# Does having the same order statistics imply the same distribution?

I'm stuck at the following problem:

Say $$X_i \overset{\text{iid}}{\sim} \operatorname{Exp}(\lambda)$$ for $$i = 1, \ldots, n$$. Denote $$X_{(1)}, \ldots, X_{(n)}$$ the order statistic from the $$n$$ samples.

And I have some random variables with the same distribution (not necessarily independent) $$Y_1, \ldots, Y_n$$, where its order statistics have the same distribution with those of $$X$$, i.e. $$Y_{(1)} \overset{d}{\equiv} X_{(1)}, \ldots, Y_{(n)} \overset{d}{\equiv}X_{(n)}$$

Does this imply $$Y \equiv X$$?

• Hint: Examine just one of the order statistics and relate its distribution to the underlying distribution. If you select a particularly nice order statistic, the comparison will be simple. – whuber Nov 22 '18 at 15:02
• You are using the word "sample" incorrectly. You don't have $n$ samples; you have one sample consisting of $n$ independent observations. – Michael Hardy Nov 22 '18 at 15:58

The distribution of $$(Y_1,\ldots,Y_n)$$ is identified by the joint distribution of the order statistic $$(Y_{(1)},\ldots,Y_{(n)})$$ and of the rank statistic $$(\sigma_1,\ldots,\sigma_n)$$. In the case of an independent sample, the later distribution is uniform over the set of all possible permutations. But it could be any other distribution over thee set of all possible permutations.
Second, while the marginal distributions of the components of the order statistic of $$Y$$ are the same as the marginal distributions of the components of the order statistic of $$X$$, this does not imply that the joint distribution is the same.