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Posterior and Predictive Probability Rules · Tutorial 158 of 696

Beta-Binomial Predictive Success for Binary Outcomes

You will learn to combine a beta posterior with future binomial outcomes, identify which future counts meet a final posterior go criterion, and sum their predictive probabilities.

Intermediate 12 min read

What You'll Learn

  • Specify the conditional response model and beta posterior that support beta-binomial prediction.
  • Derive the predictive probability of each possible number of future responses.
  • Translate a final posterior go threshold into the qualifying future response counts.
  • Calculate predictive success by summing the probabilities of qualifying counts.
  • Check calculations, assumptions, and interpretations for common errors.
  • Distinguish predictive success from current posterior evidence and plug-in calculations.

Why Use a Beta-Binomial Calculation?

At an interim analysis of a trial with a binary endpoint, the observed responses do not determine whether the trial will ultimately meet its go criterion. The remaining participants may have different numbers of responses, and each possible final count produces a different posterior distribution. Beta-binomial predictive success calculates the probability, given the interim data and model, that the final posterior decision will be “go.”

The calculation has two linked parts: first, determine which future response counts would satisfy the final posterior rule; second, calculate the predictive probability of those counts and add them. This tutorial specializes predictive averaging to a binary response rate with a beta posterior. It is not the current probability that the response rate exceeds a benchmark: it is the probability, conditional on the current data and model, that future data will make the final rule pass.

Model and Final Go Rule

Let \(p\) be the probability of a binary response for a participant in the population of interest. Suppose the prior distribution is \(p\sim\operatorname{Beta}(a_0,b_0)\), where \(a_0,b_0>0\). Conditional on \(p\), responses are independent Bernoulli outcomes with the same response probability \(p\). At the interim analysis, \(x\) responses have been observed among \(n\) participants. Conjugacy gives the interim posterior \(p\mid\text{interim data}\sim\operatorname{Beta}(a,b)\), where \(a=a_0+x\) and \(b=b_0+n-x\).

Suppose \(m\) responses remain to be observed. Let \(Y\) denote the number of responses among those remaining participants. Conditional on \(p\), \(Y\) has a binomial distribution. The final analysis updates the interim posterior using the future count \(Y=y\). If the final go rule compares the response rate with a benchmark \(p_B\), it may be written as: go if the posterior probability that \(p>p_B\) is at least \(\gamma\), where \(0<\gamma<1\) is fixed in advance.

Definition: Under this model, the predictive success probability at the interim analysis is the posterior predictive probability that the final data will meet the prespecified go rule. For future count \(y\), the final posterior is \(\operatorname{Beta}(a+y,b+m-y)\). Define the qualifying set \(G\) as the values \(y\in\{0,1,\ldots,m\}\) for which \(P(p>p_B\mid\text{interim data},Y=y)\geq\gamma\). Then predictive success is \(P(Y\in G\mid\text{interim data})\).

This definition assumes that the endpoint is recorded as a binary response, the remaining participants share the response probability \(p\) with those already observed, and responses are conditionally independent given \(p\). It also assumes the beta prior and the go rule are appropriate for the design question. The binomial model does not automatically account for clustering, changing response probabilities over time, informative missingness, or different response probabilities across treatment groups.

Deriving the Beta-Binomial Predictive Distribution

The predictive probability of a future count is obtained by averaging its binomial probability over the interim posterior for \(p\). The beta density is \(p^{a-1}(1-p)^{b-1}/B(a,b)\), where \(B(a,b)\) is the beta function. Therefore:

$$ \begin{aligned} P(Y=y\mid\text{interim data}) &=\int_0^1 P(Y=y\mid p)\,\pi(p\mid\text{interim data})\,dp\\ &=\binom{m}{y}\int_0^1 p^y(1-p)^{m-y} \frac{p^{a-1}(1-p)^{b-1}}{B(a,b)}\,dp\\ &=\binom{m}{y}\frac{B(a+y,b+m-y)}{B(a,b)}, \qquad y=0,\ldots,m. \end{aligned} $$

The last equality follows because the integrand contains \(p^{a+y-1}(1-p)^{b+m-y-1}\), whose integral from 0 to 1 is \(B(a+y,b+m-y)\). This is the beta-binomial predictive distribution. Its probabilities sum to one because they were obtained by integrating a binomial distribution that sums to one for every \(p\):

$$ \sum_{y=0}^{m}P(Y=y\mid\text{interim data}) =\int_0^1\left\{\sum_{y=0}^{m}P(Y=y\mid p)\right\} \pi(p\mid\text{interim data})\,dp =\int_0^1\pi(p\mid\text{interim data})\,dp=1. $$

For each possible \(y\), the final posterior tail probability is the beta-distribution tail \(P(p>p_B\mid Y=y,\text{interim data})\). Calculate that tail, compare it with \(\gamma\), and include the predictive probability for \(y\) in the sum only if it passes. This enumeration is exact up to numerical evaluation of beta functions and beta tail probabilities.

1
Update the interim posterior.
Record its beta parameters \(a,b\), after incorporating the interim responses.
2
Evaluate each possible final count.
For \(y=0,\ldots,m\), form \(\operatorname{Beta}(a+y,b+m-y)\) and check whether its benchmark tail reaches \(\gamma\).
3
Weight qualifying counts.
Calculate each count’s beta-binomial predictive probability and add the probabilities for all qualifying counts.

Worked Calculations

Worked Example: Correctly Identifying Both Qualifying Counts

Suppose a hypothetical trial has interim posterior \(p\mid\text{data}\sim\operatorname{Beta}(2,2)\), with two participants still to respond. The final rule is go if \(P(p>0.5\mid\text{final data})\geq0.65\). Find predictive success.

For \(y=0,1,2\), the final posteriors are respectively \(\operatorname{Beta}(2,4)\), \(\operatorname{Beta}(3,3)\), and \(\operatorname{Beta}(4,2)\). Their probabilities above 0.5 are \(0.1875\), \(0.5000\), and \(0.8125\). For example, the \(\operatorname{Beta}(4,2)\) cumulative probability at 0.5 is \(5(0.5)^4-4(0.5)^5=0.1875\), so its upper-tail probability is \(1-0.1875=0.8125\). Thus only \(y=2\) meets the 0.65 rule? No: the final posterior for \(y=1\) has tail \(0.5000\), below 0.65. The qualifying count is \(y=2\) alone.

The predictive probabilities under \(\operatorname{Beta}(2,2)\), for two future responses, are:

$$ \begin{aligned} P(Y=0)&=\frac{B(2,4)}{B(2,2)} =\frac{1/20}{1/6}=0.3000,\\ P(Y=1)&=2\frac{B(3,3)}{B(2,2)} =2\frac{1/30}{1/6}=0.4000,\\ P(Y=2)&=\frac{B(4,2)}{B(2,2)} =\frac{1/20}{1/6}=0.3000. \end{aligned} $$

Only \(Y=2\) qualifies, so predictive success is \(P(Y=2)=0.3000\). This example illustrates why the posterior tail and predictive weight must both be checked for every count; a count can be likely without satisfying the final criterion.

Worked Example: A Higher Posterior Threshold

Suppose the interim posterior is \(\operatorname{Beta}(3,2)\), two participants remain, and go requires \(P(p>0.5\mid\text{final data})\geq0.80\). The possible final posteriors are \(\operatorname{Beta}(3,4)\), \(\operatorname{Beta}(4,3)\), and \(\operatorname{Beta}(5,2)\) for \(y=0,1,2\). Their tails above 0.5 are \(0.34375\), \(0.65625\), and \(0.890625\). Therefore, only \(y=2\) meets the final rule.

The predictive mass for two future responses is:

$$ P(Y=2\mid\text{data}) =\frac{B(5,2)}{B(3,2)} =\frac{1/30}{1/12} =0.4000. $$

Predictive success is \(0.4000\). The predictive distribution is properly normalized: the masses for \(y=0,1,2\) are \(0.2000,0.4000,0.4000\), which sum to 1. This check is useful when implementing the formula.

Worked Example: One Remaining Participant and a Non-Midpoint Benchmark

Suppose the interim posterior is \(\operatorname{Beta}(2,3)\), one participant remains, the benchmark is \(p_B=0.4\), and the final go threshold is \(\gamma=0.65\). If the future participant does not respond, the final posterior is \(\operatorname{Beta}(2,4)\). Its tail above 0.4 is \((1-0.4)^4(1+4(0.4))=0.33696\), below the threshold. If the participant responds, the final posterior is \(\operatorname{Beta}(3,3)\). Its cumulative probability at 0.4 is \(10(0.4)^3-15(0.4)^4+6(0.4)^5=0.31744\), so its upper tail is \(1-0.31744=0.68256\), above the threshold.

Thus the only qualifying future count is \(y=1\). Its predictive probability is:

$$ P(Y=1\mid\text{data}) =\binom{1}{1}\frac{B(3,3)}{B(2,3)} =\frac{1/30}{1/12} =0.4000. $$

Predictive success is \(0.4000\). The calculation also illustrates that the benchmark need not be 0.5: use the benchmark in the final posterior tail, while the beta-binomial predictive mass is determined by the interim posterior and the number remaining.

Interpretation, Checks, and Failure Modes

A predictive success probability of 0.3000 means that, under the specified model and conditional on the interim data, the probability that future responses will lead the final posterior rule to say go is 30%. It does not mean that the current posterior probability of benefit is 30%, nor that the probability the response rate exceeds the benchmark is 30%. Those are different probabilities about different events.

A reliable calculation keeps three objects separate: the final posterior tail for each hypothetical future count, the predictive probability of that count, and the final sum over counts that pass. Common errors include reversing a beta tail, using the interim posterior tail as if it were predictive success, and including counts that have substantial predictive probability but fail the posterior threshold. In the first worked example, \(P(p>0.5)\) for \(\operatorname{Beta}(4,2)\) is 0.8125, not 0.6875; and the count \(y=1\) has predictive mass 0.4000 but does not meet the 0.65 final threshold. Therefore predictive success there is 0.3000, not 0.7000.

Useful diagnostics include verifying that the predictive masses are nonnegative and sum to one, confirming that the final posterior parameters add the future successes to \(a\) and future failures to \(b\), and checking that the qualifying set behaves sensibly as the go threshold changes. Raising \(\gamma\) cannot add a qualifying count. As another check, if no future count passes, predictive success is zero; if every possible count passes, it is one.

Exact enumeration is particularly convenient because the future count has only \(m+1\) possible values. A plug-in calculation that fixes \(p\) at a point estimate omits uncertainty about \(p\) and generally gives a different predictive distribution. Simulation can approximate the same posterior predictive quantity in more complex settings, but for this beta-binomial case enumeration avoids simulation error. Neither exact calculation nor simulation protects against a misspecified model: dependence among participants, substantial heterogeneity in response probabilities, or changes in response rates over time may make the common-\(p\) binomial assumption inappropriate. The result is also conditional on the chosen prior, benchmark, and posterior threshold; it does not by itself establish clinical value or determine whether the decision rule is well calibrated for a broader operating objective.

Key takeaway: First determine which hypothetical final response counts meet the posterior go rule. Then sum their beta-binomial predictive probabilities. The answer is a probability of a future decision conditional on the interim data and model, not a posterior probability about the response rate itself.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. An interim posterior is \(\operatorname{Beta}(1,1)\), and two responses remain. State the posterior after \(Y=0\), \(Y=1\), and \(Y=2\).
  2. Why does the beta-binomial predictive distribution account for more uncertainty than a binomial distribution evaluated only at the posterior mean?
  3. If the qualifying future counts are \(y=1\) and \(y=2\), and their predictive probabilities are 0.25 and 0.35, what is predictive success?
  4. For a fixed benchmark and future count, what happens to the qualifying set if the required posterior threshold \(\gamma\) is increased?

Solutions. (1) The three posteriors are \(\operatorname{Beta}(1,3)\), \(\operatorname{Beta}(2,2)\), and \(\operatorname{Beta}(3,1)\), because each future response adds one to the first beta parameter and each nonresponse adds one to the second. (2) The beta-binomial averages the binomial distribution over the entire interim posterior for \(p\), rather than fixing \(p\) at one value; this includes uncertainty in \(p\). (3) Predictive success is \(0.25+0.35=0.60\), because those and only those counts meet the final rule. (4) The qualifying set can only stay the same or shrink: each count must meet a more demanding posterior-tail threshold, so no newly qualifying count can be added.

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