Gamma Mean, Variance, And Variance Function

Why the Gamma Variance Function Matters

For a positive continuous response, the mean alone does not describe how much variation to expect. In a gamma response model, the variance grows with the square of the mean, and a separate dispersion parameter controls its scale. This relationship is useful because it tells us what pattern of variability the model assumes before we consider predictors or choose a link.

The key distinction is between the variance of a particular response distribution and the variance function, which describes how the variance depends on the mean after the dispersion factor has been separated out. We derive both from the gamma cumulant function, using the exponential-dispersion parameterization.

Gamma Exponential-Dispersion Form

Let \(Y\) have a gamma distribution with mean \(\mu>0\) and dispersion \(\phi>0\). In shape-scale notation its shape is \(1/\phi\) and its scale is \(\phi\mu\). Thus its density, for \(y>0\), is

The natural parameter is \(\theta=-1/\mu<0\). The cumulant function for the exponential-dispersion representation is \(b(\theta)=-\log(-\theta)\), and the dispersion factor is \(\phi\). Specifically, the density can be written as

where

The function \(b\) is called the cumulant function because it determines the response’s cumulants. Here the dispersion parameter scales those cumulants: the first cumulant is \(b'(\theta)\), and the second is \(\phi b''(\theta)\). We derive these relationships for this gamma distribution below rather than treating them as formulas to memorize.

Deriving the Cumulant Generating Function

The cumulant generating function of \(Y\) is \(K_Y(t)=\log E[\exp(tY)]\), wherever the expectation is finite. In the density above, multiplying by \(\exp(ty)\) changes the term \(y\theta\) to \(y(\theta+\phi t)\). The resulting kernel is integrable precisely when \(\theta+\phi t<0\), the natural-parameter domain of this gamma density. Comparing its normalizing factor with that of the original density gives

Taking logarithms yields the cumulant generating function. Substituting \(b(\theta)=-\log(-\theta)\) and \(\theta=-1/\mu\) makes its form especially clear:

The stated domain follows from \(\theta+\phi t<0\): since \(\theta=-1/\mu\), this inequality is equivalent to \(t<1/(\phi\mu)\). In particular, the domain contains an open interval around zero, so the first two derivatives of \(K_Y\) at zero give the mean and variance.

Mean, Variance, and Variance Function

Differentiate the cumulant generating function with respect to \(t\). For general \(\mu>0\) and \(\phi>0\), the first two derivatives are

Because the moment generating function is finite near zero, the standard cumulant identities apply: \(K_Y'(0)=E[Y]\) and \(K_Y''(0)=\operatorname{Var}(Y)\). Evaluating at zero gives the gamma mean and variance. Separating the dispersion factor identifies the variance function.

This calculation also verifies the general exponential-dispersion derivative relationships for this case. Differentiating the cumulant function \(b\) with respect to its natural parameter gives

Since \(\theta=-1/\mu\), it follows that \(b'(\theta)=\mu\) and \(\phi b''(\theta)=\phi\mu^2\). Thus the cumulant-function and cumulant-generating-function derivations agree.

What the Variance Function Says

The variance function \(V(\mu)=\mu^2\) is not itself the variance. It describes the mean-dependent part of the variance; the full variance includes the dispersion multiplier \(\phi\). At fixed dispersion, doubling the mean multiplies the variance by four. At fixed mean, doubling the dispersion doubles the variance.

The coefficient of variation makes the role of dispersion especially transparent. Since \(\operatorname{Var}(Y)=\phi\mu^2\), the standard deviation is \(\sqrt{\phi}\mu\) for \(\mu>0\). Dividing by the mean gives a coefficient of variation of \(\sqrt{\phi}\), independent of \(\mu\). Thus changing the mean while holding dispersion fixed changes the absolute spread but not the spread relative to the mean.

Common Mistake: Confusing the Function With the Variance

A frequent error is to report \(V(\mu)=\mu^2\) as the full variance and omit \(\phi\). For example, at \(\mu=20\) and \(\phi=0.25\), this mistake gives a variance of 400. But 400 is the variance-function value; the variance is \(\phi V(\mu)=0.25(400)=100\). The correction is to keep the two quantities distinct: \(V(\mu)\) describes how variance depends on the mean, while \(\phi\) scales that relationship.

The reverse mistake is to treat \(\phi\mu^2\) as the variance function. Under the definition used here, the variance function excludes the dispersion factor, so it is \(\mu^2\). This separation is what allows the mean-dependent variance pattern and the overall dispersion scale to be discussed separately.