Why Update Toxicity Probabilities Directly?
A CRM dose-response model represents toxicity probabilities through a low-dimensional parameter and a prespecified skeleton. An empirical approach instead estimates a probability at each dose from the observed outcomes, then imposes the clinical ordering that toxicity risk should not decrease as dose increases. This can be useful when the focus is on the observed dose levels rather than on fitting a particular curve. It also makes the role of the monotonicity constraint visible: the data provide dose-specific proportions, while the constraint may require adjacent estimates to be pooled.
The direct approach is not assumption-free. It requires comparable, adequately observed toxicity outcomes at each dose and an ordering restriction. It also cannot infer a dose-specific probability from a dose with no observations without adding further assumptions. Unlike a fixed-skeleton CRM, it does not use a prespecified parametric curve to borrow information across the entire dose grid.
Likelihood and the Monotonicity Constraint
For the direct binomial likelihood, assume that each participant has a binary toxicity outcome, participants’ outcomes are independent conditional on their assigned doses, and participants assigned dose \(j\) share a common toxicity probability \(q_j\). The toxicity assessment window and outcome definition must be consistent across participants. The observed dose assignments and outcomes then contribute the likelihood
This is a likelihood conditional on the dose assignments. In an adaptive trial, the assignment process is based on accrued information; the dose-level likelihood still uses the outcomes at the doses actually assigned. Its interpretation depends on the stated outcome and independence assumptions, and adaptive allocation does not create observations at doses that were rarely or never used.
If \(n_j>0\), the unconstrained estimate at dose \(j\) is the observed proportion \(y_j/n_j\). Those proportions need not be nondecreasing. A lower observed rate at a higher dose may arise from sampling variability, especially with small \(n_j\). The constrained estimate resolves such order violations by pooling adjacent doses and assigning the pooled block a common estimate. The pooling is weighted by the number of participants at each dose.
Why Pooling Gives the Maximum-Likelihood Estimate
Consider a consecutive block \(B\) of dose levels that are assigned a common probability \(q\). Write \(Y_B=\sum_{j\in B}y_j\) for its total toxicities and \(N_B=\sum_{j\in B}n_j\) for its total participants. Ignoring constants that do not depend on \(q\), the block log-likelihood is
When \(0<Y_B<N_B\), the maximum is interior. Differentiating and setting the score to zero gives
The derivative argument applies only when the block contains both outcomes. If \(Y_B=N_B\), then \(\ell_B(q)=N_B\log(q)\), which increases on \(0<q<1\), so its maximum over \(0\leq q\leq1\) is at \(q=1\). If \(Y_B=0\), then \(\ell_B(q)=N_B\log(1-q)\), which decreases on \(0<q<1\), so its maximum is at \(q=0\). Thus the block maximum-likelihood estimate in all cases is the observed proportion \(Y_B/N_B\), including the boundary values 0 and 1.
To enforce the order restriction, begin with one block per observed dose, each fitted at its sample proportion. If two adjacent blocks have estimates \(a>b\), assigning those decreasing estimates violates the nondecreasing constraint. The constrained maximum for those two blocks is obtained by setting their probabilities equal: the combined block likelihood is maximized at the combined observed proportion, with the sample sizes acting as weights. Merge the blocks and repeat until all adjacent block estimates are nondecreasing. The binomial log-likelihood is concave, and this pooling procedure is the weighted isotonic maximum-likelihood solution; the final ordered block estimates satisfy the order constraint and maximize the likelihood under it.
For each dose with \(n_j>0\), calculate \(y_j/n_j\) and retain its weight \(n_j\).
If a higher-dose block has a smaller estimate than the block immediately below it, those adjacent blocks violate the required ordering.
Replace the violating blocks by one block with estimate equal to total toxicities divided by total participants in the merged block.
Continue checking adjacent blocks until estimates are nondecreasing. Each member of a pooled block receives the same fitted probability.
Worked Examples
Worked Example: Pooling a Decreasing Pair
Suppose four dose levels have the following hypothetical outcomes: \(n=(3,4,5,4)\) and \(y=(0,2,1,3)\). The observed proportions are \(0/3=0\), \(2/4=0.5000\), \(1/5=0.2000\), and \(3/4=0.7500\). The second and third estimates decrease with dose, so they cannot both remain as fitted under the monotonicity constraint.
Pool those two doses. Their combined number of toxicities is \(2+1=3\), and their combined sample size is \(4+5=9\). The pooled estimate is
The fitted estimates are therefore \((0,\ 0.3333,\ 0.3333,\ 0.7500)\), which are nondecreasing. The pooled value is not the unweighted average \((0.5000+0.2000)/2=0.3500\); dose 3 contributes five outcomes and dose 2 contributes four, so the correct pooled estimate weights each dose by its sample size.
Worked Example: Repeated Pooling with Unequal Counts
Consider four hypothetical doses with \(n=(5,4,6,3)\) and \(y=(1,3,2,2)\). The observed proportions are \(0.2000,\ 0.7500,\ 0.3333,\ 0.6667\). The second estimate exceeds the third, so pool doses 2 and 3:
The resulting sequence is \(0.2000,\ 0.5000,\ 0.5000,\ 0.6667\), already nondecreasing, so no further pooling is required. This example shows why a pooled estimate is determined by the combined event and participant counts, not by averaging the two proportions equally. The estimate \(0.5000\) describes the common fitted probability for these two dose levels under the order constraint; it does not claim that the underlying probabilities must truly be equal.
Worked Example: Boundary Proportions and a Pooled Interior Estimate
Suppose hypothetical outcomes at four ordered doses are \(n=(4,3,4,2)\) and \(y=(0,3,1,2)\). The raw proportions are \(0,\ 1,\ 0.2500,\ 1\). Dose 1 has no observed toxicities, and dose 4 has toxicity for both participants. Their unconstrained binomial estimates are the boundary values 0 and 1.
Doses 2 and 3 violate monotonicity because \(1>0.2500\). Pooling gives
The final estimates are \((0,\ 0.5714,\ 0.5714,\ 1)\). They satisfy the order restriction. The fact that the first and last estimates are 0 and 1 follows from the binomial likelihood for those observed outcomes, not from assuming in advance that either dose has exactly zero or certain toxicity risk. With such small samples, those boundary estimates also illustrate a limitation: the maximum-likelihood estimate alone can give an overly stark impression of what the data establish.
What the Estimates Do—and Do Not—Say
The fitted probabilities summarize the dose-level data under a monotonicity constraint. A plateau means that the likelihood-based order restriction has assigned the same estimate to consecutive doses; it does not show that the true risks are identical. Likewise, a boundary estimate can occur with few observations and should not be read as proof that a dose is harmless or inevitably toxic. The amount of information at each dose remains important, even when the fitted sequence looks smooth.
A dose with \(n_j=0\) has no dose-specific observed proportion: \(y_j/n_j\) is undefined, and its outcomes contribute no term to the likelihood. Its probability is not identified by direct empirical estimation alone. The monotonicity inequalities may bound it relative to neighboring probabilities, but generally do not determine a unique value. Do not enter a zero for an unobserved dose or apply PAVA as if it had a positive sample size. Any rule for assigning an estimate there would require an additional convention or modeling assumption.
Useful diagnostics include checking the raw proportions and counts, listing every pooled block, and confirming that the final estimates are nondecreasing and lie between 0 and 1. A long plateau or repeated boundary estimates can signal sparse information rather than a well-determined dose-response shape. Also check that outcomes have had adequate time to be observed: treating participants with incomplete toxicity follow-up as nontoxic would distort both the counts and the fitted probabilities.
Comparison, Limitations, and a Common Mistake
| Approach | How it represents the dose response | Main implication |
|---|---|---|
| Direct empirical updating with isotonic constraints | Estimates probabilities at observed doses and pools adjacent order violations. | Few structural assumptions about curve shape, but limited extrapolation and potentially unstable estimates with sparse data. |
| Fixed-skeleton CRM | Uses a prespecified dose skeleton and updates a model parameter to relate doses. | Can share information across the dose grid through the model, but conclusions depend on the skeleton and model assumptions. |
The empirical method is more local: only adjacent doses are pooled to satisfy the order restriction, and it does not generate a smooth parametric curve between dose levels. A fixed-skeleton CRM can provide estimates at sparsely observed doses by relying on its model structure, but this comes at the cost of stronger structural assumptions. Neither approach removes the need to assess data quality, dose ordering, and the consequences of sparse observations. Empirical isotonic estimates also do not by themselves provide a posterior distribution or a complete measure of uncertainty.
Common mistake: “The observed toxicity proportions decrease between two doses, so I will average them and assign that value to both.” An unweighted average is generally wrong when the dose sample sizes differ. The correction is to pool the counts: add toxicities, add participants, and divide. Also, the score equation for a pooled block is an interior calculation only when the block contains both toxicities and nontoxicities. For an all-toxicity block the maximum is 1; for a no-toxicity block it is 0. The pooled observed proportion gives the maximum in both boundary cases as well.
Check Your Understanding
Work through the questions, then compare with the solutions below.
- At two adjacent doses, \(n=(2,4)\) and \(y=(1,1)\). What are the raw proportions, and what fitted estimates result from pooling?
- Four equally sized dose groups each have \(n_j=10\), with raw proportions \(0.1000,\ 0.6000,\ 0.4000,\ 0.8000\). Find the isotonic estimates.
- What can direct empirical estimation determine at a dose with no participants?
- Does a pooled plateau establish that the true toxicity probabilities at those doses are equal? Explain.
Solutions. 1. The raw proportions are \(1/2=0.5000\) and \(1/4=0.2500\), a violation. Pooling gives \((1+1)/(2+4)=2/6=0.3333\), so both fitted estimates are \(0.3333\). 2. The second and third estimates violate the order. Their pooled estimate is \((10(0.6000)+10(0.4000))/20=10/20=0.5000\). The fitted sequence is \(0.1000,\ 0.5000,\ 0.5000,\ 0.8000\). 3. With no participants, the dose has no observed proportion and contributes no likelihood term. Monotonicity may restrict its value relative to neighboring doses, but does not generally identify a unique estimate. 4. No. Pooling assigns a common fitted value to satisfy the constrained likelihood; it does not prove equality of the underlying probabilities.