Solving Problems Submitted to MAA Journals (Part 7i)

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

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

Motivated by the graph of $f(x)$ , I thought of a two-step method for showing $f$ must be positive: show that $f$ is an increasing function, and show that $\displaystyle \lim_{x \to -\infty} f(x) = 0$ . If I could prove both of these claims, then that would prove that $f$ must always be positive.

I was able to show the second step by demonstrating that, if $x<0$ ,

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

As discussed in the last post, the limit $\displaystyle \lim_{x \to -\infty} f(x) = 0$ follows from this equality. However, I just couldn’t figure out the first step.

So I kept trying.

And trying.

Until it finally hit me: I’m working too hard! The goal is to show that $f(x)$ is positive. Clearly, clearly, the right-hand side of the last equation is positive! So that’s the entire proof for $x<0$ … there was no need to prove that $f$ is increasing!

For $x \ge 0$ , it’s even easier. If $x$ is non-negative, then

$f(x) = 1 + \sqrt{2\pi} x e^{x^2/2} \Phi(x) \ge 1 + \sqrt{2\pi} \cdot 0 \cdot 1 \cdot \frac{1}{2} = 1 > 0$ .

So, in either case, $f(x)$ must be positive. Following the logical thread in the previous posts, this demonstrates that $\hbox{Var}(Z_1 \mid Z_1 > a+bZ_2) < 1$ , so that $\hbox{Var}(X \mid X <Y) < \hbox{Var}(X)$ , 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.

Solving Problems Submitted to MAA Journals (Part 7h)

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

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 7g)

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

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 7f)

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

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

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 7e)

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

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.

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

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

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

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

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

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

$E(Z_1 \mid I_A) = \displaystyle \frac{e^0}{\sqrt{2\pi}\sqrt{\sigma^2+1} \frac{1}{2}} = \sqrt{\frac{2}{\pi(\sigma^2+1)}}$ ,

matching what we found earlier.

In the next post, we consider the calculation of $E(Z_1^2 \mid I_A)$ .

Solving Problems Submitted to MAA Journals (Part 7d)

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

Based on the experience of the special cases, it seems likely that I’ll eventually need to integrate over the joint probability density function of $X$ and $Y$ . However, it’s a bit easier to work with standard normal random variables than general ones, and so I’d like to rewrite in terms of $Z_1$ and $Z_2$ to whatever extent is possible.

As it turns out, the usual scaling and shifting properties of variance apply to a conditional variance on any event $A$ . The event that we have in mind, of course, is $X > Y$ . As discussed in the previous post, this can be rewritten as $Z_1 > a + b Z_2$ , where $a = (\mu_2 - \mu_1)/\sigma_1$ and $b = \sigma_2/\sigma_1$ .

We are now ready to derive the scaling and shift properties for $\hbox{Var}(X \mid A)$ . We begin by using the definition

$\hbox{Var}(X \mid A) = E(X^2 \mid A) - [E(X \mid A)]^2 = \displaystyle \frac{E(X^2 I_A)}{P(A)} - \left[ \frac{E(XI_A)}{P(A)} \right]^2$ .

Let’s examine the unconditional expectations $E(XI_A)$ and $E(X^2 I_A)$ . First,

$E(XI_A) = E([\mu_1 + \sigma_1 Z_1] I_A) = \mu E(I_A) + E(\sigma_1 Z_1 I_A) = \mu_1 P(A) + \sigma_1 E(Z_1 I_A)$ ,

and so

$E(X \mid A) = \displaystyle \frac{E(X I_A)}{P(A)} = \mu_1 \frac{P(A)}{P(A)} + \sigma_1 \frac{E(Z_1 I_A)}{P(A)} = \mu_1 + \sigma_1 E(Z_1 \mid A)$ .

Next,

$E(X^2 I_A) = E([\mu_1 + \sigma_1 Z_1]^2 I_A)$

$= E([\mu_1^2 + 2\mu_1 \sigma_1 Z_1+ \sigma_1^2 Z_1^2] I_A)$

$= \mu_1^2 E(I_A) + 2\mu_1 \sigma_1 E(Z_1 I_A) + \sigma_1^2 E(Z_1^2 I_A)$ ,

and so

$E(X^2 \mid A) = \displaystyle \frac{E(X^2 I_A)}{P(A)} = \mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A)$ .

Therefore,

$\hbox{Var}(A) = E(X^2 \mid A) - [ E(X \mid A) ]^2$

$= \mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A) - [\mu_1 + \sigma_1 E(Z_1 \mid A)]^2$

$=\mu_1^2 + 2 \mu_1 \sigma_1 E(Z_1 \mid A) + \sigma_1^2 E(Z_1^2 \mid A) - \mu_1^2 - 2\mu_1 \sigma_1 E(Z_1 \mid A) - \sigma_1^2 [E(Z_1 \mid A)]^2$

$= \sigma_1^2 E(Z_1^2 \mid A) - [E(Z_1 \mid A)]^2$

$= \sigma_1^2 \hbox{Var}(Z_1 \mid A)$ .

So, not surprisingly, $\hbox{Var}(X \mid A) = \hbox{Var}(\mu_1 + \sigma_1 Z_1 \mid A) = \sigma_1^2 \hbox{Var}(Z_1 \mid A)$ .

Also, the ultimate goal is to show that $\hbox{Var}(X \mid A)$ is less than $\hbox{Var}(X) = \sigma_1^2$ , where $A$ is the event $X>Y$ or, equivalently, $Z_1 > a + bZ_2$ . We see that it will be sufficient to show that

$\hbox{Var}(Z_1 \mid a + bZ_2 ) < 1$ .

We start the calculations of this conditional variance in the next post.

Solving Problems Submitted to MAA Journals (Part 7c)

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

Not quite knowing how to start, I decided to begin by simplifying the problem and assume that both $X$ and $Y$ follow a standard normal distribution, so that $E(X) = E(Y) = 0$ and $\hbox{SD}(X)=\hbox{SD}(Y) = 1$ . After solving this special case, I then made a small generalization by allowing $\hbox{SD}(Y)$ to be arbitrary. Solving these two special cases boosted my confidence that I would eventually be able to tackle the general case, which I start to consider with this post.

We suppose that $E(X) = \mu_1$ , $\hbox{SD}(X) = \sigma_1$ , $E(Y) = \mu_2$ , and $\hbox{SD}(Y) = \sigma_2$ . The goal is to show that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 < \hbox{Var}(X) = \sigma_1^2$ .

Based on the experience of the special cases, it seems likely that I’ll eventually need to integrate over the joint probability density function of $X$ and $Y$ . However, it’s a bit easier to work with standard normal random variables than general ones, and so I wrote $X = \mu_1 + \sigma_1 Z_1$ and $Y = \mu_2 + \sigma_2 Z_2$ , where $Z_1$ and $Z_2$ are independent standard normal random variables.

We recall that

$E(X \mid X > Y) = \displaystyle \frac{E (X I_{X>Y})}{P(X>Y)}$ ,

and so we’ll have to compute $P(X>Y)$ . We switch to $Z_1$ and $Z_2$ :

$P(X>Y) = P(\mu_1 + \sigma_1 Z_1 > \mu_2 + \sigma_2 Z_2)$

$= P(\sigma_1 Z_1 > \mu_2 - \mu_1 + \sigma_2 Z_2)$

$= \displaystyle P \left( Z_1 > \frac{\mu_2 - \mu_1}{\sigma_1} + \frac{\sigma_2}{\sigma_1} Z_2 \right)$

$= \displaystyle P(Z_1 > a + b Z_2)$

$= \displaystyle P(bZ_2 - Z_1 < -a)$ ,

where we define $a = (\mu_2 - \mu_1)/\sigma_1$ and $b = \sigma_2/\sigma_1$ for the sake of simplicity. Since $Z_1$ and $Z_2$ are independent, we know that $W = bZ_2 - Z_1$ will also be a normal random variable with

$E(W) = b E(Z_2) - E(Z_1) = 0$

and

$\hbox{Var}(W) = \hbox{Var}(bZ_2) + \hbox{Var}(-Z_1) = b^2 \hbox{Var}(Z_2) + (-1)^2 \hbox{Var}(Z_1) = b^2+1$ .

Therefore, converting to standard units,

$P(W < -a) = \displaystyle \Phi \left( \frac{-a - 0}{\sqrt{b^2+1}} \right) = \Phi(c)$ ,

where $c = -a/\sqrt{b^2+1}$ and $\Phi$ is the cumulative distribution function of the standard normal distribution.

We already see that the general case is more complicated than the two special cases we previously considered, for which $P(X>Y)$ was simply equal to $\frac{1}{2}$ .

In future posts, we take up the computation of $E(X I_{X>Y})$ and $E(X^2 I_{X>Y})$ .

Solving Problems Submitted to MAA Journals (Part 7b)

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

Not quite knowing how to start, I decided to begin by simplifying the problem and assume that both $X$ and $Y$ follow a standard normal distribution, so that $E(X) = E(Y) = 0$ and $\hbox{SD}(X)=\hbox{SD}(Y) = 1$ . This doesn’t solve the original problem, of course, but I hoped that solving this simpler case might give me some guidance about tackling the general case. I solved this special case in the previous post.

Next, to work on a special case that was somewhere between the general case and the first special case, I kept $X$ as a standard normal distribution but changed $Y$ to have a nonzero mean. As it turned out, this significantly complicated the problem (as we’ll see in the next post), and I got stuck.

So I changed course: for a second attempt, I kept $X$ as a standard normal distribution but changed $Y$ so that $E(Y) = 0$ and $\hbox{SD}(Y) = \sigma$ , where $\sigma$ could be something other than 1. The goal is to show that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 < 1$ .

We begin by computing $E(X \mid X > Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)}$ . The denominator is straightforward: since $X$ and $Y$ are independent normal random variables, we also know that $X-Y$ is normally distributed with $E(X-Y) = E(X)-E(Y) = 0$ . (Also, $\hbox{Var}(X-Y) = \hbox{Var}(X) + (-1)^2 \hbox{Var}(Y) = \sigma^2+1$ , but that’s really not needed for this problem.) Therefore, $P(X>Y) = P(X-Y>0) = \frac{1}{2}$ since the distribution of $X-Y$ is symmetric about its mean of $0$ .

Next, I wrote $Y = \sigma Z$ , where $Z$ has a standard normal distribution. Then

$E(X I_{X>\sigma Z}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \int_{\sigma z}^\infty x e^{-x^2/2} e^{-z^2/2} \, dx dz$ ,

where we have used the joint probability density function for the independent random variables $X$ and $Z$ . The region of integration is $\{(x,z) \in \mathbb{R}^2 \mid x > \sigma z \}$ , matching the requirement $X > \sigma Z$ . The inner integral can be directly evaluated:

$E(X I_{X>Y}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ -e^{-x^2/2} \right]_{\sigma z}^\infty e^{-z^2/2} \, dz$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ 0 + e^{-\sigma^2 z^2/2} \right] e^{-z^2/2} \, dz$

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

At this point, I rewrote the integrand to be the probability density function of a random variable:

$E(X I_{X>Y}) = \displaystyle \frac{1}{\sqrt{2\pi} \sqrt{\sigma^2+1}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sqrt{ \frac{1}{\sigma^2+1}}} \exp \left[ -\frac{z^2}{2 \cdot \frac{1}{\sigma^2+1}} \right] \, dz$ .

The integrand is the probability density function of a normal random variable with mean 0 and variance $\sigma^2+1$ , and so the integral must be equal to 1. We conclude

$E(X \mid X>Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)} = \frac{ \frac{1}{\sqrt{2\pi(\sigma^2+1)}} }{ \frac{1}{2} } = \sqrt{\frac{2}{\pi(\sigma^2+1)}}$ .

Next, we compute the other conditional expectation:

$E(X^2 \mid X > \sigma Z) = \displaystyle \frac{E(X^2 I_{X>\sigma Z})}{P(X>\sigma Z)} = \displaystyle \frac{2}{2\pi} \int_{-\infty}^\infty \int_{\sigma z}^\infty x^2 e^{-x^2/2} e^{-z^2/2} \, dx dz$ .

The inner integral can be computed using integration by parts:

$\displaystyle \int_{\sigma z}^\infty x^2 e^{-x^2/2} \, dx = \int_{\sigma z}^\infty x \frac{d}{dx} \left( -e^{-x^2/2} \right) \, dx$

$= \displaystyle \left[-x e^{-x^2/2} \right]_{\sigma z}^\infty + \int_{\sigma z}^\infty e^{-x^2/2} \, dx$

$= \sigma z e^{-\sigma^2 z^2/2} + \displaystyle \int_{\sigma z}^\infty e^{-x^2/2} \, dx$ .

Therefore,

$E(X^2 \mid X > \sigma Z) = \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty \sigma z e^{-\sigma^2 z^2/2} e^{-z^2/2} \, dz + 2 \int_{-\infty}^\infty \int_{\sigma z}^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-z^2/2} \, dx dz$

$= \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty \sigma z e^{-(\sigma^2+1) z^2} \, dz + 2 \int_{-\infty}^\infty \int_{\sigma z}^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-z^2/2} \, dx dz$ .

We could calculate the first integral, but we can immediately see that it’s going to be equal to 0 since the integrand $z e^{-(\sigma^2+1) z^2}$ is an odd function. The double integral is equal to $P(X>\sigma Z)$ , which we’ve already shown is equal to $\frac{1}{2}$ . Therefore, $E(X^2 \mid X > Y) = 0 + 2 \cdot \frac{1}{2} = 1$ .

We conclude that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 = 1 - \displaystyle \frac{2}{\pi(\sigma^2 + 1)}$ ,

which is indeed less than 1. If $\sigma = 1$ , we recover the conditional variance found in the first special case.

After tackling these two special cases, we start the general case with the next post.

Solving Problems Submitted to MAA Journals (Part 7a)

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

I admit I did a double-take when I first read this problem. If $X$ and $Y$ are independent, then the event $X>Y$ contains almost no information. How then, so I thought, could the conditional distribution of $X$ given $X>Y$ be narrower than the unconditional distribution of $X$ ?

Then I thought: I can believe that $E(X \mid X > Y)$ is greater than $E(X)$ : if we’re given that $X>Y$ , then we know that $X$ must be larger than something. So maybe it’s possible for $\hbox{Var}(X \mid X>Y)$ to be less than $\hbox{Var}(X)$ .

Still, not quite knowing how to start, I decided to begin by simplifying the problem and assume that both $X$ and $Y$ follow a standard normal distribution, so that $E(X) = E(Y) = 0$ and $\hbox{SD}(X)=\hbox{SD}(Y) = 1$ . This doesn’t solve the original problem, of course, but I hoped that solving this simpler case might give me some guidance about tackling the general case. I also hoped that solving this special case might give me some psychological confidence that I would eventually be able to solve the general case.

For the special case, the goal is to show that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 < 1$ .

We begin by computing $E(X \mid X > Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)}$ . The denominator is straightforward: since $X$ and $Y$ are independent normal random variables, we also know that $X-Y$ is normally distributed with $E(X-Y) = E(X)-E(Y) = 0$ . (Also, $\hbox{Var}(X-Y) = \hbox{Var}(X) + (-1)^2 \hbox{Var}(Y) = 2$ , but that’s really not needed for this problem.) Therefore, $P(X>Y) = P(X-Y>0) = \frac{1}{2}$ since the distribution of $X-Y$ is symmetric about its mean of $0$ .

Next,

$E(X I_{X>Y}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \int_y^\infty x e^{-x^2/2} e^{-y^2}/2 \, dx dy$ ,

where we have used the joint probability density function for $X$ and $Y$ . The region of integration is $\{(x,y) \in \mathbb{R}^2 \mid x > y \}$ , taking care of the requirement $X > Y$ . The inner integral can be directly evaluated:

$E(X I_{X>Y}) = \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ -e^{-x^2/2} \right]_x^\infty e^{-y^2/2} \, dy$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty \left[ 0 + e^{-y^2/2} \right] e^{-y^2/2} \, dy$

$= \displaystyle \frac{1}{2\pi} \int_{-\infty}^\infty e^{-y^2} \, dy$ .

At this point, I used a standard technique/trick of integration by rewriting the integrand to be the probability density function of a random variable. In this case, the random variable is normally distributed with mean 0 and variance $1/2$ :

$E(X I_{X>Y}) = \displaystyle \frac{1}{\sqrt{2\pi} \sqrt{2}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sqrt{1/2}} \exp \left[ -\frac{y^2}{2 \cdot \frac{1}{2}} \right] \, dy$ .

The integral must be equal to 1, and so we conclude

$E(X \mid X>Y) = \displaystyle \frac{E(X I_{X>Y})}{P(X>Y)} = \frac{ \frac{1}{2\sqrt{\pi}} }{ \frac{1}{2} } = \frac{1}{\sqrt{\pi}}$ .

We parenthetically note that $E(X \mid X>Y) > 0$ , matching my initial intuition.

Next, we compute the other conditional expectation:

$E(X^2 \mid X > Y) = \displaystyle \frac{E(X^2 I_{X>Y})}{P(X>Y)} = \displaystyle \frac{2}{2\pi} \int_{-\infty}^\infty \int_y^\infty x^2 e^{-x^2/2} e^{-y^2/2} \, dx dy$ .

The inner integral can be computed using integration by parts:

$\displaystyle \int_y^\infty x^2 e^{-x^2/2} \, dx = \int_y^\infty x \frac{d}{dx} \left( -e^{-x^2/2} \right) \, dx$

$= \displaystyle \left[-x e^{-x^2/2} \right]_y^\infty + \int_y^\infty e^{-x^2/2} \, dx$

$= y e^{-y^2/2} + \displaystyle \int_y^\infty e^{-x^2/2} \, dx$ .

Therefore,

$E(X^2 \mid X > Y) = \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty y e^{-y^2/2} e^{-y^2/2} \, dy + 2 \int_{-\infty}^\infty \int_y^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-y^2/2} \, dx dy$

$= \displaystyle \frac{1}{\pi} \int_{-\infty}^\infty y e^{-y^2} \, dy + 2 \int_{-\infty}^\infty \int_y^\infty \frac{1}{2\pi} e^{-x^2/2} e^{-y^2/2} \, dx dy$ .

We could calculate the first integral, but we can immediately see that it’s going to be equal to 0 since the integrand $y e^{-y^2}$ is an odd function. The double integral is equal to $P(X>Y)$ , which we’ve already shown is equal to $\frac{1}{2}$ . Therefore, $E(X^2 \mid X > Y) = 0 + 2 \cdot \frac{1}{2} = 1$ .

We conclude that

$\hbox{Var}(X \mid X > Y) = E(X^2 \mid X > Y) - [E(X \mid X > Y)]^2 = 1 - \displaystyle \left( \frac{1}{\sqrt{\pi}} \right)^2 = 1 - \frac{1}{\pi}$ ,

which is indeed less than 1.

This solves the problem for the special case of two independent standard normal random variables. This of course does not yet solve the general case, but my hope was that solving this problem might give me some intuition about the general case, which I’ll develop as this series progresses.

Solving Problems Submitted to MAA Journals (Part 4)

The following problem appeared in Volume 96, Issue 3 (2023) of Mathematics Magazine.

I’ll admit when I first read this problem, I didn’t believe it. I had to draw a couple of Venn diagrams to convince myself that it actually worked:

Of course, pictures are not proofs, so I started giving the problem more thought.

I wish I could say where I got the inspiration from, but I got the idea to define a new random variable $N$ to be the number of events from $A_1, \dots A_n$ that occur. With this definition, $B_k$ becomes the event that $N \ge k$ , so that

$\displaystyle \sum_{k=1}^n P(B_k) = \sum_{k=1}^n P(N \ge k)$

At this point, my Spidey Sense went off: that’s the tail-sum formula for expectation! Since $N$ is a non-negative integer-valued random variable, the mean of $N$ can be computed by

$E(N) = \displaystyle \sum_{k=1}^n P(N \ge k)$ .

Said another way, $E(N) = \displaystyle \sum_{k=1}^n P(B_k)$ .

Therefore, to solve the problem, it remains to show that $\displaystyle \sum_{k=1}^n P(A_k)$ is also equal to $E(N)$ . To do this, I employed the standard technique from the bag of tricks of writing $N$ as the sum of indicator random variables. Define