From Probability Requirements to a Dose Rule
Dose selection often involves a benefit that increases with dose and a toxicity risk that also increases with dose. A decision-maker may specify both a minimum acceptable probability of efficacy and a maximum acceptable probability of toxicity. The task is then not simply to find the dose with the highest predicted efficacy: it is to maximize a stated benefit while respecting both probability requirements.
The probabilities used for these requirements must have a defined basis. They may reflect uncertainty in model parameters, future patient outcomes, or both, according to the clinical decision being made. For example, a probability may be calculated over posterior parameter draws, or over parameter draws together with simulated outcomes. The probability definition and uncertainty distribution should match the clinical meaning of the thresholds. In what follows, \(E_d\) denotes the event that the efficacy criterion is met at dose \(d\), and \(T_d\) denotes the event that the toxicity criterion is met.
The notation \(\operatorname*{arg\,max}\) means the dose, or doses, at which the largest feasible benefit is attained. If no dose meets both requirements, the feasible set is empty: the rule does not authorize choosing the “least bad” dose without a separate decision about which requirement may be relaxed. If multiple feasible doses have the same maximum benefit, additional tie-breaking criteria are needed.
Deriving the Feasible Set
Define the feasible set as \(F=\{d\in D:p_E(d)\geq q_E,\ p_T(d)\leq q_T\}\). The optimization is a comparison of benefit values only over \(F\). This distinction is important: a dose with very high benefit is irrelevant to the constrained optimum if it violates either probability requirement.
A useful special case has a continuous dose range, an increasing efficacy probability, an increasing toxicity probability, and an increasing benefit. If the efficacy probability reaches its threshold at a lower boundary \(d_E\), while toxicity reaches its limit at an upper boundary \(d_T\), then efficacy requires \(d\geq d_E\) and toxicity requires \(d\leq d_T\). The feasible interval is therefore \([d_E,d_T]\), provided \(d_E\leq d_T\) and both boundaries lie in the allowable dose range. Since benefit increases throughout that interval, its maximum is at \(d_T\). This conclusion depends on the stated monotonicity and interval conditions; it is not a general rule to select the highest dose that appears feasible.
With a finite set of candidate doses, or with nonmonotone probabilities or benefit, do not infer the optimum from a boundary argument. Calculate both probabilities at each candidate, exclude infeasible doses, and compare the specified benefit among the remaining doses.
Worked Example: Continuous Dose With Increasing Benefit
Suppose, hypothetically, that dose \(d\) is measured in milligrams and the specified probabilities and benefit are \(p_E(d)=1-e^{-0.12d}\), \(p_T(d)=1-e^{-0.012d}\), and \(B(d)=10(1-e^{-0.08d})\) benefit points. Require \(p_E(d)\geq0.85\) and \(p_T(d)\leq0.20\), with \(d\geq0\). Assume these functions are the agreed decision quantities over the dose range being considered.
The efficacy condition gives \(1-e^{-0.12d}\geq0.85\), so \(e^{-0.12d}\leq0.15\), and hence \(d\geq-\ln(0.15)/0.12=15.8093\) mg, approximately 15.81 mg.
The toxicity condition gives \(1-e^{-0.012d}\leq0.20\), so \(e^{-0.012d}\geq0.80\), and hence \(d\leq-\ln(0.80)/0.012=18.5953\) mg, approximately 18.60 mg. The feasible interval is therefore approximately 15.81 to 18.60 mg. It is nonempty because the lower boundary is less than the upper boundary.
The benefit is increasing because its derivative is \(0.8e^{-0.08d}>0\) for all \(d\geq0\). Thus the continuous constrained optimum is the exact upper boundary, \(d^*=-\ln(0.80)/0.012\), or approximately 18.60 mg. At this exact boundary, toxicity probability is 0.20 and efficacy probability is \(1-e^{-0.12(18.5953)}\), approximately 0.8926. The benefit is \(10[1-e^{-0.08(18.5953)}]\), approximately 7.741 points.
If only whole-milligram doses can be prescribed, the largest feasible integer dose is 18 mg. At 18 mg, efficacy probability is \(1-e^{-2.16}=0.8847\), toxicity probability is \(1-e^{-0.216}=0.1943\), and benefit is \(10(1-e^{-1.44})=7.631\) points. The 19 mg dose exceeds the toxicity boundary, so it is infeasible even though its benefit is greater.
Applying the Rule to Uncertainty-Based Probabilities
In model-informed decisions, \(p_E(d)\) and \(p_T(d)\) are often estimated from a distribution representing uncertainty. For example, each draw of model parameters can be used to determine whether the efficacy target and toxicity limit are met at a candidate dose. The estimated probabilities are the proportions of draws meeting the corresponding criteria. If simulated future outcomes are also included, the event indicators must be defined using those outcomes as well.
The benefit must be specified on the same decision basis as the optimization. It could be a fixed score assigned to each dose or an expected benefit calculated under the agreed uncertainty distribution. The chance constraints do not determine what benefit should be used; they restrict the set of doses over which the stated benefit is optimized.
Worked Example: Selecting From Monte Carlo Estimates
Suppose 1,000 hypothetical parameter-and-outcome draws are used to estimate probabilities at four candidate doses. The requirements are \(p_E(d)\geq0.80\) and \(p_T(d)\leq0.15\). The benefit values, in prespecified points, are also supplied.
| Dose (mg) | Efficacy events | \(\hat{p}_E(d)\) | Toxicity events | \(\hat{p}_T(d)\) | Benefit (points) |
|---|---|---|---|---|---|
| 5 | 720 of 1,000 | 0.720 | 30 of 1,000 | 0.030 | 4.2 |
| 10 | 840 of 1,000 | 0.840 | 80 of 1,000 | 0.080 | 6.8 |
| 15 | 910 of 1,000 | 0.910 | 160 of 1,000 | 0.160 | 8.0 |
| 20 | 950 of 1,000 | 0.950 | 270 of 1,000 | 0.270 | 8.6 |
For example, at 10 mg the estimates are \(\hat{p}_E(10)=840/1000=0.840\) and \(\hat{p}_T(10)=80/1000=0.080\). Both constraints hold, since \(0.840\geq0.80\) and \(0.080\leq0.15\), so 10 mg is feasible.
At 5 mg, efficacy fails because \(0.720<0.80\). At 15 mg, toxicity fails because \(0.160>0.15\); the same constraint fails at 20 mg because \(0.270>0.15\). Thus, among these candidates, 10 mg is the only feasible dose and is the selected dose, with benefit 6.8 points. The higher benefit at 20 mg cannot override the toxicity constraint.
These probabilities are Monte Carlo estimates, not exact values. For example, the approximate Monte Carlo standard error for the efficacy estimate at 10 mg is \(\sqrt{0.84(1-0.84)/1000}=0.0116\). A decision near a threshold may be sensitive to simulation error; more draws or an explicitly conservative decision procedure may then be appropriate.
Marginal Constraints Are Not Automatically Joint Constraints
The two constraints in the definition are marginal: one applies to the probability of efficacy, and the other to the probability of toxicity. Meeting both does not imply that a specified proportion of draws meet efficacy and avoid toxicity together. The events may be dependent, so the joint probability cannot generally be obtained by multiplying the marginal probabilities or by assuming independence.
If the clinical requirement is explicitly that both conditions hold in the same case with probability at least \(q_J\), that is a different constraint: \(P(E_d\cap T_d^c)\geq q_J\), where \(T_d^c\) is the event of no toxicity. The appropriate formulation must follow the stated decision requirement, not be chosen after seeing which one makes a dose feasible.
Worked Example: Checking a Joint Requirement
At a hypothetical dose of 12 mg, consider 200 draws. Efficacy occurs in 180 draws, toxicity occurs in 20 draws, and both efficacy and no toxicity occur together in 166 draws. Suppose the marginal requirements are \(p_E\geq0.85\) and \(p_T\leq0.15\), while a separate proposed joint requirement is \(P(E_d\cap T_d^c)\geq0.85\).
The marginal estimates are \(\hat{p}_E=180/200=0.900\) and \(\hat{p}_T=20/200=0.100\). Both marginal constraints pass. But the joint estimate is \(166/200=0.830\), which is below 0.85. Therefore, this dose meets the two marginal constraints but fails the stated joint requirement.
The counts also show why multiplication would be inappropriate: \(0.900(1-0.100)=0.810\), which is not the observed joint proportion, 0.830. The joint event must be evaluated directly when it is the quantity required by the decision.
Diagnostics, Alternatives, and Limitations
Before acting on a selected dose, check that the probability definitions, thresholds, and dose range match the intended clinical decision. For simulation-based estimates, inspect the number of draws and the distance from each probability estimate to its threshold. If an estimate is close to a boundary, simulation variability can change whether it appears feasible. Repeating the calculation with more draws is a useful diagnostic; it does not remove uncertainty in the underlying model or make an unsuitable probability definition valid.
Also verify that all candidate doses were evaluated consistently, that the benefit was calculated on the prespecified scale, and that rounding did not change a feasibility classification. When continuous-dose monotonicity is assumed, inspect the probability and benefit functions over the relevant range rather than relying on a few points to establish it. A coarse dose grid can miss a narrow feasible region or the best available dose.
Chance-constrained selection differs from maximizing benefit without constraints: the latter may choose a dose whose toxicity probability exceeds the stated limit. It also differs from maximizing efficacy probability alone, which ignores the specified benefit and can favor unnecessarily high doses. A weighted benefit-minus-harm score is another possible decision rule, but it trades outcomes through chosen weights rather than enforcing the two probability limits directly. None of these alternatives is interchangeable without an explicit decision rationale.
The result is conditional on the model, uncertainty distribution, event definitions, benefit measure, and allowable dose set. A feasible dose is not a guarantee that a particular patient will respond or avoid toxicity. Nor does feasibility establish that the model is correct. If no dose is feasible, the appropriate conclusion is that the stated constraints cannot be jointly met over the evaluated set; the constraints or decision basis require review rather than silent relaxation.
Common Mistake
- Wrong reasoning: “The dose has the largest benefit, so it must be optimal, even though its toxicity probability exceeds the limit.”
- Correction: The optimization is restricted to doses satisfying every stated constraint. First determine feasibility; then compare benefit only among feasible doses. A dose that fails either constraint is not a candidate for the constrained optimum.
- Wrong reasoning: “The efficacy probability and toxicity probability each pass, so the joint efficacy-and-safety probability must also pass.”
- Correction: Separate marginal conditions do not specify the joint probability. Calculate the joint event directly if the clinical requirement is joint.
Check Your Understanding
For each problem, identify the feasibility decision before stating the selected dose. Solutions are included below.
- A candidate has efficacy probability 0.88 and toxicity probability 0.14. The requirements are efficacy at least 0.85 and toxicity at most 0.15. Is it feasible?
- Suppose \(p_E(d)=1-e^{-0.10d}\), \(p_T(d)=1-e^{-0.01d}\), and \(B(d)=5d\), with \(d\geq0\). Find the feasible interval for efficacy at least 0.80 and toxicity at most 0.10. State the continuous optimum.
- Three candidate doses have estimated efficacy probabilities 0.82, 0.90, and 0.94; toxicity probabilities 0.05, 0.12, and 0.18; and benefits 6, 8, and 10 points, respectively. With thresholds 0.85 and 0.15, which dose is selected?
- Why can two marginal probability constraints pass while a joint efficacy-without-toxicity requirement fails?
Solutions. (1) Yes: \(0.88\geq0.85\) and \(0.14\leq0.15\), so both conditions hold. (2) Efficacy requires \(d\geq-\ln(0.20)/0.10=16.09\). Toxicity requires \(d\leq-\ln(0.90)/0.01=10.54\). Since \(16.09>10.54\), the feasible set is empty and there is no constrained optimum. (3) The first dose fails efficacy; the third fails toxicity; the middle dose passes both, so the selected dose is the second, with benefit 8 points. (4) The marginal probabilities count efficacy and toxicity separately. Their overlap depends on the dependence between events, so the joint probability must be calculated directly and may be below its required threshold.