Solving Problems Submitted to MAA Journals (Part 7g)

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.

The goal is to show that $\hbox{Var}(X \mid X > Y) < \hbox{Var}(X)$ . In previous posts, we showed that it will be sufficient to show that $\hbox{Var}(Z_1 \mid Z_1 > a + bZ_2) < 1$ , where $a = (\mu_2 - \mu_1)/\sigma_1$ and $b = \sigma_2/\sigma_1$ . We also showed that $P(Z_1 > a + bZ_2) = \Phi(c)$ , where $c = -a/\sqrt{b^2+1}$ and

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

is the cumulative distribution function of the standard normal distribution.

To compute

$\hbox{Var}(Z_1 \mid Z_1 > a + bZ_2) = E(Z_1^2 \mid Z_1 + a bZ_2) - [E(Z_1 \mid Z_1 > a + bZ_2)]^2$ ,

we showed in the two previous posts that

$E(Z_1 \mid Z_1 > a + bZ_2) = \displaystyle \frac{e^{-c^2/2}}{\sqrt{2\pi}\sqrt{b^2+1} \Phi(c)}$

and

$E(Z_1^2 mid Z_1 > a + bZ_2) = 1 -\displaystyle \frac{c e^{-c^2/2}}{ \sqrt{2\pi} (b^2+1) \Phi(c)}$ .

Therefore,

$\hbox{Var}(Z_1 \mid A) = 1 - \displaystyle\frac{c e^{-c^2/2}}{ \sqrt{2\pi} (b^2+1) \Phi(c)} - \left( \frac{e^{-c^2/2}}{\sqrt{2\pi (b^2+1)} \Phi(c)} \right)^2$

$= 1 - \displaystyle\frac{c e^{-c^2/2}}{ \sqrt{2\pi} (b^2+1) \Phi(c)} - \frac{e^{-c^2}}{2\pi (b^2+1) [\Phi(c)]^2}$

$= 1 - \displaystyle\frac{c}{ \sqrt{2\pi} (b^2+1) \Phi(c) e^{c^2/2}} - \frac{1}{2\pi (b^2+1) [\Phi(c)]^2e^{c^2}}$

$= 1 - \displaystyle\frac{\sqrt{2\pi} c e^{c^2/2} \Phi(c) + 1}{2\pi (b^2+1) [\Phi(c)]^2 e^{c^2}}$ .

To show that $\hbox{Var}(Z_1 \mid A) < 1$ , it suffices to show that the second term must be positive. Furthermore, since the denominator of the second term is positive, it suffices to show that $f(c) = 1 + \sqrt{2\pi} c e^{c^2/2} \Phi(c)$ must also be positive.

And, to be honest, I was stuck here for the longest time.

At some point, I decided to plot this function in Mathematica to see if I get some ideas flowing:

The function certainly looks like it’s always positive. What’s more, the graph suggests attempting to prove a couple of things: $f$ is an increasing function, and $\displaystyle \lim_{x \to -\infty} f(x) = 0$ . If I could prove both of these claims, then that would prove that $f$ must always be positive.

Spoiler alert: this was almost a dead-end approach to the problem. I managed to prove one of them, but not the other. (I don’t doubt it’s true, but I didn’t find a proof.) I’ll discuss in the next post.

Solving Problems Submitted to MAA Journals (Part 7g)

Published by John Quintanilla

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