Common Errors in Reading GLM Coefficients and Indications

Why Reading GLM Output Requires Care

A generalized linear model can produce a compact coefficient table, but the entries are not usually rating factors in the form they will be used. In a log-link model, a coefficient is an additive change in the linear predictor; exponentiating the appropriate coefficient or combination of coefficients gives a multiplicative change in the fitted mean. Reference levels and interactions determine which combination is appropriate.

This tutorial focuses on interpreting supplied fitted-model output. It assumes the predictor definitions and coding are known; the same numerical coefficient can have different meanings under different coding schemes. Interpretations below concern fitted conditional means, not causal effects.

Definition: For a response with positive conditional mean $\mu(x)$, a log-link GLM has $$ \log \mu(x)=\eta(x)=\beta_0+\sum_j \beta_j z_j(x), $$ where $\eta(x)$ is the linear predictor, $\beta_0$ is the intercept, and the $z_j(x)$ are specified predictor columns. Thus $\mu(x)=\exp(\eta(x))$. A change in predictors from profile $x_0$ to profile $x_1$ has fitted mean ratio $\exp(\eta(x_1)-\eta(x_0))$.

The response and exposure convention matter. For example, if a count $N$ is modeled with exposure $e$, a common specification is $\log E[N\mid x,e]=\log(e)+\eta(x)$. The exposure term is an offset, not an estimated rating coefficient. The coefficient comparisons below concern the fitted rate per unit of exposure, so the exposure cancels when comparing otherwise identical risks.

Coefficient, Relativity, and Percentage Change

Under a log link, an additive linear-predictor difference $b$ corresponds to a fitted mean ratio of $\exp(b)$. The relative change is $\exp(b)-1$, not $b$. For a small coefficient, $b$ may be a useful approximation to the proportional change, but the exact model implication is obtained by exponentiating.

Worked Example: Exponentiating a Reference-Level Contrast

A hypothetical insurer fits a Poisson log-link model for claim frequency per vehicle-year. Let Small be the reference vehicle class and let $I_L$ equal 1 for Large vehicles and 0 otherwise. The fitted model is $\log \mu=-2.1203+0.2231I_L$. Interpret the class coefficient and calculate both fitted rates.

Method and model conditions. The model is $\log \mu=\beta_0+\beta_L I_L$, where $\mu$ is expected claims per vehicle-year. The model uses the stated reference coding: $I_L=0$ for Small and $I_L=1$ for Large. The coefficient is therefore the Large-versus-Small difference in log fitted frequency, holding the other model terms fixed.

Step 1: Calculate the reference rate. For Small, $I_L=0$, so $\log \mu_S=-2.1203$. Therefore $\mu_S=\exp(-2.1203)\approx 0.1200$ claims per vehicle-year.

Step 2: Convert the coefficient to a factor. The Large-to-Small fitted rate ratio is $\exp(0.2231)\approx 1.2500$. The fitted frequency is 25.00% higher for Large vehicles, conditional on the model’s other predictors.

Step 3: Calculate the Large rate. Its linear predictor is $-2.1203+0.2231=-1.8972$. Thus $\mu_L=\exp(-1.8972)\approx 0.1500$ claims per vehicle-year. Checking the ratio, $0.1500/0.1200=1.2500$, consistent with the exponentiated coefficient.

Common mistake avoided. Calling 0.2231 a 22.31% increase treats a log-scale coefficient as a direct percentage. The exact percentage increase is $(1.2500-1)\times100\%=25.00\%$.

This ratio interpretation follows directly by dividing the two fitted means. If two profiles differ only by a predictor whose contribution to the linear predictor changes by $b$, then their means are $\exp(\eta+b)$ and $\exp(\eta)$. Their ratio is $\exp(\eta+b)/\exp(\eta)=\exp(b)$. The common terms, including the intercept, cancel.

Reference Levels and Interactions

Under treatment (reference) coding, a coefficient for a categorical level commonly compares that level with the omitted reference level, holding the other predictors fixed. The intercept is the fitted log mean for the profile whose coded predictors are all at their reference or zero values. Do not assume these interpretations if the output uses a different contrast convention: read the stated coding and predictor definitions.

With an interaction, a main-effect coefficient is generally a conditional comparison at the value or level where the interacting predictor is zero or at its reference. It is not automatically the overall effect across all levels. The relevant comparison is found by subtracting the full linear predictors for the two profiles; this retains any interaction contribution that changes between them.

Worked Example: Interpreting an Interaction Conditionally

A hypothetical insurer models pure premium with a log link. Let $A=1$ denote a newer vehicle and $B=1$ denote coastal territory; zero denotes each corresponding reference level. The fitted equation, in dollars, is $\log \mu=\log(600)+0.18A+0.10B-0.12AB$. Find the fitted pure premium for a newer vehicle in coastal territory, and compare the effect of coastal territory for newer vehicles with its effect for reference-age vehicles.

Method and model conditions. The fitted mean is $\mu=\exp(\eta)$; the base pure premium of $600 applies when $A=B=0$. The interaction term $AB$ is included, so all terms must be evaluated for each profile.

Step 1: Evaluate the target profile. For $A=B=1$, the interaction product is $AB=1$. The linear-predictor increment above the base is $0.18+0.10-0.12=0.16$.

Step 2: Convert that increment to a factor and premium. The factor relative to the reference profile is $\exp(0.16)\approx1.1735$. The fitted pure premium is $600(1.1735109)=\$704.11$, rounded to the nearest cent.

Step 3: Compare the territory effect at each vehicle level. For reference-age vehicles ($A=0$), changing from $B=0$ to $B=1$ changes the linear predictor by $0.10$, giving factor $\exp(0.10)\approx1.1052$. For newer vehicles ($A=1$), the change is $0.10-0.12=-0.02$, giving factor $\exp(-0.02)\approx0.9802$. Thus the fitted coastal-versus-reference territory ratio is about 0.9802 for newer vehicles, not 1.1052.

Common mistake avoided. Adding the two main effects and ignoring the interaction gives $0.18+0.10=0.28$, or factor $\exp(0.28)\approx1.3231$. That is not the fitted factor for the target profile because its interaction contribution, $-0.12$, must also be included. A main-effect coefficient alone does not describe every conditional comparison when an interaction is present.

Key calculation: For a comparison between profiles, calculate $\eta(x_1)-\eta(x_0)$ from every applicable model term, then exponentiate. This procedure works whether the comparison involves main effects, interactions, or both.

Standard Errors Are on the Coefficient Scale

A reported standard error for a log-link coefficient describes sampling uncertainty in that coefficient, on the linear-predictor scale. It is not itself the standard error of the exponentiated factor. A standard Wald confidence interval is first formed for the coefficient and then exponentiated at both endpoints. Exponentiation preserves the interval’s ordering but makes its scale asymmetric around the point estimate.

Worked Example: Turning a Coefficient Interval Into a Factor Interval

A hypothetical fitted log-link model reports a coefficient estimate of $0.30$ for a rating-level contrast, with standard error $0.10$. Using a 95% Wald interval and $1.96$ as the normal critical value, calculate the estimated factor, its approximate delta-method standard error, and its confidence interval.

Method and assumptions. Treat the coefficient estimate as approximately normal with standard error $0.10$. The factor is $R=\exp(\beta)$. The requested interval is a Wald interval on the coefficient scale, transformed to the factor scale.

Step 1: Factor estimate. $\hat R=\exp(0.30)\approx1.3499$.

Step 2: Approximate standard error of the factor. By the delta method, the derivative of $\exp(\beta)$ is $\exp(\beta)$, so $\operatorname{SE}(\hat R)\approx\exp(0.30)(0.10)=0.1350$. This is a local approximation, not an exact transformation of the coefficient’s standard error.

Step 3: Form the coefficient interval. The endpoints are $0.30\pm1.96(0.10)=0.30\pm0.196$, giving $(0.104,0.496)$.

Step 4: Exponentiate the endpoints. The factor interval is $(\exp(0.104),\exp(0.496))\approx(1.1096,1.6420)$. The point estimate $1.3499$ lies between these endpoints, as expected.

Common mistake avoided. Exponentiating the standard error, $\exp(0.10)$, does not give the standard error of the factor. The coefficient standard error is on the log scale; the delta-method approximation uses $\exp(\hat\beta)\operatorname{SE}(\hat\beta)$, while a transformed Wald interval is obtained by exponentiating the coefficient interval’s endpoints.

For a contrast involving several estimated coefficients, its uncertainty also depends on their covariances. If the contrast is $c^\mathsf{T}\hat\beta$ with coefficient covariance matrix $\widehat{\Sigma}$, its estimated variance is $c^\mathsf{T}\widehat{\Sigma}c$. Adding coefficient standard errors as if they were independent is not justified unless the relevant covariances are zero. Once a standard error for the contrast is correctly obtained, the same log-scale interval and exponentiation procedure applies.

Coefficients Are Not Complete Rating Indications

A coefficient is an estimated parameter tied to a particular model specification and coding. An exponentiated coefficient is a fitted multiplicative contrast under that specification. Neither should be mistaken for a dollar premium or for a complete indication for a policy. To obtain a fitted amount, the relevant model terms must be combined with the intercept or a specified base amount. The resulting amount is conditional on the other predictors, exposure convention, and model assumptions.

A factor may also be reported relative to a specified reference profile, while a filed or displayed rating factor may be subject to a chosen normalization or other practical constraints. Those adjustments are not implicit in the coefficient. When reading an indication, identify the base, the comparison profile, the included terms, and whether any further transformation is stated.

Worked Example: From Coefficients to a Policy-Level Indication

A hypothetical insurer’s log-link pure-premium model has a reference-profile pure premium of $800. For a particular policy, the supplied contributions are $0.30$ for its vehicle category, $0.20$ for its territory, and $-0.10$ for the interaction between those characteristics. Calculate the fitted indication for this policy relative to the reference profile. No normalization or other adjustments are specified.

Method and model conditions. The stated reference premium is the fitted amount at the reference profile. The supplied contributions are on the log scale, and the interaction applies to this policy. The policy-level factor is the exponential of the sum of the contributions.

Step 1: Add all applicable contributions. The log-scale difference is $0.30+0.20-0.10=0.40$.

Step 2: Convert to the fitted factor. The factor relative to the reference profile is $\exp(0.40)\approx1.4918$.

Step 3: Apply the factor to the base indication. The fitted policy-level pure premium is $800(1.4918247)=\$1,193.46$, rounded to the nearest cent.

Step 4: Check the result. The factor exceeds 1, so the indication must be above the $800 reference amount; $1,193.46 is consistent. The number $0.40$ is the combined log-scale increment, not the policy’s premium or a 40% increase.

Common mistake avoided. Applying only the vehicle coefficient would give a factor of $\exp(0.30)$, omitting the territory and interaction contributions. The complete fitted indication requires evaluating every applicable term for the policy profile.

Exam Traps and a Reliable Reading Sequence

The most reliable way to interpret output is to translate it in a fixed order: establish the response scale and link, identify the model terms and coding, specify the comparison or policy profile, calculate the linear-predictor difference, and only then exponentiate. For uncertainty, first calculate the standard error of the coefficient or contrast on the log scale; then form and transform the interval as required.

Exponentiating the wrong quantity. A log-scale coefficient $b$ implies factor $\exp(b)$, not $1+b$, except as a small-effect approximation.
Ignoring the reference convention. A coefficient’s comparison depends on how the predictors are coded. The intercept is not automatically a universal base rate.
Reading a main effect as unconditional. With an interaction, the conditional effect depends on the interacting predictor’s value.
Using a standard error on the wrong scale. A coefficient standard error is on the coefficient scale. Do not label it the factor’s standard error or exponentiate it by itself.
Calling a coefficient a final rate indication. A policy-level fitted amount requires the base and every applicable model contribution, and may differ from any subsequently adjusted rating display.

These distinctions are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. In an exam calculation, showing the profile comparison and the linear-predictor arithmetic makes the interpretation auditable and helps reveal missing interaction terms or mistaken reference levels.

Check Your Understanding

For each problem, identify the relevant model comparison before calculating. Answers are provided so you can check the result.

A log-link frequency model uses a reference class with fitted rate 0.08 and a coefficient of $0.1823$ for another class. Find the fitted rate for that class and its percentage change from reference. (Answer: factor $\exp(0.1823)\approx1.2000$; rate $0.0960$; 20.00% higher.)
A log-link model has terms $0.12A+0.08B-0.15AB$, where $A$ and $B$ are binary indicators. Find the fitted factor for $A=B=1$ relative to $A=B=0$. (Answer: $\exp(0.12+0.08-0.15)=\exp(0.05)\approx1.0513$.)
In the same model, calculate the factor for changing $B$ from 0 to 1 when $A=1$, and compare it with the factor when $A=0$. (Answer: for $A=1$, $\exp(0.08-0.15)=\exp(-0.07)\approx0.9324$; for $A=0$, $\exp(0.08)\approx1.0833$.)
A coefficient estimate is $0.25$ with standard error $0.05$. Using a 95% Wald interval with critical value 1.96, give the factor estimate and transformed interval. (Answer: factor $\exp(0.25)\approx1.2840$; coefficient interval $(0.152,0.348)$; factor interval approximately $(1.1641,1.4162)$.)
A reference pure premium is $500. A policy has applicable log-scale contributions of $-0.10$, $0.25$, and $0.05$, including an interaction term. Calculate the fitted policy amount. (Answer: total increment $0.20$; factor $\exp(0.20)\approx1.2214$; fitted amount $500(1.2214)=\$610.70$, to the nearest cent.)

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