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?