Solving Problems Submitted to MAA Journals (Part 7h)

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.

In previous posts, we reduced the problem to showing that if $g(x) = \sqrt{2\pi} x e^{x^2/2} \Phi(x)$ , then $f(x) = 1 + g(x)$ is always positive, where

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

is the cumulative distribution function of the standard normal distribution. If we can prove this, then the original problem will be true.

When I was solving this problem for the first time, my progress through the first few steps was hindered by algebra mistakes and the like, but I didn’t doubt that I was progressing toward the answer. At this point in the solution, however, I was genuinely stuck: nothing immediately popped to mind for showing that $g(x)$ must be greater than $-1$ .

So I turned to Mathematica, just to make sure I was on the right track. Based on the graph, the function of $f(x)$ certainly looks positive.

What’s more, the graph suggests attempting to prove a couple of things: $f$ is an increasing function, and $\displaystyle \lim_{x \to -\infty} f(x) = 0$ or, equivalently, $\displaystyle \lim_{x \to -\infty} g(x) = -1$ . If I could prove both of these claims, then that would prove that $f$ must always be positive.

I started by trying to show

$\displaystyle \lim_{x \to -\infty} g(x) = \lim_{x \to \infty} x e^{x^2/2} \int_{-\infty}^x e^{-t^2/2} \, dt = -1$ .

I vaguely remembered something about the asymptotic expansion of the above integral from a course decades ago, and so I consulted that course’s textbook, by Bender and Orszag, to refresh my memory. To derive the behavior of $g(x)$ as $x \to -\infty$ , we integrate by parts. (This is permissible: the integrands below are well-behaved if $x<0$ , so that $0$ is not in the range of integration.)

$g(x) = \displaystyle x e^{x^2/2} \int_{-\infty}^x e^{-t^2/2} \, dt$

$= \displaystyle x e^{x^2/2} \int_{-\infty}^x \frac{1}{t} \frac{d}{dt} \left(-e^{-t^2/2}\right) \, dt$

$= \displaystyle x e^{x^2/2} \left[ -\frac{1}{t} e^{-t^2/2} \right]_{-\infty}^x - x e^{x^2/2} \int_{-\infty}^x \frac{d}{dt} \left(\frac{1}{t} \right) \left( -e^{-t^2/2} \right) \, dt$

$= \displaystyle x e^{x^2/2} \left[ -\frac{1}{x} e^{-x^2/2} - 0 \right] + |x| e^{x^2/2} \int_{-\infty}^x \frac{1}{t^2} e^{-t^2/2} \, dt$

$\displaystyle = - 1 +|x| e^{x^2/2} \int_{-\infty}^x \frac{1}{t^2} e^{-t^2/2} \, dt$

$\displaystyle = -1+ |x| e^{x^2/2} \int_{-\infty}^x \frac{1}{t^2} e^{-t^2/2} \, dt$ .

This is agonizingly close: the leading term is $-1$ as expected. However, I was stuck for the longest time trying to show that the second term goes to zero as $x \to -\infty$ .

So, once again, I consulted Bender and Orszag, which outlined how to show this. We note that

$\left|g(x) + 1\right| = \displaystyle |x| e^{x^2/2} \int_{-\infty}^x \frac{1}{t^2} e^{-t^2/2} \, dt < \displaystyle |x| e^{x^2/2} \int_{-\infty}^x \frac{1}{x^2} e^{-t^2/2} \, dt = \displaystyle \frac{g(x)}{x^2}$ .

Therefore,

$\displaystyle \lim_{x \to -\infty} \left| \frac{g(x)+1}{g(x)} \right| \le \lim_{x \to -\infty} \frac{1}{x^2} = 0$ ,

so that

$\displaystyle \lim_{x \to -\infty} \left| \frac{g(x)+1}{g(x)} \right| =\displaystyle \lim_{x \to -\infty} \left| 1 + \frac{1}{g(x)} \right| = 0$ .

Therefore,

$\displaystyle \lim_{x \to -\infty} \frac{1}{g(x)} = -1$ ,

$\displaystyle \lim_{x \to -\infty} g(x) = -1$ .

So (I thought) I was halfway home with the solution, and all that remained was to show that $f$ was an increasing function.

And I was completely stuck at this point for a long time.

Until I realized — much to my utter embarrassment — that showing $f$ was increasing was completely unnecessary, as discussed in the next post.

Solving Problems Submitted to MAA Journals (Part 7h)

Published by John Quintanilla

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