Interactions Under Nonlinear Binary Links

Why the Scale of an Interaction Matters

In a binary-response model, an interaction coefficient describes how predictors combine on the model’s link scale. But researchers often want to know how a change in one predictor affects the probability of the outcome, and that effect may depend on the other predictor even when the link-scale predictor has no interaction term. The key is to calculate the probability contrast at each value of the other predictor, then compare those contrasts.

This distinction is especially clear with probit and complementary log-log regression. Their inverse links are nonlinear, so equal changes on the predictor scale need not produce equal changes in probability. We will use binary predictors coded 0 and 1, which let us define both interactions as differences between four fitted values.

Definition: Let $Y$ be binary, let $X,Z\in\{0,1\}$, and write $p_{xz}=P(Y=1\mid X=x,Z=z)$. The conditional probability contrast for changing $X$ from 0 to 1 at $Z=z$ is $\Delta_X(z)=p_{1z}-p_{0z}$. The probability-scale interaction is the difference-in-differences $\Delta_X(1)-\Delta_X(0)=p_{11}-p_{01}-p_{10}+p_{00}$.

The order of subtraction matters: this definition asks how much the probability contrast for $X$ changes when $Z$ changes from 0 to 1. Reversing the order changes the sign, but not the substantive question of whether the two conditional contrasts differ.

Predictor-Scale Interaction in a 2-by-2 Model

For the four combinations of two binary predictors, write the linear predictor as $\eta(x,z)=\beta_0+\beta_1x+\beta_2z+\beta_3xz$. The model relates this predictor to the conditional probability through an inverse link $h$: $p_{xz}=h(\eta(x,z))$. Under the probit link $h(\eta)=\Phi(\eta)$, where $\Phi$ is the standard normal cumulative distribution function. Under the complementary log-log link, $h(\eta)=1-\exp\{-\exp(\eta)\}$.

Result: In the 0/1-coded model $\eta(x,z)=\beta_0+\beta_1x+\beta_2z+\beta_3xz$, the predictor-scale difference-in-differences equals $\beta_3$: $\eta(1,1)-\eta(0,1)-\eta(1,0)+\eta(0,0)=\beta_3$. This is an algebraic statement about the predictor, not a claim that the probability-scale interaction equals $\beta_3$.

To verify the result, the four predictor values are $\eta(0,0)=\beta_0$, $\eta(1,0)=\beta_0+\beta_1$, $\eta(0,1)=\beta_0+\beta_2$, and $\eta(1,1)=\beta_0+\beta_1+\beta_2+\beta_3$. Substitution gives

\eta(1,1)-\eta(0,1)-\eta(1,0)+\eta(0,0) =(\beta_0+\beta_1+\beta_2+\beta_3)-(\beta_0+\beta_2)-(\beta_0+\beta_1)+\beta_0 =\beta_3.

Thus $\beta_3=0$ means the predictor contrast for $X$ is the same at both values of $Z$. It does not say that the probability contrast is the same, because each predictor value is passed through $h$ before probabilities are compared.

How Curvature Creates a Probability-Scale Interaction

Consider first the model with no predictor interaction, so $\beta_3=0$. Let $a=\beta_0$, $b=\beta_1$, and $c=\beta_2$. Its probability-scale interaction is $h(a+b+c)-h(a+c)-h(a+b)+h(a)$. If $h$ is twice continuously differentiable over the range of predictor values, this contrast can be written as an integral of the curvature of $h$.

Result: For twice continuously differentiable $h$, the probability-scale interaction in an additive predictor model satisfies $$ h(a+b+c)-h(a+c)-h(a+b)+h(a) =\int_0^b\int_0^c h''(a+s+t)\,dt\,ds. $$ The integrals are oriented, so the identity also holds when $b$ or $c$ is negative.

For the proof, apply the fundamental theorem of calculus first to the contrast between $h(a+s+c)$ and $h(a+s)$, and then to the resulting difference in first derivatives:

\begin{aligned} h(a+b+c)-h(a+c)-h(a+b)+h(a) &=\int_0^b \bigl[h'(a+s+c)-h'(a+s)\bigr]\,ds\\ &=\int_0^b\int_0^c h''(a+s+t)\,dt\,ds. \end{aligned}

The stated smoothness ensures these applications of the fundamental theorem are valid throughout the intervals of integration. If $b,c>0$ and $h''$ is strictly positive throughout the rectangle of predictor values, the probability interaction is positive; if $h''$ is strictly negative throughout, it is negative. If curvature changes sign across that rectangle, the integral combines positive and negative contributions, so its sign must be determined from the contrast itself. This result explains how a nonzero probability interaction can arise with $\beta_3=0$.

Worked Example: Probit with No Predictor Interaction

Suppose a probit model has $\beta_0=-1$, $\beta_1=0.8$, $\beta_2=0.7$, and $\beta_3=0$. Calculate both conditional probability contrasts and their difference.

The predictor values are $\eta(0,0)=-1$, $\eta(1,0)=-0.2$, $\eta(0,1)=-0.3$, and $\eta(1,1)=0.5$. Applying $h(\eta)=\Phi(\eta)$ gives, to six decimal places,

p_{00}=\Phi(-1)=0.158655,\quad p_{10}=\Phi(-0.2)=0.420740,\quad p_{01}=\Phi(-0.3)=0.382089,\quad p_{11}=\Phi(0.5)=0.691462.

When $Z=0$, changing $X$ from 0 to 1 changes the probability by $p_{10}-p_{00}=0.420740-0.158655=0.262085$. When $Z=1$, the corresponding contrast is $p_{11}-p_{01}=0.691462-0.382089=0.309373$. Therefore, the probability-scale interaction is $0.309373-0.262085=0.047288$, approximately 0.0473.

The predictor interaction is zero, but the probability interaction is not. The probit transformation makes the same predictor change correspond to different probability changes at different starting predictor values.

Worked Contrasts Under Probit and Complementary Log-Log Links

A coefficient should not be reported as though it were a probability difference. Instead, calculate fitted probabilities for the four predictor combinations and form contrasts that match the question. The following examples show the calculation under each link, including the direction of the probability-scale interaction.

Worked Example: Probit with a Nonzero Predictor Interaction

Suppose $\beta_0=-1$, $\beta_1=1$, $\beta_2=0.5$, and $\beta_3=0.5$. The four predictor values are $\eta(0,0)=-1$, $\eta(1,0)=0$, $\eta(0,1)=-0.5$, and $\eta(1,1)=1$. Their predictor-scale difference-in-differences is $1-(-0.5)-0-1=0.5$, as specified by $\beta_3$.

Under the probit link, the fitted probabilities are

p_{00}=\Phi(-1)=0.158655,\quad p_{10}=\Phi(0)=0.500000,\quad p_{01}=\Phi(-0.5)=0.308538,\quad p_{11}=\Phi(1)=0.841345.

At $Z=0$, the probability contrast for $X$ is $0.500000-0.158655=0.341345$. At $Z=1$, it is $0.841345-0.308538=0.532807$. The probability-scale interaction is $0.532807-0.341345=0.191462$. It is not the predictor-scale coefficient $0.5$: the two numbers are contrasts on different scales.

Worked Example: Complementary Log-Log with No Predictor Interaction

Consider $\eta(x,z)=0+0.5x+0.5z$, so $\beta_3=0$. For the complementary log-log inverse link $h(\eta)=1-\exp\{-\exp(\eta)\}$, the predictor values at $(0,0),(1,0),(0,1),(1,1)$ are $0,0.5,0.5,1$. The corresponding probabilities are

h(0)=1-e^{-1}=0.632121,\quad h(0.5)=1-e^{-e^{0.5}}=0.807704,\quad h(1)=1-e^{-e}=0.934012.

Thus $p_{00}=0.632121$, $p_{10}=p_{01}=0.807704$, and $p_{11}=0.934012$. At $Z=0$, the probability contrast for $X$ is $0.807704-0.632121=0.175583$. At $Z=1$, it is $0.934012-0.807704=0.126308$. Subtracting the first contrast from the second gives $0.126308-0.175583=-0.049275$, or approximately $-0.04928$ using unrounded values.

Equivalently, the probability interaction is $p_{11}-p_{01}-p_{10}+p_{00}=0.934012-0.807704-0.807704+0.632121$, approximately $-0.04928$. The negative sign means that the probability contrast for $X$ is smaller when $Z=1$, despite the additive predictor.

A Reliable Calculation and Interpretation

For binary predictors, a short calculation prevents confusion between the two scales. Start with the predictor values implied by the model; transform each value with the appropriate inverse link; then calculate the two conditional probability contrasts before subtracting them. Keeping the order visible makes the sign interpretable.

Evaluate the predictor.
Compute $\eta(0,0)$, $\eta(1,0)$, $\eta(0,1)$, and $\eta(1,1)$ from the fitted coefficients.

Transform to probabilities.
Apply the model’s inverse link $h$ to each of the four predictor values.

Calculate conditional contrasts.
Find $p_{10}-p_{00}$ for $Z=0$ and $p_{11}-p_{01}$ for $Z=1$.

Compare and interpret.
Subtract the $Z=0$ contrast from the $Z=1$ contrast. State the result as a difference in probability changes, not as a change in a link-scale coefficient.

Common Mistake: Treating the Coefficient as a Probability Interaction

A common error is to see $\beta_3=0$ and conclude that the effect of $X$ on the probability is identical for both values of $Z$. That conclusion follows only on the predictor scale: $\beta_3=0$ makes the predictor-scale difference-in-differences zero. With a nonlinear inverse link, the probability contrasts can still differ, as the probit and complementary log-log examples show.

The reverse interpretive error is to report a nonzero $\beta_3$ as the size of a probability-scale interaction. The probit example with $\beta_3=0.5$ had a probability-scale interaction of about $0.1915$, not $0.5$. The correction is to label the scale of every contrast and, when the question concerns probabilities, calculate the four probabilities and their difference-in-differences.

Key takeaway: With nonlinear binary links, predictor-scale and probability-scale interactions answer different questions. For the probability-scale question, calculate the conditional probability contrast at each value of the other predictor and compare those contrasts directly.

Check Your Understanding

For each question, keep the predictor scale distinct from the probability scale.

In the 0/1-coded model, what predictor-scale contrast does $\beta_3$ represent?
Why can a probit model with $\beta_3=0$ still have a nonzero probability-scale interaction?
For the complementary log-log example, which conditional probability contrast is smaller, and by approximately how much?
Write the probability-scale difference-in-differences using $p_{00},p_{10},p_{01},p_{11}$.
Under what condition on $h''$, when $b,c>0$, does the integral expression imply a positive probability-scale interaction in an additive predictor model?