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Conditional and Predictive Power · Tutorial 344 of 699

A Worked Conditional-Power Calculation

Learn to calculate and compare plug-in and design-effect conditional power from an interim result in a two-stage normal trial.

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What You'll Learn

  • Set up the two-stage normal model and its information assumptions.
  • Derive the conditional distribution of the final test statistic given interim data.
  • Calculate conditional power under a plug-in effect estimate.
  • Calculate conditional power under a prespecified design effect.
  • Interpret differences between the two calculations and identify when assumptions may fail.

Why Calculate Conditional Power From Interim Data?

At an interim analysis, a trial team may ask how likely the trial is to meet its final success criterion if it continues as planned. Conditional power answers this question by conditioning on the data already observed and specifying an effect size for the future-data model. It is not one fixed property of the trial: the answer can change substantially depending on which effect size is assumed.

This tutorial works through a two-stage normal trial with one final efficacy test. It calculates conditional power in two common ways: using the interim effect estimate as the assumed effect, and using a prespecified design effect. The first reflects the interim estimate; the second asks what success probability remains if a chosen effect, often the effect used in the design, is the true effect.

Definition: Conditional power at an interim analysis is the probability, under a specified model and assumed treatment effect, that the trial will satisfy its final success criterion, conditional on the interim data observed and on the trial continuing. The assumed effect is part of the definition of a particular conditional-power calculation.

Two-Stage Normal Model and Calculation

Let \(I_1\) be the information at the interim analysis and \(I_2\) the planned total information at the final analysis, with \(0<I_1<I_2\). Let \(\theta\) be the treatment effect, with positive values favoring treatment. Information is measured in inverse squared effect units. In the canonical normal model, the cumulative score \(S_k\) at information \(I_k\) has mean \(\theta I_k\) and variance \(I_k\), and score increments from disjoint information intervals are independent. The standardized statistic is \(Z_k=S_k/\sqrt{I_k}\).

For the calculations below, assume the effect is constant across the trial, the information values are fixed and known, and the normal independent-increments model is adequate. Suppose the final test is one-sided, rejects \(H_0:\theta=0\) in favor of \(H_a:\theta>0\) when \(Z_2>c\), and has significance level \(\alpha=0.025\). Thus \(c=z_{1-\alpha}\), approximately \(1.960\). If the interim result is \(Z_1=z_1\), the conditional power under an assumed effect \(\theta_A\) is calculated from the conditional distribution of \(Z_2\).

$$ S_2=S_1+(S_2-S_1), \qquad S_2-S_1 \sim N\!\left(\theta_A(I_2-I_1),\,I_2-I_1\right) $$

The future increment is independent of the observed interim score \(S_1=\sqrt{I_1}z_1\). Dividing the sum by \(\sqrt{I_2}\) gives the conditional mean and variance of the final statistic.

$$ Z_2 \mid Z_1=z_1 \sim N\!\left( \sqrt{\frac{I_1}{I_2}}\,z_1+ \frac{\theta_A(I_2-I_1)}{\sqrt{I_2}}, \quad 1-\frac{I_1}{I_2} \right) $$

The second argument of \(N\) is the variance. Therefore the conditional standard deviation is \(\sqrt{1-I_1/I_2}\), and the one-sided conditional power is

$$ \operatorname{CP}(\theta_A) = \Phi\!\left[ \frac{ \sqrt{I_1/I_2}\,z_1+ \theta_A(I_2-I_1)/\sqrt{I_2}-c }{ \sqrt{1-I_1/I_2} } \right], $$

where \(\Phi\) is the standard normal cumulative distribution function. The numerator is the conditional mean of \(Z_2\) minus the final boundary; the denominator expresses the remaining uncertainty from future data.

1
Fix the success rule.
Identify the final boundary \(c\) and the planned total information \(I_2\).
2
Record the interim information.
Use the observed \(z_1\) and \(I_1\) to specify what is already known.
3
Choose the assumed effect.
For plug-in conditional power, set \(\theta_A=\hat{\theta}_1=z_1/\sqrt{I_1}\). For design-effect conditional power, set \(\theta_A=\theta_D\), the prespecified effect.
4
Standardize and evaluate.
Substitute into the conditional-power formula and evaluate \(\Phi\) at the resulting standardized distance above the boundary.

Worked Calculations

Worked Example: Comparing Plug-In and Design-Effect Conditional Power

Consider a hypothetical two-stage trial of a continuous outcome measured in score units. At the interim analysis, \(I_1=50\) inverse score-squared units and \(z_1=1.40\); the planned total information is \(I_2=100\) inverse score-squared units. Use \(c=1.960\). Calculate conditional power under the interim plug-in estimate and under a design effect of \(\theta_D=0.30\) score units.

The plug-in effect estimate is \(\hat{\theta}_1=z_1/\sqrt{I_1}\). Thus

$$ \hat{\theta}_1=\frac{1.40}{\sqrt{50}}=0.19799\ \text{score units}. $$

The conditional standard deviation is \(\sqrt{1-50/100}=0.7071\). Under the plug-in effect, the conditional mean is

$$ \sqrt{\frac{50}{100}}(1.40)+ \frac{0.19799(100-50)}{\sqrt{100}} =0.9899+0.9899=1.9799. $$

The standardized distance above the boundary is \((1.9799-1.960)/0.7071=0.0281\), so plug-in conditional power is \(\Phi(0.0281)=0.511\), or about 51.1%.

Under the design effect, the conditional mean is

$$ \sqrt{\frac{50}{100}}(1.40)+ \frac{0.30(100-50)}{\sqrt{100}} =0.9899+1.5000=2.4899. $$

The standardized distance is \((2.4899-1.960)/0.7071=0.7495\), giving \(\Phi(0.7495)=0.773\), or about 77.3% conditional power. These results differ because they answer different questions: one assumes the interim estimate persists, while the other assumes an effect of 0.30 score units.

Worked Example: A Smaller Interim Estimate

In a second hypothetical trial, \(I_1=40\), \(I_2=100\), and \(z_1=0.80\). Use the same one-sided boundary \(c=1.960\) and a design effect \(\theta_D=0.25\) in the relevant effect units. Calculate conditional power under both assumptions.

The conditional standard deviation is \(\sqrt{1-40/100}=\sqrt{0.6}=0.7746\). The plug-in estimate is \(0.80/\sqrt{40}=0.1265\). Its conditional mean is

$$ \sqrt{0.4}(0.80)+\frac{0.1265(60)}{10} =0.5060+0.7590=1.2649. $$

The standardized distance is \((1.2649-1.960)/0.7746=-0.8974\), so plug-in conditional power is \(\Phi(-0.8974)=0.185\), about 18.5%.

Under the design effect, the conditional mean becomes

$$ \sqrt{0.4}(0.80)+\frac{0.25(60)}{10} =0.5060+1.5000=2.0060. $$

The standardized distance is \((2.0060-1.960)/0.7746=0.0593\), giving \(\Phi(0.0593)=0.524\), about 52.4%. The low plug-in value does not mean that the design-effect calculation is wrong: it reflects a different assumed effect. Neither number is the probability that the treatment works.

Worked Example: A Favorable Interim Estimate

Suppose a third hypothetical trial has \(I_1=75\), \(I_2=120\), and \(z_1=2.30\). Use \(c=1.960\), and compare plug-in conditional power with conditional power under \(\theta_D=0.20\) effect units.

Here \(\sqrt{1-75/120}=\sqrt{0.375}=0.6124\), and the interim estimate is \(2.30/\sqrt{75}=0.2656\) effect units. Under the plug-in effect, the conditional mean is

$$ \sqrt{\frac{75}{120}}(2.30)+ \frac{0.2656(45)}{\sqrt{120}} =1.8183+1.0910=2.9093. $$

The standardized distance is \((2.9093-1.960)/0.6124=1.550\), yielding plug-in conditional power \(\Phi(1.550)=0.939\), about 93.9%.

Under the design effect, the conditional mean is

$$ \sqrt{\frac{75}{120}}(2.30)+ \frac{0.20(45)}{\sqrt{120}} =1.8183+0.8216=2.6399. $$

The standardized distance is \((2.6399-1.960)/0.6124=1.110\), giving \(\Phi(1.110)=0.867\), about 86.7%. Both calculations indicate a high conditional chance of crossing the final boundary, but the plug-in result is higher because the interim estimate exceeds the design effect.

Why the Two Calculations Differ

The distinction can be seen directly from the assumed effect. The plug-in approach uses \(\hat{\theta}_1=z_1/\sqrt{I_1}\), the effect estimate from the interim data. The design-effect approach uses a fixed \(\theta_D\), chosen before or independently of the interim estimate. For fixed \(z_1\), the conditional mean increases with \(\theta_A\) when \(I_2>I_1\), so conditional power increases with the assumed effect. A larger plug-in result than design-effect result is therefore expected when \(\hat{\theta}_1>\theta_D\), and the reverse is expected when \(\hat{\theta}_1<\theta_D\).

As a check, substituting the plug-in estimate into the conditional mean gives \(\sqrt{I_2/I_1}\,z_1\). This follows because

$$ \sqrt{\frac{I_1}{I_2}}z_1+ \frac{(z_1/\sqrt{I_1})(I_2-I_1)}{\sqrt{I_2}} =z_1\sqrt{\frac{I_2}{I_1}}. $$

This simplified expression is useful for checking arithmetic, but it does not remove the underlying assumption that the interim effect estimate is the effect to use for future increments.

Conditional power differs from unconditional power, which averages over possible interim results before the trial begins. It also differs from a p-value: a p-value is calculated from observed data under \(H_0\), whereas conditional power is a future-success probability conditional on the current data and an assumed effect. It is not a probability that \(H_0\) is true, nor is it a guarantee that the final test will succeed.

Assumptions, Diagnostics, and Limitations

Before using the calculation, confirm that the quantities have the intended meaning. Information must be cumulative, so \(I_2>I_1>0\); if \(I_2=I_1\), there is no future information and the formula's conditional variance is zero. The interim statistic and information must correspond to the same analysis and scale. The direction of \(Z_1\), \(\theta_A\), and the rejection boundary must also agree: positive values in these examples favor treatment.

The normal model assumes independent future increments with the stated mean and variance. Dependence between increments, an effect that changes over time, or a substantial departure from the normal approximation can make the calculation unreliable. Information may also differ from the planned amount if recruitment, measurement, or follow-up changes. Replacing \(I_2\) with a revised value is not just an arithmetic update unless the revised design and its assumptions are justified.

A useful diagnostic is to check the conditional standard deviation and the direction of the result. As \(I_1/I_2\) approaches one, little information remains and the conditional variance approaches zero; a result far from the final boundary will then imply a conditional power near zero or one. If \(\theta_A\) is increased while the other inputs remain fixed, conditional power for this positive one-sided test should not decrease. A violation usually signals a sign, boundary, or substitution error.

Plug-in conditional power can be unstable because the interim estimate is noisy, particularly when \(I_1\) is small. Design-effect conditional power avoids using that estimate as the future effect, but its answer depends on whether the chosen effect is plausible. Both are conditional calculations under a fixed effect, not methods for averaging over uncertainty about the effect. If the variance is estimated, information is uncertain, or the future design changes, the simple known-information normal calculation may require modification.

Common Pitfall

Common mistake: Reporting one conditional-power value as though it were an objective probability of eventual success, without stating the assumed effect. Correction: Name the conditioning data, the effect assumption, and the final success rule. For example, “Conditional power is 77.3% given \(z_1=1.40\) and assuming \(\theta=0.30\) score units, for a final one-sided boundary of 1.960.” This makes clear that a different assumed effect can produce a different probability.

Check Your Understanding

Try each question, then compare your work with the solutions.

  1. With \(I_1=36\), \(I_2=64\), \(z_1=1.00\), and \(\theta_D=0.25\), calculate design-effect conditional power for \(c=1.960\).
  2. For \(I_1=25\), \(I_2=100\), and \(z_1=1.20\), find the plug-in effect estimate and plug-in conditional power using \(c=1.960\).
  3. For a positive assumed effect and a positive one-sided test, what should happen to conditional power if the assumed effect increases while all other inputs stay fixed? Why?
  4. State two reasons why a calculated conditional power may be unreliable in an actual trial.

Solutions. (1) The conditional standard deviation is \(\sqrt{1-36/64}=\sqrt{0.4375}=0.6614\). The conditional mean is \(\sqrt{36/64}(1.00)+0.25(28)/8=0.7500+0.8750=1.6250\). Thus the standardized distance is \((1.6250-1.960)/0.6614=-0.5065\), and conditional power is \(\Phi(-0.5065)=0.306\), about 30.6%. (2) The plug-in estimate is \(1.20/\sqrt{25}=0.2400\). The conditional standard deviation is \(\sqrt{0.75}=0.8660\), and the conditional mean is \(\sqrt{0.25}(1.20)+0.2400(75)/10=0.6000+1.8000=2.4000\). Therefore conditional power is \(\Phi((2.4000-1.960)/0.8660)=\Phi(0.5081)=0.694\), about 69.4%. (3) It should increase: a larger positive assumed effect increases the conditional mean of the final statistic, moving it farther above the fixed boundary. (4) Examples include dependent or non-normal future increments, a changing treatment effect, incorrect information values, and a future design that differs materially from the planned design. Any two, with a brief explanation, are sufficient.

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