The Gamma-Poisson Mixture Derivation

Why Mix a Poisson Rate?

A Poisson model assigns one fixed frequency rate to every risk with the same exposure. In practice, risks may differ in ways the model does not observe: for example, a hypothetical portfolio of small delivery vehicles may include routes with different traffic conditions. If each vehicle has its own rate, but those rates vary across the portfolio, a single fixed-rate Poisson model can understate the variation in claim counts.

A gamma-Poisson mixture represents this unobserved rate variation directly. Conditional on a risk's rate, its claim count is Poisson. Across risks, the rate is gamma-distributed. Averaging the conditional Poisson probabilities over the gamma distribution produces a negative binomial distribution for the count.

Definition: Let $\Lambda$ be a positive, unobserved claim frequency for a fixed exposure period. Suppose $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$, using the shape-rate convention, with density $$ f_\Lambda(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0, $$ where $\alpha>0$ and $\beta>0$. Conditional on $\Lambda=\lambda$, let $N\mid\Lambda=\lambda\sim\operatorname{Poisson}(\lambda)$. The unconditional distribution of $N$ is the gamma-Poisson mixture.

Here $\beta$ is a rate parameter, not a scale parameter. The gamma mean and variance are $E[\Lambda]=\alpha/\beta$ and $\operatorname{Var}(\Lambda)=\alpha/\beta^2$. The exposure period is fixed and absorbed into $\Lambda$, so the count $N$ and the rate parameters must refer to that same period. No interest or mortality assumptions enter this frequency model.

Deriving the Mixed Count Distribution

For any nonnegative integer $n$, the unconditional probability of $n$ claims is the average of the conditional Poisson probability over all possible values of $\Lambda$. The key is to multiply the conditional probability by the gamma density and integrate over positive rates.

Write the conditional probability.
Given $\Lambda=\lambda$, the Poisson model gives $P(N=n\mid\Lambda=\lambda)=e^{-\lambda}\lambda^n/n!$.

Average over the random rate.
Multiply that conditional probability by $f_\Lambda(\lambda)$, then integrate over $\lambda>0$.

Evaluate the gamma integral.
Combine the exponential terms and use the gamma integral with shape $\alpha+n$ and rate $\beta+1$.

\begin{aligned} P(N=n) &=\int_0^\infty P(N=n\mid\Lambda=\lambda)f_\Lambda(\lambda)\,d\lambda\\ &=\int_0^\infty \frac{e^{-\lambda}\lambda^n}{n!} \frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda\\ &=\frac{\beta^\alpha}{\Gamma(\alpha)n!} \int_0^\infty \lambda^{\alpha+n-1}e^{-(\beta+1)\lambda}\,d\lambda\\ &=\frac{\beta^\alpha}{\Gamma(\alpha)n!} \frac{\Gamma(\alpha+n)}{(\beta+1)^{\alpha+n}}\\ &=\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)n!} \left(\frac{\beta}{\beta+1}\right)^\alpha \left(\frac{1}{\beta+1}\right)^n,\qquad n=0,1,2,\ldots. \end{aligned}

The integral identity used in the fourth line is $\int_0^\infty \lambda^{c-1}e^{-r\lambda}\,d\lambda=\Gamma(c)/r^c$, for $c>0$ and $r>0$. In the present calculation, $c=\alpha+n>0$ and $r=\beta+1>0$, so the identity applies.

This probability function is the negative binomial form for the number of failures before $\alpha$ successes when $\alpha$ is a positive integer, with success probability $p=\beta/(\beta+1)$. The expression derived above also applies when $\alpha$ is any positive real number, through the gamma function. The exact probability function matters more here than the name of a parameterization: different conventions attach different meanings to “success probability” and “failure count.”

Key result: If $\Lambda\sim\operatorname{Gamma}(\alpha,\beta)$ in the shape-rate convention and $N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)$, then $$ P(N=n)=\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)n!} \left(\frac{\beta}{\beta+1}\right)^\alpha \left(\frac{1}{\beta+1}\right)^n,\qquad n=0,1,2,\ldots. $$ The count distribution is negative binomial, with $E[N]=\alpha/\beta$ and $\operatorname{Var}(N)=\alpha/\beta+\alpha/\beta^2$.

Why the Mixture Has Extra Variance

The mean and variance show precisely how rate variation affects the count. Conditional on $\Lambda$, a Poisson count has conditional mean $\Lambda$ and conditional variance $\Lambda$. The law of total expectation gives the unconditional mean:

E[N]=E\big[E[N\mid\Lambda]\big]=E[\Lambda]=\frac{\alpha}{\beta}.

For the variance, the law of total variance separates expected variation within a fixed rate from variation in the rate itself:

\begin{aligned} \operatorname{Var}(N) &=E\big[\operatorname{Var}(N\mid\Lambda)\big] +\operatorname{Var}\big(E[N\mid\Lambda]\big)\\ &=E[\Lambda]+\operatorname{Var}(\Lambda)\\ &=\frac{\alpha}{\beta}+\frac{\alpha}{\beta^2}. \end{aligned}

The first term, $\alpha/\beta$, is the average Poisson variation conditional on the rate. The second term, $\alpha/\beta^2$, is the additional variation caused by different risks having different rates. Since $\alpha,\beta>0$, this additional term is positive: the mixture variance exceeds its mean. If the rate were fixed at the same mean instead, the Poisson variance would equal $\alpha/\beta$.

This is a marginal statement about a population of risks. It does not say that a particular risk's conditional count is non-Poisson: given its rate, the model remains Poisson. The negative binomial distribution describes the count before that risk-specific rate is observed or conditioned upon.

Worked Examples

Worked Example: Calculate Probabilities and Moments From a Gamma Rate

A hypothetical book of one-period equipment policies is modeled by a latent claim rate $\Lambda\sim\operatorname{Gamma}(2,3)$, using shape-rate parameters. Conditional on $\Lambda$, the claim count $N$ is Poisson with mean $\Lambda$. Find $P(N=0)$, $P(N=1)$, the mean, and the variance. No interest or mortality assumptions are involved.

Method. Use the gamma-Poisson mixture probability function, then use the total expectation and total variance formulas. The parameters are $\alpha=2$ and $\beta=3$, so $\beta+1=4$.

For zero claims, $n=0$:

P(N=0)=\frac{\Gamma(2)}{\Gamma(2)0!} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^0 =\frac{9}{16}=0.5625.

For one claim, $n=1$:

P(N=1)=\frac{\Gamma(3)}{\Gamma(2)1!} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^1 =2\cdot\frac{9}{16}\cdot\frac{1}{4} =\frac{18}{64}=\frac{9}{32}=0.28125.

The mean is $E[N]=\alpha/\beta=2/3$. The variance is $\operatorname{Var}(N)=2/3+2/9=8/9$. Thus the mean is approximately $0.667$ claims and the variance is approximately $0.889$ claims squared.

Check. The two probabilities sum to $0.5625+0.28125=0.84375$, which is less than 1, as expected because counts of two or more remain possible. The variance exceeds the mean. The common mistake avoided here is treating $\beta=3$ as a scale and substituting it directly as the gamma mean; under the stated shape-rate convention, $E[\Lambda]=2/3$, not $2\cdot3$.

Worked Example: A Nonzero Claim Count With a Different Shape

For a hypothetical one-period warranty portfolio, suppose $\Lambda\sim\operatorname{Gamma}(3,1)$ and $N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)$. Calculate the probability of exactly two claims and the unconditional mean and variance. No interest or mortality assumptions apply.

Method. Apply the mixture probability function with shape $\alpha=3$, rate $\beta=1$, and count $n=2$. Here $\beta+1=2$.

\begin{aligned} P(N=2) &=\frac{\Gamma(5)}{\Gamma(3)2!} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2\\ &=\frac{24}{2\cdot2}\cdot\frac{1}{8}\cdot\frac{1}{4}\\ &=6\cdot\frac{1}{32} =\frac{3}{16} =0.1875. \end{aligned}

The mean is $E[N]=3/1=3$. The variance is $\operatorname{Var}(N)=3/1+3/1^2=6$.

Check. The probability is between 0 and 1, and the variance $6$ is greater than the mean $3$, consistent with rate mixing. The common mistake avoided is omitting the $2!$ in the count probability: the coefficient is $\Gamma(5)/(\Gamma(3)2!)=24/(2\cdot2)=6$, not 12.

Worked Example: Compare a Mixture With a Fixed-Rate Poisson Model

A hypothetical group of one-period risks has a latent rate $\Lambda\sim\operatorname{Gamma}(1,1)$, and $N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)$. Compare the probability of at least three claims under this mixture with that from a fixed-rate Poisson model having the same mean. No interest or mortality assumptions apply.

Mixture calculation. The mixture has $\alpha=1,\beta=1$, so its mean is $1$. For $n=0,1,2$, the probability function simplifies to $P(N=n)=(1/2)^{n+1}$, because $\Gamma(n+1)/(\Gamma(1)n!)=1$. Therefore,

P(N\ge3)=1-\left(\frac12+\frac14+\frac18\right)=\frac18=0.1250.

Fixed-rate Poisson comparison. A Poisson model with the same mean has rate $1$. Its probability of at least three claims is

P(N\ge3)=1-P(N=0)-P(N=1)-P(N=2) =1-e^{-1}\left(1+1+\frac{1}{2}\right) =1-\frac{2.5}{e} \approx 0.0803.

The mixture gives a higher probability of three or more claims: $0.1250$ compared with approximately $0.0803$. Both models have mean 1, but the mixture has variance $1+1=2$, while the fixed-rate Poisson variance is 1.

Check. The comparison uses the same mean, so the difference is not an exposure or average-frequency difference; it reflects rate variation. The common mistake avoided is comparing a gamma-Poisson mixture with a Poisson model at a different mean.

Interpretation and Exam Traps

The latent rate is a model for unobserved risk variation, not an extra claim added to the count. For each risk, a particular value of $\Lambda$ governs the conditional Poisson frequency. Mixing means that the analyst averages over the gamma distribution of possible rates. The resulting negative binomial probabilities assign relatively more variation to counts than a fixed-rate Poisson model with the same mean.

Check the gamma convention. This tutorial uses shape $\alpha$ and rate $\beta$, so $E[\Lambda]=\alpha/\beta$. A gamma distribution described using shape and scale has a different second parameter convention.
Do not substitute the average rate before mixing. A Poisson model with rate $E[\Lambda]$ has the correct mixture mean but generally not its probabilities or variance.
Keep conditional and unconditional statements separate. Given $\Lambda$, the count is Poisson. Without conditioning on $\Lambda$, the count has the negative binomial mixture distribution.
Match the period and the rate. The $\Lambda$ in the conditional Poisson model must be the expected count for the specified exposure period.

The derivation is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus because it connects conditional frequency assumptions to a marginal count model and makes the source of overdispersion explicit.

Check Your Understanding

For each problem, use the shape-rate convention and treat the stated exposure period as fixed. No interest or mortality assumptions are involved.

A hypothetical risk group has $\Lambda\sim\operatorname{Gamma}(2,4)$ and $N\mid\Lambda\sim\operatorname{Poisson}(\Lambda)$. Find $P(N=0)$. (Answer: $16/25=0.6400$)
Suppose $\Lambda\sim\operatorname{Gamma}(1,2)$ and the count conditional on the rate is Poisson. Find the unconditional probability of exactly one claim. (Answer: $2/9=0.2222$)
For $\Lambda\sim\operatorname{Gamma}(3,1)$ mixed with a conditional Poisson count, find the unconditional mean and variance. (Answer: mean $3$, variance $6$)
Let $\Lambda\sim\operatorname{Gamma}(2,2)$. Find the probability of exactly two claims in the mixed model. (Answer: $4/27\approx0.1481$)
Let $\Lambda\sim\operatorname{Gamma}(1,1)$. Compare the mixture probability of at least three claims with that of a fixed-rate Poisson model having the same mean. (Answer: mixture $1/8=0.1250$; fixed-rate Poisson $1-2.5/e\approx0.0803$)

What You'll Learn