Is this true: $\operatorname{Cov}(X,Y)=\operatorname{Cov}(E[X|Y],Y)$? Covariance of variables versus conditional expectations?

Question

Let $X$ and $Y$ be to correlated random variables. Is the following true?

$\operatorname{Cov}[E[X|Y],Y]=\operatorname{Cov}[X,Y]$

I derived this for from the standard expression for covariance, and it holds if $E[E[X|Y]*Y]=E[XY]$ is correct, but somehow this seems wrong to me.

Answer 1

Note that $$ YE(X\mid Y)=E(XY\mid Y) $$ by the pull out property of conditional expectation whence $$ E[Y(X\mid Y)]=E[E(XY\mid Y)]=EXY $$ by the tower law.

Answer 2

It is correct. There is nothing wrong.

$\begin{align}\mathsf{Cov}(\mathsf E(X\mid Y),Y)&=\mathsf E(\mathsf E(X\mid Y)~Y)-\mathsf E(\mathsf E(X\mid Y))~\mathsf E(Y)&&\text{definition of covariance}\\&=\mathsf E(\mathsf E(XY\mid Y))-\mathsf E(\mathsf E(X\mid Y))~\mathsf E(Y)&&\text{distributivity of conditional expectation}^\star\\&=\mathsf E(XY)-\mathsf E(X)~\mathsf E(Y)&&\text{law of total expectation}^\dagger\\&=\mathsf{Cov}(X,Y)&&\text{definition of covariance}\end{align}$

$\star$ : also known as the "pull out property" of conditional expectation, which is inherited from summation and integration.

$\dagger$ : also known as the "tower property".

So for a pair of continuous random variables with joint, conditional, and marginal pdf, we have: $$\small\begin{align}\mathsf {Cov}(\mathsf E(X\mid Y),Y) &=\int_\Bbb R\left[\int_\Bbb R x f_{\small X\mid Y}(x\mid y)~\mathrm d x\right] y~f_{\small Y}(y)~\mathrm d y-\int_\Bbb R\left[\int_\Bbb R x f_{\small X\mid Y}(x\mid y)~\mathrm d x\right] f_{\small Y}(y)~\mathrm d y\cdot\int_\Bbb R y f_{\small Y}(y)~\mathrm d y\\[1ex] &=\int_\Bbb R\left[\int_\Bbb R xy~ f_{\small X\mid Y}(x\mid y)~\mathrm d x\right] f_{\small Y}(y)~\mathrm d y-\int_\Bbb R\left[\int_\Bbb R x f_{\small X\mid Y}(x\mid y)~\mathrm d x\right] f_{\small Y}(y)~\mathrm d y\cdot\int_\Bbb R y f_{\small Y}(y)~\mathrm d y\\[1ex]&=\iint_{\Bbb R^2} xy~f_{\small X,Y}(x,y)~\mathrm d (x,y)-\iint_{\Bbb R^2} x~f_{\small X,Y}(x,y)\mathrm d (x,y)\cdot\int_\Bbb R y~f_{\small Y}(y)~\mathrm dy\\[1ex]&=\mathsf {Cov}(X,Y)\end{align}$$