sufficient vs. minimally sufficient vs. complete statistic and MLE for uniform distribution 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 $\theta &gt; 0$ 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?
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.
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.