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?

  • 2
    $\begingroup$ Is this either homework or self study? $\endgroup$ Dec 8, 2016 at 21:26
  • 1
    $\begingroup$ No. I would even be greatful for resources where this question is asked. $\endgroup$
    – morty
    Dec 8, 2016 at 22:53

1 Answer 1


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.


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