Conditional And Unconditional Power

Why distinguish two kinds of power?

In a trial with fixed allocation, the planned numbers assigned to each arm are known in advance. With response-adaptive randomization, the sequence of assignments can depend on earlier outcomes, and the allocation history is itself random. Two different questions then arise: How likely is rejection for this particular allocation history? And what is the rejection probability before the trial, averaging over all histories the design could generate?

The first quantity is conditional power, conditional on a specified allocation history. The second is unconditional power for the design. The distinction matters because a conditional calculation describes one realized path, while a design-level power calculation must account for every possible path and how likely it is to occur. Neither quantity can in general be recovered by treating the realized or expected allocation as if it had been fixed in advance.

Here, “conditional power” means conditioning on an allocation history, not the different quantity sometimes called conditional power at an interim analysis, which also conditions on interim outcome data. The rejection rule and the alternative model must be specified for either calculation.

Definitions and the averaging identity

Definition: Let \(H\) denote the random allocation history generated by a trial design, and let \(R\) be the event that the prespecified test rejects \(H_0\). For an alternative parameter value \(\theta\), the conditional power for history \(h\) is \(P_\theta(R\mid H=h)\). The unconditional power is \(P_\theta(R)\), the rejection probability under the alternative before the allocation history is known.

If the possible histories form a discrete set, the law of total probability gives the key relationship:

P_\theta(R)=\sum_h P_\theta(R\mid H=h)P_\theta(H=h).

Thus, unconditional power is a weighted average of conditional powers. The weights are the probabilities of the histories under the alternative and the specified design. If histories are continuous or too numerous to list, the corresponding operation is an integral or an expectation over the design’s distribution of histories.

For response-adaptive randomization, an allocation history is generated through a sequence of assignment decisions, potentially using previous outcomes. Consequently, both \(P_\theta(H=h)\) and \(P_\theta(R\mid H=h)\) must come from the joint model for outcomes and randomization. In particular, conditioning on an outcome-dependent allocation path can change the conditional distribution of the outcomes. It is not generally valid to take a fixed-allocation power formula and apply it to the counts on that path as if those counts had been set independently of outcomes.

For an allocation scheme selected independently of outcomes, this complication does not arise: conditional on a selected allocation, the specified outcome model applies as usual. The first example uses this simpler setting to make the averaging calculation transparent.

A normal-outcome illustration

Worked Example: Conditional Power for Two Allocation Histories

Hypothetical design. A trial plans four participants and selects one of two allocation histories before any outcomes are observed. With probability 0.60, it assigns two participants to each arm; with probability 0.40, it assigns three to treatment and one to control. Within the selected history, assignments are randomized independently of outcomes. Assume independent outcomes with known standard deviation \(\sigma=1\) outcome unit in both arms. Test \(H_0:\mu_T-\mu_C=0\) against a two-sided alternative using the statistic \(Z=(\bar{x}_T-\bar{x}_C)/SE\), with \(SE=\sqrt{1/n_T+1/n_C}\), rejecting when \(|Z|>1.96\). Calculate power under the specific alternative \(\mu_T-\mu_C=2.5\) outcome units.

The normal, independent outcome model and known common standard deviation are assumptions of this illustration. The allocation branch is selected independently of outcomes, so the usual fixed-allocation distribution of \(Z\) applies within either branch. Under the alternative, \(Z\) has a normal distribution with variance 1 and mean equal to the effect divided by its standard error.

For the balanced history, \(n_T=n_C=2\), so

SE=\sqrt{\frac{1}{2}+\frac{1}{2}}=1,\qquad Z\sim N(2.5,1).

The conditional power is the probability of falling in either rejection tail:

P_\theta(|Z|>1.96\mid H=\text{balanced}) =1-\Phi(1.96-2.5)+\Phi(-1.96-2.5) =\Phi(0.54)+\Phi(-4.46) \approx 0.7054.

Thus, under the stated alternative, power conditional on the balanced history is about 70.5%.

For the 3:1 history, \(n_T=3\) and \(n_C=1\). Its standard error is \(\sqrt{1/3+1}=\sqrt{4/3}\approx1.155\), so the alternative mean of \(Z\) is \(2.5/1.155\approx2.165\). Therefore,

P_\theta(|Z|>1.96\mid H=\text{3:1}) =1-\Phi(1.96-2.165)+\Phi(-1.96-2.165) \approx \Phi(0.205)+\Phi(-4.125) \approx 0.5813.

Power conditional on the 3:1 history is about 58.1%. This is lower because the allocation is less balanced and hence gives a larger standard error for the treatment-control difference.

Unconditional power and a short justification

In the illustration, the allocation history is selected before outcomes, with probabilities 0.60 and 0.40. Applying the averaging identity gives the unconditional power:

P_\theta(R) =(0.7054)(0.60)+(0.5813)(0.40) =0.6558.

The design’s power under the specified alternative is about 65.6%. This is not the power for either history alone; it averages their rejection probabilities according to how often each history is generated.

The justification is a partition argument. The events “reject and the history is balanced” and “reject and the history is 3:1” are disjoint, and together they make up the rejection event. Therefore,

P_\theta(R) =P_\theta(R,H=\text{balanced})+P_\theta(R,H=\text{3:1}) =\sum_h P_\theta(R\mid H=h)P_\theta(H=h).

This argument requires no special power theorem; it is the law of total probability. Its practical implication is important: the conditional powers alone do not determine unconditional power. Their weights are also needed.

Worked Example: Why Expected Allocation Counts Are Not a Substitute

Use the same hypothetical design. The expected treatment and control counts are \(E(n_T)=(0.60)(2)+(0.40)(3)=2.4\) and \(E(n_C)=1.6\). A tempting shortcut is to insert these expected counts into the fixed-allocation standard-error formula.

That shortcut gives \(SE=\sqrt{1/2.4+1/1.6}=\sqrt{1.0417}\approx1.021\), and hence a noncentral mean of \(2.5/1.021\approx2.449\). The resulting two-sided normal power is approximately

1-\Phi(1.96-2.449)+\Phi(-1.96-2.449) \approx\Phi(0.489)+\Phi(-4.409) \approx0.688.

This is about 68.8%, not the correct unconditional power of 65.6%. The calculation treated fractional expected counts as if they described a single fixed design. But the actual design has two possible histories with different standard errors, and its power is the weighted average of their powers. In general, applying a nonlinear power calculation to average allocation counts does not equal averaging the powers calculated for the possible histories.

Outcome-dependent histories require care

In a response-adaptive design, earlier responses can influence later assignment probabilities. A realized history may therefore be more or less likely under different outcome models. The unconditional calculation must use the history probabilities under the alternative being considered, not automatically the probabilities under \(H_0\) or the probabilities observed in a different trial.

There is a second issue: a history can carry information about the outcomes that helped generate it. A fixed-count formula may not give \(P_\theta(R\mid H=h)\) if reaching history \(h\) depended on those outcomes. The conditional rejection probability must reflect the actual joint mechanism, including the adaptive assignment rule. For design-level power, one can instead work directly with the joint distribution of assignments and outcomes, or enumerate the histories and calculate the appropriate conditional probabilities for each.

Worked Example: Use Alternative-Specific History Weights

Hypothetical calculation. Suppose a validated joint design-and-outcome model gives two possible allocation histories. Under a specified alternative, history A occurs with probability 0.70 and has conditional power 0.80; history B occurs with probability 0.30 and has conditional power 0.40. These conditional probabilities already account for the response-dependent rule and the outcome distribution conditional on each history.

The unconditional power is

P_\theta(R)=(0.80)(0.70)+(0.40)(0.30)=0.56+0.12=0.68.

Thus the design’s power under this alternative is 68%. If the same histories instead occurred with probabilities 0.50 and 0.50 under the alternative, the average would be \((0.80)(0.50)+(0.40)(0.50)=0.60\). The conditional powers have not changed, but the design-level power has, because the alternative-specific distribution of histories has changed.

This example assumes the supplied conditional powers and history probabilities come from a valid joint model. It does not imply that those probabilities can be found by treating response-dependent allocations as independent of outcomes.

Diagnostics, comparisons, and limitations

A careful power calculation should make clear what is being conditioned on, which alternative is being used, and how the histories are generated. For an adaptive design, useful checks include:

Check the weights. Confirm that history probabilities are computed under the specified alternative and sum to 1 over the possible histories.
Check the conditional model. Verify that the distribution used for each conditional power respects any dependence between earlier outcomes and later allocations.
Check the test rule. Apply the same prespecified rejection rule to every history, unless the design explicitly specifies a history-dependent test.
Check the averaging. The unconditional power must lie between the smallest and largest conditional powers receiving positive weight.
Check null calibration separately. A test with a target type I error rate overall need not have that same error rate conditional on every history. Assessing type I error is a separate calculation under \(H_0\).

The main alternative to history-by-history calculation is direct simulation of the complete trial under the alternative: repeatedly generate outcomes and assignments according to the specified model and adaptive rule, apply the test, and estimate the rejection proportion. This estimates unconditional power directly and can be practical when enumerating histories is difficult. History-based calculation can be easier to inspect and can identify which paths contribute most to power. Simulation error should be reported or controlled; a simulation estimate is not the exact probability.

Both approaches depend on the assumed outcome model, effect value, randomization rule, and analysis. Power is not a property of an allocation history alone. Nor does a high conditional power on one favorable path establish high unconditional power if that path is uncommon. The calculations also concern rejection probabilities under specified assumptions; they do not, by themselves, establish that an adaptive design is preferable on other grounds.

Common mistake

Common pitfall: “This realized allocation had high power, so the design had high power.” The conclusion confuses a history-specific rejection probability with the design’s average over histories. The correction is to calculate conditional power for each relevant history and weight those values by the probabilities of those histories under the alternative. In response-adaptive designs, also verify that the conditional outcome model and history weights reflect the response-dependent assignment mechanism.

Check Your Understanding

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

A design has two histories with probabilities 0.25 and 0.75 under a specified alternative. Their conditional powers are 0.90 and 0.50. What is unconditional power?
Why is power calculated using expected allocation counts generally not equal to unconditional power?
In a response-adaptive design, why might a fixed-allocation power formula fail to give conditional power for a realized allocation history?
A reported unconditional power is 0.72. Two histories have conditional powers 0.60 and 0.80, and both have positive probability. Is the report possible? Explain.

Solutions. (1) Average using the stated alternative-specific weights: \((0.90)(0.25)+(0.50)(0.75)=0.225+0.375=0.600\). Unconditional power is 60%. (2) Power is a nonlinear function of allocation counts, so evaluating it at the expected counts generally differs from averaging its values over possible histories. (3) Earlier outcomes may influence assignments and thus the probability of a history. Conditioning on that history can change the outcome distribution; the conditional calculation must reflect the adaptive rule and the joint outcome-assignment model. (4) Yes. A weighted average of 0.60 and 0.80 with positive weights must lie strictly between them, and 0.72 does. For example, weights 0.40 and 0.60 give \((0.60)(0.40)+(0.80)(0.60)=0.72\).