Can anyone help me compute the limit of the average of the top half +1 of order marginal order distribution of $n$ draws from $X$, as $i\to\infty$?
Specifically, the limit as $i\to\infty$ of $\frac{1}{2^{i+1}-2^i+1} \sum_{k=2^i}^{2^{i+1}}\mathbb{E}X_{k,2^{i+1}}$
where $X$ is represented by cdf $F(x) = (1-(1-x)^2)^2$ and $X_{k,n}$ is the $k$th smallest order statistic of $n$ draws from $X$.
Mathematica hangs when I try to compute it and I don't know how to reduce it.
The most simple code I've been able to generate excludes (for the time being) the divisor for the average and the "1-" in the integrand of the expectation, and still hangs.
Code as follows:
F = (1 - (1 - x)^2)^2; Gk[z_, i_, n_] := Sum[n! (z^k)*(1 - z)^(n - k)/(k!*(n - k)!), {k, i, n}]; Assuming[0 < x < 1 && Element[i, Integers] && i > 0, Limit[Integrate[ Sum[Gk[F, k, 2^(i + 1)], {k, 2^i, 2^(i + 1)}], {x, 0, 1}], i -> Infinity]]