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I think I have some confusion going on when it comes to sufficient vs. minimally sufficient vs. complete statistic. I have read some posts here (ex: Finding complete sufficient statistic) but I am still not exactly sure I understand. So I summarized all the questions I have from all the different posts I have read here into one problem/question.

Let $\theta > 0$ be a parameter and let $X_1, ..., X_n$ be a random sample with pdf $f(x|\theta)=\frac{1}{2\theta}$ if $-\theta \leq x \leq \theta$ and $0$ otherwise.

a) Find the MLE of $\theta$

$L(\theta|X)=\frac{1}{(2\theta)^n}I(\theta)_{(max(-X_{(1)},X_{(n)}), \infty)}=\frac{1}{(2\theta)^n}I(|X|_{(n)})_{(0,\theta)}$.

So my question: How do you know if the MLE is $max(-X_{(1)},X_{(n)})$ or $|X|_{(n)}?$.

b) Is the MLE sufficient for $\theta$?

From a), regardless of which is the MLE, both $max(-X_{(1)},X_{(n)})$ and $|X|_{(n)}$ are sufficient by the factorization theorem. $(X_{(1)},X_{(n)})$ would also be sufficient because of the factorization theorem as well.

c) Is the MLE minimally sufficient for $\theta$?

Regardless of which one is the MLE, wouldn't all $max(-X_{(1)},X_{(n)})$, $|X|_{(n)}$, and $(X_{(1)},X_{(n)})$ be minimally sufficient?

From Casella and Berger's book, a sufficient statistic $T(X)$ is minimally sufficient if for every two sample points $x,y$, the ratio $f(x|\theta)/f(y|\theta)$ is free of $\theta$ iff $T(x)=T(y).$

For $max(-X_{(1)},X_{(n)}):$ $\frac{f(x|\theta)}{f(y|\theta)} = \frac{\frac{1}{(2\theta)^n}I(\theta)_{(max(-X_{(1)},X_{(n)}), \infty)}}{\frac{1}{(2\theta)^n}I(\theta)_{(max(-Y_{(1)},Y_{(n)}), \infty)}}$, which is free of $\theta$ iff $max(-X_{(1)},X_{(n)})=max(-Y_{(1)},Y_{(n)})$

For $|X|_{(n)}$: $\frac{f(x|\theta)}{f(y|\theta)} = \frac{\frac{1}{(2\theta)^n}I(|X|_{(n)})_{(0,\theta)}}{\frac{1}{(2\theta)^n}I(|Y|_{(n)})_{(0,\theta)}}$, which is free of $\theta$ iff $|X|_{(n)}=|Y|_{(n)}$

For $(X_{(1)},X_{(n)})$: $ \frac{f(x|\theta)}{f(y|\theta)}=\frac{\frac{1}{(2\theta)^n}I(X_{(1)})_{(-\theta, \infty)}I(X_{(n)})_{(-\infty, \theta)}}{\frac{1}{(2\theta)^n}I(Y_{(1)})_{(-\theta, \infty)}I(Y_{(n)})_{(-\infty, \theta)}}$, which is free of $\theta$ iff $(X_{(1)},X_{(n)})=(Y_{(1)},Y_{(n)})$

d) Is the MLE complete for $\theta$?

Regardless of which is the MLE, which of $max(-X_{(1)},X_{(n)})$, $|X|_{(n)}$, and $(X_{(1)},X_{(n)})$ would be complete?

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  • Is there any difference between $\max\left(-X_{(1)},X_{(n)}\right)$ and $|X|_{(n)}$ ? – Henry Apr 07 '20 at 22:07
  • @Henry I think not intuitively, but can we say that the random variable defined as $max(-X_{(1)}, X_{(n)})$ is the same random variable as the one defined as $|X|_{(n)}$? – funmath Apr 07 '20 at 22:16
  • If they are always equal then they are the same statistic – Henry Apr 07 '20 at 22:20
  • @Henry Ok, thank you. What about minimally sufficiency and completeness for $max(-X_{(1)},X_{(n)})$ and $(X_{(1)},X_{(n)})$? – funmath Apr 07 '20 at 22:22
  • @Henry since $\theta$ is a location parameter would $(X_{(1)},X_{(n)})$ be incomplete because it would not be independent of the the range random variable? – funmath Apr 07 '20 at 22:27
  • $(X_{(1)},X_{(n)})$ is clearly not minimal sufficient, since $\max\left(-X_{(1)},X_{(n)}\right)$ is sufficient – Henry Apr 07 '20 at 22:53
  • All these questions have been answered before. Did you search the site? – StubbornAtom Apr 08 '20 at 02:46
  • @StubbornAtom I mentioned this at the beginning of my post. However, the answers provided did not address the questions I have (which are written above). Maybe there are posts that do address them, but I have not been able to find them – funmath Apr 08 '20 at 03:34
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    MLE is derived here and here among several posts. Note that $\max(-X_{(1)},X_{(n)})=\max_{1\le i\le n}|X_i|$. Minimal sufficiency (and hence sufficiency) is discussed here and here. And since $|X_i|\sim U(0,\theta)$, it follows from here that $\max_{1\le i\le n} |X_i|$ is complete. – StubbornAtom Apr 08 '20 at 07:39
  • You can also take a look at https://stats.stackexchange.com/q/354893/119261. – StubbornAtom Apr 08 '20 at 08:18
  • @StubbornAtom Thank you so much. Can you claim that $(X_{(1)},X_{(n)})$ is not complete because if it were, it would be independent of the range random variable $(X_{(n)}-X_{(1)})$ and it clearly is not (because the range is a function of the statistic)? Is there another way to show this statistic is not complete? – funmath Apr 08 '20 at 14:44
  • Since $(X_{(1)},X_{(n)})$ is not minimal sufficient, it cannot be complete. See https://stats.stackexchange.com/q/435938/119261. Alternatively, if my calculations are correct then $E_{\theta}(X_{(n)}+X_{(1)})=0$ for every $\theta$, but $X_{(n)}+X_{(1)}=0$ does not hold almost surely. – StubbornAtom Apr 08 '20 at 15:41
  • @StubbornAtom Thank you. I have one last question about the log likelihood function. Since this function is a parameter of $\theta$, should it be written as $L(\theta|X)=\frac{1}{(2\theta)^n}I(\theta){(max(-X{(1)},X_{(n)}), \infty)}$ or $L(\theta|X)=\frac{1}{(2\theta)^n}I(|X|{(n)}){(0,\theta)}$? (Just wondering what the correct way to express the indicator function would be, or if it does not matter) – funmath Apr 08 '20 at 18:10
  • Since the likelihood is a function of $\theta$, writing $I_{(\max_{1\le i\le n}|x_i|,\infty)}(\theta)$ makes this point explicit. Of course you can simply write this as $I(0<\max_{1\le i\le n}|x_i|<\theta<\infty)$. Please provide an answer below to your questions, since they have already been answered. I would vote to close this as a duplicate, but you have made it difficult to choose one suitable target post. No need to further extend this comment section. – StubbornAtom Apr 08 '20 at 21:15

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As is possible to see in the comment section, all questions in the post have been answered before. Take a look at the links provided in the comment section. They summarize everything well.

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