Question

已知X服从N(1,9),Y服从N(0,16),且X与Y的相关系数p(xy)=-1/2,设Z=1/3X+1/2Y,求方差DZ=?

Answer 1

X~N(1,9) Y~N(0,16)
EX=1 DX=9 EY=0 DY=16
DX=E(X^2)-(EX)^2 所以E(X^2)=10
DY=E(Y^2)-(EY)^2 所以E(Y^2)=16
p(XY)=Cov(X,Y)/√DX*√DY 所以Cov(X,Y)=-6
Cov(X,Y)=E(XY)-(EX)(EY)
所以E(XY)=-6
DZ=D(Z^2)-(DZ)^2
=E[(1/3X+1/2Y)^2]-[E(1/3X+1/2Y)}^2
=E[1/9X^2+1/4Y^2+1/3XY]-[1/3EX+1/2EY]^2
=(1/9EX^2+1/4EY^2+1/3EXY)-(1/3EX+1/2EY)^2
=(1/9*10+1/4*16+1/3*(-6))-(1/3*1+1/2*0)^2
=3
上面的又是复制党= =