Poisson-Gamma Credibility as a Weighted Average

Why the Posterior Mean Is a Credibility-Weighted Rate

In a Poisson-Gamma model, the claim rate is not treated as known. Instead, it is represented by a random variable with a prior distribution, and observed claims provide evidence about its value. The posterior mean combines the prior information with the observed experience. Its weighted-average form makes the role of each source of information explicit.

The key is to keep the exposure unit consistent. If \(e\) is exposure and \(N\) is the observed claim count, then \(N/e\) is the observed rate only when \(e>0\). The prior contributes information too: under the Gamma shape-rate parameterization below, its mean is \(\alpha/\beta\), and \(\beta\) acts as the prior exposure strength in the weighted-average formula.

Model and notation: Let \(\theta\) be a claim rate per unit of exposure. Conditional on \(\theta\), suppose the claim count \(N\) over fixed exposure \(e\) satisfies \(N\mid\theta\sim\operatorname{Poisson}(e\theta)\). Before observing \(N\), let \(\theta\sim\operatorname{Gamma}(\alpha,\beta)\), using the shape-rate convention with density proportional to \(\theta^{\alpha-1}e^{-\beta\theta}\), where \(\alpha>0\) and \(\beta>0\). Thus \(E[\theta]=\alpha/\beta\). Assume the exposure is known and the count process is homogeneous over the stated exposure period.

This tutorial concerns the posterior mean of the underlying rate \(\theta\). It is not itself a claim count: a future expected count also depends on the future exposure. The posterior mean rate is particularly useful when the desired premium or estimate is expressed per unit of exposure.

Deriving the Weighted-Average Formula

Given an observed count \(N=n\), the Poisson likelihood as a function of \(\theta\) is proportional to \(\theta^n e^{-e\theta}\). Multiplying by the Gamma prior kernel gives a posterior kernel with an updated shape and rate. The next derivation shows how those updates lead to the credibility-weighted form.

Write the likelihood kernel.
For fixed exposure \(e\) and observed count \(n\), the Poisson likelihood is proportional, as a function of \(\theta\), to \(\theta^n e^{-e\theta}\). Terms that do not involve \(\theta\), such as \(e^n/n!\), can be omitted from the kernel.

Multiply by the prior kernel.
The prior kernel is \(\theta^{\alpha-1}e^{-\beta\theta}\). The product is proportional to \(\theta^{\alpha+n-1}e^{-(\beta+e)\theta}\), which is a Gamma kernel.

Take the posterior mean.
The posterior is \(\operatorname{Gamma}(\alpha+n,\beta+e)\), so its mean is \((\alpha+n)/(\beta+e)\). Divide the numerator and denominator by \(\beta+e\) and express \(\alpha\) as \(\beta(\alpha/\beta)\).

\begin{aligned} \pi(\theta\mid N=n) &\propto \theta^n e^{-e\theta}\theta^{\alpha-1}e^{-\beta\theta}\\ &= \theta^{\alpha+n-1}e^{-(\beta+e)\theta},\\[4pt] \theta\mid N=n &\sim \operatorname{Gamma}(\alpha+n,\beta+e),\\[4pt] E[\theta\mid N=n] &=\frac{\alpha+n}{\beta+e}\\ &=\frac{\beta}{\beta+e}\frac{\alpha}{\beta} +\frac{e}{\beta+e}\frac{n}{e}, \qquad e>0. \end{aligned}

Theorem: For \(e>0\), the posterior mean rate is \(E[\theta\mid N=n]=(1-Z)(\alpha/\beta)+Z(n/e)\), where \(Z=e/(\beta+e)\). Thus \(1-Z=\beta/(\beta+e)\) weights the prior mean rate, and \(Z\) weights the observed rate. Both weights are between zero and one, and they sum to one.

This is a weighted average because \(\beta+e\) is the combined rate parameter: the prior contributes \(\beta\), while the observed exposure contributes \(e\). The prior mean \(\alpha/\beta\) is the rate being blended with the empirical rate \(n/e\). The parameter \(\beta\) has the same exposure scale as \(e\); comparing them is meaningful only when they use the same exposure unit.

For fixed prior parameters, increasing \(e\) increases \(Z\), shifting the posterior mean toward the observed rate. Increasing \(\beta\) while holding the prior mean fixed makes the prior more influential and reduces \(Z\). In fact, if the prior mean is held at \(m\), increasing \(\beta\) also increases \(\alpha=m\beta\), so the prior contributes more information without changing its central rate.

When \(e=0\), no count exposure has been observed. The empirical rate \(n/e\) is undefined, so the weighted-average expression using that rate does not apply. In the stated model a zero exposure yields \(N=0\) with probability one, and the posterior remains the prior; its mean is \(\alpha/\beta\). This agrees with assigning zero weight to experience, but it does not make \(0/0\) a valid observed rate.

Worked Applications

Worked Example: A High Observed Rate with Moderate Exposure

A hypothetical roadside-assistance portfolio has a prior claim rate \(\theta\sim\operatorname{Gamma}(4,2)\), where \(\theta\) is claims per policy-year and the Gamma parameters use the shape-rate convention. During \(e=3\) policy-years of exposure, \(n=9\) claims are observed. Find the posterior mean rate and show its weighted-average interpretation.

Method and assumptions. Use the Poisson-Gamma model: conditional on a fixed rate \(\theta\), the count over 3 policy-years is Poisson with mean \(3\theta\). The prior mean rate is \(\alpha/\beta=4/2=2\) claims per policy-year. The observed rate is \(n/e=9/3=3\) claims per policy-year.

The credibility weight on the observed rate is \(Z=e/(\beta+e)=3/(2+3)=3/5=0.6\). The weight on the prior mean is \(1-Z=2/5=0.4\). Therefore,

E[\theta\mid N=9] =0.4(2)+0.6(3) =0.8+1.8 =2.6000 \text{ claims per policy-year}.

Direct check. The posterior distribution is \(\operatorname{Gamma}(4+9,2+3)=\operatorname{Gamma}(13,5)\). Its mean is \(13/5=2.6000\), matching the weighted average.

Common mistake avoided. The observed rate is 3, not 9: the count must be divided by the 3 policy-years. Weighting the raw count as if it were a rate would mix incompatible quantities.

Worked Example: One Claim Over Substantial Exposure

A hypothetical equipment-insurance segment has prior rate \(\theta\sim\operatorname{Gamma}(6,3)\), measured in claims per 1,000 insured-equipment-years. It records \(n=1\) claim over \(e=1.5\) thousand insured-equipment-years. Calculate the posterior mean and explain why it remains below the prior mean.

Method and assumptions. Conditional on \(\theta\), take the count to be Poisson with mean \(e\theta\), using exposure measured in thousands of insured-equipment-years. The prior mean is \(6/3=2\) claims per 1,000 insured-equipment-years. The observed rate is \(1/1.5=2/3\) claim per 1,000 insured-equipment-years.

The experience weight is \(Z=1.5/(3+1.5)=1.5/4.5=1/3\). The prior weight is \(2/3\). The posterior mean is

\begin{aligned} E[\theta\mid N=1] &=\frac{2}{3}(2)+\frac{1}{3}\left(\frac{2}{3}\right)\\ &=\frac{4}{3}+\frac{2}{9} =\frac{14}{9} =1.5556 \text{ claims per 1,000 insured-equipment-years}. \end{aligned}

Direct check. The posterior is \(\operatorname{Gamma}(6+1,3+1.5)=\operatorname{Gamma}(7,4.5)\). Its mean is \(7/4.5=14/9=1.5556\), as above. Since the observed rate \(2/3\) is below the prior mean 2 and has positive weight, the result lies between them.

Common mistake avoided. Using \(Z=\beta/(\beta+e)\) for the observed rate reverses the roles of the weights. Here the observed-rate weight is \(e/(\beta+e)=1/3\); the prior-rate weight is \(2/3\).

Worked Example: Zero Claims Do Not Mean a Zero Posterior Rate

A hypothetical small-business insurance group has prior rate \(\theta\sim\operatorname{Gamma}(3,1.5)\), in claims per business-year. It reports zero claims over \(e=4\) business-years. Find the posterior mean and interpret the size of the experience weight.

Method and assumptions. Given \(\theta\), assume the claim count is Poisson with mean \(4\theta\). The prior mean is \(3/1.5=2\) claims per business-year. The observed rate is \(0/4=0\).

The experience weight is \(Z=4/(1.5+4)=4/5.5=8/11\), and the prior weight is \(1-Z=3/11\). Consequently,

\begin{aligned} E[\theta\mid N=0] &=\frac{3}{11}(2)+\frac{8}{11}(0)\\ &=\frac{6}{11} =0.5455 \text{ claims per business-year}. \end{aligned}

Direct check. The posterior is \(\operatorname{Gamma}(3+0,1.5+4)=\operatorname{Gamma}(3,5.5)\). Its mean is \(3/5.5=6/11=0.5455\), matching the weighted result. The mean is positive because the prior still has positive weight, even though the observed count is zero.

Common mistake avoided. Replacing the posterior mean by the observed rate would give zero. That discards the prior information. The observed rate has a large weight here, but it does not have all the weight.

Interpretation and Exam Traps

The weight \(Z=e/(\beta+e)\) depends on exposure and prior strength, not on the observed count. A surprising count changes the observed rate being blended; it does not change \(Z\) under this model. This distinction is useful when comparing two groups with the same prior but different exposure: the group with more exposure receives greater weight on its own observed rate.

A second useful check is that a weighted average with nonnegative weights summing to one must lie between the prior mean and observed rate. If the calculated posterior mean falls outside that interval for \(e>0\), check the parameterization, the exposure units, and which weight has been assigned to which rate.

Do not confuse shape and rate. Under the convention used here, the prior mean is \(\alpha/\beta\). A Gamma distribution parameterized by scale uses a different expression, so the scale cannot be substituted for \(\beta\).
Do not compare incompatible exposure units. If \(e\) is measured in policy-years, \(\theta\) must be a rate per policy-year, and \(\beta\) must use the corresponding exposure scale.
Do not use an empirical rate at zero exposure. For \(e=0\), the posterior remains the prior and \(n/e\) is not defined.
Do not assign the wrong weight. The experience rate receives \(e/(\beta+e)\); the prior mean receives \(\beta/(\beta+e)\).

The weighted-average form is useful because it turns the posterior update into an interpretable credibility statement: the prior mean and the observed rate are blended according to their relative information strengths. The posterior distribution remains important when uncertainty about the rate itself is needed, but the posterior mean gives the central rate estimate directly.

Check Your Understanding

For each problem, use the Gamma shape-rate convention and state the posterior mean claim rate. In each case, assume a fixed, known exposure and a Poisson count conditional on the rate.

A prior is \(\operatorname{Gamma}(2,1)\), in claims per unit exposure. There are 5 claims over 2 units. Find the prior mean, experience weight, and posterior mean. (Answer: prior mean \(2\), \(Z=2/3\), posterior mean \(7/3=2.3333\) claims per unit exposure.)
A prior is \(\operatorname{Gamma}(8,4)\), and 1 claim is observed over 2 policy-years. Find the observed rate and posterior mean. (Answer: observed rate \(0.5\), posterior mean \(1.5000\) claims per policy-year.)
A prior is \(\operatorname{Gamma}(3,1.5)\). No exposure is observed. What is the posterior mean, and can an observed rate be calculated? (Answer: posterior mean \(2.0000\) claims per exposure unit; no, the observed rate is undefined.)
A prior is \(\operatorname{Gamma}(5,2.5)\), with rate measured per 1,000 insured-hours. There are 6 claims over 7.5 thousand insured-hours. Find the credibility weight and posterior mean. (Answer: \(Z=0.75\); posterior mean \(1.1000\) claims per 1,000 insured-hours.)
A prior is \(\operatorname{Gamma}(4,2)\) claims per policy-year. Two independent observation periods for the same rate have counts and exposures \((3,1)\) and \((1,2)\), respectively. Combine the evidence and find the posterior mean. (Answer: total count \(4\), total exposure \(3\), \(Z=3/5\), posterior mean \(1.6000\) claims per policy-year.)

What You'll Learn