Maximum Likelihood Estimation for Poisson Rates

Why Estimate a Poisson Rate?

Count data arise when we record how many events occur in a specified period or amount of exposure. Examples include hypothetical counts of equipment failures per month, claims reported during a policy period, or arrivals at a service desk during an hour. A Poisson model provides a way to estimate the average count, or rate, from repeated observations.

The key modeling detail is what each observation represents. If all observations cover the same exposure, the maximum likelihood estimate of the Poisson parameter is the average of the observed counts. If observations cover different known exposures, the estimate is total events divided by total exposure. The distinction matters: the unweighted sample mean of the counts is generally not a rate per unit exposure when observation periods differ.

Definition: Suppose $X_1,\ldots,X_n$ are independent and have the same Poisson distribution with parameter $\lambda$, where $\lambda\geq 0$. Thus $X_i\sim\operatorname{Poisson}(\lambda)$, and $$ P(X_i=x_i)=\frac{e^{-\lambda}\lambda^{x_i}}{x_i!},\qquad x_i=0,1,2,\ldots. $$ The parameter $\lambda$ is the expected count in the observation unit, since $E[X_i]=\lambda$. The maximum likelihood estimator (MLE) is the value of $\lambda$ that maximizes the likelihood of the observed counts.

Independence means that the joint probability of the observed counts is the product of their individual probabilities. The likelihood treats the observed values $x_1,\ldots,x_n$ as fixed and regards that joint probability as a function of the unknown parameter $\lambda$. Factors that do not depend on $\lambda$, such as the factorials, do not affect which parameter value maximizes it.

Deriving the MLE for Equal Exposures

For observed counts $x_1,\ldots,x_n$, independence gives the likelihood

L(\lambda) =\prod_{i=1}^n\frac{e^{-\lambda}\lambda^{x_i}}{x_i!} =\frac{e^{-n\lambda}\lambda^{\sum_{i=1}^n x_i}}{\prod_{i=1}^n x_i!}, \qquad \lambda\geq 0.

When the total count is positive, taking logarithms turns the product into a sum and makes differentiation straightforward. Write $T=\sum_{i=1}^n x_i$. For $\lambda>0$, the log-likelihood is

\ell(\lambda)=\log L(\lambda) =-n\lambda+T\log\lambda-\sum_{i=1}^n\log(x_i!).

The final term is constant with respect to $\lambda$. Differentiating and setting the score equal to zero gives a candidate:

\frac{d\ell}{d\lambda}=-n+\frac{T}{\lambda}=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{T}{n} =\frac{1}{n}\sum_{i=1}^n x_i.

A score equation gives a candidate, so we also check that this value is a maximum. The second derivative is $-T/\lambda^2$, which is negative when $T>0$. The log-likelihood is therefore strictly concave for $\lambda>0$, and the candidate is its unique maximum there. At $\lambda=0$, the likelihood is zero when $T>0$, so the boundary cannot improve on this positive candidate.

Theorem: For independent observations $X_i\sim\operatorname{Poisson}(\lambda)$, with common exposure and $\lambda\geq0$, the MLE is $$ \widehat{\lambda}=\overline{X}=\frac{1}{n}\sum_{i=1}^n X_i. $$ If the total count is zero, this formula gives $\widehat{\lambda}=0$, which is also the likelihood maximum at the boundary.

The result has a direct interpretation. Across $n$ equal observation units, the total number of recorded events is divided by the number of units. Thus the estimate is the observed average count per unit. This interpretation follows from the likelihood derivation; it is not a rule to use without checking that the observations have equal exposure and share one Poisson parameter.

Worked Example: Estimating a Common Count Parameter

A hypothetical monitoring system records equipment alerts during five equal-length shifts. The counts are $2,0,3,1,4$. Assume the counts are independent and each has a Poisson distribution with the same parameter $\lambda$, measured in alerts per shift. Find the MLE and verify that it maximizes the likelihood.

Method. Use the independent, equal-exposure Poisson likelihood. First compute the total and the sample size:

T=2+0+3+1+4=10,\qquad n=5.

The likelihood, as a function of $\lambda$, is

L(\lambda) =\frac{e^{-5\lambda}\lambda^{10}}{2!\,0!\,3!\,1!\,4!}.

For $\lambda>0$, its log-likelihood has derivative

\ell'(\lambda)=-5+\frac{10}{\lambda}.

Setting this equal to zero gives $\widehat{\lambda}=10/5=2$ alerts per shift. The second derivative at any positive $\lambda$ is $\ell''(\lambda)=-10/\lambda^2<0$, so the log-likelihood is strictly concave and this candidate is the unique maximum. The boundary likelihood at zero is zero because the total count is positive. As an additional check, the estimated expected total over five shifts is $5(2)=10$, equal to the observed total. The common mistake to avoid is dividing by the total number of alerts rather than by the five observation periods.

Known Unequal Exposures

A rate per unit exposure can also be estimated when the observation intervals differ in known length. Suppose $X_i$ counts events over exposure $t_i>0$, and assume the observations are independent with $X_i\sim\operatorname{Poisson}(\lambda t_i)$. Here $\lambda\geq0$ is the common rate per exposure unit; the expected count in observation $i$ is $\lambda t_i$. This model assumes the rate is constant across observations and that the specified exposures are known.

The likelihood contains one factor for each observation. Collecting the terms involving $\lambda$ gives

L(\lambda) =\prod_{i=1}^n\frac{e^{-\lambda t_i}(\lambda t_i)^{x_i}}{x_i!} =e^{-\lambda\sum_i t_i}\lambda^{\sum_i x_i} \prod_{i=1}^n\frac{t_i^{x_i}}{x_i!}.

For a positive total count, the terms involving $\lambda$ in the log-likelihood are $-\lambda\sum_i t_i+(\sum_i x_i)\log\lambda$. Differentiating gives the rate estimate below. The same second-derivative argument as before verifies a unique maximum when the total count is positive.

\widehat{\lambda} =\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n t_i}.

In words, this is total events divided by total exposure. If every $t_i=1$, the denominator is $n$ and the estimate reduces to the sample mean count. If exposures differ, averaging the individual counts does not generally estimate events per exposure unit.

Worked Example: Combining Counts From Unequal Observation Periods

A hypothetical field study records event counts over four known exposure periods. The exposure lengths are $1,2,0.5,1.5$ monitoring units, and the corresponding counts are $1,5,0,6$. Assume independent counts with $X_i\sim\operatorname{Poisson}(\lambda t_i)$, where $\lambda$ is the common rate per monitoring unit. Estimate $\lambda$.

Method. Add counts and exposures separately:

\sum_i x_i=1+5+0+6=12,\qquad \sum_i t_i=1+2+0.5+1.5=5.

Therefore,

\widehat{\lambda}=\frac{12}{5}=2.4

events per monitoring unit. For a check, the fitted expected counts for the four observations are $2.4(1)=2.4$, $2.4(2)=4.8$, $2.4(0.5)=1.2$, and $2.4(1.5)=3.6$; they sum to $12$, the observed total. The unweighted mean count would be $12/4=3$, but it is not the estimate of events per unit exposure because the observation periods are not all one unit long. The common mistake is to use the number of observations as the denominator rather than total exposure.

The Zero-Total-Count Case

When every observed count is zero, the score equation involving $\log\lambda$ cannot be used at $\lambda=0$. The likelihood itself makes the answer clear. Under equal exposures,

L(\lambda)=e^{-n\lambda},\qquad \lambda\geq0.

This likelihood decreases as $\lambda$ increases, so its maximum over the allowed parameter space occurs at $\lambda=0$. At that value, each observation has probability one of being zero under the model, and the likelihood is $1$. Under known unequal exposures, the corresponding likelihood is $e^{-\lambda\sum_i t_i}$, which likewise has its maximum at zero.

The parameter space matters. With the standard model allowing $\lambda=0$, zero is an attained MLE. If a problem instead imposes the strict restriction $\lambda>0$, there is no attained maximum when all counts are zero; the likelihood approaches its supremum as $\lambda$ decreases toward zero. Always use the parameter space stated in the problem.

Worked Example: All Observed Counts Are Zero

A hypothetical sensor is monitored over four equal one-day periods, and no alerts are recorded on any day. Assume independent daily counts $X_i\sim\operatorname{Poisson}(\lambda)$, with $\lambda\geq0$. Find the MLE and explain why solving the usual interior score equation is not appropriate.

Method. The total count is $T=0+0+0+0=0$, so the likelihood is

L(\lambda)=\prod_{i=1}^4 e^{-\lambda}=e^{-4\lambda}.

For any $\lambda>0$, $e^{-4\lambda}<1$, while $L(0)=1$. Hence the maximum over $\lambda\geq0$ is attained at

\widehat{\lambda}=0

alerts per day. The formula $\widehat{\lambda}=T/n$ agrees: $0/4=0$. The interior score expression $-4+T/\lambda$ is not valid at $\lambda=0$, and substituting $T=0$ into it cannot establish the boundary maximum. Comparing the likelihood at the boundary, or observing that it decreases for positive $\lambda$, resolves the case. The common mistake is to reject zero because a logarithm or score calculation was written only for $\lambda>0$.

Exam Traps and Model Checks

Before calculating, identify the unit in which the rate is being reported and verify that the observations support a common-rate model. A count of 8 over four hours and a count of 3 over one hour should not be treated as two equal-duration counts when estimating events per hour. Likewise, if the observations are dependent or their rates differ for reasons not represented in the model, the likelihood above does not apply as stated.

A particularly common error is to use the mean of individual rates even when the desired likelihood estimate is based on unequal exposures. For example, in the unequal-exposure example, the individual rates are $1/1=1$, $5/2=2.5$, $0/0.5=0$, and $6/1.5=4$. Their unweighted mean is $1.875$, not $2.4$. The likelihood estimate weights each observation through its exposure: it is $12/5=2.4$. Use the sample mean of counts for equal exposures; use total count divided by total exposure for known unequal exposures.

Key takeaway: Under independent Poisson counts with a common parameter, the MLE is total count divided by the number of equal-exposure observations. Under a common rate and known unequal exposures, it is total count divided by total exposure. If the total count is zero and zero is allowed in the parameter space, the MLE is zero.

Check Your Understanding

For each problem, use the stated independent Poisson model and show which total belongs in the denominator.

A hypothetical lab records counts $3,1,0,4,2$ over five equal one-hour periods. Estimate the Poisson mean per hour. (Answer: $2$ events per hour.)
A hypothetical counter records zero arrivals in each of six equal intervals. The parameter space is $\lambda\geq0$. Find the MLE and state whether it is attained. (Answer: $\widehat{\lambda}=0$, attained at the boundary.)
A hypothetical inspection process has counts $4,1,9$ over exposures of $2,1,3$ machine-hours, respectively, with independent $X_i\sim\operatorname{Poisson}(\lambda t_i)$. Find the rate MLE per machine-hour. (Answer: $14/6=7/3\approx2.3333$ events per machine-hour.)
A hypothetical sensor records counts $0,2,3,7$ over exposures $0.5,1,2,1.5$ days, with independent counts and a common rate per day. Calculate the MLE and check it by comparing total fitted expected count with total observed count. (Answer: $\widehat{\lambda}=12/5=2.4$ events per day; the fitted total expected count is $2.4(5)=12$.)
Four independent equal-exposure counts have total $15$, while five other independent counts have total $15$, with all nine observations sharing the same Poisson parameter. Find the MLE using all observations, and explain why averaging the two group means equally is not the likelihood calculation. (Answer: $15/9=5/3\approx1.6667$ counts per exposure unit; the two groups contain different numbers of observations.)

What You'll Learn