文档介绍：第六章****题 6
1、在下列矩阵的空白处填上适当的数字，使矩阵 Q 成为一个转移强度矩阵。
é -5
2
3 ù
ê
ú
Q= ê
0
0
0
ú
ê
2
2
-4ú
ë
û
2、一连续时间 Markov 链有 0 和 1 两个状态，在状态 0 和 1 的逗留时间分别服从参数为 l > 0 及 m > 0 的指数分布。试求在时刻 0 从状态 0 起始， t 时刻后过程处
于状态 0 的概率 P00 (t )。
答：
设 x (t )为 t 时刻所处状态，记
P00 (t ) = P (x ( t ) = 0 x (0 ) = 0 ), P01 (t ) = P (x ( t ) = 1 x (0 ) = 0)
易知： P00 (t )+ P01 (t ) =1 ，采用无穷小分析法
P00 (t + Dt ) = P (x (t + Dt ) = 0 x (0 ) = 0)
P (x (t + Dt ) = 0, x (t ) = 0 x (0 ) = 0 )+ P (x (t + D t ) = 0, x (t ) = 1 x (0 ) = 0)
P00 (t ) P (x (t + Dt ) = 0 x (t ) = 0 )+ P01 (t ) P (x (t + Dt ) = 0 x (t ) = 1)
P00 (t ) (1 - l Dt + o ( Dt ))+ (1- P00 (t )) m Dt
P00 (t + DDtt) - P00 (t ) = - ( l + m ) P00 (t )+ m + o (DDtt )
P00¢ (t ) = - ( l + m ) P00 (t )+ m Þ P00 (t ) = l l m e-(l +m)t + l m m
++
3、在上题中如果 l = m 定义 N (t )为过程在[0, t ]中改变状态的次数，试求 N (t )的
概率分布。
答：
P01 (t ) = 1 - P00 (t )
=
l
-
l
e - (l +m)t
=
1
(1 - e -2lt ), (l = m )
l +m
l +m
2
P
t
)
=
1
1+e -2 l t
)
，P
(
t
)
=
1
1+e -2 l t
)
，P
(
t
)
=
1
1- e-2lt
)
00 (
2
(
11
2
(
10
2
(
记 Pk (t ) = P (N (t ) = k x (0 ) = 0)
k
Pk (t + t ) = P (N (t + t ) = k
x (0 ) = 0 ) = åP (N (t + t ) =