Solving Problems Submitted to MAA Journals (Part 2b)

The following problem appeared in Volume 53, Issue 4 (2022) of The College Mathematics Journal. This was the second-half of a two-part problem.

Suppose that $X$ and $Y$ are independent, uniform random variables over $[0,1]$ . Define $U_X$ , $V_X$ , $B_X$ , and $W_X$ as follows:

$U_X$ is uniform over $[0,X]$ ,

$V_X$ is uniform over $[X,1]$ ,

$B_X \in \{0,1\}$ with $P(B_X=1) = X$ and $P(B_X=0)=1-X$ , and

$W_X = B_X \cdot U_X + (1-B_X) \cdot V_X$ .

Prove that $W_X$ is uniform over $[0,1]$ .

Once again, one way of showing that $W_X$ is uniform on $[0,1]$ is showing that $P(W_X \le t) = t$ if $0 \le t \le 1$ .

My first thought was that the value of $W_X$ depends on the value of $X$ , and so it makes sense to write $P(W_X \le t)$ as an integral of conditional probabilities:

$P(W_X \le t) = \displaystyle \int_0^1 P(W_X \le t \mid X = x) f(x) \, dx$ ,

where $f(x)$ is the probability density function of $X$ . In this case, since $X$ has a uniform distribution over $[0,1]$ , we see that $f(x) = 1$ for $0 \le x \le 1$ . Therefore,

$P(W_X \le t) = \displaystyle \int_0^1 P(W_X \le t \mid X = x) \, dx$ .

My second thought was that $W_X$ really has a two-part definition:

$W_X = \displaystyle \bigg\{ \begin{array}{ll} U_X, & B_X = 1 \\ V_X, & B_X = 0 \end{array}$

So it made sense to divide the conditional probability into these two cases:

$P(W_X \le t) = \displaystyle \int_0^1 P(W_X \le t, B_X = 1 ~\hbox{or}~ B_X = 0 \mid X = x) \, dx$

$= \displaystyle \int_0^1 P(W_X \le t, B_X = 1 \mid X = x) \, dx + \int_0^1 P(W_X \le t, B_X = 0 \mid X = x) \, dx$

My third thought was that these probabilities can be rewritten using the Multiplication Rule. This ordinarily has the form $P(E \cap F) = P(E) P(F \mid E)$ . For an initial conditional probability, it has the form $P(E \cap F \mid G) = P(E \mid G) P(F \mid E \cap G)$ . Therefore,

$P(W_X \le t) = \displaystyle \int_0^1 P(B_X = 1 \mid X = x) P(W_X \le t \mid B_X = 1, X = x) \, dx$

$+ \displaystyle \int_0^1 P(B_X = 0 \mid X=x) P(W_X \le t \mid B_X = 0, X = x) \, dx$ .

The definition of $B_X$ provides the immediate computation of $P(B_X = 1 \mid X = x)$ and $P(B_X = 0 \mid X = x)$ :

$P(W_X \le t) =\displaystyle \int_0^1 x P(W_X \le t \mid B_X = 1, X = x) \, dx$

$+ \displaystyle \int_0^1 (1-x) P(W_X \le t \mid B_X = 0, X = x) \, dx$

Also, the two-part definition of $W_X$ provides the next step:

$P(W_X \le t) =\displaystyle \int_0^1 x P(U_X \le t \mid B_X = 1, X = x) \, dx$

$+ \displaystyle \int_0^1 (1-x) P(V_X \le t \mid B_X = 0, X = x) \, dx$

We split each of these integrals into an integral from $0$ to $t$ and then an integral from $t$ to $1$ . First,

$\displaystyle \int_0^1 x P(U_X \le t \mid B_X = 1, X = x) \, dx$

$= \displaystyle \int_0^t x P(U_X \le t \mid B_X = 1, X = x) \, dx + \displaystyle \int_t^1 x P(U_X \le t \mid B_X = 1, X = x) \, dx$ .

We now use the following: if $0 \le x \le 1$ and $U$ is uniform over $[0,x]$ , then

$P(U \le t) = \displaystyle \bigg\{ \begin{array}{ll} t/x, & t \le x \\ 1, & t > x \end{array}$

We observe that $t > x$ in the first integral, while $t \le x$ in the second integral. Therefore,

$\displaystyle \int_0^1 x P(U_X \le t \mid B_X = 1, X = x) \, dx = \int_0^t x \cdot 1 \, dx + \int_t^1 x \cdot \frac{t}{x} \, dx$

$= \displaystyle \int_0^t x \, dx + \int_t^1 t \, dx$ .

For the second integral involving $V_X$ , we again split into two subintegrals and use the fact that if $V$ is uniform on $[x,1]$ , then

$P(V \le t) = \displaystyle \bigg\{ \begin{array}{ll} 0, & t \le x \\ (t-x)/(1-x), & t > x \end{array}$

Therefore,

$\displaystyle \int_0^1 (1-x) P(V_X \le t \mid B_X = 0, X = x) \, dx$

$=\displaystyle \int_0^t (1-x) P(V_X \le t \mid B_X = 0, X = x) \, dx$

$+ \displaystyle \int_t^1 (1-x) P(V_X \le t \mid B_X = 0, X = x) \, dx$

$= \displaystyle \int_0^t (1-x) \cdot \frac{t-x}{1-x} \, dx + \int_t^1 (1-x) \cdot 0 \, dx$

$= \displaystyle \int_0^t (t-x) \, dx$ .

Combining, we conclude that

$P(W_X \le t) = \displaystyle \int_0^t x \, dx + \int_t^1 t \, dx + \displaystyle \int_0^t (t-x) \, dx$

$= \displaystyle \int_0^t [x + (t-x)] \, dx + \int_t^1 t \, dx$

$= \displaystyle \int_0^t t \, dx + \int_t^1 t \, dx$

$= \displaystyle \int_0^1 t \, dx$

$= t$ ,

from which we conclude that $W_X$ is uniformly distributed on $[0,1]$ .

As I recall, this took a couple days of staring and false starts before I was finally able to get the solution.

Solving Problems Submitted to MAA Journals (Part 2b)

Published by John Quintanilla

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