Poisson And Gamma Exam Cases With Common Traps

Choosing a Model for Counts and Positive Amounts

A claim count and a claim amount answer different questions. A policy can have zero, one, or several claims during an observation period, so its claim count is discrete and can be zero. A claim severity, by contrast, is an amount conditional on a claim having occurred; it is positive and continuous. These differences determine which response distribution is appropriate and how exposure enters the model.

For frequency, a Poisson model can represent counts over an exposure period. For severity, a Gamma model can represent skewed positive amounts. In a common actuarial setup, the models are fitted separately: one to counts and one to observed positive claim amounts. Their predictions can then be combined to estimate expected aggregate cost, provided the quantities and assumptions are aligned.

Definition: In a Poisson regression, the response $N_i$ is a nonnegative count and, conditional on predictors, has a Poisson distribution with mean $\mu_i$. In a Gamma regression, the response $Y_i$ is a strictly positive continuous amount and, conditional on predictors, has a Gamma distribution with mean $\mu_i$. A zero count is valid for the Poisson response; a zero amount is not a valid observation from the ordinary Gamma response model.

The examples use log links because they give multiplicative effects and keep fitted means positive. For a count model, exposure is the amount of time or risk observed, expressed in a stated unit. For a severity model, the response is the amount per claim, not the number of claims or the total amount accumulated over a policy’s exposure.

Poisson Counts: Rates, Exposure, and Zeros

Suppose $N_i$ is the number of events observed for unit $i$, and $e_i>0$ is its exposure measured in a consistent unit, such as policy-years. A Poisson regression with a log link and an exposure offset is

N_i\mid \mathbf{x}_i \sim \operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\log(e_i)+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}.

The offset $\log(e_i)$ has coefficient fixed at 1; it is not another coefficient to estimate. If $\lambda_i=\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta})$, then $\lambda_i$ is the modeled event rate per exposure unit and $\mu_i=e_i\lambda_i$ is the expected count over the actual exposure. A fitted Poisson mean can be non-integer: it is an expected count, not a claim that a fractional event was observed.

Why the offset works: If events occur at a constant rate $\lambda_i$ per exposure unit, then doubling the exposure doubles the expected count. Thus $\mu_i=e_i\lambda_i$. Taking logarithms gives $\log(\mu_i)=\log(e_i)+\log(\lambda_i)$, so the log exposure enters with a fixed coefficient of 1.

The Poisson model assigns probability to zero as well as positive counts: $P(N_i=0\mid\mathbf{x}_i)=\exp(-\mu_i)$. Consequently, policy records with no claims are part of a frequency data set. Omitting them and fitting only to policies with claims changes the question: that restricted data no longer represent the count distribution for all exposed policies.

Worked Example: Converting an Exposure-Adjusted Count Prediction

A hypothetical portfolio uses a Poisson log-link model. Its baseline claim rate is $0.20$ claims per policy-year, and a specified risk class has a rate ratio of $1.5$. A policy in that class is observed for $2.5$ policy-years. Find its expected count and the probability of at least one claim. Assume the Poisson model is appropriate, the rate is constant over the period, and the risk class remains unchanged. Interest and mortality assumptions do not apply.

Method. Apply the rate ratio to the baseline rate, multiply by exposure to obtain the Poisson mean, and then use the Poisson probability of zero. The model for the count is $N\sim\operatorname{Poisson}(\mu)$.

The class rate is

\lambda=0.20(1.5)=0.30 \quad\text{claims per policy-year}.

The expected count over $2.5$ policy-years is

\mu=e\lambda=2.5(0.30)=0.75.

Therefore $N\sim\operatorname{Poisson}(0.75)$, and

P(N\geq 1)=1-P(N=0)=1-\exp(-0.75)=0.5276.

The result checks against the setup: the expected count is below one, and the chance of at least one claim is just over one-half. Common mistake avoided: treating $0.75$ as a probability. It is the expected count; the probability is calculated from the Poisson distribution.

Gamma Amounts: Positive Responses and Multiplicative Effects

For positive claim amounts, a Gamma regression with a log link specifies a positive conditional mean:

\log(\mu_i)=\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}, \qquad \mu_i=E[Y_i\mid\mathbf{x}_i]=\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}).

One common parameterization is $Y_i\mid\mathbf{x}_i\sim\operatorname{Gamma}(\alpha,\theta_i)$, where $\alpha$ is the shape and $\theta_i$ is the scale. Under this parameterization, $E[Y_i\mid\mathbf{x}_i]=\alpha\theta_i=\mu_i$ and $\operatorname{Var}(Y_i\mid\mathbf{x}_i)=\alpha\theta_i^2=\mu_i^2/\alpha$. Equivalently, the variance is $\phi\mu_i^2$ with $\phi=1/\alpha$. The Gamma response is strictly positive; it is not the usual model for claim counts or zero-inflated amounts.

With a log link, a coefficient $\beta_j$ represents a change on the log-mean scale. Increasing predictor $x_j$ by one unit multiplies the fitted mean by $\exp(\beta_j)$, holding other predictors fixed. The coefficient itself is not the dollar change in the mean.

Worked Example: Predicting a Positive Claim Amount

A hypothetical Gamma model for individual claim severity uses a log link. The baseline mean severity is $ \$600$. For a specified territory, the fitted coefficient is $\beta_1=\log(1.5)$, with the territory indicator equal to 1. Assume the conditional Gamma shape is $\alpha=3$ in both territories. Find the territory’s mean, Gamma scale, and variance. The response is claim amount conditional on a claim; interest and mortality assumptions do not apply.

Method. Exponentiate the coefficient to obtain its multiplicative effect on the mean. Then use the stated shape to recover the Gamma scale and variance.

The territory coefficient is $\beta_1=\log(1.5)\approx0.4055$, but the multiplier is

\exp(\beta_1)=\exp(\log(1.5))=1.5.

Thus the conditional mean severity is

\mu=600(1.5)=\$900.

For shape $\alpha=3$, the scale is

\theta=\frac{\mu}{\alpha}=\frac{900}{3}=\$300.

The conditional variance is

\operatorname{Var}(Y\mid\text{territory}) =\alpha\theta^2 =3(300)^2 =270{,}000\text{ dollars}^2.

As a check, $\alpha\theta=3(300)=\$900$, matching the predicted mean. Common mistake avoided: interpreting $0.4055$ as a dollar increase or as the multiplier. It is the log-scale coefficient; the multiplier is $1.5$.

Combining Frequency and Severity

A frequency model and a severity model can be combined to estimate expected aggregate claim cost. Let $N$ be the number of claims over a specified exposure, and let $X_1,\ldots,X_N$ be the corresponding claim amounts. If the claim amounts have common conditional mean $m$, then, conditional on the risk characteristics,

E\left[\sum_{j=1}^{N}X_j\right] =E[N]\,m.

This follows by conditioning on $N$: if $N=n$, the expected sum is $nm$. Averaging over $N$ gives $E[N]m$. Independence between frequency and severity, conditional on the modeled risk characteristics, is a standard sufficient assumption for this calculation. The product is an expected cost, not a prediction that every policy will incur that amount.

Worked Example: Expected Cost With a Chance of No Claims

A hypothetical policy has $1.5$ policy-years of exposure. Its Poisson frequency rate is $0.08$ claims per policy-year before a risk factor of 2. The Gamma severity model has baseline mean $ \$1{,}600$ and a multiplicative risk factor of $1.25$. Assume conditional independence of frequency and severity, a constant frequency rate during exposure, and Gamma severity with shape $\alpha=4$. Claims are paid in full, with no expenses or limits. Find the probability of no claims and expected aggregate claim cost. Interest and mortality assumptions do not apply.

Method. First calculate the exposure-adjusted Poisson mean. Separately calculate the Gamma conditional mean. Multiply these means for expected aggregate cost.

The adjusted frequency rate is

\lambda=0.08(2)=0.16 \quad\text{claims per policy-year}.

For $1.5$ policy-years, the count mean is

\mu_N=1.5(0.16)=0.24, \qquad N\sim\operatorname{Poisson}(0.24).

The probability of no claims is

P(N=0)=\exp(-0.24)=0.7866.

The severity mean is $m=1{,}600(1.25)=\$2{,}000$. With Gamma shape 4, its scale is $\theta=2{,}000/4=\$500$, and its variance is $4(500)^2=1{,}000{,}000$ dollars squared. Expected aggregate cost is

E\left[\sum_{j=1}^{N}X_j\right] =0.24(2{,}000)=\$480.00.

The probability of no claims is high, while the expected cost is positive because the remaining outcomes can include one or more claims. Common mistake avoided: treating the Gamma severity model as though it included policies with no claims. The zero-claim probability comes from the Poisson frequency model; the Gamma model describes amounts only when claims occur.

Common Traps in Mixed Count and Amount Cases

In a mixed case, identify the response, its unit, and its observation set before using a fitted model. A count model uses one response per exposure unit and includes zero counts. A severity model uses positive amounts for observed claims. If the question asks for total expected cost, ensure the frequency prediction and severity prediction refer to the same risk characteristics and time period.

Wrong exposure unit: If exposure is supplied in months but the modeled rate is per year, convert months to years before applying the offset. Six months is $0.5$ years, not 6 years.
Wrong use of an offset: In the Poisson log-mean equation, $\log(e_i)$ is added with coefficient fixed at 1. The offset is not a fitted risk factor.
Misreading coefficients: For a log link, $\exp(\beta_j)$ is the multiplicative effect for a one-unit predictor increase. It is not generally equal to $\beta_j$.
Putting zeros in a Gamma model: A zero claim count belongs in the frequency data. An ordinary Gamma severity response cannot take the value zero.
Confusing rate and expected count: A rate is per exposure unit; the expected count over a particular exposure is rate times exposure.

For example, if a frequency rate is $0.12$ per policy-year and the policy is observed for six months, the expected count is $0.12(0.5)=0.06$, not $0.12(6)=0.72$. The latter calculation uses months as though they were years and overstates the mean by a factor of 12.

Key takeaway: Use Poisson regression for counts over exposure and Gamma regression for strictly positive amounts. Apply exposure to the frequency mean, interpret log-link coefficients through exponentiation, and combine fitted frequency and severity means only when their units and conditioning assumptions match. These distinctions are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each problem, identify the model quantity being calculated before applying the arithmetic.

A Poisson model gives a rate of $0.12$ claims per policy-year for a risk class with rate ratio $1.5$. A policy is observed for 9 months. Find its expected count and probability of zero claims. (Answer: expected count $0.135$; probability of zero claims $0.8737$.)
A Gamma log-link severity model has baseline mean $ \$800$, a group coefficient $\log(1.2)$, and shape $\alpha=5$. Find the group mean, Gamma scale, and variance. (Answer: mean $ \$960$; scale $ \$192$; variance $184{,}320$ dollars squared.)
A hypothetical portfolio has 10 policy-years of exposure and 3 claims with amounts $ \$900$, $ \$1{,}200$, and $ \$900$. Using the empirical frequency rate and mean severity, calculate expected claim cost per policy-year. State which observations enter the frequency and severity calculations. (Answer: frequency rate $0.3$ per policy-year; mean severity $ \$1{,}000$; expected cost $ \$300$ per policy-year. All 10 policy-years, including zero-claim exposures, enter frequency; the 3 positive claim amounts enter severity.)
A policy has exposure of 2 policy-years, a baseline Poisson rate of $0.06$ per policy-year, and frequency multiplier $2.5$. Its baseline Gamma severity mean is $ \$1{,}200$, with severity multiplier $0.8$. Find the expected count, probability of zero claims, and expected aggregate cost. (Answer: expected count $0.3$; probability of zero claims $0.7408$; expected aggregate cost $ \$288.00$.)
A Poisson log-link model has baseline rate $0.15$ per exposure-year and a predictor coefficient $-0.2231$, approximately $\log(0.8)$. For a unit with predictor value 1 and exposure of 4 years, find the rate, expected count, and probability of no events. (Answer: rate $0.12$ per year; expected count $0.48$; probability of no events $0.6188$.)

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