Monotone Links And One-To-One Mean Mapping

Introduction

A link function translates an admissible mean onto the scale of the linear predictor. For that translation to be unambiguous in both directions, a given mean must produce only one predictor, and a given predictor in the link’s range must correspond to only one mean. Strict monotonicity guarantees this correspondence.

The key qualification is in the link’s range. A link can have a unique inverse on its range without that range being all of the real numbers. This distinction determines whether every possible real-valued linear predictor is allowed or whether the predictor must satisfy an additional restriction.

Strict Monotonicity and the Inverse

Definition: Let \(\mathcal{M}\) be a set of admissible means and let \(g:\mathcal{M}\to\mathbb{R}\) be a link function. The function \(g\) is strictly increasing if \(a<b\) implies \(g(a)<g(b)\) for every \(a,b\in\mathcal{M}\). It is strictly decreasing if \(a<b\) implies \(g(a)>g(b)\). It is strictly monotone if it is either strictly increasing or strictly decreasing. The range of \(g\) is \(g(\mathcal{M})=\{g(\mu):\mu\in\mathcal{M}\}\).

The strict inequality matters. Strictly increasing links preserve the ordering of means; strictly decreasing links reverse it. In either case, two distinct admissible means cannot have the same link value.

Theorem: If \(g\) is strictly monotone on \(\mathcal{M}\), then it is one-to-one on \(\mathcal{M}\). Consequently, it has a unique inverse function \(g^{-1}:g(\mathcal{M})\to\mathcal{M}\), with \(g^{-1}(g(\mu))=\mu\) for every \(\mu\in\mathcal{M}\).

Proof. Take any \(\mu_1,\mu_2\in\mathcal{M}\) and suppose \(g(\mu_1)=g(\mu_2)\). If \(\mu_1\ne\mu_2\), then either \(\mu_1<\mu_2\) or \(\mu_2<\mu_1\). Strict monotonicity would then imply \(g(\mu_1)\ne g(\mu_2)\), a contradiction. Thus \(\mu_1=\mu_2\), so \(g\) is one-to-one. By definition, every \(\eta\in g(\mathcal{M})\) is \(g(\mu)\) for at least one \(\mu\in\mathcal{M}\); one-to-one-ness guarantees there is exactly one such \(\mu\). Assigning that unique \(\mu\) to each \(\eta\) defines \(g^{-1}\) on the range. \(\square\)

This result uses no continuity assumption. The inverse is defined on the range, even if that range is not all of \(\mathbb{R}\). Strict monotonicity ensures uniqueness, not that every real number is attained.

Worked Examples

Worked Example: A Log Link for Positive Means

Suppose the admissible means are \(\mathcal{M}=(0,\infty)\) and the link is \(g(\mu)=\log(\mu)\). Determine the mean corresponding to predictor \(\eta=\log(2.5)\), and check that the link and inverse give a unique match.

Strict monotonicity. The derivative of \(\log(\mu)\) is \(1/\mu\), which is positive for every \(\mu>0\). Thus the log link is strictly increasing on \((0,\infty)\).

Inverse calculation. Solving \(\eta=\log(\mu)\) for \(\mu\) gives \(\mu=\exp(\eta)\). Therefore,

\mu=g^{-1}(\eta)=\exp(\log(2.5))=2.5.

The log function maps \((0,\infty)\) onto all of \(\mathbb{R}\), so every real predictor has one positive mean under this link. For example, \(\eta=0\) gives \(\mu=\exp(0)=1\), while \(\eta=\log(2.5)\) gives \(\mu=2.5\). Strict increase rules out a second positive mean for either predictor.

Worked Example: A Logit Link for Proportions

Let the admissible mean be a probability \(\mu\in(0,1)\), and use the logit link \(g(\mu)=\log\!\left(\frac{\mu}{1-\mu}\right)\). Find the predictor for \(\mu=0.75\), then recover the mean from that predictor.

Forward calculation. The odds corresponding to \(0.75\) are \(0.75/(1-0.75)=0.75/0.25=3\). Hence \(\eta=\log(3)\).

Strict monotonicity. Differentiating the logit gives \(\frac{1}{\mu}+\frac{1}{1-\mu}=\frac{1}{\mu(1-\mu)}\), which is positive for every \(\mu\in(0,1)\). The logit is strictly increasing on its domain.

Inverse calculation. Starting from \(\eta=\log\!\left(\frac{\mu}{1-\mu}\right)\), exponentiate to obtain \(e^\eta=\frac{\mu}{1-\mu}\). Then \(e^\eta(1-\mu)=\mu\), so \(e^\eta=\mu(1+e^\eta)\) and \(\mu=\frac{e^\eta}{1+e^\eta}=\frac{1}{1+e^{-\eta}}\). Substituting \(\eta=\log(3)\) gives

g^{-1}(\log(3)) =\frac{3}{1+3} =\frac{3}{4} =0.75.

The logit range is all of \(\mathbb{R}\), so each finite real predictor corresponds to exactly one probability strictly between zero and one.

Worked Example: An Identity Link for Real-Valued Means

For a model with admissible means \(\mathcal{M}=\mathbb{R}\), use the identity link \(g(\mu)=\mu\). Find the mean for predictor \(\eta=-1.2\), and explain what happens to ordering.

The identity link is strictly increasing: if \(\mu_1<\mu_2\), then \(g(\mu_1)=\mu_1<\mu_2=g(\mu_2)\). Its inverse is also the identity, so \(\mu=g^{-1}(\eta)=\eta=-1.2\). Its range is \(\mathbb{R}\), and every real predictor is allowed. Moreover, if one predictor is larger than another, its corresponding mean is larger by the same amount, because the link does not transform the scale.

The Mean–Predictor Correspondence in a GLM

In a generalized linear model, the link relates the mean \(\mu\) to the linear predictor \(\eta\) through \(\eta=g(\mu)\). The linear predictor is formed from predictors and coefficients; the link then expresses the mean on the model’s response scale. For a strictly monotone link, the exact correspondence is between \(\mathcal{M}\), the admissible means, and \(g(\mathcal{M})\), the link’s range.

Theorem: Let \(g:\mathcal{M}\to\mathbb{R}\) be strictly monotone, and let \(\mathcal{E}=g(\mathcal{M})\). Then \(g\) is a bijection from \(\mathcal{M}\) to \(\mathcal{E}\), and \(g^{-1}\) is a bijection from \(\mathcal{E}\) to \(\mathcal{M}\). Thus \(\eta=g(\mu)\) gives a one-to-one correspondence between admissible means and predictors in \(\mathcal{E}\). If \(\mathcal{E}=\mathbb{R}\), every real-valued predictor corresponds to one admissible mean.

Proof. By definition, \(g(\mathcal{M})=\mathcal{E}\), so every \(\eta\in\mathcal{E}\) is \(g(\mu)\) for at least one \(\mu\in\mathcal{M}\); this makes \(g\) onto \(\mathcal{E}\). By strict monotonicity and the preceding theorem, distinct means cannot have the same image, so \(g\) is one-to-one. It is therefore a bijection from \(\mathcal{M}\) to \(\mathcal{E}\), with inverse \(g^{-1}\). If \(\mathcal{E}=\mathbb{R}\), every real \(\eta\) is in the domain of that inverse and has exactly one corresponding mean. \(\square\)

This qualification is important when the linear predictor can vary over the real numbers. If the link range is only a subset of \(\mathbb{R}\), then \(\eta\) must lie in that subset for the inverse link to produce an admissible mean. One cannot conclude that an inverse is available for every real predictor merely from strict monotonicity.

Order and Uniqueness

Monotonicity also tells us how comparisons on one scale translate to the other. For a strictly increasing link, larger means give larger predictors and vice versa. For a strictly decreasing link, larger means give smaller predictors and vice versa.

Result: If \(g\) is strictly increasing and \(\eta_1<\eta_2\) are in its range, then \(g^{-1}(\eta_1)<g^{-1}(\eta_2)\). If \(g\) is strictly decreasing, then \(g^{-1}(\eta_1)>g^{-1}(\eta_2)\).

Proof. Let \(\mu_1=g^{-1}(\eta_1)\) and \(\mu_2=g^{-1}(\eta_2)\). If \(g\) is strictly increasing and \(\mu_1\ge\mu_2\), then either \(\mu_1=\mu_2\), which would imply \(\eta_1=\eta_2\), or \(\mu_1>\mu_2\), which would imply \(g(\mu_1)>g(\mu_2)\), or \(\eta_1>\eta_2\). Each possibility contradicts \(\eta_1<\eta_2\). Hence \(\mu_1<\mu_2\). If \(g\) is strictly decreasing and \(\mu_1\le\mu_2\), equality again contradicts the distinct predictors, while \(\mu_1<\mu_2\) implies \(g(\mu_1)>g(\mu_2)\), or \(\eta_1>\eta_2\). Thus \(\mu_1>\mu_2\). \(\square\)

For the log and logit links above, the relationship is order-preserving. This conclusion concerns the mean and predictor values themselves; it does not by itself describe how changing a particular covariate affects the mean. That also depends on how the covariate enters the linear predictor.

Common Pitfalls and Why the Distinction Matters

Assuming weak monotonicity is enough. A nondecreasing link can have a flat section: two distinct means in that section have the same link value, so there is no unique inverse there. The guarantee in this tutorial requires strict monotonicity.
Assuming the inverse covers every real number. The inverse is defined on the link’s range. If that range is not all of \(\mathbb{R}\), a predictor outside it has no corresponding admissible mean under that link.
Confusing a unique mean with identifiable coefficients. A one-to-one link uniquely determines a mean from a given predictor. It does not ensure that the model coefficients are uniquely determined by the design matrix. That is a separate identifiability question.
Claiming strict monotonicity is necessary for every possible inverse. It is a sufficient and useful condition for the links considered here. The defining requirement for an inverse function is that the function be one-to-one; strict monotonicity is a direct way to establish that property.

The practical check is straightforward: identify the admissible mean set, verify that the link is strictly monotone there, and determine its range. These steps establish whether the mean–predictor mapping is one-to-one and exactly which predictors have admissible means. They also keep a scale-level property of the link distinct from questions about coefficient estimation or interpretation.

Check Your Understanding

Use the domain, range, and strict monotonicity of a link to answer each question.

For \(g(\mu)=\log(\mu)\) on \((0,\infty)\), what is the mean corresponding to \(\eta=0\), and why is it unique?
Why does strict monotonicity ensure that two different admissible means cannot have the same predictor?
For a strictly decreasing link, if \(\eta_1<\eta_2\), which of \(g^{-1}(\eta_1)\) and \(g^{-1}(\eta_2)\) is larger?
Does strict monotonicity alone ensure that every real predictor has an admissible mean? Explain using the link’s range.
Why does a one-to-one mean–predictor mapping not, by itself, establish that the regression coefficients are identifiable?