Covariance Identity: $\operatorname{Cov}(X,Y) = E(\operatorname{Cov}(X,Y\mid Z)) + \operatorname{Cov}(E(X\mid Z),E(Y\mid Z))$

Question

How can I show that:$\DeclareMathOperator{\Cov}{Cov}$

$\Cov(X,Y) = E(\Cov(X,Y\mid Z)) + \Cov(E(X\mid Z),E(Y\mid Z))$?

With $X, Y$ and $Z\;$ r.v with finite variances.

Answer 1

Here's a ugly looking jumble of Expectations

$$\begin{align}E[\DeclareMathOperator{\Cov}{Cov}\Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\\[2ex] &=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\\[2ex] &=E[XY]-E[E[X|Z]E[Y|Z]] \end{align}$$ Similarly, $$\begin{align} \Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\\[2ex] &=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\\[2ex] \end{align}$$

Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property