inferenceself-studysufficient-statisticsuniform distribution
So, I know that $\max(-X_{(1)},X_{(n)})$ is a sufficient statistic for the parameter $\theta$. But can I also say that $(X_{(1)},X_{(n)})$ are jointly sufficient for the parameter $\theta$ ?
In other words, can a single parameter have jointly sufficient statistics?
Let's take care of the routine calculus for you, so you can get to the heart of the problem and enjoy formulating a solution. It comes down to constructing rectangles as unions and differences of triangles.
First, choose values of $a$ and $b$ that make the details as simple as possible. I like $a=0,b=1$: the univariate density of any component of $X=(X_1,X_2,\ldots,X_n)$ is just the indicator function of the interval $[0,1]$.
Let's find the distribution function $F$ of $(Y_1,Y_n)$. By definition, for any real numbers $y_1 \le y_n$ this is
$$F(y_1,y_n) = \Pr(Y_1\le y_1\text{ and } Y_n \le y_n).\tag{1}$$
The values of $F$ are obviously $0$ or $1$ in case any of $y_1$ or $y_n$ is outside the interval $[a,b] = [0,1]$, so let's assume they're both in this interval. (Let's also assume $n\ge 2$ to avoid discussing trivialities.) In this case the event $(1)$ can be described in terms of the original variables $X=(X_1,X_2,\ldots,X_n)$ as "at least one of the $X_i$ is less than or equal to $y_1$ and none of the $X_i$ exceed $y_n$." Equivalently, all the $X_i$ lie in $[0,y_n]$ but it is not the case that all of them lie in $(y_1,y_n]$.
Because the $X_i$ are independent, their probabilities multiply and give $(y_n-0)^n = y_n^n$ and $(y_n-y_1)^n$, respectively, for these two events just mentioned. Thus,
$$F(y_1,y_n) = y_n^n - (y_n-y_1)^n.$$
The density $f$ is the mixed partial derivative of $F$,
$$f(y_1,y_n) = \frac{\partial^2 F}{\partial y_1 \partial y_n}(y_1,y_n) = n(n-1)(y_n-y_1)^{n-2}.$$
The general case for $(a,b)$ scales the variables by the factor $b-a$ and shifts the location by $a$. Thus, for $a \lt y_1 \le y_n \lt b$,
$$F(y_1,y_n; a,b) = \left(\left(\frac{y_n-a}{b-a}\right)^n - \left(\frac{y_n-a}{b-a} - \frac{y_1-a}{b-a}\right)^n\right) = \frac{(y_n-a)^n - (y_n-y_1)^n}{(b-a)^n}.$$
Differentiating as before we obtain
$$f(y_1,y_n; a,b) = \frac{n(n-1)}{(b-a)^n}(y_n-y_1)^{n-2}.$$
Consider the definition of completeness. Let $g$ be any measurable function of two real variables. By definition,
$$\eqalign{E[g(Y_1,Y_n)] &= \int_{y_1}^b\int_a^b g(y_1,y_n) f(y_1,y_n)dy_1dy_n\\ &\propto\int_{y_1}^b\int_a^b g(y_1,y_n) (y_n-y_1)^{n-2} dy_1dy_n.\tag{2} }$$
We need to show that when this expectation is zero for all $(a,b)$, then it's certain that $g=0$ for any $(a,b)$.
Here's your hint. Let $h:\mathbb{R}^2\to \mathbb{R}$ be any measurable function. I would like to express it in the form suggested by $(2)$ as $h(x,y)=g(x,y)(y-x)^{n-2}$. To do that, obviously we must divide $h$ by $(y-x)^{n-2}$. Unfortunately, for $n\gt 2$ this isn't defined whenever $y-x$. The key is that this set has measure zero so we can neglect it.
Accordingly, given any measurable $h$, define
$$g(x,y) = \left\{\matrix{h(x,y)/(y-x)^{n-2} & x \ne y \\ 0 & x=y}\right.$$
Then $(2)$ becomes
$$\int_{y_1}^b\int_a^b h(y_1,y_n) dy_1dy_n \propto E[g(Y_1,Y_n)].\tag{3}$$
(When the task is showing that something is zero, we may ignore nonzero constants of proportionality. Here, I have dropped $n(n-1)/(b-a)^{n-2}$ from the left hand side.)
This is an integral over a right triangle with hypotenuse extending from $(a,a)$ to $(b,b)$ and vertex at $(a,b)$. Let's denote such a triangle $\Delta(a,b)$.
Ergo, what you need to show is that if the integral of an arbitrary measurable function $h$ over all triangles $\Delta(a,b)$ is zero, then for any $a\lt b$, $h(x,y)=0$ (almost surely) for all $(x,y)\in \Delta(a,b)$.
Although it might seem we haven't gotten any further, consider any rectangle $[u_1,u_2]\times [v_1,v_2]$ wholly contained in the half-plane $y \gt x$. It can be expressed in terms of triangles:
$$[u_1,u_2]\times [v_1,v_2] = \Delta(u_1,v_2) \setminus\left(\Delta(u_1,v_1) \cup \Delta(u_2,v_2)\right)\cup \Delta(u_2,v_1).$$
In this figure, the rectangle is what is left over from the big triangle when we remove the overlapping red and green triangles (which double counts their brown intersection) and then replace their intersection.
Consequently, you may immediately deduce that the integral of $h$ over all such rectangles is zero. It remains only to show that $h(x,y)$ must be zero (apart from its values on some set of measure zero) whenever $y \gt x$. The proof of this (intuitively clear) assertion depends on what approach you want to take to the definition of integration.
Regarding 1., note that interpretation of a sufficient statistic is:
"no other statistic that can be calculated from the same sample provides any additional information as to the value of the parameter".
So, we need both the statistics to provide the "most information" possible about the value of the parameter. Not necessarily "full information".
Regarding 2.
Note that your likelihood can be written as
$$ f_{\theta} (x_1, \dots, x_n) = \dfrac{1}{\theta^n} I\left( \dfrac{\max x_i}{2} \leq \theta \leq \min x_i \right)\,.$$
The MLE is the statistic that maximizes the likelihood. You can show that the likelihood is a decreasing function of $\theta$. So the likelihood is maximized on the smallest value $\theta$ is allowed to take, which is $(\max x_i)/2$
Best Answer
Suppose we have a random sample $(X_1,X_2,\cdots,X_n)$ drawn from $\mathcal U(-\theta,\theta)$ distribution.
PDF of $X\sim\mathcal U(-\theta,\theta)$ is $$f(x;\theta)=\frac{1}{2\theta}\mathbf1_{-\theta<x<\theta},\quad\theta>0$$
Joint density of $(X_1,X_2,\cdots,X_n)$ is
\begin{align}f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^nf(x_i;\theta) \\&=\frac{1}{(2\theta)^n}\mathbf1_{-\theta<x_1,\cdots,x_n<\theta} \\&=\frac{1}{(2\theta)^n}\mathbf1_{0<|x_1|,\cdots,|x_n|<\theta} \\&=\frac{1}{(2\theta)^n}\mathbf1_{\max_{1\le i\le n}|x_i|<\theta} \end{align}
To clearly use the factorisation theorem, let us define $\mathbb I(x)=\begin{cases}1&,\text{ if }x>0\\0&,\text{ otherwise }\end{cases}$
Then we have
\begin{align} f_{\theta}(x_1,x_2,\cdots,x_n)&=\frac{1}{(2\theta)^n}\mathbb I\left(\theta-\max_{1\le i\le n}|x_i|\right) \\&=g\left(\theta,\max_{1\le i\le n}|x_i|\right)h(x_1,x_2,\cdots,x_n) \end{align}
where $g\left(\theta,\max_{1\le i\le n}|x_i|\right)=\frac{1}{(2\theta)^n}\mathbb I\left(\theta-\max_{1\le i\le n}|x_i|\right)$ depends on $\theta$ and on $x_1,x_2,\cdots,x_n$ through $\max_{1\le i\le n}|x_i|$, and $h(x_1,x_2,\cdots,x_n)=1$ is independent of $\theta$.
So by factorisation theorem, $\max_{1\le i\le n}|X_i|$ is a sufficient statistic for $\theta$.
In fact, it can be shown to be minimal sufficient for $\theta$.
Since $-\theta<x_i<\theta\implies \theta>\max(-x_{(1)},x_{(n)})$, you are correct that $\max(-X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$ by a similar logic. In fact, if I am not wrong, it can be shown that $$\max(-X_{(1)},X_{(n)})=\max_{1\le i\le n}|X_i|$$
It is perfectly valid for a single unknown parameter to have jointly sufficient statistics since the definition of sufficiency tells us that a set of statistics $T_1(X_1,\cdots,X_n),\cdots,T_k(X_1,\cdots,X_n)$ is jointly sufficient for $\theta$ (which may be a vector) if and only if the conditional distribution of $X_1,\cdots,X_n$ given $T_1,\cdots,T_k$ does not depend on $\theta$.
As for example, we always have two trivial choices of jointly sufficient statistics. One is the sample $(X_1,X_2,\cdots,X_n)$ itself and the other is the set of order statistics $(X_{(1)},X_{(2)},\cdots,X_{(n)})$ as mentioned by knrumsey in the comments.
Again, $(X_{(1)},X_{(n)})$ is also sufficient for $\theta$ as it has the two components $X_{(1)}$ and $X_{(n)}$ of the minimal sufficient statistic, but it is not itself minimal sufficient as mentioned in the comments.
To clarify further queries of OP in the comments:
If a statistic $T$ is sufficient for a one-dimensional parameter $\theta$, then $\mathbf T=(T,T_0)$ will also be a sufficient statistic for $\theta$ for any other statistic $T_0$. This is because $\mathbf T$ can be looked as a one-to-one function of $T$. In particular, if $T$ is minimal sufficient, then $\mathbf T$ will be sufficient in a sense that the second component $T_0$ in $\mathbf T$ will be of no use in further data condensation and data reduction without loosing any information about $\theta$. So $T$ will be preferred to $(T,T_0)$ as a sufficient statistic.
For example, consider the $\mathcal N(\theta,1)$ population. Here the sample mean $\bar X$ is minimal sufficient for $\theta$. Now if we take $\mathbf T=(\bar X,S^2)$ where $S^2$ is the sample variance, then $\mathbf T$ will remain sufficient but no longer minimal sufficient.
Thanks to @knrumsey for pointing out my error.