Дронов С.В. Задачник по теории вероятностей (второй семестр)
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g(x)
θ
p
µ
(x) =
1
2
e
−|x−µ|
.
p
θ
(x) =
θx
θ
, x ∈ [0, 1]
0, x 6∈ [0, 1]
p
θ
(x) =
2x
θ
2
, x ∈ [0, θ]
0, x 6∈ [0, θ]
p
θ
(x) =
θ
√
2πx
3
e
−θ
2
/(2x)
, x > 0
0, x ≤ 0
p
θ
(x) =
(
e
θ−x
, x ≥ θ,
0, x < θ
p
α,µ
(x) =
1
2α
exp
−
|x − µ|
α
.
θ
∗
=
¯
X
p
θ
(x) =
1
π(1 + (x − θ)
2
)
?
θ
∗
σ
2
(θ
∗
− θ)
√
n
d
−→ η,
η N(0, σ
2
)
θ
∗
= h
1
n
n
X
j=1
g(x
j
)
,
σ
2
= (h
0
(a))
2
+∞
Z
−∞
(g(x) − a)
2
dF
θ
(x),
a = Mg(x
1
)
d(θ) =
M(θ
∗
−θ)
2
K
b
= {θ
∗
| Mθ
∗
= b}.
b = 0
θ
±
1 − ε
lim
n→∞
P (θ
−
≤ θ ≤ θ
+
) ≥ 1 − ε.
X U
[0,θ]
θ
∗
1
= 2
¯
X θ
∗
2
= (1 + 1/n) X
(n)
M(θ
∗
i
− θ)
2
= Dθ
∗
i
, i = 1, 2.
d
1
(θ) = D
θ
θ
∗
1
= 4D
θ
¯
X =
1
n
2
(4
P
n
j=1
D
θ
x
j
) =
= (4nD
θ
x
1
)/n
2
= θ
2
/3n.
d
2
(θ) = D
θ
θ
∗
2
= (1 +
1
n
)
2
M
θ
X
2
(n)
− ((1 +
1
n
)M
θ
X
(n)
)
2
=
= (1 +
1
n
)
2
M
θ
X
2
(n)
− θ
2
.
X
(n)
M
θ
X
2
(n)
=
θ
Z
0
x
2
nx
n−1
θ
n
dx =
n
θ
n
θ
n+2
n + 2
=
n
n + 2
θ
2
,
d
2
(θ) =
θ
2
n(n + 2)
.
d
1
(θ) d
2
(θ) θ
∗
1
n = 1 θ
∗
2
X
λ λ
λ
∗
1
=
¯
X, λ
∗
2
=
v
u
u
t
X
2
+
1
4
−
1
2
.
λ
∗
1
g(x) = h(x) = x, a = Mx
1
= λ, h
0
(a) = 1,
σ
2
1
= Dx
1
= λ λ
∗
2
g(x) = x
2
, h(x) =
q
x + 1/4 − 1/2.
a = Mx
2
1
= λ
2
+ λ, h
0
(a) =
1
2λ + 1
,
σ
2
2
=
Dx
2
1
(2λ + 1)
2
.
x
1
Π
λ
Mx
4
1
= λ
4
+ 6λ
3
+ 6λ
2
+ λ; (Mx
1
)
2
= (λ
2
+ λ)
2
= λ
4
+ 2λ
3
+ λ
2
,
σ
2
2
=
4λ
3
+ 5λ
2
+ λ
(2λ + 1)
2
= λ
1 +
λ
(2λ + 1)
2
.
σ
2
1
< σ
2
2
λ
∗
1
θ
∗
θ σ
2
θ
1 − ε
x
P (|(θ
∗
−θ|
√
n ≤ x) → P (|η| < x) = P (
η
σ
<
x
σ
) = 2Φ(x)−1.
ε x
2Φ(x) − 1 = 1 − ε,
P (θ
∗
−
x
√
n
≤ θ ≤ θ
∗
+
x
√
n
) → 1 − ε (n → ∞).
x
1 − ε/2
θ
∗
k
=
k
r
(k + 1)X
k
U
[0,θ]
lim
k→∞
θ
∗
k
λ Γ
α,λ
α
X
α k
α
∗
k
=
k!
X
k
!
1/k
θ
∗
1
= X
(1)
θ
∗
2
= 2
¯
X − 1
U
[θ,1]
X N(α, 1) α > 0
α
∗
1
=
¯
X α
∗
2
= max{0,
¯
X}
U
[θ,θ+1]
θ
∗
= [x
1
]
θ
∗
α
αθ
α
∗
=
n − 1
n
¯
X
Γ
α,1
θ
∗
= (1 + 1/n)X
(n)
[0, θ]
L(X, θ) = h(X) exp{A(θ)T (X) + B(θ)},
h, A, T, B
N(a, 1)
N(a, σ
2
)
U
[a,b]
Γ
α,λ
[0, θ]
θ
∗
= 2
¯
X
ˆ
θ = X
(n)
lim
n→∞
P
nX
(k)
θ
< y
!
→ Γ
1,k
(y)
k, y
¯
X
m
θ
σ
2
(θ)
θ
U
[0,θ]
1−ε [X
(n)
, X
(n)
/Ψ]
Ψ
Ψ
n−1
(n − (n −1)Ψ) = ε.
X
(1)
U
[θ,θ+1]
U
[θ, 2θ]
¯
X, X
2
U
[0,θ]
H
0
H
1
δ : X → {0, 1}
S = {X ∈ X : δ(X) = 1},
α(δ) = P
0
(S)
1 − P
1
(S)
α(δ) β(δ) = P
1
(S)
δ
(
β(δ) → max
α(δ) ≤ α
α
S =
X ∈ X :
L(X, 1)
L(X, 2)
> t
,
t P
0
(S) ≤ α
H
0
= {N(0, 1)} H
1
= {Π
λ
}
H
0
= {N(0, 1)}
H
1
= {P (0 ≤ ξ ≤ 1) = 1}
N(α, 1)
H
0
= {α = 0} H
1
= {α = 1}
S = {X : X
1
≥ 3}.
H
0
= {U
[0,1]
} H
1
sup
t∈[0,1]
|F
∗
n
(t) − t| >
1
3
.
n
F G
H
0
= {F = G}
n/2
δ > 0
n
δ
ε
α H
0
=
{θ = 1} N(θ, 1) N(1, θ) E
θ
p = θ/2 Π
θ
σ
2
H
0
= {σ
2
= σ
2
1
} H
1
= {σ
2
< σ
2
1
}
H
i
= {E
α
i
}, i = 0, 1; H
i
= {Π
λ
i
}, i = 0, 1;
m p
i
, i = 1, 2
{N(a, 1)} {N(b, 2)}.