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Insurance exposure measures the amount of opportunity for claims, such as the number of policy-years observed. With known positive insurance exposures $w_j$ and a common annual claim frequency $\Theta$, a simple model iswhere the claim counts $K_j$ are conditionally independent. The insurance exposures are fixed inputs here; the conditional independence assumption concerns the claim counts.

Put $W={\sum }_j{w}_j$ and $K={\sum }_j{K}_j$. Multiplying the conditional probability generating functions gives $K\mid \Theta \sim \mathrm{Poisson}(W\Theta )$. The conditional likelihood has the formAll dependence on $\theta$ is through the total claim count $K$ and total insurance exposure $W$; with the insurance exposures fixed, $K$ is a sufficient statistic. This is the exposure aggregation used in Poisson-gamma conjugacy with unequal exposures.

The normalized claim count $K_j/w_j$ has conditional expectation $\Theta$ and conditional variance $\Theta /w_j$. Thus its conditional expectation does not change with insurance exposure, while its conditional variance decreases. For example, two groups with $2$ and $3$ policy-years at annual frequency $0.4$ have expected claim counts $0.8$ and $1.2$; together they have a Poisson distribution claim count of mean $2$. The count model and aggregation follow from the Poisson distribution formula; the likelihood and example are derived here.^[ref]

Use the shape-rate convention for the gamma distribution $\theta \sim \mathrm{Gamma}(\alpha ,\beta )$, with prior distribution probability density function proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$. The observations are independent conditional on the common parameter $\theta$. Unconditionally a common nondegenerate prior distribution induces covariance $\alpha /{\beta }^2$ between distinct observations, so the Bayesian inference model requires conditional independence.

For unit-exposure counts, the likelihood is proportional to ${\theta }^{{\sum }_i{x}_i}{e}^{-n\theta }$. Multiplying by the prior distribution givesThe future count has conditional expectation $\theta$, so posterior distribution averaging givesThis is an exact Bayesian credibility estimate, with credibility factor $Z=n/(n+\beta )$ and collective mean $\alpha /\beta$.

For the insurance exposure and claims inflation calculation put $c_j=c(1+r{)}^j$, assuming $c>0$, $1+r>0$ and positive policy counts $n_j$. Let $K_j$ be the total claim count in year $j$. Conditional independence of the policy counts gives $K_j\mid \theta \sim \mathrm{Pois}(n_j\theta )$, and the observed average amount satisfiesFor possible observations, the recovered $k_j$ are nonnegative integers. If they are not, the data have zero likelihood under the model and no conditional posterior distribution is defined for those impossible values.

Write $W={\sum }_{j=1}^n{n}_j$ and $K={\sum }_{j=1}^n{k}_j$. Since $Y_j$ and $K_j$ determine each other, conditioning on the observed average amounts is equivalent to conditioning on these counts. The independent yearly count likelihood isThe Poisson-gamma conjugacy with unequal exposures therefore givesFor the future year, $\mathbb{E}[Y_{n+1}\mid \theta ]=c_{n+1}\theta$; the number of future policies cancels because $Y_{n+1}$ is a per-policy average. HencewithThe collective mean $m$ is the prior distribution expected claim amount per policy in the future year's monetary units. The experience mean $m(\mathbf{y})$ first inflates each past per-policy amount to that same future year and then averages with policy-exposure weights. Equivalently, it estimates annual claim frequency by total observed claims divided by total past insurance exposure and converts that frequency to the future claim cost. This is inflation-adjusted Poisson-gamma credibility.

For fixed prior distribution parameters, $Z$ increases strictly with total past insurance exposure: $dZ/dW=\beta /(W+\beta {)}^2>0$, and $Z\to 1$ as $W\to \infty$. The prior distribution weight decreases to zero; $\beta$ acts as equivalent prior distribution insurance exposure. With no past insurance exposure the appropriate weight is zero and the prediction is $m$, without defining an empirical experience mean.