TITLE: Joint density of order statistics
QUESTION [9 upvotes]: 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$
$\textbf{(i)} \quad$the 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

REPLY [6 votes]: 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 by
$$F(y) := \begin{cases}
n! \prod_{i=1}^{n}{f(y_i)}, & y_1 < \ldots < y_n, \\
0, & \text{ otherwise} .
\end{cases}$$