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:

$$f_{Y_{(1)}, \ldots, Y_{(n)}}(y_1, \ldots, y_n) = n! \prod_{i=1}^n f_i(y_i),~ y_1 \leq \ldots y_n \tag{2}$$


1 Answer 1


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 $n!$.

  • $\begingroup$ Thanks for great clarification! Then how can I sample the ordered statistics realizations from $f_{Y_{(1)}, \ldots, Y_{(n)}}$? Following my notation for case of i.i.d. variables, we just need to generate $n$ values from $f(x)$ and sort them. $\endgroup$ Feb 20, 2020 at 15:14
  • $\begingroup$ @PiotrSemenov: Indeed; that's the definition of the order statistics; you don't need their PDF for that purpose. $\endgroup$
    – joriki
    Feb 20, 2020 at 15:15
  • $\begingroup$ Right. But imagine you know nothing about the underlying distributions $f_i$. You know only joint PDF $f_{Y_{(1)}, \ldots, Y_{(n)}}$. $\endgroup$ Feb 20, 2020 at 15:28
  • $\begingroup$ @PiotrSemenov: In which form do you have $f_{Y_{(1)}, \ldots, Y_{(n)}}$? $\endgroup$
    – joriki
    Feb 20, 2020 at 16:49
  • $\begingroup$ I guess it is too general problem I've just asked for. Let it be the normal multivariate distribution defined for ordered arguments. It has a non-zero mean vector and a diagonal covariance matrix. Anyway thanks for your answer! $\endgroup$ Feb 20, 2020 at 16:56

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