D(X-Y)+[E(X-Y)]^2=DX+DY-2(EXY-EXEY)+(EX-EY)^2怎么推导出来的?

我是上帝拽2022-10-04 11:39:542条回答

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sherrystt 共回答了22个问题 | 采纳率90.9%
对于D的定义
D(X)=E(X²)-【E(X)】²
于是就有
D(X-Y)=E((X-Y)²)-【E(X-Y)】²
D(Y)=E(Y²)-【E(Y)】²
从而消去
D(X-Y)+[E(X-Y)]^2=DX+DY-2(EXY-EXEY)+(EX-EY)^2所有的D,就得
E((X-Y)²)-【E(X-Y)】²+[E(X-Y)]^2=E(X²)-【E(X)】²+E(Y²)-【E(Y)】²-2(EXY-EXEY)+(EX-EY)^2
整理为E((X-Y)²)=E(X²)-【E(X)】²+E(Y²)-【E(Y)】²-2(EXY-EXEY)+【E(X)】²+【E(Y)】²-2EXEY
再整理E((X-Y)²)=E(X²)+E(Y²)-2E(XY)
分明有E((X-Y)²)=E(X²-2XY+Y²)=E(X²)+E(Y²)-2E(XY)
从而
D(X-Y)+[E(X-Y)]^2=DX+DY-2(EXY-EXEY)+(EX-EY)^2
得证
1年前
埃塞俄比亚的狐狸 共回答了40个问题 | 采纳率
从而
D(X-Y)+[E(X-Y)]^2=DX+DY-2(EXY-EXEY)+(EX-EY)^2
得证
1年前

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