Дронов С.В. Задачник по теории вероятностей (первый семестр)
Подождите немного. Документ загружается.
ξ
p(x) = e
−x
(x ≥ 0).
η = e
−ξ
.
ξ
F (x) η = F (ξ).
ξ
F (x) η
F
−1
(z) = sup{x | F (X) ≤ z}, z ∈ [0, 1]
F
−1
(η)
ξ
Mξ =
X
x
i
P(ξ = x
i
),
ξ ξ
p(t)
Mξ =
∞
Z
−∞
t p(t) dt.
α, β
M(αξ ± βη) = αMξ ± βMη;
C MC = C;
ξ ≥ η ⇒ Mξ ≥ Mη;
ξ, η ⇒ Mξη = Mξ · Mη;
Mg(ξ) = Σg(x
i
)P(ξ = x
i
),
Mg(ξ) =
∞
Z
−∞
g(x)p(x)dx
g
x ξ
ξ
ξ −x 100 − x 1000 − x 2500 − x 5000 − x
Mξ = 160 −x. Mξ = x/2 x ≈
5000+
2 × 2500 + 5 × 1000 + 1000 = 16000
ξ
n p Mξ
n = 1
Mξ = 0 × q + 1 × p = p.
n > 1 η
1
, ..., η
n
ξ = η
1
+ ... + η
n
⇒ Mξ = Mη
1
+ ... + Mη
n
= np.
s
λ
v, 0 ≤ v ≤ v
max
kv (k = 1/v
max
)
t
0
s
n
H
n
τ = s/v + ξt
0
,
ξ (0 ≤ ξ ≤ n).
H
n
ξ n
p = kv.
M(ξ/H
n
) = kvn.
P(H
n
) =
λ
n
n!
e
−λ
, n = 0, 1, 2, ...,
Mξ = kv
∞
X
n=0
n
λ
n
n!
e
−λ
= kvλ e
−λ
∞
X
n=1
λ
n−1
(n − 1)!
= λkv.
f(v) = Mτ =
s
v
+ t
0
Mξ =
s
v
+ λt
0
kv.
v = v
0
=
s
sv
max
λt
0
.
v
opt
=
v
0
v
0
≤ v
max
v
max
v
0
> v
max
∞
X
n=1
1
n
2
=
π
2
6
.
ξ
P(ξ = n) =
6
π
2
n
2
≡ p
n
, n = 1, 2, ...
Mξ =
∞
X
n=1
np
n
=
6
π
2
∞
X
n=1
1
n
p
v
λ p
ξ
Mξ < ∞
Mξ =
∞
X
n=1
P(ξ ≥ n).
ξ ≥ 0 Mξ < ∞
lim
x→∞
xP(ξ ≥ x) = 0.
ξ
Ox
x + y = 1
p(t) = At
λ−1
e
−αt
(t ≥ 0),
λ, α
A
Γ
Γ(z) =
∞
Z
0
x
z−1
e
−x
dx.
ξ
Dξ = M(ξ − Mξ)
2
= Mξ
2
− (Mξ)
2
.
Dξ ≥ 0, Dξ = 0 ⇐⇒ ξ = .
D(λξ) = λ
2
Dξ.
D(ξ + c) = Dξ.
D(ξ + η) = Dξ + Dη ξ η.
D(ξ + η) = Dξ + Dη + 2 (ξ, η)
ξ, η (ξ, η) = Mξη − MξMη
ξ η
ρ(ξ, η) =
(ξ, η)
√
Dξ Dη
=
Mξη − Mξ · Mη
√
Dξ Dη
.
|ρ(ξ, η)| ≤ 1;
|ρ(ξ, η)| = 1 ⇐⇒ ∃a, b : ξ = aη + b.
ξ, η ⇒ ρ(ξ, η) = 0.
ξ
a ∈ (0, 1) η = |ξ − a|
ξ, η
ξ
f(x) =
x ∈ [0, 1]
x
Mξ =
1
R
0
xdx =
1
2
; Mξ
2
=
1
R
0
x
2
dx =
1
3
;
Dξ = Mξ
2
− (Mξ)
2
=
1
12
;
Mη = M|ξ − a| =
1
R
0
|x − a|dx =
=
a
R
o
(a − x)dx +
1
R
a
(x − a)dx = a
2
− a +
1
2
;
Mη
2
=
1
R
0
(x − a)
2
dx =
(1−a)
3
+a
3
3
;
Dη =
(1−a)
3
+a
3
3
−
a
2
− a +
1
2
2
;
Mξη = Mξ|ξ − a| =
1
R
0
x|x − a|dx =
a
3
3
−
a
2
2
+
1
12
,
ρ(ξ, η) =
4a
3
− 6a
2
+ 1
√
1 − 12a
4
+ 24a
3
− 12a
2
.
a
(ξ, η) = M(ξ − Mξ)(η − Mη).
ξ N(a, σ
2
)
|ρ(ξ, η)| = 1. ξ = aη + b,
a ρ
n
< Ω, F, P >
ξ η (ξ, η)
F (x, y) = P(ξ < x, η < y) = P({ξ < x} ∩ {η < y})
ξ η
p(u, v)
(∀x.y) F (x, y) =
x
Z
−∞
y
Z
−∞
p(u, v)dudv,
p F
A ⊆ R
2
P((ξ, η) ∈ A) =
Z Z
{(u,v)|(u,v)∈A}
p(u, v)dudv.
(ξ, η) ξ η
f, g
f(x) =
∞
Z
−∞
p(x, v)dv; g(y) =
∞
Z
−∞
p(u, y)du.
ξ, η f, g
p
ξ+η
(z) =
∞
Z
−∞
f(x)g(z − x) dx.
(ξ, η)
p(u, v) =
(u, v) ∈ S
S ρ(ξ, η)
f
1
, f
2
ξ, η
f
1
(x) =
1
R
0
p(x, v)dv =
x/2
R
0
dv =
x
2
(0 ≤ x ≤ 2);
f
2
(y) =
2
R
0
p(u, y)du =
2
R
2y
du = 2 − 2y (0 ≤ y ≤ 1).
x, y
Mξ =
2
R
0
x
2
2
dx =
4
3
; Mη =
1
R
0
y(2 − 2y)dy =
1
3
;
Mξ
2
= 2; Mη
2
=
1
6
; Dξ =
2
9
; Dη =
1
18
.
Mξη =
2
Z
0
du
u/2
Z
0
uvdv = =
1
2
.
(ξ, η) = Mξη − MξMη =
1
18
ρ(ξ, η) =
1
2
.
p ξ η
(ξ, η)
f(x, y) =
A
π
2
(3 + x
2
)(1 + y
2
)
.
A (ξ, η)
ξ η
(ξ, η) ∈ {(x, y) : 2 ≤
q
x
2
+ y
2
≤ 3}.
F (x, y) = (1 − e
−ax
)(1 − e
−by
) (x, y ≥ 0).
(ξ, η)
f(x, y) = Axy D
x + y = 1, x = 0, y = 0 f(x, y) = 0 D A
ξ η
λ µ
ξ + η
ξ
η
ξ, η
ξ
ϕ
ξ
(t) = M exp{itξ}
ξ p(x)
p(x) =
1
2π
+∞
Z
−∞
ϕ(t)e
−itx
dt.
ϕ(0) = 1
ϕ
aξ+b
(t) = e
ibt
ϕ
ξ
(at)