Колодий Н.А., Мазепа Е.А. Сборник задач и упражнений по теории вероятностей
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ξ
1
ξ
2
p
1
= 1−q
1
p
2
= 1 − q
2
ξ
1
ξ
2
ξ
1
ξ
2
f
ξ
1
(x) = f
ξ
2
(x) =
1
√
2π
e
−
x
2
2
.
ζ
3
=
ξ
1
ξ
2
f
ζ
3
(x) =
+∞
Z
−∞
|y|f
ξ
1
(xy)f
ξ
2
(y) dy =
1
2π
+∞
Z
−∞
|y|e
−
(xy)
2
2
e
−
y
2
2
dy =
=
1
π
+∞
Z
0
ye
−
y
2
2
(x
2
+1)
dy = −
1
π(x
2
+ 1)
e
−
y
2
2
(x
2
+1)
+∞
0
=
1
π(x
2
+ 1)
.
ξ
1
ξ
2
ξ
1
+ ξ
2
ξ
1
− ξ
2
ξ η
ξ
2
+ η
2
p
ξ
2
+ η
2
ξ
ξ
z
ψ
ξ
(z) =
M
z
ξ
=
∞
X
k=0
z
k
P
{ξ = k}, |z| ≤ 1.
|z| = 1
ξ
ξ
ξ
t
φ
ξ
(t) =
M
e
itξ
, t ∈ (−∞, +∞).
P
{|ξ| < ∞} = 1
ψ
ξ
(z) φ
ξ
(t)
ψ
ξ
(1) = φ
ξ
(0) = 1.
M
|ξ|
k
< ∞ k ≥ 1
φ
(k)
ξ
(0) = i
k
M
ξ
k
.
ξ η
φ
ξ+η
(t) = φ
ξ
(t)φ
η
(t), ψ
ξ+η
(z) = ψ
ξ
(z)ψ
η
(z).
ξ
1
ξ
2
F
ξ
1
(x) F
ξ
2
(x)
φ
ξ
(t)
ξ
f
ξ
(x)
f
ξ
(x) =
1
2π
+∞
Z
−∞
e
−itx
φ
ξ
(t) dt
ξ
ψ
ξ
(z)
ξ + n nξ n
ψ
ξ
(z) =
M
z
ξ
ψ
ξ+n
(z) =
M
z
ξ+n
=
M
z
ξ
z
n
= z
n
M
ξ
n
= z
n
ψ
ξ
(z),
ψ
nξ
(z) =
M
z
nξ
=
M
(z
n
)
ξ
= ψ
ξ
(z
n
).
1
4
(1+z)
2
p(1−qz)
−1
, p, q >
0, p+q = 1
e
λ(z−1)
, λ > 0 (p+qz)
n
1 +
1−z
z
ln(1 − z)
ψ
ξ
(z) =
M
z
ξ
=
∞
P
k=0
z
k
P
{ξ = k}.
1
4
(1 + z)
2
1
4
(1 + z)
2
=
1
4
· 1 +
1
2
· z +
1
4
z
2
.
ξ 0, 1, 2
ψ
0
=
1
4
ψ
1
=
1
2
ψ
2
=
1
4
p(1 − qz)
−1
p(1 − qz)
−1
= p
∞
X
k=0
(qz)
k
=
∞
X
k=0
z
k
pq
k
.
ξ k =
0, 1, 2, . . .
ψ
k
= pq
k
ψ(z) = |z|
φ
ξ
(t)
ξ a b
aξ + b
φ
ξ
(t)
ξ
−ξ
F (x) F (x) = 1 − F (−x − 0).
ξ η
φ(t)
ξ−η
e
−i|t|
a
1
cos t+. . .+a
n
cos nt+b
1
sin t+
. . .+b
n
sin nt, b
1
·. . .·b
n
6= 0, a
i
b
i
ξ η ξ
η
ζ
ξ ξ + ζ = η
ζ
φ
η
(t) = φ
ξ+ζ
(t) = φ
ξ
(t)φ
ζ
(t)
φ
ξ
(t) φ
η
(t)
φ
ξ
(t) =
∞
X
k=0
e
itx
k
p
k
= e
it0
1
2
+ e
it1
1
2
=
1
2
(1 + e
it
).
φ
η
(t) =
∞
X
k=0
e
itx
k
p
k
= e
it0
1
8
+e
it1
1
4
+e
it2
1
2
+e
it3
1
8
=
1
8
+e
it
1
4
+e
it2
1
2
+e
it3
1
8
.
1
8
+ e
it
1
4
+ e
it2
1
2
+ e
it3
1
8
=
1
2
(1 + e
it
)φ
ζ
(t).
t = π
1
4
= 0 · φ
ζ
(t)
1
4
= 0
ζ
ξ η ξ
η
ζ ξ ξ + ζ = η
ξ η
ξ + η
ξ η
[a, b]
λ > 0
a b
λ > 0
N(a, σ
2
)
p ∈ (0, 1)
ξ
[a, b] f
ξ
(x) =
1
b−a
x ∈ [a, b]
φ
ξ
(x) =
M
e
itξ
=
b
Z
a
e
itx
1
b − a
dx =
1
b − a
1
it
e
itx
b
a
=
e
itb
− e
ita
it(b − a)
=
= e
it(a+b)
2
sin(t
b−a
2
)
t
b−a
2
.
ξ
P
{ξ = k} =
C
k
n
p
k
q
n−k
k = 0, n p + q = 1
φ
ξ
(x) =
M
e
itξ
=
n
X
k=0
e
itk
C
k
n
p
k
q
n−k
=
n
X
k=0
C
k
n
(e
it
p)
k
q
n−k
= (q + pe
it
)
n
.
ξ η
N(a
1
, σ
2
1
) N(a
2
, σ
2
2
)
ξ − η N(a
1
− a
2
, σ
2
1
+ σ
2
2
)
N(a
1
−
a
2
, σ
2
1
+ σ
2
2
)
ξ −η
φ
1
(t) φ
2
(t)
a
1
a
2
a
1
+ a
2
+
. . . = 1
φ(t) =
∞
P
i=1
a
i
φ
i
(t)
cos t cos
2
t e
−t
2
e
−|t|
1
1+t
2
1
1−it
sin t
t
e
−|t|
cos t
cos t
cos t
cos t =
e
it
+e
−it
2
=
1
2
e
−it
+
1
2
e
it
ξ
φ
ξ
(t) =
M
e
itξ
=
∞
P
k=1
e
itx
k
p
k
x
1
= −1 x
2
= 1 p
1
= p
2
=
1
2
e
−t
2
f
ξ
(x) =
1
2π
+∞
Z
−∞
e
−itx
e
−t
2
dt.
x
f
0
ξ
(x) =
1
2π
+∞
Z
−∞
(−it)e
−itx
e
−t
2
dt =
i
4π
+∞
Z
−∞
e
−itx
de
(−t
2
)
=
=
i
4π
e
−t
2
e
−itξ
+∞
−∞
+ ix
+∞
Z
−∞
e
−t
2
e
−itx
dt
=
= −
x
4π
+∞
Z
−∞
e
−t
2
e
−itx
dt = −
x
2
f
ξ
(x).
f
0
ξ
(x) = −
x
2
f
ξ
(x)
f
ξ
(x) = ce
−
x
2
4
c
c =
1
2
√
π
f
ξ
(x) =
1
2
√
π
e
−
x
2
4
ξ N(0, 2)
ξ
ξ
1
ξ
2
ξ = ξ
1
+ξ
2
ξ
1
, . . . , ξ
n
N(a, σ
2
)
1
n
n
P
j=1
ξ
j
N(a,
1
n
σ
2
)
ξ
i
N(a,
1
n
σ
2
)
n n n (n > 1)
n ≥ 3
n
k
6n
1, 2, . . . , 6 n
n m
k
r
n k
n
x y
[0,
1
2
]
x + iy
2
√
3
[−1, 2]
A
l
=
l B
l
=
l C
l
=
l A
2
, C
2
A
2
, B
3
A
4
, B
4
k− 1−p
k
k = 1, 2
A B C
C A4B
p
1
p
2
p
3
p
1
+p
2
+p
3
= 1
1
2
(1 −α)
A B
B