3
$\begingroup$

Let $X_1$,$X_2$,...,$X_n$ be i.i.d. according to Gamma($\alpha$,$\beta$). Denote the mean by $\mu := E[X_i] = \alpha/\beta$.

Can you find an unbiased and efficient estimator for $\mu$?

MLE gives unbiased and efficient estimators for $\alpha$ and $\beta$ separately. Combining them yields that $\frac{\sum_{i=1}^nX_i}{n}$ should be an unbiased estimator of $\mu$. Is it also efficient?

$\endgroup$
  • 2
    $\begingroup$ Is this either homework or self study? $\endgroup$ – Michael Chernick Dec 8 '16 at 21:26
  • 1
    $\begingroup$ No. I would even be greatful for resources where this question is asked. $\endgroup$ – morty Dec 8 '16 at 22:53
1
$\begingroup$

Since this is a regular exponential family, the pair of statistics $$\left(\sum_{i=1}^n X_i,\sum_{i=1}^n \log X_i\right)$$is minimal sufficient and complete. Since$$\frac{1}{n} \sum_{i=1}^n X_i$$is unbiased, and a function of this pair, this means this is the minimum variance unbiased estimator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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