Dose-Response Models For Efficacy

Why model efficacy across dose?

When efficacy is measured at several dose levels, the observed response proportions can suggest whether benefit rises with dose, levels off, or declines at higher doses. A dose-response model summarizes that pattern and can estimate efficacy at doses with limited data. Its shape is an assumption about how the outcome changes with dose, not a fact guaranteed by the data. In dose escalation, that distinction matters: a predicted efficacy increase is evidence about benefit, but it does not establish that a dose is safe to give.

Here the outcome is binary: each participant either meets a prespecified efficacy definition at a fixed assessment time or does not. The models below describe the probability of that efficacy outcome as a function of dose. They are not models for the joint distribution of efficacy and toxicity.

Definition: Let \(d\) denote a participant’s assigned dose and \(Y\) their binary efficacy outcome, where \(Y=1\) means the prespecified efficacy criterion is met. The dose-response function is \(p(d)=P(Y=1\mid d)\). A monotone efficacy curve is nondecreasing over the dose range under study. A plateau curve rises and approaches an upper limit. A nonmonotone curve rises and then falls, or otherwise changes direction, over that range.

Data, assumptions, and fitting

Suppose dose levels are \(d_1,\ldots,d_J\). At dose \(d_j\), let \(n_j\) participants have evaluable outcomes and let \(r_j\) meet the efficacy criterion. A basic grouped-binomial model assumes that, conditional on assigned dose, participants’ outcomes are independent Bernoulli variables with the same probability \(p(d_j)\). Thus \(r_j\) has a binomial distribution with size \(n_j\) and probability \(p(d_j)\). The assessment definition and time window should be consistent across dose groups, and missing outcomes should not create an unaddressed selection problem.

These assumptions are substantive. Independence may fail with clustering or repeated observations treated as if they were separate participants. A changing assessment window can make probabilities at different doses incomparable. If efficacy depends on baseline characteristics that differ across dose groups, an unadjusted dose curve can mix dose effects with differences in participant composition. Randomization, a suitable adjustment model, or both may be needed for the intended interpretation.

For a proposed curve \(p(d;\theta)\), where \(\theta\) denotes its parameters, the grouped-binomial likelihood is

L(\theta)=\prod_{j=1}^{J}{n_j\choose r_j}\,p(d_j;\theta)^{r_j}\{1-p(d_j;\theta)\}^{n_j-r_j}.

A maximum-likelihood fit chooses parameter values that maximize this likelihood, subject to any constraints defining the model. The binomial coefficients do not depend on \(\theta\), so maximizing the likelihood is equivalent to maximizing the log-likelihood after omitting those constants. This fitting procedure does not make an unsuitable curve suitable: model shape and assumptions still require assessment.

Monotone logistic curves

A simple monotone model uses a linear dose effect on the log-odds scale:

\operatorname{logit}\{p(d)\}=\log\!\left(\frac{p(d)}{1-p(d)}\right)=\beta_0+\beta_1 x, \qquad x=\frac{d-d_{\mathrm{ref}}}{s_d}.

Here \(d_{\mathrm{ref}}\) is a reference dose and \(s_d\) is a prespecified positive dose scale, so \(x\) is a scaled dose. The logistic transformation maps any real linear predictor to a probability strictly between 0 and 1. If \(\beta_1\geq 0\), the curve is nondecreasing in dose; if \(\beta_1>0\), it is strictly increasing. The model assumes a constant increase in log-odds per unit of scaled dose, which need not mean a constant increase in probability.

Worked Example: Interpreting a Monotone Logistic Curve

Consider a hypothetical trial with binary efficacy assessed at the same time for all participants. Doses are coded \(x=0,1,2,3\), with independent outcomes within and across dose groups. Suppose the illustrative curve has \(\beta_0=-2.00\) and \(\beta_1=0.800\). The positive slope implies a monotone increase in efficacy probability across these dose levels.

At \(x=2\), the linear predictor is \(-2.00+0.800(2)= -0.400\). Therefore,

p(2)=\frac{\exp(-0.400)}{1+\exp(-0.400)} =\frac{0.6703}{1+0.6703} \approx 0.4013.

The model assigns an efficacy probability of about 40.1% at \(x=2\). At \(x=3\), the predictor is \(-2.00+0.800(3)=0.400\), giving \(p(3)=\exp(0.400)/\{1+\exp(0.400)\}\approx 0.5987\). The odds at \(x=3\) are \(\exp(0.800)\approx 2.23\) times the odds at \(x=2\), not 2.23 times the probability.

For an illustrative check, suppose the observed counts were 12 of 100, 23 of 100, 40 of 100, and 60 of 100 across the four levels. The model-implied expected counts are approximately 11.9, 23.1, 40.1, and 59.9. These close values make the curve plausible as a summary of these hypothetical rates, but do not by themselves establish goodness of fit or prove that the displayed coefficients are maximum-likelihood estimates.

Plateau-shaped efficacy

A common plateau model is the Emax curve:

p(d)=p_0+E_{\max}\frac{d}{ED_{50}+d}, \qquad d\geq 0.

Here \(p_0\) is the probability at zero dose, \(E_{\max}\) is the increase from \(p_0\) toward the asymptotic plateau, and \(ED_{50}>0\) is the dose at which half of that increase is reached. Probability constraints require \(0\leq p_0\) and \(p_0+E_{\max}\leq 1\), with \(E_{\max}\geq 0\). The curve is nondecreasing because its derivative is \(E_{\max}ED_{50}/(ED_{50}+d)^2\), which is nonnegative. As \(d\) grows without bound, \(p(d)\) approaches \(p_0+E_{\max}\); it does not exceed that limit.

Worked Example: Calculating an Emax Plateau

Suppose hypothetical independent binary outcomes are assessed at four nonnegative doses using the same fixed efficacy definition. Consider the illustrative parameters \(p_0=0.100\), \(E_{\max}=0.600\), and \(ED_{50}=1.50\), with doses \(d=0,1,3,6\). The probability constraints hold because \(0.100+0.600=0.700\leq 1\).

At \(d=3\),

p(3)=0.100+0.600\frac{3}{1.50+3} =0.100+0.600\left(\frac{3}{4.50}\right) =0.100+0.400=0.500.

The full predictions are \(p(0)=0.100\), \(p(1)=0.100+0.600(1/2.50)=0.340\), \(p(3)=0.500\), and \(p(6)=0.100+0.600(6/7.50)=0.580\). At very high doses, the curve approaches 0.700. At \(d=1.50\), the dose equals \(ED_{50}\), so the increase above baseline is \(0.600/2=0.300\), giving \(p(1.50)=0.400\).

For concrete hypothetical data, suppose the corresponding counts are 10/100, 34/100, 50/100, and 58/100. Each observed proportion equals the model prediction at its dose. That agreement illustrates the calculations; it does not show that real dose-response data must follow an Emax curve or that these parameters are known without uncertainty.

Nonmonotone curves

A quadratic logistic model allows efficacy probability to rise and then fall:

\operatorname{logit}\{p(x)\}=\beta_0+\beta_1x+\beta_2x^2.

The derivative of the log-odds with respect to \(x\) is \(\beta_1+2\beta_2x\). Since the logistic function has a positive derivative, the probability changes direction at the same point as the log-odds. If \(\beta_2<0\), the turning point is \(x^*=-\beta_1/(2\beta_2)\); it represents a maximum when it lies within the dose range. A turning point outside the studied range does not establish a rise-then-fall pattern among the studied doses.

Worked Example: Finding a Nonmonotone Peak

Consider a hypothetical study with independent binary outcomes and a common efficacy assessment window at scaled doses \(x=0,1,2,3,4\). Use the illustrative quadratic-logistic parameters \(\beta_0=-1.50\), \(\beta_1=1.40\), and \(\beta_2=-0.350\). The negative quadratic coefficient allows a downturn; the fitted curve’s turning point must still be checked against the studied range.

The turning point is \(x^*=-1.40/\{2(-0.350)\}=2.00\), which is within the range. At \(x=2\), the predictor is \(-1.50+1.40(2)-0.350(2^2)=-0.100\), so \(p(2)=\exp(-0.100)/\{1+\exp(-0.100)\}\approx 0.4750\).

At \(x=3\), the predictor is \(-1.50+1.40(3)-0.350(3^2)=-0.450\), giving \(p(3)\approx 0.3894\). At \(x=4\), it is \(-1.50+1.40(4)-0.350(4^2)=-1.50\), giving \(p(4)\approx 0.1824\). The corresponding probabilities at \(x=0\) and \(x=1\) are also approximately 0.1824 and 0.3894. Thus this curve rises toward \(x=2\) and falls afterward.

For a numerical check, suppose hypothetical counts at \(x=0,1,2,3,4\) are 36, 78, 95, 78, and 36 among 200 participants per dose. The predicted counts are approximately 36.5, 77.9, 95.0, 77.9, and 36.5. The observed rates broadly align with the illustrative curve. The symmetry here follows from the chosen parameters and dose levels; it is not a general property of nonmonotone dose-response curves.

Justification, diagnostics, and alternatives

The Emax curve’s plateau interpretation follows directly from its limiting behavior: as \(d\) increases, \(d/(ED_{50}+d)\) approaches 1, so \(p(d)\) approaches \(p_0+E_{\max}\). Its slope is nonnegative under the stated parameter constraints. For the quadratic-logistic model, differentiating the probability gives

\frac{dp(x)}{dx}=p(x)\{1-p(x)\}\{\beta_1+2\beta_2x\}.

Because \(0<p(x)<1\), the first two factors are positive. Therefore the sign of the probability’s slope is determined by \(\beta_1+2\beta_2x\), which changes sign at \(x^*\) when that point exists. This establishes why the calculated turning point identifies the curve’s peak when \(\beta_2<0\).

Diagnostics should examine observed proportions alongside fitted probabilities, preferably with uncertainty intervals, and inspect whether residual discrepancies show a systematic pattern by dose. Sparse data, very small or very large response proportions, and few dose levels can make parameters unstable or estimates extreme. A quadratic curve can turn sharply between observed doses, so its apparent peak may be weakly supported even when the numerical fit is good. A plateau is hard to identify if no doses approach the plateau; \(E_{\max}\) and \(ED_{50}\) can then be imprecise or strongly dependent on each other. Check convergence, parameter constraints, uncertainty, and sensitivity to dose scaling and model form.

The main alternatives impose different shapes. A monotone logistic model is parsimonious and appropriate when a continuing increase is credible, but it has no plateau and imposes linearity on the log-odds scale. Emax explicitly represents diminishing gains toward an asymptote, but assumes that particular saturating shape. A quadratic-logistic model is flexible enough for a rise and fall, but can overfit sparse data and may imply implausible behavior outside the studied range. A model with one shape should not be selected solely because its curve looks clinically attractive; compare fit, uncertainty, scientific plausibility, and the consequences of misspecification.

If efficacy informs escalation, state the implication conditionally: under the chosen model and its assumptions, the estimated probability of efficacy changes in a particular way over the studied doses. A small apparent increase with wide uncertainty may not support a meaningful distinction between doses. Nor does an efficacy curve by itself establish that escalation is safe, that a dose is acceptable, or that a joint benefit-harm criterion is met. Those decisions require the separate safety and clinical rules of the design.

Common mistake and practical interpretation

Common pitfall: “The fitted efficacy curve increases with dose, so the trial should escalate.” This reasoning treats an assumed or estimated efficacy trend as a safety conclusion. The correction is to report what the efficacy model supports, with its uncertainty and dose range, and apply the design’s safety requirements separately before making an escalation decision.

Another frequent error is to call a curve “plateauing” merely because two adjacent observed proportions are similar. Sampling variability can produce a flat-looking pair even when the underlying curve is still increasing. Conversely, a noisy lower proportion at a high dose does not on its own establish a true downturn. Model-based shape claims depend on data, assumptions, and uncertainty together.

Check Your Understanding

Answer each question, then compare with the solutions.

For a monotone logistic model with \(\beta_1=0.50\), what happens to the odds when scaled dose increases by one unit?
An Emax model has \(p_0=0.15\) and \(E_{\max}=0.50\). What is its limiting efficacy probability as dose increases?
For a quadratic-logistic curve with \(\beta_1=1.20\) and \(\beta_2=-0.30\), find the turning point and state what additional check is needed to call it a peak in the studied range.
Why can an observed efficacy proportion that falls at the highest dose fail to prove a nonmonotone relationship?
Name one reason a good-looking efficacy trend alone cannot justify escalation.

Solutions. 1. The odds are multiplied by \(\exp(0.50)\approx 1.65\) per one-unit increase in scaled dose; this is not a 65-percentage-point increase in probability. 2. The limit is \(p_0+E_{\max}=0.15+0.50=0.65\), or 65%. 3. \(x^*=-1.20/\{2(-0.30)\}=2.00\). Check that \(x=2\) lies inside the studied dose range; with \(\beta_2<0\), it is then a maximum of the curve. 4. The observed proportion is subject to sampling variation, and sparse data may give substantial uncertainty about the underlying dose-response shape. 5. Efficacy models describe benefit, not safety; escalation must also meet the design’s safety requirements.