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B)(tX}),({},),({TttYTttX∈∈),()sin()(+∞−∞∈Θ+=ttAtXω()1,1−−ωπΘ)1,1(−Θπω)]([tYE)]([tYDCC−,2121)(ipCCtX−λ{}0),(≥ttX}1),({≥nnXpq−=1}0),({≥nnX31})0({==iXP3,2,1=i1232121211213)3(,2)1({==XXP}2)5(,=X)(tX}),({},),({TtYTttX∈∈tX(t)t,t,X(t),0,1,2,…,t,}0|{≥=ttT,ntX=)(},2,1,0{L=E,}0),({≥ttXY(t)t,t,Y(t),t,0≥t,}0|{≥=ttT,stX=)(}0|{≥=ssE,}0),({≥ttY.(1)A(-1,1),,()sin()()XtAttRω=+Θ∈,ωΘ()Xt2()Xt112A=±111(sin(xtω=+Θ2))21()sin()2txttω=−+ΘΘ(0,2),,UAπω()12,ππ=2()sin()cos()2sin()sin()XtAtAtXtAtAtπωωωπωΘ=+==+=−2πω2πω2)]([tYE)]([tYD(1)N(t),j()jNt123()()()()()NtNtNtNtNt⇒=++(6)tππ123232323()3),()(2),()()()[0,)()10()15()()()10()15()531021550()2()100()225()2531002225500NttNttNttYtttNtNtEYttENtENtttttDYttDNtDNtttttππ++∴=++=×+×+==++=⋅+⋅+=111Y(t)=5N5EN5DN()()()NtYtXn=∑100221()[(()/())](()/())(())5050(())(())5066(2)()(()())[[(()())/()]]500(())5006nkkEYtEEYtNtEYtNtkPNtkkPNtkENttDYtEYtEYtEEYtEYtNtENtt=∞=∞==========−=−==∑∑CC−,2121)(ipCCtX−λ}0),({≥ttX1tEX(t)=02+∞=22)(ctEX321tt})()({})()({)()(),(221222122121ctXtXPcctXtXPctXtEXttRX−=−===})({})({122122=−−=−=ttNPcttNPc)(0121220)(21221212)!12())(()!2())((ttkkkttkekttcekttc−−+∞=++∞=−−∑∑+−−−=λλλλ)(!))((122)(22)(01221212ttRcecekttcXttttkk−==−−=−−−−+∞=∑λλλ,,12tt)(),(21221ttRcttRXX−=,,)(),(122222112ttRcecttRXttX−==−−λ,,t}0),({≥ttX}1),({≥nnX:}1),({≥nnX,nnnnmL21)(),(,),(),(21nXnXnXnXmL})(,)(|)({),,,,,,,|,(112121mmmnxnXxnXxnXPnnnxxxnxF≤≤≤=LLL})(,)({})(,)(,)({1111mmmmxnXxnXPxnXxnXxnXP≤≤≤≤≤=LL})({})({})({})({})({})({})({})({1111mmmmmmmmxnXPxnXPxnXPxnXPxnXPxnXPxnXPxnXP≤≤≤=≤≤≤≤≤=LL},|,{})(|)({})({})(,)({mmmmmmmmnxnxFxnXxnXPxnXPxnXxnXP=≤≤=≤≤≤=,pq−=1N1,2,…..,NX(n)nX(n-1)X(n-1),}1),({≥nnXE={1,2,......,N},}1{≥=nT}1),({≥nnX1N((1)/())111ijpPXkjXkipjiqjiiNo=+===+⎧⎪==−⎨⎪⎩1((1)/()1)2((1)/())11jNjpPXkjXkpjqjNopPXkjXkNpjqjNo=+===⎧⎪==⎨⎪⎩=+===⎧⎪==−⎨⎪⎩{}0),(≥nnX31})0({==iXP3,2,1=i123212121121}2)5(,=X3)3(,2)1({==XXP1⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=0012/12/102/12/10P⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=2/12/104/14/12/14/14/12/10012/12/102/12/100012/12/102/12/102P2}2)5(,3)3(,2)1({===XXXP}3)3(|2)5({}2)1(|3)3({}2)1({======XXPXXPXP2412141]02121[31])0([)2(31)2(23231=××++==∑=ppppiiiA2003.1.6),()sin()(+∞−∞∈Θ+=ttAtXω)1,1(−ωΘΘπω},cos)({RttAtX∈=ω3=λ)},(sincos)({+∞−∞∈+=ttBtAtXωωω0)()(==BEAE2)()(σ==BDADA)10(ppX(n)nA1{},2,1),(L=nnX12{},2,1),(L=nnXR11234657892:MMYnM,n⎩⎨⎧−=MMXi11L,2,1=iM,21}1{}1{=−===XPXPt=n,M1nniiYX==∑,,}1,{≥=nnT,},2,1,0{L±±=E{,1}nYn≥.(1)A(-1,1),,()sin()()XtAttRω=+Θ∈,ωΘ1()Xt2()Xt112A=±111(sin(xtω=+Θ2))21()sin()2txttω=−+Θ1Θ(0,2),,UAπω()12,2()sin()cos()2sin()sin()XtAtAtXtAtAtπππωωωπωΘ==+==+=−22πω2πω23{()cos,}XtAttRωω=∈AN(0,1)t0X(t)A0AN(0,1)(tX0)0(0,(cos))Ntω212=((),())RXtXt212(coscosEAttωω)])ttDAAωω=+212coscos([E1,0,DAEA==12coscosttωω=12121(2),())((),())()()XXCXtXtRXtXtmtmt=−0Co2tX2ωω=(,()XtRmt∀∈=))(),((v1tX1ttcoscos3=λ(3)tπ773515(35((5)7N==)15)7!7!Pee−×−×=153315(3)((3)3)!nkknPPNekττ∞−=≤=≥=∑()2100001cost0cos2cos2xttωωωπ⎧⎫⎛⎞⎪⎪−≠⎨⎬⎜⎟⎝⎠⎪⎪⎩⎭11(,)expfxt=⋅()20001expcost0cos2cos2xdxttωπ⎧⎫⎛⎞⎪⎪−≠⎨⎬⎜⎟⎝⎠⎪⎪⎩⎭1011(,)xFxtωω−∞=⋅∫(2)415215151215151−−−××−×−−=eee=××++−=−152)1215151(1e99996.05.128115=×−−e)},(sincos)({+∞−∞∈+=ttBtAtXωωω0)()(==BEAE2)()(σ==BDAD1()cossinEXttEAtEBωω==0222222222(sincos)sin2ttσωω=2()(cossin2sincos)EXtEAtBtABtttEAEBωωωωωσ=++++⋅⋅=∞31211222212121212221()()(cossin)(cossin)(coscossinsin(sincoscossin))cos()EXtXtEAtBtAtBtEAttBttABttttttωωωωωωωωωωωσω=++=+++=−ωA)10(ppX(n)nA1},2,1),({L=nnX2},2,1),({L=nnXL,1=iAY2,01⎩⎨⎧=AiiiX(n)∑==niiYnX1)(L,2,1=n4321nnnn≤∀1∑+=−21112)()(nniiYnXnX=A1n2nA∑+==−43112)()(nniiYnXnX3n4n}1,{≥iYi{(),1,2}Xnn=L52X(n)nA{(),1,2}Xnn=LiYX01pi1-pp+−=+−==1)1()()(0ϕitititYYpeppepeeEtiiX(n)YitnitXnXpEEeEtnii]1[)))()(1)()(1+−==∑==∏==ϕnk−=)0nitniitYei(pee(knknkknitkppCe−=−=∑)1(0kitknppeC−=∑1()(kn{(),1,2}Xnn=L(())(1),0,1,2kknknPXnkCppkn−==−=LR1X(n)nX(n)1,2,3,…,9,E={1,2,3,4,5,6,7,8,9},,{≥=nnT}1}1),({≥nnXk+1X(k+1)kX(k)k}1),({≥nnX1100000002200110000000221100000221100000002211000000022111000000333000000010⎛⎞⎟⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠1100000002211000002200⎜⎟⎜⎟⎜⎟⎜⎜6
本文标题:随机过程习题与答案
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