Poisson Frequency Maximum Likelihood

Why Estimate a Poisson Frequency Rate?

A frequency model describes how many claims occur in a specified exposure period. In a simple Poisson model, the parameter is the expected number of claims per exposure unit. Estimating that parameter from complete claim counts is a basic likelihood calculation: the data determine a likelihood, and the rate that maximizes it is the maximum likelihood estimate (MLE).

The key distinction is between a count and a rate. A total of 12 claims is not, by itself, a frequency rate: it must be related to the exposure over which those claims were observed. With one-year observations, the MLE is the sample mean count per year. With unequal observation periods, it is total claims divided by total exposure.

Definition: Suppose $N_i$ is the number of claims observed during known exposure $t_i$, for $i=1,\ldots,n$. Under a homogeneous Poisson frequency model, the observations are independent and $$ N_i\sim\operatorname{Poisson}(\lambda t_i), $$ where $\lambda\geq 0$ is the claim rate per exposure unit. The counts are complete: each claim count over its stated exposure period is observed. The exposures are treated as fixed and known, and the same rate applies to every observation.

For example, if exposure is measured in policy-years, $\lambda$ is the expected number of claims per policy-year. The Poisson parameter for an individual observation is $\lambda t_i$, not necessarily $\lambda$: it is the expected number of claims over that observation’s actual exposure.

Likelihood and the Score Equation

For a Poisson random variable with mean $\lambda t_i$, the probability of observing $N_i=n_i$ is $e^{-\lambda t_i}(\lambda t_i)^{n_i}/n_i!$. Independence means that the likelihood for all observations is the product of these probabilities. Define total claims $S=\sum_{i=1}^n n_i$ and total exposure $T=\sum_{i=1}^n t_i$. Then

L(\lambda) =\prod_{i=1}^n \frac{e^{-\lambda t_i}(\lambda t_i)^{n_i}}{n_i!} =e^{-\lambda T}\lambda^S \prod_{i=1}^n\frac{t_i^{n_i}}{n_i!}.

The factors involving only observed counts and known exposures do not depend on $\lambda$. For $S>0$ and $\lambda>0$, take the logarithm and differentiate:

\ell(\lambda) =-\lambda T+S\log\lambda+ \sum_{i=1}^n n_i\log t_i-\sum_{i=1}^n\log(n_i!), \qquad \ell'(\lambda)=-T+\frac{S}{\lambda}.

Setting the score $\ell'(\lambda)$ equal to zero gives $\widehat{\lambda}=S/T$. The second derivative is $\ell''(\lambda)=-S/\lambda^2<0$ when $S>0$, so the log-likelihood is strictly concave and this critical point is the unique maximum.

Result: For independent complete Poisson counts with a common rate $\lambda$ and known exposures $t_i$, if at least one claim is observed, the MLE is $$ \widehat{\lambda}=\frac{\sum_{i=1}^n n_i}{\sum_{i=1}^n t_i}. $$ If all exposures equal $t$, this is $\overline{N}/t$. If every observation has one exposure unit, it is the sample mean count $\overline{N}$.

Why the Sample Mean Appears

When every observation covers one exposure unit, each $N_i$ has a Poisson distribution with mean $\lambda$. In that case, $T=n$ and $S=\sum_i n_i$, so the score equation becomes

-n+\frac{\sum_{i=1}^n n_i}{\lambda}=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{1}{n}\sum_{i=1}^n n_i=\overline{N}.

This is more than a convenient shortcut. The score balances the total observed claims against the total expected claims: at the estimate, $\widehat{\lambda}T=S$. Thus the fitted model’s expected total number of claims equals the observed total. With equal unit exposures, that balance says that the estimated expected count per observation equals the observed average count.

Worked Example: Estimating a Rate From One-Year Counts

A hypothetical insurer observes complete claim counts for six policies, each covered for exactly one policy-year. The counts are $0,1,2,0,3,2$. Assume independent Poisson counts with a common rate $\lambda$ per policy-year. Estimate the rate, then find the fitted probability that a policy-year has at least two claims and the expected number of claims over three policy-years.

Method and data summary. The observations have equal, one-year exposures, so the MLE is their sample mean. The total count is $S=0+1+2+0+3+2=8$, and the total exposure is $T=6(1)=6$ policy-years.

Therefore, $\widehat{\lambda}=S/T=8/6=4/3$, or approximately $1.333$ claims per policy-year. As a score-equation check, $-T+S/\widehat{\lambda}=-6+8/(4/3)=-6+6=0$.

Fitted probability. For a one-policy-year count $N$, the fitted distribution is $\operatorname{Poisson}(4/3)$. Hence $P(N\geq2)=1-P(N=0)-P(N=1)=1-e^{-4/3}(1+4/3)$. Using $e^{-4/3}\approx0.263597$, this is $1-(0.263597)(2.333333)\approx0.3849$, to four decimal places.

Three-year expected count. The expected number of claims over three policy-years is $\widehat{\lambda}(3)=\frac{4}{3}(3)=4$.

Check and common mistake. The fitted expected total over the six observed policy-years is $(4/3)(6)=8$, matching the observed total. Do not report the total of eight claims as the estimated annual rate; the rate divides claims by exposure.

Unequal Exposures

When observation periods differ, averaging the individual counts gives each observation equal weight even though the observations cover different amounts of exposure. The likelihood accounts for exposure through the means $\lambda t_i$, and its score uses the total exposure $T$. Thus the appropriate estimator is total claims divided by total exposure. It is not generally the unweighted average of the individual rates $n_i/t_i$.

Worked Example: Combining Different Observation Periods

A hypothetical book of policies has four complete count records. The exposures, measured in policy-years, are $0.5,1,1.5,3$, and the respective claim counts are $1,0,3,2$. Assume the counts are independent and each has a Poisson distribution with mean equal to its exposure times a common annual rate $\lambda$. Estimate the rate and compare it with the unweighted average of the four individual rates.

Method. Sum the observed counts and exposures separately: $S=1+0+3+2=6$ claims, and $T=0.5+1+1.5+3=6$ policy-years. The MLE is $\widehat{\lambda}=S/T=6/6=1$ claim per policy-year.

The score check is $-T+S/\widehat{\lambda}=-6+6/1=0$. The fitted expected counts for the four records are, respectively, $\widehat{\lambda}t_i=0.5,1,1.5,3$, which sum to $6$, the observed total.

The individual observed rates are $1/0.5=2$, $0/1=0$, $3/1.5=2$, and $2/3$. Their unweighted average is $(2+0+2+2/3)/4=7/6\approx1.167$ claims per policy-year. That is not the MLE for this common-rate model. The likelihood-based estimate uses the total count over total exposure, giving $1$ claim per policy-year.

Check and common mistake. The fitted expected total is $(1)(6)=6$, consistent with the six observed claims. Averaging the four individual rates equally is the trap: it gives the half-year record as much weight as the three-year record.

The Zero-Count Boundary Case

The score equation derivation above uses $\log\lambda$, so it applies directly only when $S>0$ and $\lambda>0$. If no claims are observed, $S=0$, the likelihood instead reduces to $L(\lambda)=e^{-\lambda T}$ times factors that do not depend on $\lambda$. For positive total exposure, this decreases as $\lambda$ increases. If the parameter space includes $\lambda=0$, the maximum is at that boundary.

Worked Example: No Claims in the Observed Exposure

A hypothetical small program records zero claims during three fully observed periods with exposures of $1$, $1.5$, and $2$ policy-years. Assume independent Poisson counts with a common rate $\lambda\geq0$ per policy-year. Find the MLE and explain why solving the usual score equation is not appropriate.

Method. The total count is $S=0+0+0=0$, and total exposure is $T=1+1.5+2=4.5$ policy-years. The likelihood is $L(\lambda)=e^{-4.5\lambda}$, since all three factorial terms equal $0!=1$ and all count powers equal one.

For every $\lambda\geq0$, $L(\lambda)=e^{-4.5\lambda}\leq1$, with equality at $\lambda=0$. Therefore $\widehat{\lambda}=0$ claims per policy-year. Equivalently, for $\lambda>0$, the log-likelihood has derivative $-4.5<0$, so it decreases throughout the positive parameter range and is maximized at the included boundary.

The formula $S/T$ also gives $0/4.5=0$, but the interior score equation $ -T+S/\lambda=0$ cannot be used at $\lambda=0$. The likelihood itself establishes the boundary maximum. If the model instead required $\lambda>0$, no positive value would attain the maximum; the likelihood would approach its supremum as $\lambda$ approaches zero.

Check and common mistake. Zero observed claims do not make the likelihood undefined. The error is to take $\log\lambda$ and apply an interior score formula at $\lambda=0$. Evaluate the likelihood at the boundary instead.

Assumptions and Exam Traps

The estimate $S/T$ relies on the stated model, not just on having a table of counts. It assumes a common rate over the exposures being combined, independent Poisson counts, known exposure amounts, and complete observations over those exposure periods. If the rate varies across groups or time, one common-rate estimate may not describe the data appropriately. If counts are incomplete or observations are subject to truncation or censoring, the complete-count likelihood used here is not the right likelihood.

A further unit check is essential. If $t_i$ is measured in policy-months, $S/T$ is claims per policy-month. To report a yearly rate, convert consistently: multiply a monthly rate by 12. Do not mix exposures recorded in years with counts or rates expressed in months.

Common exam trap: Using $S/n$ when exposure periods differ. The correct denominator is total exposure $T$, so the common-rate estimate is $S/T$. The sample mean $S/n$ is the MLE as a rate only when each observation has one exposure unit; with equal exposure $t$, the rate estimate is $\overline{N}/t$.

These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. When applying them, state the rate’s time or exposure unit, distinguish the Poisson mean for a particular observation from the common rate, and check that fitted expected claims equal observed total claims over the combined exposure.

Check Your Understanding

For each problem, assume complete counts and the common-rate Poisson model stated in the question. Determine the requested estimate and keep the exposure units consistent.

A hypothetical service contract portfolio has five independent one-year count observations: $1,2,0,3,2$. Estimate the annual claim rate. (Answer: $1.6$ claims per year)
A hypothetical equipment pool has counts $0,2,3$ over exposures of $0.5,1,2$ equipment-years. Estimate the common rate per equipment-year. (Answer: $10/7\approx1.4286$ claims per equipment-year)
Three independent observation periods have counts $2,0,5$ and exposures $1,0.5,2$ years. Find the MLE per year and verify it by comparing fitted expected total claims with the observed total. (Answer: $2$ claims per year; fitted total $2(3.5)=7$)
A hypothetical benefit plan observes zero claims in $18$ fully observed policy-months. Under a common Poisson rate, find the MLE per policy-year. (Answer: $0$ claims per policy-year)
Four complete portfolio records each cover $2$ policy-years and have counts $1,4,2,0$. Estimate the rate per policy-year and express it approximately per policy-month. (Answer: $7/8=0.875$ claims per policy-year, or $0.0729$ claims per policy-month)

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