Gamma Density In Exponential-Dispersion Form

Why Write the Gamma Density This Way?

A generalized linear model is built around a response distribution whose density can be organized into a standard form. For a gamma response, that organization makes clear which quantity plays the role of the natural parameter, which function is the cumulant function, and which terms belong to the base measure. The same probability distribution can be described using shape and scale, or using a mean and a dispersion parameter; the exponential-dispersion form makes the latter structure explicit.

The goal here is to perform that rewrite carefully. In particular, the gamma scale is not itself the dispersion parameter, and terms involving the response value must be separated from terms involving the model parameter. Keeping those distinctions straight is essential when reading or deriving a gamma likelihood.

The Exponential-Dispersion Form

A one-parameter exponential-dispersion density has the form below, with respect to Lebesgue measure for a continuous response. The functions \(a\), \(b\), and \(c\) specify the dispersion scaling, cumulant function, and remaining density term.

For the gamma form used here, \(a(\phi)=\phi\), with \(\phi>0\). The natural parameter must lie in a domain where \(b(\theta)\) is defined; the gamma response will require \(\theta<0\). The density is supported on \(y>0\), so its value is zero for \(y\leq 0\). In a formula for \(c(y,\phi)\), this support restriction can be stated separately rather than encoded in the logarithmic expression.

Deriving the Gamma Representation

Use the gamma shape-and-scale convention: a gamma variable with shape \(\alpha>0\) and scale \(\beta>0\) has density \(y^{\alpha-1}\exp(-y/\beta)/\{\Gamma(\alpha)\beta^\alpha\}\) for \(y>0\). Write its mean as \(\mu=\alpha\beta\), and set \(\phi=1/\alpha\). Equivalently, \(\alpha=1/\phi\) and \(\beta=\phi\mu\). Substituting these relations gives

Taking logarithms separates the terms involving \(y\) and \(\mu\):

The terms involving the response and the mean can be grouped as \((-y/\mu-\log\mu)/\phi\). Set \(\theta=-1/\mu\). Since \(\mu>0\), this gives \(\theta<0\), and \(\log\mu=-\log(-\theta)\). Therefore, if \(b(\theta)=-\log(-\theta)\), then \(y\theta-b(\theta)=-y/\mu-\log\mu\). The terms left over are precisely \(c(y,\phi)\) in the definition above.

Proof. Begin with the gamma density in shape-and-scale form and substitute \(\alpha=1/\phi\) and \(\beta=\phi\mu\). Its logarithm is the displayed expression above. With \(\theta=-1/\mu\), we have \(\theta<0\), \(-\log(-\theta)=\log\mu\), and hence \[ \frac{y\theta-b(\theta)}{\phi} = \frac{-y/\mu-\log\mu}{\phi}. \] The remaining terms in the log density are \((1/\phi-1)\log y-\log\Gamma(1/\phi)-(1/\phi)\log\phi\). Thus the log density equals \(\{y\theta-b(\theta)\}/\phi+c(y,\phi)\), which proves the claimed representation for \(y>0\). The gamma density is zero outside that support. \(\square\)

Normalization Confirms the Cumulant Function

The algebra identifies the required components; normalization provides an independent check. For fixed \(\phi>0\) and \(\theta<0\), define \(q=-\theta>0\), so \(b(\theta)=-\log q\). Then the exponential-dispersion expression on \(y>0\) is

To integrate this expression, use the gamma integral \(\int_0^\infty y^{r-1}e^{-sy}\,dy=\Gamma(r)/s^r\), valid for \(r>0\) and \(s>0\). Here \(r=1/\phi>0\) and \(s=q/\phi>0\). Therefore the integral of the density expression is

This calculation shows that the stated base-measure term and cumulant function produce a normalized density throughout the domain \(\theta<0\). The boundary \(\theta=0\) is not included: there \(q=0\), and the gamma-integral argument requires a strictly positive rate.

Worked Conversions

Interpreting the Components Correctly

The parameter mapping is summarized below. In particular, the scale is \(\phi\mu\), not \(\phi\); the dispersion is the reciprocal of the gamma shape.

The division between \(b\) and \(c\) is made according to the exponential-dispersion convention: \(y\theta-b(\theta)\) is divided by \(a(\phi)\), while \(c(y,\phi)\) is not. A quick check is to inspect the log density: the term \(-y/(\phi\mu)\) and the term \(-\log\mu/\phi\) belong together in \(\{y\theta-b(\theta)\}/\phi\). The remaining powers and normalizing constants form \(c\).

Common Mistake: Calling the Scale the Dispersion

A frequent error is to identify the gamma scale \(\beta\) with the exponential-dispersion parameter \(\phi\). This fails because the scale changes with the mean: under this parameterization, \(\beta=\phi\mu\), while \(\phi=1/\alpha\). For example, shape \(4\) and scale \(2.5\) give \(\phi=1/4\), not \(2.5\), and \(\mu=10\). Substituting \(2.5\) for \(\phi\) would produce the wrong shape \(1/\phi\) and the wrong powers and normalizing constants in the density.

Another useful safeguard is to check the support and parameter domain whenever using the representation: \(y\) must be positive, \(\phi\) must be positive, and \(\theta\) must be negative. A formula involving \(\log y\) is not a gamma density formula at \(y\leq0\), and \(b(\theta)=-\log(-\theta)\) is not real-valued for \(\theta\geq0\).