unofficial solutions manual to r.a gibbon's a primer in game theory

The Unofficial Solution Manual to

A Primer in Game Theoryby RA Gibbons

Unfinished Draft

Navin Kumar

Delhi School of Economics

2

This version is an unreleased and unfinished manuscript.

The author can be reached at [email protected]

Last Updated: January 20, 2013

Typeset using LATEX 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 [email protected], or tell me in person.

- Navin Kumar

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 iteratedelimination of strictly dominated strategies. The Nash Equilibriumare (T,R) and (M,L).

Answer 1.3 For whatever value Individual 1 chooses (denotedby 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 isS1 + S2 = 1.

Answer 1.4 The market price of the commodity is determined bythe formula P = a− Q in which Q is determined Q = q1 + ... + qn

The cost for an individual company is given by Ci = c · qi. Theprofit made by a single firm is

πi = (p− c) · qi = (a−Q− c) · qi = (a− q∗1 − ...− q∗n − c) · qi

Where q∗j is the profit-maximizing quantity produced by firm j inequilibrium. This profit is maximized at

dπidqi

= (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 andCramer’s rule, but a simpler method would be to observe that sinceall firms are symmetric, their equilibrium quantity will be the samei.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

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 = q2

m =qm

2=

a− c4

In such a scenario, the profits earned by Firm 1 (and, symmetrically,Firm 2) is

π1mm = (P− c) · q1

m = (a−Q− c) · q1m =

(a− a− c

2− c)· q1

m

=a− c

2· a− c

4=

(a− c)2

8= 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 − q2

cc

= a− 2 · a− c3

=a + c

2

π1cc = (P− c) · q1

c = (a− 2(a− c)3

− c) · q1c = (

a + 2c3− c) · q1

c

=(a− c)2

9= 0.11 · (a− c)2

What if one of the firms (say Firm 1) plays the Cournot quantityand the other plays the Monopoly quantity? 2 Firm 1’s profits are: 2 The price is given by

P = a−Q

= a− q1c − q2

m

= a− a− c3− a− c

4

= a− 712· (a− c)

=5a12

+7c12

π1cm = (P− c) · q1

c = (5a12

+7c12− c) · a− c

3=

536· (a− c)2

= 0.14 · (a− c)2

And Firm 2’s profits are:

π2cm = (P− c) · q2

m = (5a12

+7c12− c) · a− c

4=

548· (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 1

qm 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 ofthe other firm’s choice, both firm’s would maximize their payoff bychoosing to produce the Cournot quantity. Each firm has a strictlydominated strategy ( qm

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

7

been had they cooperated by producing qm together (which wouldhave earned them 0.13 · (a− c)2).

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

π1 = (P− c1) · q1 = (a−Q− c1) · q1 = (a− q1 − q2 − c1) · q1

At the maximum level of profit,

dπ1

dq1= (a− c1 − q1 − q2) + q1 · (−1) = 0⇒ q1 =

a− c1 − q2

2

And by a similar deduction,

q2 =a− c2 − q1

2

Plugging the above equation into it’s predecessor,

q1 =a− c1 − a−c2−q1

22

=a− 2c1 + c2

3

And by a similar deduction,

q2 =a− 2c2 + c1

3

Now,

2c2 > a + c1 ⇒ 0 > a− 2c2 + c1 ⇒ 0 >a− 2c2 + c1

3⇒ 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 toshut down.

Answer 1.7 We know that,

qi =

a− pi if pi < pj,a−pi

2 if pi = pj,

0 if pi > pj.

We must now prove pi = pj = c is the Nash Equilibrium of thisgame. To this end, let’s consider the alternatives exhaustively.

If pi > pj = c, qi = 0 and πj = 0. In this scenario, Firm j canincrease profits by charging pj + ε where ε > 0 and ε < pi − pj,raising profits. Thus this scenario is not a Nash Equilibrium.

If pi > pj > c, qi = 0 and πi = 0 and Firm i can make pos-itive profits by charging pj − ε > c. Thus, this cannot be a NashEquilibrium.

If pi = pj > c, πi = (pi − c) · a−pi2 . Firm i can increase profits by

charging pi − ε such that pj > pi − ε > 0, grab the entire market anda larger profit, provided

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

2


Therefore, this is not a Nash EquilibriumIf pi = pj = c, then πi = πj = 0. Neither firm has any reason

to deviate; if FFirm i were to reduce pi, πi would become negative.If Firm i were to raise pi, qi = πi = 0 and he would be no betteroff. Thus Firm i (and, symmetrically, Firm j) have no incentive todeviate, making this a Nash Equilibrium.

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

Si =

xi +

12 · (xj − xi) if xi < xj,

12 if xi = xj,

(1− xi) +12 · (xi − xj) if xi > xj.

We aim to prove the Nash Equilibrium is ( 12 , 1

2 ). Let us exhaustivelyconsider the alternatives.

Suppose xi = 12 and xj > 1

2 i.e. One candidate is a centristwhile the other (Candidate j) isn’t. In such a case, Candidate j canincrease his share of the vote by moving to the left i.e. reducing xj.If xj <

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

to the right.Suppose xi > xj > 1

2 ; Candidate i can gain a larger share bymoving to a point xj > xi >

12 . Thus this is not a Nash Equilibrium.

Suppose xi = xj =12 ; the share of i and j are 1

2 . If Candidate i wereto deviate to a point (say) xi >

12 , his share of the vote would de-

cline. Thus ( 12 , 1

2 ) is a unique Nash Equilibrium. This is the famousMedian Voter Theorm, used extensively in the study of politics. Itexplains why, for example, presidential candidates in the US veersharply 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 thatPlayer 2 would be indifferent between Mum and Fink. The payofffrom 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 1

Up(p) 1, 0 1, 2 0, 1

Down(1− p) 0, 3 0,1 2, 0

Figure 1.1.1.

Here, Player 1 must set p so that Player 2 is indifferent betweenLeft, Middle and Right. The payoffs from Left and Middle, forexample, have to be equal. i.e.

p · 0 + (1− p) · 3 = p · 2 + (1− p) · 1⇒ p = 0.5

Similarly, 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, 6

Figure 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.


Answer 1.11 This game can be written as

Player 2

L(q0) C(q1) R(1− q0 − q1)

Player 1

T(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 theexpected 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 − 1

Similarly,

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 =

59

∴p1 = 2 · p0 − 1 = 2 · 59− 1 =

19

∴1− p0 − p1 = 1− 59− 1

9=

39

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)T1 2, 1 0, 2

(1− p)B1 1, 2 3, 0

11

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 112 w1, 1

2 w1 w1,w2

(1− p)Apply2 to Firm 2 w2,w112 w2, 1

2 w2

There are two pure strategy Nash Equilibrium (Apply to Firm 1,Apply to Firm 2) and (Apply to Firm 2, Apply to Firm 1). In amixed-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 − w2

w1 + w2

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

2w1 − w2

w1 + w2< 1⇒ 1

2w1 < 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

Dynamic Games of Complete Information

Answer 2.1 The total family income is given by

IC(A) + IP(A)

This is maximized at

d(IC(A) + IP(A))

dA= 0⇒ dIC(A)

dA= −dIP(A)

dA

The utility function of the parents is given by

V(IP − B) + kU(IC + B)


kdU(IC + B)

dB+

dV(IP − B)db

= 0

⇒ kdU(IC + B)d(IC + B)

· d(IC + B)dB

+dV(IP − B)d(IP − B)

· IP − BdB

= 0

⇒ kU′(IC + B)−V′(IP − B) = 0

⇒ V′(IP − B∗) = kU′(IC + B∗)

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

The child’s utility function is given by U(IC(A)− B∗(A)). This ismaximized at

dU(IC(A) + B∗(A))

dA= 0

⇒ U′(IC(A) + B∗(A)) ·(

dIC(A)

dA+

dB∗(A)

dA

)= 0

⇒ dIC(A)

dA+

dB∗(A)

dA= 0⇒ I′C(A) = −B′∗(A)

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

dV(IP(A)− B∗(A))

dA= 0

14 Dynamic Games of Complete Information

⇒ V′(IP(A)∗B(A)) ·(

dIP(A)

dA− dB∗(A)

dA

)= 0

⇒ I′P(A) = B′∗(A)

Answer 2.2 The utility function of the parent is given by V(IP −B) + k[U1(IC − S) + U2(B + S)]. This is maximized at

d · {V(IP − B) + k[U1(IC − S) + U2(B + S)]}dB

= 0

⇒ −V′ + [U′1(−S′B) + U′2(S′B + 1)] = 0⇒ V′ = −kU′1S′ + U′2S′ + U′2

The utility of the child is given by U1(IC − S) + U2(S + B). This ismaximized at:

d[U1(IC − S) + U2(S + B)]dS

= 0⇒ U′1 = U′2(1 + B′)

Total utility is given by

V(Ip − B) + k(U1(IC − S) + U2(B + S)) + U1(IC − S) + U2(B + S)

= V(Ip − B) + (1 + k)(U1(IC − S) + U2(B + S))

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

V′ · (−B′S) + (1 + K) · [U′1(−1) + U′2(1 + B′S)] = 0

⇒ U′1 = U′2(1 + B′S)−V′B′S1 + k

as opposed to U′1 = U′2(1 + B′S), which is the equilibrium condition.

Since V′B′S1+k > 0, the equilibrium U′1 is ‘too high’ which means that S

- the level of savings - must be too low (since dU′dS < 0). It should be

higher.

Answer 2.3 To be done

Answer 2.4 Let’s suppose c2 = R − c1, partner 2’s payoff is V −(R − c1)

2. If c1 ≥ R, partner 2’s best response is to put in 0 andpocket V. If c1 < R and partner 2 responds with some c2 suchthat c2 < R − c1, his payoff is −c2

2. If he puts in nothing, he will,receive a payoff of zero. There is, therefore, no reason to put in sucha low amount. There is - obviously - also no reason to put in anyc2 > 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

15

⇒ δ ≥(

R√V− 1)2

If R2 ≥ 4V, δ would have to be greater than one, which is impossi-

ble. Therefore, if δ ≥(

R√V− 1)2

and 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, wherep ∈ (−∞, ∞). 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 tothe 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 thedesired equilibrium is:

yD0 − yE0 ≤ p ≤ yDS − yES

Given this condition, the worker will acquire the promotion iff heacquires the skill. He will acquire the skill iff the benefit outweighsthe cost i.e.

wD − C ≥ wE ⇒ wE + p− C ≥ wE ⇒ p ≥ c

That is, the premium paid by the company must cover the cost oftraining. The company wishes (obviously) to minimize the pre-mium, which occurs at:

wD − wE = p =

C if C ≥ yD0 − 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 yij − 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 − q3

The profit earned by a firm is given by πi = (p− c) · qi. For Firm 2,for example

π2 = (a− q∗1 − q2 − q3 − c) · q2


which is maximized at

dπ2

dq2= (a− q∗1 − q2 − q3 − c) + q2(−1) = 0

⇒ q2 =a− q∗1 − q3 − c

2

Symmetrically,

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.

∴ π1 = (a− q1 − q2 − q3 − c) · q1 = (a− q1 − 2 · a− c− q1

3− c) · q1 =

(a− q1 − c

3

)· q1


dπ1

dq1=

a− q1 − c3

+−q1

3= 0⇒ q∗1 =

a− c2

Plugging this into the previous equations, we get

q2 = q3 =a− c

6

Answer 2.7 The profit earned by firm i is

πi = (p− w) · Li

Which is maximized at

dπidLi

=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− w

a− w...

a− w

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) ·n

n + 1· (a− w) =

nn + 1

· (aw− awa − w2 + wwa)

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

dUdw

= (a− 2w + wa) ·n

n + 1= 0⇒ w =

a + wa

2

The sub-game perfect equilibrium is Li = a−wn+1 and w = a+wa

2 .Although the wage doesn’t change with n, the union’s utility (U)is an increasing function of n

n+1 , which is an increasing functionof n. This is so because the more firms there are in the market, thegreater the quantity produced: more workers are hired to producethis larger quantity, increasing employment and the utility of thelabor 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− 2tj

3= 0⇒ tj =

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. Whathappens to domestic sales?

∴ hi =a− wi + ti

3=

a− wi +a−wi

23

=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− wi2

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

πe=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

Ue=0 = (wi − wa)Li = (wi − wa)

(a− wi

2

)=

wia− wa − w2i + wiwa

2


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

dUe=0

dwi=

a− 2wi + wa

2= 0⇒ wi =

a + wa

2

Now, tariffs decline to zero. In this situation, tj = 0 and therefore

hj = hi =a− wi

3

and

ej = ei =a− wi

3

Due to this, prices fall:

Pi = Pj = a−Q = a− hi − ej = a− a− wi3− a− wi

3=

a + 2wi3

So what happens to profits?

πt=0i = (pi − wi)hi + (pj − wi)ej =

(a + 2wi

3− wi

)(a− wi

3

)+

(a + 2wi

3− wi

)(a− wi

3

)

⇒ πt=0i = 2

(a− wi

3

)2= 2 · πe=0

i

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

Lt=0i = qi = hi + ei =

23· (a− wi) =

43·(

a− wi2

)=

43· Le=0

i

Employment rises. And what happens to the wage? That dependson the payoff the union now faces:

Ut=0 = (wi − wa) · Li =23(wi − wa)(a− wi) =

23· (wia− waa− w2

i + wawi)


dUt=0

dwi=

23(a− 2wi + wa) = 0⇒ wi =

a + wa

2

Which is the same as before.

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

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 POji(X1, X2)

denote player i’s payoff in round j when player 1 plays X1 and

19

player 2 plays X2. If player 2 doesn’t deviate, he earns the sum ofpayoffs from two rounds of gaming:

PO11(Q1, Q2) + PO2

1(P1, P2) = 4 + 2 = 6

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

PO11(Q1, P2) + PO2

1(S1, S2) = 0 + 2 = 2

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

PO11(P1, Q2) + PO2

1(R1, R2) = x + 0 = x

And when they both deviate:

PO11(P1, P2) + PO2

1(P1, P2) = 2 + 2 = 4

Thus, if player 2 deviates, player 1’s best response is to deviate fora 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 whenplaying 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 toplayer 2.

Thus the condition for which the strategy is a sub-game perfectNash 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? Ifhe plays T, his playoff is: 3 3 If you do not understand the nota-

tion, see Answer 2.10

PO11(T, R) + PO2

1(M, C) = 5 + 1 = 6

If he plays B:

PO11(B, R) + PO2

1(T, L) = 4 + 3 = 7

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


To get a intuitive idea of how we constructed the strategy, notethat there are two Nash equilibrium that player 2 can play in thesecond 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 price

is, therefore:

P = a−Q = q−(

a− c2

)=

a + c2

The players can construct the following strategy.• In Round 1, play pi =

a+c2

• In Round t 6=1, if the previous Round had pj 6= a+c2 , play pi = c

(the Bertrand equilibrium). Else, play pi =a+c

2Now, if player i deviates by charging a−c

2 − ε where ε > 0, hesecures the entire market and earns monopoly profits for one roundand Bertrand profits (which are 0) for all future rounds which totalsto

πdeviate =

(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 ollow =

(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

πdeviate ≤ π f ollow

⇒ (a− c)2

4≤(

11− δ

)· (a− c)2

2

⇒ δ ≥ 12

Q.E.D.

Answer 2.14 The monopoly quantity when demand is high isaH−c

2 which makes the monopoly price

pH = aH −(

aH − c2

)=

aH + c2

21

This is the price that the firms have to maintain when demand ishigh. Conversely, when demand is low

pL =aL + c

2

Let pM be the monopoly price, defined as

pM =

pH if ai = aH

pL if ai = aL

Consider the following strategy for firm i:• In Round 1, set pi = pM.• In Round t 6= 1, if pj = pM in the previous round, play pM, else

play pi = cThe 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

πdeviate =(ai − c)2

4+ δ · 0 + δ2 · 0 + . . . =

(ai − c)2

4

If he follows the strategy, he earns:

π f ollow =(ai − c)2

8+ δ ·

(π · (aH − c)2

8+ (1− π) · (aL − c)2

8

)+ . . .

⇒ π f ollow =(ai − c)2

8+

δ

1− δ·(

π · (aH − c)2

8+ (1− π) · (aL − c)2

8

)The strategy is stable if

πdeviate ≤ π f ollow

⇒ (ai − c)2

4≤ (ai − c)2

8+

δ

1− δ·(

π · (aH − c)2

8+ (1− π) · (aL − c)2

8

)

Answer 2.15 If the quantity produced by a monopolist is a−c2 , the

quantity produced by a single company in a successful cartel is

qmn =

a− c2n

Therefore, the profit earned by one of these companies is

πm = (p− c)qmn = (a−Q− c)qm

n =

(a− a− c

2− c)(

a− c2n

)=

1n·(

a− c2

)2

The Cournot oligopoly equilibrium quantity5 is a−c1+n which means 5 See Answer 1.4

that the profit earned at this equilibrium is

πc = (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 qm

n


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

produce qmn , else produce qc

n.Now, the best response to everyone else producing qm

n 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′ − q′2

Which is maximized at

dπ′

dq′=

n + 12n· (a− c)− 2q′ = 0⇒ q′ =

a + 14n

(a− c)

The profit at q′ (cheating gain) is

π′ =

(a− c− a− c

2n· (n− 1)− n + 1

4n· (a− c)

)·(

n + 14n· (a− c)

)=

[n + 1

4n· (a− c)

]2

If the firm deviates, it earns the cheating gain for one round andCournot profits for all future rounds i.e. the gain from deviatingfrom the strategy is

πdeviate =

(n + 1

4n· (a− c)

)2+ δ

(a− c1 + n

)2+ δ2

(a− c1 + n

)2+ . . .

⇒ πdeviate =

[(n + 1

4n

)2+

δ

1 + δ

(1

1 + n

)2](a− c)2

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

π f ollow =1n

(a− c

2

)2+ δ · 1

n

(a− c

2

)2+ δ2 · 1

n

(a− c

2

)2+ . . . =

11− δ

· 1n

(a− c

2

)2

The strategy is stable if

π f ollow ≥ πdeviate

⇒ 11− δ

· 1n

(a− c

2

)2≥[(

n + 14n

)2+

δ

1− δ

(1

n + 1

)2](a− c)2

⇒ δ∗ ≤ n2 + 2n + 1n2 + 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 payoffof 1 and player 2 would get a payoff of 0 i.e. (1,0). In a two period

23

game, if the two players can’t agree, the game goes to the secondstage, 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 firstround, they go on to the second round, at which point it becomesa two period game and player 2 gets a payoff of 1− δ and player 1

gets 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 δ

1+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 δ

1+δ ,which will be accepted, leaving 1

1+δ for B. This is worth δ · 11+δ to

B in the first round (which is why B will refuse anything less thanthis 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 theinvestor does in the first round and x2 is what the investor does inthe second round. The game can be represented by the followingtable:

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 notSubgame Perfect Equilibria since there is no subgame in which bothor either player doesn’t withdraw his or her funds in the secondround.

Answer 2.23 The optimal investment is given by

d(v + I − p− I2)

dI= 0⇒ I∗ =

12

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

v + I − p− I2 ≥ −I2 ⇒ p ≤ v + I


Thus, the highest possible price that the buyer will pay at this pointis p = v + I. If, however, the buyer doesn’t invest, the buyer willbuy if

v− p ≥ 0⇒ v ≥ p

Thus the buyer would be willing to pay p = v. Thus investment isI ⊆ {0, 1

2}. The price is drawn from p ⊆ {v, v + 12}. There is no gain

from charging anything other than these prices.

v + 12 v

12 , A − 1

4 , v + 12

14 , v

12 , R − 1

4 , 0 − 14 , 0

0, A − 12 , v + 1

2 0, v

0, 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 cancollapse the above table into a simpler one:

(q)v + 12 (1− q)v

(p)I = 12 Accept − 1

4 , v + 12

14 , v

(1− p)I = 0Reject 0, 0 0, 0

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

Static Games of Incomplete Information

Answer 3.1 See text

Answer 3.2 Firm 1 aims to maximize

π1 = (p− c)q1 = (ai − c− q1 − q2)q1

Which is done by

dπ1

dq1= ai − c− q1 − q2 + q1(−1) = 0⇒ q1 =

ai − c− q2

2

Thus, the strategy for firm 1 is

q1 =

aH−c−q2

2 if ai = aHaL−c−q2

2 if ai = aL

Now, the firm 2 aims to maximize

π2 = (p− c)q2 = (a− c− q1 − q2)q2


dπ2

dq2= a− c− q1 − q2 + q2(−1) = 0⇒ q2 =

a− c− q1

2=

θaH + (1− θ)aL − c− q1

2

Plugging in q1, we get

q2 =θaH + (1− θ)aL − c−

[θaH+(1−θ)aL−c−q2

2

]2

⇒ q2 =θaH + (1− θ)aL − c

3

Now, we need to find out what firm 1’s output would be. If ai =

aH ,

qH1 =

aH − c− θaH+(1−θ)aL−c3

2=

(3− θ)aH − (1− θ)aL − 2c6

But what if ai = aL?

qL1 =

aL − c− θaH+(1−θ)aL−c3

2=

(2 + θ)aL − θaH − 2c6

26 Static Games of Incomplete Information

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

q2 ≥ 0⇒ θaH + (1− θ)aL − c ≥ 0⇒ θ ≥ c− aLaH − aL

Which also requires

θ ≤ 1⇒ 1 ≥ c− aL ⇒ aL ≥ c− 1

Furthermore,

qL1 ≥ 0⇒ (2 + θ)aL − θaH − 2c ≥

6≥ 0⇒ θ ≤ 2 · c− aL

aH − aL

Which subsumes the last-but-one result. And finally,

qH1 ≥ 0⇒ θ ≤ 2c− 3aH + aL

aH − aL

Answer 3.3 The profits earned by firm 1 is given by

π1 = (p1 − c)q1 = (p1 − c)(a− p1 − b1 p2)


dπ1

dp1= a− p1 − b1 p2 + p1(−1) = 0⇒ p1 =

a− b1 p2

2=

a− b1[θpH + (1− θ)pL]

2

Now, what if b1 = bH? To start with, p1 = pH and

pH =a− bH [θpH + (1− θ)pL]

2⇒ pH =

a− (1− θ)bH pL2 + θbH

And if b1 = bL:

pL =a− bL[θpH + (1− θ)pL]

2=

a− θbL pH2 + bL(1− θ)

Which means that,

pH =a− (1− θ)

[a−θbL pH

2+(1−θ)bL

]2 + θbH

⇒ pH =a(1− [1− θ]bH)

4 + 2(1− θ)bL + 2θbH

Similarly,

pL =a(1− θbL)

4 + 2(1− θ)bL + 2θbH

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

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

Now, if nature chooses game 1, player 1 will play T. If naturechooses game 2, player 1 will play B. Furthermore, if player 2 playsL with probability p:

π2 = p[12· 0 + 1

2· 1] + (1− p)[

12· 0 + 1

2· 2] = 1− 1

2· p

This is maximized at p = 0 i.e. player 2 will always play R. Thus thePure-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 > bj ∀j = 1, 2, . . . , i− 1, i + 1, . . . , nvi−bj

m if bi = bj

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

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

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

⇒ πi = (vi − bi) · [P(bi > aj + cjvj)]n−1 = (vi − bi) ·

[P

(vj <

bi − aj

cj

)]n−1

= (vi − bi)

(bi − aj

cj

)n−1


dπidbi

= (−1) ·(

bi − aj

cj

)n−1

+ (vi − bi)

(n− 1

cj

)(bi − aj

cj

)n−2

= 0

(bi − aj

cj

)n−2

·(

aj + (n− 1)vi − nbi

cj

)= 0

This requires that either

bi − aj

cj= 0⇒ bi = aj

Or that

aj + (n− 1)vi − nbi

cj= 0⇒ bi =

aj

n+

n− 1n· vi

28 Static Games of Incomplete Information

Now, we know that bi = ai + civi. Here, we know that ci =n−1

n and

ai =aj

n⇒ a1 =

a2

n=

a3

n= . . . =

an

n

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

bi =n− 1

n· vi

Answer 3.7

Answer 3.8

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

π2(L′) = 1 · p + 0 · (1− p) = p

The payoff from playing R′

π2(R′) = 0 · p + 1 · (1− p) = 1− p

Player 2 will always play L′ if

π2(L′) > π2(R′)⇒ p > 1− p⇒ p >12

The playoff to player 1 from playing L is

π1(L) = 4 · q + 0 · (1− q) = 4q

And the payoff from playing M is

π1(M) = 3 · q + 0 · (1− q) = 3q

Player 1 will always play L if

π1(L) > π1(M)⇒ 4q > 3q

Which is true. Thus p = 1. In which case, player 2 will always playL′. 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 Dynamic Games of Incomplete Information

The expected values of the payoffs to player 2 are:

π2(L′) = 3 · p + 0 · (1− p) = 3p

π2(M′) = 2 · p + 2 · (1− p) = 2

π2(R′) = 0 · p + 3 · (1− p) = 3− 3p

And the payoffs to player 1 are:

π1(R) = 2

π1(L) = 1 · q1 + 1 · q2 + 4 · (1− q1 − q2) = 4− 3q1 − 3q2

π1(M) = 4 · q1 + 0 · q2 + 3 · (1− q1 − q2) = 3 + q1 − 3q2

The only Nash Equilibrium is (R, M′); it is also sub-game perfect.To be a Perfect Bayesian Equilibrium, player 2 must believe that

π2(M′) > π2(L′)⇒ 2 > 3p⇒ 23> p

and

π2(M′) > π2(R′)⇒ 2 > 3− 3p⇒ p >13

Furthermore, player 1 must believe

π1(R) > π1(L)⇒ 2 > 4− 3q1 − 3q2 ⇒ q1 >23− q2

Since q1 > 0, this implies that

23− q2 > 0⇒ 2

3> q2

and

π1(R) > π1(M)⇒ 2 > 3 + q1 − 3q2 ⇒ 3q2 − 1 > q1

Which, in turn, requires

3q2 − 1 >23− q2 ⇒ q2 >

512

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

13> p >

23

, 3q2 − 1 > q1 >23− q2,

23> q2 >

512

]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

31

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

π1(L) > π1(M)⇒ 3 · q + 0 · (1− q) > 0 · q + 3 · (1− q)⇒ q >12

Thus if q > 0.5, p = 1.(b) Player 2 always picks either L′ or R′; player 2 will always play

L′ if

π2(L′) > π2(R′)⇒ 0 · p + 1 · (1− p) > 1 · p + 0 · (1− p)⇒ 12> p

Thus, if p < 0.5, q = 1. This violates the condition we uncovered inpart (a), proving that there is no PSNE.

In a mixed strategy BE, player 1 plays L with probability p andplayer 2 plays L′ with probability q. In equilibrium, player 2 isindifferent between L′ and R′:

π2(L′) = π2(R′)⇒ 0 · p + 1 · (1− p) = 1 · p + 0 · (1− p)⇒ p =12

And similarly, for player 1:

π1(L) = π1(M)⇒ 3 · q + 0 · (1− q) = 0 · q + 3 · (1− q)⇒ q =12

Thus, in a mixed strategy equilibrium, player 1 plays R with p = 0.5and player 2 plays L′ with probability q = 0.5.

Answer 4.3 (a) Let’s start with the pooling equilibrium (R, R). Inthis situation, p = 0.5. Now, the payoff to the receiver is

πR(R, u) = 0.5 · (1) + 0.5 · (0) = 0.5

πR(R, d) = 0.5 · (0) + 0.5 · (2) = 1

Thus, if the sender plays R, the receiver will play d. We have to testtwo strategies for the receiver: (u, d) and(d, d). Under the strategy(d, d)

π1(L, d) = 2 and π1(R, d) = 3

π2(L, d) = 3 and π2(R, d) = 2

There is no incentive for type 1 to deviate and play L, but there isan incentive for type 2 to do so. Under the strategy (u, d),

π1(L, u) = 1 and π1(R, d) = 3

π2(L, u) = 0 and π2(R, d) = 2

Neither type 1 nor type 2 have any reason to L instead of R. Thus,we have the following pooling equilibrium:

[(R, R), (u, d), p = 0.5, 1 ≥ q ≥ 0]


(b) We must find a pooling equilibrium in which the senderplays (L, L, L). For the receiver, the payoffs are

πR(L, u) =13· 1 + 1

3· 1 + 1

3· 1 = 1

πR(L, d) =13· 0 + 1

3· 0 + 1

3· 0 = 0

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

π1(L, u) = 1 and π1(R, u) = 0

π2(L, u) = 2 and π2(R, u) = 1

π3(L, u) = 1 and π3(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 =13

, 1 ≥ q ≥ 0]

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

0.5. πR(L, u) = πR(L, d). Thus, it doesn’t matter for the receiverwhether he/she plays u or d.• Under (u, u), π1(L, u) = 1 < π1(R, u) = 2, making it unsustain-

able.• Under (u, d), π2(L, u) = 0 < π2(R, d) = 1, making it unsustain-

able.• Under (d, d), π2(L, d) = 0 < π2(R, d) = 1, making it unsustain-

able.• Under (d, u), π2(L, u) = 0 < π2(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 πR(1, L, u) > πR(1, L, d) andπR(2, R, u) > πR(2, R, d)deviate:

• For type 1, π1(L, u) = 1 > π1(R, d) = 0 i.e. no reason to play Rinstead of L.• For type 2, π2(L, u) = 0 < π2(R, d) = 1 i.e. no reason to play L

instead of R.Let’s examine pooling equilibrium (R, R). πR(R, u) = 1 >

πR(R, d) = 0.5. Therefore, the two strategies that can be followed bythe 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 πR(1, R, u) = 2 > πR(1, R, d) =0 and πR(2, L, u) = 0 < πR(2, L, d) = 1• For type 1, π1(L, d) = 2 ≥ π1(R, u) = 2, i.e. will play L.

• For type 2, π2(L, d) = 0 ≤ π2(R, u) = 1, i.e. will play R.

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). πR(L, u) = 1.5 >

πR(L, d) = 1, therefore player 2 will respond to L with u. The twostrategies are (u, u) and (u, d).• Under (u, u), π(L, u) > π(R, u) for both types.• Under (u, d), π1(L, u) = 3 < π1(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, π1(L, d) = 1 > π1(R, u) = 0 i.e. type 1 will play L.• For type 2, π2(L, d) = 0 < π2(R, u) = 1 i.e. type 2 will play R.Let’s examine pooling equilibrium (R, R). πR(R, u) = 1 >

πR(L, d) = 0.5, therefore player 2 will respond to R with u. The twostrategies are (u, u) and (d, u).• Under (u, u), π1(L, u) = 3 > π1(R, u) = 0, making (R, R)

unsustainable.• Under (d, u), π1(L, d) = 1 > π1(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, π1(L, d) = 1 > π1(R, u) = 0, making (R, L)

unsustainable.• For type 2, π2(L, d) = 0 < π2(R, u) = 1, which doesn’t conflict

with 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.πR(L, u) = 1 > πR(L, d) = 0.5. Thus, the receiver’s response toL 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 is

no incentive for either type to play R instead of L.• Under the strategy (u, u), π2(L, u) = 0 < π2(R, u) = 1, making

(L, L) unsustainable.


Let’s examine (L, R). The best response to this is (u, d). In re-sponse to this,• For type 1, π1(L, u) = 1 < π1(R, d) = 3 i.e. type 1 will play R

which violates the equilibrium• For type 2, π2(L, u) = 0 < π2(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, π1(L, u) = 1 < π1(R, d) = 3, i.e. type 1 will play R.• For type 2, π2(L, u) = 0 < π2(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 equilibriumother 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) and6.(R, R, L).

Let’s start with pooling equilibrium (R, R, R). πR(R, u) = 23 >

πR(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), π1(L, d) < π1(R, u), making the equilibrium

unsustainable. Let’s examine the various separating the equilib-rium.

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

• For type 1, πS(1, L, u) = 1 > πS(1, R, d) = 0 i.e. type 1 will playL.• For type 2, πS(2, L, u) = 2 > πS(2, R, d) = 1 i.e. type 2 will play

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

R.Thus, this is a viable equilibrium.2. (L, R, L). The best response to this is (u, u), 9 9 since πR(2, R, u) = 1 > πR(2, R, d) =

0 and 0.5 · πR(1, L, u) + 0.5 ·πR(3, L, u) = 1 > 0.5 · πR(1, L, d) + 0.5 ·πR(3, L, d) = 0

• For type 1, πS(1, L, u) = 1 > πS(1, R, u) = 0 i.e. type 1 will playL.• For type 2, πS(2, L, u) = 2 > πS(2, R, u) = 1 i.e. type 2 will play

L, instead of R.• For type 3, πS(3, L, u) = 1 > πS(3, R, u) = 0 i.e. type 3 will play

L.Thus, this is not a viable equilibrium.3. (R, L, L). The best response to this is (u, u), 10 10 since πR(1, R, u) = 1 < πR(1, R, d) =

0 and 0.5 · πR(2, L, u) + 0.5 ·πR(3, L, u) = 1 > 0.5 · πR(2, L, d) + 0.5 ·πR(3, L, d) = 0

• For type 1, πS(1, L, u) = 1 > πS(1, R, u) = 0 i.e. type 1 will playL, instead of R.• For type 2, πS(2, L, u) = 2 > πS(2, R, u) = 1 i.e. type 2 will play

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

L.Thus, this is a not viable equilibrium.

35

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

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



L, instead of R.Thus, this is not a viable equilibrium.Let’s test (u, d)• For type 1, πS(1, L, u) = 1 > πS(1, R, d) = 0 i.e. type 1 will play

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

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

R.Thus, this is not a viable equilibrium.5. (R, L, R). The best response to this is either (u, u) or (u, d) 12 12 since πR(2, L, u) = 1 > πR(2, L, d) =

0 and 0.5 · πR(2, L, u) + 0.5 ·πR(3, L, u) = 0.5 = 0.5 · πR(2, L, d) +0.5 · πR(3, L, d) = 0.5

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



L, instead of R.Thus, this is not a viable equilibrium.Let’s test (u, d)• For type 1, πS(1, L, u) = 1 > πS(1, R, d) = 0 i.e. type 1 will play

L, instead of R.• For type 2, πS(2, L, u) = 2 > πS(2, R, d) = 0 i.e. type 2 will play

L.• For type 3, πS(3, L, u) = 1 = πS(3, R, d) = 1 i.e. type 3 can play

R.Thus, this is not a viable equilibrium.6. (R, R, L). The best response to this is (u, u), 13 13 since πR(3, L, u) = 1 > πR(3, L, d) =

0 and 0.5 · πR(1, R, u) + 0.5 ·πR(2, R, u) = 0.5 > 0.5 · πR(1, R, d) +0.5 · πR(2, R, d) = 0

• For type 1, πS(1, L, u) = 1 > πS(1, R, u) = 0 i.e. type 1 will playL, instead of R• For type 2, πS(2, L, u) = 2 > πS(2, R, u) = 1 i.e. type 2 will play

L, instead of R• For type 3, πS(3, L, u) = 1 > πS(3, R, u) = 0 i.e. type 3 will play

L.Thus, this is not a viable equilibrium.The only other perfect Bayesian Equilibrium is[

(L, L, R), (u, d), p, q0 =13

, q1 =13

]Answer 4.6 Type 2 will always play R since πS(2, R, a) > πS(2, R, u) and πS(2, R, a) >πS(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, πR(x, L, u) > πR(x, L, d) for

both types, so the Receiver will alwaysplay u


a payoff of (2, 1). This gives type 1 a higher payoff than if he playedR, which would have given him a payoff of 1. Thus, the perfectBayesian Equilibrium is

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

unofficial solutions manual to r.a gibbon's a primer in game theory

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