# Minimal Sufficient Statistic for the distribution $U(-\theta, \theta)$ [duplicate]

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. 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 $$|Xi|$$.

• Neither $X_{(n)}$ nor $(X_{(1)},X_{(n)})$ is minimal sufficient: math.stackexchange.com/q/2116770/321264 Feb 24, 2021 at 9:46
• It was max of $|Xi|$ Feb 24, 2021 at 10:19

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}
• 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. Feb 24, 2021 at 14:28