设X,Y,Z是三个随机变量,已知E(X)=E(Y)=1,E(Z)=-1;D(X)=D(Y)=D(Z)=1; ρxy=0,ρxz=1⼀2,ρyz=-1⼀2,

求E(X+Y+Z)和D(X+Y+Z)
2025-05-21 15:20:10
推荐回答(1个)
回答1:

ρ(x,y)=cov(x,y)/(√D(x)√D(y))=[E(X,Y)-E(X)E(Y)]/2=0
cov(x,y)=0
同理
cov(x,z)=1
cov(y,z)=-1
E(W)=E(X)+E(Y)+E(Z)=1
D(W)=D(X+Y+Z)=D(X)+D(Y)+D(Z)+2cov(XY)+2cov(XZ)+2cov(YZ)=6+2-2=6
E(W²)=D(W)+E²(W)=7
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