Deriving A Bayesian Credibility Premium

Why the Posterior Mean Is a Credibility Premium

A frequency-based experience rating must balance two sources of information: the risk’s observed claims and a prior expectation for risks of its type. A risk with substantial exposure should generally be rated more heavily on its own experience than a risk with only a small amount of exposure. The Bayesian Poisson-Gamma model makes that balance explicit: its posterior mean is a weighted average of the observed claim frequency and the prior mean.

This tutorial derives that result and interprets its weight. The premium developed here is a claim-frequency rate, not a claim-cost or dollar premium. To turn it into a loss-cost premium would require additional information about claim severity.

Definition: Let \(N\) be the number of claims observed over exposure \(E\), measured in a fixed exposure unit. Conditional on an unknown risk-specific frequency \(\lambda\), assume \(N\mid\lambda\sim\operatorname{Poisson}(E\lambda)\). Assign \(\lambda\) a Gamma prior with shape \(\alpha>0\) and rate \(\beta>0\), written \(\lambda\sim\operatorname{Gamma}(\alpha,\beta)\), with density \(f(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\), for \(\lambda>0\). Under this shape-rate convention, \(E[\lambda]=\alpha/\beta\).

The parameter \(\lambda\) is a frequency per exposure unit. For example, if exposure is measured in thousands of vehicle-years, \(\lambda=1.4\) means an expected 1.4 claims per thousand vehicle-years. The prior mean \(\alpha/\beta\) is also a frequency per exposure unit. The prior rate parameter \(\beta\) has units of exposure; this matters when interpreting the resulting credibility weight.

The model assumes a constant \(\lambda\) for the risk during the period being modeled, and independent claim counts conditional on \(\lambda\). Conditional on that frequency, expected claims over exposure \(E\) are \(E\lambda\). The model does not account by itself for changes in coverage, claim definitions, trend, or risk characteristics over time; data need to be made comparable before they are pooled.

Deriving the Posterior Mean

The likelihood contribution from \(N\) claims over exposure \(E\) is proportional, as a function of \(\lambda\), to \(\lambda^N e^{-E\lambda}\). Multiplying it by the Gamma prior density gives the posterior kernel. The resulting distribution is Gamma, with its shape increased by the observed claim count and its rate increased by the observed exposure.

Write the likelihood.
For \(N\mid\lambda\sim\operatorname{Poisson}(E\lambda)\), the likelihood is proportional to \(\lambda^N e^{-E\lambda}\). Factors not involving \(\lambda\) can be omitted when finding the posterior distribution.

Multiply by the prior.
The prior kernel is \(\lambda^{\alpha-1}e^{-\beta\lambda}\). Multiplication gives the posterior kernel \(\lambda^{\alpha+N-1}e^{-(\beta+E)\lambda}\).

Recognize the posterior distribution.
This is the kernel of a Gamma distribution with shape \(\alpha+N\) and rate \(\beta+E\). Therefore, \(\lambda\mid N\sim\operatorname{Gamma}(\alpha+N,\beta+E)\).

Take the posterior mean.
For a Gamma distribution with shape \(a\) and rate \(b\), the mean is \(a/b\). Hence the posterior mean frequency is \((\alpha+N)/(\beta+E)\).

To express this mean as a credibility-weighted average, let \(m=\alpha/\beta\) be the prior mean frequency and let \(\bar{X}=N/E\) be the observed frequency when \(E>0\). Define \(Z=E/(\beta+E)\). Since \(1-Z=\beta/(\beta+E)\), the posterior mean becomes:

E[\lambda\mid N] =\frac{\alpha+N}{\beta+E} =\frac{E}{\beta+E}\frac{N}{E} +\frac{\beta}{\beta+E}\frac{\alpha}{\beta} =Z\bar{X}+(1-Z)m.

This equality is more than a convenient rearrangement. The posterior numerator combines prior shape \(\alpha\) with observed claims \(N\); the denominator combines prior exposure-equivalent amount \(\beta\) with actual exposure \(E\). Dividing both contributions by the same total \(\beta+E\) gives their relative weights. Thus, when \(E\) is large compared with \(\beta\), \(Z\) approaches 1 and experience receives more weight. When \(E\) is small compared with \(\beta\), \(Z\) is closer to 0 and the prior mean receives more weight.

Theorem: Under the stated Poisson likelihood and Gamma shape-rate prior, the posterior mean frequency is \(Z(N/E)+(1-Z)(\alpha/\beta)\), where \(Z=E/(\beta+E)\), provided \(E>0\). The parameter \(\beta\) acts as the prior’s exposure-equivalent amount. If \(E=0\), no observed frequency \(N/E\) is defined; the posterior remains the prior and its mean is \(\alpha/\beta\).

The posterior mean is the frequency estimate that minimizes expected squared error under the posterior distribution. For a prospective period with exposure \(E_f\), the corresponding expected claim count is \(E_f E[\lambda\mid N]\), because conditional expected claims are \(E_f\lambda\), and averaging over the posterior replaces \(\lambda\) by its posterior mean. This is the frequency-based premium component considered here.

Applying the Credibility Formula

Worked Example: A Single Period of Claim Experience

A hypothetical fleet has 9 claims over 3 exposure units. Its prior frequency is modeled as \(\lambda\sim\operatorname{Gamma}(4,2)\), using the shape-rate convention. Calculate the posterior mean frequency, express it as a credibility blend, and find expected claims over a future exposure of 1.5 units. Name the method and identify a common mistake to avoid.

Method. Use Poisson-Gamma conjugacy and rewrite the posterior mean as a credibility-weighted average. The prior mean is \(m=\alpha/\beta=4/2=2\) claims per exposure unit. The observed frequency is \(\bar{X}=N/E=9/3=3\) claims per exposure unit.

The posterior parameters are shape \(\alpha+N=4+9=13\) and rate \(\beta+E=2+3=5\). Therefore, the posterior mean frequency is \(13/5=2.6\) claims per exposure unit.

The credibility weight is \(Z=E/(\beta+E)=3/(2+3)=3/5=0.6\). The complementary weight is \(1-Z=0.4\). The weighted average is \(0.6(3)+0.4(2)=1.8+0.8=2.6\), agreeing with the posterior calculation.

For future exposure \(E_f=1.5\), expected claims are \(1.5(2.6)=3.9\). This is an expected count, so it need not be an integer. Check: the posterior mean lies between the observed frequency 3 and prior mean 2, and multiplying by future exposure gives 3.9 expected claims.

Common mistake avoided: Treating the claim count \(N=9\) as the observed frequency and calculating \(0.6(9)+0.4(2)=6.2\). This is wrong because \(N\) is a count over 3 exposure units, not a rate per exposure unit. The blend must use \(N/E=3\), giving 2.6 claims per unit.

With several periods, the same derivation applies when they share the same underlying frequency and are conditionally independent given \(\lambda\). The likelihood depends on the total number of claims and total exposure. Consequently, period-by-period data can be reduced to \(N=\sum_j N_j\) and \(E=\sum_j E_j\), provided the exposure units and claim experience are comparable.

Worked Example: Combining Periods With Unequal Exposure

A hypothetical equipment portfolio has claim counts of 2, 3, and 1 in three successive periods, with exposures of 1, 2, and 0.5 units, respectively. Use the prior \(\lambda\sim\operatorname{Gamma}(3,1.5)\). Find the posterior mean frequency and expected claims over a future exposure of 4 units. State why averaging the three period frequencies equally would be inappropriate.

Method. Aggregate claims and exposure, then apply the Poisson-Gamma posterior formula. Total claims are \(N=2+3+1=6\). Total exposure is \(E=1+2+0.5=3.5\) units. The prior mean is \(m=3/1.5=2\) claims per unit.

The posterior shape is \(\alpha+N=3+6=9\), and the posterior rate is \(\beta+E=1.5+3.5=5\). The posterior mean frequency is \(9/5=1.8\) claims per unit.

The observed frequency is \(N/E=6/3.5=12/7\), or approximately 1.7143 claims per unit. The credibility weight is \(Z=3.5/(1.5+3.5)=3.5/5=0.7\). The blend is \(0.7(12/7)+0.3(2)=1.2+0.6=1.8\), which confirms the posterior result.

For future exposure of 4 units, expected claims are \(4(1.8)=7.2\). Check: the posterior frequency is between the observed frequency, approximately 1.7143, and the prior mean of 2; the future expected count is 7.2.

Common mistake avoided: Giving each period’s frequency equal weight would ignore unequal exposure. The periods have frequencies \(2/1=2\), \(3/2=1.5\), and \(1/0.5=2\), whose unweighted average is \(1.8333\). The correct aggregate frequency is total claims divided by total exposure, \(6/3.5\approx1.7143\), because the period with more exposure contributes more information under this model.

Units, Interpretation, and Limits

The credibility formula requires the prior rate \(\beta\) and actual exposure \(E\) to use the same exposure unit. The prior shape \(\alpha\) is dimensionless, while the prior mean \(\alpha/\beta\) has frequency units. If exposure is converted, the numerical values of both \(\beta\) and the frequency rate must be converted consistently.

Worked Example: Preserving the Exposure Unit

A hypothetical insurer measures exposure in thousands of vehicle-years. Its prior is \(\lambda\sim\operatorname{Gamma}(6,5)\), so the prior mean is 1.2 claims per thousand vehicle-years and the prior exposure-equivalent amount is 5 thousand vehicle-years. A risk produces 7 claims over 4 thousand vehicle-years. Find its posterior mean frequency and expected claims over a future 2.5 thousand vehicle-years. Show the credibility blend and the unit error that could produce a wrong result.

Method. Keep counts, rates, and exposure in thousands of vehicle-years. The observed frequency is \(7/4=1.75\) claims per thousand vehicle-years. The credibility weight is \(Z=4/(5+4)=4/9\), and the prior weight is \(1-Z=5/9\).

The posterior shape is \(6+7=13\), and the posterior rate is \(5+4=9\) thousand vehicle-years. Thus, the posterior mean is \(13/9=1.4444\) claims per thousand vehicle-years, rounded to four decimal places.

The blend gives the same result: \((4/9)(1.75)+(5/9)(1.2)=7/9+6/9=13/9\). For future exposure of 2.5 thousand vehicle-years, expected claims are \(2.5(13/9)=65/18=3.6111\), rounded to four decimal places. Check: the posterior mean lies between 1.2 and 1.75 claims per thousand vehicle-years, and the future count is the rate multiplied by 2.5 thousand vehicle-years.

Common mistake avoided: If the analyst converts the observed exposure to 4,000 vehicle-years but leaves \(\beta=5\) in thousand-vehicle-year units, the credibility calculation mixes units. The denominator \(5+4{,}000\) has no coherent exposure unit. Either retain exposure in thousands, as above, or convert the prior rate to 5,000 vehicle-years and express the frequency per vehicle-year consistently.

Several conditions limit the interpretation of the result. A shared \(\lambda\) across periods is not appropriate if underlying frequency changes materially and those changes have not been adjusted. Counts should represent the same claim definition, and exposure should be measured on a consistent basis. The credibility weight is determined by \(E\) and \(\beta\) in this model; it is not an independent adjustment selected after seeing the count. Nor does the frequency premium include severity, expenses, limits, or other components of a charged premium.

The central exam skill is to move reliably between the posterior-parameter calculation and the credibility interpretation. First identify the count and exposure in consistent units. Then calculate the posterior mean, or use the equivalent blend. These routes should agree exactly before any rounding. This derivation is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, use the Gamma shape-rate convention and keep exposure units consistent.

A risk has prior \(\operatorname{Gamma}(2,1)\) and observes 5 claims over 2 exposure units. Find the posterior mean frequency and credibility weight. (Answer: posterior mean \(7/3=2.3333\) claims per unit; \(Z=2/3\).)
A hypothetical class has prior mean frequency 1.5 claims per unit and prior exposure-equivalent amount 4 units. It observes 8 claims over 6 units. Find the prior shape, credibility weight, posterior mean frequency, and expected claims over 3 future units. (Answer: \(\alpha=6\); \(Z=0.6\); posterior mean \(2.1\) claims per unit; 6.3 expected claims.)
A risk has prior \(\operatorname{Gamma}(5,2)\). It reports 0 claims over 3 exposure units. Calculate the posterior mean and express it as a blend of the prior mean and observed frequency. (Answer: posterior mean \(1.0\) claim per unit; \(Z=0.6\); blend \(0.6(0)+0.4(2.5)=1.0\).)
A portfolio has counts 1 and 5 over exposures 0.5 and 2.5 units. Its prior is \(\operatorname{Gamma}(4,2)\). Find the aggregate observed frequency, posterior mean, and expected claims over 2 future units. (Answer: aggregate frequency \(6/3=2\) claims per unit; posterior mean \(10/5=2\) claims per unit; 4 expected claims.)
Exposure is measured in hundreds of machine-years. A prior mean of 0.8 claims per hundred machine-years has prior strength 5 hundred machine-years. A risk has 3 claims over 10 hundred machine-years. Find its credibility weight, posterior mean frequency, and expected claims over 4 hundred machine-years. (Answer: \(Z=2/3\); posterior mean \(1.6\) claims per hundred machine-years; 6.4 expected claims.)

What You'll Learn