求教一道概率证明题设x y是相互独立的随机变量,证明(1)若E(X)=E(Y)=0,则D(XY)=D(X)D(Y),(2
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求教一道概率证明题
设x y是相互独立的随机变量,证明(1)若E(X)=E(Y)=0,则D(XY)=D(X)D(Y),(2)若E(X)=0或E(Y)=0,则D(XY)>=D(X)D(Y)
设x y是相互独立的随机变量,证明(1)若E(X)=E(Y)=0,则D(XY)=D(X)D(Y),(2)若E(X)=0或E(Y)=0,则D(XY)>=D(X)D(Y)
∵X,Y相互独立, ∴X^2,Y^2也相互独立
(1) D(XY)=E[XY-E(XY)]^2
=E(XY-EXEY)^2
=E(X^2Y^2)
=E(X^2)E(Y^2)
=E[(X-EX)^2]E[(Y-EY)^2]
=D(X)D(Y)
(2)不妨设E(X)=0,E(Y)可能等于0也可能不等于0
(EY)^2≥0
由(1)可知D(XY)=E[(X-EX)^2]E(Y^2)
≥E[(X-EX)^2][E(Y^2)-(EY)^2] (等号在EY=0时成立)
=E[(X-EX)^2][E(Y^2)-2(EY)^2+(EY)^2]
=E[(X-EX)^2]E[(Y-EY)^2]
=D(X)D(Y)
(1) D(XY)=E[XY-E(XY)]^2
=E(XY-EXEY)^2
=E(X^2Y^2)
=E(X^2)E(Y^2)
=E[(X-EX)^2]E[(Y-EY)^2]
=D(X)D(Y)
(2)不妨设E(X)=0,E(Y)可能等于0也可能不等于0
(EY)^2≥0
由(1)可知D(XY)=E[(X-EX)^2]E(Y^2)
≥E[(X-EX)^2][E(Y^2)-(EY)^2] (等号在EY=0时成立)
=E[(X-EX)^2][E(Y^2)-2(EY)^2+(EY)^2]
=E[(X-EX)^2]E[(Y-EY)^2]
=D(X)D(Y)
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