AMM problem 12450

This is AMM problem 12450 that I solved with my friends Satvik Saha and Sohom Gupta.

Problem Statement -
Let $n$ be an odd positive number, and suppose that $x_1, \dots, x_n$ are chosen randomly and uniformly from the interval $[0, 1]$. For $1 \leq i \leq n$, let $y_i = x_i - x_i^2$. What is the expected value of the median of $\{y_1, \dots, y_n\}$?

Solution -
Let $g(x) = x - x^2$ on $[0, 1]$, and let

$$h(y) = \frac{1 - \sqrt{1 - 4y}}{2}$$

be the inverse of the $g$ restricted to $[0, 1/2]$.

Let $X \sim \mathcal{U}(0, 1)$, and set $Y = g(X)$.

Then, by symmetry, $Y$ has density 

$$f_Y(y) = 2 f_X(h(y)) \cdot \frac{d}{dy}h(y) = 2 h'(y)$$

from which it follows that the cdf of $Y$ on $[0, 1/4]$ is $F_Y(y) = 2 h(y)$. From an iid sample $y_1, \dots, y_n \sim F_Y$ where $n = 2k + 1$, the sample median $\xi = y_{[k + 1]}$ has density

$$f_\xi(\xi) = \frac{n!}{(k!)^2} \, F_Y(\xi)^k(1 - F_Y(\xi))^k f_Y(\xi) = \frac{n!}{(k!)^2} (2h(\xi))^k (1 - 2h(\xi))^k\, 2h'(\xi)$$

and hence, the expected value of the median is

\begin{align*}\mathbb{E}[\xi] = \int \xi f_\xi(\xi) \:d\xi&= \int_0^{1/4} \frac{n!}{(k!)^2} \, \xi\, (2h(\xi))^k (1 - 2h(\xi))^k \cdot 2h'(\xi) \:d\xi \\&= \int_0^1 \frac{n!}{(k!)^2} \, g(x/2)\, x^k(1 - x)^k \:dx \\&= \int_0^1 \frac{n!}{(k!)^2} \,\left[\frac{x}{2} - \frac{x^2}{4}\right]\, x^k(1 - x)^k \:dx \\&= \frac{n!}{(k!)^2} \left[\frac{k! (k + 1)!}{2(n + 1)!} - \frac{k!(k + 2)!}{4(n + 2)!}\right] \\&= \frac{3n + 5}{16n + 32}\end{align*}

which is the final answer.