# Proof that n-order statistics are sufficient for a sample of size n

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

$$\prod_{i=1}^nf(X_{i})$$

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

$$g(\theta)=\prod_{i=1}^nf(X_{i})$$

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

• This is not completely correct, as $g(\theta)\ne\prod_{i=1}^nf(X_{i})$. – Xi'an Apr 3 '15 at 7:43
• 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. – Hong Ooi Apr 3 '15 at 7:58
• @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. – Elvis Apr 3 '15 at 8:44

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 $$\prod_i f(x_i) = \prod_i f(y_i)$$.