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Consider the Gamma distribution:

$\Gamma(\alpha,\beta) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}$

According to many sources (e.g. wikipedia), the mean is given by $\mu=\alpha/\beta$. On the other hand, using Maximum Likelihood Estimation, we can find that (see page 11 here for example) $\hat{\beta}_{MLE}=\frac{\hat{\alpha}_{MLE}}{\bar{x}}$, where $\bar{x}$ is the average. Accordingly, $\hat{\mu}_{MLE} = \frac{\hat{\alpha}_{MLE}}{\hat{\beta}_{MLE}}=\bar{x}$.

Is this correct? If so, why do many sources go on length to do numerical estimation of the parameters? Since most of the time all what we care about is mean and variance, I don't see why we would resort to numerical estimation of individual parameters.

student1
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    but, it's not straightforward to estimate $\alpha$. To get the $\hat{\alpha}$ estimate, you need to set the first derivative with respect to $\alpha$ to zero. As it states in the reference on page 11, the resulting equation can only be solved using numerical methods. – mlofton Aug 13 '22 at 04:07
  • I think my question is more like, our ultimate goal is usually to find the mean which in this case can be found without having to numerically solve for alpha. But people still do that, why? Is it that the individual parameters are interesting in themselves for some reason? – student1 Aug 13 '22 at 04:14
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    In the gamma case, the ratio of the two MLE's is the mean. but, in general, A) you might want to calculate probabilities by using the MLE's to estimate the density or B) the mean does not come out to a nice form like $\bar{x}$. For example, the variance of the gamma I don't think has a nice form so you will need the MLE's of the parameters to estimate that. – mlofton Aug 13 '22 at 04:20
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    Re "ultimate goal:" that's unusually limited. In most cases the objective either requires estimating the distribution, not just one of its properties; or it requires estimates of multiple properties, which is tantamount to estimating the distribution anyway. Bear in mind that what makes a problem statistical (as opposed to a sterile mathematical "fitting" question) is recognition of the value of quantifying the amount of variability -- and in this case that cannot be done without estimating the shape parameter. – whuber Aug 13 '22 at 14:09

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Indeed it's isn't necessary to solve for the shape parameter, $α$ in order to estimate $\mu$. If you reparameterize the gamma to the shape-mean parameterization (as you essentially have in a gamma GLM), then the MLE of the mean can be done without having an estimate of α and indeed will be $\bar{x}$.

You can find the shape-mean parameterization here and if you do the calculations - which are quite straightforward - you'll see the shape parameter cancel out of the MLE for $\mu$. So you can certainly estimate $\mu$ by MLE without solving for the shape parameter.

However, for the variance ($\frac{\mu^2}{\alpha}$ in this parameterization), I don't think you can avoid estimating the shape parameter (or something essentially as difficult, if you replace $\alpha$ in yet another reparameterization).

For example, I don't think it helps whether you replace $\alpha$ by the coefficient of variation or even the variance itself, you'll either end up with a function of the shape parameter or you end up with a problem of similar (or slightly greater) difficulty to estimating the shape. Might as well just be done and estimate the shape itself in that case.

Glen_b
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