Solving Problems Submitted to MAA Journals (Part 7d)

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.

We suppose that $E(X) = \mu_1$ , $\hbox{SD}(X) = \sigma_1$ , $E(Y) = \mu_2$ , and $\hbox{SD}(Y) = \sigma_2$ . With these definitions, we may write $X = \mu_1 + \sigma_1 Z_1$ and $Y = \mu_2 + \sigma_2 Z_2$ , where $Z_1$ and $Z_2$ are independent standard normal random variables.

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’d like to rewrite in terms of $Z_1$ and $Z_2$ to whatever extent is possible.

As it turns out, the usual scaling and shifting properties of variance apply to a conditional variance on any event $A$ . The event that we have in mind, of course, is $X > Y$ . As discussed in the previous post, this can be rewritten as $Z_1 > a + b Z_2$ , where $a = (\mu_2 - \mu_1)/\sigma_1$ and $b = \sigma_2/\sigma_1$ .

We are now ready to derive the scaling and shift properties for $\hbox{Var}(X \mid A)$ . We begin by using the definition

$\hbox{Var}(X \mid A) = E(X^2 \mid A) - [E(X \mid A)]^2 = \displaystyle \frac{E(X^2 I_A)}{P(A)} - \left[ \frac{E(XI_A)}{P(A)} \right]^2$ .

Let’s examine the unconditional expectations $E(XI_A)$ and $E(X^2 I_A)$ . First,

$E(XI_A) = E([\mu_1 + \sigma_1 Z_1] I_A) = \mu E(I_A) + E(\sigma_1 Z_1 I_A) = \mu_1 P(A) + \sigma_1 E(Z_1 I_A)$ ,

and so

$E(X \mid A) = \displaystyle \frac{E(X I_A)}{P(A)} = \mu_1 \frac{P(A)}{P(A)} + \sigma_1 \frac{E(Z_1 I_A)}{P(A)} = \mu_1 + \sigma_1 E(Z_1 \mid A)$ .

Next,

$E(X^2 I_A) = E([\mu_1 + \sigma_1 Z_1]^2 I_A)$

$= E([\mu_1^2 + 2\mu_1 \sigma_1 Z_1+ \sigma_1^2 Z_1^2] I_A)$

$= \mu_1^2 E(I_A) + 2\mu_1 \sigma_1 E(Z_1 I_A) + \sigma_1^2 E(Z_1^2 I_A)$ ,

and so

$E(X^2 \mid A) = \displaystyle \frac{E(X^2 I_A)}{P(A)} = \mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A)$ .

Therefore,

$\hbox{Var}(A) = E(X^2 \mid A) - [ E(X \mid A) ]^2$

$= \mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A) - [\mu_1 + \sigma_1 E(Z_1 \mid A)]^2$

$=\mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A) - \mu_1^2 - 2\mu_1 \sigma_1 E(Z_1 \mid A) - \sigma_1^2 [E(Z_1 \mid A)]^2$

$= \sigma_1^2 E(Z_1^2 \mid A) - [E(Z_1 \mid A)]^2$

$= \sigma_1^2 \hbox{Var}(Z_1 \mid A)$ .

So, not surprisingly, $\hbox{Var}(X \mid A) = \hbox{Var}(\mu_1 + \sigma_1 Z_1 \mid A) = \sigma_1^2 \hbox{Var}(Z_1 \mid A)$ .

Also, the ultimate goal is to show that $\hbox{Var}(X \mid A)$ is less than $\hbox{Var}(X) = \sigma_1^2$ , where $A$ is the event $X>Y$ or, equivalently, $Z_1 > a + bZ_2$ . We see that it will be sufficient to show that

$\hbox{Var}(Z_1 \mid a + bZ_2 ) < 1$ .

We start the calculations of this conditional variance in the next post.

Solving Problems Submitted to MAA Journals (Part 7d)

Published by John Quintanilla

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