Posterior Probabilities Of Acceptable Doses

What Does It Mean for a Dose to Be Acceptable?

In dose finding, a dose can have encouraging efficacy and still be unsuitable if its toxicity risk is too high. Conversely, a dose may appear safe but offer too little benefit. A useful summary therefore concerns both characteristics at once: given the observed data and the assumed model, how probable is it that the dose exceeds a minimum efficacy level while remaining below a maximum toxicity level?

This is a posterior probability about unknown dose-specific response rates. It is not the probability that a particular patient will respond without toxicity, nor is it a rule that automatically selects a dose. The threshold values express the clinical criteria being assessed and should be specified in the context of the trial.

Definition: For dose $j$, let $p_{E,j}$ be its efficacy probability and $p_{T,j}$ its toxicity probability. Let $\eta_E$ be the minimum acceptable efficacy probability and $\eta_T$ the maximum acceptable toxicity probability. The dose is acceptable when $$ A_j=\{p_{E,j}>\eta_E,\ p_{T,j}<\eta_T\}. $$ Given observed data $D$ and a specified Bayesian model, the posterior probability of acceptability is $$ \pi_j=P(A_j\mid D)=P(p_{E,j}>\eta_E,\ p_{T,j}<\eta_T\mid D). $$

The strict inequalities above follow the stated definition. If a protocol instead defines a threshold using a non-strict inequality, use that definition consistently. For continuous posterior distributions, the probability of equality to a fixed threshold is zero, so strict versus non-strict inequalities do not change the numerical probability. The thresholds are probabilities on the same scale as the dose-specific response rates: for example, $\eta_E=0.40$ means a minimum efficacy rate of 40%, not an efficacy increase of 40 percentage points.

The calculation requires a posterior distribution for the efficacy and toxicity parameters under the model. That distribution depends on the likelihood, prior, and observed data. If efficacy and toxicity are modeled jointly, the posterior dependence between their rates matters. Even when the two marginal posterior distributions are known, they do not by themselves determine the probability that both thresholds are met.

Calculating the Joint Posterior Probability

In general, the probability is the posterior mass assigned to the region above the efficacy threshold and below the toxicity threshold. Writing the joint posterior density of the two rates as $f(p_{E,j},p_{T,j}\mid D)$, the calculation is

\pi_j=\int_{\eta_E}^{1}\int_{0}^{\eta_T} f(p_{E,j},p_{T,j}\mid D)\,dp_{T,j}\,dp_{E,j}.

If the posterior factorizes into independent efficacy and toxicity components, this joint probability is the product of two marginal probabilities. That factorization is an assumption about the posterior under the chosen model; it does not follow merely because efficacy and toxicity are recorded as separate outcomes. With dependent posterior rates, calculate the joint probability directly from the joint posterior.

Worked Example: Independent Beta Posteriors

Hypothetical data and model. At one dose, suppose efficacy is modeled as binomial with a Beta(1,1) prior. Four of four patients have an efficacy response, giving a Beta(5,1) posterior. Toxicity is modeled with a separate binomial likelihood and a Beta(1,1) prior; zero of three patients have a toxicity event, giving a Beta(1,4) posterior. Assume the model gives independent posterior distributions for the two rates. Assess acceptability using $\eta_E=0.40$ and $\eta_T=0.20$.

For a Beta(5,1) random variable, the density is $5p^4$ for $0<p<1$. Thus the posterior probability of exceeding the efficacy threshold is

P(p_{E,j}>0.40\mid D) =\int_{0.40}^{1}5p^4\,dp =1-(0.40)^5 =0.98976.

For a Beta(1,4) random variable, the cumulative distribution function at $t$ is $1-(1-t)^4$. Therefore,

P(p_{T,j}<0.20\mid D) =1-(1-0.20)^4 =1-(0.80)^4 =0.5904.

By the assumed posterior independence, multiply the two probabilities:

\pi_j=0.98976(0.5904)=0.584354304\approx 0.5844.

Under this hypothetical model and data, the posterior probability that the dose’s efficacy rate exceeds 40% while its toxicity rate is below 20% is about 58.44%. This is not the probability that a patient has both outcomes, and it does not establish that the dose should be chosen.

Why Dependence Changes the Answer

For any two events, the probability that both occur depends on how often they occur together, not just on their separate probabilities. In this setting, model structure or shared patient-level factors can induce posterior dependence between efficacy and toxicity rates. When that dependence is present, multiplying the marginal threshold probabilities can overstate or understate the joint probability.

A direct way to calculate the joint probability is to draw paired values of the efficacy and toxicity rates from their joint posterior distribution, then check both threshold conditions for each pair. The pairs must come from the joint posterior: drawing efficacy and toxicity values independently would impose posterior independence, even if the fitted model does not make that assumption.

Worked Example: Joint Posterior Simulation With Dependence

Hypothetical posterior simulation. Suppose a fitted joint model produces 1,000 paired posterior draws for one dose, using thresholds $\eta_E=0.40$ and $\eta_T=0.20$. In 760 draws, $p_{E,j}>0.40$; in 700 draws, $p_{T,j}<0.20$; and in 620 draws both conditions hold.

The joint posterior probability is estimated by the fraction of paired draws satisfying both conditions:

\widehat{\pi}_j=\frac{620}{1000}=0.620.

The marginal estimates are $760/1000=0.760$ for efficacy and $700/1000=0.700$ for toxicity. Multiplying those marginals would give $0.760(0.700)=0.532$, not 0.620. The discrepancy occurs because the simulated indicators for passing the two thresholds are associated in these draws. The calculation based on paired draws preserves the dependence represented in the model.

The approximate Monte Carlo standard error of the estimated joint probability is

\sqrt{\frac{0.620(1-0.620)}{1000}} =\sqrt{0.0002356} \approx 0.01535.

This standard error describes simulation error from using a finite number of posterior draws; it is not the posterior uncertainty in the unknown rates themselves. Increasing the number of effective draws can reduce this computational error.

Justification for the Simulation Estimate

Let $I^{(m)}_j$ equal 1 if posterior draw $m$ satisfies both acceptability conditions and 0 otherwise. If the draws are independent draws from the joint posterior, then each $I^{(m)}_j$ is a Bernoulli random variable with success probability $\pi_j$. The average of these indicators estimates that probability:

\widehat{\pi}_j=\frac{1}{M}\sum_{m=1}^{M}I^{(m)}_j, \qquad E(\widehat{\pi}_j\mid D)=\pi_j, \qquad \operatorname{Var}(\widehat{\pi}_j\mid D)=\frac{\pi_j(1-\pi_j)}{M}.

The expectation and variance follow from $E(I^{(m)}_j\mid D)=\pi_j$, $\operatorname{Var}(I^{(m)}_j\mid D)=\pi_j(1-\pi_j)$, and independence across draws. Replacing the unknown $\pi_j$ in the variance formula with $\widehat{\pi}_j$ gives the usual estimated Monte Carlo standard error. For correlated simulation output, such as draws from Markov chain Monte Carlo, the nominal number of draws overstates the independent information. Assess convergence and use an effective sample size or a suitable batch-based Monte Carlo error estimate rather than relying on the independent-draw formula alone.

Specify the event.
State the efficacy and toxicity parameters, the thresholds, and whether equality is included.

Obtain the posterior.
Use the model fitted to the data, retaining any dependence between efficacy and toxicity rates.

Calculate the joint mass.
Integrate over the acceptable region or count paired posterior draws that satisfy both conditions.

Check computational adequacy.
For simulation, check convergence and Monte Carlo error, especially when the estimated probability is near a decision-relevant boundary.

Interpret in context.
Report the probability as conditional on the data and model, with the clinical thresholds and dose identified.

Comparisons, Diagnostics, and Limitations

A plug-in calculation substitutes point estimates, such as posterior means, into the threshold conditions. It can say whether those point estimates lie in the acceptable region, but it does not quantify how much posterior probability lies there. A credible interval for each rate separately gives marginal uncertainty, but separate intervals do not generally answer the joint question either. The joint posterior probability directly summarizes uncertainty about the combined event.

Several checks help identify an unreliable calculation:

Check the event and units. Thresholds must refer to the same dose-specific probability scale as $p_{E,j}$ and $p_{T,j}$. Reversing the toxicity inequality answers a different question.
Check the dependence assumption. A product of marginal probabilities is justified only when the posterior factorizes for these rates. If dependence is modeled, use joint draws or joint integration.
Check simulation behavior. Poor chain mixing, lack of convergence, or too few effective draws can make the estimated probability unstable. Compare independent chains or longer runs and inspect Monte Carlo error.
Check model sensitivity. The probability is conditional on the chosen likelihood, prior, dose-response structure, and outcome definitions. A threshold probability cannot correct misspecification or make sparse data informative.

One limitation is that a single probability compresses a two-dimensional posterior distribution into one number. Different posterior shapes can produce the same acceptability probability while implying different clinical tradeoffs. The result also depends on the chosen thresholds, so it should be reported with them rather than as an unqualified property of the dose. Posterior acceptability is an input to dose assessment; it does not, by itself, specify how to rank acceptable doses or what action to take.

Common pitfall: “The posterior probability that efficacy exceeds its threshold is 0.8, and the probability that toxicity is below its threshold is 0.8, so the probability the dose meets both is 0.64.” This multiplication is correct only if the two threshold events are independent under the joint posterior. Otherwise, calculate the probability of both events from the joint distribution. Also, do not interpret a posterior probability of 0.64 as saying that 64% of treated patients will have both outcomes.

Worked Example: Comparing Acceptability Across Doses

Hypothetical posterior calculations. For three doses, suppose a joint model produces 2,000 paired posterior draws per dose using common thresholds $\eta_E=0.35$ and $\eta_T=0.25$. The numbers of draws meeting both conditions are 1,450, 1,100, and 360 for doses 1, 2, and 3, respectively.

The estimated probabilities are calculated as the joint passing count divided by 2,000:

\widehat{\pi}_1=\frac{1450}{2000}=0.725,\qquad \widehat{\pi}_2=\frac{1100}{2000}=0.550,\qquad \widehat{\pi}_3=\frac{360}{2000}=0.180.

For these hypothetical data and this model, dose 1 has the largest estimated posterior probability of meeting both thresholds, at 72.5%; dose 2 has 55.0%; and dose 3 has 18.0%. This comparison describes threshold acceptability only. It does not establish that dose 1 is the best dose, because other considerations and a prespecified decision rule may matter.

For example, the estimated Monte Carlo standard error for dose 1, under independent posterior draws, is

\sqrt{\frac{0.725(1-0.725)}{2000}} =\sqrt{0.0000996875} \approx 0.009984.

Thus the simulation estimate has Monte Carlo error of about one percentage point under the stated independent-draw calculation. If the draws are autocorrelated, the effective sample size is smaller than 2,000 and the appropriate Monte Carlo error will be larger.

Check Your Understanding

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

If $p_{E,j}\mid D$ has a Beta(4,1) posterior, what is $P(p_{E,j}>0.50\mid D)$?
If $p_{T,j}\mid D$ has a Beta(1,3) posterior, calculate $P(p_{T,j}<0.20\mid D)$.
In 1,000 paired posterior draws, efficacy passes its threshold 800 times, toxicity passes its threshold 650 times, and both pass 540 times. What is the estimated joint acceptability probability? Why is multiplying the two marginal estimates not justified by these counts alone?
For an estimated joint probability of 0.40 from 1,000 independent posterior draws, calculate the approximate Monte Carlo standard error.

Solutions. (1) For a Beta(4,1) distribution, the cumulative distribution function at $x$ is $x^4$. Therefore $P(p_{E,j}>0.50\mid D)=1-(0.50)^4=0.9375$. (2) For Beta(1,3), the cumulative distribution function at $x$ is $1-(1-x)^3$. Thus $P(p_{T,j}<0.20\mid D)=1-(0.80)^3=0.4880$. (3) The estimated joint probability is $540/1000=0.540$. The marginal estimates are 0.800 and 0.650, but their product assumes posterior independence; the joint count shows the observed paired-draw proportion directly. (4) The standard error is $\sqrt{0.40(0.60)/1000}=\sqrt{0.00024}\approx 0.01549$, or about 1.55 percentage points of Monte Carlo error.