From a Log Coefficient to a Rating Relativity
A rating model with a log link represents differences between categories on a logarithmic scale. The fitted coefficient may therefore look unlike the rating factor an actuary would place in a multiplicative rating structure. Exponentiating the coefficient converts that log-scale difference into a factor: a value greater than 1 indicates a higher modeled amount than the reference category, and a value below 1 indicates a lower amount.
The key is to distinguish the coefficient from the relativity. A coefficient such as \(0.20\) is not a 20% increase. Its corresponding multiplicative change is \(\exp(0.20)\), or about 1.2214, which is about a 22.14% increase. The conversion is essential whenever a fitted log-scale category effect must be interpreted in the original units of a rating model.
This interpretation assumes the modeled mean is positive and the model uses the log link, so the logarithm of the mean is modeled. It also assumes that \(\beta_j\) is indeed the log-scale contrast with the stated reference category. If a coefficient is defined using a different contrast, its exponent describes that contrast instead; it should not automatically be called a factor relative to a particular category.
Why Exponentiation Gives the Factor
To justify the conversion, compare the fitted means for the category of interest and its reference category, at the same values of all other rating variables. Let \(\mu_j\) be the modeled mean for category \(j\), \(\mu_{\mathrm{ref}}\) the modeled mean for the reference category, and \(\beta_j\) the log-scale contrast between them. By the definition of that contrast,
Subtracting \(\log(\mu_{\mathrm{ref}})\) from both sides gives \(\log(\mu_j)-\log(\mu_{\mathrm{ref}})=\beta_j\). The logarithm of a ratio is the difference of the logarithms, so the left side is \(\log(\mu_j/\mu_{\mathrm{ref}})\). Exponentiating both sides then gives
Thus the exponentiated coefficient is a ratio, not an additive change in dollars. Once the reference mean is known, multiplying it by the ratio gives the modeled mean for the category. The absolute dollar difference depends on the reference mean; the relativity itself does not.
Worked Conversions
In practice, coefficients are often reported as rounded estimates, so an exponentiated factor will also be approximate. Keep adequate precision during the calculation, then round the reported money amount to the cent. The examples use hypothetical fitted rating quantities.
Worked Example: A Positive Coefficient and a Higher Modeled Loss Cost
A hypothetical log-link model gives a category coefficient of \(0.2231436\) for a delivery-vehicle category relative to the reference category. At the same values of the other rating variables, the reference category’s modeled annual loss cost is $640. Find the relativity, the category’s modeled loss cost, and the percentage change from the reference.
Step 1: Exponentiate the coefficient. The log-scale coefficient is the natural logarithm of the multiplicative factor, so
The category’s relativity is approximately 1.25. It is 1.25 times the reference modeled loss cost, not 0.2231436 times that amount.
Step 2: Calculate the modeled loss cost.
The modeled difference is \(\$800.00-\$640.00=\$160.00\).
Step 3: State the percentage change.
The fitted category mean is 25.0000% higher than the reference mean. Check: 25% of $640 is $160, and $640 plus $160 is $800. This method avoids reading the coefficient 0.2231436 as a 22.31436% increase.
Worked Example: A Negative Coefficient and a Lower Modeled Rate
A hypothetical rating model assigns a category coefficient of \(-0.1053605\) to a type of small commercial equipment relative to the reference category. The reference category’s fitted annual rate is $1,250, with the other modeled variables held fixed. Find the category relativity, fitted rate, dollar reduction, and percentage reduction.
Step 1: Convert the coefficient to a relativity.
The factor is below 1, so the fitted amount is lower than the reference amount.
Step 2: Apply the factor to the reference rate.
Step 3: Calculate the reduction and its percentage. The dollar reduction is \(\$1{,}250.00-\$1{,}125.00=\$125.00\). The percentage change is
The category’s fitted rate is therefore 10.0000% lower than the reference rate, a reduction of $125.00. Check: 10% of $1,250 is $125, and subtracting that amount gives $1,125. The negative coefficient indicates the direction of change, but its magnitude is not itself a percentage reduction.
Worked Example: Avoiding Direct Subtraction of a Negative Coefficient
For a hypothetical specialty-service category, the fitted coefficient relative to the reference is \(-0.3566749\). The reference modeled annual amount is $950. Calculate the relativity and fitted amount, then diagnose the error in calculating the rate as \(\$950(1-0.3566749)\).
Step 1: Exponentiate the coefficient.
The category has a relativity of approximately 0.70, or 70% of the reference modeled amount.
Step 2: Calculate the fitted amount using the relativity.
The reduction from the reference amount is \(\$950.00-\$665.00=\$285.00\), which is 30% of $950.
Step 3: Identify and correct the mistake. Directly subtracting the coefficient from 1 gives
That calculation is arithmetically consistent, but it uses the wrong factor. The coefficient is on the log scale, so the factor must be \(\exp(-0.3566749)\), not \(1-0.3566749\). The correct fitted amount is \(\$950\exp(-0.3566749)\approx \$665.00\). Check: a 0.70 relativity means a 30% reduction, and 30% of $950 is $285; $950 minus $285 is $665.
Interpreting the Result Precisely
A clear interpretation names both the comparison and the scale. For example, if the coefficient is \(-0.1053605\), it is not enough to say “the coefficient is negative” or “the rate falls by 0.1053605.” Exponentiating gives a factor of 0.90, so the modeled amount for that category is 90% of the reference amount, or 10% lower, with other modeled variables held fixed.
The distinction between relativity and percentage change is useful in both directions. A factor of 1.08 means the modeled amount is 108% of the reference amount, hence 8% higher. A factor of 0.92 means it is 92% of the reference amount, hence 8% lower. In general, subtract 1 from the factor to get the signed proportional change. Do not subtract the coefficient from 1.
| Log-scale coefficient | Exponentiated factor | Interpretation relative to reference |
|---|---|---|
| \(\beta_j=0\) | \(\exp(\beta_j)=1\) | No modeled change |
| \(\beta_j>0\) | \(\exp(\beta_j)>1\) | Higher modeled amount |
| \(\beta_j<0\) | \(0<\exp(\beta_j)<1\) | Lower modeled amount |
A relativity describes a ratio of fitted means, not a guarantee about every individual observation. Actual insurance losses can vary around the modeled mean. Further, the category comparison is conditional on the other quantities in the model being held fixed; it does not by itself describe the difference between two risks that may differ in several ways.
Common Mistakes and Exam Checks
- Using the coefficient as the factor. If \(\beta_j=0.20\), the factor is \(\exp(0.20)\), not 0.20.
- Treating the coefficient as a percentage. A coefficient of 0.20 does not mean a 20% increase. Exponentiate it first; \(\exp(0.20)\approx1.2214\), about a 22.14% increase.
- Subtracting a coefficient from 1. For a negative coefficient, \(1-\beta_j\) is generally not the relativity. Use \(\exp(\beta_j)\).
- Reversing the comparison. A factor of 0.80 means the modeled amount is 80% of the reference, or 20% lower. It does not mean 80% lower.
- Omitting the reference comparison. State which category is the reference and that other modeled variables are held fixed.
- Rounding too early. Keep the coefficient and exponentiation at available precision, calculate the fitted amount, and round money only in the final reported result.
Check Your Understanding
For each problem, interpret the coefficient relative to the stated reference category. Keep other modeled variables fixed, and round money to the cent.
- A hypothetical log-link model gives a coefficient of \(0.0769610\) for a category whose reference modeled annual loss cost is $700. Find the relativity, modeled loss cost, and percentage change. (Answer: relativity \(\approx1.08\); modeled loss cost $756.00; 8.0000% higher.)
- A fitted coefficient is \(-0.2231436\), and the reference modeled rate is $840. Find the factor, fitted rate, and dollar reduction. (Answer: relativity \(\approx0.80\); fitted rate $672.00; reduction $168.00, or 20.0000%.)
- A category coefficient is \(0.4054651\), and the reference mean is $460. Calculate the category factor, modeled mean, and dollar increase. (Answer: relativity \(\approx1.50\); modeled mean $690.00; increase $230.00, or 50.0000%.)
- A coefficient of \(-0.0512933\) applies to a category with a reference modeled amount of $2,400. Find the relativity and fitted amount, and explain the percentage interpretation. (Answer: relativity \(\approx0.95\); fitted amount $2,280.00; 5.0000% lower, a $120.00 reduction.)
- A hypothetical fitted category mean is $1,080, while the reference category mean at the same other predictor values is $900. Find the relativity, the implied log-scale coefficient, and the percentage change. (Answer: relativity \(1.20\); coefficient \(\log(1.20)\approx0.1823\); 20.0000% higher.)