Solving Problems Submitted to MAA Journals (Part 7e)

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 the previous two 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 begin with

$E(Z_1 \mid Z_1 + abZ_2) = \displaystyle \frac{E(Z_1 I_{Z_1 > a+b Z_2})}{P(Z_1 > a + bZ_2)} = \frac{E(Z_1 I_{Z_1 > a+b Z_2})}{\Phi(c)}$ .

The expected value in the numerator is a double integral:

$E(Z_1 I_{Z_1 > a+b Z_2}) = \displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty z_1 I_{z_1 > a + bz_2} f(z_1,z_2) \, dz_1 dz_2 = \displaystyle \int_{-\infty}^\infty \int_{a+bz_2}^\infty z_1 f(z_1,z_2) \, dz_1 dz_2$ ,

where $f(z_1,z_2)$ is the joint probability density function of $Z_1$ and $Z_2$ . Since $Z_1$ and $Z_2$ are independent, $f(z_1,z_2)$ is the product of the individual probability density functions:

$f(z_1,z_2) = \displaystyle \frac{1}{\sqrt{2\pi}} e^{-z_1^2/2} \frac{1}{\sqrt{2\pi}} e^{-z_2^2/2} = \frac{1}{2\pi} e^{-z_1^2/2} e^{-z_2^2/2}$ .

Therefore, we must compute

$E(Z_1 I_A) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \int_{a+bz_2}^\infty z_1 e^{-z_1^2/2} e^{-z_2^2/2} \, dz_1 dz_2$ ,

where I wrote $A$ for the event $Z_1 > a + bZ_2$ .

I’m not above admitting that I first stuck this into Mathematica to make sure that this was doable. To begin, we compute the inner integral:

$E(Z_1 I_A) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ - e^{-z_1^2/2} \right]_{a+bz_2}^\infty e^{-z_2^2/2} \, dz_2$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{(a+bz_2)^2}{2} \right] \exp\left[-\frac{z_2^2}{2} \right] dz_2$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{(b^2+1)z_2^2+2abz_2+a^2}{2} \right]$ .

At this point, I used a standard technique/trick of completing the square to rewrite the integrand as a common pdf.

$E(Z_1 I_A) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{b^2+1}{2} \left( z_2^2 + \frac{2abz_2}{b^2+1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right) \right] \exp \left[ -\frac{1}{2} \left(a^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right) \right] dz_2$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{b^2+1}{2} \left( z_2^2 + \frac{2abz_2}{b^2+1} + \frac{a^2b^2}{(b^2+1)^2} \right) \right] \exp \left[ -\frac{1}{2} \left(a^2 - \frac{a^2b^2}{b^2+1} \right) \right] dz_2$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{b^2+1}{2} \left( z_2 + \frac{ab}{b^2+1} \right)^2 \right] \exp \left[ -\frac{1}{2} \left( \frac{a^2}{b^2+1} \right) \right] dz_2$

$= \displaystyle \frac{e^{-c^2/2}}{2\pi} \int_{-\infty}^\infty \exp \left[ -\frac{b^2+1}{2} \left( z_2 + \frac{ab}{b^2+1} \right)^2 \right] dz_2$ .

I now rewrite the integrand so that has the form of the probability density function of a normal distribution, writing $2\pi = \sqrt{2\pi} \sqrt{2\pi}$ and multiplying and dividing by $\sqrt{b^2+1}$ in the denominator:

$E(Z_1 I_A) = \displaystyle \frac{e^{-c^2/2}}{\sqrt{2\pi}\sqrt{b^2+1}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sqrt{ \displaystyle \frac{1}{b^2+1}}} \exp \left[ - \frac{\left(z_2 + \displaystyle \frac{ab}{b^2+1} \right)^2}{2 \cdot \displaystyle \frac{1}{b^2+1}} \right] dz_2$ .

This is an example of making a problem easier by apparently making it harder. The integrand is equal to $P(-\infty < V < \infty)$ , where $V$ is a normally distributed random variable with $E(V) = -ab/(b^2+1)$ and $\hbox{Var}(V) = 1/(b^2+1)$ . Since $P(-\infty < V < \infty) = 1$ , we have

$E(Z_1 I_A) = \displaystyle \frac{e^{-c^2/2}}{\sqrt{2\pi}\sqrt{b^2+1}}$ ,

and so

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

We note that this reduces to what we found in the second special case: if $\mu_1=\mu_2=0$ , $\sigma_1 = 1$ , and $\sigma_2 = \sigma$ , then $a = 0$ , $b = \sigma$ , and $c = 0$ . Since $\Phi(0) = \frac{1}{2}$ , we have