Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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]]
share|improve this question
    
"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. –  Jand Aug 16 '11 at 7:09

1 Answer 1

up vote 4 down vote accepted

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 $L_a=a^{-1}E(X;X\ge m_a)$.

share|improve this answer
1  
@JandR Just to supplement Didier's excellent answer (+1), approximations of order statistics by quantiles of the parent distribution is discussed at length in "A First Course in Order Statistics" by Arnold et al. –  Sasha Aug 16 '11 at 12:24
    
@Sasha-Thanks! All my probability and statistics books gloss over order statistics in a few pages. This will be really helpful! –  Jand Aug 16 '11 at 13:48
    
@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 $a=\frac{1}{2}$ ? –  Jand 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? –  Jand Aug 16 '11 at 17:33
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.