# The two estimators of mean of Gamma distribution and the estimators' variances

In Casella Example 10.1.18, the author says it is not easy to calculate the mean of gamma distribution. It seems that we CAN use the easy way $$\bar X=\frac{\sum X_i}n$$, but the variance of the mean we thus get is big, especially when the mean is small and $$\beta$$ is big (and $$\alpha$$ is small). This indicates when mean is small, our estimate of it by this method is very unreliable. Is my understanding correct?

We know $$\mu=\alpha\beta$$ by the integration of $$\int xf(x|\alpha, \beta)$$. (This integration can be easily done by noticing inside the integration the function is similar to $$f(x|\alpha', \beta)$$.)

It seems that we can use MLE of $$\mu$$ (i.e. value of $$\alpha\beta$$ that maximize $$l(\mu, \beta|\mathbf{x})$$, we can calc this value with the usual trick of changing the variables to a pair one of which is the variable we wanna study; this value is difficult to calculate since the zero of partial derivative of this log likelihood is difficult to calculate) to estimate the mean instead, this method will produce a smaller variance.

My question is that why MLE of $$\alpha\beta$$ is the mean?

(Updated: My thought is that $$\alpha\beta$$ is the mean, and its MLE is the most likely value of $$\alpha\beta$$, and therefore, that of the mean. So what the author tries to say is just so, that is, it's somehow reasonable to regard this MLE is the mean.)

ps: we can calculate the variance of mean gotten by the first method by $$\mathrm{Var}(\frac{\sum X_i}n)=\frac{\sum\mathrm{Var}( X_i)}{n^2}$$, we can calculate the variance of mean gotten by the second method by $$\frac{h'(\mu)}{I_n(\mu)}=\frac1{-\frac{\partial^2}{\partial \theta^2} \log l(\mu, \beta|\mathbf{X})}$$.