电子科大 应用随机过程及应用 (陈良均 朱庆棠)第三章作业

(ii) 分解 对于参数为λ 对于参数为λ的Poisson过程， 过程，假设发生的每一个事件 独立的以概率做了记录， 独立的以概率做了记录，未做记录的概率为1-p。令 N1(t)是到t为止做了记录的事件数， 为止做了记录的事件数，而N2(t)是未做记录 的事件数， 的事件数，则{N1(t)；t ≥0}和 {N2(t)；t ≥0}分别是具 有参数pλ 和(1-p)λ的独立Poisson过程。 过程。
相互独立。 相互独立。而且
P ( N (t ) = k ) = ∑ P ( N 1 (t ) = j, N 2 (t ) = k − j ) = ∑ P ( N 1 (t ) = j )P ( N 2 (t ) = k − j )
j=0 j=0 j k− j k k
(λ t ) (λ t ) = ∑ 1 e − λ1 t 2 e −λ2t j! ( k − j )! j=0
[
]
( )
( )
(
)
ρ=
(
)(
)
一维概率密度函数
一维特征函数 二维概率函数 f (s , t , x , y ) = −
[X − m (t )]2 t ∈ T 1 exp − 2 D (t ) 2 λ D (t ) x∈ R t∈T ϕ (t , u ) = exp im (t )u − 1 D (t )u 2 2 x∈ R f (t , x ) =
i i i =1
n
X (t )为正态分布 m X (t ) = E [X (t )] = E [ξ t + W (t )] = E (t )E (ξ ) + E [W (t )] = 0
(t > s ) E [X 2 (t )] = E [ξ 2 t 2 + W (t )W (s ) + W (t )ξ s + W (s )ξ t ] = ts + s σ 2 D (t ) = t 2 + t 2σ 2 D (s ) = s 2 + s 2 σ 2 C (s , t ) = C (t , s ) = R (t , s ) = ts + s σ 2
(
习题十
习题十三
E [N (2 )N (3 )] = E N (2 )( N (3 ) − N (2 )) + N = E [N (2 )]E [N (3 ) − N (2 )] + E N E N
[
2
[
2
(2 )]
(2 )]
X (t ) = W (t + a ) − W (t ) ~ N 0,σ 2a m X (t ) = mX (s ) = 0 R(s, t ) = C (s, t ) = E[X (t )X (s )]
P ( N 1 (t ) = j, N 2 (t ) = k ) = P ( N 1 (t ) = j, N 2 (t ) = k | N (t ) = k + j )P ( N (t ) = k + j )
j+k k + j j k (λt ) − ( 1 ) p p = j ( k + j )! j ( p λ t ) − pλ t [(1 − p )λ t ]k − ( 1− p ) λ t e e = k! j!
s = 2 k = 2
t = 3 j =1
P {N (3 ) = 2 , N (2 ) = 1} P {N (3 ) − N (2 ) = 1 , N (2 ) = 1} = P {N (2 ) = 1} P {N (2 ) = 1} k − j =1
证明： 证明：
N1(t) N2(t) N(t)
Poisson过程的性质
n维特征函数
ϕ (u ) = ϕ (t1 , t 2 ...t n ; x1 , x 2 .. x n ) = exp{iµ T u − 1 2 uT cu}
二维特征函数
ϕ (s , t ; u , v ) = exp {i [um (s ) + vm (t )] −
二维特征函数的向量形 iµ T u − ϕ (u ) = ex ρ {
c (t1 , t1 ), c (t1 , t 2 )...c (t1 , t n ) c (t , t ),......... .......... ...... 2 2 c= . . ( ) ( ) , .......... ..... , c t t c t t n n n 1
2
E [W (t )W (s )] = min (s , t )σ
= sσ
2
s2 + σ 2s C = ts + s σ 2
ts + s σ 2 t 2 + σ 2t
ρ =
ts + s σ
2
(s
2
+ σ 2 s t 2 + σ 2t
)(
)
习题八
习题九
D( X + Y ) = D(W (4) + W (3) − W (2) − W (1)) = D(W (4) − W (3) + 2W (3) − 2W (2) + W (2) − W (1)) D(W (4) − W (3)) + 4D(W (3) − W (2)) + D(W (2) − W (1)) = 6σ 2 = 24 1.D( X + Y ) = D( X ) + D(Y ) + 2 cov( X ,Y ) D( X ) = D(Y ) = 2σ 2 = 8 cov( X ,Y ) = 4 2. cov( X ,Y ) = E( XY ) − E( X )E(Y ) = E(W (4)W (3) − W (4)W (1) + W (1)W (2) − W (2)W (3)) = 3σ 2 − σ 2 + σ 2 − 2σ 2 = σ 2 = 4
(
2
)
X − m (s )2 D (s )
(
)
=
1
(2π )
n
2
c
1
2
( x − µ ) c −1 ( x − µ ) exp − 1 2
T
{
}
2 ρ ( X − m (s ))( X − m (t )) + D (s )D (t )
(y
2 − m (t )) D (t )
证明： 证明：
N(t) N1(t) N2(t)
令s<t<u，因为N1(t)-N1(s) 与N1(u)-N1(s)分别是N(t)-N(s) 与N(u)-N(s)的子集合 的子集合， 集合，由N(t)-N(s) 与N(u)-N(s)的独立性 可知N1(t)-N1(s) 与N1(u)-N1(s)相互独立。 相互独立。
(
)
= E (W (t + a )W (s + a ) + W (t )W (s ) − W (s )W (t + a ) − W (t )W (s + a )) 令s < t = σ 2 (s + a ) − σ 2 min(s + a, t ) − σ 2 s + σ 2 s = σ 2 max(0, s + a − t )
X (t ) = W (t + a ) − W (a ) ~ N 0, σ 2 t m X (t ) = m X (s ) = 0 R(s, t ) = C (s, t ) = E [X (t )X (s )]
Байду номын сангаас
(
) )
= E W (t + a )W (s + a ) + W 2 (a ) − W (t + a )W (a ) − W (s + a )W (a ) = (min(s, t ) + a )σ 2 + aσ 2 − 2aσ 2 = σ 2 min(t , s )
1 2
1 2
[u

2
D (s ) + 2 uvc
(s , t ) +
v 2 D (t )
]}
式 u T cu
}
u u = v
µ =
m ( s ) m (t )
µ T = (m (s ), m (t ))
m (t1 ) m (t 2 ) µ = . . m (t n )
2 2
[
P {N (2 ) = 1 , N (3 ) = 2 } = P {N (2 ) = 1 , N (3 ) − N (2 ) = 1} = = 2 2 × 2 1 × 11 × e − 2 × 3 = 8 e − 6 P {N (3 ) = 2 N (2 ) = 1 } = = 2e −2 t = 3 s = 2
k
k 1 j k− j = e − ( λ1 + λ 2 ) t ( λ1 t ) ( λ 2 t ) j k!
λ1 λ2 λn λ1+λ2+…+λn
=
[( λ1 + λ 2 ) t ]k − ( λ1 + λ 2 ) t e . k!
⇒ 参数为λ1+λ2的Poisson分布。
{
}
二维概率密度函数 1 T −1 1 (x − µ ) f (x ) = 1 exp − 2 ( x − µ ) c 2π c 2
{
}
n维概率密度函数 f ( x ) = f (t1 , t 2 ...t n , x1 , x 2 .. x n )
1 2λ D (t )D (s ) 1 − ρ