Gamma GLMs for Claim Severity

Why Model Severity Variation?

A claim-severity model describes the size of a loss conditional on a claim being observed. Predicting the average severity is only part of the task: the model also specifies how the spread of claim amounts changes as the expected severity changes. That spread matters when assessing uncertainty and whether a proposed distribution is plausible for the claim-level data.

A Gamma generalized linear model (GLM) is one standard option for positive, continuous severities. Its defining variance pattern is especially useful to recognize: at a fixed dispersion, severity variance is proportional to the square of the conditional mean. This differs from the constant-variance and Poisson variance patterns and should be checked against the type of data being modeled.

Definition: A Gamma GLM for claim severity models each positive claim-level response $Y_j$, conditional on its covariates $x_j$, as Gamma distributed with conditional mean $\mu_j>0$ and dispersion $\phi>0$. In the mean-dispersion parameterization, $Y_j\mid x_j\sim\operatorname{Gamma}(1/\phi,\phi\mu_j)$, where the two arguments are the shape and scale. Thus $E[Y_j\mid x_j]=\mu_j$ and $\operatorname{Var}(Y_j\mid x_j)=\phi\mu_j^2$. The GLM specifies the mean through a link, $g(\mu_j)=x_j^\mathsf{T}\beta$.

The response here is a positive continuous claim amount, not a claim count. As established in the earlier material on claim-level severity targets and sampling units, the observational unit and definition of the response must be clear before choosing a distribution. The model described here assumes the observed severity values are positive, complete, and measured on a consistent basis.

Parameterization and the Mean–Variance Relationship

The Gamma distribution is often introduced using a shape parameter $k>0$ and a scale parameter $\theta>0$. Under this convention, $E[Y]=k\theta$ and $\operatorname{Var}(Y)=k\theta^2$. To express the distribution in terms of a desired mean $\mu$ and dispersion $\phi$, set $k=1/\phi$ and $\theta=\phi\mu$. These choices give the required mean and variance:

E[Y]=k\theta=\frac{1}{\phi}(\phi\mu)=\mu,\qquad \operatorname{Var}(Y)=k\theta^2 =\frac{1}{\phi}(\phi\mu)^2 =\phi\mu^2.

The parameter $\phi$ is the dispersion, not the variance and not the standard deviation. It determines relative spread: the conditional coefficient of variation is $\sqrt{\phi}$, since $\sqrt{\operatorname{Var}(Y\mid x)}/\mu=\sqrt{\phi\mu^2}/\mu=\sqrt{\phi}$. For a Gamma distribution, the shape is $1/\phi$, so the same dispersion also means the same shape across modeled risk groups. A smaller $\phi$ corresponds to a larger shape and less relative variability.

Key result: For a Gamma GLM with common dispersion $\phi$, the conditional variance and standard deviation are $\operatorname{Var}(Y_j\mid x_j)=\phi\mu_j^2$ and $\operatorname{SD}(Y_j\mid x_j)=\sqrt{\phi}\mu_j$. If the mean is multiplied by $c>0$, the variance is multiplied by $c^2$, while the coefficient of variation stays constant.

This quadratic variance function is not an arbitrary rule added after fitting the mean. It follows from the Gamma distribution’s shape-scale parameterization when the shape (equivalently, the dispersion) is held fixed. Consequently, a Gamma GLM is plausible when larger expected severities tend to have larger absolute spreads in roughly the same proportion to their means. It is less plausible when the data show a substantially different variance pattern, after allowing for sampling variability and the fitted mean structure.

Choosing the Mean Structure

A GLM relates the conditional mean to explanatory variables by choosing a link function $g$ and writing $g(\mu_j)=x_j^\mathsf{T}\beta$. The mean must remain positive because a Gamma response has positive support. The log link is a common choice: it maps any real linear predictor to a positive mean. Other links can be used when their resulting fitted means remain positive over the covariate range of interest.

The link determines how covariates describe the mean; the Gamma family determines the conditional variance pattern. These are separate modeling choices. For example, a log link does not itself imply a Gamma variance function, and specifying a Gamma variance function does not force the use of a log link. This distinction is useful when evaluating whether a model’s mean specification and distributional assumptions are each reasonable.

Define the response.
Identify the claim-level severity being modeled and ensure the target is positive, continuous, and consistently defined.

Specify the conditional mean.
Choose covariates and a link so that $g(\mu_j)=x_j^\mathsf{T}\beta$ and fitted means are positive.

Assess the variance assumption.
For a proposed dispersion, compare severity variation with $\phi\mu_j^2$, recognizing that individual sample variances are noisy.

Worked Example: Interpreting Dispersion for One Risk Group

A hypothetical commercial equipment portfolio uses a Gamma severity model for complete positive claim amounts. For a particular covariate profile, the fitted mean is $\$2{,}000$ and the dispersion is $\phi=0.25$. Find the implied Gamma shape and scale, variance, standard deviation, and coefficient of variation. State the method and a common error it avoids.

Method. Use the mean-dispersion parameterization, $k=1/\phi$ and $\theta=\phi\mu$, then apply the Gamma mean and variance formulas.

Shape and scale. The shape is $k=1/0.25=4$. The scale is $\theta=0.25(\$2{,}000)=\$500$. The resulting distribution is $\operatorname{Gamma}(4,\$500)$ in shape-scale notation; its mean is $4(\$500)=\$2{,}000$, as required.

Variance and standard deviation. The conditional variance is $\phi\mu^2=0.25(\$2{,}000)^2=0.25(\$4{,}000{,}000)=\$1{,}000{,}000$ squared dollars. Its standard deviation is $\sqrt{\$1{,}000{,}000}=\$1{,}000$.

Coefficient of variation. It is $\$1{,}000/\$2{,}000=0.5$, or 50%; equivalently, $\sqrt{0.25}=0.5$.

Check. The shape-scale parameters give mean $\$2{,}000$, variance $\$1{,}000{,}000$ squared dollars, and the same 50% coefficient of variation. A common error is to use $\phi\mu$, which gives $0.25(\$2{,}000)=\$500$; that is not the Gamma variance and does not even have variance units. The Poisson variance function would give $\mu=\$2{,}000$, also not the Gamma variance.

How the Variance Changes Across Risk Groups

The square in the variance formula has a practical consequence: proportional changes in the mean produce larger proportional changes in variance. If two groups have the same $\phi$ and one group’s mean severity is twice the other’s, its variance is four times as large and its standard deviation is twice as large. The coefficient of variation remains unchanged. Therefore, comparing raw variances across groups without accounting for their means can give a misleading impression of relative variability.

Worked Example: Comparing Two Severity Levels

A hypothetical insurer models positive, complete repair claims under a Gamma distribution with common dispersion $\phi=0.16$. For two risk profiles, the fitted mean severities are $\$500$ and $\$1{,}000$. For each profile, calculate the Gamma shape, scale, variance, and standard deviation. Explain the comparison and name the error the calculation avoids.

Method. Apply $k=1/\phi$, $\theta=\phi\mu$, and $\operatorname{Var}(Y)=\phi\mu^2$ separately to each mean.

Profile 1, mean $\$500$. The shape is $1/0.16=6.25$, and the scale is $0.16(\$500)=\$80$. The mean check is $6.25(\$80)=\$500$. The variance is $0.16(\$500)^2=0.16(\$250{,}000)=\$40{,}000$ squared dollars, so the standard deviation is $\sqrt{\$40{,}000}=\$200$.

Profile 2, mean $\$1{,}000$. The shape remains $6.25$, and the scale is $0.16(\$1{,}000)=\$160$. The mean check is $6.25(\$160)=\$1{,}000$. The variance is $0.16(\$1{,}000)^2=0.16(\$1{,}000{,}000)=\$160{,}000$ squared dollars, so the standard deviation is $\sqrt{\$160{,}000}= \$400$.

Comparison and check. The second mean is double the first, its variance is four times as large, and its standard deviation is double. Both coefficients of variation are $200/500=0.4$ and $400/1{,}000=0.4$. The method avoids the common mistake of scaling variance in direct proportion to the mean: the Gamma variance scales with the square of the mean.

When Is the Gamma Variance Pattern Plausible?

A Gamma model is a candidate when observations are positive, continuous amounts and the data support a mean-dependent spread with an approximately stable coefficient of variation. Under the specified parameterization, the conditional skewness is $2\sqrt{\phi}$, so a common dispersion also imposes a common positive skewness. These implications are part of the model, not optional features that can be ignored while retaining the same distributional assumption.

A useful preliminary check is to group claims by fitted mean or by a relevant risk segment, calculate a sample mean and sample variance for each group, and compare the ratios of variance to squared mean. If the Gamma pattern is reasonable and one common dispersion is appropriate, those ratios should be broadly similar. They need not be identical: finite samples produce random variation, and the mean structure may be misspecified. A systematic trend in the ratios, however, is a reason to investigate rather than to assume a common $\phi$.

This comparison is a diagnostic, not a proof that the Gamma distribution fits. The full distribution also implies positive support and a particular shape. In practice, conclusions should consider the amount and quality of data, the fitted mean, and whether the severity target has been altered by selection or limits. A zero-valued response, a negative response, or a response that is censored or selected is not an ordinary complete positive Gamma observation; its data structure needs to be addressed rather than silently treated as if the stated model applied unchanged.

Worked Example: Checking a Common Dispersion Across Segments

A hypothetical insurer divides complete, positive claim-level losses into three groups based on a rating characteristic. The group summaries are: Group A, sample mean $\$300$ and sample variance $\$27{,}000$ squared dollars; Group B, sample mean $\$600$ and sample variance $\$108{,}000$ squared dollars; Group C, sample mean $\$900$ and sample variance $\$243{,}000$ squared dollars. Assess whether the summaries are consistent with a common Gamma dispersion, and give the corresponding shape and scales. State the method and a common error it avoids.

Method. For each group, compute the moment ratio $\widehat{\phi}=s^2/\bar{y}^{\,2}$. Under a Gamma model this ratio estimates the common dispersion, although it will vary from sample to sample.

Group A. The squared mean is $(\$300)^2=\$90{,}000$. Thus the ratio is $27{,}000/90{,}000=0.30$.

Group B. The squared mean is $(\$600)^2=\$360{,}000$. Thus the ratio is $108{,}000/360{,}000=0.30$.

Group C. The squared mean is $(\$900)^2=\$810{,}000$. Thus the ratio is $243{,}000/810{,}000=0.30$.

Interpretation and parameter check. All three ratios equal 0.30, so these summaries are exactly consistent with a common dispersion of $\phi=0.30$, though summaries alone cannot establish a full distributional fit. The shape is $1/0.30=3.\overline{3}$. The scales are $0.30(\$300)=\$90$, $0.30(\$600)=\$180$, and $0.30(\$900)=\$270$. Multiplying each scale by the common shape gives the stated group means. The method avoids comparing raw variances alone: the variances differ, but each is 0.30 times its group’s squared mean.

Common Mistakes and Model Boundaries

Using a linear variance function. The Gamma GLM variance is $\phi\mu^2$, not $\phi\mu$ and not $\mu$. The latter is the Poisson variance function for a count response.
Calling dispersion the variance. $\phi$ is dimensionless in the mean-dispersion parameterization; the response variance has squared response units and depends on $\mu^2$.
Assuming the link sets the variance. The link specifies the conditional mean structure. The selected family supplies its variance function.
Applying the model to values it cannot represent. An ordinary Gamma response is strictly positive and continuous. Zero or negative observations are outside its support, and censoring or selection changes what is observed.
Treating a matching moment ratio as proof. Similar values of $s^2/\bar y^{\,2}$ are evidence about the variance pattern, not confirmation that the entire conditional severity distribution is Gamma.

Key takeaway: A Gamma GLM is appropriate to consider for positive continuous severities when a mean-based GLM structure is sensible and the conditional variance is plausibly $\phi\mu^2$. Its constant coefficient of variation is a specific, testable modeling implication.

Check Your Understanding

Work each problem using the Gamma shape-scale convention stated above. These are hypothetical severity-model scenarios.

A positive claim severity has fitted mean $\$750$ and Gamma dispersion $0.20$. Find its variance, standard deviation, shape, and scale. (Answer: variance $=\$112{,}500$ squared dollars; standard deviation $\approx\$335.41$; shape $=5$; scale $=\$150$.)
Two risk profiles have means $\$400$ and $\$1{,}200$, with common Gamma dispersion $0.09$. Find both variances and standard deviations, and state the variance ratio. (Answer: variances $=\$14{,}400$ and $\$129{,}600$ squared dollars; standard deviations $=\$120$ and $\$360$; the second variance is 9 times the first.)
Two groups of complete positive claims have summaries $(\bar y,s^2)=(\$250,\$15{,}625)$ and $(\$500,\$62{,}500)$. Calculate the two moment ratios and the implied common Gamma shape if the ratios agree. (Answer: both ratios are $0.25$, implying shape $=4$.)
A fitted Gamma model has mean severity $\$1{,}500$ and dispersion $0.40$. Calculate the variance, standard deviation to the nearest cent, and scale parameter. (Answer: variance $=\$900{,}000$ squared dollars; standard deviation $=\$948.68$; scale $=\$600$.)
A severity data set includes a point mass of recorded zero payments along with positive continuous claim amounts. Can an ordinary Gamma GLM for all recorded responses represent this data structure without modification? Give the model-based reason. (Answer: No. An ordinary Gamma distribution has strictly positive support and assigns no probability to an exact zero; the zero-valued component requires separate treatment or a different response model.)

What You'll Learn