Posterior Probability of Hitting the Target Interval

Why Calculate a Target-Interval Probability?

In dose escalation, the target toxicity interval describes a range of dose-limiting toxicity (DLT) rates considered to have the desired toxicity level. For each candidate dose, the model does not supply a known true DLT rate; it supplies a posterior distribution that represents uncertainty about that rate after combining the observed data with the model and prior. A useful summary is the posterior probability that the dose’s true DLT rate falls inside the specified interval.

This probability is dose-specific and conditional on the observed trial data, the model, and its prior. It is not the probability that a patient will experience a DLT at that dose, nor is it by itself a dose recommendation. The calculation summarizes uncertainty about the dose-level rate. The target interval must be defined in advance and in the clinical context; calculating its probability does not establish that the interval is clinically appropriate.

Definition: Let $p_j$ be the true DLT rate at candidate dose $d_j$, and let $(p_L,p_U)$ be the prespecified target toxicity interval, with $0<p_L<p_U<1$. Given the observed data $D$, the posterior target-interval probability at dose $d_j$ is $$ q_j=P(p_L\leq p_j\leq p_U\mid D). $$ For a continuous posterior distribution, including either endpoint instead of excluding it does not change this probability: the posterior probability of any single endpoint is zero.

Model and Calculation

Use the two-parameter Bayesian logistic regression model (BLRM) for the dose-toxicity relationship. Write $x_j=\log(d_j/d_{\mathrm{ref}})$, where $d_{\mathrm{ref}}$ is a chosen reference dose, and let $a$ and $b$ be the intercept and dose slope. The DLT rate at dose $d_j$ is

\operatorname{logit}(p_j)=a+b x_j,\qquad p_j=\frac{\exp(a+b x_j)}{1+\exp(a+b x_j)}.

For dose group $j$, suppose $n_j$ patients have evaluable DLT outcomes and $y_j$ experience a DLT. The usual cohort likelihood assumes conditionally independent Bernoulli outcomes with a common DLT probability within each dose group; equivalently, $y_j\mid p_j\sim\operatorname{Binomial}(n_j,p_j)$. The BLRM places a prior on its parameters and updates it with these outcomes to obtain a posterior distribution for $a$, $b$, and hence each $p_j$. This model assumes that the chosen logistic curve is an adequate representation of the dose-toxicity relationship over the dose range being considered. A positive dose slope enforces a nondecreasing toxicity curve as dose increases. The prior and the model’s dose scaling also matter, particularly when data are sparse.

The target-interval event can be expressed directly in terms of the linear predictor. Since the logistic function is strictly increasing, its inverse, $\operatorname{logit}(p)=\log(p/(1-p))$, preserves order. Therefore,

p_L\leq p_j\leq p_U \quad\Longleftrightarrow\quad \operatorname{logit}(p_L)\leq a+b x_j\leq\operatorname{logit}(p_U).

This equivalence is useful for checking an implementation: a posterior draw is inside the target interval on the probability scale exactly when its linear predictor is between the transformed limits. In practice, posterior draws are a flexible way to calculate $q_j$, without assuming that the dose-specific posterior distribution has a particular shape.

Calculation from posterior draws: For each posterior draw $m$, calculate $p_j^{(m)}=\operatorname{logit}^{-1}(a^{(m)}+b^{(m)}x_j)$. Set $I_j^{(m)}=1$ if $p_L\leq p_j^{(m)}\leq p_U$, and set it to 0 otherwise. With $M$ draws, estimate the target-interval probability by $$ \widehat{q}_j=\frac{1}{M}\sum_{m=1}^{M} I_j^{(m)}. $$

Worked Calculations

Worked Example: Normal Approximation at One Dose

Suppose the target interval is 0.20 to 0.30. At a hypothetical candidate dose, a normal approximation to the posterior distribution of the BLRM linear predictor is $\eta_j=a+b x_j\mid D\sim N(-1.10,0.25^2)$. This is an approximation for this calculation, not a claim that every BLRM posterior is exactly normal. Find the approximate posterior probability that the DLT rate is in the target interval.

The transformed interval endpoints are $\operatorname{logit}(0.20)=\log(0.20/0.80)=-1.3863$ and $\operatorname{logit}(0.30)=\log(0.30/0.70)=-0.8473$. Thus the event is $-1.3863\leq\eta_j\leq-0.8473$. Standardizing both endpoints using the stated posterior mean and standard deviation gives

P(-1.3863\leq\eta_j\leq-0.8473\mid D) =\Phi\left(\frac{-0.8473-(-1.10)}{0.25}\right) -\Phi\left(\frac{-1.3863-(-1.10)}{0.25}\right)

The upper standardized value is $1.0108$, and the lower is $-1.1452$. Using $\Phi(1.0108)\approx0.8439$ and $\Phi(-1.1452)\approx0.1261$, the probability is $0.8439-0.1261=0.7178$. Under this approximation, the posterior probability that the DLT rate at this dose lies between 0.20 and 0.30 is about 0.718, or 71.8%.

Worked Example: Comparing Candidate Doses Using Posterior Draws

Consider hypothetical trial data from three candidate doses: at 2 mg, 0 of 3 patients had a DLT; at 4 mg, 1 of 3 did; and at 8 mg, 2 of 3 did. Suppose a BLRM fit, using its specified prior and these data, produced 1,000 independent posterior draws. For the target interval $(0.20,0.30)$, the draw counts inside the interval were 248 at 2 mg, 631 at 4 mg, and 102 at 8 mg. Calculate and compare the estimated probabilities.

At 2 mg, $\widehat q=248/1000=0.248$, or 24.8%. At 4 mg, $\widehat q=631/1000=0.631$, or 63.1%. At 8 mg, $\widehat q=102/1000=0.102$, or 10.2%. For independent draws, the estimated Monte Carlo standard error is $\sqrt{\widehat q(1-\widehat q)/M}$. For example, at 4 mg it is $\sqrt{0.631(1-0.631)/1000}=\sqrt{0.000232839}\approx0.0153$, or about 1.53 percentage points.

The estimate is largest at 4 mg in this hypothetical analysis. These probabilities are summaries of posterior uncertainty conditional on the fitted model and the stated data; they do not say that 63.1% of patients treated at 4 mg would experience DLTs. The small cohorts also mean the analysis depends on the model and prior, so these figures should not be read as precise, model-free rates.

Worked Example: A Posterior Mean Is Not an Interval Probability

Suppose a hypothetical BLRM analysis at 6 mg gives 400 posterior draws of the DLT rate. Of these, 143 lie between 0.20 and 0.30. The mean of all 400 rate draws is 0.302. What is the estimated target-interval probability, and what does the posterior mean tell us?

The estimated probability is $143/400=0.3575$, or 35.8%. The posterior mean rate is $0.302$, just above the target interval’s upper limit. These summaries answer different questions: the mean averages the rates across all draws, while the interval probability counts the fraction of draws inside the specified range. A mean just outside the interval does not imply that few or no posterior draws are inside it. Conversely, a mean inside the interval would not establish that most of the posterior probability lies there.

Why the Draw-Based Formula Works

The target-interval probability is the posterior expectation of an indicator. An indicator is 1 when its event is true and 0 otherwise, so its expectation equals the probability of that event. Specifically,

E[I_j\mid D] =1\cdot P(p_L\leq p_j\leq p_U\mid D) +0\cdot P(p_j\notin[p_L,p_U]\mid D) =q_j.

Posterior draws approximate this expectation by an average. If draws were independent, the indicator draws would be Bernoulli with probability $q_j$, giving variance $q_j(1-q_j)$. Replacing the unknown $q_j$ with $\widehat q_j$ yields the usual estimated Monte Carlo standard error:

\widehat{\operatorname{MCSE}}(\widehat q_j) =\sqrt{\frac{\widehat q_j(1-\widehat q_j)}{M}}.

This derivation explains both the method and its limitation: the standard error describes numerical uncertainty from finite simulation, not all uncertainty about the true DLT rate. In Markov chain Monte Carlo (MCMC), successive draws are usually correlated. The independent-draw formula then understates simulation error if applied as though all $M$ draws were independent. Use an effective sample size for the indicator sequence, or an appropriate MCMC standard-error estimate, and assess chain convergence before relying on the estimate.

Diagnostics, Alternatives, and Common Pitfalls

A target-interval probability should be accompanied by checks that it is being estimated from a trustworthy posterior computation. Review the chains for mixing and convergence in the BLRM parameters and in the dose-specific target indicators. Check the effective sample size of the indicators: a chain can explore parameter values adequately in general yet spend long periods on one side of an interval boundary. If few draws fall inside the interval, the estimated probability may be especially sensitive to simulation error; if all or nearly all do, check that the chains have explored the relevant tails. Increase simulation or investigate the fit when estimates vary materially across chains or runs.

Also check the model inputs and transformations. Confirm that the target bounds are the prespecified values in the intended units, that the same DLT definition and follow-up window were used for all relevant patients, and that the dose scaling and reference dose match the fitted BLRM. A coding error in either endpoint or a mismatch between a dose and its scaled value can change the answer. When data are limited, compare reasonable prior specifications and inspect whether the posterior curve is being driven mainly by the prior. The output is conditional on those choices, not a direct empirical proportion.

There are several tempting alternatives, but each answers a different question. The posterior mean DLT rate is a useful summary of central tendency; it does not give the probability of lying in a range. A credible interval describes posterior uncertainty using a chosen coverage level and endpoints; whether it overlaps the target interval does not determine how much posterior probability is inside that interval. The draw-based indicator calculation answers the target-interval question directly. A normal approximation to the linear predictor can be a convenient check or a quick calculation, but it can be inaccurate when the posterior is skewed, unusually shaped, or near a boundary. Prefer posterior draws when available and verify that a convenient approximation agrees closely enough for the intended summary.

Because the fitted dose-response curve may be monotone, posterior DLT rates at higher doses are ordered within a draw when the slope is constrained positive. However, the posterior probability of being in the target interval need not increase with dose: it can rise as rates approach the interval and fall after rates tend to exceed it. Each dose’s probability is a separate summary of its posterior distribution. This calculation alone does not account for every clinical or design consideration involved in choosing a dose.

Common mistake: “The posterior mean rate is inside the target interval, so the probability of hitting the target must be high.” This conclusion does not follow: a mean gives no direct information about the fraction of posterior mass inside a particular interval. Calculate the fraction of posterior draws inside the interval, or integrate the dose-specific posterior distribution over that interval.

Key takeaway: For each candidate dose, transform the posterior uncertainty in the BLRM into draws of its DLT rate, count the draws inside the prespecified target interval, and report the resulting conditional posterior probability with appropriate simulation and model checks.

Check Your Understanding

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

For a target interval of 0.10 to 0.25, write the event on the linear-predictor scale.
Out of 400 posterior draws at a dose, 96 DLT rates lie inside the target interval. What is the estimated target-interval probability?
A posterior mean DLT rate is 0.24 for a target interval of 0.20 to 0.30. Does this establish that the posterior probability of being in the interval exceeds 0.5? Explain.
For 250 independent posterior draws, $\widehat q=0.50$. Calculate the estimated Monte Carlo standard error.

Solutions. 1. The event is $\operatorname{logit}(0.10)\leq a+b x_j\leq\operatorname{logit}(0.25)$, or approximately $-2.1972\leq a+b x_j\leq-1.0986$. 2. $96/400=0.240$, so the estimated probability is 24.0%. 3. No. The mean is only an average; it does not specify how much posterior mass lies between 0.20 and 0.30. Count posterior draws in the interval or integrate the posterior distribution over it. 4. $\sqrt{0.50(1-0.50)/250}=\sqrt{0.001}=0.03162$, or about 3.16 percentage points.