Worked Inverse Gaussian Severity Model

Why Use an Inverse Gaussian Severity GLM?

Claim severities are positive and often have substantial variation. A generalized linear model (GLM) allows the expected severity to vary with rating characteristics while specifying how the variance changes with that expected severity. The inverse Gaussian model is one option when variability grows more rapidly with the mean than it does under a Gamma model.

The distinction is not simply a different label for the same fit. Given a log link and a saturated model for groups, both models can produce the same fitted group means. Their conditional variances, however, grow at different rates as those means increase. That difference matters when interpreting uncertainty and assessing whether a severity model is appropriate.

Definition: For a positive claim severity $Y$ with conditional mean $\mu$, an inverse Gaussian GLM specifies $$ Y\mid \boldsymbol{x}\sim \operatorname{IG}(\mu,\lambda), \qquad E[Y\mid \boldsymbol{x}]=\mu,\qquad \operatorname{Var}(Y\mid \boldsymbol{x})=\phi\mu^3, $$ where $\lambda=1/\phi$ is the inverse Gaussian shape parameter and $\phi>0$ is the dispersion parameter. In a GLM, the mean is linked to the predictors by $g(\mu)=\eta=\boldsymbol{x}^{\mathsf T}\boldsymbol{\beta}$. With a log link, $\log\mu=\eta$, so the fitted mean is positive for every finite linear predictor.

For clarity, the inverse Gaussian density under this parameterization is

f(y\mid\mu,\lambda)= \left(\frac{\lambda}{2\pi y^3}\right)^{1/2} \exp\left\{-\frac{\lambda(y-\mu)^2}{2\mu^2y}\right\}, \qquad y>0.

The tutorial uses individual, fully observed claim severities as responses. The claims are assumed conditionally independent given their covariates, and the specified severity target is the same for every observation being modeled. If observations are truncated, censored, or have different severity targets, the likelihood and fitting procedure must reflect that data structure; applying this ordinary density as though every response were a complete observation would not be appropriate.

The Log Link and the Variance Function

Under a log link, each coefficient describes a multiplicative effect on the conditional mean. If a predictor increases by one unit while the other predictors remain fixed, the mean is multiplied by $\exp(\beta_j)$. A binary indicator with coefficient $\beta_j$ therefore multiplies the expected severity for the indicated group by $\exp(\beta_j)$, relative to its reference group.

The inverse Gaussian variance function is cubic in the mean. In particular, if the mean is multiplied by $c$, while dispersion is held fixed, then

\operatorname{Var}(Y\mid \mu\text{ changed to }c\mu) =\phi(c\mu)^3 =c^3\operatorname{Var}(Y\mid\mu).

A Gamma GLM with the same mean and a log link instead has $\operatorname{Var}(Y\mid\boldsymbol{x})=\phi\mu^2$. Its variance is multiplied by $c^2$ when the mean is multiplied by $c$. Dispersion values have different units and interpretations across these two variance functions, so compare the models through their stated variance formulas rather than treating their dispersion parameters as interchangeable.

Why the Fitted Mean Is the Group Average in a Saturated Fit

A useful special case is a saturated model with one separate mean for each group. For a group with observations $y_1,\ldots,y_n$, the inverse Gaussian log likelihood, omitting terms that do not depend on $\mu$, is

\ell(\mu)=-\frac{\lambda}{2} \sum_{i=1}^n\left(\frac{y_i}{\mu^2}-\frac{2}{\mu}+\frac{1}{y_i}\right).

Differentiating with respect to $\mu$ gives the score

\frac{\partial\ell}{\partial\mu} =\frac{\lambda}{\mu^3}\sum_{i=1}^n(y_i-\mu) =\frac{\lambda}{\mu^3}\left(\sum_{i=1}^n y_i-n\mu\right).

At an interior maximum the score is zero. Since $\lambda>0$ and $\mu>0$, this requires $\sum_i y_i-n\mu=0$, or $\hat{\mu}=\sum_i y_i/n$. Thus, for a saturated group fit, the estimated inverse Gaussian mean is the arithmetic average of the observed severities. The log link changes how group means are represented through coefficients; it does not change this group-average result.

Worked Example: Fit a Two-Group Inverse Gaussian Model

A hypothetical equipment insurer observes six complete claim severities, in dollars. Three claims are from standard equipment, with severities $80, $100, and $120. Three are from high-complexity equipment, with severities $160, $200, and $240. Fit an inverse Gaussian GLM with a log link and a separate mean for each group. Assume conditional independence and a common positive dispersion parameter. The dispersion is not needed to estimate the saturated group means.

Method. Use the saturated-group score result: calculate each group’s arithmetic mean, then express the fitted means as a log-link intercept and group coefficient. This avoids incorrectly treating the group means as geometric averages merely because the link is logarithmic.

For the standard group, the total severity is $80+100+120=300$, so

\hat{\mu}_{\mathrm{standard}}=\frac{300}{3}=100.

For the high-complexity group, the total is $160+200+240=600$, so

\hat{\mu}_{\mathrm{complex}}=\frac{600}{3}=200.

Let $z=0$ denote standard equipment and $z=1$ denote high-complexity equipment. The fitted log-link model is $\log\mu=\beta_0+\beta_1z$. At $z=0$, $\log 100=\beta_0$, giving $\hat{\beta}_0=\log 100\approx4.6052$. At $z=1$, $\log 200=\hat{\beta}_0+\hat{\beta}_1$, so

\hat{\beta}_1=\log(200)-\log(100)=\log 2\approx0.6931.

The high-complexity group’s fitted mean is therefore multiplied by $\exp(\hat{\beta}_1)=2$ relative to the standard group. As a check, $\exp(\hat{\beta}_0)=100$, and $\exp(\hat{\beta}_0+\hat{\beta}_1)=100(2)=200$, matching both arithmetic group means. The fitted group residuals also sum to zero: $80+100+120-3(100)=0$, and $160+200+240-3(200)=0$.

Calculating Conditional Means and Variances

After fitting a model, first evaluate the linear predictor for the risk being rated. Exponentiating that predictor gives the fitted conditional mean. To calculate a conditional variance, use the inverse Gaussian variance function and the model’s dispersion estimate or specified dispersion. The variance describes claim-to-claim variability around the conditional mean; it is not itself a predicted claim severity.

Worked Example: Predict Means and Variances From GLM Coefficients

Suppose a hypothetical severity model for equipment repairs has fitted predictor $\log\mu=\log(100)+\log(1.5)x+\log(2)z$, where $x=1$ indicates a reinforced component and $z=1$ indicates high repair complexity. Let the inverse Gaussian dispersion be $\phi=0.0001$ per dollar. Assume the model applies to positive, complete individual claim severities and that the stated dispersion is common across these covariate combinations. Calculate the fitted mean and variance for all four combinations.

Method. Exponentiate the linear predictor to obtain each mean, then calculate $\phi\mu^3$. The common mistake to avoid is multiplying the mean by a coefficient on the log scale instead of exponentiating that coefficient.

When $x=0,z=0$, the mean is $100$, and the variance is $0.0001(100^3)=0.0001(1{,}000{,}000)=100$ dollars squared.

When $x=1,z=0$, the mean is $100(1.5)=150$. Since $150^3=3{,}375{,}000$, the variance is $0.0001(3{,}375{,}000)=337.50$ dollars squared.

When $x=0,z=1$, the mean is $100(2)=200$. Since $200^3=8{,}000{,}000$, the variance is $0.0001(8{,}000{,}000)=800$ dollars squared.

When $x=1,z=1$, both multiplicative effects apply, so the mean is $100(1.5)(2)=300$. Since $300^3=27{,}000{,}000$, the variance is $0.0001(27{,}000{,}000)=2{,}700$ dollars squared.

The results are internally consistent: adding the reinforced-component indicator multiplies each mean by 1.5, while adding high complexity multiplies it by 2. For the inverse Gaussian model, those mean changes multiply the variances by $1.5^3=3.375$ and $2^3=8$, respectively. For example, the variance rises from 100 to 337.50 when the mean rises from 100 to 150.

Comparing Inverse Gaussian and Gamma Fits on the Same Claims

The following comparison uses a saturated two-group model, so both distributions estimate each group’s mean by its arithmetic average. To estimate dispersion for this example, use the Pearson estimate, defined as the sum of squared Pearson residuals divided by the residual degrees of freedom. This is a moment-based dispersion estimate; it is not the maximum-likelihood estimate of dispersion for either distribution.

Result: For fitted mean $\hat{\mu}_i$, an inverse Gaussian Pearson residual is $(y_i-\hat{\mu}_i)/\hat{\mu}_i^{3/2}$; a Gamma Pearson residual is $(y_i-\hat{\mu}_i)/\hat{\mu}_i$. With $n$ observations and $p$ fitted mean parameters, the Pearson dispersion estimate is the sum of squared residuals divided by $n-p$. The corresponding fitted variance at mean $\mu$ is $\hat{\phi}\mu^3$ for inverse Gaussian and $\hat{\phi}\mu^2$ for Gamma.

Worked Example: Compare Two Variance Structures on the Same Claims

Use the six hypothetical severities from the earlier fit: $80, $100, and $120 in the standard group; $160, $200, and $240 in the high-complexity group. Fit separate group means under both an inverse Gaussian model and a Gamma model, each with a log link. Assume complete, independent claim severities. Estimate each model’s dispersion using the Pearson estimate, with residual degrees of freedom $n-p=6-2=4$, and compare fitted variances at the two group means.

Method. First use the group averages, which are 100 and 200 in both models. Then calculate the Pearson dispersion separately under each model’s variance function. The common mistake is to use $\mu^2$ for the inverse Gaussian residual or $\mu^3$ for the Gamma residual.

For the inverse Gaussian fit, the squared Pearson residual sum in the standard group is $(80-100)^2/100^3+(100-100)^2/100^3+(120-100)^2/100^3$. This equals $400/1{,}000{,}000+0+400/1{,}000{,}000=0.0008$. In the high-complexity group it is $1600/8{,}000{,}000+0+1600/8{,}000{,}000=0.0004$. Thus the total is $0.0012$, and

\hat{\phi}_{\mathrm{IG}}=\frac{0.0012}{4}=0.0003.

The fitted inverse Gaussian variances are $0.0003(100^3)=300$ dollars squared and $0.0003(200^3)=2{,}400$ dollars squared.

For the Gamma fit, the standard-group squared Pearson residual sum is $400/100^2+0+400/100^2=0.08$. The high-complexity-group sum is $1600/200^2+0+1600/200^2=0.08$. The total is $0.16$, so

\hat{\phi}_{\mathrm{Gamma}}=\frac{0.16}{4}=0.04.

The fitted Gamma variances are $0.04(100^2)=400$ dollars squared and $0.04(200^2)=1{,}600$ dollars squared. Both models fit the same means, 100 and 200, but their estimated dispersions and fitted variances differ. In this example, doubling the mean multiplies the inverse Gaussian variance by $2^3=8$, from 300 to 2,400; it multiplies the Gamma variance by $2^2=4$, from 400 to 1,600. The arithmetic is checked directly by substituting each mean and dispersion into its model’s variance function.

Practical Interpretation and Exam Traps

The choice between these severity distributions is not made by looking only at fitted means. In a saturated group model, identical means are compatible with different assumptions about how spread changes across groups. Model selection and validation should consider whether the conditional variance pattern is plausible for the claim population, along with the adequacy of the mean structure.

Do not confuse the link with the distribution. A log link ensures positive fitted means; it does not determine whether the response is Gamma or inverse Gaussian.
Do not use the wrong power of the mean. The inverse Gaussian variance is $\phi\mu^3$; the Gamma variance is $\phi\mu^2$.
Do not interpret a log-link coefficient as an additive change. A coefficient $\beta$ corresponds to a multiplicative mean change of $\exp(\beta)$.
Do not compare dispersion numbers without their variance functions. The dispersion parameters are tied to different models and have different units.
Do not use an ordinary severity density for a different observation mechanism. Truncation, censoring, or a changed definition of the response requires an appropriate likelihood and target.

These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. A reliable solution states the response distribution and mean-variance relationship, evaluates the log-link mean, and only then applies the appropriate variance function.

Check Your Understanding

For each problem, assume independent, fully observed positive claim severities and use the distribution and dispersion stated. Give variance answers in dollars squared.

A saturated inverse Gaussian log-link model has a reference-group mean of $80 and a second-group mean multiplier of 1.25. What are the fitted means for the two groups? (Answer: $80 and $100.)
An inverse Gaussian model has fitted mean $\mu=240$ and dispersion $\phi=0.00005$ per dollar. Calculate the conditional variance. (Answer: $691.20 squared.)
At the same fitted mean of $500, an inverse Gaussian model has $\phi=0.00002$ per dollar and a Gamma model has $\phi=0.03$. Calculate each conditional variance and state which is larger. (Answer: Inverse Gaussian $2,500; Gamma $7,500; the Gamma variance is larger.)
A saturated inverse Gaussian model has two groups with severities $90 and $110 in group A, and $180 and $220 in group B. Calculate the fitted group means and the log-link group coefficient for B relative to A. (Answer: Means $100 and $200; coefficient $\log 2\approx0.6931$.)
An inverse Gaussian log-link model has $\log\mu=\log(120)+\log(1.25)x+\log(0.5)z$. For $x=2,z=1$, with $\phi=0.00008$ per dollar, calculate the fitted mean and conditional variance. (Answer: Mean $93.75; variance $65.92 dollars squared.)

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