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 $\frac{h'(\mu)}{I_n(\mu)}=\frac1{-\frac{\partial^2}{\partial \theta^2} \log l(\mu, \beta|\mathbf{X})}$.



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