随机过程第一章

mX(t)=E[X(t)]=xdFX(x;t

[x-mX(t)]2 (x;t二1自相关函数："s,t˛T RX(s,t)=E[X(s)X(t CX(s,t)=E{[X(s)-mX(s)][X(t)-mXCX(s,t)=RX(s,t)-mX(s)mX）mX(t)”0CX(s,t)=RX(s,t)2）s=ts2(t)=C(t,t)=R(t,t)-m2 YP1例1已知g(t)为确定信YP1X(t)Yg(t

s2(t

(s,t

(s,tXXXX g2(t g(s)g(t g(s)g(t例2已知X(t)acos(wtqa0,wmX(t RX(s,t

0

+q)1dq=

(s,t)

0

2cos[w(st，只与tst例

0X(t)

-

0已知随机点在(ab]内出现的次数N(ae-ltP{N(t0,t0+t]=k} ˆpk(t) mX RX(s,t). mX(t)=1P{X(t)=1}+(-1)P{X(t)=-=1(1

+

+)e-

+)e-

不妨设s£tRX(s,t)=E[X(s)X(t=P{X(s)=1,X(t)=1}+P{X(s)=-1,X(t)=--P{X(s)=1,X(t)=-1}-P{X(s)=-1,X(t)==P{X(s)=1,X(t)-X(s)=+P{X(s)=-1,X(t)-X(s)=-P{X(s)=1,X(t)-X(s)=--P{X(s)=-1,X(t)-X(s)=-=P{X(t)-X(s)=1}-P{X(t)-X(s)=-PN(st偶数次

PN(st奇数次X去掉条件s£t， (s,t)=e-2X

三

˛T RXY(s,t)=E[X(s)Y(t"s,t˛T,CXY(s,t)=E{[X(s)-mX()]Yt)-mY定 对于两个随机过程{X(t),t˛T}和{Y(t),t˛T如果对"s,t˛T 都RXY(s,t)”则称X(t)与Y(t)为正交过程X(t)与Y(t)正 "s,t˛T RXY(s,t)”X(t)与Y(t)不相 "s,t˛T CXY(s,t)”"m,n˛N,"ti,tj˛t)X(t)与Y(t)独立s,t˛TE[X(s)Y(t)]=mX(s)mYRW(s,t)=RX(s,t)+RXN(s,t)+RNX(s,t)+RN(s,t)特别地，当X(t)与N(t)RW(s,t)=RX(s,t)+RN(s,t)四 复随机过程Z(t)为Z(t)=X(t)+jY(t其中X(t、Y(t)mZ(t)=E[Z(t)]ˆmX(t)+jmYZs2(t)ˆZ

Z(t)-

(t)=E{[Z(t)-mZ(t)][Z(t)-mZ(t)}RZ(s,t)ˆE{[Z(s)][Z(tCZ(s,t)ˆE{[Z(s)-mZ(s)][Z(t)-mZ(t

RZ(s,t)=RZ(t, CZ(s,t)=CZ(t,CZ(s,t)=RZ(s,t)-mZ(s)mZ(t 满mX(t)=常数 (b)RX(s,t)DRX(s-t

£t<t˛T,"t1<t (s,t)=R(s,t)=s2(min(s,t XX记mminst其中s2(minsts2(mX2独立增量过程.t1<t2£t3<t4££t2k<t2k+1£˛T,XX (s,t)=s2(min(s,tXX2XCX(s,t)=RX(s,t)=s(min(s,tXt1,t2,t1+s,t2s˛T, 1E离散型：设随机过程{X(t),t˛T}的状态空间是可数集E,若对于任意正整数 n及t1<t2 <<tn <t˛T, P{X(t1)=i1,X(t2)=i2,,X(tn)=in}>P{X(t)=i/X(t1)=i1,X(t2)=i2,,X(tn)=in=P{X(t)=i/X(tn)=in ，t E连续型：设{X(t),t˛T}为连续型随机过程 ,若对于任意正整数n及t1 <t2 <<tn <t˛T,

n及 << <t˛T,因为X(0)=另一方面，PX(ti/X(tninXt)所以：P{X(tiX(t1i1X(t2i2,X(tnin=P{X(t)=i/X(tn)=in即X(t，t˛T}为Markov设X(t(t0)A在(0t次数且满足X(0)=0£st(l(t- -lP{X(t)-X(s)=k}

=特别地，当s0

P{X(t)-X(0)=k} t,

=(tEX(t)=DX(t)=RX(s,t)=l2st+l

(s,t)=l

(tX(00; s,t0X(tX(s)~Ns2ts(tEX(t)=0,DX(t)=s

(s,t)=

(s,t)=s2 的证明: 任取正整数n及"0£t1 £t2 ££tn，则X(t1)-X(0),X(t2)-X(t1),,X(tn)-X(tn-1)X(t

0X(t1)-X(0) 因为 =

0X(t2)-X(t1) X

) 1X(tn)-X(tn- X(t1)-X(0) 0 X

)-X(t1)

N

)