Comparing Frequency-Severity and Tweedie Predictions

Why Compare the Two Model Structures?

A frequency-severity model predicts how often claims occur and how large they are. An aggregate Tweedie model predicts total loss directly. Both can produce an expected loss for the same risk, but a matching expected loss does not by itself imply that the models give the same probability of no loss, variance, or overall distribution. Those distinctions matter when interpreting predictions and checking whether two model structures are comparable.

The comparison is meaningful only after aligning the risk, coverage period, and exposure basis. In this tutorial, $\lambda$ is the expected number of claims per unit of exposure, $m$ is mean claim severity, and $e$ is the risk's exposure measured in those units. Let $X$ denote aggregate loss over that exposure. The frequency-severity model assumes a claim count $N$ and severities $Z_1,Z_2,\ldots$, with $X=\sum_{j=1}^{N}Z_j$.

Definition: Under a frequency-severity model, the expected aggregate loss per unit of exposure is the expected claim frequency multiplied by expected claim severity: $\mu=\lambda m$. For exposure $e$, the expected aggregate loss is $E[X]=e\lambda m=e\mu$. This expectation does not require frequency and severity to be independent, but the product form does require $m$ to be the appropriate expected severity conditional on the frequency structure; the usual independent frequency-severity model assumes the count is independent of the claim severities.

For an aggregate Tweedie model, $\mu$ is the expected loss per unit of exposure. In the compound Poisson-gamma case, the model describes a Poisson number of positive gamma-distributed losses. It can therefore represent both a point mass at zero and a continuous distribution of positive aggregate losses. A Tweedie model with a variance power between 1 and 2 has this compound Poisson-gamma interpretation under the standard parameterization.

Key comparison: If the frequency-severity model has Poisson frequency and independent gamma severities, its aggregate loss distribution is a compound Poisson-gamma distribution and can be represented by a Tweedie model with $1<p<2$. The mapping requires compatible parameters and exposure. Outside those assumptions, the two structures may still predict the same mean, but their full distributions need not match.

When the Distributions Match

Suppose frequency per unit of exposure is Poisson with mean $\lambda$, and individual claim severity $Z$ has a gamma distribution with shape $\alpha$ and scale $\theta$. Then $E[Z]=m=\alpha\theta$, and $E[Z^2]=\alpha(\alpha+1)\theta^2$. If the number of claims is independent of severities, the aggregate is compound Poisson-gamma.

A Tweedie model with mean $\mu=\lambda\alpha\theta$, variance function $\operatorname{Var}(X)=\phi\mu^p$ for unit exposure, and matching compound Poisson-gamma parameters has

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

Since $\alpha>0$, this gives $1<p<2$. The same compound process has probability of zero claims, and therefore zero aggregate loss, equal to $e^{-\lambda e}$ over exposure $e$. A Tweedie model representing that same process must give the same zero-loss probability. The variance expression above applies at unit exposure; for total loss over exposure $e$, the compound process has variance $e\phi\mu^p$.

Align the target.
Confirm that both predictions refer to the same risk, coverage period, and exposure units. Distinguish total expected loss $e\mu$ from expected loss per exposure unit $\mu$.

Compare the means.
Calculate $e\lambda m$ from frequency and severity, then compare it with the aggregate model's total expected loss.

Compare distributional assumptions.
Check whether frequency is Poisson, severities are gamma, and frequency is independent of severity before claiming a Tweedie distributional match.

Check more than the mean.
When a full distributional comparison is needed, compare quantities such as zero-loss probability and variance, not just the pure premium.

The parameter mapping follows from matching the compound process to the Tweedie mean and variance structure. For unit exposure, the compound process has mean $\lambda E[Z]$ and variance $\lambda E[Z^2]$. With gamma severity, these become $\mu=\lambda\alpha\theta$ and $\lambda\alpha(\alpha+1)\theta^2$. Matching the power of the mean in the Tweedie variance function gives $p=(\alpha+2)/(\alpha+1)$; matching the variance coefficient then gives the stated $\phi$.

Worked Comparisons

Worked Example: Matching a Frequency-Severity Model to Tweedie

A hypothetical risk has annual claim frequency $\lambda=0.20$. Claim severity is gamma with shape $\alpha=2$ and scale $\theta=\$2{,}500$. Compare the annual expected loss and distributional parameters with a matching Tweedie model. Use one year of exposure.

Method and assumptions. Use the compound Poisson-gamma mapping. Claims occur according to a Poisson count, severities are independent gamma variables, and the same annual exposure applies to both representations.

First, the mean severity is $m=\alpha\theta=2(\$2{,}500)=\$5{,}000$. The frequency-severity expected annual loss is

E[X]=\lambda m=0.20(\$5{,}000)=\$1{,}000.

The Tweedie mean per year must therefore be $\mu=\$1{,}000$. Its power is

p=\frac{\alpha+2}{\alpha+1} =\frac{2+2}{2+1} =\frac{4}{3}.

For the dispersion calculation, $\alpha(\alpha+1)\theta^2=2(3)(\$2{,}500)^2=6(\$6{,}250{,}000)=\$37{,}500{,}000$. Also, $\mu^p=(\$1{,}000)^{4/3}=10{,}000$ in the corresponding currency-power units. Thus

\phi=\frac{0.20(\$37{,}500{,}000)}{10{,}000}=750.

As a check, gamma severity has $E[Z^2]=\$37{,}500{,}000$, so the compound Poisson aggregate variance is $0.20E[Z^2]=\$7{,}500{,}000\;(\text{dollars})^2$. The Tweedie variance is $\phi\mu^p=750(10{,}000)=\$7{,}500{,}000\;(\text{dollars})^2$, in agreement. The probability of zero annual loss is $e^{-0.20}=0.8187$ to four decimal places.

Conclusion and mistake avoided. The models have the same annual mean, variance, and zero-loss probability because the Poisson-gamma assumptions and parameters match. Do not compare $\phi$ itself with a frequency or severity parameter; it is a dispersion parameter in the Tweedie variance representation.

Worked Example: Equal Pure Premiums, Different Distributions

Two hypothetical risk classes each have one year of exposure. Class A has Poisson frequency $\lambda_A=0.10$; Class B has Poisson frequency $\lambda_B=0.20$. In both classes, severity is gamma with shape $\alpha=2$, but the mean severity is $\$10{,}000$ for A and $\$5{,}000$ for B. Compare their expected losses, variances, and probabilities of zero loss.

Method and assumptions. Calculate the frequency-severity mean and compound Poisson variance for each class. The count is Poisson, severities are independent gamma variables, and frequency and severity are independent within each class.

For Class A, $\mu_A=0.10(\$10{,}000)=\$1{,}000$. Its gamma scale is $\theta_A=m_A/\alpha=\$5{,}000$, so

E[Z_A^2]=\alpha(\alpha+1)\theta_A^2 =6(\$5{,}000)^2 =\$150{,}000{,}000.

Thus $\operatorname{Var}(X_A)=0.10(\$150{,}000{,}000)=\$15{,}000{,}000\;(\text{dollars})^2$, and $P(X_A=0)=e^{-0.10}=0.9048$.

For Class B, $\mu_B=0.20(\$5{,}000)=\$1{,}000$. Its scale is $\theta_B=\$2{,}500$, giving $E[Z_B^2]=6(\$2{,}500)^2=\$37{,}500{,}000$. Therefore $\operatorname{Var}(X_B)=0.20(\$37{,}500{,}000)=\$7{,}500{,}000\;(\text{dollars})^2$, and $P(X_B=0)=e^{-0.20}=0.8187$.

Conclusion and mistake avoided. Both classes have the same expected loss, but Class A has twice the variance and a higher probability of no claims. The higher severity in A produces a more variable aggregate loss. The common mistake is to infer that equal pure premiums imply equal loss distributions; the calculations show that they do not.

Worked Example: Reconciling Model Outputs With a Shared Mean

For one year of exposure, a hypothetical frequency-severity model predicts Poisson frequency $\lambda=0.08$ and exponential severity with mean $\$7{,}812.50$. An aggregate Tweedie model predicts mean $\mu=\$625$, power $p=1.5$, and dispersion $\phi=500$. Compare the means, implied frequency and severity from the Tweedie parameters, variances, and zero-loss probabilities.

Method and assumptions. Use the compound Poisson-gamma parameter mapping for the Tweedie model. An exponential severity is gamma with shape $\alpha=1$. Both predictions use the same one-year exposure; the comparison checks whether the distributions agree, not merely whether their means do.

The frequency-severity mean is $0.08(\$7{,}812.50)=\$625$, matching the Tweedie mean. For $p=1.5$, the corresponding gamma shape is $\alpha=(2-p)/(p-1)=0.5/0.5=1$. The Tweedie-implied frequency is

\lambda_{\text{Tw}}=\frac{\mu^{2-p}}{\phi(2-p)} =\frac{\sqrt{625}}{500(0.5)} =\frac{25}{250}=0.10.

The implied mean severity is $\$625/0.10=\$6{,}250$. Thus the aggregate model corresponds to a Poisson frequency of 0.10 and exponential severity with mean $\$6{,}250$, not to the frequency-severity model's 0.08 and $\$7{,}812.50$.

For the frequency-severity model, exponential severity has second moment $E[Z^2]=2m^2=2(\$7{,}812.50)^2=\$122{,}070{,}312.50$. Its aggregate variance is $0.08E[Z^2]=\$9{,}765{,}625\;(\text{dollars})^2$, and its zero-loss probability is $e^{-0.08}=0.9231$.

For the Tweedie-implied model, $E[Z^2]=2(\$6{,}250)^2=\$78{,}125{,}000$. Its aggregate variance is $0.10E[Z^2]=\$7{,}812{,}500\;(\text{dollars})^2$, also equal to $\phi\mu^p=500(625^{1.5})=500(15{,}625)$. Its zero-loss probability is $e^{-0.10}=0.9048$.

Conclusion and mistake avoided. The means match at $\$625$, but the implied frequencies differ, so the variances and zero-loss probabilities differ. Do not declare the distributions equivalent solely because their expected losses agree.

Interpreting Differences and Avoiding Traps

A practical comparison begins by establishing whether the outputs represent the same target. A frequency-severity model may report frequency per vehicle-year and severity per claim, while an aggregate model may report total loss per policy-year. Convert to a common exposure basis before multiplying or comparing. If a risk has $e$ exposure units, frequency-severity expected loss is $e\lambda m$, not just $\lambda m$, unless $e=1$.

A second distinction is between a prediction of the conditional mean and a prediction of the full loss distribution. Pure premium comparisons use expected loss. Distributional comparisons also require assumptions about count variability, severity shape, and dependence. Even when the expected loss formulas coincide, differences in those assumptions can change the likelihood of a zero, the variance, and the probability of large losses.

The following calculation illustrates a common exposure trap. If $\lambda=0.12$ claims per year, $m=\$4{,}000$, and a policy has $e=2$ years of exposure, the expected total loss is $2(0.12)(\$4{,}000)=\$960$. Multiplying $\lambda m$ by exposure a second time would incorrectly give $\$1{,}920$. Conversely, comparing the annual pure premium of $\$480$ directly with the two-year aggregate prediction of $\$960$ also compares different targets.

Key takeaway: Frequency-severity and Tweedie models can be equivalent at the distributional level under the compound Poisson-gamma assumptions and a consistent parameter mapping. Otherwise, compare their predictions on the same exposure basis and treat agreement in expected loss as agreement of means only.

Check Your Understanding

For each problem, use the stated exposure basis and assumptions. Probabilities are rounded to four decimal places where requested.

A risk has Poisson annual frequency $0.15$ and mean severity $\$4{,}000$. Find its annual expected loss and probability of zero claims. (Answer: $\$600$; $0.8607$.)
Annual frequency is Poisson with mean $0.20$, and severity is gamma with shape $3$ and mean $\$3{,}000$. Find the annual expected loss, $E[Z^2]$, and aggregate variance. (Answer: $\$600$; $\$12{,}000{,}000$; $\$2{,}400{,}000\;(\text{dollars})^2$.)
A policy has exposure $e=2$ years and Poisson frequency $0.06$ per year. Mean severity is $\$5{,}000$. Find total expected loss over the policy period and the probability of zero claims during that period. (Answer: $\$600$; $0.8869$.)
A unit-exposure Tweedie prediction has $\mu=\$900$, $p=1.5$, and $\phi=300$, and is assumed to have a compound Poisson-gamma representation. Find the implied frequency, gamma shape, mean severity, aggregate variance, and zero-loss probability. (Answer: $0.20$; shape $1$; $\$4{,}500$; $\$8{,}100{,}000\;(\text{dollars})^2$; $0.8187$.)
A frequency-severity model has Poisson frequency $0.12$ and exponential severity with mean $\$5{,}000$. A Tweedie model predicts the same unit-exposure mean but has $p=1.5$ and $\phi=250$. Compute the frequency-severity variance and zero-loss probability. Then find the Tweedie-implied frequency and state whether the full distributions match. (Answer: frequency-severity variance $\$6{,}000{,}000\;(\text{dollars})^2$, zero-loss probability $0.8869$; Tweedie-implied frequency $\sqrt{600}/125\approx0.1960$, so the distributions do not match.)

What You'll Learn