Maximum likelihood for a normal sample
Section of Estimation, part of Statistical inference
Maximum likelihood for a normal sample
| Type | Section |
| Taught in | Estimation |
| Part | Statistical inference |
For X_1,\ldots,X_n independent and \operatorname{N}(\mu,\sigma^2),
\log L(\mu,\sigma^2)=-\frac1{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2-\frac n2\log\sigma^2-\frac n2\log(2\pi).
The two likelihood equations are
\frac{\partial\log L}{\partial\mu}=\frac1{\sigma^2}\sum_{i=1}^n(x_i-\mu)=0, \qquad \frac{\partial\log L}{\partial\sigma^2}=\frac1{2\sigma^4}\sum_{i=1}^n(x_i-\mu)^2-\frac n{2\sigma^2}=0.
The first gives \widehat\mu=\overline x whatever the value of \sigma^2. Substituting into the second gives
\widehat\sigma^2=\frac1n\sum_{i=1}^n(x_i-\overline x)^2.
So the normal mean is estimated by the sample mean, and the maximum likelihood estimator of the variance divides by n, not by n-1. The two estimators coincide with the moment estimators. The derivative was taken with respect to \sigma^2, treated as the parameter; the next result shows that working with \sigma instead gives the same answer.
Used in
- Taught in Estimation
- All 6 questions on Estimation
See also
Estimator · Method of moments · Sample correlation coefficient · Likelihood function · Maximum likelihood · Invariance of maximum likelihood · Unbiased estimator · Mean squared error