Deriving Poisson-Gamma Forecasts for Claim Counts

From an Uncertain Claim Rate to a Count Forecast

A Poisson model is useful when a claim rate is fixed: conditional on that rate, it gives a distribution for the number of claims in a specified exposure period. In ratemaking, however, the rate for a particular risk may not be known exactly. A Bayesian model represents that uncertainty with a posterior distribution for the rate. Forecasting then requires averaging the conditional Poisson probabilities over the posterior—not simply substituting one rate estimate into a Poisson formula.

For a Poisson likelihood and a Gamma posterior, this averaging has a closed-form result: the future count follows a negative binomial distribution. The forecast horizon matters directly, so exposure units must be consistent with the units used to express the rate.

Definition: Let $\lambda$ be a risk's claim rate per exposure unit. Suppose the posterior distribution is Gamma with shape $\alpha>0$ and rate $\beta>0$, written $\lambda\mid\text{data}\sim\operatorname{Gamma}(\alpha,\beta)$, with density $$ f(\lambda\mid\text{data})=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0. $$ The second parameter is the rate, not the scale; the posterior mean rate is $\alpha/\beta$. For a future period of $h$ exposure units, assume the future claim count $N$, conditional on $\lambda$, is Poisson with mean $h\lambda$. The resulting marginal distribution of $N$ is the Poisson-Gamma predictive distribution, also called a negative binomial distribution.

The Poisson assumption is conditional on a fixed $\lambda$: claims occur independently at a constant rate over the exposure being modeled. The Gamma distribution describes uncertainty about that rate after the observed experience. Conditional on the rate, future counts are independent of the observed counts. If the claim process has changing rates within the forecast period, or dependence not represented by the model, this mixture does not by itself account for those features.

Deriving the Predictive Distribution

Use the Gamma shape-rate convention above. Conditional on $\lambda$, the probability of $k$ claims over $h$ exposure units is $e^{-h\lambda}(h\lambda)^k/k!$. The unconditional forecast probability averages this expression over the posterior density of $\lambda$.

\begin{aligned} P(N=k\mid\text{data}) &=\int_0^\infty P(N=k\mid\lambda)f(\lambda\mid\text{data})\,d\lambda\\ &=\int_0^\infty \frac{e^{-h\lambda}(h\lambda)^k}{k!} \frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda\\ &=\frac{h^k\beta^\alpha}{k!\,\Gamma(\alpha)} \int_0^\infty \lambda^{\alpha+k-1}e^{-(\beta+h)\lambda}\,d\lambda. \end{aligned}

For $c>0$ and $r>0$, the Gamma integral is $\int_0^\infty x^{r-1}e^{-cx}\,dx=\Gamma(r)/c^r$. Applying it with $r=\alpha+k$ and $c=\beta+h$ gives the predictive probability mass function:

P(N=k\mid\text{data}) = \frac{\Gamma(\alpha+k)}{\Gamma(\alpha)\,k!} \left(\frac{\beta}{\beta+h}\right)^\alpha \left(\frac{h}{\beta+h}\right)^k, \qquad k=0,1,2,\ldots

This is a negative binomial probability mass function with shape parameter $\alpha$ and probability parameter $p=\beta/(\beta+h)$, using the convention that the random variable counts failures before $\alpha$ successes when $\alpha$ is an integer. The formula remains valid for any positive real $\alpha$, using the Gamma function. The complementary parameter is $1-p=h/(\beta+h)$. In particular, $P(N=0\mid\text{data})=p^\alpha$.

The posterior predictive mean is $h\alpha/\beta$. One way to check it is to take the conditional mean $E[N\mid\lambda]=h\lambda$ and average over the posterior, whose mean rate is $\alpha/\beta$. This mean is a useful check, but it does not make the predictive distribution Poisson: the distribution accounts for uncertainty in $\lambda$.

Worked Forecasts

Worked Example: One-Year Forecast After an Observed Claim

A hypothetical insured's annual claim rate $\lambda$ has prior distribution $\operatorname{Gamma}(1,2)$, with shape 1 and rate 2 per year. The insured produces 1 claim over 1 year of exposure. Forecast the probability of 0, 1, or 2 claims during the next year, and the probability of at most 2 claims. Assume a constant rate conditional on $\lambda$, and conditionally Poisson claim counts.

Method and posterior. With a Gamma shape-rate prior and a Poisson count likelihood, $n$ observed claims over exposure $e$ update the parameters to shape $a+n$ and rate $b+e$. Here $a=1$, $b=2$, $n=1$, and $e=1$, so the posterior is $\operatorname{Gamma}(2,3)$. The rate is per year, and the future horizon is $h=1$ year.

Thus $p=\beta/(\beta+h)=3/(3+1)=3/4$ and $1-p=1/4$. For shape $\alpha=2$, the coefficient in the mass function is $\Gamma(2+k)/(\Gamma(2)k!)=k+1$.

For zero claims, $$ P(N=0)=\left(\frac{3}{4}\right)^2=\frac{9}{16}=0.5625. $$

For one claim, $$ P(N=1)=2\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)=\frac{18}{64}=0.28125. $$

For two claims, $$ P(N=2)=3\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^2=\frac{27}{256}=0.10546875. $$

Therefore $P(N\leq 2)=0.5625+0.28125+0.10546875=0.94921875$. Rounded to four decimal places, the probabilities are $0.5625$, $0.2813$, $0.1055$, and $P(N\leq2)=0.9492$. The check is that the cumulative probability is less than 1, leaving $1-0.94921875=0.05078125$ for three or more claims. The common mistake avoided here is using the prior parameters after observing experience instead of updating them first.

Worked Example: A Six-Month Forecast With Annual Rate Units

A hypothetical commercial vehicle risk has posterior annual claim rate $\lambda\sim\operatorname{Gamma}(2,4)$, with shape 2 and rate 4 per year. Forecast the probability of at least 2 claims in the next six months. Assume the same conditional Poisson model and that six months represents $h=0.5$ years of exposure.

Method. Use the predictive negative binomial formula with $\alpha=2$, $\beta=4$, and $h=0.5$. Then $p=4/(4+0.5)=8/9$ and $1-p=1/9$.

First calculate the probabilities below 2: $$ P(N=0)=\left(\frac{8}{9}\right)^2=\frac{64}{81}=\frac{576}{729}=0.790123\ldots, $$ $$ P(N=1)=2\left(\frac{8}{9}\right)^2\left(\frac{1}{9}\right) =\frac{128}{729}=0.175583\ldots. $$

The requested probability is the complement: $$ P(N\geq2)=1-P(N=0)-P(N=1) =1-\frac{576}{729}-\frac{128}{729} =\frac{25}{729}=0.034293\ldots. $$

Rounded to four decimal places, the probability is $0.0343$. The common mistake avoided is treating six months as one full year: doing so would use $h=1$ and forecast for twice the requested exposure. The rate is annual, so the six-month horizon must be entered as 0.5 years.

Worked Example: A Fractional Shape Parameter

For a hypothetical small artisan business, the prior annual rate is $\operatorname{Gamma}(1.5,2)$, with shape 1.5 and rate 2 per year. The business has 0 claims over 1 year of exposure. Find the probability of at most 2 claims over the next 2 years. Assume a constant conditional Poisson rate.

Method and posterior. Updating the Gamma shape-rate prior gives posterior shape $1.5+0=1.5$ and rate $2+1=3$. Thus the posterior is $\operatorname{Gamma}(1.5,3)$. The future exposure is $h=2$ years, so $p=3/(3+2)=3/5$ and $1-p=2/5$. This non-integer shape is allowed; the Gamma function supplies the coefficients.

For $k=0$, $$ P(N=0)=\left(\frac{3}{5}\right)^{1.5}=0.464758\ldots. $$

For $k=1$, the coefficient is $\Gamma(2.5)/\Gamma(1.5)=1.5$, so $$ P(N=1)=1.5\left(\frac{3}{5}\right)^{1.5}\left(\frac{2}{5}\right) =0.278855\ldots. $$

For $k=2$, the coefficient is $\Gamma(3.5)/(\Gamma(1.5)2!)=1.5(2.5)/2=1.875$, so $$ P(N=2)=1.875\left(\frac{3}{5}\right)^{1.5}\left(\frac{2}{5}\right)^2 =0.139427\ldots. $$

Adding the three probabilities gives $P(N\leq2)=0.464758\ldots+0.278855\ldots+0.139427\ldots=0.883040\ldots$. Rounded to four decimal places, $P(N\leq2)=0.8830$. The common mistake avoided is assuming the negative binomial shape must be a whole number; the formula is valid for every $\alpha>0$.

Computing More Probabilities Reliably

When several consecutive probabilities are needed, the mass function also gives a recurrence. For $k\geq0$, divide the probability at $k+1$ by the probability at $k$. The common powers of $p$ cancel, and the ratio of Gamma functions is $\Gamma(\alpha+k+1)/\Gamma(\alpha+k)=\alpha+k$. Therefore:

\frac{P(N=k+1\mid\text{data})}{P(N=k\mid\text{data})} = \frac{\alpha+k}{k+1}\frac{h}{\beta+h}.

This recurrence is useful for checking calculations and generating a sequence of probabilities once $P(N=0)$ is known. It also makes clear that each step uses the same forecast horizon and posterior parameters; the count index changes, but the model does not.

Confirm the posterior convention.
Write the posterior as Gamma shape $\alpha$, rate $\beta$. Do not mistake the rate $\beta$ for the scale $1/\beta$.

Express the horizon in matching exposure units.
If $\lambda$ is claims per year, enter a half-year forecast as $h=0.5$, not $h=6$ or $h=1$.

Calculate the two negative binomial parameters.
Set $p=\beta/(\beta+h)$ and $1-p=h/(\beta+h)$.

Evaluate the requested count probabilities.
Use the Gamma-function mass formula or its recurrence, then sum or take a complement for cumulative or threshold probabilities.

Common Exam Trap: Plugging In the Posterior Mean

A frequent error is to replace the uncertain rate by its posterior mean and then use a single Poisson distribution. That calculates a conditional forecast at one selected rate, not the posterior predictive distribution. For example, with posterior $\operatorname{Gamma}(2,3)$ and a one-year horizon, the mean rate is $2/3$. A Poisson plug-in calculation gives $P(N=0)=e^{-2/3}=0.5134$, whereas integrating over the posterior gives $P(N=0)=(3/4)^2=0.5625$. The two answers differ because the predictive calculation averages over all plausible rates, rather than treating the posterior mean as known without uncertainty.

A second check is parameterization: a Gamma shape-scale prior must be converted before using the shape-rate formulas here. If the scale is $\theta$, then the equivalent rate is $\beta=1/\theta$. Applying the scale directly as $\beta$ changes the forecast probabilities.

Key takeaway: With posterior rate $\operatorname{Gamma}(\alpha,\beta)$ in shape-rate form and future exposure $h$, the forecast count has a negative binomial distribution with $P(N=k)=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)k!}\left(\frac{\beta}{\beta+h}\right)^\alpha\left(\frac{h}{\beta+h}\right)^k$. Its use depends on the conditional Poisson model and consistent exposure units.

Check Your Understanding

For each problem, use a Gamma shape-rate posterior and a conditionally Poisson future count. Give probabilities to four decimal places.

A risk has posterior $\operatorname{Gamma}(2,3)$, with rate per year. For a 2-year forecast, find $P(N=0)$ and $P(N=1)$. (Answer: $P(N=0)=0.3600$, $P(N=1)=0.2880$)
A risk has posterior $\operatorname{Gamma}(3,6)$, with rate per year. Find the probability of at least one claim in the next year. (Answer: $1-(6/7)^3=0.3703$)
A risk has posterior $\operatorname{Gamma}(1,2)$, with rate per year. Find the probability of exactly 2 claims over the next 18 months. (Answer: $(4/7)(3/7)^2=0.1050$)
A risk has posterior $\operatorname{Gamma}(0.5,1)$, with rate per year. Find the probability of at most 1 claim over the next year. (Answer: $0.5^{0.5}+0.5(0.5^{0.5})(0.5)=0.8839$)
A risk has posterior $\operatorname{Gamma}(4,2)$, with rate per year. Find the probability of at most 1 claim over the next 3 years. (Answer: $(2/5)^4[1+4(3/5)]=0.0870$)

What You'll Learn