Method of moments for the normal distribution

Question

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

Derive the MoM estimators for the mean $\mu$ and variance $\sigma^2$ of a Normal $(\mu, \sigma^2)$ distribution. Do they match the MLE?

Accepted Answer

The recipe. With sample moments m1 = \frac1n\sum Xi and m2 = \frac1n\sum Xi^2, match theory to data. Match the two moments. \mathbb{E}[X] = \mu = m1, \qquad \mathbb{E}[X^2] = \mu^2 + \sigma^2 = m2. The first gives \hat\mu{	ext{MoM}} = m1 = \bar X. Substituting into the second: \hat\sigma^2{	ext{MoM}} = m2 - m1^2 = \frac{1}{n}\sum (Xi - \bar X)^2. Do they match the MLE? Yes, exactly. The normal MLE is also \hat\mu = \bar X and \hat\sigma^2 = \frac1n\sum(Xi - \bar X)^2. So for the normal, MoM and MLE give identical estimators, another case (like the exponential and Poisson) where the two reci…

Method of moments for the normal distribution

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