# Asymptotic distribution of the average of $p$ sample-percentile

Here is a question which seems to be intuitively simple, yet I am not able to show it rigorously.

Any help, whether it is carrying out my attempt or to suggest any other approach - would be greatly appreciated.

Also, please, see the note on a similar question.

The problem

Let $\{x_i\}_{i=1}^n$ iid samples drawn from uniform distribution $X_i \sim U[0,1]$. What is the asymptotic distribution of sum of the top $0<p<1$ sample percentile?

That is, what is the asymptotic distribution of $S_p(n)$, where

$$S_p(n) := \frac{1}{p\cdot n}\sum_{i = p \cdot n}^n X_{(i)}\,.$$

(with $X_{(i)}$ is the i-th order statistic. Additionally, to spare the clutter in the notation, I have ignored rounding to the integer in the summation over index $i= \lfloor p \cdot n \rfloor$)

Intuitively, I would argue that when $n$ grows high the top $p$ percentile is uniformly distributed on the interval $[p ,1]$ therefor, the sum is asymptotically equivalent to the sum of uniform random samples drawn from $U[p, 1]$. Which leads to

$$\sqrt{n}\left(\frac{S_p(n) - \mathbb{E}S_p(n)}{\sqrt{\text{var}(S_p(n))}}\right) = \sqrt{n}\left(\frac{S_p(n) - \frac{(p+1)}{2}}{\frac{(1-p)}{\sqrt{12}}}\right) \to N(0, 1)\,.$$

Note:

There is a similar question asked here, I was not sure whether to post a comment there or ask a new question, eventually I did both and put a reference in both places.

Additionally, since one can always go from general distribution to uniform one by inverse cdf transform $F^{-1}(x)$, what I ask is a in a more general setting.

Outline of my Attempt to solve.

(As already noted, this is in the spirit of the answer to this question)

$\mathbf{1)}$ Look at the random vector $$(y, x_{m+1}, \ldots, x_{n}) := \left(\sum_{j=m}^n X_{(j)}, X_{(m+1)}, \ldots, X_{(n)}\right)\,.$$ $\mathbf{2)}$ Find the joint distribution of the vector from $\mathbf{1)}$. By change of variable for distribution of $$\left(X_{(m)}, X_{(m+1)}, \ldots, X_{(n)}\right)\,.$$

$\mathbf{3)}$ Integrate out $x_{m+1}, \ldots, x_n$ to get distribution for $\sum_{j=m}^n X_{(j)}$. (Stuck here)

$\mathbf{4)}$ See if this helps to find the asymptotic distribution. (Actually, not sure if the previous step really helps here)

Details

In general, the joint probability density of order statistics of iid samples drawn form continuous distribution $F(x)$ (see here): $$f(x_m ,\ldots, x_n) = n! f(x_n) \cdots f(x_{m})\frac{F(x_m)^{m-1}}{(m-1)!} \mathbb{I}\{0 <x_m<x_{m+1}< \cdots< x_n\}\,,$$

where $\mathbb{I}\{A\}$ is indicator of the set $A$.

For the uniform distribution the above yields $$f(x_m ,\ldots, x_n) = n! \frac{(x_m)^{m-1}}{(m-1)!} \mathbb{I}\{0 <x_m<x_{m+1}< \cdots< x_n\}\,.$$

Let change of variables $$\tau: \left(X_{(m)}, \ldots, X_{(n)}\right) \to \left(\underbrace{\sum_{j=m}^n X_{(j)}}_{Y}, X_{(m+1)}, \ldots, X_{(n)}\right)\,.$$ $$\tau^{-1}(Y, X_{(m+1)}, \ldots, X_{(n)}) \to \left(Y - \sum_{j=m+1}^n X_{(j)}\,, X_{(m+1)}\,, \ldots\,, X_{(n)}\right).$$ The determinant of the Jacobian of the inverse $\tau^{-1}$ is one. The distribution after the change of variables is $$f(y, x_{m+1}, \ldots, x_n) = n! \frac{(y - \sum_{j=m+1}^n x_{j})^{m-1}}{(m-1)!} \mathbb{I}\left\{0 <y - \sum_{j=m+1}^n x_{j}< x_{m+1}< \cdots< x_n\right\}\,.$$

Basically, at this point I want to be able to integrate

$$\int \cdots \int n! \frac{(y - \sum_{j=m+1}^n x_{j})^{m-1}}{(m-1)!} \mathbb{I}\left\{0 <y - \sum_{j=m+1}^n x_{j}< x_{m+1}< \cdots< x_n\right\} dx_{m+1}\cdots dx_{n}$$

$$= \int_{\frac{y}{2}}^y dx_{m+1} \int_{\frac{y-x_{m+1}}{2}}^y dx_{m+2} \cdots \int_{\frac{y- \sum_{j=1}^kx_j}{2}}^y d{x_{k+1}} \cdots \int_{\frac{y-\sum_{j=1}^{n-1}x_j}{2}}^y n! \frac{(y - \sum_{j=m+1}^n x_{j})^{n-m-1}}{(n-m-1)!} d{x_{n}} \,.$$

I am stuck, can't see an easy way to disentangle the resulting integral .

• For the case of exp. distribution, i.e., $X_i \sim \exp(1)$, one can show that $(X_{(m+1)} - X_{(m)}, \ldots, X_{(n)} - X_{(m)})| X_{m}$ is distributed as ord. statistic of $Y_i, n-m$ i.i.d samples from $exp(1)$, therefor $\frac{1}{m-n}\sum_{i=m+1}^n X_{(i)} - X_{(m)}| X_{(m)} = \sum_{i=1}^{n-m} Y_i - (n-m)X_{m} |X_{m}$. At the limit $n\to \infty$, $X_{(m}) \to F^{-1}(p)$ while $\sum_{i=1}^{n-m} Y_i$ has asymptotic normal distribution..., does this extends to the unconditioned version? Is this development only works for exponential distribution, and cannot help with uniform distribution? – them Aug 28 '16 at 13:26
• For problems like this, I would suggest doing some numerical experiments before going too deep into the math. That way you have some guidance on where the math should be headed. – GeoMatt22 Aug 28 '16 at 18:44
• @GeoMatt22 thanks for pointing out. I have also asked a similar question on exponential distribution here. The question with the exponential distribution is from a practice exam, but no solution is given there. So I know the other question should have relatively straightforward solution. – them Aug 28 '16 at 19:17