We can say a random sample means that $ X_1, X_2, \ldots , X_n $ are iid from a same distribution. But Can we use a same analogue to order statistics? I mean, Can we convert $ X_1, X_2, \ldots , X_n $ to order statistics $ Y_1, Y_2, \ldots, Y_n $ without losing properties of random sample? As I write this question, it is get in my head that because order statistics are relevant to each other, So they never be a random sample. Is that right? Could you explain more sophisticated answer to me?
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What is your definition of "random sample"? – David K Jul 14 '21 at 02:02
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I mean a random sample is a bunch of variables that are iid. – Minho Kang Aug 04 '21 at 02:19
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@MinhoKang After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Jose Avilez Apr 27 '22 at 01:46
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If by "random sample" you mean an iid collection of random variables, then no, the order statistics do not constitute a random sample. In particular, the order statistics are not identical because you are guaranteed that $$Y_1 \leq Y_2 \leq \ldots \leq Y_n$$
Moreover, if you assume that the $X_i$'s admit densities, then you can argue that the $Y_i$'s are not independent by observing that their joint densities do not factor into the product of their marginal densities (e.g. see here).
Jose Avilez
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