Logistic Dose-Toxicity Models for Ordered Doses

Why Model Toxicity Across Ordered Doses?

In an escalation trial, toxicity is recorded as a binary outcome at each administered dose. The earlier material on binary DLT likelihoods establishes how to describe such outcomes at a fixed dose. A dose-toxicity model adds a structure across doses: it uses their ordering to represent how the probability of toxicity changes as dose increases. This can be useful when observations at individual dose levels are sparse, but it also makes an important assumption about the shape of that change.

A logistic model expresses the log odds of toxicity as a linear function of a dose covariate. If that covariate increases with the ordered dose and its slope is nonnegative, the model gives nondecreasing toxicity probabilities across dose levels. Its simplicity is useful, but monotonicity alone does not guarantee that the model describes the observed pattern well.

Definition: Let $Y_{ij}=1$ if participant $i$ treated at dose level $j$ experiences a dose-limiting toxicity (DLT), and $Y_{ij}=0$ otherwise. Let $x_j$ be a numerical covariate assigned to dose level $j$, with $x_1<x_2<\cdots<x_J$. A one-covariate logistic dose-toxicity model is $$ P(Y_{ij}=1\mid x_j)=p_j,\qquad \operatorname{logit}(p_j)=\log\left(\frac{p_j}{1-p_j}\right)=\alpha+\beta x_j, $$ or, equivalently, $p_j=\operatorname{logit}^{-1}(\alpha+\beta x_j)=1/\{1+\exp[-(\alpha+\beta x_j)]\}$. The monotone version requires $\beta\geq 0$.

Assumptions and Parameter Interpretation

The model is defined for a binary DLT endpoint over a specified assessment window. Conditional on dose, each participant's outcome is treated as a Bernoulli observation with probability $p_j$; for the grouped-binomial likelihood, outcomes are conditionally independent within and across dose levels. Participants assigned the same dose share the same modeled toxicity probability. The model also assumes that the log odds—not the probability itself—change linearly with the chosen dose covariate.

The covariate must respect the dose ordering. It might be the actual dose in mg, a transformed dose, or a numerical rank assigned to dose levels. The choice has consequences. With rank coding $x_j=0,1,2,\ldots$, a constant slope represents the same change in log odds at every adjacent step. If the physical dose increments are unequal, that is not the same assumption as a constant change in log odds per mg.

The intercept $\alpha$ is the log odds at $x=0$, and $\operatorname{logit}^{-1}(\alpha)$ is the corresponding toxicity probability. If zero is outside the observed dose range, this is an extrapolated reference value, not necessarily the risk at a dose used in the trial. The slope $\beta$ is the change in log odds for a one-unit increase in $x$. For an increase of $\Delta x$, the odds ratio is $\exp(\beta\Delta x)$. This is a multiplicative change in odds, not a constant increase in risk.

When $\beta>0$, higher values of $x$ produce higher toxicity probabilities. When $\beta=0$, the modeled probability is constant across doses. A negative slope would produce decreasing risks, so it is excluded when specifying a nondecreasing toxicity model. The model does not force risks to increase by equal amounts: the logistic curve can rise slowly near its lower and upper extremes and more steeply in between.

Worked Example: Reading Risk and Odds Changes Across Ranked Doses

Suppose three ordered dose levels are coded $x=0,1,2$, and a hypothetical model has $\alpha=-2$ and $\beta=0.8$. Calculate each toxicity probability and interpret the slope.

At $x=0$, the linear predictor is $-2+0.8(0)=-2$, so $p_1=1/\{1+\exp(2)\}\approx0.1192$. At $x=1$, it is $-1.2$, giving $p_2=1/\{1+\exp(1.2)\}\approx0.2315$. At $x=2$, it is $-0.4$, giving $p_3=1/\{1+\exp(0.4)\}\approx0.4013$.

Because $\beta=0.8>0$, the modeled risks increase across the ordered levels. A one-step increase multiplies the odds by $\exp(0.8)\approx2.226$. Using the unrounded model values, the odds at $x=1$ are approximately $0.3012$, and the odds at $x=2$ are approximately $0.6703$; multiplying the unrounded odds at $x=1$ by $\exp(0.8)$ gives the odds at $x=2$, corresponding to risk $0.4013$. The risk increases are not equal: approximately 0.1123 from $x=0$ to $x=1$, and 0.1698 from $x=1$ to $x=2$.

Likelihood for Escalation Data

Let $n_j$ be the number treated at dose level $j$, and $y_j$ the number with DLT. Under the assumptions above, $Y_j=\sum_iY_{ij}$ has a binomial distribution with size $n_j$ and probability $p_j$. The grouped likelihood is the product of the dose-level binomial probabilities. Its log form is useful for estimating model parameters or checking a proposed parameter pair.

L(\alpha,\beta)=\prod_{j=1}^{J}{n_j\choose y_j}p_j^{y_j}(1-p_j)^{n_j-y_j}, \qquad \ell(\alpha,\beta)=\sum_{j=1}^{J}\left[ \log{n_j\choose y_j}+y_j\log(p_j)+(n_j-y_j)\log(1-p_j) \right].

The binomial coefficient is part of the likelihood for the grouped count. It does not depend on $\alpha$ or $\beta$, so it can be omitted when finding parameter values that maximize the likelihood; it must be included if reporting the probability of the observed grouped counts. Substituting $p_j=\operatorname{logit}^{-1}(\alpha+\beta x_j)$ gives another useful expression:

\ell(\alpha,\beta)= \sum_{j=1}^{J}\left[ \log{n_j\choose y_j} +y_j(\alpha+\beta x_j) -n_j\log\{1+\exp(\alpha+\beta x_j)\} \right].

This expression follows because $\log(p_j)=\eta_j-\log(1+\exp(\eta_j))$ and $\log(1-p_j)=-\log(1+\exp(\eta_j))$, where $\eta_j=\alpha+\beta x_j$. Combining the $y_j$ toxicity terms and $n_j-y_j$ non-toxicity terms leaves $y_j\eta_j-n_j\log(1+\exp(\eta_j))$, plus the binomial coefficient. This also shows why each likelihood contribution depends on both the number of toxicities and the number without toxicity.

Worked Example: Evaluating the Likelihood for Hypothetical Cohorts

Suppose six participants are treated at each of three ranked dose levels $x=0,1,2$. Hypothetical DLT counts are $y=(1,2,4)$. Evaluate the grouped likelihood at $\alpha=-2,\ \beta=0.8$, using the probabilities from the previous example.

The likelihood is $L={6\choose1}p_0(1-p_0)^5{6\choose2}p_1^2(1-p_1)^4{6\choose4}p_2^4(1-p_2)^2$, where $p_0\approx0.1192,\ p_1\approx0.2315,\ p_2\approx0.4013$. Substituting the unrounded logistic probabilities gives

L= 6p_0(1-p_0)^5 \times15p_1^2(1-p_1)^4 \times15p_2^4(1-p_2)^2 \approx 0.01482.

For the log likelihood, the binomial-coefficient terms sum to $\log(6)+\log(15)+\log(15)\approx7.2079$. The remaining contributions are $\log(p_0)+5\log(1-p_0)\approx-2.7616$, $2\log(p_1)+4\log(1-p_1)\approx-3.9797$, and $4\log(p_2)+2\log(1-p_2)\approx-4.6781$. Therefore, $\ell\approx7.2079-2.7616-3.9797-4.6781=-4.2115$, consistent with $\exp(-4.2115)\approx0.01482$. This is the likelihood of these grouped counts under the specified parameter values, not a probability that the parameters are correct.

Why the Monotonicity Constraint Works

The model's monotonicity can be verified directly rather than inferred from a plotted curve. Differentiating the logistic probability with respect to the dose covariate gives a derivative whose sign is the sign of $\beta$.

p(x)=\frac{1}{1+\exp[-(\alpha+\beta x)]}, \qquad \frac{dp(x)}{dx} =\beta\,p(x)\{1-p(x)\}.

For finite $\alpha+\beta x$, $0<p(x)<1$, and therefore $p(x)\{1-p(x)\}>0$. Thus $dp(x)/dx\geq0$ exactly when $\beta\geq0$. For a finite increase $\Delta x>0$, the corresponding odds ratio is $\exp(\beta\Delta x)$, which is at least one under the same condition. A positive slope gives strictly increasing risks; a zero slope gives equal risks at all modeled doses.

Worked Example: Unequal Physical Dose Increments

Consider hypothetical doses of 5, 10, and 20 mg, modeled with $x$ equal to dose in mg. Let $\alpha=-2.4$ and $\beta=0.04$ per mg. Calculate the risks and compare the odds across the dose range.

At 5 mg, the predictor is $-2.4+0.04(5)=-2.2$, so the risk is $\operatorname{logit}^{-1}(-2.2)\approx0.09975$. At 10 mg, the predictor is $-2.0$, giving risk approximately $0.1192$. At 20 mg, it is $-1.6$, giving risk approximately $0.1680$. The positive per-mg slope yields increasing risk.

The odds ratio for 5 to 10 mg is $\exp\{0.04(5)\}=\exp(0.2)\approx1.221$. For 10 to 20 mg it is $\exp\{0.04(10)\}=\exp(0.4)\approx1.492$. The odds ratio for 5 to 20 mg is $\exp\{0.04(15)\}=\exp(0.6)\approx1.822$. Unequal dose increments therefore imply unequal adjacent odds ratios, even though the log-odds change is constant per mg.

Model Choice, Diagnostics, and Limitations

A linear dose covariate is compact and borrows information across dose levels, but its shape assumption should be examined. A categorical-dose logistic model instead assigns a separate toxicity probability to each dose, without imposing a linear relationship across levels. It is more flexible but uses more parameters, which can be difficult to estimate with few participants per level. An isotonic approach imposes an ordering on the probabilities without requiring a straight line in log odds; it is another option when monotonicity is credible but the logistic shape is not. These approaches answer related but not identical modeling questions.

Useful checks include comparing observed DLT proportions $y_j/n_j$ with fitted probabilities $p_j$, plotting both against ordered dose, and checking whether residual discrepancies are concentrated at particular dose levels. Observed proportions can fluctuate out of order by chance, especially in small cohorts; a local reversal alone does not establish that the underlying risk is nonmonotone. Conversely, a fitted increasing curve can hide systematic mismatch, such as a sharp increase between two levels or a plateau that the linear log-odds form does not capture.

Sparse data can also make parameter estimates unstable. If all observed outcomes are non-toxic or all are toxic, or if outcomes are strongly separated by dose, an unconstrained logistic estimate may diverge rather than yield finite parameter estimates. A monotonicity constraint does not automatically solve sparse-data or separation problems. Estimates may also land at the boundary $\beta=0$, especially when the data provide little evidence for a trend. Such a result means the constrained likelihood is maximized at a flat dose relationship; it is not proof that true toxicity risks are identical.

The model describes a population-average risk pattern under its assumptions. It does not account by itself for participant-level predictors, dependence between outcomes, incomplete toxicity follow-up, or changes in DLT definition. Those features require additional modeling choices. Nor does an increasing fitted risk establish that dose causes toxicity in every setting: interpretation depends on trial conduct and the adequacy of the model. In particular, monotonicity is a substantive assumption to examine, not a feature guaranteed by the word “logistic.”

Common Pitfall

Common mistake: Treating $\beta$ as the increase in toxicity probability for each dose step. In this model, $\beta$ is the increase in log odds per one covariate unit; the odds are multiplied by $\exp(\beta)$. The probability difference depends on the starting risk and can vary across the dose range.

A related calculation error is substituting the linear predictor $\alpha+\beta x_j$ for $\log(p_j)$ in the binomial likelihood. The linear predictor is the log odds, not the log probability. For example, when the predictor is $-2$, $\log(p)\approx-2.1269$, not $-2$. Use $y_j\log(p_j)+(n_j-y_j)\log(1-p_j)$, or the equivalent simplified log-likelihood expression, to keep toxicity and non-toxicity contributions correct.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For $x=0,1,2$, what happens to the modeled toxicity risks if $\alpha=-1.5$ and $\beta=0$?
If $\beta=0.3$ per rank step, what is the odds ratio for a two-step increase?
Why can a model with positive $\beta$ still have unequal changes in risk across adjacent dose levels?
For hypothetical data $n=5,\ y=0$ at one dose, which log-likelihood contribution contains the toxicity term, and what is that term?
Name one advantage and one limitation of treating dose as categorical rather than as a single linear covariate.

Solutions. (1) The modeled risks are equal at all three levels because $\alpha+\beta x=-1.5$ for every $x$; each risk is $\operatorname{logit}^{-1}(-1.5)\approx0.1824$. (2) The odds ratio is $\exp(0.3\times2)=\exp(0.6)\approx1.822$. (3) The slope makes log odds change linearly, but the logistic transformation is nonlinear, so probability differences depend on the starting risk. (4) The contribution is $y\log(p)+(n-y)\log(1-p)$; with $y=0,n=5$, it is $5\log(1-p)$, and there is no toxicity term because no DLT was observed. (5) A categorical model can represent irregular risk patterns without a linear log-odds assumption, but it requires more parameters and may be poorly determined with sparse observations.