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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- Oct 4th 2012, 04:20 PMnoblewhalecov(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 - Oct 4th 2012, 05:29 PMchiroRe: 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]. - Oct 5th 2012, 07:08 PMnoblewhaleRe: 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!!