Tweedie Moments From Compound Components

Why Component Moments Matter

A compound loss combines two sources of uncertainty: how many claims occur and how large each claim is. For ratemaking, its expected aggregate loss is not enough to describe its variability. The count distribution and severity distribution together determine the aggregate variance, and that variance can be written in the Tweedie form $\operatorname{Var}(S)=\phi\mu^p$ for a compound Poisson-Gamma model.

The connection is useful because it gives an interpretation to the Tweedie power: it is linked to the shape of the Gamma claim severity. The derivation also makes clear a qualification that is easy to miss. A fixed Poisson rate and a fixed severity distribution do not, by themselves, give a general power variance function as the mean changes. To describe a Tweedie family across different means, the component parameters must vary according to the Tweedie parameterization.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$ be the claim count, and let $Z_1,Z_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}Z_j$, with $S=0$ when $N=0$. For the compound Poisson-Gamma model, assume $Z_j\sim\operatorname{Gamma}(\alpha,\theta)$, using shape $\alpha>0$ and scale $\theta>0$. Thus $E[Z_j]=\alpha\theta$ and $\operatorname{Var}(Z_j)=\alpha\theta^2$.

Here $\lambda$ is the expected claim count for the period being modeled. If the period or exposure changes, the count parameter must be defined for that period. All component distributions and independence assumptions above apply to the aggregate whose moments are being calculated.

Deriving the Aggregate Mean and Variance

The count affects the aggregate mean directly: conditional on $N=n$, the aggregate is a sum of $n$ severities. For the variance, it is important to account for both the variation in severity at a given count and the variation in the count itself.

E[S\mid N]=N E[Z],\qquad \operatorname{Var}(S\mid N)=N\operatorname{Var}(Z)

Taking expectations gives $E[S]=E[N]E[Z]=\lambda E[Z]$. For the variance, the law of total variance gives $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. Because a Poisson variable has both mean and variance $\lambda$, the two terms are $\lambda\operatorname{Var}(Z)$ and $\lambda(E[Z])^2$, respectively.

E[S]=\lambda E[Z],\qquad \operatorname{Var}(S)=\lambda\left(\operatorname{Var}(Z)+(E[Z])^2\right) =\lambda E[Z^2]

For Gamma severity, $E[Z]=\alpha\theta$, $\operatorname{Var}(Z)=\alpha\theta^2$, and therefore $E[Z^2]=\alpha(\alpha+1)\theta^2$. Substitution yields the component-moment formulas.

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

Derivation: The term $\lambda\operatorname{Var}(Z)$ is the expected conditional severity variance. The term $\lambda(E[Z])^2$ is the variance of the conditional mean $\lambda$-count sum, because $\operatorname{Var}(N)=\lambda$. Their sum is $\lambda E[Z^2]$, not merely $\lambda\operatorname{Var}(Z)$. The additional squared-mean term is the contribution of random claim count.

Worked Example: Aggregate Moments From Gamma Severity

A hypothetical coverage has a claim count $N\sim\operatorname{Poisson}(3)$ for the policy period. Claim severities are independent of the count and independently distributed as $\operatorname{Gamma}(2,500)$, where the parameters are shape and scale in dollars. Calculate the aggregate mean, variance, and standard deviation.

Method. First calculate the severity moments, then substitute them into the compound Poisson formulas. The mean severity is $E[Z]=\alpha\theta=2(500)=\$1{,}000$. The severity variance is $\operatorname{Var}(Z)=\alpha\theta^2=2(500^2)=\$500{,}000$.

The second moment is $E[Z^2]=\operatorname{Var}(Z)+(E[Z])^2 =500{,}000+1{,}000{,}000=\$1{,}500{,}000$, where the units are dollars squared.

The aggregate mean is $E[S]=3(1{,}000)=\$3{,}000$. The aggregate variance is $\operatorname{Var}(S)=3(1{,}500{,}000)=\$4{,}500{,}000$, and the standard deviation is $\sqrt{4{,}500{,}000}=\$2{,}121.32$, rounded to the nearest cent.

Check. The variance is also $3(500{,}000)+3(1{,}000)^2=1{,}500{,}000+3{,}000{,}000=\$4{,}500{,}000$. The common mistake is to use only the first term and omit the variance from the random count.

Connecting the Moments to Tweedie Power

A Tweedie variance function has the form $\operatorname{Var}(S)=\phi\mu^p$, where $\mu=E[S]$, $\phi>0$ is the dispersion parameter, and $p$ is the variance power. For a compound Poisson-Gamma Tweedie model, $1<p<2$. The component parameters can be expressed in terms of $\mu,\phi,p$ as follows.

Theorem: Under the compound Poisson-Gamma Tweedie parameterization, the Gamma shape, Gamma scale, and Poisson mean are $\alpha=(2-p)/(p-1)$, $\theta=\phi(p-1)\mu^{p-1}$, and $\lambda=\mu^{2-p}/\{\phi(2-p)\}$, respectively. These relations assume $\mu>0$, $\phi>0$, and $1<p<2$.

The mapping follows directly from the component moments. With these parameters, the aggregate mean is $\lambda\alpha\theta$. Substituting gives $\{\mu^{2-p}/[\phi(2-p)]\}\{(2-p)/(p-1)\}\{\phi(p-1)\mu^{p-1}\}=\mu$. For the variance, use $E[Z^2]=\alpha(\alpha+1)\theta^2$. Since $\alpha+1=1/(p-1)$, the second moment is $\phi^2(2-p)\mu^{2p-2}$. Multiplying by $\lambda$ gives $\operatorname{Var}(S)=\phi\mu^p$.

\alpha=\frac{2-p}{p-1} \quad\Longleftrightarrow\quad p=\frac{\alpha+2}{\alpha+1}

This relationship shows why the Tweedie power is tied to severity shape: a given Gamma shape determines $p$. The scale and Poisson mean then adjust with $\mu$ and $\phi$ to produce the required aggregate mean and variance.

Worked Example: Recovering the Tweedie Power From Severity Shape

For a hypothetical compound Poisson-Gamma aggregate, the severity distribution has shape $\alpha=2$ and scale $\theta=2$ in thousands of dollars. The Poisson claim count has mean $\lambda=2$. Calculate the aggregate mean and variance, then identify the Tweedie power and dispersion.

Method. The Gamma severity mean is $E[Z]=2(2)=4$ thousand dollars. Its variance is $2(2^2)=8$ thousand-dollars squared, so its second moment is $8+4^2=24$ thousand-dollars squared.

The aggregate mean is $2(4)=8$ thousand dollars. Its variance is $2(24)=48$ thousand-dollars squared. The power corresponding to $\alpha=2$ is $p=(\alpha+2)/(\alpha+1)=4/3$.

To find the dispersion, use $\operatorname{Var}(S)=\phi\mu^p$: $\phi=48/8^{4/3}=48/16=3$. Thus the aggregate moments are $\mu=8$, $p=4/3$, and $\phi=3$, with monetary units as specified.

Check. The component mapping gives $\theta=3(1/3)8^{1/3}=2$ and $\lambda=8^{2/3}/[3(2/3)]=2$, recovering the stated severity scale and count mean. The trap is to report $p=1/\alpha$; the correct relation is $p=(\alpha+2)/(\alpha+1)$.

Worked Example: Constructing Components From Tweedie Parameters

A hypothetical aggregate loss, measured in thousands of dollars, is specified by a Tweedie mean $\mu=100$, dispersion $\phi=0.02$, and power $p=1.5$. Find its compound Poisson-Gamma parameters and verify its mean and variance.

Method. The Gamma shape is $\alpha=(2-1.5)/(1.5-1)=0.5/0.5=1$. The Gamma scale is $\theta=0.02(0.5)(100^{0.5})=0.01(10)=0.1$ thousand dollars. The Poisson mean is $\lambda=100^{0.5}/[0.02(0.5)]=10/0.01=1{,}000$.

A severity therefore has mean $\alpha\theta=0.1$ thousand dollars and second moment $\alpha(\alpha+1)\theta^2=1(2)(0.1^2)=0.02$ thousand-dollars squared. The aggregate mean is $1{,}000(0.1)=100$ thousand dollars. Its variance is $1{,}000(0.02)=20$ thousand-dollars squared.

Check. The Tweedie formula gives $\phi\mu^p=0.02(100^{1.5})=0.02(1{,}000)=20$, as required. The large expected count paired with small severities is consistent with the specified aggregate mean; do not mistake the severity scale for the aggregate mean.

What the Power Does—and Does Not—Say

For one fixed compound Poisson model, $\lambda$, $\alpha$, and $\theta$ are fixed. In that model, the aggregate moments are fixed as well. If the count rate alone changes while the severity distribution stays fixed, then $E[S]=\lambda E[Z]$ and $\operatorname{Var}(S)=\lambda E[Z^2]$, so variance is proportional to mean. This is a linear relationship, corresponding to power $p=1$, not a general compound Poisson-Gamma Tweedie variance function with $1<p<2$.

The Tweedie mapping instead describes a family of distributions indexed by $\mu$: as the mean changes, both the Gamma scale and Poisson mean change according to the formulas above, while the power fixes the Gamma shape. This distinction prevents an incorrect conclusion that any fixed frequency-severity model automatically has variance proportional to an arbitrary power of its mean.

Key takeaway: For a compound Poisson aggregate, $E[S]=\lambda E[Z]$ and $\operatorname{Var}(S)=\lambda E[Z^2]$. In the compound Poisson-Gamma Tweedie parameterization, the Gamma shape and Tweedie power satisfy $p=(\alpha+2)/(\alpha+1)$, and the component parameters vary with the aggregate mean and dispersion.

Check Your Understanding

For each problem, state the component moments or parameter conversion needed before giving the requested aggregate result.

A hypothetical risk has $N\sim\operatorname{Poisson}(5)$ and independent severities $Z\sim\operatorname{Gamma}(2,10)$, with scale in dollars. Find the aggregate mean and variance. (Answer: mean $\$100$; variance $\$3{,}000$)
A hypothetical policy period has $N\sim\operatorname{Poisson}(3)$ and severity $Z\sim\operatorname{Gamma}(4,5)$, with scale in dollars. Find the aggregate mean and variance. (Answer: mean $\$60$; variance $\$1{,}500$)
A compound Poisson-Gamma model has Gamma severity shape $\alpha=5$. What Tweedie power corresponds to that shape? (Answer: $p=7/6$)
A Tweedie aggregate has $\mu=16$, $\phi=0.5$, and $p=1.5$. Find its Poisson mean and Gamma shape and scale, then verify the aggregate variance. (Answer: $\lambda=16$, $\alpha=1$, $\theta=1$; variance $32$)
A hypothetical three-unit exposure has a Poisson claim rate of $2$ per unit. Severity is $\operatorname{Gamma}(2,4)$, with scale in thousands of dollars. Find the aggregate mean and variance over the full exposure. (Answer: mean $48$ thousand dollars; variance $576$ thousand-dollars squared)

What You'll Learn