设随机变量X~N(0,4),Y~N(0,4),并且X和Y相互独立，求D(X+Y)和D(2X-3Y)

Question

设随机变量X~N(0,4),Y~N(0,4),并且X和Y相互独立，求D(X+Y)和D(2X-3Y)

Answer 1

D(X+Y)=E[(X+Y)^2]-E^2(X+Y).因为X与Y相互独立，所以E(XY)=E(X)E(Y);故上式等于: E(X^2+2XY+Y^2)-[E^2(X)+E^2(y)]=E(X^2)+E(Y^2)+2E(XY)-[E^2(X)+E^2(y)]; 又因为E(X^2)=E(Y^2)=D(X)+E^2(X)=4,所以结果是D(X+Y)=4+4+...