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?

  • $\begingroup$ Is there any difference between $\max\left(-X_{(1)},X_{(n)}\right)$ and $|X|_{(n)}$ ? $\endgroup$
    – Henry
    Apr 7 '20 at 22:07
  • $\begingroup$ @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 $|X|_{(n)}$? $\endgroup$
    – funmath
    Apr 7 '20 at 22:16
  • $\begingroup$ If they are always equal then they are the same statistic $\endgroup$
    – Henry
    Apr 7 '20 at 22:20
  • $\begingroup$ @Henry Ok, thank you. What about minimally sufficiency and completeness for $max(-X_{(1)},X_{(n)})$ and $(X_{(1)},X_{(n)})$? $\endgroup$
    – funmath
    Apr 7 '20 at 22:22
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 8 '20 at 7:39

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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