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Let $X_1,\dots,X_n$ be a sample from uniform distribution on $(-\theta,\theta)$ with parameter $\theta>0$.

It is easy to show that $T(X) = (X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$ where $X_{(1)}$ and $X_{(n)}$ stands for the minimum and the maximum from the sample $X_1,\dots,X_n$ respectively.

I want to show that it is also minimal sufficient. To do so I look at the ratio of the densities

$$ \frac{f(x_1,\dots,x_n;\theta)}{f(y_1,\dots,y_n;\theta)} = \frac{1_{[-\theta<x_{(1)}\leq x_{(n)}<\theta]}}{1_{[-\theta<y_{(1)}\leq y_{(n)}<\theta]}} $$ We want to show that this ratio is a constant as a function of $\theta$ iff $(x_{(1)},x_{(n)})=(y_{(1)},y_{(n)})$.

1. question: how should I understand the ratio if it is not defined (e.g. $\frac{0}{0}$)?

It is easy to show that if $(x_{(1)},x_{(n)})=(y_{(1)},y_{(n)})$ than the ratio is constant as a function of $\theta$ (if I neglect the problem of understanding the $\frac{0}{0}$ case). But:

2. question: how to show that if the ratio is constant as a function of $\theta$ then $(x_{(1)},x_{(n)})=(y_{(1)},y_{(n)})$?

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so I realised there was a mistake in the assignment.

The statistic $(X_{(1)},X_{(n)})$ is NOT minimal sufficient for $\theta$. The minimal statistic is $\max\{ -X_{(1)}, X_{(n)} \}$ which follows easily from the fact that the density of $X_1,\dots,X_n$ can be expressed as

$$ \frac{1}{(2\theta)^n} \mathbb{1}_{[\max\{ -X_{(1)}, X_{(n)} \} < \theta]} .$$

As for the first question, I found a reference - Theorem 2.29, Mark J. Schervish, Theory of Statistics, 1995. One needs to check if one density is a multiple of the other and the multiplicative constant does not depend on $\theta$.

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    $\begingroup$ Your intuition should show that $(X_{(1)}, X_{(n)})$ is not minimal, because the support is symmetric: if you observed $\boldsymbol x = (3, -1, 4, -5, 1, 0)$, you immediately know that $\theta \ge 5$, and that this is the best you can do in the sense that the other data contribute no additional information about the parameter. $\endgroup$ – heropup Nov 1 '16 at 20:18
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For the case where we have the undefined ratio $\frac{0}{0}$, I believe we can ignore it in the context of the problem. We seek a minimally sufficient statistic for the purposes of estimating $\theta$. Thus, the case where \begin{equation} \frac{ \mathbb{I}(-X_{(n)}, X_{(1)}) }{ \mathbb{I}(-Y_{(n)}, Y_{(1)}) } = \frac{0}{0} \end{equation}

implies we are using an estimate of $\theta$ that is not wide enough to span observed values in both sets of observations (i.e., the estimate has zero likelihood). Though mathematically possible in the pure sense, it is quite unreasonable in the realm of statistical inference.

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