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

[x-mX(t)]2 (x;t二1自相關(guān)函數(shù)："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)為確定信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，只與tst例

0X(t)

-

0已知隨機(jī)點(diǎn)在(ab]內(nèi)出現(xiàn)的次數(shù)N(ae-ltP{N(t0,t0+t]=k} ?pk(t) mX RX(s,t). mX(t)=1P{X(t)=1}+(-1)P{X(t)=-=1(1

+

+)e-

+)e-

不妨設(shè)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偶數(shù)次

PN(st奇數(shù)次X去掉條件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 對于兩個(gè)隨機(jī)過程{X(t),t?T}和{Y(t),t?T如果對"s,t?T 都RXY(s,t)”則稱X(t)與Y(t)為正交過程X(t)與Y(t)正 "s,t?T RXY(s,t)”X(t)與Y(t)不相 "s,t?T CXY(s,t)”"m,n?N,"ti,tj?t)X(t)與Y(t)獨(dú)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)特別地，當(dāng)X(t)與N(t)RW(s,t)=RX(s,t)+RN(s,t)四 復(fù)隨機(jī)過程Z(t)為Z(t)=X(t)+jY(t其中X(t、Y(t)mZ(t)=E[Z(t)]?mX(t)+jmYZs2(t)?Z

Z(t)-

(t)=E{[Z(t)-mZ(t)][Z(t)-mZ(t)}RZ(s,t)?E{[Z(s)][Z(tCZ(s,t)?E{[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 滿mX(t)=常數(shù) (b)RX(s,t)DRX(s-t

￡t<t?T,"t1<t (s,t)=R(s,t)=s2(min(s,t XX記mminst其中s2(minsts2(mX2獨(dú)立增量過程.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離散型：設(shè)隨機(jī)過程{X(t),t?T}的狀態(tài)空間是可數(shù)集E,若對于任意正整數(shù) 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連續(xù)型：設(shè){X(t),t?T}為連續(xù)型隨機(jī)過程 ,若對于任意正整數(shù)n及t1 <t2 <<tn <t?T,

n及 << <t?T,因?yàn)閄(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}為Markov設(shè)X(t(t0)A在(0t次數(shù)且滿足X(0)=0￡st(l(t- -lP{X(t)-X(s)=k}

=特別地，當(dāng)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 的證明: 任取正整數(shù)n及"0￡t1 ￡t2 ￡￡tn，則X(t1)-X(0),X(t2)-X(t1),,X(tn)-X(tn-1)X(t

0X(t1)-X(0) 因?yàn)? =

0X(t2)-X(t1) X

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

)-X(t1)

N

)