Tutorials › Bayesian Clinical Trials › Predictive Success When Control Response Is Uncertain

Posterior and Predictive Probability Rules · Tutorial 171 of 696

Predictive Success When Control Response Is Uncertain

Learn how uncertainty in the control response rate changes the predictive distribution of future outcomes and the probability that a prespecified final success rule will be met.

Intermediate 13 min read

What You'll Learn

  • Set up posterior and future-data models for binary treatment and control responses.
  • Derive the beta-binomial predictive distribution by integrating over an uncertain response rate.
  • Calculate predictive success using a prespecified final decision rule.
  • Compare full posterior integration with a plug-in control-rate approximation.
  • Check predictive distributions, identify modeling failures, and handle a zero remaining control sample.

Why Control-Rate Uncertainty Matters

In a binary treatment-control trial, predictive success is the probability, given the data observed so far, that the final analysis will meet a prespecified success rule. The future control response rate is often not known exactly. Treating it as a fixed number, such as its current posterior mean, can understate the variation in future control outcomes and can misrepresent how those outcomes will update the final comparison.

The key is to average over the current posterior distribution of the control response rate as well as over future control outcomes. This is not the same as setting the rate equal to its posterior mean and simulating a binomial count. The difference is especially important when the posterior is diffuse or when several control participants remain.

Definition: Let \(p_T\) and \(p_C\) be the treatment and control response probabilities. Given current data, predictive success is the posterior predictive probability that the final data satisfy a specified success rule. It averages over both the unknown response probabilities and the outcomes of the remaining participants.

Model and Predictive-Success Calculation

Suppose the treatment arm has observed \(x_T\) responses among \(n_T\) participants, and the control arm has observed \(x_C\) responses among \(n_C\) participants. Use independent beta priors, \(p_T\sim\operatorname{Beta}(a_{T0},b_{T0})\) and \(p_C\sim\operatorname{Beta}(a_{C0},b_{C0})\). Conditional on each response probability, outcomes are Bernoulli and independent within and between arms. These assumptions give independent current posteriors: \(p_j\mid D\sim\operatorname{Beta}(a_j,b_j)\), where \(a_j=a_{j0}+x_j\) and \(b_j=b_{j0}+n_j-x_j\), for \(j=T,C\). Here \(D\) denotes the current data.

Let \(m_T\) and \(m_C\) be the numbers of remaining participants. Conditional on the response probabilities, future response counts \(Y_T\) and \(Y_C\) follow independent binomial distributions. A final decision rule might declare success when the final posterior probability of superiority reaches a threshold \(\gamma\). For each possible pair of future counts, the final posteriors are \(\operatorname{Beta}(a_T+Y_T,b_T+m_T-Y_T)\) and \(\operatorname{Beta}(a_C+Y_C,b_C+m_C-Y_C)\).

Predictive-success formula: For the rule that declares success when the final posterior probability of superiority is at least \(\gamma\), define \(G(y_T,y_C)=1\) if \(P(p_T>p_C\mid D,Y_T=y_T,Y_C=y_C)\geq\gamma\), and \(G(y_T,y_C)=0\) otherwise. Then predictive success is the sum of \(G(y_T,y_C)\) times the joint posterior predictive probability of those future counts. Under independent arm posteriors, that joint probability is the product of their beta-binomial probabilities.

For a current \(\operatorname{Beta}(a,b)\) posterior and \(m\) remaining participants, integrating the binomial likelihood over the posterior gives

$$ P(Y=y\mid D) =\binom{m}{y}\frac{B(a+y,b+m-y)}{B(a,b)},\qquad y=0,\ldots,m. $$

Here \(B(\cdot,\cdot)\) is the beta function. This beta-binomial distribution is the predictive distribution for the remaining response count. Substituting the arm-specific beta-binomial probabilities into the sum gives an exact finite calculation of predictive success. The final success rule can also be another prespecified rule; the predictive averaging is the same, but the indicator \(G\) must represent that rule exactly.

Why the Beta-Binomial Is More Variable Than a Plug-In Binomial

Write \(s=a+b\) and \(\mu=a/(a+b)\) for the mean of the current beta posterior. Conditional on \(p\), \(Y\sim\operatorname{Binomial}(m,p)\). The law of total variance gives

$$ \begin{aligned} E(Y\mid D)&=m\mu,\\ \operatorname{Var}(Y\mid D) &=E\{\operatorname{Var}(Y\mid p,D)\mid D\} +\operatorname{Var}\{E(Y\mid p,D)\mid D\}\\ &=mE\{p(1-p)\mid D\}+m^2\operatorname{Var}(p\mid D)\\ &=m\mu(1-\mu)\left(\frac{s+m}{s+1}\right). \end{aligned} $$

For the last equality, a beta distribution has variance \(\mu(1-\mu)/(s+1)\), so \(E\{p(1-p)\}=\mu(1-\mu)-\operatorname{Var}(p)\). Compared with the plug-in binomial variance \(m\mu(1-\mu)\), the beta-binomial variance is larger when \(m\geq 2\) and the posterior has nonzero variance. For \(m=1\), the count is Bernoulli with probability \(\mu\), so the two predictive distributions agree. If \(m=0\), the future count is identically zero and its variance is zero; the variance ratio is not defined. In particular, the factor \((s+m)/(s+1)\) is at least one when \(m\geq1\), but it is less than one when \(m=0\), when it is not a variance multiplier for a nonzero binomial variance.

Worked Example: Exact Prediction With One Remaining Participant per Arm

Suppose the current posteriors are \(p_T\mid D\sim\operatorname{Beta}(1,1)\) and \(p_C\mid D\sim\operatorname{Beta}(1,1)\), with one participant remaining in each arm. Use the final rule that declares success if \(P(p_T>p_C\mid\text{final data})\geq0.80\). Each future response count is zero or one, and each has predictive probability \(0.5\); independence makes each of the four pairs have probability \(0.5\times0.5=0.25\).

If \(Y_T=1\) and \(Y_C=0\), the final posteriors are \(\operatorname{Beta}(2,1)\) and \(\operatorname{Beta}(1,2)\). Their densities are \(2x\) and \(2(1-y)\) on \([0,1]\), respectively. Thus

$$ P(p_T>p_C\mid D,Y_T=1,Y_C=0) =\int_0^1 2x\{1-(1-x)^2\}\,dx =\frac{5}{6}\approx0.8333. $$

This pair meets the 0.80 rule. Reversing the responses gives a probability of superiority of \(1/6\), which does not meet the rule. If both participants respond or neither responds, both final posteriors are identical, so the probability of superiority is \(0.5\), also below the threshold. Therefore only one of the four future-outcome pairs is a success, and predictive success is \(0.25\). This calculation integrates over the uncertain response rates when evaluating both the future data and the final posterior decision.

Worked Calculations With a Posterior-Mean Rule

The examples below use a different, deliberately simple final rule: declare success when the final posterior mean of \(p_T\) exceeds the final posterior mean of \(p_C\). This is a specified decision rule, not a synonym for requiring a particular posterior probability of superiority. It makes the predictive averaging transparent. In an actual trial analysis, use the rule that was selected for that trial.

Worked Example: Integrating an Uncertain Control Rate

Suppose the current posteriors are \(p_T\mid D\sim\operatorname{Beta}(4,2)\) and \(p_C\mid D\sim\operatorname{Beta}(3,3)\), with one remaining treatment participant and two remaining control participants. The current posterior means are \(4/6=0.6667\) and \(3/6=0.5000\). The treatment future response probability is \(2/3\), so \(P(Y_T=0\mid D)=1/3\) and \(P(Y_T=1\mid D)=2/3\).

For control, the beta-binomial probabilities are

$$ \begin{aligned} P(Y_C=0\mid D)&=\frac{3\cdot4}{6\cdot7}=\frac{2}{7},\\ P(Y_C=1\mid D)&=\frac{2\cdot3\cdot3}{6\cdot7}=\frac{3}{7},\\ P(Y_C=2\mid D)&=\frac{3\cdot4}{6\cdot7}=\frac{2}{7}. \end{aligned} $$

After the future data, the treatment posterior mean is \(4/7\) if \(Y_T=0\), and \(5/7\) if \(Y_T=1\). The control posterior mean is \(3/8\), \(4/8\), or \(5/8\) when \(Y_C\) is zero, one, or two. If \(Y_T=1\), the treatment posterior mean exceeds the control mean for all three control counts. If \(Y_T=0\), it exceeds the control mean for \(Y_C=0\) and \(Y_C=1\), but not for \(Y_C=2\), where both means are \(5/8\) and the rule requires a strict excess.

By independence of the predictive counts, predictive success is therefore

$$ \begin{aligned} P(\text{success}\mid D) &=P(Y_T=1\mid D) +P(Y_T=0\mid D)P(Y_C\leq1\mid D)\\ &=\frac{2}{3}+\frac{1}{3}\left(\frac{2}{7}+\frac{3}{7}\right) =\frac{19}{21}\approx0.9048. \end{aligned} $$

If instead the control rate were fixed at its current posterior mean of \(0.5\), a binomial model would give control-count probabilities \(0.25,0.50,0.25\). The same rule would then yield \(2/3+(1/3)(0.75)=11/12\approx0.9167\). The plug-in calculation is not the posterior predictive calculation: it does not average over uncertainty in \(p_C\), and it gives a different probability of success.

Worked Example: A Zero Remaining Control Sample

Suppose the current treatment posterior mean is \(0.40\), and the current control posterior is \(\operatorname{Beta}(2,5)\), with three control participants remaining and none remaining in treatment. Use the posterior-mean rule from the previous example. The control predictive probabilities for \(Y_C=0,1,2,3\) are

$$ \left( \frac{5\cdot6\cdot7}{7\cdot8\cdot9}, \frac{3\cdot2\cdot5\cdot6}{7\cdot8\cdot9}, \frac{3\cdot2\cdot3\cdot5}{7\cdot8\cdot9}, \frac{2\cdot3\cdot4}{7\cdot8\cdot9} \right) = \left(\frac{5}{12},\frac{5}{14},\frac{5}{28},\frac{1}{21}\right). $$

These probabilities sum to one. With \(Y_C=k\), the final control posterior mean is \((2+k)/10\). It is below the fixed treatment mean \(0.40\) for \(k=0\) or \(k=1\), equal to it for \(k=2\), and above it for \(k=3\). Thus

$$ P(\text{success}\mid D) =P(Y_C\leq1\mid D) =\frac{5}{12}+\frac{5}{14} =\frac{65}{84}\approx0.7738. $$

If no control participants remained instead, \(Y_C=0\) with probability one and \(\operatorname{Var}(Y_C\mid D)=0\). The uncertainty in \(p_C\) would still be present in the current posterior and in any final posterior-probability criterion, but it would not create variation in a future control count when there are no future control outcomes.

Practical Checks, Alternatives, and Limitations

For modest remaining sample sizes, the beta-binomial probabilities can be enumerated and the final rule evaluated at every possible pair of counts. For larger problems, simulation draws response probabilities from their current posteriors, draws future counts conditional on those probabilities, applies the final analysis, and estimates predictive success by the fraction of simulated trials meeting the rule. Simulation approximates the same integral; it does not justify replacing an uncertain response rate by a fixed value.

Several checks help reveal errors or model mismatch:

  • For each arm, verify that the predictive probabilities over \(y=0,\ldots,m\) sum to one and that the predictive mean is \(m a/(a+b)\).
  • When \(m=1\), confirm that the beta-binomial prediction reduces to a Bernoulli distribution with response probability \(a/(a+b)\). When \(m=0\), the count is zero with probability one.
  • Check that the final success indicator uses the stated rule, including whether its inequality is strict and which threshold applies.
  • Compare predictions across plausible priors or posterior concentrations. A diffuse posterior can produce a substantially more variable future count than a concentrated posterior with the same mean.
  • Check whether the assumed arm independence is defensible. If a shared parameter or hierarchical model induces dependence between \(p_T\) and \(p_C\), the joint predictive distribution does not factor into two beta-binomial probabilities; integrate over the joint posterior instead.

A plug-in binomial calculation is a convenient approximation, but it discards posterior uncertainty and may change the probability of crossing a final decision boundary. Direct enumeration is exact under the stated model, while simulation is flexible for complicated rules and models but has Monte Carlo error. Both approaches still depend on the likelihood, prior, and assumptions being reasonable. Neither predictive success nor a high probability of meeting the rule guarantees that the treatment effect is clinically important, that the model is correct, or that the trial procedure has acceptable performance across repeated trials.

Common mistake: “The posterior mean control response rate is 0.5, so the future control count is binomial with response probability 0.5.” That calculation conditions on a fixed rate. The posterior predictive distribution instead averages the binomial likelihood over the posterior for \(p_C\), giving a beta-binomial count when the posterior is beta. The two distributions have the same mean, but can have different variances and different probabilities of meeting the final success rule.

Check Your Understanding

Answer each question, then compare with the solutions.

  1. Under a beta posterior with parameters \(a=3,b=5\), what is the predictive mean of the number of responses among \(m=4\) remaining participants?
  2. Why can the beta-binomial predictive variance exceed the variance from a binomial model fixed at the posterior mean?
  3. If no control participants remain, what is the predictive distribution of the future control response count?
  4. In the second worked example, why does the posterior-mean decision rule not equal the rule based on \(P(p_T>p_C\mid\text{final data})\geq\gamma\)?

Solutions. (1) The posterior mean response rate is \(3/(3+5)=3/8\), so the predictive mean is \(4(3/8)=1.5\) responses. (2) Multiple future outcomes share the same unknown \(p\); integrating over its posterior adds variation, specifically the term \(m^2\operatorname{Var}(p\mid D)\) in the law of total variance. (3) It is zero with probability one, so its variance is zero, even though uncertainty about the underlying response probability remains. (4) Comparing posterior means checks whether the posterior expected difference is positive. A posterior-probability rule checks how much posterior probability lies above zero and requires that probability to exceed a threshold; these are different summaries and need not give the same decision.

Clinical Biostats