# Joint density of order statistics

I need some help to understand the following proposition (mainly to understand how it is proven):

Let $Y_1,Y_2...,Y_n$ be $n$ random variables which are independent, identically distributed random variables with probability density function $f$. The joint density of the order statistics $Y_{(1)},Y_{(2)},..,Y_{(n)}$ is given by:

\textbf{(1)}\quad$$f(y_1,y_2,...,y_n)= n!\prod\limits_{i=1}^{n}f(y_1) \qquad y_1 <y_2...<y_n \textbf{(i)} \quadthe preceding follows since: (Y_{(1)},Y_{(2)},..,Y_{(n)}) will equal (y_1,y_2...,y_n) if (Y_1,Y_2,...Y_n) is equal to any of the n! permutations of (y_1,y_2...,y_n) ** this part i get; they are saying that the ordered statistics is equal to the tuple (y_1,y_2...,y_n) only if the unordered (Y_1,Y_2,...Y_n) is equal to a permutation of (y_1,y_2...,y_n). But what does have for consequence in \textbf{(1)} ?? ** \textbf{(ii)}\quad the probability density that (Y_1,Y_2,...,Y_n) is equal to y_{i_{1}},...,y_{i_{n}} is \prod_{j=1}^{n}f(y_{i_{j}}) = \prod_{j=1}^{n}f(y_j) when i_1,...i_n is a permutation of 1,2...,n ** here Iam completely lost; the probability density that (Y_1,Y_2,...,Y_n) is equal to y_{i_{1}},...,y_{i_{n}} what do they mean? probability density (function)? I don't know what they mean at all with this statement • For your second point, they basically mean that the joint density of your sample (not the ordered statistics) is independent of the order in which you observe (y_1,\ldots,y_n) – M Turgeon May 15 '14 at 15:07 ## 2 Answers By hypothesis, all the Y_i are identically distributed with density function f. This implies that$$ \forall 1 \leq i,j \leq n: P(Y_i = Y_j) = 0 $$This implies for the order statistics that$$ Y_{(1)} < \ldots < Y_{(n)}$$almost surely. Consequently, (Y_{(1)}, \ldots, Y_{(n)}) \in \mathbb{R}^n_{\Delta} a.s., where$$ \mathbf{R}^n_{\Delta} := \{(y_1, \ldots, y_n) \in \mathbb{R}^n \mid y_1 < \ldots < y_n\}. $$Consequently, it suffices to calculate the joint density on a generator of the Borel algebra of \mathbb{R}^n_{\Delta} with is given by sets of the type$$ ]t_0, t_1] \times \ldots \times ]t_{n-1} \times t_n], \; t_0 < \ldots < t_n. We denote by \mathfrak{S}_n the group of all permutations of \{1, \ldots, n\} and calculate \begin{align*} P(Y_{(1)} \in ]t_0, t_1], \ldots, Y_{(n)} \in ]t_{n-1}, t_n]) &= P(\dot \bigcup_{\pi \in \mathfrak{S}_n}{ \{Y_{\pi(1)} \in ]t_0, t_1], \ldots, Y_{\pi(n)} \in ]t_{n-1}, t_n] \} }) \\ &= \sum_{\pi \in \mathfrak{S}_n}{ P(\{Y_{\pi(1)} \in ]t_0, t_1], \ldots, Y_{\pi(n)} \in ]t_{n-1}, t_n] \}) } \\ &= \sum_{\pi \in \mathfrak{S}_n}{ P(Y_{\pi(1)} \in ]t_0, t_1]) \ldots P(Y_{\pi(n)} \in ]t_{n-1}, t_n]) } \\ &= \sum_{\pi \in \mathfrak{S}_n}{\prod_{i=1}^{n}\int_{]t_{i-1},t_i]}{f(x_i)dx_i}} \\ &= n! \prod_{i=1}^{n}\int_{]t_{i-1},t_i]}{f(x_i)dx_i} \\ &= \int_{]t_0, t_1] \times \ldots ]t_{n-1},t_n]}{n! \prod_{i=1}^{n}{f(x_i)}d(x_1, \ldots, x_n)} \end{align*} For the second equality it is crucial that these sets are disjoint for the various \pi \in \mathfrak{S_n} and for the third that the Y_i are independent. Alltogether this implies that the joint densiy of the order statistics is given byF(y) := \begin{cases} n! \prod_{i=1}^{n}{f(y_i)}, & y_1 < \ldots < y_n, \\ 0, & \text{ otherwise} . \end{cases}$It may help to see a simple example of small size. Suppose$n = 2$. Then our order statistics consist of the smaller of the two observations, and the larger. So if, say, we observed$Y_1 = 3, Y_2 = 5$, then$Y_{(1)} = Y_1 = 3$and$Y_{(2)} = Y_2 = 5$; nothing has changed. But if the first observation was$5$and the second was$3$, we would have to interchange them to write$Y_{(1)} = 3, Y_{(2)} = 5$as we did before. Notice then that for a given realization of the ordered set of random variables, there is in general two ways to have observed the original variables so as to obtain the ordered result (we can ignore the case where the observations are equal because for continuous random variables, such an outcome has infinitesimal probability). Therefore, in general, for any given realization of order statistics of a sample of size$n$, there are$n!$possible ways that we could have observed a sample that, upon ordering, would yield such a realization. That's basically why the joint density is scaled the way it is. • This illustrates that$(Y_{(1)}, \ldots, Y_{(n)}$will equal$(y_1, \ldots, y_n)$if$(Y_1, \ldots, Y_n)$is equal to any of the$n!$permutations of$(y_1, \ldots, y_n)\$. But how exactly does one prove the claim concerning the joint probability density? Such a permutation will in general not be unique. – Meneldur Nov 20 '14 at 16:30
• Yes the permutation is unique (except on a null event). – Did Dec 1 '14 at 9:22