Problem:
Let $(X_1,X_2, \ldots, X_n) $ be a random sample from Gamma Distribution. Prove that $ \bar X $ is the UMVUE of $ f(a,b) = \frac{a}{b}$.
My work so far: It is easy to see that $ \mathbb{E}(\bar X) = \mathbb{E}(X)$. Since $ \mathbb{E}(X) =\frac{a}{b} $ for Gamma Distribution,
we conclude that $ \mathbb{E}(\bar X) = \frac{a}{b}$, so $ \bar X $ is an unbiased estimator of $ f(a,b) = \frac{a}{b}$. In addition, we observe
that Gamma is included in the Exponential Family and we can see that $ T=(\sum_{i=1}^{n} X_i, \sum_{i=1}^{n} \log X_i $)
is a sufficient and complete statistic for $ (a,b)$. Now can we infer that since $ \bar X $ is an unbiased
estimator and a function of a sufficient and complete statistic, it must be the UMVUE of $ f(a,b) =\frac{a}{b}$
or we have to prove that $ \mathbb{E}(\bar X|T) = \bar{X} $ ?
Best Answer
Indeed, this is the Lehman-Scheffe Theorem.
(https://en.wikipedia.org/wiki/Lehmann%E2%80%93Scheff%C3%A9_theorem)
Also for your very last statement, note that $\overline{X} \mid T$ is equal to $\overline{X}$, since if you know $T$ i.e. you know $\sum_{i=1}^n X_i$, then you immediately know $\overline{X}$.