Multiplicative Interaction in Logistic Models

Why the Scale Matters

A genotype and an exposure can jointly relate to a binary outcome in ways that depend on the scale used to compare outcomes. In logistic regression, the interaction coefficient describes whether the genotype odds ratio changes across exposure levels. It does not directly say whether the increase in disease risk due to the genotype changes across those levels. Those are different questions, even when they concern the same four genotype-exposure groups.

This tutorial considers a binary genotype indicator $G$, coded 0 or 1, a binary exposure indicator $E$, also coded 0 or 1, and a binary outcome $Y$, with $Y=1$ denoting the outcome. The interpretation depends on these coding choices: $G=1$ and $E=1$ must represent the genotype and exposure categories being compared.

Definition: A logistic regression with a genotype-by-exposure product term can be written as $$ \operatorname{logit}\{P(Y=1\mid G,E,C)\} =\beta_0+\beta_GG+\beta_EE+\beta_{GE}GE+\boldsymbol{\gamma}^{T}C, $$ where $\operatorname{logit}(p)=\log\{p/(1-p)\}$, and $C$ denotes any included covariates. The coefficient $\beta_{GE}$ is the multiplicative interaction coefficient on the odds scale. With $G$ and $E$ coded 0/1, $\exp(\beta_{GE})$ is the ratio of the genotype odds ratio at $E=1$ to the genotype odds ratio at $E=0$, conditional on the covariates in the model.

The model assumes that, conditional on the included variables, the log odds follow the specified linear predictor. For the standard coefficient-based inference described below, observations must also be independent under the sampling design, the model must be identifiable, and the estimated standard error must be adequate for the approximation used. A product term does not, by itself, establish a causal interaction.

Deriving the Ratio of Odds Ratios

For a fixed value of the covariates $C$, compare the outcome odds for $G=1$ with those for $G=0$, first at $E=0$. The exposure and product terms vanish when $E=0$, so the log odds difference is $\beta_G$. Exponentiating gives the genotype odds ratio at $E=0$.

\operatorname{OR}_{G\mid E=0}=\exp(\beta_G).

At $E=1$, the genotype terms in the log odds are $\beta_GG+\beta_{GE}G$. The log odds difference comparing $G=1$ with $G=0$ is therefore $\beta_G+\beta_{GE}$, giving:

\operatorname{OR}_{G\mid E=1}=\exp(\beta_G+\beta_{GE}).

Dividing the genotype odds ratio at $E=1$ by the one at $E=0$ cancels the main-effect term $\beta_G$. This establishes the interpretation of the product coefficient:

\frac{\operatorname{OR}_{G\mid E=1}}{\operatorname{OR}_{G\mid E=0}} = \frac{\exp(\beta_G+\beta_{GE})}{\exp(\beta_G)} =\exp(\beta_{GE}).

Null hypothesis: No multiplicative interaction on the odds scale means the genotype odds ratio is the same at both exposure levels. In this model, that is equivalent to $$ H_0:\beta_{GE}=0 \quad\Longleftrightarrow\quad \frac{\operatorname{OR}_{G\mid E=1}}{\operatorname{OR}_{G\mid E=0}}=1. $$ A two-sided alternative is $H_a:\beta_{GE}\ne 0$, or equivalently a ratio of odds ratios different from 1.

Worked Example: Interpreting a Fitted Interaction Coefficient

Suppose a hypothetical study fits the stated logistic model and estimates $\hat{\beta}_G=0.4055$ and $\hat{\beta}_{GE}=0.2624$. Interpret the genotype odds ratios at the two exposure levels and their ratio, holding included covariates fixed.

At $E=0$, the estimated genotype odds ratio is $$ \exp(\hat{\beta}_G)=\exp(0.4055)\approx 1.500. $$ Thus, at exposure level 0, the estimated outcome odds for $G=1$ are 1.5 times the odds for $G=0$, conditional on the model covariates.

At $E=1$, the genotype log odds ratio is $0.4055+0.2624=0.6679$, so the odds ratio is $$ \exp(0.4055+0.2624)=\exp(0.6679)\approx 1.950. $$ The ratio of the two genotype odds ratios is $$ \frac{1.950}{1.500}=1.300, $$ which agrees with $\exp(\hat{\beta}_{GE})=\exp(0.2624)\approx1.300$. The estimated genotype odds ratio is 30% larger at $E=1$ than at $E=0$. This is a comparison of odds ratios, not a statement that genotype increases risk by 30%, nor a causal conclusion.

Estimating and Testing Interaction From Stratum Counts

With binary $G$ and $E$, four genotype-exposure strata define four outcome odds. For a simple model with no additional covariates, a logistic model containing the intercept, both main effects, and the product term can represent these four stratum-specific log odds. If each stratum has both outcome values, its observed odds are the number with $Y=1$ divided by the number with $Y=0$. The ratio of odds ratios can then be calculated directly.

For inference, a common large-sample Wald procedure estimates the standard error of $\hat{\beta}_{GE}$, calculates $z=\hat{\beta}_{GE}/SE(\hat{\beta}_{GE})$, and compares the absolute value with a standard normal distribution under $H_0$. A two-sided p-value is $2P(Z\ge |z|)$ for $Z\sim N(0,1)$. This procedure relies on an adequate normal approximation; small or empty cells can make it unreliable.

Worked Example: Ratio of Odds Ratios and Its Confidence Interval

Consider hypothetical independent observations in four strata. The table gives the number with the outcome and the number without it. Assume the simple four-parameter logistic model is used, the observations are independent, and all cells are positive.

Exposure $E$	Genotype $G$	$Y=1$	$Y=0$
0	0	20	80
0	1	30	70
1	0	40	60
1	1	60	40

At $E=0$, the genotype odds ratio is $(30/70)/(20/80)=12/7\approx1.714$. At $E=1$, it is $(60/40)/(40/60)=9/4=2.250$. Hence the estimated ratio of odds ratios is $$ \widehat{\operatorname{ROR}} =\frac{9/4}{12/7} =\frac{21}{16} =1.3125, \qquad \hat{\beta}_{GE}=\log(21/16)\approx0.2719. $$

For these independent stratum counts, the estimated variance of the log ratio is the sum of the reciprocals of the eight cell counts: $$ \widehat{\operatorname{Var}}(\hat{\beta}_{GE}) =\frac1{20}+\frac1{80}+\frac1{30}+\frac1{70} +\frac1{40}+\frac1{60}+\frac1{60}+\frac1{40} \approx0.193452. $$ Therefore $SE(\hat{\beta}_{GE})=\sqrt{0.193452}\approx0.4398$. The Wald statistic is $z=0.2719/0.4398\approx0.618$, giving a two-sided p-value of approximately $0.536$. At significance level $\alpha=0.05$, these data do not provide evidence against $H_0:\beta_{GE}=0$; this is not proof that the null is true.

A 95% Wald interval on the log scale is $$ 0.2719\ \pm\ 1.96(0.4398)\approx(-0.5901,\ 1.1340). $$ Exponentiating its endpoints gives an approximate 95% confidence interval for the ratio of odds ratios: $$ \bigl(\exp(-0.5901),\ \exp(1.1340)\bigr)\approx(0.554,\ 3.108). $$ The interval includes 1, consistent with the two-sided test. In repeated samples under the model and its assumptions, intervals constructed this way would cover the true ratio of odds ratios about 95% of the time. The interval is wide: the data are compatible with a substantially smaller or larger genotype odds ratio at $E=1$ than at $E=0$.

Odds-Scale Interaction Is Not Risk-Scale Interaction

For each genotype-exposure combination, let $p_{ge}=P(Y=1\mid G=g,E=e)$, with covariates held fixed if they are included. Additive interaction on the risk scale is commonly summarized by the contrast $p_{11}-p_{10}-p_{01}+p_{00}$. A value of zero means the risk difference for genotype is the same at both exposure levels. By contrast, $\beta_{GE}=0$ means the genotype odds ratio is the same at both exposure levels. The two statements are not equivalent because odds $p/(1-p)$ are a nonlinear transformation of risk $p$.

Worked Example: No Odds Interaction but Nonzero Risk Interaction

Suppose a hypothetical population has $p_{00}=0.10$, and the genotype odds ratio is 2 at both exposure levels while the exposure odds ratio is 3 at both genotype levels. This corresponds to a logistic model with $\beta_{GE}=0$, $\beta_G=\log(2)$, and $\beta_E=\log(3)$.

The baseline odds are $0.10/0.90=1/9$. Multiplying odds by 2 for $G=1$, by 3 for $E=1$, and by 6 for both gives the four risks: $$ p_{00}=0.10,\quad p_{10}=\frac{2/9}{1+2/9}=\frac{2}{11}\approx0.1818,\quad p_{01}=\frac{3/9}{1+3/9}=0.25,\quad p_{11}=\frac{6/9}{1+6/9}=0.40. $$ The genotype odds ratio is 2 at $E=0$ and also 2 at $E=1$, so the ratio of odds ratios is $2/2=1$. But the additive risk interaction is $$ p_{11}-p_{10}-p_{01}+p_{00} =0.40-\frac{2}{11}-0.25+0.10 \approx0.0682. $$ Thus, there is no interaction on the odds scale in this model, while the additive risk contrast is about 0.0682, or 6.82 percentage points. Neither result contradicts the other; each answers a question on a different scale.

Worked Example: Zero Risk Interaction but Nonzero Odds Interaction

In a separate hypothetical setting, suppose the four risks are $p_{00}=0.10$, $p_{10}=0.20$, $p_{01}=0.30$, and $p_{11}=0.40$. The additive interaction contrast is $$ 0.40-0.20-0.30+0.10=0. $$ The genotype risk difference is $0.20-0.10=0.10$ at $E=0$ and $0.40-0.30=0.10$ at $E=1$.

The genotype odds ratio at $E=0$ is $(0.20/0.80)/(0.10/0.90)=2.25$. At $E=1$, it is $(0.40/0.60)/(0.30/0.70)=14/9\approx1.556$. The ratio of odds ratios is therefore $$ \frac{14/9}{2.25}=\frac{14/9}{9/4}=\frac{56}{81}\approx0.6914, $$ not 1. These risks have no additive interaction, yet they do have interaction on the odds scale. The examples show why the scale must be named when reporting an interaction result.

Model Interpretation and Common Pitfalls

A product coefficient answers a specific, conditional question: does the genotype odds ratio differ between the two exposure levels in the fitted logistic model, at fixed values of included covariates? It does not report a difference in probability or risk. To discuss risks, one must examine predicted probabilities or other suitable risk estimates and compute a risk-scale contrast. The intercept and main-effect coefficients matter for those probabilities; the interaction coefficient alone cannot supply them.

Reading $\beta_{GE}$ as an odds ratio. The coefficient is on the log-odds scale. Its exponential, $\exp(\beta_{GE})$, is the ratio of odds ratios.
Calling $\exp(\beta_{GE})$ a risk ratio. It compares odds ratios, not risks or risk ratios. Odds and risks are distinct quantities.
Assuming no interaction is scale-free. A null product coefficient means no interaction on the odds scale in the specified model. It does not imply a zero additive risk contrast.
Ignoring coding and reference groups. The main effects refer to the groups coded 0, and changing coding can change coefficient signs and reference comparisons. State which genotype and exposure categories are coded 1.
Overstating a non-significant result. A p-value above $\alpha$ does not establish equal odds ratios. Consider the estimate and its confidence interval, as well as the precision of the data.

The distinction matters because different scientific questions call for different scales. A logistic product term is appropriate when the target is effect modification of the odds ratio under the stated model. Questions about excess cases, absolute risk differences, or public-health burden require risk-scale quantities. A complete report names the scale, describes the comparison and coding, and gives uncertainty rather than relying on the word “interaction” alone.

Key takeaway: In a logistic model with binary 0/1 genotype and exposure, $\exp(\beta_{GE})$ equals the genotype odds ratio at exposure 1 divided by the genotype odds ratio at exposure 0. Its null value is 1. This odds-scale null is distinct from the null of no additive interaction in risks.

Check Your Understanding

Answer each question using the stated outcome, genotype, exposure, and scale.

In a 0/1-coded logistic model, what quantity does $\exp(\beta_{GE})$ compare?
Write the null hypothesis of no multiplicative interaction on the odds scale, both in terms of $\beta_{GE}$ and the ratio of odds ratios.
If $\hat{\beta}_{GE}<0$, what does that imply about the estimated genotype odds ratios at $E=1$ and $E=0$?
Why does a zero additive risk interaction not necessarily imply $\beta_{GE}=0$?
A confidence interval for a ratio of odds ratios includes 1. What does that say about the evidence against the null, and what does it not prove?

Exposure \(E\)	Genotype \(G\)	\(Y=1\)	\(Y=0\)
0	0	20	80
0	1	30	70
1	0	40	60
1	1	60	40