Frequency and Severity Models

Claim Count Models

Model claim frequency with exposure, overdispersion, heterogeneity, and excess zeros.

Claim Count Models

A count without an exposure period is rarely comparable. Let NiN_i be claims from exposure eie_i—for example, earned vehicle-years—during a stated period.

1. Begin with an observed rate

PortfolioVehicle-yearsClaimsRaw countClaims per vehicle-year
A10,000400lower0.040
B15,000525higher0.035

Portfolio B has more claims but lower observed frequency. Modelling counts without the exposure denominator reverses the comparison.

2. Poisson: the benchmark process

If NPoisson(λ)N\sim\operatorname{Poisson}(\lambda),

P(N=n)=eλλnn!,E[N]=Var(N)=λ.P(N=n)=e^{-\lambda}\frac{\lambda^n}{n!}, \qquad E[N]=\operatorname{Var}(N)=\lambda.

For risk ii, write λi=eiri\lambda_i=e_i r_i, where rir_i is expected frequency per unit exposure. A log-link regression becomes

logE[Ni]=logei+β0+xiβ.\log E[N_i]=\log e_i+\beta_0+x_i^\top\beta.

logei\log e_i is an offset: its coefficient is fixed at one. If a rating factor has coefficient 0.200.20, its multiplicative effect on expected frequency is e0.201.22e^{0.20}\approx1.22, holding other variables and exposure fixed.

Why Poisson appears

Poisson is plausible when many opportunities each have small event probability and events are approximately independent over a fixed exposure. These are process claims to investigate, not consequences of integer-valued data.

3. Binomial: fixed opportunities

If exactly mm policies can each produce one event with probability pp independently,

NBinomial(m,p),E[N]=mp,Var(N)=mp(1p).N\sim\operatorname{Binomial}(m,p), \qquad E[N]=mp, \qquad \operatorname{Var}(N)=mp(1-p).

Binomial is appropriate for “at least one claim” per policy. It is not appropriate when one policy can generate several claims unless the response is deliberately reduced to a binary indicator.

4. Negative binomial: extra-Poisson variation

A useful parameterisation has

E[Ni]=μi,Var(Ni)=μi+αμi2,α>0.E[N_i]=\mu_i, \qquad \operatorname{Var}(N_i)=\mu_i+\alpha\mu_i^2, \qquad \alpha>0.

It can arise from Poisson counts with a Gamma-distributed latent rate. Different policyholders then have different unobserved propensities, even after measured rating factors are included.

Diagnose, do not merely switch families

Overdispersion—sample or residual variance exceeding the Poisson mean—can arise from:

  • unobserved heterogeneity;
  • common weather or event shocks;
  • omitted seasonality or trend;
  • excess zeros;
  • exposure errors;
  • serial or spatial dependence.

Negative binomial handles one variance pattern but does not identify which mechanism caused it.

5. Zero-inflated and hurdle models

Two distinct questions can generate excess zeros:

  • Zero-inflated model: some observations are in a structural-zero state; the remaining state can also produce zeros.
  • Hurdle model: one process determines zero versus positive; a zero-truncated count model determines positive counts.

Use them only when the zero mechanism has operational meaning. A flexible model can fit zeros while giving misleading interventions.

6. Time and dependence

For monthly or regional counts, condition on known structure:

logE[Nit]=logeit+β0+xitβ+s(t)+bi,\log E[N_{it}] =\log e_{it}+\beta_0+x_{it}^{\top}\beta+s(t)+b_i,

where s(t)s(t) can represent seasonality/trend and bib_i a group effect. Event clustering may require random effects, event-level models, or correlated processes; multiplying independent Poisson probabilities is then invalid.

7. A current data-literacy example

The Association of British Insurers reported £11.9bn of motor claim payouts across 2.5m claims in 2025. Dividing gives £4,760 per reported claim, but that number should not be presented as market severity without checking:

  • whether payout and count scopes match;
  • claim versus coverage definitions;
  • whether figures are paid during year or ultimate accident-year cost;
  • mix of damage, theft, injury, and third-party claims;
  • the ABI's warning that improved 2025 coverage limits direct year-on-year comparison.

The example illustrates why a quotient is not automatically a statistical estimand.

8. Validation checklist

CheckWhat failure suggests
predicted vs observed counts by exposure bandoffset or non-linearity problem
Pearson/deviance residual patternmisspecified mean or variance
variance-to-mean by homogeneous segmentoverdispersion or omitted grouping
zero calibrationexcess-zero mechanism
rolling-time validationtrend/seasonality instability
event-level residual clusteringdependence

Practice

  1. Expected annual rate is 0.08 per policy. What is Poisson expected count for 2,500 policy-years?
  2. Under μ=2\mu=2 and α=0.5\alpha=0.5, find negative-binomial variance.
  3. A portfolio's exposure doubles while rate is unchanged. What happens to expected count and expected rate?
Answers
  1. 2,500×0.08=2002{,}500\times0.08=200.
  2. 2+0.5(22)=42+0.5(2^2)=4.
  3. Expected count doubles; expected count per exposure remains unchanged.

Source for the current example

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