# When do maximum likelihood and method of moments produce the same estimators?

I was asked this question the other day and had never considered it before.

My intuition comes from the advantages of each estimator. Maximum likelihood is preferably when we are confident in the data generating process because, unlike the method of moments, it makes use of the knowledge of the entire distribution. Since MoM estimators only use information contained in the moments, it seems like the two methods should produce the same estimates when the sufficient statistics for the parameter we are attempting to estimate are exactly the moments of the data.

I've checked this result with a few distributions. Normal (unknown mean and variance), exponential, and Poisson all have sufficient statistics equal to their moments and have MLEs and MoM estimators the same (not strictly true for things like Poisson where there are multiple MoM estimators). If we look at a Uniform$(0,\theta)$, the sufficient statistic for $\theta$ is $\max(X_1,\cdots,X_N)$ and the MoM and MLE estimators are different.

I thought perhaps this was a quirk of the exponential family, but for a Laplace with known mean the sufficient statistic is $\frac{1}{n} \sum |X_i|$ and the MLE and the MoM estimator for the variance are not equal.

I've so far been unable to show any sort of result in general. Does anybody know of general conditions? Or even a counter example would help me refine my intuition.

• MM and MLE coincide for the canonical parameter in exponential families. But making a transformation would generally mean you lose this equivalence (as also suggested by Xi'an's answer). – hejseb Feb 17 '17 at 19:03

Furthermore, as stated in the question, there are many MoM estimators. An infinity of them, actually. But they are all based on the empirical distribution, $\hat{F}$, which may be seen as a non-parametric MLE of $F$, although this does not relate to the question.