Model-Based Class Relativities

Why Model-Based Relativities Need a Reference

Classification models estimate how expected claim frequency, severity, or cost varies with rating characteristics. A fitted coefficient is not automatically a rate factor: its meaning depends on the model’s link function and on which class is the reference. For a log-link model, exponentiating the appropriate coefficient or coefficient contrast converts a difference on the log scale into a multiplicative rate relativity.

The method complements the class-level experience and relativity calculations used elsewhere in classification ratemaking. It is especially useful when several rating variables are modeled together, because the model can estimate each variable’s effect conditional on the others. The resulting factors are model indications; a model estimate by itself does not determine the final selected rates.

Definition: In a log-link model, the conditional mean $m$ is modeled as $\log(m)=\eta$, where $\eta$ is a linear predictor containing an intercept and applicable model terms. If class $r$ is the reference, its class coefficient is set to zero or omitted. The relativity for class $c$ compared with $r$, with other modeled characteristics held fixed, is the ratio of their fitted means: $R_{c:r}=m_c/m_r$.

For a categorical variable with one reference class and indicator variables for the other classes, the linear predictor can be written as $\eta_c=\beta_0+\beta_c$. The reference class has $\eta_r=\beta_0$, because $\beta_r=0$. Therefore the ratio of fitted means is determined by the difference in linear predictors, not by the coefficient as a direct percentage.

R_{c:r}=\frac{m_c}{m_r} =\frac{\exp(\beta_0+\beta_c)}{\exp(\beta_0)} =\exp(\beta_c)

When both classes have estimated coefficients—for example, because the model’s parameterization does not make either class coefficient zero—the relevant factor is the exponential of their coefficient difference. With several model terms, the contrast includes every term that changes between the two class profiles. Terms that are the same in both profiles cancel.

From a Fitted Model to a Class Rate

A common frequency model uses a Poisson count response and a log link. Let $N_i$ be the claim count for observation $i$, and let $e_i$ be its exposure measured in the model’s exposure unit. Conditional on the modeled characteristics, assume $N_i$ is Poisson with mean $e_i\lambda_i$, where $\lambda_i$ is the expected frequency per exposure unit. The model is

N_i\mid x_i \sim \operatorname{Poisson}(e_i\lambda_i), \qquad \log(\lambda_i)=\beta_0+\sum_j \beta_j x_{ij}.

Equivalently, the model for the count has $\log(e_i)$ as an offset, a term whose coefficient is fixed at one. An offset accounts for the amount of exposure; it is not an additional class factor. The frequency relativities come from the classification terms in the linear predictor. The Poisson assumption specifies the count distribution and its mean; if the count variation is greater than the Poisson model allows, uncertainty estimates may need adjustment, but the exponential transformation of the fitted log mean remains the way to obtain a mean ratio.

Worked Example: Converting Frequency Coefficients to Factors

A hypothetical insurer fits a Poisson frequency model for a small commercial equipment policy. The response is annual claim count, exposure is measured in policy-years, and the model assumes independent Poisson counts conditional on the rating variables. The fitted model has reference territory North, with annual frequency $0.10$ claims per policy-year. Territory South has estimated coefficient $\hat{\beta}_{S}=\log(1.25)=0.2231$, and territory West has $\hat{\beta}_{W}=\log(0.80)=-0.2231$. Calculate the frequency indications and expected claim counts for 1,000 policy-years in each territory.

Method and reference. The log-link Poisson model is $\log(\lambda)=\beta_0+\beta_{\text{territory}}$, with $\beta_{\text{North}}=0$. Thus North’s fitted frequency is $0.10$, and each other territory’s factor is the exponential of its coefficient.

South. Its factor is $\exp(0.2231)=1.25$. Its fitted frequency is $0.10\times1.25=0.125$ claims per policy-year. For 1,000 policy-years, the expected claim count is $1{,}000\times0.125=125$.

West. Its factor is $\exp(-0.2231)=0.80$. Its fitted frequency is $0.10\times0.80=0.080$ claims per policy-year. For 1,000 policy-years, the expected claim count is $1{,}000\times0.080=80$.

Check. Relative to North, South’s expected frequency is 25% higher and West’s is 20% lower. The expected counts are the fitted frequencies multiplied by exposure, as required by the Poisson mean $e_i\lambda_i$. The common mistake avoided is treating a log-scale coefficient as the factor itself.

The same transformation applies to other response distributions with a log link. For example, a Gamma severity model can be used for positive claim severities when a positive, right-skewed response is suitable. If its conditional mean severity is $s$, then $\log(s)$ is modeled as a linear predictor, and exponentiating a coefficient contrast gives a ratio of conditional mean severities. The distribution and link determine the model assumptions; the coefficient-to-relativity calculation follows from the link.

Worked Example: Combining Frequency and Severity Effects

A hypothetical insurer models claim frequency and claim severity separately for a new equipment coverage. For frequency, claim count is Poisson with mean $e\lambda$, where exposure $e$ is in policy-years; the reference frequency is $0.08$ claims per policy-year. A territory coefficient is $\log(1.20)$. For severity, individual claim amount is Gamma with fixed shape parameter $4$ and conditional mean $s$; its scale parameter is therefore $s/4$. Reference mean severity is $5,000, and a machinery class coefficient is $\log(1.10)$. Both models use log links. Calculate the expected loss cost per policy-year for the reference profile and for a policy in the indicated territory and machinery class.

Reference profile. Expected loss cost per policy-year is frequency multiplied by mean severity: $0.08\times \$5{,}000=\$400$. This multiplication gives expected aggregate claim cost when the modeled frequency and severity apply to the same policy profile.

Frequency effect. The territory factor is $\exp(\log(1.20))=1.20$. The indicated frequency is $0.08\times1.20=0.096$ claims per policy-year.

Severity effect. The machinery factor is $\exp(\log(1.10))=1.10$. The indicated mean severity is $\$5{,}000\times1.10=\$5{,}500$. Under the stated Gamma model, the corresponding scale parameter is $\$5{,}500/4=\$1{,}375$; the shape remains 4.

Combined loss cost and relativity. The indicated loss cost is $0.096\times\$5{,}500=\$528$ per policy-year. Its relativity to the reference profile is $\$528/\$400=1.32$, which also equals $1.20\times1.10$.

Check. The frequency factor changes expected claim count, while the severity factor changes expected dollars per claim. Multiplying both factors gives the loss-cost factor only under this separate frequency-severity setup, with the modeled means applying to the same profile. The common mistake avoided is applying just one of the two effects to the pure premium.

Why Exponentiation Gives a Rate Ratio

The key justification is the algebra of the log link. For any two profiles $a$ and $b$, suppose their modeled means are $m_a=\exp(\eta_a)$ and $m_b=\exp(\eta_b)$. Dividing the means cancels any linear-predictor terms common to both profiles. The remaining contrast is exponentiated, yielding their fitted mean ratio.

Write each profile’s linear predictor.
For example, $\eta_a=\beta_0+\beta_A$ and $\eta_b=\beta_0+\beta_B$, with the same other variables held fixed.

Use the log-link definition of the fitted mean.
Then $m_a=\exp(\eta_a)$ and $m_b=\exp(\eta_b)$.

Take the ratio.
$m_a/m_b=\exp(\eta_a)/\exp(\eta_b)=\exp(\eta_a-\eta_b)=\exp(\beta_A-\beta_B)$. Thus the model-based relativity is the exponential of the coefficient contrast.

This derivation also explains what “holding other characteristics fixed” means: their terms are identical in $\eta_a$ and $\eta_b$, so they cancel from the ratio. If the model contains an interaction, the relevant interaction term may differ between the profiles and must be included in the contrast. A factor cannot be read from a single main-effect coefficient when the modeled contrast also changes an interaction term.

Worked Example: Rebasing Coefficients to a New Reference

A hypothetical insurer fits a Gamma log-link model for positive claim severity. The distribution has fixed shape parameter $5$, so the conditional scale is the conditional mean divided by 5. For a fixed set of other characteristics, the estimated mean for Standard use is $720. Business use has coefficient $\hat{\beta}_{B}=\log(1.40)=0.3365$, and Delivery use has coefficient $\hat{\beta}_{D}=\log(1.80)=0.5878$, both relative to Standard. Find the fitted means, the Delivery-to-Business relativity, and the coefficients if Business is instead chosen as the reference.

Fitted means under the original reference. Standard has mean $720. Business has factor $\exp(0.3365)=1.40$, so its fitted mean is $\$720\times1.40=\$1{,}008$. Delivery has factor $\exp(0.5878)=1.80$, so its fitted mean is $\$720\times1.80=\$1{,}296$. The Gamma scale parameters are respectively $\$720/5=\$144$, $\$1{,}008/5=\$201.60$, and $\$1{,}296/5=\$259.20$.

Delivery compared with Business. The mean ratio is $\$1{,}296/\$1{,}008=1.2857$ to four decimal places. Equivalently, the coefficient contrast is $0.5878-0.3365=0.2513$, and $\exp(0.2513)=1.2857$ to four decimal places.

Rebase to Business. The new intercept corresponds to the Business mean, $1,008. Standard’s coefficient becomes $0-0.3365=-0.3365$, giving factor $\exp(-0.3365)=1/1.40=0.7143$. Delivery’s coefficient becomes $0.5878-0.3365=0.2513$, giving factor $1.2857$. Business has coefficient zero and factor 1.

Check. Under the rebased model, Standard’s mean is $\$1{,}008\times0.7143=\$720$, and Delivery’s is $\$1{,}008\times1.2857=\$1{,}296$, up to rounding. Changing the reference changes the coefficient representation, not the fitted means or pairwise relativities.

Interpretation and Exam Traps

A coefficient such as $0.20$ is a difference in log means, not a 20% rate increase. Its implied factor is $\exp(0.20)$, approximately $1.2214$, or a 22.14% increase. A coefficient of $-0.20$ gives factor $\exp(-0.20)$, approximately $0.8187$, or an 18.13% decrease. The increases and decreases are not symmetric in percentage terms because the factors are multiplicative.

A frequent direction error is to use the wrong class in the denominator. If a model estimates $\beta_A$ and $\beta_B$, the factor for A relative to B is $\exp(\beta_A-\beta_B)$; reversing the comparison gives its reciprocal, $\exp(\beta_B-\beta_A)$. State the requested comparison explicitly before calculating.

Do not exponentiate the intercept alone to get every class factor. The intercept sets the fitted mean for the reference profile, subject to the other model terms.
Do not treat an offset as a class coefficient. Exposure adjusts the expected count for the amount insured or time at risk; class terms produce the modeled relativities.
Do not compare coefficients without specifying the reference. A coefficient is interpreted relative to the omitted class or parameterization. A pairwise relativity requires the relevant contrast.
Do not assume separate factors combine in every model. Multiplication follows when the log-linear predictor contains the corresponding additive terms and the comparison has no changing interaction term.

These interpretations are aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus. In practice, a model-based factor describes the fitted response mean under the model and data used. It should be distinguished from an overall rate indication, a selected final factor, or a comparison with a currently used factor.

Check Your Understanding

For each problem, identify the reference or comparison carefully before translating coefficients into relativities.

A hypothetical Poisson frequency model uses policy-years as exposure and has reference frequency $0.06$ claims per policy-year. A class coefficient is $0.1823$, approximately $\log(1.20)$. What are the class factor, indicated frequency, and expected count for 500 policy-years? (Answer: factor $1.20$; frequency $0.072$; expected count $36$.)
In a log-link model, class A has coefficient $\log(1.30)$ and class B has coefficient $\log(0.90)$, each measured against the same omitted reference. What is A’s relativity to B? (Answer: $1.30/0.90=1.4444$ to four decimal places.)
A Poisson frequency model has reference frequency $0.04$ per policy-year and a class factor of $1.50$. A separate Gamma severity model has reference mean $2,400, fixed shape 4, and a class factor of $0.75$ for the same profile. Find reference and indicated loss costs and their relativity. (Answer: reference loss cost $96; indicated frequency $0.06$, severity $1,800, and loss cost $108; relativity $1.125$.)
With Class A as reference, estimated coefficients for B and C are $\log(1.20)$ and $\log(0.75)$. Rebase to B as reference and state the factors for A, B, and C. (Answer: A $1/1.20=0.8333$; B $1.0000$; C $0.75/1.20=0.6250$.)
A log-link model has reference mean $600. Class X has coefficient $\log(1.25)$. For a different profile, the same class coefficient applies, but an interaction term adds $\log(0.80)$ to the linear predictor. Find the second profile’s factor relative to the reference and its fitted mean. (Answer: factor $1.25\times0.80=1.00$; fitted mean $600.)

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