Integration Using Schwinger Parametrization

I recently read the terrific article Integration Using Schwinger Parametrization, by David M. Bradley, Albert Natian, and Sean M. Stewart in the American Mathematical Monthly. I won’t reproduce the entire article here, but I’ll hit a couple of early highlights.

The basic premise of the article is that a complicated integral can become tractable by changing it into an apparently more complicated double integral. The idea stems from the gamma integral

$\Gamma(p) = \displaystyle \int_0^\infty t^{p-1} e^{-t} \, dt$ ,

where $\Gamma(p) = (p-1)!$ if $p$ is a positive integer. If we perform the substitution $t = \phi u$ in the above integral, where $\phi$ is a quantity independent of $t$ , we obtain

$\Gamma(p) = \displaystyle \int_0^\infty (\phi u)^{p-1} e^{-\phi t} \phi \, du = \displaystyle \int_0^\infty \phi^p u^{p-1} e^{-\phi u} \, du$ ,

which may be rewritten as

$\displaystyle \frac{1}{\phi^p} = \displaystyle \frac{1}{\Gamma(p)} \int_0^\infty t^{p-1} e^{-\phi t} \, dt$

after changing the dummy variable back to $t$ .

A simple (!) application of this method is the famous Dirichlet integral

$I = \displaystyle \int_0^\infty \frac{\sin x}{x} \, dx$

which is pretty much unsolvable using techniques from freshman calculus. However, by substituting $\phi = x$ and $p=1$ in the above gamma equation, and using the fact that $\Gamma(1) = 0! = 1$ , we obtain

$I = \displaystyle \int_0^\infty \sin x \int_0^\infty e^{-xt} \, dt \, dx$

$= \displaystyle \int_0^\infty \int_0^\infty e^{-xt} \sin x \, dx \, dt$

after interchanging the order of integration. The inner integral can be found by integration by parts and is often included in tables of integrals:

$I = \displaystyle \int_0^\infty -\left[ \frac{e^{-xt} (\cos x + t \sin x)}{1+t^2} \right]_{x=0}^{x=\infty} \, dt$

$= \displaystyle \int_0^\infty \left[0 +\frac{e^{0} (\cos 0 + t \sin 0)}{1+t^2} \right] \, dt$

$= \displaystyle \int_0^\infty \frac{1}{1+t^2} \, dt$ .

At this point, the integral is now a standard one from freshman calculus:

$I = \displaystyle \left[ \tan^{-1} t \right]_0^\infty = \displaystyle \frac{\pi}{2} - 0 = \displaystyle \frac{\pi}{2}$ .

In the article, the authors give many more applications of this method to other integrals, thus illustrating the famous quote, “An idea which can be used only once is a trick. If one can use it more than once it becomes a method.” The authors also add, “We present some examples to illustrate the utility of this technique in the hope that by doing so we may convince the reader that it makes a valuable addition to one’s integration toolkit.” I’m sold.

A vivid illustration of a discontinuous function

The essay Singular Limits in the May 2002 issue of Physics Today has a vivid illustration of a discontinuous function $F(x)$ which measures the ickiness one feels after eating an apple but observing that proportion $x$ of a maggot is still inside the apple. For this function, $\displaystyle \lim_{x \to 0^+} F(x) \ne F(0)$ .

Biting into an apple and finding a maggot is unpleasant enough, but finding half a maggot is worse. Discovering one-third of a maggot would be more distressing still: The less you find, the more you might have eaten. Extrapolating to the limit, an encounter with no maggot at all should be the ultimate bad-apple experience. This remorseless logic fails, however, because the limit is singular: A very small maggot fraction ( $f \ll 1$ ) is qualitatively different from no maggot ( $f=0$ ).

Higher derivatives in ordinary speech

Just about every calculus student is taught that the first derivative is useful for finding the slope of a curve and finding velocity from position, and that the second derivative is useful for finding the concavity of a curve and finding acceleration from position.

I recently came across a couple of quotes that, taken literally, are statements about third and fifth derivatives.

Per Wikipedia, President Nixon announced in 1972 that the rate of increase of inflation was decreasing. Taken literally, this claims that “the second derivative of inflation is negative, and so the third derivative of purchasing power [since inflation is the derivative of purchasing power] is negative.” As dryly stated in the Notices of the American Mathematical Society, “[t]his was the first time a sitting president used the third derivative to advance his case for reelection”; the article then ponders the implications of the abuse of mathematics.

More recently, the popular blog Math With Bad Drawings had some fun analyzing a clause that appeared in a 2013 op-ed piece: “As the rate of acceleration of innovation increases…” Taken literally, the words rate, innovation and increases all refer to a first derivative (innovation would be the rate at which technology changes), while the word acceleration refers to a second derivative. Therefore, taken literally and not rhetorically (which was clearly the authors’ intent), this brief clause is a claim that the fifth derivative of technology is positive.

Horrible False Analogy

I had forgotten the precise assumptions on uniform convergence that guarantees that an infinite series can be differentiated term by term, so that one can safely conclude

$\displaystyle \frac{d}{dx} \sum_{n=1}^\infty f_n(x) = \sum_{n=1}^\infty f_n'(x)$ .

This was part of my studies in real analysis as a student, so I remembered there was a theorem but I had forgotten the details.

So, like just about everyone else on the planet, I went to Google to refresh my memory even though I knew that searching for mathematical results on Google can be iffy at best.

And I was not disappointed. Behold this laughably horrible false analogy (and even worse graphic) that I found on chegg.com:

Suppose Arti has to plan a birthday party and has lots of work to do like arranging stuff for decorations, planning venue for the party, arranging catering for the party, etc. All these tasks can not be done in one go and so need to be planned. Once the order of the tasks is decided, they are executed step by step so that all the arrangements are made in time and the party is a success.

Similarly, in Mathematics when a long expression needs to be differentiated or integrated, the calculation becomes cumbersome if the expression is considered as a whole but if it is broken down into small expressions, both differentiation and the integration become easy.

Pedagogically, I’m all for using whatever technique an instructor might deem necessary to to “sell” abstract mathematical concepts to students. Nevertheless, I’m pretty sure that this particular party-planning analogy has no potency for students who have progressed far enough to rigorously study infinite series.

Solving Problems Submitted to MAA Journals (Part 7i)

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 $f(x) = 1 + \sqrt{2\pi} x e^{x^2/2} \Phi(x)$ , 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.

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 7g)

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

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

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.

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.

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

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.

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.

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.

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.

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.