Method of Moments

Study guide: Exam MAS-I

The Method of Moments (MoM) estimates parameters by setting the sample moments equal to the theoretical moments and solving. With $k$ unknown parameters, the first $k$ moment equations are used. It is the simplest estimation method and often supplies the starting values for Maximum Likelihood Estimation.

$\frac{1}{n}\sum_{i=1}^{n} X_i^{\,r} = E[X^{\,r}], \qquad r = 1, 2, \ldots, k$

$\bar{X} = E[X \mid \theta], \qquad \frac{1}{n}\sum X_i^2 = E[X^2 \mid \theta]$

One parameter needs one equation ( $\bar{X} = \mu(\theta)$ ); two parameters need the first two moments, usually solved as mean and variance
MoM estimators are consistent by the law of large numbers, but they are generally not efficient and not always unbiased — MLE beats them asymptotically
They can produce impossible values (a negative variance, a shape parameter below zero) when the sample moments are unusual; MLE is constrained by the likelihood and cannot
MoM is not built from the Sufficient Statistic, so in some models it leaves information on the table — for the gamma (both parameters unknown), beta and Pareto, the moment estimators are not functions of the sufficient statistics. In others it lands on one anyway: $\bar{X}$ for the Bernoulli and Poisson, and $\big(\bar{X}, \tfrac{1}{n}\sum (X_i - \bar{X})^2\big)$ for the Normal, where the MoM estimators are also the MLEs
It handles awkward likelihoods well: for a mixture or a compound distribution, the moments are simple even when the density is not
Censored or truncated data breaks the plain method — the sample moments no longer estimate the population moments, so the theoretical moments must be recomputed under the Censoring or Truncation scheme

Gamma Parameters by Moments A sample of claim amounts has $\bar{x} = 500$ and $s^2 = 125{,}000$ (using the $n$ divisor). Fit a Gamma distribution with shape $\alpha$ and scale $\theta$ by the method of moments.

Answer For a Gamma, $E[X] = \alpha\theta$ and $\text{Var}(X) = \alpha\theta^2$ . Setting these to the sample values: $\alpha\theta = 500, \qquad \alpha\theta^2 = 125{,}000$ Dividing the second by the first gives $\theta = 250$ , and then $\alpha = \frac{500}{250} = 2$ The fitted distribution is $\text{Gamma}(\alpha = 2, \theta = 250)$ .

Method of Moments vs MLE for the Uniform A sample from $\text{Uniform}(0, \theta)$ has $\bar{x} = 6$ with largest observation $x_{(n)} = 14$ . Compare the MoM and ML estimates of $\theta$ .

Answer MoM: $E[X] = \theta/2$ , so $\hat\theta_{\text{MoM}} = 2\bar{x} = 12$ . MLE: the likelihood is positive only when $\theta \ge x_{(n)}$ and decreasing in $\theta$ , so $\hat\theta_{\text{MLE}} = x_{(n)} = 14$ . The MoM estimate is impossible — it is smaller than an observed value. This is the standard illustration that matching moments ignores the structure of the likelihood.