Solving Problems Submitted to MAA Journals (Part 7f)

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

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

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 previous post that

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

We now turn to the second conditional expectation:

$E(Z_1^2 \mid Z_1 + abZ_2) = \displaystyle \frac{E(Z_1^2 I_{Z_1 > a+b Z_2})}{P(Z_1 > a + bZ_2)} = \frac{E(Z_1^2 I_{Z_1 > a+b Z_2})}{\Phi(c)}$ .

The expected value in the numerator is a double integral:

$E(Z_1 I_{Z_1 > a+b Z_2}) = \displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty z_1^2 I_{z_1 > a + bz_2} f(z_1,z_2) \, dz_1 dz_2 = \displaystyle \int_{-\infty}^\infty \int_{a+bz_2}^\infty z_1^2 f(z_1,z_2) \, dz_1 dz_2$ ,

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:

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

Therefore, we must compute

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

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:

we begin by using integration by parts on the inner integral:

$\displaystyle \int_{a+bz_2}^\infty z_1^2 e^{-z_1^2/2} \, dz_1 = \int_{a+bz_2}^\infty z_1 \frac{d}{dz_1} \left(-e^{-z_1^2/2} \right) \, dz_1$

$=\displaystyle \left[ -z_1 e^{-z_1^2/2} \right]_{a+bz_2}^\infty + \int_{a+bz_2}^\infty e^{-z_1^2/2} \, dz_1$

$= (a+bz_2) \displaystyle \exp \left[-\frac{(a+bz_2)^2}{2} \right] + \int_{a+bz_2}^\infty e^{-z_1^2/2} \, dz_1$

Therefore,

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

The second term is equal to $\Phi(c)$ since the double integral is $P(Z_1 > a+bZ_2)$ . For the first integral, we complete the square as before:

$E(Z_1^2 I_A) = \Phi(c) + \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty (a+bz_2) \exp \left[-\frac{(b^2+1)z_2^2 + 2abz_2 + a^2}{2} \right] \, dz_2$

$= \Phi(c) + \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty (a + bz_2) \exp \left[ -\frac{b^2+1}{2} \left( z_2^2 + \frac{2abz_2}{b^2+1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right) \right] \exp \left[ -\frac{1}{2} \left(a^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right) \right] dz_2$

$= \Phi(c) +\displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty (a + bz_2)\exp \left[ -\frac{b^2+1}{2} \left( z_2^2 + \frac{2abz_2}{b^2+1} + \frac{a^2b^2}{(b^2+1)^2} \right) \right] \exp \left[ -\frac{1}{2} \left(a^2 - \frac{a^2b^2}{b^2+1} \right) \right] dz_2$

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

$= \Phi(c) +\displaystyle \frac{e^{-c^2/2}}{2\pi} \int_{-\infty}^\infty (a + bz_2)\exp \left[ -\frac{b^2+1}{2} \left( z_2 + \frac{ab}{b^2+1} \right)^2 \right] dz_2$ .

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:

$E(Z_1^2 I_A) = \Phi(c) + \displaystyle \frac{e^{-c^2/2}}{\sqrt{2\pi}\sqrt{b^2+1}} \int_{-\infty}^\infty (a+bz_2) \frac{1}{\sqrt{2\pi} \sqrt{ \displaystyle \frac{1}{b^2+1}}} \exp \left[ - \frac{\left(z_2 + \displaystyle \frac{ab}{b^2+1} \right)^2}{2 \cdot \displaystyle \frac{1}{b^2+1}} \right] dz_2$ .

This is an example of making a problem easier by apparently making it harder. The integrand has the probability density function of a normally distributed random variable $V$ with $E(V) = -ab/(b^2+1)$ and $\hbox{Var}(V) = 1/(b^2+1)$ . Therefore, the integral is equal to $E(a + bV)$ , so that

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

$= \Phi(c) + \displaystyle \frac{e^{-c^2/2}}{\sqrt{2\pi (b^2+1)}} \left( a -\frac{ab^2}{b^2+1} \right)$

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

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

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

Therefore,

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

We note that this reduces to what we found in the second special case: if $\mu_1=\mu_2=0$ , then $a = 0$ and $c = 0$ , so that $E(Z_1^2 \mid Z_1 > a + bZ_2) = 1$ , matching what we found earlier.

In the next post, we consider the calculation of $\hbox{Var}(Z_1^2 \mid I_A)$ .

Solving Problems Submitted to MAA Journals (Part 7f)

Published by John Quintanilla

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