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Operating Characteristics and Controversies · Tutorial 329 of 501

Power Under Response-Adaptive Randomization

Learn to connect adaptive allocation counts to treatment-contrast variance and calculate power without confusing a conditional calculation with the operating characteristics of the full design.

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What You'll Learn

  • Define power for a specified alternative and a fully specified adaptive design.
  • Derive the variance contribution from random two-arm allocation counts under explicit assumptions.
  • Calculate conditional power for a two-sided test using arm-specific sample sizes.
  • Explain why averaging power over allocation outcomes differs from using an average variance.
  • Identify when adaptive outcome-dependent allocation requires full-design simulation.
  • Diagnose common errors in power calculations for adaptive trials.

Power Is a Property of the Full Design

In a response-adaptive trial, the number assigned to each arm is random: assignments depend on information accumulated during the trial. Power therefore cannot generally be read off from a single fixed sample-size formula. It is the probability that the prespecified analysis rejects the null hypothesis when a specified alternative is true, with that probability taken over the trial’s random assignments and outcomes.

This definition matters because an adaptive design includes both an allocation rule and an analysis rule. To evaluate power, one must use the same allocation rule, stopping rule (if any), estimator, and rejection rule that the trial will use. A calculation conditional on a particular allocation is useful, but it answers a narrower question than the unconditional power of the design.

Definition: For a specified treatment-effect value \(\delta\) under the alternative and a prespecified test, power is \(P_{\delta}(\text{reject }H_0)\). In an adaptive trial, this probability is over the joint distribution of the allocation process and the outcomes. Conditional power given an allocation history is the rejection probability conditional on that history; it is not, by itself, the trial’s unconditional power.

For a two-arm comparison, let \(N_1\) and \(N_0\) denote the final numbers assigned to arms 1 and 0, with \(N_1+N_0=N\) when the total sample size is fixed. Let the estimand be the mean contrast \(\delta=\mu_1-\mu_0\). Under a two-sided test of \(H_0:\delta=0\), power depends on how often the test statistic crosses either rejection boundary when the true contrast is \(\delta\). In an adaptive design, that probability averages over the possible allocation histories, not just over outcome noise for one chosen pair of sample sizes.

How Allocation Counts Enter Contrast Variance

Start with the familiar two-arm sample-mean contrast \(\hat{\delta}=\bar{X}_1-\bar{X}_0\). For fixed arm counts and independent outcomes with arm-specific variances \(\sigma_1^2\) and \(\sigma_0^2\), its variance is the sum of the two arm-mean variances. If the counts are random, this expression becomes a function of the realized counts.

$$ \operatorname{Var}(\hat{\delta}\mid N_1,N_0) = \frac{\sigma_1^2}{N_1}+\frac{\sigma_0^2}{N_0}. $$

This conditional formula requires care in an adaptive trial. It holds when, conditional on the counts being considered, the observations used in the contrast have the stated independent outcome variances and the conditional mean of the contrast is \(\delta\). Those conditions are natural for fixed allocation, and can be reasonable for a working approximation or a count mechanism independent of the outcome residuals. They do not automatically hold when the allocation rule uses earlier trial outcomes: conditioning on the final counts can then select histories with different outcome patterns.

Under the stated conditions, if the contrast has conditional mean \(\delta\) for every count pair, the law of total variance gives a direct result. Since \(E(\hat{\delta}\mid N_1,N_0)=\delta\), the variance of this conditional expectation is zero. Thus the unconditional variance is the expected conditional variance:

$$ \operatorname{Var}(\hat{\delta}) = E\!\left[ \frac{\sigma_1^2}{N_1}+\frac{\sigma_0^2}{N_0} \right] = \sigma_1^2 E\!\left(\frac{1}{N_1}\right) +\sigma_0^2 E\!\left(\frac{1}{N_0}\right). $$

The reciprocal counts, rather than the reciprocals of the average counts, appear in this expression. When the conditional mean is not constant, the general decomposition also includes variation in that mean:

$$ \operatorname{Var}(\hat{\delta}) = E\{\operatorname{Var}(\hat{\delta}\mid N_1,N_0)\} + \operatorname{Var}\{E(\hat{\delta}\mid N_1,N_0)\}. $$

For outcome-adaptive assignment, even this count-based decomposition may not capture all relevant dependence. The allocation history can matter beyond the final counts. The decomposition remains true, but the conditional variance and conditional mean must describe the actual adaptive data-generating process; they cannot simply be assumed to equal the fixed-count expressions.

Key result: Under a valid conditional fixed-count model with constant conditional mean, random counts contribute through \(E(1/N_1)\) and \(E(1/N_0)\). For a genuinely response-dependent allocation rule, this is a variance benchmark, not a universal formula. The joint allocation-and-outcome distribution determines the actual power.

From Contrast Variance to Conditional Power

Suppose for illustration that the outcome variances are known and that, conditional on the counts, the contrast is normally distributed with mean \(\delta\) and variance \(\sigma_1^2/N_1+\sigma_0^2/N_0\). A two-sided \(z\)-test at significance level \(\alpha\) rejects when the absolute standardized contrast exceeds \(z_{1-\alpha/2}\). Conditional on the counts, the standardized statistic has mean \(\lambda=\delta/SE\) and variance 1, where \(SE\) is the conditional standard error.

$$ \operatorname{Power}(N_1,N_0) = \Phi\!\left(-z_{1-\alpha/2}-\lambda\right) + 1-\Phi\!\left(z_{1-\alpha/2}-\lambda\right), \qquad \lambda=\frac{\delta}{SE}. $$

The first tail is rejection in the lower direction and the second is rejection in the upper direction. For a design with several possible allocation outcomes, unconditional power is the average of the conditional rejection probabilities over those outcomes only when those probabilities correctly condition on the relevant allocation history. For outcome-dependent allocation, simulation of the complete trial is often the clearest way to obtain that average.

Worked Example: Power at a Specified Allocation

Consider a hypothetical two-arm trial with 100 participants, 30 assigned to arm 1 and 70 to arm 0. Suppose the outcome standard deviation is 10 units in each arm, the true mean difference is \(\delta=4\) units, and the planned test is two-sided at \(\alpha=0.05\). Assume for this conditional calculation that the normal model and known variances apply.

The conditional standard error is \(SE=\sqrt{10^2/30+10^2/70}=\sqrt{4.7619}=2.182\) units. The noncentrality parameter is \(\lambda=4/2.182=1.833\), and \(z_{0.975}=1.960\). Therefore,

$$ \begin{aligned} \operatorname{Power}(30,70) &=\Phi(-1.960-1.833)+1-\Phi(1.960-1.833)\\ &=\Phi(-3.793)+1-\Phi(0.127)\\ &\approx 0.00007+0.4495\\ &\approx 0.4496. \end{aligned} $$

The conditional power is approximately 0.4496, or 45.0%. It is the rejection probability under the specified 4-unit difference for this allocation and this test, under the stated model. It is not yet the unconditional power of a response-adaptive design that could produce other allocations.

Averaging Over Random Counts

Even when conditional normal calculations are justified, averaging the conditional powers is not generally equivalent to calculating power from an average variance. Power is a nonlinear function of the standard error. The next example isolates this feature in a count-mixture benchmark. It assumes the count outcome is independent of outcome residuals, so it illustrates the mathematics of random counts rather than claiming that outcome-dependent response adaptation satisfies this assumption.

Worked Example: A Two-Point Count Mixture

Suppose a hypothetical 100-person design has \(N_1=30,N_0=70\) with probability 0.5 and \(N_1=50,N_0=50\) with probability 0.5. Assume both arm standard deviations are 10 units, the true difference is 4 units, and the same two-sided 5% known-variance \(z\)-test is used. Under the benchmark assumptions, unconditional power is the probability-weighted average of the two conditional powers.

For counts 30 and 70, the preceding calculation gives power approximately 0.4496. For counts 50 and 50, \(SE=\sqrt{100/50+100/50}=2\) units and \(\lambda=4/2=2\). Thus,

$$ \begin{aligned} \operatorname{Power}(50,50) &=\Phi(-1.960-2)+1-\Phi(1.960-2)\\ &=\Phi(-3.960)+1-\Phi(-0.040)\\ &\approx 0.00004+0.51595\\ &\approx 0.51599. \end{aligned} $$

The unconditional power is therefore \(0.5(0.4496)+0.5(0.51599)=0.4828\), or about 48.3%. This calculation averages the rejection probabilities, rather than substituting an average count into a fixed-allocation formula.

The count-variance benchmark also shows why inverse counts matter. Here, \(E(1/N_1)=0.5(1/30+1/50)=0.02667\), and \(E(1/N_0)=0.5(1/70+1/50)=0.01714\). The benchmark variance is \(100(0.02667+0.01714)=4.381\) square units. A normal power calculation using only this average variance would not reproduce the mixture power exactly, because the power function is nonlinear. In a true response-adaptive trial, outcome-dependent allocation adds another reason to evaluate the full design rather than rely on this benchmark.

Evaluating Power for the Actual Adaptive Procedure

When assignment probabilities depend on observed responses, a defensible power evaluation must reproduce that dependence. For each simulated trial, generate outcomes under the chosen alternative, apply the allocation rule using only the information available at each assignment, complete the trial, and apply the prespecified analysis. The estimated power is the fraction of simulated trials in which the analysis rejects \(H_0\). The outcome model, adaptation rule, sample-size rule, and test must all match the proposed design.

1
Specify the alternative and the complete design.
State the treatment effect and other outcome-model parameters, the allocation rule, the total sample size or stopping rule, and the final test.
2
Generate a complete trial.
Simulate each outcome under the alternative and update assignment probabilities according to the adaptive rule at the correct time.
3
Apply the planned analysis.
Calculate the prespecified statistic and record whether its rejection rule is met. Do not substitute an analysis chosen after seeing the simulated allocation.
4
Repeat and quantify simulation error.
With \(M\) independent simulated trials and \(R\) rejections, estimate power by \(R/M\); its Monte Carlo standard error is approximately \(\sqrt{\hat{p}(1-\hat{p})/M}\).

Worked Example: Monte Carlo Power for an Adaptive Trial

Suppose a hypothetical response-adaptive design is evaluated under a specified alternative using 2,000 simulated trials. Each simulated trial follows the same response-based allocation rule and final two-sided test. If 1,134 trials reject \(H_0\), the estimated power is \(\hat{p}=1,134/2,000=0.5670\).

The Monte Carlo standard error is \(\sqrt{0.5670(1-0.5670)/2,000}\approx\sqrt{0.0001227555}\approx 0.01108\). An approximate 95% interval describing simulation uncertainty is \(0.5670\mathbin{\pm}1.96(0.01108)=(0.5453,0.5887)\). This interval quantifies the finite number of simulation replicates; it is not a confidence interval for a real treatment effect. The estimated power is about 56.7%, with Monte Carlo error of roughly 1.1 percentage points in standard-error units.

This estimate reflects the simulated adaptive procedure only if each replicate generated the assignment sequence from its own evolving outcomes and then applied the planned test. Holding the allocation fixed across replicates would answer a conditional question instead.

Diagnostics, Alternatives, and Limitations

A useful power analysis reports more than one number. Check the distribution of \(N_1\) and \(N_0\), including how often either arm receives few observations, and inspect whether allocation histories with different counts also have different outcome patterns. For simulation, verify that assignment probabilities are updated at the intended times, that every replicate reaches the specified sample size or stopping rule, and that the final test is implemented exactly as planned. Monte Carlo uncertainty should be small enough for the design decision; increase \(M\) if the estimated power is too imprecise.

The main alternatives answer different questions. A fixed-count normal calculation is quick and interpretable for a particular allocation. A count-mixture calculation averages such conditional results when the conditional model is valid and the count distribution is available. Full-design simulation can account for response-dependent allocation and a complicated test, but its accuracy depends on the outcome model and the correctness of the code. These methods are complementary: the simple formula helps explain the role of counts, while simulation evaluates the specified adaptive procedure.

A count-only variance calculation has important limitations. It can miss dependence between allocation and observed outcomes, and an average standard error does not determine the average rejection probability. Moreover, a power estimate is specific to the alternative and analysis used; it does not establish power uniformly over all possible effects. The earlier material on type I error and confidence interval coverage concerns other operating characteristics: high power under an alternative does not by itself establish correct error control or coverage.

Common Mistake

Wrong reasoning: “The adaptive trial is expected to assign 40 participants to arm 1 and 60 to arm 0, so I can calculate power as though those counts were fixed.” That replaces random counts by their averages and ignores both the nonlinear power calculation and the possibility that outcome-dependent allocation changes the distribution of the estimator.

Correction: A fixed-count calculation is conditional on those counts. For random counts, average valid conditional rejection probabilities over the allocation outcomes. When assignments respond to trial outcomes, evaluate the allocation and analysis together—often by simulating complete trials—and report the alternative and assumptions used.

Check Your Understanding

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

  1. In one sentence, distinguish conditional power at fixed arm counts from unconditional power for an adaptive design.
  2. Under the stated constant-conditional-mean assumptions, what count quantities enter the unconditional variance of the two-arm contrast?
  3. Why can power calculated from an average variance differ from the average of conditional powers?
  4. A simulation produces 800 rejections in 1,000 trials. Give the estimated power and approximate Monte Carlo standard error.

Solutions. (1) Conditional power is the rejection probability for a specified allocation or history; unconditional power averages rejection over the random allocation and outcome process. (2) The variance uses \(\sigma_1^2E(1/N_1)+\sigma_0^2E(1/N_0)\), provided the conditional variance formula and constant conditional mean hold. (3) The rejection probability is nonlinear in the standard error, so applying the power formula to an average variance need not equal averaging the separate power values. (4) The estimate is \(800/1,000=0.800\); the approximate Monte Carlo standard error is \(\sqrt{0.8(0.2)/1,000}=0.01265\), or about 1.265 percentage points.

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