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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)?

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  • $\begingroup$ Is $Y$ a random variable? $\endgroup$
    – Greenparker
    Commented May 5, 2016 at 19:02
  • $\begingroup$ Yes but it may depend on $X$. $\endgroup$
    – Aaron Voelker
    Commented May 5, 2016 at 19:08
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    $\begingroup$ 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 $Y^2$.) $\endgroup$
    – whuber ♦
    Commented May 5, 2016 at 19:35
  • $\begingroup$ Thanks whuber. I believe this leads to the correct solution (see the answer I made)! $\endgroup$
    – Aaron Voelker
    Commented May 6, 2016 at 0:03

2 Answers 2

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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).

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  • $\begingroup$ Also note that we can conclude $|E(XY)| \le \sqrt{\text{Var}(X) + E(X)^2}$, since $\text{Var}(XY) \ge 0 \implies E(XY)^2 \le E((XY)^2)$. $\endgroup$
    – Aaron Voelker
    Commented May 6, 2016 at 0:41
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    $\begingroup$ 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. $\endgroup$
    – Greenparker
    Commented May 6, 2016 at 2:12
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    $\begingroup$ 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 $\text{Var}(Y), E(Y^2) \le \sup(Y^2)$. $\endgroup$
    – Aaron Voelker
    Commented May 6, 2016 at 2:30
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    $\begingroup$ 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. $\endgroup$
    – Greenparker
    Commented May 6, 2016 at 2:37
  • $\begingroup$ I changed my answer to reflect the changes. $\endgroup$
    – Greenparker
    Commented May 6, 2016 at 2:41
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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.

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    $\begingroup$ 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? $\endgroup$
    – Aaron Voelker
    Commented May 5, 2016 at 19:23
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    $\begingroup$ @Aaron Voelker: There is no need for independence in the calculations. $\endgroup$
    – kjetil b halvorsen ♦
    Commented May 6, 2016 at 7:59
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    $\begingroup$ @kjetilbhalvorsen $\mathbb{E}(XY \mid Y=y) = \mathbb{E}(yX)$ does not hold without some assumptions (such as independence). $\endgroup$
    – Juho Kokkala
    Commented May 6, 2016 at 11:15
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    $\begingroup$ @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. $\endgroup$
    – whuber ♦
    Commented Nov 9, 2017 at 15:07
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    $\begingroup$ @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... . $\endgroup$
    – whuber ♦
    Commented Nov 9, 2017 at 19:55

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