Solutions Manual

for

! Microeconomic Theory Mas-Colell, Wbinston, and Green

Prepared by :

Chiaki Hara i Cambridge Universiry

Ilya Segal

Un iversity oj California, Berkeley • ,

Steve Tadelis

Harvard Un iverSity

I I , ,

New York Oxford OXFORD UNIVERSITY PRESS 1997

•

x - y a nd hence x ~ y. If, on the contrary. u(x) > u.(y), then X > y, and

hence x :: y. Thus, if u.(x) ~ u{y), then x ~ y. So u{·) represents r. -

l.B.S First, we shall pr-ove by induction on the number r..; of the elements of X

that, if there is no indifference between any two different elements of X,

then the:-e exists a utility function . If N =: 1, there is nothing t o prove: •

Just ass ign any number to the unique element. So let N > 1 and s uppose that ·

the above assertion is true for N - 1. We will show that it is st ill true for

N. Write X = {x1' ... . xN_l'xN}. By the induction hypothesis. :: can be

represented by a utility function u(·) on the subset {xl' ...• xN

_1}. Without

loss of generality we can assume that u(x11 > u{xZ) > ... > u(xN_1 1.

Consider the following three cases:

Case L For eve,y , < N. xN >- x .. , Case 2, For every , < N. x. ~ xN· ,

N such that x. >- xN

>- x, , 1 • ' J

Case 3, Tnere exist , < N and j <

Since there is no indifference between two different elements, these three

c ases are are exhaus~jve and mutually exclusive. We shall now show how the

value of u(xN

) should be detednined. in each of the three cases, fo:- u(· ) t o

r eor esent >- on the whole X. . -If Case 1 a pplies, t hen take U(X

N) to be larger than u (x

1). If Case 2

aoolies. • • ·a" " \,. "'-

applies. Le:

XN" ... I >-• • •

x .}. J

u {x.. ,) t o be smaller than N

I = {i e {l, ... , N - 1}: X. , Suppose now that Case 3

and J = {j EO, ... , r-.: - l}:

Campletene!Os and the assumption that there is no indifference

impli es that I v J = 0 •. .. • N - 1). The transitivity implies that both I and

J are " . ." I nT.e 'Y8!S , in the sense that if i E I and i' < i, then i' E l ; and if

J E J and j' > j, then j' E J. Let i- = max I, then j- + 1 = min J. Take

1-2

i

r

r r •

• •

•

I

•

the open interval (u(x .• I),U{X'.»' Then it is t o s~e 1 • 1

that u(· ) rep,eser.ts ?: on the whole X.

Suppose next that there may be indifference bet weer. som~ two elements of

x = { xl •... ' x " J. Fa. each n = 1 •. .. • N, define X ;: {x e X: x - x}. Then, 1', nm mn

bo" the reflex ivity of - (Proposition l.B.Hiill. v

N X = X. Also, by the

n=l n

transitivity of - (Proposition l.8.Hiill, if X ~ . n x . m then n X

m = 0 . So

let M be a subset of O •... ,N} such that x = U MX and X ;: X fo:-mE m m n

any m E M

and any n E M with m ;: n. Define an relation >-* on {X : m e M} by letting X - m m

>-- X if and only if xm ~ x - n - n

In fact, by the definition of M, there is no

indifference between two different elements of {X : m EM}. m

Thus, bv the o

preceding result. there exists a utility function u*(') that represents ~ .. -= u·(X ) if m e M and x

m n EX .

m It is easy t.o

show t.hat, b y t.h~ tranS itivity, u(·J represents ?:.

I.e,! Ii y E C({x,Y,z}l. then the \VA would imply that Y E C({x,y}). B~t

contradicts the equality C({x,y}) = {x}. Hence Y E: C({x,y,z») . Th'JS

C« X,y ,Z}) E {{x},{z},{x, z}}.

I.C.2 The p:-ope .... t y in t he q uestion are equiva lent t o the f ollow ir.g prope:-ty ;

If B E ~ , B' E $ , X E B. 'i E 8, x E 8', Y E 8'. x E C(E ), and Y E C(S' ), t hen

X E C(E') and Y E G(E). We shall thus prove the equivalence betweer. this

property and the Weak Axiom.

. S up;lOse :~irs·. '-hat t he Weak Axiom IS sati sfied, Assume t ha', 8 e 2, S ' E

~ , X E 8. Y E B , XES' , Y E 8' . X E C(Sl , and y E G{E' ). If we app ly the

Weak Axiom t\ ... ·ice, we obtain x E e(B') and y E G(S ). Hence t he above Drooertv • • 0

1-3

.. ,- ".,- -,,_.. --.'"'

r Suppose conversely that the above property is satisfied. Let B E :8, x Ii:

B, Y E B, XES', and x E C(S). Furthermore, let 8' E 'E, x E 8', Y E B', and

Y E C(S'). Tnen the above condition implies that x E CIS' ) (and Y E em )). r Thus the Weak Axiom is satisfied.

f i.,

l.C.3 (a) Suppose that x >-- y, then there is some E E :B such that x E 8, y E

r B, x. E C(B), and Y E CeB). Thus x ~. y. Suppose that y ~- x, then there

exists B E :a such that x E B, Y E B and x E C(B]' But the Weak Axiom implies

•

that Y E e(B), which is a contradiction. Hence if x. ).- y, then we cannot have ;

Y )-- x. He~ce x )-•• y . -Converseiy, SUppose that x )-•• Y. then x ~. y but not y Hence

• •

there is some B E 1l such that x E B, Y E B, x E erB) and if x E E' and YES'

fa, any E' E ~ , ther, Y E em' }. In particular, x E GtE) and y E e(E )' Thus •

x )-. y

The equality of the two relation is not guaranteed without the WA. As

car. be seen from the above proof, the WA is not necessary to guarantee tha: if"

x )-•• y, the:! x )- . y . But the converse need not be true, as shown by the

fcUo·.ving example. Define X· = {x·,Y.z}, :B = {{x,y},(x,y ,z}}, C({x ,y} ) = {x},

and C({x, y,z}) = {y}. Then x )-. y and Y )-. x. But neither x >- - y nor y >-- x .

•

(b) The re lation )- . need not be transitive, as shown by the following example.

Define X = {x,y,z}, ~ = {{x,Y},{y,z}}, C({x,y» = {x} and C({y,z} ) = {y}.

The:l. x >-* y and y )-- 2. But we do not have x )-- z (because neither of the tWO -sets m :B incl udes {x,z}} and hence we do not have x )-. z either.

(e) According to the proof of Proposition l.D.2, if 'B includes all three-

e:er.1er,t subset of X, then ::* is transitive. By Proposition l.B.Hil, F·· is

1-4

transit ive . Since)-·)s equal to )-•• , )-. is also transit ive.

An alternative proof is as foll ows; Let x e X. y e X, z e X. x }-W y, and

y )-- z. Then tx.,y,z} E. !B and. by (a). x )-.- y, a nd y )-•• z. Hence we have

nelthe:- y )-- x nor Z >-- y. Since >-- rationalizes (~,C(·)}, this imolies that - - - . y ~ C{(x ,Y.z}) and z f. C({x,y,z}). Since C«x,y.z}) ~ 0, C({x,y,Z} ) = {x}.

Thus x >-. z.

l.D.l The si!Iplest example is X = {x,y}, !8 =- {{xl, {y}}, C({x}} "" {x}, C({y})

= {y}. The!"! at]" rational preference relation of X rationa lizes C( · l.

1.0.2 By Ex~cise 1.B.S. let u{·) be a utility representation of >-. Since X -is finite. for any 8 c X with B ~ 0, there exists x E 8 such that u(x) ~ u(yl

for al l Y E B. Then x E CW(B.~) and h~ce C·(B.t1 ~ 0. (A direct p:-oof with

no use 0: utility representation is possible, but it is essentially the same

as the proof of Exercise 1.8.5.)

1.0.3 Su?pose that the Weak Axiom holds.· I f x E C(X ), then x E C({x ,z}), • •

which cor.tradlcts t he equality C({x,z} ) = {z}. If Y E C(X), then), e

C({X,y»), wn: d contradicts C({x,y}) =- {x}. If z e C(X ), ther.. z e C({y,z} )'

which comracrts C{{x.z}) = {y}. Thus (!B,C(')) must vio late the Weak Axiom.

1.0.4 Let ~ ntionali~ ct·) relative to 1l. Let x E C(Bl

v B2

) and Y E C(Bll

U C(E2

). t hec:l :1. ~ Y b~a~ B1 v Bz

:l e(E I) v C(S21. Tnus x E C(C(S I) v

CIB2

11.

Le: x € ~C(81) u C{B2)l and y E B1 v 82

, then there a,e f our cases:

Case 1. x e C(Bl), y e B

1.

1-5

--_ . . .

r Case 2 . X'. e (SI). y e 8

2,

Case 3. x E C(8zJ. y e B)"

x e CCB21. y • H2· r Case 4.

If either Case 1 or 4 is true, then x >- y follows directly fror:: -rationalizability . r If Case 2 is true, then pick any z e C(Bz l. Then z ).- y. -

Hence, -by the transitivity, x >- y. - If ~ z. -f .

Case 3 is true, then pick any Z E C(BI' and do the same argurne:1.t as for Case

2. , . ! •

1.D.5 (a) Assign probability 1/6 to each of the SIX possible p,eferences,

which are x >- y >- z, x >- z >- y, y >- x >- z, y >- z ,.. x. z >- x >- y. and z >- y >-

r X. ,

(b) If the gIven stochastic choice function were rationalizab!e , ther. the _. p robability that at least one of x >- y, y >- Z, and z >- x holds wou!c be at

mas: J x 0/4) = 3/4. But, in fact, at least one of the three relations

a lways hol ::is , because, if the first two do not. hold. t hen y >- x a:J.d z >- y.

Hence the transitivity implies the third. Thus. the given stccha.stic choice

•

function is not: rationalizable.

(c) T3e same argu ment as 10 (b ) can be used to sho w t hat a. ~ 1/3. Since

C(x.y}) = C({y,z}) - C({z.x}) :::: (a.. 1 - 0:) is equivalent to C{{y,x}) ::::

C({z.y}) = C({ x ,Z}) (I - 0'., 0'. ) , if we apply the same argumer.t as in (b) to y

)- x, z )- y. a :1d X )- z . t hen we can establish 1 - a ?! 1/3. that I S . 0'. -::: 2/3.

Thus, in order f o. the given stochastic choice function is rationalizable, it

is necessary that ex. e [1I3.2I3 J. Moreover . this condition 15 actuaIly

s ufficie:1t : Fo. anv 0:. E [1/3,2/3J. assign probability a. - 1/3 t o each of x )-•

}-6

-- ... , .. .

',"

y ;- z, y ;- z >- x , and z >- x >- y; assign probability 2/3 - a: t o each of x ;- z >­

y, y >- x >- z , and z >- y >- x. Then we obtain t he given stochastic choic~

functi on.

•

1-7

--------

CHAPTER 1

l.B.l S ince y >- z implies y >- Z, the transi tiv ity implies tha t x ;- z . - -Suppose that z >- x. - Since y ~ z , the transitivity t hen implies that y >- x. -But this contradicts x >- y. Thus we cannot have z >- x. - Hence x >- z.

1.B.2 By the completeness , x >- x f or - every x E X. Hence the,e IS no x e X

such that x >- x. Suppose that x >- y and y >- Z, then x >- y ;: z. By (jii) of

Proposition LB. 1, which was proved in Exercise L8.1, we have x >- z . Hence >-

IS trans itive. Property (jJ 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 reflexive. Suppose that x - y and y - z . Tnen x t y, Y t z. y = x,

and z >- y. - By t he transitivity, this implies that x t z and z >- x. - Thus x -

z . Hence - is t r ansitive. Suppose x that - y. Then x >- y and y >- x. - - Thus Y

?; x and x ~ y. Hence y - x . Thus - is symmetric Property Oil is now

prnved.

,

LB.3 Le;: x e X and Y E X. Since u(') r epresents >-, x }- v - -' . ," ""d o"lv ,'c l. _.. " . '

u( x } ~ u(yl. Since f { ' ) is strictly increasing, u (x) it u ( y) if and only if

v { x ) ~ v ( y ). Hence x ~ y if and only if vex) ::!: v(y). Theref 0:'1": v ( . )

represents ~.

1.BA Suppose first that x ?::' y. If. furthermore , y ?::' x , then x - y and hence

u tx ) u ty). If , on the contrary, we do not have y }- x, t h1":n x }- y. Hence -u{ x ) > u (y} . Thus , if x ?::' y, then u(x) ~ u(y).

Suppose conversely that u(x ) ~ u(y ). If , furthermore, u (x) = u (y), then

- H

•

x - y a nd hence x ~ y. If, on the contrary. u(x) > u.(y), then X > y, and

hence x :: y. Thus, if u.(x) ~ u{y), then x ~ y. So u{·) represents r. -

l.B.S First, we shall pr-ove by induction on the number r..; of the elements of X

that, if there is no indifference between any two different elements of X,

then the:-e exists a utility function . If N =: 1, there is nothing t o prove: •

Just ass ign any number to the unique element. So let N > 1 and s uppose that ·

the above assertion is true for N - 1. We will show that it is st ill true for

N. Write X = {x1' ... . xN_l'xN}. By the induction hypothesis. :: can be

represented by a utility function u(·) on the subset {xl' ...• xN

_1}. Without

loss of generality we can assume that u(x11 > u{xZ) > ... > u(xN_1 1.

Consider the following three cases:

Case L For eve,y , < N. xN >- x .. , Case 2, For every , < N. x. ~ xN· ,

N such that x. >- xN

>- x, , 1 • ' J

Case 3, Tnere exist , < N and j <

Since there is no indifference between two different elements, these three

c ases are are exhaus~jve and mutually exclusive. We shall now show how the

value of u(xN

) should be detednined. in each of the three cases, fo:- u(· ) t o

r eor esent >- on the whole X. . -If Case 1 a pplies, t hen take U(X

N) to be larger than u (x

1). If Case 2

aoolies. • • ·a" " \,. "'-

applies. Le:

XN" ... I >-• • •

x .}. J

u {x.. ,) t o be smaller than N

I = {i e {l, ... , N - 1}: X. , Suppose now that Case 3

and J = {j EO, ... , r-.: - l}:

Campletene!Os and the assumption that there is no indifference

impli es that I v J = 0 •. .. • N - 1). The transitivity implies that both I and

J are " . ." I nT.e 'Y8!S , in the sense that if i E I and i' < i, then i' E l ; and if

J E J and j' > j, then j' E J. Let i- = max I, then j- + 1 = min J. Take

1-2

i

r

r r •

• •

•

I

•

CHAPTER 8 •

• •

8.B.l Finll i chooses h. to . 1

ize CU:.h. + ~m.h.) - w'(h.)2. The JJ JJ 11

F.O.C. is: Cl + ~(.TT.h') - 2w.h. = O. The best response function for firm J~ 1 J 1 1

• • 1 IS

1 J 1 J Wi has a strictly dominant

strategy iff f3 = 0, i.e., if the best response function of i is not dependent

on the action of the other firms. If f3 = O. firm j's strictly dominant

strategy is h. = 1

Cl

2w .. I

•

8.B.2 Ca) Suppose s~ E s. and s~ E s. are two weakly dominant strategies for I 1 1 1

player i. This implies that u.(s~,s .) i!! u.(s~,s .) V S~E S. and V s .E S ., 1 1 -1 I I -1 I 1 -1-1

and u.(s~, s .) i!! u.(s~, s .1 1 1 -1 1 1 -1

V S~E S. a~d .' 5 • E S .. In particular, u.(s~, 1 1 -1 -1 1 1

2 2 s .) i!! u.(s., s .) and u.(s., -1 1 1 -1 1 1

s .) -I

1 i!! u.(s., s.) V s . E S .,

1 1 -I -1 -1 1 . 2

u.(s., s.l = u.(s., s.) V 5. 1 1 -1 1 1 -1 -I

E S .. -1

•

(b)

Player 2 L R

• u 1 , 4 2,5 •

Player 1

D 1 , 2 • 2,3

Figure 8.B.2

Therefore,

Both of player 1'5 strategies (U) and (0) are weakly dominant. However,

player 2 prefers that player 1 uses strategy (Ul.

•

•

8-1 •

------ --- . . . .. . __ .- ~--------- . ._ .. _--------- - - ---- -~

•

,. ,

•

.-

•

• •

r .......

•

8.B.3 Suppose not. Assume bidder i bids b. > v.. Then if some other bidder 1 1

bids something larger than bi, bidder i is just as well of as if he would have

bid v.. If all other players bid lower than v., then bidder i obtains the 1 . 1

object pays the amount of the second highest bid. If the second highest

bid is b. < v., this results in the same payoff for player j as if he bid v .. J 1 1

However. suppose that the second highest bid of the other is b / v r Then, by

bidding bi

bidder i will win the object and obtain a negative payoff. By

bidding v. he will not win the object and obtain a payoff of zero. Therefore, I

bidding b. > v. is weakly dominated by bidding v .. I 1 1

Suppose bidder i bids b. < v.. Then if all other bidders bid something 1 J .

.

smaller than b., bidder i is just as well of as if he would have bid v.. He 1 1

•

will win the object and pay the the second highest bid. If some other player

bids higher than v., then bidder i does not win the object regardless Whether 1

he bids b. or v.. However, suppose that nobody bids higher than v. and the 111

highest bid of the other players is b. with b.< b.< v.. Then by bidding b. J 1 J 1 1

bidder i will not win the object, therefore getting a payoff of O. By bidding

•

v .• he would win the object, pay b. < v., and thus obtain a payoff of v. - b. 1 J 1 1 J

> O. Therefore. bidding b. < v. is weakly dominated by bidding v.. This 111

argument implies that bidding v. is a weakly dominant strategy. 1

8.S.4 Call the set of strategies for player i that remain after N rounds of

deletion of strictly dominated strategies Suppose S. E . is a strictly 1 1

dominated strategy given the strategies . of the other players Therefore. -I

there exists a strategy s! E .• which is not a strictly dominated strategy 1 1

• gIven " which strictly dominates s,. Suppose further that s. will not be -1 1 I

deleted in the N+l round . •

•

8-2

------ - - - - -- - --- .. - - -- - - -_ .. ------- - - ------: - ------- - - - - -

•

•

Since s. was strictly dominated by s! I I

• given ., it -1

will still be

strictly dominated by si given +~ S; -1

d • +1 (. h . ., an s. E. with t e given -1 1 1

assumptions). Thus the strategy s. will 1

be deleted in the next round.

(Note: If s. is only weakly dominated by s! given 1 1

.• then s. may no longer -1 1

be weakly dominated given +1 •

• Since -1

+1 . may no longer include the strategies -1

of the opponent relative to which some other strategy of player i will •

strictly be better. Thus the order of deletion does matter for the set of

strategies surviving a process of iterated deletion of weakly dominated •

strategies).

8.B.S (a) Suppose, player j produces q.. Player i's best response can be J

calculated by maximizing (this is symmetric for both players): •

max (a - b(q. .. q.l - cJq. 1 J 1

•

which yields the F.O.C: (a - b(2q. + q.) - cJ = 0, so the best response is: 1 J

b.(q.) = 1 J

a-c 2b

-q.

J 2' •

•

•

Now, since ql ~ 0, q2 ~ (a-c)/2b (all other strategies would be strictly

•

dominated by q2 = (a-c) I 2b). Therefore, since q2 :s (a-c)/2b, we have that •

ql ~ (a-c)/2b - (a-c)/4b = (a-c)/4b. Thus, since ql ~ (a-c) I 4b, then

q2 ~ (a-c)/2b - (a-c)/Sb = 3(a-c)/Sb. Continuing in this fashion we will

obtain: q = (a-c)/2b - q/2. Thus, after successive elimination of strictly

dominated strategies, ql = q2 = (a-c)/3b.

(b) Suppose, player j produces qj and player h produces qh' Player 1's best

response can be calculated by maximizing (a - b(q. + q. + qh) - clq., which 1 J 1

yields the F.O.C [a - b(2q. + q. + qh) - cl ::: 0, implying the best response: 1 J

b. (q .,qh) = (a-c)/2b - (q. + qh)/2. 1 J J

Now, since q2' q3 ~ 0, ql :s (a-c)/2b (all other strategies would

strictly be dominated by ql = (a-c)/2b). Thus, since ql ~ (a-c)/2b and

•

S-3

•

•

•

•

similarly q3 :s {a-c}/2b. we have q2 ~ {a-c)/2b - [2(a-c}/2b)/2 ... o. •

Therefore, successive elimination of strictly dominated strategies, implies

that ql' q2' q3 ~ 0 and ql' Q2' Q3 :s (a-c)/2b. However, a unique prediction

cannot be obtained. •

•

8.B.6 Suppose s! is strictly dominated by the strategy cr!. Suppose further .11

that cr i is a mixed strategy.in which s1 is played with strictly positive

probability cr.(s!»O. We claim that cr. is strictly dominated by the mixed 1 1 1

strategy cr~, which is equivalent to cr. except that instead of playing s! with 1 1 1

•

probability cr.(s!) it plays cr! with probability cr.(s!). This follows since: 11111

u.(cr., cr.) := 1: cr.(s,} u.(s., cr .) 1 1 -1 1 1 1 1 -1

= ~. cr.(s.l U.(s., cr .) + cr.(s!) u.(s!, cr .) s. s. 1 1 1 1 -1 1 1 1 1 -1

1 1

< ~. cr.(s.l u.(s., cr .) + :.(S!) u.(cr!, cr .) s. S. 1 1 1 1 -1 1 1 . 1 1 -1 1 1

= u. (cr~. cr .1 1 1 -1

for all cr . E A(S .). -1 -1

•

8.B.7 Suppose in negation that cr. is a strictly dominant mixed strategy of 1

player i. and suppose, s~ ..... s~ are the pure strategies that are played 1 1

•

with positive probability in the mixed strategy cr .. 1

Since cr. is a strictly 1

dominant strategy: u.(cr., s .J > u.(s!, s.) 't:I S!E S. and 't:I s . E S .. In 1 1 -1 1 1 -1 1 1 -1-1

• J particular. u.(cr .• s .J > u.(s .• s .) 't:I j = 1 ..... N. This implies that

1 1 -1 1 1 -1

u.(cr .• s .1 1 1 -1

N . > .1: l[cr. (s~)

J= 1 1

•

u.(s~,s .)) = 1 1 -1

u.(cr .• s .) , a contradiction. 1 1 -1

B.C.l 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 never •

a best-response. This contradicts the assumption that the strategy is

rationalizable. Therefore, both procedures must yield the same prediction.

8.C.2 Call the set of strategies for p}::iyer i that remain after N rounds of

deletion of never best-response strategies

.

Suppose s. is never a 1

best-response to any strategy in • • -1

Suppose further that s. will not be 1

deleted in the N+1 round. Since s. was ne~er a best response to a strategy in 1

+1 • .' ., it will clearly not be a best-response to a strategy in

-1 -1 -1

Thus this strategy will be deleted in the next round.

8.C.3 Suppose that sl is a pure strategy of player 1 that is never a best

response for any mixed strategy of player 2. Suppose in negation that

not strictly dominated. Construct the following correspondence for any

0'. e fl(S.) for i = I, 2: 1 1

A A A

(0'1,0'2)" {O'll 0'1 e argmax g 1 (0' 1'0'2)} x {0'2 1

The first part of this correspondence is the best response function for

player I 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

•

. --------------------------------------------------------~

o

strategies of player 2, ror ,!"hich sl is not strictly dominated (it is a

non-empty set since 51 is not a strictly dominated strategy. i.e., • • It IS not

strictly dominated by err)' Therefore, the second part of the correspondence

is convex valued and upper hemicontinuous due to the usual assumptions. Thus,

by Kakutani's theorem there exists a fixed point (cri ,cri ) of this

correspondence such that glCsl' cri) ~ gl(cri' cr2) from the second part of the

correspondence, and gl(1. cri} ~ gl(cr1, cri} for all crl

E ,d(Sl)" Therefore,

gl (sl' cr2.) ~ gl (cr1, cr2.) for all cr 1 E AeS l ), which contradicts the assllmption •

that sl is a pure strategy of player 1 that is never a best response for any

mixed strategy of player 2. Therefore, if sl is a pure strategy of player 1

that is never a best response for any mixed strategy of player 2, then •

strictly dominated by some mixed strategy of player 1. . •

8.C.4 (First Printing Ell ata: a typo appears in the lower left box of the

payoff matrix. Player 1's payoff should be n + 4c and not 1) + 4c.]

For the continuation of this answer, a strategy for player 2 is to play u with

probability 0: and D with probability 1-0:, and for player 3 is to play l with

probability and r with probability 1-(3. Denote by P A the expe.~ted payoff of

player 1 when action A € {L,M,R} is taken given 0: and (3. Direct calculation •

•

and simple algebra yield:

P = n + 30: + 3{3 - 30:{3 - 1 M 2

P L = n + (213 - 1)c

• P

R = n + H - 2(3)c

(a) To show that M is never a best response to any pair of strategies of

players 2 and 3, (0:,(3>, we have three cases:

Case 1: (3 > 112

Note that in this case --;;-- = 11[3/2 - 3(3) < O. Thus the highest payoff for lh

8 - 6

•

•

player 1 if he plays M is obtained when ex = 0, and his payoff will be P M(ex=O) •

3 3 = If + 11[~ - 11 < If + 4c[~ - 11 < If + 4c[~ -

.

11 = PL' Further note that P L •

•

is independent of ex, so that these inequalities hold for all ex. Therefore, M

cannot be a best respo~se in this case . •

2: 13< 1/2

8PM

Now, 8ex > 0, the highest payoff for player 1 if he plays M is obtained when

ex = I,

Further note

1 3 If + 11(- - ~) <

2 2

that PR

is

independent of ex. so that these inequalities hold for all ex. Therefore, M

cannot be a best response in this case. .

Case 3: 13= 1/2

In this case PM = If 11 - -4 < If This concludes that M can never be a

best response.

(b) Suppose in negation that there exists a mixed strategy. in which player 1

plays R with probability 7 and L with probability 1-7. that strictly dominates

M. • •

Case 1: 7 ~ liZ.

If 13 = 0 and ex= 1 then PM = Tl + 1)/2 > Tl. The mixed strategy will give a

payoff of lr - 4c(1-27) ~ Tl. Therefore, M cannot be a strictly dominated by •

the mixed strategy in this case.

Case 2: 7 > liZ.

•

then PM = Tl + 1)/2 > Tl • The mixed strategy will give a If f3 = 1 and ex= 0

payoff of Tl- 4d27-1) ~ If. Therefore. M cannot be a strictly dominated by .

the mixed strategy in this case. This implies a contradiction, so that M

cannot be strictly dominated.

!c) Suppose players correlate in the following way: Players 2 and 3 play •

•

•

8-7

•

(U, r) with probability 1/2 and (0, l) with probability 1/2. Any mixed •

strategy for player 1 involving only Land R will give him a payoff of ll'.

However, playing M will yield him a payoff of ll' + 1)/2. Thus M is a

best-response to the above correlat~d strategy of player 2 and 3.

8.0.1 We know already from section 8.C that a4and b4 are not rationalizable

strategies. Thus, these strategies cannot be played with positive probability

in a mixed strategy Nash equilibrium. Suppose that there exists a mixed

stratev equilibrium in which a1

and a3 are both played with a strictly

positive probability. Then the expected payoff from playing either one of

them has to be equal (see exercise 8.0.2). This implies that the probability •

that player 2 plays bi

has to be equal to the probability that he plays b3

. •

Now, suppose that player 2 plays b. and b_ with probability ex and b2

with 1 ,

probability 1-2ex. The expected payoff for player 1 obtained by playing either

al

or a3

equals: 1ex + 0-2ex)2. The expected payoff for player 1 when playing

a2

equals: Sc + Sex + 0-2ex)3 = lOex + 0-2cx)3 > 1ex + 0-2cx)2. Therefore, in a

mixed strategy equilibrium al

and a3

cannot both be played with positive

probability since playing a2

would give the player a larger payoff.

Suppose, there exists a mixed .strategy equilibrium in which player 1

plays a I and a2

with strictly positive probability. Clearly, player 2' s best

response to this strategy of player I does not involve playing b3

with

strictly positive probability (given the strategy of player I, playing b2

is

strictly better for player 2). Thus player 2 will play bl

with probability

and b2

with probability 1-/3. The payoff for player 1 from playing a l

equals: O-f3}2, playing a2

yields: 5/3 ... 0-/3)3 > 0-/3)2. Therefore, in a

mixed strategy equilibrium a1

and a2

cannot both be played with strictly

positive probability since playing a2

is always better. ,

•

8-8

•

•

•

Similarly, it can be shown that there exists no mixed strategy

equilibrium in which a2 and a3 are both played with strictly positive

probability. Therefore, player 1 always plays a2

in a Nash equilibrium.

Player 2 will then play his best resp.mse b2. Thus (a2 • b

2) being played with

certainty is the unique mixed strategy equilibrium.

• -8.D.2 We will show that any Nash equilibrium (NE) must be in

CIIl S • 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.si •...• sj) be a (mixed) NE and suppose in negation that it

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 CT. and a. such that 1 1

k-l U.(CT.,!'; .) '" u.(a.,s .) '<I s .E S . , and a.

1 1 -1 1 1 -1 -1 -1 1

is played with positive probability s!(a.). Since k is the first round at 1 1

which any of the NE strategies, (si,sz, ... ,sj). are ruled out, we must have

k-l that S·.E S ., Hence. U.(CT .• S·.) > u.(a .• s·.). Let the strategy s~ be

-1 -1 1 1 -1 1 1 -1 1

derived from s! 1

execpt that any probability of playing

playing CT •• We thus have that: 1

a. is replaced by 1 .

( ' .)- ( .. ) u.s.,sl -u.s.,sl 1 1 - 1 1 -

+ s!(a,)·(u.(cr.,s·l) - u.(a.,s·l)] > u.(s!,s·l) 1111- 11- 11-

which contradicts the assumption that (si,sz, .... sj) is a NE.

•

8.D.3 First of all, notice that the first auction bid is a simultaneous move •

game where a strategy for a player consists of a bid. Let bl

be the bid of

Player 1, and b2

be the bid of Player 2.

(j) If bl

> b2

, Player I gets the object and pays b1

for it;

Player 2 does not get the object. Thus. in this case:

•

8-9 •

,

,

U 1 (bi, bl)=v l-bl 11: s va 1 uation of the object

minus what he has to pay for itJ.

•

,

Therefore, we have for i,j E {t,l}, i ;e j:

0, b. < b. 1 J

I 2: (Vi - bi) • b i = b2

(v.-b.). b.>b. 1 • 1 J

,

•

/", (a) We claim that no strategy for player I is strictly dominated. Suppose in

negation that bi

is strictly dominated by bit Le., for any bi u1(bi,b

2) >

u 1(bl ,bl )· Take bi = max {b1,bi} + I,then: bi > bI' and bi > bi. Hence,

,

•

f '

u 1 (bi,bi) = u l (bl,bi) = 0, a contradiction. Therefore. no strategy for player

1 is strictly dominated. 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 bi

for Player

weakly dominated by v l' Note that, if bl

> v I:

•

Oi)

8-10

-- ------------ - - - - - ------ -•

8.£.2 (a) Suppose that all the bidders, use the bidding function b(v), that

is if their valuation is v they bid bey). The expected payoff for a bidder

whose valuation is v. · is given by: 1

(v.-b(v.»·Pr{b(v.) > b(v.)} + O·Pr{b(v.) < b(v.H. (Note that we ignore a tie 1 1 1 J 1 J

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.n II: Pr{v. > v.> = v.l v (since the valuations are . 1 J IJ 1

-uniformly distributed on [0, vJ. bey) 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 FOC is: 1

(v. - b(v))/v - b' (v) vi';', · = O. bey) is an equilibrium bidding function if 1

it is optimal for the bidder not to pretend to have a different valuation,

that is, if v = v. is the optimal solution to · the above FOC, i.e., if . 1 .

•

- -(v. - b(v. »/v - b' (v.) v.lv = O. This is a differential equation that has to 1 1 1 1

be satisfied by the bidding function b( v) in order to be an equilibriu!l1

bidding function. The solution to this differential equation is b(v) = v/2.

Thus a bidder whose valuation is v will bid vIZ (a monotonic linear function).

•

(b) We can proceed as above by assuming that all bidders use the same bidding

function b( v). Now, Pr{b(v.) > b(v') 't/ j ~ j} = = Pr{v.> v. 't/ j ~ i} = 1 J 1 J

1-1 -(v,) I v. Proceeding as in a) above, we get the following differential 1

equation:

(I-IHv. - b(v. »(v. )1-2/~ -III

differential equation is bey)

b'(v.) (v,)I-l/v = O. The solution to this 1 1

1-1 = I . v.

I-I . As I ... CIO, bey) = I ·V ... v, l.e., as

the number of players goes to infinity each player will bid his valuation . •

•

8-21

8.E.3 A firm of type i = H or 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,U firm 1 will maximize:

Max 1 q. 1

1 2 1 O-",)[(a - b(q.+ qH) - c.)~.] + ",[(a 111

•

111 _. . - .

1 c.)q. I 1 1

- c.] = O. In 1

1 2 qL = qL = qL'

Plugging this into the F.O.C we get the following two eqations:

(l-",)(a - 3b qH - cHI + Il(a ~ b(2qH+ qL) - cHI =- 0,

Cl-",)[a - b(qH+ 2qL) - cL] + ",(a - 3b qL - cLJ • o.

Therefore, we obtain that

q = H

q = L

a - c H

+ ~(c 2 L

a - c + L

8.F.l For the proof of Proposition 8.F.l, 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 Equilibrium 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

. ,

•

•

•

•

• J - ....

CHAPTER 9 •

• •

9.8.1 There are 5 subgames, each t"lne beginning at a different node of the

game (this includes the whole game itself).

•

9.8.2 (a) Clearly if (1"0 is •

a Nash equilibrium . of r E' and rE

is the only •

•

proper subgame of r E' then (1" 0 induces a NE in every proper subgame of the game

r E. Thus, by definition (1"0 is a subgame perfect NE of r E . .

(b) Assume in negation that (1"0 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. nE) of rEin which the

o restriction of (1" to TIE is not a SPNE. This implies that there exists a

proper subgame (say. 0El of nE

in which the restriction of (1"0 to 0E is not a

NE. Since 0E is a proper subgame of nE

and nE

is a proper subgame of r E' then

~ is also a proper subgame of r E. Therefore, (1"0 does not induce a NE in a

proper subgame of r E - contradiction.

9.B.3 ~et player 1's pure strategy be sle{L,R}, player 2's be s2e{a,b}, and

•

player 3's be s3= (x,y,z) where x,y,z e{l.r}, x is what 3 does after player 1

played L, y is what 3 does after 1 played Rand 2 played a, and z is what 3

does after 1 . played . Rand 2 played b. The pure strategy SPNE identified in the

example is (R,a,(r,r,l)), which is easily seen to be a NE. Three other NE

which are not SPNE but yield the same outcome are (R,a,O,r.I), (R,a,O,r,r»

and (R,a,(r,r,r». For each of these NE, player 3 is not choosing a rational

move for some of his nodes. If player 3 is reached, he will always do the best

thing for himself, therefore if 1 plays L, ;3 will play r. To support this as a

9 - 1 •

. '"

•

•

•

•

CHAPTER 9 •

• •

.

9.B.l There are 5 subgames, each "ne beginning at a different node of the

game (this includes the whole game itself).

•

o . 9.B.2 Ca) Clearly if (T is a Nash equilibrium . of

•

•

•

and rE

is the only

proper subgame of r E' then (T 0 induces a NE in every proper subgame of the game

o r E' Thus, by definition (T is a subgame perfect NE of r E' .

•

(b) o

Assume in negation that (T 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, TIE) of rEin which the

restriction of (TOto TIE is not a SPNE. This implies that there exists a

proper subgame (say, 0E) of ITE in which the restriction of (TOto ~ is not a

NE. Since 0E is a proper subgame of ITE and TIE is a proper subgame of r E' then

'1: is also a proper subgame of r E'

proper subgame of r E - contradiction.

o Therefore, (T does not induce a NE in a

9.B.3 Let player l's pure strategy be sle{L.R}, player 2's be s2e{a,b}, and . .

player 3's be 53= (x,y,z) where x,y,z eO,r}, x is what 3 does after player 1

played L, y is what 3 does after 1 played Rand 2 played a. and z is what 3

does after 1 played Rand 2 played b. The pure strategy SPNE identified in the

example is (R,a,(r,r,l)). which is easily seen to be a NE. Three other NE

which are not SPNE but yield the same outcome are (R,a,(I,r,I), (R,a,(I,r,r»

and (R,a,(r,r,r». for each of these NE, player 3 is not choosing a rational

move for some of his nodes. If player 3 is reached, he will always do the best

thing for himself, therefore if 1 plays L, 3 will play r. To support this as a

•

9 - 1

(0.-0.0.) Player j will not accept, and offer __ l_l--=J_. v to

(1-0.0.) player i in the next .

1 J

period. Player i will then accept and obtain a (0.-0.0.)

payoff of 0.' 1 1 J 'v = 1

(1-0.0.) 1 J

2 (1-0 i ) (I-o . ) o. . . v < __ ....:J"--. v.

1 Thus he will choose not to deviating from the

0-0 . 0 .) 0-0 . 0 .) 1 J 1 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. (982) "Perfect Equilibria in a Bargaining Model,"

Econometri.ca, 50:97-109 (pp. 107-108).

9.8.14 Ca) The extensive form ot the game is depicted in figure 9.B.14(a)

below (the letter "d" is used instead of 0). 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.14(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 In. 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 I:

2 2 z + oy + 0 x = z + o(y + ox) > z + 00 + o)z = Z + OZ + 0 Z .

This causes firm E to choose Out in t=2. Working backward we get that at t=I,

•

firm I chooses Accommodate and firm E chooses In. However, the choice of firm •

E at t=O depends on the value of k. If k > 1 then firm E will choose not to

enter, and if k < 1 then E will enter. For k = 1 both are part of the

9 - 11

,

••

•

(unique) continuation SPNE. so there are two SPNE in this case . •

,

•

Not Enter

Out In

1+d -k I

+dz+d2

Ac.

• •

Enter t 0

In

Ac. Ft.

Out In •

I 1- d-k

•

Out In

I -1 +d -k

T I I

t 1 I I

Out

-1-<1 -k

In

+dy+d2

y+dz+d\ y+dy+d 2

Ft. Ac. Ft. Ac. Ft.

2 2 1- d+d -k 1- d- d -k 2 2

1 +d+d -k -1 +d- d -Ie 2 2

+dy+d z z+ y+ ~ y+dz+d2

+dz+d2y

re 9.B.14 a •

9 - 12

Ac.

2 1- d+d-

+dy+d2

•

•

T I I

t

I I

Ft.

2 -1- d- d

+dy+d

3

------ --- - --- - - - ----~---------,-, - ' -•

•

•

•

r ....... ·

Not Enter Enter

? •

•

Out In

I-k

+dx+d2x Ac. Ft.

1 +d+d2-k 1 +d_d2_k z+dz+cfz z+dz+d1y

,

t , 0

In

Ac. Ft.

E

Out

-I - k •

y+dx+d2

Pi re 9.B.14 b

9 - 13

T I I

t 1 I I

t

•

T •

I I

2

I I

------- - -- ."."".,- --- ,---,. " - ,. _ ------- - - ' .. ------

I

9.C.l First. if r1" is a NE then (i) and (ij) 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 r1" is a NE. If Oi) 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 CO and (m are satisfied, then r1" 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.

9.C.2 Let r1"r be the probability that firm I fights after entry, let III be

firm I's belief that In 1 was E's strategy if entry has occurred and let r1" 0'

r1" I' CT.2

denote the probabilities with which firm E actually chooses Out, In1, .

and In2

respectively. •

As in example 9.C.3. we cannot have 2 III < 3 (same analysis). We could, however.

have 2

Il > -1 3 : In this case firm I will play "Fight" with probability 1. and

having 7<0 implies that firm E will choose "Out", which supports any beliefs

in the information set of firm I. This concludes that one class of weak

perfect Bayesian I

playing "Fight".

• '"'1= ~, and the analysis is as in example The second class of weak PBE is where

9.C.3 with crr = l/(r+2), and (0"0''''1''''2)

equilibrium only if '1 is "large enough" in the interval (-1.0). When '1 is

close to -I, then r1" F is close to 1. and the expected payment for firm E of the

•

• •

9 - 14

•

•

•

-".

.

strategy (0"0,0"1,0"2) = (O.~.~) is close to -I, therefore E would choose "Out" •

with probability 1. If, on the other hand, 7 is close to 0, then 0" F is close

1 to 2' and the expected payment for firm E of the strategy (er O,er

1,O" 2) =

The set of 7'S for which the equilibrium of example 9.C.3 will still be an

equilibrium can be calculated.

•

9.C.3 (a) For given values of A and ~, there exists the following weak

perfect Bayesian equilibrium (WPBE):

•

The seller will demand a price of (l-~)vH + ~.vL in the first period and

vL in the second period. A buyer of type L will buy the good if the price is -

at most vL

. A buyer of type H will buy the good in the first period if the

price is at most O-o)vH

+ oVL

and in the second period if the price is at

most vH

.

To show that these strategies constitute an equilibrium suppose the high

type buyer deviates and does not buy in the first period. Then he will buy

the good in the second period and obtain a payoff of o(vH

- vL

). Buying in

the first period gives him a payoff of vH

- ll-o)vH

- OVL

= o(vH

- vL

). Thus he

does not benefit by deviating. A player of type L does not benefit by

deviating and accepting in the first period since he will obtain a negative

payoff. The seller will not benefit by demanding a higher price in the first

period since nobody will buy at that price (both types will then buy in the •

second period), Suppose he offers a lower price. If this price is higher

than vL' then only the high type will buy, and the seller will obtain a lower

price without increasing his sales volume. Suppose he offers the good for vL

'

then both types will buy the good and the seller will make a profit of vL

. If ,

,

• •

9 - 15

r "

•

strategy •

with probability 1. If, on the other hand, '1 is close to 0, then (1' f is close

1 to 2' and the expected payment for firm E of the strategy ((1'0,(1'1,(1'2) =

The set of '1'S for which the equilibrium of example 9.C.3 will still be an

equilibrium can be calculated.

•

9.C.3 (a) for given values of A and ~, there exists the following weak

perfect equilibrium (WPBE):

•

The seller will demand a price of (l-~)vH + ~.vL in the first period and

vL

in the second period. A buyer of type L will buy the good if the price is • -

at most vL

. A buyer of type H will buy the good in the first period if the

price is at most (1-~)vH + ~vL and in the second period if the price is at

most vH

.

To show that these strategies constitute an equilibrium suppose the high

type buyer deviates and does not buy in the first period. Then he will buy ,

the good in the second period and obtain a payoff of ~(vH - vL

). Buying in

the first period gives him a payoff of vH

- (l-~)vH- ~vL= ~(vH - vL

). Thus he

does not benefit by deviating. A player of type L does not benefit by

deviating and accepting in the first period since he will obtain a negative

payoff. The seller will not benefit by demanding a higher price in the first

period since nobody will buy at that price (both types will then buy in the

second period); Suppose he offers a lower price. If this price is higher

than vL' then only the high type will buy. and the seller will obtain a lower

price without increasing his sales volume. Suppose he offers the good for vL'

then both types will buy the good and the seller will make a profit of vL

. If ,

• ,

9 - 15

•

•

,

he uses the above strategy he will make a profit of A[ (l-<3)v H + <3v L I +

U-A)OVL

~ vL

' therefore he will not deviate. ,

~: A[U-<3)VH

+ <3VL

) + O-A)<3VL

< vL

·

The seller will offer the good for vL

in every period, and both types buy

the good in period 1.

Clearly, both types of buyers will not benefit by deviating, since they - •

obtain the, good for the "lowest possible price" vL

' and the analysis of Case 1

shows that the seller will not deviate with the assumption of case 2.

(b) Suppose the buyer offers the good for the price x in the first period and

-y in the second period (clearly, x,y E [V, v)). Then there exists a unique -type v· who is indifferent between buying the good in the first period or in

the second period, v· - x = <3(v· - y), or t'. = ex - <3y) • Given that all

( 1-<3 )

types v !: v· will buy the good in period I, in eqUilibrium we must have that

the price y was set to maximize the second period profits: ,

into the expression for v·, we obtain: v· = -r( l:--x_--:~" 0/2) . The expressions for y

and v· obtained imply that if the seller charges x in the first period he will •

, •

x charge 2 _ o in the second period, all buyers of

• _ . x type v ~ v - (1 _ <3/2)

will buy in the first period, and all x types with (l _ <3/2)

x ~ -:::'2----=0 will

buy in the second period. The seller's profits will thus be:

IT = - x v - .,...,.---=-=-<"

(I - 0/2) x . x + o· .

2 - 0 x - x

(1 - (5/2) 2 -

2 -- - x

v - (1 - 0/2) 'x + o· x 2 -

The optimal selling price for the first period can be obtained by maximizing

,

9 - 16

, •

---_._---- ---- - - - -------_ .. . --_ .... _ -----------------

r "

•

•

r". o

Summing up, the weak perfect Bayesian equilibrium of this game· is

as follows: The seller offers the good for v(2 _ ~)2

x = 8 - 615 in the first period

v(2 - ~) and for y = 8 _ 6~ in period 2. All bUyers of type v E

-[ V(2 - d) -)

4 - 3d ,v

will buy in period 1, and all buyers -v(2 - a)

of type v E o[ 8 _ 6a -

buy in period 2.

o

o

9.C.4 [First Printing Errata: in part (e) it should read "Now allow Ms. P,

after ... "]

For an analysis of this problem see:

Nalebuff, B. (1987) "Credible Pretrial Negotiation," Rand Journal

of Economics, 18(2):198-210.

•

9.C.S For an analysis of this problem see:

Nalebuff, B. (1987) "Credible Pretrial Negotiation," Rand Journal

of Economics, 18(2):198-210.

o

•

9.C.6 Example 9.C.3: Since the example restricts attention to "( > 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 SEe

sequences of strategies be: For firm I, (IT~,IT~)

firm E, (CT~, n n 12111 IT I' IT 2) = (ii' 3 - 2n ' 3 - 2n ) ,

= ( 1 2 +

1 1 - - -) n ' n '

where n = 1,2, ....

Let the

and for

Clearly,

these strategies converge to the WPBE strategies described in the textbook.

o •

The sequence of strategies for firm E generates a unique sequence of beliefs

that firm E played In1

as follows:

2 1 - -n 3 2n 4n - 3 n 2 J.l 1

-- 2 1 - -3 2n

- ) -- 3' 1 1 6n - 6 110 - -3 2n +

Thus, the WPBE described is a SEe

9-17

--------------------- " ---- - -----_ ._ ..... __ ._----

•

•

,

•

• •

•

U (S.A) = p

AW-Cp

for s~Aw-cd

s for s~Aw-cd

-AW-C d

-s for S~AW-a

Therefore. Ms. P's expected payoff for an offer s will be given by (recall •

that teliefs for Ms. P are determined b) f(A):

1

(AW-C )f(A)dA + P

SfCA)dA = (AW-C )fCA)dA P

s-c +Sl-F __ d .

o o •

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:

1 - F s-C

d 1l

.'

(b) Implicit differentiation of (j) above yields

•

•

(ii) -s-C

d s-c

+ !...f __ d 1 s-cd

ds + -·f -- dc s-C

d W 1l 1l

•

s +c s-cd -'""'2~P . f -1l-

1l

+

1l 1l P

.f' S-c

d

-W

d1l = 0

which will determine the change in the optimal value of s given changes in any

of the parameters. For example. if A were uniformly distributed on the

interval [ 0.11 then the expression • (ii) reduces to that In s = 1l - C so we p

ds ds ds have 1. -1. and O. - - -7 • - . - -d1l dc dC

d •

P

(c) For an analysis of this part see:

Nalebuff, B. (1987) "Credible Pretrial Negotiation," Rand Journal

of Economics. 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

•

,

• L

; , •

, • • i ..

• • , • , , • ,

,--------------------------------------------------- ------------------------------

of Economics, J8(2):1~8-21O. •

•

9.C.6 9.C.3: Since the example restricts attention to '1 > 0, there is

a unique Weak Perfect Bayesian Equilibrium (WPBEl. Since every Sequential

Equilibrium (SE) is a WPBE, this is the only candidate for a SE.

1 1 :--- - - -)

1' . n'n' . n n n 12111 ·

flr,m E, (er 0' er I' er 2) = (ii' 3' - in ' 3 - 2n ) , where n = 1,2, ....

Let the

and for

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 In1

as follows:

2 1 - -0 3 20 4n - 3 n 2 III - - ) - - - -2 1 1 1 6n - 6 3' - - + --3 2n 3 2n

Thus, theWPBE d~scribed is a SE.

.

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 (4,~) in player

2's information set. 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. This

causes him to play "r" if

5), which in turn induces player 1 to play "y". This is the uniql.,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 (er O.er F,er A)' the probabilities of playing (Out).

(In,Fight), and On,Accommodate) respectively. Let a strategy of firm I be

(er f.er a l. the pr·obabilities of playing Fight, and Accommodate respectively (if

reached). Consider the following sequences of str'ategies:For firm E.

n n n 1 n-l 1 (er O.er F,er A) = ( 1 - n' 2' 2

n n a n .!. ). n

These sequences converge to the strategies described, and furthermore generate •

9 - 19

the belief sequence:

n-l 2

• n n n •

~ - ) 1. - n-l 1 CD +

2 2 n n

•

This supports the described strategies as a SE .

Another SE ( . WPBE) is where firm E plays Un, Accommodate) and firm I

plays Accommodate. This is supported by the sequences of strategies: For firm

n n n E, (er O,er F,er A) = (~l 1 1

2n ' 2n ' 1 - -) . n '

1 = (- I­n '

9.C.7 Cal The extensive form game is given in figure 9.C.7(a).

•

•

•

•

D

4 2

Player 1

Player 2

u

1 1

•

5 1

D

T

Figure 9.C.7(a)

u

2 2

•

1.). n

•

The set of pure strategies for player 1 is 51= {B,n, and for player 2 is 5

2=

<DD,DU,UD,UU} where playing AB means playing A at node v2

and B at node v3

.

By backward induction it is easy to see that the unique 5PNE is (B,DUL There

are two more classes of NE: (i) Player 1 plays T, and player 2 plays UU with

probability p and DU with probability l-p, with p , and (ji) Player 1

plays B, and player 2 plays DU with probability p and DO with probability I-p,

. h 1 Wlt P ~ 3" .

•

• •

•

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 D U

T S,1 2,2

Player 1 .

B 4,2 1,1 D

4 2

r

•

Figure 9.C.7(b)

B T

_P!!yer 2

1 1

5 1

u

2 2

Since playing T is a strictly dominant strategy for player I, we have a unique •

NE: (T,U).

(cl The modified game is depicted in figure 9.C.7(c).

•

n

4 2

r ,

P

r

r

I I

~

~

f----- Nature ----

I-p r

~ ~

~ r r

n

4 2

u

1 1

2

2

Is

IT 1

o

5 I

- ........

Figure 9.C.7(c)

P

........ ........

........

2 2

........

........ ~

~

5 I

lop

o u

2 2

•

•

Ik denotes player 2's information set after she observes k E {B,n, r is the

probability she assigns to the event that player 1 played B after she finds

herself in information set IB' and similarly q is the probability she assigns

to the event that player 1 played B after she finds herself in information set

• •

9 - 21

•

IT' Let s E [0,1) denote the probability that player 1 plays B. We can

have three. possible situations in a WPBE: First, player 1 playing s = 1;

second, player 1 playing s z: 0; and third, player 1 playing s E (0,1). Player

1 playing s = 1 cannot be part of a WPPE. 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 0, player 1 will prefer to deviate and play T. Second, •

•

,

player 1 playing s = 0 is part of a WPBE. 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 5 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 e(,lJilibrium we must have:

s'p sO-p) r = '( ~l ---5-'-) -.-:O,...----'--p') -+-5-' -p ,and q = s (1 - p) + pO - 5) ,

Simple algebra shows that > s = p < if and only if > 1

q < 2 • ,

, and that > s = <

(1 - p)

if and only if > 1 r = -< 2' This observation allows us to concentrate on 4 cases

,

as follows:

(0 s > p and s > O-p): In this case we must have

implies that player 2 will always play D, which in turn implies that player

l's best response is s = 0, Therefore there cannot be a WPBE in this case.

(ii) s < p and s < (l-p): In this case we must have

implies that player 2 will always play U, which in turn implies that player

1'5 best response is 5 = 0, This coincides with the pure strategy WPBE

described earlier, •

(iii) (l-p) < s < p: (which implies p > ~) In this case we must have 1

q < 2 and

1 Th' . I' r > 2' IS Imp les that player 2 will play U in information set IT and will

play D in information set IB' Player l's best response will now depend on p. ,

9 - 22

•

.----------------------------------------------

,

•

•

! •

• i • ,

• •

I

, , , · , •

..

Playing B will give 1 an expected payoff of 4p + lCl-p). and playing T will •

give him 2p + 50-pl. 2 If p. 3' then player 1 will have a unique best

response which rules out such WPBE. mixed

and player 2 will play U in information set IT ~d will play D in infol'JJlation •

set lB'

(Iv) p < s < O-pl: (which implies p < ~) This case is symmetric to

1 case CHi) above. If p. 3' then player 1 will have a unique best response

which rules out such WPBE. However. if 1 P = 3' the we have a mixed strategy

WPBE as follows: player 1 plays B with probability 1 2 s E ( 3' ' 3' >, and player

2 will play D in information set IT and will play U in information

To conclude, there exists a unique pure strategy WPBE as described

• earlier, and if p is randomly drawn from the interval (0,1) then the pure

strategy WPBE is the unique WPBE with probability 1. However, if p = ~ or

exists a mixed strategy WPBE as described in

cases (Hi) and (iv) above.

•

9.D.l In a similar manner to the proof of proposition 9.B.l, let N be the

maximal number of moves from a terminal node to the root of the tree. The

statement is clearly true for N = 1 because only one player will have to play .

•

at this node, and since no two terminal nodes give him the same payoffs, he

will have a unique optimal strategy that is not weakly dominated. Now suppose

that the result is true for N = n -I, and consider the case where N = n. Let

the set of final decision nodes (those that are before the terminal nodes) be

Va. Since for every player no two terminal nodes yield the same payoffs, then

o at each node vk

E V there is a unique optimal strategy for the player at that

node. Let (T k be this unique strategy. We can therefore associate with each

9 - 23 •

node vk E o V a payoff to all players given cr k. These payoffs induce a game

•

with N = n-l, and by our hypothesis this game has a unique prediction obtained

through iterated removal of weakly dominated strategies. Playing the optimal •

strategy in the n'th node given the unique prediction of the N = n-1 game will

give us the unique prediction obtained through iterated removal of weakly

dominated strategies. •

•

•

•

• •

•

•

•

,

•

•

•

9 - 24 ••

•

•

:

x solves I

Max, u.(x,) x E X

s. t. the old budget constraint.

1 I

J • (iii) Market clearing: r x = Co)

11 I 1=1

depend on prices at aU.

IO.C.I. (a) The consumer solves L

Max r log Xl

1=1

J • + r y for each t z: 1, •.. , L - does not

j ::1 IJ

L'

s.t. r PIX. :s w. 1=1

The first-order condition for the Lagrangean of this program can be written as

X = "Alp. t=l •...• L. where "A > O. Substituting in the budget constraint. we I I

,

written as X (p,w) = L w . 1 P

find "A a w/L. therefore the demand function can be

The wealth eff ect is ax (p,w)/aw = 1

(b) As L ) CIO, the wealth effect ax1(p,w>jaw ) o.

IO.C.2. (a) The consumer solves

Max a + f3 In x + m (x,m)

s. t. P x + m ~ w . The m

first-order condition (assuming interior solution) yields x(p) = f3/p.

The firm solves Max pq - rrq. - q~O

The firm's first-order condition (assuming interior solution) is p = fF.

I

-

-

(b) From the two first-order conditions and the consumer's budget constraint •

• • the competitive equilibrium is p = fF, X

,

• = (3/fF. m = ,

Co) m

- (3.

IO.C.3. (a) Assuming interior solution, the first-order condition is

• c'(q ) = "A > 0 for all j. J J

, ,

,

• ,

10-3

,

------- - ---- ------- ------- ---------------- ------- ----- .. -- -

•

J • • J • J • (b) C'(q) = r c' (q ) dq /dq =

J=1 J J J r ~ dq /dq • ~

J=1 J d( r q ) /dq

J =1 J . = ~ dq/dq = ~.

• Therefore. C'(q) = c'(q ) for all j. J J

(el Each firm j solves Max pq - c (q ). q j J J

J

The first-order condition (assuming interior solution) is c;(qJ) = p.

-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 (q) - the solution to the central J j

.

authority's program in part (a) for total output q. Therefore, J J. r q = r q (q) = q. In other words, if the market price is C'(q), then the

J =1 1 J =1 J

industry produces q. Therefore. C'(') is the inverse of the industry supply

function .

IO.CA. (a) The central authority's problem can be written as

Max (x I •• I X )~O

1 I

I

s. t. r x :s x. , J = 1

Assuming interior solution. the first-order condition is

• ¢I'(x ) = ~ > 0 for all i. i J

I 1 • • • (b) 1'(X) = L ¢I' (x ) dx /dx = LA dx/dX

I i , i = 1 I ., 1

• Therefore, 1'(X) - ¢'(x ) for all i. -

.i I

(c) Each consumer solves Max A. (x ) - Px . 'f'J i J

X .,

1

- A d( LX -1=1

• ) /dx

J

•

--

The first-order condition (assuming interior solution) is ¢I'(x) = P. I J

•

A dx/dx

If P = 1'(X), then we have ¢I'(x) = 1'(xl for every i. If ¢'(.) is strictly . I J I

•

•

decreasing for all i, we must have x = x (xl - the solution to the central I I

authority's program in part (al above for total consumption x. Therefore, I I.

--

LXI = L XI (x) = x. In other words, if the market price is 1'(X), then the 1=1 1=1

•

•

10-4

•

A.

~.--.---------------- - ­-------------------------

aggregate demand is x. Therefore, r'{'} is the inverse of the aggregate demand

function.

IO.C.S. The system of equations UO.C.4)-OO.C.6} here takes the following

form: •

• • 4>:(x 1) = p + t, i = 1 ....... 1, •

• •• c'(q ) = p ~ J I

• J.

LXI = Lqj' 1=1 J =1

j = 1, ••• , J,

• • These equations describe the equilibrium • (x I • q I p ) as an implicit function

of t. Differentiating with respect to t, we get

•• • 4>"(x ) x '(t) = p 'et) + 1, 1 I I

i = 1, ... , 1,

•• • c"(q ) q '(t) = p '(t), -l J J J

• • j = 1, ... , J,

•

LX '(t) = L q '(t). 1 j

1=1 J =1

• • • 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 •

• (p '(t) + I

1) L (¢;'(x: )]-1 = 1 =1

J • • -I P '(t) [(c"(q)] .

J =1 J J

• From here we can express p 'It): •

•

1

L {¢;'(X:)]-I

• p '(t) = I = 1

I J r.{¢~'(X~)]-I- L (c:'(q~)]-I

I I J J 1::1 J=1

•

Compare to the expression on page 324 of the textbook.

IO.C.6. Ca) 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 •

•

,

• • •

c The equilibrium market price p is determined from equalizing demand and

supply:

c c x(p + tl = 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-tL The equilibrium market price pP is determined from

•

equalizing demand and supply:

x(/) = q(pp - tl. •

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 •

P C x(p ) = x(p + tl.

•

•

(b) If the ad valorem tax T is levied on the consumer. then he pays

.

(l+T)p for every unit of the good, and the demand at market price p becomes

X«(l+T)P). The equilibrium market price pC is determined from equalizing

demand and supply:

c c x((l+T)p ) = q(p ). (0

On the other hand, If the ad valorem tax T is levied on the producer,

collects (l-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-T)p ). (2)

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 0) we have

• •

10-6

•