随机过程答案-西交大

精品文档【第一章】1.1证明：∵A,,∴FF,F,F,且F11111∵A,AF,F,A,AcFA222∴AF,AFF,,AcAF,AF22c222且AAc,AAcF2∴F是事件域。且FF。∵212∴F23∴F是事件域。且FF∴F,F,F皆为事件域且FF32312312是事件域。F3。1.2一次投掷三颗均匀骰子可能出现的点数ω为i,j,k,iR,jR,kR,ji,kj,1i6,j6精品文档放心下载∴样本空间n6i,j,ki1

jikj事件Ai,j,k|i,j,kj6,k6，iR,jR,kR,ji,kj,i6,1事件域F2概率测度1欢迎下载。精品文档PAi,j,k167i7j6,j6,k6，iR,jR,kR,ji,kj,1i则,F,P为所求的概率空间。1.3证明：（1）由公理可知P0（2）有概率测度的可列可加性可得PkPAknnAk1k1（3）∵A,BF,AB∴BAF,ABA由概率测度的可列可加性可得：P(B)PABAP(A)P(BA)谢谢阅读即PBAPBPA有概率测度的非负性可得PBPAPBA0，即PBPA精品文档放心下载（4）若B,由（3）则有PA1PA感谢阅读（5）∵PA1A2PAPAPA12A12假设mkmPAk1立，则k1PAkPAPAAiji1ijm1ijkmAjAkPA1m1A12Am成2欢迎下载。精品文档m1mmm谢谢阅读m1PPA

kPAkPA

kPAm1A谢谢阅读Ak1k1m1k1k1精品文档放心下载kPAm1PAAPAmAAkijijkk11ijm1ijkmAmm1PA1PAAAk12mAk1n1PAA12Amk1m1PAPAPAm1AAkijijk11ijm1ijkm1ijmPAiPAAAAAAjijkm1m11ijkmPA1m1PAAAA12mm1Am1PAPAAAAm1ijijkkk11ijm11ijkm1谢谢阅读1PAm2A12也成立由数学归纳法可知nknPAk1k1PAkPAPA1ijn1ijkniiAjAAjkPA1n1AnA12nnnkPPAnkPAAkPAPAkAAk111k2k21k2nnP1PA2nPAkPAkPAkAAk321k2k3P1PA2PAnkk3PAnk1k1.4（1）0PPAPBPP124精品文档放心下载0PAPBPPAPBPAP感谢阅读PAPBPPAPABPA1PA14PA谢谢阅读3欢迎下载。精品文档（2）PPPPPPABCPPPP谢谢阅读PABCPPPABC1精品文档放心下载elseifPP=PP感谢阅读可由这个式子的轮换对称性证明这种情况（3）nAnAAAkkk1k1PAPPA1PAnknAPAnnnk1kkkk1k1k11PAnPAnk1kPAPAnn1k1k1.5PXkkn!n!A，FXPXx1PXx1nnnk!nk!kkk1.6由全概率公式PYXPY0PX0PY1PX1PY2PX2感谢阅读1111PY01PY0PY1=1e14241.7证明：显然Fxx,n11Fx,x1nFy,xPx1nX1y,xX112,,xnX2n0假设4欢迎下载。精品文档12成立iFx,xPx1nX1y,x112Xx2y,,X2iy,xX,,xXiiiinn0从而FxPx,xXy,xXy,12i+11111222in,xXy,xX,iiii1i1,xnXnPxXy,xXy,,xXy,yX,111222i1i1iiiPxXy,xXy,,xXy,xX111222i1i1i1i2（分布函数对于每一变元单调不减）也成立,xXnn,,xXi2nn0由数学归纳法可知12xnnFx,Px1X1y,x112X2xy,,X2nynn01.8hx,yex'yexyxxy'ehx,yeex'y'ex'yxyxyexex'eyey'0xx',yy'所以h是二元单调不减函数。hx,yhx,y

0xy谢谢阅读但对每一个变元是单调减的。gx,y0or10,gx,y0or10yx对于每一变元单调不减；gx,ygx',y'x,ygx',ygx,y'感谢阅读xy当gx',y'=1，gx,y=0gx',y=1gx,y'=1时精品文档放心下载gx,y0,是单调不增x的。y1.95欢迎下载。精品文档XXXX1EYE132EXE3EXE0精品文档放心下载2122233322XXDYEYEYEYE22222X2XXX1312321X2321.10n:n1/pn1EX1/pi1/pp1pi1/pp1pin1niii01+2n1pn11.11+gxxdFxgxxudFxgxxudFx+g+xudFxgEx01.12（1）X~exp,fxex,FXXx1ex,x0YminX,t,t0fYy|0Xtfx|0xtXfy,0XtexYFy,0Xt1e1eextxWhenXt,YtPYt|XtPXt|Xt1感谢阅读PYt,XtPYt|XtPXtet感谢阅读（2）fXtx|xexetext1.13（1）6欢迎下载。精品文档PB12Bn10PA0PPAPBBBB12n112side=rightside|APAtherightoftheBn1（2）k1PA|APB|AAkkPAB|Ak1kPk|APB|AAB|APBAk1i1

k1.14证明：PAB|CPA|PB|CPAB|CPA|CPB|CPA|CPA|谢谢阅读证毕1.15xxeffx|x1,x1x1X21FX1EX|X1xfx|x1dx5精品文档放心下载11.16CovX,EY|XEXEY|XEEY|X感谢阅读EEXY|XY|XEXEYEEXEYCovX,Y精品文档放心下载1.17fXx|yf1x,y2XYfYy21y1y7欢迎下载。精品文档EX|yxfx|ydx1X|Y

yx1y,0y11y1y2fY|Xxy|fx,y1XYxxfXEY|xyfy|xxxY|X00yxx21.18证明DX|YEXEX|Y|Y精品文档放心下载222XEX|YEX|YEX感谢阅读2|YEX2|Y2EX|YEX|YEEX|Y|Y2EX2|YEX|Y2EDX|YDEX|Y谢谢阅读EEX2|Y谢谢阅读EX|YEEX|YEEX|Y222精品文档放心下载221.19fy1,0y1fx,yfx精品文档放心下载XZzFXYzFzxdxdyfXxyzdyxdx1zf0Xz1Fzydy10XzF

zFfZXX1.208欢迎下载。精品文档（1）PXYzPXYz|Y0PY0,0z1感谢阅读PXYz|Y1PY1,1z2PXzPY0,0z1精品文档放心下载PXz1PY1,1z2z1p,0z1精品文档放心下载z1p,1z2（2）Pz1p,z0pz0z11.21XX1EXEX1XX12121EX1X2=X11E2EX2X1X1X12XXEXEXXE精品文档放心下载11XXXX121212121.22（1）fXX12x,x121122expxx222121Y,XXY1Y122212111111y

f,yYYexpy21212y2y242221212（2）9欢迎下载。精品文档YY2X,11222221covY,Y1221221.23（1）xy211yyx22xy2212y23322y1J2y2yy18yyf212y,yYYY2y1231223y38y,yyyey312323（2）xy211yyx22221J8yyy12yf,yYY121281y2ey221.24（1）PXk|ke00k!PXk|PXdFkececdk00k!0cc1k1？未算，请验证0（2）PXk|k0dF00k!ePXkPX

k|0010欢迎下载。精品文档1.25Gzpn0znnPYm|XnCp1pnmmmn谢谢阅读PYmPYm|Xnpnn0Cp1pGnn!0mmnn0nm0nmz1ppmpz1pGCnznmmmm0n0Ynnm01.26GXpnzn143n15znn1n04315n1.27设:YXX12YXX21tttYXX21ttXXt211YXX无关。121.2811欢迎下载。精品文档PXkCp1p1kkk

n1keikt1peiktptCkpk1p精品文档放心下载nk0nEX'0''0p,EX2ii22np1pEX2感谢阅读1.301PlimY0PX1nnnn3211.311.32A~N,B~N&A,B独立X~Nt精品文档放心下载X,X12~N0,1t;0;0,1t121.34证明：EXtmtCYtmt谢谢阅读1X12Y2EXtEYtmtmt1X12Y2精品文档放心下载tttt1X12Y2谢谢阅读m0tmttmtmX1X1Y2Y2X,Y不相关。1.35证明：12欢迎下载。精品文档EXtmtC1X1EXtEYt120sint1m2d2X0YtYt2mY2t21mycost20t1,d；X,Y不相关。【第二章】2.11.NtnSnt2.NtnSnt3.NtnStn13欢迎下载。精品文档2.2证明：记定义2.1.2的条件（），（），（）为；记定义2.1.4的条件（），（2），

（3）为E,F,G精品文档放心下载AE,G&FB有定理2.1.3，G&F对G的等式右边进行Taylor展开，分别令n1&2可得C，即GC感谢阅读A&B&CE&F&G2.3tENtmtNtt-tmt,C,t,t谢谢阅读Rmtt=EN112XN

2122ttN12NtNt1212ENtNtNtNt22tt122212精品文档放心下载ENtNttt2tt2122212tt2ttttt222122212谢谢阅读t2tt2t2ttt2122212t22.4证明：PSSnn10由定理2.1.8TnPTn0~PSnS0S1Sn10S2PTn0a.s.2.514欢迎下载。精品文档PN1tsN2tsN1tN2tnPnm0N1tsN1tmN2tsN2tnmssnmnm1e2essm!nm!12m0nsemnmns1212m!nm!m0sntheoremsbinomial12精品文档放心下载e12n!2.61.PnNtntPNt/2nmm,NtNt/2n!etP

nNt/2m,NtttPNt/mPNNt/2mte22nmt2tnme2nm!2m!PnNt/m|Nt2PnNt/2m,NtPttmtnmnNte21tnmn!22e22m!nm!m!n

m!n!ntet215欢迎下载。精品文档tPkPnNtnnn!etPnN2tk,NtNtn,NNtnPNtntk!nPN2tettknetkn!NtknPnN2tk|NtPnN2tk,NtPnNtnttknn!etetkn!.n!tnettknkn!et2.7PNt8/60Nt0780ee78602.8ES5S2ET3T4T5T332

252.9PT1PT1022511ee842.1116欢迎下载。精品文档（1）covT,NTETNTETENT2（2）DNTDEEDNT|TNTTT222.12LetXNXii1EENttX=NtX|ENt2t2t谢谢阅读covNt,NtXtXtttt22ti1i2.13Pns|NtS1n1PNsnPNtNs0|NtnP0|Ns1|N

tPNstnPNs0PNtNsnPnNtPnNtesn!tsnets1sntPS1tnetn!s|Ntn11snt2.1417欢迎下载。精品文档PMmEPYSYSm|SmTSTEe2Tm!212122TTm=emmeT2T2e2T22m!m!

22

00mT0m!22eT2mTTm1|e=Te222m!0m1!2020Tm1em1!222T02e2T1PMmm2.17PN11123PTPN10PTPN321PN33PN31PN33PT1PN3122,T22319827112131627PT12.18LfsF2xex2sLmsLfs2s2s22s1Lfsmtte

11142t222.1918欢迎下载。精品文档Tn,xex~fxxT,,T1n1n,mintnttn,,tgs,1~ft,,t0,mint1n,,t1n1n,mint

nesn

,,t,sn0,mint1n,,tgnns1ngs,,s1ndsn1nnsn1est,,tnn1n1!PNtnPStntnsn1esn=n0n1!nen1n1ktketk!k0mint，,1,tn2.20证明：必要性：mtENtt充分性：=LtLmss2sfsL1LsLLmsfss1LT~f指数分布fsetNt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：12对于2k1的情况与k1的情况等价k1k1显然成立k=1k1证明：PXmpi|X0i,,Xmp1i

mp0

pm1PXmpi,

|Xmp1i,kpmpmp1kk1时的结论32k12显然成立23mt,Pm1Xti|Xtim1mt1i|Xt1m1PtX0i,,Xm1i|Xti=m1m0m1mtmiSi,,1inXti|Xt,Xt1iPXtiXt

1i|Xti,m10m110m11tttttmmm10m10m1iSPPiSnXm1i|Xt,,Xt1i,Xt1iXti,,Xt1im10m11m110

m1tttt0m1m10iPSiSiStm11iSn1Xm1iXt1i|Xti,Xti,m1m110ttmtm10mtm11Xm1i|Xti,Xti,m10tmt0m证毕3.2证明：pijPXm1j|Xmi，jPXnjrightside=PX0iPX1|X0PXni|Xn1in1side0n20欢迎下载。精品文档3.3证明PYn1j,j'|Y0i,,Yni,j,i01n感谢阅读PXn,Xn1j感谢阅读1j,Xn2j'|X0i,0谢谢阅读PXn,Xn1j感谢阅读1j,Xn2j'|Xnin谢谢阅读PYn,j1j,j'|Ynin3.4证明：假设第n次投掷硬币得到正面的事件为，那么第n次投币后得到正面的次数为谢谢阅读ni。X(n)Ynnk1其中，YX(nX(n)，因为。。。Yn之间相互独n1立，所以，Yn+1和（），。。。Xn）相互独立。谢谢阅读P{X(ni|Xi,X(2)i,...,X(n)i}n112nii|Xi,...,X(n)i谢谢阅读}n1n1n1nii}n1n1n精品文档放心下载同理P{X(ni|X(n)i}n1nn1in1i}n所以P{X(ni|Xi,X(2)n1i12,...,X(n)i}P{X(n|X(n)i}innn1所以，X(n)是markov而pijXj|Xmm1Pm1ji0ji)ji1其他21欢迎下载。精品文档0.50.5所以转移矩阵为P0.50.5....0.50.50.53.5p1,1111,p,pp1,0p0,12221,10,012p,p1,101,1111,p,p0,141,024PXn10|Xn1,Xn11精品文档放心下载PXn10|Xn1,Xn11PXn10|Xn1感谢阅读不是MC22欢迎下载。精品文档Xnnn12Pnn1j,nn1iPXn1j|Xni精品文档放心下载22Pnn1i21122jjj22212nnnn111

11121

2122n

nG1|1122jXi111G4241122Counterevidence:PX31|X20,X11,PX31|X0201322

2ThusitdisqualifiesMC.精品文档放心下载3.6S1,211,0P0.650.350.5551P40.61115,0.38885谢谢阅读PX523.7（1）PX03,X111111,X23（2）48823欢迎下载。精品文档p13231pp481rr3r13.10123456d11,d2doesn't，d3d4d5d61谢谢阅读3.11（1）11A是随机矩阵A感谢阅读Anm1m11111AAn1m1n1A111m1m1m1An（2）A是双重随机矩阵T是一般的随机矩阵n,T也是随机矩阵n由1的结论可知AAA=

nT

TnA,A是一般的随机矩阵TnnAn是双重随机矩阵3.12PX02112142111042,20P2209/57637/96145/576精品文档放心下载谢谢阅读EX2PX212PX223PX23谢谢阅读1103/5843.1324欢迎下载。精品文档PX1X01

1214812f3,f,f333521142114484f3235353521,f2f034f

f3121212123.14pz111f1nP

z1F1fjjn0z1z1n0n1p当j为非常返态，即fnn03.15充分性链所有状态都相通pi,j使得0由题意npp0nn

pn1pnp2n1pnn10pp精品文档放心下载n1|di，2n1|di精品文档放心下载di1是非周期的3.16（1）123正常返，d14（2）1234d33.17（1）SNC225欢迎下载。精品文档（2）SNC1C23（3）SNC7精品文档放心下载3.181jSpn1jCnp0njSC则CppnnjSjCp1jCn13.19413.21（1）S0,1,2，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精品文档放心下载有关，与过去无关。a122a1a1a1b1aa22P1ab22b21aa21bb1a1bababa21bbab1a1b22222（2）=P++012=,,26欢迎下载。精品文档3.22nn1nPnnnnnn1123nn,,n感谢阅读3.231是吸收态，2，,N)limjlimPXnjlimnj1精品文档放心下载NNnkkkjnnnk1k1nj1limnlim0Pn0limPnnnn0，0|1n精品文档放心下载3.24=P&1解得11111=，，，，555553.25=P,1123563

解得=，，

感谢阅读27欢迎下载。精品文档【第四章】4.1E2nn

nEaXXbYY2感谢阅读22nn

nEa2XXb2Y

Y2abXXYYnnnn精品文档放心下载n2ablimEXXYYnnn2ablimEXXEY精品文档放心下载0nY,CauchySchwarzInequality22nn精品文档放心下载Enn结论成立n024.2XtXt,limgtgt00tttt感谢阅读由定理4.1.6,gtXtgt精品文档放心下载0tt00Xt004.3（1）28欢迎下载。精品文档XttXt由题：存在00t0tXtXtXtXt感谢阅读E存在均方收敛准则Cauchy明精品文档放心下载tt010020t0t0t1t2不妨设t=t1212XtXtt2E010存在t0t211XttXtE2XtXtt=Et22010t0t0t01012111Xt2Xt感谢阅读XttXtt2t2E010t2E010=0tt1122t0t0t0精品文档放心下载11111XttXt00t0精品文档放心下载均方连续（2）Xttt0XtEX'tE00t0tXtXttt

tE000t0tt0tdEXt（3）aXtttbYttbYtaXbY'感谢阅读0000tttXttXtYttYt精品文档放心下载ab=aX''0000tttt（4）29欢迎下载。精品文档fttXttftXt谢谢阅读fX'ttt0ftXttXtfttftXt感谢阅读tt0fttftXttXt精品文档放心下载tt0XttXtfttftftXtttt0t0谢谢阅读f'tXtftX'tXttXtfttftt0

0X'tt04.4Xt'sinAt'EsinAt|A'E精品文档放心下载sinAt'|AEAcosAt|AAcosAt感谢阅读4.7均方不连续均方不可导解：{X（,t是泊松过程，所以有P{X(t)-X(s)k}e-s)[s)]kk!,k0,1,...,t现在令s=0，那么有X。当st有C)(s)s][X(t)X谢谢阅读(s)s]{[X(s)s][X(t)t][X(s)s]}}谢谢阅读(s)s]}2D[X(s)]2s当s>t时可得C(s,t)t。所以C(s,t)XXmin(s,t)，所以R(s,t)C(s,t)m(t)XX22因为Rst)在（ttt大于等于0处二元连续，泊松过程在t大于等于0是均方谢谢阅读X连续。讨论其均方可导性30欢迎下载。精品文档R(t,tl)(t,t)h,tl)Rt)

RRXXXXh0,l0h)2h)(tl)h)2h)tt2t(tl)t感谢阅读2t2h0,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：Xh)X)R(h)RE[XX'E[X]XR'(t)(0)h0Xhhh0（2）证明：E[X'dE[X(t)]0X(0)Xh)X(t)Xl)X)E[X'(t)X')]E[*]h0,l0hlXh)Xl)X(t)Xl)Xh)X)X(t)X)E[]h0,l0hl[R(lh)R(l)][R(h)R()]1h0,l0hlR'(l)R'()limR''()''()l0l得4.9EZtatEcosatisiniatsinEcosatcossinatsinsinatcoscos精品文档放心下载cosatisinaticosEsin=0atEat

cossinateEet,tEZtZt谢谢阅读tEeiiatRZZt4.1031欢迎下载。精品文档tlimClimt1Xt00Ct0XtC0limCXXt0tR0limRXXt02tmtR2C=tX000CXXCXX0limtlimCmtlimRt0XttXXtt04.11000CCXX2EXXEXX2EXX2XXnnnnn1n1n1n1E24EXX22222n2n2n2n22n2n1

222n221EX2n2X0011122EXEXEXnEXnE0nn00

n1n1XEX

EXXX

EXnnnnmnm1nmnm1m2EXX2mnn1是平稳过程4.12ENtN1ttt0感谢阅读ENtN1tNtN1t感谢阅读ENtNtN11tNtN1tNtN1tENtNttt1tENtN1ttN12精品文档放心下载ttttt2tt12t2精品文档放心下载2t1只看含有t的项,可以完全相消是平稳过程4.13证明：（X）0E[X(n)X(m)]cos(mj)]a)aNNcos(nkkkjk1j1j32欢迎下载。精品文档当k不等于j时，有E[aa)]cos(n)cos(mkjkkjj谢谢阅读aa)]E[cos(n)cos(mkjkkjj感谢阅读aaj20k20cos(n(k)mjj)kjcos(n()mjkjjk)*1dd42kj(kj)当k=j时，E[aakkcos(n)cos(m)]kkkkcos（*jda221d2k2004kk2a2cos（）k2kaN2E[X(n)X(m)]cos（m-n）

与n无关，因此为平稳过程。感谢阅读2k4.14证明：设（t）N(tL)N(t),t0谢谢阅读那么，E(X(t))E(X)X)][N)]E[N))E[NL)N(tL)N(t)N12121122E[NL)NL)]E[NL)]E[NL)N)N121212L)L)L))]22t2t2tt22