Log-Link Coefficients as Multiplicative Relativities

Why a Log-Link Coefficient Is Not a Rating Factor

A log-link generalized linear model represents a positive expected quantity on a logarithmic scale. Its fitted coefficients therefore do not usually describe dollar changes directly. In ratemaking, the practical question is often how a rating characteristic changes the expected rate relative to a comparison category. Exponentiating the coefficient translates that log-scale change into a multiplicative factor.

That translation is simple, but interpretation requires care. A positive coefficient means a factor greater than one; a negative coefficient means a factor between zero and one. Neither the coefficient itself nor its sign alone gives the percentage change on the response scale. The factor must be calculated first, and the percentage change is measured relative to one.

Definition: In a log-link GLM, a coefficient $\beta$ attached to a binary indicator that is zero for a comparison level and one for another level corresponds to the multiplicative relativity $\exp(\beta)$, provided the other modeled predictors are held fixed. The relativity is the modeled mean for the indicated level divided by the modeled mean for the comparison level. A factor above one indicates a higher modeled mean; a factor below one indicates a lower modeled mean.

From a Log-Scale Difference to a Relativity

Suppose a GLM models a positive mean $\mu$ using the natural log link. For a binary indicator $x$, where $x=0$ denotes the comparison level and $x=1$ denotes the level being evaluated, write the predictor as $\eta=\alpha+\beta x+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}$. Here, $\mathbf{z}$ denotes the other predictors. The model is $\log(\mu)=\eta$. This setup assumes that the effect of $x$ is represented by this coefficient and that no interaction changes its effect across the values of $\mathbf{z}$.

\mu(x=0)=\exp\left(\alpha+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}\right), \qquad \mu(x=1)=\exp\left(\alpha+\beta+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}\right)

Dividing the second mean by the first cancels the terms common to both levels. This gives the justification for exponentiating the coefficient: the coefficient is a difference in log means, and exponentiating a log difference produces a ratio of means.

\frac{\mu(x=1)}{\mu(x=0)} = \frac{\exp\left(\alpha+\beta+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}\right)} {\exp\left(\alpha+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}\right)} = \exp(\beta)

Thus, the factor applies to the modeled mean, not directly to the log of the mean. If the model is a claim-frequency model with an exposure offset, this ratio compares expected claim counts at the same exposure. For unequal exposures, expected counts also reflect the exposure difference; the coefficient alone remains a ratio of rates per unit of exposure, not a ratio of counts regardless of exposure.

Key result: For a log-link coefficient $\beta$, the corresponding multiplicative factor is $r=\exp(\beta)$. If $r>1$, the modeled mean is $100(r-1)\%$ higher than the comparison mean. If $0<r<1$, it is $100(1-r)\%$ lower. If $\beta=0$, then $r=1$, so the coefficient represents no difference in the modeled mean between the two levels, conditional on the other modeled predictors.

Worked Interpretations

In the examples below, the stated coefficients are fitted log-link coefficients in hypothetical models. The comparison is made with the other modeled predictors held fixed. Dollar quantities are rounded to the nearest cent when shown.

Worked Example: A Positive Coefficient and a Higher Expected Rate

A hypothetical property insurer fits a log-link model for annual pure premium. A binary indicator is one for buildings with a specified roof feature and zero for the comparison roof category. The fitted coefficient is $\beta=0.2623643$. For a fixed set of other predictors, the modeled annual pure premium for the comparison category is $500. Find and interpret the relativity, then calculate the modeled pure premium for the indicated category.

Step 1: Exponentiate the coefficient. The factor is $r=\exp(0.2623643)\approx1.3000$. The indicator level has a modeled mean 1.3000 times the comparison level’s mean.

Step 2: Convert the factor to a percentage change. Since $r>1$, the indicated level is higher by $(1.3000-1)\times100\%=30.00\%$. This is a 30.00% increase, not a 0.2623643% increase.

Step 3: Apply the factor to the supplied comparison mean. The modeled pure premium is $500\times1.3000=\$650.00$.

Check. The increase is $\$650.00-\$500.00=\$150.00$, and $\$150.00/\$500.00=0.30$, or 30.00%. This confirms the interpretation. The method is exponentiation followed by comparison with one; it avoids treating a log-scale coefficient as a percentage.

Worked Example: A Negative Coefficient and a Lower Expected Rate

A hypothetical insurer models annual expected claim frequency per vehicle-year with a log link and an exposure offset. For a particular vehicle-use category, the fitted indicator coefficient relative to its comparison category is $\beta=-0.1625189$. At the same values of all other predictors, the comparison category has an annual expected frequency of $0.0800$ claims per vehicle-year. Find the relativity and the indicated category’s expected frequency.

Step 1: Convert the coefficient to a factor. $r=\exp(-0.1625189)\approx0.8500$. The indicated category has 0.85 times the comparison category’s expected frequency per vehicle-year.

Step 2: State the direction and magnitude. Because the factor is below one, the decrease is $(1-0.8500)\times100\%=15.00\%$. A negative coefficient does not mean a negative frequency; the exponentiated factor and the modeled mean remain positive.

Step 3: Calculate the indicated rate. $0.0800\times0.8500=0.0680$ expected claims per vehicle-year.

Check. The difference is $0.0800-0.0680=0.0120$ claims per vehicle-year, which is $0.0120/0.0800=0.15$, a 15.00% decrease. This comparison is a rate comparison for equal exposure. For two vehicles with different exposures, expected counts would also depend on their respective exposure amounts. The method avoids interpreting the coefficient as a negative percentage or ignoring the exposure unit.

Worked Example: Accurate Exponentiation and Percentage Rounding

A hypothetical rating analysis reports a fitted log-link coefficient of $0.2231$ for one level relative to its comparison level. Calculate the factor and the percentage increase, retaining enough precision to round the final answers to four decimal places.

Step 1: Exponentiate the reported coefficient. Using the coefficient as stated, $\exp(0.2231)\approx1.24994556$. Rounded to four decimal places, the factor is $1.2499$.

Step 2: Calculate the percentage increase using the unrounded factor. The percentage increase is $(1.24994556-1)\times100\%\approx24.994556\%$, which rounds to 24.9946%.

Step 3: Check the rounding convention. The factor is slightly less than 1.2500, so it is not exactly a 25.0000% increase. Reporting the factor to four decimals gives 1.2499; using the unrounded factor for the percentage avoids introducing an extra rounding error.

Check. The factor exceeds one, and the percentage increase is positive. The method is to exponentiate the given coefficient, then subtract one and convert to a percentage. It avoids replacing the reported coefficient with a nearby value that would imply a different result.

Worked Example: A Negative Coefficient With a Close-to-One Factor

For a different hypothetical rating characteristic, a log-link coefficient is $-0.2231$. Calculate the relativity and percentage decrease, rounding both to four decimal places at the end.

Step 1: Exponentiate the coefficient. $\exp(-0.2231)\approx0.80003484$, which rounds to a factor of $0.8000$ to four decimal places.

Step 2: Calculate the percentage decrease. Because the factor is below one, the decrease is $(1-0.80003484)\times100\%\approx19.996516\%$. Rounded to four decimal places, this is 19.9965% lower.

Step 3: Interpret the result. The factor compares the indicated level with its comparison level, holding the other predictors fixed. It means the modeled mean is about 80.00% of the comparison mean, or 19.9965% lower using the unrounded factor.

Check. $0.80003484+0.19996516=1$, so the factor and the proportional decrease are consistent. The method avoids calculating the percentage decrease from the rounded factor, which would give 20.0000% and lose the precision requested.

Reading the Factor Correctly

A relativity is always interpreted against a particular comparison level. The coefficient does not, by itself, identify a dollar premium; it describes a ratio between modeled means. To turn the ratio into a mean for a particular risk, a comparison mean must also be available. A factor of 1.30, for example, says “30% higher than the comparison mean,” not “add 30 dollars” or “charge 30% of the comparison amount.”

The interpretation is conditional on the model specification. In the binary-indicator setup used here, $\exp(\beta)$ is the ratio when other predictors are held fixed. If the model includes an interaction involving that indicator, the effect can depend on the other variable’s value, so the single coefficient alone may not describe the factor for every risk. Likewise, for a model with an offset, compare rates at the same exposure when interpreting the coefficient as a rate relativity.

Factors also behave differently from percentage changes when moving in opposite directions. A factor of 1.25 means 25% higher. Its reciprocal, 0.80, means 20% lower, not 25% lower. The percentage decrease is measured against the original comparison mean: $1-0.80=0.20$. This asymmetry is a consequence of using different denominators when expressing the two comparisons.

Common Mistakes and an Exam Check

Identify the model scale.
Confirm that the model uses a log link. Exponentiation gives the multiplicative mean ratio for the coefficient under the stated comparison and model conditions.

Calculate the factor.
Use $r=\exp(\beta)$. Do not report $\beta$ itself as the relativity.

Interpret direction and size.
For $r>1$, calculate $(r-1)\times100\%$. For $r<1$, calculate $(1-r)\times100\%$.

Check the comparison.
State which level is being compared with which, and hold other predictors fixed. If the model has an exposure offset, distinguish rates from expected counts.

Round at the end.
Use adequate precision in the exponentiation and percentage calculation, then round the final quantities as requested.

Common exam trap. A coefficient of $0.2231$ is not a 22.31% increase. That calculation reads a log-scale coefficient as though it were already a response-scale percentage. The correction is to compute $\exp(0.2231)\approx1.24994556$, then calculate $(1.24994556-1)\times100\%\approx24.9946\%$. Similarly, for $-0.2231$, exponentiate first: the factor is approximately $0.80003484$, so the decrease is approximately $19.9965\%$, not 22.31%. Always interpret the factor relative to one and specify the comparison level.

Key takeaway: In a log-link GLM, the fitted coefficient is a change in log mean. Exponentiating it gives the multiplicative relativity. Interpret that factor against its comparison level, convert it to a percentage using its distance from one, and keep exposure and other predictors consistent in the comparison.

Check Your Understanding

For each problem, exponentiate the coefficient before interpreting it. Round only where the question requests it.

A hypothetical log-link model has a binary-indicator coefficient $\beta=\ln(1.10)$. What is the relativity, and what percentage change does it represent against the comparison level? (Answer: relativity $1.10$; 10% higher.)
A model coefficient is $\beta=\ln(0.75)$. Calculate the relativity and state the percentage change. (Answer: relativity $0.75$; 25% lower.)
For equal exposure and fixed other predictors, a hypothetical comparison category has expected frequency $0.060$ claims per vehicle-year. The indicated category has coefficient $\beta=\ln(0.90)$. Find its expected frequency and percentage change. (Answer: $0.054$ claims per vehicle-year; 10% lower.)
A fitted coefficient is $0.2231$. Using four decimal places for the final factor and percentage, find the relativity and percentage increase. (Answer: relativity $1.2499$; 24.9946% higher.)
A hypothetical frequency model uses an exposure offset. The indicated category has coefficient $\beta=\ln(1.25)$. One risk in that category has exposure $0.40$ vehicle-years. If the comparison rate is $0.080$ expected claims per vehicle-year, calculate the indicated rate and expected claim count for that exposure. (Answer: relativity $1.25$; indicated rate $0.100$ claims per vehicle-year; expected count $0.040$ claims.)

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