Method of moments, derive and compare

Question

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

Derive the MoM estimators for (a) the rate $\lambda$ of an exponential distribution and (b) both parameters of a Gamma $(\alpha, \theta)$ distribution. How does MoM compare to maximum likelihood?

Accepted Answer

The recipe. With sample moments m1 = \frac{1}{n}\sum Xi and m2 = \frac{1}{n}\sum Xi^2, solve \mathbb{E}[X] = m1, \qquad \mathbb{E}[X^2] = m2, \dots for the unknown parameters, as many moment equations as parameters. (a) Exponential(\lambda). One parameter, one equation: \mathbb{E}[X] = 1/\lambda = m1 = \bar{X}, so \hat{\lambda}{	ext{MoM}} = \frac{1}{\bar{X}}. (Here MoM and MLE coincide, that happens for many one-parameter exponential-family models.) (b) Gamma(\alpha, 	heta) (shape–scale, mean \alpha	heta, variance \alpha	heta^2). Match mean and variance, using \hat{v} = m2 - m1^2 for the …

Method of moments, derive and compare

Your answer

More Statistics questions