Method of moments
Definition taught in Estimation, part of Statistical inference
Method of moments
| Type | Definition |
| Taught in | Estimation |
| Part | Statistical inference |
| Worked examples | 1 |
Suppose the first k population moments exist and determine \theta=(\theta_1,\ldots,\theta_k). The method of moments matches them to the corresponding sample moments,
\mathbb{E}_\theta(X^r)=\frac1n\sum_{i=1}^n x_i^r,\qquad r=1,\ldots,k,
and solves the k equations for \theta.
Examples
Two estimators of a uniform endpoint
Worked example from Estimation
Let X_1,\ldots,X_n be a random sample from the uniform distribution on (0,\theta), and let M=\max_i X_i.
Question. Compare the mean squared errors of the moment estimator 2\overline X and the maximum likelihood estimator M.
Solution. 2\overline X is unbiased, and \operatorname{Var} X_i=\theta^2/12, so
\operatorname{MSE}(2\overline X)=4\cdot\frac{\theta^2}{12n}=\frac{\theta^2}{3n}.
For the maximum, \mathbb{P}(M\le m)=(m/\theta)^n on (0,\theta), so M has density nm^{n-1}/\theta^n. Then \mathbb{E} M=n\theta/(n+1) and \mathbb{E} M^2=n\theta^2/(n+2), and
\operatorname{MSE}(M)=\mathbb{E} M^2-2\theta\mathbb{E} M+\theta^2=\frac{2\theta^2}{(n+1)(n+2)}.
The maximum is biased low, because it can never exceed \theta, but its mean squared error falls like 1/n^2 rather than 1/n. With n=10 the two are 0.033\,\theta^2 and 0.015\,\theta^2. The rescaled maximum (n+1)M/n is unbiased and has mean squared error \theta^2/\{n(n+2)\}, smaller still.
Used in
- Taught in Estimation, where the derivation and examples are.
- All 5 questions on Estimation
See also
Estimator · Sample correlation coefficient · Likelihood function · Maximum likelihood · Invariance of maximum likelihood · Unbiased estimator · Mean squared error · Consistent estimator