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Population Enrichment and Subgroup Selection · Tutorial 429 of 799

Power Under Uncertain Biomarker Prevalence

Evaluate subgroup and all-comer power over plausible biomarker prevalence values, and distinguish average power from power at a single planning estimate.

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What You'll Learn

  • Define average power by integrating conditional power over a distribution of plausible biomarker prevalence values.
  • Calculate subgroup and all-comer power under stated treatment-effect and outcome-variance assumptions.
  • Explain why power at mean prevalence need not equal average power.
  • Distinguish uncertainty about prevalence from random variation in the trial’s observed subgroup count.
  • Compare average-power planning with plug-in and conservative planning approaches.
  • Identify diagnostics and assumptions that can make prevalence-averaged power misleading.

Why Prevalence Uncertainty Changes a Power Calculation

A power calculation for a biomarker-defined subgroup depends on how many trial participants are expected to be biomarker-positive. If the prevalence is uncertain, calculating power at one chosen estimate can conceal a meaningful range of possible performance. The same uncertainty can also affect an all-comer claim when treatment effects differ between biomarker-positive and biomarker-negative participants: the population-average effect then depends on the prevalence.

The aim is not to replace a design assumption with a more elaborate calculation for its own sake. It is to describe how trial performance changes across plausible prevalence values, and to make clear which summary is being reported. In particular, average power over a distribution of plausible prevalences is not the same quantity as power evaluated at the average prevalence.

Definition: Let \(\pi\) denote the biomarker-positive prevalence in the population from which trial participants are recruited. If uncertainty about \(\pi\) is represented by a probability distribution \(g(\pi)\), the prevalence-averaged power for a specified test is the average of its power conditional on \(\pi\), weighted by \(g\). It summarizes performance across the stated distribution; it does not assert that a fixed population prevalence is randomly changing during the trial.

Conditional Power Under a Planning Model

Begin with a simple model to make the role of prevalence explicit. Suppose the outcome \(Y\) is continuous, larger values are beneficial, and the treatment-control contrast is estimated using independent randomized groups. Assume a common within-group standard deviation \(\sigma\), balanced allocation within the population being tested, and a large enough sample for a normal approximation. The test is two-sided at significance level \(\alpha\), and the treatment effect under the planning alternative is \(\delta\). These assumptions are planning approximations; departures from them require a different standard error or power calculation.

For a total of \(m\) participants in the population being tested, balanced allocation gives approximately \(m/2\) per arm. The standard error of the difference in means is \(SE=2\sigma/\sqrt{m}\). If the estimated contrast is approximately normal with mean \(\delta\) and standard error \(SE\), the rejection region is outside \([-z_{1-\alpha/2}SE,z_{1-\alpha/2}SE]\). Its power is

$$ \operatorname{Power}(m,\delta) = \Phi\!\left(-z_{1-\alpha/2}-\frac{\delta}{SE}\right) + 1-\Phi\!\left(z_{1-\alpha/2}-\frac{\delta}{SE}\right), \qquad SE=\frac{2\sigma}{\sqrt{m}}. $$

Here \(\Phi\) is the standard normal cumulative distribution function. The formula includes both tails of the two-sided test. For a subgroup analysis in an all-comer trial with total sample size \(N\), the expected subgroup size under a fixed prevalence \(\pi\) is \(N\pi\). A first-order planning calculation substitutes \(m=N\pi\). An all-comer analysis uses \(m=N\), but its treatment effect may depend on \(\pi\).

Suppose the treatment effect is \(\delta_+\) in biomarker-positive participants and \(\delta_-\) in biomarker-negative participants. If the subgroup effects are defined on the same outcome scale and the target is the population-average treatment contrast, then

$$ \delta_A(\pi)=\pi\delta_+ +(1-\pi)\delta_-. $$

Thus the all-comer power at prevalence \(\pi\) is calculated using \(m=N\) and \(\delta=\delta_A(\pi)\), under the stated common-variance approximation. If the outcome variance differs by biomarker stratum, or the analysis uses a different estimand or standard error, the all-comer power calculation must reflect that model rather than blindly using this formula.

From Conditional to Prevalence-Averaged Power

Let \(\operatorname{Power}_+(\pi)\) denote subgroup power conditional on prevalence \(\pi\), and let \(\operatorname{Power}_A(\pi)\) denote all-comer power conditional on \(\pi\). With \(g(\pi)\) representing a defensible distribution of plausible prevalence values, the corresponding average powers are

$$ \overline{\operatorname{Power}}_+ =\int_0^1 \operatorname{Power}_+(\pi)g(\pi)\,d\pi, \qquad \overline{\operatorname{Power}}_A =\int_0^1 \operatorname{Power}_A(\pi)g(\pi)\,d\pi. $$

A short justification follows directly from conditioning. Let \(R\) be the event that the specified test rejects \(H_0\). Averaging the conditional rejection probability over the prevalence distribution gives

$$ P(R)=\int_0^1 P(R\mid \pi)g(\pi)\,d\pi. $$

This identity is useful only after the target distribution and the conditional power model have been specified. The distribution might represent uncertainty about which recruitment population will be encountered, or uncertainty in an estimate derived from external data. It should not be treated as an objective feature of the trial unless its source and interpretation are justified.

In practice, the integral can be evaluated numerically. For a discrete set of scenarios \(\pi_1,\ldots,\pi_K\), with weights \(w_k\) that sum to one, use

$$ \overline{\operatorname{Power}}\approx \sum_{k=1}^{K}w_k\operatorname{Power}(\pi_k). $$

For a continuous distribution, numerical integration or simulation can be used. When using simulation, draw prevalence values from the specified distribution, calculate conditional power for each draw, and average those powers. Report the distribution and the calculation method so the result can be reproduced.

Worked Example: Averaging Subgroup Power Across Scenarios

Suppose a hypothetical all-comer trial plans to enroll \(N=400\) participants. The biomarker-positive treatment effect is \(\delta_+=5\) outcome units, the common within-group standard deviation is \(\sigma=10\) units, and the subgroup test is two-sided with \(\alpha=0.05\). For this planning calculation, treat \(N\pi\) as the subgroup total and use balanced allocation. Consider prevalence values \(0.20\), \(0.40\), and \(0.60\), with respective scenario weights \(0.25\), \(0.50\), and \(0.25\).

At \(\pi=0.20\), the expected subgroup size is \(m=400(0.20)=80\). Therefore \(SE=2(10)/\sqrt{80}=2.236\) units, and the standardized effect is \(5/2.236=2.236\). Using \(z_{0.975}=1.96\), the power is approximately \(\Phi(-4.196)+1-\Phi(-0.276)=0.609\).

At \(\pi=0.40\), \(m=160\), \(SE=20/\sqrt{160}=1.581\) units, and the standardized effect is \(5/1.581=3.162\). The power is approximately \(\Phi(-5.122)+1-\Phi(-1.202)=0.885\).

At \(\pi=0.60\), \(m=240\), \(SE=20/\sqrt{240}=1.291\) units, and the standardized effect is \(5/1.291=3.873\). The power is approximately \(\Phi(-5.833)+1-\Phi(-1.913)=0.972\).

The weighted average is \(0.25(0.609)+0.50(0.885)+0.25(0.972)=0.838\), or about 83.8%. By contrast, power at the mean prevalence \(0.40\) is 88.5%. Both figures are correct, but answer different questions: one averages performance across the stated scenario distribution, while the other conditions on prevalence being 0.40.

All-Comer Power When Effects Differ by Biomarker

An all-comer estimate can have a different relationship with prevalence. Under the common-variance planning model, its standard error depends on \(N\), but its population-average treatment effect changes with the proportion in each stratum. Consequently, a prevalence distribution can change all-comer power even when the planned total sample size is fixed.

Worked Example: Averaging All-Comer Power

Use the hypothetical trial above with \(N=400\), \(\sigma=10\), and \(\alpha=0.05\). Suppose the treatment effect is \(\delta_+=5\) units in biomarker-positive participants and \(\delta_-=0\) units in biomarker-negative participants. Under the stated common-variance approximation, the all-comer standard error is \(SE=20/\sqrt{400}=1.000\) unit.

At \(\pi=0.20\), the population-average effect is \(\delta_A(0.20)=0.20(5)+0.80(0)=1\) unit. Its standardized effect is \(1/1=1\), giving power \(\Phi(-2.96)+1-\Phi(0.96)\approx0.170\).

At \(\pi=0.40\), \(\delta_A(0.40)=2\) units. The standardized effect is \(2\), and power is \(\Phi(-3.96)+1-\Phi(-0.04)\approx0.516\).

At \(\pi=0.60\), \(\delta_A(0.60)=3\) units. The standardized effect is \(3\), giving power \(\Phi(-4.96)+1-\Phi(-1.04)\approx0.851\).

With the same scenario weights as before, the all-comer average power is \(0.25(0.170)+0.50(0.516)+0.25(0.851)=0.513\), or about 51.3%. This differs from 51.6%, the power at mean prevalence \(0.40\). The small difference here is specific to these values; nonlinear power functions can produce larger differences. If \(\delta_+=\delta_-\), prevalence does not alter this population-average effect under this model, although it still affects the subgroup’s expected size.

Prevalence Uncertainty Is Not the Same as Count Variation

The calculations so far use the expected subgroup size \(N\pi\) at each prevalence. In an actual trial, the observed number of biomarker-positive participants can vary around that expectation. If each enrolled participant’s biomarker status is an independent Bernoulli outcome with probability \(\pi\), then the subgroup count \(M\), conditional on \(\pi\), follows \(M\sim\operatorname{Binomial}(N,\pi)\). This additional sampling variation is distinct from uncertainty represented by \(g(\pi)\).

A more complete average-power calculation can average first over possible subgroup counts and then over prevalence:

$$ \overline{\operatorname{Power}}_+ = \int_0^1 \left[ \sum_{m=0}^{N} P(M=m\mid \pi)\operatorname{Power}_+(m) \right]g(\pi)\,d\pi. $$

Here \(\operatorname{Power}_+(m)\) uses the actual subgroup size \(m\), the planned treatment allocation, and the specified analysis. For \(m=0\), a subgroup test cannot be performed, so its rejection probability is zero. In real calculations, small or uneven arm counts may also require methods other than the simple normal approximation. The formula makes clear why using \(N\pi\) is a convenient approximation rather than an exact account of trial-level power.

Worked Example: Planning to a Lower Prevalence Scenario

A hypothetical design team wants at least 80% subgroup power for a two-sided test with \(\alpha=0.05\), \(\delta_+=5\) units and \(\sigma=10\) units. Using the large-sample normal approximation, a standardized effect of about \(z_{0.975}+z_{0.80}=1.96+0.842=2.802\) gives approximately 80% power; the lower-tail probability is negligible at this effect size.

For subgroup size \(m\), the standardized effect is \(\delta_+\sqrt{m}/(2\sigma)\). Solving \(\delta_+\sqrt{m}/(2\sigma)\ge 2.802\) gives \(m\ge(2\sigma(2.802)/\delta_+)^2=(20(2.802)/5)^2=125.6\). Round up to 126 expected subgroup participants for this approximate calculation.

If the planning target is the lower scenario \(\pi=0.25\), the corresponding total is \(N\ge126/0.25=504\). At \(\pi=0.40\), the same expected subgroup count requires \(N\ge126/0.40=315\). The 504-participant figure meets the expected-count criterion at prevalence 0.25; it does not guarantee 80% power in every trial at that prevalence, because the realized subgroup count may be below its expectation. A design requiring a specified probability of meeting a power target must include count variation in its calculation.

Choosing a Summary and Checking the Model

Prevalence-averaged power is one useful summary, not a universal replacement for other planning summaries. Its interpretation depends on the decision being made and on the credibility of the prevalence distribution.

ApproachQuestion answeredMain caution
Plug-in powerWhat is power at a specified prevalence?A single estimate may hide sensitivity to prevalence.
Prevalence-averaged powerWhat is mean power across the stated distribution?The average can obscure low-power portions of that distribution.
Conservative scenarioWhat is power at a selected low or otherwise adverse prevalence?The scenario must be justified; it is not automatically a guarantee.
Count-integrated powerWhat is power after allowing subgroup counts to vary at each prevalence?Requires a count and analysis model that reflect the design.

Several checks make the calculation more informative. Report the assumed prevalence distribution, its source, its range, and the conditional power at meaningful points such as lower and upper quantiles. Check whether the distribution places substantial weight on prevalences where subgroup power is below the design target. For an all-comer claim, also check how the assumed effects in each biomarker stratum determine \(\delta_A(\pi)\). If the result changes sharply under small changes to the distribution or treatment-effect assumptions, present that sensitivity rather than only one average.

The normal approximation, common variance, treatment-effect specification, and allocation assumptions should be checked against the planned design. If allocation differs by stratum, the subgroup standard error must reflect the actual numbers assigned to each arm. If outcome variance varies across strata, a common-\(\sigma\) calculation may misstate all-comer precision. If prevalence differs between the planning source and trial recruitment population, an otherwise correct integration can still answer the wrong question. Simulation can accommodate a more detailed design, but does not repair unsupported assumptions: it should implement a clearly stated data-generating and analysis model.

Common Mistake

A common mistake is to calculate power once at the mean prevalence and label it “average power.” That substitution is generally invalid because power is a nonlinear function of both the effect and the standard error: in general, \(E\{\operatorname{Power}(\pi)\}\ne\operatorname{Power}\{E(\pi)\}\). The correction is to calculate conditional power across the prevalence distribution and average those values with the specified weights. Report power at mean prevalence separately if it is also of interest.

A second interpretive pitfall is to treat 80% prevalence-averaged power as meaning that every prevalence scenario has 80% power. It means only that the weighted average of conditional powers is 80%. The conditional powers may lie both above and below that value, so their range or selected quantiles should accompany the average when design decisions depend on avoiding poor-performing scenarios.

Key takeaway: State whether a reported power is conditional on one prevalence, averaged across a specified prevalence distribution, or additionally integrated over random subgroup counts. Those quantities answer different planning questions.

Check Your Understanding

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

  1. If a prevalence distribution has mean 0.35, is power at prevalence 0.35 necessarily equal to prevalence-averaged power? Explain.
  2. In a trial of 500 participants, what is the expected subgroup size at prevalence 0.30? Why is this not necessarily the realized subgroup size?
  3. Suppose \(\delta_+=4\) units and \(\delta_-=1\) unit. Write the all-comer effect as a function of prevalence \(\pi\).
  4. What does a prevalence-averaged power of 75% say, and what does it not say?

Solutions. (1) No. Power is nonlinear, so the weighted average of conditional powers generally differs from power evaluated at the mean prevalence. (2) The expected size is \(500(0.30)=150\) participants. The realized count varies across trials; under independent biomarker status it has a binomial distribution conditional on prevalence. (3) \(\delta_A(\pi)=4\pi+1(1-\pi)=1+3\pi\) units. (4) It says that conditional powers average to 75% under the specified prevalence distribution and power model. It does not say that power is 75% at every prevalence, or that every trial has a 75% probability of success under all possible assumptions.

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