The limit of the expectation of the top half+1 order stats of $n$ draws of $X$ as $n\to\infty$?

Question

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]]

Answer 1

The empirical distribution of the sample $(X_{k,n})_{1\le k\le n}$ converges to $F$ when $n\to\infty$ (and some call this result the fundamental theorem of statistics). The distributions you are considering are uniformly bounded (everything happens in $[0,1]$) hence, when $n\to\infty$ (or, if $n=2^{i+1}$, when $i\to\infty$), the empirical mean of the upper half of the sample $$ \frac2n\sum_{k=n/2}^nX_{k,n}, $$ with or without the expectations, converges and its limit is $L=E(X\mid X\ge m)$, for any median $m$ of $F$. Thus, $F(m)=\frac12$ and $$ L=2E(X; X\ge m)=2E(X)-2E(X;X\le m). $$ Numerically, $m=1-\sqrt{1-1/\sqrt2}$ and, for every $x$ in $[0,1]$, $$ E(X;X\le x)=(4/3)x^3-(3/2)x^4+(2/5)x^5, $$ from which the value of $L$ should be easy to compute.

Likewise, for any $a$ in $(0,1)$, the empirical mean of the top $a$ part of the sample $$ \frac1{an}\sum_{k=(1-a)n}^nX_{k,n}, $$ converges and its limit is $L_a=E(X\mid X\ge m_a)$, for any $(1-a)$ quantile $m_a$ of $F$. Thus, $F(m_a)=1-a$ and $L_a=a^{-1}E(X;X\ge m_a)$.