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Normal distribution - Maximum Likelihood Estimation

This lecture deals with maximum likelihood estimation of the parameters of the normal distribution.

Before continuing, you might want to revise the basics of maximum likelihood estimation (MLE).

Table of contents

Assumptions
The likelihood function
The log-likelihood function
The maximum likelihood estimators
Asymptotic variance
Other examples

Assumptions

Our sample is made up of the first terms of an IID sequence of normal random variables having mean $mu _{0}$ and variance $sigma _{0}^{2}$ .

The probability density function of a generic term of the sequence is [eq2]

The mean $mu _{0}$ and the variance $sigma _{0}^{2}$ are the two parameters that need to be estimated.

Main steps leading to the solution of the maximization problem: compute the likelihood, take the log, calculate the gradient of the log-likelihood function and set it equal to zero.

The likelihood function

The likelihood function is [eq3]

Proof

Given the assumption that the observations from the sample are IID, the likelihood function can be written as [eq4]

The log-likelihood function

The log-likelihood function is [eq5]

Proof

By taking the natural logarithm of the likelihood function, we get [eq6]

The maximum likelihood estimators

The maximum likelihood estimators of the mean and the variance are [eq7]

Proof

We need to solve the following maximization problem The first order conditions for a maximum are [eq9] The partial derivative of the log-likelihood with respect to the mean is [eq10] which is equal to zero only if [eq11] Therefore, the first of the two first-order conditions implies [eq12] The partial derivative of the log-likelihood with respect to the variance is [eq13] which, if we rule out $sigma ^{2}=0$ , is equal to zero only if [eq14] Thus, the system of first order conditions is solved by [eq15]

Thus, the estimator is equal to the sample mean and the estimator is equal to the unadjusted sample variance.

Asymptotic variance

The vectoris asymptotically normal with asymptotic mean equal toand asymptotic covariance matrix equal to [eq19]

Proof

The first entry of the score vector is [eq21] The second entry of the score vector is [eq22] In order to compute the Hessian [eq23] we need to compute all second order partial derivatives. We have [eq24] and [eq25] Finally, [eq26] which, as you might want to check, is also equal to the other cross-partial derivative . Therefore, the Hessian is [eq28] By the information equality, we have that [eq29] As a consequence, the asymptotic covariance matrix is [eq30]

In other words, the distribution of the vector can be approximated by a multivariate normal distribution with mean and covariance matrix [eq33]

Other examples

StatLect has several pages that contain detailed derivations of MLEs. Learn how to find the estimators of the parameters of the following distributions and models.

	Type	Solution
Exponential distribution	Univariate distribution	Analytical
Poisson distribution	Univariate distribution	Analytical
T distribution	Univariate distribution	Numerical
Multivariate normal distribution	Multivariate distribution	Analytical
Gaussian mixture	Mixture of distributions	Numerical (EM)
Normal linear regression model	Regression model	Analytical
Logistic classification model	Classification model	Numerical
Probit classification model	Classification model	Numerical

How to cite

Please cite as:

Taboga, Marco (2021). "Normal distribution - Maximum Likelihood Estimation", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/normal-distribution-maximum-likelihood.