# Joint pdf of functions of order statistics

Let $Y_1 < Y_2 <\ldots <Y_{10}$ be the order statistics of a random sample from a continuous type distribution with cdf $F(x)$. How would I begin to show that the joint distribution of $V_1=F(Y_4)-F(Y_2)$ and $V_2=F(Y_{10})-F(Y_6)$ is given by:

$$h(v_1,v_2)= \frac{10!}{1!3!4!} v_1 v_2^3 (1-v_1-v_2)^4$$

Any hints? Thank you.

• @wolfies This is all the information I am given. – JohnK Dec 4 '13 at 16:20
• @wolfies This is precisely what I am suggesting. This is a general condition that holds and people who have seen order statistics, tolerance intervals and coverages before might be able to help me derive it. – JohnK Dec 4 '13 at 16:28
• And that's all the information you need: $F$ is the probability integral transform so indeed $F(Y_i)$ is distributed as the $i^\text{th}$ order statistic from a uniform distribution. For a non-rigorous (but nevertheless accurate) heuristic, think of this order statistic as dividing the interval $[0,1]$ into three bins of length $F(Y_i)$, $dY_i$ (an infinitesimal), and $1-F(Y_i)$, into which precisely $i-1$, $1$, and $n-i$ of the values fall (that's where the multinomial coefficients come from). – whuber Dec 4 '13 at 16:31
• @whuber It's the functions that I having trouble comprehending. And by functions I mean the difference between the cdf values$v_1,v_2$. Why is the unit interval split like that precisely? – JohnK Dec 4 '13 at 16:34
• Second time this has come up - lol - and I learn not from my errors :) – wolfies Dec 4 '13 at 16:36

This is a nonrigorous demonstration (at least as far as those who don't use nonstandard analysis are concerned) that provides some intuition for working with order statistics.

The function $F$ makes the distribution uniform. Consider four positions $0\lt x_2\lt x_4\lt x_6\lt x_{10}\lt 1$ and four corresponding "infinitesimal" lengths $dx_2, \ldots, dx_{10}$. These determine eight locations within the unit interval which thereby divide it into nine bins corresponding to the colored inequality symbols:

$$0\color{red}\le x_2\color{red}\lt x_2+dx_2\color{red}\le x_4\color{red}\lt x_4+dx_4\color{red}\le x_6\color{red}\lt x_6+dx_6\color{red}\le x_{10}\color{red}\lt x_{10}+dx_{10}\color{red}\le 1.$$

By ignoring events of zero probability, we readily deduce that the first bin, $[0, x_2),$ is occupied by precisely one value, $x_1,$ because $0\le x_1\lt x_2$ (by definition of order statistics). The use of "$\lt$" instead of "$\le$" ignores the zero chance that $x_1=x_2$; this is where continuity is needed.

Similarly, the bin $[x_2+dx_2, x_4)$ is occupied only by $x_3$; the bin $[x_4+dx_4, x_6]$ is occupied only by $x_5$, and $[x_6+dx_6, x_{10})$ is occupied by $x_7, x_8,$ and $x_9$. The four infinitesimal bins $[x_2, x_2+dx_2)$, $[x_4, x_4+dx_4)$, $[x_6, x_6+dx_6),$ and $[x_{10}, x_{10} + dx_{10}),$ are occupied by $x_2, x_4, x_6,$ and $x_{10},$ respectively. Finally, the bin $[x_{10}+dx_{10}, 1]$ has no occupants.

The chance that a random uniform value occupies any bin is equal to the bin's length. The chance of any particular collection of occupancy numbers is given by a multinomial distribution; in this case it equals (up to the lowest order of infinitesimals)

$$\binom{10}{1,1,1,1,1,1,3,1,0}x_2^1(x_4-x_2)^1(x_6-x_4)^1(x_{10}-x_6)^3(1-x_{10})^0dx_2dx_4dx_6dx_{10}.$$

(By definition, the multinomial coefficient $\binom{10}{1,1,1,1,1,1,3,1,0}=\frac{10!}{1!1!1!1!1!1!3!1!0!}=\frac{10!}{3!}.$)

Changing variables to $x_4 = x_2+v_1$ and $x_6 = x_{10}-v_2$ in order to represent the ranges $v_1=x_4-x_2$ and $v_2=x_{10}-x_6$ turns this into

$$\left(\frac{10!}{3!}v_1v_2^3dv_1dv_2\right)\left(x_2(x_{10}-x_2-v_1-v_2)) dx_2dx_{10}\right).$$

Performing the integrations for $x_2$ and $x_{10}$, subject to the order restrictions $0\le x_2\le 1-v_1-v_2$ and $x_2+v_1+v_2\le x_{10}\le 1$, produces the joint PDF of $(v_1,v_2)$ given in the question. Its natural support is where $0\le v_1\le 1,$ $0\le v_2\le 1$, and $v_1+v_2\le 1.$

A computer algebra system can make the calculations less painful. Here is a Mathematica 9 solution:

Integrate[
PDF[OrderDistribution[{UniformDistribution[], 10}, {2, 4, 6, 10}]][{x2, x4, x6, x10}]
/. {x4 -> x2 + v1, x6 -> x10 - v2}, {x2, 0, 1}, {x10, x2, 1}]


$\begin{array}{cc} \{ & \begin{array}{cc} 25200 \text{v1} \text{v2}^3 (\text{v1}+\text{v2}-1)^4 & 0<\text{v1}<1\land \text{v2}\geq 0\land \text{v1}+\text{v2}<1 \\ 0 & \text{True} \\ \end{array} \\ \end{array}$

• Reviewing your answer, the only trouble I am having is justifying the order restrictions for $x_2$ and $x_{10}$. Would you mind giving me the intuition there? – JohnK Dec 4 '13 at 22:16
• The restrictions are immediate from the definitions of order statistics and the $v_i$: we know that two of the bins between $x_2$ and $x_{10}$ have widths of $v_1$ and $v_2$ (while the rest can be arbitrarily small), so $x_2+v_1+v_2$ cannot exceed $x_{10}$. – whuber Dec 4 '13 at 22:31
• Statistics is fascinating, isn't it? – JohnK Dec 4 '13 at 22:36