Selecting A Claim Frequency Model From A Scenario

Start With the Claim-Generating Mechanism

A claim-count model should be chosen from the way claims can occur, not simply from the fact that a count is being modeled. A contract covering a fixed number of independent items may lead to a binomial count. Events occurring at a stable rate over exposure may lead to a Poisson count. A negative binomial model can represent a count of failures before a fixed number of successes, or extra variation among risks with different underlying rates. The right choice depends on the assumptions the scenario supports.

The first checks are practical: identify the observational unit and exposure period, determine whether the number of opportunities is fixed, and establish whether the count can exceed that number. Then check the model’s assumptions before calculating the requested probability. These distinctions are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: A binomial count $N\sim\operatorname{Bin}(n,q)$ is the number of successes in $n$ independent trials, each with the same success probability $q$. Its support is $0,1,\ldots,n$. A Poisson count $N\sim\operatorname{Poisson}(\lambda)$ has parameter $\lambda>0$, which is its mean, and support $0,1,2,\ldots$. A negative binomial count $N$ may be defined as the number of failures before the $r$th success in independent trials with success probability $p$; its support is $0,1,2,\ldots$, and $$ P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k,\qquad k=0,1,2,\ldots. $$

A Practical Model-Selection Sequence

Use the mechanism to narrow the choices. A fixed number of opportunities with independent, equally likely claim outcomes points to a binomial model. A count over time or exposure with a stable event rate points to a Poisson model. A failure-count process with a specified target number of successes points directly to the negative binomial model. If risks have different underlying claim rates, a negative binomial model may also arise after those rates are averaged out.

Define the count.
State what one event is, which risks or opportunities are included, and the period of observation. A count per policy-year is not automatically a count per policy.

Check whether the maximum is fixed.
If there are exactly $n$ independent opportunities and each can produce at most one claim, the count cannot exceed $n$. This is consistent with a binomial support.

Check the event mechanism.
A stable rate over exposure supports a Poisson count. A number of failures before a fixed number of successes supports the negative binomial failure-count model.

Check assumptions and parameters.
Look for independence, equal probabilities, rate stability, or risk heterogeneity as appropriate. Convert the stated period and exposure to the units required by the model.

Calculate only after selecting the model.
Use the model’s support and probability formula to evaluate the requested event. A formula cannot repair an incorrect choice of mechanism.

Fixed Opportunities: The Binomial Model

Suppose a portfolio contains $n$ specified items, each of which either has a claim during the period or does not. The binomial model is appropriate if each item has the same claim probability $q$, the outcomes are independent, and no item can generate more than one counted claim during that period. If the scenario allows multiple claims per item, this binomial setup does not apply to the total claim count without a different definition or model.

For $N\sim\operatorname{Bin}(n,q)$, the probability of exactly $k$ claims is $P(N=k)=\binom{n}{k}q^k(1-q)^{n-k}$, for $k=0,\ldots,n$. The factor $\binom{n}{k}$ counts which $k$ opportunities have claims; independence gives the probability $q^k(1-q)^{n-k}$ for each such arrangement.

Worked Example: Claims From a Fixed Set of Devices

A hypothetical warranty portfolio covers 8 identical devices for one year. Each device has probability $0.10$ of producing at least one covered claim during the year, independently of the others, and each device can contribute at most one count. Find the probability of exactly 2 claims.

Method and model. There are $n=8$ fixed opportunities, each has the same probability $q=0.10$, and outcomes are independent. Thus $N\sim\operatorname{Bin}(8,0.10)$, with support $0,\ldots,8$.

Substitute $k=2$: $$ P(N=2)=\binom{8}{2}(0.10)^2(0.90)^6. $$

The combination count is $\binom{8}{2}=8\cdot7/(2\cdot1)=28$. Also, $(0.10)^2=0.01$ and $(0.90)^6=0.531441$. Therefore $$ P(N=2)=28(0.01)(0.531441)=0.14880348. $$

Rounded to four decimal places, the probability is 0.1488. The result is between 0 and 1 and is compatible with the stated support. This method avoids treating the 8 devices as an unbounded event process: the fixed maximum and one-claim-per-device condition matter.

Stable Event Rate: The Poisson Model

For a Poisson count, the parameter is the expected count over the stated exposure. If a process has a constant rate $\lambda_0$ per unit of exposure and is observed for $t$ units, the Poisson parameter is $\lambda=\lambda_0t$. The model assumes a stable rate and independent occurrence in disjoint exposure intervals; it does not impose a fixed upper bound on the count. If the rate changes materially over time, a single constant-rate Poisson model may not be appropriate.

For $N\sim\operatorname{Poisson}(\lambda)$, $P(N=k)=e^{-\lambda}\lambda^k/k!$, for nonnegative integers $k$. The exposure multiplication is essential: a rate per year must be multiplied by years, not by a number of months treated as years.

Worked Example: Converting an Annual Rate to a Two-Year Count

A hypothetical class of policies experiences claims at a constant rate of $0.75$ claims per policy-year. For one policy observed for 2 years, assume the Poisson process conditions hold. Find the probability of at least 2 claims.

Method and model. The exposure is 2 policy-years, so the Poisson parameter is $\lambda=(0.75\text{ claims per policy-year})(2\text{ policy-years})=1.5$ claims. Treating claim counts as dimensionless in the probability calculation, $N\sim\operatorname{Poisson}(1.5)$.

“At least 2” is the complement of 0 or 1. Thus $$ P(N\geq2)=1-P(N=0)-P(N=1) =1-e^{-1.5}-1.5e^{-1.5} =1-2.5e^{-1.5}. $$

Using $e^{-1.5}\approx0.22313016$, the probability is $1-(2.5)(0.22313016)=1-0.55782540=0.44217460$. Rounded to four decimal places, the answer is 0.4422. It lies between 0 and 1. The complement calculation avoids summing an unbounded set of probabilities, and the exposure conversion avoids using $0.75$ as the two-year parameter.

Failure Counts and Heterogeneity: The Negative Binomial Model

In the failure-count definition, $N$ records how many failures occur before the $r$th success. This model is suitable when the scenario genuinely describes repeated independent trials with a fixed success probability and asks for failures up to a target success. It is not automatically the right model merely because a claim count has variance greater than its mean.

Another useful mechanism is unobserved heterogeneity: risks may have different underlying claim rates even if each risk’s count is Poisson conditional on its own rate. Mixing over those rates can produce a negative binomial marginal count. The following derivation makes the parameter conversion explicit.

Derivation: Suppose, conditional on a risk’s annual rate $\Lambda=\lambda$, its count over $t$ years is Poisson with mean $t\lambda$. Across risks, let $\Lambda$ have a gamma distribution with shape $\alpha>0$ and rate parameter $\beta>0$, with density $$ f_\Lambda(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0. $$ Here $\lambda$ is measured in claims per year, so $\beta$ has units of years when claim counts are treated as dimensionless. Integrating the conditional Poisson probability over rates gives $$ P(N=k)=\int_0^\infty \frac{e^{-t\lambda}(t\lambda)^k}{k!} \frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda =\frac{\Gamma(k+\alpha)}{\Gamma(\alpha)k!} \left(\frac{\beta}{\beta+t}\right)^\alpha \left(\frac{t}{\beta+t}\right)^k. $$ The last equality follows by collecting $\lambda^{k+\alpha-1}e^{-(\beta+t)\lambda}$ in the integrand and using $\int_0^\infty x^{s-1}e^{-cx}\,dx=\Gamma(s)/c^s$, for $s,c>0$. This is the negative binomial probability formula with $r=\alpha$ and $p=\beta/(\beta+t)$. The mixture has an unbounded count support, consistent with its Poisson components.

Worked Example: Claim Counts With Different Latent Rates

In a hypothetical portfolio, conditional on a risk’s annual claim rate $\Lambda$, its one-year count $N$ is Poisson with mean $\Lambda$. Across risks, assume $\Lambda$ has a gamma distribution with shape $\alpha=2$ and rate parameter $\beta=1$ year. Find the probability that a randomly selected risk has at least 2 claims in one year.

Method and model. The period is $t=1$ year. The gamma exponent is $-\beta\lambda$, so $\beta$ must have units reciprocal to $\lambda$; since $\lambda$ is claims per year, $\beta=1$ year is dimensionally consistent. By the mixture derivation, the marginal count is negative binomial with $r=\alpha=2$ and $p=\beta/(\beta+t)=(1\text{ year})/(1\text{ year}+1\text{ year})=1/2$.

For this negative binomial parameterization, the probability of zero claims is $$ P(N=0)=\binom{1}{0}(1/2)^2(1/2)^0=1/4. $$ The probability of exactly one claim is $$ P(N=1)=\binom{2}{1}(1/2)^2(1/2)^1 =2(1/4)(1/2)=1/4. $$ Therefore $$ P(N\geq2)=1-P(N=0)-P(N=1)=1-\frac14-\frac14=\frac12. $$

The answer is 0.5000, to four decimal places. It is valid for the unbounded support. This method accounts for variation in rates across risks; treating every risk as if it had one common fixed rate would discard the stated heterogeneity.

Common Selection Traps

A model can produce a plausible number while still being inconsistent with the scenario. Check the mechanism and support before trusting a calculation.

Wrong: selecting binomial just because the portfolio has $n$ policies. A policy can have multiple claims, so the total count may exceed $n$. Correction: use a binomial count only if each of the $n$ specified independent opportunities can contribute at most one counted success.
Wrong: using the annual rate as the count parameter over a longer period. For a constant rate of $0.75$ per year over 2 years, using $\lambda=0.75$ ignores half the exposure. Correction: use $\lambda=0.75(2)=1.5$.
Wrong: selecting negative binomial solely because a count is overdispersed. Overdispersion alone does not identify the mechanism or parameterization. Correction: check for a failure-count process or a specified source of heterogeneity, and state how $r,p$ are defined.
Wrong: overlooking support. A binomial model cannot assign positive probability to $n+1$ claims from $n$ one-claim opportunities. Correction: compare the requested event with the model’s support before applying a formula.
Wrong: mixing gamma rate and scale conventions. Under the density containing $e^{-\beta\lambda}$, $\beta$ is a rate parameter with units reciprocal to $\lambda$. Correction: state the density and units, rather than importing a scale parameter into the rate formula.

Key takeaway: Identify the count mechanism, verify its assumptions and support, and convert rates using the stated exposure. Only then select a probability formula. A fixed number of independent one-claim opportunities, a stable event rate, and a failure-count or heterogeneous-rate mechanism lead to different models.

Check Your Understanding

For each scenario, identify the model and parameters before calculating. Give probabilities to four decimal places.

A hypothetical set of 12 independent components each has probability $0.05$ of at least one failure claim in a year, and each component can contribute at most one count. Find the probability of no claims. (Answer: $0.95^{12}=0.5404$, binomial with $n=12,\ q=0.05$)
A hypothetical risk has a constant Poisson claim rate of $0.4$ claims per year and is observed for 3 years. Find the probability of exactly one claim. (Answer: $1.2e^{-1.2}=0.3614$, Poisson with $\lambda=1.2$)
Six independent opportunities each produce a claim with probability $0.20$, at most one claim per opportunity. Find the probability of at least one claim. (Answer: $1-(0.80)^6=0.7379$, binomial with $n=6,\ q=0.20$)
A process counts failures before the third success in independent trials, each with success probability $p=0.60$. Find the probability of exactly two failures. (Answer: $\binom{4}{2}(0.60)^3(0.40)^2=0.2074$, negative binomial with $r=3,\ p=0.60$)
Conditional on a risk’s annual claim rate $\Lambda$, its one-year count is Poisson with mean $\Lambda$. Across risks, $\Lambda$ is gamma with shape $\alpha=3$ and rate parameter $\beta=2$ years. Identify the marginal model and find the probability of zero claims in one year. (Answer: negative binomial with $r=3,\ p=2/(2+1)=2/3$, so $P(N=0)=(2/3)^3=0.2963$)

What You'll Learn