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 $\displaystyle E[f(X)E[Y|X]]=E[f(X)Y]$ for any function $\displaystyle f$ (that's a characterization of the conditional expectation).
You can use the above equality with $\displaystyle f(X)=X-E[X]$ : it gives $\displaystyle E[(X-E[X])E[Y|X]]=E[(X-E[X])Y]$. Since additionaly $\displaystyle E[E[Y|X]]=E[Y]$, 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!