Why the Slope Sign Matters
In dose escalation, candidate doses have a natural order. A dose-toxicity model is easier to interpret and use when it respects that order: if a patient receives a higher dose, the modeled probability of a dose-limiting toxicity (DLT) should not be lower. In the two-parameter Bayesian logistic regression model (BLRM), this ordering follows from the sign of the slope on log dose. The result is a mathematical property of the curve for fixed parameter values; it is not a claim that observed DLT proportions must rise at every dose.
The derivation depends on the model’s dose scale and on the ordinary properties of the logistic function. It also clarifies what the model does not guarantee: if the slope is allowed to be negative, the curve decreases, and if its sign is uncertain, the model does not impose increasing toxicity for every possible parameter value.
Model and Assumptions
Let \(d\) be a positive dose and let \(d_{\mathrm{ref}}\) be a fixed, positive reference dose on the same scale. For fixed model parameters \(\alpha>0\) and \(\beta\), the dose-only BLRM toxicity probability is
Here \(\operatorname{logit}^{-1}(z)=1/(1+\exp(-z))\). The model uses a linear predictor on the log-dose scale, with \(\beta\) as the coefficient multiplying log dose. The reference dose makes the dose ratio dimensionless; the logarithm therefore applies to a ratio, not to a dose with units. The probability is conditional on the model parameters and on the dose-only model being appropriate for the patients and DLT definition under consideration.
This conclusion concerns the modeled probability at a dose, not the DLT status of an individual or the observed proportion in a particular cohort. It assumes the doses being compared use the same DLT definition and follow-up window, and that the dose-only curve is intended to describe comparable patients. Patient characteristics or other predictors, if included in an expanded model, can change which probabilities are being compared.
Deriving the Monotonicity Result
Write the linear predictor as \(\eta(d)=\log(\alpha)+\beta\log(d/d_{\mathrm{ref}})\), so \(p(d)=\operatorname{logit}^{-1}\{\eta(d)\}\). The derivative of the inverse-logit function at \(\eta\) is \(p(1-p)\), and the derivative of \(\eta(d)\) with respect to dose is \(\beta/d\). Applying the chain rule gives the dose-scale derivative.
For every finite positive dose and finite parameter values, \(0<p(d\mid\alpha,\beta)<1\). Thus \(p(d)\{1-p(d)\}>0\), and \(d>0\). The derivative’s sign is therefore exactly the sign of \(\beta\). A positive slope implies a positive derivative at every positive dose, which establishes strict increase; a zero slope gives zero derivative throughout; and a negative slope gives a negative derivative throughout.
The same conclusion can be shown directly for any two ordered doses. If \(d_2>d_1>0\), then \(\log(d_2/d_1)>0\), and the difference between their linear predictors is
When \(\beta>0\), this difference is positive. Since the inverse-logit function is strictly increasing, \(p(d_2)>p(d_1)\). This pairwise argument makes explicit that the result applies to any two positive doses, not just adjacent levels in a dose grid.
The curve is not linear in dose. Although the derivative has the sign of \(\beta\), its size also depends on dose and on \(p(d)\{1-p(d)\}\). For fixed \(\beta\), the probability changes most rapidly with the linear predictor when the probability is near one-half; changes are more gradual when the probability is near zero or one. A positive slope guarantees ordering, not a constant increase in probability per unit of dose.
Worked Example: Checking the Dose Derivative
Suppose, hypothetically, a dose-only BLRM has \(\alpha=0.20\), \(\beta=1.4\), and \(d_{\mathrm{ref}}=100\) mg. Calculate the modeled probabilities at 50, 100, and 200 mg, and evaluate the derivative at 100 mg.
At 50 mg, the modeled odds are \(0.20(50/100)^{1.4}\). Since \(0.5^{1.4}\approx0.3789\), the odds are \(0.07579\), giving \(p(50)=0.07579/(1+0.07579)\approx0.07045\).
At 100 mg, the dose ratio is 1, so the odds are \(0.20\) and \(p(100)=0.20/(1+0.20)\approx0.1667\). At 200 mg, \(2^{1.4}\approx2.639\), so the odds are \(0.20(2.639)\approx0.5278\) and \(p(200)=0.5278/(1+0.5278)\approx0.3455\). The probabilities increase in dose order, as the positive slope predicts.
At 100 mg, the derivative is \(\beta p(1-p)/d=1.4(0.1667)(0.8333)/100\approx0.001944\) per mg. This is the local rate of change in modeled probability at that dose, not a constant change that applies across the entire dose range.
What the Ordering Does—and Does Not—Say
The ordering result is useful because it provides a coherent structural assumption for toxicity across candidate doses. When \(\beta>0\), the model cannot assign a lower toxicity probability to a higher dose for the same fixed parameter values. The result applies regardless of whether the candidate doses are equally spaced: it depends on their ordering, not on a constant dose increment.
However, a monotone probability curve does not require observed DLT fractions to be monotone. Cohort outcomes are random. A lower-dose cohort can happen to have a larger observed fraction than a higher-dose cohort, even when the underlying modeled probabilities increase. The model concerns the probabilities generating outcomes, not a rule that the observed data must satisfy.
Worked Example: Nonmonotone Observed Fractions
Consider a hypothetical escalation with six patients at each of 50, 100, and 200 mg. Suppose the observed DLT counts are 1, 3, and 2, respectively. The observed fractions are \(1/6\approx0.1667\), \(3/6=0.5000\), and \(2/6\approx0.3333\). They rise and then fall.
Use the parameters from the preceding example, which give modeled probabilities of approximately 0.07045, 0.1667, and 0.3455 at these doses. These probabilities are strictly increasing, but the observed fractions are not. There is no contradiction: the counts are realizations of random outcomes, and the model does not require their fractions to follow the probability curve exactly. A nonmonotone sequence of observed fractions is a reason to examine the data and model fit, not by itself a mathematical refutation of monotonicity in the modeled probabilities.
Worked Example: A Negative Slope Reverses the Ordering
For contrast, consider a hypothetical curve with \(\alpha=0.25\), \(\beta=-0.8\), and \(d_{\mathrm{ref}}=100\) mg. At 50 mg the odds are \(0.25(0.5)^{-0.8}\approx0.4353\), so \(p(50)\approx0.4353/1.4353=0.3033\). At 200 mg the odds are \(0.25(2)^{-0.8}\approx0.1436\), so \(p(200)\approx0.1436/1.1436=0.1255\).
The higher dose has the lower modeled toxicity probability. This is precisely the decreasing ordering implied by \(\beta<0\); the logistic form alone does not enforce an increasing dose-toxicity relationship. Whether parameter values with negative slope are possible depends on the parameter space or prior specification used for the model.
Assumptions, Diagnostics, and Alternatives
The derivative result itself is exact under the stated model, but using the result in a clinical trial requires attention to the modeling assumptions. The doses must be positive for the log-dose term and derivative to be defined. The curve also treats dose as the driver of toxicity in the stated model; if risk differs systematically across cohorts for reasons not represented in that model, the fitted dose curve may not capture those differences. Monotonicity is an assumption about the model’s dose-response structure, not proof of a causal effect of increasing dose.
A practical check is to calculate or plot the modeled probability at every candidate dose for representative parameter values and verify the ordering. For a positive slope, numerical values should increase with dose; a computed decrease can indicate a coding error, an inconsistent dose scale, or a formula that differs from the stated model. A zero dose cannot be inserted into the log-dose formula. Nor should doses expressed on different scales be compared without first expressing them consistently.
A key distinction in a Bayesian analysis is between the slope of one parameter draw and uncertainty about the slope. Each draw with \(\beta>0\) produces an increasing curve; a draw with \(\beta<0\) produces a decreasing one. If the model allows both signs, the model does not enforce an increasing curve for every draw. If increasing toxicity is a scientific requirement, that requirement must be reflected in the parameter space or prior specification. This tutorial’s derivative result does not decide whether such a constraint is scientifically justified.
The BLRM offers a smooth curve with an explicit dose ordering when the slope is positive. In contrast, assigning a separate probability to each dose without an ordering constraint does not itself guarantee monotonicity. An isotonic approach can impose an ordered sequence of probabilities without using this particular logistic curve, while a more flexible dose-response model may allow shapes that the two-parameter curve cannot represent. These alternatives trade off structural simplicity against flexibility; none removes the need to choose assumptions suited to the trial.
The central result is narrow but important: for the stated log-dose logistic curve, positivity of the slope is both necessary and sufficient for strict increase across positive doses. The conclusion follows from the derivative or, equivalently, from the ordering of the linear predictors. It does not assert that risk increases by a fixed amount, that every cohort will show more DLTs at higher doses, or that the model is adequate for every trial.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For the stated BLRM, what is the sign of the derivative with respect to dose when \(\beta=0\), and what does that imply for the curve?
- Why does \(d_2>d_1>0\) imply \(p(d_2)>p(d_1)\) when \(\beta>0\)?
- Can observed DLT fractions of 2/8 at a lower dose and 1/8 at a higher dose occur under a positive-slope curve? Explain.
- What happens to the modeled ordering if the slope is negative?
Solutions. (1) The derivative is \(p(d)\{1-p(d)\}\beta/d=0\), so the probability is constant across positive doses. (2) The linear-predictor difference is \(\beta\log(d_2/d_1)>0\); the inverse-logit function is strictly increasing, so the higher dose has the higher probability. (3) Yes. The observed fractions are random outcomes and need not be ordered like the underlying probabilities. (4) The derivative is negative at every positive dose, so the modeled probability decreases as dose increases.