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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:

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

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

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Note that likelihood function depends on the sample $X_1,\ldots,X_n$. Therefore, there can be no $x$ in the argument. $$f(X_1,\ldots,X_n\mid \theta)=\prod_{i=1}^n\left[\left(\frac{1}{2\theta}\right)^n I(|X_i|<\theta)\right]=\prod_{i=1}^n\left(\frac{1}{2\theta}\right)^n \cdot I\left(\max_{1\leqslant i\leqslant n}|X_i|<\theta\right).$$

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 $\theta$.

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  • $\begingroup$ Thank you very much! I appreciate the helpful links :) $\endgroup$ – Student Jun 25 at 17:30

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