Log-Link Relativities for Claim Severity

Why a Log Link Produces Severity Relativities

Claim-severity rating often needs to express how the expected cost of a claim changes across risk characteristics. A multiplicative model is useful for this purpose: each characteristic contributes a factor to a reference severity. The log link gives a direct way to fit such a model while ensuring that fitted mean severities remain positive.

Here, $Y$ denotes a positive claim-severity response and $x$ denotes the claim's observed rating characteristics. The quantity being modeled is the conditional mean $\mu(x)=E[Y\mid x]$, not a promise about the amount of any individual claim. The log link states that the logarithm of this mean is a linear function of the predictors.

Definition: A log-link severity model has the form $\log \mu(x)=\eta(x)$, where $\mu(x)=E[Y\mid x]$ and $\eta(x)$ is a linear predictor. Equivalently, $\mu(x)=\exp(\eta(x))$. A coefficient's exponent is a multiplicative factor in the conditional mean when the associated predictor changes by one unit, holding the other model terms fixed.

For a categorical predictor, the reference category is represented by zero in the model's indicator coding. Its coefficient compares a non-reference category with that reference. For a numerical predictor, a one-unit change is defined by the predictor's measurement scale: one year, one point, or one dollar, for example. Always identify that unit before interpreting the coefficient.

The positive fitted mean follows directly because the exponential of every finite real number is positive. If $\eta(x)=\beta_0+\beta_1x_1+\cdots+\beta_px_p$, then

\mu(x)=\exp(\beta_0)\prod_{j=1}^{p}\exp(\beta_jx_j).

This is a multiplicative expression. The intercept determines the reference mean when all coded predictors are zero, and the exponentiated predictor terms multiply that mean. In a severity model, a fitted mean of $1,000 does not imply that every claim costs $1,000; it describes the model's average response for claims with those characteristics.

Interpreting a Coefficient as a Conditional Ratio

Suppose two claims are alike on all predictors except one numerical predictor $x_j$, and the second claim's value is one unit larger. Their linear predictors differ by $\beta_j$. Taking the ratio of their fitted means cancels the common terms:

\frac{\mu(x_j+1)}{\mu(x_j)} = \frac{\exp(\eta(x_j)+\beta_j)}{\exp(\eta(x_j))} = \exp(\beta_j).

Thus, $\exp(\beta_j)$ is the conditional severity ratio for a one-unit increase, not an additive dollar change. The corresponding percentage change in fitted mean is $100\{\exp(\beta_j)-1\}\%$. For a change of $c$ units in a numerical predictor with no interaction, the ratio is $\exp(c\beta_j)$.

For example, a coefficient of $0.10$ gives a one-unit ratio of $\exp(0.10)$, approximately 1.1052, or a 10.52% increase. It does not mean an increase of ten cents, ten dollars, or 10% exactly. The exact percentage is obtained by exponentiating first and then subtracting one.

Worked Example: Two Categorical Severity Relativities

A hypothetical insurer fits a claim-severity model for positive claim payments. The reference claim is in vehicle class Standard and territory Inland, with fitted mean severity of $800. The log-link coefficients are $\beta_{\text{Heavy}}=\log(1.25)$ and $\beta_{\text{Coastal}}=\log(0.80)$, with no interaction. Calculate the fitted mean for a Heavy-class claim in Coastal territory, and interpret each factor. Assume conditional severity is Gamma with shape $4$ and scale $\mu/4$, so its conditional mean is $\mu$. Interest and mortality assumptions do not apply to this claim-severity calculation.

Method. Start from the reference fitted mean and multiply by the exponentiated coefficient for each applicable category. The Heavy factor is $\exp\{\log(1.25)\}=1.25$. Holding territory fixed, Heavy claims have a fitted mean 1.25 times that for Standard claims. The Coastal factor is $\exp\{\log(0.80)\}=0.80$; holding vehicle class fixed, Coastal claims have a fitted mean 0.80 times the Inland mean.

The combined prediction is

\mu_{\text{Heavy, Coastal}} =800(1.25)(0.80) =800(1.00) =\$800.00.

The two factors exactly offset in this particular combination. As a check, the fitted mean is 25% higher from the vehicle-class factor and 20% lower from the territory factor; the combined change is multiplicative, not the arithmetic sum $25\%-20\%=5\%$. The correct combined ratio is $1.25\times0.80=1.00$. This method avoids treating categorical coefficients as dollar additions.

Numerical Predictors and Changes Across Several Units

A numerical predictor's coefficient is tied to its unit. If the coefficient is per year of building age, a five-year comparison requires five coefficient increments. If the predictor was rescaled—for example, recorded in decades rather than years—the coefficient and its one-unit interpretation would change accordingly, even though the model's fitted predictions need not change.

For a model with no interactions involving $x$, comparing values $x$ and $x+c$, while holding other predictors fixed, gives

\frac{\mu(x+c)}{\mu(x)}=\exp(c\beta_x).

This ratio is conditional on the other modeled characteristics. It describes an association in the fitted mean, not necessarily a causal effect. Other differences between claims, model specification, and data limitations can affect the interpretation.

Worked Example: A Five-Year Increase in Building Age

A hypothetical property insurer models positive claim severity using building age in years. At age 40, for the stated reference territory and construction class, fitted mean severity is $1,200. The age coefficient is $0.04$ per year, with no interactions. Find the fitted means at ages 45 and 50 and the percentage increase from age 40 to age 45. Assume conditional severity is Gamma with shape $4$ and scale $\mu/4$. Interest and mortality assumptions do not apply.

Method. For five additional years, the log mean increases by $5(0.04)=0.20$. The fitted-mean ratio is therefore $\exp(0.20)=1.2214028$. The increase as a percentage is $(1.2214028-1)\times100\%=22.14028\%$.

The age-45 fitted mean is

\mu(45)=\$1{,}200\exp\{(45-40)(0.04)\} =\$1{,}200(1.2214028) =\$1{,}465.68.

For age 50, the change from age 40 is ten years, so the log-scale change is $10(0.04)=0.40$. Thus, $\exp(0.40)=1.4918247$, and

\mu(50)=\$1{,}200\exp(0.40) =\$1{,}200(1.4918247) =\$1{,}790.19.

As a check, the age-50 ratio is the square of the age-45 ratio because the comparison spans twice as many years: $1.2214028^2\approx1.4918247$. The stated values are rounded to the nearest cent only at the final step. The common mistake is to multiply the coefficient $0.04$ by 100 and call the result an exact 4% annual increase; the exact annual factor is $\exp(0.04)$, or about 1.0408.

Interactions Change the Meaning of a Relativity

An interaction allows the effect of one predictor to depend on the value of another. In that case, a coefficient for one predictor alone is not generally the full comparison between two risk groups. All terms that change between the groups must be included in the linear-predictor difference and then exponentiated.

For a binary location indicator $L$, binary construction indicator $M$, and interaction $LM$, consider

\log\mu=\beta_0+\beta_L L+\beta_M M+\beta_{LM}LM.

When $M=0$, the location ratio for $L=1$ versus $L=0$ is $\exp(\beta_L)$. When $M=1$, that ratio is $\exp(\beta_L+\beta_{LM})$. The interaction coefficient adjusts the location effect on the log scale; its exponent is the factor by which the location ratio itself changes between the two construction groups.

Worked Example: Comparing Risk Groups With an Interaction

A hypothetical commercial-property severity model uses Inland as the reference location ($L=0$) and Wood as the reference construction ($M=0$). The reference fitted mean is $950. Coefficients are $\beta_L=\log(1.10)$, $\beta_M=\log(0.90)$, and $\beta_{LM}=\log(1.05)$. Calculate the fitted means for Inland Wood, Coastal Wood, Inland Masonry, and Coastal Masonry. Also find the Coastal-to-Inland ratio for masonry claims. Assume conditional severity is Gamma with shape $5$ and scale $\mu/5$. Interest and mortality assumptions do not apply.

Method. Exponentiating the coefficients gives factors 1.10 for Coastal, 0.90 for Masonry, and 1.05 for the interaction when both indicators equal one. For Inland Wood, all indicators are zero, so the prediction is the reference mean, $950.00.

For Coastal Wood, $L=1$ and $M=0$, so the interaction term is zero:

\mu_{\text{Coastal, Wood}} =\$950(1.10) =\$1{,}045.00.

For Inland Masonry, $L=0$ and $M=1$:

\mu_{\text{Inland, Masonry}} =\$950(0.90) =\$855.00.

For Coastal Masonry, both indicators equal one, so all three factors apply:

\mu_{\text{Coastal, Masonry}} =\$950(1.10)(0.90)(1.05) =\$950(1.0395) =\$987.525 \approx \$987.53.

The Coastal-to-Inland masonry ratio is $(1.10)(1.05)=1.155$, or a 15.5% higher fitted mean. A direct check gives $\$987.525/\$855=1.155$. The interaction changes the Coastal factor from 1.10 for Wood to 1.155 for Masonry. Using 1.10 alone for the masonry comparison would omit the interaction and understate the modeled ratio.

Common Interpretation Trap

Suppose a log-link coefficient is $\log(1.20)$. A frequent error is to call this a $0.20 increase or to add 0.20 to a reference severity. The coefficient is on the log-mean scale. Exponentiating gives the ratio $1.20$, which means a 20% increase in fitted mean, conditional on the other model terms.

For a reference fitted mean of $700, the incorrect additive calculation would give $\$700+\$0.20=\$700.20$. The correct calculation is

\$700\exp\{\log(1.20)\} =\$700(1.20) =\$840.00.

The correct dollar difference in this particular comparison is $140, but it depends on the starting mean. A 20% relativity applied to a different reference mean produces a different dollar difference. The factor, rather than a fixed dollar increment, is what the coefficient represents.

Key takeaway: In a log-link severity model, exponentiate coefficient changes to obtain fitted-mean ratios. Multiply the reference mean by all applicable factors, including interaction terms. Interpret ratios as conditional comparisons of expected severity, not as fixed dollar changes or claims about every individual claim.

Check Your Understanding

For each hypothetical model, calculate the requested ratio or fitted mean. Assume there are no interactions unless the question specifies one. Round dollar amounts to the nearest cent.

A reference risk has fitted mean severity $600. A categorical indicator has coefficient $\log(1.15)$. What is the fitted mean when the indicator is one, and what is its ratio to the reference? (Answer: $690.00; ratio 1.15.)
A numerical predictor is measured in years and has coefficient $0.03$ per year. Find the fitted-mean ratio and percentage change for a five-year increase. (Answer: Ratio $\exp(0.15)=1.1618$; increase 16.18%.)
A reference fitted mean is $1,100. Two applicable categorical effects have coefficients $\log(0.85)$ and $\log(1.20)$. Find the combined ratio and fitted mean. (Answer: Ratio $0.85(1.20)=1.02$; fitted mean $1,122.00.)
In a model with location indicator $L$, construction indicator $M$, and interaction $LM$, the coefficients are $\log(1.08)$, $\log(0.95)$, and $\log(1.10)$, respectively. What is the location ratio for $M=1$, comparing $L=1$ with $L=0$? (Answer: $1.08(1.10)=1.188$, or an 18.8% increase.)
A risk with score 10 has fitted mean severity $2,000. The score coefficient is $-0.12$ per point, with no interactions. Calculate the fitted mean at score 13 and the percentage change from score 10. (Answer: Ratio $\exp(-0.36)=0.6977$; fitted mean $1,395.35; decrease 30.23%.)

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