Deriving Mean-Scale Effects With The Chain Rule

From a Link-Scale Coefficient to a Mean-Scale Derivative

In a generalized linear model, the linear predictor describes the mean after applying the link function: \(g(\mu)=\eta\). The inverse link therefore gives \(\mu=g^{-1}(\eta)\). When a predictor is continuous, we can ask a more local question than how its coefficient changes the linear predictor: at a given predictor value, how quickly does the modeled mean change as the covariate changes?

The answer follows from the chain rule. The covariate changes the linear predictor, and the inverse link translates that change onto the mean scale. Consequently, a coefficient that is constant on the link scale need not correspond to a constant change in the mean. The derivative of the inverse link determines how the mean-scale effect varies with the current linear predictor.

In the common case \(\eta(x)=\beta_0+\beta_1x+\sum_{j=2}^p\beta_j z_j\), with the \(z_j\) held fixed, \(\frac{d\eta}{dx}=\beta_1\). That coefficient is the change in the linear predictor per unit change in \(x\). To obtain the mean-scale derivative, it must be multiplied by the rate at which the inverse link changes at the current predictor.

The Chain-Rule Result

The assumptions are local: the predictor needs a derivative at the covariate value of interest, and the inverse link needs a derivative at the resulting predictor value. No assumption that the link is canonical is required. For the usual linear predictor, the result becomes \(\frac{d\mu}{dx}=h'(\eta)\beta_1\). If the predictor includes a transformed covariate, its derivative must also be included.

Proof. Set \(\eta_0=\eta(x)\), and let \(\Delta\eta=\eta(x+\Delta x)-\eta(x)\). Differentiability of \(\eta\) gives \(\Delta\eta=\eta'(x)\Delta x+o(\Delta x)\). Differentiability of \(h\) at \(\eta_0\) gives \(h(\eta_0+\Delta\eta)-h(\eta_0)=h'(\eta_0)\Delta\eta+o(\Delta\eta)\). If \(\eta'(x)\ne 0\), dividing by \(\Delta x\) and taking the limit yields \(h'(\eta_0)\eta'(x)\). If \(\eta'(x)=0\), then \(\Delta\eta=o(\Delta x)\); the second expansion, divided by \(\Delta x\), tends to zero, including its remainder because \(o(\Delta\eta)/\Delta x\) tends to zero. Thus the derivative has the stated value in either case. \(\square\)

How the Mean-Scale Derivative Changes

The first derivative tells us the mean-scale effect at a particular point. To describe how that effect itself varies with \(x\), differentiate once more. A useful special case has a linear predictor in \(x\), so that \(\eta(x)=a+\beta x\) for fixed \(a\) and \(\beta\).

Proof. The chain-rule result gives \(\frac{d\mu}{dx}=h'(\eta)\eta'(x)=\beta h'(\eta)\), since \(\eta'(x)=\beta\). Differentiating this expression with respect to \(x\), using the chain rule again, gives \(\frac{d^2\mu}{dx^2}=\beta h''(\eta)\eta'(x)=\beta^2 h''(\eta)\). Since \(\beta^2\geq0\), the sign of this second derivative is the sign of \(h''(\eta)\) when \(\beta\ne0\); when \(\beta=0\), the mean is constant in \(x\). This proves the stated monotonicity of the first derivative. \(\square\)

For a nonlinear predictor, the second derivative has an additional term. If \(\eta\) and \(h\) are twice differentiable, differentiating \(h'(\eta)\eta'\) yields \(\frac{d^2\mu}{dx^2}=h''(\eta)(\eta')^2+h'(\eta)\eta''\). The second term accounts for curvature in the predictor itself. Thus the way a mean-scale effect varies can reflect both the inverse link and the covariate’s contribution to the predictor.

Transformed Covariates and Finite Changes

If a predictor contains a transformed covariate, the chain rule applies at both stages. For example, if \(\eta(x)=\beta_0+\beta_1x^2\), then \(\eta'(x)=2\beta_1x\), not \(\beta_1\). The mean-scale derivative is therefore \(h'(\eta)2\beta_1x\). A coefficient multiplying a transformation describes change per unit of that transformed quantity; it does not by itself give the derivative per unit of the original covariate.

A derivative is a local rate, whereas a finite change compares means at two distinct covariate values. If \(h'\) and \(\eta'\) are continuous between \(x_0\) and \(x_1\), the fundamental theorem of calculus gives \(\mu(x_1)-\mu(x_0)=\int_{x_0}^{x_1}h'(\eta(x))\eta'(x)\,dx\). Thus a derivative at one endpoint generally does not equal the finite change divided by \(x_1-x_0\). A derivative-based approximation can be useful for a small change, but its accuracy depends on how much the derivative varies across that change.

Common Pitfall and Why the Derivative Matters

• Wrong reasoning: “The coefficient of \(x\) is \(0.8\), so increasing \(x\) by one increases the mean by \(0.8\).” This treats a link-scale change as though it were already on the mean scale.
• Correction: For a linear predictor, the mean-scale derivative is \(0.8h'(\eta)\), evaluated at the current predictor. It equals \(0.8\) only if the inverse-link derivative at that predictor equals \(1\).
• Another common error: When \(\eta\) contains \(x^2\), using only the coefficient of \(x^2\). For a derivative with respect to \(x\), differentiate the predictor too: \(\frac{d}{dx}(\beta_1x^2)=2\beta_1x\).

Mean-scale derivatives are useful when a question concerns the rate of change in the modeled mean near a specified covariate value. They also make clear why the covariate’s effect can depend on where in the predictor range it is evaluated, even when the linear predictor has a constant coefficient. To report such an effect responsibly, specify the covariate value and the other covariates held fixed, and distinguish a local derivative from a finite contrast.