Modeling Exposure-Response for a Binary Clinical Outcome

Why Model a Binary Clinical Outcome?

Many clinical outcomes are recorded as a response or no response, an event or no event, or success or failure by a defined time. The central exposure-response question is then not how much a continuous measurement changes, but how the probability of the binary outcome varies across exposure. A model can summarize that relationship and estimate the exposure corresponding to a clinically chosen response probability. Those estimates are useful only when the endpoint, model assumptions, and uncertainty are made explicit.

The case below is entirely hypothetical. It considers a binary clinical response by a prespecified assessment visit and exposure measured as area under the concentration-time curve (AUC), in mg·h/L. The analysis describes the association between exposure and response in the modeled population. By itself, it does not establish that changing an individual patient's exposure would cause the modeled change in response probability.

The Binomial Logistic Exposure-Response Model

Definition: For individual $i$, let $Y_i=1$ denote response and $Y_i=0$ denote no response. Let $x_i$ be exposure and let $p_i=P(Y_i=1\mid x_i)$. A two-parameter logistic exposure-response model is $$ Y_i\mid x_i\sim \operatorname{Bernoulli}(p_i),\qquad \operatorname{logit}(p_i)=\log\left(\frac{p_i}{1-p_i}\right)=\beta_0+\beta_1x_i. $$ Equivalently, $p_i=1/\{1+\exp[-(\beta_0+\beta_1x_i)]\}$. The parameters $\beta_0$ and $\beta_1$ describe the intercept and exposure slope on the log-odds scale.

The model assumes the response is correctly coded as binary, observations are conditionally independent given the modeled predictors, and the log-odds of response are linear in exposure over the range being modeled. It also assumes the exposure and response are measured and defined consistently. If relevant baseline predictors are included, they must be specified in the model, and the exposure relationship is interpreted conditional on them. For the simple model here, the exposure coefficient is an association; unmeasured differences between people at different exposures can confound that association.

The slope has a precise interpretation: for an increase of one exposure unit, the modeled odds of response are multiplied by $\exp(\beta_1)$, holding other included predictors fixed. This is an odds ratio, not a probability difference. The probability change depends on the starting probability. A positive slope means a positive modeled association between exposure and response probability, but does not imply that every person's outcome improves as exposure rises.

Fitting and Checking the Model

For independent observations, the likelihood is the product of their Bernoulli probabilities. If observations are grouped at exposure $x_j$, with $r_j$ responses among $n_j$ individuals, the corresponding binomial likelihood is proportional to $$ L(\beta_0,\beta_1)=\prod_j p_j^{r_j}(1-p_j)^{n_j-r_j}, \qquad p_j=\frac{1}{1+\exp[-(\beta_0+\beta_1x_j)]}. $$ The maximum-likelihood estimates choose coefficients that make the observed response counts plausible under this likelihood. Standard errors usually come from the estimated covariance matrix, obtained from the curvature of the log-likelihood near its maximum. They quantify sampling uncertainty under the model; they do not account automatically for model misspecification or unmeasured confounding.

Before interpreting a fitted curve, inspect observed response proportions across exposure, fitted probabilities, and the range and distribution of exposure. A systematic pattern in residuals or observed-versus-predicted responses can indicate that a straight line on the log-odds scale is inadequate. Sparse data at the low or high end, extreme fitted probabilities, and very large coefficient standard errors are warnings. Complete or quasi-complete separation can occur when exposure values or covariate patterns perfectly distinguish responders from nonresponders; then a finite maximum-likelihood estimate may not exist, or estimates may be unstable.

Worked Example: Fitting the Two-Parameter Curve

Suppose a hypothetical trial has 100 participants at AUC 1 mg·h/L, of whom 30 respond, and 100 participants at AUC 3 mg·h/L, of whom 60 respond. Assume independent responses conditional on exposure and apply the logistic model above.

The observed response proportions are $0.30$ and $0.60$, both strictly between zero and one. With two distinct exposure values and two model parameters, the fitted curve can reproduce these two group proportions exactly. The group logits are $$ \operatorname{logit}(0.30)=\log(0.30/0.70)=-0.84730,\qquad \operatorname{logit}(0.60)=\log(0.60/0.40)=0.40547. $$

Solving the two linear equations gives $$ \hat\beta_1=\frac{0.40547-(-0.84730)}{3-1}=0.62638,\qquad \hat\beta_0=-0.84730-(0.62638)(1)=-1.47368. $$ Thus, to three significant figures, the fitted logit is $-1.47+0.626x$. Each 1 mg·h/L increase in AUC corresponds to an estimated odds multiplier of $\exp(0.626)=1.87$. This is an estimated 87% increase in odds per unit, not an 87 percentage-point increase in response probability.

For grouped binomial data, the observed information matrix at this fit is $X^\mathsf{T}WX$, where each group contributes $n_jp_j(1-p_j)$ to the weight and its exposure to the design row. Here the weights are $100(0.30)(0.70)=21$ and $100(0.60)(0.40)=24$, so $$ X^\mathsf{T}WX= \begin{pmatrix} 45&93\\ 93&237 \end{pmatrix}, \qquad \widehat{\operatorname{Cov}}(\hat\beta_0,\hat\beta_1)= \frac{1}{2016} \begin{pmatrix} 237&-93\\ -93&45 \end{pmatrix} \approx \begin{pmatrix} 0.1176&-0.04613\\ -0.04613&0.02232 \end{pmatrix}. $$ The determinant is $45(237)-93^2=2016$, so the displayed inverse is well-defined. Since both observed proportions are interior, this fit has finite coefficients. If either group proportion were zero or one, an exact fitted probability of zero or one would require an infinite logit; a finite maximum-likelihood estimate need not exist.

Exact agreement at these two exposure groups is not evidence that the log-odds relationship is linear between those values or beyond them. More exposure levels and adequate data across the range are needed to assess the shape. In this example, a plot cannot reveal departures at unobserved intermediate exposures.

Exposure for a Target Response Probability

A target probability $q$, with $0<q<1$, corresponds to log-odds $\log\{q/(1-q)\}$. When $\beta_1\ne0$, solving the logistic model for the exposure associated with that probability gives $$ x_q=\frac{\log\{q/(1-q)\}-\beta_0}{\beta_1}. $$ For $q=0.50$, the target log-odds are zero, so $x_{0.50}=-\beta_0/\beta_1$. This is the modeled exposure at which the response probability is 50%, not a guarantee that any particular patient will respond at that exposure.

Worked Example: Exposure Associated With a 50% Response Probability

Using the fitted coefficients above, the estimated exposure corresponding to a 50% response probability is $$ \hat x_{0.50}=\frac{-\hat\beta_0}{\hat\beta_1} =\frac{1.47368}{0.62638}=2.35269\approx2.353\ \text{mg·h/L}. $$

At that exposure, the fitted logit is approximately zero and the fitted probability is $1/(1+\exp(0))=0.50$. This is an estimate of a population response probability under the fitted model. It is not an individual-level threshold separating certain responders from certain nonresponders.

A confidence interval for this target exposure can be approximated with the delta method. For $g(\beta_0,\beta_1)=-\beta_0/\beta_1$, the gradient is $$ \nabla g= \left(-\frac{1}{\beta_1},\frac{\beta_0}{\beta_1^2}\right). $$ The first-order variance approximation is $\nabla g^\mathsf{T}\widehat{\operatorname{Cov}}(\hat\beta_0,\hat\beta_1)\nabla g$. Using the rounded estimates and covariance entries gives gradient $(-1.596,-3.755)$, and $$ \widehat{\operatorname{Var}}(\hat x_{0.50}) =(-1.596)^2(0.1176)+(-3.755)^2(0.02232) +2(-1.596)(-3.755)(-0.04613) \approx0.06135. $$ Thus $SE(\hat x_{0.50})=\sqrt{0.06135}=0.2477$ mg·h/L. A 95% Wald interval is $2.35269\pm1.96(0.2477)$, or approximately $1.867$ to $2.838$ mg·h/L. Under the model and the large-sample approximation, this procedure has 95% coverage over repeated samples. It does not mean there is a 95% probability that the fixed target exposure lies in this particular interval.

Uncertainty in Predicted Response and Exposure Effect

Uncertainty in a fitted probability at a specified exposure can be assessed first on the logit scale, where the model is linear. At $x$, the estimated linear predictor is $\hat\eta(x)=\hat\beta_0+\hat\beta_1x$, with estimated variance $$ \widehat{\operatorname{Var}}\{\hat\eta(x)\} =\widehat{\operatorname{Var}}(\hat\beta_0) +x^2\widehat{\operatorname{Var}}(\hat\beta_1) +2x\widehat{\operatorname{Cov}}(\hat\beta_0,\hat\beta_1). $$ A Wald interval can be calculated on this scale and transformed at both endpoints using the logistic function. The transformed interval remains between zero and one. It is an interval for the modeled response probability, not for an individual outcome, which remains binary.

Worked Example: Probability Estimate and Test at a Specified Exposure

At AUC $2.5$ mg·h/L, the fitted logit is $$ \hat\eta(2.5)=-1.474+(0.6264)(2.5)=0.0920, \qquad \hat p(2.5)=\frac{1}{1+\exp(-0.0920)}=0.5230. $$ The model therefore estimates a response probability of about 52.3% at that exposure.

Using the covariance estimates from the fit, the logit variance and standard error are $$ \widehat{\operatorname{Var}}\{\hat\eta(2.5)\} =0.1176+(2.5)^2(0.02232)+2(2.5)(-0.04613) =0.02645, \qquad SE\{\hat\eta(2.5)\}=\sqrt{0.02645}=0.1626. $$ A 95% Wald interval for the logit is $0.0920\pm1.96(0.1626)=[-0.2267,0.4107]$. Transforming the endpoints gives an approximate 95% interval of $0.444$ to $0.601$ for the modeled response probability.

To test whether the exposure slope is zero, use $H_0:\beta_1=0$ against $H_a:\beta_1\ne0$. The estimated slope standard error is $\sqrt{0.02232}=0.1494$, so the Wald statistic is $z=0.6264/0.1494=4.19$. The two-sided normal-reference p-value is approximately $0.000028$. Under the model, this is strong evidence against a zero linear exposure slope. It is not the probability that $H_0$ is true, and it does not establish causation.

A different question is whether the response probability at AUC 2.5 differs from 50%. Testing $H_0:\eta(2.5)=0$ gives $z=0.0920/0.1626=0.566$, with a two-sided p-value of about $0.57$. The data do not provide strong evidence that the modeled probability at this exposure differs from 50%. This test is distinct from the test of whether the exposure slope is zero.

Justification and Limits of the Uncertainty Calculations

The delta method approximates the variance of a smooth function of estimated coefficients by propagating their covariance through the function's local slope. If $\hat\theta=(\hat\beta_0,\hat\beta_1)$ has covariance matrix $\widehat V$, a first-order Taylor expansion around the estimates gives $$ g(\hat\theta)\approx g(\theta)+\nabla g(\theta)^\mathsf{T}(\hat\theta-\theta). $$ Taking variances of the right-hand side gives $\operatorname{Var}\{g(\hat\theta)\}\approx\nabla g^\mathsf{T}V\nabla g$, which yields the target-exposure calculation above. The approximation is most credible when the coefficient estimates are approximately normal and the target is not highly sensitive to small coefficient changes.

The target $x_q$ divides by $\beta_1$. If the slope is close to zero or poorly estimated, the target can be extremely uncertain or effectively unbounded; a symmetric Wald interval may then be misleading. Profile-likelihood intervals or a suitable bootstrap can better represent an asymmetric or weakly identified target, though they still rely on a defensible model and data. A target outside the observed exposure range is an extrapolation, not a data-supported estimate. Even a narrow interval does not validate the chosen target probability as clinically appropriate.

The simple logistic model is attractive because it gives probabilities within zero and one, has interpretable log-odds coefficients, and is straightforward to fit. Alternatives include modeling observed proportions at exposure groups, which avoids imposing a single curve but can be unstable with sparse groups and cannot naturally predict between groups; a probit model, which uses a different link and may yield similar fitted probabilities but changes coefficient interpretation; and a more flexible exposure function, which may capture curvature but requires more data and can overfit. Model choice should follow the question, the exposure range, and diagnostic evidence rather than a preference for one link function.

A practical pitfall is to treat a statistically significant exposure slope as proof that higher exposure causes greater response. The wrong reasoning is: “The slope test has a small p-value, therefore increasing a patient's exposure will improve their outcome.” The correction is that the test supports an exposure association under the fitted model and its assumptions. Causal interpretation additionally depends on study design and control of confounding, including factors that influence both exposure and outcome. Similarly, excellent fit at a few exposure groups is not proof that the curve is correct elsewhere.

Define the endpoint and exposure.
Specify the response coding, assessment time, exposure metric, units, and analysis population.

Fit and inspect the model.
Check exposure coverage, observed and fitted response patterns, separation, and whether a linear logit is plausible.

Answer the clinical question with uncertainty.
Report target probabilities or exposures with units and intervals, distinguish tests that address different hypotheses, and limit conclusions to what the design supports.

Key takeaway: Logistic exposure-response modeling translates a binary outcome into a probability curve and can estimate an exposure associated with a chosen probability. The estimate is meaningful only over a defensible exposure range, with uncertainty and model assumptions reported, and without confusing association with causation.

Check Your Understanding

Work through these questions, then compare your reasoning with the brief solutions.

If the estimated exposure slope is positive, what does $\exp(\hat\beta_1)$ represent?
Why can two exposure groups with interior response proportions be fitted exactly by an intercept and linear exposure term, and what does that fail to establish?
For $\hat\beta_0=-1.0$ and $\hat\beta_1=0.5$, find the exposure associated with a 50% modeled response probability.
At an exposure of $x$, why must the covariance between $\hat\beta_0$ and $\hat\beta_1$ be included in the estimated variance of the fitted logit?
What does a small p-value for the slope test establish, and what does it not establish?

Solutions. (1) It is the estimated multiplicative change in response odds for a one-unit increase in exposure, holding other modeled predictors fixed. (2) Two interior proportions have finite logits, and the two coefficients solve the two logit equations; this exact fit does not assess curvature between or beyond the observed exposures. (3) $x_{0.50}=-(-1.0)/0.5=2$ exposure units. (4) The fitted logit is a sum of two estimated coefficients, which are generally correlated; its variance therefore includes $2x\widehat{\operatorname{Cov}}(\hat\beta_0,\hat\beta_1)$. (5) It is evidence against a zero slope under the model and test assumptions; it does not prove causation or give the probability that the null hypothesis is true.