# Sufficient statistics for Uniform $(-\theta,\theta)$

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?

• Yes you can have jointly sufficient statistics. In fact the order statistics $T=(X_{(1)}, X_{(2)}, \cdots X_{(n)})$ are always sufficient (somewhat trivially). What you lack however, is minimal sufficiency (and in this case completeness), a rather important property for many applications. – knrumsey Jul 6 '18 at 21:03
• Your question is not very clear, but note that if $(T_1, T_2)$ is sufficient, then so will be $g(T_1, T_2)$ when $g$ is a one-to-one function. – kjetil b halvorsen Jul 6 '18 at 23:42
• @knrumsey comments that $(X_{(1)}, X_{(n)})$ will not be minimally sufficient (unless $n=1$). That is because in this peculiar model, we have another sufficient statistic, $\max\left( |X_{(1)}|, |X_{(n)}| \right)$. – kjetil b halvorsen Jul 6 '18 at 23:52
• In fact, in some families like $N(\theta,\theta^2)$, the dimension of the sufficient statistic is two while the parameter $\theta$ is one-dimensional. But that is not the case here. In the Uniform family, we often get $(X_{(1)},X_{(n)})$ as the two-dimensional sufficient statistic, like in $U(a,b)$ where both $a,b$ are unknown and even in $U(\theta,\theta+1)$ where $\theta$ is the single parameter. – StubbornAtom Jul 7 '18 at 8:05
• No, the argument is more involved! But note that in $U(-\theta,\theta)$ there is a symmetry not present in $U(a,b)$ or $U(\theta,\theta+1)$. – kjetil b halvorsen Jul 7 '18 at 8:12

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.

• So, Imagine that I know that $\sum X_{i}$ is sufficient for a certain paramether $\theta$. Can I also say that $(\sum X_{i}, X_{(a)})$ (where $X_{(a)}$ is any order statistics) is jointly sufficient for the same paramether? In other words,can I manipulate the factorisation theorem in such way that any order statistics or even the sample itself appears within the part of the function that depends on both the sample and $\theta$ (meaning they will be sufficient for $\theta$)?. I'm not sure if the second part is very clear, since english is not my primary language. Thanks for the answers! – Tricolor Jul 7 '18 at 22:06
• @Tricolor I don't think that's true in general. Say we consider the Bernoulli$(\theta)$ distribution. Then $T(X_1,\cdots,X_n)=\sum_{i=1}^n X_i$ is a minimal sufficient statistic for $\theta$. In other words, $T$ is itself a function of every other set of sufficient statistics. As in, $(X_{(1)},\cdots,X_{(n)})$ is jointly sufficient for $\theta$, and $T=\sum_{i=1}^nX_i=\sum_{i=1}^n X_{(i)}$ is a function of $X_{(1)},\cdots,X_{(n)}$. – StubbornAtom Jul 8 '18 at 8:10
• @Tricolor When you factor the joint density, do you get a $g$ as in my answer where $g$ depends on $x_1,\cdots,x_n$ through $(T,X_{(i)})$? If you don't, then you can be certain that your claim does not hold. In the above answer, the particular pair $(X_{(1)},X_{(n)})$ is a sufficient statistic because the minimal sufficient statistic $\max(-X_{(1)},X_{(n)})$ contains the components $X_{(1)}$ and $X_{(n)}$. That doesn't mean we can incorporate any particular order statistic along with a sufficient statistic to get a jointly sufficient statistic. – StubbornAtom Jul 8 '18 at 8:10
• If $X_{(n)}$ is sufficient for $b$, then so is $(T_0, X_{(n)})$ for any statistic $T_0$. As pointed out by @kjetilbhalvorsen, this is because $g(x) = (a,x)$ is a one to one function from $\mathbb R$ to $\mathbb R^2$. Again, what you lose is not sufficiency, but minimal sufficiency. – knrumsey Jul 9 '18 at 4:30
• @Tricolor Please see the above comments and my edit. – StubbornAtom Jul 9 '18 at 10:03