Interpreting Frequency Regression Coefficients

Why Interpret Coefficients on the Frequency Scale?

A frequency regression model describes how expected claim frequency varies with measured risk characteristics. In a log-link Poisson model, the regression equation is written on the logarithmic scale, but its practical interpretation is usually on the original frequency scale: a coefficient corresponds to a multiplicative change in expected frequency. The distinction matters because a coefficient such as 0.20 is not a 20% increase.

The interpretation also depends on what is held fixed, how a predictor is coded, and—when predictors interact—which other predictor values are being compared. Exposure must be handled separately from frequency: a policy with twice the exposure has twice the expected count at the same modeled frequency, not twice the frequency.

Definition: Let $N_i$ be the claim count for observation $i$, $e_i$ its exposure, and $x_i$ its vector of risk characteristics. In a log-link Poisson regression model, conditional on the predictors, $$ N_i\mid x_i,e_i\sim\operatorname{Poisson}(\mu_i),\qquad \log(\mu_i)=\log(e_i)+\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}. $$ The term $\log(e_i)$ is an exposure offset: its coefficient is fixed at one. The modeled expected frequency per exposure unit is $$ \lambda_i=\frac{\mu_i}{e_i} =\exp\left(\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}\right). $$

This model assumes that, conditional on the predictors and exposure, each count has a Poisson distribution with mean $\mu_i$. In particular, its conditional variance equals its conditional mean. The usual regression setup also treats observations as conditionally independent. The interpretation of the fitted coefficients as frequency effects follows from the log link; it does not by itself establish that a predictor causes a change in claims.

For a predictor that enters the linear predictor without interactions, increasing it by $\Delta x$, while holding other predictors fixed, changes expected frequency by the ratio $\exp(\beta\Delta x)$. For a one-unit increase, the ratio is $\exp(\beta)$. For an indicator variable coded zero or one, $\exp(\beta)$ is the frequency ratio for the category coded one compared with the reference category coded zero.

Why Exponentiating a Coefficient Gives a Frequency Ratio

The multiplicative interpretation is a direct consequence of the log link. Suppose two otherwise identical risks differ only in a predictor $x_j$, which increases by $\Delta x_j$. Their log frequencies differ by $\beta_j\Delta x_j$. Exponentiating both log frequencies and taking their ratio gives:

\frac{\lambda_{\mathrm{new}}}{\lambda_{\mathrm{old}}} = \frac{\exp(\eta_{\mathrm{old}}+\beta_j\Delta x_j)} {\exp(\eta_{\mathrm{old}})} = \exp(\beta_j\Delta x_j),

where $\eta$ denotes the part of the linear predictor excluding the exposure offset. The other terms cancel because they are held fixed. This is why the coefficient itself is a difference in log frequency, while its exponential is a ratio of frequencies. The exact percentage change is $\bigl(\exp(\beta_j\Delta x_j)-1\bigr)\times100\%$, not $\beta_j\Delta x_j\times100\%$, except as a small-effect approximation.

Interpreting Reference Categories and Exposure

A categorical predictor is commonly represented by indicator variables. One category is omitted and serves as the reference; its effect is included in the intercept. A coefficient for another category compares that category with the reference, holding other modeled characteristics fixed. Changing the reference category changes the reported coefficients, but not the fitted frequencies or comparisons between risks.

For an observation with exposure $e_i$, first calculate its modeled frequency from the risk-characteristic part of the linear predictor. Then multiply by exposure to obtain expected claim count. A frequency ratio between two risks with different exposures compares their rates per exposure unit; it does not compare their expected counts unless their exposures are equal.

Worked Example: A Two-Level Vehicle-Age Indicator

A hypothetical insurer fits a Poisson frequency model with annual policy-years as the exposure unit. Vehicle age is represented by an indicator that equals one for age seven and zero for age two, the reference category. The fitted model is

\log(\mu_i)=\log(e_i)-1.44+0.40I(\text{age seven}).

Find the fitted annual frequencies for both vehicle ages, their frequency ratio, and the expected counts for 1.5 policy-years at each age.

Method. The intercept gives the age-two log frequency, and the age-seven indicator coefficient gives the change in log frequency relative to age two.

For age two, the indicator is zero. Its annual frequency is

\lambda_2=\exp(-1.44)=0.2369277587\ldots.

For age seven, the indicator is one, so the log frequency is $-1.44+0.40=-1.04$. Its annual frequency is

\lambda_7=\exp(-1.04)=0.3534546819\ldots.

The age-seven to age-two frequency ratio is obtained directly from the indicator coefficient:

\frac{\lambda_7}{\lambda_2}=\exp(0.40)=1.4918246976\ldots.

Thus the fitted frequency for age-seven vehicles is about 1.4918 times the age-two frequency, or about 49.18% higher. For 1.5 policy-years, expected count equals exposure times the unrounded modeled frequency. For age two,

\mu_2=1.5\exp(-1.44)=0.3553916380\ldots\approx0.3553916.

For age seven,

\mu_7=1.5\exp(-1.04)=0.5301820229\ldots\approx0.5301820.

The expected counts are rounded to seven decimal places. As a check, their ratio is $0.5301820229\ldots/0.3553916380\ldots=1.4918246976\ldots$, matching the frequency ratio because both risks have the same exposure. Common mistake avoided: multiplying exposure by already-rounded frequencies can change the last reported digit; use the unrounded model values until the final rounding.

Continuous Predictors and the Meaning of a Unit

For a continuous predictor, the coefficient describes a one-unit increase in the units used to fit the model. If the predictor is measured in thousands of miles, a one-unit increase means 1,000 miles, not one mile. For a change of several units, multiply the coefficient by the number of units changed before exponentiating. A negative coefficient gives a ratio below one for an increase in the predictor; it does not mean that frequency becomes negative.

Worked Example: Comparing Mileage and Garaging Characteristics

A hypothetical personal auto model uses annual policy-years of exposure. Let $x$ be annual mileage in thousands of miles, and let $g=1$ indicate garaging in a secured facility, with $g=0$ otherwise. The fitted frequency equation is

\log(\mu_i)=\log(e_i)-2.00+0.10x_i-0.20g_i.

Compare the fitted annual frequencies for a vehicle driven 3,000 miles and not garaged in a secured facility, and a vehicle driven 8,000 miles and garaged in one. Calculate the frequency ratio and the exact percentage difference.

Method. Evaluate the linear predictor for both risks, then exponentiate the difference to obtain their frequency ratio. The mileage difference is five model units because $8{,}000-3{,}000=5{,}000$ miles.

For the first risk, $x=3$ and $g=0$, so its log frequency is $-2.00+0.10(3)-0.20(0)=-1.70$. Its fitted annual frequency is

\lambda_1=\exp(-1.70)=0.1826835241\ldots.

For the second risk, $x=8$ and $g=1$, so its log frequency is $-2.00+0.10(8)-0.20(1)=-1.40$. Its fitted annual frequency is

\lambda_2=\exp(-1.40)=0.2465969639\ldots.

The log-frequency difference is $-1.40-(-1.70)=0.30$. Equivalently, the mileage contribution to the difference is $0.10(5)=0.50$, and the secured-garaging contribution is $-0.20$, giving $0.50-0.20=0.30$. Therefore,

\frac{\lambda_2}{\lambda_1}=\exp(0.30)=1.3498588076\ldots.

The second risk has a fitted frequency about 1.3499 times the first, or about $(1.3498588076\ldots-1)\times100\%=34.99\%$ higher. The ratio also checks against the fitted frequencies: $0.2465969639\ldots/0.1826835241\ldots=1.3498588076\ldots$. Common mistake avoided: treating the five-unit mileage difference as five miles, or treating the coefficient 0.10 as a 10% increase, ignores the predictor’s units and the log link.

Interactions Change the Interpretation of Main Effects

An interaction allows the effect of one predictor to depend on another predictor. With two zero-one indicators $A$ and $B$, a model containing $A$, $B$, and their product $AB$ has linear predictor

\eta=\beta_0+\beta_AA+\beta_BB+\beta_{AB}AB.

The coefficient $\beta_A$ is the effect of changing $A$ from zero to one when $B=0$. When $B=1$, that change in log frequency is $\beta_A+\beta_{AB}$. Thus an interaction coefficient is not, by itself, the frequency ratio for either group; exponentiate the relevant sum to get the ratio for the specified comparison.

Worked Example: Frequency Ratios With an Interaction

A hypothetical insurer models annual claim frequency using $A=1$ for urban territory and $A=0$ for rural territory, and $B=1$ for a newly registered vehicle and $B=0$ otherwise. The fitted model is

\log(\mu_i)=\log(e_i)-1.50+0.20A_i-0.10B_i+0.30A_iB_i.

Calculate the fitted frequency for each of the four combinations, the urban-to-rural ratio for each registration group, and the expected count for 0.75 policy-years for an urban newly registered vehicle.

Method. Substitute each indicator combination into the linear predictor. For each urban-to-rural comparison, keep registration status fixed and subtract the corresponding log frequencies.

For a rural vehicle that is not newly registered, $A=0,B=0$, so $\eta=-1.50$ and $\lambda=\exp(-1.50)=0.2231301601\ldots$. For an urban vehicle that is not newly registered, $A=1,B=0$, so $\eta=-1.50+0.20=-1.30$ and $\lambda=\exp(-1.30)=0.2725317930\ldots$.

For a rural newly registered vehicle, $A=0,B=1$, so $\eta=-1.50-0.10=-1.60$ and $\lambda=\exp(-1.60)=0.2018965180\ldots$. For an urban newly registered vehicle, $A=1,B=1$, so $\eta=-1.50+0.20-0.10+0.30=-1.10$ and $\lambda=\exp(-1.10)=0.3328710837\ldots$.

When $B=0$, the urban-to-rural log-frequency difference is $0.20$, so the ratio is $\exp(0.20)=1.2214027582\ldots$. When $B=1$, the difference is $0.20+0.30=0.50$, so the ratio is $\exp(0.50)=1.6487212707\ldots$. The interaction therefore makes the urban comparison larger for newly registered vehicles.

For 0.75 policy-years of exposure in the urban, newly registered group, expected count is

\mu=0.75\exp(-1.10)=0.2496533128\ldots\approx0.2497.

The count is rounded to four decimal places. The four fitted frequencies and both ratios are consistent: $\exp(-1.10)/\exp(-1.60)=\exp(0.50)$. Common mistake avoided: interpreting $0.20$ as the urban effect for both registration groups ignores the interaction; for $B=1$, the relevant log effect is $0.20+0.30$.

A Practical Interpretation Checklist

For a fitted coefficient, an accurate interpretation should identify the comparison, specify the units, and state whether it concerns frequency or count. Apply the following sequence:

Identify the coding.
For a categorical variable, determine the reference category. For a continuous variable, state the unit in which the predictor was entered.

Specify the comparison.
State which predictor changes and which other characteristics are held fixed. With an interaction, calculate the combined coefficient for the particular comparison.

Convert to a ratio.
Exponentiate the coefficient for a one-unit change, or exponentiate the coefficient times the change in predictor value.

Translate the ratio.
A ratio $r$ means the expected frequency is $r$ times as large. The exact percentage change is $(r-1)\times100\%$.

Apply exposure only for counts.
Multiply modeled frequency by exposure to calculate expected claim count. Do not include exposure in a risk-characteristic frequency ratio unless the question asks about counts and exposures differ.

The most common exam trap is reporting a log-scale coefficient as a percentage change. For example, if $\beta=0.20$, the incorrect calculation is “frequency increases by 20%.” The correct one-unit frequency ratio is $\exp(0.20)=1.2214$, corresponding to an increase of about 22.14%. For a negative coefficient such as $-0.20$, the ratio is $\exp(-0.20)=0.8187$, corresponding to a decrease of about 18.13%, not a decrease of 20%.

These interpretations describe fitted associations conditional on the model’s included predictors. They do not imply that changing a characteristic would cause claims to change by that amount. Nor does the coefficient interpretation establish that the model is suitable: it is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus to interpret the fitted log-link effects, while model adequacy is a separate question.

Check Your Understanding

For each model, calculate the requested frequency ratio or expected count. Interpret ratios as comparisons of expected frequency unless the question specifically asks for expected count.

A fitted model has $\log(\mu_i)=\log(e_i)-1.80+0.25U_i$, where $U=1$ denotes a coastal territory and $U=0$ is the reference territory. Find the coastal-to-reference frequency ratio and exact percentage increase. (Answer: $\exp(0.25)=1.2840$, an increase of 28.40%.)
A continuous predictor $x$ is measured in units of 100 miles and has coefficient $-0.06$. For two otherwise identical risks differing by 500 miles, find the frequency ratio for the risk with more mileage compared with the other. (Answer: $\exp(-0.06\times5)=\exp(-0.30)=0.7408$, a decrease of 25.92%.)
A model is $\log(\mu_i)=\log(e_i)-2.20+0.15x_i-0.20g_i$. For $x=4$, $g=1$, and exposure $e=3$ policy-years, calculate the log frequency, frequency, and expected count. (Answer: log frequency $-1.80$, frequency $\exp(-1.80)=0.1653$ per policy-year, expected count $3\exp(-1.80)=0.4959$.)
A model has $\log(\mu_i)=\log(e_i)-2.00+0.30A_i+0.20B_i-0.10A_iB_i$. Find the frequency ratio for changing $A$ from zero to one when $B=0$, and then when $B=1$. (Answer: when $B=0$, $\exp(0.30)=1.3499$; when $B=1$, $\exp(0.30-0.10)=\exp(0.20)=1.2214$.)
A coefficient of $0.04$ applies per $1,000 increase in a continuous rating variable. Find the frequency ratio and percentage increase for a $6,000 increase, holding other variables fixed. (Answer: $\exp(0.04\times6)=\exp(0.24)=1.2712$, an increase of 27.12%.)

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