Solving Problems Submitted to MAA Journals (Part 7i)

The following problem appeared in Volume 131, Issue 9 (2024) of The American Mathematical Monthly.

Let and be independent normally distributed random variables, each with its own mean and variance. Show that the variance of conditioned on the event is smaller than the variance of alone.

In previous posts, we reduced the problem to showing that if , then is always positive, where

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

Motivated by the graph of , I thought of a two-step method for showing must be positive: show that is an increasing function, and show that . If I could prove both of these claims, then that would prove that must always be positive.

I was able to show the second step by demonstrating that, if ,

.

As discussed in the last post, the limit follows from this equality. However, I just couldn’t figure out the first step.

So I kept trying.

And trying.

And trying.

Until it finally hit me: I’m working too hard! The goal is to show that is positive. Clearly, clearly, the right-hand side of the last equation is positive! So that’s the entire proof for … there was no need to prove that is increasing!

For , it’s even easier. If is non-negative, then

.

So, in either case, must be positive. Following the logical thread in the previous posts, this demonstrates that , so that , thus concluding the solution.

And I was really annoyed at myself that I stumbled over the last step for so long, when the solution was literally right in front of me.