I am trying to get an equation that will show the rate of change of the expected value of $\frac{2^n}{2}$th lowest of $2^n$ draws from $X$ as $n$ increases (where n >1). Let's call the order statistic in question $X_n$ for shorthand. Let's define function $g(n) = X_{n+1} -X_n$ as the function I'm looking for. So, for example, $g(2) = \mathbb{E}X_{4,8} - \mathbb{E}X_{2,4}$ , and $g(3) = \mathbb{E}X_{8,16} - \mathbb{E}X_{4,8}$
$X$ is represented by cdf $F(x) = (1-(1-x)^2)^2$. The median of this is $m=1-\sqrt{1-1/\sqrt2}$. I know that as $n\to\infty$, $\mathbb{E}X_n \to m$, and $g(n) \to 0$. I can manually calculate the values of $g(2) . . . g(6)$ using Mathematica, but the equations I use hang Mathematica after that. The equation I use is:
$g(n) = \int_0^1 (\sum_{k=\frac{2^{n}}{2}}^{2^{n}} \frac{(2^{n})!}{k!(2^{n}- k)!}F(x)^k(1-F(x))^{2^{n}- k} - \sum_{k=\frac{2^{n+1}}{2}}^{2^{n+1}} \frac{(2^{n+1})!}{k!(2^{n+1}- k)!}F(x)^k(1-F(x))^{2^{n+1}- k}) \, dx $
Is there any way I can create a manageable $g(n)$ (or otherwise managebley calculate $\mathbb{E}X_n$) for high values of $n$? In terms of underlying problem, what I'm really trying to ultimately calculate is $h(n) = (\mathbb{E}X_{2^n,2^n} -\mathbb{E}X_{\frac{2^n}{2},\frac{2^n}{2}})- (\mathbb{E}X - \mathbb{E}X_{\frac{2^n}{2}, 2^n}) $. Doing manual caluclations for the first 6 values, this function starts positive, increases, then starts decreasing, with its limit being negative (note the amount of initial increasing increases when I change $F(x)$ to $F(F(x))$ and so on). My first priority, I suppose, is to know when $h(n) = 0$, but having a manageable way to calculate its value at any $n$ would be fantastic.
Any ideas?