Frequency Likelihoods with Unequal Exposures

Why Unequal Exposures Change the Likelihood

Frequency data often come from policies observed for different lengths of time. One policy may contribute a full year of experience, while another contributes only a few months because it began partway through the study period. A count of two claims over six months does not represent the same observed frequency as two claims over two years. A likelihood must therefore connect each observed count to its own exposure.

The key modeling choice is to specify how the expected count changes with exposure. For a constant event rate, doubling the exposure doubles the expected count. This proportionality belongs in the probability model itself; it is not an adjustment to make after fitting an equal-exposure model. The following results assume exposures are known and fixed, and that observations are independent conditional on the model parameters.

Definition: Let \(e_i>0\) be the exposure for observation \(i\), measured in a consistent unit such as policy-years, and let \(N_i\) be its observed event count. In a Poisson rate model with common rate \(\lambda\) events per exposure unit, \(N_i\sim\operatorname{Poisson}(\lambda e_i)\). Thus \(E[N_i]=\lambda e_i\), and the likelihood is the product of the observation-specific Poisson probabilities.

Poisson Likelihood and Its Rate Estimate

For independent counts \(N_1,\ldots,N_m\), the probability mass function for a Poisson random variable with mean \(\lambda e_i\) gives the likelihood

L(\lambda)=\prod_{i=1}^{m}\frac{(\lambda e_i)^{n_i}e^{-\lambda e_i}}{n_i!}.

Here \(n_i\) denotes the observed value of \(N_i\). Taking logarithms turns the product into a sum. Terms that do not depend on \(\lambda\) can be set aside when maximizing:

\ell(\lambda) =\sum_{i=1}^{m}\left[n_i\log(\lambda e_i)-\lambda e_i-\log(n_i!)\right] =\left(\sum_i n_i\right)\log\lambda-\lambda\left(\sum_i e_i\right)+C,

where \(C=\sum_i[n_i\log e_i-\log(n_i!)]\) is constant with respect to \(\lambda\). Differentiating and setting the score equal to zero yields the maximum likelihood estimate.

\frac{d\ell}{d\lambda} =\frac{\sum_i n_i}{\lambda}-\sum_i e_i=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{\sum_i n_i}{\sum_i e_i}.

This is total observed count divided by total exposure—not the unweighted average of the policy-level counts or rates. If every observed count is zero, the likelihood is maximized at the boundary \(\widehat{\lambda}=0\). Otherwise, the score equation gives a positive estimate.

Derivation: The exposure enters the expected count as \(\lambda e_i\). In the log-likelihood, each observation contributes \(n_i\log\lambda-\lambda e_i\), apart from terms not involving \(\lambda\). Summing produces \((\sum_i n_i)\log\lambda-\lambda(\sum_i e_i)\). Its derivative balances total observed events against total exposure, which explains why the estimate is the pooled count divided by pooled exposure.

Worked Example: Estimating a Poisson Rate from Policy-Year Counts

A hypothetical portfolio has four independently observed policies. Their exposures, in policy-years, are \(0.5, 1, 1.5,\) and \(2\), and their respective claim counts are \(0,1,2,\) and \(4\). Assume a common constant Poisson claim rate per policy-year. Estimate the rate by maximum likelihood and calculate the fitted expected count for the policy with exposure \(1.5\) years.

Method. Use the independent Poisson likelihood with mean \(\lambda e_i\) for each policy. The exposure values are fixed, and the rate is common across policies.

The total count is \(0+1+2+4=7\). The total exposure is \(0.5+1+1.5+2=5\) policy-years. Therefore,

\widehat{\lambda}=\frac{7}{5}=1.4 \quad\text{claims per policy-year}.

The fitted expected count for the \(1.5\)-year policy is \(\widehat{\lambda}e=1.4(1.5)=2.1\) claims. As a likelihood check, the log-likelihood at the estimate is

\ell(1.4) =7\log(1.4)+2\log(1.5)+4\log(2) -1.4(5)-\log(2!)-\log(4!) \approx -4.9324.

The terms \(\log(0!)\) and \(\log(1!)\) are both zero. The calculation uses the original observation-specific exposures; the pooled estimate also satisfies the score equation because \(7/1.4-5=0\). Common mistake avoided: the unweighted average count is \(7/4=1.75\), but that is not a rate per policy-year. Dividing by total exposure, not number of policies, gives the correct estimate.

Exposure in Related Count Models

Unequal observation periods do not automatically imply a Poisson model. The count distribution must match the data-generating assumptions. In particular, a binomial model counts successes among a known number of opportunities, whereas a Poisson rate model relates events to an amount of exposure. If the number of trials differs by observation, it belongs in each binomial probability.

Binomial model: If observation \(i\) has \(m_i\) independent opportunities, each with the same success probability \(p\), then \(N_i\sim\operatorname{Binomial}(m_i,p)\). For independent observations, the likelihood is \(L(p)=\prod_i {m_i\choose n_i}p^{n_i}(1-p)^{m_i-n_i}\), and its maximum likelihood estimate is \(\widehat p=\sum_i n_i/\sum_i m_i\), provided there is at least one opportunity.

The formula resembles the Poisson rate estimate but answers a different question: it is a proportion of successes per opportunity, not events per unit of time. A duration can be substituted for \(m_i\) only if the model actually specifies discrete, countable opportunities and \(m_i\) is their number.

Worked Example: Unequal Numbers of Binomial Opportunities

In a hypothetical inspection program, three machines are checked over different numbers of independent operating cycles. The first has \(3\) cycles and \(1\) detected defect, the second has \(4\) cycles and \(0\) defects, and the third has \(5\) cycles and \(4\) defects. Assume each cycle independently produces a detected defect with a common probability \(p\). Estimate \(p\) by maximum likelihood.

Method. Treat each cycle as a Bernoulli opportunity and use the binomial likelihood for each machine. The total number of defects is \(1+0+4=5\), and the total number of cycles is \(3+4+5=12\). The estimate is

\widehat p=\frac{5}{12}=0.4167

to four decimal places. As a check, the fitted expected defect counts for the three machines are \(3(5/12)=1.25\), \(4(5/12)=1.6667\), and \(5(5/12)=2.0833\). They sum to \(5\), the total observed count, as expected at the binomial maximum likelihood estimate. Common mistake avoided: using \(5\) divided by the number of machines would not estimate a success probability per cycle; the denominator must be total opportunities.

A related model for overdispersed counts is the negative binomial distribution. One useful exposure-scaled specification gives each observation a shape proportional to its exposure. Let \(\kappa>0\) be the shape per exposure unit, \(\lambda>0\) the mean event rate per unit, and set \(r_i=\kappa e_i\). Use the parameterization in which a negative binomial count with shape \(r_i\) and success probability \(p\) has mean \(r_i(1-p)/p\). Setting \(p=\kappa/(\kappa+\lambda)\) makes the mean \(\lambda e_i\). The variance is \(\lambda e_i+(\lambda e_i)^2/(\kappa e_i)\), which exceeds the mean for finite positive \(\kappa\).

P(N_i=n_i) =\frac{\Gamma(n_i+\kappa e_i)}{\Gamma(\kappa e_i)n_i!} \left(\frac{\kappa}{\kappa+\lambda}\right)^{\kappa e_i} \left(\frac{\lambda}{\kappa+\lambda}\right)^{n_i}.

This is a particular exposure-scaled negative binomial model: its shape increases with exposure, and it uses a common rate and dispersion parameter per unit. It is not interchangeable with every possible negative binomial regression specification. If \(\kappa\) is fixed, the terms involving \(\lambda\) in the joint log-likelihood are \((\sum_i n_i)\log\lambda-(\sum_i n_i+\kappa\sum_i e_i)\log(\kappa+\lambda)\), up to constants. Differentiation again gives \(\widehat\lambda=\sum_i n_i/\sum_i e_i\). Estimating \(\kappa\) as well generally requires solving an additional likelihood equation; the simple rate formula alone does not estimate dispersion.

Worked Example: Fitting an Exposure-Scaled Negative Binomial Rate

Suppose three independent hypothetical policy records have exposures \(0.5,1,\) and \(1.5\) policy-years, with counts \(0,2,\) and \(3\). Assume the exposure-scaled negative binomial model just defined, with known \(\kappa=2\) per policy-year and unknown rate \(\lambda\). Estimate \(\lambda\), then calculate the probability of observing exactly two claims in the one-year record at the fitted parameters.

Method. For fixed \(\kappa\), maximize the exposure-scaled negative binomial likelihood for the common rate. Total exposure is \(0.5+1+1.5=3\) policy-years, and total count is \(0+2+3=5\). Thus

\widehat\lambda=\frac{5}{3}=1.6667 \quad\text{claims per policy-year}.

For the one-year record, the shape is \(r=\kappa e=2(1)=2\), and the fitted probability parameter is \(p=\kappa/(\kappa+\widehat\lambda)=2/(2+5/3)=6/11\). Hence \(1-p=5/11\). The probability of two claims is

P(N=2) ={2+2-1\choose 2}\left(\frac{6}{11}\right)^2 \left(\frac{5}{11}\right)^2 =3\left(\frac{36}{121}\right)\left(\frac{25}{121}\right) =\frac{2700}{14641} \approx 0.1844.

The mean for that one-year observation is \(r(1-p)/p=2(5/6)=5/3\), agreeing with the fitted rate times one year. The likelihood calculation uses \(\kappa\) as fixed; it does not claim that these data estimate the dispersion. Common mistake avoided: using the Poisson probability for exactly two claims would discard the stated negative binomial dispersion assumption.

Model Checks Before Maximizing

Likelihood calculations are only as sound as their exposure definitions and assumptions. Before differentiating, confirm that counts and exposures refer to the same observational units and that every observation uses the parameterization intended by the model. In a Poisson rate model, an exposure of \(0.25\) means one quarter of the stated exposure unit, not one full observation with a smaller weight applied afterward.

Check the units. A rate estimated per policy-year must use exposures in policy-years. Converting all exposure values to months changes the numerical rate by a factor of twelve, though not the fitted expected counts.
Check independence and common rate. The product likelihood assumes independent observations conditional on the model and a common rate. If rates differ systematically among groups, a single common-rate estimate may not represent the intended model.
Check the count mechanism. Poisson counts assume the stated Poisson model; binomial counts require known numbers of opportunities and a common success probability here. Use a negative binomial likelihood only when its dispersion structure is part of the assumed model.
Do not confuse count with rate. The pooled count is \(\sum_i n_i\), while the pooled rate divides that count by \(\sum_i e_i\). They have different units and answer different questions.

The central practical point is that unequal exposure belongs inside each observation’s expected count or trial count. For a Poisson rate, this produces an estimate based on all events over all exposure. For a binomial model, the denominator is the number of opportunities. For an exposure-scaled negative binomial model, the same pooled-rate estimate follows when its dispersion parameter is fixed, but the probability model and resulting predictive probabilities remain different from Poisson.

Check Your Understanding

For each problem, use the model and exposure definition stated in the question.

Three independent records follow a common Poisson rate per policy-year. Their exposures are \(0.25,0.75,\) and \(1.5\) years, and their counts are \(1,0,\) and \(3\). Find the maximum likelihood rate. (Answer: \(1.6\) claims per policy-year.)
Three batches have \(3,4,\) and \(5\) independent test opportunities, with observed success counts \(1,0,\) and \(4\). Under a common binomial success probability, find its maximum likelihood estimate. (Answer: \(5/12=0.4167\), to four decimal places.)
Independent Poisson observations have exposures \(1,2,\) and \(3\) policy-years, with counts \(2,1,\) and \(6\). Find the common rate estimate and the fitted mean count for the third observation. (Answer: \(\widehat\lambda=1.5\) per policy-year; fitted mean \(4.5\).)
Use the exposure-scaled negative binomial model with known \(\kappa=1.5\) per exposure unit. Two records have exposures \(0.5\) and \(2\), and counts \(1\) and \(5\). Find the maximum likelihood rate. (Answer: \(6/2.5=2.4\) events per exposure unit.)
Independent Poisson observations have exposures \(1,0.5,\) and \(1.5\) years and counts \(2,1,\) and \(3\). Estimate the common rate, then find the fitted probability that a new record with exposure \(0.75\) years has zero events. (Answer: \(\widehat\lambda=2\) per year; \(P(N=0)=e^{-1.5}\approx0.2231\).)

What You'll Learn