Inverse-Gaussian and Gamma Variance Compared

Why Compare Variance Functions?

Two regression models can describe the same conditional mean while assigning different amounts of variability to responses at that mean. For continuous positive responses, the inverse-Gaussian and gamma models are useful to compare because their variance functions have different powers of the mean. As the mean changes across covariate values, that difference changes the comparison: neither model is uniformly more variable over every possible mean.

The comparison is about conditional variance at a given covariate value, not about a sample variance pooled across different covariate values. We will first compare the models with the same mean and dispersion values in a fixed response scale, then show why matching the variance at a reference mean is often a more interpretable comparison.

Variance Functions and the Comparison Setup

These are conditional statements: for each fixed \(x\), they relate the conditional mean to the conditional variance. The gamma parameterization is equivalent to a gamma distribution with shape \(1/\phi_{\mathrm{G}}\) and scale \(\phi_{\mathrm{G}}\mu(x)\). The inverse-Gaussian parameterization uses the usual mean parameter \(\mu(x)\) and shape parameter \(1/\phi_{\mathrm{IG}}\). Each gives the stated mean and variance.

To compare the variance functions, suppose first that both models have the same positive conditional mean \(\mu\). If we also specify dispersion values, their variance ratio is

For fixed positive dispersions, the ratio grows linearly with the mean. The crossover mean is \(\mu_*=\phi_{\mathrm{G}}/\phi_{\mathrm{IG}}\): below it the inverse-Gaussian variance is smaller, at it the variances agree, and above it the inverse-Gaussian variance is larger. This comparison presumes a fixed response scale. Because the dispersion parameters have different units in the two variance functions, equating their numerical values is not by itself a scale-independent way to match models.

Matching at a Reference Mean

For a comparison that does not depend on assigning the same numerical value to two differently scaled dispersion parameters, choose a positive reference mean \(\mu_0\) and require that the two models have the same variance there. The required relation follows directly from the variance functions:

This says how to choose one dispersion once the other and the reference mean are specified. It does not assert that the models have the same variance away from \(\mu_0\).

Proof. Matching at \(\mu_0\) means \(\phi_{\mathrm{IG}}\mu_0^3=\phi_{\mathrm{G}}\mu_0^2\). Since \(\mu_0>0\), division by \(\mu_0^2\) gives \(\phi_{\mathrm{G}}=\phi_{\mathrm{IG}}\mu_0\). At any other \(\mu>0\), the ratio is therefore

The ratio is below, equal to, or above one precisely when \(\mu\) is below, equal to, or above \(\mu_0\), respectively. \(\square\)

The result is especially useful when the mean function varies with \(x\). At covariate values where \(\mu(x)<\mu_0\), the matched inverse-Gaussian model has smaller conditional variance; where \(\mu(x)>\mu_0\), it has larger conditional variance. The point of agreement is set by the chosen reference mean.

How Variability Changes as the Mean Grows

The powers of the mean describe how quickly variance changes when the mean changes. With dispersion held fixed, multiplying the mean by a positive factor \(c\) multiplies inverse-Gaussian variance by \(c^3\), but gamma variance by \(c^2\). In particular, doubling the mean multiplies these variances by eight and four, respectively.

To derive this, write either variance function as \(V(\mu)=\phi\mu^p\), with \(\phi>0\), \(\mu>0\), and \(p=3\) or \(p=2\). Its elasticity is the proportional change in variance per proportional change in the mean, expressed as \(\mu V'(\mu)/V(\mu)\). Differentiating gives \(V'(\mu)=p\phi\mu^{p-1}\), so

Thus, locally, a one-percent increase in the mean corresponds to a three-percent increase in inverse-Gaussian variance or a two-percent increase in gamma variance, with dispersion held fixed. Equivalently, over any finite multiplication by \(c>0\), \(V(c\mu)/V(\mu)=c^p\). This is a statement about the variance functions, not about how the mean must behave in a particular regression.

A related measure is the conditional coefficient of variation, defined here as the conditional standard deviation divided by the conditional mean. Taking square roots of the two variance functions and dividing by \(\mu>0\) gives

At fixed inverse-Gaussian dispersion, relative variability increases with the mean; at fixed gamma dispersion, the coefficient of variation is constant across means. This complements, rather than replaces, a comparison of variances: the coefficient of variation divides out the mean, while conditional variance is measured in squared response units.

Response Units and a Scale-Invariant Comparison

The numerical dispersions cannot be interpreted independently of response units. If a response is changed from \(Y\) to \(Y'=aY\), where \(a>0\), its mean becomes \(\mu'=a\mu\) and its variance becomes \(a^2\operatorname{Var}(Y\mid x)\). The dispersion values must transform differently under the two variance functions.

Derivation. For the inverse-Gaussian model, the transformed variance is \(a^2\phi_{\mathrm{IG}}\mu^3\). Since \(\mu' = a\mu\), writing this as \(\phi'_{\mathrm{IG}}(\mu')^3\) requires \(\phi'_{\mathrm{IG}}a^3\mu^3=a^2\phi_{\mathrm{IG}}\mu^3\), hence \(\phi'_{\mathrm{IG}}=\phi_{\mathrm{IG}}/a\). For the gamma model, the transformed variance is \(a^2\phi_{\mathrm{G}}\mu^2=\phi_{\mathrm{G}}(a\mu)^2\), so \(\phi'_{\mathrm{G}}=\phi_{\mathrm{G}}\). \(\square\)

This transformation preserves the relative comparison. The crossover mean changes to \(a\mu_*\), as it should when the unit changes, and a reference mean changes from \(\mu_0\) to \(a\mu_0\). The variance ratio at the correspondingly transformed mean remains \(\mu/\mu_0\).

Common Pitfall: Treating Dispersion Values as Universally Matched

A tempting argument is: “Both models have dispersion \(0.1\), so compare their variances at mean \(2\); the inverse-Gaussian variance must be larger because it uses \(\mu^3\) and the gamma variance uses \(\mu^2\).” This overlooks that the ratio with equal numerical dispersion values is \(\mu\), and so is not always greater than one. It also overlooks that the numerical dispersion values are tied to the response scale and have different units in these two parameterizations.

The correction is to state the comparison being made. For fixed numerical dispersion values in a specified response scale, compute the ratio \((\phi_{\mathrm{IG}}/\phi_{\mathrm{G}})\mu\) and identify its crossover. For a comparison that matches variability at a meaningful mean, choose \(\mu_0\), set the dispersions so the variances agree there, and use the ratio \(\mu/\mu_0\). Neither comparison establishes which model assigns more probability to a particular extreme outcome: variance alone does not determine tail probabilities.