I need to find the sufficient statistic for the parameter $\theta$ for a uniform distribution $U(-\theta, \theta)$ for a sample of size $n$.

The joint density of the sample can be written as:

$f(x_1, x_2,..,x_n|\theta) = (\frac{1}{2\theta})^n,-\theta<x_i<\theta \text{ for all i = 1,2,3,..,n}$. I can express it as :

$f(x|\theta) = (\frac{1}{2\theta})^n \prod_{i=1}^{n} I_{(-\theta, \theta)}(x_i) = (\frac{1}{2\theta})^n I_{(X_{(n)}, \infty)}(\theta) I_{(-X_{(1)}, \infty)}(\theta) \prod_{i=1}^{n}I_{(-\infty,\infty)}$

Now, to prove a statistic, $ T(x)$, as minimally sufficient, we need to prove that the ratio $\frac{f(x|\theta)}{f(y|\theta)}$ to be constant as a function of $\theta$.

The ratio is computed as follows:

$\frac{f(x|\theta)}{f(y|\theta)} = \frac{(\frac{1}{2\theta})^n I_{(X_{(n)}, \infty)}(\theta) I_{(-X_{(1)}, \infty)}(\theta) \prod_{i=1}^{n}I_{(-\infty,\infty)}}{(\frac{1}{2\theta})^n I_{(Y_{(n)}, \infty)}(\theta) I_{(-Y_{(1)}, \infty)}(\theta) \prod_{i=1}^{n}I_{(-\infty,\infty)}}$

We can only get constant as a function of $\theta$ after substituting $X_{(n)} = Y_{(n)}$ and $X_{(1)} = Y_{(1)}$. Hence, $(X_{(1)}, X_{(1)})$ will be minimally sufficient. Please take a moment to review my steps and let me know where exactly I went wrong because the correct answer is shown as max of $|Xi|$.


1 Answer 1


From the range of your uniform distribution, you can see that $T(\mathbf{x}) = \max_{i=1,...,n} |X_i|$ is going to be the minimal sufficient statistic. To demonstrate sufficiency formally, we note that the likelihood function reduces to:

$$\begin{align} L_\mathbf{x}(\theta) &= \prod_{i=1}^n \text{U}(x_i| -\theta, \theta) \\[6pt] &= \prod_{i=1}^n \frac{1}{2 \theta} \cdot \mathbb{I}(|x_i| \leqslant \theta) \\[6pt] &= \frac{1}{2^n \theta^n} \prod_{i=1}^n \mathbb{I}(|x_i| \leqslant \theta) \\[6pt] &= \frac{1}{2^n \theta^n} \cdot \mathbb{I} \Big( \max_{i=1,...,n} |x_i| \leqslant \theta \Big) \\[6pt] &= \frac{1}{2^n \theta^n} \cdot \mathbb{I}( T(\mathbf{x}) \leqslant \theta ). \\[6pt] \end{align}$$

Using the Fisher–Neyman factorisation theorem this demonstrates that this statistic is sufficient. Minimal sufficiency follows from the fact that there is no sufficient statistic from which this statistic cannot be obtained.

  • $\begingroup$ Specially the last line made everything clear to me. Plus +1 for that. This topic has been causing a lot of trouble for me since last week. $\endgroup$
    – userNoOne
    Feb 24, 2021 at 14:19
  • $\begingroup$ Hello @Ben, I was wondering about the completeness of this minimal sufficient estimator. So, here is what I am thinking. Since $|X|$ follows $U(0,\theta)$, I will have $\frac{|X|}{\theta}$ following $U(0,1)$ which means that it is ancillary. I once read that if a function of sufficient statistic is ancillary, then it cannot be complete. Hence, I will concluyde that this found minimally sufficient statistic will not be complete. Now, I was wondering whether there can be a situation when a family of distribution is not complete? If yes, is there exist a formal way to prove that. $\endgroup$
    – userNoOne
    Feb 24, 2021 at 14:28
  • 1
    $\begingroup$ @userNoOne: May I recommend that you pose that as a new question on the forum, and link back to this question for context. $\endgroup$
    – Ben
    Feb 25, 2021 at 0:07

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