Choosing a Frequency Model for Covered Losses

Why the Frequency Model Depends on the Count

A frequency model describes the random number of events in a specified period or portfolio. Before choosing one, define what counts as an event. A reported loss, an incurred claim, and a claim producing a positive payment are not necessarily the same event. For a contract with a deductible, for example, an incurred loss may result in no payment. Whether that loss belongs in the frequency count depends on the question being modeled.

The exposure structure is equally important. A fixed set of policies may allow at most one claim per policy during a period; another portfolio may have no fixed maximum and may be modeled by a rate per unit of exposure. A negative binomial model can be useful when counts are more variable than a simple Poisson model predicts, but extra variation is a diagnostic, not by itself proof that the negative binomial assumptions are appropriate.

Definition: A frequency variable $N$ is the number of events meeting a stated definition during a stated period and for a stated exposure set. A frequency model is appropriate only when its assumptions reasonably describe those event opportunities, their probabilities or rates, and their dependence. Always specify whether zero-payment claims are included in the event definition.

Three Frequency Models and Their Assumptions

Use the binomial model when there are $n$ identifiable opportunities, each opportunity can produce at most one counted event, and the event indicators are independent with the same probability $p$. For example, the opportunities could be policies in a fixed policy-period, if each policy can have at most one event counted in that period.

N\sim\operatorname{Bin}(n,p),\qquad E[N]=np,\qquad \operatorname{Var}(N)=np(1-p).

The Poisson model describes a count with parameter $\lambda$, where $\lambda$ is both the mean and the variance. It is often suitable when events occur independently at a stable rate over exposure, there is no relevant fixed maximum count, and event clustering or unmodeled rate variation is not material. If the rate is $\lambda_0$ per unit of exposure and total exposure is $e$, then the model parameter is $\lambda=\lambda_0e$.

N\sim\operatorname{Poisson}(\lambda),\qquad P(N=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad E[N]=\operatorname{Var}(N)=\lambda.

Use the negative binomial model when the count has meaningful overdispersion relative to a Poisson model and a negative binomial mechanism is plausible. Here use the convention that $N$ counts failures before the $r$th success, with success probability $p$. The parameter $r$ may be a positive non-integer in the usual actuarial model.

P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k,\quad k=0,1,\ldots, \qquad E[N]=\frac{r(1-p)}{p},\quad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}.

A useful mechanism for this model is a Poisson count whose rate varies across otherwise similar exposure groups. In that setting, a group with a high underlying rate tends to produce more events, and variation in rates adds count variability. The negative binomial is not automatically suitable whenever a data set has variance above its mean: dependence, changing exposure, recording differences, or an unsuitable event definition may also explain the pattern.

Justifying the Variance Diagnostics

The moment formulas help explain why the models can be distinguished, while also showing why moments alone cannot settle the choice. For a binomial count, write $N=I_1+\cdots+I_n$, where each $I_j$ is 1 if opportunity $j$ produces an event and 0 otherwise. Independence gives zero covariance between distinct indicators, so the variance is the sum of their variances.

E[I_j]=p,\qquad \operatorname{Var}(I_j)=p(1-p), \qquad \operatorname{Var}(N)=\sum_{j=1}^{n}\operatorname{Var}(I_j)=np(1-p).

For a Poisson count, mean and variance are both $\lambda$. For the stated negative binomial parameterization, the variance exceeds the mean whenever $0<p<1$, since $\operatorname{Var}(N)=E[N]/p$. That pattern can be consistent with heterogeneous risk, but an observed sample variance is subject to sampling variation and does not identify its cause.

The rate-mixture explanation can be justified directly. Suppose that, conditional on a random rate $\Lambda$, the count is Poisson with mean $\Lambda$. The law of total expectation gives $E[N]=E[\Lambda]$. The law of total variance gives

\operatorname{Var}(N) =E[\operatorname{Var}(N\mid\Lambda)] +\operatorname{Var}(E[N\mid\Lambda]) =E[\Lambda]+\operatorname{Var}(\Lambda).

Thus variation in the underlying rate increases the count variance above its mean when $\operatorname{Var}(\Lambda)>0$. A gamma distribution for $\Lambda$ yields a negative binomial count under standard parameterizations. This derivation explains a possible reason for overdispersion; it does not establish that every overdispersed portfolio has gamma-distributed rates.

A Practical Model-Selection Sequence

Define the event and period.
State whether an event means a reported claim, an incurred loss, or a positive payment, and specify the exposure period.

Check for a fixed number of opportunities.
If there are $n$ independent opportunities with a common event probability and at most one event per opportunity, consider the binomial model.

Check whether a stable event rate is credible.
If a count arises over exposure with no relevant fixed maximum and approximately independent events at a stable rate, consider the Poisson model.

Investigate excess variability.
Compare the mean and variance, then examine exposure differences, heterogeneity, dependence, and the event definition. Consider a negative binomial model only when its assumptions are defensible.

Recheck the count after coverage terms.
A zero-payment claim remains in the count if the modeled event is an incurred or reported claim. If the modeled event is a positive payment, it is excluded.

These are model checks, not a mechanical ranking. For instance, the Poisson distribution can approximate a binomial count when there are many opportunities and small event probabilities, but that approximation does not erase a real fixed maximum. If the portfolio structure is known, it should inform the choice even when two models give similar numerical results.

Worked Example: A Fixed Number of Policy Opportunities

A hypothetical group has 40 policies observed for one quarter. Each policy can have at most one event meeting the stated claim definition, and the policies are assumed independent with the same event probability, $p=0.05$. Select a frequency model, calculate its mean and variance, and find the probabilities of zero and one event. State the method and the mistake it avoids.

Method. There are 40 fixed opportunities, each with at most one event, a common probability, and independence. Therefore $N\sim\operatorname{Bin}(40,0.05)$.

The mean is $np=40(0.05)=2$. The variance is $np(1-p)=40(0.05)(0.95)=1.9$.

P(N=0)=\binom{40}{0}(0.05)^0(0.95)^{40} =0.95^{40}\approx0.1285.

For exactly one event,

P(N=1)=\binom{40}{1}(0.05)(0.95)^{39} =40(0.05)(0.95)^{39}\approx0.2706.

The probabilities sum to $0.3991$, so the probability of two or more events is $1-0.1285-0.2706=0.6009$. The check is that the mean of 2 is consistent with substantial probability above one event. Common mistake avoided: choosing Poisson solely because the expected count is small. The binomial model reflects the known 40-opportunity structure; a Poisson approximation would be a separate approximation, not the exact structural model.

Worked Example: Scaling a Poisson Rate by Exposure

A hypothetical coverage program has 1,200 exposure units. Assume events occur independently at a constant rate of $0.004$ per exposure unit, with no relevant fixed maximum or material rate heterogeneity. Find the count model, its mean and variance, and the probabilities of zero and exactly three events. State the method and the mistake it avoids.

Method. Multiply the rate per unit by total exposure: $\lambda=1{,}200(0.004)=4.8$. Under the stated rate assumptions, $N\sim\operatorname{Poisson}(4.8)$. Its mean and variance are both $4.8$.

P(N=0)=e^{-4.8}\approx0.0082.

For exactly three events,

P(N=3)=e^{-4.8}\frac{4.8^3}{3!} =e^{-4.8}\frac{110.592}{6} \approx0.1517.

The probabilities are stated to four decimal places. Common mistake avoided: using $0.004$ as the Poisson parameter for the whole program. It is a rate per exposure unit; the total-exposure parameter is $4.8$.

Worked Example: Diagnosing Overdispersion and Fitting a Negative Binomial

In a hypothetical data review, a consistently defined quarterly claim count has empirical mean 6 and empirical variance 15 across comparable groups. Suppose a negative binomial model is being considered, with $N$ counting failures before $r$ successes. Use the two moments to estimate $r$ and $p$, then check the fitted moments and calculate $P(N=0)$. State the method and the mistake it avoids.

Method. Equate the negative binomial mean and variance to the empirical moments. Dividing the variance by the mean gives $1/p$, so $p=6/15=0.4$. Then $6=r(1-p)/p$, giving $r=6(0.4)/(0.6)=4$.

The fitted mean is $4(0.6)/0.4=6$, and the fitted variance is $4(0.6)/(0.4)^2=15$, matching the two input moments. The zero-count probability is

$$ P(N=0)=p^r=0.4^4=0.0256. $$

The moment fit is internally consistent, but it does not prove that the negative binomial is the correct model. A modeler should investigate whether heterogeneous rates or another plausible mechanism explains the extra variation. Common mistake avoided: treating variance greater than mean as conclusive proof of a negative binomial distribution.

Worked Example: A Deductible and the Definition of a Counted Event

A hypothetical portfolio has 100 independent policies in one period. Each policy has probability $0.10$ of an incurred claim. Conditional on an incurred claim, the loss is \$500 with probability $0.30$, or \$2,000 with probability $0.70$. The policy has a \$500 ordinary deductible, so the payment is the amount by which the loss exceeds \$500. Find the mean and variance of (a) the number of incurred claims and (b) the number of claims producing positive payments. State the method and the mistake it avoids.

Method. For incurred claims, each policy is one independent opportunity with probability $0.10$. The incurred-claim count is $\operatorname{Bin}(100,0.10)$, with mean $100(0.10)=10$ and variance $100(0.10)(0.90)=9$.

A \$500 loss produces payment $\max(500-500,0)=\$0$; a \$2,000 loss produces payment $\max(2{,}000-500,0)=\$1{,}500$. Thus the probability that any one policy produces a positive payment is $0.10(0.70)=0.07$. The positive-payment count is $\operatorname{Bin}(100,0.07)$, with mean $100(0.07)=7$ and variance $100(0.07)(0.93)=6.51$.

The two counts answer different questions: an incurred claim at exactly the deductible remains in the first count but not the second. Common mistake avoided: assuming that every incurred claim generates a positive payment, or silently switching between incurred-claim frequency and positive-payment frequency.

Common Traps When Choosing a Model

Ignoring the exposure structure. A fixed number of independent opportunities with at most one event each points to a binomial model, even if a Poisson approximation might be numerically close.
Using the wrong Poisson parameter. A rate per unit of exposure must be multiplied by total exposure before it is used as $\lambda$.
Assuming overdispersion identifies its cause. A variance above the mean can be consistent with a negative binomial model, but changing rates, dependence, or data issues may also account for it.
Leaving zero-payment claims undefined. State whether the count includes incurred losses that produce no payment. The deductible alone does not determine the frequency variable.
Changing the event definition mid-calculation. A model fitted to reported claims should not be used without adjustment as if it described positive payments, unless those counts are in fact the same under the contract and data definition.

Frequency selection matters because it describes how many claims enter the later calculation chain. An incorrect count definition can change the fitted frequency model even when the severity or payment calculation is correct. A clear event definition, followed by structural checks and dispersion diagnostics, is more reliable than choosing a familiar distribution from the mean alone. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the appropriate model or model parameter and complete the requested calculation.

A hypothetical portfolio has 60 independent policy-period opportunities, with at most one counted claim per opportunity and claim probability $0.02$. Find the model, mean, and variance. (Answer: $\operatorname{Bin}(60,0.02)$; mean $1.2$; variance $1.176$.)
A hypothetical book has 750 exposure units and a constant Poisson claim rate of $0.003$ per unit. Find $\lambda$ and $P(N=0)$, giving the probability to four decimals. (Answer: $\lambda=2.25$; $P(N=0)=e^{-2.25}\approx0.1054$.)
A negative binomial count uses the failures-before-$r$-successes convention. Its mean is 8 and variance is 20. Find the moment estimates of $r$ and $p$. (Answer: $p=8/20=0.4$; $r=8(0.4)/(0.6)=16/3\approx5.333$.)
There are 200 independent policy opportunities. Each has a $0.04$ probability of an incurred claim; conditional on a claim, the loss exceeds the deductible with probability $0.75$. Find the mean and variance of the positive-payment count. (Answer: positive-payment probability $0.04(0.75)=0.03$; binomial model; mean $6$; variance $200(0.03)(0.97)=5.82$.)
A portfolio has exactly 50 independent policy-period opportunities, at most one event per opportunity, and a common event probability of $0.08$. An observed sample variance is slightly above its sample mean. Which model is structurally indicated, and does that sample variance alone justify replacing it with a negative binomial model? (Answer: the binomial model, $\operatorname{Bin}(50,0.08)$, is structurally indicated; no, a sample variance slightly above the mean alone does not establish a negative binomial mechanism.)

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