From Link-Scale Coefficients to Mean Changes
An inverse-Gaussian regression coefficient describes a change in the linear predictor, not generally a fixed change in the response mean. To interpret it on the response scale, first apply the inverse link. The resulting change may be additive, multiplicative, or nonlinear, and it can depend on the starting covariate values.
Write \(x\) for a covariate vector and \(\beta\) for the coefficient vector. The linear predictor is \(\eta(x)=x^\top\beta\), and the conditional mean is \(\mu(x)=E[Y\mid x]=g^{-1}(\eta(x))\). Here \(Y\mid x\) has an inverse-Gaussian distribution with positive mean \(\mu(x)\). All interpretations below concern this conditional mean; they do not say that an individual observed response changes by the same amount.
For a one-unit increase in covariate \(x_j\), holding all other covariates fixed, the linear predictor changes by \(\beta_j\). For a change of \(\delta\) units, it changes by \(\beta_j\delta\). The inverse link determines how that predictor change translates into a mean difference or ratio.
Exact Contrasts Under Four Links
For the log link, \(g(\mu)=\log\mu\), so \(\mu=\exp(\eta)\). For the identity link, \(g(\mu)=\mu\), so \(\mu=\eta\). For the reciprocal link, \(g(\mu)=1/\mu\), so \(\mu=1/\eta\). Under the canonical inverse-Gaussian link, the natural parameter is \(\eta=-1/(2\mu^2)\), giving \(\mu=(-1/(2\eta))^{1/2}\). The canonical convention here uses this natural parameter itself, including its negative sign and factor of \(1/2\).
- Log link: \(\mu_B/\mu_A=\exp(d)\), so \(\mu_B-\mu_A=\mu_A\{\exp(d)-1\}\).
- Identity link: \(\mu_B-\mu_A=d\), so \(\mu_B/\mu_A=(\eta_A+d)/\eta_A\).
- Reciprocal link: \(\mu_B/\mu_A=\eta_A/\eta_B\), and \(\mu_B-\mu_A=-d/(\eta_A\eta_B)\).
- Canonical link: \(\mu_B/\mu_A=(\eta_A/\eta_B)^{1/2}\), and \(\mu_B-\mu_A=\mu_A\{(\eta_A/\eta_B)^{1/2}-1\}\).
Proof. Under the log link, \(\mu_A=\exp(\eta_A)\) and \(\mu_B=\exp(\eta_A+d)\). Dividing gives \(\mu_B/\mu_A=\exp(d)\); subtracting gives \(\mu_A\{\exp(d)-1\}\). Under the identity link, \(\mu_A=\eta_A\) and \(\mu_B=\eta_A+d\), so subtraction gives \(d\), and division gives \((\eta_A+d)/\eta_A\).
Under the reciprocal link, \(\mu_A=1/\eta_A\) and \(\mu_B=1/\eta_B\), where \(\eta_B=\eta_A+d\). Thus \(\mu_B/\mu_A=\eta_A/\eta_B\), and \[ \mu_B-\mu_A=\frac{1}{\eta_B}-\frac{1}{\eta_A} =\frac{\eta_A-\eta_B}{\eta_A\eta_B} =-\frac{d}{\eta_A\eta_B}. \] Under the canonical link, \(\mu_A=(-1/(2\eta_A))^{1/2}\) and \(\mu_B=(-1/(2\eta_B))^{1/2}\). Their ratio is \((\eta_A/\eta_B)^{1/2}\); this is real and positive because both predictors are negative. Subtracting the means gives the stated difference. The domain conditions ensure the displayed means are positive and defined. \(\square\)
These formulas apply to any finite contrast, including a change in several covariates at once. In particular, for a one-unit change in \(x_j\), set \(d=\beta_j\); for a \(\delta\)-unit change, set \(d=\beta_j\delta\). For a categorical indicator, the coefficient compares the group coded 1 with the reference group coded 0, conditional on the other covariates.
Worked Examples
Worked Example: A Log-Link Coefficient Gives a Mean Ratio
Suppose a log-link model for mean response time has \(\eta=\beta_0+0.08x\), where \(x\) is years of exposure. Compare two otherwise identical cases whose exposure differs by five years. Suppose the lower-exposure case has fitted mean \(\mu_A=12\).
The predictor change is \(d=0.08(5)=0.40\). The exact mean ratio is \[ \frac{\mu_B}{\mu_A}=\exp(0.40)\approx1.4918. \] Thus the fitted conditional mean is about 49.18% higher for the case with five more years of exposure. Its fitted mean is \(12(1.4918)\approx17.90\), and the mean difference is approximately \(17.90-12=5.90\). The coefficient \(0.08\) is not an increase of 0.08 time units per year; it is a change in log mean per year.
Worked Example: An Identity-Link Coefficient Gives a Mean Difference
Suppose an identity-link model has \(\eta=\beta_0-0.6x\), where \(x\) is a dose measured in units. At dose \(x_A\), let the fitted mean be \(8\). Compare that case with one at dose \(x_B=x_A+3\).
The predictor change is \(d=(-0.6)(3)=-1.8\). The identity-link formula gives \(\mu_B-\mu_A=d=-1.8\), so \(\mu_B=8-1.8=6.2\). Both means are positive, as required. The ratio is \(6.2/8=0.775\): the second fitted mean is 77.5% of the first, or 22.5% lower. The difference of \(-1.8\) is constant for this three-unit dose change, while the ratio depends on the starting mean.
Worked Example: A Reciprocal-Link Coefficient Changes the Mean Nonlinearly
Suppose the reciprocal-link predictor is \(\eta=1/\mu=0.05+0.01x\), with \(x\) a treatment-intensity measure. Compare a case with \(x=0\) to one with \(x=2\).
At \(x=0\), \(\eta_A=0.05\), so \(\mu_A=1/0.05=20\). At \(x=2\), \(\eta_B=0.05+0.01(2)=0.07\), so \(\mu_B=1/0.07=100/7\approx14.286\). Both predictors are positive. The exact ratio is \(\mu_B/\mu_A=0.05/0.07=5/7\approx0.7143\), and the difference is \(100/7-20=-40/7\approx-5.714\). A positive coefficient raises the reciprocal mean predictor and, in this example, lowers the conditional mean.
Worked Example: A Canonical-Link Coefficient and Its Domain
Suppose the canonical predictor is \(\eta=-1/(2\mu^2)=-0.02+0.005x\). Compare \(x=0\) with \(x=2\).
At \(x=0\), \(\eta_A=-0.02\), and \(\mu_A=(-1/(2(-0.02)))^{1/2}=(25)^{1/2}=5\). At \(x=2\), \(\eta_B=-0.02+0.005(2)=-0.01\), and \(\mu_B=(-1/(2(-0.01)))^{1/2}=(50)^{1/2}\approx7.071\). Both predictors remain negative, so the canonical inverse link is valid at both points. The ratio is \((\eta_A/\eta_B)^{1/2}=((-0.02)/(-0.01))^{1/2}=\sqrt{2}\approx1.414\). The mean difference is about \(7.071-5=2.071\). Here a positive coefficient moves the negative predictor toward zero and increases the mean.
Local Slopes and Domain Checks
A coefficient can also be translated into a local rate of change in the conditional mean. This derivative describes the effect of an infinitesimally small covariate change at a specified point. It is not a replacement for the exact finite-contrast formulas when the covariate change is substantial.
- Log link: \(\partial\mu/\partial x_j=\beta_j\mu\).
- Identity link: \(\partial\mu/\partial x_j=\beta_j\).
- Reciprocal link: \(\partial\mu/\partial x_j=-\beta_j\mu^2\).
- Canonical link: \(\partial\mu/\partial x_j=\beta_j\mu^3\).
Derivation. Since \(\eta=x^\top\beta\), its derivative with respect to \(x_j\) is \(\beta_j\). The chain rule therefore gives \(\partial\mu/\partial x_j=(dg^{-1}/d\eta)\beta_j\). For the log link, \(d\exp(\eta)/d\eta=\mu\); for identity, the derivative is 1. For the reciprocal link, \(d(1/\eta)/d\eta=-1/\eta^2=-\mu^2\). For the canonical link, \(\mu=(-2\eta)^{-1/2}\), so \[ \frac{d\mu}{d\eta}=(-2\eta)^{-3/2}=\mu^3, \] where the equality follows from \(\mu=(-2\eta)^{-1/2}>0\). Multiplying each derivative by \(\beta_j\) proves the result. \(\square\)
The derivative signs help explain direction, but the predictor domain must still be checked over the contrast. The reciprocal link requires \(\eta>0\), and the canonical link requires \(\eta<0\), at every covariate value being interpreted. The identity link requires a positive fitted mean. A coefficient can point in a direction that eventually reaches an invalid predictor; a local slope alone does not guarantee that a large finite extrapolation is valid.
Common Interpretation Pitfalls
- Reading every coefficient as a mean difference. This is correct for the identity link, but not for log, reciprocal, or canonical links. Apply the inverse link before interpreting a response-scale effect.
- Calling a log-link coefficient a percentage directly. The exact percentage change for a one-unit increase is \(100\{\exp(\beta_j)-1\}\%\), not \(100\beta_j\%\). The latter is only an approximation when \(\beta_j\) is small.
- Using a single ratio for every link. The log-link ratio is independent of the starting covariates. For identity, reciprocal, and canonical links, the mean ratio generally depends on the starting predictor or mean.
- Ignoring the link domain. A reciprocal predictor must be positive and a canonical predictor must be negative. An algebraic contrast formula is not a valid model interpretation if either endpoint violates its domain.
- Making a causal claim from a regression coefficient. These formulas describe model-based conditional means. A causal interpretation requires additional design or identification assumptions.
Check Your Understanding
For each question, distinguish the predictor-scale change from the conditional-mean change.
- Under a log link, a coefficient is \(0.12\). What is the exact mean ratio for a three-unit covariate increase?
- Under an identity link, why is the mean difference for a fixed covariate contrast constant across starting means, while the ratio need not be?
- A reciprocal-link model has predictor \(0.08\) at one covariate value and a predictor change of \(0.02\). Calculate the two means and their ratio.
- For the canonical link, explain why both endpoint predictors must be negative before interpreting a finite contrast.
- Which of the four links gives a local mean slope proportional to \(\mu^3\), and what is that slope for coefficient \(\beta_j\)?