习题三新.doc

****题三1、对Markov链,0,?nxn试证条件P??ixixixjxnnnn??????,,...,1001=P??ixjxnn???1(1)等价于对所有时刻n,m及所有状态,,...,,,...,10mnjjii有P??nnmmnnixixjxjx??????,...,,...,0011=P??nnmmnnixjxjx?????,...,11(2)证明:???P??nnmmnnixjxjx?????,....,11=P????nnmmnnnnixixPjxjxixix????????,...,,...,,,...,001100=P??0011,....,,.....,ixixjxjxnnmmnmmn????????.P??0011,...,,...,ixixjxnnmmn??????÷P??nnixix??,....,00利用(1)=P??11??????mmnmmnjxjx……P??nnnixjx?????nnixix??,...,00÷P??nnixix??,....,00=P??11????????2211????????mmnmmnjxjx….P??nnnixjx???11=P??11??????mmnmmnjxjx….P??nnnnixjxjx?????1122,=……..=P??nnnmmnixjxjx?????11,......???在(2)中取m=1,即得(1)。2、考虑状态0,1,2上的一个Markov链,0,?nxn它有转移阵P=??????????,初始分布为,,,???ppp试求概率P??2,1,0210???xxx解:P??2,1,0210???xxx=P??1212????1101????00?x=0××=0注:??110111???xxPP=,??121212???xxPP=。信号只有0和1两种,分为多个阶段传输,在每一步上出错的概率为?。0X=0是送出的信号,而nX是在第n步接收到的信号。假定nX为一Markov链,它有转移概率矩阵P=11? ?? ??? ?? ??? ?,0<?<1试求:(a)两步均不出错的概率P(0X=0,1X=0,2X=0)(b)试求两步传送后收到正确信号的概率。(c)试求5步之后传送无误的概率P5 0( 0| 0)X X? ?解:(a)P(0X=0,1X=0,2X=0)=P(0X=0)P(1X=0|0X=0)P(2X=0|1X=0)=(0X=0)=P(0X=0)2(1 )??(b)P(2X=0|0X=0)=P(2X=0,1X=0|0X=0)+P(2X=0,1X=1|0X=0)=P1X=0|0X=0)P(2X=0|1X=0)+P(1X=1|0X=0)P(2X=0|1X=1)=2(1 )??+2?(c)5 0 5 1 1 0 5 1 1 0( 0| 0) ( 0| 0) ( 0| 0) ( 0| 1) ( 1| 0)P X X P X X P X X P X X P X X? ???????????=(1-?)[5 2 2 1 5 2 2 1( 0| 0) ( 0| 0) ( 0| 1) ( 1| 0)P X X P X X P X X P X X? ????????]+?[5 2 2 1 5 2 2 1( 0| 0) ( 0| 0) ( 0| 1) ( 1| 1)P X X P X X P X X P X X? ????????]=2(1 )??25 2 5 2 5 2 5 2( 0| 0) (1 ) ( 0| 1) ( 0| 0) (1 ) ( 0| 1)P X X P X X P X X P X X? ? ???? ????????????=[2(1 )??+2?].{5 3 3 2( 0| 0) ( 0| 0)P X X P X X? ???+5 3 3 2( 0| 1) ( 1| 0)P X X P X X? ???}+2?(1-?){5 3 3 2 5 3 3 2( 0| 0) ( 0| 1) ( 0| 1) ( 1| 1)P X X P X X P X X P X X?