# Find best unbiased estimator for $\theta$ when $X_i\sim U(-\theta,\theta)$

I am having an issue finding a best unbiased estimator for $$\theta$$. Any help is appreciated.

Let $$X_1, ..., X_n$$ be a random sample from a population with pdf: $$f(x\mid\theta)=\frac{1}{2\theta}$$ $$-\theta0$$.

I understand that $$T(x) := |X_n|$$ is the sufficient statistic since by the factorization theorem we have:

$$f(x\mid \theta) = \left(\frac{1}{2\theta}\right)^n \prod_{i=1}^n I[|x_i|<\theta]$$

I think my main issue is showing that this sufficient statistic is also a complete sufficient statistic. Can somebody please aid me in this?

• Jun 25 '20 at 16:22

Note that likelihood function depends on the sample $$X_1,\ldots,X_n$$. Therefore, there can be no $$x$$ in the argument. $$f(X_1,\ldots,X_n\mid \theta)=\prod_{i=1}^n\left[\left(\frac{1}{2\theta}\right) I(|X_i|<\theta)\right]=\left(\frac{1}{2\theta}\right)^n \cdot I\left(\max_{1\leqslant i\leqslant n}|X_i|<\theta\right).$$
Then the sufficient statistics is $$T(X_1,\ldots,X_n) = \max_{1\leqslant i\leqslant n}|X_i| = |X|_{(n)}$$. This is the last order statistics of the sample $$|X_1|,\ldots, |X_n|$$. Note that $$|X_i|$$ are uniformly distributed on $$(0,\theta)$$.
The statistics $$T=|X|_{(n)}$$ is also complete. So by Lehmann–Scheffé theorem, if some function of $$T$$ is unbiased then it is UMVUE.
Then you can find pdf of last order statistics, then calculate its expected value $$\mathbb E\bigl[|X|_{(n)}\bigr]=\frac{n}{n+1}\theta$$ and correct the sufficient statistics to be unbiased estimator for $$\theta$$. Finally, $$\theta^*=\frac{n+1}{n}\cdot |X|_{(n)}=\frac{n+1}{n} \cdot T$$ is the best unbiased estimator for $$\theta$$.