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

increased.
网 2.4 Solving the game by backward induction: 案 In Period two: given c1 , partner 2 choose c2 , the minimize of c2 to complete 后答 the project is R − c1 . 课 If V ≥ (R − c1)2, c2 = R − c1 , both receive V.
S)
+
U
′
2
(
S
+
B)i V
′′(I
p
V ′′(I p − B)
−
B)
+
kU
′′
2
(
S
+
B)
=0
Hence, in the game, child chooses S* such that:
−U1′ ( I c
−
S*)
+
U
′
2
(
S
*
+
B)i V
′′(I
p
V ′′(I p − B)
−
B)
+
kU
′′
2
U
′
2
(S
+
B)(1
+
∂B ∂S
)
=
0
后 From (*), we have
课 ∂B = −
kU
′′
2
(S
+
B)
∂S
V
′′(I
p
−
B)
+
kU
′′
2
(
S
+
B)
−U1′ ( I c
−
S)
+
U
′
2
(
S
+
B)(1 −
V
′′(I
p
kU
′′
2
(S
+
B)
−
B)
+
kU
′′
2
(
S
+
) B)
So
=
−U1′ ( I c
−
2
If the firms use trigger strategies, then if there is no firm deviate, both get
案网 1i(a − c)2
on every stage game, and the total discounted profit is
1i( a − c)2 22 .
(S
*
+
B)
=
0
On the other hand, if child chooses S′ to maximize U1(Ic − S) +U2 (S + B) ,
where B is exogenous, then S′ satisfies:
−U1′ ( I c
−
S ′)
+
U
′
2
(
S
′
+
B)
=
0
)2
+
(1 −
π
)
1 i( 2
aL − 2
c
)2
)
.
22
1−δ
In
period
with
demand aH , payoff from deviating is
(aH − c)2 ; 2
in period with
demand
aL , payoff from deviating is
( aL − c )2 . 2
课 1−δ
2
2
2.14
The monopoly price is
pH
=
aH + c , 2
pL
=
aL + c 2
If the firms use trigger strategies, and there is no firm deviate, in period with
demand aH , the total payoff is
1i( aH
−
c )2
+
δ (π i1i( aH − c )2 22
+
(1− π
)
1i(aL − c)2 ) 22
,
22
1−δ
in period with demand aL , the total discounted payoff is
1 i( aL
−
c )2
+
δ
(π
i1i( aH − 22
c
=
Li
=
1 (a n +1
−
w)
.
m At stage 1, the union chooses w to maximize its utility:
co max . w
n
n +1
(a
−
w)(w
−
wa
)
aw FOC:
a + wa
− 2w =
0⇒
w
=
a + wa 2
.
d Then payoffs of the union are n (a − wa )2 , which is increasing with n. If n h n +1 2
p
−
B)
i(
I
′
c
(
A)
+
I
′
p
(
A))
=
0
⇒
Ic′
(
A)
+
I
′
p
(
A)
=
0(
Since U ′(i) > 0, V ′′(i) < 0 and U ′′(i) < 0)
Another approach: Solving the game by backward induction:
At stage 2, given Ic and I p , parent chooses B to maximize his payoff:
22
1−δ
答 The payoff from deviating on an stage is ( a − c)2 . For the trigger strategies to 2
后 be SPNE, we must have
1i(a − c)2 22
≥ ( a − c )2 , that is
δ ≥1.
If V < (R − c1)2, c2 = 0 , both receive 0.
In Period one: partner 1 choose c1 . Consider four cases: (1) if R ≤ (1− δ )V , that is V − R2 ≥ δV , c1 = R , partner 1 will complete the
max B
V
(I
p
−
B)
+
kU
(Ic
+
B)
m FOC: kU ′(Ic + B) −V ′(I p − B) = 0 *
o At stage 1, child chooses A to maximize his payoff:
.c max A
U
(
Ic
(
A)
+
B)
daw FOC:
U
′(
I
c
(
A)
+
B)i(
( A)
+
B)i(
I
′
c
(
A)
+
kU
−kU ′′(Ic + B) ′′(Ic + B) +V ′′(I p
−
B
)
i
I
′
c
(
A)
+
kU
V ′′(I p − B) ′′(Ic + B) +V ′′(I
p
−
iI B)
′
p
(
A))
=
U
′(Ic
(
A)
+
B)i kU
V ′′(I p − B) ′′(Ic + B) +V ′′(I
0
m ⇒
V
′′(
I
p
(
A)
−
B)(
I
′p
(
A)
−
∂B ∂A
)
=
kU
′′(
I
c
(
A)
+
B)(
I
c′
(
A)
+
∂B ∂A
)
=
0
(Since
(Ic′(A)+
∂B) ∂A
=0)
.co Because V is strictly concave, so
V
′′(I
p
(
A)
−
B)
>
0
⇒
(I
′p
(
A)
−
∂B ) ∂A
+
B)i V
′′(I
p
V ′′(I p − B)
−
B)
+
kU
′′
2
(S
*
+
B)
+
U
′
2
(S
*
+
B)i V
′′(I
p
kU
′′
2
(S
*
+
B)
−
B
)
+
kU
′′
2
(
S
*
+
B)
hda =
U
′
2
(S
*
+
B)i V
′′(I
p
kU
′′
2
(S
*
+
B)
−
B)
+
kU
′′
2
(
S
*
+
B)
.k > 0 = f (S′)
www So S* < S′ . If child save more, i.e. S′ , both the parent’s and child’s payoffs could be
2.1 Proof: we should show that A maximizes Ic ( A) + I p ( A) , this is equivalent to
show
Ic′
(
A)
+
I
′
p
(
A)
=
0
.
Solving the game by backward induction:
At stage 2, given Ic and I p , parent chooses B to maximize his payoff:
=
0
(2)
w (1)+(2) ⇒ Ic′( A) + I ′p (A) = 0
hda 2.2 Proof: k At, first, solving the game by backward induction: . At stage 2, given S, parent chooses B to maximize his payoff:
project himself. So c2 = 0 .
(2) If (1− δ )V < R ≤ V , that is V − R2 < δV and V ≥ R2 , c1 = 0, c2 = R .
(3) If (1+ δ ) V ≥ R > V , that is R2 > V , and (R − V )2 ≤ δV , the minimize of c1 to complete the project satisfies: V = (R − c1)2 , c1 < R , So c1 = R − V , c2 = V .
We need to show S* < S′ .
Denote
f
(S) =
−U1′ ( I c
−
S
)
+
U
′
2
(S
+
B),
f
′(S )
= U1′′(Ic
−
S
)
+
U
′′
2
(
S
+
B) <
0
om f
(S*)
=
−U1′ ( I c
−
S*)
+
U
′
2
(
S
*
+
B)
w.c =
−U1′ ( I c
−
S*)
+
U
′
2
(S
*
aH − 2
c
)2
+
(1 −
π
)
1 i( 2
aL − 2
c )2
)
≥
( aL
−
c
)2
22
1−δ
2
.c 1i(aH − c)2
aw ⇒ δ
≥
(π i1i( aH
− c)2
2 + (1− π )
2 1i( aL
− c)2) +
1i( aH
− c)2
d 2 2
22
22
h The lowest of δ such that the firms can use trigger strategies to sustain these
k monopoly price levels in a SPNE is:
(4) If R > (1+ δ ) V , that is (R − V )2 > δV , it is not worth completing the project, hence c1 = 0, c2 = 0 .
2.7 In the subgame, the equilibrium is
qi
)
=
0
Because U is increasing and strictly concave, so
U
′(
I
c
(
A)
+
B)
>
0
⇒
(
Ic′
(
A)
+
∂B ∂A
)
=
0
（1）
(*)
对 A 求偏导：
kU
′′(Ic
(
A)
+
B)(Ic′
(
A)
+
∂B ∂A
)
−V
′′(I
p
(
A)
−
B)(I
′
p
(
A)
−
∂B ∂A
)
=
max B
V
(I
p
−
B)
+
kU
(Ic
+
B)
FOC: kU ′(Ic + B) −V ′(I p − B) = 0 *
At stage 1, child chooses A to maximize his payoff:
max A
U
(
Ic
(
A)
+
B)
FOC:
U
′(Ic
(
A)
+
B)i(
I
′
c
(
A)
+
∂B ∂A
I
′
c
(
A)
+
∂B ∂A
)
=
0
h From (*), we have:
.k ∂B w ∂A
=
∂B ∂Ic
i
I
′
c
(
A)
+
∂B ∂I p
iI
′
p
(
A)
ww =
−
kU
kU ′′(Ic + ′′(Ic + B) +V
B) ′′(I
p
−
B)
i
I
′
c
(
A)
−
kU
−V ′′(Ic +
′′(I p − B) +V
For the trigger strategies to be SPNE, we must have
1 i( aH
−
c )2
+
δ (π
i1i(aH − c)2 22
+
(1− π
)
1i(aL − c)2 ) 22
≥
( aH
− c)2
22
1−δ
2
And
om 1i(aL
−
c )2
+
δ
(π
i 1 i( 2
k increases, the total output increases, so does the demand for labor, so the union’ utility
. increases.
www 2.13 Proof: The monopoly price is p = a + c .
ww max w B
V
(I
p
−
B)
+
k (U1 ( I c
−
S)
+
U2
(S
+
B))
FOC:
−V
′(I
p
−
B)
+
wenku.baidu.com
kU
′
2
(
S
+
B)
=
0
(*)
网 At stage 1, child maximizes his payoff:
案 max S
U1
(
Ic
−
S
)
+U2
(S
+
B)
答 FOC:
−U1′
(Ic
−
S
)
+
B) ′′(I
p
−
iI B)
′
p
(
A)
案网 =
−
kU
kU ′′(Ic + ′′(Ic + B) +V
B) ′′(I
p
−
B)
i
I
′
c
(
A)
+
kU
V ′′(I p − B) ′′(Ic + B) +V ′′(I
p
−
B)
iI
′
p
(
A)
答 U
′(
I
c
(
A)
+
B)i(
Ic′
(
A)
+
∂B ∂A
)
课后 =
U
′(Ic