Response Rescaling and Dispersion

Why Response Units Matter

A response measured in grams has different numerical values from the same response measured in milligrams. In a regression model, this change affects more than the displayed fitted means: the inverse-Gaussian dispersion parameter changes as well. The coefficient transformation depends on the link, so there is no single rule that coefficients always stay fixed or always scale by the same amount.

Let \(Y\) have an inverse-Gaussian distribution with mean \(\mu>0\) and dispersion \(\phi>0\). Its variance is \(\operatorname{Var}(Y)=\phi\mu^3\). We will find the exact parameter transformation when the recorded response is multiplied by \(c>0\), and then use the link equation to determine what happens to regression coefficients and predictions.

Definition: A positive response rescaling replaces \(Y\) by \(Y^*=cY\), where \(c>0\) is a fixed constant. Here \(c\) multiplies the numerical response values. If the new unit is smaller, the numerical values increase and \(c>1\); if the new unit is larger, the numerical values decrease and \(c<1\).

How the Inverse-Gaussian Parameters Transform

The inverse-Gaussian density in the mean-dispersion parameterization 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.

For \(Y^*=cY\), the transformed variable has density \(f_{Y^*}(z)=c^{-1}f_Y(z/c)\) for \(z>0\). Substituting \(z/c\) into the density gives

\begin{aligned} f_{Y^*}(z) &=\frac{c^{-1}}{\sqrt{2\pi\phi(z/c)^3}} \exp\left\{-\frac{(z/c-\mu)^2}{2\phi\mu^2(z/c)}\right\}\\ &=\frac{1}{\sqrt{2\pi(\phi/c)z^3}} \exp\left\{-\frac{(z-c\mu)^2}{2(\phi/c)(c\mu)^2z}\right\}. \end{aligned}

The final expression is the inverse-Gaussian density with mean \(c\mu\) and dispersion \(\phi/c\). This density calculation verifies the full distributional transformation, not just a match between moments.

Theorem: If \(Y\) is inverse-Gaussian with mean \(\mu>0\) and dispersion \(\phi>0\), then \(Y^*=cY\), for \(c>0\), is inverse-Gaussian with \( \mu^*=c\mu \) and \( \phi^*=\phi/c \).

The variance relation gives a short check of the dispersion result. Since \(\operatorname{Var}(cY)=c^2\operatorname{Var}(Y)\), the transformed variance is \(c^2\phi\mu^3\). Writing it in inverse-Gaussian form using \(\mu^*=c\mu\) requires

\operatorname{Var}(Y^*)=\phi^*(\mu^*)^3 =\phi^*c^3\mu^3=c^2\phi\mu^3, \qquad\text{so}\qquad \phi^*=\frac{\phi}{c}.

The density argument establishes the distributional result; this variance calculation makes the inverse power of \(c\) in dispersion especially easy to check.

Worked Example: Converting Milligrams to Micrograms

Suppose a positive response is recorded in milligrams per liter and has inverse-Gaussian mean \(0.8\) and dispersion \(0.06\). Convert the numerical response to micrograms per liter.

One milligram is \(1000\) micrograms, so each numerical response value in micrograms per liter is \(1000\) times its value in milligrams per liter. Thus \(c=1000\), and the transformed parameters are

\mu^*=1000(0.8)=800,\qquad \phi^*=\frac{0.06}{1000}=0.00006.

A variance check gives the same answer. Originally, \(\operatorname{Var}(Y)=0.06(0.8)^3=0.03072\). After conversion, the variance of the numerical response is \(1000^2(0.03072)=30720\). The transformed model gives \(0.00006(800)^3=30720\).

The new unit, a microgram, is one-thousandth the size of a milligram; that is why the numerical response is multiplied by \(1000\). The dispersion is divided by \(1000\), not multiplied by it.

Transforming Regression Coefficients

For a generalized linear model with design vector \(x\), the linear predictor is \(\eta=x^\mathsf{T}\beta\), and the mean satisfies \(g(\mu)=\eta\). After response rescaling, the mean is \(\mu^*=c\mu\). Therefore the transformed link-scale prediction must be \(g(c\mu)\). Whether this is still a linear predictor in the same design variables depends on the link.

Theorem: Suppose the original model has \(g(\mu)=x^\mathsf{T}\beta\), and for the relevant means the link satisfies \(g(c\mu)=a(c)g(\mu)+b(c)\), where \(a(c)\) and \(b(c)\) do not depend on \(x\). Then the transformed linear predictor is \(a(c)x^\mathsf{T}\beta+b(c)\). If the design matrix has a coefficient vector \(\delta\) such that \(x^\mathsf{T}\delta=1\) for every observation, the transformed model has coefficient vector \(\beta^*=a(c)\beta+b(c)\delta\).

To verify the result, start with \(\eta=x^\mathsf{T}\beta=g(\mu)\). By the assumed link identity,

g(\mu^*)=g(c\mu)=a(c)g(\mu)+b(c) =a(c)x^\mathsf{T}\beta+b(c) =x^\mathsf{T}\{a(c)\beta+b(c)\delta\}.

The last equality uses \(x^\mathsf{T}\delta=1\). This proves the stated coefficient transformation. If the model has an intercept, \(\delta\) is the vector with value \(1\) in the intercept position and \(0\) in every slope position. Without an intercept, the constant \(b(c)\) need not be representable by the design, so the transformed predictor may not belong to the same linear-predictor model. \(\square\)

Several commonly used links obey this identity. The table gives their link-scale transformation and, when an intercept is present, the corresponding coefficient rule. For the canonical inverse-Gaussian link we use \(g(\mu)=-1/(2\mu^2)\).

Link	Transformed link value \(g(c\mu)\)	Coefficient transformation
Log: \(g(\mu)=\log\mu\)	\(g(\mu)+\log c\)	Intercept increases by \(\log c\); slopes are unchanged
Identity: \(g(\mu)=\mu\)	\(c\,g(\mu)\)	Every coefficient is multiplied by \(c\)
Reciprocal: \(g(\mu)=1/\mu\)	\(g(\mu)/c\)	Every coefficient is divided by \(c\)
Canonical: \(g(\mu)=-1/(2\mu^2)\)	\(g(\mu)/c^2\)	Every coefficient is divided by \(c^2\)

These are parameter-level transformations of the regression equation. They describe how a model written in one response unit corresponds to the same mean relationship written in another. They do not imply that the coefficients have the same numerical values across links.

Worked Example: Log-Link Coefficients and Predictions

Suppose \(\log\mu=0.4+0.3x\), and express the response in units that multiply each numerical value by \(10\). Then \(c=10\), and

\log\mu^*=\log(10\mu)=\log\mu+\log 10 =(0.4+\log 10)+0.3x.

Thus the transformed intercept is \(0.4+\log 10\), approximately \(2.7026\), and the slope remains \(0.3\). At \(x=2\), the original predictor is \(0.4+0.3(2)=1\), so the original fitted mean is \(\mu=e^1\approx2.7183\). The transformed predictor is \(1+\log 10\approx3.3026\), and its inverse-link prediction is \(e^{1+\log 10}=10e\approx27.1828\). This is exactly \(10\) times the original fitted mean.

If the original dispersion is \(0.08\), the transformed dispersion is \(0.08/10=0.008\), irrespective of the log-link coefficient calculation.

Worked Example: Identity-Link Coefficients

Consider the identity-link model \(\mu=2+0.5x\) and multiply the response by \(c=3\). The transformed mean is

\mu^*=3\mu=3(2+0.5x)=6+1.5x.

The transformed intercept is \(6\) and the transformed slope is \(1.5\). At \(x=4\), the original fitted mean is \(2+0.5(4)=4\); the transformed fitted mean is \(6+1.5(4)=12\), which equals \(3(4)\). If the original dispersion is \(0.12\), then \(\phi^*=0.12/3=0.04\).

The identity link also requires fitted means to be positive for an inverse-Gaussian model. The coefficient algebra above applies where the original and transformed fitted means are in the model's valid positive range; multiplying a positive mean by \(3\) preserves positivity.

Worked Example: Reciprocal-Link Coefficients

Suppose \(1/\mu=2+0.5x\). After multiplying the response by \(3\), the transformed reciprocal-link predictor is

\frac{1}{\mu^*}=\frac{1}{3\mu} =\frac{1}{3}\left(2+0.5x\right) =\frac{2}{3}+\frac{1}{6}x.

At \(x=1\), the original predictor is \(2+0.5=2.5\), so the original mean is \(1/2.5=0.4\). The transformed predictor is \(2/3+1/6=5/6\), whose reciprocal is \(6/5=1.2\). This is \(3(0.4)\), as required. The original mean is \(0.4\), not \(2\): the value \(2\) is the intercept on the reciprocal link scale, not a fitted mean. With original dispersion \(0.09\), the transformed dispersion is \(0.03\).

Why Dispersion Depends on Units

The inverse-Gaussian variance function is cubic in the mean, while variance itself scales quadratically when the response is multiplied by \(c\). The dispersion must therefore absorb the remaining factor: \(\phi^*=\phi/c\). In this parameterization, dispersion is not a unit-free measure of spread. It has reciprocal response units: if numerical responses are recorded in a unit made \(c\) times smaller, the numerical values increase by \(c\), and the numerical dispersion decreases by \(c\).

This is why dispersion values should not be compared without accounting for response units. Two analyses can describe the same underlying measurements and the same relative model fit, yet report different numerical dispersions because one response is in milligrams and the other in micrograms. The fitted means and link coefficients also have link-specific unit transformations, as the examples show.

Common Mistakes

Reversing the conversion factor. If a new unit is one-thousandth as large, each numerical response is multiplied by \(1000\). The parameter rule then gives \(\phi^*=\phi/1000\).
Leaving dispersion unchanged. The transformed variance must be \(c^2\operatorname{Var}(Y)\), while the new mean cubed is \(c^3\mu^3\). Matching them requires division of dispersion by \(c\).
Applying one coefficient rule to every link. The log link shifts the intercept, whereas identity, reciprocal, and canonical links multiply the entire linear predictor by a link-specific factor.
Confusing an intercept with a fitted mean. Under a reciprocal link, an intercept of \(2\) is a link-scale value. At \(x=0\), if the model contains only that intercept, the mean is \(1/2\), not \(2\).
Assuming a missing intercept causes no issue. For the log link, the added constant \(\log c\) must be representable by the design. Without an intercept or another column combination that equals a constant, the transformed predictor need not stay in the same model.

Key takeaway: Under \(Y^*=cY\), an inverse-Gaussian mean becomes \(c\mu\) and its dispersion becomes \(\phi/c\). Regression coefficients transform according to \(g(c\mu)\), so the link determines how the change in response units appears on the linear-predictor scale.

Check Your Understanding

Use the parameter and link transformations to answer each question.

A response is converted to units whose numerical values are four times larger. If the original mean is \(1.5\) and dispersion is \(0.2\), what are the transformed mean and dispersion?
For a log-link model with an intercept, how do the intercept and slopes change when \(Y^*=cY\)?
A reciprocal-link model has predictor \(1/\mu=1.5+0.25x\). If \(c=2\), write the transformed predictor and calculate the original and transformed means at \(x=2\).
Why does the variance relation imply \(\phi^*=\phi/c\), rather than \(\phi^*=c\phi\)?
Why might a log-link model without an intercept fail to represent the transformed means using the same design variables?