Binary Efficacy Outcomes and Exposure

Why Binary Efficacy Outcomes Need Their Own Exposure Model

Some efficacy endpoints are recorded as a response or nonresponse at a defined assessment time: for example, whether a patient meets a prespecified response criterion by week 12. The outcome is binary, so a model for a continuous outcome is not an appropriate probability model. In particular, a fitted linear equation can predict values below 0 or above 1. Logistic and probit regression instead model the probability of response while keeping predictions between 0 and 1.

The interpretation requires care. An exposure coefficient is generally a conditional effect on a model scale, not a change in response probability that applies equally to every patient. The probability change depends on the starting probability and, when included, on covariates. For decision-making, it is often useful to translate the model into probabilities at specified exposures and to average those probabilities over a clearly defined covariate distribution.

Definition: Let \(Y_i=1\) denote that patient \(i\) meets the defined efficacy-response criterion in the specified assessment window, and let \(Y_i=0\) denote that the patient does not. For exposure \(E_i\) and covariates \(Z_i\), the binary-outcome model specifies \(p_i=P(Y_i=1\mid E_i,Z_i)\). A logistic model uses the logit of this probability; a probit model uses the standard normal quantile of this probability.

Model Form and Interpretation

For a simple logistic model, let \(x_i\) be exposure expressed in a stated unit or scale, such as increments of 10 mg/L in an exposure metric. Let \(Z_i\) denote covariates included in the model. The linear predictor and response probability are

\operatorname{logit}(p_i)=\log\left(\frac{p_i}{1-p_i}\right) =\beta_0+\beta_E x_i+\boldsymbol{\gamma}^{\mathsf T}Z_i, \qquad p_i=\frac{1}{1+\exp[-(\beta_0+\beta_E x_i+\boldsymbol{\gamma}^{\mathsf T}Z_i)]}.

At fixed covariate values, increasing \(x_i\) by one unit multiplies the odds of response by \(\exp(\beta_E)\), provided the model has a linear exposure term and no exposure-by-covariate interaction. The odds are \(p_i/(1-p_i)\); they are not the probability itself. A positive \(\beta_E\) means higher conditional odds under this model, but the probability increase is not constant across exposure values.

A probit model replaces the logit with the standard normal cumulative distribution function \(\Phi\):

\Phi^{-1}(p_i)=\beta_0+\beta_E x_i+\boldsymbol{\gamma}^{\mathsf T}Z_i, \qquad p_i=\Phi(\beta_0+\beta_E x_i+\boldsymbol{\gamma}^{\mathsf T}Z_i).

A probit coefficient is a change in the normal-quantile scale, not a log-odds change. Consequently, \(\exp(\beta_E)\) is not an odds ratio for a probit model. Both links give probabilities in the interval from 0 to 1, but their coefficients are on different scales and should not be compared directly.

Assumptions and scope: The outcome must be defined consistently as binary for a specified assessment window. The chosen model must represent the conditional response probability adequately over the exposure range of interest, including the link and exposure functional form. Observations must have an appropriate independence or dependence structure for the analysis; any repeated responses or clustering require a model that accounts for that structure. Exposure, covariates, and the analysis population must be defined in a way consistent with the estimand. A fitted association does not by itself establish a causal effect of changing exposure; the causal assumptions discussed in the earlier material on causal assumptions in individual-level exposure-response remain relevant.

Calculating Conditional Response Probabilities

Worked Example: Logistic Probability at Two Exposures

Hypothetical data and model. Suppose a study defines response at week 12 as \(Y=1\), and exposure \(x\) is the average concentration in increments of 10 mg/L. A fitted logistic model, at the specified covariate values, is \(\operatorname{logit}(p)=-1.2+0.8x\). Calculate the predicted response probabilities at \(x=0\) and \(x=1\), and interpret the exposure coefficient.

At \(x=0\), the linear predictor is \(-1.2+0.8(0)=-1.2\). Therefore,

p(0)=\frac{1}{1+\exp(1.2)} =\frac{1}{1+3.3201} =0.2315.

At \(x=1\), corresponding to 10 mg/L on this scale, the linear predictor is \(-1.2+0.8(1)=-0.4\). Thus,

p(1)=\frac{1}{1+\exp(0.4)} =\frac{1}{1+1.4918} =0.4013.

The modeled probability increases by \(0.4013-0.2315=0.1698\), or about 17.0 percentage points, between these two exposure values at the specified covariate values. The conditional odds ratio for a 10 mg/L increase is \(\exp(0.8)=2.226\). This is a multiplication of odds, not a multiplication of response probability. These are model-based predictions, not proof that increasing exposure would cause this change.

Worked Example: Probit Probability at Two Exposures

Hypothetical model. Suppose a probit model for the same type of binary endpoint is \(\Phi^{-1}(p)=-0.5+0.6x\), where \(x\) is exposure in 10 mg/L increments. Calculate the predictions at \(x=0\) and \(x=1\).

At \(x=0\), the linear predictor is \(-0.5\), so \(p(0)=\Phi(-0.5)=0.3085\). At \(x=1\), it is \(-0.5+0.6=0.1\), so \(p(1)=\Phi(0.1)=0.5398\). The predicted difference is \(0.5398-0.3085=0.2313\), or 23.13 percentage points.

The coefficient \(0.6\) is a change in the probit scale for a 10 mg/L exposure increase. It is not a log odds ratio. The probability predictions are obtained by applying \(\Phi\) to each linear predictor; reporting the coefficient alone would not convey the response probabilities at these exposures.

Conditional Effects and Marginal Effects

A conditional probability fixes covariates at specified values. A marginal probability averages predictions over a stated covariate distribution. For a target population represented by patients \(i=1,\ldots,n\) with weights \(w_i\), where \(w_i\geq0\) and \(\sum_i w_i=1\), the standardized probability at exposure \(x\) is

\bar p(x)=\sum_{i=1}^{n}w_i\,P(Y_i=1\mid x,Z_i).

The exposure is set to the same value \(x\) for each patient in this calculation, while each patient's covariates are retained. This answers a population-specific question: what average response probability does the fitted model predict at exposure \(x\) for this covariate distribution? The marginal contrast \(\bar p(x_1)-\bar p(x_0)\) is a difference in probabilities. It is distinct from a conditional odds ratio.

Justification: Averaging conditional probabilities follows from the law of total probability. For discrete covariate patterns, the standardized probability at exposure \(x\) is \(\sum_z P(Y=1\mid x,Z=z)P_{\text{target}}(Z=z)\), whereas the observational conditional probability is \(P(Y=1\mid x)=\sum_z P(Y=1\mid x,Z=z)P(Z=z\mid x)\). These are equal when the conditional covariate distribution given \(x\) matches the target distribution. Replacing the population weights \(P(Z=z)\) with the specified standardization weights gives the formula above. In general, applying the inverse-logit to an average linear predictor does not give this average probability, because the inverse-logit is nonlinear.

Worked Example: Standardizing Over Two Covariate Groups

Hypothetical model predictions. Consider a logistic model with a prognostic covariate defining two groups, each assigned standardization weight \(0.5\). At \(x=0\), the model predicts response probabilities \(0.2000000000\) and \(0.4284951334\). At \(x=1\), it predicts \(0.3181217\) and \(0.5831906\), respectively. The exposure scale is one 10 mg/L increment. The unrounded model predictions for the second exposure give a standardized probability of \(0.4506561699\).

At \(x=0\), the standardized probability is

\bar p(0)=0.5(0.2000000000)+0.5(0.4284951334) =0.3142475667.

At \(x=1\), average the two group-specific probabilities using their unrounded model predictions. The resulting value is \(\bar p(1)=0.4506561699\). Calculate the marginal difference before rounding the component probabilities:

\bar p(1)-\bar p(0) =0.4506561699-0.3142475667 =0.1364086032 \approx 0.1364.

Thus, the standardized predicted response probability is higher by approximately 13.64 percentage points at \(x=1\) than at \(x=0\) for this specified two-group covariate distribution. The standardization weights define the population represented by this contrast; changing those weights can change the marginal result.

Probability Changes, Diagnostics, and Alternatives

For the logistic model with a linear exposure term, differentiating the inverse-logit probability at fixed covariates gives

\frac{\partial p}{\partial x} =\beta_E p(1-p).

Since \(p(1-p)\) changes with \(p\), the probability slope varies across exposures even when \(\beta_E\) is constant. Its absolute value is largest at \(p=0.5\), where \(p(1-p)=0.25\), and becomes smaller as \(p\) approaches 0 or 1. For a probit model, the corresponding derivative is \(\beta_E\phi(\eta)\), where \(\eta\) is the linear predictor and \(\phi\) is the standard normal density. This also varies with exposure through \(\eta\). For finite exposure changes, calculate the two predicted probabilities and subtract them rather than treating a local derivative as the full change.

The model should be checked against the data and the intended prediction task. Warning signs include fitted probabilities close to 0 or 1 for exposure regions with little information, very large or unstable coefficients, sparse response or nonresponse groups, and systematic differences between observed response rates and predictions across exposure ranges. Inspect predictions across the observed exposure range and relevant covariate patterns; avoid relying on extrapolation where data are scarce. Separation, in which exposure or covariates perfectly or nearly perfectly distinguish responders from nonresponders, can make ordinary maximum-likelihood coefficient estimates unstable or unbounded. A numerically returned estimate does not remove that problem.

Logistic and probit links are common alternatives for a binary response, and often give similar fitted probabilities over the data-supported range. Their coefficients differ in scale, so compare their predicted probabilities or contrasts on a common scale rather than comparing coefficient magnitudes. A linear probability model is another possible approximation, but its unrestricted fitted values can lie outside 0 and 1, and its error variance depends on the response probability. None of these link choices resolves confounding, exposure measurement error, or an unsuitable assessment window. In particular, a conditional model association should not be described as causal without the necessary identification assumptions.

Common Mistakes

Reading an odds ratio as a probability ratio. An odds ratio of 2 does not mean that response probability doubles. Convert model predictions to probabilities at the exposures of interest.
Reporting a conditional coefficient as a population probability change. State the covariate values for a conditional prediction, or standardize over an explicitly defined covariate distribution.
Subtracting rounded values. Retain adequate precision in both probabilities, calculate the contrast, and round only the final difference. In the standardization example, the unrounded difference is 0.1364086032, which rounds to 0.1364, not 0.1365.
Calling a fitted exposure association causal. A model of observed exposure and response alone does not guarantee that the exposure contrast is unconfounded or that changing exposure would produce the predicted response change.
Comparing logistic and probit coefficients as if they shared a scale. They do not; compare predicted probabilities for the same patients and exposures instead.

Key takeaway: Define the binary response and assessment window, specify the link and exposure scale, and translate fitted predictions into probabilities. Keep conditional effects distinct from standardized marginal effects, and interpret each contrast for its stated population and covariate distribution.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

A logistic model has exposure coefficient \(0.5\) per 5 mg/L. What is the conditional odds ratio for a 5 mg/L increase, and what does it not mean?
For a logistic model, the predicted response probability at one exposure is 0.20. If the exposure coefficient is 0.7 per unit, calculate the probability at one unit higher, holding covariates fixed.
Two covariate groups have equal standardization weights and predicted probabilities at a specified exposure of 0.30 and 0.50. What is the standardized probability?
Why can the marginal probability difference change when the covariate distribution used for standardization changes?
A probit exposure coefficient is positive. Can it be interpreted as a conditional odds ratio? Explain.

Solutions. (1) The odds ratio is \(\exp(0.5)=1.649\), meaning the conditional odds are multiplied by about 1.649 for a 5 mg/L increase under the model; it does not mean the response probability increases by that factor. (2) The starting odds are \(0.20/0.80=0.25\). The new odds are \(0.25\exp(0.7)=0.5034\), giving probability \(\frac{0.25\exp(0.7)}{1+0.25\exp(0.7)}\approx0.3349\). (3) The standardized probability is \(0.5(0.30)+0.5(0.50)=0.40\). (4) The covariate-specific probabilities differ, so changing the weights changes their weighted average and can change a contrast between exposure values. (5) No. The probit coefficient changes the normal-quantile scale; obtain probabilities using \(\Phi\), and do not interpret its exponent as an odds ratio.