# 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 $$\theta > 0$$ 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?

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