三.有限維概率馬氏鏈{X(t),tt0t1t2}在初始時(shí)pj(t0)P{X(t0)j},j稱為初始分xxO 馬爾可夫鏈在任何時(shí)刻tn的一維概率分pj(tn)P{X(tn)j},j0,1,2,xxj)Ottn-t設(shè)：已知初始pj(t0P{X(t0j},j0,12,以及n步轉(zhuǎn)移概率p(n)( ) i,j0,1,2求：瞬時(shí)概

ij pj(tn)P{X(tn)j},j0,1,2,P{X(tn)j} P{X(tn1)i}P{X(tn)j|X(tn1)i pi(tn1)pij(tn1),j0,1,2,i于 pj(tn) pi(tn1)pij(tn1),j同樣的方法p(t)p(t)p(n)(t),j pj(tn) pi(tn1)pij,j0,1,2,i p(t) p(t)p(n),j0,1,2,i結(jié)論：齊次馬爾可夫鏈在時(shí)n的瞬(p0(tn),p1(tn),,pj(tn),

0p1jp(p0(t0),p1(t0),,pj(t0),)

pppppp

(p(t),p(t),,p(t),)P(n) n步轉(zhuǎn)移概率矩陣P(n) (p(n))例7在例3(一維隨機(jī)游動(dòng))中,設(shè)質(zhì)點(diǎn)在初始時(shí)0505/902/04/04/90004/1/902/92/3P(2)

1/001/0000

11p2(t0)0(t0)p1(t0)p3(t0)p4(t0)4p(t)p(t)p(2)

11/91/ p(t)p2(t2)

44i

p(t)p i i4

1014/94/p3(t2)4

i

pi(t0)

(2i

10ip4(t2)i

pi(t0)

14/94/方法二按題意有00,1020040

(p0(t2),p1(t2),p2(t2),p3(t2),p4(t2(p(t),p(t),p(t),p(t),p(t)) 1/ 2/ 000 5/9 4/9 00

01/ 4/ 4/0 1/9 2/9 2/30 1 4/ 4/ O k2 求P{X(k1j1X(k2j2,X(knjn}定理態(tài)空間S={01,2,i,},則對(duì)任意n個(gè)非負(fù)整數(shù)<2kn和S內(nèi)的任意n個(gè)狀態(tài),2,,,有P{X(k1)j1,X(k2)j2,,X(kn)jnp(0)p(k1

p(k2k1

(knkn1ii

j1

jn1復(fù)習(xí)乘法公式的推廣形P(A1A2AnP(1)P(A2|1)P(A3|A1A2)P(An|A1其中P(A1A2 An1)證P{X(k1)j1,X(k2)j2,,X(kn)jnP{X(k1)j1}P{X(k2)j2|X(k1)j1}P{X(k3)j3|X(k1)j1,X(k2)P{X(kn)jn|X(k1)j1,X(k2)j2,,X(kn1)P{X(k1)j1}P{X(k2)j2|X(k1)j1}P{X(k3)j3|X(k2)P{X(kn)jn|X(kn1) j j P{X(k)j}p(k2k1)p(k3k2) j j 1 2 n1P{X(k1)j1} P{X(0)i}P{X(k1)j1|X(0)p(0)p(k1i

于是P{X(k1)j1X(k2j2X(knjnp(0)p(k1

p(k2k1

p(k3k2

p(knkn1i

j1

j2

jn1證的概率分別為1/3和求第2、3、6步都傳輸出1的概 的概率是多01P01Pp0(0)1/3,p1(0)2/一步轉(zhuǎn)移概率P 11P{X21,X31,X

0.1i

(0)p(2) i

(3p1 1p(p(0)p(2)p(0)p(2)) )先 因 0 qpP 1q p有相異的特征 11,2p所以P相似于對(duì)角 0 2 2 P(HH1)HHPn (HH1)nHnH 011(p 11(pq)n 1

1 (pq) (pq) 公式,當(dāng)系統(tǒng)經(jīng)n級(jí)傳輸 P{X01|XnP{X01,Xn1}P{X01}P{Xn1|X0P{Xn p(0)P(n p(0)P(n)p(0)

P{Xn(2/3)(1(pq)n)1(1/3)(pq)定義 對(duì)于齊次馬爾可夫{X(t),tt0,t1,t2,P(pij),pijP{X(tm1)j|X(tm) (0,1,2,,j,

0,j

1 ipij,j0,1,2,i則稱(0,1,2,,j,)為平穩(wěn)分布,稱改寫(xiě)成向量:平穩(wěn)分布(0,1,2,,j,(0,1,2,,j,) 顯然(P)PP2(P)P2P3Pn定理如果齊次馬爾可夫鏈{X(t),tt0,t1,t2的初始分布pj(t0P{X(t0jj0,12,)pj(tn)P{X(tn)j}pj(t0),j0,1,2,證明由平穩(wěn)分布的性質(zhì),得(p0(tn),p1(tn),,pj(tn),(p(t),p(t),,p(t),)P(n) (p0(t0),p1(t0),,pj(t0),.設(shè)轉(zhuǎn)移概率矩1/ 2/ 0P1/ 1/ 1/3 2/ 1/3pj(0)P{X0j}1/3,j1,2,3求P{X11X22,X3求P{X求平穩(wěn)分布

解P{X11,X22,X3P{X1}{X22|X1}{X33|X22,X1P{X1}{X22|X1}{X33|X2P{X11}p123P{X0j}P{X11|X0j}p123j3P{X0j}pj1p123j211(110) 3

23P{X0j}P{X23|X033jP{Xjj

j}p(21(223) p1 1p 1 2 p2 1 1 2 3 p0 1 解之

2 3p1p2p3p1, 2, 例帶一個(gè)反射壁的一維隨機(jī)游動(dòng),以(j表示在時(shí)刻tn粒子處于狀態(tài)j,狀態(tài)空間S={0,1,2,,j,pp00pj,jpj,j

p j0,1,2,q j1,2,3,pq求平穩(wěn)分布j,j0,1,2,解根 P(p)

j jj即應(yīng)滿

j1pj1,jj1pj1, pj1qj

(2) j

由(1由(2)

1(p/j1j(p/q)(jj1(p/q)j1

0[(p/q)j1(p/q)j[(p/q)j(p/j得到j(luò)pq)j0

j

(p/q)j 1由(3),(4)得

01p/ 當(dāng)p/q<1時(shí),即當(dāng)p<q時(shí),解得0=1p/q,j(p/q)j(1p/ j0,1,2,, 0P p p 求粒子從狀態(tài)1經(jīng)二步、經(jīng)三步轉(zhuǎn)移回到狀態(tài)的轉(zhuǎn)移概率求過(guò)程的平穩(wěn)分 p(2

P{X

)1|X ) n 2p1kpkk

qp0pq2P(2)

2

p2pq q2q2pP(3)P3q32pq3

2p2qp3 q3 2pq2 2p2qp3 p

P{X )1|X(t)1} p0p0qp1qp20p1p0pp10p2qp2p00p1pp2pp0p1p2解之

p01pq,p11pq,p21設(shè)氏鏈,一步轉(zhuǎn)移概率矩陣為0 04 4 初始pi0PXi01,2