Solving Problems Submitted to MAA Journals (Part 7a)

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.

I admit I did a double-take when I first read this problem. If $X$ and $Y$ are independent, then the event $X>Y$ contains almost no information. How then, so I thought, could the conditional distribution of $X$ given $X>Y$ be narrower than the unconditional distribution of $X$ ?

Then I thought: I can believe that $E(X \mid X > Y)$ is greater than $E(X)$ : if we’re given that $X>Y$ , then we know that $X$ must be larger than something. So maybe it’s possible for $\hbox{Var}(X \mid X>Y)$ to be less than $\hbox{Var}(X)$ .

Still, 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 also hoped that solving this special case might give me some psychological confidence that I would eventually be able to solve the general case.

For the special case, 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, $\hbox{Var}(X-Y) = \hbox{Var}(X) + (-1)^2 \hbox{Var}(Y) = 2$ , 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,

$E(X I_{X>Y}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \int_y^\infty x e^{-x^2/2} e^{-y^2}/2 \, dx dy$ ,

where we have used the joint probability density function for $X$ and $Y$ . The region of integration is $\{(x,y) \in \mathbb{R}^2 \mid x > y \}$ , taking care of the requirement $X > Y$ . The inner integral can be directly evaluated:

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

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

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty e^{-y^2} \, dy$ .

At this point, I used a standard technique/trick of integration by rewriting the integrand to be the probability density function of a random variable. In this case, the random variable is normally distributed with mean 0 and variance $1/2$ :

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

The integral must be equal to 1, and so we conclude

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

We parenthetically note that $E(X \mid X>Y) > 0$ , matching my initial intuition.

Next, we compute the other conditional expectation:

$E(X^2 \mid X > Y) = \displaystyle \frac{E(X^2 I_{X>Y})}{P(X>Y)} = \displaystyle \frac{2}{2\pi} \int_{-\infty}^\infty \int_y^\infty x^2 e^{-x^2/2} e^{-y^2/2} \, dx dy$ .

The inner integral can be computed using integration by parts:

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

$= \displaystyle \left[-x e^{-x^2/2} \right]_y^\infty + \int_y^\infty e^{-x^2/2} \, dx$

$= y e^{-y^2/2} + \displaystyle \int_y^\infty e^{-x^2/2} \, dx$ .

Therefore,

$E(X^2 \mid X > Y) = \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty y e^{-y^2/2} e^{-y^2/2} \, dy + 2 \int_{-\infty}^\infty \int_y^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-y^2/2} \, dx dy$

$= \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty y e^{-y^2} \, dy + 2 \int_{-\infty}^\infty \int_y^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-y^2/2} \, dx dy$ .

We could calculate the first integral, but we can immediately see that it’s going to be equal to 0 since the integrand $y e^{-y^2}$ is an odd function. The double integral is equal to $P(X>Y)$ , 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

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

which is indeed less than 1.

This solves the problem for the special case of two independent standard normal random variables. This of course does not yet solve the general case, but my hope was that solving this problem might give me some intuition about the general case, which I’ll develop as this series progresses.

Solving Problems Submitted to MAA Journals (Part 7a)

Published by John Quintanilla

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