gibbons & kumar - the unofficial solution manual to a primer in game theory - unfinished draft

8/20/2019 GIBBONS & KUMAR - The Unofficial Solution Manual to a Primer in Game Theory - Unfinished Draft

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This version is an unreleased and unfinished manuscript.

The author can be reached at [email protected]

Last Updated: January 20, 2013

Typeset using L AT EX and the Tufte book class.

This work is not subject to copyright. Feel free to reproduce, distributeor falsely claim authorship of it in part or whole.


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This is strictly a beta version. Two thirds of it are missing and there are er-

rors aplenty. You have been warned.

On a more positive note, if you do find an error, please email me at

[email protected], or tell me in person.

- Navin Kumar


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Static Games of Complete Information

Answer 1.1 See text.

Answer 1.2 B is strictly dominated by T. C is now strictly dom-

inated by R. The strategies (T,M) and (L,R) survive the iterated

elimination of strictly dominated strategies. The Nash Equilibrium

are (T,R) and (M,L).

Answer 1.3 For whatever value Individual 1 chooses (denoted

by S1), Individual 2’s best response is S2 = B2(S1) = 1 − S2.Conversely, S1 = B1(S2) = 1 − S1. We know this because if S2 <1 − S1, then there is money left on the table and Individual 2 couldincrease his or her payoff by asking for more. If, however, S2 >

1 − S1, Individual 2 earns nothing and can increase his payoff byreducing his demands sufficiently. Thus the Nash Equilibrium is

S1 + S2 = 1.

Answer 1.4 The market price of the commodity is determined by

the formula P = a − Q in which Q is determined Q = q1 + ... + qnThe cost for an individual company is given by Ci = c · qi. Theprofit made by a single firm is

π i = ( p − c) · qi = (a − Q − c) · qi = (a − q∗1 − ... − q∗n − c) · qiWhere q∗ j is the profit-maximizing quantity produced by firm j inequilibrium. This profit is maximized at

dπ i

dqi

= (a

−q∗1

−...

−q∗i

−...

−q∗n

−c)

−q∗i = 0

⇒ a − q∗1 − ... − 2 · q∗i − ... − q∗n − c = 0⇒ a − c = q∗1 + ... + 2 · q∗i + ... + q∗n

For all i=1, ... ,n. We could solve this question using matrices and

Cramer’s rule, but a simpler method would be to observe that since

all firms are symmetric, their equilibrium quantity will be the same

i.e.

q∗1 = q∗2 = ... = q∗i = ... = q∗n

Which means the preceding equation becomes.

a − c = (n + 1)q∗i ⇒ q∗i = a − cn + 1


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6 Static Games of Complete Information

A similar argument applies to all other firms.

Answer 1.5 Let qm be the amount produced by a monopolist.

Thus, if the two were colluding, they’d each produce

q1m = q2m = qm2 = a − c

4

In such a scenario, the profits earned by Firm 1 (and, symmetrically,

Firm 2) is

π 1mm = (P − c) · q1m = (a − Q − c) · q1m =

a − a − c

2 − c

· q1m

= a − c

2 · a − c

4 =

(a − c)28

= 0.13 · (a − c)2

If both are playing the Cournot equilibrium quantity, than the prof-

its earned by Firm 1 (and Firm 2) are: 1 1 The price is determined by

P = a − Q= a − q1cc −q2cc= a − 2 · a − c

3

= a + c

2

π 1cc = (P − c) · q1c = (a −

2(a − c)3

− c) · q1c = (a + 2c

3 − c) · q1c

= (a − c)2

9

= 0.11 · (a − c)2

What if one of the firms (say Firm 1) plays the Cournot quantity

and the other plays the Monopoly quantity? 2 Firm 1’s profits are: 2 The price is given by

P = a − Q= a − q1c − q2m

= a − a

−c

3 − a

−c

4

= a − 712

· (a − c)

= 5a

12 +

7c

12

π 1cm = (P − c) · q1c = (

5a

12 +

7c

12 − c) · a − c

3 =

5

36 · (a − c)2

= 0.14 · (a − c)2And Firm 2’s profits are:

π 2cm = (P − c) · q2m = (

5a

12 +

7c

12 − c) · a − c

4 =

5

48 · (a − c)2

= 0.10 · (a − c)2

For notational simplicity, let

α ≡ (a − c)2

Their profits are reversed when their production is. Thus, the pay-

offs are:

Player 2

qm qc

Player 1qm 0.13α, 0.13α 0.10α, 0.14α

qc 0.14α, 0.10α 0.11α, 0.11α

As you can see, we have a classic Prisoner’s Dilemma: regardless of

the other firm’s choice, both firm’s would maximize their payoff by

choosing to produce the Cournot quantity. Each firm has a strictly

dominated strategy ( qm2 ) and they’re both worse off in equilibrium(where they make 0.11 · (a − c)2 in profits) than they would have


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been had they cooperated by producing qm together (which would

have earned them 0.13 · (a − c)2).

Answer 1.6 Price is determined by P = a − Q and Q is determined by Q = q1 + q2. Thus the profit generated for Firm 1 is given by:

π 1 = (P − c1) · q1 = (a − Q − c1) · q1 = (a − q1 − q2 − c1) · q1At the maximum level of profit,

dπ 1dq1

= (a − c1 − q1 − q2) + q1 · (−1) = 0 ⇒ q1 = a − c1 − q22And by a similar deduction,

q2 = a − c2 − q1

2

Plugging the above equation into it’s predecessor,

q1 = a − c1 − a−c2−q12

2 =

a − 2c1 + c23

And by a similar deduction,

q2 = a − 2c2 + c1

3

Now,

2c2 > a + c1 ⇒ 0 > a − 2c2 + c1 ⇒ 0 > a − 2c2 + c13 ⇒ 0 > q2⇒ q2 = 0

Since quantities cannot be be negative. Thus, under certain condi-

tions, a sufficient difference in costs can drive one of the firms to

shut down.

Answer 1.7 We know that,

qi =

a − pi if pi < p j,a− pi

2 if pi = p j,

0 if pi > p j.

We must now prove pi = p j = c is the Nash Equilibrium of this

game. To this end, let’s consider the alternatives exhaustively.

If pi > p j = c, q i = 0 and π j = 0. In this scenario, Firm j can

increase profits by charging p j + ε where ε > 0 and ε < pi − p j,raising profits. Thus this scenario is not a Nash Equilibrium.

If pi > p j > c, q i = 0 and π i = 0 and Firm i can make pos-

itive profits by charging p j − ε > c. Thus, this cannot be a NashEquilibrium.

If pi = p j > c, π i = ( pi − c) · a− pi2 . Firm i can increase profits bycharging pi − ε such that p j > pi − ε > 0, grab the entire market anda larger profit, provided

( pi − ε − c) · (a − pi − ε) > ( pi − c) · (a − pi)2


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Therefore, this is not a Nash Equilibrium

If pi = p j = c, then π i = π j = 0. Neither firm has any reason

to deviate; if FFirm i were to reduce pi, π i would become negative.

If Firm i were to raise pi, q i = π i = 0 and he would be no better

off. Thus Firm i (and, symmetrically, Firm j) have no incentive to

deviate, making this a Nash Equilibrium.

Answer 1.8 The share of votes received by a candidate is given by

Si =

xi + 12 · (x j − xi) if x i < x j,

12 if x i = x j,

(1 − xi) + 12 · (xi − x j) if x i > x j.

We aim to prove the Nash Equilibrium is ( 12 , 12 ). Let us exhaustively

consider the alternatives.Suppose x i =

12 and x j >

12 i.e. One candidate is a centrist

while the other (Candidate j) isn’t. In such a case, Candidate j can

increase his share of the vote by moving to the left i.e. reducing x j.

If x j < 1

2 , Candidate j can increase his share of the vote by moving

to the right.

Suppose x i > x j > 1

2 ; Candidate i can gain a larger share by

moving to a point x j > x i > 12 . Thus this is not a Nash Equilibrium.

Suppose x i = x j = 12 ; the share of i and j are

12 . If Candidate i were

to deviate to a point (say) x i > 1

2 , his share of the vote would de-

cline. Thus ( 12 , 12 ) is a unique Nash Equilibrium. This is the famous

Median Voter Theorm, used extensively in the study of politics. Itexplains why, for example, presidential candidates in the US veer

sharply to the center as election day approaches.


Answer 1.10 (a) Prisoner’s Dilemma

Player 2

( p) Mum (1 − p)Fink Player 1

(q) Mum −1, −1 −9, 0(1 − q)Fink 0, −9 −6, −6

Prisoner’s Dilemma

In a mixed strategy equilibrium, Player 1 would choose q such that

Player 2 would be indifferent between Mum and Fink. The payoff

from playing Mum and Fink must be equal. i.e.

−1 · q + −9 · (1 − q) = 0 · q + −6 · (1 − q) ⇒ q = −3.5

This is impossible.


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(b)

Player 2

Le f t(q0) Middle(q1) Right(1 − q0 − q1)

Player 1U p( p) 1, 0 1, 2 0, 1

Down(1 − p) 0, 3 0,1 2, 0Figure 1.1.1.

Here, Player 1 must set p so that Player 2 is indifferent between

Left, Middle and Right. The payoffs from Left and Middle, for

example, have to be equal. i.e.

p · 0 + (1 − p) · 3 = p · 2 + (1 − p) · 1

⇒ p = 0.5Similarly, the payoffs from Middle and Right have to be equal

2 · p + 1 · (1 − p) = 1 · p + 0 · (1 − p)⇒ p = −0.5

Which, besides contradicting the previous result, is quite impossi-

ble.

(c)

Player 2

L(q0) C(q1) R(1 − q0 − q1)

Player 1

T ( p0) 0, 5 4, 0 5, 3

M( p1) 4, 0 0, 4 5, 3

B(1 − p0 − p1) 3, 5 3, 5 6, 6Figure 1.1.4.

In a mixed equilibrium, Player 1 sets p0 and p1 so that Player 2

would be indifferent between L, C and R. The payoffs to L and Cmust, for example, be equal i.e.

4 · p0 + 0 · p1 + 5 · (1 − p0 − p1) = 0 · p0 + 4 · p1 + 5 · (1 − p0 − p1)⇒ p0 = p1

Similarly,

0 · p0 + 4 · p1 + 5 · (1 − p0 − p1) = 3 · p0 + 3 · p1 + 6 · (1 − p0 − p1)⇒ p1 = 2.5 − p0

Which violates p0 = p1.


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Answer 1.11 This game can be written as

Player 2

L(q0) C(q1) R(1 − q0 − q1)

Player 1T ( p0) 2, 0 1, 1 4, 2

M( p1) 3, 4 1, 2 2, 3

B(1 − p0 − p1) 1, 3 0, 2 3, 0

In a mixed Nash Equilibrium, Player 1 sets p0 and p1 so that the

expected payoffs from L and C are the same. i.e.

E2(L) = E2(C)

⇒ 0 · p0 + 4 · p1 + 3 · (1 − p0 − p1) = 1 · p0 + 2 · p1 + 2 · (1 − p0 − p1)

⇒ p1 = 2 · p0 − 1Similarly,

E2(C) = E2(R)

⇒ 1 · p0 + 2 · p1 + 2 · (1 − p0 − p1) = 2 · p0 + 3 · p1 + 0 · (1 − p0 − p1)⇒ p1 = 23 − p0

Combining these,

2 · p0 − 1 = 23 − p0 ⇒ p0 = 5

9

∴ p1 = 2 · p0 − 1 = 2 · 59 − 1 = 1

9

∴1 − p0 − p1 = 1 − 59 − 1

9 =

3

9

Now we must calculate q0 and q1. Player 2 will set them such that

E1(T ) = E1( M)

⇒ 2 · q0 + 1 · q1 + 4 · (1 − q0 − q1) = 3 · q0 + 1 · q1 + 2 · (1 − q0 − q1)⇒ q1 = 1 − 1.5 · q0

And

E1( M) = E1(B)

⇒ 3 · q0 + 1 · q1 + 2 · (1 − q0 − q1) = 1 · q0 + 0 · q1 + 3 · (1 − q0 − q1)

Qed

Answer 1.12

(q)L2 (1 − q)R2( p)T 1 2, 1 0, 2

(1 − p)B1 1, 2 3, 0


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Player 1 will set p such that

E2(L) = E2(R)

⇒ 1

· p + 2

·(1

− p) = 2

· p + 0

·(1

− p)

⇒ p = 23

Player 2 will set q such that

E1(T ) = E1(B)

⇒ 2 · q + 0 · (1 − q) = 1 · q + 3 · (1 − q)

⇒ q = 34

Answer 1.13

(q)Apply1 to Firm 1 (1 − q)Apply1 to Firm 2( p)Apply2 to Firm 1 12 w1,

12 w1 w1,w2

(1 − p)Apply2 to Firm 2 w2,w1 12 w2, 12 w2There are two pure strategy Nash Equilibrium (Apply to Firm 1,

Apply to Firm 2) and (Apply to Firm 2, Apply to Firm 1). In a

mixed-strategy equilibrium, Player 1 sets p such that Player 2 isindifferent between Applying to Firm 1 and Applying to Firm 2.

E2(Firm 1) = E2(Firm 2)

⇒ p · 12

w1 + (1 − p) · w1 = p · w2 + (1 − p) · 12 w2

⇒ p = 2w1 − w2w1 + w2

Since 2w1 >

w2, 2w1 − w2 is positive and p > 0. For p < 1 to betrue, it must be the case that

2w1 − w2w1 + w2

< 1 ⇒ 12

w1 < w2

Which is true. And since the payoffs are symmetric, a similar analy-

sis would reveal that

q = 2w1 − w2

w1 + w2

Answer 1.14


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Dynamic Games of Complete Information

Answer 2.1 The total family income is given by

I C( A) + I P( A)

This is maximized at

d(I C( A) + I P( A))dA

= 0 ⇒ dI C( A)dA

= −dI P( A)dA

The utility function of the parents is given by

V (I P − B) + kU (I C + B)


k dU (I C + B)

dB +

dV (I P − B)db

= 0

⇒ k dU (I C + B)

d(I C + B) · d(I C + B)

dB +

dV (I P

−B)

d(I P − B) · I P

−B

dB = 0

⇒ kU (I C + B) − V (I P − B) = 0

⇒ V (I P − B∗) = kU (I C + B∗)

Where B∗ is the maximizing level of the bequest. We know it exists because a) there are no restrictions on B and b) V() and U() are

concave and increasing

The child’s utility function is given by U (I C( A) − B∗( A)). This is

maximized atdU (I C( A) + B

∗( A))dA

= 0

⇒ U (I C( A) + B∗( A)) ·

dI C( A)

dA +

d B∗( A)dA

= 0

⇒ dI C( A)dA

+ d B∗( A)

dA = 0 ⇒ I C( A) = −B∗( A)

We now have only to prove B ∗( A) = I P( A)

dV (I P( A) − B∗( A))dA

= 0


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14 Dynamic Games of Complete Information

⇒ V (I P( A)∗B( A)) ·

dI P( A)

dA − d B

∗( A)dA

= 0

⇒ I P( A) = B ∗( A)

Answer 2.2 The utility function of the parent is given by V (I P −B) + k [U 1(I C − S) + U 2(B + S)]. This is maximized at

d · {V (I P − B) + k [U 1(I C − S) + U 2(B + S)]}dB

= 0

⇒ −V + [U 1(−SB) + U 2(SB + 1)] = 0 ⇒ V = −kU 1S + U 2S + U 2The utility of the child is given by U 1(I C − S) + U 2(S + B). This ismaximized at:

d[U 1(I C − S) + U 2(S + B)]dS

= 0 ⇒

U 1

= U 2

(1 + B)

Total utility is given by

V (I p − B) + k (U 1(I C − S) + U 2(B + S)) + U 1(I C − S) + U 2(B + S)

= V (I p − B) + (1 + k )(U 1(I C − S) + U 2(B + S))

This is maximized (w.r.t S) at:

V · (−BS) + (1 + K ) · [U 1(−1) + U 2(1 + BS)] = 0

⇒ U 1 = U 2(1 + BS) − V BS1 + k as opposed to U 1 = U 2(1 + BS), which is the equilibrium condition.Since

V BS1+k > 0, the equilibrium U

1 is ‘too high’ which means that S

- the level of savings - must be too low (since dU

dS R − c1. He will put in R − c1 if it’s better than putting innothing i.e.

V − (R − c1)2 ≥ 0 ⇒ c1 ≥ R −√

V

For player 1, any c1 > R −√

V is dominated by c1 = R −√

V . Now,

player 1 will do this if the benefit exceeds the cost:

δV ≥ (R − √ V )2


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⇒ δ ≥

R√ V

− 12

If R2 ≥ 4V , δ would have to be greater than one, which is impossible. Therefore, if δ

≥ R√

V −1

2and R2

≥ 4V (i.e. the cost is not

‘too high’), c1 = R − √ V and c2 = √ V . Otherwise, c2 = 0 andc1 = 0.

Answer 2.5 Let the ‘wage premium’ be p = wD − wE, where p ∈ (−∞,∞). In order to get the worker to acquire the skill, thefirm has to credibly promise to promote him if he acquires the skill

- and not promote him if he doesn’t.

Let’s say that he hasn’t acquired the skill. The firm will not pro-

mote him iff the returns to the firm are such that:

yD0 −

wD ≤

yE0 −

wE ⇒

yD0 −

yE0 ≤

wD −

wE

= p

If he does acquire the skill, the firm will promote if the returns to

the firm are such that:

yDS − wD ≥ yES − wE ⇒ yDS − yES ≥ wD − wE = p

Thus the condition which the firm behaves as it ought to in the

desired equilibrium is:

yD0 − yE0 ≤ p ≤ y DS − yES

Given this condition, the worker will acquire the promotion iff heacquires the skill. He will acquire the skill iff the benefit outweighs

the cost i.e.

wD − C ≥ wE ⇒ wE + p − C ≥ wE ⇒ p ≥ c

That is, the premium paid by the company must cover the cost of

training. The company wishes (obviously) to minimize the pre-

mium, which occurs at:

wD − wE = p =

C if C ≥ y D0 − yE0, yD0

− yE0 if C < yD0

− yE0

A final condition is that the wages must be greater than the alterna-

tive i.e. wE ≥ 0 and wD ≥ 0. The firm seeks to maximize y ij − wi,which happens at wE = 0 and wD = p.

Answer 2.6 The price of the good is determined by

P(Q) = a − q1 − q2 − q3The profit earned by a firm is given by π i = ( p − c) · qi. For Firm 2,for example

π 2 = (a − q∗1 − q2 − q3 − c) · q2


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which is maximized at

dπ 2dq2

= (a − q∗1 − q2 − q3 − c) + q2(−1) = 0

⇒ q2 = a − q∗1 − q3 − c2Symmetrically,

q3 = a − q∗1 − q2 − c

2

Putting these two together,

q2 = a − q∗1 −

a−q∗1−q2−c2 − c

2 ⇒ q2 =

a − c − q∗13

Which, symmetrically, is equal to q3.

∴ π 1 = (a − q1 − q2 − q3 − c) · q1 = (a − q1 − 2 · a − c − q13 − c) · q1 =

a − q1 − c3

· q1


dπ 1dq1

= a − q1 − c

3 +

−q13

= 0 ⇒ q∗1 = a − c

2

Plugging this into the previous equations, we get

q2 = q3 = a − c

6

Answer 2.7 The profit earned by firm i is

π i = ( p − w) · LiWhich is maximized at

dπ idLi

= d( p − w) · Li

dLi=

d(a − Q − w)LidLi

= 0

⇒ d(a − L1 − . . . − Li − . . . − Ln − w) · LidLi

= 0

⇒ (a − L1 − . . . − Li − . . . − Ln − w) + Li(−1) = 0

⇒ L1 + . . . + 2Li + . . . + Ln = a − w ∀ i = 1, . . . , n

This generates a system of equation similar to the system in Ques-

tion (1.4)

2 1 . . . 1

1 2 . . . 1...

... . . .

...

1 1 . . . 2

·

L1

L2...

Ln

=

a − wa − w

...

a − w


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17

Which resolves to

Li = a − wn + 1

Thus, total labor demand is given by

L = L1 + L2 + . . . + Ln = a − wn + 1

+ . . . + a − wn + 1

= n

n + 1 · (a − w)

The labor union aims to maximize

U = (w − wa) · L = (w − wa) · nn + 1

· (a − w) = nn + 1

· (aw − awa − w2 + wwa)

The union maximize U by setting w. This is maximized at

dU

dw = (a − 2w + wa) · n

n + 1 = 0 ⇒ w = a + wa

2

The sub-game perfect equilibrium is Li =

a

−w

n+1 and w =

a+wa

2 .Although the wage doesn’t change with n, the union’s utility (U )

is an increasing function of nn+1 , which is an increasing function

of n. This is so because the more firms there are in the market, the

greater the quantity produced: more workers are hired to produce

this larger quantity, increasing employment and the utility of the

labor union.

Answer 2.8

Answer 2.9 From section 2.2.C, we know that exports from coun-

try i can be driven to 0 iff

e∗i =a − c − 2t j

3 = 0 ⇒ t j = a − c2

which, symmetrically, is equal to ti. Note that in this model c =

wi and we will only be using wi from now, for simplicity. What

happens to domestic sales?

∴ hi = a − wi + ti

3 =

a − wi + a−wi23

= a − wi

2

Which is the monopoly amount. Now, in the monopoly-union bar-

gaining model, the quantity produced equals the labor demanded.Thus,

Le=0i = a − wi

2

The profit earned by firm i in this case is given by

π e=0i = ( pi − wi)hi = (a − wi − hi)hi =

a − wi −

a − wi

2

a − wi

2

=

a − wi

2

2

The union function’s payoff is defined by

U e=0 = (wi − wa)Li = (wi − wa)

a − wi2

= wi a − wa − w

2i + wiwa

2


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The union sets wages to maximize utility with the following condi-

tion:

dU e=0

dwi=

a − 2wi + wa2

= 0 ⇒ w i = a + wa2Now, tariffs decline to zero. In this situation, t j = 0 and therefore

h j = h i = a − wi

3

and

e j = ei = a − wi

3

Due to this, prices fall:

Pi = P j = a − Q = a − hi − e j = a − a − wi3 − a − wi

3 =

a + 2wi3

So what happens to profits?

π t=0i = ( pi − wi)hi + ( p j − wi)e j =

a + 2wi

3 − wi

a − wi

3

+

a + 2wi

3 − wi

a − wi

3

⇒ π t=0i = 2

a − wi3

2= 2 · π e=0i

Thus, profits are higher at zero tariffs. What happens to employ-

ment?

Lt=0i = q i = h i + ei = 2

3 ·(a

−wi) =

4

3 ·a − wi

2 =

4

3 ·Le=0i

Employment rises. And what happens to the wage? That depends

on the payoff the union now faces:

U t=0 = (wi − wa) · Li = 23 (wi − wa)(a − wi) = 2

3 · (wi a − waa − w2i + wawi)


dU t=0

dwi=

2

3(a − 2wi + wa) = 0 ⇒ wi = a + wa2

Which is the same as before.

Answer 2.10 Note that (P1, P2), (R1, R2) and (S1, S2) are Nash

Equilibrium.

P2 Q2 R2 S2

P1 2, 2 x, 0 -1, 0 0,0

Q1 0, x 4, 4 -1, 0 0, 0

R1 0, 0 0, 0 0, 2 0, 0

S2 0, -1 0, -1 -1, -1 2, 0

So what are player 1’s payoff from playing the strategy? Let PO ji (X 1, X 2)denote player i’s payoff in round j when player 1 plays X 1 and


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19

player 2 plays X 2. If player 2 doesn’t deviate, he earns the sum of

payoffs from two rounds of gaming:

PO11(Q1, Q2) + PO21(P1, P2) = 4 + 2 = 6

And if player 2 deviates from the strategy, player 1 earns:

PO11(Q1, P2) + PO21(S1, S2) = 0 + 2 = 2

Now, let’s look at his payoff from deviating, when 2 doesn’t:

PO11(P1, Q2) + PO21(R1, R2) = x + 0 = x

And when they both deviate:

PO11(P1, P2) + PO21(P1, P2) = 2 + 2 = 4

Thus, if player 2 deviates, player 1’s best response is to deviate for

a payoff of 4

(as opposed to the 2

he’d get when not deviating). If,however, player 2 doesn’t deviate, player 1 gets a payoff of 6 when

playing the strategy and x when he doesn’t. Thus, he (player 1)

will play the strategy i f f x < 6. A symmetric argument applies to

player 2.

Thus the condition for which the strategy is a sub-game perfect

Nash Equilibrium is

4 < x < 6

Answer 2.11 The only pure-strategy subgame-perfect Nash Equi-

librium in this game are (T , L) and ( M, C).

L C R

T 3,1 0,0 5,0

M 2,1 1,2 3,1

B 1,2 0,1 4,4

Unfortunately, the payoff (4,4) - which comes from the actions (B,R)

- cannot be maintained. Player 2 would play R, if player 1 plays B -

player 1 however would deviate to T to earn a payoff of 5. Consider,

however, the following strategy for player 2:

• In Round 1, play R.• In Round 2, if Round 1 was (B,R), play L. Else, play M.Player 1’s best response in Round 2 is obviously T or M, depend-

ing on what player 2 does. But what should he do in Round 1? If

he plays T, his playoff is: 3 3 If you do not understand the nota-tion, see Answer 2.10

PO11(T , R) + PO21( M, C) = 5 + 1 = 6

If he plays B:

PO11(B, R) + PO21(T , L) = 4 + 3 = 7

Thus, as long as player 2 is following the strategy given above, wecan induce (B,R).


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To get a intuitive idea of how we constructed the strategy, note

that there are two Nash equilibrium that player 2 can play in the

second round: a "reward" equilibrium in which player 1 gets 3 and

"punishment" equilibrium in which player 1 gets 1. By (credibly)

threatening to punish player 1 in Round 2, player 2 induces "good"

behavior in Round 1.


Answer 2.13 The monopoly quantity is a−c2 . The monopoly priceis, therefore:

P = a − Q = q −

a − c2

=

a + c

2

The players can construct the following strategy.

• In Round 1, play pi = a+c

2• In Round t=1, if the previous Round had p j = a+c2 , play pi = c(the Bertrand equilibrium). Else, play pi =

a+c2

Now, if player i deviates by charging a−c2 − ε where ε > 0, hesecures the entire market and earns monopoly profits for one round

and Bertrand profits (which are 0) for all future rounds which totals

to

π deviate =

a − a − c

2 − c

·

a − c2

+ δ · 0 + δ2 · 0 + . . . = (a − c)

2

4

The payoff from sticking to the strategy (in which both firms pro-

duce a

−c

4 ) the payoff is

π f ol low =

a − a − c

2 − c

·

a − c4

+ δ ·

a − a − c

2 − c

·

a − c4

+ . . .

=

1

1 − δ

·

a − c2

·

a − c4

=

1

1 − δ

· (a − c)2

8

The strategy is stable if

π deviate ≤ π f ol low

⇒ (a − c)2

4 ≤

1

1 − δ

· (a − c)2

2

⇒ δ ≥ 12

Q.E.D.

Answer 2.14 The monopoly quantity when demand is high isa H −c

2 which makes the monopoly price

p H = a H −

a H − c2

= a H + c

2


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21

This is the price that the firms have to maintain when demand is

high. Conversely, when demand is low

pL = aL + c

2

Let p M be the monopoly price, defined as

p M =

p H if a i = a H pL if a i = a L

Consider the following strategy for firm i:

• In Round 1, set pi = p M.• In Round t = 1, if p j = p M in the previous round, play p M, else

play pi = c

The payoff received from deviating is the monopoly profit for

one round 4 and then zero profits in all future rounds: 4 See the previous question for thederivation of one round monopolyprofits

π deviate = (ai − c)2

4 + δ · 0 + δ2 · 0 + . . . = (ai − c)

2

4

If he follows the strategy, he earns:

π f ol low = (ai − c)2

8 + δ ·

π · (a H − c)

2

8 + (1 − π ) · (aL − c)

2

8

+ . . .

⇒ π f ol low = (ai − c)2

8 +

δ

1 − δ ·π · (a H − c)

2

8 + (1 − π ) · (aL − c)

2

8


π deviate ≤ π f ol low

⇒ (ai − c)2

4 ≤ (ai − c)

2

8 +

δ

1 − δ ·π · ( a H − c)

2

8 + (1 − π ) · (aL − c)

2

8

Answer 2.15 If the quantity produced by a monopolist is a−c2 , thequantity produced by a single company in a successful cartel is

qmn = a − c

2n

Therefore, the profit earned by one of these companies is

π m = ( p − c)qmn = (a − Q − c)qmn =

a − a − c2

− c

a − c2n

=

1

n ·

a − c2

2

The Cournot oligopoly equilibrium quantity5 is a−c1+n which means 5 See Answer 1.4that the profit earned at this equilibrium is

π c = (a − Q − c)qcn =

a − n

a − c1 + n

− c

·

a − c1 + n

=

a − c1 + n

2

A grim trigger strategy for a single company here is• In Round t=1, produce qmn


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• In Round t > 1, if the total quantity produced in t − 1 is n · qmn ,produce qmn , else produce q

cn.

Now, the best response to everyone else producing qmn is deter-

mined by finding q which renders profit

π = (a − Q − c)q = a − a − c2n

· (n − 1) − q − c q = n + 12n

· (a − c)q − q2

Which is maximized at

dπ

dq =

n + 1

2n · (a − c) − 2q = 0 ⇒ q = a + 1

4n (a − c)

The profit at q (cheating gain) is

π =

a − c − a − c

2n · (n − 1) − n + 1

4n · (a − c)

·

n + 1

4n · (a − c)

=

n + 1

4n · (a − c)

2

If the firm deviates, it earns the cheating gain for one round and

Cournot profits for all future rounds i.e. the gain from deviating

from the strategy is

π deviate =

n + 1

4n · (a − c)

2+ δ

a − c1 + n

2+ δ2

a − c1 + n

2+ . . .

⇒ π deviate =

n + 1

4n

2+

δ

1 + δ

1

1 + n

2(a − c)2

If the firm follows the strategy, it’s payoff is π m for all rounds:

π f ol low = 1n

a − c

2

2

+ δ · 1n

a − c

2

2

+ δ2 · 1n

a − c

2

2

+ . . . = 11 − δ ·

1n

a − c

2

2


π f ol low ≥ π deviate

⇒ 11 − δ ·

1

n

a − c

2

2≥

n + 1

4n

2+

δ

1 − δ

1

n + 1

2(a − c)2

⇒ δ∗ ≤ n2 + 2n + 1

n2 + 6n + 1

Thus as n rises, δ∗ falls. If you want to know more, see Rotemberand Saloner (1986).

Answer 2.16

Answer 2.17


Answer 2.19 In a one period game, player 1 would get a payoff of 1 and player 2 would get a payoff of 0 i.e. (1,0). In a two period


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23

game, if the two players can’t agree, the game goes to the second

stage, at which point, player 2 gets 1 and player 1 gets 0. This pay-

off of 1 in the second round is worth δ to player 2 in the first round.

If player 1 offers δ to player 2 in the first round, player 2 will accept,

getting a payoff of 1

−δ i.e. (1

−δ, δ).

In a three period game, if player 2 rejects the offer in the first

round, they go on to the second round, at which point it becomes

a two period game and player 2 gets a payoff of 1 − δ and player 1gets δ. This payoff (δ,1 − δ) is worth (δ2,δ[1 − δ]) to the players inthe first round. Thus, if player 1 makes an offer of (1 − δ[1 − δ],δ[1 −δ])=(1 − δ + δ2,δ − δ2), player 2 would accept and player 1 wouldsecure a higher payoff.

Answer 2.20 In round 1, player A offers ( 11−δ , δ

1+δ ) to player B,

which player B accepts since δ1+δ ≥ δs∗ = δ 11−δ .What if A deviates from the equilibrium, offers less and B re-

fuses? The game then goes into the next round and B offers A δ1+δ ,

which will be accepted, leaving 11+δ for B. This is worth δ · 11+δ toB in the first round (which is why B will refuse anything less than

this amount) and δ2

1+δ to A in the first round. This is less than the1

1+δ A would have made had he not deviated.

Answer 2.21

Answer 2.22 In the first round, investors can either withdraw (w)

or not (d). A strategy can represented as x1x2 where x1 is what the

investor does in the first round and x2 is what the investor does inthe second round. The game can be represented by the following

table:

ww wd dd dw

ww r,r r,r D,2r-D D,2r-D

wd r,r r,r D,2r-D D,2r-D

dd 2r-D,D 2r-D,D R,R D,2R-D

dw 2r-D,D 2r-D,D 2R-D,D R,R

There are 5

Nash Equilibria: (dw,dw) , (ww,ww) , (ww,wd) , (wd,ww)and (wd,wd). Of these, (ww,wd), (wd,ww) and (wd,wd) are not

Subgame Perfect Equilibria since there is no subgame in which both

or either player doesn’t withdraw his or her funds in the second

round.

Answer 2.23 The optimal investment is given by

d(v + I − p − I 2)dI

= 0 ⇒ I ∗ = 12

The boost in the value added is If the buyer had played ‘Invest’,

buyer will buy if

v + I − p − I 2 ≥ −I 2 ⇒ p ≤ v + I


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Thus, the highest possible price that the buyer will pay at this point

is p = v + I . If, however, the buyer doesn’t invest, the buyer will

buy if

v

− p

≥ 0

⇒ v

≥ p

Thus the buyer would be willing to pay p = v. Thus investment is

I ⊆ {0, 12}. The price is drawn from p ⊆ {v, v + 12 }. There is no gainfrom charging anything other than these prices.

v + 12 v

12 , A − 14 , v + 12 14 , v12 , R − 14 , 0 − 14 , 00, A − 12 , v + 12 0, v0, R 0, 0 0, 0

As you can see from this (complicated) table, if I = 12 , A weakly

dominates R. And if I = 0 R weakly dominates A. Thus we can

collapse the above table into a simpler one:

(q)v + 12 (1 − q)v( p)I = 12 Accept − 14 , v + 12 14 , v

(1 − p)I = 0Reject 0, 0 0, 0

The only pure Nash Equilibrium is for the buyer to not invest and

the


25/36

Static Games of Incomplete Information

Answer 3.1 See text

Answer 3.2 Firm 1 aims to maximize

π 1 = ( p − c)q1 = (ai − c − q1 − q2)q1Which is done by

dπ 1dq1

= a i − c − q1 − q2 + q1(−1) = 0 ⇒ q1 = ai − c − q22Thus, the strategy for firm 1 is

q1 =

a H −c−q22 if a i = a H

aL−c−q22 if a i = a L

Now, the firm 2 aims to maximize

π 2 = ( p − c)q2 = (a − c − q1 − q2)q2This is maximized at

dπ 2dq2

= a − c − q1 − q2 + q2(−1) = 0 ⇒ q2 = a − c − q12 = θa H + (1 − θ)aL − c − q1

2

Plugging in q1, we get

q2 =θa H + (1 − θ)aL − c −

θa H +(1−θ)aL−c−q2

2

2

⇒ q2 = θa H + (1 − θ)aL − c3Now, we need to find out what firm 1’s output would be. If a i =

a H ,

q H 1 = a H − c − θa H +(1−θ)aL−c3

2 =

(3 − θ)a H − (1 − θ)aL − 2c6

But what if a i = a L?

qL1 = aL − c −

θ

a H +(1−θ

)aL−c32

= (2 + θ)aL − θa H − 2c6


26/36

26 Static Games of Incomplete Information

Now, based on these results, the constraints for non-negativity are:

q2 ≥ 0 ⇒ θa H + (1 − θ)aL − c ≥ 0 ⇒ θ ≥ c − aLa H − aL

Which also requires

θ ≤ 1 ⇒ 1 ≥ c − aL ⇒ a L ≥ c − 1

Furthermore,

qL1 ≥ 0 ⇒ (2 + θ)aL − θa H −2c ≥

6 ≥ 0 ⇒ θ ≤ 2 · c − aL

a H − aLWhich subsumes the last-but-one result. And finally,

q H 1 ≥ 0 ⇒ θ ≤ 2c − 3a H + aL

a H − aL

Answer 3.3 The profits earned by firm 1 is given by

π 1 = ( p1 − c)q1 = ( p1 − c)(a − p1 − b1 p2)


dπ 1dp1

= a − p1 − b1 p2 + p1(−1) = 0 ⇒ p1 = a − b1 p22 = a − b1[θ p H + (1 − θ) pL]

2

Now, what if b1 = b H ? To start with, p1 = p H and

p H = a − b H [θ p H + (1 − θ) pL]

2 ⇒ p H =

a − (1 − θ)b H pL2 +

θb H

And if b1 = bL:

pL = a − bL[θ p H + (1 − θ) pL]

2 =

a − θbL p H 2 + bL(1 − θ)

Which means that,

p H =a − (1 − θ)

a−θbL p H 2+(1−θ)bL

2 + θb H

⇒ p H = a(1 − [1 − θ]b H )4 + 2(1 − θ)bL + 2θb H Similarly,

pL = a(1 − θbL)

4 + 2(1 − θ)bL + 2θb H

Answer 3.4 Game 1 is played with 0.5 probability:

L R

(q)T 1,1 0,0

(1-q)B 0,0 0,0


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27

If nature picks game 2, which 0.5 probability:

L R

(q)T 0,0 0,0

(1-q)B 0,0 2,2

If nature picks game 2, player 1 will always play B, since it weakly

dominates T and player 2 will play R, since it weakly dominates L.

Now, if nature chooses game 1, player 1 will play T. If nature

chooses game 2, player 1 will play B. Furthermore, if player 2 plays

L with probability p:

π 2 = p[1

2 · 0 + 1

2 · 1] + (1 − p)[ 1

2 · 0 + 1

2 · 2] = 1 − 1

2 · p

This is maximized at p = 0 i.e. player 2 will always play R. Thus the

Pure-strategy Bayesian Nash equilibrium is

PSNE = {(1, T , R), (2, B, R)}

Answer 3.5

Answer 3.6 The payoff is given by

ui =

vi − bi if bi > b j ∀ j = 1, 2, . . . , i − 1, i + 1 , . . . , nvi−b j

m if bi = b j

0 if bi < b j for any j = 1, 2, . . . , i − 1, i + 1 , . . . , n

The beliefs are: v j is uniformly distributed on [0,1]. Actions aregiven by bi ⊆ [0, 1] and types are given by v i ⊆ [0, 1]. The strategy isgiven by bi = a i + civi. Thus, the aim is to maximize

π i = (vi − bi) ·P(bi > b j∀ j = 1, . . . , i − 1, i + 1 , . . . , n) = (vi − bi) · [P(bi > b j)]n−1

⇒ π i = (vi − bi) · [P(bi > a j + c jv j)]n−1 = (vi − bi) ·P

v j <

bi − a jc j

n−1= (vi − bi)

bi − a j

c j

n−1


dπ i

dbi = (−1) ·bi

−a j

c j

n−1

+ (vi − bi)n

−1

c j bi −

a j

c j

n−2

= 0

bi − a j

c j

n−2·

a j + (n − 1)vi − nbic j

= 0

This requires that either

bi − a jc j

= 0 ⇒ bi = a j

Or that

a j + (n − 1)vi − nbic j

= 0 ⇒ bi = a jn + n − 1n · vi


28/36

28 Static Games of Incomplete Information

Now, we know that b i = a i + civi. Here, we know that c i = n−1

n and

ai =a j

n ⇒ a1 = a2n =

a3n

= . . . = an

n

Which is only possible if a1 = a2 = . . . = an = 0. Thus,

bi = n − 1

n · vi

Answer 3.7

Answer 3.8


29/36

Dynamic Games of Incomplete Information

Answer 4.1 (a)

(q)L’ (1-q)R’

L 4,1 0,0

M 3,0 0,1

R 2,2 2,2

The Nash Equilibria are (L, L), (R, R) and they are both sub-gameperfect equilibria. Now, the payoff to player 2 from playing L is

π 2(L) = 1 · p + 0 · (1 − p) = p

The payoff from playing R

π 2(R) = 0

· p + 1

·(1

− p) = 1

− p

Player 2 will always play L if

π 2(L) > π 2(R) ⇒ p > 1 − p ⇒ p > 12

The playoff to player 1 from playing L is

π 1(L) = 4 · q + 0 · (1 − q) = 4q

And the payoff from playing M is

π 1( M) = 3 · q + 0 · (1 − q) = 3q

Player 1 will always play L if

π 1(L) > π 1( M) ⇒ 4q > 3q

Which is true. Thus p = 1. In which case, player 2 will always play

L. Thus the outcome (R, R) violates Requirements 1 and 2.(b)

L M R

L 1,3 1,2 4,0

M 4,0 0,2 3,3

R 2,4 2,4 2,4


30/36

30 Dynamic Games of Incomplete Information

The expected values of the payoffs to player 2 are:

π 2(L) = 3 · p + 0 · (1 − p) = 3 p

π 2( M) = 2

· p + 2

·(1

− p) = 2

π 2(R) = 0 · p + 3 · (1 − p) = 3 − 3 p

And the payoffs to player 1 are:

π 1(R) = 2

π 1(L) = 1 · q1 + 1 · q2 + 4 · (1 − q1 − q2) = 4 − 3q1 − 3q2

π 1( M) = 4 · q1 + 0 · q2 + 3 · (1 − q1 − q2) = 3 + q1 − 3q2The only Nash Equilibrium is (R, M); it is also sub-game perfect.To be a Perfect Bayesian Equilibrium, player 2 must believe that

π 2( M) > π 2(L) ⇒ 2 > 3 p ⇒ 2

3 > p

and

π 2( M) > π 2(R) ⇒ 2 > 3 − 3 p ⇒ p > 1

3

Furthermore, player 1 must believe

π 1(R)

> π 1(L) ⇒ 2 > 4 − 3q1 − 3q2 ⇒ q1 >

2

3 − q2Since q1 > 0, this implies that

2

3 − q2 > 0 ⇒ 23 > q2

and

π 1(R) > π 1( M) ⇒ 2 > 3 + q1 − 3q2 ⇒ 3q2 − 1 > q1Which, in turn, requires

3q2 − 1>

2

3 − q2 ⇒ q2 >

5

12

The pure Bayesian Nash equilibrium is(R, M),

1

3 > p >

2

3, 3q2 − 1 > q1 > 23 − q2,

2

3 > q2 >

5

12

Answer 4.2

(q)L (1 − q)R( p)L 3,0 0,1

(1

− p) M 0,1 3,0

R 2,2 2,2


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31

As you can see, there is no Pure Nash Equilibrium. But, we need

rigorous proof: A pure strategy Nash Equilibrium exists if

(a) Player 1 always picks either L or M. For example, player 1

will always play L if

π 1(L) > π 1( M) ⇒ 3 · q + 0 · (1 − q) > 0 · q + 3 · (1 − q) ⇒ q > 12Thus if q > 0.5, p = 1.

(b) Player 2 always picks either L or R ; player 2 will always playL if

π 2(L) > π 2(R) ⇒ 0 · p + 1 · (1 − p) > 1 · p + 0 · (1 − p) ⇒ 12 > p

Thus, if p


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(b) We must find a pooling equilibrium in which the sender

plays (L, L, L). For the receiver, the payoffs are

π R(L, u) = 1

3 · 1 + 1

3 · 1 + 1

3 · 1 = 1

π R(L, d) = 1

3 · 0 + 1

3 · 0 + 1

3 · 0 = 0

There are two strategies: (u, u) and (u, d). Under (u, u):

π 1(L, u) = 1 and π 1(R, u) = 0

π 2(L, u) = 2 and π 2(R, u) = 1

π 3(L, u) = 1 and π 3(R, u) = 0

None of the three types have an incentive to send R instead of L.

Thus, we have the following equilibrium:

[(L, L, L), (u, u), p = 1

3, 1 ≥ q ≥ 0]

Answer 4.4 (a) Let’s examine pooling equilibrium (L, L). p =

0.5. π R(L, u) = π R(L, d). Thus, it doesn’t matter for the receiver

whether he/she plays u or d.

• Under (u, u), π 1(L, u) = 1 < π 1(R, u) = 2, making it unsustain-able.

• Under (u, d), π 2(L, u) = 0 < π 2(R, d) = 1, making it unsustain-able.

• Under (d, d), π 2(L, d) = 0 < π 2(R, d) = 1, making it unsustain-able.

• Under (d, u), π 2(L, u) = 0 < π 2(R, d) = 1, making it unsustain-able.

Thus, (L, L) is not a sustainable equilibrium.

Let’s examine separating equilibrium (L, R). The best response

to this is (u, d)6. Let’s see if either of the types have an incentive to 6 Since π R(1, L, u) > π R(1, L, d) andπ R(2, R, u) > π R(2, R, d)deviate:

• For type 1, π 1(L, u) = 1 > π 1(R, d) = 0 i.e. no reason to play Rinstead of L.

• For type 2, π 2(L, u) = 0 < π 2(R, d) = 1 i.e. no reason to play Linstead of R.

Let’s examine pooling equilibrium (R, R). π R(R, u) = 1 >

π R(R, d) = 0.5. Therefore, the two strategies that can be followed by

the receiver are (u, u) and (d, u).

• Under (u, u), π (L, u) < π (R, u) for both types.• Under (d, u), π (L, d) ≤ π (R, u) for both types.Let’s examine pooling equilibrium (R, L). The best response to

this is (d, u)7. 7 Since π R(1, R, u) = 2 > π R(1, R, d) =

0 and π R(2, L, u) = 0 < π R(2, L, d) = 1• For type 1, π 1(L, d) = 2 ≥ π 1(R, u) = 2, i.e. will play L.• For type 2, π 2(L, d) = 0 ≤ π 2(R, u) = 1, i.e. will play R.


33/36

33

Thus the perfect Bayesian equilibrium are:

[(L, R), (u, d), p, q]

[(R, R), (u, u), p, q = 0.5]

[(R, R), (d, u), p, q = 0.5]

[(R, L), (d, u), p, q]

(b) Let’s examine pooling equilibrium is (L, L). π R(L, u) = 1.5 >

π R(L, d) = 1, therefore player 2 will respond to L with u. The two

strategies are (u, u) and (u, d).

• Under (u, u), π (L, u) > π (R, u) for both types.

• Under (u, d), π 1(L, u) = 3 < π 1(R, d) = 4, making it unsustain-able.

Let’s examine separating equilibrium (L, R). The best response to

this is (d, u).

• For type 1, π 1(L, d) = 1 > π 1(R, u) = 0 i.e. type 1 will play L.• For type 2, π 2(L, d) = 0 < π 2(R, u) = 1 i.e. type 2 will play R.Let’s examine pooling equilibrium (R, R). π R(R, u) = 1 >

π R(L, d) = 0.5, therefore player 2 will respond to R with u. The two

strategies are (u, u) and (d, u).

• Under (u, u), π 1(L, u) = 3 > π 1(R, u) = 0, making (R, R)unsustainable.

• Under (d, u), π 1(L, d) = 1 > π 1(R, u) = 0, making (R, R)unsustainable.

Let’s examine separating equilibrium (R, L). The best response to

this is (d, u).

• For type 1, π 1(L, d) = 1 > π 1(R, u) = 0, making (R, L)unsustainable.

• For type 2, π 2(L, d) = 0 < π 2(R, u) = 1, which doesn’t conflictwith the equilibrium.

The perfect Bayesian Equilibria are

[(L, L), (u, u), p = 0.5, q]

[(L, R), (d, u), p = 1, q = 0]

Answer 4.5 Let’s examine 4.3(a). We’ve already tested equilib-

rium (R, R). Let’s try another pooling equilibrium (L, L). q = 0.5.

π R(L, u) = 1 > π R(L, d) = 0.5. Thus, the receiver’s response to

L will always be u. We have to test two strategies for the receiver:

(d, u) and(u, u).

• Under the strategy (d, u), π (L, d) = 2 > π (R, u) = 0 i.e. there isno incentive for either type to play R instead of L.

• Under the strategy (u, u), π 2(L, u) = 0 < π 2(R, u) = 1, making(L, L) unsustainable.


34/36


Let’s examine (L, R). The best response to this is (u, d). In re-

sponse to this,

• For type 1, π 1(L, u) = 1 < π 1(R, d) = 3 i.e. type 1 will play Rwhich violates the equilibrium

• For type 2, π 2(L, u) = 0 < π 2(R, d) = 2 i.e. type 2 will play R

which doesn’t violate the equilibrium.

Let’s examine (R, L). The best response to this is (u, d). In re-

sponse to this,

• For type 1, π 1(L, u) = 1 < π 1(R, d) = 3, i.e. type 1 will play R.• For type 2, π 2(L, u) = 0 < π 2(R, d) = 2, i.e. type 2 will play R,

violating the equilibrium.

The perfect Bayesian Equilibrium is

[(L, L), (d, u), p, q = 0.5]

Now, let’s examine 4.3(b). There is one pooling equilibrium

other than the (L, L, L): (R, R, R). There are six pooling equilib-ria: 1.(L, L, R), 2.(L, R, L), 3.(R, L, L), 4.(L, R, R), 5.(R, L, R) and

6.(R, R, L).

Let’s start with pooling equilibrium (R, R, R). π R(R, u) = 2

3 >

π R(R, d) = 13 . Thus, receiver will play the strategy (u, u) or (d, u).

• For strategy (u, u), π (L, u) < π (R, u) for all types, making itunsustainable.

• For strategy (d, u), π 1(L, d) < π 1(R, u), making the equilibriumunsustainable. Let’s examine the various separating the equilib-

rium.

1. ( L, L, R). The best response to this is (u, d), 8 8 Since π R(3, R, u) = 0 < π R(3, R, d) =1 and 0.5 · π R(1, L, u) + 0.5 ·π R(2, L, u) = 1 > 0.5 · π R(1, L, d) + 0.5 ·π R(1, L, d) = 0

• For type 1, π S(1, L, u) = 1 > π S(1, R, d) = 0 i.e. type 1 will playL.


• For type 3, π S(3, L, u) = 1 < π S(3, R, d) = 2 i.e. type 3 will playR.

Thus, this is a viable equilibrium.

2. ( L, R, L). The best response to this is (u, u), 9 9 since π R (2, R, u) = 1 > π R(2, R, d) =0 and 0.5 · π R(1, L, u) + 0.5 ·π R(3, L, u) = 1 > 0.5 · π R(1, L, d) + 0.5 ·π R(3, L, d) = 0

• For type 1, π S(1, L, u) = 1 > π S(1, R, u) = 0 i.e. type 1 will playL.

• For type 2, π S(2, L, u) = 2 > π S(2, R, u) = 1 i.e. type 2 will playL, instead of R .


Thus, this is not a viable equilibrium.

3. ( R, L, L). The best response to this is (u, u), 10 10 since π R (1, R, u) = 1 < π R (1, R, d) =0 and 0.5 · π R(2, L, u) + 0.5 ·π R(3, L, u) = 1 > 0.5 · π R(2, L, d) + 0.5 ·π R(3, L, d) = 0



• For type 3, π S(3, L, u) = 1 > π S(3, R, u) = 0 i.e. type 3 will play

L.Thus, this is a not viable equilibrium.


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4. ( L, R, R). The best response to this is (u, u) and (u, d) 11 11 since π R (1, L, u) = 1 > π R(1, L, d) =0 and 0.5 · π R(2, R, u) + 0.5 ·π R(3, R, u) = 0.5 = 0.5 · π R(2, R, d) +0.5 · π R(3, R, d) = 0.5

Let’s test (u, u):


• For type 2, π S(2, L, u) = 2 > π S(2, R, u) = 1 i.e. type 2 will play

L, instead of R .



Let’s test (u, d)


• For type 2, π S(2, L, u) = 2 > π S(2, R, d) = 1 i.e. type 2 will playL, instead of R .

• For type 3, π S(3, L, u) = 1 < π S(3, R, d) = 2 i.e. type 3 will playR.


5. ( R, L, R). The best response to this is either (u, u) or (u, d) 12 12 since π R (2, L, u) = 1 > π R(2, L, d) =0 and 0.5 · π R(2, L, u) + 0.5 ·π R(3, L, u) = 0.5 = 0.5 · π R(2, L, d) +0.5 · π R(3, L, d) = 0.5

Let’s test (u, u).





Let’s test (u, d)• For type 1, π S(1, L, u) = 1 > π S(1, R, d) = 0 i.e. type 1 will play

L, instead of R .


• For type 3, π S(3, L, u) = 1 = π S(3, R, d) = 1 i.e. type 3 can playR.


6. ( R, R, L). The best response to this is (u, u), 13 13 since π R (3, L, u) = 1 > π R(3, L, d) =0 and 0.5 · π R(1, R, u) + 0.5 ·π R(2, R, u) = 0.5 > 0.5 · π R(1, R, d) +0.5

·π

R(2, R, d) = 0

• For type 1, π S(1, L, u) = 1 > π S(1, R, u) = 0 i.e. type 1 will playL, instead of R

• For type 2, π S(2, L, u) = 2 > π S(2, R, u) = 1 i.e. type 2 will playL, instead of R



The only other perfect Bayesian Equilibrium is(L, L, R), (u, d), p, q0 =

1

3, q1 =

1

3

Answer 4.6 Type 2 will always play R since π S(2, R, a) > π S(2, R, u) and π S(2, R, a) >

π S(2, R, d). Thus if the Receiver gets the message L, he knows thatit can only be type 1. In such a case, the Receiver plays u14, creating 14 in fact, π R (x, L, u) > π R(x, L, d) for

both types, so the Receiver will alwaysplay u


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a payoff of (2, 1). This gives type 1 a higher payoff than if he played

R, which would have given him a payoff of 1. Thus, the perfect

Bayesian Equilibrium is

[(L, R), (u, a), p = 1, q = 0]

gibbons & kumar - the unofficial solution manual to a primer in game theory - unfinished draft

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