The Method of Moments: Two-Parameter Distributions

For a distribution with a single unknown parameter, the method of moments solves the equation $E[X] = \bar X$ for the parameter. When a distribution depends on two unknown parameters $\theta_1$ and $\theta_2$, one equation is no longer enough — we add a second moment equation.

The method of moments estimates $(\theta_1, \theta_2)$ by equating the first two population moments to the corresponding sample moments:

$$E[X] = \bar X, \qquad E[X^2] = m_2,$$

where

$$\bar X = \dfrac{1}{n}\sum\limits_{i=1}^n X_i, \qquad m_2 = \dfrac{1}{n}\sum\limits_{i=1}^n X_i^2.$$

Since $\text{Var}(X) = E[X^2] - (E[X])^2$, we may equivalently equate the mean and variance:

$$E[X] = \bar X, \qquad \text{Var}(X) = \hat\sigma^2,$$

where

$$\hat\sigma^2 = m_2 - \bar X^2 = \dfrac{1}{n}\sum\limits_{i=1}^n (X_i - \bar X)^2$$

is the (uncorrected) sample variance. Note that the divisor is $n$, not $n-1$.

Either form yields a system of two equations in the two unknowns. Solving the system gives the MoM estimates $(\hat\theta_1, \hat\theta_2)$.