Thread: 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

Hey noblewhale.

Hint: The expectation is distributive where E[(X-a)(Y-b)] = E[XY] - E[bX] - E[aY] + E[ab].

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!!