Worked Gamma Model For Skewed Severity

From a Fitted Gamma Model to a Severity Prediction

Positive insurance severities are often right-skewed: many claims may be moderate, while a smaller number are much larger. A Gamma regression model can describe positive, continuous responses while allowing their variability to increase with their mean. With a log link, the fitted mean remains positive and covariate effects act multiplicatively, making predictions interpretable as relative changes in expected severity.

This tutorial works through fitted Gamma log-link models, including the calculation of predicted means, comparisons between profiles, and the role of dispersion in describing response variation. The examples concern hypothetical severity data; they do not represent an actual insurer or industry data. The material is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: For a positive response $Y$ and covariates $x$, a Gamma regression model in the mean-dispersion parameterization specifies $Y\mid x$ to have conditional mean $\mu(x)>0$ and conditional variance $\operatorname{Var}(Y\mid x)=\phi\mu(x)^2$, where $\phi>0$ is the dispersion parameter. Equivalently, $Y\mid x$ has a Gamma distribution with shape $1/\phi$ and scale $\phi\mu(x)$. A Gamma log-link model specifies $\log(\mu(x))=\eta(x)=\beta_0+\beta_1x_1+\cdots+\beta_px_p$, so $\mu(x)=\exp(\eta(x))$.

The Gamma distribution assigns probability to positive values, not zero or negative values. A Gamma model is therefore not suitable for a response that can be zero as recorded, without first specifying a model or data treatment appropriate to those zeros. The examples below assume strictly positive severity observations and use a common dispersion parameter. They focus on the fitted mean relationship; they do not attempt to estimate regression coefficients from raw observations.

For a given profile, the linear predictor $\eta(x)$ is on the log-mean scale. The prediction on the severity scale is its exponential, $\mu(x)=\exp(\eta(x))$. A coefficient is therefore not generally a dollar change in the mean. For a one-unit increase in a continuous covariate $x_j$, holding the other covariates fixed, the ratio of the new mean to the old mean is $\exp(\beta_j)$.

Worked Predictions and Multiplicative Effects

Worked Example: A Risk Class and Years of Experience

A hypothetical model for positive claim severity, measured in dollars, is $\log(\mu)=\log(900)+\log(1.3)X+0.075Z$. Here $X=1$ indicates a specified risk class and $X=0$ is the reference class; $Z$ is years above the reference experience level. Predict the mean for $X=1,Z=4$, compare it with the mean for $X=0,Z=4$, and calculate the conditional variance for the first profile when $\phi=0.16$. Assume a Gamma response with shape $1/\phi$ and scale $\phi\mu$. No interest-rate or mortality assumption enters this severity prediction.

Method and setup. The specified Gamma shape is $1/0.16=6.25$. The scale for any profile is $0.16\mu$, so it depends on the profile’s predicted mean. First calculate the linear predictor for $X=1,Z=4$:

\eta=\log(900)+\log(1.3)+0.075(4) =\log(900)+\log(1.3)+0.30.

Exponentiating gives $\mu=900(1.3)\exp(0.30)=1170\exp(0.30)$. Using $\exp(0.30)\approx1.3498588$, the fitted mean is $1170(1.3498588)\approx\$1{,}579.33$, rounded to the nearest cent. The model’s conditional scale for this profile is $0.16\mu$, approximately $0.16(1{,}579.3348)=\$252.69$.

For the reference class at the same experience level, set $X=0$: $\mu_0=900\exp(0.30)\approx\$1{,}214.87$. The class effect is clearer as a ratio: $\mu/\mu_0=1.3$. Thus the model predicts a 30% higher mean for the specified class, holding experience fixed.

The conditional variance for the first profile uses the unrounded mean: $\operatorname{Var}(Y\mid X=1,Z=4)=0.16[1170\exp(0.30)]^2$. Since $[1170\exp(0.30)]^2=1{,}368{,}900\exp(0.60)\approx2{,}494{,}298.43$, the variance is approximately $0.16(2{,}494{,}298.43)=399{,}087.75$ dollars squared. The standard deviation is $\sqrt{0.16}\mu=0.4\mu$, approximately $\$631.73$. These values describe variation in individual severities conditional on the profile, not uncertainty in the estimated coefficients.

Check and common mistake. The class ratio is exactly 1.3 because the profiles differ only in $X$. A common error is to add 0.3 to the mean, or to treat $\log(1.3)$ as a dollar increment. The coefficient is on the log scale; exponentiating it produces the multiplicative factor 1.3.

Worked Example: Comparing Two Different Risk Profiles

A hypothetical severity model, with amounts measured in dollars, is $\log(\mu)=\log(800)+\log(1.2)R+0.025A$. Here $R=1$ marks a specified region, $R=0$ is the reference region, and $A$ is years above a reference age. Compare profile A, $(R=1,A=6)$, with profile B, $(R=0,A=10)$, by calculating both predicted means, their dollar difference, and their ratio. Assume Gamma responses with $\phi=0.16$, hence shape 6.25 and profile-specific scale $0.16\mu$. No interest-rate or mortality assumption is needed.

Profile A. Its linear predictor is $\log(800)+\log(1.2)+0.025(6)$. Therefore $\mu_A=800(1.2)\exp(0.15)=960(1.16183424)\approx\$1{,}115.36$.

Profile B. Its linear predictor is $\log(800)+0.025(10)$. Therefore $\mu_B=800\exp(0.25)=800(1.28402542)\approx\$1{,}027.22$.

The predicted difference is $\mu_A-\mu_B\approx1{,}115.3609-1{,}027.2203=\$88.14$, rounded to the nearest cent. For the ratio, cancel the common baseline: \[ \frac{\mu_A}{\mu_B} =\frac{800(1.2)\exp(0.15)}{800\exp(0.25)} =1.2\exp(-0.10) \approx1.0858. \] Profile A’s fitted mean is about 1.0858 times profile B’s, or approximately 8.58% higher. The region effect by itself is 20%, and the four-year difference in $A$ offsets some of that increase.

Check and common mistake. The direct ratio of the two calculated means is approximately $1{,}115.36/1{,}027.22=1.0858$, consistent with the ratio from the model. A common mistake is to compare profiles by reporting only the region’s 20% effect; that ignores the difference in $A$.

Why the Effects Are Multiplicative

The multiplicative interpretation follows directly from the log link. Suppose two profiles differ only in covariate $x_j$, which increases by one unit. Their linear predictors differ by $\beta_j$. Exponentiating each predictor and taking the ratio removes all terms shared by the profiles:

\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 mean ratio for a one-unit increase, with other covariates held fixed. The corresponding percentage change is $100[\exp(\beta_j)-1]\%$, not $100\beta_j\%$ exactly. The approximation $100\beta_j\%$ can be useful for a small coefficient, but the exact percentage should be calculated by exponentiating.

For a binary indicator, changing from 0 to 1 multiplies the fitted mean by $\exp(\beta_j)$, conditional on the other covariates. For a change of $c$ units in a continuous covariate, the ratio is $\exp(c\beta_j)$. These statements concern fitted conditional means; they do not assert that every individual response changes by that factor.

Worked Example: Predicting Mean and Variation for a New Profile

A hypothetical model for positive repair severities in dollars is $\log(\mu)=\log(1200)+\log(1.15)D+0.04K$. Here $D=1$ identifies a specified repair category, and $K$ is distance in units of 10 kilometers above a reference distance. Predict the mean and standard deviation for $D=1,K=3$, given $\phi=0.09$. Assume a Gamma response with shape $1/\phi$ and scale $\phi\mu$. No interest-rate or mortality assumption enters the model.

The shape parameter is $1/0.09=11.111\ldots$. The linear predictor for this profile is $\log(1200)+\log(1.15)+0.04(3)=\log(1200)+\log(1.15)+0.12$. Hence $\mu=1200(1.15)\exp(0.12)=1380(1.12749685)\approx\$1{,}555.95$, rounded to the nearest cent. The profile-specific scale is $0.09\mu\approx\$140.04$.

The conditional variance is $0.09\mu^2$, and the standard deviation is $\sqrt{0.09}\mu=0.3\mu\approx\$466.78$. Using the unrounded mean, the variance is approximately $0.09[1380\exp(0.12)]^2=\$217{,}887.02$ dollars squared. As a check, squaring the standard deviation gives approximately $466.7837^2=217{,}887.02$.

Interpretation and common mistake. The standard deviation is 30% of the conditional mean because the coefficient of variation is $\sqrt{\phi}=0.3$. The response is variable even if the fitted mean is known exactly. It would be incorrect to call this standard deviation the standard error of the fitted mean: coefficient-estimation uncertainty is a separate source of uncertainty.

Reading a Fitted Model Carefully

A reliable calculation keeps the model’s scales and conditioning explicit. Begin with the covariate values, evaluate the linear predictor, and only then exponentiate to obtain the mean. If comparing profiles, distinguish a dollar difference from a ratio: the difference is $\mu_A-\mu_B$, while the relative effect is $\mu_A/\mu_B$. A ratio is often the more direct expression of a log-link effect, but it does not replace a dollar prediction when the question asks for an amount.

The dispersion parameter controls conditional response variability under this Gamma parameterization. Since $\operatorname{Var}(Y\mid x)=\phi\mu^2$, the conditional coefficient of variation is $\sqrt{\operatorname{Var}(Y\mid x)}/\mu=\sqrt{\phi}$. For a fixed $\phi$, larger fitted means come with proportionally larger standard deviations. This mean-variance relationship is part of the model assumption; it is not a claim that all positive severity data necessarily follow it.

Several checks help prevent interpretation errors:

Check positivity: exponentiating any finite linear predictor gives a positive fitted mean.
Check the units: a prediction inherits the units of the response, while its variance has squared response units.
Check the covariate scale: a coefficient for distance per 10 kilometers applies to a one-unit increase in those 10-kilometer units, not to one kilometer.
Check which uncertainty is requested: $\phi\mu^2$ describes response variance conditional on covariates, not uncertainty in coefficient estimates.
Check that comparisons hold other covariates fixed: otherwise a single coefficient cannot describe the full profile difference.

A fitted model can therefore produce three distinct answers from the same parameter set: a predicted conditional mean, a multiplicative comparison between means, and a conditional variance for individual responses. Naming which quantity is being calculated is as important as performing the arithmetic.

Key takeaway: In a Gamma log-link model, calculate the fitted mean by exponentiating the complete linear predictor. A coefficient produces a multiplicative mean effect through its exponential, while dispersion determines response variance through $\phi\mu^2$.

Check Your Understanding

For each problem, use the stated Gamma mean-dispersion parameterization. Round monetary amounts to the nearest cent where applicable.

A model is $\log(\mu)=\log(700)+0.20X+0.03Z$, where $X$ is binary and $Z$ is measured in years above a reference. Find the fitted mean for $X=1,Z=5$. (Answer: $\$993.35$)
A severity model has $\log(\mu)=\log(1000)+\log(1.25)R+0.02A$. Compare profile A, $R=1,A=3$, with profile B, $R=0,A=8$. Give both means and their difference. (Answer: profile A: $\$1{,}327.30$; profile B: $\$1{,}173.51$; difference: $\$153.78$)
For a profile, the fitted mean is $\$1{,}800\exp(0.10)$ and $\phi=0.12$. Find the coefficient of variation and conditional standard deviation. (Answer: coefficient of variation $=\sqrt{0.12}\approx0.3464$; standard deviation $\approx\$689.12$)
A model is $\log(\mu)=\log(600)+0.30D+0.04K$. Compare two profiles that have the same $D$, but whose $K$ values are 2 and 7. What is the ratio of the second fitted mean to the first? (Answer: $\exp[0.04(7-2)]=\exp(0.20)\approx1.2214$, or about 22.14% higher)
At $X=0$, a fitted model predicts a mean of $\$500$; increasing $X$ from 0 to 1 multiplies the mean by 1.15. Under a log link with this single covariate effect, find the mean at $X=4$ and the standard deviation if $\phi=0.25$. (Answer: mean $=500(1.15)^4\approx\$874.50$; standard deviation $=0.5(874.5031)\approx\$437.25$)

What You'll Learn