Poisson Frequency Estimation by Moments and Likelihood

Why Exposure Matters in Frequency Estimation

A claim count has meaning only in relation to the time or amount of risk that produced it. Four claims observed over ten policy-years suggest a different frequency from four claims observed over two policy-years. A Poisson frequency model makes this relationship explicit: it describes the expected number of claims as a rate multiplied by exposure.

This tutorial estimates that rate using the method of moments and maximum likelihood. The key practical result is simple: under a common Poisson rate, add the observed counts and divide by the total exposure. The reasoning matters, however, because averaging unweighted rates or using the wrong exposure unit can produce an incorrect estimate.

Definition: Suppose $N_j$ is the number of claims observed in record $j$, and $e_j$ is its exposure measured in a specified unit, such as policy-years. Under a homogeneous Poisson frequency model, the records are independent and $$ N_j\sim\operatorname{Poisson}(\lambda e_j), $$ where $\lambda\geq 0$ is the claim frequency rate per exposure unit. Thus $E[N_j]=\lambda e_j$, and $\operatorname{Var}(N_j)=\lambda e_j$. The model assumes a constant rate over the observed exposures, independent counts across records, and consistently recorded claims and exposure.

The exposure unit is part of the interpretation of $\lambda$. If exposure is measured in policy-years, $\lambda$ is claims per policy-year. If exposure is measured in vehicle-years, it is claims per vehicle-year. A rate cannot be interpreted correctly unless its denominator is stated.

For example, a count observed over one-half of a policy-year has mean $\lambda/2$, not $\lambda$. The model does not say that every policy produces a fractional number of claims; it says that the expected count over a specified exposure is proportional to that exposure.

Moment Estimation

For observations with a common exposure $e$, the expected count for each record is $\lambda e$. The first sample moment, the observed average count, can therefore be matched to its model expectation. This gives an estimate of the rate by dividing the average count by the exposure per record.

When exposures differ, the total count $C=\sum_j N_j$ has expectation $\lambda E$, where $E=\sum_j e_j$ is the total exposure. Matching the observed total count to its expected value gives the pooled moment estimate. This formulation also works when all exposures are equal.

Moment estimator: For independent Poisson counts with a common rate $\lambda$ and total exposure $E=\sum_j e_j>0$, the exposure-adjusted method-of-moments estimate is $$ \widehat{\lambda}_{\mathrm{MM}}=\frac{C}{E} =\frac{\sum_j N_j}{\sum_j e_j}. $$ When every record has exposure $e$, this is equivalently the sample mean count divided by $e$.

Worked Example: Converting a Mean Count to an Annual Rate

A hypothetical portfolio has four independent policy records, each with exposure of $0.5$ policy-year. The recorded claim counts are $0,1,2,1$. Estimate the Poisson claim rate per policy-year by moments, and check the implied mean count for one record.

Method. Use the observed average count and divide by the exposure per record. Equivalently, use total claims divided by total policy-year exposure.

The total number of claims is $$ C=0+1+2+1=4. $$ The total exposure is $$ E=4(0.5)=2\text{ policy-years}. $$ Therefore, $$ \widehat{\lambda}_{\mathrm{MM}}=\frac{C}{E}=\frac{4}{2}=2 \text{ claims per policy-year}. $$ As a check using the common-exposure form, the sample mean count is $4/4=1$ claim per record. Each record has $0.5$ policy-year of exposure, so the estimated rate is $1/0.5=2$ claims per policy-year. At that rate, the fitted mean count for a half-year record is $2(0.5)=1$, matching the observed average.

Common mistake avoided: The average count, $1$, is not itself the annual rate because each record represents only half a year. Dividing by the exposure per record gives the correct rate.

Maximum Likelihood Estimation

The likelihood uses the probability of the observed counts as a function of the unknown rate. Under the independent Poisson model, the probability of count $N_j=n_j$ is $e^{-\lambda e_j}(\lambda e_j)^{n_j}/n_j!$. Multiplying these probabilities across records gives the likelihood.

L(\lambda)=\prod_j\frac{e^{-\lambda e_j}(\lambda e_j)^{n_j}}{n_j!} \ \propto\ e^{-\lambda E}\lambda^C, \qquad C=\sum_j n_j,\quad E=\sum_j e_j.

The omitted factor in the proportional expression depends on the observed counts and exposures, but not on $\lambda$, so it does not change which value maximizes the likelihood. For $C>0$, the log-likelihood, apart from terms independent of $\lambda$, is $C\log\lambda-\lambda E$. Differentiating and setting the derivative to zero gives $C/\lambda-E=0$, hence the estimate $C/E$.

Maximum-likelihood estimate: Assume independent counts $N_j\sim\operatorname{Poisson}(\lambda e_j)$, with $\lambda\geq 0$ and total exposure $E=\sum_j e_j>0$. If $C=\sum_j n_j>0$, the maximum-likelihood estimate is $$ \widehat{\lambda}_{\mathrm{MLE}}=\frac{C}{E}. $$ If $C=0$, the likelihood is maximized at the boundary $\widehat{\lambda}_{\mathrm{MLE}}=0$. The positive-total-exposure condition is essential: if $E=0$, the data provide no exposure-based information about the rate, and this result does not apply.

For $C>0$, the second derivative of the log-likelihood is $-C/\lambda^2<0$, so the stationary point is a maximum. When $C=0$, the likelihood is proportional to $e^{-\lambda E}$. If $E>0$, this decreases as $\lambda$ increases over $\lambda\geq 0$, so its maximum is at zero. This also explains why the zero-count case must be treated separately: the log-likelihood expression involving $C\log\lambda$ is not a useful way to evaluate the boundary at $\lambda=0$.

Why the Two Estimates Agree

The moment and likelihood estimates coincide here because the observed total count is the statistic that carries the rate information, and its expectation is $\lambda E$. The moment method equates the observed total to that expectation. The likelihood method finds the rate that makes the observed total most probable, accounting for the same total exposure.

Write the count model.
For each record, use $N_j\sim\operatorname{Poisson}(\lambda e_j)$, with independent records and a common rate per exposure unit.

Combine the data.
Set $C=\sum_j n_j$ and $E=\sum_j e_j$. The likelihood depends on $\lambda$ through $e^{-\lambda E}\lambda^C$.

Estimate and interpret.
For $E>0$, use $C/E$. State the exposure unit in the answer, and check whether the fitted expected total count, $\widehat{\lambda}E$, equals $C$.

The last check follows directly from the estimate: $(C/E)E=C$. It is an arithmetic check, not proof that the Poisson model is appropriate. The estimate summarizes frequency under the assumed model; it does not by itself verify that the rate is constant or that records are independent.

Worked Example: Combining Different Exposure Amounts

A hypothetical insurer observes three independent risks under the same claim rate. Their exposures are $1$, $2$, and $0.5$ policy-years, and their claim counts are $2$, $1$, and $0$, respectively. Estimate the rate by moments and maximum likelihood. Also calculate the fitted mean count for the second risk.

Method. Pool the counts and exposures, rather than giving each risk’s count-per-exposure ratio equal weight.

The total count is $$ C=2+1+0=3, $$ and total exposure is $$ E=1+2+0.5=3.5\text{ policy-years}. $$ The moment estimate is $$ \widehat{\lambda}_{\mathrm{MM}}=\frac{3}{3.5} =\frac{6}{7}\approx 0.8571 \text{ claims per policy-year}. $$ The likelihood estimate is the same, since $C>0$ and $E>0$: $$ \widehat{\lambda}_{\mathrm{MLE}}=\frac{3}{3.5} \approx 0.8571 \text{ claims per policy-year}. $$ For the second risk, with $2$ policy-years of exposure, the fitted mean count is $$ \widehat{\lambda}(2)=\frac{6}{7}(2)=\frac{12}{7}\approx 1.7143 \text{ claims}. $$ As a check, the fitted mean total count over all records is $\widehat{\lambda}E=(3/3.5)(3.5)=3$, the observed total.

Common mistake avoided: The individual observed rates are $2/1=2$, $1/2=0.5$, and $0/0.5=0$. Their unweighted average is $2.5/3\approx0.8333$, not the pooled estimate. Records with more exposure represent more opportunity for claims; the rate estimate must weight them through their exposure.

The Zero-Claim Boundary Case

A zero observed count does not create a division problem as long as total exposure is positive. The moment estimate $C/E$ is zero, and the likelihood also has its maximum at zero. This is a boundary estimate, not evidence that future claims are impossible. It reflects only the fitted rate under the specified model and the data observed.

Worked Example: No Claims in Positive Exposure

A hypothetical group of policies produces zero claims over a total of $4$ policy-years. Estimate the common Poisson rate by moments and maximum likelihood. Then calculate the likelihood of observing zero claims if the rate were instead $0.25$ claims per policy-year.

Method. First estimate using the total count and positive total exposure. Then use the Poisson probability for the total count under the stated alternative rate.

Here $C=0$ and $E=4>0$, so $$ \widehat{\lambda}_{\mathrm{MM}}=\frac{0}{4}=0, \qquad \widehat{\lambda}_{\mathrm{MLE}}=0 \text{ claims per policy-year}. $$ At the alternative rate $\lambda=0.25$, the total count over $4$ policy-years has a Poisson distribution with mean $\lambda E=(0.25)(4)=1$. Therefore, the probability of zero claims is $$ P(C=0)=e^{-1}\frac{1^0}{0!}=e^{-1}\approx0.3679. $$ The zero-rate estimate is consistent with the observed count as a maximum-likelihood fit: with $E=4>0$, the likelihood $e^{-4\lambda}$ decreases as $\lambda$ increases from zero. At $\lambda=0$, the model assigns probability $1$ to zero claims; the estimate does not establish that the true rate is literally zero.

Common mistake avoided: The conclusion that the likelihood is maximized at zero depends on positive total exposure. If $E=0$, the likelihood is constant in $\lambda$, and the data cannot uniquely determine the rate. Here $E=4$, so the boundary conclusion is justified.

Checks Before Reporting an Estimate

A frequency estimate is only as interpretable as its exposure definition and data. Before reporting $C/E$, verify that the numerator and denominator refer to the same covered risks and time period. For example, claims recorded for a set of policies should not be divided by exposure from a different set of policies.

Check the unit. Report whether the rate is per policy-year, vehicle-year, or another exposure unit.
Check the model’s scope. The formula assumes one common rate across the records. If risk characteristics or time periods imply different rates, a common-rate estimate may not represent each group well.
Check the records. Claim counts and exposures must use compatible definitions and reporting periods. Missing claims or inconsistent exposure measurement can distort the estimate.
Check total exposure. The estimator $C/E$ requires $E>0$. With no exposure, a rate cannot be estimated by this method.

These checks distinguish a correct calculation from a defensible interpretation. The shared estimate $C/E$ is a direct consequence of the homogeneous Poisson model; it is not a general formula for every frequency model. The material here is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated exposure unit and assume the independent homogeneous Poisson model unless the question asks otherwise.

Four independent risks each have $0.25$ vehicle-year of exposure. Their claim counts are $0,1,0,2$. Find the moment estimate of the rate per vehicle-year. (Answer: $3$ claims per vehicle-year.)
Three records have exposures of $1$, $0.5$, and $2$ policy-years, with counts $1$, $0$, and $4$. Find the moment and likelihood estimates of the common rate. (Answer: $5/3.5\approx1.4286$ claims per policy-year.)
Five independent policies each contribute $2$ policy-years, and their counts are $3,1,0,2,4$. Find the estimated rate and the fitted expected total count. (Answer: $1$ claim per policy-year; fitted expected total count $=10$.)
A group has zero observed claims over $2.5$ policy-years. Find the moment and likelihood estimates, and calculate the probability of zero claims at an alternative rate of $0.4$ per policy-year. (Answer: both estimates are $0$; alternative-rate probability $e^{-1}\approx0.3679$.)
Three records have exposures $0.5$, $1.5$, and $1$ policy-years, with counts $1$, $2$, and $0$. Find the pooled rate estimate and the fitted mean count for a new risk with $2.5$ policy-years of exposure. (Answer: $1$ claim per policy-year; fitted mean count $=2.5$.)

What You'll Learn