Compound Poisson-Gamma Construction

From Claim Frequency and Severity to Aggregate Loss

An aggregate pure premium model describes the total loss from a policy or risk over a specified period, including the possibility of no claims. One way to construct that distribution is to combine a model for the number of claims with a model for each claim’s severity. When the count is Poisson and severities are independent Gamma random variables, the resulting aggregate distribution is a compound Poisson-Gamma model. This construction is also the probabilistic form of a Tweedie model with variance power strictly between one and two.

The construction is useful because it makes the sources of aggregate risk explicit: the claim count varies, and each claim amount varies. It also provides a single aggregate-loss model with a point mass at zero and positive losses otherwise. The result depends on the assumptions below; it is not a claim that every portfolio’s count and severity distributions have these forms.

Definition: For one fixed observation period, let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$. Let $Z_1,Z_2,\ldots$ be independent and identically distributed claim severities, independent of $N$, with $Z_j\sim\operatorname{Gamma}(\alpha,\theta)$ in the shape-scale convention: $E[Z_j]=\alpha\theta$ and $\operatorname{Var}(Z_j)=\alpha\theta^2$, where $\alpha>0$ and $\theta>0$. Define aggregate loss by $S=\sum_{j=1}^{N}Z_j$, with $S=0$ when $N=0$. This is a compound Poisson-Gamma model.

The period and monetary units must be consistent throughout. For example, if $Z_j$ is measured in thousands of dollars, then $S$ is also in thousands of dollars and $\operatorname{Var}(S)$ is in squared thousands of dollars. The parameters $\lambda$, $\alpha$, and $\theta$ are not interchangeable: $\lambda$ controls the expected claim count, $\alpha$ controls Gamma shape, and $\theta$ is the Gamma scale.

Aggregate Mean, Variance, and Transform

Conditioning on the number of claims gives the aggregate moments. Given $N=n$, the aggregate is a sum of $n$ independent severities, so $E[S\mid N=n]=n\alpha\theta$ and $\operatorname{Var}(S\mid N=n)=n\alpha\theta^2$. Applying the laws of total expectation and total variance yields

E[S]=\lambda\alpha\theta,\qquad \operatorname{Var}(S)=\lambda\alpha(\alpha+1)\theta^2.

For the variance, the total-variance calculation is

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])\\ &=E[N]\alpha\theta^2+\operatorname{Var}(N)(\alpha\theta)^2\\ &=\lambda\alpha\theta^2+\lambda\alpha^2\theta^2\\ &=\lambda\alpha(\alpha+1)\theta^2. \end{aligned}

The first variance contribution is the variation in severities conditional on the count; the second is the variation in the count itself. Omitting either contribution understates aggregate variance.

A transform also follows directly by conditioning. For values of $t$ for which the Gamma moment-generating function exists, $M_Z(t)=(1-\theta t)^{-\alpha}$, with $t<1/\theta$. Given $N=n$, independence gives $E[e^{tS}\mid N=n]=M_Z(t)^n$. Averaging over the Poisson count,

\begin{aligned} M_S(t) &=E[M_Z(t)^N]\\ &=\exp\{\lambda(M_Z(t)-1)\}\\ &=\exp\left\{\lambda\left[(1-\theta t)^{-\alpha}-1\right]\right\}, \qquad t<1/\theta. \end{aligned}

This derivation shows why the aggregate is compound: its transform is obtained by inserting the severity transform into the Poisson count’s probability-generating function. In general, the positive aggregate $S\mid S>0$ is not itself Gamma; it can contain one, two, or more Gamma severities added together.

Why the Tweedie Variance Power Lies Between One and Two

A Tweedie model in the relevant range is an exponential dispersion model with mean $\mu>0$, dispersion parameter $\phi>0$, and variance function $\operatorname{Var}(S)=\phi\mu^p$, where $1<p<2$. To connect it to the compound construction, write $\mu=E[S]=\lambda\alpha\theta$ and use the compound-model variance. Dividing variance by mean gives

\frac{\operatorname{Var}(S)}{\mu} =\frac{\lambda\alpha(\alpha+1)\theta^2}{\lambda\alpha\theta} =(\alpha+1)\theta.

If the variance is represented as $\phi\mu^p$, then $\operatorname{Var}(S)/\mu=\phi\mu^{p-1}$. Matching these expressions and the compound-model parameters gives the following standard conversion:

Theorem: A compound Poisson-Gamma model with shape $\alpha>0$, scale $\theta>0$, and mean $\mu=\lambda\alpha\theta$ has Tweedie variance power and dispersion $p=(\alpha+2)/(\alpha+1)$ and $\phi=(\alpha+1)\theta\mu^{-1/(\alpha+1)}$. Equivalently, given Tweedie parameters $\mu,\phi,p$ with $1<p<2$, its compound parameters are $\alpha=(2-p)/(p-1)$, $\theta=\phi(p-1)\mu^{p-1}$, and $\lambda=\mu^{2-p}/[\phi(2-p)]$.

For the first direction, the definition of $p$ implies $p-1=1/(\alpha+1)$. Thus $\phi\mu^{p-1}=(\alpha+1)\theta$, which reproduces the variance-to-mean ratio and hence $\operatorname{Var}(S)=\phi\mu^p$. For the reverse direction, substitute $\alpha=(2-p)/(p-1)$ and the stated $\theta$ and $\lambda$ into the compound mean:

\lambda\alpha\theta = \frac{\mu^{2-p}}{\phi(2-p)} \frac{2-p}{p-1} \phi(p-1)\mu^{p-1} =\mu.

The compound variance then equals $\phi\mu^p$ by the same variance-to-mean identity. Finally, $\alpha>0$ implies $p-1=1/(\alpha+1)$ is positive and less than one, so $1<p<2$. Conversely, every $p$ strictly between one and two gives a positive Gamma shape. The conversion assumes the compound Poisson and Gamma specification above; it does not establish that this model is appropriate for every loss portfolio.

Identify the parameterization.
Check whether the given Gamma parameters are shape and scale, and keep the observation period and loss units fixed.

Calculate the aggregate mean.
Use $\mu=\lambda\alpha\theta$, not the mean severity alone.

Convert shape to variance power.
Use $p=(\alpha+2)/(\alpha+1)$; then verify that $1<p<2$.

Find or verify dispersion.
Use $\phi=(\alpha+1)\theta\mu^{-(p-1)}$, or check that $\phi\mu^p$ equals the compound variance.

Worked Conversions

Worked Example: Constructing the Aggregate and Tweedie Parameters

A hypothetical class of risks has a Poisson claim count with $\lambda=4$ claims per period. Each claim severity is Gamma with shape $\alpha=2$ and scale $\theta=1$ thousand dollars. Find the aggregate mean and variance, then identify its Tweedie parameters and probability of zero aggregate loss.

Method and calculation. The severity mean is $\alpha\theta=2(1)=2$ thousand dollars. Therefore the aggregate mean is $\mu=\lambda\alpha\theta=4(2)(1)=8$ thousand dollars.

The aggregate variance is $\lambda\alpha(\alpha+1)\theta^2=4(2)(3)(1^2)=24$ squared thousands of dollars. The Tweedie power is $p=(\alpha+2)/(\alpha+1)=(2+2)/(2+1)=4/3$.

The dispersion is $\phi=(\alpha+1)\theta\mu^{-1/(\alpha+1)} =3(1)8^{-1/3}=3/2=1.5$. Checking the variance-function form gives $\phi\mu^p=1.5(8^{4/3})=1.5(16)=24$, as required.

The aggregate is zero exactly when $N=0$, so $P(S=0)=e^{-4}=0.0183$ to four decimals. This check confirms the mass at zero is inherited from the count model. The method avoids treating the aggregate as one Gamma claim: its positive values can be sums of several severities.

Worked Example: Computing the Power and Dispersion From Frequency and Severity

For a hypothetical risk group, the annual claim count is Poisson with $\lambda=5$. Individual severities are Gamma with shape $\alpha=3$ and scale $\theta=0.4$ thousand dollars. Calculate the aggregate mean and variance, Tweedie power and dispersion, and probability of no annual loss.

Method and calculation. The expected severity is $3(0.4)=1.2$ thousand dollars. Thus $\mu=5(3)(0.4)=6$ thousand dollars.

The aggregate variance is $5(3)(3+1)(0.4)^2=5(3)(4)(0.16)=9.6$ squared thousands of dollars. The power is $p=(3+2)/(3+1)=5/4=1.25$, which lies between one and two.

For dispersion, $\phi=(3+1)(0.4)6^{-1/(3+1)}=1.6/6^{1/4}$. Since $6^{1/4}\approx1.5651$, $\phi\approx1.0223$. The verification is $\phi\mu^p=(1.6/6^{1/4})6^{5/4}=1.6(6)=9.6$, matching the compound variance. The zero-loss probability is $e^{-5}=0.0067$ to four decimals.

This calculation uses scale $0.4$ thousand dollars, not a Gamma rate of $0.4$. Confusing scale with rate would change the severity mean and all subsequent aggregate quantities.

Worked Example: Recovering Frequency and Severity From Tweedie Parameters

A hypothetical aggregate-loss model for one period has Tweedie mean $\mu=4$ thousand dollars, dispersion $\phi=0.5$, and power $p=1.5$. Recover its compound Poisson-Gamma parameters and verify the aggregate moments.

Method and calculation. First calculate the Gamma shape: $\alpha=(2-p)/(p-1)=(2-1.5)/(1.5-1)=0.5/0.5=1$.

The Gamma scale is $\theta=\phi(p-1)\mu^{p-1}=0.5(0.5)4^{0.5}=0.25(2)=0.5$ thousand dollars. The Poisson mean is $\lambda=\mu^{2-p}/[\phi(2-p)]=4^{0.5}/[0.5(0.5)]=2/0.25=8$.

Check the aggregate mean: $\lambda\alpha\theta=8(1)(0.5)=4$ thousand dollars. Check the variance: $\lambda\alpha(\alpha+1)\theta^2=8(1)(2)(0.5^2)=4$ squared thousands of dollars. The Tweedie expression gives $\phi\mu^p=0.5(4^{1.5})=0.5(8)=4$, agreeing with the compound calculation. This reverse conversion avoids using $p$ itself as the Gamma shape; here the correct shape is one.

Zero Loss, Interpretation, and Common Traps

Because severities are positive, $S=0$ if and only if $N=0$, giving $P(S=0)=e^{-\lambda}$. This is a point mass, not part of a positive continuous density. For $s>0$, the aggregate distribution combines the possibilities of one or more claims. In particular, conditional on at least one claim, the aggregate is a mixture over the possible positive claim counts; it is generally incorrect to assign it the original single-claim Gamma distribution.

A second useful check is dimensional. The mean aggregate loss has the same money unit as $\theta$, while the aggregate variance has its square. The shape $\alpha$, power $p$, and count rate $\lambda$ are dimensionless for a fixed period. The dispersion’s units must make $\phi\mu^p$ a variance. A numerically correct calculation with inconsistent money units or time periods is still an incorrect model specification.

Using the wrong Gamma convention. In shape-scale form, the severity mean is $\alpha\theta$. If a Gamma distribution is instead specified by shape and rate, its mean is shape divided by rate; do not insert a rate into a scale formula.
Dropping frequency variation. The aggregate variance includes both conditional severity variance and variance from the random count.
Confusing severity and aggregate means. The mean severity is $\alpha\theta$; the aggregate mean is $\lambda\alpha\theta$.
Using the wrong shape-power conversion. The relationship is $p=(\alpha+2)/(\alpha+1)$, not $p=\alpha$ or $p=1+1/\alpha$.

The compound construction gives a direct interpretation of the Tweedie range: small Gamma shape corresponds to a power closer to two, while increasing shape moves the power toward one. The model’s appeal is that count variation, severity variation, and the zero-loss probability are represented within one aggregate distribution. Its assumptions should nevertheless be assessed for the application, including whether the count is suitably Poisson, severities are adequately Gamma, and count and severity are independent.

Key takeaway: Independent Gamma severities with a Poisson claim count produce a compound Poisson-Gamma aggregate with mean $\lambda\alpha\theta$, variance $\lambda\alpha(\alpha+1)\theta^2$, and Tweedie power $p=(\alpha+2)/(\alpha+1)$ strictly between one and two. The zero-loss probability is $e^{-\lambda}$.

Check Your Understanding

For each problem, use shape-scale Gamma notation and keep the stated period and units consistent.

A hypothetical portfolio has $\lambda=2$ and Gamma severity parameters $\alpha=4$, $\theta=\$250$. Find the aggregate mean, variance, Tweedie power, and probability of zero loss. (Answer: mean $\$2{,}000$; variance $\$2{,}500{,}000$; $p=1.2$; $P(S=0)=0.1353$.)
A risk has Poisson frequency mean $6$ and severity $Z\sim\operatorname{Gamma}(2,0.25)$, with severity measured in thousands of dollars. Find the aggregate mean, variance, and Tweedie power. (Answer: mean $3$ thousand dollars; variance $2.25$ squared thousands of dollars; $p=4/3$.)
A Tweedie compound Poisson-Gamma model has $p=1.25$. What is the corresponding Gamma shape? (Answer: $\alpha=(2-1.25)/(1.25-1)=3$.)
A Tweedie model has $\mu=9$ thousand dollars, $\phi=0.2$, and $p=1.5$. Find the compound Poisson mean and Gamma shape and scale. (Answer: $\lambda=30$, $\alpha=1$, and $\theta=0.3$ thousand dollars.)
A hypothetical risk has $\lambda=1.5$. Find the probabilities of zero aggregate loss, positive aggregate loss, exactly one claim, and at least two claims. (Answer: $P(S=0)=0.2231$; $P(S>0)=0.7769$; $P(N=1)=0.3347$; $P(N\ge2)=0.4422$.)

What You'll Learn