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 $1\le i\le n$, 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 

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

from which it follows that the cdf of $Y$ on $[0,1/4]$ is $F_Y(y)=2h(y)$. 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

$$f_{\xi }(\xi )=\frac{n!}{(k!{)}^2}{F}_Y(\xi {)}^k(1-F_Y(\xi ){)}^k{f}_Y(\xi )=\frac{n!}{(k!{)}^2}(2h(\xi ){)}^k(1-2h(\xi ){)}^{k}2h^{\prime }(\xi )$$

and hence, the expected value of the median is

$$\begin{array}{rl}\mathbb{E}[\xi ]=\int \xi f_{\xi }(\xi )d\xi & ={\int }_0^{1/4}\frac{n!}{(k!{)}^2}\xi (2h(\xi ){)}^k(1-2h(\xi ){)}^k\cdot 2h^{\prime }(\xi )d\xi \\ & ={\int }_0^1\frac{n!}{(k!{)}^2}g(x/2)x^k(1-x{)}^{k}dx\\ & ={\int }_0^1\frac{n!}{(k!{)}^2}[\frac{x}{2}-\frac{x^2}{4}]x^k(1-x{)}^{k}dx\\ & =\frac{n!}{(k!{)}^2}[\frac{k!(k+1)!}{2(n+1)!}-\frac{k!(k+2)!}{4(n+2)!}]\\ & =\frac{3n+5}{16n+32}\end{array}$$

which is the final answer.