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I am not able to find the minimal sufficient statistic for the following density function:

$$f(x_i;\theta) = 2(1+\theta-x_i)I_{\theta \le x_i \le \theta+1}$$

The function does not belong to the exponential family distribution and so I apply the the Lehmann Scheffé Theorem, from which I get that a minimal sufficient statistic should be:

$$T(\mathbf x) = (\min(x_i), \max(x_i), \prod(1+\theta+x_i))$$

But since a statistic cannot depend on the parameter, it is wrong for sure.

StubbornAtom
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Lucas
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  • Do you mean the Factorisation theorem when you say Lehmann-Scheffe? – StubbornAtom Aug 18 '18 at 19:37
  • No, with the Factorisation theorem we just find a sufficient statistic; here instead I need mininality. – Lucas Aug 18 '18 at 22:27
  • @Lucas : The Lehmann–Scheffé theorem draws conclusions from completeness, and say the conditional expected value of an unbiased estimator, given a statistic that is both complete and sufficient, is the UMVUE. The theorem does not show minimality of a sufficient statistic. – Michael Hardy Aug 19 '18 at 19:03
  • I think you need to tell us what you thought the Lehmann–Scheffé theorem says, since otherwise we can't tell what you did. – Michael Hardy Aug 19 '18 at 19:06
  • Yes, I know. There is another result by Lehmann-Scheffé (1950) in which they show how to find a minimal sufficient statistic, by using the ratio of two likelihood functions. – Lucas Aug 19 '18 at 19:09
  • I used the same approach of this example: https://math.stackexchange.com/questions/2885935/proving-minimal-sufficiency-using-lehmann-scheffe-approach-when-joint-pdf-is-onl – Lucas Aug 19 '18 at 19:22
  • The first step is to find a sufficient statistic, then look for minimal sufficiency of that statistic. For this problem, I think we only have the trivial sufficient statistics $(X_1,X_2,\ldots,X_n)$ or $(X_{(1)},\ldots,X_{(n)})$ for $\theta$. Either of them could be minimal sufficient. The $T$ you found (using Factorization theorem presumably) is not a statistic in the first place, as you say. – StubbornAtom Aug 19 '18 at 19:27
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    Actually, it is the sample $(X_1,\ldots,X_n)$ which could be minimal sufficient, not the order statistics. – StubbornAtom Aug 19 '18 at 19:33

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