Question

设随机变量X1,X2,X3,X4相互独立,且具有相同的分布,数学期望为0,方差为B^2,令 X=X1+X2+X3,

Answer 1

E(X)=E(X1+X2+X3)=E(X1)+E(X2)+E(X3)=0,同理E(Y)=0

E(XY)=E(X2^2)+E(X3^2)=2B^2

Cov(X,Y)=E(XY)-E(X)*E(Y)=2B^2

D(X)=D(X1+X2+X3)=D(X1)+D(X2)+D(X3)=3B^2

D(Y)=D(X2+X3+X4)=D(X2)+D(X3)+D(X4)=3B^2

Pxy=Cov(X,Y)/√D(X)*√D(Y)=2B^2/3B^2=2/3