Question

概率论问题....

设随机变量X,Y相互独立,且E(X)=E(Y)=1,D(X)=2,D(Y)=3,求D(XY)

Answer 1

因为随机变量X，Y相互独立
所以E(XY)=E(X)E(Y)
D(X)=E(X^2)-E^2(X)
D(Y)=E(Y^2)-E^2(Y)
因为E(X)=E(Y)=1，D(X)=2，D(Y)=3
所以E(X^2)=2+1^2=3，E(Y^2)=3+1^2=4
所以D( XY )
= E[ XY - E(XY) ]^2
= E[ (XY)^2 - 2XYE(XY) + E^2(XY) ]
= E[ (XY)^2 ] - E[ 2XYE(XY) ] + E[ E^2(XY) ]
= E[ (X^2Y^2) ] - 2E(XY)E[ XY ] + E^2(XY)
= E(X^2)E(Y^2) - 2E^2(XY) + E^2(XY)
= E(X^2)E(Y^2) - E^2(X)E^2(Y)
= 3*4 - 1*1
= 11

Answer 2

D(XY)=E{[XY-EXEY]^2}=E(X^2)*E(Y^2)-(EX)^2*E(Y)^2
=3*4-1=11
其中：E(X^2)=D(X)+[E(X)]^2=2+1=3
E(Y^2)=D(Y)+[E(Y)]^2*=3+1=4