Дронов С.В. Задачник по теории вероятностей (второй семестр)
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6
ω
2
-
ω
1
A B
x
B
y
D
y
E
C
x
C
y
C
y
0
D
y
0
C
y
0
ξ η
(ξ, η) x y
y < x/2 {ξ < x, η < y}
AB
x
ED
y
E
(x, y) (x − y)y
y
0
≥ x/2 {ξ < x, η < y
0
} = {ξ < x}
F (x, y) = F
ξ
(x)
F (x, y) =
x < 0 y < 0
(x − y)y 0 ≤ x ≤ 2, 0 ≤ y < x/2
x
2
/4 0 ≤ x ≤ 2, y ≥ x/2
(2 − y)y x > 2, 0 ≤ y ≤ 1
x > 2, y > 1.
F (1, 1/2) = 1/4 F (1) = 1/4
F (1/2) = 3/4 F (1, 1/2) 6= F (1)F (1/2) ξ η
ξ η
ξ η ξ
0 ≤ η ≤ ξ/2
P (ξ < 1/4, 1/3 ≤ η < 1) = 0,
P (ξ < 1/4) P (1/3 ≤ η < 1)
P (ξ < 1/4, 1/3 ≤ η < 1) 6= P (ξ < 1/4)P (1/3 ≤ η < 1),
ξ η
Ω = [0, 1], F = B[0, 1]. σ(ξ)
ξ(ω) Ω ξ(ω) = ω ξ(ω) = (ω −
1
2
)
2
;
ξ(ω) =
(
0 ≤ ω ≤ 1/2,
ω
2
, 1/2 < ω ≤ 1;
ξ(ω) =
(
ω ∈ [0, 1/2)
ω ∈ [1/2, 1].
σ(ξ) = {ξ
−1
(B), B ∈ B}.
ξ, η
f η = f(ξ)
σ(η) ⊂ σ(ξ)
Ω = [0, 1], F = B[0.1], P
ξ ξ(ω) = ω ξ(ω) =
ω
2
ξ(ω) = ω
−α
(α > 0)
ξ(ω) =
(
2ω, 0 ≤ ω ≤ 1/2,
2(1 − ω), 1/2 < ω ≤ 1;
ξ(ω) =
ω, 0 ≤ ω ≤ 1/3,
1/3 < ω < 2/3
−ω
3
, 2/3 ≤ ω < 1;
ξ(ω) =
(
ω ∈ [0, 1/2)
ω ∈ [1/2, 1].
Ω
F σ Ω P
ξ + η
ξ = ω
1
+ ω
2
η = ω
1
− ω
2
ξ = ω
1
η = ω
2
ξ =
(
ω
1
= ω
2
ω
1
6= ω
2
η = ω
1
ω
2
.
ξ η
OX
R
2
R
3
α
OX
ξ η
E
α
f(x) = αe
−αx
, x ≥ 0.
ξ
3
ξ −η max{ξ, η
2
} |ξ −η|
ξ, η
ξ η σ(ξ) ⊂ σ(η)
ξ
ξ, η
F, G P (ξ > η) =
1 (∀x)(F (x) < G(x))
ξ, η ϕ, ψ
ϕ(ξ), ψ(η)
σ(ξ) σ(η)
σ {∅, Ω}
Ω = [0, 1], F = B[0, 1] P
ξ(ω) = ω
2
η(ω) = 1 −ω
2
ξ(ω) = ω
2
ζ(ω) =
1 − ω
Ω = {1, 2, 3}, F = ρ(Ω), P (1) = P (2) = P (3) =
1/3 < Ω, F, P >
Ω = [0, 1], F = B[0, 1], P
ξ(ω) = ω
ξ
x
1
, ..., x
n
Mξ =
n
X
j=1
x
j
P (ξ = x
j
).
ξ
(ξ
n
)
{n∈N}
(∀ω) (ξ
n
(ω) % ξ(ω)).
ξ Mξ = lim Mξ
n
Mξ = ∞
ξ
ξ
+
= max{0, ξ}, ξ
−
= max{0, −ξ}.
Mξ = Mξ
+
− Mξ
−
.
ξ
+
, ξ
−
• M(ξ ± η) = Mξ ± Mη
• M(λξ) = λMξ
• ξ, η Mξη = MξMη
• f(x)
Mf(ξ) =
+∞
Z
−∞
f(x)dF (x).
Z
Ω
f(ξ) dP =
+∞
Z
−∞
f(x)dF (x)
Mξ =
+∞
Z
−∞
xp(x) dx,
p(x)
Mξ =
+∞
R
−∞
xdF (x), F (x)
k ξ
Mξ
k
Dξ = M(ξ − Mξ)
2
= Mξ
2
− (Mξ)
2
.
ξ
x
1
, ..., x
m
lim
n→∞
n
q
Mξ
n
.
n
Mξ
n
=
m
X
j=1
x
n
j
p
j
,
p
j
= P (ξ = x
j
)
P
m
j=1
p
j
= 1
x ξ
Mξ
n
= x
n
m
X
j=1
α
j
p
j
, α
j
= (x
j
/x)
m
≤ 1.
m
X
j=1
α
j
p
j
≤
m
X
j=1
p
j
= 1.
k α
k
= 1
m
X
j=1
α
j
p
j
≥ p
k
.
p
1/n
k
x ≤
n
q
Mξ
n
≤ x.
M|ξ|
α
< ∞
lim
t→+∞
t
α
P (|ξ| ≥ t) = 0.
M|ξ|
α
=
+∞
Z
−∞
|x|
α
dF (x).
lim
t→∞
Z
|x|≥t
|x|
α
dF (x) = 0,
R
|x|≥t
|x|
α
dF (x) ≥ t
α
R
|x|≥t
dF (x) = M1
{|ξ|≥t}
=
= t
α
P (|ξ| ≥ t).
0 ≤ t
α
P (|ξ| ≥ t) ≤
Z
|x|≥t
|x|
α
dF (x),
ξ
F Mξ
+
Mξ
−
ξ Ω = [0, 1] F = B[0.1] P
ξ(ω) = ω ξ(ω) = ω − 1/2 ξ(ω) = sin πω
Ω
F = B(Ω)
ξ(ω) =
ω
1
< ω
2
ω
1
= ω
2
ω
1
> ω
2
ξ
ξ ≥ 0 Mξ < ∞
Mξ =
+∞
Z
0
P (ξ ≥ x) dx.
M|ξ| < ∞
Mξ = −
0
Z
−∞
P (ξ < x) dx +
+∞
Z
0
P (ξ ≥ x) dx.
F M|ξ|
α
< ∞ (α > 0)
|x|
α−1
(1 − F (x) + F (−x))
ξ ≥ 0, Mξ < ∞
1
Mξ
≤ M
1
ξ
!
.
M|ξ|
γ
< ∞ (γ > 0) 0 < β ≤ γ
M|ξ|
β
< ∞
ξ
x
1
, ..., x
m
lim
n→∞
Mξ
n+1
Mξ
n
.
+∞
R
−∞
f(x) dF (x)
F (x) = 1
(1, +∞)
(x) f(x) = x
2
F (x) =
1
2
+
1
π
(x) f(x) = (x)
F (x) =
0, x ≤ 0
sin x
2
0 < x ≤
π
2
π
2
< x ≤ 2
1, x > 2
f(x) = x.
ξ
Mξ = a P (ξ = n)
ξ P (0 < ξ < 1) =
1 Dξ < Mξ
ξ
η
(
q
Dξ −
q
Dη)
2
≤ D(ξ + η) ≤ (
q
Dξ +
q
Dη)
2
.
ξ X = (x
1
, ..., x
n
) n
ξ n
x
1
, ..., x
n
ξ
ξ
∗
ξ
∗
x
1
x
n
1/n 1/n
ξ
Mξ
∗
=
1
n
n
X
j=1
x
j
≡
¯
X−
Dξ
∗
=
1
n
n
X
j=1
(x
j
−
¯
X)
2
=
1
n
n
X
j=1
x
2
j
− (
¯
X)
2
≡ S
2
−