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Prior Predictive Checks · Tutorial 84 of 696

Rare Safety Events And Zero-Event Trials

Translate rare-event priors into prior predictive counts and zero-event probabilities, and assess whether those predictions are plausible for the planned trial.

Intermediate 11 min read

What You'll Learn

  • Calculate the chance of no observed adverse events under a fixed event probability and under a prior distribution.
  • Derive the beta-binomial predictive probability of zero events.
  • Compare priors with the same mean but different uncertainty.
  • Relate planned sample size to the prior predictive chance of observing at least one event.
  • Identify model assumptions and practical reasons zero-event predictions may be unreliable.

Why Zero-Event Predictions Need Attention

For a rare adverse event, a trial can observe no events even when the event probability is not zero. That makes a predicted zero count clinically important: it helps reveal whether a proposed prior and planned sample size jointly imply that “no events observed” should be routine or surprising. This check does not establish that a safety risk is absent. It describes what the prior model predicts before trial outcomes are observed.

The calculation depends on both the event probability and the number of participants observed. A fixed-probability calculation gives the chance of zero events at one specified probability. A Bayesian prior predictive calculation averages that chance over the prior distribution for the probability. These are not generally the same calculation, even if the fixed probability is set equal to the prior mean.

Definition: Let \(p\) be the probability that one participant experiences a specified adverse event during a defined observation window, and let \(K\) be the number of participants with that event among \(n\) participants. A zero-event trial, for this endpoint and window, has \(K=0\). Its prior predictive probability is \(P(K=0)\), calculated by averaging over the prior distribution for \(p\) as well as the participant outcomes conditional on \(p\).

Model and Assumptions

For a simple calculation, suppose each participant contributes one binary outcome: the specified event either occurs or does not occur during the defined window. Conditional on a common event probability \(p\), assume the participant outcomes are independent and have the same probability \(p\). Then \(K\mid p\) has a binomial distribution. To represent uncertainty about the event probability before seeing trial data, use a beta prior with positive shape parameters \(a\) and \(b\).

$$ K\mid p \sim \operatorname{Binomial}(n,p), \qquad p\sim \operatorname{Beta}(a,b). $$

This model requires more than a small event rate. The endpoint and observation window must be defined consistently for every participant; event ascertainment must be sufficiently reliable; and the common-probability, conditional-independence assumption must be reasonable for the intended calculation. For example, this setup treats each participant as contributing one event indicator, not a recurrent-event count. Differences in follow-up or substantial dependence between participants may require a different model. The calculation is a prior predictive check under stated assumptions, not a guarantee about the trial.

The beta prior has mean \(a/(a+b)\). Conditional on a particular \(p\), the probability of no events is \((1-p)^n\). Integrating this conditional probability over the prior gives the prior predictive probability of zero events:

$$ P(K=0) = \int_0^1 P(K=0\mid p)\,\pi(p)\,dp = \int_0^1 (1-p)^n \frac{p^{a-1}(1-p)^{b-1}}{B(a,b)}\,dp = \frac{B(a,b+n)}{B(a,b)}. $$

Here \(B(\cdot,\cdot)\) is the beta function, and \(\pi(p)\) is the beta prior density. More generally, integrating the binomial probability for any count \(k\) gives the beta-binomial predictive distribution:

$$ P(K=k) = {n\choose k}\frac{B(a+k,b+n-k)}{B(a,b)}, \qquad k=0,1,\ldots,n. $$

The predictive mean count is \(E[K]=n\,a/(a+b)\), but a mean does not determine how much probability lies at zero. The probability of at least one event is \(P(K\geq 1)=1-P(K=0)\). Both quantities are useful: the mean describes the average count over prior predictive repetitions, while the zero-event probability directly describes how often the model predicts no observed events.

Worked Prior Predictive Calculations

Worked Example: A Rare Event in a 500-Participant Trial

Hypothetical setup. Suppose a trial plans to observe 500 participants for a specified safety endpoint. The elicited prior is \(p\sim\operatorname{Beta}(1,1999)\), with one event indicator per participant and the conditional independence and common-probability assumptions described above. The prior mean event probability is \(1/2000=0.0005\), or 0.05%.

The expected number of participants with the event is \(E[K]=500(0.0005)=0.25\) participants. For a \(\operatorname{Beta}(1,b)\) prior, \(B(1,b)=1/b\), so:

$$ P(K=0) = \frac{B(1,1999+500)}{B(1,1999)} = \frac{1/2499}{1/1999} = \frac{1999}{2499} \approx 0.7999. $$

Thus the prior predictive probability of zero participants with the event is about 80.0%, and the probability of at least one is about \(1-0.7999=0.2001\), or 20.0%. Under this prior, a zero-event result in a trial of this size is expected to be common. It is not, by itself, evidence that the event probability is zero.

For comparison, inserting the prior mean into the fixed-probability formula gives \((1-0.0005)^{500}\approx 0.7787\). That is a calculation at \(p=0.0005\), not the prior predictive probability. The distinction matters because the prior predictive calculation averages across possible values of \(p\).

Worked Example: Same Prior Mean, Different Uncertainty

Hypothetical setup. Consider the same 500-participant trial and compare the prior above with \(p\sim\operatorname{Beta}(2,3998)\). The second prior also has mean \(2/4000=0.0005\), so both priors imply the same expected event count, 0.25 participants.

For a \(\operatorname{Beta}(2,b)\) prior, \(B(2,b)=1/[b(b+1)]\). Therefore, under the second prior:

$$ P(K=0) = \frac{B(2,4498)}{B(2,3998)} = \frac{3998(3999)}{4498(4499)} \approx 0.7901. $$

This prior predicts a zero-event trial with probability about 79.0%, compared with about 80.0% under \(\operatorname{Beta}(1,1999)\). The two probabilities are similar here, but they are not identical: equal prior means and equal expected counts do not make the full prior predictive distributions equal. The shape of the prior also matters. A safety review should therefore examine the predicted zero-event probability directly rather than infer it from the mean alone.

How Prior Uncertainty Affects the Zero Probability

For \(n\geq 2\), the function \(f(p)=(1-p)^n\) is convex on \(0\leq p\leq 1\), since \(f''(p)=n(n-1)(1-p)^{n-2}\geq 0\). Jensen’s inequality therefore gives:

$$ E[(1-p)^n] \geq (1-E[p])^n. $$

The left side is the prior predictive probability of zero events; the right side is the fixed-probability calculation at the prior mean. Thus, for this model and \(n\geq2\), substituting the prior mean generally understates the prior predictive chance of a zero count. The inequality is strict when the prior has nonzero uncertainty and \(n\geq2\). When \(n=1\), the probability of zero is linear in \(p\), so the two calculations are equal.

This result helps explain why uncertainty in a rare-event probability can make zero-event trials more probable than a calculation at the average probability suggests. It does not say that every prior with greater variance has a larger zero probability than every other prior: the full distribution matters, and comparisons should use the actual prior predictive calculation.

Worked Example: Choosing a Sample Size for a Predictive Check

Hypothetical setup. Retain the prior \(p\sim\operatorname{Beta}(1,1999)\), and ask how many participants would be needed for the prior predictive probability of observing at least one event to reach 90%. This is a check on what the prior and sample size jointly predict; it is not a guarantee that a trial will detect an event.

For \(n\) participants, the zero-event probability is \(1999/(1999+n)\). Requiring a probability of at least 0.90 for one or more events means:

$$ 1-\frac{1999}{1999+n}\geq 0.90 \quad\Longleftrightarrow\quad \frac{1999}{1999+n}\leq 0.10 \quad\Longleftrightarrow\quad n\geq 17991. $$

At \(n=17991\), the zero-event probability is \(1999/19990=0.1000\), so the probability of one or more events is 0.9000. The expected count is \(17991(0.0005)=8.996\) participants. A 500-participant trial, by contrast, has only about a 20.0% prior predictive chance of observing at least one event under this prior. Whether either sample size is clinically or operationally appropriate requires considerations beyond this calculation.

Diagnostics, Alternatives, and Limitations

A prior predictive check is most useful when it is tied to a specific safety question. Before interpreting a zero probability, verify the endpoint definition, denominator, follow-up window, and event ascertainment. Then check whether the prior predictive probability of zero events, and the expected count, are plausible to the clinical team for that trial size. If simulations are used, draw \(p\) from the prior and then draw \(K\mid p\); fixing \(p\) at its mean skips the uncertainty the prior predictive check is meant to include.

The binomial calculation is a useful baseline, not a universal safety-event model. It may be unsuitable if participants have materially different exposure times, event risks vary in ways not represented by the shared \(p\), outcomes are dependent, or participants can have multiple events and the target is a recurrent-event count. In those settings, the probability of zero events should be calculated from a model that represents the relevant follow-up and dependence structure. A zero count can also arise from limited follow-up or incomplete ascertainment, neither of which is resolved by choosing a different prior for \(p\).

A fixed-\(p\) calculation is simpler and appropriate when the question is explicitly conditional on one specified event probability. A Poisson approximation may be convenient for a binomial count when \(p\) is small and \(n\) is sufficiently large, with mean \(np\); at fixed \(p\), it approximates the zero probability by \(\exp(-np)\). It is an approximation, and substituting a prior mean into it still does not integrate over prior uncertainty. The beta-binomial expression gives the exact prior predictive calculation for the beta-binomial model. Simulation can check the calculation or handle extensions, but the simulation must reproduce the intended model and prior.

A common mistake is to say, “The prior mean implies 0.25 events, so a zero-event trial is unlikely,” or to calculate zero probability only at the prior mean and call it prior predictive. An expected count of 0.25 does not mean that 0.25 events will occur in a trial, and the plug-in calculation does not average across plausible probabilities. Calculate \(P(K=0)\) under the full prior predictive distribution, then interpret it in the context of the event definition, sample size, and assumptions.

Key takeaway: For \(K\mid p\sim\operatorname{Binomial}(n,p)\) and \(p\sim\operatorname{Beta}(a,b)\), the prior predictive probability of a zero-event trial is \(B(a,b+n)/B(a,b)\). Examine that probability alongside the expected count, and treat both as implications of the specified model—not as proof that an event is absent or present.

Check Your Understanding

Use the beta-binomial model where a question specifies it. Each problem is followed by a solution.

  1. Suppose \(p\sim\operatorname{Beta}(1,999)\) and \(n=100\). What is the prior predictive probability of zero events? Solution: For a \(\operatorname{Beta}(1,b)\) prior, \(P(K=0)=b/(b+n)\). Thus \(P(K=0)=999/1099\approx0.9090\), or 90.90%.
  2. For the prior and sample size in question 1, what is the probability of at least one event and the expected event count? Solution: \(P(K\geq1)=1-0.9090=0.0910\), or about 9.10%. The prior mean is \(1/1000=0.001\), so \(E[K]=100(0.001)=0.1000\) participants.
  3. Why is \((1-E[p])^n\) generally not the prior predictive zero-event probability when \(n\geq2\)? Solution: The prior predictive probability is \(E[(1-p)^n]\), which averages the conditional zero probability over \(p\). Since \((1-p)^n\) is convex for \(n\geq2\), Jensen’s inequality gives \(E[(1-p)^n]\geq(1-E[p])^n\).
  4. Name one situation in which the binomial model may not be appropriate for a safety-event count. Solution: If participants have substantially different follow-up times, a common event probability for one shared window may not represent their risks. A model accounting for exposure or follow-up differences would be needed.
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