Prove that $\operatorname{cov}\left[x - E(x\mid y), E(x\mid y)\right] = 0$

Question

Contents below are what I have done.

\begin{align} \operatorname{cov}\left[E(x\mid y),x - E(x\mid y)\right] &= E[xE(x\mid y)] - E(E^2(x\mid y)) - E(x)E(x - E(x\mid y)) \\&=E\left[xE(x\mid y)\right] - E(E^2(x\mid y)) - E(x) [E(x) - E(x)] \\&= E\left[xE(x\mid y)\right] - E(E^2(x\mid y)) \end{align}

and then I don't know how to deal with this equation. Please tell me how to do next and the reason. Thanks.

Answer 1

Conditioning on $y$ we get $E[x(E(x|y)]=E(E(x(E(X|y)|y)))$. But $E(x(E(X|y)|y))=E((E(x|y)(E(x|y))=E[(E(x|y)]^{2}$. Just use this in the first line of your proof.