概率与随机过程习题答案

(2)、令Z (t) W (t) Xt Z (t) E[W (t) Xt] 0
CZ (t1, t2 ) E[Z (t1)Z (t2 )] E[(W (t1) Xt1)(W (t2 ) Xt2 )] E[W (t1)W (t2 )] E[W (t1) Xt2 ] E[ Xt1W (t2 )] E[ Xt1Xt2 ]
Q X (t),Y (t)相互独立
P{S (t) S (t0 ) k} P{X (t) Y (t) X (t0 ) Y (t0 ) k}
k
P{X (t) X (t0 ) i,Y (t) Y (t0 ) k i} i0
k
P{X (t) X (t0 ) i}P{Y (t) Y (t0 ) k i} i0
解 (1)、令Z (t) W (t) At Z (t) E[W (t) At] At CZ (t1, t2 ) E[(Z (t1) Z (t1))(Z (t2 ) Z (t2 ))]
E[(W (t1) At1 At1)(W (t2 ) At2 At2 )]
E[W (t1)W (t2 )] 2 min{t1, t2}, t1, t2 0
解 (1)、X (t),Y (t){t (0, )}是独立增量过程,X (t)和 Y (t)相互独立 {S(t),t (0, )}是独立增量过程
(2)、S(0) X (0) Y (0) 0
(3)、t t0 0, X (t) X (t0 ) : ((t t0 )) Y (t) Y (t0 ) : ((t t0 ))
t1E[(Y Y )( X X )] t1t2E[(Y Y )(Y Y )]
CXX
t2CXY
t1CYX
t1t2CYY
12
(t1 t2 )1 2
t1t2
2 2
例4 设X (t)和Y (t){t (0, )}是两个相互独立的、分
别具有强度和的泊松过程.试证明S(t) X (t) Y (t) 是具有强度为 的泊松过程.
随机过程补充例题
例 1 随机过程 {X (t),t T}, x是任一实数,定义另一随机
过程Y
(t)
1, 0
X (t) x ,t T.试将Y (t)的均值函数和自 X (t) x
相关函数用X (t)的一维和二维分布函数表示.
解 Y (t) E[Y (t)] 1 P{Y (t) 1} 0 P{Y (t) 0}
k [ (t t0 )]i e (tt0 ) g[ (t t0 )]ki e (tt0 )
i0
i!
(k i)!
e( )(tt0 ) k!
k
Cki [ (t t0 )]i
i0
[ (t t0 )]ki
[( )(t t0 )]k
k!
e( )(tt0 ) S (t) S (t0 ) :
[( )(t t0 )]
{S (t), t 0}是强度为 的泊松过程
例5 设{W (t)，t 0}是以 2为参数的维纳过程，求下列
过程的协方差函数： (1)W (t) At, ( A为常数); (2)W (t) Xt, X为与{W (t)，t 0}相互独立的标准正态变量;
(3)aW ( t a2 ), a为正常数;
例 2 已知随机过程 {X (t),t T}的均值函数X (t)和协 方差函数CX (t1,t2 ),(t)是普通函数.试求随机过程Y ( t) X (t) (t)的均值函数和协方差函数.
解 Y (t) E[ X (t) (t)] X (t) (t)
CY (t1, t2 ) E[(Y (t1) Y (t1))(Y (t2 ) Y (t2 ))] E[( X (t1) (t1) X (t1) (t1))
( X (t2 ) (t2 ) X (t2 ) (t2 ))] E[( X (t1) X (t1))( X (t2 ) X (t2 ))]
CX (t1, t2 )
例 3 假定Z (t) X Yt,t R.若已知二维随机变量
(
X
,
Y
)的协方差矩阵为
12 1
2
的协方差函数.
1
2 2
且P[ X (1) 0] P[ X (1) 1] 1
2
2
2
P[ X (1) 0] P[ X ( 1) 1] 1
2
2
2
0, x 0
F
( x;
1) 2
1 2
,
1,
0 x 1 x 1
1,
X
(1)
2,
出现H 出现T
2
,试求Z
(t
)
解
CZ (t1,t2 ) E[( X Yt1 (X Yt1))( X Yt2 (X Yt2 ))]
E[(( X X ) (Yt1 Yt1))(( X X ) (Yt2 Yt2 ))]
E[( X X )( X X )] t2E[( X X )(Y Y )]
P{X (t) x} FX (x,t)
RY (t1,t2 ) E[Y (t1)Y (t2 )] 1 PY (t1)Y (t2 ) 1 0 PY (t1)Y (t2 ) 0 PY (t1)Y (t2 ) 1 PX (t1) x1, X (t2 ) x2 FX (x1, x2;t1,t2 )
例6 利用抛掷硬币的试验定义一随机过程:
X
(t
)
cos
2t,
t,
出现H 出现T
，
t
，
假设P(H )
P(T )
1 2
,
试确定X
(t)的(1)一维分布函数F
( x;
1 ), 2
F
(Fra Baidu bibliotek
x;1);
(2)二维分布函数F
(
x1,
x2
;
1 2
,1);
解
由X (t)的定义
X (1) 2
0, 1,
出现H 出现T ,
2 min{t1, t2} t1t2 , t1, t2 0
(3)、令Z (t) aW ( t )a2 Z (t) aE[W ( t a2)] 0
CZ (t1, t2 ) E[aW (t1 a2 )aW (t2 a2 )]
a2 E[W ( t1 a2 )W (t2 a2 )] a2 2 min{t1 a2 , t2 } a2 2 min{t1, t2}, t1, t2 0