From Model Output to Decision-Facing Outcomes
A dose-response model does not always report its predictions in the units needed to describe outcomes at a dose. A fitted model may produce a linear predictor, a log mean, or a hazard ratio, while a decision maker needs a probability of response, an average outcome in clinical units, or an absolute event risk over a specified period. Translating between these quantities is part of interpreting the model, not a cosmetic change in presentation.
The key distinction is between a model scale, on which a model is specified or estimated, and a decision-facing outcome scale, on which the predicted outcome is expressed. A link function connects the two. In addition, predictions may be conditional on particular covariates or summarized over a population. Those summaries answer different questions.
The estimand defines which outcome, population, and conditions the prediction is intended to describe. Before translating a model output, therefore, make sure that the prediction target matches the estimand: an individual’s conditional response, a population average, and a risk by a fixed time are not interchangeable. A transformation cannot repair a mismatch in the outcome or population being predicted.
Model Scales, Link Functions, and Prediction Targets
For a generalized linear model, a typical form is \(g\{E(Y\mid d,x)\}=\eta(d,x)\), where \(Y\) is the outcome, \(d\) is dose, \(x\) denotes covariates, \(g\) is a link function, and \(\eta\) is the linear predictor. The outcome-scale prediction is obtained by applying the inverse link \(g^{-1}\): \(E(Y\mid d,x)=g^{-1}\{\eta(d,x)\}\). The exact interpretation depends on the outcome and model. For example, in a binary-outcome logistic model the inverse link gives a probability; in a log-link model it gives a conditional mean.
A prediction should also name its time horizon when the outcome changes over time, and its covariate condition when it is conditional. If the objective is a population summary, the conditional predictions generally need to be averaged over the target population’s covariate distribution. Because the inverse link is generally nonlinear, evaluating the model at average covariates is not usually the same as averaging outcome-scale predictions.
Worked Example: Converting a Logistic Prediction to Response Probability
Suppose a hypothetical binary efficacy outcome is modeled as \(\operatorname{logit}\{p(d)\}=-2+0.8\ln(d/10)\), where \(p(d)\) is the probability of response at dose \(d\), with dose measured in mg. Assume the model form is appropriate, the dose is within its supported range, and the prediction is for the population represented by the model. Find the predicted probability at 20 mg.
The linear predictor at 20 mg is \(\eta(20)=-2+0.8\ln(20/10)=-2+0.8(0.6931)=-1.4455\). The logit model specifies \(\operatorname{logit}(p)=\ln\{p/(1-p)\}\); its inverse is the logistic function:
The model predicts a response probability of about 0.191, or 19.1%, at 20 mg. The value \(-1.4455\) is a log-odds prediction, not a probability; reporting it as “negative 1.45% response” would be incorrect. This calculation alone does not establish that 20 mg is preferable to another dose.
For a binary outcome, probability and risk are often used for the predicted proportion experiencing the outcome. When risk is reported, specify the event and, for a time-dependent outcome, the time period. A probability of response by the end of treatment and a hazard ratio during treatment are different quantities even if they come from related models.
Transforming Log-Scale Predictions Carefully
A log-scale model can make positive, right-skewed outcomes easier to model, but exponentiating its predicted log mean does not generally produce the arithmetic mean on the original scale. If \(\ln(Y)\mid d,x\sim N\{\mu(d,x),\sigma^2\}\), then \(Y\) has a lognormal distribution. Its median is \(\exp\{\mu(d,x)\}\), while its arithmetic mean is \(\exp\{\mu(d,x)+\sigma^2/2\}\). Thus, the appropriate back-transformation depends on the quantity the decision requires.
The mean formula follows directly from the normal moment-generating result. If \(Z=\ln(Y)\sim N(\mu,\sigma^2)\), then \(Y=\exp(Z)\), and \(E(Y)=E\{\exp(Z)\}=\exp(\mu+\sigma^2/2)\). The variance term matters: since \(\sigma^2\) is nonnegative, the arithmetic mean is at least as large as the median, and is larger whenever \(\sigma^2>0\).
Worked Example: Back-Transforming a Lognormal Outcome
In a hypothetical dose group, a continuous biomarker \(Y\), measured in mg/L, is modeled as lognormal with \(\ln(Y)\mid d\sim N\{\mu(d),\sigma^2\}\). At the dose of interest, suppose \(\mu(d)=\ln(10)\) and \(\sigma=0.40\). Assume the lognormal model is adequate and that the intended summary is the arithmetic mean, not the median.
The predicted median is \(\exp\{\mu(d)\}=\exp\{\ln(10)\}=10.00\) mg/L. For the arithmetic mean, first calculate the log-scale variance: \(\sigma^2=0.40^2=0.16\). Then
The model therefore predicts an arithmetic mean of 10.83 mg/L and a median of 10.00 mg/L. Reporting 10.00 mg/L as the mean would omit the distributional adjustment and answer a different question. The calculation also relies on the assumed lognormal distribution; a log-scale regression alone does not guarantee that its residual distribution is normal with constant variance.
Absolute Risk from Time-to-Event Predictions
In a proportional-hazards model, a hazard ratio describes the relative instantaneous event rate among individuals still event-free. It is not an absolute probability. To report an absolute event risk by a fixed time, a baseline cumulative hazard and an appropriate survival-model prediction are needed. In the absence of competing risks, survival through time \(t\) is \(S(t\mid d,x)=\exp\{-H(t\mid d,x)\}\), so the probability of the event by \(t\) is \(1-S(t\mid d,x)\).
Worked Example: Converting a Hazard Ratio to Risk by a Fixed Time
Consider a hypothetical time-to-event model with proportional hazards and no competing risks. For a specified reference covariate profile, suppose the baseline cumulative hazard by 12 weeks is \(H_0(12)=0.10\). At a dose of interest, the model predicts a hazard ratio of 1.8 relative to that profile’s reference dose. Assume the proportional-hazards model and baseline cumulative hazard apply through 12 weeks.
Under proportional hazards, the cumulative hazard at the dose is \(H(12)=1.8(0.10)=0.18\). The predicted event risk by 12 weeks is
The predicted risk is 16.5% by 12 weeks for the specified profile, under the stated assumptions. The hazard ratio of 1.8 is not a risk of 180%, nor does it mean the 12-week risk is 1.8 times a baseline risk in every setting. A risk ratio would require the corresponding absolute risks. For \(n=100\) otherwise comparable individuals, the model-based expected number of events is \(100(0.1647)=16.47\); this is an expectation, not a claim that a fraction of an event will occur.
Conditional Predictions and Population Summaries
A model may give a prediction at a covariate profile \(x\), but a dose decision may concern a target population with varying covariates. A standardized population prediction averages outcome-scale predictions over that population. For strata \(x_1,\ldots,x_K\) with target-population weights \(w_1,\ldots,w_K\), where \(w_k\geq0\) and \(\sum_k w_k=1\), the standardized outcome is \(\sum_k w_k m(d,x_k)\). The weights should represent the intended population, not automatically the sample composition if that composition differs from the target.
Worked Example: Standardizing Response Probability Across Covariate Groups
Suppose a hypothetical logistic model predicts \(\operatorname{logit}\{p(d,x)\}=-2+0.5\ln(d/10)+2x\), where \(x=0\) or \(1\) denotes two covariate groups. At 10 mg, the target population is 75% group 0 and 25% group 1. Assume these group-specific predictions are calibrated for the target population and that the given weights are appropriate. Calculate the standardized response probability.
At 10 mg, \(\ln(10/10)=0\). For group 0, \(\eta=-2\), so \(p(10,0)=\exp(-2)/\{1+\exp(-2)\}=0.1192\). For group 1, \(\eta=-2+2=0\), so \(p(10,1)=\exp(0)/\{1+\exp(0)\}=0.5000\). The standardized probability is the weighted average of these probabilities:
The standardized prediction is 21.4% for the specified target population. It is not obtained by inserting the mean covariate \(x=0.25\) into the model: that gives \(\eta=-1.5\) and \(p=0.1824\), a different result. The difference arises because the inverse-logit transformation is nonlinear. The target population and its weights are part of the interpretation.
Uncertainty, Diagnostics, and Common Pitfalls
A predicted value is not known without uncertainty. Parameter uncertainty, residual variability, and—when relevant—variation in covariates can affect a dose-specific quantity. Uncertainty intervals should be calculated for the quantity being reported, respecting the model and its prediction target. For a nonlinear transformation, simply transforming the endpoints of a symmetric interval on the model scale may not give an appropriate interval on the outcome scale. Simulation from the fitted parameter distribution, followed by transformation and summarization, is one general way to propagate parameter uncertainty. A predictive distribution for an individual outcome includes additional outcome variability and is not the same as an interval for a population mean or response probability.
Useful checks include confirming that probabilities lie between 0 and 1, risks refer to a stated horizon, predicted continuous outcomes have sensible units and ranges, and group weights sum to one. Examine whether the dose lies within the range supporting the data, whether covariate profiles are represented, and whether the model’s dose-response pattern is plausible. Calibration, residual, and model-fit diagnostics should be considered for the outcome and model being used. Implausible transformed predictions can signal extrapolation, a link-scale misunderstanding, or a model that does not describe the data adequately.
Common mistake: “The logistic model predicts 0.7, so the response probability is 70%,” when 0.7 is actually the linear predictor. The correction is to identify the model’s link and apply its inverse; for the logit link, the probability is \(\exp(0.7)/\{1+\exp(0.7)\}=0.6682\). Similarly, a hazard ratio is not an absolute risk, and exponentiating a log-scale mean may yield a median rather than an arithmetic mean. Always label the quantity, scale, units, population, and time horizon.
Alternatives depend on the intended summary. A conditional prediction describes a specified covariate profile; standardization summarizes predictions over a target covariate distribution. A mean on the original scale and a geometric mean answer different questions for skewed outcomes. A hazard ratio summarizes relative instantaneous rates, whereas an absolute risk over a horizon is often easier to interpret in event counts. None is universally preferable: select and report the quantity that corresponds to the estimand and decision context. These transformations and summaries describe outcomes; combining outcomes into an explicit preference or tradeoff requires an additional decision framework.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- A logistic model gives a linear predictor of \(-0.4\) at a dose. What response probability does it imply?
- A lognormal outcome has log-scale mean \(\ln(5)\) and log-scale standard deviation \(0.30\). Find its median and arithmetic mean, in the original outcome units.
- A model gives cumulative hazard 0.25 by 8 weeks, with no competing risks. What is the event risk by 8 weeks?
- Why can predictions for two covariate groups not generally be averaged by first averaging their covariate values and then applying a nonlinear inverse link?
Solutions. (1) Apply the inverse logit: \(\exp(-0.4)/\{1+\exp(-0.4)\}=0.4013\), or 40.1%. (2) The median is \(\exp\{\ln(5)\}=5.00\). The variance on the log scale is \(0.30^2=0.09\), so the arithmetic mean is \(\exp\{\ln(5)+0.09/2\}=5\exp(0.045)=5.230\) outcome units. (3) Risk is \(1-\exp(-0.25)=0.2212\), or 22.1% by 8 weeks. (4) A nonlinear inverse link does not generally preserve averages: in general, \(g^{-1}\{\sum_k w_k\eta_k\}\neq\sum_k w_k g^{-1}(\eta_k)\). For a population response probability, calculate group-specific probabilities first and then average them using the target-population weights.