Solving Problems Submitted to MAA Journals (Part 2a)

The following problem appeared in Volume 53, Issue 4 (2022) of The College Mathematics Journal. This was the first problem that was I able to solve in over 30 years of subscribing to MAA journals.

Suppose that $X$ and $Y$ are independent, uniform random variables over $[0,1]$ . Now define the random variable $Z$ by

$Z = (Y-X) {\bf 1}(Y \ge X) + (1-X+Y) {\bf 1}(Y<X)$ .

Prove that $Z$ is uniform over $[0,1]$ . Here, ${\bf 1}[S]$ is the indicator function that is equal to 1 if $S$ is true and 0 otherwise.

The first thing that went through my mind was something like, “This looks odd. But it’s a probability problem using concepts from a senior-level but undergraduate probability course. This was once my field of specialization. I had better be able to get this.”

My second thought was that one way of proving that $Z$ is uniform on $[0,1]$ is showing that $P(Z \le t) = t$ if $0 \le t \le 1$ .

My third thought was that $Z$ really had a two-part definition:

$Z = \bigg\{ \begin{array}{ll} Y-X, & Y \ge X \\ 1-X+Y, & Y <X \end{array}$

So I got started by dividing this probability into the two cases:

$P(Z \le t) = P(Z \le t, Y \ge X ~\hbox{or}~ Y < X)$

$= P(Z \le t, Y \ge X ~\hbox{or}~ Z \le t, Y < X)$

$= P(Z \le t, Y \ge X) + P(Z \le t, Y < X)$

$= P(Y-X \le t, Y \ge X) + P(1 - X + Y \le t, Y < X)$

$= P(Y \le X + t, Y \ge X) + P( Y \le X + t - 1, Y < X)$

$= P(X \le Y \le X + t) + P(Y \le X + t - 1)$ .

In the last step, since $0 \le t \le 1$ , the events $Y \le X + t - 1$ and $Y < X$ are redundant: if $Y \le X + t - 1$ , then $Y$ will automatically be less than $X$ . Therefore, it’s safe to remove $Y < X$ from the last probability.

Ordinarily, such probabilities are computed by double integrals over the joint probability density function of $X$ and $Y$ , which usually isn’t easy. However, in this case, since $X$ and $Y$ are independent and uniform over $[0,1]$ , the ordered pair $(X,Y)$ is uniform on the unit square $[0,1] \times [0,1]$ . Therefore, probabilities can be found by simply computing areas.

In this case, since the area of the unit square is 1, $P(Z \le t)$ is equal to the sum of the areas of

$\{(x,y) \in [0,1] \times [0,1] : x \le y \le x + t \}$ ,

which is depicted in green below, and

$\{(x,y) \in [0,1] \times [0,1] : y \le x + t - 1 \}$ ,

which is depicted in purple.

First, the area in green is a trapezoid. The $y-$ intercept of the line $y = x+t$ is $(0,t)$ , and the two lengths of $t$ and $1-t$ on the upper left of the square are found from this $y-$ intercept. The area of the green trapezoid is easiest found by subtracting the areas of two isosceles right triangles:

$latex \displaystyle \frac{(1)^2}{2} – \frac{(1-t)^2}{2} = \frac{1-1+2t-t^2}{2} = \frac{2t-t^2}{2}.

Second, the area in purple is an isosceles right triangle. The $y-$ intercept of the line $y=x+t-1$ is $(0,t-1)$ , so that the distance from the $y-$ intercept to the origin is $1-t$ . From this, the two lengths of $1-t$ and $t$ are found. Therefore, the area of the purple right triangle is $\displaystyle \frac{t^2}{2}$ .

Adding, we conclude that

$P(Z \le t) = \displaystyle \frac{2t-t^2}{2} + \frac{t^2}{2} = \frac{2t}{2} = t$ .

Therefore, $Z$ is uniform over $[0,1]$ .

A closing note: after going 0-for-4000 in my previous 30+ years of attempting problems submitted to MAA journals, I was unbelievably excited to finally get one. As I recall, it took me less than an hour to get the above solution, although writing up the solution cleanly took longer.

However, the above was only Part 1 of a two-part problem, so I knew I just had to get the second part before submitting. That’ll be the subject of the next post.

Solving Problems Submitted to MAA Journals (Part 2a)

Published by John Quintanilla

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