随机过程习题解析.pdf

1、随机过程习题解析（未完成版）8/31/2015121.令X(t)为二阶矩存在的随机过程.试证它是宽平稳的当且仅当EX(t)与EX(t)X(s+t)都不依赖s.证：充分性：若EX(s)与EX(s)X(s+t)都不依赖s则EX(s) =常数m, EX(s)X(s+t) = f(t)令s = s+t,)EX(s)X(s) = f(ss)RX(s,s) = EX(s)X(s)EX(s)EX(s)= f(ss)m2)X(t)是宽平稳的必要性：若X(t)宽平稳则EX(S)为常数m，即EX(S)与s无关则RX(s,s) = EX(s)X(s)EX(s)EX(s)= g(ss)令s = s+t则EX(s)X(

2、s+t) = m2 +g(t)与s无关2.记U1,Un为在(0,1)中均匀分布的独立随机变量.对0 t,并记X(t) = 1n n k=1I(t,Uk),06t 61,这是U1,Un的经验分布函数.试求过程X(t)的均值和协方差函数.解：EX(t) = E1nn k=1I(t,Uk)= EI(t,U1)= t01dx =t3Rx(s,t) = EX(s)X(t)EX(s)EX(t)= E 1n2n i=1I(s,Ui)n j=1I(s,Uj)st= 1n2 (n2n)E(I(s,U1)I(t,U2)+nE(I(s,U1)I(t,U1)st= 1n2 (n2n)st +nmin(s,t)st=

3、1nmin(s,t)st3.令Z1,Z2为独立的正态随机变量,均值为0,方差为2, 为实数.定义过程X(t) = Z1 cost+Z2 sint.试求X(t)的均值函数和协方差函数.它是宽平稳的吗?解：EX(t) = costEZ1 +sintEZ2 = 0RX(s,t) =Cov(Z1 coss+Z2 sins,Zzcost +Z2 sint)= cosscostCov(Z1,Z1)+sinssintCov(Z2,Z2)= 2 cos(st)只与st有关, )是宽平稳的4. Poisson过程X(t),t 0满足(i) X(t) = 0; (ii)对t s, X(t)X(s)服从均值为(ts

4、)的Possion分布; (iii)过程是有独立增量的.试求其均值函数和协方差函数.它是宽平稳的吗?解：EX(t) = EX(t)X(0) = tRX(s,t) =Cov(X(t),X(s)=Cov(X(s)X(t)+X(t)X(0),X(t)X(0)=Cov(X(t)X(0),X(t)X(0)（独立增量）= t (st)非宽平稳5. X(t)为第4题中的Possion过程.记Y(t) = X(t +1)X(t),试求过程Y(t)的均值函数和协方差函数,并研究其平稳性.4解：EY(t) = EX(t +1)EX(t) = RX(s,t) =Cov(X(s+1)X(s),X(t +1)X(t)=

5、Cov(X(s+1),X(t +1)+Cov(X(s),X(t)Cov(X(s),X(t +1)Cov(X(s+1),X(t)= min(s+1,t +1)+min(s,t)min(s,t +1)min(s+1,t)令 = st,当 1或 0是独立增量过程.证：XnXn1 = Zn)X1X0,X2X1,XnXn1即Z1,Z2,Zn独立同分布)Xn,n0为独立增量过程又Xt1+hXt1 = Zt1+1 +Zt1+h5Xt2+hXt2 = Zt2+1 +Zt2+h*Zn独立同分布)Zt1+1 +Zt1+h d= Zt2+1 +Zt2+h)是平稳的逆命题：已知Xn = n i=0Zi,过程Xn,n0

6、是平稳独立增量过程)XiXi1 = Zi与XjXj1 = Zj独立同分布即Zi,i = 0,1,是一串独立同分布的随机变量8.若X1,X2,为独立随机变量,还要添加什么条件才能确保它是严平稳的随机过程?解：若X1,X2,严平稳,则对任意正整数m和n , Xm和Xn的分布都相同,从而X1,X2,是一列同分布的随机变量.而当X1,X2,是一列独立同分布的随机变量时.对任意正整数k及n1,nk,k维随机向量(Xn1,Xnk)的分布函数为(记X1,X2,的共同分布函数为F(x)F(Xn1,Xnk)(x1,xk) = FXn1 (x1)FXnk (xn)= F(x1)F(xk). 34X Y).解：易见

7、答案为14.12.气体分子的速度V有三个垂直分量Vx,Vy,Vz,它们的联合分布密度依MaxwellBoltzman定律为fVx,Vy,Vz(v1,v2,v3) = 1(2kT)3/2 exp(v21 +v22 +v232kT),其中k是Boltzman常数, T为绝对温度,给定分子的总动能为e.试求x方向的动量的绝对值的期望值.解：由题中所给分布律知分子质量为单位质量,即有e = 12(V2x +V2y +V2z ).则所求为E|Vx|12(V2x +V2y +V2z )= e6作(Vx,Vy,Vz)的球坐标变换Vx = RcosVy = RcossinVz = Rsinsin,则(R,)的

8、联合概率密度为fR,(r,) = fVx,Vy,Vz(rcos,rcos sin,rsin sin)r2 sin=2r2(kT)3/2er2/2kT 12 12 sin= fR(r)f()f()由此可知R,相互独立.) E|Vx|12(V2x +V2y +V2z )= e= ER|cos|12R2 = e= ER|cos|R =2e=2eER|cos|=2e 012 sin|cos|d=e213.若X1,X2,Xn独立同分布.它们服从参数为的指数分布.试证ni=1Xi是参数为(n,)的分布,其密度为f(t) = expt(t)n1/(n1)! , t 0.证：令Y = n i=1XigY(t)

9、 = gnX(t) =( t)n)Y服从参数为(n,)的分布,其密度函数如题所述14.设X1和X2为相互独立的均值为1和2的Possion随机变量.试求X1 +X2的分布,并计算给定X1 +X2 = n时X1的条件分布.解：令Y = X1 +X27gY(t) = gX1(t)gX2(t)= e1(et1)e2(et1)= e(1+2)(et1)Y P(1 +2)给定X1 +X2 = n时X1服从参数为p = 11+2,n = n的二项分布15.若X1,X2,独立且有相同的以为参数的指数分布, N服从几何分布,即P(N = n) = (1)n1,n = 1,2,0 1独立且服从参数为的几何分布,

10、 0 0为一强度是的Possion过程.对s 0试计算EN(t)N(t +s).解：原式= EN(t)(N(t +s)N(t)+N(t)= E(N(t)E(N(t +s)N(t)+E(N2(t)= ts+VarN(t)+E2N(t)= 2ts+t +(t)2= 2t(s+t)+t3.电报依平均速率为每小时3个的Possion过程到达电报局,试问:(i)从早上八时到中午没收到电报的概率;(ii)下午第一份电报到达时间的分布是什么?解:(i)令t的计时单位为小时,并以早上8:00为起始时刻所求事件即N(4) = 0其概率为e121200! 6.1106(ii)取中午12:00为起始时刻, T表示下

11、午第一份电报到达时间F(t)=P(T 6t) = P(N(t)1) = 1e3t) f(t) = 3e3t,即T服从参数为3的指数分布4. N(t),t 0为一 = 2的Possion过程,试求：(i)PN(1)62;(ii)PN(1) = 1且N(2) = 3;(iii)PN(1)2|N(1)1.解：11(i)原式= PN(1)N(0) = 0+PN(1)N(0) = 1+PN(1)N(0) = 2= e2 + 2e21! +22e22!= 5e2(ii)原式= PN(1)N(0) = 1且N(2)N(1) = 2= PN(1)N(0) = 1PN(2)N(1) = 2= 2e22e2= 4

12、e4(iii)原式= PN(1)2且N(1)1PN(1)1= PN(1)2PN(1)1= 1PN(1)N(0)611PN(1)N(0) = 0= 13e21e26.一部600页的著作总共有240个印刷错误,试利用Possion过程近似求出某连续三页无错误的概率.解：设N(t)到第t页为止的印刷错误数,则N(t)为一Possion过程观点一：该Possion过程的参数未知则所求概率为PN(t +3)N(t) = 0|N(600)N(0) = 240= PN(t +3)N(t) = 0;N(600)N(0) = 240/PN(600)N(0) = 240= PN(3)N(0) = 0PN(600)

13、N(3) = 240/PN(600)N(0) = 240= e3 (597)240240! e597/(600)240240! e600=(597600)2400.30030.312观点二：该Possion过程的参数已知, = 240/600 = 0.4,则所求概率为PN(3) = 0= e0.43 0.30120.37. N(t)是强度为的Possion过程.给定N(t) = n,试求第r个事件(r6n)发生的时刻Wr的条件概率密度fWr|N(t)=n(wr|n).解：fWr|N(t)=n(wr|n)wr= PN(wr)N(0) = r1,N(wr +wr)N(wr) = 1|N(t) =

14、n= PN(wr)N(0) = r1PN(wr +wr)N(wr) = 1PN(t)N(wr +wr) = nr/PN(t) = n= (wr)r1(r1)! ewr (wr +o(wr) (twrwr)nr(nr)! e(twrwr)/(t)nn! et两边除以wr并令wr 0得fWr|N(t)=n(wr|n) = n!(r1)!(nr)! (wr)r1(twr)nrtn8.令Ni(t),t 0,i = 1,2,n为n个独立的有相同强度参数的Possion过程.记T为在全部n个过程中至少发生了一件事的时刻,试求T的分布.解：FT(t) = PT 6t= 1P(T t)dt解：令Wn = n

15、k=1Yk法一：P(T t) = PX(t)6= PN(t)k=1Yk 6=+ n=0PN(t)k=1Yk 6|N(t) = nPN(t) = n=+ n=0Pn k=1Yk 6|N(t) = nPN(t) = n=+ n=0PWn 6PN(t) = n求和式中当n = 0时认为PWn 6|N(t) = n= 1*Yk exp(), )Wn = n k=1Yk (n,) P(Wn 6) = n(n1)! 0sn1esds (n1)PN(t) = n= (t)nn! et)P(T t) = et +et+ n=1(t)nn!(n1)! 0sn1esds14)ET = +0P(T t)dt= 1

16、+ n=1()nn!(n1)! +0tnet dt 0sn1esds= 1 + 1+ n=1n(n+1)n!(n1)! 0sn1esds= 1 + 1 0+n=1(s)n1es(n1)!d(s)= 1+从结果看,若越大(系统所受冲击越频繁), 越小(每次冲击所造成的平均损害越大), 越小(系统所能承受的的损害极限越小),则系统平均寿命越短,且当等于0时系统的平均寿命即为第一次冲击到来的平均时间,符合常识.法二：*Gn() = PY1 +Yk 6= PWn 6= PN1()n=+ k=n()nk! e其中N1(t)是强度为的Possion过程)P(T t) = PX(t)6=+ n=0(t)ne

17、tn! Gn() (例2.4)=+ n=0+ k=n(t)nn! et ()nk! e=+ k=0()nk! ek n=0(t)nn! et15)ET = +0P(T t)dt=+ k=0()kk! ek n=0 +0(t)nn! et dt=+ k=0()kk! e (k+1)(n+1)n!= 1+ k=0()kk! e + + k=1()k1(k1)! e= 1+12.令N(t)是强度函数为(t)的非齐次Possion过程, X1,X2,为事件间的时间间隔. (i)Xi是否独立;(ii)Xi是否同分布;(iii)试求X1与X2的分布.解：记m(t) =t0 (u)du法一：等待时间W1,W

18、2的联合分布函数为FW1,W2(t1,t2) = P(W1 6t1,W2 6t2), 06t1 1,N(t2)2)=+ k=2k =1P(N(t1) = ,N(t2) = k)=+ k=2k =1P(N(t1) = ,N(t2)N(t1) = k)=+ k=2k =1P(N(t1) = )P(N(t2)N(t1) = k)=+ k=2k =1(m(t1)! em(t1) m(t2)m(t1)k(k)! em(t2)m(t1)=+ k=2k =1(m(t1)m(t2)m(t1)k!(k)! em(t2)= em(t2)+ k=21k!k =1Ck(m(t1)m(t2)m(t1)k= em(t2)

19、+ k=21k! k=0Ck(m(t1)m(t2)m(t1)km(t2)m(t1)k= em(t2)em(t2)em(t2)m(t1)m(t1)= 1em(t1)m(t1)em(t2) fW1,W2(t1,t2) = 2FW1,W2(t1,t2)t1t2 = (t1)(t2)em(t2)16*W1 = X1W2 = X1 +X2) fX2,X2(t1,t2) = (t1)(t1 +t2)em(t1+t2)不能写为g1(t1)g2(t2)形式)X1,X2不独立又有fX1(t1) = (t1) +0(t1 +t2)em(t1+t2)dt2= (t1)em(t1)em(+), t1 0下面确定em(

20、):1 = +0fX1(t1)dt1 = +0(t1)em(t1)em(+)dt1 = 1m(+)+1em(+)em(+) = 0) fX1(t1) = (t1)em(t1) (t1 0)fX2(t2) = +0(t1)(t1 +t2)em(t1+t2)dt1 (t2 0)X1,X2不同分布且其概率密度函数如上.法二：PX2 t|X1 = s= limssPN(s+t)N(s) = 0|N(s) = 0,N(s)N(s) = 1= PN(s+t)N(s) = 0= em(s)m(s+t)与s有关,故X1,X2不独立.PX1 t= PN(t) = 0= em(t)PX2 t= +0PX2 t|X

21、1 = sfX1(s)ds= +0em(s+t)(s)ds= em(t) +0em(t)m(s+t)(s)ds= PX1 t +0em(t)m(s+t)(s)dsPX1 t)X1,X2不同分布(PX1 t和PX2 t给出X1和X2的分布).1713.考虑对所有t,强度函数(t)均大于0的非齐次Possion过程N(t),t 0.令m(t) =t0 (u)du,m(t)的反函数为(t),记N1(t)N(t).试证N1(t)是通常的Possion过程,试求N1(t)的强度参数.解：(i)N1(0) = N(0) = N(0) = 0(ii)*m()单增, )(t)单增)对任意06t1 t;(ii)

22、N(t)6kWk t;(iii)N(t) kWk k=Wk 6t=Wk t18(ii)N(t)6k=N(t) k+1=Wk+1 6t=Wk+1 t=Wk t(iii)N(t) k=N(t)k+1=Wk+1 6t=Wk 0)解：(a)gX1(t)+X2(t)(v) = gX1(t)(v)gX2(t)(v)= exp1t(ev1)exp2t(ev1)= exp(1 +2)t(ev1)X1(t)+X2(t)是强度为1 +2的Possion过程.(b)gX1(t)X2(t)(v) = gX1(t)(v)gX2(t)(v)= exp1t(ev1)E(evX2(t)= exp1t(ev1)exp2t(ev

23、1)不能化为Possion过程的矩母函数形式,)X1(t)X2(t)不是Possion过程.(c)X1(t)的相邻两个事件时间间隔服从参数为1的指数分布,其概率密度函数为f() = 1e119)Pk = +0f()PX2(t +)X2(t) = kd= +01e1 (2)kk! e2 d= 1k2(1 +2)k+1k! +0(1 +2)ke(1+2) d(1 +2)= 11 +2( 21 +2)k201.对Markov链Xn,n0,试证条件PXn+1 = j|X0 = i0,Xn1 = in1,Xn = i= PXn+1 = j|Xn = i等价于对所有时刻n,m及所有状态i0,in, j1,

24、 jm有PXn+1 = j1,Xn+m = jm|X0 = i0,Xn = i= PXn+1 = j1,Xn+m = jm|Xn = in.证：只需令m = 1由P27(3.4)可知PXn+1 = j1,Xn+m = jm|X0 = i0,Xn = in= PXn+1 = j1,Xn+m = jm,X0 = i0,Xn = in/PX0 = i0,Xn = in= PX0 = i0Pj1 j2 Pjm1 jmPi0i1 Pi1 j1/PX0 = i0Pi0i1 Pin1in= PXn = iPj1 j2 Pjm1 jmPin j1/PXn = in= PXn+1 = j1,Xn+m = jm|

25、Xn = in2.考虑状态0,1,2上的一个Markov链Xn,n0,它有转移概率矩阵P =0.1 0.2 0.70.9 0.1 00.1 0.8 0.1,初始分布为p0 = 0.3,p1 = 0.4,p2 = 0.3,试求概率PX0 = 0,X1 = 1,X2 = 2.解：PX0 = 0,X1 = 1,X2 = 2= p0P01P12 = 0.30.10 = 03.信号传送问题.信号只有0,1两种,分为多个阶段传输.在每一步上出错的概率为 . X0 = 0是送出的信号,而Xn是在第n步接收到的信号.假定Xn为一Markov链,它有转移概率矩阵P00 = P11 = 1,P01 = P10 =

26、,0 0为一M.C.其转移概率矩阵为P =0 1 20 12 12 01 12 0 122 0 0 122) E(T|X0 = 0)=2 k=0E(T|X0 = 0,X1 = k)P(X1 = k|X0 = 0)=2 k=0E(T|X1 = k)P(X1 = k|X0 = 0)= (1+v) 12 +E(T|X1 = 1) 12及E(T|X1 = 1)=2 k=0E(T|X1 = 1,X2 = k)P(X2 = k|X1 = 1)=2 k=0E(T|X2 = k)P(X2 = k|X1 = 1)= (2+v) 12 +0+ 12 2解得E(T|X0 = 0) = 6,平均需掷6次.6.迷宫问题

27、.将小家鼠放入迷宫内作动物的学习试验,如下图所示.在迷宫的第7号小格内放有美味食物而第8号小格内则是电击捕鼠装置.假定当家鼠位于某格时有k个出口可以离去,则它总是随机地选择一个,概率为1/k.并假定每一次家鼠只能跑到相邻的小格去.令过程Xn为家鼠在时刻n时所在小格的号码,试写出这一Markov过程的转移概率阵,并求出家鼠在遭到电击前能找到食物的概率.解：23P =0 1 2 3 4 5 6 7 80 0 12 12 0 0 0 0 0 01 13 0 0 13 0 0 0 13 02 13 0 0 13 0 0 0 0 133 0 14 14 0 14 14 0 0 04 0 0 0 13 0

28、 0 13 13 05 0 0 0 13 0 0 13 0 136 0 0 0 0 12 12 0 0 07 0 0 0 0 0 0 0 1 08 0 0 0 0 0 0 0 0 1设uk为家鼠从k出发在遭到电击前能找到食物的概率,显然u7 = 1,u8 = 0,设T为进入吸收态时刻,则当06k66时,uk = PXT = 7|X0 = k=8 i=0PXT = 7,X1 = i|X0 = k=8 i=0PXT = 7,X1 = iPX1 = i|X0 = k)u0 = 12(u1 +u2)u1 = 13(u0 +u3 +u7)u2 = 13(u0 +u3 +u8)u3 = 14(u1 +u2

29、 +u4 +u5)u4 = 13(u3 +u6 +u7)u5 = 13(u3 +u6 +u8)u6 = 12(u4 +u5)u7 = 1u8 = 0u0 = 12u1 = 23u2 = 13u3 = 12u4 = 23u5 = 13u6 = 12u7 = 1u8 = 07.记Zi,i = 1,2,为一串独立同分布的离散随机变量. PZ1 = k =pk 0,k = 0,1,2, k=0= 1.记Xn = Zn,n = 1,2,.试求过程Xn的转移概率矩阵.解：*PXn+1 = in+1|X1 = i1,Xn = in= PXn+1 = in+1PXn+1 = in+1|Xn = in= PXn

30、+1 = in+1)PXn+1 = in+1|X1 = i1,Xn = in= PXn+1 = in+1|Xn = in24)Xn是一M.C.其转移概率矩阵为P =p0 p1 p2 p0 p1 p2 p0 p1 p2 . . . .8.对第7题中的Zi,令Xn = maxZ1,Zn,n= 1,2,并约定X0 = 0.Xn是否为Markov链?如果是,其转移概率阵是什么?解：Xn+1 = maxZ1,Zn,Zn+1= maxmaxZ1,Zn,Zn+1= maxXn,Zn+1)Xn是M.C.Pij =0 , j iik=0pk , j = i0 ,其他其转移概率矩阵为P =p0 p1 p2 0 p

31、0 + p1 p2 0 0 p0 + p1 + p2 . . . .9.设f(n)ij表示从i出发在n步转移时首次到达j的概率,试证明P(n)ij =n k=0f(k)ij P(nk)jj .证：设Tj = minn : n0且Xn = j25)P(n)ij = PXn = j|X0 = i=n k=0PXn = j,Tj = k|X0 = i=n k=0PTj = k|X0 = iP(nk)jj=n k=0PXk = j,Xs = j(s = 0,1,k1)|X0 = iP(nk)jj=n k=0f(k)ij P(nk)jj10.对第7题中的Zi,若定义Xn = n i=1Zi,n = 1,

32、2,X0 = 0,试证Xn为Markov链.并求其转移概率矩阵.证：对n0有PXn+1 = in+1|X0 = i0,Xn = in (i0 = 0)= PZn+1 = in+1in|X0 = i0,Xn = in= PZn+1 = in+1in=Pin+1in ,in+1in = 0,1,2,0 ,owPXn+1 = in+1|Xn = in= PXn +Zn+1 = in+1|Xn = in= PZn+1 = in+1in)Xn是M.C.P =p0 p1 p2 0 p0 p1 0 0 p0 . . . .11.一Markov链有状态0,1,2,3和转移概率矩阵P =0 12 0 120 0

33、 1 00 0 0 112 0 01226试求f(n)00 ,n = 1,2,3,4,5,其中f(n)00由PXn = i,Xk = i,k = 1,n1|X0 = i定义.解：f(1)00 = P00 = 0, f(2)00 =(12 012)(0 0 12)T= 14对n2有f(n)00 =(12 012)0 1 00 0 10 0 12n20012当n = 3时, f(3)00 = 18当n4时,f(n)00 =(12 012)0 0 12n20 0 12n10 0 12n10012=12n +12n+212.在成败型的重复试验中,每次试验结果为成功（S）或失败（F）.同一结果相继出现称

34、为一个游程（run）,比如一结果FSSFFFSF中共有两个成功游程,三个失败游程.设成功概率为p,失败概率为q = 1p.记Xn为第n次试验后成功游程的长度（若第n次试验,则Xn = 0）.试证Xn,n = 1,2,为一Markov链,并确定其转移概率阵.记T为返回状态0的时间,试求T的分布及均值.并由此对这一Markov链的状态进行分类.证：Xn+1 =Xn+1 ,第n+1次试验成功0 ,第n+1次试验失败)Xn是M.C.P(Xn+1 = j|Xn = i) =p , j = i+1q , j = 0P =0 1 2 3 0 q p 0 0 1 q 0 p 0 2 q 0 0 p . . .

35、 . . .27T = minn : Xn = 0,Xs = 0(s = 1,2,n1)P(T = k) = pk1q k = 1,2,ET =+ k=1pk1qk) pET =+ k=1pkqk)(1p)ET = qET = q+ pq+ p2q+= q1p = 1)ET = 1q所有状态互通,为一类13.试证各方向游动的概率相等的对称随机游动在二维时是常返的证：为使n步游动后回到原位置,向左移动的步数应等于向右移动的步数,向上移动的步数应等于向下移动的步数, )P(2k+1)00 = 0,k = 0,1,2,再考虑2k步的情况,P(2k)00 =k k1=0Ckk12k2k1Ck2k1C2

36、k12k(14)2k=k k1=0(2k)!(k!)2(Ck1k)2( 122k)2断言k k1=0(Ck1k)2 = (2k)!(k!)2 ,)P(2k)00 =(2k)!(k!)2 122k)2当n充分大时,由Stiring公式P(2k)00 (2k)2k+ 12 e2k2(kk+ 12 ek2)2 122k)2= 1k28) + n=1P(n)ii = + (i代表任一格点)二维对称随机游动是常返的断言的证明：(1+x)k =k k1=0Ck1k xk1(1+x)2k =(k k1=0Ck1k xk1)(k k2=0Ck2k xk2)比较两边xk的系数,有Ck2k = k k1=0Ck1

37、k Ckk1k = k k1=0(Ck1k)2 = (2k)!(k!)214.某厂对该厂生产的同类产品的三种型号调查顾客的消费习惯.并把它们归结为Markov链模型.记顾客消费习惯在A,B,C三种型号间的转移概率矩阵分别为下列四种.请依这些转移阵所提供的信息对厂家提出关于A,B两种型号的咨询意见.(1)1 0 00 1 00 0 0, (2)0 12 1212 0121212 0,(3)12 01213131312 012, (4)0 1 00 0 11 0 0.解：(1)不是概率转移矩阵,第三行行和不为1.(2)P2 =0 12 1212 0121212 02=121414141214141

38、412)A,B,C三状态互通,所有状态可遍历.设(A,B,C)为经过长时间后三个产品的市场占有额,则(A,B,C)0 12 1212 0121212 0= (A,B,C)A +B +C = 1 A = B = C = 1)三个品牌竞争力差不多,可以都生产.29(3)由归纳法可知Pn =12 01212(1 13n) 13n12(1 13n)12 012可见B = limn(13)n= 0,(A,C)12121212= (A,C)A +C = 1 A = C = 12)B将逐渐淡出市场,建议停止生产B,扩大对A的生产.(4)A,B,C三状态互通,所有状态可遍历.(A,B,C)0 1 00 0 1

39、1 0 0= (A,B,C)A +B +C = 1 A = B = C = 13)A,B,C市场占有率相同,可维持现状.15.考虑一有限状态的Markov链.试证明(a)至少有一个状态是常返的,(b)任何常返状态必定是正常返的.证：(a)反设所有状态均为瞬过或零常返(加强结论),则对iS,有limn+P(n)ii = 0 ()考虑P(n)ij = + k=1f(k)ij P(nk)jj ,则有 k=1f(k)ijP(nk)jj 6P(n)ij 6 k=1f(k)ij P(nk)jj + k=lf(k)ij ()30固定,令n+,则由()得06 limn+P(n)ij 60+ k=f(k)ij

40、()在()中令+,由于+ k=1f(k)ij 61收敛) limn+P(n)ij = 0 (4)若此有限状态M.C.有N个状态,则N j=1P(n)ij = 1 (5)(5)中令n+,由(4)得0 = 1,矛盾)至少有一个状态是(正)常返的(b)若存在零常返状态i,可构造C(i) = j|i j,则C(i)为原M.C.的一不可约子M.C.(有限状态),于是C(i)中所有状态均为零常返,与有限状态M.C.至少有一个正常返状态矛盾, )任何常返状态均为正常返16.考虑一生长与灾害模型.这类Markov链有状态0,1,2,当过程处于状态i时它既可能以概率pi转移到i+1(生长)也能以概率qi = 1pi落回到状态0(灾害).而从状态“0”又必然“无中”生有.即P01 1.(a)试证所有状态为常返的条件是limn(p1p2p3pn) = 0.(b)若此链为常返,试求其为零常返的条件.证：(a)其概率转移阵为P =0 1 2 3 cdots0 0 1 0 0 1 q1 0p1 02 q2 0 0p23 q3 0 0 0. . .易知此M.C.是不可约的, )只需证状态0常返 limn+ p1p2pn = 0