Factor-by-Continuous Interactions

Why a Factor-by-Continuous Interaction Matters

A GLM without an interaction between a factor and a continuous predictor uses the same link-scale slope for every factor group. That assumption may be inappropriate. For example, the association between a policyholder’s years of experience and expected claim count might differ by vehicle type. A factor-by-continuous interaction allows the fitted slope to vary by group.

Interpreting the interaction requires keeping two scales distinct. The model’s linear predictor describes changes on the link scale; the response-scale mean is obtained by applying the inverse link. A group-specific link-scale slope is therefore not generally the same as a constant change in the expected outcome.

Definition: Let $A$ be a binary factor, coded $0$ for the reference group and $1$ for a comparison group, and let $x$ be a continuous predictor. A GLM with a factor-by-continuous interaction has conditional mean $\mu=E[Y\mid A,x]$, link $g$, and linear predictor $$ g(\mu)=\eta=\beta_0+\beta_1x+\beta_2A+\beta_3Ax. $$ The coefficient $\beta_3$ is the change in the link-scale slope when moving from the reference group to the comparison group.

This expression assumes that, with other predictors held fixed, the linear predictor is linear in $x$ and includes the displayed main effects and interaction. The GLM also specifies a response distribution and link; the model’s conditional mean must be compatible with that family and link. The formula does not describe a nonlinear relationship in $x$ unless additional terms, such as transformations or polynomial terms, are included.

Finding Group-Specific Slopes and Predictions

Set the factor indicator to each group’s value. For the reference group, $A=0$, so the predictor is $\eta_0(x)=\beta_0+\beta_1x$. For the comparison group, $A=1$, so it is $\eta_1(x)=(\beta_0+\beta_2)+(\beta_1+\beta_3)x$. Thus, the slopes with respect to $x$ on the link scale are $\beta_1$ and $\beta_1+\beta_3$, respectively.

\frac{d\eta_0}{dx}=\beta_1,\qquad \frac{d\eta_1}{dx}=\beta_1+\beta_3,\qquad \eta_1(x)-\eta_0(x)=\beta_2+\beta_3x.

The main-effect coefficient $\beta_2$ is the groups’ link-scale difference specifically at $x=0$, not necessarily their difference at every $x$. The interaction changes that difference as $x$ changes. If $x=0$ is outside the meaningful range, centering $x$ at a useful value can make the main-effect interpretation more relevant; it does not remove the interaction or change fitted predictions when the model is re-expressed consistently.

Worked Example: Different Claim-Count Slopes by Vehicle Type

Model and assumptions. In a hypothetical portfolio, let $Y$ be a policy’s claim count over one year. Assume $Y\mid A,x$ has a Poisson distribution with parameter $\mu$, so $E[Y\mid A,x]=\mu$, and use a log link. Let $A=0$ denote one vehicle type and $A=1$ another. The fitted model is $\log(\mu)=\log(2)+\log(2)x+\log(1.5)A+\log(1.25)Ax$. Calculate each group’s link-scale slope, then compare predicted means at $x=2$ and the effect of increasing $x$ by one unit from there.

Method. Substitute each factor value, identify its slope, and exponentiate the resulting linear predictor to obtain the Poisson mean.

For $A=0$, $\log(\mu_0)=\log(2)+\log(2)x$. Its link-scale slope is $\log(2)$, corresponding to a one-unit multiplicative factor of $e^{\log(2)}=2$ in the mean.

For $A=1$, $\log(\mu_1)=[\log(2)+\log(1.5)]+[\log(2)+\log(1.25)]x=\log(3)+\log(2.5)x$. Its link-scale slope is $\log(2.5)$, and its one-unit mean multiplier is $e^{\log(2.5)}=2.5$.

At $x=2$, the reference-group mean is $\mu_0=e^{\log(2)+2\log(2)}=2(2^2)=8$. The comparison-group mean is $\mu_1=e^{\log(3)+2\log(2.5)}=3(2.5^2)=18.75$.

Increasing $x$ from 2 to 3 multiplies the reference mean by 2, giving $8(2)=16$, an increase of $16-8=8$. For the comparison group, the multiplier is 2.5, giving $18.75(2.5)=46.875$, an increase of $46.875-18.75=28.125$.

Check. The means at $x=3$ also follow directly from the model: $\mu_0=2(2^3)=16$, and $\mu_1=3(2.5^3)=46.875$. The ratio between consecutive means in each group matches its stated one-unit multiplier. The common mistake avoided here is treating the log-scale slope as an additive change in claim count.

From Link-Scale Slopes to Response-Scale Changes

For a differentiable inverse link $g^{-1}$, the response mean is $\mu=g^{-1}(\eta)$. Holding the factor group and other predictors fixed, the chain rule gives the response-scale derivative as the derivative of the inverse link multiplied by the group’s link-scale slope. This is a local rate of change at the specified predictor value, not a finite change over an interval.

\frac{d\mu}{dx} = \frac{d\,g^{-1}(\eta)}{d\eta}\frac{d\eta}{dx} = (g^{-1})'(\eta)\,s_A, \qquad s_A= \begin{cases} \beta_1,&A=0,\\ \beta_1+\beta_3,&A=1. \end{cases}

This follows because $\mu$ depends on $x$ through $\eta$: a small change in $x$ first changes $\eta$ at rate $s_A$, and the inverse link converts that link-scale change into a change in $\mu$. Unless $(g^{-1})'(\eta)$ is constant, the response-scale derivative also depends on the current fitted value.

For a log link, $\mu=e^\eta$, so $(g^{-1})'(\eta)=e^\eta=\mu$ and $d\mu/dx=\mu s_A$. Over a finite change $\Delta x$, the mean ratio within a group is $e^{s_A\Delta x}$, and the additive change is $\mu(x)\{e^{s_A\Delta x}-1\}$. For a logit link, $p=\mu=1/(1+e^{-\eta})$, and $(g^{-1})'(\eta)=p(1-p)$. The probability derivative is therefore $p(1-p)s_A$; a finite probability change must instead be found by calculating both endpoint probabilities.

Worked Example: Probability Changes Under a Logit Interaction

Model and assumptions. In a hypothetical study, $Y$ indicates whether a submitted application is approved. Assume $Y\mid A,x$ is Bernoulli with probability $p$, and use the logit link: $\operatorname{logit}(p)=-2+0.5x+1A-0.2Ax$. Calculate both groups’ link-scale slopes and their predicted probabilities at $x=2$. Then calculate each probability’s local response-scale derivative there and the comparison group’s finite probability change from $x=2$ to $x=3$.

Method. Substitute the group indicator into the linear predictor. Apply the inverse-logit function for predictions and $p(1-p)s_A$ for the local derivative.

For $A=0$, the predictor is $\eta_0=-2+0.5x$, with slope $s_0=0.5$. At $x=2$, $\eta_0=-2+0.5(2)=-1$, so $p_0=1/(1+e^1)=0.2689$.

For $A=1$, the predictor is $\eta_1=-2+1+(0.5-0.2)x=-1+0.3x$, with slope $s_1=0.3$. At $x=2$, $\eta_1=-1+0.3(2)=-0.4$, so $p_1=1/(1+e^{0.4})=0.4013$.

At the reference-group prediction, the local derivative is $0.5(0.2689)(1-0.2689)=0.0983$. At the comparison-group prediction, it is $0.3(0.4013)(1-0.4013)=0.0721$. These are local probability changes per one unit of $x$ at $x=2$, not exact one-unit changes.

For the comparison group at $x=3$, $\eta_1=-1+0.3(3)=-0.1$, so $p_1(3)=1/(1+e^{0.1})=0.4750$. Its finite increase from $x=2$ is $0.4750-0.4013=0.0737$, to four decimal places. This is different from the local derivative $0.0721$.

Check. The comparison group’s link-scale predictor increases by $0.3$ from $-0.4$ to $-0.1$, as its slope requires. The probability remains between 0 and 1. The common mistake avoided is reporting $0.3$ as a probability increase; it is a log-odds slope.

More Than Two Factor Levels

With a factor having $K$ levels, treatment coding chooses one reference level and represents each other level with an indicator. The model includes an indicator and its interaction with $x$ for each non-reference level. If level $j$ has main-effect coefficient $\gamma_j$ and interaction coefficient $\delta_j$, its predictor is $(\beta_0+\gamma_j)+(\beta_1+\delta_j)x$. Its link-scale slope is $\beta_1+\delta_j$; the reference-level slope remains $\beta_1$.

In R, a formula such as outcome ~ group * x includes the main effects of group and x, as well as their interaction. With a factor, the numerical coefficients depend on the chosen reference level and coding. Predictions for a given group and $x$ do not depend on which reference level is used, provided the model is represented consistently.

Worked Example: Slopes Across Three Service Zones

Model and assumptions. A hypothetical study models annual service expense $Y$ in dollars by service zone and $x$, a workload measure. Assume $Y\mid\text{zone},x$ is normally distributed with mean $\mu$ and variance $400$ dollars squared, and use an identity link, so $\mu=E[Y\mid\text{zone},x]$. Zone A is the reference. The model is $\mu=100+4x+20B-3Bx-10C+2Cx$, where $B=1$ for Zone B and $C=1$ for Zone C; both are zero for Zone A. Find each zone’s slope and predicted mean at $x=5$.

Method. Substitute the appropriate indicator values, then read the coefficient of $x$ in each group-specific mean. Under the identity link, that coefficient is both the link-scale slope and the response-scale slope.

For Zone A, $B=C=0$, so $\mu_A=100+4x$. Its slope is $4$ dollars per workload unit, and at $x=5$, $\mu_A=100+4(5)=120$ dollars.

For Zone B, $B=1,C=0$, so $\mu_B=(100+20)+(4-3)x=120+x$. Its slope is $1$ dollar per workload unit, and at $x=5$, $\mu_B=120+5=125$ dollars.

For Zone C, $B=0,C=1$, so $\mu_C=(100-10)+(4+2)x=90+6x$. Its slope is $6$ dollars per workload unit, and at $x=5$, $\mu_C=90+6(5)=120$ dollars.

Check. Substitution into the original expression gives Zone B $100+20+20-15=125$, and Zone C $100+20-10+10=120$. The equal predicted means for A and C at $x=5$ do not imply equal slopes: their slopes are 4 and 6. The common mistake avoided is inferring equal group effects from predictions that happen to match at one predictor value.

Interpretation Traps

Omitting the interaction coefficient. The comparison-group slope is $\beta_1+\beta_3$, not $\beta_3$ alone. The interaction coefficient is the difference between group slopes.
Using $\beta_2$ as the group difference at every $x$. The link-scale difference is $\beta_2+\beta_3x$; $\beta_2$ applies at $x=0$.
Confusing scales. Under a log link, a slope is a change in log mean per unit, and its exponential is a mean multiplier. Under a logit link, a slope is a change in log odds, not a probability change.
Treating a derivative as a finite change. The derivative is local. For a finite interval, calculate predictions at both endpoints and subtract them (or take their ratio when the target is a ratio).
Ignoring units. A slope is per one unit of $x$. If $x$ is measured in hundreds of units, that is not the same increment as one individual unit.

Key takeaway: First calculate the slope for each group on the link scale. Then use the specified inverse link to obtain predicted means and response-scale changes. The interaction changes the link-scale slope; the link determines how that change appears in the response.

Check Your Understanding

For each problem, calculate on the stated scales and check that your answer respects the model’s link.

A Poisson GLM with log link has $\log(\mu)=\log(3)+\log(1.2)x+\log(2)A+\log(1.5)Ax$. Find both groups’ link-scale slopes and the mean for $A=1$ at $x=2$. (Answer: slopes $\log(1.2)$ and $\log(1.8)$; comparison-group mean $19.44$.)
A Bernoulli GLM with logit link has $\operatorname{logit}(p)=-1+0.4x-0.5A+0.2Ax$. Find the comparison-group slope, probability at $x=1$, and local probability derivative there. (Answer: slope $0.6$; $p=0.2891$; derivative $0.1233$.)
An identity-link model has mean $\mu=50+2x+8A-0.5Ax$. Find both group slopes and the group means at $x=4$. (Answer: slopes $2$ and $1.5$; means $58$ and $64$.)
A log-link model has $\log(\mu)=\log(4)+0.1x+\log(1.5)A-0.05Ax$. For $A=1$, find the slope, mean at $x=2$, and exact mean change from $x=2$ to $x=4$. (Answer: slope $0.05$; mean $6e^{0.1}\approx6.6310$; change $6e^{0.1}(e^{0.1}-1)\approx0.6974$.)
A logit model has $\operatorname{logit}(p)=-2+0.8x+0.6A-0.3Ax$. At $x=1$, calculate the link-scale group difference, both probabilities, and their probability difference (comparison minus reference). (Answer: link-scale difference $0.3$; probabilities $0.2315$ and $0.2891$; probability difference $0.0576$.)

What You'll Learn