Recently saw this on an interview prep guide. Let $X \sim \mathcal{N}(0,1)$ and $Y \sim Unif(0,1)$. What is $Var(XY)$?
If $X$ and $Y$ were independent, then the calculation is easy since whatever the joint pdf is factors into the marginal pdfs: \begin{align*} Var(XY) & = E((XY)^2) - (E(XY))^2 \\ & = \int_{-\infty}^\infty \int_0^1 x^2 y^2 \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dy dx - \left(\int_{-\infty}^\infty \int_0^1 x y \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dy dx\right)^2 \\ & = \ldots. \end{align*}
It seems I have no insight into calculating $Var(XY)$ if $X$ and $Y$ are not independent, though. Any thoughts? Or perhaps the question was missing this independence assumption?
Yes, the question was missing the independence assumption. When you make $Y$ large when $|X|$ is large you get a greater variance than when you make $Y$ small when $|X|$ is large.
Edit:
Since the answer was apparently not clear enough, here's some more detail: Let $X\sim\mathcal N(0,1)$, with cumulative distribution function $G(x)=\frac12\left(1+\operatorname{erf}\left(x/\sqrt2\right)\right)$, and let $Y=|2G(X)-1|$. Then $Y\sim U(0,1)$. Likewise, if $Y=1-|2G(X)-1|$, then $Y\sim U(0,1)$. In the first case, $Y$ is large when $|X|$ is large, whereas in the second case $Y$ is large when $|X|$ is small, so the variance of $XY$ is greater in the first case.