方兆本_随机过程_答案.pdf

nX i=1 X i) 3] = nX j=1 nX i=1 nX k=1 E(X iX jX k) = 0, E(Y 4|N=n) =E[( NX i=1 X i) 4|N=n] =E[( nX i=1 X i) 4|N=n] =E[( nX i=1 X i) 4] = nX j=1 nX i=1 nX k=1 nX l=1 E(X iX jX kX l) = nX i=1 E(X 4i) + nX i,j=1,i6=j E(X 2iE(X 2j) =n 2, Y E(Y) =E[E(Y|N)] = 0, E(Y 2) =E[E(Y 2|N)] =E(N) 10 : =β∞ X n=1 n(1?β) n?1= 1β, E(Y 3) =E[E(Y 3|N)] = 0, E(Y 4) =E[E(Y 4|N)] =E(N 2) =β∞ X n=1 n 2(1?β) n?1= 1β. 2 , n, E(e tY|N=n) =E(exp{t NX i=1 X i} |N=n) =E(exp{t nX i=1 X i} |N=n) =E(exp{t nX i=1 X i}) = [E(e tX 1)] n = ?12 (e ?t+e t) ? n, E(e tY|N) = ?12 (e ?t+e t) ? N, Y g Y(t) =E(e tY) =E[E(e tY|N)] =β∞ X n=1 ?12 (e ?t+e t) ? n(1?β) n?1 = β(e ?t+e t) 2?(1?β)(e ?t+e t) . Y E(Y) = dg Y(t) dt | t=0= 0, E(Y 2) = d 2g Y(t) dt 2 | t=0= 1β, E(Y 3) = d 3g Y(t) dt 3 | t=0= 0, E(Y 4) = d 4g Y(t) dt 4 | t=0= 1β.