Inverse-Gaussian Mean-Dispersion Representation

Why this representation matters

A probability density can look very different from the exponential-dispersion form used to organize generalized linear models. Rewriting a familiar density in that form is not just a change of notation: it makes clear which part depends on the response, which part carries the mean parameter, and where the dispersion parameter enters. Here we derive that representation for the inverse-Gaussian distribution, keeping track of the support and every normalization factor.

We use \(Y\) for the response, \(\mu>0\) for its mean parameter, and \(\phi>0\) for its dispersion. The inverse-Gaussian density is defined on \(y>0\). In the exponential-dispersion convention, we seek to write it as \(\exp\{[y\theta-b(\theta)]/\phi+c(y,\phi)\}\). The task is to identify each component from the density, rather than infer them by analogy with another distribution.

Deriving the components

The essential step is to expand the quadratic expression in the exponent. Since \(y>0\) and \(\mu>0\), division by \(y\) and by \(\mu^2\) is valid:

The first two terms depend on the response through the term proportional to \(y\), and on the parameter through \(\mu\). To match the specified exponential-dispersion form, set \(\theta=-1/(2\mu^2)\). Then the response-dependent term is \(y\theta/\phi\). The remaining mean-dependent term must equal \(-b(\theta)/\phi\), so \(b(\theta)=-1/\mu\). Because \(-2\theta=1/\mu^2\) and \(\mu>0\), this is equivalently \(b(\theta)=-\sqrt{-2\theta}\).

The density prefactor and the remaining term involving \(y\) do not depend on \(\mu\), so they belong in the carrier. In particular, the factor \(1/\sqrt{2\pi\phi y^3}\) contributes its logarithm to \(c(y,\phi)\). This gives the following complete factorization.

The algebra can be checked by expanding the proposed representation back into the original density. Since \(\sqrt{-2\theta}=1/\mu\), its exponent apart from the logarithmic prefactor is

Also, exponentiating the carrier’s logarithmic term gives \((2\pi\phi y^3)^{-1/2}\). These two checks recover both the exponent and the prefactor of the stated density.

Parameter domain and normalization

The finite-mean inverse-Gaussian family has \(\theta<0\), since \(\theta=-1/(2\mu^2)\) and \(\mu\) is finite and positive. Conversely, every \(\theta<0\) determines a unique positive finite mean parameter:

There is a boundary distinction worth making. If one asks for the largest set of \(\theta\) values for which the normalizing integral with this carrier is finite, the endpoint \(\theta=0\) is included. It gives a normalized boundary density, but not a member with finite \(\mu\): the inverse relation would require \(\mu\) to diverge. For \(\theta>0\), the integral diverges because the factor \(e^{y\theta/\phi}\) grows exponentially as \(y\) increases, whereas the carrier decreases only as a power of \(y\). Thus \(\theta<0\) is the parameter range corresponding to finite positive means, while \(\theta\leq 0\) is the finite-normalizer range for this carrier.

The factorization also has to integrate to one. The following calculation verifies the normalizer directly and shows why the square-root expression in \(b(\theta)\) is required. For \(a>0\) and \(d>0\), use the integral identity

To verify the identity, substitute \(y=d/t^2\). Reversing the bounds after substitution gives $$ \int_0^\infty y^{-3/2}e^{-ay-d/y}\,dy =\frac{2}{\sqrt d}J(\sqrt{ad}),\qquad J(r)=\int_0^\infty e^{-t^2-r^2/t^2}\,dt. $$ For \(r>0\), differentiation under the integral is valid on every compact interval of positive \(r\): the integrand and its derivative are bounded near \(t=0\) by a constant times a power of \(1/t\) times \(e^{-r_0^2/t^2}\), and near infinity by a polynomial times \(e^{-t^2}\), where \(r_0\) is the interval’s positive lower endpoint. Differentiating and then substituting \(u=r/t\) in the resulting integral gives

As \(r\) decreases to zero, dominated convergence applies because \(0\leq e^{-t^2-r^2/t^2}\leq e^{-t^2}\), an integrable function on \((0,\infty)\). Hence \(J(0+)=\int_0^\infty e^{-t^2}\,dt=\sqrt{\pi}/2\). Solving \(J'(r)=-2J(r)\) with this limit yields \(J(r)=(\sqrt{\pi}/2)e^{-2r}\), which proves the identity.

Now set \(a=-\theta/\phi\) and \(d=1/(2\phi)\), both positive when \(\theta<0\). The density in the proposed representation is $$ \frac{e^{\sqrt{-2\theta}/\phi}}{\sqrt{2\pi\phi}}\, y^{-3/2}\exp\!\left\{-\frac{-\theta}{\phi}y-\frac{1}{2\phi y}\right\}. $$ The integral identity says that the integral of the \(y\)-dependent factor \(y^{-3/2}\exp\{-ay-d/y\}\) is \(\sqrt{2\pi\phi}\exp\{-\sqrt{-2\theta}/\phi\}\). Multiplying by the displayed leading factor gives one. Therefore the proposed log-partition term does normalize the density for every \(\theta<0\).

Worked checks and a common pitfall

A frequent mistake is to omit the dispersion factor when expanding the exponent. The constant term is \(1/(\phi\mu)\), not \(1/\mu\); omitting \(\phi\) prevents the expression from matching \(-b(\theta)/\phi\). Another mistake is to put \(-1/(2\phi y)\) into the log-partition function. That term depends on the observed response \(y\), so it belongs in the carrier, which may depend on \(y\) and \(\phi\) but not on the mean parameter. Finally, the factor \(y^{-3/2}\) in the density is part of the carrier’s logarithmic term; dropping it changes the density and destroys normalization.