Dispersion in Gamma and Inverse Gaussian Severity Fits

Why Dispersion Matters in Severity Models

A severity model estimates an expected claim amount for each risk class. Two classes can have the same fitted mean but different variability around that mean. The dispersion parameter describes this variability in Gamma and inverse Gaussian generalized linear models (GLMs). It is therefore important for uncertainty measures and prediction, even when it does not change the fitted mean relativities.

This tutorial considers independent claim-level severity observations, with a correctly specified conditional mean model and a single positive dispersion parameter shared across observations. No interest or mortality assumption is involved: these are claim severity models. The claim distribution and the fitted mean determine the variance relationship; dispersion sets its scale.

Definition: In a severity GLM, let $Y_j$ be the observed severity for claim $j$, and let $\mu_j=E[Y_j\mid x_j]$ be its fitted conditional mean given predictors $x_j$. The dispersion parameter $\phi>0$ scales the conditional variance. For a Gamma GLM, $\operatorname{Var}(Y_j\mid x_j)=\phi\mu_j^2$. For an inverse Gaussian GLM, $\operatorname{Var}(Y_j\mid x_j)=\phi\mu_j^3$.

The power of the mean differs between the models, so dispersion should not be interpreted without naming the distribution. For Gamma severity, the conditional coefficient of variation is constant across risk classes: $\sqrt{\operatorname{Var}(Y_j\mid x_j)}/\mu_j=\sqrt{\phi}$. For inverse Gaussian severity, it is $\sqrt{\phi\mu_j}$, so it depends on the fitted mean as well as on dispersion.

These variance formulas use the standard GLM parameterizations. A Gamma response with mean $\mu$ and dispersion $\phi$ has shape $1/\phi$ and scale $\phi\mu$. An inverse Gaussian response has mean $\mu$ and shape parameter $\lambda=1/\phi$, giving variance $\mu^3/\lambda=\phi\mu^3$. The parameters describe conditional severity given the predictors; they are not parameters of claim frequency.

Estimating Dispersion from Residuals

For a fitted model, the Pearson residual for claim $j$ is the observed-minus-fitted severity divided by the square root of the variance function without its dispersion factor. Thus the residual measures the discrepancy in units of the model's expected standard deviation when $\phi=1$.

r_{P,j}=\frac{Y_j-\hat{\mu}_j}{\sqrt{V(\hat{\mu}_j)}}

Here $V(\mu)=\mu^2$ for Gamma and $V(\mu)=\mu^3$ for inverse Gaussian. A common residual estimate of dispersion is the Pearson chi-square statistic divided by residual degrees of freedom:

\hat{\phi}_{P}=\frac{\sum_{j=1}^{n}r_{P,j}^{\,2}}{n-p} =\frac{1}{n-p}\sum_{j=1}^{n}\frac{(Y_j-\hat{\mu}_j)^2}{V(\hat{\mu}_j)}

In this expression, $n$ is the number of observations and $p$ is the number of estimated mean-model coefficients, including the intercept when one is present. The denominator $n-p$ is the residual degrees of freedom. This Pearson estimate is a widely used estimate, not an identity for every possible dispersion estimator. A likelihood-based estimate may use a different calculation, and the particular estimation method should be identified when interpreting a reported value.

Worked Example: Pearson Dispersion for a Gamma Severity Fit

A hypothetical insurer fits a Gamma log-link model to five independent claim severities, measured in dollars. The model has $p=2$ estimated mean coefficients. The fitted means and observed severities are $100$ and $110$, $120$ and $108$, $150$ and $180$, $200$ and $160$, and $250$ and $250$, respectively. Estimate dispersion by the Pearson method. For this Gamma model, assume shape $1/\phi$ and scale $\phi\mu_j$. Interest and mortality assumptions do not apply.

Method. For a Gamma model, $V(\mu_j)=\mu_j^2$, so $r_{P,j}=(Y_j-\hat{\mu}_j)/\hat{\mu}_j$. The five residuals are $10/100=0.10$, $-12/120=-0.10$, $30/150=0.20$, $-40/200=-0.20$, and $0/250=0$.

Their squared values are $0.0100$, $0.0100$, $0.0400$, $0.0400$, and $0$. Therefore, the Pearson chi-square statistic is $0.0100+0.0100+0.0400+0.0400=0.1000$. The residual degrees of freedom are $n-p=5-2=3$.

\hat{\phi}_{P}=\frac{0.1000}{3}=0.033333\ldots

Thus $\hat{\phi}_{P}$ is approximately $0.0333$, to four decimal places. Under the stated Gamma parameterization, the estimated shape is $1/\hat{\phi}_{P}=30$, and at a fitted mean of $150$ dollars the estimated scale is $\hat{\phi}_{P}(150)=5$ dollars. The estimated conditional coefficient of variation is $\sqrt{0.033333\ldots}=0.1826$, to four decimals. As a check, the residual sum of squares divided by the stated three degrees of freedom gives the reported estimate.

Common mistake avoided: Dividing by $n=5$ would give $0.0200$, but the stated Pearson estimate uses residual degrees of freedom, $n-p=3$, after estimating two mean coefficients.

Worked Example: Pearson Dispersion for an Inverse Gaussian Fit

A hypothetical portfolio has four independent claim severities, measured in thousands of dollars. An inverse Gaussian model with an intercept only fits $\hat{\mu}_j=1$ for each claim; the observed values are $1.2,\ 0.9,\ 1.1,\ 0.8$. Estimate dispersion by the Pearson method. Use mean $\mu_j$, shape $\lambda=1/\phi$, and variance $\mu_j^3/\lambda=\phi\mu_j^3$. Interest and mortality assumptions do not apply.

Method. There is one estimated mean coefficient, so $n-p=4-1=3$. For the inverse Gaussian, $V(1)=1^3=1$, so the four Pearson residuals are $0.2,\ -0.1,\ 0.1,\ -0.2$. Their squared values sum to $0.04+0.01+0.01+0.04=0.10$.

\hat{\phi}_{P}=\frac{0.10}{3}=0.033333\ldots,\qquad \hat{\lambda}=\frac{1}{\hat{\phi}_{P}}=30

The estimated dispersion is $0.0333$, to four decimal places, and the corresponding shape parameter is $30$. At the fitted mean of $1$ thousand dollars, the estimated variance is $0.033333\ldots(1)^3=0.033333\ldots$ in squared thousands of dollars, and the standard deviation is $0.1826$ thousand dollars. At a mean of $4$ thousand dollars with the same dispersion, the variance would instead be $0.033333\ldots(4)^3=2.133333\ldots$, and the coefficient of variation would be $\sqrt{0.033333\ldots(4)}=0.3651$. The different coefficient of variation confirms that inverse Gaussian relative variability depends on the mean. The calculation checks because the residual sum of squares, $0.10$, divided by three residual degrees of freedom yields $1/30$.

Common mistake avoided: Using the Gamma variance $\phi\mu^2$ here would give the wrong residuals and dispersion estimate. The inverse Gaussian variance function is $\mu^3$, not $\mu^2$.

Why Dispersion Changes Uncertainty, Not Log-Link Relativities

Suppose the mean is modeled with a log link, $\log(\mu_j)=x_j^\mathsf{T}\beta$, where $x_j$ is the predictor vector and $\beta$ is the coefficient vector. A shared dispersion parameter appears as a common scale factor in the likelihood score for $\beta$. Consequently, in an unpenalized fit with the same observations and mean specification, changing the estimate of that common factor does not change the coefficient estimates or fitted means.

The cancellation can be seen directly for both distributions. For Gamma severity, the score contribution for $\beta$ under the log link is proportional to $x_j(Y_j-\mu_j)/(\phi\mu_j)$. For inverse Gaussian severity it is proportional to $x_j(Y_j-\mu_j)/(\phi\mu_j^2)$. In either case, setting the sum of score contributions to zero cancels the common factor $1/\phi$. The remaining equation determines the fitted mean coefficients without using the numerical value of $\phi$.

Write the variance function.
Gamma uses $V(\mu)=\mu^2$; inverse Gaussian uses $V(\mu)=\mu^3$. The full variance is $\phi V(\mu)$.

Form the log-link information.
For a GLM, the information contribution before the common $1/\phi$ factor is proportional to $(d\mu/d\eta)^2/V(\mu)$, where $\eta=\log(\mu)$ and $d\mu/d\eta=\mu$.

Identify how dispersion enters.
For Gamma, $\mu^2/V(\mu)=1$; for inverse Gaussian, $\mu^2/V(\mu)=1/\mu$. Thus the information matrix is $I(\beta)=X^\mathsf{T}WX/\phi$, with the appropriate model-specific weights in $W$.

Invert the information.
The approximate covariance matrix is $\operatorname{Cov}(\hat{\beta})\approx\phi(X^\mathsf{T}WX)^{-1}$. Larger dispersion therefore increases coefficient variances and standard errors, even though the fitted means need not change.

The weight pattern in the information matrix is not the focus here: for Gamma log-link severity it is constant before accounting for any observation-level factors, while for inverse Gaussian log-link severity it depends on $1/\mu_j$. In both cases the scalar $\phi$ multiplies the covariance matrix. This is the key distinction: the mean model determines the relativities; dispersion calibrates the uncertainty around their estimates.

A multiplicative relativity for a one-unit change in a predictor with coefficient $\beta_k$ is $\exp(\beta_k)$. Since $\hat{\beta}_k$ does not change when only the common dispersion estimate changes under these assumptions, the point relativity does not change. Its standard error on the coefficient scale does change. Approximate confidence intervals for a relativity must therefore account for dispersion before transforming back to the multiplicative scale.

Worked Example: Dispersion and a Relativity Confidence Interval

In a hypothetical Gamma severity model with a log link, a factor coefficient is $\hat{\beta}=\log(1.10)$. The corresponding diagonal element of the estimated coefficient covariance matrix before multiplying by dispersion is $0.81$. The Pearson dispersion estimate is $0.04$. Find the coefficient standard error and a 95% Wald confidence interval for the factor relativity. Assume a Gamma conditional distribution with variance $\phi\mu^2$, and no interest or mortality component.

Method. Multiply the unscaled covariance element by the dispersion estimate to obtain the coefficient variance: $0.04(0.81)=0.0324$. The standard error is $\sqrt{0.0324}=0.18$. The point relativity remains $\exp(\hat{\beta})=1.10$.

The 95% Wald interval on the coefficient scale is $\hat{\beta}\pm1.96(0.18)$. Since $\log(1.10)=0.09531018$ and $1.96(0.18)=0.3528$, its endpoints are $-0.25748982$ and $0.44811018$. Exponentiating gives the relativity interval:

[\exp(-0.25748982),\ \exp(0.44811018)] \approx [0.7730,\ 1.5654]

The endpoints are rounded to four decimal places. The interval contains the point estimate $1.10$, as expected for a symmetric Wald interval on the coefficient scale transformed by the increasing exponential function. The upper endpoint is $1.5654$, not $1.5655$. If dispersion were smaller while the mean-model fit stayed the same, the coefficient standard error and interval width would decrease, but the point relativity would remain $1.10$.

Common mistake avoided: Using $0.81$ directly as the coefficient variance gives a standard error of $0.9$, omitting the dispersion scale. The covariance element must first be multiplied by $\hat{\phi}$.

Interpretation, Assumptions, and Exam Traps

Dispersion has two related but distinct interpretations. First, it scales the conditional variance of individual severities around their fitted means. Second, it scales the sampling uncertainty of estimated mean-model coefficients through the covariance matrix. It does not itself increase or decrease the fitted mean in the standard setup. Confusing these roles can lead to incorrect rate indications or misleading statements about relativity changes.

The result that dispersion leaves fitted means unchanged requires a common scalar dispersion parameter and the same mean specification and data. It applies to the standard unpenalized likelihood fit for the Gamma and inverse Gaussian models described here. It need not hold if the model changes in another way—for example, if different observations have different dispersion structures, or if a procedure uses dispersion-dependent penalties or changes the estimating equations. A changed dispersion estimate should not be treated as evidence that the fitted mean has changed unless the mean fit itself has been recomputed under a changed specification.

Do not swap variance functions. Gamma variance grows as $\mu^2$; inverse Gaussian variance grows as $\mu^3$. The appropriate function belongs in the Pearson residual.
Do not equate dispersion with the mean. A larger $\phi$ means greater conditional variability at the same fitted mean, not a higher expected severity.
Do not ignore degrees of freedom. The Pearson estimate divides by residual degrees of freedom, not automatically by the raw number of claims.
Do not report unchanged relativities as unchanged uncertainty. The point estimates may be the same while standard errors, confidence intervals, and prediction variability differ.
Do not assume constant relative variability for inverse Gaussian claims. Its coefficient of variation is $\sqrt{\phi\mu}$, so it increases with the conditional mean when $\phi$ is fixed.

Key takeaway: For Gamma and inverse Gaussian severity GLMs, dispersion scales variance and coefficient uncertainty. Under a common dispersion parameter and standard unpenalized log-link fitting, it does not change the fitted mean relativities. Always use the model's own variance function when estimating or interpreting dispersion.

Check Your Understanding

For each problem, use the stated model and parameterization. Round standard errors and probabilities, if any, to four decimal places.

A Gamma severity model has fitted mean $\mu=200$ dollars and dispersion $\phi=0.0225$. Find the conditional variance, standard deviation, and coefficient of variation. (Answer: variance $900$ dollars squared; standard deviation $\$30.00$; coefficient of variation $0.1500$.)
An inverse Gaussian severity model has mean $\mu=4$ thousand dollars and dispersion $\phi=0.05$. Find its conditional variance, standard deviation, and coefficient of variation. (Answer: variance $3.2$ squared thousands of dollars; standard deviation $1.7889$ thousand dollars; coefficient of variation $0.4472$.)
A Gamma model has four observations, one fitted mean coefficient, and Pearson residuals $0.1,\ -0.2,\ 0.1,\ 0$. Estimate dispersion using residual degrees of freedom. (Answer: $(0.01+0.04+0.01+0)/(4-1)=0.0200$.)
In a log-link severity model, a coefficient's covariance element before dispersion scaling is $0.64$, and $\hat{\phi}=0.0625$. The coefficient estimate is $\log(1.25)$. Find its standard error and point relativity. (Answer: standard error $\sqrt{0.64(0.0625)}=0.2000$; point relativity $1.25$.)
At the same fitted mean $\mu=2$ and dispersion $\phi=0.04$, compare Gamma and inverse Gaussian conditional variances and coefficients of variation. (Answer: Gamma variance $0.16$, coefficient of variation $0.2000$; inverse Gaussian variance $0.32$, coefficient of variation $0.2828$.)

What You'll Learn