Interpreting Poisson Frequency Relativities

From Log-Scale Coefficients to Frequency Relativities

In a Poisson frequency model with a log link, model coefficients are expressed on the log scale, while ratemaking decisions are often discussed as multiplicative changes in claim frequency. Interpreting a coefficient therefore requires a conversion: exponentiate it to obtain the corresponding frequency relativity. The comparison is always conditional on the other model predictors and depends on which class is the reference.

This distinction matters when combining rating characteristics. A positive coefficient does not mean that the fitted frequency increases by that coefficient as a percentage; it means the fitted frequency is multiplied by the coefficient’s exponential. Likewise, when a model includes an interaction, the joint relativity is not generally the product of the main-effect relativities alone.

Definition: Suppose the conditional claim count $N$, for exposure $e>0$, follows a Poisson distribution with mean $\mu$, and the model uses a log link: $$ N\mid \mathbf{x}\sim\operatorname{Poisson}(\mu),\qquad \log(\mu)=\log(e)+\beta_0+\mathbf{x}^{\mathsf T}\boldsymbol{\beta}. $$ The offset $\log(e)$ has coefficient fixed at one. The fitted frequency rate per unit of the specified exposure basis is $$ \lambda(\mathbf{x})=\frac{\mu}{e} =\exp\!\left(\beta_0+\mathbf{x}^{\mathsf T}\boldsymbol{\beta}\right). $$ For two rating profiles, the frequency relativity of profile 1 to profile 0 is $\lambda(\mathbf{x}_1)/\lambda(\mathbf{x}_0)$. This interpretation assumes the same exposure basis and compares profiles under the same fitted model.

The Poisson assumption describes the conditional count distribution: given the predictors and exposure, the count has mean equal to its variance, $\mu$. The log link ensures a positive fitted mean. The coefficients describe fitted associations in this specified model; by themselves, they do not establish causal effects. If the Poisson assumptions or model specification are inadequate, exponentiating the coefficients still gives the model’s fitted multiplicative contrasts, but those contrasts may not be reliable descriptions of the underlying risk.

Why Exponentiating a Coefficient Gives a Relativity

Consider two profiles that differ only in a predictor $x_j$, which increases by one unit. If its coefficient is $\beta_j$, the log fitted rate for the second profile is the log fitted rate for the first profile plus $\beta_j$. Taking the ratio of the rates gives the reason the exponential is the relevant conversion.

\log\lambda_1-\log\lambda_0=\beta_j \quad\Longrightarrow\quad \log\left(\frac{\lambda_1}{\lambda_0}\right)=\beta_j \quad\Longrightarrow\quad \frac{\lambda_1}{\lambda_0}=\exp(\beta_j).

Thus, a coefficient of $\log(1.30)$ corresponds to a relativity of $1.30$: the fitted rate is 30% higher than that of the comparison profile, holding other predictors fixed. A coefficient of $\log(0.80)$ corresponds to a relativity of $0.80$, or a fitted rate 20% lower. A negative coefficient can therefore imply a relativity greater than zero and less than one; it does not imply a negative frequency.

Key result: With reference-level coding and no interaction involving a predictor, the exponentiated coefficient for a nonreference level is that level’s fitted frequency ratio to the reference level, with the other predictors held fixed. The intercept is the log fitted rate for the profile whose predictors are all at their reference or zero values.

Combining Class Effects and Interactions

For categorical predictors coded relative to reference levels, each included coefficient contributes to the linear predictor for the selected profile. If there is no interaction, exponentiating the sum of the applicable coefficients is equivalent to multiplying their individual relativities. This gives a convenient way to calculate a fitted rate from a reference rate.

\lambda(\text{class}) =\lambda(\text{reference})\, \exp(\beta_1+\beta_2+\cdots+\beta_k) =\lambda(\text{reference})\, \exp(\beta_1)\exp(\beta_2)\cdots\exp(\beta_k).

An interaction adds a coefficient for a particular combination of predictor levels. Its exponential multiplies the joint relativity. Consequently, a main-effect coefficient describes its contrast at the reference level of the interacting predictor, not necessarily the contrast at every level.

Worked Example: Combining Two Relativities

A hypothetical insurer fits a Poisson log-link frequency model for policy-year claim counts. Its reference class has a fitted rate of $0.0250$ claims per policy-year. Relative to that class, the target has a territory coefficient $\log(1.40)$ and a vehicle-type coefficient $\log(0.80)$. There is no interaction. Calculate the target fitted rate and determine whether it is above or below the reference rate.

Method. Exponentiate each coefficient to obtain its relativity. With no interaction, multiply the relativities and the reference rate.

The territory relativity is $$ \exp(\log(1.40))=1.40, $$ and the vehicle-type relativity is $$ \exp(\log(0.80))=0.80. $$ The combined relativity is $$ 1.40(0.80)=1.12. $$ Therefore the target fitted rate is $$ 0.0250(1.12)=0.0280\text{ claims per policy-year}. $$ The rate is above the reference: the combined relativity $1.12$ is 12% greater than one.

Check. The separate percentage changes are not added. Multiplication gives $1.12$, so the calculated rate is $0.0030$ claims per policy-year higher than $0.0250$, consistent with a 12% increase.

Common mistake avoided. Treating a territory relativity of $1.40$ and a vehicle relativity of $0.80$ as a net increase of $40\%-20\%=20\%$ is incorrect. The factors multiply, giving a net increase of 12%.

Reference Coding and the Intercept

With treatment (reference-level) coding, one level of each categorical predictor is omitted and serves as that predictor’s reference. The intercept is the log rate for the profile at all reference levels, provided there are no other nonzero predictors in that profile. Exponentiating the intercept gives that profile’s fitted rate. A coefficient for another level is a log rate ratio relative to the omitted level, not a fitted rate by itself.

A different coding scheme can change the individual coefficient interpretations. For example, a model using effect coding may not assign the intercept to a single observed reference class. In every case, predictions come from the full linear predictor; interpreting a coefficient requires knowing how the predictors were coded.

Worked Example: Using an Intercept and an Interaction

A hypothetical policy-frequency model has reference rate $0.0400$ claims per policy-year for reference territory and reference vehicle type. The territory-B coefficient is $\log(1.25)$, the vehicle-V coefficient is $\log(0.80)$, and the territory-B-by-vehicle-V interaction coefficient is $\log(1.10)$. Calculate the fitted rate for a policy in territory B with vehicle type V.

Model setup. Assume conditional Poisson counts and a log link, with the stated policy-year exposure basis. The fitted log rate for the target includes the intercept and all three applicable coefficients: $$ \log\lambda_{B,V} =\log(0.0400)+\log(1.25)+\log(0.80)+\log(1.10). $$

Exponentiating converts the sum into a product. The main-effect product is $$ 1.25(0.80)=1.00. $$ Including the interaction gives the joint relativity $$ 1.00(1.10)=1.10. $$ Thus the fitted rate is $$ \lambda_{B,V}=0.0400(1.10)=0.0440\text{ claims per policy-year}. $$

Check. The main effects cancel in this combination, so without the interaction the fitted rate would be $0.0400$. The interaction multiplies it by $1.10$, producing $0.0440$, as calculated.

Common mistake avoided. Ignoring the interaction would yield $0.0400$, while applying the interaction without the main effects would also miss part of the specified linear predictor. The target prediction requires every coefficient whose terms apply to that profile.

Continuous Predictors and the Size of a Change

For a continuous predictor, the coefficient’s relativity corresponds to a one-unit increase in the predictor, with other model terms fixed. The unit is whatever the model uses: a coefficient per one mile is not the same interpretation as a coefficient per 1,000 miles. If the predictor changes by $c$ units, its rate ratio is $\exp(c\beta)$, not merely $\exp(\beta)$.

Worked Example: Applying a Continuous-Predictor Relativity

A hypothetical Poisson model predicts claims per policy-year. The reference profile has rate $0.0500$. A continuous vehicle-age predictor is measured in years, with coefficient $\log(0.97)$ per additional year. The target vehicle is four years older than the reference vehicle and is also in a territory with relativity $0.90$. There are no interactions. Calculate the target rate, rounding the final rate to four decimal places.

Method. The four-year difference contributes relativity $\exp(4\log(0.97))=(0.97)^4$. Multiply that by the territory relativity and the reference rate.

The continuous-predictor factor is $$ (0.97)^2=0.9409,\qquad (0.97)^4=(0.9409)^2=0.88529281. $$ Combining it with the territory factor gives $$ 0.88529281(0.90)=0.796763529. $$ The unrounded fitted rate is $$ 0.0500(0.796763529)=0.03983817645 \text{ claims per policy-year}. $$ Rounded only at the final step to four decimal places, the rate is $0.0398$ claims per policy-year.

Check. The four-year age difference reduces the rate by a factor below one, as does the territory factor. Their product, $0.796763529$, is below one, so the target rate must be below $0.0500$.

Common mistake avoided. Applying $0.97$ only once would treat a four-year difference as a one-year difference. The coefficient is per year, so the correct factor is $(0.97)^4$.

A Reliable Calculation Process

A disciplined sequence helps prevent confusion between coefficients, relativities, and fitted rates. In particular, identify the comparison profile before interpreting any coefficient and check whether any applicable interaction terms are present.

Identify the model scale and rate basis.
Confirm that the model uses a log link and note the stated exposure basis for the fitted rate.

Identify the reference profile.
Read the predictor coding and determine which levels correspond to omitted or baseline categories.

Build the target linear predictor.
Include the intercept and every main-effect, continuous-predictor, and interaction term that applies to the target.

Exponentiate and compare.
Exponentiating the full predictor gives the fitted rate; exponentiating its difference from the reference predictor gives the relativity.

Common Interpretation Traps

Using a coefficient as if it were a relativity. The relativity is $\exp(\beta)$, not $\beta$. A coefficient of $\log(1.40)$ gives a factor of $1.40$.
Adding percentage changes instead of multiplying factors. With no interaction, two relativities combine by multiplication. A factor of $1.40$ and a factor of $0.80$ combine to $1.12$.
Applying a one-unit continuous effect to a multi-unit change. For a $c$-unit change, use $\exp(c\beta)$.
Forgetting an interaction. A joint profile’s fitted rate must include every applicable interaction coefficient.
Confusing a coefficient with a class rate. Under reference coding, a nonreference coefficient is a log ratio; the rate requires the intercept and all applicable terms.

A useful numerical check is to calculate the target-to-reference rate ratio directly from the difference in linear predictors. If the target has a positive log-predictor difference, its fitted rate must be higher; if the difference is negative, its fitted rate must be lower. This check catches sign errors and implausible interpretations before rounding.

Key takeaway: In a log-link frequency model, differences in linear predictors become multiplicative rate ratios after exponentiation. Identify the reference profile, use the correct predictor units, include applicable interactions, and compare fitted rates on the same exposure basis.

Check Your Understanding

For each problem, assume a Poisson count model with a log link and the stated exposure basis. Use the specified reference coding and include all listed terms.

A reference class has a fitted rate of $0.0320$ claims per policy-year. A target differs by territory relativity $1.15$ and vehicle relativity $0.90$, with no interaction. Find the target rate and say whether it is above or below the reference. (Answer: $0.03312$ claims per policy-year; above.)
The intercept corresponds to a reference rate of $0.0180$ claims per policy-year. A nonreference territory has coefficient $\log(0.75)$. Find its rate and state the percentage change from reference. (Answer: $0.0135$ claims per policy-year; 25% lower.)
A reference rate is $0.0600$. The target has territory relativity $1.20$, vehicle relativity $0.85$, and an interaction relativity of $1.10$. Calculate the joint relativity and fitted rate. (Answer: joint relativity $1.122$; fitted rate $0.06732$ claims per stated exposure unit.)
A model has reference rate $0.0400$, a continuous-predictor coefficient $\log(1.02)$ per 1,000 miles, and territory relativity $0.95$. A target is 3,000 miles above the reference value. Find the fitted rate, rounding to five decimal places. (Answer: $0.04033$ claims per stated exposure unit.)
A reference class has rate $0.0250$. Under reference-level coding, class B has coefficient $\log(1.30)$ and class C has coefficient $\log(0.70)$. Find the fitted rates for B and C and the rate ratio of B to C. (Answer: B rate $0.0325$, C rate $0.0175$ claims per stated exposure unit, and B-to-C ratio $1.8571$.)

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