概率论问题....
2个回答
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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
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
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