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