博弈论基础 (罗伯特.吉本斯 著) 中国社会科学出版社 课后答案

博弈论

2.1 Proof: we should show that A maximizesIc(A)+Ip(A), this is equivalent to showIc′(A)+Ip′(A)=0.

Solving the game by backward induction:

At stage 2, givenIcandIp, parent chooses B to maximize his payoff: maxV(Ip B)+kU(Ic+B)

B

At stage 1, child chooses A to maximize his payoff: maxU(Ic(A)+B)

A

=

答

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U′(Ic(A)+B)i(Ic′(A)+

V′′(Ip B)

(Ic′(A)+Ip′(A))=0=U′(Ic(A)+B)i

kU′′(Ic+B)+V′′(Ip B) Ic′(A)+Ip′(A)=0(SinceU′(i)>0,V′′(i)<0andU′′(i)<0)Another approach:

Solving the game by backward induction:

At stage 2, givenIcandIp, parent chooses B to maximize his payoff: maxV(Ip B)+kU(Ic+B)

B

课

=U′(Ic(A)+B)i(Ic′(A)+

案

V′′(Ip B)kU′′(Ic+B)′= iIc(A)+iIp′(A)

kU(Ic+B)+V(Ip B)kU(Ic+B)+V(Ip B)

B

A

V′′(Ip B) kU′′(Ic+B)′iIp′(A))Ic(A)+

kU(Ic+B)+V(Ip B)kU(Ic+B)+V(Ip B)

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V′′(Ip B)kU′′(Ic+B)

iIc′(A) iIp′(A)

kU′′(Ic+B)+V′′(Ip B)kU′′(Ic+B)+V′′(Ip B)

w

ww

B B′ B′=Ic(A)+iIp(A) A Ic Ip

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B

FOC: U′(Ic(A)+B)i(Ic′(A)+=0

A

From (*), we have:

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FOC: kU′(Ic+B) V′(Ip B)=0 *

博弈论

FOC: kU′(Ic+B) V′(Ip B)=0 *

At stage 1, child chooses A to maximize his payoff: maxU(Ic(A)+B)

A

′(A)+I′(1)+(2) Icp(A)=0

B

BkU2′′(S+B)

=

SV′′(Ip B)+kU2′′(S+B)

U1′(Ic S)+U2′(S+B)(1

kU2′′(S+B)

)

V′′(Ip B)+kU2′′(S+B)

V′′(Ip B)V′′(Ip B)+kU2′′(S+B)

So = U1′(Ic S)+U2′(S+B)i

=0

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B

FOC: U1′(Ic S)+U2′(S+B)(1+)=0

S

From (*), we have

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答

maxU1(Ic S)+U2(S+B)

S

案

At stage 1, child maximizes his payoff:

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FOC: V′(Ip B)+kU2′(S+B)=0 (*)

w

maxV(Ip B)+k(U1(Ic S)+U2(S+B))

ww

2.2 Proof: At, first, solving the game by backward induction:

At stage 2, given S, parent chooses B to maximize his payoff:

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that: Hence, in the game, child chooses S* such

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.

B

FOC: U′(Ic(A)+B)i(Ic′(A)+=0

A

Because U is increasing and strictly concave, so

B

U′(Ic(A)+B)>0 (Ic′(A)+=0 （1）

A

B B

′(A)+) V′′(Ip(A) B)(Ip′(A) =0 (*) 对A求偏导： kU′′(Ic(A)+B)(Ic

A A

B B

′′′()(())(())=0(Since(Ic′(A)+ B=0) V′′(Ip(A) B)(I′A =kUIA+BIA+pcc

A A A

B

Because V is strictly concave, so V′′(Ip(A) B)>0 (I′A ())=0 (2) p

A

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博弈论

U1′(Ic S*)+U2′(S*+B)i

V′′(Ip B)

V′′(Ip B)+kU2′′(S*+B)

=0

On the other hand, if child chooses S′ to maximizeU1(Ic S)+U2(S+B), where B is exogenous, then S′ satisfies: U1′(Ic S′)+U2′(S′+B)=0 We need to showS*<S′.

Denote f(S)= U1′(Ic S)+U2′(S+B),f′(S)=U1′′(Ic S)+U2′′(S+B)<0

f(S*)= U1′(Ic S*)+U2′(S*+B)

*

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aw.

′′(S*+B)kU2

= U1′(Ic S)+U2′(S+B)i+U2′(S+B)i

*

V′′(Ip B)+kU2′′(S+B)V′′(Ip B)+kU2′′(S*+B)

V′′(Ip B)

*

If V<(R c1)2,c2=0, both receive 0. In Period one: partner 1 choose c1. Consider four cases:

(1)

if R≤, that is V R2≥δV, c1=R, partner 1 will complete the project himself. So c2=0. (2)

If

<R≤V R2<δV and V≥R2, c1=0,c2=R.

课

If V≥(R c1)2,c2=R c1, both receive V.

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the project is R c1.

答

In Period two: given c1, partner 2 choose c2, the minimize of c2 to complete

案

increased.

2.4 Solving the game by backward induction:

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w

SoS*<S′. If child save more, i.e.S′, both the parent’s and child’s payoffs could be

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>0=f(S′)

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′′(S*+B)kU2

=U2′(S+B)i

V′′(Ip B)+kU2′′(S*+B)

*

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博弈论

(3)

If (1≥R>, that is R2>

V, and (R 2≤δV, the minimize of c1 to complete the project satisfies: V=(R c1)2,c1<

R, So

c1=R

c2=.

(4)

If R>(1

, that is (R2>δV, it is not worth completing the project, hence c1=0,c2=0.

1

(a w). n+1

At stage 1, the union chooses w to maximize its utility:

n

max(a w)(w wa)

wn+1

a+wa

. FOC: a+wa 2w=0 w=2

na wa2

Then payoffs of the union are (, which is increasing with n. If n

2n+1

increases, the total output increases, so does the demand for labor, so the union’ utility increases.

a+c

2.13 Proof: The monopoly price is p=.

2

If the firms use trigger strategies, then if there is no firm deviate, both get

1a c2()

1a c2. ) on every stage game, and the total discounted profit is (

1 δ22

a c2

The payoff from deviating on an stage is (. For the trigger strategies to

2

1a c2()

a c21≥(, that is δ≥. be SPNE, we must have 1 δ22a+ca+c

2.14 The monopoly price is pH=H ,pL=L

22

If the firms use trigger strategies, and there is no firm deviate, in period with

1a c21a c2

δ(πii(H)+(1 π)(L)

1aH c2, demandaH, the total payoff is (+221 2.7 In the subgame, the equilibrium is qi=Li=

in period with demand aL

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答

案

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, the total discounted payoff is

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博弈论

1aL c2(+22

δ(πii(

1aH c21a c2

)+(1 π)i(L))

.

1 aH c2

); in period with 2

In period with demandaH, payoff from deviating is (demand aL, payoff from deviating is (