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Math Help - cov(x,y)=E[(x-Ex)(y-Ey)]=E[(x-Ex)y]=E[X(Y-EY)]

  1. #1
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    cov(x,y)=E[(x-Ex)(y-Ey)]=E[(x-Ex)y]=E[X(Y-EY)]

    Help anyone:

    How to prove

    cov(x,y)=E[(x-Ex)(y-Ey)]=E[(x-Ex)y]=E[X(Y-EY)]

    Thanks
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  2. #2
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    Re: cov(x,y)=E[(x-Ex)(y-Ey)]=E[(x-Ex)y]=E[X(Y-EY)]

    Hey noblewhale.

    Hint: The expectation is distributive where E[(X-a)(Y-b)] = E[XY] - E[bX] - E[aY] + E[ab].
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  3. #3
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    Re: cov(x,y)=E[(x-Ex)(y-Ey)]=E[(x-Ex)y]=E[X(Y-EY)]

    ok...

    E[(x-Ex)(y-Ey)]
    = E(xy-xEy-yEx+ExEy)
    = E(xy)-ExEy-EyEx+ExEy
    =E(xy)-ExEy

    E[(x-Ex)Y]
    =E(xy-yEx)
    =E(xy)-EyEx

    got it! Thank you Chiro!!
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