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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} $ ?

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Now can we infer that since $\overline{X}$ is an unbiased estimator and a function of a sufficient and complete statistic, it must be the UMVUE of $f(a,b)=a/b$?

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 $\overline{X}$.

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  • $\begingroup$ Both in my textbook and in this wikipedia article, Lehmann–Scheffé theorem seems to be used only in the estimation of a single parameter. This is why I have doubt if it is applicable here. $\endgroup$ – Greg Nov 22 '16 at 20:25
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    $\begingroup$ @Greg. It's OK with multiple parameters. Fix $n$ our sample size. Define a function $g$ by $g(x,y) = \dfrac{x}{n}$. Then, for our complete & sufficient statistic $T = \left( \sum_{i=1}^n X_i, \sum_{i=1}^n \log X_i \right)$, we see $g(T) = \overline{X}$ is unbiased for $a/b$, & done by Lehman Scheffe. $\endgroup$ – user365239 Nov 22 '16 at 20:33

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