Deriving Conditional Power in the Normal Model

Why Condition on the Interim Result?

At an interim analysis, the observed statistic contains information about the trial’s final statistic. Conditional power quantifies the chance of crossing the final rejection boundary given that interim result, under a specified model for the outcomes still to be observed. To calculate it, we need the distribution of the final statistic conditional on the interim statistic—not merely the final statistic’s distribution before any data are observed.

This tutorial derives that conditional distribution in the canonical normal model. The derivation makes clear why the interim result, the information accrued, and the assumed effect all matter. It also shows why conditional power is a model-based probability for future data, rather than a probability that the treatment effect is true.

Canonical Normal Model and Assumptions

Let \(I_1\) be the information available at the interim analysis and \(I_2\) the planned total information, with \(0<I_1<I_2\). Let \(S(I)\) denote the score-scale statistic accumulated by information level \(I\). Under the normal model with drift \(\theta\), assume \(S(I)\sim N(\theta I,I)\), and assume increments over disjoint information intervals are independent. Thus the future increment from \(I_1\) to \(I_2\) has distribution \(S(I_2)-S(I_1)\sim N(\theta(I_2-I_1),I_2-I_1)\), independently of \(S(I_1)\). The standardized statistic at information \(I\) is \(Z(I)=S(I)/\sqrt I\).

Definition: In this model, \(\theta\) is the assumed drift per unit of information. It determines the mean of the accumulated score, \(\theta I\). Conditional power at a specified \(\theta\) is the probability, under this model for the future increment, that the final statistic crosses its rejection boundary, conditional on the observed interim statistic.

The model is exact, for example, when the relevant score has normal independent increments with known variance and information is correctly measured. Its use requires that \(I_1\) and \(I_2\) refer to the same analysis and that the future information and test boundary are specified. It also requires a coherent effect scale: \(\theta\) is measured per information unit, so changing the absolute information while keeping an information fraction fixed can change the conditional distribution. In applications where variance is estimated or increments are not well approximated by independent normal increments, the formula may be an approximation rather than an exact result.

Deriving the Conditional Distribution

Write \(Z_1=Z(I_1)\) and \(Z_2=Z(I_2)\). Once \(Z_1=z_1\) is observed, the interim score is fixed at \(S(I_1)=z_1\sqrt{I_1}\). The final score is this observed score plus the independent future increment. Dividing their sum by \(\sqrt{I_2}\) gives the conditional distribution directly.

\begin{aligned} Z_2 &=\frac{S(I_1)+\{S(I_2)-S(I_1)\}}{\sqrt{I_2}},\\ Z_2\mid Z_1=z_1 &\sim N\left( \frac{z_1\sqrt{I_1}+\theta(I_2-I_1)}{\sqrt{I_2}}, \frac{I_2-I_1}{I_2} \right). \end{aligned}

The second argument of \(N\) is the variance. The conditional mean can also be written using the information fraction \(t=I_1/I_2\):

\mu_{2\mid 1} =\sqrt{t}\,z_1+\theta\frac{I_2-I_1}{\sqrt{I_2}} =\sqrt{t}\,z_1+\theta\sqrt{I_2}(1-t), \qquad \sigma^2_{2\mid 1}=1-t.

The first term in the mean is the interim score carried forward, rescaled to the final statistic. The second is the expected contribution from the remaining information under the assumed effect. The conditional variance is the fraction of total information that remains. In particular, conditional power depends on \(t\), \(z_1\), and the standardized effect scale \(\theta\sqrt{I_2}\); equivalently, the absolute total information \(I_2\) matters as well as the information fraction.

This derivation uses only the independent-increment assumption and the stated normal distributions. It does not treat the observed interim statistic as a random quantity after conditioning: once \(z_1\) is given, only the future increment remains random.

From the Distribution to Conditional Power

Let \(c\) be the final upper rejection boundary. For an upper-tailed test, the final rejection event is \(Z_2\ge c\). If \(\Phi\) denotes the standard normal cumulative distribution function, standardizing the conditional normal distribution gives the conditional power.

CP_{\theta}(z_1) =P_{\theta}(Z_2\ge c\mid Z_1=z_1) =1-\Phi\left( \frac{c-\mu_{2\mid 1}}{\sqrt{1-t}} \right).

For a two-sided test with final boundaries \(-c\) and \(c\), rejection occurs in either tail. The two conditional tail probabilities must therefore be added:

CP_{\theta,\,2s}(z_1) = \Phi\left(\frac{-c-\mu_{2\mid 1}}{\sqrt{1-t}}\right) + 1-\Phi\left(\frac{c-\mu_{2\mid 1}}{\sqrt{1-t}}\right).

These expressions calculate power under the specified value of \(\theta\), conditional on the observed \(z_1\). The effect used for the calculation must be identified: it could be a design alternative or an effect estimate chosen for a plug-in calculation. The formula itself does not select that effect or establish that it is correct.

Worked Example: Upper-Tailed Conditional Power

Suppose a hypothetical trial has \(I_1=50\), planned \(I_2=100\), observed \(z_1=1.20\), and an assumed drift \(\theta=0.20\) per information unit. The final upper boundary is \(c=1.96\). The independent-increment normal model is assumed, the information values are fixed, and the specified \(\theta\) is on the score scale used to define \(I\).

The remaining information is \(I_2-I_1=50\), and the conditional mean and standard deviation are

\mu_{2\mid 1} =\frac{1.20\sqrt{50}+0.20(50)}{\sqrt{100}} =0.8485+1.000 =1.8485, \qquad \sigma_{2\mid 1} =\sqrt{\frac{50}{100}} =0.7071.

Thus the conditional power is

CP_{\theta}(1.20) =1-\Phi\left(\frac{1.96-1.8485}{0.7071}\right) =1-\Phi(0.1576) \approx 0.4374.

Under the assumed drift and the observed interim result, the model assigns about a 43.74% probability to a final statistic at or above 1.96. This is not the unconditional power of the trial, which averages over possible interim results.

Worked Example: Two-Sided Conditional Power

Consider a hypothetical two-sided test with final boundaries \(-1.96\) and \(1.96\). Let \(I_1=40\), \(I_2=100\), \(z_1=1.50\), and assume \(\theta=0.1184\). The canonical normal model and independent future increments are assumed, and the same information scale is used for the specified drift and observed statistic.

Here \(t=0.40\), and the conditional mean and standard deviation are

\mu_{2\mid 1} =\sqrt{0.40}(1.50)+0.1184\frac{60}{10} =0.9487+0.7104 =1.6591, \qquad \sigma_{2\mid 1}=\sqrt{0.60}=0.7746.

Using the displayed rounded values, the upper-tail standardized value is \((1.960-1.659)/0.7746=0.3886\), and the lower-tail value is \((-1.960-1.659)/0.7746\approx-4.672\). Therefore,

\begin{aligned} CP_{\theta,\,2s}(1.50) &=\Phi(-4.672)+1-\Phi(0.3886)\\ &\approx 0.0000015+0.3488\\ &\approx 0.3488. \end{aligned}

The lower-tail probability is negligible at the reported precision, so almost all of this conditional rejection probability comes from crossing the upper boundary. Both tails are included because the test is two-sided.

Worked Example: How the Assumed Effect Changes the Answer

For a third hypothetical calculation, let \(I_1=25\), \(I_2=100\), \(z_1=0.50\), and use an upper boundary \(c=1.645\). Assume the same fixed-information normal model. First calculate using \(\theta=0.10\); then compare with the calculation at \(\theta=0\).

At \(\theta=0.10\), \(t=0.25\), so

\mu_{2\mid 1} =\sqrt{0.25}(0.50)+0.10\frac{75}{10} =0.25+0.75 =1.00, \qquad \sigma_{2\mid 1}=\sqrt{0.75}=0.8660.

The conditional power is

1-\Phi\left(\frac{1.645-1.00}{\sqrt{0.75}}\right) =1-\Phi(0.7448) \approx 0.2282.

If instead \(\theta=0\), the conditional mean is \(0.25\), while the standard deviation remains \(0.8660\). The resulting conditional rejection probability is \(1-\Phi\{(1.645-0.25)/0.8660\} =1-\Phi(1.611)\approx0.0536\). The interim statistic and information values are unchanged, but the assumed future drift differs. This illustrates why a conditional-power value must always be reported with its assumed effect.

Interpretation, Diagnostics, and Alternatives

Conditional power is useful because it isolates the model’s prediction for the remaining data after the interim result is known. The calculation is straightforward, but its interpretation depends on assumptions that should be checked rather than taken for granted.

Check the information scale. Confirm that \(I_1\) and \(I_2\) refer to the same statistic and that the assumed \(\theta\) is expressed per unit of that information. Keeping \(I_1/I_2\) fixed does not make the absolute information irrelevant: the effect term includes \(\theta\sqrt{I_2}\).
Check the remaining information. If \(I_1\) is close to \(I_2\), the conditional standard deviation \(\sqrt{1-t}\) is small. A small change in the conditional mean relative to the boundary can then produce a large change in the tail probability. At \(I_1=I_2\), there is no future increment and the conditional distribution becomes degenerate; the formula with a positive denominator is not applicable.
Check the assumed effect and boundary. Replacing a two-sided rejection event with only an upper tail, or using a boundary from a different testing procedure, changes the probability being calculated. Report whether the calculation uses a design effect or a plug-in estimate.
Check the model’s fit to the design. Materially estimated variance, changing information definitions, dependence not captured by the canonical model, or departures from normal independent increments can undermine the calculation. In such cases, the stated normal tail probability may be only an approximation.

A plug-in conditional-power calculation uses an interim estimate to supply an effect for the future increment; an assumed-effect calculation instead fixes the effect to a prespecified value. Both use the conditional distribution derived here, but they answer different model-based questions. Unconditional power is another alternative: it averages rejection probability over the possible interim results rather than conditioning on one observed \(z_1\). Simulation can be useful when the design’s data-generating process or decision rules are too complex for this simple normal calculation, but its accuracy then depends on correctly representing those features.

A central limitation is that conditional power does not, by itself, guarantee that a design with interim-driven changes retains its original type I error rate. It is a conditional probability under a specified model, not an error-control procedure. When interim results influence later design choices, the final testing rule and its operating characteristics require appropriate evaluation. Nor should a high conditional power be read as evidence that the assumed effect is true; it is the probability of a specified future rejection event conditional on the observed interim statistic and the model assumptions.

Common Pitfall

Common mistake: Treating conditional power as a function only of the information fraction and observed interim statistic, or interpreting it as an unconditional probability that the study will succeed. The correction is to include the assumed effect on the correct information scale: the conditional mean contains \(\theta\sqrt{I_2}(1-t)\). The result is conditional on \(z_1\), and its value depends on both the assumed effect and the planned total information.

Check Your Understanding

Work through each question, then compare with the solutions below.

Under the canonical normal model, what are the conditional mean and variance of \(Z_2\) given \(Z_1=z_1\)?
For an upper-tailed test, why is conditional power \(1-\Phi\{(c-\mu_{2\mid1})/\sqrt{1-t}\}\), rather than \(\Phi\{(c-\mu_{2\mid1})/\sqrt{1-t}\}\)?
With \(t=0.50\), \(z_1=0\), and \(\theta\sqrt{I_2}=1\), calculate the conditional mean and variance of \(Z_2\).
State one reason that fixing the information fraction alone does not determine conditional power.

Solutions. (1) The mean is \(\sqrt{t}\,z_1+\theta\sqrt{I_2}(1-t)\), and the variance is \(1-t\). (2) Rejection is the upper-tail event \(Z_2\ge c\), so standardizing gives one minus the standard normal cumulative probability at the standardized boundary. (3) The mean is \(\sqrt{0.50}(0)+1(1-0.50)=0.50\), and the variance is \(0.50\). (4) At the same fraction, changing the absolute \(I_2\) changes the standardized effect \(\theta\sqrt{I_2}\) when \(\theta\) is held fixed, and can therefore change the conditional mean and tail probability.