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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.

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    $\begingroup$ 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? $\endgroup$ – Glen_b Jan 28 '14 at 6:28
  • $\begingroup$ If that's hard, try $n=3$ and $k=2$ and list the whole thing out. $\endgroup$ – Glen_b Jan 28 '14 at 6:34
  • $\begingroup$ Oh I see thanks. The key is that for discrete rvs some order statistics may take on the same values. $\endgroup$ – DanRoDuq Jan 28 '14 at 15:55
  • $\begingroup$ 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. $\endgroup$ – Glen_b Jan 28 '14 at 21:57
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As mentioned in comments, it's clearly not true for discrete random variables.

The problem is, as the original poster suggested in comments, that we can get ties.

The nonzero probability of ties make the equality

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

- which works in the continuous case - untrue in general.

(This is a familiar problem when dealing with nonparametric tests.)

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