Method of moments for the beta distribution

Question

The method of moments (MoM) estimates parameters by matching theoretical moments to sample moments.

Derive the MoM estimators for both shape parameters $\alpha$ and $\beta$ of a Beta $(\alpha, \beta)$ distribution (support $[0,1]$ ). Why is MoM especially attractive here?

Accepted Answer

The recipe. With sample mean m1 and sample variance \hat v = \frac1n\sum Xi^2 - m1^2, match theory to data. Set up the two equations. Write s = \alpha + \beta. The mean and variance are \mu = \frac{\alpha}{s}, \qquad 	ext{Var} = \frac{\mu(1-\mu)}{s + 1}. Match them to m1 and \hat v. From the variance equation, solve for s: \hat v = \frac{m1(1 - m1)}{s + 1} \implies s + 1 = \frac{m1(1 - m1)}{\hat v} \implies \hat s = \frac{m1(1 - m1)}{\hat v} - 1. Then split \hat s into the two shapes using the mean: \hat\alpha{	ext{MoM}} = m1 \,\hat s, \qquad \hat\beta{	ext{MoM}} = (1 - m1)\,\hat s. (This …

Method of moments for the beta distribution

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