When method of moments equals maximum likelihood

Question

For a Poisson(\lambda) sample, the method of moments gives \hat\lambda = \bar X. Show that the maximum-likelihood estimate is also \bar X, and explain why the two methods coincide here while they often disagree elsewhere.

Accepted Answer

The likelihood estimate. The log-likelihood for Poisson data is, dropping constants, \ell(\lambda) = \Big(\sumi xi\Big)\log\lambda - n\lambda. Set the derivative to zero: \frac{\sumi xi}{\lambda} - n = 0, which gives \hat\lambda{	ext{MLE}} = \bar X. This is exactly the method-of-moments estimate. Why they agree here. The Poisson mean equals its parameter, \mathbb{E}[X] = \lambda. So "match the mean" and "maximize the likelihood" are solving for the same thing through the same summary, the sum of the data. More generally, for a one-parameter exponential family written in its natural form, the…

When method of moments equals maximum likelihood

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