Let $X_1, \ldots, X_n$ be $n$ i.i.d. variables with probability density $f(x)$. Let $X_{(1)}, \ldots, X_{(n)}$ be the ordered statistics. Then the joint probability density of all $n$ order statistics is:
$$f_{X_{(1)}, \ldots, X_{(n)}}(x_1, \ldots, x_n) = n! \prod_{i=1}^n f(x_i),~ x_1 \leq \ldots, x_n \tag{1}$$
Assume now that $Y_1, \ldots, Y_n$ are not identically distributed but still independent variables with probability densities respectively $f_i(x)$. What is the joint probability density $f_{Y_{(1)}, \ldots, Y_{(n)}}(y_1, \ldots, y_n)$ of all the order statistics for $Y_1, \ldots, Y_n$?
I wonder if eq. (1) can be generalized in intuitive way:
$$f_{Y_{(1)}, \ldots, Y_{(n)}}(y_1, \ldots, y_n) = n! \prod_{i=1}^n f_i(y_i),~ y_1 \leq \ldots y_n \tag{2}$$
Best Answer
No, you can’t generalize in this manner, since each of the $Y_{(i)}$ could come from each of the distributions whereas your formula arbitrarily assigns each of them to some particular distribution.
First taking one step back, we should correct $(1)$; the correct form is
$$ f_{X_{(1)}, \ldots, X_{(n)}}(x_1, \ldots, x_n) = [x_1\le x_2\le\cdots\le x_n]n! \prod_{i=1}^n f(x_i)\;, $$
where $[c]$ is the indicator function for the condition $c$. The corresponding expression in the case where the $Y_i$ are not identically distributed (but still independent; you didn’t mention that) is
$$ f_{Y_{(1)}, \ldots, Y_{(n)}}(y_1, \ldots, y_n) = [y_1\le y_2\le\cdots\le y_n]\sum_{\sigma\in S_n} \prod_{i=1}^n f_i(y_{\sigma(i)})\;, $$
which is readily seen to reduce to the specific form for the case of identical distributions, and explains why it contains a factor of $n!$.