Planning After Selection
At a phase transition, the trial team may select one treatment for further study and decide how much confirmatory information to collect. A natural planning question is: given the phase 2 data already observed, how much phase 3 information would provide a chosen probability of success? Conditional power answers this question under a specified model for the treatment effect. It is a planning calculation, not by itself a guarantee of success or a replacement for a valid final testing procedure.
The calculation below uses an independent-increments combination test. The phase 2 result contributes a fixed observed stage statistic; the phase 3 statistic is random, with its distribution determined by the assumed effect and planned information. A critical value \(c\) represents the rejection threshold for the confirmatory claim under the chosen testing plan. If treatment selection or multiplicity requires a different calibrated threshold, that threshold must be used instead of an unadjusted one.
Model and Conditional-Power Formula
Suppose the selected treatment is tested against control on a continuous endpoint. Let the treatment effect be measured in units for which the estimator has variance \(1/I\), where \(I\) is information. For example, with independent normally distributed outcomes having known variance \(\sigma^2\), equal allocation with \(n\) participants per group gives information \(I=n/(2\sigma^2)\) for the difference in means. Define \(Z_1\) and \(Z_2\) as the stage 1 and stage 2 standardized statistics. Assume the stages use independent participants and that, under a specified effect \(\delta\), \(Z_2\sim N(\delta\sqrt{I_2},1)\). The observed phase 2 statistic is \(z_1\), and the planned phase 3 information is \(I_2\).
For the illustration, use information weights \(w_1=\sqrt{I_1/(I_1+I_2)}\) and \(w_2=\sqrt{I_2/(I_1+I_2)}\), so \(w_1^2+w_2^2=1\). The combination statistic is \(Z_C=w_1z_1+w_2Z_2\), and the claim is rejected when \(Z_C>c\). These formulas assume the weights and critical value correspond to the design being evaluated, the stage 2 statistic is approximately standard normal with the stated mean, and the stage increments are independent. The phase 2 statistic is held fixed in this conditional calculation.
The denominator is \(w_2\), not 1, because the stage 1 contribution is fixed once \(z_1\) is observed. Only the future stage statistic contributes sampling variation to the conditional distribution of \(Z_C\). The assumed effect \(\delta\) is a planning input; changing it can change conditional power substantially.
Deriving the Calculation
Under the assumed effect, \(Z_2\) has mean \(\delta\sqrt{I_2}\) and variance 1. Therefore \(Z_C\), conditional on \(z_1\), has mean \(w_1z_1+w_2\delta\sqrt{I_2}\) and variance \(w_2^2\). Standardizing the event \(Z_C>c\) gives the formula above:
This derivation also shows why sample-size planning is generally a numerical problem: the weights depend on \(I_2\), and \(I_2\) appears inside the square roots. To plan for target conditional power \(q\), solve \(\operatorname{CP}(I_2;\delta)=q\), then round the resulting sample size upward as required by the allocation and design. A numerical root should be checked by substituting it back into the target equation.
Worked Example: Conditional Power at a Proposed Stage Size
Assume a single selected-treatment test has a valid one-sided critical value \(c=1.96\). Suppose \(I_1=25\), the observed statistic is \(z_1=2.00\), and the planning effect is \(\delta=0.270\) standardized effect units. Evaluate a proposed \(I_2=36\) information units.
The weights are \(w_1=5/\sqrt{61}=0.6402\) and \(w_2=6/\sqrt{61}=0.7682\). Under the planning effect, the stage 2 statistic has mean \(\delta\sqrt{I_2}=0.270(6)=1.620\). Thus the conditional standard-normal argument is \((0.6402(2.00)+0.7682(1.620)-1.96)/0.7682=0.7353\).
Therefore the conditional power is \(\Phi(0.7353)\approx0.769\), or about 76.9%. This is below an 80% target for the specified effect and observed stage 1 statistic. It is not the probability that the null hypothesis is false, nor a probability statement about the unknown true effect.
Solving for Information and Participants
A conditional-power target specifies an equation, not a universal sample-size rule. The result depends on the observed phase 2 statistic, assumed effect, critical value, information weights, and target. A numerical solver can be used, but the solution should be verified by recomputing the weights, the normal argument, and the resulting probability.
Worked Example: Finding Information for 80% Conditional Power
Use \(I_1=25\), \(z_1=2.00\), \(\delta=0.270\), and \(c=1.96\) as above. Find \(I_2\) giving target conditional power \(q=0.80\). Since \(\Phi^{-1}(0.80)=0.8416\), solve \((w_1z_1+w_2\delta\sqrt{I_2}-1.96)/w_2=0.8416\), with weights recalculated for each candidate \(I_2\).
Numerical solution gives approximately \(I_2=43.8\) information units. Check the result: \(\sqrt{I_2}\approx6.618\), \(\sqrt{I_1+I_2}=\sqrt{68.8}\approx8.295\), so \(w_1\approx0.6028\) and \(w_2\approx0.7980\). The argument is \((0.6028(2.00)+0.7980(0.270)(6.618)-1.96)/0.7980\approx0.841\). Hence \(\Phi(0.841)\approx0.800\), confirming the rounded solution.
If outcomes have known variance \(\sigma^2=1\) and equal allocation, \(I_2=n/2\) where \(n\) is the number per group. The information solution corresponds to \(n=2(43.8)=87.6\), so at least 88 participants per group, or 176 total, are needed to attain at least that information. Rounding upward changes the information slightly; the conditional power can be recomputed at \(I_2=44\).
Accounting for Uncertainty in the Effect
The plug-in calculation treats the assumed effect as fixed. If that effect was estimated from a modest phase 2 sample, this can give a falsely reassuring impression of precision. There are two distinct sources of uncertainty: future phase 3 sampling variability, which appears in conditional power, and uncertainty about which effect value is a reasonable planning input. A single plug-in effect does not represent the second source.
One way to summarize effect uncertainty is to specify a planning distribution \(G\) for \(\delta\), based on an explicitly stated model or prior information. The average of conditional power over this distribution is often called assurance in Bayesian planning, or expected conditional power when the distribution is used only as a planning device. It is not a frequentist probability that the fixed treatment effect is random. The distribution must be justified; it does not arise automatically from observing a phase 2 estimate.
For a normal planning distribution \(\delta\sim N(m,\tau^2)\), with stage 2 independent of stage 1 conditional on \(\delta\), averaging over effect uncertainty yields
To see the added variance, conditional on \(\delta\), the random part of \(Z_C\) is \(w_2Z_2\), with variance \(w_2^2\). Variation in \(\delta\) adds variance \(w_2^2I_2\tau^2\), because the stage 2 mean contains \(\delta\sqrt{I_2}\). The total predictive variance is consequently \(w_2^2(1+I_2\tau^2)\). If \(\tau=0\), the formula reduces to plug-in conditional power at \(\delta=m\).
Worked Example: Planning With Effect Uncertainty
Continue with \(I_1=25\), \(z_1=2.00\), and \(c=1.96\). Suppose the planning distribution is \(\delta\sim N(0.270,0.120^2)\). At \(I_2=43.8\), the plug-in conditional power at the mean effect is about 80.0%, as in the preceding example. For assurance, the effect-uncertainty factor is \(\sqrt{1+43.8(0.120^2)}=\sqrt{1.6307}=1.277\).
The numerator of the assurance argument is \(w_1z_1+w_2m\sqrt{I_2}-c\), which is approximately \(w_2(0.8416)\). Thus the assurance argument is approximately \(0.8416/1.277=0.659\), and assurance is \(\Phi(0.659)\approx0.745\). In this planning model, a design with 80% plug-in conditional power has about 74.5% assurance after averaging over the specified effect uncertainty.
To target 80% assurance under the same distribution, solve the assurance equation numerically. A solution is approximately \(I_2=72.7\) information units: its weights are about \(w_1=0.5059\) and \(w_2=0.8626\), and the assurance argument is approximately 0.841. With variance 1 and equal allocation, this corresponds to \(n=2(72.7)=145.4\), rounded up to 146 participants per group. This larger sample size reflects the stated uncertainty distribution; it is not universally required whenever a phase 2 estimate is used.
Diagnostics, Alternatives, and Limitations
Before relying on a planned size, check that the model describes the actual design. Stage independence can fail if participants or outcomes are reused in ways not represented by the stagewise model. The normal approximation may be poor for sparse binary outcomes or limited event counts. Information may not increase proportionally with enrollment when follow-up, variance, or allocation differs from the assumed model. A selected phase 2 estimate can also be exaggerated because it was chosen from several candidates. Treating that estimate as an unbiased, precise effect may inflate plug-in conditional power.
Useful checks include evaluating conditional power over a plausible range of effects, comparing plug-in power with assurance or another uncertainty analysis, confirming the actual stage 2 information implied by integer allocation, and checking that the critical value reflects the planned treatment-selection and multiplicity procedure. The assumption distribution for assurance should include relevant uncertainty; simply centering it at a selected estimate may fail to address winner's-curse bias. If selection changes the distribution of the stage 1 statistic, a simple calculation conditional only on the observed \(z_1\) does not automatically resolve that issue.
| Planning approach | What it answers | Main limitation |
|---|---|---|
| Plug-in conditional power | Success probability conditional on observed \(z_1\) and one assumed effect | Can be sensitive to an optimistic or noisy effect estimate |
| Assurance or expected conditional power | Average success probability over a stated effect distribution | Depends on the suitability of that distribution |
| Fixed sample-size planning | Power under a prespecified effect and design before phase 2 data are observed | Does not tailor confirmatory information to the observed phase 2 result |
Conditional power is a design aid, not a device for rescuing a weak result by choosing an arbitrary effect or sample size. Any data-dependent change in sample size must be considered together with the test that will ultimately be used. The earlier material on combination tests and conditional error explains why the final testing rule must remain valid under the adaptation. A high conditional-power target does not itself establish type I error control, and a low value does not prescribe a futility decision.
Common Pitfall
- Wrong reasoning: “The phase 2 estimate is the effect, so the calculated 80% conditional power means the trial has an 80% chance of success.” This treats a noisy estimate as known truth and ignores the assumptions behind the future-stage model.
- Correction: State the effect value used, condition on the observed stage statistic, and describe the result as a model-based probability of rejection. If effect uncertainty matters, assess it with a justified planning distribution or sensitivity analysis; do not interpret assurance as a probability about a fixed parameter unless a probability model for that parameter has been specified.
- Another check: Do not round a numerical information root down and assume the target is still met. Recalculate power at the rounded, implementable sample size.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- Why is the denominator in the conditional-power argument \(w_2\), rather than 1?
- For \(I_1=25\), \(z_1=2.00\), \(I_2=36\), and \(c=1.96\), what conditional power results if the assumed effect is \(\delta=0\)?
- Under equal allocation with known outcome variance \(\sigma^2=2\), what information does 100 participants per group provide for a difference in means?
- What additional assumption is needed to calculate assurance from a phase 2 effect estimate?
Solutions. 1. Given the observed stage 1 statistic, only \(Z_2\) is random, so the variance of the combination statistic is \(w_2^2\), and its standard deviation is \(w_2\). 2. The weights remain \(w_1=0.6402\) and \(w_2=0.7682\); the argument is \((0.6402(2.00)-1.96)/0.7682=-0.8847\), giving \(\Phi(-0.8847)\approx0.188\), or 18.8%. 3. \(I_2=n/(2\sigma^2)=100/(2(2))=25\) information units. 4. Specify and justify a distribution or other uncertainty model for the effect, such as a planning distribution based on a stated analysis; the estimate alone does not define a probability distribution for a fixed effect.