Why Frequency Estimation Requires an Exposure Model
Claim counts do not have a useful frequency interpretation without knowing how much risk was observed. One claim from a half-year policy and one claim from a five-year policy are not equivalent evidence about an annual rate. Similarly, a binomial count must be interpreted relative to its number of possible claim opportunities. This tutorial develops moment and likelihood estimates for three standard count models, making each model’s exposure and parameterization explicit.
Suppose there are \(n\) independent observations. For observation \(j\), let \(X_j\) be the observed claim count, and let \(e_j\) be its known exposure measured in a specified unit, such as policy-years. For binomial data, write \(m_j\) for the known number of opportunities and \(X_j\) for the number of successes. The observations are assumed to follow the stated model independently; estimates are only as appropriate as those assumptions.
Poisson Frequency With Unequal Exposure
For a Poisson frequency model, assume independent counts with \(X_j\sim\operatorname{Poisson}(\lambda e_j)\), where \(e_j\) is exposure and \(\lambda\) is the expected number of claims per exposure unit. Thus \(E[X_j]=\lambda e_j\) and \(\operatorname{Var}(X_j)=\lambda e_j\). This model assumes a constant rate per exposure unit and does not account for extra variation between observations beyond Poisson variation.
The total count \(T=\sum_{j=1}^n X_j\) has expected value \(\lambda E\), where \(E=\sum_{j=1}^n e_j\) is total exposure. Equating observed and expected totals gives the moment estimate. The likelihood gives the same estimate.
To see why the likelihood result holds, the Poisson likelihood is proportional, as a function of \(\lambda\), to
For \(T>0\), the log-likelihood has derivative \(-E+T/\lambda\), which is zero at \(\lambda=T/E\); its second derivative is \(-T/\lambda^2<0\), so this is a maximum. If \(T=0\), the likelihood is maximized at the boundary estimate \(\widehat{\lambda}=0\).
Worked Example: A Poisson Rate With Different Policy Terms
A hypothetical portfolio records 0 claims over 0.5 policy-years, 1 claim over 1 policy-year, 2 claims over 1.5 policy-years, and 3 claims over 2 policy-years. Assume independent Poisson counts with a constant annual claim rate. Estimate the rate by moments and likelihood, then estimate the expected count for a new 2-year policy.
Method and setup. The model is \(X_j\sim\operatorname{Poisson}(\lambda e_j)\), with exposure measured in policy-years. Total claims are \(0+1+2+3=6\). Total exposure is \(0.5+1+1.5+2=5\) policy-years.
Estimate. Both methods give \(\widehat{\lambda}=6/5=1.2\) claims per policy-year. For a 2-year policy, the fitted mean count is \(1.2(2)=2.4\) claims.
Check and trap avoided. The fitted total expected count across the observed portfolio is \(1.2(5)=6\), matching the observed total. Dividing by four policies instead would give 1.5 claims per policy, which ignores that the policy terms differ.
Binomial Frequency With Different Numbers of Opportunities
Suppose \(X_j\) counts the number of claims among \(m_j\) specified opportunities, with independent outcomes and the same claim probability \(p\) for every opportunity. Then \(X_j\sim\operatorname{Binomial}(m_j,p)\), so \(E[X_j]=m_jp\). Here an opportunity could be a covered item or a defined exposure period in which at most one claim is possible. The binomial model is not appropriate if an opportunity can generate multiple claims or if the opportunity outcomes do not share the assumed probability.
Let \(M=\sum_jm_j\) be the total number of opportunities and \(T=\sum_jX_j\) the total number of claims. Equating observed and expected totals gives \(T=Mp\), so the moment estimate is \(T/M\). The likelihood is proportional to \(p^T(1-p)^{M-T}\), which is maximized at the same estimate when \(0<T<M\); the boundary estimates are 0 if \(T=0\), and 1 if \(T=M\).
Worked Example: Estimating a Claim Probability From Unequal Opportunity Counts
A hypothetical insurer observes three groups of covered devices. The groups have 4, 6, and 5 devices, and the respective numbers of devices with a claim are 1, 2, and 4. Assume each device independently has at most one claim during the observation period, with the same probability \(p\). Estimate \(p\) by moments and likelihood.
Method and setup. Use \(X_j\sim\operatorname{Binomial}(m_j,p)\). The total number of opportunities is \(M=4+6+5=15\), and total claims are \(T=1+2+4=7\).
Estimate. The moment estimate is \(T/M=7/15=0.4667\) to four decimal places. The likelihood is proportional to \(p^7(1-p)^8\); its log-likelihood derivative is \(7/p-8/(1-p)\). Setting this to zero gives \(7(1-p)=8p\), hence \(p=7/15=0.4667\).
Check and trap avoided. The estimate predicts \(15(7/15)=7\) claims across these opportunities, as observed. Averaging the three group-specific proportions, \(1/4\), \(2/6\), and \(4/5\), would give a different result because it assigns equal weight to groups of different sizes.
Negative Binomial Frequency: Mean and Size
A negative binomial model can represent count variation greater than the Poisson model allows. To avoid ambiguity between conventions, define \(X\) here as the number of failures before \(r\) successes, with success probability \(p\), where \(r>0\) need not be an integer. Its probability function, mean, and variance are
For estimation, it is often clearer to use the equivalent mean-size parameterization: \(\mu=E[X]\) and size \(r\). Then \(p=r/(r+\mu)\), and \(\operatorname{Var}(X)=\mu+\mu^2/r\). Smaller \(r\) means greater variance at a fixed mean. In the following equal-exposure model, independent counts have this same \(\mu\) and \(r\); if exposure changes the mean, set \(\mu_j=\lambda e_j\). The model must specify how the size behaves across exposures: assuming one common \(r\) is not the same as assuming size increases with exposure.
For equal exposures, let \(\bar X\) be the sample mean and let \(m_2=n^{-1}\sum_j(X_j-\bar X)^2\) be the empirical central second moment. Equating the model mean and variance to these empirical moments yields
This formula follows by rearranging \(m_2=\bar X+\bar X^2/r\). If \(m_2\leq\bar X\), it does not give a positive finite size estimate: the sample does not exhibit the overdispersion required by this negative binomial model-of-moments calculation. Do not force a negative \(r\) or interpret the formula as a valid fit.
For independent observations with common \(\mu\) and \(r\), MLE generally requires numerical optimization for \(r\). Once \(r\) is fixed, maximizing over \(\mu\) gives \(\widehat{\mu}=\bar X\); the remaining shape estimate solves a likelihood score equation. It need not equal the moment estimate, and an approximate numerical solution should be identified as such.
Worked Example: Negative Binomial Moments and Likelihood
A hypothetical unit of exposure has claim counts \(0,0,0,1,2,3\) in six independent observations. Assume the equal-exposure negative binomial model with common mean \(\mu\) and size \(r\). Estimate both parameters by moments, then obtain the likelihood estimates.
Moments method. The total is 6, so \(\bar X=6/6=1\). The sum of squared counts is \(0+0+0+1+4+9=14\). Thus the empirical second central moment, using divisor \(n\), is \(m_2=14/6-1^2=4/3\). Therefore \(m_2-\bar X=4/3-1=1/3\), and \(\widehat r_{\mathrm{MOM}}=1^2/(1/3)=3\). The moments estimates are \(\widehat\mu=1\), \(\widehat r=3\), and the corresponding probability is \(\widehat p=3/(3+1)=0.75\).
Likelihood method. In this equal-exposure model the maximum-likelihood estimate of \(\mu\) is the sample mean, 1. For the counts in this example, the score equation for \(r\), after substituting \(\mu=1\), is
This equation comes from differentiating the negative binomial log-likelihood with respect to \(r\); the three nonzero counts contribute the terms involving reciprocal sums, and the six observations contribute the logarithm term. Numerical maximization gives \(\widehat r_{\mathrm{MLE}}\approx1.700\). Hence \(\widehat\mu_{\mathrm{MLE}}=1\) and \(\widehat p_{\mathrm{MLE}}=1.700/(1.700+1)=0.6296\), with \(p\) rounded to four decimal places.
Check and trap avoided. Under the moment fit, the variance is \(1+1/3=4/3\), equal to the empirical central second moment. The likelihood fit need not reproduce that variance: MLE maximizes the probability of the observed sample rather than matching its first two moments. Also, using the unbiased sample variance with divisor \(n-1\) in the moment formula would change the estimate; here the method of moments explicitly uses the empirical moment with divisor \(n\).
Choosing and Checking an Estimator
Moment estimates are often direct and interpretable, but they may not exist in the parameter space or may be sensitive to the empirical moments. Likelihood estimates use the full specified probability model and obey its parameter restrictions, but can require numerical optimization. Neither method repairs a poorly chosen model.
| Model | Exposure information | Typical estimate | Important condition |
|---|---|---|---|
| Poisson | Exposure \(e_j\) for each count | Total claims divided by total exposure | Assumes a constant rate and Poisson variance |
| Binomial | Opportunity count \(m_j\) for each group | Total claims divided by total opportunities | Each opportunity has at most one success |
| Negative binomial | Specify how mean and size relate to exposure | Mean and size can be estimated by moments or likelihood | Size convention and overdispersion must be explicit |
A frequent error is to treat exposure as though it were merely a label. For Poisson counts, the mean is \(\lambda e_j\), not \(\lambda\) for every observation regardless of term. For binomial counts, the mean is \(m_jp\), not \(p\) claims per group. A second trap is to quote a negative binomial estimate without defining its parameterization: some sources count failures before successes, while others use mean and size directly. State the convention and translate parameters using \(p=r/(r+\mu)\) when using the convention defined here.
These estimates connect observed claim records to a fitted frequency model. In any application, verify the observation unit, exposure scale, support of the count, and independence assumption before interpreting the result. The estimation techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, state the model and parameterization before calculating.
- Independent Poisson counts are 2, 1, and 3 over exposures of 1, 2, and 3 policy-years. Find the MLE of the claim rate per policy-year. (Answer: 1.000 claim per policy-year)
- Three binomial groups have 5, 8, and 12 opportunities, with 2, 3, and 5 claims. Find the common-probability moment and likelihood estimate. (Answer: \(10/25=0.4000\))
- Equal-exposure negative binomial counts are \(0,1,1,2,4\). Using the empirical central second moment with divisor \(n\), find the moments estimates of \(\mu\) and \(r\). (Answer: \(\widehat\mu=1.6,\ \widehat r=10.6667\))
- For a negative binomial distribution defined as failures before \(r\) successes, let \(\mu=2\) and \(r=4\). Find \(p\) and the variance. (Answer: \(p=2/3,\ \operatorname{Var}(X)=3\))
- Two independent Poisson observations have counts 1 and 4 at exposures 0.5 and 1.5 policy-years. What is the fitted expected count over a further 2 policy-years? (Answer: \(\widehat\lambda=2.5\) per policy-year; fitted expected count \(=5\))