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Sample Size, Power and Assurance · Tutorial 246 of 696

Hybrid Bayesian Success Criteria and Frequentist Constraints

Build and evaluate a hybrid trial design by translating its Bayesian success rule into a decision threshold, then checking frequentist error-rate and power constraints.

Advanced 11 min read

What You'll Learn

  • Translate a posterior-probability success rule into an observable-data cutoff.
  • Calculate type I error and power under a known-variance normal model.
  • Check Bayesian and frequentist criteria jointly rather than assuming one guarantees the other.
  • Recognize how the analysis prior can affect frequentist operating characteristics.
  • Identify when a proposed hybrid design is infeasible and how changing the rule can help.

Why Combine Bayesian and Frequentist Criteria?

A trial may define success using a posterior probability, such as requiring at least a 97.5% posterior probability that the treatment effect is positive. A sponsor or regulator may also require that the design control the probability of a false positive at a prespecified level and attain adequate power at a clinically relevant effect. These requirements answer different questions: the posterior rule describes how the trial decision is made from the observed data, while frequentist constraints describe how often that rule would succeed across repeated trials under specified true effects.

A hybrid design must satisfy both requirements for the same decision rule. It is not enough to quote the posterior threshold, or to choose a sample size that meets a power target under a different test. The analysis model, prior, success rule, sample size, and frequentist operating characteristics must all be specified consistently.

Model and Decision Rule

Consider a hypothetical, balanced, two-arm trial with a continuous endpoint. Let \(\theta\) be the treatment-minus-control difference in population means, measured in endpoint units. Assume independent normally distributed outcomes in the two arms with a common, known standard deviation \(\sigma\). With \(n\) participants per arm, the observed difference in sample means, denoted \(\bar{D}\), has sampling distribution

$$ \bar{D}\mid\theta\sim N(\theta,s_D^2),\qquad s_D^2=\frac{2\sigma^2}{n}. $$

For the Bayesian analysis, use the proper prior \(\theta\sim N(0,\tau^2)\), independent of the trial data. This prior is centered at no treatment difference and has variance \(\tau^2\). Under this model, the posterior distribution is normal. Define \(w_n=\tau^2/(\tau^2+s_D^2)\). Then

$$ \theta\mid\bar{D}\sim N\left(w_n\bar{D},\,w_ns_D^2\right). $$

Suppose the prespecified Bayesian success rule is \(P(\theta>0\mid\bar{D})>\gamma\), where \(\gamma\) is the chosen posterior-probability threshold. Write \(z_\gamma\) for the \(\gamma\) quantile of the standard normal distribution. Since the posterior probability is \(\Phi\bigl(w_n\bar{D}/\sqrt{w_ns_D^2}\bigr)\), the rule is equivalent to

$$ \bar{D}>c_n,\qquad c_n=z_\gamma\frac{s_D}{\sqrt{w_n}}. $$

Thus, under this particular normal model, the Bayesian rule is an observable-data cutoff. The cutoff depends on the sample size and on the analysis prior through \(w_n\). The formula assumes the stated endpoint model, known variance, proper normal prior, and a success criterion based on a positive effect. If success instead requires exceeding a clinically important margin, the posterior probability and cutoff must be defined relative to that margin.

Definition: A hybrid design uses a prespecified Bayesian decision rule and requires that its frequentist operating characteristics also meet prespecified constraints. For example, it may require a posterior probability above \(\gamma\), a type I error no greater than \(\alpha^*\), and power of at least \(1-\beta^*\) at a specified treatment effect.

Deriving the Frequentist Constraints

For the decision rule \(\bar{D}>c_n\), the probability of declaring success when the true effect is \(\theta\) is

$$ P_\theta(\bar{D}>c_n) =1-\Phi\left(\frac{c_n-\theta}{s_D}\right) =\Phi\left(\frac{\theta}{s_D}-\frac{z_\gamma}{\sqrt{w_n}}\right). $$

This is the operating-characteristic function for this design: it gives the success probability at each specified true effect. For the one-sided null \(H_0:\theta\leq 0\), the probability of success is largest at the boundary \(\theta=0\), because the displayed expression increases as \(\theta\) increases. The type I error is therefore

$$ \alpha_n=\sup_{\theta\leq0}P_\theta(\bar{D}>c_n) =1-\Phi\left(\frac{z_\gamma}{\sqrt{w_n}}\right). $$

At a prespecified alternative \(\theta=\delta>0\), the power is

$$ \operatorname{Power}_n(\delta) =1-\Phi\left(\frac{c_n-\delta}{s_D}\right) =\Phi\left(\frac{\delta}{s_D}-\frac{z_\gamma}{\sqrt{w_n}}\right). $$

A design with target type I error \(\alpha^*\) and target power \(1-\beta^*\) must satisfy \(\alpha_n\leq\alpha^*\) and \(\operatorname{Power}_n(\delta)\geq1-\beta^*\) for the same \(n\), prior, endpoint model, and success rule. If there are nuisance parameters, the required error-rate or power criterion must also state how they are handled, for example by taking a worst case over a specified range.

Worked Examples

Worked Example: Applying the Posterior Success Rule

Suppose a hypothetical balanced trial has \(n=25\) participants per arm, known outcome standard deviation \(\sigma=10\) points, and analysis prior \(\theta\sim N(0,20^2)\) points. The observed treatment-minus-control difference is \(\bar{D}=6.5\) points. The success rule requires \(P(\theta>0\mid\bar{D})>0.975\).

First, the sampling variance and weight are

$$ s_D^2=\frac{2(10^2)}{25}=8,\qquad w_n=\frac{400}{400+8}=\frac{400}{408}\approx0.9804. $$

The posterior mean is \(w_n\bar{D}\approx0.9804(6.5)=6.373\) points, and the posterior standard deviation is \(\sqrt{w_ns_D^2}=\sqrt{(400/408)(8)}\approx2.801\) points. Therefore,

$$ P(\theta>0\mid\bar{D}) =\Phi\left(\frac{6.373}{2.801}\right) \approx\Phi(2.275) \approx0.9886. $$

The posterior probability exceeds 0.975, so this hypothetical trial meets its stated Bayesian success rule. Equivalently, the data cutoff is \(c_n=z_{0.975}\sqrt{8/w_n}\approx5.599\) points, calculated using the unrounded quantities. The observed difference of 6.5 points exceeds that cutoff.

Worked Example: Finding an Infeasible Combination of Constraints

Use the same hypothetical outcome model and prior, with \(\sigma=10\) points and \(\tau=20\) points. Suppose the proposed rule is \(P(\theta>0\mid\bar{D})>0.975\), while the design requirements are type I error at most 0.01 and power at least 0.80 when \(\theta=5\) points.

At \(n=1\) per arm, \(s_D^2=200\), \(w_n=400/(400+200)=2/3\), and \(z_{0.975}/\sqrt{w_n}\approx2.400\). The type I error at the null boundary is

$$ \alpha_1=1-\Phi(2.400)\approx0.0082. $$

So the error-rate constraint is met at \(n=1\). But the standardized alternative effect is \(5/\sqrt{200}\approx0.3536\), giving power

$$ \operatorname{Power}_1(5) =\Phi(0.3536-2.400) \approx0.0203. $$

This is far below 0.80. Nor can a larger \(n\) solve both requirements under this rule. As \(n\) increases, \(s_D^2=200/n\) decreases and \(w_n=400/(400+200/n)\) increases. Consequently, \(z_{0.975}/\sqrt{w_n}\) decreases, so the type I error increases. At \(n=2\), for example, \(w_n=0.8\), and \(\alpha_2=1-\Phi(1.959964/\sqrt{0.8})\approx0.0142\), already above 0.01. Thus the error-rate constraint holds at \(n=1\) but not for \(n\geq2\), while the power target fails at \(n=1\). There is no sample size satisfying both proposed constraints.

Worked Example: Adjusting the Posterior Threshold to Meet Both Targets

Retain the same hypothetical model and power target of 0.80 at \(\theta=5\) points, but use the stricter posterior rule \(P(\theta>0\mid\bar{D})>0.99\). For this rule, \(z_\gamma=z_{0.99}\approx2.32635\). Since \(w_n<1\) at finite \(n\), \(z_{0.99}/\sqrt{w_n}>z_{0.99}\), so the type I error is less than 0.01.

At \(n=81\) per arm, \(s_D=\sqrt{200/81}\approx1.571\), \(w_n=400/(400+200/81)\approx0.9939\), and \(z_{0.99}/\sqrt{w_n}\approx2.3335\). The type I error is approximately \(1-\Phi(2.3335)=0.0098\). The power at 5 points is

$$ \operatorname{Power}_{81}(5) =\Phi\left(\frac{5}{\sqrt{200/81}}-2.3335\right) =\Phi(3.1820-2.3335) \approx0.802. $$

At \(n=80\), the corresponding power is \(\Phi(3.1623-2.3336)\approx0.796\), below target. Because the alternative effect divided by \(s_D\) increases with \(n\), while \(z_{0.99}/\sqrt{w_n}\) decreases, power increases with \(n\) in this model. Thus 81 participants per arm is the smallest balanced allocation meeting both stated targets under this rule.

What the Joint Check Does—and Does Not—Guarantee

These examples show why the two criteria must be evaluated together. A posterior probability threshold does not, by itself, equal a frequentist type I error rate. In the normal model above, the prior affects the relationship: for a fixed \(\gamma\), \(w_n<1\) makes the data cutoff more demanding than the large-sample cutoff \(z_\gamma s_D\). As \(n\) grows, \(w_n\) approaches 1, and the type I error approaches \(1-\gamma\). With \(\gamma=0.975\), that limit is 0.025, not 0.01. At small sample sizes, however, the error can be below 0.01; it is incorrect to claim it always exceeds 0.01.

A hybrid design differs from two common alternatives. A posterior-only design chooses the Bayesian rule without imposing a repeated-sampling error-rate or power requirement. A conventional frequentist design specifies a test and its error rates, without making a posterior-probability threshold its success definition. A hybrid design retains the stated posterior rule but checks frequentist behavior for that same rule. Its constraints may lead to a stricter posterior threshold, a different sample size, or a conclusion that the requested combination is infeasible under the assumed model.

The calculation also has limits. It relies on a known variance, normal sampling model, balanced allocation, and the specified normal analysis prior. Unknown variance, time-to-event outcomes, missing data, nonadherence, multiple endpoints, or a different analysis can change the operating characteristics. In such settings, the design must evaluate the actual planned analysis and decision rule; substituting the formulas here without checking their assumptions is not justified.

Diagnostics and Common Pitfalls

  • Check that the estimand and success event match. The examples define \(\theta\) as treatment minus control and success as evidence that \(\theta>0\). Reversing the contrast or using a positive clinical margin changes the rule.
  • Check the null being protected. For \(H_0:\theta\leq0\), the boundary \(\theta=0\) gives the maximum false-positive probability in this model. If the null or nuisance-parameter space differs, evaluate the appropriate supremum rather than only one convenient case.
  • Do not assume a posterior threshold guarantees a chosen error rate. Calculate the type I error for the actual sample size and prior. In particular, a threshold of 0.975 does not generally impose a 0.01 type I error bound.
  • Recheck all constraints after changing the design. Changing \(n\), \(\gamma\), allocation, or the prior changes the cutoff and can change both power and type I error. A design that meets one target may fail another.
  • Inspect the model and computation. Check that variance and effect units agree, that the planned sample size is per arm or total as stated, and that numerical calculations use unrounded values until the final report.
Key takeaway: A hybrid design is feasible only when the same prespecified Bayesian success rule meets every required frequentist constraint. Derive or compute the rule's operating characteristics under the planned model, verify the null and alternative conditions, and report infeasibility when no allowed design satisfies them.

Check Your Understanding

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

  1. In the normal model, why is the largest type I error for \(H_0:\theta\leq0\) attained at \(\theta=0\)?
  2. For the posterior threshold \(\gamma=0.975\), what is the limiting type I error as \(n\) tends to infinity in the stated model?
  3. With \(\sigma=10\), \(\tau=20\), and \(\gamma=0.975\), calculate \(w_n\) and \(z_\gamma/\sqrt{w_n}\) at \(n=2\). Does the type I error meet a 0.01 limit?
  4. Why is it not sufficient to calculate power for a conventional test and then claim that the Bayesian decision rule meets the same power target?

Solutions. (1) The success probability is \(\Phi(\theta/s_D-z_\gamma/\sqrt{w_n})\), which increases with \(\theta\). Its supremum over \(\theta\leq0\) is therefore at \(\theta=0\). (2) As \(n\) grows, \(s_D^2\) tends to zero and \(w_n\) tends to 1. Hence the type I error tends to \(1-\Phi(z_{0.975})=0.025\). (3) At \(n=2\), \(s_D^2=200/2=100\), so \(w_n=400/(400+100)=0.8\). The ratio is \(1.959964/\sqrt{0.8}\approx2.191\), giving type I error \(1-\Phi(2.191)\approx0.0142\), which exceeds 0.01. (4) Power belongs to a particular decision rule under a particular model. A conventional test can have a different cutoff from the posterior rule, so its power does not establish the power of the Bayesian rule; evaluate the rule that will actually determine success.

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