Bayesian Poisson-Gamma Experience Rating

Why Use a Poisson-Gamma Update?

An experience rating needs to balance a risk’s own claim history against information available before that history is observed. A short history can be especially misleading: a low count may reflect a genuinely low frequency, or simply random variation. The Bayesian Poisson-Gamma model makes this balance explicit by starting with a prior distribution for the risk’s claim frequency and updating it with the risk’s observed claims and exposure.

The result is a posterior distribution for the risk’s frequency. Its mean is a natural estimate for the expected claim count per unit of exposure. This tutorial focuses on the update and the resulting frequency premium; it does not estimate claim severity or a dollar loss cost.

Definition: Let $\lambda$ be a risk’s claim frequency per unit of exposure. Assume the prior distribution is Gamma with shape $\alpha>0$ and rate $\beta>0$, written $\lambda\sim\operatorname{Gamma}(\alpha,\beta)$. Its density is $$ \pi(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0, $$ and its mean is $\alpha/\beta$. Here $\beta$ is a rate parameter, not a scale parameter. If exposure is measured in years, $\lambda$ is claims per year and $\beta$ is measured in years.

Model, Assumptions, and Posterior Update

For observed exposure $E$, let $N$ be the risk’s claim count. The model assumes that, conditional on its frequency $\lambda$, the count is Poisson with mean $E\lambda$:

N\mid\lambda,E\sim\operatorname{Poisson}(E\lambda).

The exposure must be expressed in the same unit used to define $\lambda$. The model treats exposure as known, assumes a constant frequency over the observed period, and assumes claims arise independently conditional on $\lambda$. The Gamma prior represents uncertainty about the risk’s underlying frequency across risks; it does not remove the conditional Poisson assumption.

Under these assumptions, the posterior distribution is also Gamma. The observed count adds to the prior shape, while the observed exposure adds to the prior rate:

Theorem: If $\lambda\sim\operatorname{Gamma}(\alpha,\beta)$ and $N\mid\lambda,E\sim\operatorname{Poisson}(E\lambda)$, then $$ \lambda\mid N=n,E\sim\operatorname{Gamma}(\alpha+n,\beta+E), \qquad E[\lambda\mid N=n,E]=\frac{\alpha+n}{\beta+E}. $$ The posterior mean is a frequency estimate per unit of exposure, not a dollar premium.

Why the Update Has This Form

The Poisson likelihood for a count $n$, conditional on $\lambda$, is proportional to $\lambda^n e^{-E\lambda}$; the factor $E^n/n!$ does not depend on $\lambda$. Multiplying this likelihood by the Gamma prior density gives the part of the posterior density that depends on $\lambda$:

\begin{aligned} \pi(\lambda\mid n,E) &\propto \lambda^n e^{-E\lambda} \lambda^{\alpha-1}e^{-\beta\lambda}\\ &=\lambda^{\alpha+n-1}e^{-(\beta+E)\lambda}. \end{aligned}

This has the kernel of a Gamma density with shape $\alpha+n$ and rate $\beta+E$. The normalizing constant is therefore the one that makes it a $\operatorname{Gamma}(\alpha+n,\beta+E)$ density. For a Gamma distribution with shape $a$ and rate $b$, integrating its density gives mean $a/b$: the numerator integral is proportional to $\Gamma(a+1)/b^{a+1}$, and $\Gamma(a+1)=a\Gamma(a)$. Thus the posterior mean is $(\alpha+n)/(\beta+E)$, as stated.

Counts and exposure play distinct roles. Each observed claim increments the shape by one; each unit of exposure increments the rate by one. With observations from several periods that share the same frequency, use the total count and total exposure: $n=\sum_j n_j$ and $E=\sum_j E_j$. This follows because the independent Poisson likelihoods multiply to a likelihood proportional to $\lambda^{\sum_j n_j}e^{-\lambda\sum_j E_j}$.

From Posterior Frequency to a Frequency Premium

Suppose the next rating period has exposure $E_{\text{new}}$, and the risk’s frequency remains constant. Conditional on $\lambda$, the expected claim count for that period is $E_{\text{new}}\lambda$. Averaging over the posterior gives

E[N_{\text{new}}\mid n,E,E_{\text{new}}] =E_{\text{new}}E[\lambda\mid n,E] =E_{\text{new}}\frac{\alpha+n}{\beta+E}.

Consequently, $(\alpha+n)/(\beta+E)$ is the expected claim count per unit of future exposure: the frequency premium. Multiplying it by future exposure gives an expected claim count, not a dollar amount. A dollar pure premium would also require an appropriate expected claim severity, which is outside this frequency-only update.

Worked Example: Updating a Frequency Rate After Five Claims

Method and trap to avoid. Use the Gamma-Poisson conjugate update. The common mistake is to divide the posterior shape by the posterior rate and present a rounded decimal as an exact equality.

A hypothetical contractor program assigns a prior frequency $\lambda\sim\operatorname{Gamma}(3,1)$, where $\lambda$ is measured in claims per contractor-year. The prior mean is $3/1=3$ claims per contractor-year. The risk has 5 claims over 5 contractor-years. Calculate its posterior mean frequency and the expected claim count for the next 2 contractor-years.

The observed count is $n=5$, and exposure is $E=5$ contractor-years. The posterior parameters are

\alpha+n=3+5=8,\qquad \beta+E=1+5=6.

Therefore, the posterior distribution is $\operatorname{Gamma}(8,6)$, and its mean is

E[\lambda\mid N=5,E=5]=\frac{8}{6}=\frac{4}{3} \approx 1.3333\text{ claims per contractor-year}.

The fraction $4/3$ is exact; the decimal is rounded to four places. For 2 future contractor-years, the expected count is $2(4/3)=8/3\approx2.6667$ claims, with the decimal rounded to four places. As a check, the observed frequency was $5/5=1$ claim per contractor-year, and the posterior mean of $1.3333$ lies between that experience and the prior mean of 3.

Worked Example: Combining Unequal Period Exposures

Method and trap to avoid. Add counts and exposures separately, then apply the update once. Do not average the period frequencies without accounting for their unequal exposures.

A hypothetical vehicle portfolio has prior frequency $\lambda\sim\operatorname{Gamma}(2,8)$, in claims per vehicle-year. It records 1 claim in a period with 4 vehicle-years of exposure and 3 claims in a later period with 8 vehicle-years. Find the posterior mean frequency and the expected count for 20 future vehicle-years.

The total observed count and exposure are

n=1+3=4,\qquad E=4+8=12\text{ vehicle-years}.

The posterior parameters are $\alpha+n=2+4=6$ and $\beta+E=8+12=20$. Thus the posterior is $\operatorname{Gamma}(6,20)$, with mean

E[\lambda\mid n=4,E=12]=\frac{6}{20}=0.30 \text{ claims per vehicle-year}.

For 20 future vehicle-years, the expected claim count is $20(0.30)=6$. The observed aggregate frequency is $4/12=1/3\approx0.3333$, while the prior mean is $2/8=0.25$; the posterior mean of 0.30 is between them. As a further check, the two periods could be updated sequentially: the first gives shape $2+1=3$, rate $8+4=12$; the second then gives shape $3+3=6$, rate $12+8=20$, the same posterior.

Worked Example: Zero Claims Still Update the Frequency

Method and trap to avoid. Add exposure to the posterior rate even when the observed count is zero. Zero claims do not mean zero frequency when exposure is finite.

A hypothetical small equipment account has prior frequency $\lambda\sim\operatorname{Gamma}(4,12)$, in claims per account-year. It has no claims during 6 account-years. Find the posterior mean frequency and the expected count for 3 future account-years.

Here $n=0$ and $E=6$. The posterior shape remains $4+0=4$, while the posterior rate is $12+6=18$. Thus the posterior is $\operatorname{Gamma}(4,18)$, and

E[\lambda\mid n=0,E=6]=\frac{4}{18}=\frac{2}{9} \approx0.2222\text{ claims per account-year}.

The decimal is rounded to four places. The expected count in 3 future account-years is $3(2/9)=2/3\approx0.6667$, also rounded to four places. The result is below the prior mean $4/12=1/3$, as zero-claim experience should be, but remains positive because the prior shape is positive. The update is consistent with the model: the zero-count likelihood is $e^{-6\lambda}$, which contributes exposure to the rate but no $\lambda$ factor to the shape.

Common Mistakes and Limits of the Model

Using a Gamma scale instead of a rate. In this tutorial’s parameterization, the mean is shape divided by rate. A source using shape-scale notation requires a conversion before applying these formulas.
Adding exposure to the shape. The count increments the shape; exposure increments the rate. The posterior is $\operatorname{Gamma}(\alpha+n,\beta+E)$, not $\operatorname{Gamma}(\alpha+E,\beta+n)$.
Mixing exposure units. If the frequency is defined per 100 vehicle-years, exposure must be counted in 100-vehicle-year units. For example, 250 vehicle-years is 2.5 such units, not 250.
Calling a frequency rate a dollar premium. The posterior mean is claims per exposure unit. It does not include severity, expenses, or other components of a dollar rate.
Ignoring model mismatch. The update may be unsuitable if frequency changes materially over time, exposure is not proportional to expected claims, or conditional counts are not plausibly Poisson (for example, because of unmodeled dependence or clustering).

The method matters because it makes the role of prior information and experience transparent: the prior contributes shape $\alpha$ and rate $\beta$, while the data contribute count $n$ and exposure $E$. It is a compact, interpretable way to update a frequency estimate when the stated assumptions are reasonable, and is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Key takeaway: With a Gamma shape-rate prior and a Poisson count over known exposure, update to $\operatorname{Gamma}(\alpha+n,\beta+E)$. The posterior mean frequency is $(\alpha+n)/(\beta+E)$; keep the count, exposure, and frequency units consistent.

Check Your Understanding

For each problem, use the Gamma shape-rate parameterization. Treat all stated exposures as known and frequencies as constant over the relevant periods.

A risk has prior $\operatorname{Gamma}(2,6)$, with frequency measured per policy-year, and observes 3 claims over 4 policy-years. What is the posterior mean frequency? (Answer: $5/10=0.5$ claims per policy-year.)
A risk has prior $\operatorname{Gamma}(3,9)$ per account-year and records zero claims in 3 account-years. What is its posterior mean frequency? (Answer: $3/12=0.25$ claims per account-year.)
A risk has prior $\operatorname{Gamma}(1.5,4)$ per unit of exposure and observes 5 claims over 8 units. Calculate the posterior mean frequency, to four decimal places. (Answer: $6.5/12\approx0.5417$ claims per exposure unit.)
A fleet has prior $\operatorname{Gamma}(2,5)$, with frequency measured per 100 vehicle-years. It observes 6 claims over 400 vehicle-years. Find the posterior mean frequency and expected count over the next 250 vehicle-years. (Answer: posterior mean $8/9\approx0.8889$ claims per 100 vehicle-years; expected count $2.5(8/9)=20/9\approx2.2222$.)
A risk has prior $\operatorname{Gamma}(2,4)$ per policy-year. It observes 2 claims in 3 policy-years, then 1 claim in 2 policy-years, with the same underlying frequency throughout. Find the final posterior mean and verify the totals used. (Answer: total count $3$, total exposure $5$, posterior $\operatorname{Gamma}(5,9)$, mean $5/9\approx0.5556$ claims per policy-year.)

What You'll Learn