Solving Problems Submitted to MAA Journals (Part 7b)

The following problem appeared in Volume 131, Issue 9 (2024) of The American Mathematical Monthly.

Let $X$ and $Y$ be independent normally distributed random variables, each with its own mean and variance. Show that the variance of $X$ conditioned on the event $X>Y$ is smaller than the variance of $X$ alone.

Not quite knowing how to start, I decided to begin by simplifying the problem and assume that both $X$ and $Y$ follow a standard normal distribution, so that $E(X) = E(Y) = 0$ and $\hbox{SD}(X)=\hbox{SD}(Y) = 1$ . This doesn’t solve the original problem, of course, but I hoped that solving this simpler case might give me some guidance about tackling the general case. I solved this special case in the previous post.

Next, to work on a special case that was somewhere between the general case and the first special case, I kept $X$ as a standard normal distribution but changed $Y$ to have a nonzero mean. As it turned out, this significantly complicated the problem (as we’ll see in the next post), and I got stuck.

So I changed course: for a second attempt, I kept $X$ as a standard normal distribution but changed $Y$ so that $E(Y) = 0$ and $\hbox{SD}(Y) = \sigma$ , where $\sigma$ could be something other than 1. The goal is to show that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 < 1$ .

We begin by computing $E(X \mid X > Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)}$ . The denominator is straightforward: since $X$ and $Y$ are independent normal random variables, we also know that $X-Y$ is normally distributed with $E(X-Y) = E(X)-E(Y) = 0$ . (Also, $\hbox{Var}(X-Y) = \hbox{Var}(X) + (-1)^2 \hbox{Var}(Y) = \sigma^2+1$ , but that’s really not needed for this problem.) Therefore, $P(X>Y) = P(X-Y>0) = \frac{1}{2}$ since the distribution of $X-Y$ is symmetric about its mean of $0$ .

Next, I wrote $Y = \sigma Z$ , where $Z$ has a standard normal distribution. Then

$E(X I_{X>\sigma Z}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \int_{\sigma z}^\infty x e^{-x^2/2} e^{-z^2/2} \, dx dz$ ,

where we have used the joint probability density function for the independent random variables $X$ and $Z$ . The region of integration is $\{(x,z) \in \mathbb{R}^2 \mid x > \sigma z \}$ , matching the requirement $X > \sigma Z$ . The inner integral can be directly evaluated:

$E(X I_{X>Y}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ -e^{-x^2/2} \right]_{\sigma z}^\infty e^{-z^2/2} \, dz$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ 0 + e^{-\sigma^2 z^2/2} \right] e^{-z^2/2} \, dz$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty e^{-(\sigma^2+1) z^2/2} \, dz$ .

At this point, I rewrote the integrand to be the probability density function of a random variable:

$E(X I_{X>Y}) = \displaystyle \frac{1}{\sqrt{2\pi} \sqrt{\sigma^2+1}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sqrt{ \frac{1}{\sigma^2+1}}} \exp \left[ -\frac{z^2}{2 \cdot \frac{1}{\sigma^2+1}} \right] \, dz$ .

The integrand is the probability density function of a normal random variable with mean 0 and variance $\sigma^2+1$ , and so the integral must be equal to 1. We conclude

$E(X \mid X>Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)} = \frac{ \frac{1}{\sqrt{2\pi(\sigma^2+1)}} }{ \frac{1}{2} } = \sqrt{\frac{2}{\pi(\sigma^2+1)}}$ .

Next, we compute the other conditional expectation:

$E(X^2 \mid X > \sigma Z) = \displaystyle \frac{E(X^2 I_{X>\sigma Z})}{P(X>\sigma Z)} = \displaystyle \frac{2}{2\pi} \int_{-\infty}^\infty \int_{\sigma z}^\infty x^2 e^{-x^2/2} e^{-z^2/2} \, dx dz$ .

The inner integral can be computed using integration by parts:

$\displaystyle \int_{\sigma z}^\infty x^2 e^{-x^2/2} \, dx = \int_{\sigma z}^\infty x \frac{d}{dx} \left( -e^{-x^2/2} \right) \, dx$

$= \displaystyle \left[-x e^{-x^2/2} \right]_{\sigma z}^\infty + \int_{\sigma z}^\infty e^{-x^2/2} \, dx$

$= \sigma z e^{-\sigma^2 z^2/2} + \displaystyle \int_{\sigma z}^\infty e^{-x^2/2} \, dx$ .

Therefore,

$E(X^2 \mid X > \sigma Z) = \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty \sigma z e^{-\sigma^2 z^2/2} e^{-z^2/2} \, dz + 2 \int_{-\infty}^\infty \int_{\sigma z}^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-z^2/2} \, dx dz$

$= \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty \sigma z e^{-(\sigma^2+1) z^2} \, dz + 2 \int_{-\infty}^\infty \int_{\sigma z}^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-z^2/2} \, dx dz$ .

We could calculate the first integral, but we can immediately see that it’s going to be equal to 0 since the integrand $z e^{-(\sigma^2+1) z^2}$ is an odd function. The double integral is equal to $P(X>\sigma Z)$ , which we’ve already shown is equal to $\frac{1}{2}$ . Therefore, $E(X^2 \mid X > Y) = 0 + 2 \cdot \frac{1}{2} = 1$ .

We conclude that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 = 1 - \displaystyle \frac{2}{\pi(\sigma^2 + 1)}$ ,

which is indeed less than 1. If $\sigma = 1$ , we recover the conditional variance found in the first special case.

After tackling these two special cases, we start the general case with the next post.

Solving Problems Submitted to MAA Journals (Part 7b)

Published by John Quintanilla

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