0
$\begingroup$

The expectation of the product of two independent random variables $X$ and $Y$ is the product of the expectations: \begin{align} E(XY) = E(X)E(Y) \end{align} Let's add another random variable $Z$ in the mix, do we have the following equality: \begin{align} E(XY \vert Z) = E(X \vert Z)E(Y \vert Z) \end{align}

$\endgroup$
  • $\begingroup$ $E(XY) = E(X)E(Y)$ is, generally speaking, not true. $\endgroup$ – Viktor Mar 2 at 9:09
  • $\begingroup$ Yes of course, I forgot to write the independent hypothesis, just edited the post $\endgroup$ – Victor Mar 2 at 9:13
  • $\begingroup$ The short answer is: you need conditional independence of $X$ and $Y$ given $Z$. Independence of $X$ and $Y$ is insufficient. See stats.stackexchange.com/a/184933/80704 $\endgroup$ – Viktor Mar 2 at 9:24
3
$\begingroup$

Definitely not true (in general). For example, let's say, $X,Y$ are the two different coin tosses. And, $Z$ is the number of heads you get after these tosses. Apparently, $X$ and $Y$ are independent; but given $Z$, they're not. Also, independence means $E[XY]=E[X]E[Y]$, but having it holds doesn't mean they're independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.