Logit Links As Bounded-Mean Transformations

From a Real Predictor to a Bounded Mean

A linear predictor can take any real value, but some quantities of interest must lie within a restricted range. A mean representing a proportion, for example, must be between zero and one. The logit link connects these scales: it takes a mean in that open interval to a real number, and its inverse maps a finite real predictor back to a mean strictly inside the interval.

This tutorial focuses on that inverse transformation and its behavior. We will not fit a binary-response model: the calculations concern the link as a mathematical map, without estimating coefficients or specifying a response likelihood.

The logit is built from the odds \(\mu/(1-\mu)\): it takes their logarithm. The inverse-logit reverses those operations. If \(\eta\) is a linear predictor, applying the inverse-logit gives the corresponding mean on the original, bounded scale. The formula is defined for every finite \(\eta\in\mathbb{R}\), because \(e^\eta>0\) and therefore \(1+e^\eta>0\).

Deriving the Inverse-Logit Formula

The formula follows by solving the defining equation of the logit link for \(\mu\). This algebra also shows why the input to the logit must be strictly between zero and one.

Dividing the numerator and denominator by \(e^\eta\) gives the equivalent expression \(\mu=1/(1+e^{-\eta})\). Each form is useful: the first makes positivity and the upper bound easy to see, while the second makes the limit as \(\eta\) grows large especially transparent.

Why the Output Is Always Between Zero and One

It is not enough to observe that the inverse-logit formula looks like a fraction. To establish its range, we must show both that its value is positive and that it is less than one for every finite real input.

Proof. For finite \(\eta\), \(e^\eta>0\), so the numerator and denominator of \(e^\eta/(1+e^\eta)\) are positive. Hence \(g^{-1}(\eta)>0\). Also \(1+e^\eta>e^\eta\), so dividing the positive numerator by the larger denominator gives \(g^{-1}(\eta)<1\).

To verify that every \(\mu\in(0,1)\) is attained, note that \(\mu/(1-\mu)>0\), so \(\eta=\log(\mu/(1-\mu))\) is a finite real number. Exponentiating gives \(e^\eta=\mu/(1-\mu)\). Substitution then yields \[ \frac{e^\eta}{1+e^\eta} = \frac{\mu/(1-\mu)}{1+\mu/(1-\mu)} = \frac{\mu/(1-\mu)}{1/(1-\mu)} = \mu, \] where \(1-\mu>0\). Finally, if the inverse-logit of \(\eta\) is \(\mu\), rearranging the defining equation gives \(e^\eta=\mu/(1-\mu)\), and taking logarithms gives \(\eta=\operatorname{logit}(\mu)\). Thus the real number is unique. \(\square\)

The theorem establishes that the inverse-logit is a one-to-one correspondence between the real line and the open interval \((0,1)\). In particular, a finite predictor cannot produce exactly zero or exactly one. The limiting behavior near those endpoints is a related but distinct fact.

Odds Interpretation and Concrete Values

The odds identity follows directly from the inverse-logit. If \(\mu=e^\eta/(1+e^\eta)\), then \(1-\mu=1/(1+e^\eta)\), so \(\mu/(1-\mu)=e^\eta\). Taking logarithms recovers \(\eta\). Consequently, exponentiating a predictor gives the odds associated with its inverse-logit mean. This is an algebraic interpretation of the transformation, not a claim that a particular model has been estimated.

The formula also makes the direction of the transformation precise. If \(\eta_1<\eta_2\), then \(e^{\eta_1}<e^{\eta_2}\). For positive numbers \(a<b\), \(a/(1+a)<b/(1+b)\), because multiplying by the positive denominators reduces that inequality to \(a<b\). Setting \(a=e^{\eta_1}\) and \(b=e^{\eta_2}\) shows that the inverse-logit is strictly increasing. Therefore, larger predictors correspond to larger transformed means. Its shape and local rate of change are separate questions; here the essential point is the ordering and bounded range.

Limiting Behavior for Very Large Predictors

Although every finite predictor gives an interior value, the transformed mean can get arbitrarily close to either endpoint. The limits follow directly from the formula and do not require a model-fitting argument.

Proof for the upper limit. For any \(\eta\), \[ 1-g^{-1}(\eta)=\frac{1}{1+e^\eta}. \] Let \(\varepsilon\) satisfy \(0<\varepsilon<1\). Choose \(M=\log(1/\varepsilon-1)\), which is finite because \(1/\varepsilon-1>0\). For \(\eta>M\), \(e^\eta>1/\varepsilon-1\), and hence \[ 0<1-g^{-1}(\eta)=\frac{1}{1+e^\eta}<\varepsilon. \] This is the definition of convergence to one as \(\eta\to+\infty\).

Proof for the lower limit. Write \(g^{-1}(\eta)=e^\eta/(1+e^\eta)\). Given \(0<\varepsilon<1\), choose \(M=\log(\varepsilon/(1-\varepsilon))\), which is finite. If \(\eta<M\), then \(e^\eta<\varepsilon/(1-\varepsilon)\), so \[ 0<g^{-1}(\eta)=\frac{e^\eta}{1+e^\eta}<\varepsilon. \] The last inequality follows by multiplying by the positive denominator and rearranging to \((1-\varepsilon)e^\eta<\varepsilon\). This proves convergence to zero as \(\eta\to-\infty\). \(\square\)

Common Mistake and Why the Transformation Matters

Common mistake: “The predictor is positive, so its inverse-logit must be greater than one,” or, “a very large finite predictor gives a mean of exactly one.” Both claims confuse the unbounded predictor scale with the bounded mean scale. The correction is to substitute into the inverse-logit: its positive denominator is always greater than its numerator, so the output stays strictly below one; it is also positive. A large finite predictor may yield a value extremely close to one, but the limit does not make the value equal to one at any finite input.

This distinction matters whenever the quantity being represented must remain a proportion. An unrestricted linear predictor can be transformed into a valid interior mean without restricting the predictor itself. The inverse-logit also provides a coherent way to read the transformed value: take its odds to recover \(e^\eta\), or take the logarithm of those odds to recover \(\eta\). Keeping these scales distinct prevents interpreting a predictor as though it were already a mean.