Tweedie Exposure Scaling Derivation

Why Exposure Scaling Matters

A portfolio’s expected aggregate loss generally grows when the amount of insured exposure grows. Its variance and probability of no loss also change, but not in the same way as the expected loss. Keeping these quantities distinct is important when comparing policies or periods with different exposure amounts, and when interpreting a Tweedie model for aggregate losses.

This tutorial derives the scaling rules from a compound Poisson-Gamma model. The exposure multiplier $c$ is relative to one baseline exposure unit: for example, $c=3$ could represent three vehicle-years if one unit is a vehicle-year. The model assumes the same claim frequency rate and severity distribution apply across those units.

Definition: Let $N_c$ be the number of claims during exposure $c$, with $N_c\sim\operatorname{Poisson}(c\lambda)$, where $\lambda>0$ is the expected claim count per exposure unit. Let the claim severities $X_1,X_2,\ldots$ be independent and identically distributed Gamma random variables, independent of $N_c$, with shape $\alpha>0$ and scale $\theta>0$. Define aggregate loss by $S_c=\sum_{j=1}^{N_c}X_j$, with $S_c=0$ when $N_c=0$. The Gamma scale convention gives $E[X_j]=\alpha\theta$ and $\operatorname{Var}(X_j)=\alpha\theta^2$.

This model requires claims to occur according to a Poisson count with rate proportional to exposure, and requires positive, identically distributed severities that are independent of the count and of one another. It does not describe situations in which frequency or severity changes with exposure, claims are dependent, or a non-claim source of loss can produce a zero or positive aggregate. In particular, the zero-loss result below relies on each claim severity being strictly positive, as it is for a Gamma distribution.

Deriving the Aggregate Mean and Variance

Write $m_1=E[X_j]$ and $m_2=E[X_j^2]$. For Gamma severity with shape $\alpha$ and scale $\theta$,

m_1=\alpha\theta,\qquad m_2=\operatorname{Var}(X_j)+E[X_j]^2 =\alpha\theta^2+\alpha^2\theta^2 =\alpha(\alpha+1)\theta^2.

The aggregate mean follows by conditioning on the claim count. Given $N_c=n$, the total is a sum of $n$ severities, so its conditional mean is $nm_1$. Since $E[N_c]=c\lambda$, the unconditional mean is $c\lambda m_1$.

Result: Under the stated compound Poisson-Gamma assumptions, aggregate loss has mean $c\lambda\alpha\theta$ and variance $c\lambda\alpha(\alpha+1)\theta^2$. Both aggregate moments are proportional to exposure $c$.

For the variance, the law of total variance separates variation in severities from variation in the number of claims. Conditional on $N_c$, the variance is $N_c\operatorname{Var}(X_j)$; the conditional mean $N_cm_1$ also varies because $N_c$ varies.

\begin{aligned} \operatorname{Var}(S_c) &=E[\operatorname{Var}(S_c\mid N_c)] +\operatorname{Var}(E[S_c\mid N_c])\\ &=E[N_c]\operatorname{Var}(X_j) +\operatorname{Var}(N_c)m_1^2\\ &=c\lambda\operatorname{Var}(X_j)+c\lambda m_1^2\\ &=c\lambda\bigl(\operatorname{Var}(X_j)+m_1^2\bigr) =c\lambda m_2. \end{aligned}

The last equality uses $\operatorname{Var}(X_j)+E[X_j]^2=E[X_j^2]$. It also explains why aggregate variance uses the second moment of severity, not just severity variance: variation in the claim count contributes the additional $m_1^2$ term.

Because every severity is positive, $S_c=0$ exactly when $N_c=0$. The Poisson probability of no claims therefore gives the zero-loss probability directly.

P(S_c=0)=P(N_c=0)=e^{-c\lambda}.

Thus the aggregate mean and variance grow linearly with exposure, while the zero-loss probability declines exponentially with exposure. Doubling exposure doubles both aggregate moments, but it does not halve the zero-loss probability.

Aggregate Loss Versus Loss Per Unit of Exposure

The aggregate total $S_c$ and the average loss per exposure unit $S_c/c$ answer different questions. Under the model, the latter has an exposure-invariant mean and a variance that decreases as exposure increases.

E\left[\frac{S_c}{c}\right]=\lambda m_1, \qquad \operatorname{Var}\left(\frac{S_c}{c}\right) =\frac{\lambda m_2}{c}.

The mean is invariant because both the expected total and the amount of exposure are multiplied by $c$. The variance is divided by $c^2$ when a total is divided by $c$, while the total variance itself grows by $c$; the net effect is a $1/c$ decrease. This is a model-based averaging result, not a claim that the risk of an individual policy changes when its exposure is re-expressed.

Takeaway: For $c$ exposure units, $E[S_c]$ and $\operatorname{Var}(S_c)$ each scale by $c$. For the exposure-normalized amount $S_c/c$, the mean stays constant and the variance scales by $1/c$. The probability of zero aggregate loss is $e^{-c\lambda}$.

Connection to Tweedie Exposure Scaling

For the conventional compound Poisson-Gamma form of a Tweedie model, let $\mu$ denote mean loss per exposure unit, $\phi>0$ dispersion, and $p$ the variance power, with $1<p<2$. For aggregate loss over exposure $c$, the Tweedie mean and variance are $c\mu$ and $c\phi\mu^p$. These linear exposure relationships agree with the compound Poisson derivation.

In this parameterization, the Poisson rate per exposure unit and the Gamma severity parameters can be written as follows. The Gamma parameters use the shape-scale convention defined above.

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

Substitution gives $\lambda\alpha\theta=\mu$ and $\lambda\alpha(\alpha+1)\theta^2=\phi\mu^p$, so the compound model yields $E[S_c]=c\mu$ and $\operatorname{Var}(S_c)=c\phi\mu^p$. Its zero-loss probability is

P(S_c=0)=\exp\left(-c\frac{\mu^{2-p}}{\phi(2-p)}\right).

This relation makes clear that the zero probability depends on the model’s implied Poisson rate, not on the aggregate mean alone. Two models with the same mean can have different zero probabilities if their dispersion or variance power differs.

Worked Applications

Worked Example: Scaling a Portfolio From One to Three Exposure Units

A hypothetical book has a claim rate of $\lambda=0.4$ per exposure unit. Claim severity is Gamma with shape $\alpha=2$ and scale $\theta=\$300$. Calculate aggregate mean, aggregate variance, and zero-loss probability at $c=1$ and $c=3$.

Method and severity moments. First calculate the severity mean and second moment:

E[X_j]=2(300)=\$600,\qquad E[X_j^2]=2(3)(300^2)=6(90{,}000)=\$540{,}000.

At $c=1$, expected claim count is $1(0.4)=0.4$. Therefore,

E[S_1]=0.4(600)=\$240,\qquad \operatorname{Var}(S_1)=0.4(540{,}000)=\$216{,}000, \qquad P(S_1=0)=e^{-0.4}=0.6703.

At $c=3$, expected claim count is $3(0.4)=1.2$. Therefore,

E[S_3]=1.2(600)=\$720,\qquad \operatorname{Var}(S_3)=1.2(540{,}000)=\$648{,}000, \qquad P(S_3=0)=e^{-1.2}=0.3010.

Check. The mean and variance at $c=3$ are each three times their values at $c=1$, as required. The zero probability is $e^{-1.2}$, not three times $e^{-0.4}$. The calculation avoids treating the exposure multiplier as a multiplier on the probability itself.

Worked Example: Comparing Aggregate Totals and Pure Premiums

A hypothetical risk has $\lambda=0.5$ claims per exposure unit, with severity Gamma of shape $\alpha=2$ and scale $\theta=\$250$. Find the aggregate mean and variance for $c=1$ and $c=4$, then find the mean and variance per exposure unit at $c=4$.

Method and severity moments.

E[X_j]=2(250)=\$500,\qquad E[X_j^2]=2(3)(250^2)=6(62{,}500)=\$375{,}000.

For $c=1$, expected claim count is $0.5$:

E[S_1]=(0.5)(500)=\$250,\qquad \operatorname{Var}(S_1)=(0.5)(375{,}000)=\$187{,}500.

For $c=4$, expected claim count is $4(0.5)=2$:

E[S_4]=2(500)=\$1{,}000,\qquad \operatorname{Var}(S_4)=2(375{,}000)=\$750{,}000,\qquad P(S_4=0)=e^{-2}=0.1353.

The exposure-normalized amount at $c=4$ is $S_4/4$. Its mean is $\$1{,}000/4=\$250$, and its variance is $\$750{,}000/4^2=\$46{,}875$. The mean matches the one-unit mean, while the normalized variance is one quarter of the one-unit variance.

Check. Multiplying the $c=1$ aggregate moments by four gives $\$1{,}000$ and $\$750{,}000$. Dividing the $c=4$ variance by sixteen gives $\$46{,}875$. This distinction prevents confusing the variance of a total with the variance of a per-unit average.

Worked Example: Converting an Aggregation Period Correctly

A hypothetical insurer models claims per vehicle-year with $\lambda=0.4$. Severity is Gamma with shape $2$ and scale $\$250$. One policy covers 18 months. Calculate the expected aggregate loss, aggregate variance, probability of zero loss, and the mean and variance per vehicle-year. State the exposure multiplier relative to one vehicle-year.

Method and exposure conversion. Eighteen months is $18/12=1.5$ years, so $c=1.5$, not $18$. The severity moments are

E[X_j]=2(250)=\$500,\qquad E[X_j^2]=2(3)(250^2)=\$375{,}000.

Expected claim count over the policy period is $1.5(0.4)=0.6$. Thus,

E[S_{1.5}]=0.6(500)=\$300,\qquad \operatorname{Var}(S_{1.5})=0.6(375{,}000)=\$225{,}000, \qquad P(S_{1.5}=0)=e^{-0.6}=0.5488.

The mean and variance per vehicle-year are obtained by dividing the total by $1.5$ and the total variance by $1.5^2$, respectively:

E[S_{1.5}/1.5]=\$300/1.5=\$200,\qquad \operatorname{Var}(S_{1.5}/1.5)=\$225{,}000/2.25=\$100{,}000.

Check. Directly, the per-year mean is $\lambda E[X_j]=0.4(500)=\$200$, and its variance is $\lambda E[X_j^2]=0.4(375{,}000)=\$150{,}000$. Wait: this does not match the computed normalized variance, which signals an error. Recheck the division: $\$225{,}000/1.5^2=\$225{,}000/2.25=\$100{,}000$, but direct scaling from the model is $\operatorname{Var}(S_c/c)=\lambda E[X_j^2]/c=\$150{,}000/1.5=\$100{,}000$. The one-unit variance is $\$150{,}000$; the normalized variance for 1.5 units is correctly $\$100{,}000$. The apparent discrepancy is resolved by applying the $1/c$ scaling to the per-unit variance.

Common Scaling Errors

A frequent mistake is to multiply aggregate variance by $c^2$ because the amount of exposure is multiplied by $c$. That is not the compound Poisson result: the expected number of independent claims grows by $c$, so the variance of the aggregate total grows by $c$. A square appears only when scaling a random variable itself: $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. In $S_c$, increasing exposure adds claims; it does not multiply every claim amount by $c$.

For example, in the preceding worked example the correct variance at $c=4$ is $4(\$187{,}500)=\$750{,}000$. The incorrect calculation $4^2(\$187{,}500)=\$3{,}000{,}000$ would treat the total as four times one fixed random loss, rather than as the sum over four exposure units. Conversely, for a per-unit average, the correct variance is the aggregate variance divided by $c^2$, yielding a net $1/c$ relationship.

Do not use severity variance alone. Aggregate variance is $c\lambda E[X_j^2]$, because claim-count variability contributes.
Do not scale the zero probability linearly. Use $e^{-c\lambda}$; the probability is determined by a Poisson zero count.
Keep exposure units consistent. If $\lambda$ is per vehicle-year, convert months to years before calculating $c\lambda$.
Keep totals and rates separate. Divide an aggregate total by $c$ for a per-unit mean, and divide its variance by $c^2$.

Key result: Under a homogeneous compound Poisson-Gamma model, exposure multiplication by $c$ gives $E[S_c]=c\lambda E[X_j]$, $\operatorname{Var}(S_c)=c\lambda E[X_j^2]$, and $P(S_c=0)=e^{-c\lambda}$. The corresponding per-unit mean is unchanged, while the per-unit variance is proportional to $1/c$.

Check Your Understanding

For each problem, treat exposure units as homogeneous and use Gamma shape-scale parameters. Round probabilities to four decimals and dollar amounts to cents.

A hypothetical portfolio has $\lambda=0.3$ claims per exposure unit and severity Gamma with shape $2$ and scale $\$100$. At $c=5$, find aggregate mean, aggregate variance, and zero-loss probability. (Answer: mean $\$300.00$, variance $\$90{,}000.00$, zero probability 0.2231.)
A hypothetical risk has $\lambda=0.6$, with severity Gamma shape $4$ and scale $\$50$. For $c=2$, calculate aggregate mean, aggregate variance, and zero-loss probability. (Answer: mean $\$240.00$, variance $\$60{,}000.00$, zero probability 0.3010.)
A model uses $\lambda=0.25$ per exposure unit and Gamma severity with shape $3$ and scale $\$200$. At $c=4$, find the aggregate mean and variance, zero-loss probability, and variance of loss per exposure unit. (Answer: aggregate mean $\$600.00$, aggregate variance $\$480{,}000.00$, zero probability 0.3679, per-unit variance $\$30{,}000.00$.)
A hypothetical book has $\lambda=0.2$ per exposure unit and Gamma severity with shape $2$ and scale $\$100$. Find the exposure $c$ for which the probability of zero aggregate loss is $0.1000$, then calculate the aggregate mean and variance at that exposure. (Answer: $c=11.5129$ exposure units, mean $\$460.52$, variance $\$138{,}155.10$.)
A risk has $\lambda=0.5$ and Gamma severity with shape $2$ and scale $\$100$. For $c=9$, calculate aggregate mean, aggregate variance, zero-loss probability, and variance per exposure unit. (Answer: aggregate mean $\$900.00$, aggregate variance $\$270{,}000.00$, zero probability 0.0111, per-unit variance $\$3{,}333.33$.)

What You'll Learn