Posterior Mean Toxicity Is Not an Overdose Rule

Why the Posterior Mean Is Not an Overdose Rule

A posterior mean toxicity probability can summarize the model’s central estimate at a dose. But a dose-escalation rule usually asks a different question: how much posterior probability is assigned to toxicity exceeding a prespecified limit? Those two summaries are not interchangeable. In particular, the mean alone does not reveal how much posterior mass lies above the limit.

This distinction matters when applying the overdose boundary and admissibility rules used in Bayesian dose-escalation designs. The boundary is based on posterior probability of exceeding the toxicity limit, not on whether the posterior mean toxicity happens to be below that limit. The introductory material on posterior overdose probabilities provides the broader decision context; here the focus is the error of substituting a mean for a tail probability.

Definition: At dose \(d\), let \(p_d\) be the dose’s true probability of a dose-limiting toxicity (DLT), and let the posterior distribution of \(p_d\), conditional on the observed trial data and the fitted model, be denoted by \(p(p_d\mid\text{data})\). For an upper toxicity limit \(\phi\), the posterior overdose probability is \(P(p_d>\phi\mid\text{data})\). If the design specifies an overdose-probability cutoff \(\xi\), the dose satisfies this particular overdose-control criterion when \(P(p_d>\phi\mid\text{data})\leq\xi\). The values of \(\phi\) and \(\xi\) must be specified by the design; neither is a universal constant.

The posterior mean, \(E(p_d\mid\text{data})\), averages toxicity probabilities across the posterior distribution. The overdose probability instead counts the posterior mass in the region above \(\phi\). A mean can be pulled down by substantial posterior mass at low toxicity even when a concerning amount of mass remains above the limit. Conversely, some posterior mass above the limit can coexist with a mean above the limit while the probability of exceeding it remains below a chosen cutoff.

The probability-based rule is applied to the dose-specific posterior distribution. In a BLRM, dose-specific distributions come from a joint model of toxicity across doses, so the posterior at one dose reflects the fitted dose–toxicity relationship and the data. The decision at that dose nevertheless depends on its posterior probability of exceeding \(\phi\), not on the posterior mean alone.

Why the Mean Does Not Determine the Tail

A brief inequality shows why substituting the mean cannot generally produce the required decision. Since \(p_d\) is nonnegative, whenever \(p_d>\phi\), its value is at least \(\phi\). Therefore the posterior expectation must be at least \(\phi\) times the posterior probability of that event:

E(p_d\mid\text{data}) \geq \phi\,P(p_d>\phi\mid\text{data}), \qquad P(p_d>\phi\mid\text{data}) \leq \frac{E(p_d\mid\text{data})}{\phi}.

The second inequality is Markov’s inequality. It provides an upper bound, not the overdose probability itself. For example, if the posterior mean is \(0.20\) and \(\phi=0.30\), this bound is \(0.20/0.30=0.667\). A bound of \(0.667\) cannot establish that the overdose probability is below a cutoff of \(0.25\). More importantly, many distinct posterior distributions can have the same mean and very different probabilities above \(\phi\). The following examples make that distinction explicit.

Worked Example: The Same Mean, Different Overdose Decisions

Suppose the upper toxicity limit is \(\phi=0.30\), and the design’s overdose-probability cutoff is \(\xi=0.25\). For illustration, consider two hypothetical sets of 1,000 equally weighted posterior draws of \(p_d\). The listed values stand in for the draws; actual model-based posterior simulations would generally have more varied values.

At dose A, 700 draws equal \(0.05\) and 300 equal \(0.55\). The posterior mean estimated from the draws is \(700(0.05)/1000+300(0.55)/1000=0.035+0.165=0.200\). Of the 1,000 draws, 300 exceed \(0.30\), so the estimated posterior overdose probability is \(300/1000=0.300\). Since \(0.300>0.25\), dose A fails the stated overdose criterion.

At dose B, all 1,000 draws equal \(0.20\). Its draw-based posterior mean is also \(0.200\), while none of the draws exceed \(0.30\). Its estimated posterior overdose probability is \(0/1000=0\), so dose B satisfies the criterion.

The posterior means are identical, but the overdose decisions differ. A rule based only on the mean would have no way to distinguish these two distributions. The posterior tail probability does.

Worked Decisions: Mean Below or Above the Limit

The same problem appears whether a mean-based shortcut is used to declare safety or to declare overdose. A mean below the limit does not guarantee that the posterior overdose probability is acceptable. A mean above the limit does not, by itself, establish that the overdose probability exceeds the design’s cutoff.

Worked Example: A Mean Below the Limit Can Still Fail

Use \(\phi=0.30\) and \(\xi=0.25\). At a hypothetical dose, suppose 600 of 1,000 equally weighted posterior draws of \(p_d\) equal \(0.10\), and 400 equal \(0.45\).

The posterior mean estimated from these draws is \(600(0.10)/1000+400(0.45)/1000=0.060+0.180=0.240\). That is below the upper limit of \(0.30\). However, all 400 draws at \(0.45\) exceed the limit, so the estimated posterior overdose probability is \(400/1000=0.400\). Because \(0.400>0.25\), the dose fails the stated overdose criterion.

Calling the dose safe because \(0.240<0.30\) would confuse the posterior mean with a posterior probability of exceeding the limit. The correct calculation uses the fraction of posterior draws in the overdose region.

Worked Example: A Mean Above the Limit Does Not Set the Cutoff Decision

Again take \(\phi=0.30\) and \(\xi=0.25\). At another hypothetical dose, suppose 800 of 1,000 posterior draws equal \(0.225\), while 200 equal \(0.70\).

The draw-based posterior mean is \(800(0.225)/1000+200(0.70)/1000=0.180+0.140=0.320\). The mean exceeds the upper limit. But the estimated posterior overdose probability is \(200/1000=0.200\), which is below the stated cutoff of \(0.25\). Thus, this dose satisfies the overdose-probability criterion as defined here.

This conclusion is deliberately limited: satisfying this one criterion does not by itself make the dose the recommended dose, nor does it settle any other design requirement. It shows only that “posterior mean above \(\phi\)” and “posterior probability above \(\phi\) exceeds \(\xi\)” are different statements.

Calculation and Diagnostics in Practice

With posterior simulations, the dose-specific probability is usually estimated by the proportion of draws above the limit. If \(M\) posterior draws are available at dose \(d\), and \(K\) of them satisfy \(p_d^{(m)}>\phi\), then the estimate is \(K/M\). The comparison with \(\xi\) should use the rule and threshold specified by the design. If values are rounded for display, the calculation should use the unrounded draws, particularly for draws close to \(\phi\).

Worked Example: Checking Simulation Precision Near the Cutoff

Suppose \(\phi=0.30\), \(\xi=0.25\), and 46 of 200 hypothetical posterior draws at a dose exceed \(0.30\). The estimated overdose probability is \(46/200=0.230\). Taken at face value, this estimate is below the cutoff.

The estimate can also have Monte Carlo error because it is computed from a finite simulation. Treating the exceedance indicators as draws with probability \(q=0.230\), the approximate Monte Carlo standard error is \(\sqrt{q(1-q)/M}=\sqrt{0.230(0.770)/200}=\sqrt{0.0008855}\approx0.0298\). A rough two-standard-error range for simulation error is \(0.230\pm2(0.0298)\), or approximately \(0.170\) to \(0.290\). This range is a diagnostic for computational precision, not a credible interval for \(p_d\) or for the overdose probability itself. It shows that 200 draws may not estimate the tail probability precisely enough to make a stable comparison with \(0.25\).

At an estimated exceedance probability near \(0.23\), the approximate number of independent draws needed for Monte Carlo standard error at most \(0.005\) is \(M\geq0.230(0.770)/(0.005)^2=7084\). Using roughly 7,100 or more draws would therefore improve precision under this approximation. The simulation method and its mixing must also be adequate: a large nominal draw count does not guarantee the same precision if draws are highly dependent or the chain has not explored the posterior well.

Several practical checks help detect a mistaken or unreliable decision:

Check the event being counted. The overdose event is \(p_d>\phi\), not \(p_d\geq\xi\), and not \(E(p_d\mid\text{data})>\phi\). Keep the toxicity limit and probability cutoff distinct.
Check which dose and posterior are used. Calculate the exceedance proportion from the posterior draws for the dose under consideration, using the fitted model and current data.
Inspect precision near the cutoff. If the estimated probability is close to \(\xi\), quantify Monte Carlo error and use enough well-behaved posterior draws to support the comparison.
Check the model and data inputs. Implausible posterior summaries, sensitivity to modeling choices, or a poorly explored simulation can undermine the calculation. A precise estimate from a poorly fitting model is not automatically a trustworthy risk assessment.

How This Rule Differs From Other Summaries

The posterior mean is useful for summarizing expected toxicity under the posterior, but it answers an average-value question. The posterior overdose probability answers a threshold-risk question. A posterior interval for \(p_d\) is another summary: comparing an interval endpoint with \(\phi\) can provide information about uncertainty, but it is not automatically the same as comparing \(P(p_d>\phi\mid\text{data})\) with \(\xi\). An interval-based rule needs its own stated coverage level and decision definition.

These approaches can agree, but they need not. A mean-only rule is simple but discards the distribution’s shape around the toxicity limit. An interval rule summarizes a selected range of posterior values and depends on how the interval is constructed. A posterior-probability rule directly targets the specified event and cutoff, but depends on a suitable posterior model and adequate computation. The rule should therefore be chosen and specified by the design rather than selected after inspecting which summary gives the preferred decision.

Common Mistake and Limits of Interpretation

Wrong reasoning: “The posterior mean DLT probability is \(0.24\), below the upper limit \(0.30\), so the dose is admissible.” This substitutes an average for a tail probability. Correction: calculate \(P(p_d>0.30\mid\text{data})\) from the dose-specific posterior and compare it with the prespecified cutoff. In the example above, the mean was \(0.24\), but the estimated overdose probability was \(0.40\), so the dose failed the criterion with \(\xi=0.25\).

A posterior overdose probability is conditional on the observed data and the model used to generate the posterior. It is not a guarantee about the toxicity outcomes of future patients, and it does not remove uncertainty about model assumptions. The threshold \(\phi\), cutoff \(\xi\), dose range, and other decision requirements should be prespecified and clinically justified. Passing the overdose criterion means only that the posterior probability is on the allowed side of that particular boundary under the specified analysis.

Key takeaway: A posterior mean summarizes the center of the toxicity distribution; it does not determine the posterior mass above the toxicity limit. To apply an overdose rule, calculate \(P(p_d>\phi\mid\text{data})\) and compare it with the prespecified cutoff \(\xi\), while checking that the posterior computation is reliable.

Check Your Understanding

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

With \(\phi=0.25\) and \(\xi=0.20\), 25 of 100 posterior draws exceed \(\phi\). What is the estimated overdose probability, and does it satisfy the stated criterion?
Why can two posterior distributions with the same mean have different overdose decisions?
For \(\phi=0.30\), posterior draws give an estimated mean of \(0.18\) and an estimated overdose probability of \(0.28\). Does the mean establish that the dose satisfies a rule with \(\xi=0.25\)?
What does a Monte Carlo standard error describe when estimating the posterior overdose probability from simulated draws?

Solutions. (1) The estimate is \(25/100=0.25\). Since \(0.25>0.20\), it does not satisfy the stated criterion. (2) The mean gives an average, whereas the overdose probability gives the posterior mass above the threshold; distributions with the same average can allocate different mass to that region. (3) No. The mean does not establish the tail probability. The stated estimate \(0.28\) exceeds \(0.25\), so the dose fails the criterion, subject to checking simulation precision. (4) It describes the approximate simulation error in the estimated tail probability due to using a finite number of posterior draws; it is not a credible interval for toxicity or a measure of the model’s clinical uncertainty.