A Binary Responder Analysis for an Antimicrobial Regimen

Why Model Antimicrobial Success as a Binary Outcome?

An antimicrobial regimen is often evaluated by whether a patient meets a prespecified definition of treatment success, such as clinical improvement without rescue therapy and microbiologic eradication at a fixed assessment time. This outcome is useful for decisions, but it compresses several aspects of response into two categories. A patient-level exposure-response analysis can relate the probability of success to a pharmacokinetic exposure summary while keeping the interpretation tied to that binary endpoint.

The example below uses cumulative unbound time above the pathogen's minimum inhibitory concentration, denoted $T$, measured in hours over the treatment course. The model estimates an association between $T$ and the probability of success; by itself, it does not prove that changing exposure causes the estimated change in success probability. The outcome definition, assessment time, exposure summary, and analysis population must be specified before analysis.

Definition: Let $Y_i=1$ if patient $i$ meets the prespecified treatment-success definition and $Y_i=0$ otherwise. Let $T_i>0$ be that patient's exposure summary in hours. A logistic exposure-response model specifies the conditional success probability $p_i=P(Y_i=1\mid T_i)$ through $$ \operatorname{logit}(p_i)=\log\left(\frac{p_i}{1-p_i}\right)=\beta_0+\beta_1x_i,\qquad x_i=\log_2\left(\frac{T_i}{100\text{ h}}\right). $$ Here $\beta_0$ is the log odds of success at $T_i=100$ h, and $\beta_1$ is the change in log odds for each doubling of exposure. Equivalently, $p_i=1/\{1+\exp[-(\beta_0+\beta_1x_i)]\}$.

Assumptions and Interpretation

The binary outcome is treated as a Bernoulli observation: conditional on the modeled predictors, $Y_i$ has probability $p_i$ of being 1 and probability $1-p_i$ of being 0. The standard likelihood assumes patient outcomes are conditionally independent. If observations are clustered, for example because patients share a site-level factor not represented in the model, that independence assumption may fail and uncertainty estimates can be too small.

The model also assumes that the log odds change linearly with the chosen log-base-2 exposure predictor across the exposure range being analyzed. A positive $\beta_1$ means that higher exposure is associated with higher odds of success under this model. It does not mean that the probability increases by a fixed amount for each exposure doubling: the probability change depends on the starting probability. The exposure summary should be measured consistently, and important determinants of both exposure and outcome should be considered when interpreting the association.

A coefficient can be translated into an odds ratio. For a doubling of $T$, the odds ratio is $\exp(\beta_1)$. For a change from $T_a$ to $T_b$, the odds ratio is $\exp\{\beta_1\log_2(T_b/T_a)\}$. These are ratios of odds, not ratios of probabilities.

Worked Example: Interpreting the Exposure Slope

Suppose a hypothetical development dataset yields $\hat\beta_0=-0.400$ and $\hat\beta_1=1.200$. The estimated odds ratio for each doubling of $T$ is $$ \exp(\hat\beta_1)=\exp(1.200)=3.320. $$

Thus, within the fitted model, the odds of success at twice the exposure are estimated to be 3.320 times the odds at the original exposure. At $T=100$ h, $x=0$, so the estimated probability is $$ \hat p=\frac{1}{1+\exp[-(-0.400)]}=0.4013. $$ At $T=200$ h, $x=1$ and the estimated log odds are $-0.400+1.200=0.800$, giving $\hat p=1/(1+\exp[-0.800])=0.6900$. The probability rises by about 0.2887, or 28.87 percentage points, between these exposures; that difference is not itself the odds ratio.

From Exposure to a Patient-Level Probability

Once estimates are available, a patient's exposure summary is first expressed relative to the reference exposure of 100 h and then entered into the fitted equation. Extrapolation deserves particular caution: a formula can return a probability at any positive exposure, but predictions far outside the observed exposure range are not supported merely because the equation produces a number.

Worked Example: Predicting Success at 70 Hours

For a hypothetical patient with $T=70$ h, the transformed exposure is $$ x=\log_2(70/100)=\log_2(0.70)=-0.5146. $$ Using the illustrative estimates above, the fitted log odds are $$ \hat\eta=-0.400+1.200(-0.5146)=-1.0175. $$ The estimated success probability is therefore $$ \hat p=\frac{1}{1+\exp(-\hat\eta)} =\frac{1}{1+\exp(1.0175)} \approx 0.2655. $$

The model estimates about a 26.55% success probability at 70 h, conditional on the model and its included predictors. This is not a guarantee for an individual patient. It is an estimated probability for patients represented by the model at that exposure.

As a second check, at $T=280$ h, $x=\log_2(2.8)=1.4854$, so $\hat\eta=-0.400+1.200(1.4854)=1.3825$. This gives $\hat p=1/\{1+\exp(-1.3825)\}\approx0.7994$.

Calibration in a Hypothetical Validation Sample

Calibration concerns whether predicted probabilities agree with observed outcome frequencies. A prediction of 0.70 is well calibrated in a group of comparable patients if roughly 70% succeed over repeated comparable samples. Calibration is distinct from discrimination: predictions may rank patients by risk well but still be systematically too high or too low.

Suppose the model is evaluated in a hypothetical validation sample of 70 patients, grouped here by exposure for a compact calculation. The model predictions come from the development estimates above. The observed outcome counts are assumed to be recorded independently of fitting the model.

Exposure $T$ (h)	Patients	Observed successes	Model probability	Expected successes
50	10	2	0.1680	1.680
100	20	8	0.4013	8.026
200	20	14	0.6900	13.800
280	20	16	0.7994	15.988

Worked Example: Summarizing Validation Calibration

The total observed successes are $2+8+14+16=40$. The total expected successes from the model are $$ 10(0.1680)+20(0.4013)+20(0.6900)+20(0.7994) =1.680+8.026+13.800+15.988=39.494. $$ The observed-to-expected ratio is $40/39.494=1.013$. This is close to 1, suggesting little overall calibration bias in this sample, but it does not establish good calibration at every exposure.

A Brier score is the mean squared error of predicted probabilities for binary outcomes: $n^{-1}\sum_i(Y_i-\hat p_i)^2$, with lower values indicating better probabilistic predictions on the same outcome scale. For the 280 h group, the contribution to the sum of squared errors is $$ 4(0-0.7994)^2+16(1-0.7994)^2 =4(0.7994)^2+16(0.2006)^2 =3.200. $$ Across all four groups, the corresponding squared-error contributions are approximately 1.610, 4.800, 4.202, and 3.200. Thus the Brier score is $(1.610+4.800+4.202+3.200)/70=0.1973$. This single score summarizes probability error; it does not diagnose which exposure range is miscalibrated.

Calibration checks should also consider observed and predicted frequencies across clinically meaningful exposure ranges, with uncertainty made visible. Small groups can have noticeably variable observed proportions even when predictions are reasonable. A plot of observed proportions and mean predicted probabilities by group can reveal patterns such as underprediction at low exposure and overprediction at high exposure. Grouping is a diagnostic display, not a substitute for patient-level assessment, and the grouping boundaries should not be selected solely to make the model look favorable.

Estimating an Exposure Target and Its Uncertainty

A target exposure can be defined as the exposure at which the model predicts a prespecified success probability $p^*$. This is a model-based target, not automatically a clinical recommendation. It is meaningful only if the fitted exposure-response relationship is appropriate and the target lies in a range with adequate data.

Derivation: Set the fitted probability equal to $p^*$ and apply the logit transformation: $$ \operatorname{logit}(p^*)=\beta_0+\beta_1x^*. $$ If $\beta_1\ne0$, solve for $x^*$: $$ x^*=\frac{\operatorname{logit}(p^*)-\beta_0}{\beta_1}. $$ Since $x^*=\log_2(T^*/100\text{ h})$, the corresponding exposure is $$ T^*=100\text{ h}\times 2^{x^*}. $$ If $\beta_1>0$, predicted probability increases with exposure and this target has the intended direction. A zero slope provides no finite exposure target for a probability different from the intercept probability.

Worked Example: Targeting an 80% Success Probability

For $p^*=0.80$, $\operatorname{logit}(0.80)=\log(0.80/0.20)=1.3863$. Using $\hat\beta_0=-0.400$ and $\hat\beta_1=1.200$, $$ \hat x^*=\frac{1.3863-(-0.400)}{1.200}=1.4886, \qquad \hat T^*=100\text{ h}\times2^{1.4886}\approx280.6\text{ h}. $$ The fitted model therefore associates approximately 281 h of cumulative exposure with an 80% success probability.

Uncertainty in the coefficients creates uncertainty in the target. For illustration, suppose the estimated coefficient covariance matrix is diagonal, with $SE(\hat\beta_0)=0.250$, $SE(\hat\beta_1)=0.300$, and estimated covariance zero. The delta method approximates the variance of $\hat x^*$ by propagating coefficient uncertainty through $g(\beta_0,\beta_1)=[\operatorname{logit}(p^*)-\beta_0]/\beta_1$. Its derivatives are $\partial g/\partial\beta_0=-1/\beta_1$ and $\partial g/\partial\beta_1=-x^*/\beta_1$. Therefore, $$ \widehat{\operatorname{Var}}(\hat x^*)= \frac{0.250^2}{1.200^2}+ \frac{(1.4886)^2(0.300^2)}{1.200^2} =0.1819, \qquad SE(\hat x^*)\approx0.4265. $$ An approximate 95% Wald interval for $x^*$ is $1.4886\pm1.96(0.4265)=[0.6527,2.3245]$. Transforming both endpoints gives $$ 100\text{ h}\times2^{0.6527}\approx157.2\text{ h}, \qquad 100\text{ h}\times2^{2.3245}\approx500.9\text{ h}. $$ Thus the approximate target interval is 157.2 to 500.9 h. This interval reflects the stated coefficient uncertainty and approximation; it is not a 95% probability that a fixed true target lies within this particular interval.

The interval is wide relative to the point estimate, illustrating why reporting only a target value can be misleading. The calculation also assumes the coefficient covariance matrix is estimated appropriately and that the local normal approximation is adequate. If the slope is weakly estimated or close to zero, a symmetric Wald interval for the transformed target can perform poorly; profile-likelihood or bootstrap methods may better represent the uncertainty, provided the model is refitted appropriately in the resampling procedure.

Diagnostics, Alternatives, and Limitations

Check that the observed exposure range supports the claimed prediction and target. Inspect the exposure distribution, success proportions, fitted probabilities, and calibration patterns. Look for a lack of overlap, sparse outcome counts at the extremes, influential observations, and systematic departures from the assumed linear trend on the log-odds scale. A curved pattern in calibration residuals or observed proportions can signal that a single straight-line logit relationship is inadequate. Any proposed change in functional form should be guided by scientific plausibility and adequate data, not by isolated fluctuations in small groups.

A threshold-only analysis, such as declaring success likely above a single exposure cutoff, is simpler to communicate but discards the graded information in exposure and can make patients just above and below the cutoff appear very different. Categorizing exposure into bins can help with descriptive displays but generally loses information and yields unstable estimates in small groups. A logistic model with a continuous predictor retains the exposure ordering and gives probabilities, but relies on its functional-form assumption. Alternative probability links or a more flexible exposure curve can be considered when justified by the data and the scientific question; added flexibility also increases the need for sufficient observations and validation.

Common failure modes include defining success after looking at outcomes, using a different endpoint definition across patients, ignoring missing or unevaluable outcomes, and treating estimated exposure as error-free when it is not. Confounding is another concern: severity, pathogen susceptibility, adherence, and clinical care can influence both exposure and success. Adjusting for relevant prespecified covariates may help, but does not guarantee causal identification. The estimated target is conditional on the studied population, regimen, organism, endpoint, exposure calculation, and model.

Common pitfall: A fitted probability of 0.80 is the model-estimated conditional probability of success for a patient with the specified predictors, subject to model and parameter uncertainty; it does not guarantee that patient's outcome or an 80% observed success rate in every future sample. The model estimates a conditional probability under its assumptions. Validate calibration in relevant patients, report uncertainty, and distinguish an exposure associated with a target probability from a causal or dosing recommendation.

Check Your Understanding

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

In the model, what does $\beta_1$ represent, and what is the odds ratio associated with a doubling of exposure?
If $\hat\beta_0=-0.400$ and $\hat\beta_1=1.200$, calculate the predicted probability at 100 h.
What does an observed-to-expected success ratio of 1.013 suggest, and what can it not establish?
Give one reason a target exposure estimated by inverting the fitted model may be unreliable.

Solutions. (1) $\beta_1$ is the change in log odds for a doubling of exposure; the odds ratio is $\exp(\beta_1)$. With $\hat\beta_1=1.200$, this is $\exp(1.200)=3.320$. (2) At 100 h, $x=0$, so $\hat p=1/[1+\exp(0.400)]=0.4013$. (3) The ratio is close to 1, indicating that total predicted successes are close to total observed successes in this validation sample. It cannot establish good calibration within exposure ranges or in other populations. (4) For example, the target may be outside the observed exposure range, the slope may be weakly estimated, the logit relationship may be misspecified, or relevant confounding may remain.