Дронов С.В. Задачник по теории вероятностей (второй семестр)
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×
Ω
ρ(Ω) F ⊆ ρ(Ω)
Ω ∈ F
A ∈ F ⇒
¯
A ∈ F
A, B ∈ F ⇒ A ∪ B ∈ F
(A
n
)
n∈N
⊂ F ⇒ ∪
n∈N
A
n
∈ F,
F σ A ⊆ ρ(Ω)
σ(A) σ
A σ(A) σ
A
Ω
A = {A ∈ ρ(Ω) | A }.
σ(A)
σ(A)
Ω A
¯
A
¯
A ∈ A ⊂ σ(A)
A ∈ σ(A) σ(A) Ω
σ(A) = ρ(Ω).
F σ Ω
F
F
n n
F Ω
σ Ω
n = 2 A
1
∈ F, A
1
6= ∅, A
1
6= Ω
A
2
=
¯
A
1
A
1
, ..., A
n
n
F
Ω n
F A
k
B ∈ F
B ∩ A
k
6= ∅, B ∩A
k
6= A
k
A
k
A
k
= B ∩A
k
A
n+1
=
¯
B ∩A
k
F σ ∅ ∈ F A, B ∈
F ⇒ AB ∈ F A, B ∈ F ⇒ A \ B ∈ F (A
n
)
n∈N
⊂ F ⇒
∩
n∈N
A
n
∈ F.
Ω σ
σ Ω
F, B σ ρ(Ω)
σ F ∪B F ∩B
F \B
σ
(A) σ
A σ(A) = ∩{F, F ∈ (A)}
Ω = {0, 1, 2} σ
A = {0, 1} B = {1, 2}
σ
{0}, {1}
∅
Ω σ
Ω
Ω
Ω
σ F
Ω
A
1
, ...A
n
F Ω F = σ(A
1
, ...A
n
)
n
F
n σ
2
n
Ω σ
σ
Ω σ F ⊆ ρ(Ω)
ξ : Ω → R
F (∀x ∈ R) ({ω : ξ(ω) < x} ∈ F
ξ
σ B = B(R)
B = σ{(−∞, x), x ∈ R}
A ⊂ R
B(A) = {AB : B ∈ B(R)}.
F
R ξ F
(−∞, x)
{ω : ξ(ω) < x} = ξ
−1
(−∞, x)
B
(−∞, x)
F
ξ
−1
(−∞, x)
C ∈ F
C = ξ
−1
(B)
ξ η
A = {ω ∈ Ω : ξ(ω) < η(ω)} ∈ F
A = ∪
α∈Q
{ω ∈ Ω : ξ(ω) < α ≤ η(ω)},
{ξ(ω) < α ≤ η(ω)} = {ξ(ω) < α} ∩ {η(ω) ≥ α} =
= {ξ(ω) < α} ∩ {η(ω) < α}.
{ξ < α} ∈ F; {η < α} ∈ F.
Q
A ∈ F
{η ≥ α} = η
−1
([α, ∞)) ∈ F
[a, b) [a, b] (a, b) (−∞, ∞) a, b
b > a
ξ, η
B = {ω ∈ Ω : ξ(ω) = η(ω)}; C = {ω ∈ Ω : ξ(ω) ≥ η(ω)}.
ξ, η
ξ + η ξη |ξ|
max{ξ, η} min{ξ, η} ξ
η
P (η > 0) = 1
(ξ
n
)
n∈N
inf ξ
n
, sup ξ
n
ξ f
η = f(ξ)
f (∀B ∈
B)(f
−1
(B) ∈ B).
A ⊆ R
B(A) = σ(A ∩ (−∞, x), x ∈ R).
(ξ
n
)
n∈N
ω ∈ Ω
(ξ
n
)
n∈N
ω ∈ Ω
(ξ
n
)
n∈N
< Ω, F >
ξ
< Ω, F > ξ
< Ω, F > ξ :
Ω → R
< Ω, F, P > ξ
x {ξ < x}
F (x) = P (ξ < x),
ξ (ξ, η)
F (x, y) = P (ξ < x, η < y)
ξ η
(∀x, y) (F (x, y) = P (ξ < x)P (η < y) = F (x)F (y)),
ξ η
ξ
ξ σ
ξ
σ(ξ) = σ({ξ < x}, x ∈ R).
ξ η
σ(ξ) σ(η)
Ω = [0, 1]; F = B([0, 1])
σ(ξ)
ξ =
ω ∈ [0, 1/4)
ω ∈ [1/4, 3/4)
1 ω ∈ [3/4, 1]
ξ
x ≤ 1/4 {ξ < x} 1/4 < x ≤ 1/2
{ξ < x} = {ξ = 1/4} = [0, 1/4).
1/2 < x ≤ 1
{ξ < x} = {ξ = 1/4}∪{ξ = 1/2} = [0, 3/4).
x > 1 {ξ < x}
ξ
σ(ξ) = σ([0, 1/4), [0, 3/4)) = {∅, [0, 1/4),
[0, 3/4), [3/4, 1], [0, 1/4) ∪ [3/4, 1], [1/4, 3/4), [0, 1]}.
B ∈ B B
ξ
−1
(B) = ∅ ξ
B
ξ
−1
(B) = [0, 1/4)
σ(ξ)
σ(ξ)
Ω
Ω
A(0, 0) B(2, 0) C(2, 1) F σ
Ω σ Ω
ω ∈ Ω (ω
1
, ω
2
) ξ(ω) =
ω
1
, η(ω) = ω
2
ξ η
{ξ < x} 0 ≤ x ≤ 2
A, B
x
(x, 0), C
x
(x, x/2)
AB
x
×B
x
C
x
/2 = x
2
/4 x <
0 {ξ < x} {ξ < x} = Ω x > 2
F
ξ
(x) =
0, x < 0;
x
2
/4, 0 ≤ x ≤ 2;
1, x > 2.
{η < y} 0 < y < 1
A, B, C
y
(2, y), D
y
(2y, y)
(AB + D
y
C
y
) × y/2 = (2 + 2 − 2y) × y/2 = (2 − y)y.
F
η
(y) =
0, y < 0;
(2 − y)y, 0 ≤ y ≤ 1;
1, y > 1.