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.

  • $\begingroup$ If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html $\endgroup$
    – Anvit
    Jan 26, 2019 at 16:36

1 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


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