If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$? I have a variable $X$ that I know has finite variance (and therefore also finite mean). Is it always true that its variance remains finite after scaling by $0 \le Y \le 1$?
Note that $X$ and $Y$ are not necessarily independent.
Edit: I believe the "worst-case" $Y$ is $0$ whenever $X < c$ and $1$ whenever $X \ge c$, for some $c$ (and the mirrored case)?
 A: I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:
\begin{align}
\text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\
&\le E(X^2Y^2) \\
&\le E(X^2)\sup(Y^2) \\
&= E(X^2) \\
&= \text{Var}(X) + E(X)^2 \\
&< \infty 
\end{align}
Note that the result also holds for any bounded $Y$ (since $\sup(Y^2)$ will be finite).
A: You need to use the formula 
$$  \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}}
   \Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))
$$
where $\Var $ is the variance operator. Take it term for term, write $\mu=\E X, \sigma^2=\Var X$, $\E (XY | Y= y) = \E (yX) = y \E (X) =\mu y$ with variance (over $Y$) $\Var (\mu Y) $ which is finite since $Y$ is bounded. 
Then the other term, $\Var (XY | Y=y) = \Var (yX) = y^2 \Var (X) = \sigma^2 y^2$ which again has a finite expectation since $Y$ is bounded. So the answer is yes.
