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Exponential distribution - Maximum Likelihood Estimation

by Marco Taboga, PhD

In this lecture, we derive the maximum likelihood estimator of the parameter of an exponential distribution.

The theory needed to understand the proofs is explained in the introduction to maximum likelihood estimation (MLE).

Table of contents

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

Assumptions

We observe the first terms of an IID sequence of random variables having an exponential distribution.

A generic term of the sequence $X_{j}$ has probability density function [eq2] where:

is the support of the distribution;
the rate parameter $lambda _{0}$ is the parameter that needs to be estimated.

Main steps leading to the solution of the maximum likelihood estimation 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 [eq4]

Proof

Since the terms of the sequence are independent, the likelihood function is equal to the product of their densities: [eq5] Because the observed values can only belong to the support of the distribution, we can write [eq7]

The log-likelihood function

The log-likelihood function is [eq8]

Proof

This is obtained by taking the natural logarithm of the likelihood function: [eq9]

The maximum likelihood estimator

The maximum likelihood estimator of is [eq10]

Proof

The estimator is obtained as a solution of the maximization problem The first order condition for a maximum is The derivative of the log-likelihood is [eq13] By setting it equal to zero, we obtain [eq14] Note that the division by is legitimate because exponentially distributed random variables can take on only positive values (and strictly so with probability 1).

Therefore, the estimator is just the reciprocal of the sample mean [eq17]

Asymptotic variance

The estimator is asymptotically normal with asymptotic mean equal to $lambda _{0}$ and asymptotic variance equal to [eq19]

Proof

The score is [eq20] The Hessian is [eq21] By the information equality, we have that [eq22] Finally, the asymptotic variance is [eq23]

This means that the distribution of the maximum likelihood estimator can be approximated by a normal distribution with mean $lambda _{0}$ and variance $lambda _{0}^{2}/n$ .

Other examples

StatLect has several pages like this one. Learn how to derive the MLEs of the parameters of the following distributions and models.

	Type	Solution
Normal distribution	Univariate distribution	Analytical
Poisson distribution	Univariate distribution	Analytical
T distribution	Univariate distribution	Numerical
Multivariate normal distribution	Multivariate distribution	Analytical
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). "Exponential distribution - Maximum Likelihood Estimation", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/exponential-distribution-maximum-likelihood.