Scaling Claim-Count Predictions by Future Exposure

Why Future Exposure Changes a Claim-Count Forecast

A posterior claim rate summarizes uncertainty about the frequency of claims per unit of exposure. To forecast a future count, that rate must be applied to the amount of future exposure. A forecast for 2,000 policy-years is not the same as a forecast for 200 policy-years, even when both use the same posterior information about the rate.

Exposure changes more than the forecast mean. If the rate itself is uncertain, predictions for disjoint future periods using that same rate are dependent: a high underlying rate raises the expected count in each period. Correctly reflecting this uncertainty matters when calculating forecast variability for a longer period or combining forecasts across periods.

Model and assumptions: Let $\lambda$ be a claim rate per unit of exposure, with posterior distribution $\lambda\mid\text{data}\sim\operatorname{Gamma}(a,b)$, using the shape-rate convention. Thus $a>0$, $b>0$, $E[\lambda\mid\text{data}]=a/b$, and $\operatorname{Var}(\lambda\mid\text{data})=a/b^2$. For a specified future exposure $e$, assume $N_e\mid\lambda\sim\operatorname{Poisson}(e\lambda)$. Exposure is known and measured in the same units used for $\lambda$; the rate is constant over the forecast period, conditional on $\lambda$.

The conditional Poisson assumption means that, if the rate were known, the expected claim count and its variance would both equal $e\lambda$. But the rate is not known: its posterior uncertainty adds an additional source of predictive variation. The resulting unconditional forecast is a posterior predictive distribution, formed by averaging the conditional Poisson distribution over the posterior distribution of $\lambda$.

Deriving the Predictive Distribution

For a nonnegative integer $n$, the conditional probability of $n$ claims is $P(N_e=n\mid\lambda)=\exp(-e\lambda)(e\lambda)^n/n!$. The Gamma posterior density, under the shape-rate convention, is $b^a\lambda^{a-1}\exp(-b\lambda)/\Gamma(a)$ for $\lambda>0$. Multiplying these expressions and integrating over all possible rates gives the predictive probability:

\begin{aligned} P(N_e=n\mid\text{data}) &=\int_0^\infty \frac{\exp(-e\lambda)(e\lambda)^n}{n!} \frac{b^a\lambda^{a-1}\exp(-b\lambda)}{\Gamma(a)} \,d\lambda\\ &=\frac{e^n b^a}{n!\,\Gamma(a)} \int_0^\infty \lambda^{a+n-1}\exp\bigl(-(b+e)\lambda\bigr)\,d\lambda\\ &=\frac{\Gamma(a+n)}{\Gamma(a)n!} \left(\frac{b}{b+e}\right)^a \left(\frac{e}{b+e}\right)^n. \end{aligned}

For the last equality, the integral is the Gamma integral $\Gamma(a+n)/(b+e)^{a+n}$. This is a negative binomial probability mass function, with shape parameter $a$ and probability parameter $b/(b+e)$ in the convention shown. In particular, the probability of no claims is $(b/(b+e))^a$. The predictive distribution is generally not Poisson: mixing over an uncertain rate adds variation beyond the conditional Poisson variation.

Predictive mean and variance: Given $\lambda$, the conditional mean and variance of $N_e$ are both $e\lambda$. Averaging over the posterior gives $$ E[N_e\mid\text{data}]=e\frac{a}{b}, \qquad \operatorname{Var}(N_e\mid\text{data}) =e\frac{a}{b}+e^2\frac{a}{b^2}. $$ The first variance term is the expected conditional Poisson variance. The second is the variance in the conditional mean caused by uncertainty about $\lambda$.

The variance formula follows from the law of total variance: $\operatorname{Var}(N_e\mid\text{data})=E[\operatorname{Var}(N_e\mid\lambda,\text{data})\mid\text{data}]+\operatorname{Var}(E[N_e\mid\lambda,\text{data}]\mid\text{data})$. The first term is $E[e\lambda\mid\text{data}]=ea/b$; the second is $\operatorname{Var}(e\lambda\mid\text{data})=e^2a/b^2$. This explains why simply using a Poisson variance equal to the predictive mean understates uncertainty whenever the posterior rate has positive variance.

Worked Examples

Worked Example: Scaling a Forecast to 2.5 Policy-Years

A hypothetical insurer has a posterior claim rate $\lambda\mid\text{data}\sim\operatorname{Gamma}(3,6)$, where $\lambda$ is claims per policy-year. Forecast the count for $2.5$ future policy-years. Find its predictive distribution, mean, variance, and probability of exactly two claims.

Method and units. Use the Gamma-Poisson predictive distribution, with $a=3$, $b=6$, and $e=2.5$. The future exposure is already in policy-years, so no unit conversion is needed.

The predictive probability mass function is $P(N_e=n\mid\text{data})=\frac{\Gamma(3+n)}{\Gamma(3)n!}(6/8.5)^3(2.5/8.5)^n$. Equivalently, the negative binomial shape is $3$, and its probability parameter is $6/8.5=12/17$.

The posterior mean rate is $a/b=3/6=0.5$ claims per policy-year. Therefore, $E[N_e\mid\text{data}]=2.5(0.5)=1.25$ claims.

The posterior rate variance is $a/b^2=3/36=1/12$. Thus, $\operatorname{Var}(N_e\mid\text{data})=2.5(0.5)+(2.5)^2(1/12)=1.25+6.25/12=1.770833\ldots$, or $1.7708$ claims squared to four decimal places.

For exactly two claims, the coefficient is $\Gamma(5)/(\Gamma(3)2!)=24/(2\cdot2)=6$. Hence $P(N_e=2\mid\text{data})=6(12/17)^3(5/17)^2=259200/1419857=0.1826$, rounded to four decimal places.

Check. The mean is 1.25, and the variance exceeds the mean by $1.770833\ldots-1.25=0.520833\ldots$, exactly the added rate-uncertainty term $6.25/12$. The common mistake avoided here is treating the mixture as Poisson and setting its variance equal to 1.25.

Worked Example: Converting Six Months of Exposure

For a hypothetical coverage, the posterior rate is $\operatorname{Gamma}(4,8)$ claims per policy-year. Calculate the predictive mean, variance, and probability of zero claims over six months of future exposure. Assume six months is exactly one-half policy-year.

Method and unit conversion. Apply the mixed Poisson formulas after converting the future exposure to $e=0.5$ policy-year. Using $e=6$ with a rate measured per policy-year would be a unit error.

The posterior mean rate is $4/8=0.5$, so the predictive mean is $E[N_e\mid\text{data}]=0.5(0.5)=0.25$ claims.

The posterior rate variance is $4/8^2=4/64=0.0625$. Therefore, $\operatorname{Var}(N_e\mid\text{data})=0.5(0.5)+(0.5)^2(0.0625)=0.25+0.015625=0.265625$, or $0.2656$ to four decimal places.

The probability of zero claims is $P(N_e=0\mid\text{data})=(8/(8+0.5))^4=(16/17)^4=65536/83521=0.7847$, rounded to four decimal places.

Check. The zero-claim probability is high because the predictive mean is only 0.25; it is not obtained by substituting the posterior mean rate into a Poisson model. That plug-in calculation would give $\exp(-0.25)=0.7788$, a different probability because it ignores rate uncertainty.

Combining Future Periods Under a Shared Rate

Suppose two disjoint future periods have exposures $e_1$ and $e_2$, and both use the same uncertain rate $\lambda$. Conditional on $\lambda$, their counts are independent Poisson variables with means $e_1\lambda$ and $e_2\lambda$. Marginally, however, the counts are dependent. A high value of the shared rate makes both counts more likely to be large.

Shared-rate covariance: If $N_1\mid\lambda\sim\operatorname{Poisson}(e_1\lambda)$ and $N_2\mid\lambda\sim\operatorname{Poisson}(e_2\lambda)$ are conditionally independent, then $$ \operatorname{Cov}(N_1,N_2\mid\text{data}) =e_1e_2\operatorname{Var}(\lambda\mid\text{data}) =e_1e_2\frac{a}{b^2}. $$ Their combined count has exposure $e_1+e_2$ and follows the same Gamma-Poisson predictive model for that total exposure.

To justify the covariance, conditional independence makes $\operatorname{Cov}(N_1,N_2\mid\lambda,\text{data})=0$. The law of total covariance then leaves the covariance of the conditional means: $\operatorname{Cov}(e_1\lambda,e_2\lambda\mid\text{data})=e_1e_2\operatorname{Var}(\lambda\mid\text{data})$. Consequently, the variance of the combined count includes twice the covariance: $\operatorname{Var}(N_1+N_2)=\operatorname{Var}(N_1)+\operatorname{Var}(N_2)+2\operatorname{Cov}(N_1,N_2)$.

Worked Example: Two Periods With One Shared Posterior Rate

A hypothetical risk has posterior claim rate $\lambda\mid\text{data}\sim\operatorname{Gamma}(2,4)$ claims per exposure unit. Forecast claims in two disjoint periods with exposures $e_1=1$ and $e_2=3$. Find each period's predictive mean and variance, their covariance, and the mean and variance of the combined count.

Method. Use the shared-rate formulas; the two counts are conditionally independent given $\lambda$, but not independent after integrating out $\lambda$. The posterior mean rate is $2/4=0.5$, and its variance is $2/4^2=2/16=0.125$.

For the first period, the mean is $1(0.5)=0.5$ and the variance is $1(0.5)+1^2(0.125)=0.625$.

For the second period, the mean is $3(0.5)=1.5$ and the variance is $3(0.5)+3^2(0.125)=1.5+1.125=2.625$.

Their covariance is $1(3)(0.125)=0.375$. The combined mean is $0.5+1.5=2$. Its variance is $0.625+2.625+2(0.375)=3.25+0.75=4$.

As a direct check, the combined exposure is $1+3=4$, so the variance formula gives $4(2/4)+4^2(2/4^2)=2+2=4$, agreeing with the covariance calculation.

Common mistake avoided. Adding the separate variances alone gives $0.625+2.625=3.25$, which incorrectly assumes the marginal forecasts are independent. The missing covariance contribution is $2(0.375)=0.75$; the correct combined variance is $4$.

How Exposure Affects Predictive Uncertainty

For fixed posterior parameters, the predictive mean grows linearly with exposure: $E[N_e\mid\text{data}]=e(a/b)$. The predictive variance has both a linear term and a quadratic term: $\operatorname{Var}(N_e\mid\text{data})=e(a/b)+e^2(a/b^2)$. The quadratic term reflects uncertainty about the underlying rate. It becomes especially important when forecasting across substantial exposure while retaining the same uncertain rate.

This does not mean that relative uncertainty must increase with exposure. Dividing the variance by the squared mean gives $\operatorname{Var}(N_e\mid\text{data})/E[N_e\mid\text{data}]^2=b/(ae)+1/a$. For fixed $a$ and $b$, this ratio decreases as $e$ increases. Thus the absolute variance can grow while uncertainty relative to the squared forecast mean falls.

The model has limits. It assumes a constant rate over the forecast exposure and a Poisson count conditional on that rate. If exposure is uncertain, rates vary systematically over time, or the conditional count process is not Poisson, these formulas need an appropriate model change. A unit mismatch also invalidates the calculation: a rate per policy-year must be multiplied by policy-years, not by months or policies without conversion.

Key takeaway: First express future exposure in the posterior rate's units. Then scale the predictive mean by exposure and use both terms in the predictive variance. If multiple future counts share an uncertain rate, include their positive covariance when combining them. These methods are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

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

A posterior rate is $\operatorname{Gamma}(3,9)$ claims per exposure unit. For future exposure $e=3$, find the predictive mean and variance. (Answer: mean $1$; variance $4/3=1.3333$.)
A posterior rate is $\operatorname{Gamma}(5,10)$ claims per policy-year. For $0.4$ future policy-year, find the predictive mean and variance. (Answer: mean $0.2000$; variance $0.2080$.)
A shared posterior rate is $\operatorname{Gamma}(2,3)$. Two disjoint periods have exposures 2 and 1. Find the covariance of their counts and the variance of their combined count. (Answer: covariance $4/9=0.4444$; combined variance $4$.)
A posterior rate is $\operatorname{Gamma}(4,12)$ claims per exposure unit. For future exposure $e=2$, find the predictive probability of zero claims. (Answer: $(6/7)^4=0.5398$.)
A posterior rate is $\operatorname{Gamma}(2,8)$ claims per year. Using a 365-day year, find the predictive mean and variance for 90 days of exposure. (Answer: mean $9/146=0.0616$; variance $0.0635$.)

What You'll Learn