# Gamma distribution parameters estimation [closed]

I have a set of samples taken from a population distributed with a Gamma distribution, so $$\begin{equation} f_X(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} \end{equation}$$ I should estimate $$\alpha$$ and $$\beta$$, so knowing that $$\begin{equation} \alpha=\frac{E[X]^2}{\text{Var}[X]},\quad \beta=\frac{E[X]}{\text{Var}[X]} \end{equation}$$ My idea is to compute $$\hat \mu, \hat \sigma^2$$ from the set as estimators of the mean and average of the population, then find $$\hat \alpha,\hat \beta$$ as follows $$\begin{equation} \hat \alpha=\frac{\hat \mu^2}{\hat \sigma^2},\quad \hat \beta=\frac{\hat \mu}{\sigma^2} \end{equation}$$

I'm trying to apply this logic to a task with a set of 10000 samples, but approximations seems to be not good enough. Is there any flaw in this reasoning?

EDIT: I've also tried the Newton-Raphson method to find alpha. This method leads to approximations very similar to the ones that I get from the method of moments.

EDIT 2: The logic is correct. Problem was a wrong understanding of the results of the implementation of this logic: in my task there was just half of the data needed to get all the estimators required. That is true for any possible method. Also approximation was fine.

## closed as unclear what you're asking by Stephan Kolassa, Michael Chernick, user158565, COOLSerdash, whuber♦Apr 23 at 14:37

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• You're using the method-of-moments estimators, afaik. In what way to you find them unsatisfying? Another possibility would be to use the maximum likelihood estimators. – COOLSerdash Apr 20 at 17:04
• What @COOLSerdash says. The gamma can be parameterized in two different ways, are you sure you are not confusing something there? That could be a simple explanation for why your results may be bad. – Stephan Kolassa Apr 20 at 17:07
• @COOLSerdash yes it is. I assume that for 10000 samples the method should work fine to have a good approximation. – ezy Apr 20 at 18:35
• FWIW, your expression for $f$ isn't quite correct. It's difficult to determine what you mean by "each letter with a unique ... combination" and it's impossible to determine what you mean by "unsatisfying." Could you clarify your question? – whuber Apr 23 at 4:00
• Without explanation of, or evidence for, "unsatisfying" results this question remains unclear. – Nick Cox Apr 24 at 8:16