Solving Problems Submitted to MAA Journals (Part 7c)

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 $\hbox{SD}(X)=\hbox{SD}(Y) = 1$ . After solving this special case, I then made a small generalization by allowing $\hbox{SD}(Y)$ to be arbitrary. Solving these two special cases boosted my confidence that I would eventually be able to tackle the general case, which I start to consider with this post.

We suppose that $E(X) = \mu_1$ , $\hbox{SD}(X) = \sigma_1$ , $E(Y) = \mu_2$ , and $\hbox{SD}(Y) = \sigma_2$ . The goal is to show that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 < \hbox{Var}(X) = \sigma_1^2$ .

Based on the experience of the special cases, it seems likely that I’ll eventually need to integrate over the joint probability density function of $X$ and $Y$ . However, it’s a bit easier to work with standard normal random variables than general ones, and so I wrote $X = \mu_1 + \sigma_1 Z_1$ and $Y = \mu_2 + \sigma_2 Z_2$ , where $Z_1$ and $Z_2$ are independent standard normal random variables.

We recall that

$E(X \mid X > Y) = \displaystyle \frac{E (X I_{X>Y})}{P(X>Y)}$ ,

and so we’ll have to compute $P(X>Y)$ . We switch to $Z_1$ and $Z_2$ :

$P(X>Y) = P(\mu_1 + \sigma_1 Z_1 > \mu_2 + \sigma_2 Z_2)$

$= P(\sigma_1 Z_1 > \mu_2 - \mu_1 + \sigma_2 Z_2)$

$= \displaystyle P \left( Z_1 > \frac{\mu_2 - \mu_1}{\sigma_1} + \frac{\sigma_2}{\sigma_1} Z_2 \right)$

$= \displaystyle P(Z_1 > a + b Z_2)$

$= \displaystyle P(bZ_2 - Z_1 < -a)$ ,

where we define $a = (\mu_2 - \mu_1)/\sigma_1$ and $b = \sigma_2/\sigma_1$ for the sake of simplicity. Since $Z_1$ and $Z_2$ are independent, we know that $W = bZ_2 - Z_1$ will also be a normal random variable with

$E(W) = b E(Z_2) - E(Z_1) = 0$

and

$\hbox{Var}(W) = \hbox{Var}(bZ_2) + \hbox{Var}(-Z_1) = b^2 \hbox{Var}(Z_2) + (-1)^2 \hbox{Var}(Z_1) = b^2+1$ .

Therefore, converting to standard units,

$P(W < -a) = \displaystyle \Phi \left( \frac{-a - 0}{\sqrt{b^2+1}} \right) = \Phi(c)$ ,

where $c = -a/\sqrt{b^2+1}$ and $\Phi$ is the cumulative distribution function of the standard normal distribution.

We already see that the general case is more complicated than the two special cases we previously considered, for which $P(X>Y)$ was simply equal to $\frac{1}{2}$ .

In future posts, we take up the computation of $E(X I_{X>Y})$ and $E(X^2 I_{X>Y})$ .

Solving Problems Submitted to MAA Journals (Part 7c)

Published by John Quintanilla

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