Integrated Bayesian Forecasting Workshop for Claim Experience

From Observed Claims to a Forecast

A claim-count forecast for an existing risk must account for two sources of uncertainty: future random variation in claims and uncertainty about the risk’s underlying claim frequency. A plug-in Poisson forecast that substitutes one estimated frequency accounts only for the first source. A Bayesian predictive forecast averages over the posterior distribution of the latent frequency, incorporating both.

This workshop follows one hypothetical risk from prior assumptions through observed experience to probabilities for future claim counts. It is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. We use a Poisson frequency model with a Gamma prior, with the Gamma distribution parameterized by shape and rate.

Definition: Let \(\lambda\) denote a risk’s latent claim frequency per exposure unit. Assume \(\lambda\sim\operatorname{Gamma}(\alpha,\beta)\), with density \(f(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\), for \(\lambda>0\). Conditional on \(\lambda\), the count \(N_j\) in an observation period with exposure \(e_j\) follows \(\operatorname{Poisson}(e_j\lambda)\), independently across periods. Here \(\alpha>0\) is the shape and \(\beta>0\) is the rate; \(E[\lambda]=\alpha/\beta\) and \(\operatorname{Var}(\lambda)=\alpha/\beta^2\).

If the observed counts are \(n_1,\ldots,n_J\), with total count \(n=\sum_j n_j\) and total exposure \(E=\sum_j e_j\), the posterior distribution is \(\operatorname{Gamma}(\alpha+n,\beta+E)\). This update uses the total count and exposure because the likelihood, apart from factors not involving \(\lambda\), is proportional to \(\lambda^n e^{-E\lambda}\).

Worked Case: Updating and Forecasting

Worked Example: A One-Unit Forecast After Two Observation Periods

A hypothetical risk has prior \(\lambda\sim\operatorname{Gamma}(2,2)\). It produces 1 claim over 0.5 exposure units in one period and 2 claims over 1.5 exposure units in another. Find the posterior distribution, then forecast the count over the next 1 exposure unit. Calculate the predictive mean, variance, and probabilities of zero and exactly two claims.

Method and update. The observed total is \(n=1+2=3\), and total exposure is \(E=0.5+1.5=2\). Conjugate updating gives

\lambda\mid\text{data}\sim\operatorname{Gamma}(\alpha+n,\beta+E) =\operatorname{Gamma}(2+3,2+2) =\operatorname{Gamma}(5,4).

The posterior mean frequency is \(5/4=1.25\) claims per exposure unit, and the posterior variance of the frequency is \(5/4^2=5/16=0.3125\). The forecast horizon is \(t=1\), so the future count \(M\), conditional on \(\lambda\), is \(\operatorname{Poisson}(\lambda)\).

Predictive mean and variance. The posterior predictive mean is \(E[M\mid\text{data}]=tE[\lambda\mid\text{data}]=1(5/4)=1.25\). By the law of total variance, the predictive variance is \(tE[\lambda\mid\text{data}]+t^2\operatorname{Var}(\lambda\mid\text{data})\), so

\operatorname{Var}(M\mid\text{data}) =1\left(\frac54\right)+1^2\left(\frac5{16}\right) =\frac{25}{16}=1.5625.

Predictive probabilities. The predictive distribution is negative binomial with shape \(5\) and success probability \(p=4/(4+1)=4/5\). Thus \(P(M=m\mid\text{data})=\binom{5+m-1}{m}(4/5)^5(1/5)^m\). For zero claims,

P(M=0\mid\text{data})=(4/5)^5=1024/3125=0.32768.

For exactly two claims,

P(M=2\mid\text{data}) =\binom{6}{2}(4/5)^5(1/5)^2 =15\left(\frac{1024}{3125}\right)\left(\frac1{25}\right) =\frac{15360}{78125}=0.196608.

The mean is 1.25 claims, and the variance exceeds the mean by 0.3125 because the latent rate remains uncertain. The calculation avoids the common mistake of treating the posterior mean as a known frequency and thereby discarding parameter uncertainty.

The Predictive Distribution and Its Uncertainty

For a posterior \(\lambda\mid\text{data}\sim\operatorname{Gamma}(a,b)\) and future exposure \(t\), the predictive probability of \(m\) claims is obtained by integrating the conditional Poisson probability over the posterior. This is a mixture: each possible latent rate supplies a Poisson distribution, and the posterior weights those distributions.

\begin{aligned} P(M=m\mid\text{data}) &=\int_0^\infty \frac{e^{-t\lambda}(t\lambda)^m}{m!} \frac{b^a}{\Gamma(a)}\lambda^{a-1}e^{-b\lambda}\,d\lambda\\ &=\frac{\Gamma(a+m)}{\Gamma(a)m!} \left(\frac{b}{b+t}\right)^a \left(\frac{t}{b+t}\right)^m,\qquad m=0,1,2,\ldots . \end{aligned}

The second equality follows by collecting the integrand’s powers and exponential terms: the integral is proportional to \(\int_0^\infty \lambda^{a+m-1}e^{-(b+t)\lambda}\,d\lambda=\Gamma(a+m)/(b+t)^{a+m}\). This is the negative binomial probability mass function in the shape-probability parameterization, with shape \(a\) and \(p=b/(b+t)\). Its mean and variance are

E[M\mid\text{data}]=\frac{at}{b}, \qquad \operatorname{Var}(M\mid\text{data})=\frac{at}{b}+\frac{at^2}{b^2}.

The variance also follows directly from the law of total variance. Conditional on \(\lambda\), \(E[M\mid\lambda]=\operatorname{Var}(M\mid\lambda)=t\lambda\). Therefore,

\begin{aligned} \operatorname{Var}(M\mid\text{data}) &=E[\operatorname{Var}(M\mid\lambda,\text{data})\mid\text{data}] +\operatorname{Var}(E[M\mid\lambda,\text{data}]\mid\text{data})\\ &=tE[\lambda\mid\text{data}]+t^2\operatorname{Var}(\lambda\mid\text{data}) =\frac{at}{b}+\frac{at^2}{b^2}. \end{aligned}

The first term is expected Poisson process variance; the second is the additional variance due to uncertainty in the latent frequency. This decomposition is useful when a forecast’s dispersion matters as much as its average. It applies under the stated Gamma-Poisson model; it is not a claim that every claim-count process is conditionally Poisson or that every risk has a Gamma-distributed frequency.

Worked Example: A Discrete Predictive Interval

Use the posterior \(\operatorname{Gamma}(5,4)\) from the hypothetical case to construct an equal-tail 90% predictive interval for the next 1 exposure unit. State the actual coverage, taking discreteness into account.

Method. Here \(p=4/5\), and the predictive probability at count \(m\) is \(\binom{m+4}{m}(4/5)^5(1/5)^m\). Starting with \(P(M=0)=0.32768\), successive probabilities can be computed using the ratio \(P(M=m)/P(M=m-1)=((m+4)/m)(1/5)\).

The probabilities and cumulative probabilities through the counts needed to locate the upper quantile are:

\begin{array}{c|c|c} m & P(M=m\mid\text{data}) & P(M\le m\mid\text{data})\\ 0 & 0.32768000 & 0.32768000\\ 1 & 0.32768000 & 0.65536000\\ 2 & 0.19660800 & 0.85196800\\ 3 & 0.09175040 & 0.94371840\\ 4 & 0.03670016 & 0.98041856 \end{array}

Define the discrete \(q\)-quantile as the smallest integer \(m\) for which \(P(M\le m\mid\text{data})\ge q\). Since the cumulative probability at 0 is already above 0.05, the lower 5% quantile is 0. Since the cumulative probability through 3 is 0.94371840, below 0.95, but through 4 is 0.98041856, the upper 95% quantile is 4. The equal-tail interval is therefore \([0,4]\), with actual coverage \(P(0\le M\le4\mid\text{data})=0.98041856\), or 0.9804 to four decimals. The coverage exceeds 90% because a discrete distribution cannot generally attain the nominal tail probabilities exactly.

A common exam trap is to report a continuous-style interval with tail probabilities exactly 0.05 each. Claim counts are integers; quantiles must be found from the cumulative probabilities, and the resulting interval’s coverage should be checked.

Forecasting Across a Different Exposure Horizon

Worked Example: Two Exposure Units and a Claim-Count Tail

For the same posterior \(\operatorname{Gamma}(5,4)\), forecast claims over the next 2 exposure units. Find the predictive mean and variance, the probability of exactly 2 claims, and the probability of at least 3 claims.

Method. For \(t=2\), \(p=b/(b+t)=4/6=2/3\), and \(1-p=1/3\). The predictive mean and variance are

E[M\mid\text{data}]=\frac{5(2)}4=2.5, \qquad \operatorname{Var}(M\mid\text{data}) =\frac{5(2)}4+\frac{5(2^2)}{4^2} =2.5+1.25=3.75.

The probability of exactly two claims is

P(M=2\mid\text{data}) =\binom{6}{2}(2/3)^5(1/3)^2 =\frac{480}{2187} =\frac{160}{729} \approx0.2195.

For the upper tail, \(P(M=0)=(2/3)^5=32/243=96/729\). Also, \(P(M=1)=5(1/3)(2/3)^5=160/729\), and \(P(M=2)=160/729\). Thus

P(M\ge3\mid\text{data}) =1-\frac{96+160+160}{729} =\frac{313}{729} \approx0.4294.

The check is that the events \(M=0,1,2\) are disjoint, so subtracting their total probability from 1 gives the probability of \(M\ge3\). The forecast horizon changes both the negative binomial probability parameter and the mean; using the one-unit value of \(p\) here would be a time-unit error.

Prior Sensitivity and a Bühlmann Comparison

The prior affects the posterior forecast even when two candidate priors have the same mean. A more concentrated prior generally pulls the posterior mean more strongly toward that common mean. In a Poisson-Gamma model, the prior mean is \(\alpha/\beta\), while its variance, \(\alpha/\beta^2\), controls the prior’s dispersion.

Worked Example: Prior Sensitivity and the Bühlmann Estimate

Keep the observed experience fixed at 3 claims over 2 exposure units. Compare the posterior forecast under \(\operatorname{Gamma}(2,2)\) and \(\operatorname{Gamma}(4,4)\), which both have prior mean 1 claim per exposure unit. Then compare the first posterior mean with a Bühlmann estimate based on the same count experience.

Bayesian comparison. Under \(\operatorname{Gamma}(2,2)\), the posterior is \(\operatorname{Gamma}(5,4)\), so its mean is \(5/4=1.25\). Under \(\operatorname{Gamma}(4,4)\), the posterior is \(\operatorname{Gamma}(4+3,4+2)=\operatorname{Gamma}(7,6)\), so its mean is \(7/6=1.1667\) claims per exposure unit. The second prior has the same mean but is more concentrated: its variance is \(4/16=0.25\), compared with \(2/4=0.5\) for the first. The more concentrated prior consequently gives more weight to its mean of 1.

For a one-unit forecast under the second prior, the predictive variance is \(7/6+7/36=49/36=1.3611\). Under the first prior it is \(5/4+5/16=25/16=1.5625\). Thus the prior choice affects not only the posterior mean but also the predictive uncertainty.

Bühlmann comparison. For this model, the conditional claim-count mean per exposure unit is \(\lambda\). The expected process variance is \(\operatorname{EPV}=E[\lambda]=1\), and the variance of hypothetical means is \(\operatorname{VHM}=\operatorname{Var}(\lambda)=0.5\) under the \(\operatorname{Gamma}(2,2)\) prior. Hence \(K=\operatorname{EPV}/\operatorname{VHM}=2\). The observed average count is \(3/2=1.5\) over \(n=2\) exposure units, giving credibility \(Z=n/(n+K)=2/(2+2)=0.5\). The Bühlmann estimate is

$$ Z(1.5)+(1-Z)(1)=0.5(1.5)+0.5(1)=1.25. $$

This equals the Bayesian posterior mean under the Gamma prior. The agreement is expected here: the Poisson-Gamma posterior mean is a credibility-weighted average of the observed average and prior mean, with weight \(E/(\beta+E)\). For this case that weight is \(2/(2+2)=0.5\). This comparison concerns the estimated claim frequency; a posterior predictive distribution additionally describes the spread of future counts.

The sensitivity calculation avoids another common mistake: equal prior means do not imply equal posterior forecasts. Prior concentration matters, so a responsible comparison states both the prior mean and its dispersion.

Practical Checks for an Integrated Forecast

A reliable calculation keeps the exposure units consistent from the likelihood through the forecast. The update adds total claims to the Gamma shape and total exposure to the Gamma rate. The forecast then uses its own horizon \(t\), which enters the predictive probability parameter and the variance decomposition. Before interpreting a result, check the following:

Summarize the data.
Record total claims \(n\) and total exposure \(E\), using the same exposure unit as the claim frequency.

Update the prior.
For a Gamma shape-rate prior, set posterior shape to \(\alpha+n\) and posterior rate to \(\beta+E\).

Match the future horizon.
For future exposure \(t\), use \(p=b/(b+t)\), not a probability parameter from a different horizon.

Interpret the spread.
Separate \(at/b\), the expected process variance, from \(at^2/b^2\), the contribution from latent-rate uncertainty.

These formulas rely on conditional independence and a constant latent frequency over the observation and forecast periods. If exposure changes the underlying frequency, claim counts are dependent beyond the shared latent rate, or the count distribution has additional structure, the Gamma-Poisson model may not be appropriate without extension or validation.

Key takeaway: Bayesian claim forecasting updates the latent frequency using both counts and exposure, then averages future Poisson outcomes over the posterior. The resulting predictive distribution carries process variation and parameter uncertainty, supports count probabilities and discrete intervals, and can be compared with a Bühlmann estimate of the frequency mean.

Check Your Understanding

For each problem, use a Gamma shape-rate prior and a conditionally Poisson claim-count model. Report probabilities to four decimals.

A risk has prior \(\operatorname{Gamma}(3,2)\), then produces 2 claims over 1 exposure unit. Find the posterior and the probability of zero claims over the next 1 exposure unit. (Answer: \(\operatorname{Gamma}(5,3)\); \(P(M=0)=(3/4)^5=0.2373\).)
The posterior is \(\operatorname{Gamma}(4,3)\). For a future horizon of 2 exposure units, find the probability of exactly 1 claim. (Answer: \(\binom41(3/5)^4(2/5)=648/3125=0.2074\).)
The posterior is \(\operatorname{Gamma}(6,2)\), and the future horizon is 1 exposure unit. Find the predictive mean and variance. (Answer: mean \(=3\); variance \(=6/2+6/2^2=4.5\).)
A risk has prior \(\operatorname{Gamma}(2,1)\) and records 1 claim over 1 exposure unit. Find the posterior and the probability of zero claims over the next 2 exposure units. (Answer: \(\operatorname{Gamma}(3,2)\); \(P(M=0)=(2/4)^3=0.1250\).)
For 3 claims over 2 exposure units, compare the posterior mean under priors \(\operatorname{Gamma}(2,2)\) and \(\operatorname{Gamma}(4,4)\). What is the difference between the two posterior means? (Answer: \(5/4\) versus \(7/6\); the difference is \(1/12=0.0833\) claims per exposure unit.)

What You'll Learn