自动化所2004“应用随机过程”讲义--第六讲习题参考答案.pdf

文档介绍：S S S S
(3) ò A B
¤ p Ø ƒ   8 Ü  ¿ µ A B = A (B − A) = A B − AB§ 2 d
V Ç  Œ \ 5 Œ  "
S
Ï  § ¤ ± § d d Œ  "
(4) TB ⊂ A A = B A − B T
 A B 6= φž § σ(A)
k 16‡  ƒ "  A B = φž § σ(A)k 8‡  ƒ " Ø J  Ñ
K 8 ¥ A3 = {(a, b) : a, b ∈ R}
\∞
(1) Ï  ∀n, A ⊂ An¤ ± k A ⊂ An"
n=1
\∞\∞
e ¡ y ² " ‡ y { µ b  § @ o §  1 §
An ⊂ A ∃x ∈ An, x∈/ A x − a > 0 n = d x−a e + 1
n=1 n=1
\∞\∞
K 1 § d d Œ  § g ñ " ¤ ± " d d Œ  " a q Œ
x > a + n x∈/ An An ⊂ A A = An
n=1 n=1
[∞
 B = Bn"
n=1
[∞
(2) Ï  ∀a ∈ R, (−∞, a] = (−n, a]§ ¤ ± A1 ⊂σ(A2)" Ï  ∀a, b ∈ R, (a, b] = (−∞, b] −
n=1
(−∞, a]§ ¤ ± A2 ⊂σ(A1)" n þ Œ  σ(A1) = σ(A2)
(3)
[∞
∀a ∈ R, (−∞, a] = (−n, a]
n=1
\∞ 1
= (−∞, a + )
m
m=1
\∞[∞ 1
= (−n, a + )
m
m=1 n=1
\∞[∞ 1
= [−n, a + )
m
m=1 n=1
[∞
= [−n, a]
n=1
∀a, b ∈ R, (a, b] = (−∞, b] −(−∞, a]
[∞ 1
(a, b) = (−∞, b −] −(−∞, a]
n
n=1
[∞ 1 [∞ 1
[a, b) = (−∞, b −] −(−∞, a −]
n m
n=1 m=1
[∞ 1
[a, b] = (−∞, b] −(−∞, a −]
m
m=1
d d Œ  σ(A1) = σ(Ai), 2 ≤ i ≤ 5
(3) Ï  A2k+1 ⊂ A2k§ ¤ ±
½ S
[∞∞ 1 1 1  ó ê
k=n/2 A2k = ( 4 , 2 + n ] n
Ak = S∞ 1 1 1
A2k = ( , + ] n Û ê
k=n k=(n+1)/2 4 2 n+1
1
\∞[∞\∞ 1 1 1 1 1
⇒ A = ( , + ] = ( , ]
k 4 2 n 4 2
n=1 k=n n=1
à !
\∞[∞ 1 1
d d Œ  P A = F ( ) − F ( )"
k 2 4
n=1 k=n
\∞[∞\∞
Ak = φ⇒ Ak = φ
k=n n=1 k=n
à !
[∞\∞
d d Π P Ak = 0"
n=1 k=n
(1) d 4  ½ Â
ω∈{ω: lim Xn = X}
n→∞
⇔∀m ≥ 1 ∃n ≥ 1 ∀k ≥ n |Xk(ω) − X| < 1/m
\
⇔∀m ≥ 1 ∃n ≥ 1 ω∈{ω: |Xk − X| < 1/m}
k≥n
[ \
⇔∀m ≥ 1 ω∈{ω: |Xk − X| < 1/m}
n≥1 k≥n
\ [ \
⇔ω∈{ω: |Xk − X| < 1/m}
m≥1 n≥1 k≥n

X∞ X∞ Xn
EN = nP (N = n) = P (N = n)
n=1 n=1 m=1
X∞ X∞ X∞
= P (N = m) = P (N > n)
n=1 m=n n=1
Z ∞ Z ∞ Z x
EX = xdF (x) = ( dy)dF (x)
0 0 0
Z ∞ Z ∞ Z ∞
= dF (x)dy = (1 − F (x))dx
0 y 0
Z ∞ Z ∞ Z x
E(Xn) = xndF (x) = ( nyn−1dy)dF (x)
0 0 0
Z ∞ Z ∞ Z ∞
= dF (x)nyn−1dy = nxn−1(1 − F