Головань С., Гуриев С., Макрушин А. Теория контрактов. Сборник задач с решениями
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α =
10
−γ 1
{rCΩ + I}
−1
1
1+γ
CΩ =
c
11
(σ
2
1
+ γ
2
σ
2
2
) − c
12
γσ
2
2
σ
2
2
(c
12
− γc
11
)
c
12
(σ
2
1
+ γ
2
σ
2
2
) − c
22
γσ
2
2
σ
2
2
(c
22
− γc
12
)
det{rCΩ + I} =(rσ
1
σ
2
)
2
det C + rσ
2
2
(γ
2
c
11
+ c
22
−2γc
12
)+rc
11
σ
2
1
→ rσ
2
1
(c
11
+ rσ
2
2
det C)
σ
2
1
→∞
10
−γ 1
{rCΩ + I}
−1
→
1
c
11
+ rσ
2
2
det C
00
−c
12
c
11
.
α
1
=0,α
2
=
(1 + γ)c
11
− c
12
c
11
+ rσ
2
2
det C
.
σ
2
2
→∞ det{rCΩ + I}→rσ
2
2
(B + rσ
2
1
det C) B ≡ γ
2
c
11
+
c
22
− 2γc
12
10
−γ 1
{rCΩ + I}
−1
→
1
B + rσ
2
1
det C
c
22
− γc
12
γc
11
− c
12
00
.
α
1
=
B + γc
11
− c
12
B + rσ
2
1
det C
,α
2
=0.
θ
a
θ
b
[0, 1]
b
i
= α
i
+ β
i
θ
i
i = a, b
f(θ
i
) i 1 θ
i
∈ [0, 1]
0 i = a, b
θ
a
>θ
b
y
a
=2 y
b
=0 y
i
i
θ
b
>θ
a
y
b
=2 y
a
=0
θ
b
= θ
a
b
i
(θ
i
)=α
i
+ β
i
θ
i
i = a, b
i b
j
(θ
j
) <
b
i
(θ
i
) i y
i
=2 b
i
b
j
(θ
j
) >b
i
(θ
i
)
i y
i
=0 b
j
b
j
(θ
j
)<b
i
(θ
i
)
(2θ
i
− b
i
(θ
i
))f(θ
j
)dθ
j
+
b
j
(θ
j
)>b
i
(θ
i
)
(b
j
(θ
j
))f(θ
j
)dθ
j
→ max
α
i
,β
i
.
b
j
(θ
j
) <b
i
(θ
i
)
b
j
(θ
j
) <b
i
(θ
i
) ⇐⇒ θ
j
<
b
i
(θ
i
) − α
j
β
j
=
α
i
− α
j
+ β
i
θ
i
β
j
.
b
i
(θ
i
)−α
j
β
j
∈ [0, 1] θ
i
∈ [0, 1]
b
j
(θ
j
)>b
i
(θ
i
)
(b
j
(θ
j
))f(θ
j
)dθ
j
=
1
b
i
(θ
i
)−α
j
β
j
(α
j
+ β
j
θ
j
)dθ
j
=
1
2β
j
[(α
j
+ β
j
)
2
− b
2
i
(θ
i
)],
b
j
(θ
j
)<b
i
(θ
i
)
(2θ
i
− b
i
(θ
i
))f(θ
j
)dθ
j
=(2θ
i
− b
i
(θ
i
))
b
i
(θ
i
) − α
j
β
j
.
1
β
j
2θ
i
b
i
(θ
i
) −
3
2
b
2
i
(θ
i
)
→ max
α
i
,β
i
.
b
i
(θ
i
) b
∗
i
(θ
i
)=
2
3
θ
i
i = a, b α
∗
i
=0 β
∗
i
=
2
3
y
a
=0 y
b
=2
θ
a
<θ
b
y
a
=2 y
b
=0 θ
a
>θ
b
θ
i
U
i
(θ
i
)=
θ
i
0
2θ
i
−
2
3
θ
i
dθ
j
+
1
θ
i
2
3
θ
j
dθ
j
=
1
3
+ θ
2
i
>θ
i
θ
i
∈ [0, 1]
J
θ
i
[0, 1]
J =3
b
i
= α
i
θ
i
i =1,...,n
n
αθ
b
(n−1)
=max
1≤j≤n−1
b
j
,b
(1)
=min
1≤j≤n−1
b
j
.
F
(n−1)
(x)=P {b
(n−1)
≤ x} =
x
α
n−1
,f
(n−1)
(x)=
n − 1
α
x
α
n−2
,
f
(n−1)
b
(n−1)
F
(1)
(y)=P {b
(1)
≤ y} =1−
1 −
y
α
n−1
,f
(1)
(y)=
n − 1
α
1 −
y
α
n−2
.
x
P {b
(1)
≤ y} =1−
1 −
y
x
n−2
,f
(1)
(y)=
n − 2
x
1 −
y
x
n−3
,
x
[0,x]
x/(n −1) α
n
α
n
θ
n
>x n
1
x/α
n
θ −
x
n − 1
dθ =
θ
2
2
−
θx
n − 1
1
x/α
n
=
1
2
−
x
n − 1
−
x
2
2α
2
n
+
x
2
(n − 1)α
n
.
α
n
α
∗
n
= n −1
b
∗
(θ)=(n − 1)θ
n [0,n−1]
(n − 1)/(n +1)
θ
r
R
[
¯
R − θ,
¯
R + θ]
¯
R>1
θ
C
max{R −r, −C} min{r, R + C}
θ θ
1
π θ
2
1 − π
¯
R
C π
θ
r
u
F
(r, R)=max{R − r, −C}
u
B
(r, R)=min{r, R + C} u
F
R u
B
R u
F
+ u
b
≡ R
R
u
B
(r, R)dF
θ
(R) → max
r
R
u
F
(r, R)dF
θ
(R) ≥ 0.
R
R
u
F
(r, R)dF
θ
(R)=0
¯
R − θ ≥ r − C
R
R
u
F
(r, R)dF
θ
(R)=
1
2θ
¯
R+θ
¯
R−θ
(R − r)dR =
¯
R − r.
¯
R + θ ≤ r −C
R
R
u
F
(r, R)dF
θ
(R)=−C.
¯
R − θ<r− C<
¯
R + θ
R
u
F
(r, R)dF
θ
(R)=
1
2θ
r−C
¯
R−θ
(−C)dR +
¯
R+θ
r−C
(R − r)dR
=
1
4θ
(C +
¯
R + θ −r)
2
− C.
E
u
F
=0
r =
¯
R,
θ ≤ C
¯
R +(
√
θ −
√
C)
2
, θ>C
¯
R
θ
r
u
F
=
−C,
θ ≤ r − C −
¯
R
¯
R − r,
¯
R − θ ≥ r − C
1
4θ
(C −
¯
R + θ − r)
2
− C,
¯
R − θ<r− C<
¯
R + θ
r ≤
¯
R θ
r>
¯
R θ
u
F
= −C u
F
=
¯
R − r u
F
≥ 0
θ ≥ (
√
C +
√
r −
¯
R)
2
θ
1
>θ
2
(r
1
,p
1
) (r
2
,p
2
) r
1
r
2
p
1
p
2
πp
1
E
1
u
B
(r
1
)+(1− π)p
2
E
2
u
B
(r
2
) → max
r
1
,r
2
,p
1
,p
2
p
1
E
1
u
F
(r
1
) ≥ 0,
p
2
E
2
u
F
(r
2
) ≥ 0,
p
1
E
1
u
F
(r
1
) ≥ p
2
E
1
u
F
(r
2
),
p
2
E
2
u
F
(r
2
) ≥ p
1
E
2
u
F
(r
1
),
E
1
E
2
p
1
πp
1
E
1
u
B
(r
1
)+(1− π)p
2
E
2
u
B
(r
2
) → max
r
1
,r
2
,p
1
,p
2
p
2
E
2
u
F
(r
2
)=0,
E
1
u
F
(r
1
)=p
2
E
1
u
F
(r
2
).
r
i
i =1, 2 E
i
u
F
(r
i
)=0
p
2
=0 p
2
> 0 θ
1
p
2
> 0
u
F
+ u
B
≡ R
π
¯
R + p
2
((1 − π)
¯
R − π E
1
u
F
(r
2
)) → max
p
2
∈(0,1]
.
(1 − π)
¯
R>πE
1
u
F
(r
2
) p
2
=1 r
1
= r
2
= r
2
(1−π)
¯
R<πE
1
u
F
(r
2
) p
2
=0 r
1
= r
1
(1 −π)
¯
R = π E
1
u
F
(r
2
) p
2
r
1
E
1
u
F
(r
1
)=p
2
E
1
u
F
(r
2
)
θ
1
θ
2
θ
1
[0, 1] θ
2
[a, 1+a] a ∈ (0, 1/3)
E[u
1
(θ
1
,θ
2
)+u
2
(θ
1
,θ
2
)] → max
y
1
(θ
1
,θ
2
)+y
2
(θ
1
,θ
2
) ≤ 1,
t
1
(θ
1
,θ
2
)+t
2
(θ
1
,θ
2
) ≤ 0,
u
1
(θ
1
,θ
2
)=θ
1
y
1
(θ
1
,θ
2
)+t
1
(θ
1
,θ
2
) u
2
(θ
1
,θ
2
)=θ
2
y
2
(θ
1
,θ
2
)+t
2
(θ
1
,θ
2
)
t
1
+ t
2
=0,y
1
=
0,
θ
1
≤ θ
2
1, θ
1
>θ
2
,y
2
=
1,
θ
1
≤ θ
2
0, θ
1
>θ
2
u
1
(θ
1
,θ
2
)+u
2
(θ
1
,θ
2
)=max{θ
1
,θ
2
} F
max{θ
1
,θ
2
}
(x)=
F
θ
1
(x)F
θ
2
(x)
F
max{θ
1
,θ
2
}
(x)=
0,
x<a
x(x − a), a ≤ x<1
x − a, 1 ≤ x<a+1
1, x ≥ a +1
F
max{θ
1
,θ
2
}
(x)=
2x − a,
a ≤ x<1
1, 1 ≤ x ≤ a +1
0,
E[max{θ
1
,θ
2
}]=
1
a
x(2x − a) dx +
1+a
1
xdx
=
2x
3
3
−
ax
2
2
1
a
=
2
3
+
a
2
+
a
2
2
+
a
3
6
.
θ
1
θ
1
θ
2
θ
2
y
2
(θ
1
,θ
2
)
θ
2
−
1 − Φ
2
(θ
2
)
ϕ
2
(θ
2
)
−
θ
1
+
Φ
1
(θ
1
)
ϕ
1
(θ
1
)
ϕ
1
(θ
1
)ϕ
2
(θ
2
) dθ
1
dθ
2
,
θ
1
θ
1
θ
2
θ
2
[θ
1
+(θ
2
− θ
1
)y
2
(θ
1
,θ
2
)]ϕ
1
(θ
1
)ϕ
2
(θ
2
) dθ
1
dθ
2
→ max
,y
2
∈ [0, 1].
ϕ
1
(θ
1
)=I
[0,1]
ϕ
2
(θ
2
)=I
[a,1+a]
Φ
1
(θ
1
)=θ
1
Φ
2
(θ
2
)=θ
2
−a
1
0
dθ
1
1+a
a
dθ
2
[y
2
(θ
1
,θ
2
)(2θ
2
− a − 1 − 2θ
1
)] ≥ 0,
1
0
dθ
1
1+a
a
dθ
2
[θ
1
+(θ
2
− θ
1
)y
2
(θ
1
,θ
2
)] → max
1
0
1+a
a
θ
1
dθ
1
dθ
2
=const,
1
0
dθ
1
1+a
a
dθ
2
[(θ
2
− θ
1
)y
2
(θ
1
,θ
2
)] → max
1
0
dθ
1
1+a
a
dθ
2
[y
2
(θ
1
,θ
2
)(2θ
2
− a − 1 − 2θ
1
)] ≥ 0.
L =(θ
2
− θ
1
)y
2
+ λy
2
(2(θ
2
− θ
1
) − (1 + a)) → max
y
2
∈[0,1]
y
2
=
0,
(1 + 2λ)(θ
2
− θ
1
) − λ((1 + a)) < 0
1, (1 + 2λ)(θ
2
− θ
1
) − λ((1 + a)) ≥ 0
x
θ
2
= θ
1
+ x
y
2
= 0
y
2
= 1
θ
1
θ
2
1
a
1 + a
0
λ =0
λ>0
y
2
=
0,
θ
2
− θ
1
<x= x(a)
1, θ
2
− θ
1
≥ x
x>0
1
0
dθ
1
1+a
a
dθ
2
[y
2
(θ
1
,θ
2
)(2(θ
2
− θ
1
) − (a + 1))] = 0.
x ≥ a x<a
x ≥ a
1+a
x
dθ
2
θ
2
−x
0
dθ
1
[2(θ
2
− θ
1
) − (a +1)]
=
1+a
x
dθ
2
(2θ
2
θ
1
− θ
2
1
)
θ
2
−x
0
−
1
2
(1 + a)(1 + a − x)
2
=
1+a
x
dθ
2
[2θ
2
(θ
2
− x) − (θ
2
− x)
2
] −
1
2
(1 + a)(1 + a − x)
2
=
2
θ
3
2
3
− xθ
2
2
−
(θ − x)
3
3
1+a
x
−
1
2
(1 + a)(1 + a − x)
2
=
2
3
(1 + a)
3
− x(1 + a)
2
−
(1 + a − x)
3
3
−
2
3
x
3
+ x
3
−
1
2
(1 + a)(1 + a − x)
2
=
2
3
x
3
− x
2
3
2
(1 + a)+(1+a)
2
x −
1
6
(1 + a)
3
=
1
6
(x − (1 + a))
2
(4x − (1 + a)).
(x − (1 + a))
2
(4x − (1 + a)) = 0 x =1+a x =(1+a)/4
a ≤ 1/3 (1 + a)/4 ≥ a
a ≤ 1/3 x<a
x<a
1
0
dθ
1
1+a
a
dθ
2
[2(θ
2
− θ
1
) − (a +1)]
=
1+a
a
dθ
2
(2θ
2
θ
1
− θ
2
1
)
1
0
− (1 + a)
=
1+a
a
dθ
2
[2θ
2
− 1] − (1 + a)=(θ
2
2
− θ
2
)
1+a
a
= a − 1.
1
a−x
dθ
1
x+θ
1
a
dθ
2
[2(θ
2
− θ
1
) − (a +1)]
=
1
a−x
dθ
1
(θ
2
2
− 2θ
1
θ
2
)
x+θ
1
a
− (1 + a)
1
2
(1 + a − x)
2
=
1
a−x
dθ
1
(x + θ
1
)
2
− 2θ
1
(x + θ
1
) − a
2
+2θ
1
a
− (1 + a)
1
2
(1 + a − x)
2
=
(x + θ
1
)
3
3
− xθ
2
1
−
2
3
θ
3
1
− a
2
θ
1
+ aθ1
2
1
a−x
− (1 + a)
1
2
(1 + a − x)
2
=
2
3
x
3
+
1
2
−
3
2
a
x
2
+(a
2
− 1)x +
−
a
3
6
−
a
2
2
+
3
2
a −
5
6
,
− (1 − a) −
2
3
x
3
+
1
2
−
3
2
a
x
2
+(a
2
− 1)x +
−
a
3
6
−
a
2
2
+
3
2
a −
5
6
= −
2
3
x
3
+
3a
2
−
1
2
x
2
+(1−a
2
)x +
a
3
6
+
a
2
2
−
a
2
−
1
6
.
P (x, a)
x =0
a ∈ [0, 1/3]
P
x
(x, a)=−2x
2
+(3a − 1)x +(1− a
2
)
= −(2x − (a + 1))(x −(a − 1)).
aP(x, a)
a − 1 (a +1)/2
[0,a)
a
x = a P (x, a)
P (a, a)=−
2
3
a
3
+
3
2
a −
1
2
a
2
+(1−a
2
)a +
a
3
6
+
a
2
2
−
a
2
−
1
6
=
a
2
−
1
6
.
P(0,a)
1 a
-1/6
2 − 1
−
P(x,a)
x
a − 1
a + 1
2