Дронов С.В. Задачник по теории вероятностей (второй семестр)
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F
∗
n
X
X
(1)
≤ X
(2)
≤ ... ≤ X
(n)
.
X
(k)
k k
x
j
F
F
∗
n
(x) =
1
n
n
X
j=1
1
(−∞,x)
(x
j
),
MF
∗
n
(x) =
1
n
n
X
j=1
P (x
j
∈ (−∞, x)) =
1
n
n
X
j=1
P (ξ < x) = F (x).
M(F
∗
n
(x))
2
=
1
n
2
M
P
n
j=1
1
(−∞,x)
(x
j
)
2
=
=
1
n
2
P
n
j=1
P
n
i=1
M1
{x
j
<x}
1
{x
i
<x}
.
(i 6= j) ⇒ M1
{x
j
<x}
1
{x
i
<x}
= M1
{x
j
<x}
M1
{x
i
<x}
= F
2
(x),
i = j F (x)
n
F (x) n(n −1) F
2
(x)
M(F
∗
n
(x))
2
=
1
n
F (x) +
n − 1
n
F
2
(x).
DF
∗
n
(x) = M(F
∗
n
(x))
2
− (MF
∗
n
(x))
2
=
F (x)(1 − F (x))
n
.
F (x) =
1−e
−x
x > 0
0 x ≤
0
X
(1)
P (X
(1)
≥ t) = P (x
1
≥ t, x
2
≥ t, ..., x
n
≥ t),
x
1
, ..., x
n
ξ
P (X
(1)
≥ t) = P
n
(x
1
≥ t) = (1 − F (x))
n
= e
−nx
x > 0 X
(1)
G(x) =
(
0, x ≤ 0
1 − e
−nx
, x > 0
X
X
B
n,p
F
P
F
∗
n
(x) =
k
n
!
= C
k
n
F
k
(x)(1 − F (x))
n−k
, k = 0, ..., n.
P (X
(k)
< y), P (X
(k)
< y, X
(k+1)
≥ y).
X
[0, θ] k
P (ξ = k) = p
k
, k = 0, ..., N.
k
P (F
∗
n
(a) < F
∗
n
(b))?
n
2/n
X
F
(y
1
, ..., y
n
) y
i
= F (x
i
)
ξ
X P
θ
, θ ∈ Θ
θ
S(X) θ
∗
= S(X)
θ
Mθ
∗
= θ,
θ
∗
θ θ
∗
θ n → ∞
θ
∗
θ = G(F ) G
F
θ
θ
∗
= G(F
∗
n
).
ξ
P
θ
f
θ
L(X, θ) =
n
Y
j=1
f
θ
(x
j
).
ˆ
θ
θ
L
l(X, θ) = ln L(X, θ) =
n
X
j=1
ln f
θ
(x
j
).
d
θ
θ
∗
d(F
θ
∗
, F
∗
n
) = min{d(F
θ
, F
∗
n
), θ ∈ Θ}.
X
B
n,p
n p
M
n,p
x
1
= np, D
n,p
x
1
= np(1 − p).
¯
X = np
S
2
= np(1 − p),
p
∗
= 1 −
S
2
¯
X
, n
∗
=
(
¯
X)
2
¯
X − S
2
−
¯
X −→ np, S −→ npq.
S
2
¯
X
−→ q p
∗
−→ 1 − q = p,
p
∗
n
∗
θ
[0, θ] θ
∗
= X
(n)
θ
∗
P (θ
∗
< x) = P
θ
(x
1
< x, ..., x
n
< x) =
=
Q
n
j=1
P (x
j
< x) = P
n
(x
1
< x).
x
1
x ∈ [0, θ] P (x <
x) = x/θ
P (θ
∗
< x) = (x/θ)
n
(0 ≤ x ≤ θ),
θ
∗
p
θ
∗
(x) =
nx
n−1
θ
n
(0 ≤ x ≤ θ).
Mθ
∗
=
θ
Z
0
nx
n
θ
n
dx =
n
θ
n
θ
n+1
n + 1
=
nθ
n + 1
,
θ
∗∗
= (n + 1)θ/n = (1 + 1/n)X
(n)
.
p
d(F, G) = sup
x
|F (x) − G(x)|.
F (x) =
0, x ≤ 0;
1 − p, 0 < x ≤ 1
1, x > 1
F
∗
n
(x) =
0, x ≤ 0;
1 −
¯
X, 0 < x ≤ 1
1, x > 1
d(F, F
∗
n
) = |p −
¯
X| p
∗
=
¯
X
F α
F (x) = α
N(a, 1) N(0, σ
2
) U
[0,θ]
U
[−θ, θ]
U
[θ−1,θ+1]
f(x) =
λ
2
e
−λ|x|
.
U
[a,b]
Γ
α,λ
P (ξ = k) =
λ
k
e
−λ
+ µ
k
e
−µ
2k!
, k = 0, 1, 2, ...
θ
F
θ
(x) = θF
1
(x) + (1 − θ)F
2
(x)
0 < θ < 1 F
1
, F
2
k
N(a, 1), a ≥ 0 U
[0,θ]
E
α
α
f(x) = e
α−x
, (x ≥ α)
α
∗
= X
(1)
X U
[0,θ]
θ
∗
1
= 2
¯
X, θ
∗
2
= X
(1)
+ X
(n)
?
X
p
∗
= (
¯
X)
2
θ
∗
= X
(n)
U
[0,θ]
τ(p) = 1/p
d(F, G) =
+∞
Z
−∞
(F (x) − G(x))
2
dF (x),
F
p
θ
(x) = f(x − θ) f
θ
θ
n [X
(1)
, X
(n)
]
N(a, 1) N(0, σ
2
) p
Π
λ
U
[−θ,0]
U
[−θ,θ]
U
[a,b]
Γ
α,λ
p
θ
(x) =
g
0
(x)
√
2π
exp{−
1
2
(g(x) − θ)
2
},