Now I know that the variance of the product of two independent variables $Y$ and $Z$ is:$\DeclareMathOperator{\Var}{Var}$

$\Var(YZ) = \Var(Y)\Var(Z) + \Var(Y)E(Z)^2+\Var(Z)E(Y)^2$

However I would like to know what the variance of the product of $Y\mid X$ and $Z\mid X$ is (assuming they are independent as well)? Is it the same as above? Which in turns give me:

$\Var(YZ\mid X) = \Var(Y\mid X) \Var(Z\mid X) + \Var(Y\mid X) E(E[Z\mid X])^2 + \Var(Z\mid X) E(E[Y\mid X])^2$

where I know that applying total expectation law by $E(E[Z\mid X]) = E(Z)$ such that I get:

$\Var(YZ\mid X) = \Var(Y\mid X) \Var(Z\mid X) + \Var(Y\mid X)E(Z)^2 + \Var(Z\mid X)E(Y)^2$

or do I have to apply the total variance law somehow?

$\Var(Y) = E[\Var(Y\mid X)] + \Var[E(Y\mid X)]$

Or is it some other method I have use to get the desired equation? Any help or input is appreciated!


1 Answer 1


We can follow a similar derivation, assuming conditional independence of $Y$ and $Z$ given $X$: $$\begin{align}\operatorname{var}(YZ|X)&=E[Y^2|X]E[Z^2|X]-E[Y|X]^2E[Z|X]^2\\&=(\operatorname{var}(Y|X)+E[Y|X]^2)(\operatorname{var(Z|X)}+E[Z|X]^2)-E[Y|X]^2E[Z|X]^2\\&=\operatorname{var}(Y|X)\operatorname{var}(Z|X)+\operatorname{var}(Y|X)E[Z|X]^2+\operatorname{var}(Z|X)E[Y|X]^2\end{align}$$

  • 1
    $\begingroup$ I was so close, but thank you for the clarification! :) $\endgroup$
    – arezaie
    Sep 27, 2021 at 14:45
  • 1
    $\begingroup$ @arezaie you’re welcome $\endgroup$
    – gunes
    Sep 27, 2021 at 14:46

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