# Sufficiency of order statistics

I am told the following proof is incorrect, but I cannot understand why.

Consider $X_{(1)}, \ldots, X_{(n)}$ are the order statistics of a random sample of size $n$. I want to show that the order statistics are sufficient. So I wrote down:

$$P(X_1, \ldots, X_n|X_{(1)}, \ldots, X_{(n)})= \tfrac{1}{n!}$$

as given the vector of order statistics, there are $n!$ possibilities for the sample $X_1, \ldots, X_n$.

As we are in an i.i.d. case, then each vector is equiprobable, and so the equality follows. I am told this is not true specially in the case of discrete random variables. I don't see how it is wrong though. Any explanation would be great.

• Try taking something simple. Perhaps start with a collection of Bernoulli random variables with $p=0.9$. Imagine you observe $k$ 1's (and $n-k$ 0's); what does the LHS work out to be? – Glen_b Jan 28 '14 at 6:28
• If that's hard, try $n=3$ and $k=2$ and list the whole thing out. – Glen_b Jan 28 '14 at 6:34
• Oh I see thanks. The key is that for discrete rvs some order statistics may take on the same values. – DanRoDuq Jan 28 '14 at 15:55
• Exactly. Would you mind writing an answer to your fine question? If you don't want to, I could write something, but I feel your answer will probably explain the issue better. – Glen_b Jan 28 '14 at 21:57

$P(X_1, \ldots, X_n|X_{(1)}, \ldots, X_{(n)}) = \frac{1}{n!}$