Continuous-Predictor Coefficients as Relativities Per Unit

From a Coefficient to a Relativity

A continuous rating variable can describe quantities such as vehicle age, annual mileage, or a building’s distance from a reference point. In a log-link model, its coefficient describes how the modeled mean changes multiplicatively when the predictor changes. The key is to interpret the coefficient in the units in which the predictor was entered—not as a direct percentage change.

The interpretation applies while other predictors in the model are held fixed. It concerns the modeled mean, such as an expected claim frequency or loss cost; it does not mean that each individual claim or policy outcome changes by that amount.

Definition: Suppose a model with a log link includes a continuous predictor $x$ through the linear term $\beta x$, with coefficient $\beta$. For two observations that differ in $x$ by $\Delta x=x_1-x_0$, and are otherwise identical in the model, the relativity of the modeled mean at $x_1$ to the modeled mean at $x_0$ is $\exp(\beta\Delta x)$. For a one-unit increase, the relativity is $\exp(\beta)$.

A relativity is a ratio. A value greater than 1 indicates an increase in the modeled mean; a value below 1 indicates a decrease. To express the change as a percentage, subtract 1 from the relativity and multiply by 100%. A coefficient is a change in the log of the mean per unit, not a percentage change per unit.

Why the Increment Is Multiplied by the Coefficient

For a log-link model, write the modeled mean as $\mu(x)=\exp(\eta(x))$, where $\eta(x)$ is the linear predictor. If the continuous predictor enters linearly and all other terms are held fixed, its contribution to the linear predictor is $\beta x$. The ratio for changing $x$ from $x_0$ to $x_1$ follows by cancellation:

\frac{\mu(x_1)}{\mu(x_0)} = \frac{\exp(C+\beta x_1)}{\exp(C+\beta x_0)} = \exp\!\left(\beta(x_1-x_0)\right) = \exp(\beta\Delta x),

Here $C$ collects the unchanged terms in the linear predictor. The equation shows why the coefficient must be multiplied by the predictor difference: $\beta$ describes the change for one unit, while $\Delta x$ gives the number of units between the two values. It also shows why the intercept and other fixed terms do not affect this ratio.

Important distinction: For an increment of $\Delta x$, the exact relativity is $\exp(\beta\Delta x)$, and the exact percentage change is $[\exp(\beta\Delta x)-1]\times100\%$. For small $\beta\Delta x$, the approximation $[\exp(\beta\Delta x)-1]\approx\beta\Delta x$ can be useful, but it is not the exact calculation.

Worked Examples

Worked Example: A Multi-Year Increase in Vehicle Age

A hypothetical insurer fits a claim-frequency model with independent Poisson claim counts conditional on the predictors. For policy $i$, the count has mean $e_i\lambda_i$, where $e_i$ is exposure in policy-years, and the model for the annual frequency is $\log(\lambda_i)=\alpha+0.08x_i$, with $x_i$ equal to vehicle age in years. Compare otherwise identical vehicles aged 2 and 7 years. Also find the age increase corresponding to a 1.50 relativity. No interest or mortality assumption is used: this is a claim-frequency comparison, not a discounted or life-contingent calculation.

Method. Use the log-link coefficient times the change in age, then exponentiate. The age difference is $7-2=5$ years, so the log difference is $0.08(5)=0.40$.

\frac{\lambda(7)}{\lambda(2)} =\exp(0.40) =1.491824697\ldots

Rounded to four decimal places, the relativity is 1.4918. The percentage change is $(1.491824697\ldots-1)\times100\%=49.1824697\ldots\%$, or 49.18% higher to two decimal places.

For the second part, let $\Delta x$ be the age increase that gives a relativity of 1.50. Take natural logarithms of both sides of $1.50=\exp(0.08\Delta x)$:

\Delta x=\frac{\ln(1.50)}{0.08} =\frac{0.4054651081\ldots}{0.08} =5.068313851\ldots

Rounded to four decimal places, the required increase is 5.0683 years, or about 5.07 years. The result is consistent: an increase of approximately 5.0683 years gives a log difference of approximately 0.405465 and thus a relativity of 1.50. The common mistake this method avoids is treating $0.08$ as an exact 8% increase per year; the model specifies a log change of 0.08 per year, so the exact one-year relativity is $\exp(0.08)$, not 1.08.

Worked Example: A Negative Coefficient and a Predictor Measured in Thousands

For a hypothetical portfolio, an insurer models annual claim frequency using independent Poisson counts with mean $e_i\lambda_i$, where $e_i$ is exposure in policy-years. The log of frequency includes the term $-0.012x_i$, where $x_i$ is annual mileage measured in thousands of miles. Compare otherwise identical risks with 4.0 and 6.5 thousand miles. Assume no interest or mortality basis; neither is needed for this frequency model.

Method. The predictor change is $6.5-4.0=2.5$ units. Since each unit represents 1,000 miles, this is an increase of 2,500 miles. The log change is $(-0.012)(2.5)=-0.030$.

\frac{\lambda(6.5)}{\lambda(4.0)} =\exp(-0.030) =0.9704455335\ldots

Rounded to four decimal places, the relativity is 0.9704. The modeled frequency is therefore $(1-0.9704455335\ldots)\times100\%=2.95544665\ldots\%$, or 2.96% lower, for the higher-mileage risk in this fitted model. The negative coefficient gives a factor below 1 for an increase in mileage.

For an absolute check, suppose the modeled frequency at 4.0 thousand miles is 0.1200 claims per policy-year. At 6.5 thousand miles the model gives $0.1200\times0.9704455335\ldots=0.1164534640\ldots$, or 0.1165 claims per policy-year to four decimal places. Dividing the unrounded new rate by 0.1200 returns the same relativity. The common mistake is to read the negative coefficient as a decrease of 1.2% for every 1,000 miles. The exact one-unit factor is $\exp(-0.012)$; the coefficient itself is a log change, not a percentage change.

Worked Example: Converting Months to a Multi-Year Severity Comparison

A hypothetical insurer models claim severity with a Gamma distribution whose conditional mean is $\mu_i$ and variance is $0.25\mu_i^2$ (equivalently, shape 4 and scale $\mu_i/4$). The log-link model includes $0.015x_i$, where $x_i$ is the age of a piece of equipment in months. At 18 months, the modeled mean severity is $640. Calculate the modeled mean at 48 months, holding all other predictors fixed. No interest or mortality assumption is used; the calculation compares conditional claim severities.

Method. The age difference is $48-18=30$ months, not 2.5 units, because the fitted predictor is measured in months. The log difference is $0.015(30)=0.45$, giving the relativity:

\frac{\mu(48)}{\mu(18)} =\exp(0.45) =1.568312185\ldots

The mean severity at 48 months is the starting mean times that relativity:

\mu(48) =640(1.568312185\ldots) =1003.7197984\ldots

Rounded to the nearest cent, the modeled mean severity is $1,003.72. It is $56.8312185\ldots\%$, or about 56.83%, higher than $640. As a distributional check, the Gamma mean is the scale times the shape: at 48 months the scale is approximately $1003.7198/4=$250.93$, and $4($250.93)$ is approximately $1,003.72 (the unrounded scale gives the unrounded mean). The common mistake is to treat 30 months as 2.5 predictor units and use $\exp(0.015(2.5))$. That would mismatch the increment with the predictor’s stated monthly units.

Units, Direction, and Interpretation

The coefficient’s numerical value depends on how the predictor is measured. A coefficient stated per mile cannot be applied directly to an increment measured in thousands of miles. Before calculating a factor, write the predictor values in the model’s units, or convert their difference into those units. The product $\beta\Delta x$ must be dimensionless.

The comparison’s direction also matters. If $\beta>0$, increasing $x$ produces a relativity above 1; reversing the comparison produces its reciprocal. If $\beta<0$, increasing $x$ produces a relativity below 1, and reversing the comparison produces a factor above 1. In either case, swapping the two values changes $\Delta x$ to $-\Delta x$:

\exp\bigl(\beta(x_0-x_1)\bigr) =\exp\bigl(-\beta(x_1-x_0)\bigr) =\frac{1}{\exp\bigl(\beta(x_1-x_0)\bigr)}.

A positive relativity does not by itself say which risk is larger: always state the comparison order, such as “the mean at $x_1$ divided by the mean at $x_0$.” Also distinguish a percentage decrease from the reverse-direction percentage increase. For example, a factor of 0.90 means a 10% decrease from the starting value; the reverse factor is $1/0.90=1.1111\ldots$, or an 11.11% increase.

Identify the predictor units.
Read the model definition to determine what one unit of $x$ represents.

Find the signed increment.
For a comparison from $x_0$ to $x_1$, calculate $\Delta x=x_1-x_0$. Keep the order of the comparison explicit.

Calculate the factor.
Evaluate $\exp(\beta\Delta x)$; do not substitute the coefficient for a percentage.

Translate and check.
For a percentage change, calculate $[\exp(\beta\Delta x)-1]\times100\%$. Check that its direction agrees with the sign of $\beta$ and $\Delta x$.

Common Exam Traps

Using the coefficient as a percentage. A coefficient of 0.08 is not exactly an 8% increase per unit. Its exact one-unit factor is $\exp(0.08)=1.0833\ldots$, an increase of about 8.33%.
Forgetting the size of the increment. A four-unit change uses $\exp(4\beta)$, not $\exp(\beta)$.
Ignoring the sign of the comparison. For a decrease in $x$, $\Delta x$ is negative. Reversing the comparison takes the reciprocal of the original factor.
Mixing units. If $x$ is measured in months, a 30-month increase is $\Delta x=30$, not 2.5. If the model instead uses years, convert the increment before applying the coefficient.
Confusing a factor with a percentage change. A factor of 1.20 represents a 20% increase; a factor of 0.80 represents a 20% decrease. Neither factor is itself the percentage.

Key takeaway: In a log-link model with a linear continuous term, the relativity for a specified change is $\exp(\beta\Delta x)$. A sound interpretation names the comparison order, uses the predictor’s actual units, and converts the resulting factor to a percentage only after exponentiating.

Check Your Understanding

For each problem, hold all other model terms fixed and use the stated units. Give the relativity in the comparison order requested.

A Poisson frequency model has log-frequency coefficient $0.04$ per year of roof age. What is the relativity for a roof-age increase of 3 years, and what percentage increase does it represent? (Answer: relativity $1.1275$, or a 12.75% increase.)
A log-link model has coefficient $-0.025$ per 1,000 kilometres of annual travel. Find the frequency relativity for an increase from 2,200 to 4,000 kilometres and state the percentage change. (Answer: relativity $0.9560$, or a 4.40% decrease.)
A severity model uses equipment age in days and has coefficient $0.006$ per day. Find the relativity for comparing 120 days with 60 days. (Answer: $\exp(0.006(120-60))=1.4333$, so the modeled mean is 1.4333 times as large.)
A log-link model has coefficient $0.03$ per unit of a continuous predictor. How many units of increase produce a 20% increase in the modeled mean? (Answer: $\ln(1.20)/0.03=6.0774$ units.)
A model has coefficient $0.012$ per $1,000 of insured equipment value. Compare a risk valued at $5,000 with one valued at $8,000, in that order. (Answer: $\exp(0.012(5-8))=0.9646$, so the $5,000 risk has a modeled mean 3.54% lower.)

What You'll Learn