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: $f(x\mid\theta)=\frac{1}{2\theta}$ $-\theta<x<\theta,\, \theta>0$.

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?


1 Answer 1


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 $\theta$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.