
(a) Let F( ·) be a distribution function, then
Ju(xldF(x} = j((3x
2
+ 7XlclF(x)
= (3(mean of F>
2
+
2
= tJJx dF(x) + 7IxdF(x) tJ(variance of F)
+
7(mean of F).
(b) We prove by contradiction that U( ·) is not compatible with any Be:-noulli
utility function.
So suppose that there Is a Be:-noulli utility function u( · )
such that U(F) = Ju(x)dF(x) for every distribution function F( ·).
Let ·x and y
be two amounts of money, G( ·) be the distribution that puts probability one at x. and H( • l be the distribution that puts probability one at y. u(x)
= U(G)
u(yl = U(H)
Thus, x
:!:::
= (mean of G) - (variance of G)
=x
- 0
= x.
= (mean of H) - (variance of Hl = y - 0 = y.
y if and only if u(xl
2:
u(y).
Hence u( ·) is strict:'·• ;;·onatone.
let F ( ·) be the distribution that puts probability one on 0 and 0 distribution t."'!at puts probability 1/2 on 0 and on 4/r > 0. and the va:-ia..."'l.ce of F ( ·) are zero, U(F 1 = 0. 0 0 u( ·) thus implies that U(F) > 0. 2
vai"iance is 4/r . contradiction. ..... l
Then
F( ·)
1\iow
be the
Since the mean
The strict monotonicity of
However, the mean of F( ·) is 2/r and the
Hence U(F) = 2/r - r(4//) = - 2/r < 0, which is a
Hence U( ·} is not compatible with any Bernoulli utility
.. . tmc .. lon. An example of two lotteries with the property requested in the exercise
was given in the above proof of incompatibility.
6-6
(Note that if all we need to
show were the incompatibility of U( ·) and any Bernoulli utility function, the equality u(x) = x obtained above would be sufficient to complete the proof, because this implies the risk neutrality, which contradicts the fact that the variance of F( • l is subtracted in the definition of U( • ). )
6.C.3
Suppose first that condition (i) ·holds.
Let x
and
E IR
> 0.
E:
Let F( ·)
be the dist:;ibution that puts probability 1/2 on x - c and on x + c, and F ( ·) be the distribution that puts probability 1/2 - n(x,c,u) on x - c and
c
1/2 + n(x,c,u) on x + c.
That is,
F(z) =
0
if
l/2
if
c
(z)
=
1/2 -
X -
£ ::S Z
1
or Ju (x}dF(x)
1
Suppose next that
i!:
-x.
If Ju Cx)dF(x) 2
Cx),
i!:
u (c(F,u 1L 1 2
Since
u (c(F,u J) and hence c(F,u J 2
1
1
c(F,u ). 2
6.C.7
Sucoose first that condition (iii) holds. ••
Let x e IR and
~
> 0.
Denote
by F{ ·) the ciis!ribution function that puts probability 1/2 - n(x,c,u l on 2 x -
~
and 112 +
n(x.~.u >
2
~-
on x +
That is,
0
F(z) =
if
1/2 ... n(x,c,u l 2
By (iii), c(F,u ) 1
if i!:
x.
x.
=-
::5
+
E[v(x
+ y)J.
and t/J"(x) = u"(w - x) + E:v"(x
+ y)).
0 for every x, which impiies that if
Now, since E[v'Cx
0
u.'(w - x ) + E[v'Cx + y)) 0 0
6-12
+ y)]
=-
> v'Cx ), 0
v'Cx ) 0
+ E!v'Cx
0
+ y))
> 0.
•
(b) Define two functions 'rl ( ·) and 'rl £ ·) by 'rl {x) = - vi(xl and TJ (x) = 1
Tnen 71 £ ·) and 11
- vz.(x).
1
2
2
1
2
are increasing and the coefficients of absolute
C·)
prudence of v ( ·) and of v £ ·) are equal to the coefficients of absolute risk 1 2 aversion of 11 ( · ) and of 11 £ • ).
Thus, if the coefficient of absolute prudence
2
1
of v ( · ) is not larger than that of v { • ) , then the coefficient of absolute 1 2 risk aversion of 1'1 ( ·) is not larger than that of 1'1 £·). . Moreover, since 1 2 E(viCx
0
+ y)]
> vi(x }, we have Ei71 Cx0 0
1
Proposition 6.C.2 to 11 £·) and 1
Eivi(zo
+
1)
2
+
y)} < lJ (x J. 1
0
£ • ), we obtain EflJ tx
2
0
Thus, by applying + y)]
< 11 Cx l. 2
0
Hence
yll > v2(zol.
The implication of this fact to part (a) is that, if the coefficient of absolute prudence of the first is not larger than that of the second, and if .
the risk y induces the first individual to save more, then it also induces the second to do so.
Hence coefficients of absolute prudence measure how much
individuals are willing to save when faced with a risk in the future.
> 0, then
(c) If .,.."'(x) •
aver-s1on.
Tl"(x)
= - v'(x)
0.
+
a-(x)z)x)u'((l - ;(x) + 7(x)z)x)
Note also that if z > 1, then (1 - ;(x) + ;(x)z)x >
Since the coefficient of relative risk aversion is increasing,
this implies that rR((l - a-(x) + ;(x)z)x) > r R(x).
Hence
- u"((l - 7Cxl + a-(x)z)x)(I - a-(x} + ;(x)z)x
> rR(x)u'((l - a-(x) By
a-Cx)z}x).
+
z - 1 > 0, - u"((l - a-(x) + a-(x)z)x)(l - ;(x) + a-(x)z)x(z - 1} > r R (x)u' Ul - a-(x) + J(X)z)x)(z - 1). •
We can sim.Ea!"ly show that this last inequality also holds for every z < 1.
- -· . .
Th ,._,.'o'"e .
- Ju .. (,t
- 7\XI +
> Jr R{x)u'((l - 7{x)
7(X)z)x)(l - a-(x) + 7(x)z)x(z - l}dF{z) +
a-(x)z)x)(z - l)dF(z)
= rR(x)Ju'((I - a-(x) + 7(X)z)x)(z- lldF(z) = 0 by the first-order condition.
(a) {First printing
6.C.l2
< 1 ar..d negative if p > 1. ... na ?. . ,
• •
:..,. .a
u ( x) = 1Qx .~
1-p
+
relative risk aversion p.
· The coefficient {3 should be positive if p This makes u( ·) increasing.) with p
~ 1
It is easy to check
and ; e IR, then u( ·) exhibits constant
Suppose conversely that u( ·) exhibits constant risk
aversicr! p, then u"(x)/u'(x) = - pix.
Thus ln (u'(x)) = - pln x + c 1 for some •
6-15
Hence u(x) = (exp c )x
1-p
1
for
2
Letting f3 = ( exp c )/(1 - p} and 7 = c , we complete the proof. 1 2
i?..
E
/(1 - p) + c
if u(x) =
(b) It is easy to check that,
~ln
x + '1 with f3 > 0 and 7 e ~. then
u( · J exhibits constant relative risk aversion one.
The other dir-ection can be
•
shown in the same way as in (a).
(c) By L'hopital's rule, [' 1-p . 1ll:lp-+l \X - 1)/(1 - p}]
Let
6.C.13
rt( • )
=
= ln x.
be the profit function and F( ·) be the distribution function
of the random price.
Since
Jensen's inequality.
n( ·)
is convex, Jn(p)dF(p}
?!
n(JpdF(p)) by
But the left-hand side is the expected payoff from the
uncertain p:-ices and the right-hand side is the utility of the vector.
price
Tnus the firm prefers the uncertain prices.
Define a functior! g( • ) by g(o:) = ko: + v(u
6.C.l4
ex;>e::~ed
+ v(x) = u•(x).
-1
(et)},
then g(u(x}) = ku(x)
It is thus sufficient to show that g( ·) is concave.
For
tnis, in tu:-n, it is sufficient to prove that (vou -l)( ·} is concave. Let c::, ,'3 e !R and A e [ 0.11. is convex.
Thus u
Since
v( •
v{ ·
-1
(Aet + ( 1 - i\.){3) :s i\.u -l(o:) + (1 - i\.)u
-1
(8).
l is nonincr-easing, this implies v(u
Since
Since u( ·) is increasing and concave, u
-1
~(i\.et + (l
- i\.)(l)) ~ v(i\.u
-1
(o:.) + (I - i\.)u
-1
((l)}.
l is concave,
•. 1 is a necessary
condition for the demand for the risky asset to be strictly positive.
In the following answers, we assume that the demands for both assets are always positive.
(c) Since the prices of the two assets are equal to one, their marginal utilities must be equal. tru'(x
Thus
+ x a> + (1 - u]u'(x + x bl = uau'(x + x al + (1 - n)bu'(x + x bl. 2 2 1 1 1 1 2 2
That is, n(l -
This and x
1
+
x
2
a)u'(x + x al + (1 - wH1 - b)u'(x + x b) = 0. 1 2 2 1
= 1 constitute the first-order condition.
(d) Taking b as constant, define
then
< 0, + (l -
< 0.
Thus. by the implicit function theorem (Theorem M.E.l), dx Ida = 1
84>/Ba Bf/>1Bx
< 0. 1
(e) It follows from the condition of (b) that b > 1, that is, that a is the worse outcome of the risky asset.
Thus, if the probability n of the worse
outcome is increased, then it is anticipated that the demand for the riskless
6-18
asset is increased.
(f} Since b
> l,
Bf/>/Brr = (1 - a)u'(x + (1 - x)a) - (1 - b}u'(x + (l - x)b)
= (1 - a}u'(x because a < 1 < b.
6.C.16
+ (1 -
x)a) + (b - l)u'(x +
Thus dx/dn = -
8(/>/Bn
x)b) > 0,
(l -
> 0, as anticipated.
Throughout the answer, we assume that u( ·} is continuous, so that the
urn and the minimum are attained. (a) If the individual owns the lottery, his random wealth is (w + G, w Thus the minimal selling price R
s
+
B).
is defined by
pu(w + G) + (1 - p)u(w + B) = u(w + R ). s (b) If he buys the lotter-y at price R, his random wealth is (w -
R + G, w - R + B).
The maximal buying price Rb is defined by
pu(w - Rb +
p}u(w - Rb +
G) + (1 -
= u(w}.
B)
However. if u( · } exhibits
(c) In general, these two prices are different.
constant absolute risk aversion. then they are the same. •
two eauat1ons car. be restated as c •
w
= w + R
s
and c
In
R
w- .
~~a.-:t.
the above
= w, where c
D
w
•
are defined as in (iii) of Proposition 6.C.3.
According to the proposition,
the constant absolute risk aversion implies that w - cw = (w - Rb) - c w-R. . 0
T."l.is is equiva!er:.t to Rs = Rb.
(d) By a direct calculation,
R
is one
o~
s
= 5[(7 - 4v'3lp
2
+
(4v'3 - 6)p + 11.
the solutions to the quadratic equation •
6-19
and -
""w-R
t
3
(1 -
6.C.l7
- 10(2p
+
7p
2
- 8p + l)Rb - 25(23p
2
- 54p + 29) = 0.
According to Exercise 6.C.l2, if u( ·) exhibits constant relative risk
aversion p, then u(x) = {3x 1 assume u(x) = ax -P. argument.
1-p
+ 0
or u[x) = /3ln(x} + 0 .
In this answer, we
The case of {3ln(x) + 'r can be proven by the same
Let's first consider the portfolio problem of the individual in
period t = 1. after a realization of the random return has generated wealth
wr
level
Denoting the distribution function of the return by F[ ·). his
problem is
As
dis~ussed
in Example 6.C.2 continued (and also in Exercise 6.C.ll), we can
show that the solution does not depend on the value of w . 1
If he chooses portfolio o:
solution by a;•. at t = 1 is w
1
=
((1 - o: )R 0
0
Denote the
at t = 0, then his random wealth
Given the solution o:• at t = 1, his
problem in period t = 0 is
1-p
Since the distributions of ..x and x2 are independent and u(x) = ax· 1
• we can
rewrite the obje:tive function as
Since the firs:
int~gral
does not depend on the choice of a. , the solution is 0
This completes the proof. For the case of a utility function exhibits constant absolute risk aversions, the absolute amounts of wealth invested on the risky asset may vary over the twc pe:-iods t = 0,1, but those in period t = 1 do not depenc! or, the realization cf
a • • - .&• L
._
-
To see this, let u(x) = - f3e-P.
1.5
6-20
The individual's p:-oblem
The solution t~:-ns out to be independent of the value of w , and hence of x . 1 0 Denote the solution by a•.
If he chooses portfolio a
random wealth at t = 1 is w
1
= (w
at t = 0, then his
0
•
- o: )R + o: xl" 0 0 0
Given the solution o:• at t
= l, his problem in period t = 0 is
Since the distributions of x
1
and x
are independent and u(x) = - {3e -p, we can
2
rewrite the objective function as
Since the first integral does not depend on the choice of o: , the solution of 0
this maximization problem is the same as the solution of the p;oblem of •
•
•
max1m1zmg
But the la';te:- is the same as what the consumer would choose at t = 1 if his coefficient of absolute risk aversion is equal to Rp. Now. if R = 1, then the consumer invests a constant absolute a:::1cunt of wealth over two periods.
Thus, their proportions out of the total wealths are •
larger i!' the total wealths are smaller.
on
\\r ..
.!.
•
ana her.ce on the realization
So, the proportion
IX
/\\' now depends 1 1
Hence the proportions can no longer be
constant.
6.C.l8
(a} A cii:-ect calculation shows that the coefficient of absolute risk
aversicr: at w = 5 is 0.1.
Exercise 6.C.l2(a) shows that the coe!'ficie:1t o:
relative :-isk ave:-sion is 0.5, which is constant over w.
(b} By a direct calculation, the certainty equivalent is 9 and the probability
premium is (v'TQ - 3}/2.
6-21
(c) By a direct calculation, the certainty equivalent is 25 and the probability premium is
em -
5)/2.
For each of these two lotteries, the difference between the mean of the lottery and the certainty equivalent is equal to one.
However, the
probability premium for the first lottery is larger.
This is because
u( ·)
exhibits constant relative risk aversion and hence decreasing absolute risk •
avers1on.
6.C.l9
Foe each n, denote by f3
n
the wealth invested in risky asset n.
wealth invested in the riskless asset is then w - Eo.Sn·
The
If the individual
random consumption is x =
13 lr + l' (3 z , where z denotes the "'"'n n ""n n n n
return of asset n.
(w - l'
By
linearity of normal distributions, x is a normal distribution with mean (w -
Lr,.Snlr
+
Lni3nf.ln and variance ,S·V,S.
£[- exp (- ax)].
The expected utility from x is
But this is equal to the value, multiplied by - l, at - c: of
the moment-generating function of the normal distribution with mean (w -
t_.s k ,_ n
f3 f.l and variance '-r: n n
+ l'
(3 • V,S.
·Therefore,
By appiying the monotone transformation u -t (- li1Xlln(- ul to this utility
function, we obtain ({w - )
f3
'nn
)r + ) (3 Jl
'"'nnn
l + (,S·Vf3)a/2.
The first-orde!"' condition for a maximum of this objective function with respect to
,a
gives the optimal portfolio
,a•
= a -lV-l(J..L - re), whe:-e e is the
vector of iRr-,; whose components are all equal to one.
6.C.20
For each e:
~ 0,
let F [; ( ·) be the distribution function of the lotterv•
6-22
that pays x
+
c with probability 1/2 and x - c with probability 112.
Then,
c(F ,u) is defined as the solution to the equation £
(I/2)u(x + c) + 0/2)u(x - c) - u(c) = 0
with respect to c.
Hence, by the implicit function theorem (Theorem M.E.l),
c(F ,u) is a differentiable function of c and c (l/2lu' (x +
c) -
(l/2)u'(x - c} - u'(c{F ,u))(oc(F ,u)/8cl = 0.
c
By putting c = 0, we obtain 8c(F ,u)/8c = 0. 0
E:
Also, by further differentiating
the left-hand side of this equality with respect to c. we obtain (112)u"(x + £) + (l/2)u"(x - c) - u"(c(F ,u))(oc{F ,u)/8d c £
2
2
2
- u'(c{F ,u))(a cCF ,u}/8c ) = o. c £
Thus, by putting c = 0 and substituting 8c(F ,u)/8c = 0, we obtain 0
2 2 u"(x) - u'(c[F ,u))(o c£F ,u)/8c ) = 0.
c
c
G( · ) be thei:- distribution functions. (a) If a Be:-noulli utility function is increasing, then there exists p e (0,1) .
su:::h that the decision maker· is indifferent between the sure outcome of $2 and the lotte:-y tha-: pays $1 with probability p and $3 with probability 1 - p. Thus, the indifference line that goes through the $2-vertex must hit some poir.t en the (Sl.$3)-face (excluding the vertices) and all indifference lines must be par-allel to it.
Conversely, this condition implies that the Bernoulli
utility functio!! increasing.
By varying p vary from 0 to 1, we can identify
the area of the lotteries that are above all indifference curves going through L.
The a:-ea is shaded in the following figure:
6-23
$3
L
•
$1
$2 Figure 6.D.l(a)
Thus, G( ·) first-order stochastically dominates F( ·) if and only if L' is •
located above the segment that goes through L and is parallel to the ($1,$2)-face and also above the segment that goes through L and parallel to the ($2,$3)-face.
(b) The distribution G( · ) first-order stochastically dominates F( • ) if and only if p
1
2:
Pi
equivalent to p
3
and p
1
+ p
2
2:
Pi
+
P2·
Since the second inequality is
:s Pj· G( ·} first-order stochastically dominates F( ·) if and
only if L' is located in the shaded area in the figure below:
$3
L
$2
$1
Figure 6.D.l(b) 6-24
(Fi:-st printing
6.0.2
a
: The phrase "the mean of x unde:- F( · ), fxdF(x),
a
•
exceeds that under G( · ), JxdG(x)" should be "the mean of x unde; G( ·), JxdG(x), cannot exceed that under F( · }, JxdF(x)". the two means should be allowed.)
x and apply Definition 6.0.1.
That is, the equality of
For the first assertion, sir.tply put
u(x)
=
As for the second, let p e (0,1/2) and consider
the following two distributions: 0 F(z)
=
G(z) =
p 1
0 1
if if 0 if 2
~
z < 0, z < 2,
~ Z,
if z < 1, if 1 :5 z.
Then F(l/2} = p > 0 = G(l/2) and JxdF(x} = 20 - p) > 1 = JxdG(x).
Hence F( ·)
does not first-order stochastically dominate G( • ), but the mea."'l of F( ·) is larger than that of G( ·).
Any e1ementa..-y increase in risk from a distribution F( ·) is a mean-
6.0.3
prese:-ving spread of F( · }.
In Example 6.0.2, we saw that any mean-preserving
spread of F( ·) is second-order stochastically dominated by F( • ).
Hence the
asse::tion follows.
Let L
6.D.4
lotteries.
•
(a) Bv a cirec: ca.lc'.liation, the means of L and L' are 2 - p
-
+
p'
Pj·
3"
1
~ p
3
and 2 - Pi
"f p - p = p• T.'1Us the two lotteries have an equal mean if and only t 1
3
1
Hence they have an equal mean if and only if they are beth on a segment
that is pa:-alle! to the segment connecting the $2-vertex and the middle point of the ($1,$3)-fa::e, as depicted below:
6-25
$3
··-.....••
•••
•• •
••
••
• ••
••
L
·...... •. _
··-..... •• •
••
$2
$1
Figme 6D.4(a)
(b) If the decision-maker exhibits risk aversion, then he prefers getting $2
with probability one to the lottery yielding $1 with probability 112 and $3 with probability 1/2.
Hence the indifference lines are steeper than the
segment connecting the $2-vertex and the middle point of the ($1,$3)-face. Hence, when L and L' have an equal mean, L is preferred to L' if and only if L is located on the right of L'.
Therefore, L second-order stochastically
dominates L' if and only if L is located on the right of L', as depicted in the figure below:
$3
•
.... ... ... ••
L
l
•• •
••
•. •
L •• •
• ••
.........
•
• ••
• ••
•.
$1
$2 Figure 6.D.4(b) 6-26
(c) The distribution of L' is a mean preserving spread of that of L if and only if they are both on a segment that is parallel to the segment connecting the $2-vertex and the middle point of the ($1,$3)-face, and L' is closer to the ($1,$3)-face than L.
This is depicted below:
$3
•• ••
•• ••
L' ·-..•
••
••
·.,_ L $2
$1
Figure 6.D.4(c) (d) Inequality (6.0.1) holds if and only if Bu~.
Pj - p
3
Pi
iii:
P1 and Pi
+
(pi
+
Pz)
?!
P + 1
since L and L' are assumed to have an equal
and hence these two inequalities are equivalent to
Pi
iii:
p
1
alone.
Thus, (6.0.1) holds if and only if L is located in the right of L', as depicted belO\\':
$3
•
•• •
•• •
L' .....
••
\
·.••
•
'·.,_ L
$2
$1
Figure 6.D.4(d)
6-27
Denote by R(x,x') the expected regret associated with lottery x
6.E.l
relative to x', and similarly for the other lotteries.
A direct calculation
yields: R(x.x') = 2/3, R(x' .x) = 13/3, R(x' ,x") = (,.12 + 1)13. R(x",x') = l's/3, • R(x",x} = Cv'2 + 1}/3, R(x,x") = v'213. Thus, x' is preferred to x·, x" is preferred to x', but x is preferred to x" .
(a) Denote the probability of .state s by tr
6.E.2
5
and the expected utility
from the contingent commodity vector (x ,x l by U(x ,x l, then U£x ,x ) = 1 2
1 2
n u(x l 1
1
+
n (1- w)u(x ). 2
2
1 2
Since u[·) is concave by the assumption of risk
aversion, U[ ·) is also concave.
Thus the preference ordering on Cx ,x ) is 1 2 •
convex. (b) According to Exercise 6.C.S(a), the concavity of U( ·) implies the risk
aversion for the lotteries on (xl'x l. 2
(c) By the additive separability of U( ·) and Exercise 3.G.4(c}, both x
1
and x
are normal goods.
6.E.3
Since g•(s) = 1 + a:(g(s) - 1) for every s, we have ~(s)
>
g(s)
If g(s)
1. Thus G*(x} :s G(xl for every x < 1 and c•(x) G( ·) and
~ G(x)
for every x > 1.
c•( · l are continuous from the right, we have
c•(l) ~ G(ll.
Since Hence
property (6.D.2} holds and thus G•( ·) second-order stochastically dominates G( · l weakly.
(If g(s)
:;c
1 for some s, t}_len G•(x) < G(x) for some x < 1 and
6-28
2
G•(x) > G(x} for some x > 1. stochastically dominates
Hence, in this case, G•( ·) second-order strictly.)
G( ·)
We shall first prove the uniqueness of the utility function on money up
6.F.l
A
Suppose that two utility .function u( · l and u( ·} satisfy
to origin and scale.
the condition of the theorem.
Since the state preferences
-
...
...
~
-s
are represented
by both J(n u{x l + (3 )dF ·ex ) and J(n u(x ) + f3 )dF (x ), by apolying s 5 s ss s 5 s ss . Proposition 6.B.2 to the set of all the lotteries in some state s. we know that n u( • ) + f3
s
...
s
-
~
and n u( ·) + 13
s
s
are the same up to origin and scale.
Hence
so are u( · ) and u( • ).
It remains to verify the uniqueness of subjective probability.
...
that both 1 n (Ju(x )dF (x )) and ) x L-ss s ss '"'ss preference relation on !f..
(Ju(x
Suppose
)dF (x )) represents the same s ss
-u( ·)
Now that we have shown that u( ·) and
are the
same up to c:-igin and scale, without loss of generality, we can assume that u( ·)
= u( · J.
We can normalize u( ·) so that u(O)
=0
and u(l)
= 1.
Note here
that if a cistribution function F ( ·) puts probability p on 1 and probability s s 1 - p
s
on 0, then the expected utility is p .
Thus, by choosing p
s
•
s
suitably
for ea:::h s. any point in [ O,l}s_ can be represented in the foim
.
'
A
Hence, if (n:, ... ,n l 5
~ (n
-
, ... ,ns), then there would exist CF ..... F l e 5 1 1
!f.
S
and (Fi·· .. ,F l e ~ such that 1 n (Ju (x )dF (x )) > ) n (Ju (x )dF' (x )),
L-ss
ss
ss
L-ss
A
[ n
ss
This
contraC.i::~s -
ss
ss
A
(Ju (x )dF (x ))
ss
ss
= Cp + p )tr(l), Pr(T > = (p + p ) tr(r), Pr(T > 1 8 5 2 3 6 4
= Cp5
= Cp7
+ p ) tr(l), 6 + p ) tr(r),
8
Pr(T ) = (p + p ) tr(L}, Pr(T ) == (pll + p l tr(L), Pr(T l = (p + p l tr(r), 9 5 10 6 7 9 10 12
PrCT > = Cp + p J cr(r). 11 12 8 The following behavioral strategy for player 1 is realization equivalent: At the root of the game, player 1 plays L,
~·
R with probabilities of (p
+
1
P + P + P l. Cp + p + p + p l and Cp + p + p + p J respectively; at 4 5 1 6 3 2 8 9 12 10 11 information set 2, player 1 plays x, y with probabilities of (p
5
+ p ll(p
6
+
5
p + P + P > and Cp + p )/(p + p + p + p ) respectively; at information set 3, 8 7 8 7 6 5 6 8 7
player 1 plays x, y with probabilities of (p (p
11
(d)
+ p
12
l/(p
9
+ p
10
+
p
11
+
p
12
>
9
+
p
10
)/(p
9
+ p
10
+
p
11
+ P
12
>
and
respectively.
Note that if player 1 reaches his (only) information set after player 2
moves, he will not remember whether he chose M or R.
Thus, the game is not of
perfect recall. The result of part (b) still holds: there exists a mixed strategy for player 1 which is realization equivalent to any behavior strategy.
7 - 3
Suppose
player 1 uses the following behavior strategy: At information set 1, player 1 plays L, M, R with probabilities of p , p 1 2 and p
respectively; at information set 2, player 1 plays x, y with
3
probabilities of q
1
and q
2
respectively.
If player 2 is using the mixed
strategy tr, then the probability that we reach each terminal node will be: Pr(T ) = p , PrCT ) = p trU) q , Pr(T > = p _cr(l) q , Pr(T ) = Pz cr(r) q , 0 2 3 1 1 2 2 1 2 1 PrCT ) = p cr(r) q , PrCT ) = p tr(l) q , PrCT ):::; p cr(l) q , PrCT l = p 2 4 2 5 3 1 6 3 2 7 3 cr(r) q , Pr(T ) = p tr(r) q . 8 3 1 2 The following mixed strategy for player 1 is realization equivalent: (L, x) with probability p , (M, x) with probability p q , 1 2 1 (M, y) with probability p q , (R, x) with probability p q , (M, y) with 2 2 3 1 probability p q . 3 2 However, there does not always exist
behavior strategy that is
O. l
strategy
cr~,
We claim that CT. is strictly dominated by the mixed
1
1
which is equivalent to CT. except that instead of playing s! with
l
1
l
probability cr.(s!} it plays cr! with probability CT.(s!). l
l
1
u.(CT., cr .) 1
l
This follows since:
I
=I: CT (s.) u.(s., CT .)
1
-1
=s . :r:i!s.• l
u.(s!, s .) 1
1
-1
1
1
-1
I
\:1 s!e S. and \:1 s . I
I
-1
E
S .. In -1
•
particular, u.(CT.,s .) > u.(s~.s .) \:1 j = 1, ... , N. This implies that I
N
1
-1
I
.
I
-1
.
u.(CT.,s .) > .I: [CT.(s~) u.(s~.s .)) = u.(CT.,s .) , a contradiction. 1 1 -1 J= l 1 1 I -1 1 1 -1
S.C.l
1
Notice that the elimination of strategies that are never a
best-response is more demanding than strictly dominated strategy elimination. Thus, in every round of elimination, the deletion of never a best response deletes more strategies than the deletion of strictly dominated strategies.
8-4
Therefore, if the elimination of strictly dominated strategies yields a unique prediction in a game, then the elimination of strategies that are never a best-response cannot yield more than one strategy.
Since a rationalizable
strategy always exist, the elimination of strategies that are never a best-response will then also yield a unique prediction. If the unique rationalizable strategy is not the unique prediction after
elimination of strictly dominated strategies, then there exist a round of elimination in which this unique rationalizable strategy was strictly dominated.
However, if this strategy was strictly dominated it was also
a best-response. rationalizable.
8.C.2
n~ver
This contradicts the assumption that the strategy is Therefore, both procedures must yield the same prediction.
Call the set of strategies for pl"lyer i that remain after N rounds of
deletion of never best-response strategies best-response to any strategy in deleted in the N+l round.
w£:.1
Suppose s. is never a 1
~ -1.. Suppose further that s.1 will not be
Since s. was never a best response to a strategy in 1
.
-N+l
., it will clear 1y not b e a best-response to a strategy m J.. -1
..N
. s= J.. -1
••
-1
Thus this strategy will be deleted in the next round.
8.C.3
Suppose that
s
1
is a pure strategy of player 1 that is never a best
response for any mixed strategy of player 2. not strictly dominated. tr.
1
Suppose in negation that s
1
is
Construct the following correspondence for any
e t.(S.) for i = 1, 2: 1
A
gl(sl, ,.2) ~ gl(,.l' The first part of this correspondence is the best response function for player 1 and therefore satisfies all the conditions of the Kakutani fixed point theorem.
The second part of the correspondence is the set of mixed
8-5
strategies of player 2, for which s
1
is not strictly dominated (it is a
non-empty set since s is not a strictly dominated strategy. i.e., it is not 1
strictly dominated by cr l.
Therefore, the second part of the correspondence
1
is convex valued and upper hemicontinuous due to the usual assumptions.
Thus,
by Kakutani's theorem there exists a fixed point (cri ,cr2 ) of this correspondence such that g (s , cr2) a: g (cri, cr2l from the second part of the 1 1
correspondence, and g
Ccrt• cr2)
1
1
!:
g (cr , cr2,) 1 1
g (s , cr2) a: g Ccr , cr2,) for all cr 1 1 1 1 1 that
s
1
E ~cs
1 ),
for all cr
1
e
~(S ) ..
1
Therefore,
which contradicts the assumption
is a pure strategy of player 1 that is never a best response for any
mixed strategy of player 2.
Therefore, if s
1
is a pure strategy of player 1
that is never a best response for any mixed strategy of player 2, then s
1
is
strictly dominated by s011e mixed strategy of player 1.
8.C.4
(First Printing En-ata: a typo appears in the lower left box of the
payoff matrix.
Player l's payoff should be
1t
+ 4c and not 11 + 4£.]
For the continuation of this answer, a strategy for player 2 is to play u with probability a and D with probability 1-a, and for player 3 is to play l with probability f3 and r with probability 1-(3.
Denote by P A the
expe.~ted
payoff of
player 1 when action A e {L,M,R} is taken given a and (3. Direct calculation .
and simple algebra yield:
p M = n + ( 3cx ; 3{3-
3a{3 - 1 ) TJ
PL = n + (2(3 - l)c PR =
n +
(1 - 2(3)c
(a} To show that M is never a best response to any pair of strategies of players 2 and 3, (cx,(3), we have three cases: Case
t.
{3 > 112
apll Note that in this case
aa
= 1J[3/2 - 3(3) < 0.
8- 6
Thus the highest payoff for
and his payoff will be p (a.=O) M
player I if he plays M is obtained when a. = 0, =
1t
3
+ 11[~ - ll
7« + (1-2«)2.
mixed strategy equilibrium a probability since playing a
2
1
and
~
Therefore, in a
cannot both be played with positive
would give the player a larger payoff.
Suppose, there exists a mixed .strategy equilibrium in which player 1 plays a
1
and a
2
with strictly positive probability.
Clearly, player 2's best
response to this strategy of player 1 does not involve playing b
3
with
strictly positive probability (given the strategy of player 1, playing b strictly better for player 2). and b
2
2
The payoff for player 1 from playing a
yields: 5(3 + (1-(3)3 > (1-(3)2.
mixed strategy equilibrium a
is
Thus player 2 will play b with probability f3 1
with probability 1-(3.
equals: (l-tJ)2, playing a
2
1
and a
positive probability since playing a
2
2
1
Therefore, in a
cannot both be played with strictly is always better.
8-8
Similarly, it can be shown that there exists no mixed strategy equilibrium in which a
2
and a
are both played with strictly positive
3
probability. Therefore, player 1 always plays a Player 2 will then play his best resp.mse b . 2
2
in a Nash equilibrium.
Thus Ca , b ) being played with 2 2
certainty Is the unique mixed strategy equilibrium.
We will show that
8.D.2
any
Nash equilibrium (NE) must be in S
00 ,
the set.
of strategies which survive iterated strict dominance. Since it is assumed that this set contains one element, this will prove the required result. Let (si,sz·····sjl be a (mixed) NE and suppose in negation that it does not survive iterated strict dominance.
Let i be the player whose strategy is
first ruled out in the iterative process (say in the kth round). Therefore, there exists
and a. such that
fT. 1
1
k-1 u.(cr.,r. .) ,.. u.(a.,s .) 'tJ s .e S . , and a. 1
1
-1
is played with positive probability s!(a.). 1
1
1
1
-1
-1
-1
1
Since k is the first round at
which any of the NE strategies, Csi,s2·····sjl. are ruled out, we must have k-1 that s•.e S . . -1
derived from playing u.(s~
Hence, u.(fT'.,s•.) > u.(a.,s•.). Let the strategy
-1
fT •• 1
1
s~
1
-1
1
1
-1
s~ 1
be
execpt that any probability of playing a. is replaced by
1
1
We thus have that:
,s• ) = u.(s!,s• ) + 1 1 -1
1 1 -1
s~(a.) 11
· [u.(cr. ,s• ) - u.(a.,s• JI > 1 1 -1 1 1 -1
u.(s~.s• ) 11-
1
which contradicts the assumption that (si,s2·····si) is a NE.
First of all, notice that the first auction bid is a simultaneous move
8.D.3
game where a strategy for a player consists of a bid. Player 1, and b
( i)
2
Let b be the bid of 1
be the bid of Player 2.
lf b > b , Player 1 gets the object and pays b for it; 2 1 1 Player 2 does not get the object.
8-9
Thus, in this case:
.
u lb ,b l=v -b (l's valuation of the object 1 1 2 1 1 minus what he has to pay for it].
If b = b , each player gets the object with probability 1 2 '1 1 (vl-bJ) 0 = ul(bl + b2) = 2(vl - bl)+ 2 .
(iii)
~:
z·
Similarly:
u (b , b ) 2 1 2
=
(v2 - b2) 2
Therefore, we have for i,j e {1,2}, i 0,
J
I
1
2 (vi - bi) • bl (v.- b.),
(a)
j:
< b.
b.
1
:;e
•
= b2
b.> b. 1 J
We claim that no strategy for player 1 is strictly dominated.
negation that b
1
Suppose in
is strictly dominated by bl. i.e., for any b : u (bl ,b l > 2 2 1
u Cb ,b ). Take b2 =max {b ,bl} + l,then: b2 > b , and b2 > b}· 1 1 2 1 1 u (hi ,b2l = u (b ,b2) = 0, a contradiction. 1 1 1 1 is strictly dominated.
Hence,
Therefore, no strategy for player
Similarly, one can prove that no strategy for player
2 is strictly dominated, and thus no strategies are strictly dominated.
(b)
We now claim that any strategy b for Player 1 such that b > v , is 1 1 1
weakly dominated by v r Note that, if b > v l: 1
(i)
If b
2
2, u (l,O) - u (0,0) = (v -l) 1 1 1 1 (ii)
If b
= 1:
2
u C0,1) = 1
(iii)
If b
Thus, in all cases ui(l' b ) 2
!:
2
(v
1
- 1) > 0
0
u 0,. b ) = u (0,b l 1 1 2 2
> 1:
2
I
u Cl, 1) = 1
1 > 0
= 0.
u (0,b ), with strict inequality in some cases. 2 1
Finally, suppose that v e {1,2} . We claim that, in this case, 1
b = v - 1 weakly dominates b = v : 1
1
1
(i)
If b
(ij)
If b
(iii)
If b
2 2 2
1
< v - 1:
u Cv -1,b ) ::: 1 > 0 = u Cv ,b ) 1 1 1 1 2 2
= v -1: 1
uCv -1, b l = l/2> 0 1 2
> v -1:
u(v -l, b ) ::: 0 = u Cv ,b ) 1 1 1 2 2
1
1
= ui (vi,b2 l inequality in (i)
and (ii).
Similarly, it can be shown that: -if v
2
-if v2
> 2, b :!!::
2
= 1 weakly dominates bl = 0
1, b2::: vz-1 weakly dominates b' z= v2
8-11
(c)
Define the best response correspondence for Player 1 as the set of
maximizers of l's utility, given the strategy for 2.
We already have the
expression for u Cb , b ); in order to find out the best response 1 1 2 ce~rrespondence,
all we have to do is maximize this function.
Denoting this
best response by R (b ), direct maximization of u(b , b } yields: 1 2 1 2 if b
( (b2+ 1),
If
{b2' b2+1}.
if {b2}' {O,l.2, ... v }, if 1 {0. 1 • 2 •...• b2-1 ) if
< v -2
1 b = v -2 2 1 b = v -1 2 1 b v1 2 b > v 2 1 2
=
Similarly, if R Cb ) is the best response correspondence for Player 2: 2 1 {bl + R Cb ) = 2 1
< v2 - 2
•
if
b
{bl' b1+1}.
if
{bl}
if
b = v - 2 1 2 b v2 - 1 1 bl = v2
1}
,
(0,1, ... , v }, if 2 {0,1, ... bl-1}. if
1
=
bl > v2
A Nash equilibrium is a pair (bi,b2l where b~ e R Cb2)• and b2 e R Cbi)· It 2 1
can be verified that a NE always exists, and, for sufficiently large of v and v 1
2
Nash
values
equilibria are not unique.
More explicitly, it can be shown that, for v , v 1 2
2 the following are Nash
equilibria:
(i) (ii) (iii)
(iv)
If v = v : (vl' v ), Cv -1, v -l), (v -2, v -2) 1 2 2 1 2 1 2 If v = v +l: (v -1, v -2), Cv -2, v -2) 1 2 1 1 1 1 If v = v +1: (v -2, v -1), Cv -2, v -2l 2 2 2 1 2 2 If v > v +1: (v -x, v -x-ll, with 1 1 1 1 2
x
:=s
v - v
:=s
1
2
Note: if v = v +2: (v -2, v -2) is also a NE. 1 2 1 1 (v)
> v +l: (v -x-l, v -x), with 1 1 2 2 2
If v
8-12
:=s
x
:=s
v
2
- v
1
Note: if v
2
= v +2: Cv -2, v -2) is also a NE. 1 2 2
Thus, generally, uniqueness of NE does not hold, although existence does.
(a) If player i demands y
8.0.4 x
~
0 is payoff equivalent.
~
100, then any strategy of player j with
Therefore, there exists no strictly dominated
strategy.
(b)
Any strategy demanding more than $100 is weakly dominated,
Case
t.
player 2 demands y
payoff equivalent. Case
~
!:
100. Then any strategy of player 1 with x
~
0 is
player 2 demands 0 ::s y < 100. Then, player 1 could
demand x = 100 - y and would obtain a payoff of 100 - y.
Demanding x > 100
will give player 1 a payoff of 0. Therefore, any strategy demanding more than $100 is weakly dominated.
(c) Any pair (x, 100 - x) with 100
~
x
!:
0 is a pure strategy Nash equilibrium
of this game. Proof: Suppose, player 1 demands x with 100
!:
y = 100 - x, his payoff will equal 100 - x
0.
2:
x
!:
0.
If player 2 demands
If player 2 demands y > 100 -
x, the demands sum to more than $100 and both players get 0.
If player 2
demands 0 ::s y < 100 - x, he will obtain his demand and therefore be· worse off .
than if he would have demanded 100 - x.
Thus, if player 1 demands x, player
2's best response is to demand y = 100 - x.
Similarly if player 2 demands 100
- x, player 1's best response is to demand x.
8.0.5
(a)
Vendor 2.
Let x
1
be the location of Vendor 1 and x
2
be the location of
Thus, we can associate a strategy for Player i with x. e [0,1]. 1
First, let us find out the payoff function for each of the vendors.
Since the
price of the ice cream is regulated, we can identify the profit of each vendor
8-13
Suppose that x < x • In this case, 2 1 xl + x2 all consumers located to the left of (below) will purchase from 2 XI + X2 Vendor 1, while all customers located to the right of will buy ice 2 with the number of customers s/he gets.
cream from Vendor 2. ul (xi. x2)
Thus: xl + x2
-
xl + X 2]) ( ., length of [0, 2
2
XI + x2 U2(Xl' x 2 > = I 2
We can derive a similar result for x
= 21.
XI + x2 ,1]) ( • length of [ 2
2
<x : 1
• • Thus, summar1zmg:
1
-2 ' 1 - xl+ x2 ' xl
< x2
1 2
=X
"2
u2(xl' x2)
-
--
• xl
2
xl + x2 2
' xl > x2
It is straightforward to check that x.l = x firm can do better by deviating). x
1
= x2
< 1/2.
2
= 1/2 constitutes a NE (no
To show uniqueness, suppose first that
Then any firm can do better by moving by
since it will sell almost 1 - x it can-be shown that x
1
= x
2
1
£
> 0 to the right,
> 1/2 units rather than l/2 units.
> 1/2 does not constitute
8-14
'i
Similarly
NE. Suppose now that
x
1
< x 2 . Then firm 1 can do better by moving to x 2
this could not have been a NE.
£,
with
> 0, therefore
£
Similarly It can be shown that x
1
>x
2
does
not constitute a NE.
(b)
Suppose that an equilibrium (xi, x2, xjl exists.
xi • x2 = xj.
Then each firm will sell 1/3.
Suppose, first, that
But any firm can increase its
sales by moving to the right (~f xi = x2 = xj < 1/2) or. the left (if ·xi = ~ = xj
!:
1/2), a contradiction.
let's say xi = x2. xi + c.
If xi = x2 < xj. then firm 3 can do better by moving to
If xi = x2 > xj. then firm 3 can do better by moving to xi -
contradiction. points.
Suppose that two firms locate at the same point,
£,
Finally, suppose that all 3 firms are located at different
But then the firm that is located the farthest on the right will be
able to increase its sales by moving to the left by c > 0, a contradiction. Thus, there exists no pure strategy NE in this game.
Case !: u > w and 1 > y.
8.0.6
In this case player 1 always plays his dominant strategy a . 1
Player 2 will
play his best response to this strategy, i.e. if v > m he will play b , if 1 v < m he will play b Case
~:
2
and otherwise he will be indifferent.
u < w and 1 < y.
In this case player 1 always plays his dominant strategy a . 2
Player 2 will
play his best response to this strategy, i.e. if x > z he will play b , if 1 x < z he will play b
2
and otherwise he will be indifferent.
""C"'"a..=s.: :.e J,: v > m and x > z. In this case player 2 always plays his dominant strategy br
Player 2 will
play his best response to this strategy, i.e. if u > w he will play a , if 1 u < w he will play a Case 4: ==-
2
and otherwise he will be indifferent.
v < m and x < z.
8-15
a
In this case player 2 always plays his dominant strategy b . 2
Player 2 will
play his best response to this strategy, i.e. if 1 > y he will play a , if 1 1 u.(s., ,. .) = u.(s.,CT .) for all« e (0, I
I
-1
I
I
-1
1).
1 2 This contradicts the fact that s. and s. are each best response to CT .. I
I
-1
Therefore, any mixed strategy NE of this game must be degenerate.
8.1;).9 (a)
Playing L or R is quite risky, since we do not know what player 1
will be playing.
The risk of obtaining a payoff of -49 is very large compared
to the payoff of 1 if player 2 played L, and the risk of obtaining a payoff of -100 is very large compared to the payoff of 2 if player
~
played LL or R.
Therefore, it seems "reasonable" to play M.
(b)
The two pure Nash equilibria of this game are (U,LL) and (O,R).
To
check for mixed strategy Nash equilibrium, player 1 must mix between U (with probability p) and D (with probability 1-p).
Player 2 then has 11 possible
mixing combinations: {LL,U, {LL,M}, {LL,R}, {L,M}, {L,R}, {M,R}, {M,L,R}, {LL,M,R}, (LL,L,R}. {LL.L,M}, and {LL,L,M.R}. We will show that only the first
combination. (LL,L}, is part of a mixed strategy NE. For player 2 to mix between LL and L (with probabilities q c;:1.d 1-q p·(2)+(1-p)·(-l00) = p·(l)+{l-p)·(-49) which
respecti.vely), we must have that
The utility of player 2 from each strategy is then: u CLU =
2
Then, for player 1 to mix between U and 0, we must have: q =
~·
and
q·(IOO)+(l-q)·(-100) = q·(-100)+(1-q)·(IOO) which yields
u (U) = u CD> = 0. Therefore, p = 1 1
'2i
and q =
strategy NE. For the rest of the answer we call this "NE•".
~
is a mixed We now show that
no other mixing combination of player 2 can be part of a mixed strategy NE. (i} If player 2 mixes with the combination (LL,M}, we must have
p =
~~
, which gives utilities
u (LU = ui(M) = 0, and
2
8-18
1
= Si > 0, so
this cannot be part of a mixed strategy NE . .
(ii) If player 2 mixes with the combination {LL,R), we must have
p
= 4. which gives
utilities
u CLU 2
= u 2 (R) = - 49.
and
u (M) 2
= o.
so this
cannot be part of a mixed strategy NE. (iii) If player 2 mixes with one of the combinations {L,M}, {L,R}. {M.R).
{M,L,R), then player 1 will have 0 as a strict best response, which in turn has R as player 2's strict best response. (iv) If player 2 mixes with one of the combinations {LL,M,R}, {LL,L,R}, or (LL,L,M,R), then the analysis of (ii) above implies that this cannot be part of a mixed strategy NE. (v) If player 2 mixes with the combination {LL,L,M} then the analysis of (i) above implies that this cannot be part of a mixed strategy NE.
(c)
The choice in part a) is not part of auy NE described above.
to see that strategy M is rationalizable: If plyer 1
play~
p =
It is easy
1
2 then M is
the unique best response of player 2.
(d)
If preplay communication is possible, the players can agree to play one
of the pure strategy NE, which are payoff equivalent and Pareto dominant for both players. Therefore, player 2 will play either LL or R depending on the agreed upon equilibrium.
8.E.l
There are four pure strategies contingent on the type of player:
AA: Attack if either weak or strong type, AN: Attack if strong and Not Attack if weak, NA: Not Attack if strong and Attack if weak, NN: Never attack. The expected payoff of each pair of strategies can be easily computed and are
8-19
given in Figure B.E.l:
Player 2
AN
AA AA
NA
M 3M s+w -w ----2 4 2 -4 4 2 4
M s+w M s+w M s+w
-4 2 • -4 2
AN
M s+w M --4 2 • -2 4
Player 1 NA
-w 3M s+w 2 •4 4
- --
I
M-s
M-s I
4
'
M s --2 4
I
M 2
4
M-w
O,M
4
0
M 10
I
'
I
NN
NN
M --2 4
M-w
I
•
I
4
M-w M-w 4 4 I
0
M 1 2
-
M 210
M 210
0,0
Figure 8.E.l Any NE of this normal form game is a Baysian NE of the original game. Case 1: M > w >
-
S1
and w > M/2 > s
From the above payoffs we can see that (AA AN) and (AN~AA) are both pure 1
strategy Bayesian Nash equilibria. Case 2: M > w > s, and M/2 < s
-
From the above payoffs we can see that (AA,NN) and (NN,AA) are both pure strategy Bayesian Nash equilibria. Case 3: w > M > s, and M/2 < s
-
From the above payoffs we can see that (AN,AN), (AA,NN) and (NN,AA) are pure strategy Bayesian Nash equilibria. Case i: w > M > s, and M/2 > s From the above payoffs we can see that (AA,AN), (AN,AA) and (AN,AN) are pure strategy Bayesian Nash equilibria.
8-20
8.£.2
(a) Suppose that all the bidders, use the bidding function b(v). that
is if their valuation is v they bid b(v). The expected payoff for a bidder whose valuation is v. is given by: 1
Cv -bCvi))·Pr(b(v ) > b(vj)} + O·Pr{bCv ) < b(vj)}. (Note that we ignore a tie 1 1 1 since it is a zero probability event given that b(v.) is a monotonic linear 1
function.)
Since both players use the same monotonic linear bidding function
then Pr(b(v.) > b(v .)} = Pr(v. > v.} = v./ 1 J 1 J 1
-
uniformly distributed on [O,v].
v (since the
valuations are
b(v) will in fact be the equilibrium bidding
function if it is not better for a player to pretend that his valuation is different.
To check this let us solve a bidders problem whose valuation is v.
1
and who has to decide whether he wants to pretend to have a different
-
valuation v'. The bidder maximizes: (v. - b(v))·(v I v), and the F"OC is:
-
(v. - b(v))lv 1
1
- = 0.
-
b' (v) vi·•
b(v) is an equilibrium bidding function if
it is optimal for the bidder not to pretend to have a different valuation, that is, if v
= v.1
is the optimal solution to· the above FOC, i.e., If
(v. - b(v. ))lv - b' (v.) v ./v = 0. 1
1
1
This is a differential equation that has to
1
be satisfied by the bidding function b(v) in order to be an bidding function.
equilibriu~
The solution to this differential equation is b(v) = v/2.
Thus a bidder whose valuation is v will bid vl2 (a monotonic linear function).
(b)
We can proceed as above by assuming that all bidders use the same bidding
function b(v). ( v.) 1
1-1 1 v. -
Now, Pr{b(v.) > b(v.)"' j 1 J
Proceeding as in· a) above,
~ i}
= = Pr{v.> 1
v."' j J
~ i}
=
we get the following differential
equation: (I -1 )( v.
- b( v. ) )( v.) 1 1 1
1-2 /v
1v
b'(v.) (v.) 1- 1
differential equation is b(v) =
I
1-1
1
1
· v.
As I ...
= 0. The solution to this m,
b(v) =
1-1
1
.
• v ... v, 1.e., as
the number of players goes to infinity each player will bid his valuation.
8-21
8.E.3
= H or
A firm of type I
L, will maximize its expected profit, taken
as given that the other firm will supply qH or qL depending whether this firm is of type H or L.
A type i e {H,L} firm 1 will maximize: 1
2
1 c.)q.]
1
Max (1-IJ)[{a - b(q.+ qH) - c.).:(.) + IJ.Ha 1 1 1 1
1
1
qi
The FOC yields:{l-IJ)[a -
2 2 b(2q~+ qH ) - c.) + IJ[a - b(2q~+ qL ) - c.) 1 1 1 1
'l'b .
. Bayes1an ' Nas h equ1 1 r1um: qH1 a symmetr1c
=
2 qH
= QH
-
an d
QLl
= 0.
In
2 = QL = qL ·
Plugging this into the F.O.C we get the following two eqations: (1-IJ}[a - 3b qH - cH] + IJ[a ~ b(2qH+ qL) - cH) = 0, {1-IJ)[a - b(qH+ 2qL) - cL] + IJ[a - 3b qL - cL) • 0. Therefore, we obtain that qH = [a - cH +;eeL - cH>]
·it; ,
qL = [a - cL + l;IJ(cH - cL))
8.F.l
·it; .
For the proof of Proposition 8.F.1, we refer to: Selten, R. (1975) "Reexamination of the Perfectness Concept for Equilibrium Points in Extensive Games," InternationaL JournaL of Game Theory.
Another good source is Section 8.4 in Fudenberg & Tirole, (1991) Game Theory, MIT press.
8.F.2
For the solution of this question we refer to: van Damme. E. (1983). Refinements of the Nash Equili.brium Concept. Berlin: Springer-Verlag (pp. 28-31).
8-22
8.F.3
For the proof of this statement, we refer to: Selten, R. (1975) "Reexamination of the Perfectness Concept for Equilibrium Points in Extensive Games," International Journal of Game Theory.
Another good source Is Section 8.4 in Fudenberg & Tirole, (1991) Game Theory, MIT press.
8-23
CHAPTER 9
9.8.1
There are 5 subgames, each "ne beginning at a different node of the
game (this includes the whole game itself).
9.8.2
(a)
Clearly if
fT" o
r E'
is a Nash equilibrium of
and
rE
is the only
•
proper subgame of
r E'
then 0
r E"
Thus, by definition
(b)
Assume in negation that
t1'
induces a NE in every proper subgame of the game
fT" o
is a subgame perfect NE of 0
t1'
r E"
is a subgame perfect equilibrium
of
r E'
but
it does not induce a subgame perfect Nash equilibrium in every proper subgame of
r E"
Then there exists a proper subgame (say, HE) of
restriction of
t1' o
to
nE
is not a SPNE.
rE
This implies that there exists a
proper subgame (say, nE) of nE in which the restriction of NE. ~
in which the
fT" o
to ~ is not a
Since nE is a proper subgame of nE and ITE is a proper subgame of is also a proper subgame of
proper subgame of
9.8.3
rE
r E'
Therefore,
0
t1'
r E'
then
does not induce a NE in a
- contradiction.
Let player l's pure strategy be s e 20, even if it has a monopoly position. not benefit by staying in the market if t > 25. the last period in which A could still be in.
Similarly, firm B will
Consider period 20.
This is
If 8 continues to stay in, it
will obtain monopoly profits during periods 21 through 25 of: (51-2·21) + (51-2·22) + (5!-2·23) + (51-2·24) + (51-2·25) = 25. If firm A were to stay in at period 20, then 8 will obtain its duopoly profits of 10.5 - 20 = -9.5 in period.
The net profit for firm 8 of staying in at
period 19 is therefore 25 - 9.5 = 15.5 > 0.
Thus, at period 20 firm B will
stay in no matter if firm A is in the market or not.
9 - 9
This implies that if
firm A stays in at period 20 it will make a loss of (lOS - 10 · 20) = - 95, which ln turn implies that firm A will stay out from period 19 on. The above argument can be applied backward and we will get the (unique) SPNE of this game.
However, the following reasoning applies: The analysis
above shows that with rational play firm 2 will exit after firm 1. Therefore, as far as firm 1 is concerned, it will never make monopoly profits. It follows that we only need calculate the last period in which firm 1 makes positive duopoly profits and this will be the last period in which both firms enter the market. In later periods, firm 2 will be in the market alone (until period 25). The unique SPNE is: For t = 1,2, ... , 10 12, ... , 25 A is out an B stays in; for t
9.8.12
i!::
both firms stay in; in t = 11,
26 both firms are out.
The unique SPNE strategies are as follows:
The offering player i offers (
ca .-a.a .l
o-a. > J
o-a.a.l I J
·v
J
'
'J·v].
ct-a.a.> 1 J
accepts if and only if he is given a share of at least
The offered player j
ca .-o.o .) 1
J · v. To show
J
o-a.a .l I J that these strategies form a SPNE, suppose that the offered player j does not
0-o. l accept the offer.
1
Then he will offer - - --·v in the next period and the
o-a.a .) 1 J
current offered player i will accept (if j offers less then i won't accept
given the strategies). Player j's payoff will be
a• J
Therefore he did not benefit by deviating. deviates and uses a different strategy. offering more than
c 1-a.a I
I
I
o-a.I aJ.>
v=
J
1
J ·v.
c1-o.a .) 1 J
Suppose the offering player i
He clearly does not benefit by
ca .-a.a .) 1 J
ca .-a.a .l
o-a.1 >
J • v to player j.
.) J
9 - 10
Suppose then that he offers less.
ca.-a.a .) 1 1
Player j will not accept, and offer - -- --=J:... • v to player i in the next
0-a.a .l 1 J Player i will then accept and obtain a payoff of
period.
a.2 · 1
c1-a i )
c1-a . > · v < ___,J"-- • v. o-a .a.) c1-a. a .) 1
J
1
Thus he will choose not to deviating from the
J
above strategy.
9.8.13
In the exercise with the cost of delay c being equal for both
players, there will generally be many SPNE of this game. For an analysis of this case we refer to: Rubinstein, A. (1982) "Perfect Equilibria in a Bargaining Model,"
Econometrica, 50:97-109 (pp. 107-108).
9.8.14
(a)
The extensive form ot the game is depicted in figure 9.B.l4(a)
below (the letter "d" is used instead of
a).
Simple backward induction leads
to the unique SPNE which is shown by arrows in the figure: Firm E enters at t=O, and always plays In thereafter. Firm I plays Accommodate for all t=l,2,3.
(b) The extensive form of the modified game is depicted in figure 9.B.l4(b). Using Backward induction, firm I will always play Accommodate in period t=3, and therefore if t=3 is reached, firm E will play ln. This causes firm I to choose Fight in t=2 since this causes firm E to exit the market, and due to condition A.2 we have that for firm 1: Z
+
ay
+
a2X = Z
+
cS(y
+ ox)
>
Z
+
a(l
+
a)z = Z
+
az
+
a2Z
•
This causes firm E to choose Out in t=2. Working backward we get that at t=l, firm I chooses Accommodate and firm E chooses ln. E at t=O depends on the value of k. enter, and if k < 1 then E will enter.
However, the choice of firm
If k > 1 then firm E will choose not to For k = 1 both are part of the
9 - 11
(unique) continuation SPNE, so there are two SPNE in this case.
T
Enter
t=O
l
T I I t=l
I I
i
T
I I t=2
I I
-1- k 2 y+dx+d
i
T I I t=3
1+d -k 2 +dz+d
1- d -k
+dy+d
-1-d -k
-1 +d -k y+dz+d\
2
y+dy+d
2
I I
Ft.i 2
2
2
2
2 l+d+d-k -1+d-d-k 1- d+d- -1-d-d-k 2 2 2 2 2 +dy+d z z+ y+ "i y+dz+d 2 +dz+d y +dy+d +dy+d
1- d + d -k 1- d- d -k
Figure 9.B.l4(a)
9 - 12
2
T
t=O
l
T I I t=l
I I
'
E Out
T I I t=2
I I
-1- k 2 y+dx+d
'
-
E Out
T I I t=3
I I
1- d -k
l+d -k 2 z+dz+d x
2
z+dy+d x
i 2
2
1+d+d-k 1+d-d-k z+dz+cfz z+dz+d)
Figure 9.B.l4(b)
9 - 13
9.C.I
First, if tr is a NE then (i) and (ii) must hold.
If (i) weren't
satisfied, then some player's information set is reached with positive probability in which he is not playing a best response to his opponents' strategies, contradicting the fact that tr is a NE.
If (ii) were not satisfied
then some player's information set is reached with positive probability in which his beliefs are not correct given his, and his opponents' strategies. Therefore. this cannot be an equilibrium. Second, if (i) and Cii) are satisfied, then tr is clearly a NE since at each information set which is reached with positive probability, sequential rationality implies that each player is playing a best response to his opponents' strategies, with correct beliefs at each such information set.
Let tr f" be the probability that firm I fights after entry. let
9.C.2
firm I's belief that ln tr , cr
1
1
be
was E's strategy if entry has occurred and let tr , 0 1
denote the probabilities with which firm E actually chooses Out, In , 1
2
and In
~-t
2
respectively.
2 As in example 9.C.3, we cannot have ,.,. < 3 (same analysis). We could, however, 1 2 have JJ. > j : In this case firm I will play "Fight" with probabi.iity 1, and 1 having 7 types v
!:
v• will buy the good in period 1, in equilibrium we must have that
the price y was set to maximize the second period profits: y • Pr{v!:y I v 0, there is
a unique Weak Perfect Bayesian Equilibrium (WPBE). Since every Sequential Equilibrium (SE) is a WPBE, this is the only candidate for a SE. sequences of strategies be: For firm I, (CT~,cr~) = ( . n n n 121 farm E. (CTo· CT 1' CT 2) = (ii, 3 - 2n
11
• 3 - 2n ) •
2
1 + 1'
Let the
1 1 -) and for n ' n '
--
where n = 1,2, ....
Clearly,
these strategies converge to the WPBE strategies described in the textbook. The sequence of strategies for firm E generates a unique sequence of beliefs that firm E played In as follows: 1 2
n J.ll
=
1 3 2n 4n - 3 ~2--;1;--';'""1--:-1- = 6n - 6 + 3 2n 3 2n
-
-
--
Thus, the WPBE described is a SE.
9 - 17
n
2
CD
3"
u (s,).)
p
~n-cp
for
s~~n-cd
s
for
s~~n-cd
-~n-c
=
d
-s
for
s~~n-cd
for
s~~n-a
Therefore, Ms. P's expected payoff for an offer s will be given by (recall that beliefs for Ms. P are determined b)
1
(s-cd)/n
E~up(s.~)
=
f(~)):
(s-cd)/n
I (~n-cp)f(~)d~ I sf().)d~ I (~n-cp)f(~)d~ =
+
0
(s-cd)/n
+ s(l - Fr:cdH
0
Ms. P will choose s to maximize this expression, and the FOC (assuming an interior solution) yields the pure strategy of Ms. P who solves:
(i)
(b)
Implicit differentiation of (i) above yields
(ii) -
which will determine the change in the optimal value of s given changes in any of the parameters.
For example, if
~
were uniformly distributed on the
interval [0,1] then the expression in (ii) reduces to s = n - c
p
so that we
ds ds have-= 1, dc = -1. and dn p
(c)
For an analysis of this part see: Nalebuff, B. (1987) "Credible Pretrial Negotiation," Rand Journal of £conomi.cs, 18(2): 198-210.
9.C.S
For an analysis of this problem first see exercise 9. C. 4 above (for a
guideline on how to solve such a problem), and then consult: Nalebuff, B. (1987) "Credible Pretrial Negotiation," Rand JournaL
9 - 18
of EconomLcs. 18(2): 198-210.
9.C.6
Example 9.C.3: Since the example restricts attention to 7 > 0, there is
a unique Weak Perfect Bayesian Equilibrium (WPBE). Since every Sequential Equilibrium (SE) is a WPBE. this is the only candidate for a SE.
Let the
1 sequences of strategies be: For firm I. (crFn"tr'An) • c - .!. , .!.). and for 2 + 7 .n n . n n n 121 11 ftrm E. (t1' , crl" t1' ) = (n. n , j - Zn ) , where n • 1,2..... Clearly. 3 2 0 2 these strategies converge to the WPBE strategies described in the textbook. The sequence of strategies for firm £ generates a unique sequence of beliefs that firm E played In as follows: 1 2 n Ill
1
3-2rl
4n - 3
n
2
- 6
CD
J"
= ..2-~1..::..._.......;:;1~--:1- = 6n 3-2ri+3- -2n
Thus, the WPBE d"!scribed is a SE. Example 9.C.4: The WPBE described In the textbook cannot be a SE since the only beliefs consistent with any mixed strategy of player 1 is 2's information set.
(~.~)
in player
Thus the unique SE must have player 2 believe that if
his information set is reached then each node is as likely as the other. causes him to play "r" if his information set is reached (since
~·10
+
This
~· 2 >
5), which in turn induces player 1 to play "y". This is the uniq\.e '::[ (which is also a WPBE). Example 9.C.S: The WPBE described in the text can be supported as a SE. Let a strategy of firm E be (tr'
0
,t1' F"tr'A)'
the probabilities of playing (Out),
(ln,Fight). and (In,Accommodate) respectively. (c:r f'tr' a),
Let a strategy of firm I be
the probabilities of playing Fight, and Accommodate respectively (if
reached).
Consider the following sequences of str·ategies:For firm E,
Z1 ),
. n n 1 and for fu·m I, (tr' f'tr' a) = ( 1 -
n , .!.n ).
n
These sequences converge to the strategies described, and furthermore generate
9 - 19
the belief sequence:
IT - n-1 n
n-1 2 n
n
1 +2 2 n n
CIO
1.
This supports the described strategies as a SE. Another SE (and WPBE) is where firm E plays Cln,Accommodate) and firm 1 plays Accommodate. This is supported by the sequences of strategies: For firm
.!._
' 2n '
9.C.7
(a)
1 -
n
1 n 1 - !). and for firm I, (CT f'CT a) = (n' n ' n
!)
The extensive form game is given in figure 9.C. 7(a).
R
Player 2 D 4 2
1
5
1
1
2 2
Figure 9.C.7(a)
The set of pure strategies for player 1 is S
= (B,T),
1
and for player 2 is
(DD,DU,UD,UU} where playing AB means playing A at node v
s 2=
and B at node v . 3 2
By backward induction it is easy to see that the unique SPNE is (B,DU).
There
are two more classes of NE: (i) Player 1 plays T, and player 2 plays UU with probability p and DU with probability 1-p, with p
?:
~
,
and (ii) Player 1
plays B, and player 2 plays DU with probability p and DD with probability 1-p, . h p
Wlt
~
J1 .
9 - 20
(b)
In this case the extensive form game, and its equivalent normal form
game, are depicted in figure 9. C. 7(b).
Player 2
T
Player
D
U
5,1
2,2
I~----~----~
B
4,2
1,1 4 2
5
1 1
2 2
1
Figure 9.C.7(b) Since playing T is a strictly dominant strategy for player 1, we have a unique NE: (T,U).
(c)
The modified game is depicted in figure 9.C. 7(c). I
1-p
....- ..- ..-
-4
2
1 1
4
2
_....,- '--=' -
-
2 Io
-
-
-
..........
-----2
Ir
u 1
1
........ ........
........ ........
u
D
s 1
2 2
s 1
2 2
Figure 9.C.7(c) Ik denotes player 2's information set after she observes k e {B,T}, r is the probability she assigns to the event that player 1 played B after she finds herself in information set 1 . and similarly q is the probability she assigns 8 to the event that player 1 played B after she finds herself in information set
9 - 21
lr·
Let s e [0,1] denote the probability that player 1 plays B.
We
can
three possible situations in a WPBE: First, player 1 playing s = 1;
have
second, player 1 playing s = 0; and third, player 1 playing s e (0,1). 1 playing s
=1
cannot be part of a WPPE.
Player
Indeed, if this were the case we
must have q = r = 1, which implies that player 2 will always play D. But given that 2 always plays D, player 1 will prefer to deviate and play T. player 1 playing s = 0 is part of a WPBE.
Second,
Indeed, if this is the case we must
have q = r = 0, which implies that player 2 will always play U, and given that 2 always plays U, player 1 will prefer to to play T.
Thus, player 1 playing T
and player 2 playing U in each of her information sets is a WPBE. To consider the possibility of a WPBE with s e (0,1), we first note that this will induce a unique pair of probability beliefs q and r derived by Bayes rule.
In particular, in such an eC'•Jilibrium we must have: r
=
s·p -o-(1::--_-s..... )(,....,l--:....._p.,....)_+_s_·_p ' and
Simple algebra shows that if and only if
> 1 r = < 2'
> s = p
u (x ) k
k
k
k
2:
u (O),and therefore k
x' "' 0. Therefore, we must have x' > 0 for at least one commodity s. k
ks
Let
x" = x'i I for i "' k, l "' s·, II x"
Is
x" kl
= x'is = x'k I
x' ~ = x' ks
ks
+ lj(l-1)
£
for i'ltk;
for l "' s; - e.
Intuitively, we are redistributing a small amount c of commodity s from consumer k to all the other consumers, so that they become better off than under the original allocation (x,y). Observe that as long as 0 < e < (x", y') will be a feasible allocation. I
5
By strong monotonicity of
preferences,· we must have u (x") > u (x') I
x~ ,
I
!:
I
u Cx ) for i. I
I
;t
k, for any
£
> 0.
By continuity of u ( · ), we must have u (x") > u (x ) for c > 0 small enough. k
Therefore, we can find an
k
£
k
k
k
> 0 small enough such that the corresponding
10-1
allocation (x", y') is feasible and makes every consumer strictly better off. Therefore, the original allocation (x,y) is not weakly Pareto efficient. Together with part (a) above, this implies that the notions of strong and weak Pareto efficiency are equivalent.
(c) [First printing errata: omit the assumption of interiority, it plays no role given that X = X I
Let 1
= 2•
= 1•
L
X1
= IR for all t.]
2
+
= X 2 = IR +
(no production). Then (x
1
,
x
2
)
1
u (x ) 1
11
1
0, u 2 (x2 )
= x 2,
w > 0, Y
=-
IR +
= (wj2, wj2) is a weakly Pareto efficient
allocation, but it is not strongly Pareto efficient. Since consumer l's preferences are not strongly monotone, we cannot reallocate utility from consumer 1 to consumer 2 to make every weak Pareto improvement a strong Pareto improvement, as we did in part (b).
10.8.2.
Suppose that the allocation
vector p
•
e IR
L
(X
• ... , X,• 1 I
and price
I
-
constitute a competitive equilibrium. Then
•
•
•
•
1
I
1
J
the allocation (x , ... , x , y , ... , y ) and price vector a.p
•
(a. > 0) also constitute a competitive equilibrium:
(i) Profit maximization: solves
Max p Y e y
•
•
solves
J
j
Max a.p • ·y J y e y J
a. Max
Y j e YJ
J
(ii) Utility maximization: Consumer i's new budget constraint is •
•
a.p · x
I
J
~ a.p · w
L e lj
+
I
J =I
•
•
•
(a.p · y ) ~ p · x j
i
~
• p ·w
J I
+ "e L. lj
J =I
which is consumer i's old budget constraint. Therefore,
x
I
solves
Max
x e X I
u (x) s. t. the new budget constraint I
I
I
10-2
~
•
p ·y
J
x solves
Max . u I (x) s.t. the old budget constraint. I
I
X
1
EX
1
I
•
(iii) Market clearing: t"' l.. XII 1=1
=wI +
J • t"' y l.. lJ j=l
for each l = 1, •••• L - does not
depend on prices at all.
lO.C.l.
(a) The consumer solves L'
L
I:
Max
log x
1=1
rpx
s. t.
1
1•1
l
l
:s w.
The first-order condition for the Lagrangean of this program can be written as
x • A/P , l=l, ... ,L, where A > 0. Substituting in the budget constraint, we 1
I
find A
= wfL,
therefore the demand function can be written as x (p,w) = Lw. 1
The wealth effect is
(b) As
L ---+
10.C.2.
(a)
m,
ax (p,w)jaw l
the wealth effect
p
1
=
ax1(p,w)jaw
-+ 0.
The consumer solves s. t. p x + m s w . The
Max a + (:J ln x + m
m
(x,m)
first-order condition (assuming interior solution) yields x(p) = fJjp. The firin solves Max pq - trq. q~O
The firm's first-order condition (assuming interior solution) is p = tr.
(b) From the two first-order conditions and the consumer's budget constraint,
the competitive equilibrium is p
10.C.3.
•
•
::: ,., x
•
= (:Jjtr, m • = w
m
- (:J.
(a) Assuming interior solution, the first-order condition is
c'(q ) = A J J
> 0 for all j.
10-3
J
L c'
(b) C'(q) =
J
J=l
•
J
(q ) dq fdq •
d(
J
J
J
• • r ;\ dq fdq = ;\ r q > fdq = ;\ dqfdq = ;\. J J J J
•
=1
=l
•
Therefore, C'(q) = c'(q ) for all j. j
j
(c) Each firm j solves Max pq - c (q ). q j j j J
The first-order condition (assuming interior solution) is c'(q ) = p. J j If p = C'(q), then we have c'(q ) = C'(q) for every j. If c'( ·) is strictly . J J J •
increasing for all j, we must have q • q j
• j
(q) -
the solution to the central
authority's program in part (a) for total output q. Therefore, J
J
Lq
Lq
=
J=l l
•
(q)
J=l
= q. In other words, if the market price is C'(q), then the
j
industry produces q. Therefore, C'( ·) is the inverse of the industry supply function.
(a) The central authority's problem can be written as
lO.C.4.
I
Max
EXI
s. t.
(X , • • , X )i!:.O I
::s
X.
I= 1
I
Assuming interior solution, the first-order condition is
•) = ;•ex I I
A > 0 for all i. I
(b) 7'(x)
= L ;·I
•
•
i
I
(x ) dx fdx
I
•
L i\
=
1=1
dx 1/' ldx = i\ d(
1:1
I
•
Lx I
) /''dx
= ;\
dxldx /'
= ;\.
1=1
•
Therefore, r'(x) = ¢'(x ) for aJI i. .i
I
(c) Each consumer solves Max X
A.
¥'1
(x ) - Px. I
i
I
The first-order condition (assuming interior solution) is
;·ex) = I I
P.
If P = r'(x), then we have ;'(x ) = 7'(x) for every i. If ;'( ·) is strictly .
I
I
decreasing for all i, we must have x
I
I
= x • (x) - the solution to the central I
authority's program in part (a) above for total consumption x. Therefore, I
I
•
L x 1 = Ex 1
1=1
(x)
= x. In other words, if the market price is 7'(x), then the
1=1
10-4
aggregate demand is x. Therefore, 7'{ ·) is the inverse of the aggregate demand function.
lO.C.S.
The system of equations (lO.C.4)-(lO.C.6) here takes the following
form:
•
9'>'(x ) 1
= p
I
•
•
+
t,
i = 1....... I,
•
•
j = 1, •••• J.
c'(q ) = p ' ~ J. J •
rx
I =1
=rq.J
I
J =1
•
•
•
These equations describe the equilibrium (x • q , p ) as an implicit function of t. Differentiating with respect to t, we get 9'>"(x • ) x •'(t)
= p •'(t) + 1,
i = 1, ••. , I,
c"(q • ) q • '(t) ~ J J J
= p • '(t),
J = 1, ••.• J'
1
I
I
• r x 1' =
1=1
• r q '(t).
J=1
J
• • • This system of linear equations should be solved for (x '(t), q '(t), p '(t). I
•
J
•
-
This can be easily done, for e.:ample, by expressing dx /dt and dq /dt from the 1
J
first two sets of equations and substituting into the third equation.
We
obtain I
•
(p '(t) + 1)
r {¢;'(x:>r
1
= p
•
]
'(t)
r {c"(q J =1
I =1
J
•
-1
)J .
J
•
From here we can express p '(t): I
• p '(t) =
[ r¢;'(x:>r
1
I =I
I
[ 1=1
•
J
f¢~'(x ~ n-1- [ fc:'(q ~JJ-1 I
I
J=l
J
J
Compare to the expression on page 324 of the textbook.
10.C.6.
(a) If the specific tax t is levied on the consumer, then he pays p+t
for every unit of the good, and the demand at market price p becomes x(p+t).
10-5
The equilibrium market price p
c
is determined from equalizing demand and
supply: x(p
c
+ t)
c
= q(p ).
On the other hand, if the specific tax t is levied on the producer, then he collects p-t from every unit of the good sold, and the supply at market price p becomes q(p-t). The equilibrium market price pp is determined from
equalizing demand and supply: x(pp)
= q(pp
- t).
It is easy to see that p solves the first equation if and only if p+t solves the second one. Therefore, pp = pc + t, which is the ultimate cost of the good to consumers in both cases. The amount purchased in both cases Is x(pP) = x(pc + t).
(b) If the ad valorem tax
Y
is levied on the consumer, then he pays
(1+-r)p for every unit of the good, and the demand at market price p becomes x((l+-r)p). The equilibrium market price pc is determined from equalizin-g demand and supply: c
c
x((l+-r)p ) = q(p ).
(1)
On the other hand, If the ad valorem tax T is levied on the producer, collects (1-T)p from then he pays (l+T)p for every unit of the good sold, and the supply at market price p becomes q((l-T)p). The equilibrium market price pP is determined from equalizing demand and supply: p
p
x(p ) = q((l--r)p ).
Consider the excess demand function for this case: z(p) = x(p) - q((l-T)p). Since x( ·) is non-increasing and q( ·) is non-decreasing, z(p) must be non-increasing. From (1) we have
10-6
(2)
c
z((I+-r)p )
= x{(l+T)pc) - q((I-T 2 )pc)l!:
c
!: x((l+T)p ) -
c
q(p )
= 0,
taking into account that q( • ) is non-decreasing and using (I). Therefore, z((I+y)pc)
!:
0 and z(pP)
= 0.
Since z( ·) is non-increasing, this
implies that (l+T)pc s pp. In words, levying the ad valorem tax on consumers leads to a lower cost on consumers than levying the same tax on producers. (In the same way it can be shown that levying the ad valorem tax on
con~umers
leads to a higher price for producers than levying the same tax to producers). If q( • ) is strictly increasing, the argument can be strengthened to obtain the strict inequality: Cl+T)pc < pp. On the other hand, when the supply
-
is perfectly inelastic, i.e. q(p) = q = const, then (1) and (2) combined yield c
x((I+T)p ) •
-
p
q • x(p ), and therefore p
p
c
=
.
Cl+T)p • Here both taxes result m
the same cost to consumers. However, the producers still bear a higher burden p
when the tax is levied directly on them: (1--r)p = (1--:){l+T)p
c
c
1/2
1
1
1
)
+ 1/Z q,icx;> for every i. - k, and
1/Jk(xk) +
1/2 t/Jk(x~).
Adding up all the inequalities, we obtain
10-12
J
I
I: tP 1(x") 1
I: c
-
I
J =1 J
1 =1
1/2 ( l>t~ (x')
>
(q")
J
I =1 1
J
I: c
-
J =I
1
l
1/2 [ r f/dx 1
+
l =1
Cq')] + J J
J
r c (q )]. J•l J J
) I
which contradicts the assumption that (x , •. , x , q , ... , q ) and (x', .. , 1
I
J
1
1
x', q', ... , q') are both solutions to (10.0.2). I
I
J
Therefore, Individual consumption levels x 's at a solution to (10.0.2) are l
uniquely determined. I
J
I: q J
( ii) The optimal aggregate production level is
I: x 1 ,
=
and it is
l =1
J=1
uniquely determines since all the x 's are uniquely determined. I
(iii) Now suppose that the cost functions c ( ·) are strictly convex. J Suppose in negation that (x , .. , x , q , ... , q ) and (x', .. , . x', q', ... , 1
J
11
1
q~
q;) are both solutions to (10.0.2), and that qk •
x " = 1/2 x 1
q" = 1/2 q J
+ 1/2
I j
x' for every i =
11
for some k.
Take
1, •.. , I,
1
+ 1/2 q~ for every j = 1, ... , J. J
Clearly, (x", .. , x", q", ... , q") satisfies the constraint in (10.0.2), 1
I
1
J
since both (x , •. , x , q , ... , q ) and (x', •. , x', q', ... , q') do. As cost 1
I
1
1
J
I
1
J
functions are strictly convex, we must have
c (q") ::s 1/2 c .J
j
(q ) +
J
j
1/2 c (q') for every J. J
c (q") < 1/2 c (q ) + 1/2 c k
k
k
k
k
k, and
#
j
(q'). k
As utility unctions ¢ 's are convex, we must have 1
; (x") z:: 1
1
1/2 ¢ (x ) + 1/2 ¢ (x') for every i = 1, ... , I. I
I
1
1
Adding up all the inequalities, we obtain
r ¢ (x") I =1 '
1
-
r c (q") j =1
J
I
J
I
>
1/2 (
J J
r ¢ (x') I =1 I
+ 1/2 (
1
1
r ¢ (x
l =1
I
I: c
j
(q')] + =1 J J J
I
) -
r c (q J=1 j
J
>J.
which contradicts the assumption that (x , .. , x , q , ... , q ) and (x', .. , I
10-13
I
1
J
1
x', q', ... , q') are both solutions to (10.0.2). I
1
J
Therefore, individual production levels q 's at a solution to
'
(10.0.2) are uniquely determined.
10.D.2.
The program (10.0.2) for the economy in Exercise 10.C.2 can be
written as follows: f/>(x) - c(x) + w
Max
m
x!:O
The first-order condition yields x
= « +
• ""' f3jtr.
x -
(3 In
tr
x + w
m
This is the same consumption (and
production) level as that obtained in Exercise IO.C.2.
• • I • • J (i) Suppose in negation that (x , m } , {z , q } is a solution I I 1=1 J J J=l
10.0.3.
to (10.0.6), but not a Pareto optimal allocation. This means that there is an allocation {x', m'J I
I
1
1=1
which is feasible (i.e. satisfies (21), {zJ', q' / J j=1
,
(2m), (3)), such that m' + ; I
I
(x') I
~
m • + ¢ (x • ) for all i = 1, ••• , I, and
m' + ¢ (x') > m k
k
I
I
k
• k
I
•
+ f/> (x ) for some k. k
k
lI I=1'
If k = 1, this means that the allocation {x', m' 1
(z'
j'
J
q'} J J=l
satisfies all the constraints in (10.0.6) and yields a higher value of the 1 . . f unction . o bJectlve t han { x • , m •} , i i 1=1
, q •}J , w h'1c h contra d.1cts the J j j=l
{ z•
assumption that the latter allocation is a solution to (10.0.6). Suppose, therefore, that k > 1. Let A = {m' + f/> (x')J k
allocation (x", m"} I
i
1 1=1
,
fz:', J
q"/ J j=1
k
k
• {m k
+
•
¢ (x )]. Take an k
k
which has some numeraire redistributed
from consumer k to consumer 1: m" = m'1 + A • 1 m" = m'k - A• k 1
and which otherwise coincides with the allocation { x' m' J 1'
10-14
I 1=1
, {z
, j'
q '}J J J=l'
It is easy to s~e that the new allocation { x .. , m .. }1 l
1
1=1
, { z'',
J
satisfies all the constraints in (10.0.6} and yields a higher value of the objective function than { x •, m •l 1
I 1•1
, {z •,J
q• /
J J=1
, which contradicts the
assumption that the latter allocation is a solution to (10.0.6). • • I m J , {z J' I 1=1
• {x , 1
(ii) Let
• J
q/
J=l
be a Pareto optimal allocation and let
-
u = m • + '(x• ) for i = 2, ... , I. We will prove that (x • , m • }I • (z• , I I I l l 1 1•1 J • J
q }
J J=l
solves the program (10.0.6) with these utility levels.
First of all, it easy to see that {x • , m • }I , (z • , q • }J I
I 1=1
J
satisfies
J J=l
all the constraints in (10.0.6). Suppose in negation that {x
• 1
,
• I , 1 1 1
m J=
•
• J
{zJ, q/J= 1 does not solve the program, i.e. that there is an
allocation {x', m'J 1
1
I 1=1
, (z', j
q•/ which J J= 1
satisfies all the constraint and
yields a higher value of the objective function. Since (x'
1'
m'}
I
I
{z' 1 q'} J J
1=1'
satisfies (21), (2m), (3), it is feasible. The
J=l
j
constraint ( 1) implies that m' + I
•
I
I 1=1
I
I
i
Since {x', m' l
,
•
m + ' ( x ) for every i = 1, ... , I.
t/) (x') ~
1
(z', q' / J
j
I
yields a higher value of the objective
j=l
function, we have m'. + ¢ (x') > m I
1
1
Therefore, {x', m') •
I
• I
m ) I
1=1
J
1 1=1 • J
I
(x •
•
•
, (z , q ) j
j
+
I
•
'(x ). 1
, {z'
I
q'}
J'
J
j J=1
is a Pareto improvement over
.
j=l
, which contradicts the assumption that the latter
allocation is Pareto optimal.
10.0.4.
The Lagrangean for he program (10.0.6) can be written as J
I
L = m 1+ '
1
(x ) + 1
L 0: IX 1
+
I =1 I
+
L '1.
I
l=Z
fm.I
r ~Jq J J =I
-
I
+ ,"' i (x I ) - u 11 -IJ{Lx I I =1
10-15
J
-rq j =1
1J
I
J
I: m1
- v{
, 1
13
J.l., v,
, 1
t
KJ,
I:z
.._
j =1
I =1
where «
J
- w 1 + [K{Z m J =1 J
J
c(q)],
J
j
J
are non-negative Lagrange multipliers.
1
The first-order conditions with respect to original variables are: BLjBm
1
•
BLjam = 1
1 - v == 0 • v = 1;
r1
-
BL/Bz . J
=- v
8L/Bx 1
= «1 +
v = + K
for L
=0
J
;;ex 1 )
-
JJ
J
!!::
2 • 1 = v for all t 1
K
J.L
= 0 for all i.;
J
J
!!::
2.
= v • 1 for all j;
•
K c~(q
BL/Bq • 13 + J.1. J
o
) = f3 + J.l. - c'(q ) = 0 for all j. J
J
j
Besides, there are first-order conditions with respect to dual variables and complementary slackness conditions. Complementary slackness conditions say that o: x I
1
=0
for all i and f3 q
J J
= 0·
for all j. Using this, (•) and (.. )
become equivalent to conditions (10.0.3) and (10.0.4) in the textbook respectively.
The first-order condition with respect to J.L yields condition
(10. 0.5) in the textbook. The remaining conditions serve to pin dow the variables
lO.E.l. tariff
n
I
and z , which are not present in program (10.0.2); j
(a) To determine how the domestic market price p depends on the T,
observe that the demand x(p), the domestic supply is J q (p), and d
the foreign supply is J q (p f
f
T),
d
where q ( ·) and q ( ·) are the supply f
d
functions of the domestic and foreign firms correspondingly.
The market
clearing condition can then be written as x(p) = J q (p) + J q (p - "T).
r r
d d
Differentiating this condition with respect to
T
and solving for p'(-r), we
obtain J q'(p) p
'( >I T
r r
-r=O = J q'(p) + J q'(p) - x'(p) d
d
•
r r
As c ( · ) are strictly convex, we have c"( ·) > 0, and therefore f
f
q'( ·) < m. The expression (•) then implies p'(-r)
r
10-16
< 1.
Domestic welfare is the sum of consumer surplus, domestic profits, and tax revenue (which can be, for example, distributed lump-sum to domestic consumers): DW(T)
= CS(p(T))
+ J
n (p(T))
c1c1
+ TJ
q (p(T)-T).
rr
We will differentiate this expression at T=O, taking into account that CS'(p) = - x(p) (Roy's identity) and
n~(p)
= qcl(p)
(Hotelling's lemma):
=
DW'(TlfT=O = -x(p)p'(O) + Jclqiplp'(O) + J,q,(p)
= f-x(p) + J clqd(p)]p'(O) + Jrqr(p) • = - J,q,(p) p'(O) + J,q,(p) • = J rqr(p) (1 - p'(O)}.
Since we have established that p'(O) < 1, we have DW'(O) > 0. i.e. the imposition of a small tariff raises domestic welfare. (b) Now we have c"( ·) = 0, and therefore q'( ·) = r r
111.
From (•) we see that
p'(T) = l, and from the result of part (a) above, DW'(O) = J q (p) [1 - r'(O)] = 0. Thus, there is no first-order effect on
r r
domestic welfare from a small tariff, and we need to look at the second derivative to determine how domestic welfare· is affected by a small tariff. However, instead of computing the second derivative of DW(-r), one can answer the question with a simple observation. Observe that since foreign firms produce at constant returns to scale, they always have zero profits, and domestic welfare equals total welfare. Imposition of a tariff distorts the global economy away from the competitive equilibrium, and therefore lowers total welfare, .which is equal to domestic welfare.
10.£.2. A
CS(p) =
Substitute s = x(p):
rJ,.. cA>p ·
A
{P(s) - p] ds
0
((x(~))
=
J,.,
[P(x(p)) -
P(OJ
10-17
~]
dx(p)
=
1\
-
(p - A p] dx(p)
= [p
1\ - p] x(p)
1\
'p J p
a.
-
x(p)dp
=
p
Cll
Cll
J"
x(p)dp.
(The second line carries out integration by parts.}
lO.E.J.
(a) [First Qrinting errata: The first sentence should read: "Show
that if t' = 0 and t' 1
I
=t1
c
1
t I for all l "' 1, then tax vector t' raises
+
the same amount of good 1 as does tax vector t." ] Given constant returns to scale, in equillbrium a firm should receive c
1
units of numeraire for each unit of good l. Under the tax vector t, a
consumer pays c units of numeraire directly to the firm for each unit of I
good L. For this transaction, the consumer has to pay the government
t units of numeraire as a tax on good 1, plus c t units of numeraire as a I
.
I I
tax on his transfer of numeraire to the firm. Thus, the total tax consumers pay on each unit of good l is t
1
c + t . This is the same amount as under the I
I
tax vector t' , and we know that t' = 0. Therefore, for each tax scheme there 1
~eaves
is an equivalent tax scheme which
numeraire untaxed.
(b) The government's problem is L
Max L CS I (c I + t I ) (t • • • • t ) 1=2 2 L L
L. t I x 1(c I
s. t.
+ t) I
1=2
>
R,
where x (p ) is consumer demand in market l, and CS (p ) is consumer surplus I
I
I
I
in market l given price p . The Lagrangean of this program is
C
L =
L cs.(cl
L + tl) + A[
1=2
where ;\
~
L tlxl(cl
+ tl) -
R},
1=2
0 is the Lagrange multiplier with the constraint.
Differentiating
with respect to t and using Roy's identity (CS'(p ) :: - x (p ) ) , we obtain I
I
the following first-order condition:
10-18
I
I
1
(for brevity we have substituted p = c + t ). • 1 I I
From here we can express t 1:
tl • u-~)/~ x.(pl)/x;cpl).
Denoting the elasticity of demand for good l by .: • p x;Cp )jx (p ), the last 1 1 1 1 1 expression can be rewritten as
This result is known as the tnverse elastictty rule: optimal rates of taxation are inversely proportional to demand elasticities. (c) From the inverse elasticity rule, optimal tax rates on two goods should be equal if demand elasticities for the two goods are equal; and markets with less elastic demands should be taxed more heavily than markets with more elastic demands. Using the argument in part (a), if t = Ct , 0 .... 0) is a tax scheme which 1
= (0,
taxes only numeraire, then t'
t c ... , t c ) is an equivalent tax scheme 1 2
1 L
leaving numeraire untaxed. Such a scheme sets equal tax rates on all (non-numeraire) goods, which is only optimal \'/hen all markets have equally elastic demands.
lO.F.t:
(i)
Suppose that p > c'(O). Using Taylor expansion of ..:·(q! near
zero, for a small c > 0 we have c(d = c(O) + c'(O)c + o(c) = c'(O)c + o(c). Therefore, n(p)
= Max pq - c(q)
2:
pc - {c'(O)c + o(c)] = {p-c'(O)Jc + o(c) > 0
q'l:O
for c > 0 sufficiently small.
(ii)
Suppose that p s c'(O). Since c( ·) is convex, we have c(c) s
cfq
c(q) + (1-c/ql c(O)
for every c,q > 0. Therefore,
10-19
= cfq
c(q)
c(q)jq
and consequently c(q)
2:
!:
c(c)/e
-+ c'(O)
c'(O)q for all q
!:
as c -+ 0,
O.But then profits
at any output q can be bounded: pq - c(q) s pq - c'(O)q = (p-c'(O)]q s 0.
Therefore, n(p) = Max pq - c(q) :s 0. q!:O
(i) Profit maximization ~ p • c'(q)
10.F.2. (a)
(ii) Market clearing (iii) Free entry
"*
A - Bp
=«
+ 2.f3q.
= Jq.
~ pq - c(q) = pq - K - o:q - (Jq = 0. 2
The solution to this system of equations is
q. =
/kf;.
p. =
Cl
+ 2
J• = (A -
Aggregate output is p
vk;,
o:B)~ - 2 a.(:3. • • • Q = J q = A - o:B
•
and q
•
and· q
•
are increasing in A.
•
r-:-'
- 2a.(3-1Kj(3.
are independent of A. This is not surprising, because
• are entirely determined by technological parameters - q • is the • • optimal scale of production, and p is the minimum average cost. J and p
Q
(b) In the short run the number of firms J • is fixed, p and q are determined by profit maximization (i) and market clearing (ii). Since we are interested in market responses to short-run changes in demand, we can differentiate (i) and (ii) treating p and q as functions of A:
= 2(3q'(A),
p'(A)
J • q'(A).
1 - Bp'(A)
We can now solve for p'(A) and q'(Al: p'(A)
= lj(J
•
+ 2(3),
10-20
q'(A) =
213/<J•
+
213).
rxB)~ rx8)~ -
As A Increases, J• = (A -
= (A
When A -+ m, J•
-
- 2 CC/3 increases, and p'(A) and q'(A) fall. 2 a.f3 -+ m, and p'(A) -+ O.The intuition is
simple: in a large market the equilibrium number of firms is large, each firm's production needs to change only slightly to accommodate a short-run shift in demand, and the market price becomes insensitive to short-run demand shifts.
lO.F.J.
[First printing errata: You need to assume that taxes are small,
which is necessary for a definite comparison. Also, the condition f/>' in the third line of the exercise should instead be· ;• and J • denote
respec~ively
1 ( ·)
1
( ·)
0.1 Let p •, q •,
the market price, each firm's production, and the
number of producing firms at the initial equilibrium.
These variables
should satisfy conditions (i)-(iii) on p.335 in the textbook. Introduction of an ad valorem tax demand function x(p) with x(p(l+-r)). (iii) intact.
T
can be represented by replacing the
This change leaves conditions (i) and
Therefore, the new long-run market price and firm output have
to satisfy (i) and (iii), and thus have to coincide with p
•
•
and q .
The new
long-run equilibrium number of firms J is determined from the modified (ii): •
A
•
x(p 0+-r)) = Jq
Differentiating this expression with respect to
T
and substituting
T
= 0, we
obtain
•
•
J' (0) = p x'(p ljq ,..
•
.
(1)
•"'""' The resulting tax revenue is R(T) = -rp q J.
of this function around ~
~
T
= 0 can be computed as
~
R(-r) ::: R' (0) T + R' I (0) •
•
A
The second-order Taylor expansion
T:
2
/2 =
""'
= P q .n-rJ' (-rl + J(-rll 1-r=o -r + (-rJ'
10-21
2 (-r> + ZJ' (T)) IT=O T /2 ~
"""
I
1•
•
•
•
•
•
•t
A
• p q J T + p q
J' (T)T.
(2)
Introduction of a per firm tax T, on the other hand, modifies the free 0
entry condition (iii), so that now it can be written as p q The profit maximization condition (i) yields p
0
0
c(q0 )
-
-
T = 0.
= c' (q0 ). Combining with the
previous equation and substituting our cost function we can write ,, (q 0 )
= ('(q0 ) + K + T)1, /q 0 .
.
0
This equation determines q . Differentiating it with respect to T, we obtain
qo' (T) The number of firms J
0
(¢'' (qo)qofl.
•
can be determined from (ii) and the profit maximization
condition:
Differentiating this equation with respect to T and evaluating at T • 0, we obtain
I I
= [x'C¢'Cq 0 )) ¢' '(q 0 ) - J OJ q 0 1 •
•
•
(T) q 0 •
= [x' (p )¢" (q ) - J 1/C¢" (q )q
The tax revenue from the per firm tax is R
0
•2
T=o
(3)
). 0
= T J (T).
(T)
The second-order
Taylor expansion of this function around T = 0 can be computed as 0
R (T) z R
0 '
(0) T + R
1
2
(0) T /2 =
(T) + J (T))
'
= J•T
'
0
0
= (TJ
0
+
IT=O
T + (TJ
0
1 '
(T) +
I
1
2
2f (T)) T=O T /2 =
.f' (O)T 2
Since at the initial equilibrium the two taxes raise the same revenue, we must
• •
have T = Tp q .
Substituting in the last Taylor expansion, we obtain 0
=p
R (T)
•
•
•
•
•
2
q J T + (p q ) J
0 I
(O)T
2
(4)
Comparing (2) and (4), we see that the first-order terms in and the second-order difference can be computed using A
0
•
•
•
•
R (T) - R(T) ::: p q [p q J •
•
= p q [ p
• X
1
•
(p ljq
•
•
A
0 '
•
(0) -
J' (O)]T •
2
(1)
•
•
•
- p J /(¢' '(q )q ) - p x' (q ljq }T
10-22
2
coincide,
and (3):
=
•
T
=
.. - p
•zT z J •I•' Cq• ) < 0. I
Therefore. a small ad valorem tax raises more revenue than the corresponding per firm tax.
10.F.4.
[First printing errata: you should assume that c(w, q) is a twice
continuously differentiable function, and that most efficient scale is uniquely defined. I (i) The long-run equilibrium price and output of each firm are given by the
following conditions: Profit maximization .. p = c (w, q), q
Free entry
"*
pq - c(w, q) = 0.
Let f(w,q) = c(w,qlfq be a firm's average cost. These equations Imply that 2
f (w,q) = c (w,qlfq - c<w,qljq = (p - pljq = 0. Intuitively, this is the q
q
first-order condition for a firm's operating at the minimum average cost in a long-run equilibrium; Now the above system can be rewritten as
-
- = f(w,q(w)) p(w)
(1)
-
f (w,q(w)) = 0,
(2)
q
-
- are equilibrium price and firm output as a function of w. where p(w) and q(w) To find out how equilibrium price depends on w, differentiate
(l)
with respect
to the input price vector w and use (2):
-
Vp(w) =
-
'1'J
w
-
f(w,q(w)) + f (w,q(w)) q
'1'J
-
w
q(w) =
'1'J
-
w
f(w,q(w))
(•)
Now, using the definition of f(w,q(w)) and Shepard's Lemma (Proposition 5.C.2 in the textbook), we obtain Vp(w) = where z(w,
q)
'I'J
f(w,q(w))
w
= Vwc(w,q(w))jq(w)
= z(w,q(w))jq ~ 0,
is a firm's input demand function. Therefore, the long-run
equilibrium price is non-increasing in factor prices.
10-23
(ii) Take any a:
> 0, and substitute a:w instead of w in equations
-
-
p(a:w) = l(a:w,q(a:w))
(1 ),
(2):
(1')
-
f (a:w,q(atw)) = 0.
(2')
q
Using the definition of f(w,q) and homogeneity of c(w,q) in w (Proposition 5.C.2 in the textbook), we obtain f(a:w,q)
= c(a.w,q)jq = a:
= a:
c(w,q)jq
f(w,q),
I (a:w,q) = « I (w,q). q
q
-
Therefore, equation (2') can be rewritten as f (w,q(a.w)) = 0, which together
-
q
-
with equation (2) implies that q(a.w) = q(w). Using this fact, the above derivations, and (1), equation (1') can be rewritten as p(a:w)
=
a. f(w,q(a:w))
= a.
.
f(w,q(w))
= a.
p(w),
-
which means that p(w) is homogeneous of degree one in w.
(iii) Differentiating (•) with respect to the input price vector
w and using
(2), we obtain 2-
2 w
-
-
-
D p(w) = D f(w,q(w)) + ajaq [V f(w,q(w))] 'I'J q(w) = w
= D2 f(w,q(w)) + w
'1'J
w
f (w,q(w))
w q
'I'J q(w) w
= D2 f(w,q(w)). w
(The order of differentiation has been changed, which we can do due to the twice continuous differentiability of
f( ·, ·).)
Now, using the definition
of f(w,q(w)), we obtain 2-
2
-
2
-
-
D p(w) = D f(w,q(w)) = D c(w,q(w))jq(w). w
w
According to Proposition 5.C.2 in the textbook, c(w,q) is a concave function 2
of w, and thus D c(w,q(w)) is a negative semidefinite matrix. Therefore, w
2
D p(w) is negative semidefinite, too, which implies that p(w) is a concave
10-24
function of w.
(a) Let q (p) be a firm's long-run choice of output as a function of
JO.F.S.
L
its price, with p • being the initial price and q • = q Cp • ) the initial output. L
Let z (p) k
p, with z
= z k <w.qL (p)) • k
be the optimal choice of factor k when output price is
= z k (p • ). Denote by q L (pI z k ) the firm's short-run choice of output ·
as a function of p, the output price, and zk the fixed input of factor Jt. Let 1t
s
(pI z
k
) denote the short-run profits as a function of output price p and
factor It's input z . k
expressed as
KL (p)
Then the long-run profit function can be
= ns(p I zk(p)). Since the firm can always keep its
short-run demand of factor k in the long run, we must have n (p) L
!:
n Cplz•) S
k
for all p. Moreover, we know that the initial situation was a long-run equilibrium, which implies that n (p • ) = n (pI z (p •)) L s k
•
n (pI z • ). Therefore, S
k
the function h(p) = n (p) - n (pI z • ) achieves its minimum value of zero at L
S
k
•
p = p . For this to hold, the following second-order condition has to be
satisfied: h"(p • ) = n"(p • ) - n"(p • I z • ) L s k
2:
0.
By Hotelling's lemma, q (p) = n'(p) and q (p) = n'(plz • ). Ther ~fer~. the last L l s S k inequality can be rewritten as •
•
q'(p ) L
!:
•
q'(p ). S
(b) We will hold the number of firms J fixed both in the long run and in the short run. (If free entry occurs in the long run, then the equilibrium price will go back to its original level which is the minimum average cost, and the question becomes trivial.) If each firm's supply function is q( • ),
the market-clearing condition can be written as x(p,cx) = Jq(p(cx)l.
10-25
Differentiating with respect to a, we obtain
x (p,a) + x (p,a)p'(a) = Jq'(p(a))p'(a) P
a
From here we can express p'(cx.):
• = xu
p'(a }
•
/
[Jq'(p ) - xp},
where a • is the initial value of a, p
x
p
• x
•
p
(p ,
a •).
•
•
p(a ),
x
a
• x
a
• a •),
(p ,
From the exercise's assumptions, both the numerator and
denominator of the fraction are positive. Denoting by ps(a) and pL (a) the short-run and long-run price responses as functions of ex respectively and using (•), we can write
•
p~(a ),
i.e. the long-run response of the market price to a marginal increase in a is smaller than the short-run response. The aggregate consumption in the market is Q(a)
= x(p(o:),a).
Differentiating
with respect to a and using the result from part (a), we obtain
•
Q'(o: ) =
X
a
+ X
•
P
p'(a )
Denoting by p (a) and p (o:) the short-run and long-run aggregate S
L
consumption as functions of a respectively and using (•) and x'(p) < 0, we obtain Q'(o: • ) =x +X p, (a: • ) p s 0: s
~X
ex
p'(o: • ) =
+X P
L
•
Q~(a. ),
i.e. the long-run response of the aggregate consumption to a marginal increase in a is larger than the short-run response.
IO.F.6.
First we derive the long-run cost function for the Cobb-Douglas
technology:
c(q) =
wz +wz
Min (z
1 I
,Z ) I 2
s. t. z
0: I
1-0:
z2
2 2
i!!:
q.
10-26
The first-order condition for this program can be written as
w.Jw2 = cxz/U-o:)zl" Together with the binding constraint in the program, this gives us factor demands:
z z
1
2
[o:w
(q) • (q)
=
I
(1-o:)w
{(1-o:)w
1-Cl
1
1
q.
fo:w ~ q.
1
2
•
Therefore, we can write c(q) • w z +w z 1 1
A = w [o:w /(1-o:)w 1 zl I
i-a.
2 2
= Aq. where
+ w [(1-o:)w 'o:w 2
1/ '
2
1« = w« w (l-«) lo:«(l-4 ) 1-«. 2
I '
- A)q
= 0.
1
The long-run equilibrium is given by (i) Profit maximization .. p = c'(q) • A. Market clearing
(ii) (iii)
"*
a - bp = Jq.
= (p
Free entry -. pq - c(q) = pq - Aq
Observe that since technology exhibits constant return to scale, profit maximization (condition (i)) automatically implies zero profits (condition (iii). Also, the number of firms and the scale of every firm are
indeterminate. Condition (i) gives the equilibrium price p • = A, and condition (ii)
gives the aggregate output Q• = J •q • = a - bp • • a - bA . In the short run capital z is fixed at z •, and the number of firms J is 1
• fixed at J . From f(z . z ) = 1 2
= q 1/tl-Cl) fz •Cl/(1-a) . I •
c(q I z l 1
I
.
q we get the short-run demand for labor •
. IS . 1/!1-al
The short-run cost funct1on •
= w 11 z +
s
•
22
I
w z Cq I z l
= w 1z
•
w q
+
2
•a/(1-al
fz 1
The first-order condition for the firm's profit maximization is
Now, z • can be computed as the long-run demand for capital: 1
z
•
1
= z I (q
•
) =
1-Cl
[o:w /(1-a)w 1 zl
I
and the first-order condition can be rewritten as p =
wo: w c1 -« l / a a( 1-a )1-o:. · ( q/ q • )Cl/(1-al 1
2
10-27
•
q ,
.
From here we can solve for the short-run supply function of a firm: s( )
•( .IA)U-«l/«
q P = q Pt
·
The short-run supply function of the industry is Qs(p)
lO.G.l.
= J•qs(p)
=
J•q•(pjA)(I-4)/Cl = Q•(pjA)U-Cl)/«.
(i) Firm j solves
M
Max
L p mqmj
t.t
q E IR J +
•
c}qlj, ... ,
-
qMJ ).
m=l
The first-order conditions for all firms are p
• m
:S
• ljaq , acj (q •IJ •••• , qMJ mj m
= I.
> 0,
with equality
(1)
.. , M, j = 1, •• , J.
The firms' first-order conditions give. JM equations. Consumer i solves M XE I
Max IRM • me IR +
s. t. m + I
~
I:
m + 1
1/J(x I
m=l
11'
M
J
•
c.J+[B LPX.::S rn ma 1 lj m= 1 J =I
... , X
(
Ml
)
•
• •
L pmqm1
- c/qlJ'
... '
m=l
q • )). Mj
The first-order conditions for all consumers are
• , ..., x • >fax a;cx Mi mi I II
::s p • , with equality if x m
• > 0, mi
(2)
m = 1, .. , M, i = 1, .. , I.
The consumers' first-order conditions give IM equations. The market-clearing conditions are I
•
'x L mi 1=1
J
•
= 'L q mj , m = 1, .. , M.
(3)
j=l
(By Lemma 10. 8.1, we do not need to check that the market for numeraire
clears.) Thus, we have the total of JM + IM + M = (J + I + l)M equations and (J + I + l)M unknowns - prices, output vectors, and consumption vectors.
Since the equations (1), (2), and (3) do not contain consumers' wealth levels
10-28
w1, the set of solutions to these equations is independent of consumer wealth. (ii) Consider the following welfare-maximization problem: I
L • I ex11 , ...,
Max
1=1
1
• L X ml
s. t.
J
x
J
-
1=1
• L q mJ
Ml .
l -
L c J(q1 J ,
••• ,
q
J =I
MJ
) + w
• 0, m = 1, •• , M.
J=l
By examining the first-order conditions to (•), it is easy to see that an allocation (x•, •• , x •, q •, .. , q • ) solves (•) if and only (p • X •' 1 I 1 I I
•
.• , x •, q •, •. , q • ) solves (1)-(3) for some price vector p •. I
1
I
(For conciseness, every variable here· has M components corresponding to the non-numeraire goods.) Suppose in negation that ep, x , .. , x , q , ... , q ) and I
I
J
1
ep• , x'1, .. , x', q', ... , q') are both solutions to (1) - (3), and therefore I
e•>.
to
I
J
and that x
x" I
q" J
Cl ear l y,
*
k
- 1/2 x 1 + - 1/2 q J +
·
for some k. Take
k
1/2 x; for every i = 1,
... ,
I,
Ijz q' for every
...,
J.
j • 1,
J
-
ex l" ,
since both
x
.. , x " , q " , ... , q "l sat'1sf'1es th e const ram · t m · I
(x , .. , l
c•>.
J
I
x, q , ... , q ) and (x', .. , x', q', ... , q') do. As cost 1 I I I I J 1
functions are convex, we must have c (q") j
j
:5
1/2 c (q ) + 1/2 c (q') for every J. = 1.... , J. J
j
j
j
As utility unctions ¢. 's are strictly convex, we must have I
~
¢ (x") I
I
1/2 ; 1ex 1 l + 1/2 ; 1ex:> for every i
*
k, and
¢ (x ") > 1/2 ¢ (x ) + 1/2 ¢ (x'). k
k
k
k
k
k
Adding up all the inequalities, we obtain
r ¢ (x") - r c Jeq:·>
I :1
I .
J
I
I
I
. 1 J=
>
1/2 [
J
r ¢ (x')
....
1
I
I
+ 112
-
r cJ
(q')] •
j=l J
J
c r ¢ I (x I > - r c (q I =1
which contradicts the assumption that
J
exI , .., x I ,
10-29
J =1 J
H. J
q , ... , q ) and (x ', .. , 1 l J
x', q', ... , q') are both solutions to (•). I
1
J
Therefore, individual consumption levels x 's at a solution to (•), and thus I
at a solution to (1)-(3), are uniquely determined. I
J
r qJ = r x
(iii} The optimal aggregate production levels are
I =1
j:1
I
• and they
are uniquely determines since all the x 's are uniquely determined. I
(iv) Now suppose that the cost functions c'( ·) are strictly convex. J Suppose in negation that (x , .. , x , q , .•• , q ) and (x', •• , x ', q ', 1
I
1
1
J
1
••• ,
1
q') are both solutions to (1)-(3), and therefore to (•), and that q J
k
"' q'
k
some k. Take
x;•
= 1/2
x
q" = 1/2 q J
x;
+ 1/2
1
for every i = 1, .•. , I,.
+ 1/2 q' for every j = 1, ... , J.
J
.
J
Clearly, (x", .. , x", q", ... , q") satisfies the constraint in (10.0.2), 1
I
J
1
since both (x, .. , x, q, ... , q ) and (x', .. , x', q', ... , q') do. As cost 1
1
J
I
I
I
J
1
functions are strictly convex, we must have
c (q") J
J
:s 1/2
c
(q ) +
J
j
1/2 c (q') for every J. J
k, and
#
j
ck(q~') < 1/2 ck(qk) + 1/2 ck(q/ As utility unctions ¢. 's are convex, we must have I
¢.(x~') ~ II
1/2 tf>.(x.) + 1/2 ¢.(x') for every i = 1, ... , I. II
•
I)
Adding up all the inequalities, we obtain
r ¢i(x;') I =I
-
r c.(q:'l j =1 J
J
I
J
I
>
1/2 {
J
r ¢ (x') I =1 I
+ 1/2 {
I
I
L t/J I (x I )
-
r c (q:ll J=1 J
-
J
[c
J
(q )],
J =I J
I =I
+
j
which contradicts the assumption that (x , · .. , x , q , ... , q l and (x', .. , I
I
I
J
x', q', ... , q') are both solutions to (•). I
I
J
Therefore, individual production levels q 's at a solution to I
(•). and thus at a solution to (1)-(3). are uniquely determined.
10-30
I
for
IO.G.2.
The first-order conditions (l) in Exercise IO.G.1 give for output of
good l:
p• c;/q•JJ) :S
1
, with equality if
q• J
> 0, j = 1, ... , J.
1
(1)
The first-order conditions (2) in Exercise 10.G.2 give for consumption of good l: •• (x 11
•)
s p • • with equality if
11
X
I
• 11
> 0, L = 1, •• , I.
(2)
The market clearing condition for good l is I
• Jell
I:
J
•
-I: qlJ.
(3)
J =I
1=1
(1)-(3) is a system of J+I+l equations with J+I+l unknowns -
•
(p , x I
•,
.. , x • • q • , ... q • ). These equations do not contain prices and
11
II
11
1J
quantities in the other markets.
From these equations, the price and
quantities in market l can be determined independently of other markets. Total welfare can be computed as I
S(x , .. , x , q , ;., q ) I
I
J
I
I
L" t/Jii(xli)
= l
J ~
-
L. j =I
1=1
= S
(X
1
11
,
.. , X
II
,
=
L t/J I (xI )
1=1
c (q lj
Ij
I
)J
+
l
J
L t/J -1.1 (x -1,1 ) - I: c -1 J(q -1
q , .. , q ) + 5 IJ
11
-1
•
•
1=1
J =I
(X
-1,1
, .. , X
-1,1
,
q
-1,1
•J
)}
, .. , q
= -I,J
).
The first term describes the surplus in market l, and the second term describes the surplus in all the other markets. The first term takes the form I
5 (x , .. , x I
II
II
, q , .. , q ) = 11
lJ
L t/J II Cx 11 )
1=1
J
-
L c lj (q I j ),
j=l
which is exactly equation (lO.E.l) in the textbook. The welfare analysis .
of market l can proceed in the same way as in Section 10. E, without consideration of other markets. If all goods are separable, every market l can be solved independently of others using equations (1)-(3). Moreover, total surplus can be written as
10-31
I
I
L
=
L' (x )
q , .. , q ) =
S(x , •. , X ,
I
J
l=l l
I
-
I
j=l L
J
L { L ' 11 (X II }
L C lj (q lJ )]
-
=
J •I
1•1 1=1
Lc
(q )
J
=
j
L 5 I(X ll ,
•• , X
l =I
II
,
q , .. , q ). IJ
U
Therefore, welfare analysis of every market can be performed independently of other markets.
IO.G.J.
•
•
2
3
(a) Let p , p
th~
be
market prices of goods 2 and 3, ·Jet x
•, x • 2
3
be
consumption of these goods by each consumer (since all consumers are identical. the will consume the same amounts). and let q •• q • be production of these goods by the firm. Assuming (x• • x • ) 2
with taxes t , t 1
2
3
2
3
» 0, the market equilibrium
on goods 1 and 2 is a solution to the following
equations, which are a modification of conditions (1)-(3) of Exercise IO.G.l:
•
P2 =c 2 +t· 2'
(1.1)
•= P3
c
(1.2)
• = P2
• • B'*'(x ,x >fax
3
+
t . 3
2
Y'
1
3
2
(2.1)
:
(2.2) Ix
•
Ix
•
(3.1}
2
3
•
= q.
(3.2)
3
Equations ( l.l) and ( 1. 2) pin down equilibrium prices, and equations (2.1), (2.2) pin down consumption of goods 2 and 3. Equations (3.1) and (3.2) pin down aggregate production levels. Since production exhibits constant returns to scale, aggregate production can be allocated across firms in an arbitrary fashion. Assuming that ·tax revenue is redistributed lump-sum across consumers, .
aggregate surplus can be computed as •
•
2
3
S = I¢(x ,x ) - (c q
•
•
f
•
•
•
•
+ c q ) = ltp(x ,x ) - c Ix - c Ix. 3 3 22 33 2 3 2 2
10-32
•
Let x (pz,p 2
3
and
)
•
.tC (p ,p ) 3 2 3
be consumer demand functions, i.e. the solutions
to (2.2), (2.3). Then aggregate surplus can be expressed as a function of taxes:
•
S(t ,t )
c +t ),x (c +t , c +t )) -
2 3
3
3
3
2
2
•
3
3
•
- c x (c +t , c +t )) - c x (c +t , c +t ) )] 2222
33
3322
33
t 3 = 0 and differentiate with respect to t 2, using (2.1); (2.2);
When we set
(1.1), (1.2), we obtain
BSCt ,O)jBt 2
2
= I[(p• - c )8x•(c +t , c 1/Bp + (p • - c )Bx•(c +t , c >/Bt J •
• +t = tlaxz(c 2
2
2
,
2
7.22
3
2
3
3
322
3
2
c 3 ljap2•
Thus. the marginal welfare change is the same as the marginal change in the area between supply and demand curves for good 2 (the derivative of expression (10.E.6) in the textbook), holding the price of good 3 constant at c . 3
(b) When we differentiate (•) with respect to t , using (2.1), 2
(2.2), (1.1), (1.2), we obtain as(t ,t >fat 23
2
= I[t ax
•(c
+t • c +t lf8p
2222
33
•
2
+ t 8x (c +t , c 3322
3
+ t )j8p ]. 3
2
The first term in the expression is the change in the area between supply and demand curves for good 2 (the. derivative of expression (lO.E.6) in the textbook), holding the price of good 3 constant at c +t . The sign of the 3
3
•
second term is the sign of ax (p ,p ljap . This derivative is positive when 3
2
3
2
goods 2 and 3 are substitutes, then the area measure would overstate the welfare loss from taxation. When goods 2 and 3 are complements, this derivative is negative, and the area measure would understate the welfare loss from taxation.
lO.G.4.
Differentiating (•) in Exercise 10.G.3 with respect to t
obtain
10-33
2
and t , we 3
•
•
8S(t ,t lfBt = I[t Bx Cc +t , ' +t )fBp + t Bx Cc +t , 3 3 2 2 2 2 2 2 3 3 2 2 2 3 BS(t
,t lfat
23
3
c +t )fapi; 3
3
= I(t ax• (c +t , c +t >fap + t ax•Cc +t , c +t >fap ]. 2222
33
3
3322
33
3
Now, S(t ,t ) - S(O,O) can be expressed, for example, as 2 3
S(t ,t ) - S(O,O) = S(t ,0) - S(O,O) + S(t ,t ) 23t 2 t 23 2
=
-
S(t ,0) 2
=
3
J
J
~
0
BSCs,Olfap2 ds +
BS(t ,slfBp ds • 2 3
2
=
J
I sax ·cc +s, c >fap ds 2
2
3
+
2
F 3
+
JIlt
ax·cc +t ,
2222
c3+slfap3
+ sax•cc +t , c +slfBp 322
3
3
1 ds.
0
lO.G.S.
(a) Let p
.. 2
,
p
·•
be the market prices of goods 2 and 3, let x
3
2
,
x
. 3
be
consumption of these goods by each consumer (since all consumers are
•
•
2j
3J
identical, the will consume the same amounts), and let q , q
•
•
2
3
of these goods by firm j. Assuming (x , x ) » 0, with taxes t , t 1
2
be production
the market equilibrium
on goods 1 and 2 is a solution to the following
equations, which are a modification of conditions (1 )-(3) of Exercise 10. G.l:
• = P2 • = P3
•
t ·
c'(q ) + 2 2
•
c'(q ) 3
3
( 1.1)
2'
+
t 3'·
(1.2)
(2.1)
(2.2)
Ix
• = q; • 2
• Ix 3
(3.1)
2
• = q . 3
(3.2)
Eliminating prices and outputs from the equations, we obtain
• •
a;cx ,x >fax 2
3
2
= c'2 (Ix •2 )
+
t2
a;cx • ,x• >fax = c'(Ix • ) + t 2
3
3
3
3
(l) (2)
3
Now, we differentiate these two equations with respect to t , 2
10-34
treating x • and x • as functions of t : 2
3
2
"" x'2 + "" x'3 = I c'' x'2 + 1 "'22 "'22 2 ~2 3x' + ' 2
x' = Ic" x'3 3
333
(To· avoid cumbersome expressions, we have defined ¢
•
•
c" • c"(lx ) x' = dx ldt ). I
I
I '
I
1/'
= 8¢Cx•,x• )/Bx Bx ,
II
2
3
I
1
•
2
From the above two equations we can express
x,3 • .,.. ""7:JfI["'.,.. 232
+ I c""" 2 .,..33 +
""22.,.. ""33 .,..
-
x;: Ic"' 3 22
Differentiating equation (1.2), we now obtain: •
2
dp-3'ldt 2 • Jc"x' • lc''• I[• - ' 3 3 3 23'' Z3
'
22 33
+ Ic"' 2
+
33
2
Ic"• - I c"c"]. 3 22 2 3
2
Since u( • ) is concave, D u( ·) must be negative semidefinite, which implies, in particular, that ~
22
it follows that ~
22
s 0, ~33 :s 0, '
2 23
- '
(/1
22 33
:s 0. Since
the cost functions are strictly convex, we must have c" > 0, c" > 0. These 2
3
inequalities together imply that the denominator of the above expression for
• dpjdt
2
sign of '
• ldt is negative the the sign of dpis always negative. Therefore, . y 2 23
•
When '
23
< 0, goods 1 and 2 are substitutes, and an increase in
t raises p • . Conversely, when 2 3 increase in t
2
.
4>
> 0, goods 1 and 2 are complements, and an
23
reduces p •. 3
(b) [First printing errata: Assume that c(q , q ) = c q 2
3
2 2
+c (q ). J 3
3
Assuming that tax revenue is redistributed lump-sum across consumers, aggregate surplus can be computed as
• •
S = lf/J(x ,x ) - (c q 2
3
•
22
• Let x (p ,p ) and 2
2
•
• •
3
2
+ c (q )) = ltf>(x ,x ) - c Ix 3
3
2
• 2
- c (Ix • ). 3
3
be consumer demand functions, i.e. the solutions
3
to (2.2), (2.3). Then aggregate surplus can be expressed as a function of taxes:
•
•
3
3
•
p ),x {c +t , p )) -
- c x • (c +t , p • ))] 2222
3
2
c
2
3
• (lx • (c +t , p3))J 3 322
10-35
Differentiating the above expression with respect to t
2
and
using (2.1), (2.2), (1.1), and (1.2), we obtain as(t ,t )jat 23
2
= It ax•(c +t , p • ljap + It ax •(c +t , p •)jap + 2
222
+ [It
2
3
2
•
ax •Cc +t ,p ) lap 2
2
2
3/'
3
3
322
+ It
3
•
3
(•)
2
ax (c +t ,p •>jap 1 dp-• ldt . 3 2 2 3 3 y 2
Now, in Exercise 10.G.3 we calculate this derivative assuming that p • is fixed 3
at c
3
+ t , i.e. dp-• ldt = 0. This is fine when technology· exhibits constant 3 y 2
returns to scale, because then the equilibrium price of good 3 is unaffected by t . 2
But in the present case, an increase In t
substitutes and lowers p
3
2
raises p • when goods are 3
when goods are complements (as shown in part (a)).
How would ignoring this effect bias are calculation of the welfare loss? We always know that we know from part
ax•3 ( ·, ·lf8p3 < 0. With complements, 8x•2 ( •• ·lj8p3 < 0, and
(~)
• that dpjdt
2
< 0. Thus,· ignoring using the formula in
Exercise lO.G.J (b) ignoring the correction terms in (•) would result in an overstatement of the welfare loss. With substitutes, Bx• ( ·, ·lj8p 2
•
know from part (a) that dpjdt
2
If t
3
> 0, and we
> 0. Thus, the first correction term is is
negative and the second correction term is positive. unclear.
3
The net effect is
= 0, the formula in Exercise lO.G.J (b) would be overstating
the welfare loss.
But if t
3
is large and t
2
is close to zero, the formula in
Exercise lO.G.3 (b) would be understating the welfare loss (which would actually be a gain).
10-36
CHAPTER 11
ll.B.l
Note first that the technologies are CRS so that producers will
either choose no production or will choose producing at full capacity .. Assume that the prices of apples and honey in the market are such that producing both products is efficient.
Since technologies are CRS, maximum profits are
equivalent to minimum costs, and generically Jones will choose only one of the possible technologies. and only if
In particular, he will choose artificial production if
H(pm+ w) < H(bpb+ kw), or, if and only if pm < bpb + (k - l)w .
Now, artificial production is socially optimal if and only if the total cost of producing both capacities is cheaper under artificial production of honey.
i.e., if and only if Hb H(p + w) + Aw < H(bpb+ kw) + (A - c)w , m
or, if and only if p < bpb+ (k m
So, if
*
~c :s:
~)w c
.
l, Jones' choice of honey production will be socially efficient. If
> 1 then for
bpb + (k -
~)w < pm
(h) + w 1
1
- T
denote the Kuhn-Tucker multiplier, and assuming an interior
solution, the FOCs are:
= 0
(l)
(2)
aq, ch. w +T) 2 2
-1 + ;\ • __...:=---.8:--w___;;~-
=0
•
which together yield, 0
0
0
0
aq, Ch •w +T )lah 2 2 •
B,P (h ,w +T )/Bw 2 2
(b)
Consider consumer 2's problem:
Letting h(ph, w ) denote his demand for h, the FOC satisfies: 2 aq,2(h(ph,w2).w2- ph ·h(ph,w2)) 8h
Differentiating this expression with respect to w
11 - 2
2
we get:
82¢2 8h(ph • w2) 8h
2
•
= 0 .
Under very general conditions (which we usually assume) we have that the cross 82¢ 82¢ 2 2 partials are equivalent, i.e., that w h = Bh w , which changes the above to 8 8 28 2 (we discard the subscript 2 since only consumer 2 is relevant):
Bh(ph, w)
(•) 8w
-- 82¢
82¢ ph. 2 aw 2
-
. Bh 2
a2;
-
Bh8w
2
2 a ; 2 Ph • 8h8w + ph· 2 8w2 82¢
•
, and substituting this into the
From the FOC
expression (•) above will express the wealth effect in the required terms. 8¢ /Bh 2 0 From part (a) above we have that ¢1 (h ) = - a; ;aw · 2 2 .
(c)
this with respect to w
(
..
)
Bh aw
0
2
2
and rearranging terms we get: 8¢2 82¢ 2 • 2
B¢2 82¢ 2 • aw
Bh aw2
8¢2 aw
We need to show that
Differentiating
c••)
2
2 ·¢"(ho) 1
+
2
BhBw
8¢2 82¢2 • aw2
Bh
2
2
-
a;2 a ;2 •
Bh Bh8w
is positive (for a positive externality.
2 To
proceed, we define consumer 2' s indirect utility function in the usual way, denoted by:
Now, consumer 2's demand for the externality, h(ph,w l, can be obtained using 2
11 - 3
Roy's identity:
and differentiating this with respect to w
-----=-
we get:
2
a2 .;2 t8ph aw2
2
8 .; 2 + z' 2 ca.; taw ) 8w 2 2 2 8.; taph 2
•
It is given that consumer 2's demand for the externality is normal, so that 8h(ph,w )18w > 0, which together with (3) above (after multiplying all terms 2 2 2 • In (3) by ca.; taw ) > 0) implies that: 2 2
2
(4)
a.;2 8 .;2 •
2 aph _aw 2
a2.;
>
2
8Pjl2 •
•
8ph8w2 aw2
Since the externality is a normal good, it is not a Giffen good, which implies
aph
or In inverse terms, Bh
< 0.
Multiplying both sides of (4) by
this we get:
---·
(5)
8ph •
•
From consumer theory (chapter 3) we know that the following ider.:ities relate the indirect utility function with the utility function:
-
•
•
---· aphaw
2
=
•
'
Bh
•
Substituting these identities into (5) above we obtain: 82¢ Bt/>2 a¢2 a2¢ 2 2• • < 2 ' 8h8w aw ah aw2 2 which implies that the numerator of the expression given in the model
0, and
11 - 4
c••) > 0.
is negative.
It is
Furthermore, strict
82¢ 2 quasiconcavity of ¢ Ch, w l implies that < 0, and 2 2 2 Bw 2
0 for a positive externality.
The reverse can be shown
for the case of a negative externality using the same methods.
With a Pigovian tax, bargaining will change the level of the
11.8.3
externality.
To see this consider the situation where an optimal tax is
chosen such that consumer 1 chooses the efficient level h (w.l.o.g.) that h is a negative externality.
0
•
and assume
An infinitesimal reduction in h
has a zero first-order effect on agent 1, and a positive first-order effect on agent 2.
Therefore, there are gains from reduction of h below h A
agents can bargain to obtain some h < h
0
0
,
and the
which will be the final choice of h
given the Pigovian tax and the ability to bargain. On the other hand, this argument will not hold for quotas. At the point h an infinitesimal reduction in the level of h by 9'>]. (h
0
)
and will increase 2's utility by
will -9'>2(h
0
,
reduce consumer l's utility 0
),
which cancel out given
the optimal choice of h
0
•
Therefore there are no gains from trade and the optimal
level will not be bargained down (the· opposite holds for a positive externality).
11.8.4
Assume that ¢ Ch,e) is such that given any level of h chosen by 2
consumer l, there is a unique optimal level of e that is consumer 2's "best response".
That is, there exists a function e•(h).
Since e will be optimally
•
chosen by 2, and it has no external effect on consumer 1, there is no reason to tax or subsidize this activity - it is not an externality.
It may,
however, still be optimal to impose some tax/subsidy on h, even though
consumer 2 can choose an e in response.
ll.B.S
(b)
(a)
The firm solves
(1)
p
(2)
0 s
$
Max p · q - c(q,h), and the FOCs are: qlh
Bc(q• h•) aq t
with equality if q• ) 0
1
1
1
with .equality if h• > 0 .
Since the consumer's utility function is quasilinear with respect to
money, the utility posibility frontier is a linear line with slope -1, and therefore we can find the Pareto optimal level of h and q
by maximizing the
sum of the profit function and the utility function without wealth, i.e., Max q,h
p·q - c(q h) + cJ>(h) , 1
which yields the FOCs, 0
(3)
(4)
(c)
0
acCq lh ) p s aq
, with equality if q
8c(q 0 ,h0 )
t/>' (h) s ---:::~Bh
I
0
with equality if h
> 0 ,
0
>0 .
Let t denote the tax rate on output, the firm solves: Max q,h
p·q - dq,h) - t·q ,
which yields the FOes, (5)
p s
(6)
0 s
Bc(q, h)
Bq
+ t
Bc(q,h) , with equality if h ah
is imposea on h, say T, the firm solves: p·q- c(q,h) - T·h ,
11 - 6
I
>0.
This gives the same level of h = h• as in part (a) above.
Max q,h
> 0
, with equality if q
If, however, a tax
wh1ch yields
t~e
F'OCs,
1 (h
0
then this gives the level h
)
= h0
1
as in part (b)
above, and efficiency is restored.
(d)
Let t denote the tax rate on output, the firm solves: Max q
p·q - c(q,o:q) - t•q ,
which yields the F'OC, 8c(q, o:q) p 8q
'.(q ). LJ""l J
The analysis of the textbook regarding private provision of a public
good yields two FOCs. (numbered as in the textbook): 9'>~
I
(x•)
~
p•
with equality if x! > 0 I
with equality
if q > 0
(ll.C.3) ,
(ll.C.4)
If we have a subsidy s for the firm then (ll.C.4) becomes: p• + s
~
Let k • Argmax {9'>~ (x •
1
0
c' (q) )},
with equality if q > 0 .
i.e., k is the index of the consumer with the highest
I
marginal utility at the optimal level x
0
(there will generically be only one .
such consumer, if there are more then simple modifications will do). subsidy in the following way: s = L;ek9'>jCx
0
11 - 7
o· o
).
Set the
The pair (x ,p }, where
p
0
= f/>k(x
0
),
will constitute a competitive equilibrium where consumer· k will
purchase exactly x. = x
0
1
,
and all other consumers will have f/>'.(x 0 ) < p 0 so J
they will not purchase any additional amount of the public good. firm will produce q
0
=x
0
Clearly, the
since it faces a gross price of p + s = ~fl>i (x 0 ). 0
First, we refer to the solution of exercise 10.E.3 for the basic
U.C.J
.
setup and the results of the Ramsey taxation problem without a public good. Second, We modify the setup of this exercise so that the consumers have identical utility functions so that we can view the problem as if we had a representative consumer. The more general case is a straightforward extension, yet the algebra is messier. are not affected.
Furthermore, the qualitative results
Let each consumer's utility function be given by: L
uCx 0 ,xl' ... ,xL) = x 1 +
L ¢ 1Cxl,x 0 ).
l=2
The government budget constraint becomes (the cost is divided by the number
of consumers since we are looking t.t a representative consumer): c(x )
L
0
E tlxL = - -
1=2
I
The Lagrangian of the problem becomes: L L
= xl
+
L
f/>l (xl,xO)
1=2
which yields the FOCs (1)
= l - ;\,l = 0,
(2)
-
11 - 8
-
(3)
(4) Note that the first, second and fourth of these equations are the equivalent to those in the regular Ramsey taxation problem, and the conclusions of that model apply.
The difference is that here the level of taxes to be generated
is endogenous and is chosen to maximize the utility of the representative consumer.
The condition for this is in (3) above, which states that the
marginal social benefit from a marginal unit of the public good must equal the marginal cost of that marginal unit.
11.0.1
Suppose that h
= h2 1
= ... hi
=h•
is a symmetric Nash equilibrium of
The it must be that for every i, h• solves
the simultaneous move game.
h.
;
Max h.
-
1
1
1-lh. Thi+ I
1
h~1 ··2
which yields the FOC 1
h.
;'
1
1
•
1-lh. yht I
1 I-lh. yhi+ I - hi - h. 1 2 l -h + 1-lh" I i I
·r
For h• to be a NE this should hold for h. = h", 1
which yields
h" =
0
- h· = 0,
be a symmetric Pareto optimum. 0
of each student i will be 0
0.
(o: fl·~'(l) )·
Now, let h = h = • · · h • h 1 2 1
h
•
1. e.,
-
ui =
The utility
2
;(1) - (h ) , which immediately implies that 2
= 0 is the Pareto optimal outcome.
The intuition is simple; by studying,
every student imposes a negative externality on others, and the competitive outcome has too much studying.
ll - 9
11.0.2
For
th~
optim~l
Pareto
which yields the FOCs
=ah--.
+
Max
{h.}
r.!_ (t/>.(h.,I:jh .) 1 L11 1 J
+ w.), 1
1
a4,.•
a;.1
r
s 0 with equality if h ~ > 0
a[.h. ·Lk-i
I
J J
I
for all 1=1, ... 1.
outcome we solve
On the other hand, in a competitive equilibrium each
individual solves
Max
f/>.(h.,h.+I: .... h.) + w. , which yields the FOC (for I
h.
I
J
J~l
I
I
I
eveFy i)
8¢. 1 Bh + 1
Bf/>. I .h. :s 0 J J
ar
with equality if hi > 0.
Bt/>k .h. for all k), then if J J
at
externality is negative (i.e,
h~
1
necessarily have h! > h~ (due to the concavity of f/>). I
> 0 and the
If, e.g.•
> 0 .we will
To restore the Pareto
I
optimal outcome in a competitive equilibrium, we must set an individual tax 0
for each i of
11.0.3
0
Bt/>k ( hk I I: .h . ) t. = h J J . 1 k~i j j
r
at
In a competitive equilibrium a price-taking firm j solves Max q. J
where Q = qj + Lk:;ejqk .
n.(q.,q .) = p·q.- c(q.,Q) J J -J J J
1
The FOC for this program is
p- c (q. Q) - c (q.,Q) q J 0 J 1
c
0,
and the SOC is satisfied by the assumptions in the exercise, - c
(q .,Q) - 2c cq . 1 Q) - c cq .,Q) < 0 . qq J q0 J 00 J
Therefore, this program has a unique solution q•, independent of j.
o•
= Jq•, then a competitive equilibrium ( 1)
_ P - cq
o•
•
will be determined
Let
by
co· r· o•l • c0 co· r· o•l .
which has a solution for p > 0 because cq > 0, c
o•
o•
0
< 0, and cq + Jc
together imply that cq(T,Q•) + c Cy-,Q•) > 0.
0
In contrast, the program for maximizing total surplus is
Q
Max S .: ql''''lqJ
J p(x)dx - \ c(q .,Q) , . .L J 0 J=1
J
11 - 10
0
>0
and the FOCs for this program are
:s
=0
= p(Q) - c (q .,Q) q. q J J
'
and we can check that the assumptions ensure that the SOC is satisfied. q
0
Let
denote the solution to this program (again, independent of j), and let
0
Q
0
= Jq .
The (2)
(Again, by cq > 0, c CQ
0
< 0, and cq + Jc
< 0, (2) implies that
and
}o[cq(~,Q)
+
p(Q
c 0 (~,Q)]
0
> 0, a solution will exist.)
0
0 )
< cqcf-.0°)
j(cqq(~,Q)
=
0
assumption), then we must have that Q
+
+
co(~
0
,Q
0
) •
(J+l)cqQ(~,Q)
> o•.
Since
Also, since p' (Q)~O. +
Jc 00 c~,Q))
> 0 (by
This is intuitive since firms
ignore the positive externality that they create, and we have an under-production competitive equilibrium.
To restore efficiency the
government can subsidize production with a subsidy of Firm j's FOC will then be 0
easy to see that Q = Q
11.0.4
p -
and p
0
0
(J-l)c (~ ,Q
=
0
s
o
Qo
= -(J-l)c0 c1 .o ).
= cq(qj,Q) + c (qj,Q), and it is 0 Qo 0 p(Q ) will cause q. = to solve this FOC. J J )
For the Pareto optimal outcome we solve
Max \ __: ¢.(h , ... ,h ) + \ .: n .(h.) , 3 -. cA
so all individuals in neighborhood A cannot be an equilibrium - contradiction.
(b)
Let an equilibrium be a pair
(El A'e ),
8
- -
where eis {8 : type 8 locates in
neighborhood i}, and let e A'eB be the average prestige levels associated with such an equilibrium.
Claim: 9 A must take on the form
A
A
[8,1), for some 9.
Proof: Assume 9' prefers A to B: (l + 8' )(1 + SA) - cA> (1 + 8' HI + SB) - cB. Rearranging gives us: (1 + 9')
CA - CB !:
-eA
11 - 14
-
- eB
,
which implies that all types
locates in which neighborhood, and it is calculated by solving:
.
which yields,
.e
,.
- cA
= o + em
+
... e -) 2
= 2(c A - c l -I. 8
.... (c)
Starting at the equilibrium with e as given above, if a small group of
individuals from the lower end of neighborhood A move to neighborh
