Deriving the Canonical Inverse-Gaussian Link

From the Inverse-Gaussian Density to Its Natural Parameter

The canonical link of a generalized linear model is determined by the distribution’s natural parameter, not chosen independently of it. For the inverse-Gaussian response, the natural parameter is negative and proportional to the reciprocal of the squared mean. This tutorial derives that relationship and shows exactly what happens to regression coefficients when the same mean model is written using a positive-valued inverse-square link.

Write $Y$ for a response with mean $\mu>0$ and dispersion $\phi>0$. Its inverse-Gaussian density is

f(y;\mu,\phi) = \frac{1}{\sqrt{2\pi\phi y^3}} \exp\left\{ -\frac{(y-\mu)^2}{2\phi\mu^2y} \right\}, \qquad y>0.

The exponential-dispersion representation uses a natural parameter $\theta$, a cumulant function $b(\theta)$, a dispersion function $a(\phi)$, and a base density factor $h(y,\phi)$:

f(y;\theta,\phi) = h(y,\phi) \exp\left\{ \frac{y\theta-b(\theta)}{a(\phi)} \right\}.

For this inverse-Gaussian parameterization, $a(\phi)=\phi$. Expand the density’s exponent before identifying the terms that depend on $y$ and $\mu$:

-\frac{(y-\mu)^2}{2\phi\mu^2y} = -\frac{y}{2\phi\mu^2} +\frac{1}{\phi\mu} -\frac{1}{2\phi y}.

The term $-1/(2\phi y)$ can be included in the base factor along with the original normalizing factor. The remaining terms determine the natural parameter and cumulant function:

h(y,\phi) = \frac{1}{\sqrt{2\pi\phi y^3}} \exp\left\{-\frac{1}{2\phi y}\right\}, \qquad \theta=-\frac{1}{2\mu^2}, \qquad b(\theta)=-\frac{1}{\mu}.

Since $\theta=-1/(2\mu^2)$, the mean can be written as $\mu=1/\sqrt{-2\theta}$. Therefore $b(\theta)=-\sqrt{-2\theta}$, with natural-parameter domain $\theta<0$. In particular, the natural parameter is not $1/\mu^2$: it is negative, and it includes the factor $1/2$.

Definition: The canonical link is the link function that maps the mean $\mu$ to the natural parameter $\theta$ in the exponential-dispersion representation. For the inverse-Gaussian distribution, $$ g(\mu)=\theta=-\frac{1}{2\mu^2}, \qquad \mu>0. $$

Why This Is the Canonical Link

In the exponential-dispersion form, the response-dependent term in the exponent is $y\theta/a(\phi)$. Comparing it with the expanded inverse-Gaussian density identifies the coefficient of $y$ as $-1/(2\phi\mu^2)$. Because $a(\phi)=\phi$, the natural parameter multiplying $y$ is $-1/(2\mu^2)$. This coefficient is what makes the link canonical.

A useful check is that the cumulant function gives the stated mean. For an exponential-dispersion model, $b'(\theta)$ is the mean. Here,

b'(\theta) = \frac{d}{d\theta}\left[-\sqrt{-2\theta}\right] = \frac{1}{\sqrt{-2\theta}} = \mu.

The second derivative is $b''(\theta)=(-2\theta)^{-3/2}=\mu^3$. Thus the variance relation $\operatorname{Var}(Y)=a(\phi)b''(\theta)$ gives $\operatorname{Var}(Y)=\phi\mu^3$, consistent with the inverse-Gaussian variance function. These derivative checks confirm the parameterization; they do not replace identifying the natural parameter from the exponent.

Theorem: For the inverse-Gaussian density above, the natural parameter is $\theta=-1/(2\mu^2)$, and the canonical link is $g(\mu)=-1/(2\mu^2)$.

Proof. Expanding the density’s exponent gives

-\frac{(y-\mu)^2}{2\phi\mu^2y} = \frac{y(-1/(2\mu^2))-(-1/\mu)}{\phi} -\frac{1}{2\phi y}.

The last term is part of $h(y,\phi)$, while the first term has the exponential-dispersion form $\{y\theta-b(\theta)\}/\phi$. Hence $\theta=-1/(2\mu^2)$ and $b(\theta)=-1/\mu$. By definition, the canonical link equals the natural parameter as a function of the mean, so $g(\mu)=\theta=-1/(2\mu^2)$. $\square$

Worked Example: Checking the Exponential-DUsuallyersion Terms

Take $\mu=2$, $\phi=0.25$, and $y=1$. The density’s exponent is

-\frac{(1-2)^2}{2(0.25)(2^2)(1)} =-\frac{1}{2} =-0.5.

The natural parameter is $\theta=-1/(2(2^2))=-1/8$, and $b(\theta)=-1/2$. The exponential-dispersion contribution is

\frac{y\theta-b(\theta)}{\phi} = \frac{(-1/8)-(-1/2)}{0.25} = \frac{3/8}{1/4} = 1.5.

The remaining base-factor contribution to the exponent is $-1/(2\phi y)=-1/(2(0.25)(1))=-2$. Adding the two pieces gives $1.5-2=-0.5$, exactly the exponent calculated directly from the density. This check also shows why the base factor must retain the term involving $1/y$.

Equivalent Inverse-Square Links and Coefficient Scaling

A link does not have to be written in the exact canonical scale to describe the same set of mean values. Multiplying a link by a nonzero constant changes the numerical values of its linear predictor; the regression coefficients must be multiplied by that same constant to preserve the predictor relationship. For the inverse-Gaussian canonical link, the positive-valued version $1/\mu^2$ is obtained by multiplying the canonical link by $-2$:

g(\mu)=-\frac{1}{2\mu^2}, \qquad g_+(\mu)=\frac{1}{\mu^2}=-2g(\mu).

Thus this particular conversion is a negative rescaling of the canonical link, not a positive multiple of it. The positive-valued link is often convenient to write, but its sign and scale must be carried through to the linear-predictor coefficients.

To state the coefficient relationship precisely, consider a model with the same design vector $x$, no offset, and linear predictor $x^\mathsf{T}\beta$. If the canonical-link model is

-\frac{1}{2\mu^2}=x^\mathsf{T}\beta,

then the model written with $g_+(\mu)=1/\mu^2$ must use coefficients $\beta_+=-2\beta$:

\frac{1}{\mu^2} = -2\left(-\frac{1}{2\mu^2}\right) = -2x^\mathsf{T}\beta = x^\mathsf{T}(-2\beta) = x^\mathsf{T}\beta_+.

This is an algebraic change of scale, not a change in the fitted means. A more general nonzero multiplier $c$ gives $g_c(\mu)=c\,g(\mu)$ and $\beta_c=c\beta$, since $g_c(\mu)=c x^\mathsf{T}\beta=x^\mathsf{T}(c\beta)$. For a positive-valued inverse-square link $g_a(\mu)=a/\mu^2$, where $a>0$, the relevant multiplier is $c=-2a$, so $\beta_a=-2a\beta$.

Theorem: For a linear predictor $x^\mathsf{T}\beta$, replacing a link $g$ by $g_c=cg$, for a fixed nonzero constant $c$, leaves the mean model unchanged when the coefficient vector is replaced by $\beta_c=c\beta$. In particular, $g_+(\mu)=1/\mu^2$ corresponds to $\beta_+=-2\beta$.

Proof. Suppose $g(\mu)=x^\mathsf{T}\beta$. Under the rescaled link, $g_c(\mu)=c g(\mu)$, so

g_c(\mu)=c x^\mathsf{T}\beta =x^\mathsf{T}(c\beta) =x^\mathsf{T}\beta_c.

Because $c\ne0$, the inverse transformation is $g(\mu)=g_c(\mu)/c$. Therefore the two equations specify the same $\mu$ for each design vector, with coefficient vectors related by $\beta_c=c\beta$. Taking $c=-2$ gives $g_c(\mu)=1/\mu^2$ and $\beta_c=-2\beta$. $\square$

Worked Example: Converting Canonical Coefficients to the Positive Link

Suppose the canonical-link predictor is $\eta=-0.12$ at a particular covariate value. The mean is determined by

-\frac{1}{2\mu^2}=-0.12 \quad\Longrightarrow\quad \mu^2=\frac{1}{0.24} \quad\Longrightarrow\quad \mu=\sqrt{\frac{1}{0.24}}\approx2.041.

With the positive-valued link, the predictor at that same covariate value must be $\eta_+=-2\eta=0.24$. It yields

\frac{1}{\mu^2}=0.24 \quad\Longrightarrow\quad \mu=\frac{1}{\sqrt{0.24}}\approx2.041.

For example, if $x^\mathsf{T}\beta=-0.12$, then $x^\mathsf{T}\beta_+=0.24$ follows from $\beta_+=-2\beta$. The coefficients change sign and double in magnitude, while the fitted mean remains the same.

Worked Example: A General Positive Multiple of the Inverse Square

Suppose a canonical-link model has $\beta=(-0.10,\ 0.03)^\mathsf{T}$, with design vector $x=(1,\ 2)^\mathsf{T}$. Its predictor is

\eta=x^\mathsf{T}\beta =(1)(-0.10)+(2)(0.03) =-0.04.

The canonical link then gives $-1/(2\mu^2)=-0.04$, hence $\mu^2=1/0.08=12.5$ and $\mu=\sqrt{12.5}\approx3.536$. Now choose the positive link $g_a(\mu)=3/\mu^2$. This is $-6$ times the canonical link, so its coefficient vector is $\beta_a=-6\beta=(0.60,\ -0.18)^\mathsf{T}$. Its predictor at the same $x$ is

x^\mathsf{T}\beta_a =(1)(0.60)+(2)(-0.18) =0.24.

Solving the positive-link equation confirms the same mean: $3/\mu^2=0.24$, so $\mu^2=3/0.24=12.5$ and $\mu\approx3.536$. The factor of $-6$, rather than $3$, is required because $3/\mu^2=-6[-1/(2\mu^2)]$.

Common Mistake and Why the Scale Matters

Common mistake: “The canonical link is $1/\mu^2$, because that is the reciprocal-square expression in the inverse-Gaussian model, so I can use the same coefficients with either link.” This reasoning drops both the negative sign and the factor $1/2$. The correction is to identify the natural parameter from the coefficient of $y$ in the exponential-dispersion form: it is $-1/(2\mu^2)$. If using $1/\mu^2$ instead, transform the coefficients by $-2$ as well. Keeping the old coefficients while changing the link generally changes the fitted means.

The scale convention matters when reporting or comparing coefficients: their numerical values depend on the link’s scale, even when the underlying fitted mean relationship is identical. It also matters if a model imposes coefficient restrictions. A transformation $\beta_c=c\beta$ carries unrestricted coefficient vectors into unrestricted coefficient vectors, but any imposed restrictions must be transformed consistently; otherwise, the two specifications need not describe the same allowed models.

Check Your Understanding

Use the density expansion and coefficient-scaling rules to answer the following.

Expand the inverse-Gaussian exponent and identify which term belongs in the base factor $h(y,\phi)$.
What are the natural parameter and canonical link when the mean is $\mu$?
If the canonical-link coefficient vector is $\beta$, what coefficient vector gives the same mean model under the link $1/\mu^2$?
For the link $a/\mu^2$, where $a>0$, find the multiplier relating its coefficient vector to the canonical-link coefficients.
Explain why using $1/\mu^2$ with unchanged canonical-link coefficients does not generally preserve fitted means.

Deriving the Canonical Inverse-Gaussian Link

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