# $\bar X$ is the UMVUE of $\frac{a}{b}$ for Gamma Distribution

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 $\mathbb{E}(\bar X|T) = \bar{X}$ ?

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 $f(a,b)=a/b$?
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 $\overline{X}$.
• @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. – user365239 Nov 22 '16 at 20:33