Reciprocal-Link Gaussian Regression and Domain Constraints

Introduction

A reciprocal link makes the reciprocal of the mean, rather than the mean itself, a linear function of the predictors. This can be useful when a scientific relationship is naturally additive on a reciprocal scale, but it also imposes a domain constraint: a linear predictor equal to zero makes the fitted mean undefined.

The main interpretive discipline is to keep the two scales separate. Coefficients describe additive changes in reciprocal means. Changes in the means themselves are generally nonlinear, and a predictor value can take a fitted mean outside the scientifically meaningful range even when the arithmetic is otherwise straightforward.

The Reciprocal-Link Gaussian Model

Let \(Y_i\) be an independent Gaussian response with mean \(\mu_i\) and variance \(\sigma^2\), and let \(\mathbf{x}_i\) be a vector of predictors, including an intercept if the model has one. Write the linear predictor as \(\eta_i=\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}\). The reciprocal link is \(g(\mu)=1/\mu\), so the mean model is defined by \(g(\mu_i)=\eta_i\).

The Gaussian distribution itself allows any real mean, so mathematical validity of this link only requires a nonzero linear predictor. A scientific model may require more. For example, when the response is a duration or concentration, a negative mean may be meaningless. Under the reciprocal link, requiring \(\mu_i>0\) is equivalent to requiring \(\eta_i>0\).

This constraint is about the mean, not the support of the Gaussian response. Even when \(\mu_i>0\), a Gaussian distribution assigns some probability to negative response values. Whether that feature is acceptable is a separate question about the suitability of the Gaussian response model.

What a Coefficient Changes

For a predictor \(x_j\), hold the other predictors fixed and increase \(x_j\) by \(\Delta\). If its coefficient is \(\beta_j\), the linear predictor changes from \(\eta\) to \(\eta+\beta_j\Delta\). Both values must be nonzero for the corresponding means to exist; if positive means are required, both must be positive.

To verify the result, the model gives \(1/\mu_{\mathrm{old}}=\eta\) and \(1/\mu_{\mathrm{new}}=\eta+\beta_j\Delta\). Subtracting the first equality from the second gives the reciprocal-mean contrast. For the mean contrast, subtract the two reciprocals: \[ \mu_{\mathrm{new}}-\mu_{\mathrm{old}} =\frac{1}{\eta+\beta_j\Delta}-\frac{1}{\eta} =\frac{\eta-(\eta+\beta_j\Delta)}{\eta(\eta+\beta_j\Delta)} =-\frac{\beta_j\Delta}{\eta(\eta+\beta_j\Delta)}. \] Thus the coefficient is not a fixed change in the mean. The mean difference depends on the starting predictor \(\eta\) as well as on the coefficient and the size of the predictor change.

More locally, for a continuous predictor, differentiating \(\mu=1/\eta\) with respect to \(x_j\) gives \[ \frac{\partial\mu}{\partial x_j} =-\frac{\beta_j}{\eta^2}. \] Where the mean is positive, \(\eta^2>0\), so a positive coefficient corresponds to a decreasing mean as that predictor increases, and a negative coefficient corresponds to an increasing mean. The size of the local change also depends on \(\eta\); it is not constant across profiles.

Checking the Predictor Domain

A fitted reciprocal mean must be checked before it is reported. For an individual profile, calculate \(\eta=\mathbf{x}^\mathsf{T}\boldsymbol{\beta}\), then check whether it is nonzero. If the scientific setting requires positive means, check that it is positive. A zero predictor gives no finite fitted mean; a negative predictor gives a negative mean.

For the final claim, each condition \(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}>0\) defines an open half-space in the space of coefficient vectors. If \(\boldsymbol{\beta}_1\) and \(\boldsymbol{\beta}_2\) satisfy every condition, then for \(0<t<1\), \[ \mathbf{x}_i^\mathsf{T}\{t\boldsymbol{\beta}_1+(1-t)\boldsymbol{\beta}_2\} =t\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}_1 +(1-t)\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}_2>0 \] for each \(i\). Their convex combination therefore satisfies every condition, proving convexity. This describes the coefficient vectors that give positive fitted means at the specified rows; it does not guarantee positive means at every possible covariate value.

The restriction can apply to a whole range of predictor values, not just the observed sample. If a model is to be used for predictions across a specified range, check the sign of the linear predictor over that range. A valid fitted mean at every observed profile does not by itself establish validity for extrapolated profiles.

Why the Domain Check Matters

As \(\eta\) approaches zero through positive values, \(1/\eta\) grows without bound. This creates an asymptote: small changes in the predictor near zero can correspond to very large changes in the fitted mean. On the other side of zero, the fitted mean is negative. Consequently, a model can produce predictions that are technically defined but unsuitable for the scientific question, or predictions that are not defined at all.

A practical check is to calculate the linear predictor, rather than judging only from a coefficient table or from fitted values at a few convenient profiles. For continuous predictors, inspect the predictor over the intended range. For categorical predictors, check each relevant category and covariate combination. If the model is used for positive means, verify \(\eta>0\) throughout the range of intended use.

Common Mistake

Wrong reasoning: “The coefficient is \(0.3\), so increasing the predictor by one unit increases the mean by \(0.3\).”

Correction: The coefficient increases the reciprocal mean by \(0.3\) per unit. If the old predictor is \(\eta\), the mean changes to \(1/(\eta+0.3)\), provided the new predictor is nonzero. For example, if \(\eta=1.5\), the old mean is \(2/3\); after a one-unit increase the new mean is \(1/1.8=5/9\), a change of \(5/9-2/3=-1/9\), not \(0.3\). Always interpret the coefficient on the reciprocal-mean scale and check that the new predictor remains in the required domain.