Worked Gamma-Poisson Rate Updates

Updating a Claim-Rate Estimate

A claim count is informative only in relation to the exposure that produced it. Nine claims over 300 exposure units suggest a different rate from nine claims over 30 units. Bayesian credibility combines that experience with a prior distribution for the unknown risk rate. The calculation produces a full posterior distribution as well as a posterior mean rate.

This tutorial uses the Gamma distribution in the shape-rate parameterization. In particular, the second parameter is a rate, not a scale. That distinction affects both the posterior update and the units of the result.

Definition: Let $\lambda$ be a risk’s claim rate per exposure unit. Write $\lambda\sim\operatorname{Gamma}(\alpha,\beta)$ in the shape-rate convention, with density $$ f(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0. $$ Here $\alpha>0$ is the shape parameter, $\beta>0$ is the rate parameter, $E[\lambda]=\alpha/\beta$, and $\operatorname{Var}(\lambda)=\alpha/\beta^2$. If the observed claim count is $N=n$ over exposure $e$, the model is $N\mid\lambda\sim\operatorname{Poisson}(e\lambda)$.

The units must agree: if $\lambda$ is measured in claims per 1,000 machine-years, then $e$ must be measured in thousands of machine-years. Under the model, the prior and the observation concern the same constant underlying rate. Conditional on that rate, claim counts follow the Poisson model; the prior represents uncertainty about which rate applies. These examples involve no interest or mortality assumptions: they update claim frequency, not discounted amounts or survival probabilities.

The Posterior Update

For a count $n$ and exposure $e$, the Poisson likelihood as a function of $\lambda$ is proportional to $\lambda^n e^{-e\lambda}$. Multiplying by the Gamma prior density gives a posterior kernel proportional to $\lambda^{\alpha+n-1}e^{-(\beta+e)\lambda}$. Those are the powers and exponential term of a Gamma density.

\lambda\mid N=n\sim\operatorname{Gamma}(\alpha+n,\beta+e), \qquad E[\lambda\mid N=n]=\frac{\alpha+n}{\beta+e}.

The update adds the observed count to the prior shape and the observed exposure to the prior rate parameter. The posterior mean has the same units as $\lambda$, since its denominator is exposure units.

Derivation: Bayes’ rule says the posterior density is proportional to the likelihood times the prior density. The likelihood is $$ P(N=n\mid\lambda)=\frac{(e\lambda)^n e^{-e\lambda}}{n!}\ \propto\ \lambda^n e^{-e\lambda}. $$ Multiplying by the prior kernel $\lambda^{\alpha-1}e^{-\beta\lambda}$ yields $$ \lambda^n e^{-e\lambda}\lambda^{\alpha-1}e^{-\beta\lambda} =\lambda^{\alpha+n-1}e^{-(\beta+e)\lambda}. $$ This is the kernel of a shape-rate Gamma distribution with shape $\alpha+n$ and rate $\beta+e$. Its mean is therefore $(\alpha+n)/(\beta+e)$.

The same posterior mean can be expressed as a weighted average of the prior mean and the observed rate. When $e>0$,

\frac{\alpha+n}{\beta+e} =\frac{\beta}{\beta+e}\frac{\alpha}{\beta} +\frac{e}{\beta+e}\frac{n}{e}.

The weights sum to one. The prior mean $\alpha/\beta$ receives weight $\beta/(\beta+e)$, and the observed rate $n/e$ receives weight $e/(\beta+e)$. This is a useful check on an update, but the posterior distribution itself still matters: its parameters describe more than its mean.

Worked Gamma-Poisson Updates

Worked Example: Updating With a Moderate Claim Count

A hypothetical book of small commercial policies has a claim rate measured in claims per policy-year. The prior is $\lambda\sim\operatorname{Gamma}(3,120)$, and the risk produces 14 claims over 400 policy-years. Find the posterior distribution and its mean. Assume a constant rate during the observation period and a conditional Poisson count. No interest or mortality assumption is involved.

Method. Use the Gamma-Poisson conjugate update, with exposure expressed in policy-years. The prior mean is $$ \frac{\alpha}{\beta}=\frac{3}{120}=0.0250 $$ claims per policy-year. The observed rate is $14/400=0.0350$ claims per policy-year.

Posterior parameters. Add the count to the prior shape and exposure to the prior rate: $$ \alpha_{\text{post}}=3+14=17,\qquad \beta_{\text{post}}=120+400=520. $$ Thus $\lambda\mid N=14\sim\operatorname{Gamma}(17,520)$.

Posterior mean. $$ E[\lambda\mid N=14]=\frac{17}{520}=0.0326923\ldots $$ To four significant figures, the posterior mean is 0.03269 claims per policy-year.

Check and common mistake avoided. The weighted-average check gives $(120(0.0250)+400(0.0350))/520=(3+14)/520=17/520$. The posterior mean lies between the prior mean, 0.0250, and the observed rate, 0.0350. A common mistake is to add the exposure to the shape; the correct update adds the count to the shape and exposure to the rate parameter.

Worked Example: Converting Exposure Units Before Updating

A hypothetical equipment portfolio has a prior rate measured in claims per 1,000 machine-years, with $\lambda\sim\operatorname{Gamma}(4,160)$. It experiences three claims over 48,000 machine-years. Find the posterior distribution and mean. Assume a constant claim rate and a conditional Poisson count; no interest or mortality assumption is used.

Method. First express exposure in the units matching the rate. Since one unit is 1,000 machine-years, $$ e=\frac{48{,}000\text{ machine-years}}{1{,}000\text{ machine-years per unit}}=48\text{ units}. $$ The prior mean is $4/160=0.0250$ claims per 1,000 machine-years, and the observed rate is $3/48=0.0625$ claims per 1,000 machine-years.

Posterior parameters. $$ \alpha_{\text{post}}=4+3=7,\qquad \beta_{\text{post}}=160+48=208. $$ Therefore, $\lambda\mid N=3\sim\operatorname{Gamma}(7,208)$.

Posterior mean. $$ E[\lambda\mid N=3]=\frac{7}{208}=0.0336538\ldots $$ To four significant figures, the posterior mean is 0.03365 claims per 1,000 machine-years.

Check and common mistake avoided. The answer is between 0.0250 and 0.0625, as expected for a weighted average. If one incorrectly inserts $48{,}000$ into the update without converting units, the calculated mean is $7/(160+48{,}000)=0.0001453$ claims per 1,000 machine-years. That calculation mixes machine-years with units of 1,000 machine-years, so it is invalid.

Worked Example: Combining Counts From Two Periods

A hypothetical risk has prior rate $\lambda\sim\operatorname{Gamma}(6,300)$, measured in claims per exposure unit. In the first period it has seven claims over 100 exposure units; in the second period it has eight claims over 150 exposure units. Find its posterior distribution and mean. Assume the rate is unchanged across both periods and the counts are conditionally independent Poisson variables given $\lambda$. No interest or mortality assumption is involved.

Method. Multiply the two conditional likelihoods. Their count-dependent parts are proportional to $$ \lambda^7e^{-100\lambda}\lambda^8e^{-150\lambda} =\lambda^{15}e^{-250\lambda}. $$ Thus the data contribute 15 claims and $100+150=250$ exposure units. The prior mean is $6/300=0.0200$; the combined observed rate is $15/250=0.0600$.

Posterior parameters. $$ \alpha_{\text{post}}=6+(7+8)=21,\qquad \beta_{\text{post}}=300+(100+150)=550. $$ Hence $\lambda\mid\text{both periods}\sim\operatorname{Gamma}(21,550)$.

Posterior mean. $$ E[\lambda\mid\text{both periods}]=\frac{21}{550}=0.0381818\ldots $$ To four significant figures, the posterior mean is 0.03818 claims per exposure unit.

Check and common mistake avoided. The weighted-average form gives $(300(0.0200)+250(0.0600))/550=(6+15)/550=21/550$. Treating either period’s rate as the only observation would discard valid information. Counts and exposures add across periods only under the stated common-rate and conditional-independence assumptions.

Interpretation and Conditions

For a collection of conditionally independent Poisson counts $N_j$ with exposures $e_j$, all sharing one unknown rate $\lambda$, the joint likelihood is proportional to $\lambda^{\sum_j n_j}e^{-\lambda\sum_j e_j}$. Consequently, the posterior parameters are $\alpha+\sum_j n_j$ and $\beta+\sum_j e_j$. This aggregation is appropriate only when the rate is common across the observations, the exposure units match, and the conditional count model is Poisson.

A different rate by period, dependence in counts, or a different count-generating process would require a different model. Nor should the posterior mean be confused with the observed rate: the former incorporates both the data and the prior. If exposure is positive, the posterior mean is between those two rates, with the relative weights shown above.

Confirm the rate unit.
Express every exposure in the unit used to define $\lambda$.

Identify the data contribution.
For a Poisson count, add counts to the prior shape and exposures to the prior rate parameter.

Report the distribution and mean.
Give both Gamma posterior parameters and calculate shape divided by rate for the posterior mean.

Check the result.
Compare the posterior mean with the prior and observed rates, and use the weighted-average identity when exposure is positive.

The most persistent exam trap is a silent mismatch between a Gamma parameterization or an exposure unit and the formula being used. Under the shape-rate convention, the prior mean is $\alpha/\beta$, and exposure is added to $\beta$. Under the stated Poisson model, a count contributes to $\alpha$. Confirming these facts before substituting numbers prevents errors that may otherwise produce a plausible-looking but incorrectly scaled answer. This calculation practice is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, use the shape-rate Gamma convention and state the posterior mean in the specified rate units.

A hypothetical fleet has prior $\lambda\sim\operatorname{Gamma}(2,80)$, where $\lambda$ is claims per vehicle-year. It has six claims over 120 vehicle-years. Find the posterior distribution and mean. (Answer: $\operatorname{Gamma}(8,200)$; 0.04000 claims per vehicle-year)
A hypothetical property group has prior $\lambda\sim\operatorname{Gamma}(3,150)$, measured in claims per 100 building-years. Four claims occur over 5,000 building-years. Convert exposure before updating. (Answer: $\operatorname{Gamma}(7,200)$; 0.03500 claims per 100 building-years)
A hypothetical service plan has prior $\lambda\sim\operatorname{Gamma}(8,400)$, measured in claims per account-year. It produces ten claims over 100 account-years. Find both posterior parameters and the posterior mean. (Answer: $\operatorname{Gamma}(18,500)$; 0.03600 claims per account-year)
A hypothetical agricultural portfolio has prior $\lambda\sim\operatorname{Gamma}(5,250)$, measured in claims per exposure unit. In two conditionally independent periods with a common rate, it records five claims over 100 units and seven claims over 200 units. Find the combined posterior mean. (Answer: $\operatorname{Gamma}(17,550)$; 0.03091 claims per exposure unit)
A hypothetical specialty book has prior $\lambda\sim\operatorname{Gamma}(4,100)$, measured in claims per 1,000 insured-item-years. It has nine claims over 200,000 insured-item-years. Find the posterior distribution and mean, converting exposure to the stated unit. (Answer: $\operatorname{Gamma}(13,300)$; 0.04333 claims per 1,000 insured-item-years)

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