Game Thoery Gibbons Unofficial Solution Manual PDF

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Unofficial solution manual to Game Theory Gibbons textbook...

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The Unofficial Solution Manual to

A Primer in Game Theory by RA Gibbons Unfinished Draft

Navin Kumar Delhi School of Economics

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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 ATEX and the Tufte book class. Typeset using L

This work is not subject to copyright. Feel free to reproduce, distribute or 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 errors 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

Static Games of Complete Information

Answer 1.1 See text. Answer 1.2 B is strictly dominated by T. C is now strictly dominated 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 could increase his or her payoff by asking for more. If, however, S2 > 1 − S1 , Individual 2 earns nothing and can increase his payoff by reducing 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 + ... + q n The cost for an individual company is given by Ci = c · q i . The profit made by a single firm is πi = ( p − c) · q i = ( a − Q − c) · q i = ( a − q1∗ − ... − qn∗ − c) · q i Where qj∗ is the profit-maximizing quantity produced by firm j in equilibrium. This profit is maximized at dπi = ( a − q ∗1 − ... − qi∗ − ... − qn∗ − c) − q ∗i = 0 dq i

⇒ a − q1∗ − ... − 2 · qi∗ − ... − qn∗ − c = 0

⇒ a − c = q1∗ + ... + 2 · qi∗ + ... + qn∗

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 = q2∗ = ... = qi∗ = ... = qn∗ Which means the preceding equation becomes. a − c = (n + 1)qi∗ ⇒ q ∗i =

a−c n+1

Static Games of Complete Information

6

A similar argument applies to all other firms. Answer 1.5 Let q m be the amount produced by a monopolist. Thus, if the two were colluding, they’d each produce 1 qm = q2m =

a−c qm = 4 2

In such a scenario, the profits earned by Firm 1 (and, symmetrically, Firm 2) is   a−c 1 1 π 1mm = ( P − c) · qm1 = ( a − Q − c) · qm = a− − c · qm 2 a−c a−c ( a − c )2 = = = 0.13 · ( a − c)2 · 4 8 2 If both are playing the Cournot equilibrium quantity, than the profits earned by Firm 1 (and Firm 2) are: 1 1 πcc = ( P − c) · qc1 = ( a −

1

P= a−Q

a + 2c 2( a − c ) − c) · q1c − c) · q1c = ( 3 3

2 = a − q1cc − qcc a−c = a−2· 3 a+c = 2

( 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: 1 = ( P − c) · q 1 πcm c = (

a−c 5 5a 7c · ( a − c )2 − c) · = + 36 3 12 12

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

a−c 5 5a 7c · ( a − c )2 − c) · = + 48 12 4 12

= 0.10 · ( a − c)2 For notational simplicity, let α ≡ ( a − c )2 Their profits are reversed when their production is. Thus, the payoffs are: Player 2

Player 1

qm

qc

qm

0.13α, 0.13α

0.10α, 0.14α

qc

0.14α, 0.10α

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

The price is determined by

2

The price is given by P= a−Q

2 = a − q1c − qm a−c a−c = a− − 4 3 7 · ( a − c) = a− 12 7c 5a + = 12 12

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been had they cooperated by producing q m 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 ) · q 1 = ( a − Q − c1 ) · q 1 = ( a − q 1 − q 2 − c1 ) · q 1 At the maximum level of profit, a − c1 − q 2 dπ1 = ( a − c1 − q1 − q2 ) + q1 · (−1) = 0 ⇒ q1 = 2 dq 1 And by a similar deduction, q2 =

a − c2 − q 1 2

Plugging the above equation into it’s predecessor, q1 =

a − c1 − 2

a − c2 − q 1 2

=

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 ⇒ 0 > q2 3

⇒ q2 = 0 Since quantities cannot be be negative. Thus, under certain conditions, a sufficient difference in costs can drive one of the firms to shut down. Answer 1.7 We know that,    a − pi q i = a −2 pi    0

if pi < pj , if pi = pj , if pi > pj .

We must now prove pi = pj = c is the Nash Equilibrium of this game. To this end, let’s consider the alternatives exhaustively. If pi > pj = c, q i = 0 and π j = 0. In this scenario, Firm j can increase 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, q i = 0 and πi = 0 and Firm i can make positive profits by charging pj − ε > c. Thus, this cannot be a Nash Equilibrium. a− p If pi = pj > c, πi = ( pi − c) · 2 i . Firm i can increase profits by charging pi − ε such that pj > pi − ε > 0, grab the entire market and a larger profit, provided

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

( a − pi ) 2

Static Games of Complete Information

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Therefore, this is not a Nash Equilibrium If pi = pj = c, then πi = π j = 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 , q i = πi = 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 =

 1    xi + 2 · ( x j − xi ) 1 2

  (1 − x ) + i

1 2

if xi < x j , if xi = x j ,

· ( xi − x j ) if xi > x j .

We aim to prove the Nash Equilibrium is (21, 12 ). Let us exhaustively consider the alternatives. Suppose xi = 12 and x j > 21 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 < 21 , Candidate j can increase his share of the vote by moving to the right. Suppose xi > x j > 21; Candidate i can gain a larger share by moving to a point x j > xi > 21. Thus this is not a Nash Equilibrium. Suppose xi = x j = 21; the share of i and j are 21 . If Candidate i were to deviate to a point (say) xi > 21 , his share of the vote would decline. Thus (21, 12 ) is a unique Nash Equilibrium. This is the famous Median Voter Theorm, used extensively in the study of politics. It explains why, for example, presidential candidates in the US veer sharply to the center as election day approaches. Answer 1.9 See text. Answer 1.10 (a) Prisoner’s Dilemma Player 2

( p) Mum Player 1

(q ) Mum

−1, −1

(1 − q ) Fink

0, −9

(1 − p) Fink −9, 0

−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

Player 1

Le f t(q0 )

Middle (q1 )

Right(1 − q0 − q1 )

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 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 = 0.5

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 impossible. (c) Player 2

Player 1

L(q0 )

C (q1 )

R (1 − q 0 − q 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 C must, 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 .

Static Games of Complete Information

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Answer 1.11 This game can be written as Player 2 L(q0 )

C (q1 )

R (1 − q 0 − q 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

Player 1

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 − 1

Similarly, E2 (C ) = E2 ( R )

⇒ 1 · p0 + 2 · p1 + 2 · (1 − p0 − p1 ) = 2 · p0 + 3 · p1 + 0 · (1 − p0 − p1 ) 2 ⇒ p1 = − p0 3 Combining these, 5 2 − p0 ⇒ p0 = 3 9 1 5 ∴ p1 = 2 · p0 − 1 = 2 · − 1 = 9 9 5 1 3 ∴1 − p0 − p1 = 1 − − = 9 9 9 2 · p0 − 1 =

Now we must calculate q0 and q1 . Player 2 will set them such that E1 ( T ) = E1 ( M )

⇒ 2 · q 0 + 1 · q 1 + 4 · (1 − q 0 − q 1 ) = 3 · q 0 + 1 · q 1 + 2 · (1 − q 0 − q 1 ) ⇒ q1 = 1 − 1.5 · q0

And E1 ( M ) = E1 ( B)

⇒ 3 · q 0 + 1 · q 1 + 2 · (1 − q 0 − q 1 ) = 1 · q 0 + 0 · q 1 + 3 · (1 − q 0 − q 1 ) Qed Answer 1.12

( q ) L2

(1 − q ) R 2

( p) T1

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=

2 3

Player 2 will set q such that E1 ( T ) = E1 ( B)

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

3 4

Answer 1.13

(q )Apply1 to Firm 1 ( p)Apply2 to Firm 1

1 1 2 w1 , 2 w1

(1 − p)Apply2 to Firm 2

w2 ,w1

(1 − q )Apply1 to Firm 2 w1 ,w2 1 1 2 w2 , 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 a mixed-strategy equilibrium, Player 1 sets p such that Player 2 is indifferent between Applying to Firm 1 and Applying to Firm 2. E2 (Firm 1) = E2 (Firm 2 )

⇒ p·

1 1 w1 + (1 − p) · w1 = p · w2 + (1 − p) · w2 2 2

⇒ p=

2w1 − w2 w1 + w2

Since 2w1 > w2 , 2w1 − w2 is positive and p > 0. For p < 1 to be true, it must be the case that 2w1 − w2 1 < 1 ⇒ w1 < w2 2 w1 + w2 Which is true. And since the payoffs are symmetric, a similar analysis would reveal that q= Answer 1.14

2w1 − w2 w1 + w2

Dynamic Games of Complete Information

Answer 2.1 The total family income is given by IC ( A ) + IP ( A ) This is maximized at dI ( A ) dI ( A ) d( IC ( A ) + IP ( A )) =− P =0⇒ C dA dA dA The utility function of the parents is given by V ( IP − B) + kU ( IC + B) This is maximized at k

⇒k

dU ( IC + B) dV ( IP − B) + =0 dB db

dU ( IC + B) d( IC + B) dV ( IP − B) IP − B · + · =0 d ( IP − B ) dB d( IC + B) dB

⇒ 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 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 ( IC ( A ) − B∗ ( A )) . This is maximized at dU ( IC ( A ) + B∗ ( A )) =0 dA

⇒ U ′ ( IC ( A) + B∗ ( A)) · ⇒



dIC ( A ) dB∗ ( A ) + dA dA



=0

dIC ( A ) dB∗ ( A ) + = 0 ⇒ IC′ ( A) = − B′∗ ( A) dA dA

We now have only to prove B′∗ ( A ) = IP′ ( A ) dV ( IP ( A ) − B∗ ( A )) =0 dA

Dynamic Games of Complete Information

14

⇒ V ′ ( IP ( A)∗B ( A)) ·



dIP ( A ) dB∗ ( A ) − dA dA



=0

⇒ IP′ ( 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)]} =0 dB

⇒ −V ′ + [U1′ (−S′B ) + U2′ (S′B + 1)] = 0 ⇒ V ′ = −kU1′ S′ + U2′ S′ + U2′ The utility of the child is given by U1 ( IC − S) + U2 (S + B). This is maximized at: d[U1 ( IC − S) + U2 (S + B)] = 0 ⇒ U ′1 = U2′ (1 + B′ ) dS Total utility is given by V ( I p − B) + k (U1 ( IC − S) + U2 ( B + S)) + U1 ( IC − S) + U2 ( B + S)

= V ( I p − B) + (1 + k )(U1 ( IC − S) + U2 ( B + S)) This is maximized (w.r.t S) at: V ′ · (− BS′ ) + (1 + K) · [U1′ (−1) + U ′2 (1 + BS′ )] = 0

⇒ U ′1 = U2′ (1 + BS′ ) −

V ′ B′S 1+k

as opposed to U1′ = U2′ (1 + B′S ), which is the equilibrium condition. V ′ BS′ 1+ k

> 0, the equilibrium U1′ is ‘too high’ which means that S ′ - the level of savings - must be too low (since dU dS < 0). It should be higher.

Since

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 and pocket V. If c1 < R and partner 2 responds with some c2 such that c2 < R − c1 , his payoff is −c22 . If he puts in nothing, he will, receive a payoff of zero. There is, therefore, no reason to put in such a low amount. There is - obviously - also no reason to put in any c2 > R − c1 . He will put in R − c1 if it’s better than putting in nothing 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 √ −1 V

2

If R2 ≥ 4V, δ would have to be greater than one, which is impossi 2 ble. Therefore, if δ ≥ √R − 1 and R2 ≥ 4V (i.e. the cost is not √ V √ ‘too high’), c1 = R − V and c2 = V. Otherwise, c2 = 0 and c1 = 0. Answer 2.5 Let the ‘wage premium’ be p = w D − w E , where p ∈ (−∞, ∞). In order to get the worker to acquire the skill, the firm 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 promote him iff the returns to the firm are such that: y D0 − w D ≤ y E0 − w E ⇒ y D0 − y E0 ≤ w D − w E = p If he does acquire the skill, the firm will promote if the returns to the firm are such that: y DS − w D ≥ y ES − w E ⇒ y DS − y ES ≥ w D − w E = p Thus the condition which the firm behaves as it ought to in the desired equilibrium is: y D0 − y E0 ≤ p ≤ y DS − y ES Given this condition, the worker will acquire the promotion iff he acquires 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 premium, which occurs at:  C if C ≥ y D0 − y E0 , wD − wE = p =  y D0 − y E0 if C < y D0 − y E0

A final condition is that the wages must be greater than the alternative i.e. w E ≥ 0 and w D ≥ 0. The firm seeks to maximize y ij − wi , which happens at w E = 0 and w D = 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) · q i . For Firm 2, for example π2 = ( a − q ∗1 − q2 − q3 − c) · q2

Dynamic Games of Complete Information

16

which is maximized at dπ2 = ( a − q1∗ − q2 − q3 − c) + q2 (−1) = 0 dq 2

⇒ q2 =

a − q1∗ − q3 − c 2

Symmetrically, q3 =

a − q1∗ − q2 − c 2

Putting these two together, q2 =

a − q1∗ −

a − q1∗− q2 − c 2

2

−c

⇒ q2 =

a − c − q 1∗ 3

Which, symmetrically, is equal to q3 . a − c − q1 − c) · q 1 = ∴ π1 = ( a − q 1 − q 2 − q 3 − c ) · q 1 = ( a − q 1 − 2 · 3 This is maximized at a − q 1 − c −q 1 dπ1 a−c = + = 0 ⇒ q1∗ = 2 3 dq 1 3 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 dπi d ( a − Q − w ) Li d ( p − w ) · Li = = =0 dL i dL i dL i

⇒

d ( a − L1 − . . . − L i − . . . − L n − w ) · L i =0 dL i

⇒ ( 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 Question (1.4)       a−w 2 1 ... 1 L1             1 2 . . . 1  L2  a − w  ·   =    .. .. . . ..   ..   ..  . .     . .  .   .    1

1

...

2

Ln

a−w



a − q1 − c 3



· q1

17

Which resolves to Li =

a−w n+1

Thus, total labor demand is given by L = L1 + L2 + . . . + L n =

a−w n a−w +...+ = · (a − w) n+1 n+1 n+1

The labor union aims to maximize U = (w − w a ) · L = ( w − w a ) ·

n n · (a − w) = · ( aw − aw a − w2 + ww a ) n+1 n+1

The union maximize U by setting w. This is maximized at dU a + wa n =0⇒w= = ( a − 2w + w a ) · 2 n+1 dw The sub-game perfect equilibrium is Li = na+−1w and w = a +2wa . Although the wage doesn’t change with n, the union’s utility (U ) is an increasing function of n+n1 , 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 country i can be driven to 0 iff ei∗ =

a − c − 2t j a−c = 0 ⇒ tj = 3 2

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 =