Since E[Y] = E[E[Y/X]], can i infer that Y = E[Y/X]? Therefore i conclude that:
Cov(X, E[Y/X])
= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
= E[(X - E[X])(Y- E[Y])]
= Cov(X, Y)
Thanks in advance!
No you can't. But, it is true that for any function (that's a characterization of the conditional expectation).
You can use the above equality with : it gives . Since additionaly , this justifies your second equality.Therefore i conclude that:
Cov(X, E[Y/X])
= E[(X - E[X])(E[Y/X] - E[E[Y/X]])]
= E[(X - E[X])(Y- E[Y])]
= Cov(X, Y)
Thanks in advance!