Why Log-Link Coefficients Need Translation
A coefficient in a log-link model is measured on the logarithmic scale, not directly as a dollar change or a percentage change in expected loss. To use the model in a rating comparison, translate the coefficient into a ratio of fitted means. That translation is exponential: a coefficient difference of \(c\) corresponds to a mean ratio of \(\exp(c)\), with percentage change \(\bigl(\exp(c)-1\bigr)\times100\%\).
The size of a coefficient alone can be misleading. A coefficient of \(0.10\), for example, is not exactly a 10% increase. It gives a ratio of \(\exp(0.10)\), or about \(1.1052\), which is a 10.52% increase. The distinction matters when interpreting estimates, communicating relativities, or comparing risks with multiple predictor differences.
Interpreting a Single Coefficient
Suppose the model includes a continuous predictor \(x\) with coefficient \(\beta\), and all other terms are held fixed. For two values \(x_1\) and \(x_2\), their linear predictors differ by \(\beta(x_2-x_1)\). Exponentiating gives the ratio of their fitted means. Thus a one-unit increase in \(x\) multiplies the mean by \(\exp(\beta)\); a change of \(\Delta x\) units multiplies it by \(\exp(\beta\Delta x)\).
The phrase one unit must be read from the predictor’s definition. If mileage is recorded in thousands of miles, one unit means 1,000 miles. If a predictor is recorded in dollars, a one-unit increase means one dollar, even if that is not a useful practical comparison. A larger change can be more informative, but the coefficient must be multiplied by that change before exponentiating.
A percentage change is directional. If \(c>0\), then \(\exp(c)>1\), indicating an increase. If \(c<0\), then \(0<\exp(c)<1\), indicating a decrease. For a decrease, the percentage change formula produces a negative result. For example, a ratio of \(0.90\) is a 10% decrease, not a 90% decrease.
Why the Exponential Gives the Ratio
The conversion follows directly from the log-link model. Let the original linear predictor be \(\eta_1\), and suppose a predictor change adds \(c\), making the new linear predictor \(\eta_2=\eta_1+c\). The corresponding means are \(\mu_1=\exp(\eta_1)\) and \(\mu_2=\exp(\eta_2)\). Therefore:
The baseline terms cancel, which is why a coefficient contrast gives a relative change even when the intercept and other predictors are not known. To express that ratio as a percentage change, subtract 1 and multiply by 100%. The result describes a change in the modeled conditional mean, holding the other included predictors fixed. It does not by itself establish that changing the predictor causes a change in loss.
Worked Example: A Continuous Mileage Predictor
In a hypothetical personal auto model, expected annual loss has a log link. The predictor \(x\) is annual mileage measured in thousands of miles, and its estimated coefficient is \(0.12\). At a particular set of other predictor values, the fitted mean for a vehicle with \(x=2\) is $800. Find the fitted mean at \(x=5\) and the percentage change. The model specifies the conditional mean only; no interest, mortality, or claim-count distribution assumption is needed for this calculation.
Method. Use the continuous-predictor contrast \(\beta(x_2-x_1)\), then exponentiate. This method avoids treating the coefficient as a direct percentage.
The mileage change is \(5-2=3\) units. Since each unit is 1,000 miles, this is an increase of 3,000 miles. The change in the linear predictor is:
The fitted-mean ratio is \(\exp(0.36)\approx1.433329\). Thus the percentage change is \((1.433329-1)\times100\%\approx43.33\%\). The new fitted mean is \(800(1.433329)\approx\$1{,}146.66\), rounded to the nearest cent.
Check. The new fitted mean exceeds $800, as expected from a positive coefficient and a positive predictor change. The ratio \(1{,}146.66/800\) is approximately \(1.4333\), consistent with the calculated factor. The common mistake would be to report a 36% increase by treating \(0.36\) as the percentage; the log-link calculation gives about 43.33%.
Categorical Predictors and Reference Levels
A categorical predictor is represented by indicator variables relative to a chosen reference level. The reference level’s coefficient is conventionally zero within that factor’s coding; each estimated category coefficient describes the difference from that reference, with other model terms held fixed. Exponentiating a category coefficient gives the category’s fitted-mean ratio to the reference category.
Comparing two nonreference levels requires a contrast: subtract the coefficient for the first level from the coefficient for the second, using the order “new minus comparison.” Exponentiate that difference to get the ratio of the new level’s mean to the comparison level’s mean. Do not interpret each nonreference coefficient as if it were a comparison with every other category.
Worked Example: Comparing Two Territory Levels
A hypothetical property insurer fits a log-link model with territory as a categorical predictor. The reference territory has coefficient 0, the Coastal level has coefficient \(0.25\), and the Inland level has coefficient \(-0.10\). For the same values of all other predictors, the fitted mean in the reference territory is $640. Find the fitted means for Coastal and Inland, and compare Coastal directly with Inland. No interest, mortality, or claim-count distribution assumption is specified or used.
Method. First compare each level with the reference using its own coefficient. Then form the direct contrast between Coastal and Inland. This avoids incorrectly treating a nonreference coefficient as a direct comparison with another nonreference level.
For Coastal, the ratio to the reference is \(\exp(0.25)\approx1.2840254\). Its fitted mean is \(640(1.2840254)\approx\$821.78\). Relative to the reference, the percentage change is \((1.2840254-1)\times100\%\approx28.40\%\).
For Inland, the ratio to the reference is \(\exp(-0.10)\approx0.9048374\). Its fitted mean is \(640(0.9048374)\approx\$579.10\). The percentage change is \((0.9048374-1)\times100\%\approx-9.52\%\), a decrease.
For Coastal compared directly with Inland, the coefficient contrast is \(0.25-(-0.10)=0.35\). Therefore, the ratio is \(\exp(0.35)\approx1.4190675\), or a 41.91% increase. The dollar difference is approximately \(\$821.78-\$579.10=\$242.68\).
Check. The ratio of the unrounded fitted means is \(\exp(0.25)/\exp(-0.10)=\exp(0.35)\), as required. Coastal has a higher fitted mean than Inland, consistent with the positive contrast. The common mistake is to compare Coastal with Inland using \(0.25\) alone; that coefficient compares Coastal with the reference, not with Inland.
Contrasts Across Multiple Predictors
When two risks differ in more than one predictor, add their contributions to the linear-predictor difference before exponentiating. For a model with continuous predictors \(x_j\) and category indicators \(z_k\), the contrast is the sum of each coefficient multiplied by its predictor difference. If \(c\) is that total contrast, the new-to-old mean ratio is \(\exp(c)\). This is the log-link reason that multiplicative rating factors combine by multiplication on the mean scale.
It is generally incorrect to add percentage changes calculated separately for each predictor. Each separate change is a ratio, and the combined effect is the product of those ratios. Equivalently, add the log-scale contributions and exponentiate once. This remains true when some contributions are positive and others negative.
Worked Example: Combining a Mileage Change and a Class Change
A hypothetical commercial vehicle model has a log link. Annual mileage \(x\) is measured in thousands of miles and has coefficient \(-0.04\). The Premium class coefficient is \(0.18\) relative to the Base class. Risk A has \(x=4\) and is in Base class, with fitted mean $1,500. Risk B has \(x=6.5\) and is in Premium class. Find Risk B’s fitted mean and its percentage change from Risk A. The fitted-mean model does not specify an interest, mortality, or claim-count distribution assumption.
Method. Add the continuous and categorical contributions to the linear-predictor contrast, then exponentiate. This avoids adding separately calculated percentage changes.
The mileage difference is \(6.5-4=2.5\) thousand miles, contributing \((-0.04)(2.5)=-0.10\). The class change from Base to Premium contributes \(0.18\). The total contrast is:
The fitted-mean ratio is \(\exp(0.08)\approx1.0832871\). Therefore, Risk B’s fitted mean is \(1{,}500(1.0832871)\approx\$1{,}624.93\), and the percentage change is approximately \(8.33\%\).
As a check, the mileage factor is \(\exp(-0.10)\approx0.9048374\), and the Premium factor is \(\exp(0.18)\approx1.1972174\). Their product is \(\exp(0.08)\approx1.0832871\). The common mistake is to add the separate percentage changes, about \(-9.52\%+19.72\%=10.20\%\). The correct combined change is about \(8.33\%\), because the factors multiply rather than their percentages adding.
Interpretation Checks and Common Traps
Before reporting a coefficient interpretation, identify the modeled mean, the predictor’s units, the comparison being made, and which other terms are held fixed. Then state the ratio and the percentage change distinctly. For example, a ratio of \(1.08\) means the new mean is 1.08 times the comparison mean, or an 8% increase; it does not mean an 108% increase.
- Calling \(\beta\) the percentage change. A coefficient of \(0.12\) gives a one-unit ratio of \(\exp(0.12)\), not an exact 12% increase.
- Forgetting the predictor units. A coefficient per thousand miles must be multiplied by the difference in thousands of miles, not by the difference in individual miles.
- Using the wrong category comparison. A category coefficient is relative to its reference level. For two nonreference levels, subtract their coefficients in the correct order.
- Adding percentages across predictors. Combine coefficients on the log scale, or multiply their mean-scale factors. Do not add the separate percentage changes.
- Reversing the comparison. The contrast for A relative to B has the opposite sign from B relative to A; their ratios are reciprocals.
Check Your Understanding
For each problem, calculate the requested comparison on the fitted-mean scale. Round percentage changes to two decimal places and dollar amounts to the nearest cent.
- A log-link model has a coefficient of \(0.06\) for a continuous predictor measured in units of $10,000. What is the mean ratio and percentage change for an increase of $25,000? (Answer: ratio \(1.1618\); increase \(16.18\%\).)
- A categorical level has coefficient \(0.22\) relative to the reference level, whose fitted mean is $900 at fixed values of the other predictors. Find the categorical level’s fitted mean and percentage change. (Answer: \(\$1{,}121.47\); increase \(24.61\%\).)
- A factor has three levels: Base, Urban, and Rural. Base is the reference; Urban’s coefficient is \(0.14\), and Rural’s is \(-0.09\). Compare Urban directly with Rural. (Answer: ratio \(\exp(0.23)\approx1.2586\); Urban is \(25.86\%\) higher.)
- A continuous predictor measured in thousands of miles has coefficient \(-0.03\). A second profile has 4,000 more miles and changes from the reference category to a category with coefficient \(0.17\). Find the total mean ratio and percentage change. (Answer: ratio \(\exp(0.05)\approx1.0513\); increase \(5.13\%\).)
- A continuous predictor has coefficient \(0.08\) per $5,000, and a categorical change contributes \(-0.11\). For a profile whose continuous predictor increases by $12,500, find the combined mean ratio and percentage change. (Answer: total contrast \(0.09\); ratio \(\exp(0.09)\approx1.0942\); increase \(9.42\%\).)