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

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]]

-
"Mathematica hangs when I try to compute it" - how about pasting in what you fed to Mathematica? – J. M. Aug 16 '11 at 7:01
Thanks, J.M. I just did. – OctaviaQ Aug 16 '11 at 7:09

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 $L_a=a^{-1}E(X;X\ge m_a)$.
@Didier: I guess the problem I'm running into is that the fraction $a$ changes as $n$ increase. When $n = 4$, $a=3/4$ , when $n = 8$, $a = 5/8$, etc. But wait: since $a = \frac{1}{2} + \frac{1}{n},$, the $\frac{1}{n}$ become 0 at the limit, so I guess I can treat this the same as $a=\frac{1}{2}$ ? – OctaviaQ Aug 16 '11 at 14:59
@Didier: I'm really stuck on something. You write $L=E(X\mid X\ge m)$ I thought $F_{X|X\geq m}(x) =\frac{F(x)}{(1-F(m)}$ When I take the derivative wrt $x$ to get $f_{X|X\geq m}(x)$ and then multiply by $x$ and take the integral from $m$ to $1$ to get $E$, I get a different answer than using your equation. I don't understand why, and I also don't understand what the semi-colon notation means (and Google is no help on that one, I've spent hours!). Would you be willing to offer a little more clarification? – OctaviaQ Aug 16 '11 at 17:33
@JandR: OK, consider the event $A=[X\ge m]$, then $L=E(X\mid A)=E(X\mathbf{1}_A)/P(A)$ (and $E(X;A)$ is just a shorthand for $E(X\mathbf{1}_A)$). But the distribution of $X$ conditionally on $A$ is not what you write. Rather, its CDF $G$ is defined by $G(x)=P(X\le x\mid A)=P(X\le x,X\ge m)/P(A)$ for every $x$. Hence $G(x)=0$ if $x<m$ and $G(x)=(F(x)-F(m))/F(m)=2F(x)-1$ if $x\ge m$. The PDF $g$ is defined by $g(x)=0$ if $x<m$ and $g(x)=2f(x)$ if $x\ge m$. Hope this clarifies things (just tell me if some problems remain). – Did Aug 16 '11 at 21:19