Probability Question About Uniform Random Variables and Median

Question

Let U, V, W ∼ Uniform(0, 1) be independent. Find the probability that the median (i.e., the second smallest) of these three random variables lies in the interval (1/4, 3/4).

I cannot figure out what to do. I am also horrible at probability. Thanks for any help.

Answer 1

Bruce Trumbo's answer is good, but if you want a more basic approach then note that the probability of a single observation of being between $0.25$ and $0.75$ is $\frac12$, being low is $\frac14$, and being high is $\frac14$. The middle value will be in the central interval if any of the following patterns happen:

• all three observations central, with probability $\frac12 \times \frac12 \times \frac12$
• two central and one low, with probability $3 \times \frac12 \times \frac12 \times \frac14$
• two central and one high, with probability $3 \times \frac12 \times \frac12 \times \frac14$
• one central and one low and one high, with probability $6 \times \frac12 \times \frac14 \times \frac14$

Add these up and you get $\frac{11}{16}=0.6875$.

Answer 2

I don't know what you have covered so far, so I won't try to give you an analytic solution. You may get more help on that if you would edit your question to give some topics in the chapter preceding the problem, speculate on which ones might be related, and say what you think is keeping you from putting the pieces together.

I will give you simulation results that will give you a very good clue whether anything you decide to try has provided the right answer. Simulation shows that the answer is about 0.687.

 m = 10^6;  n = 3;  x = runif(m*n)
 MAT = matrix(x, nrow=m)
 med = apply(MAT, 1, median)
 mean(med > 1/4 & med < 3/4)
 ## 0.687118

Addendum: No other answers and no response yet. So I'll provide the outline to one analytic approach to an exact solution. Let $X_{(2)}$ denote the second order statistic (the median) and $F_2(x)$ denote its CDF.

A general formula for order statistics can be simplified in this instance [three UNIF(0, 1) observations] to the following:

$$F_2(x) = \sum_{i=2}^3 C(3,i) x^i (1 - x)^{n-i},$$ for $0 < x < 1.$ Then $F_2(3/4) - F_2(1/4) = 0.6875,$ which was approximated by the simulation above about as well as can be expected.

If $U \sim Unif(0,1),$ then $F_U(x) = x$, for $0 < x < 1.$ In the equation above, factor $x^i$ results from $[F_U(x)]^i.$ If the formula used here is not in your textbook you can look on the Internet for 'order statistics'.