隨機(jī)過程答案-西交大

精品文檔【第一章】1.1證明：∵A,,F1,F1,F1,且F1∴F1是事件域?！逜,AF2,F2,AF2,AcA∴AcF2,AF2,AcF2,AF2,AF2且AAc,AAcF2∴F2是事件域。且F1F2?！?∴F3∴F3是事件域。且F2F3∴F1,F2,F3皆為事件域且F1F2F3。1.2一次投擲三顆均勻骰子可能出現(xiàn)的點(diǎn)數(shù) ω為i,j,k,i R,j R,k R,j i,k j,1 i 6,j 6,k 6n6∴樣本空間 = i,j,k1jikj事件i,j,k|i,j,k，iR,jR,kR,ji,kj,i6,1j6,k6A事件域F 2概率測(cè)度。1歡迎下載精品文檔1，iR,jR,kR,ji,kj,1i6,j6,k6PAi,j,k67i7j則,F,P為所求的概率空間。1.3證明：（1）由公理可知P0nn（2）有概率測(cè)度的可列可加性可得PAkPAkk1k1（3）∵A,BF,AB∴F,ABABA由概率測(cè)度的可列可加性可得：P(B)PABAP(A)P(BA)即PBAPBPA有概率測(cè)度的非負(fù)性可得PBPAPBA0，即PBPA（4）若B,由（3）則有PA1PA（5）∵PA1A2PA1PA2PA1A2假設(shè)mmm1PAkPAkPAiAjPAiAjAk1PA1A2Am成k1k11ijm1ijkm立，則。2歡迎下載精品文檔m1 m m mPAkPAk+Am1PAm1PAkPAm1Akk1k1k1k1mPAm1PAkPAiAjPAiAjAkk11ijm1ijkmm1mAmPAm1Ak1PA1A2k1m1n1PAkPAiAjPAiAjAkAm1PA1A2k11ijm1ijkmPAiAjAm1PAiAjAkAm1m1AmAm11PA1A21ijm1ijkmm1m2PAkPAiAjPAiAjAk1PA1A2Am1k11ijm11ijkm1也成立由數(shù)學(xué)歸納法可知n nPAkPAkPAiAjPAiAjAkk1k11ijn1ijknnnnnPAkPA1AkPA1PAkPA1Akk1k2k2k2nnnPA1PA2PAkPA1AkPA2Akk3k2k3nPA1PA2PAkk3nPAkk11.41）021PABPAPBPABPAB40PAPBPABPAPBPAPABPAPBPABPAPABPA1PA1PA4

n11 PA1A2 An。3歡迎下載精品文檔（2）if PAB PBC=PAB PBCPAB PAC PAB C PABCPAB PBC PACPAB C PABC PBCPAB C 1elseifPAB PBC = PAB PBC可由這個(gè)式子的輪換對(duì)稱性證明這種情況3）n nAkAAkAk1k1nnnnPAPAkPAk1PAknPAkk1k1k1k1n1PAnPAkk1nPAk1PAkn11.5PXkAnkn!，∴PXx1PXx1n!nknkk!FXnkk!1.6由全概率公式PYXPY0PX0PY1PX1PY2PX2111PY011PY0PY1=1e14241.7證明：顯然1Fx1, xn F x1, xn F y1, xn Px1 X1 y1,x2 X2, ,xn Xn 0假設(shè)。4歡迎下載精品文檔1 2 iF x1, xn Px1 X1 y1,x2 X2 y2, ,xi Xi yi,xi Xi, ,xn Xn 0成立從而12ii+1Fx1,xnPx1X1y1,x2X2y2,,xiXiyi,xi1Xi1,,xnXnPx1X1y1,x2X2y2,,xiXiyi,yi1Xi1,,xnXnPx1X1y1,x2X2y2,,xi1Xi1yi1,xi2Xi2,,xnXn0（分布函數(shù)對(duì)于每一變?cè)獑握{(diào)不減）也成立由數(shù)學(xué)歸納法可知12nFx1,xnPx1X1y1,x2X2y2,,xnXnyn01.8xhx,yex'yexyxyhx,yex'y'exy'ex'yexyexex'eyey'0x x',y y'所以h是二元單調(diào)不減函數(shù)。hx,y hx,y0, 0x y但對(duì)每一個(gè)變?cè)菃握{(diào)減的。xgx,y0or10,ygx,y0or10對(duì)于每一變?cè)獑握{(diào)不減；xygx,ygx',y'+gx,ygx',ygx,y'當(dāng)gx',y'=1，，，時(shí)gx,y=0gx',y=1gx,y'=1ygx,y0,是單調(diào)不增的。1.9。5歡迎下載精品文檔EYX1X3X2EX1EX3EX2E10E1X321X321X32DY22EY2EX12X322X1X2X3X222EYEY1X321.10n:n1/pn1iniEX1/pi1/pp1pi1/pp1pin1i01+2n1pn11.11+dFxgxxudFx+gxxgxxudFx+xudFxgEx0g1.12（1）X~exp ,fX x e x,FX x 1 e x,x 0Y min X,t,t 0fYy|0XtfXex|0xt1e

xtfYy,0XtexFy,0Xt1exWhenXt,YtPYt|XtPXt|Xt1PYt,XtPYt|XtPXtete（2）fX x|x te

xt

e

xt1.131）。6歡迎下載精品文檔PAB1B2Bn10PA0PAB1B2Bn1PAPB1B2Bn1|APAtherightsideoftheequationleftside=rightside（2）PAk|APB|AAkPAkB|Ak1k1PAkB|APBAk|APB|Ai1k11.14證明：PAB|CPA|BCPB|CPAB|CPA|CPB|CPA|CPA|BC證畢1.15fx|x1fXxxex1FX1,x112EX|X11xfx|x1dx51.16CovX,EY|XEXEXEY|XEEY|XEEXY|XEXEY|XXEYEXEYEXYEXEYCovX,Y1.17fX|Yx|yfXYx,y21fYy21y1y。7歡迎下載精品文檔EX|y1xfX|Yx|ydxy1xdx1y,0y1y1y2fY|Xy|xfXYx,y1fXxxEY|xxyfY|Xy|xxyxdy=dy200x1.18證明DX|Y2EXEX|Y|YEX22XEX|YEX|Y2|YEX2|Y2EX|YEX|YEEX|Y2|YEX2|YEX|Y2EDX|YDEX|YEEX2|Y222EX|YEEX|YEEX|YEX2EX2DX1.19f y 1,0 y 1f x,y fX xFzZ z FXY X Y zfX xdxdyx yz1 z y0dy fX xdx10FX z ydyfZ z FX z FX z 11.20。8歡迎下載精品文檔（1）PX Y zPXYz|Y0PY0,0z1PXYz|Y1PY1,1z2PXzPY0,0z1PXz1PY1,1z2z1p,0z1z1p,1z2（2）PXYz1p,z0pz0 z 11.21EX1X2X1EX11X1X2EX1X2EX12EX11=1X1X22EX1X2X1EX1XX1X21.221）fX1X2x112exp1,x2x122X11Y1+Y2,X11Y1Y222

X12EX1X22 2x222fYYy1,y212exp11y1+y21y1y21242222）。9歡迎下載精品文檔Y1Y22X2,1DY2DY122covY1,Y2DY2DY1221.231）x1 y12x2y22y12x3y32y222y1J2y12y28y1y2y3y22y3fYYYy1,y2,y3y28y1y2y3e31232）x1y12x2y22y12J8y1y2fYYy1,y2y28y1y2e2121.241）kPXk|0k!e0PXkPXk|00c？未算，請(qǐng)驗(yàn)證k1c2）kPXk|0k!e0PXkPXk|00

kdF0ececd0k!dF0。10歡迎下載精品文檔1.25Gzpnznn0nmPYm|XnCnmpm1pPYmn0PYm|XnpnCnmpm1pnG0nmn0n!nm0GYzCnmpm1pnmpnzm=Gpz1pm0n0nm01.26GXz43n15n1znn0p43n15n1n1.27設(shè):YX1X2YX2X1YtX2tX1t矩母函數(shù)與分布函數(shù)一一對(duì)應(yīng)X2tX1tYt1X1 X2的概率密度與 無關(guān)。1.28。11歡迎下載精品文檔PXkCnkpk11kpnCnkpk1k1peiktt1ppeiktk0EX'0p,EX2''0ii2DXEX2EX2np1p1.30PlimYn0limPX11nnn3211.311.32A~N0,1,B~N0,1&A,B獨(dú)立X~N0,1 tX1,X2~N0,1 t1;0,1 t2;01.34證明：CXY E Xt1 mX t1 Yt2 mYt2EXt1 mX t1 EYt2 mYt2EXt1 EmXt1 EYt2 EmYt2mXt1 mXt1 mYt2 mYt20∴X,Y不相關(guān)。1.35證明：。12歡迎下載精品文檔CXYEXt1mXt1Yt2mYt2EXt1EYt2021dmXsint0,02my21dcost0；02t0,1,2,∴X,Y不相關(guān)?！镜诙隆?.11.NtnSnt2.NtnSnt3.NtnSnt。13歡迎下載精品文檔2.2證明：記定義 的條件（1），（2），（3）為A,B,C；記定義 的條件（1），（2），3）為E,F,GAE,G&FB有定理2.1.3，A&B&CG&F對(duì)G的等式右邊進(jìn)行Taylor展開，分別令n1&2可得C，即GC∴A&B&CE&F&G2.3mNtENttCNt1,t2RXt1,t2-mNt1mNt2,t1t2=ENt1Nt22t1t2ENt1Nt2Nt222t1t2Nt2ENt1Nt2ENt222t1t2ENt2t2ENt1t2t22t222t1t2t2t1t2t22t222t1t2t22.4證明：PSn Sn1 0 PTn 0由定理 可得Tn~expPSn Sn1 0 PTn 0=1S0 S1 S2 a.s.2.5。14歡迎下載精品文檔PN1t s N2t s N1t N2t nnPN1t s N1t mPN2t s N2t n mm0n1sm2snm1se2sem0m!nm!nmnmns12se12m!nm!m0n12se12sbinomialtheoremn!2.61.nPNtntetn!PNt/2m,NtnPNt/2m,NtNt/2nmtmtnmt22PNt/2mPNtNt/2nme2em!nm!

t2PNt/2m,NtnPNt/2m|NtnnPNttmtnmtt22e2e2m!nm!netn!

nm1n!m!n m!2。15歡迎下載精品文檔tnPNtnetn!PN2tk,NtnPNtn,N2tNtknPNtnPN2tNtkntntknetn!etk!kPN2tk|NtPN2tk,NtnnNtnPtntknetn!etn!knt e t. n!k netkn!2.7780781460PNt8/60Nt0e60e150!2.8ES5S2ET3T4T53ET33152252.9PT2101PT21010511e8e42.11。16歡迎下載精品文檔1）covT,NTETNTETENT22）DNTDENT|TEDNT|T=DT+ET222.12NtLetXXii1ENtX=EENtX|Nt22tENttNt22covNt,XiENtXEXENttttti12.13PS1s|NtnPNs1|Ntn1PNs0|NtnPNs0|NtnPNs0,NtnPNs0PNtNsnPNtnPNtntsnestsn!ensn1tettn!nPS1s|Ntn11st2.14。17歡迎下載精品文檔PMmT2mEPYS2YS1m|S2S1T2EeT2m!mmT2m=T2eTeTeT2dT22dT20m!20m!T2m0eT2dT2m!mm1=T2eT2|00T2eT2dT2m!m1!T2m1eT2dT2eT2dT210m1!0mPMm2.17PN11PT1213PN10PT112316PN3 2 1 PN3 3 PN3 1278PN3 3 PT1 1,T2 1,T3 1271PN3 1 PT1 2,T2 292.182xex2LfsFs2Lfs2Lmss1Lfs2s2smt1t1e2t12242.19。18歡迎下載精品文檔ex,xTn~fx0,xnet1tnnt1,,tnT1,,Tn~ft1,,tn,min0,mint1,,tngs1,,snnesn,mint1,,tn0,mint1,,tnnn1snngnsngs1,,snds1dsn1se，t1,,tnnmin1!PNtnPSntnn1esnn=tsdsn0n1!nn1ktket，e1t1,,tnk!mink02.20證明：必要性：mt ENt t充分性：Lms=Lts2LLfsLfsL1Letfsfsmss1LfssT~fs指數(shù)分布Nt為參數(shù)的泊松分布。19歡迎下載精品文檔【第三章】3.1證明：由定義可知：1 2對(duì)于21的情況與k1的情況等價(jià)k1k顯然成立1k=1k1證明：PXmpimp|X0i0,,Xmp1ipm1PXmpimp|Xmp1imp1,p1,,k把這k個(gè)式子兩邊相乘，就是k時(shí)的結(jié)論132k12顯然成立23letmtm1,PXtm1itm1|Xtmitm=PX0i0,,Xm1im1|XtmitminSi1SPXtm1itm1|Xt0it0,,Xtm11itm11PXt0it0,,Xtm11itm11|XtmiminSi1SPXm1im1|Xt0it0,,Xtm11itm11,Xtm11itm11PXt0it0,,Xtm11itm11|Xitm11SinSi1SPXm1im1，Xtm11itm11|Xt0it0,,Xtmitmtm11S=PXm1im1|Xt0it0,,Xtmitm證畢3.2證明：pijPXm1j|Xmi，jPXnjrightside=PX0i0PX1|X0PXnin|Xn1in1leftside。20歡迎下載精品文檔3.3證明PYn1PXn1PXn1PYn1

j,j'|Y0j,Xn2j,Xn2j,j'|Yn

i0,i1, ,Yn in,jj'|X0 i0, ,X n 1 jj'|Xn in,Xn 1 jin,j3.4證明：假設(shè)第 n次投擲硬幣得到正面的事件為 Yn=1，那么第n次投幣后得到正面的次數(shù)為nX(n) Yn in。k1其中，Yn1 X(n 1) X(n)，因?yàn)閅1。。。Yn之間相互獨(dú)立，所以，Yn+1和X（1），。。。X（n）相互獨(dú)立。P{X(n1)in1|X(1)i1,X(2)i2,...,X(n)in}P{Yn1in1in|X(1)i1,...,X(n)in}P{Yn1in1in}同理P{X(n1)in1|X(n)in}P{Yn1in1in}所以P{X(n1)in1|X(1)i1,X(2)i2,...,X(n)in}P{X(n1)in1|X(n)in}所以，X(n)是markov鏈。0.5,ji0而pijP{Xm1j|Xmi}P(Ym1ji)0.5,ji10,其他。21歡迎下載精品文檔0.50.50.50.5所以轉(zhuǎn)移矩陣為P....0.50.50.53.5p1,11,p1,11,p0,01222p1,01,p1,10,p1,102p0,11,p1,01,p0,11424PXn10|Xn1,Xn11不存在PXn10|Xn1,Xn11PXn10|Xn1不是MC。22歡迎下載精品文檔nn1Xn2Pn2PXn1j|XniP

n1j,nn1i2nn1i21Pn12j1,n121Pn2LetGijPXn11122G11142411

2j11Pn12j1,n12j22i11Pn2i12j|Xni22Counterevidence:32PX31|X20,X1011,PX31|X2022ThusitdisqualifiesasMC.3.6S1,211,0P0.650.350.550.451P40.61115,0.38885PX520.388853.7（1）PX03,X11,X2311114882562）。23歡迎下載精品文檔312p13p1rpr3r1483.10，2，3，4，5，6d1 1,d2 doesn'texist，d3 d4 d5 d6 13.111）1 1A是隨機(jī)矩陣 A111AnAn1A1m11

m1 1m11 1An1m1 1m1 1m1An是隨機(jī)矩陣（2）是雙重隨機(jī)矩陣A,AT是一般的隨機(jī)矩陣A由1的結(jié)論可知An,ATn也是隨機(jī)矩陣TnnTA=ATAn,An 是一般的隨機(jī)矩陣An是雙重隨機(jī)矩陣3.12PX021114221110,,42420P2209/57637/96145/576EX2PX212PX223PX231103/5843.13。24歡迎下載精品文檔f111PX11|X011,f1120,f1132148,333545f1142114211448203235353545175175f1212,f122f123f124033.14nlimPjjzlim111pjjz1fjjn0z1z11Fjj1nfjjn0當(dāng)j為非常返態(tài)，即fjj1時(shí)，等價(jià)于pjjnn03.15充分性有限不可約Markov鏈所有狀態(tài)都相通i,j使得pij0由題意n使得：pjin0,piin0piin1pijpjin0,pii2n1piinpiin10n 1|di，2n 1|didi 1是非周期的3.161）1，2，3正常返，d 1；4，非常返2）1，2，3，4正常返，d 33.171）S N C 2 1,3,4。25歡迎下載精品文檔2）S N C1 C2 2,4 3 1,53）SNC4，2，5，71,3,63.18n1pijjSiC,假設(shè)pijn1Cpijn0jSC則C不為閉集，與題設(shè)矛盾npijnpij1jSjCpijn1jC3.191，2，3，4，3，13.211）0,1,2，因?yàn)樵O(shè)備的好壞與否與設(shè)備能否被修復(fù)彼此獨(dú)立，所以設(shè)備的好壞只與現(xiàn)在有關(guān)，與過去無關(guān)。122a1aa2aP122ba1a1b1a1baba2abb222b21aa21bb1ab2a221a21bba1b2）= P0+1+ 2=1=,,。26歡迎下載精品文檔3.22limnlimn1limnPnnnlimn1n2n31nlimn,,n3.23是吸收態(tài)，，,N是非常返態(tài)(反證法證明)12NnN0,j1limjlimPXnjlimnk0pkjk0limpkj1,j1nnnnk1k1limnlim0Pn0limPnnnn1，0，，0|1n3.24P&各項(xiàng)元素和為1解得11111= ，，，，555553.25由轉(zhuǎn)移矩陣可知，這是一個(gè)有限不可約遍歷鏈每一個(gè)狀態(tài)相通且是正常返的其極限分布是平穩(wěn)分布且唯一= P,1 2 3 15 6 3解得 = ，，。27歡迎下載精品文檔【第四章】4.1limEaXnbYnaX2bYnlimEaXnX2bYnYnlimEa2Xn2b2YnY2X2abXnXYnYn2ablimEXnXYnYn2ablimEXnX2Y2EYn,CauchySchwarzInequalityn0limEaXnbYnaXbY20n結(jié)論成立4.2由題：m.s.limXtXt0,limgtgt0tt0tt0由定理4.1.6,m.s.limgtXtgt0Xt0tt04.3（1）。28歡迎下載精品文檔由題：m.s.limXt0ttXt0存在t0limEXt0t1Xt0Xt0t2Xt0存在均方收斂準(zhǔn)則收斂準(zhǔn)則證明t1t2t10t20不妨設(shè)t1=t2Xt0t1Xt02存在limE2t10t12limEXt0t1Xt02=limEt12Xt0t12Xt0t10t10t1Xt0t12Xt0t1Xt02limt12EXt0limt12limE=022t10t1t10t10t1m.s.limXt0tXt0t0均方連續(xù)2）EX'tEXt0tXt0m.s.limtt0limEXt0tXt0limEXt0tEXt0t0tt0tEXtdt3）aXbY'taXt0taXt0bYt0tbYt0m.s.limttm.s.alimXt0tXt0m.s.blimYt0tYt0=aX'bY'tttt4）。29歡迎下載精品文檔fX'tm.s.limfttXttftXttt0m.s.limftXttXtfttftXttt0m.s.limfttftXttXttt0m.s.ftlimXttXtm.s.Xtfttfttlimtt0t0f'tXtftX'tm.s.limfttftXttXt0X't0tt04.4Xt'sinAt'EsinAt|A'EsinAt'|AEAcosAt|AAcosAt4.7 均方不連續(xù) 均方不可導(dǎo)解：{X（）0}是泊松過程，所以有t,tP{X(t)-X(s)-(t-s)[(ts)]k0,1,...,ts.k}e,kk!現(xiàn)在令s=0，那么有E[X(t)]t。當(dāng)st時(shí),有CX(s,t) E{[X(s) s][X(t) t]}E{[X(s) s]{[X(s) s] [X(t) t] [X(s) s]}}E{[X(s) s]2}D[X(s)]2 s當(dāng)s>t時(shí)可得CX(s,t) t。所以CX(s,t) min(s,t)，所以RX(s,t) CX(s,t) m(s)m(t) min(s,t) 2st因?yàn)镽X(s,t)在（t，t）t大于等于0處二元連續(xù)，泊松過程在 t大于等于0是均方連續(xù)。討論其均方可導(dǎo)性。30歡迎下載精品文檔limRX(th,tl)RX(th,t)RX(t,tl)RX(t,t)hlh0,l0lim(th)2(th)(tl)(th)2(th)tt2t(tl)t2t2h0,l0hl2泊松分布在 t大于等于0的時(shí)候是均方可導(dǎo)。4.8（1）證明：E[X(t),X'(t)]limE[X(t)X(th)X(t)]limRX(h)RX(0)R'X(0)h0hh0h（2）證明：E[X'(t)]dE[X(t)]0dtE[X'(t)X'(t)]limE[X(th)X(t)*X(tl)X(t)]h0,l0hllimX(th)X(tl)X(t)X(tl)X(th)X(t)X(t)X(t)E[hlh0,l0hlim[R(lh)R(l)][R(h)R()]10,l0hl

]得limR'( l) R'() R''()l0l4.9EZtEcosatisinatEcosatcossinatsinisinatcoscosatsincosatisinatEcossinaticosatEsin=0RZt,tEZtZtEeiateiatEeia=eia不依賴于tZt是寬平穩(wěn)過程4.10。31歡迎下載精品文檔limCXtlimtt01CX0t0limCXtCX0t0limRXtRX0t0limCXtlimRXt2mXRXt02CXt0limtt0tt0mXt0=tt0CX0CX0CX0CX04.11EXnXnEXn1nXn1n2EXn1Xn1222EXn2Xn2224EXn2Xn21+222n1212n22nEX0X01+222n2221211EXnEXn1nEXn1nEX0nE00EXnXnmEXnXnm1nmEXnXnm1mEXnXnm212是平穩(wěn)過程4.12ENtN1ttt0ENtN1tNtN1tENtNtN1tN1tNtN1tNtN1tENtNttt2tENtN1tENtN1EN1ttt22tt12tt12tt只看含有t的項(xiàng),可以完全相消是平穩(wěn)過程4.13證明：E（X）0N NE[X(n)X(m)] E[ akcos(kn k) ajcos(jm j)]k1 j1。32歡迎下載精品文檔當(dāng)k不等于j時(shí)，有E[akajcos(knk)cos(jmj)]akajE[cos(knk)cos(jmj)]akaj22j)mj)cos(n(jk)mjjk)*10cos(n(kjk42dkdj0，j)0(k當(dāng)k=j時(shí)，E[akakcos(kn k)cos(km k)]ak2 22 02akcos2

2cos（）10km-n*42dkdj（m-n）k2E[X(n)X(m)] akNcos（m-n）與n無關(guān)，因此為平穩(wěn)過程