This is problem 1.5.8 in Mathematical Statistics by Bickel and Doksum. It seems straightforward, but I am not sure if my proof is lacking in some way. It doesn't seem quite correct.

Question Let $X_1, X_2,..., X_n$ be a sample from some continuous distribution $F$ with density $f$, which is unknown. Treating $f$ as a parameter, show that the order statistics $X_{(1)},...X_{(n)}$ (cf. Problem B.2.8) are sufficient for $f$.

My Solution Attempt The sample density is simple the product of the $f(x_i)$. This yields


We apply the factorization theorem where $h(x)=1$


(here $\theta=X_{(1)},...X_{(n)}$). And so because we have factorized the sample density, the statistics are sufficient.

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    $\begingroup$ This is not completely correct, as $g(\theta)\ne\prod_{i=1}^nf(X_{i})$. $\endgroup$
    – Xi'an
    Commented Apr 3, 2015 at 7:43
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    $\begingroup$ I'm probably completely off-base here, but aren't the $n$ order statistics just a permutation of the original sample? Thus no information is lost, so by definition they must be sufficient. $\endgroup$
    – Hong Ooi
    Commented Apr 3, 2015 at 7:58
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    $\begingroup$ @HongOoi yep, basically what’s said here is that loosing the drawing order doesn’t matter. This is kind of obvious if the $X_i$'s are independent... however it is an interesting exercise to be able to prove this intuitive fact. $\endgroup$
    – Elvis
    Commented Apr 3, 2015 at 8:44

1 Answer 1


The joint distribution off all order statistics $X_{(1)}, \dots, X_{(n)}$ is $$f_o(y_1, \dots, y_n) = n! f(y_1) \times \cdots \times f(y_n)$$ for $y_1 \le \cdots \le y_n$.

Thus, the joint distribution of $X_1 ,\dots, X_n$, given $X_{(1)}, \dots, X_{(n)}$ does not depend of the density $f$! We have $$ Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!} $$ when the multisets of the $x_i$'s and of the $y_i$'s are equal.

The intuition is that loosing the drawing order doesn’t matter, for the $X_i$'s are independent.

PS Re-reading this answer long after, I tend to think that the conclusion is clear in itself: given $y_1 < \dots < y_n$, $Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!}$ means that all re-orderings of the $y_i$'s are equally probable, whatever the density $f$ is. In fact I don't see how to prove the starting result I used (the joint distribution of order statistics) without proving this at the same time... This just comes from the fact that all the points $x = (x_1, \dots, x_n)$ obtained from a permutation of the $y_i$ have the same density $\prod_i f(x_i) = \prod_i f(y_i)$.


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