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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 $(x_{(1)},x_{(n)})=(y_{(1)},y_{(n)})$?

Michael Hardy
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1 Answers1

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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 $\theta$.

Michael Hardy
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student
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    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. – heropup Nov 01 '16 at 20:18