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}

  • $\begingroup$ $E(XY) = E(X)E(Y)$ is, generally speaking, not true. $\endgroup$ – Viktor Mar 2 '19 at 9:09
  • $\begingroup$ Yes of course, I forgot to write the independent hypothesis, just edited the post $\endgroup$ – Victor Mar 2 '19 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 '19 at 9:24

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.

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