随机过程答案-西交大

精品文档【第一章】1.1证明：∵A,,F1,F1,F1,且F1∴F1是事件域。∵A,AF2,F2,AF2,AcA∴AcF2,AF2,AcF2,AF2,AF2且AAc,AAcF2∴F2是事件域。且F1F2。∵2∴F3∴F3是事件域。且F2F3∴F1,F2,F3皆为事件域且F1F2F3。1.2一次投掷三颗均匀骰子可能出现的点数 ω为i,j,k,i R,j R,k R,j i,k j,1 i 6,j 6,k 6n6∴样本空间 = i,j,k1jikj事件i,j,k|i,j,k，iR,jR,kR,ji,kj,i6,1j6,k6A事件域F 2概率测度。1欢迎下载精品文档1，iR,jR,kR,ji,kj,1i6,j6,k6PAi,j,k67i7j则,F,P为所求的概率空间。1.3证明：（1）由公理可知P0nn（2）有概率测度的可列可加性可得PAkPAkk1k1（3）∵A,BF,AB∴F,ABABA由概率测度的可列可加性可得：P(B)PABAP(A)P(BA)即PBAPBPA有概率测度的非负性可得PBPAPBA0，即PBPA（4）若B,由（3）则有PA1PA（5）∵PA1A2PA1PA2PA1A2假设mmm1PAkPAkPAiAjPAiAjAk1PA1A2Am成k1k11ijm1ijkm立，则。2欢迎下载精品文档m1 m m mPAkPAk+Am1PAm1PAkPAm1Akk1k1k1k1mPAm1PAkPAiAjPAiAjAkk11ijm1ijkmm1mAmPAm1Ak1PA1A2k1m1n1PAkPAiAjPAiAjAkAm1PA1A2k11ijm1ijkmPAiAjAm1PAiAjAkAm1m1AmAm11PA1A21ijm1ijkmm1m2PAkPAiAjPAiAjAk1PA1A2Am1k11ijm11ijkm1也成立由数学归纳法可知n nPAkPAkPAiAjPAiAjAkk1k11ijn1ijknnnnnPAkPA1AkPA1PAkPA1Akk1k2k2k2nnnPA1PA2PAkPA1AkPA2Akk3k2k3nPA1PA2PAkk3nPAkk11.41）021PABPAPBPABPAB40PAPBPABPAPBPAPABPAPBPABPAPABPA1PA1PA4

n11 PA1A2 An。3欢迎下载精品文档（2）if PAB PBC=PAB PBCPAB PAC PAB C PABCPAB PBC PACPAB C PABC PBCPAB C 1elseifPAB PBC = PAB PBC可由这个式子的轮换对称性证明这种情况3）n nAkAAkAk1k1nnnnPAPAkPAk1PAknPAkk1k1k1k1n1PAnPAkk1nPAk1PAkn11.5PXkAnkn!，∴PXx1PXx1n!nknkk!FXnkk!1.6由全概率公式PYXPY0PX0PY1PX1PY2PX2111PY011PY0PY1=1e14241.7证明：显然1Fx1, xn F x1, xn F y1, xn Px1 X1 y1,x2 X2, ,xn Xn 0假设。4欢迎下载精品文档1 2 iF x1, xn Px1 X1 y1,x2 X2 y2, ,xi Xi yi,xi Xi, ,xn Xn 0成立从而12ii+1Fx1,xnPx1X1y1,x2X2y2,,xiXiyi,xi1Xi1,,xnXnPx1X1y1,x2X2y2,,xiXiyi,yi1Xi1,,xnXnPx1X1y1,x2X2y2,,xi1Xi1yi1,xi2Xi2,,xnXn0（分布函数对于每一变元单调不减）也成立由数学归纳法可知12nFx1,xnPx1X1y1,x2X2y2,,xnXnyn01.8xhx,yex'yexyxyhx,yex'y'exy'ex'yexyexex'eyey'0x x',y y'所以h是二元单调不减函数。hx,y hx,y0, 0x y但对每一个变元是单调减的。xgx,y0or10,ygx,y0or10对于每一变元单调不减；xygx,ygx',y'+gx,ygx',ygx,y'当gx',y'=1，，，时gx,y=0gx',y=1gx,y'=1ygx,y0,是单调不增的。1.9。5欢迎下载精品文档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欢迎下载精品文档PAB1B2Bn10PA0PAB1B2Bn1PAPB1B2Bn1|APAtherightsideoftheequationleftside=rightside（2）PAk|APB|AAkPAkB|Ak1k1PAkB|APBAk|APB|Ai1k11.14证明：PAB|CPA|BCPB|CPAB|CPA|CPB|CPA|CPA|BC证毕1.15fx|x1fXxxex1FX1,x112EX|X11xfx|x1dx51.16CovX,EY|XEXEXEY|XEEY|XEEXY|XEXEY|XXEYEXEYEXYEXEYCovX,Y1.17fX|Yx|yfXYx,y21fYy21y1y。7欢迎下载精品文档EX|y1xfX|Yx|ydxy1xdx1y,0y1y1y2fY|Xy|xfXYx,y1fXxxEY|xxyfY|Xy|xxyxdy=dy200x1.18证明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欢迎下载精品文档（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欢迎下载精品文档Y1Y22X2,1DY2DY122covY1,Y2DY2DY1221.231）x1 y12x2y22y12x3y32y222y1J2y12y28y1y2y3y22y3fYYYy1,y2,y3y28y1y2y3e31232）x1y12x2y22y12J8y1y2fYYy1,y2y28y1y2e2121.241）kPXk|0k!e0PXkPXk|00c？未算，请验证k1c2）kPXk|0k!e0PXkPXk|00

kdF0ececd0k!dF0。10欢迎下载精品文档1.25Gzpnznn0nmPYm|XnCnmpm1pPYmn0PYm|XnpnCnmpm1pnG0nmn0n!nm0GYzCnmpm1pnmpnzm=Gpz1pm0n0nm01.26GXz43n15n1znn0p43n15n1n1.27设:YX1X2YX2X1YtX2tX1t矩母函数与分布函数一一对应X2tX1tYt1X1 X2的概率密度与 无关。1.28。11欢迎下载精品文档PXkCnkpk11kpnCnkpk1k1peiktt1ppeiktk0EX'0p,EX2''0ii2DXEX2EX2np1p1.30PlimYn0limPX11nnn3211.311.32A~N0,1,B~N0,1&A,B独立X~N0,1 tX1,X2~N0,1 t1;0,1 t2;01.34证明：CXY E Xt1 mX t1 Yt2 mYt2EXt1 mX t1 EYt2 mYt2EXt1 EmXt1 EYt2 EmYt2mXt1 mXt1 mYt2 mYt20∴X,Y不相关。1.35证明：。12欢迎下载精品文档CXYEXt1mXt1Yt2mYt2EXt1EYt2021dmXsint0,02my21dcost0；02t0,1,2,∴X,Y不相关。【第二章】2.11.NtnSnt2.NtnSnt3.NtnSnt。13欢迎下载精品文档2.2证明：记定义 的条件（1），（2），（3）为A,B,C；记定义 的条件（1），（2），3）为E,F,GAE,G&FB有定理2.1.3，A&B&CG&F对G的等式右边进行Taylor展开，分别令n1&2可得C，即GC∴A&B&CE&F&G2.3mNtENttCNt1,t2RXt1,t2-mNt1mNt2,t1t2=ENt1Nt22t1t2ENt1Nt2Nt222t1t2Nt2ENt1Nt2ENt222t1t2ENt2t2ENt1t2t22t222t1t2t2t1t2t22t222t1t2t22.4证明：PSn Sn1 0 PTn 0由定理 可得Tn~expPSn Sn1 0 PTn 0=1S0 S1 S2 a.s.2.5。14欢迎下载精品文档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欢迎下载精品文档tnPNtnetn!PN2tk,NtnPNtn,N2tNtknPNtnPN2tNtkntntknetn!etk!kPN2tk|NtPN2tk,NtnnNtnPtntknetn!etn!knt e t. n!k netkn!2.7780781460PNt8/60Nt0e60e150!2.8ES5S2ET3T4T53ET33152252.9PT2101PT21010511e8e42.11。16欢迎下载精品文档1）covT,NTETNTETENT22）DNTDENT|TEDNT|T=DT+ET222.12NtLetXXii1ENtX=EENtX|Nt22tENttNt22covNt,XiENtXEXENttttti12.13PS1s|NtnPNs1|Ntn1PNs0|NtnPNs0|NtnPNs0,NtnPNs0PNtNsnPNtnPNtntsnestsn!ensn1tettn!nPS1s|Ntn11st2.14。17欢迎下载精品文档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欢迎下载精品文档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证明：必要性：mt ENt t充分性：Lms=Lts2LLfsLfsL1Letfsfsmss1LfssT~fs指数分布Nt为参数的泊松分布。19欢迎下载精品文档【第三章】3.1证明：由定义可知：1 2对于21的情况与k1的情况等价k1k显然成立1k=1k1证明：PXmpimp|X0i0,,Xmp1ipm1PXmpimp|Xmp1imp1,p1,,k把这k个式子两边相乘，就是k时的结论132k12显然成立23letmtm1,PXtm1itm1|Xtmitm=PX0i0,,Xm1im1|XtmitminSi1SPXtm1itm1|Xt0it0,,Xtm11itm11PXt0it0,,Xtm11itm11|XtmiminSi1SPXm1im1|Xt0it0,,Xtm11itm11,Xtm11itm11PXt0it0,,Xtm11itm11|Xitm11SinSi1SPXm1im1，Xtm11itm11|Xt0it0,,Xtmitmtm11S=PXm1im1|Xt0it0,,Xtmitm证毕3.2证明：pijPXm1j|Xmi，jPXnjrightside=PX0i0PX1|X0PXnin|Xn1in1leftside。20欢迎下载精品文档3.3证明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证明：假设第 n次投掷硬币得到正面的事件为 Yn=1，那么第n次投币后得到正面的次数为nX(n) Yn in。k1其中，Yn1 X(n 1) X(n)，因为Y1。。。Yn之间相互独立，所以，Yn+1和X（1），。。。X（n）相互独立。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链。0.5,ji0而pijP{Xm1j|Xmi}P(Ym1ji)0.5,ji10,其他。21欢迎下载精品文档0.50.50.50.5所以转移矩阵为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欢迎下载精品文档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欢迎下载精品文档312p13p1rpr3r1483.10，2，3，4，5，6d1 1,d2 doesn'texist，d3 d4 d5 d6 13.111）1 1A是随机矩阵 A111AnAn1A1m11

m1 1m11 1An1m1 1m1 1m1An是随机矩阵（2）是双重随机矩阵A,AT是一般的随机矩阵A由1的结论可知An,ATn也是随机矩阵TnnTA=ATAn,An 是一般的随机矩阵An是双重随机矩阵3.12PX021114221110,,42420P2209/57637/96145/576EX2PX212PX223PX231103/5843.13。24欢迎下载精品文档f111PX11|X011,f1120,f1132148,333545f1142114211448203235353545175175f1212,f122f123f124033.14nlimPjjzlim111pjjz1fjjn0z1z11Fjj1nfjjn0当j为非常返态，即fjj1时，等价于pjjnn03.15充分性有限不可约Markov链所有状态都相通i,j使得pij0由题意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欢迎下载精品文档2）S N C1 C2 2,4 3 1,53）SNC4，2，5，71,3,63.18n1pijjSiC,假设pijn1Cpijn0jSC则C不为闭集，与题设矛盾npijnpij1jSjCpijn1jC3.191，2，3，4，3，13.211）0,1,2，因为设备的好坏与否与设备能否被修复彼此独立，所以设备的好坏只与现在有关，与过去无关。122a1aa2aP122ba1a1b1a1baba2abb222b21aa21bb1ab2a221a21bba1b2）= P0+1+ 2=1=,,。26欢迎下载精品文档3.22limnlimn1limnPnnnlimn1n2n31nlimn,,n3.23是吸收态，，,N是非常返态(反证法证明)12NnN0,j1limjlimPXnjlimnk0pkjk0limpkj1,j1nnnnk1k1limnlim0Pn0limPnnnn1，0，，0|1n3.24P&各项元素和为1解得11111= ，，，，555553.25由转移矩阵可知，这是一个有限不可约遍历链每一个状态相通且是正常返的其极限分布是平稳分布且唯一= P,1 2 3 15 6 3解得 = ，，。27欢迎下载精品文档【第四章】4.1limEaXnbYnaX2bYnlimEaXnX2bYnYnlimEa2Xn2b2YnY2X2abXnXYnYn2ablimEXnXYnYn2ablimEXnX2Y2EYn,CauchySchwarzInequalityn0limEaXnbYnaXbY20n结论成立4.2由题：m.s.limXtXt0,limgtgt0tt0tt0由定理4.1.6,m.s.limgtXtgt0Xt0tt04.3（1）。28欢迎下载精品文档由题：m.s.limXt0ttXt0存在t0limEXt0t1Xt0Xt0t2Xt0存在均方收敛准则收敛准则证明t1t2t10t20不妨设t1=t2Xt0t1Xt02存在limE2t10t12limEXt0t1Xt02=limEt12Xt0t12Xt0t10t10t1Xt0t12Xt0t1Xt02limt12EXt0limt12limE=022t10t1t10t10t1m.s.limXt0tXt0t0均方连续2）EX'tEXt0tXt0m.s.limtt0limEXt0tXt0limEXt0tEXt0t0tt0tEXtdt3）aXbY'taXt0taXt0bYt0tbYt0m.s.limttm.s.alimXt0tXt0m.s.blimYt0tYt0=aX'bY'tttt4）。29欢迎下载精品文档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 均方不连续 均方不可导解：{X（）0}是泊松过程，所以有t,tP{X(t)-X(s)-(t-s)[(ts)]k0,1,...,ts.k}e,kk!现在令s=0，那么有E[X(t)]t。当st时,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当s>t时可得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因为RX(s,t)在（t，t）t大于等于0处二元连续，泊松过程在 t大于等于0是均方连续。讨论其均方可导性。30欢迎下载精品文档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的时候是均方可导。4.8（1）证明：E[X(t),X'(t)]limE[X(t)X(th)X(t)]limRX(h)RX(0)R'X(0)h0hh0h（2）证明：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不依赖于tZt是宽平稳过程4.10。31欢迎下载精品文档limCXtlimtt01CX0t0limCXtCX0t0limRXtRX0t0limCXtlimRXt2mXRXt02CXt0limtt0tt0mXt0=tt0CX0CX0CX0CX04.11EXnXnEXn1nXn1n2EXn1Xn1222EXn2Xn2224EXn2Xn21+222n1212n22nEX0X01+222n2221211EXnEXn1nEXn1nEX0nE00EXnXnmEXnXnm1nmEXnXnm1mEXnXnm212是平稳过程4.12ENtN1ttt0ENtN1tNtN1tENtNtN1tN1tNtN1tNtN1tENtNttt2tENtN1tENtN1EN1ttt22tt12tt12tt只看含有t的项,可以完全相消是平稳过程4.13证明：E（X）0N NE[X(n)X(m)] E[ akcos(kn k) ajcos(jm j)]k1 j1。32欢迎下载精品文档当k不等于j时，有E[akajcos(knk)cos(jmj)]akajE[cos(knk)cos(jmj)]akaj22j)mj)cos(n(jk)mjjk)*10cos(n(kjk42dkdj0，j)0(k当k=j时，E[akakcos(kn k)cos(km k)]ak2 22 02akcos2

2cos（）10km-n*42dkdj（m-n）k2E[X(n)X(m)] akNcos（m-n）与n无关，因此为平稳过程