随机过程第1章习题参考答案

第1章解：2200020()cos0,()coscos()~(0,cos)EXttEXDXttDXtXtNt解：12t,1,1()20,X出现正面出现反面,1()~(1,)2XtB，0,0111(,){()},022211,2xFxPXtxxx1,1,()2,tXt出现正面出现反面，0,11(,1),1221,2xFxxx解：112233()(,)()(,)()(,)()1(1sincos)3EXtXtPXtPXtPtt12121112121222313231212(,)()()()(,)(,)()(,)(,)1()(,)(,)(1sinsincoscos)3XRttEXtXtPXtXtPXtXtPXtXttttt解：利用随机变量函数的分布密度的方法。()(),()XtxtXtegXgxe，其反函数为1lnxyt。()Xt的分布密度为()11()(ln)(0,0)Xtfyfytytty解：00111()|(1)TxtxtTTtEXtedxeeTTtTt，因为120,0tt，因此120tt1212121212120()()()0012121(,)()().111|(1)()()TxtxtXTxttxttTttTRttEXtXteedxTedxeeTTttTtt解：（1）12(),((),()XtXtXt均为离散随机变量，以下求离散分布律。()~(1,)XtBp。对于任意的的12,tt，12(),()XtXt的取值为0,1，且2121221{()0,()0}(1){()0,()1}{()0,()1}PXtXtpPXtXtPXtXtpq212{()1,()1}PXtXtp2121212(),(,)()()11{()1,()1}XEXtpRttEXtXtPXtXtp也可以求分布函数，但较为复杂。求()Xt的一维分布。0,0()~(1,),(,){()}1,011,1xXtBpFxtPXtxpxx求二维分布。12(),()XtXt是独立的，因此1212112211221122(,,,){(),()){()}.{()}(,).(,)FxxttPXtxXtxPXtxPXtxFxtFxt解：（1）11YX，分布列同1X（2）212YXX，可能取值为2,0,2，其概率如下2121{2}{1,2}4PYPXX，21212{0}{1,1}{1,1}2PYPXXPXX2121{2}{1,1}4PYPXX（3）120(...)0nnniiEYEXXXEX（4）,ijXX相互独立，且21iEX。1111min(,)2111(.)()min(,)nnmnmYnmijijijijnmnmijiijiREYYEXXEXXEXXEXmn解：()(()())()()()()XEYtEXttEXttmtt1212112212122112(,)()()(()())(()())()()()()()()()()YRttEYtYtEXttXttEXtXttEXttEXttt1212122112(,)()()()()()()()()XXXXXCttmtmttmttmttt12121122(,)(,)(()())(()())YYXXCttRttmttmtt=12(,)XCtt证明：()1{()}(,)XEYtPXtxFxt12121212(,)()(){(),()}(,,,)YXRttEYtYtPXtxXtxFxxtt解：解：200011()cos()cos()022EXtEAEttd0212102220102020120120012012(,)cos()cos()1cos()cos()211[cos(()2)cos()]34112cos()cos()126XRttEAttEAttttttdtttt解：为随机变量，根据定义有，当0，()1tXt时，1EX，当0t时0000001122112212100201()cos()cos1111sin|[sin()sin()]2221cossin()2tttttEXttfdtdttttttt00000000121121221211212211221211211212221212012121(,)coscos1[cos()cos()]2111{sin()|sin()|}211(){sin()cos21tXtttttttRttttdttttdttttttttttttttttt12120()sin()cos}2ttttt上式假设12120,0tttt。若120tt，121212012121212121201212121111()(,)sin()cos220,0，111()(,)sin()cos220,0,(0,0)1XXXtttRtttttttttttttRtttttttttttR2时有当即=t=0时00001211210122021212112012222021212(,)(coscossin)22(coscossin)211coscos[coscoscossin()]24cossin2tXttttCttttttdttttttdttttt解()EXtEXa。根据协方差的定义有1211222(,)[()()][()(()]()()XCttEXtEXtXtEXtEXaXaDX解：由定义知，121122121212(,)[()][()][()()][()()]()(,)XCttEXtYEXtYXtYEXtYEXEXtYEYXEXtYEYDXttCovXYttDY解：与14题类似，根据定义计证明：00101()()()[()][()l.i.m]1lim{()[()()]}1(,)lim[(,)(,)]ttXXXtdXtXttXtEXtEXtdttEXtXttXttRttRtttRtttt解：111001100[()][()]102EZEXtdtEXtYdtEXtdtEYdtEXEY11122220002222[()]21142323EZEXtYdtEXtdtEXYtdtEYEXEXYEY解解：（1）00()()()()()[][..]limttdXtXttXtEXttEXtEElimdttt()dEtdt（2）利用均方收敛的性质。0000''()()()()..[()()][()()]..()()()()......()()tttaXttbYttaXtbYtlimtaXttXtbYttYtlimtXttXtYttYtalimblimttaXtbYt证明：利用均方极限的性质，有101101()()..()()()lim()()()()()nbkkkkaknbkkkkakEftXtdtElimfuXuttfuEXuttftEXtdt证明：根据均方极限的定义，即要证明00020020002222000lim|()()()()|lim|()(()()()(()()|2{lim()|()()|()|()()|0ttttttEgtXtgtXtEgtXtXtXtgtgtgtEXtXtEXtgtgt