Exponential-Dispersion Form of the Inverse-Gaussian

Why the Exponential-Dispersion Form Matters

A probability density can be written in several algebraically equivalent ways, but the form used in a generalized linear model makes particular structure visible. The exponential-dispersion form separates the response, a natural parameter, a cumulant function and a dispersion parameter. For the inverse-Gaussian distribution, this separation also reveals the variance function and the part of the density that must remain in the base measure.

Use the inverse-Gaussian parameterization \(Y\sim\operatorname{IG}(\mu,\lambda)\), with mean \(\mu>0\) and shape \(\lambda>0\). Its density with respect to Lebesgue measure on \((0,\infty)\) is

The exponential-dispersion representation will use \(\phi=1/\lambda\), so larger shape corresponds to smaller dispersion. In general notation, its density has the form \(\exp\{[y\theta-b(\theta)]/a(\phi)+c(y,\phi)\}\). For this distribution, \(a(\phi)=\phi\). The derivation below identifies the remaining components directly from the density.

Deriving the Components

First expand the quadratic expression in the exponent. Since \((y-\mu)^2=y^2-2\mu y+\mu^2\), division by \(2\mu^2y\) gives three terms:

The first two terms involve \(y\) and \(\mu\), and can be collected in the exponential-dispersion numerator. The final term depends on the response and shape parameter, so it belongs in \(c(y,\phi)\). Set

Because \(\theta<0\), we can recover \(\mu=(-2\theta)^{-1/2}\). Consequently, \(b(\theta)=-\sqrt{-2\theta}\). The natural-parameter domain is the open interval \((-\infty,0)\); the restriction is essential because the mean must be positive and finite.

To check the match, substitute the definitions into the part involving \(y\) and \(\theta\):

The exponential of \(c(y,\phi)\) is

Multiplying these factors gives the expanded density: the exponent combines to \(-\lambda(y-\mu)^2/(2\mu^2y)\), while the leading factor is \(\sqrt{\lambda/(2\pi)}\,y^{-3/2}\). Thus the representation matches the specified inverse-Gaussian density, including its normalization factor.

The Base Measure and Its Role

The response-dependent base measure is often displayed as \(h(y,\phi)=\exp\{c(y,\phi)\}\), the factor in the density that is not part of \(\exp\{[y\theta-b(\theta)]/\phi\}\). For this model,

This factor depends on the response \(y\), as well as on \(\phi\). In the exponential-dispersion notation, that dependence is represented by \(c(y,\phi)\). It is not a constant that can be dropped when writing the density: the \(y^{-3/2}\) and \(\exp\{-1/(2\phi y)\}\) terms are needed to reproduce the inverse-Gaussian law. The density remains defined with respect to Lebesgue measure on the positive real line; \(h(y,\phi)\) is its base factor, not a different choice of sample-space support.

What the Cumulant Function Implies

A useful check on an exponential-dispersion representation is that its cumulant function recovers the mean and variance. For a density of the stated form, if differentiation under the integral is valid, then \(E[Y]=b'(\theta)\) and \(\operatorname{Var}(Y)=\phi b''(\theta)\). Here the required differentiations are valid: on any compact interval of negative \(\theta\) values, the density and the derivatives of its unnormalized integrand are bounded in the tails by integrable functions with exponential decay in \(y\) and in \(1/y\). More explicitly, the exponent contains a negative multiple of \(y\) as \(y\) grows and a negative multiple of \(1/y\) as \(y\) approaches zero; differentiation introduces only factors bounded by constants times \(1+y^2\) on that interval. These bounds justify differentiating the normalization integral twice.

For completeness, these identities follow by differentiating the normalization rather than by relying only on a parameter-matching check. Write the density as \(f(y;\theta,\phi)=\exp\{[y\theta-b(\theta)]/\phi+c(y,\phi)\}\). Its derivative with respect to \(\theta\) is \(f(y;\theta,\phi)(y-b'(\theta))/\phi\). Since the density integrates to one, differentiating that integral once gives

Hence \(E[Y]=b'(\theta)\). For the second derivative, the derivative of the log density is \((y-b'(\theta))/\phi\), and its second derivative is \(-b''(\theta)/\phi\). Differentiating the normalized density twice therefore yields

It follows that \(\operatorname{Var}(Y)=\phi b''(\theta)\). Finally, from \(b(\theta)=-\sqrt{-2\theta}\), direct differentiation gives \(b'(\theta)=(-2\theta)^{-1/2}\) and \(b''(\theta)=(-2\theta)^{-3/2}\). Substituting \(\mu=(-2\theta)^{-1/2}\) gives the stated mean and variance. \(\square\)

Common Mistakes and a Practical Check

The most common errors arise from confusing the shape parameter with dispersion or assigning a response-dependent term to the wrong part of the representation.

• Using \(\lambda\) as the dispersion. In the density parameterized by shape \(\lambda\), the exponential-dispersion value is \(\phi=1/\lambda\). For example, shape \(4\) means dispersion \(1/4\), not \(4\).
• Giving \(b(\theta)\) the wrong sign. Since \(b(\theta)=-1/\mu\), it is negative. Then \(y\theta-b(\theta)=-y/(2\mu^2)+1/\mu\), which produces the positive \(+\lambda/\mu\) term in the expanded density.
• Dropping the response-dependent base term. Keeping only \(y\theta-b(\theta)\) omits both the inverse-\(y\) term and the \(y^{-3/2}\) factor. The result is not the inverse-Gaussian density.
• Allowing nonnegative natural parameters. The inverse-Gaussian mean parameterization requires \(\theta<0\). At \(\theta=0\), the corresponding mean is not finite, and \(\theta>0\) cannot be written as \(-1/(2\mu^2)\) for a positive real \(\mu\).