Method of moments for the Poisson, and a dispersion check

Question

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

Derive the MoM estimator for the rate $\lambda$ of a Poisson distribution. The Poisson has only one parameter but two moments equal to $\lambda$ , so what does comparing the two estimators tell you?

Accepted Answer

The recipe. Match theoretical moments to sample moments, one equation per parameter. One parameter, mean-based estimator. For Poisson(\lambda), \mathbb{E}[X] = \lambda, so setting it equal to the sample mean m1 = \bar X gives \hat\lambda{	ext{MoM}} = \bar X. Here MoM and MLE coincide, as they do for many one-parameter exponential-family models. The extra moment is a diagnostic. The Poisson is special because its variance also equals \lambda. So a second, equally valid estimator is the sample variance: \hat\lambda{	ext{var}} = \hat v = \frac{1}{n}\sum (Xi - \bar X)^2. If the Poisson model is…

Method of moments for the Poisson, and a dispersion check

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