随机过程(林元烈)第六讲习题参考答案

∀n, A ⊂ An
∞
A ⊂ An"
n=1
e¡y² y{µb §@o § § ∞ An ⊂ A"
∞
∃x ∈ An, x ∈/ A
x−a > 0
n=
1 x−a
+1
n=1
n=1
Kx
>
a+
1 n
§
d
d

x
∈/
§ An gñ"¤± ∞
An
⊂
A"ddA =
∞
An" a q
∞ B = Bn"
n=1
n=1
n=1
Ï §¤± Ï ∞
(2)
∀a ∈ R, (−∞, a] = (−n, a]
A1 ⊂ σ(A2)" ∀a, b ∈ R, (a, b] = (−∞, b] −
§¤± nþ (−∞, a]
n=1
A2 ⊂ σ(A1)"
σ(A1) = σ(A2)
(3)
∞
∀a ∈ R, (−∞, 源自文库] = (−n, a]
n≥1 k≥n
⇔ ω∈
{ω : |Xk − X| < 1/m}
m≥1 n≥1 k≥n
1.14
∞
∞n
EN =
nP (N = n) =
P (N = n)
n=1
n=1 m=1
∞∞
∞
=
P (N = m) = P (N > n)
n=1 m=n
n=1
EX = =
∞
∞x
xdF (x) = ( dy)dF (x)
0 ∞∞
=( A2k
1 4
,
=
1 2
+
1 n
]
(
1 4
,
1 2
+
1 n+1
]
n n
1
⇒
∞ n=1
∞ k=n
Ak
=
∞
(
n=1
1 4
,
1 2
+
1 n
]
=
(
1 4
,
1 2
]
ddP
∞∞
Ak
=
F
(
1 2
)
−
F
(
1 4
)"
n=1 k=n
∞
∞∞
Ak = φ ⇒
Ak = φ
k=n
n=1 k=n
dd ∞ ∞
P
Ak = 0"
1.24 X, Y éÜVÇݼêµ
f(X,Y )(x, y) = e−(x+y)I(x>0,y>0)
d Ý 1 ªµ X = UV/(1 + V ), Y = U/(1 + V ) Jacobi
ddµ ©OÈ©>
J = −u/(1 + v)2
f(U,V
)(u,
v)
=
e−u
k=0
=
n
kCnk
k=0
λ1
k
λ1 + λ2
λ2
n−k
λ1 + λ2
=
nλ1 λ1 + λ2
n k=1
Cnk−−11
λ1
k−1
λ1 + λ2
λ2
n−k
λ1 + λ2
=
nλ1 λ1 + λ2
3
¤±E(N1|N1
+
N2)
=
λ1
λ1 +
λ2
(N1
+
N2)"
E(N1 + N2|N1) = E(N1|N1) + E(N2|N1) = N1 + E(N2) = N1 + λ2
(−∞, b
−
1 n
]
−
∞
(−∞, a −
1 m
]
n=1
m=1
[a, b]
=
(−∞, b] −
∞
(−∞, a −
1 m
]
m=1
ddσ(A1) = σ(Ai), 2 ≤ i ≤ 5
Ï §¤± 1.6 (3)
A2k+1 ⊂ A2k
∞
Ak =
k=n
∞
óê ∞ k=n/2
A2k
Ûê k=(n+1)/2
0
0
∞
dF (x)dy = (1 − F (x))dx
0y
0
E(Xn) = =
∞
∞x
xndF (x) = ( nyn−1dy)dF (x)
0
0
0
∞∞
∞
dF (x)nyn−1dy =
nxn−1(1 − F (x))dx
0y
0
1.18
∞
P (ξ = k) =
P (N = n)P (ξ = k|N = n)
=
e−λ λk pk k!
∞
λn−k (n − k)!
(1
−
p)n−k
n=k
=
∞
λn k!(n
e−λ − k)!
pk
(1
−
p)n−k
n=k
=
e−λλk k!
pk
eλ(1−p)
=
(λp)k k!
e−λp
∼
P
o(λp)
D(ξ) = λp"
n
P (N1 + N2 = n) =
P (N1 + N2 = n, N1 = k)
k=0 n
=
P (N2 = n − k)P (N1 = k)
k=0
=
n λk1 e−λ1 λn2 −ke−λ2 k! (n − k)!
k=0
=
(λ1
+ λ2)n n!
e−(λ1 +λ2 )
P (N1 = k|N1 + N2 = n)
=
P (N1 = k)P (N2 = n − k) P (N1 + N2 = n
1V(114...134)ÇÏ(K3A)8B\ò¥B5⊂AA=3A=§φB¤{" (§a±¤,σbAp)(A:=Øa) ,Bbk∈ 1AR6}−8Bܧd" ¿dµAA" BB
= =
A (B − A) = A
φ§σ(A)k8
B − AB§2d "ØJÑ
Ï ¤±k (1)
n=1
=
∞
(−∞, a
m=1
+
1 m
)
=
∞
∞
(−n, a
+
1 m
)
m=1 n=1
=
∞
∞
[−n, a
+
1 m
)
m=1 n=1
∞
= [−n, a]
n=1
∀a, b ∈ R, (a, b] = (−∞, b] − (−∞, a]
(a, b) =
∞
(−∞, b
−
1 n
]
−
(−∞,
a]
n=1
[a, b) =
∞
n=k
∞
n
=
P (N = n)P ( Xi = k|N = n)
n=k
i=1
∞
n
=
P (N = n)P ( Xi = k)
n=k
i=1
2
¤±E(ξ) = λp,
1.19 (1) (2)
=
∞
λne−λ n!
Cnkpk(1 − p)n−k
n=k
=
∞
λn k!(n
e−λ − k)!
pk
(1
−
p)n−k
n=k
=
Cnk
λ1
k
λ1 + λ2
λ2
n−k
λ1 + λ2
(3)
P (N1 + N2|N3) = P (N1|N3) + P (N2|N3) = P (N1) + P (N2) = P (N1 + N2)
¤± Õá N1 + N2 N3
"
(4)
n
E(N1|N1 + N2 = n) =
kP (N1 = k|N1 + N2 = n)
1.20 ùpÑ1n Ky²§cüaq"
(3)
E(E(IA|IB)|IB, IC ) = E(P (A|B)IB + P (A|B)IB|IB, IC ) = E(P (A|B)IB|IB, IC ) + E(P (A|B)IB|IB, IC ) = P (A|B)IB + P (A|B)IB = E(IA|IB)
d4½Â n=1 k=n
1.10 (1)
ω
∈
{ω
:
lim
n→∞
Xn
=
X}
⇔ ∀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}