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Lehmann and Casella (http://www.springerlink.com/content/978-0-387-98502-2) state it as an example (Example 1.6.10) that for arbitrary continuous distributions, the order statistics are always sufficient. Is this a theorem, or are pathological examples known to exist? Are there necessary conditions for the order statistics to constitute a set of sufficient statistics?

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Sufficient for what parameter(s)? – onestop Dec 21 '11 at 20:28
every order statistic? If you know every order statistic then you know the entire data set. – Macro Dec 21 '11 at 20:46
@Macro Yes, but knowing the entire data set is not the same as knowing all the information in the sample. – fg nu Dec 21 '11 at 21:09
@onestop Good point. For the parameters of the prototypical pdf of an iid sample. – fg nu Dec 21 '11 at 21:11

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The order statistics are just the sorted data values, so for any case where the data is univariate iid, the order statistics have the exact same information as the original data (just in a different order). If the order in the data matters (not iid, e.g. time series) then the order statistics don't have that information and that would be one case where they were not sufficient. Another case would be non-univariate cases, the order statistics of X and the order statistics of Y would be sufficient for the X and Y distributions seperately, but not for covariance or correlation parameters.

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Thanks. That makes complete sense. I guess I was looking more for weird probabilistic constructions where the full set of order statistics is not sufficient for the parameters of the distribution of iid data. – fg nu Dec 21 '11 at 21:12

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