# 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)?

• Is $Y$ a random variable? May 5, 2016 at 19:02
• Yes but it may depend on $X$. May 5, 2016 at 19:08
• An utterly trivial inequality sometimes is quite useful in such situations: $\mathbb{E}(X^2Y^2) \le \mathbb{E}(X^2)\sup(Y^2)$. (This is perhaps the simplest special case of Hölder's inequality for $p=1,q=\infty$ applied to $X^2$ and $Y^2$.)
– whuber
May 5, 2016 at 19:35
• Thanks whuber. I believe this leads to the correct solution (see the answer I made)! May 6, 2016 at 0:03

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).

• Also note that we can conclude $|E(XY)| \le \sqrt{\text{Var}(X) + E(X)^2}$, since $\text{Var}(XY) \ge 0 \implies E(XY)^2 \le E((XY)^2)$. May 6, 2016 at 0:41
• I don't believe kjetil assumed independence between $X$ and $Y$. The law of total variation holds in general, and not assuming independence. So I can find nothing in his statement that assumes independence. Also notice that your conclusion is exactly the same as my conclusion that is based off of kjetil's answer. May 6, 2016 at 2:12
• Independence must have been used somewhere (I guess when factoring out the expectation), otherwise the first equation from your answer (as shown in my comment) is stating that $\text{Var}(XY)$ is the same whether or not $X$ and $Y$ are independent, which is a contradiction. The fact that we came to the same conclusion is kind of a "coincidence", because we are both giving upper bounds. Mine comes from dropping the $E(XY)^2$ term and $Y^2 \le \sup(Y^2)$, and yours comes from $\text{Var}(Y), E(Y^2) \le \sup(Y^2)$. May 6, 2016 at 2:30
• I think I figured out where kjetil is using independence. By the law of total variance $Var(XY) = Var(E(XY|Y)) + E(Var(XY|Y))$. If we just look at the first term $Var(E(XY \mid Y)) = Var(Y^2E(X \mid Y))$ which is not the same as $Var(Y^2E(X))$. It is the same only when $X$ and $Y$ are independent. May 6, 2016 at 2:37
• I changed my answer to reflect the changes. May 6, 2016 at 2:41

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.

• Nice. To check my understanding this is using the law of total variance? Also this seems to prove something more general: that the variance $XY$ is finite as long as the variance of both $X$ and $Y$ are finite? May 5, 2016 at 19:23
• @Aaron Voelker: There is no need for independence in the calculations. May 6, 2016 at 7:59
• @kjetilbhalvorsen $\mathbb{E}(XY \mid Y=y) = \mathbb{E}(yX)$ does not hold without some assumptions (such as independence). May 6, 2016 at 11:15
• @Juho $\mathbb{E}(1 X) = \mathbb{E}(X)=0.5$, too. The relation $\mathbb{E}(XY\mid Y=y)=\mathbb{E}(X)y$ is an example of a very general theorem called "taking out what is known." It does not require independence of $(X,Y)$. See en.wikipedia.org/wiki/Conditional_expectation#Basic_properties.
– whuber
Nov 9, 2017 at 15:07
• @Juho Sorry, my comment was stupid. Of course I needed to write the conditional expectation in $\mathbb{E}(XY\mid Y=y)=\mathbb{E}(X\mid Y=y) y$. For some reason I just automatically understood these expectations as being conditional even when they weren't... .
– whuber
Nov 9, 2017 at 19:55