Worked Normal-Endpoint Conditional-Power Calculation

Why Condition on the Interim Result?

At an interim analysis, the observed Z statistic is already known. Conditional power answers a forward-looking question: if a specified effect governed the data collected after this interim analysis, what is the probability that the final test would cross its efficacy boundary? It is a probability for future data conditional on the interim result and on an explicitly chosen effect assumption. It is not the probability that the assumed effect is true.

This tutorial treats a normally distributed endpoint and a one-sided upper efficacy boundary. The final information level and boundary are inputs to the calculation. As in the earlier material on target conditional power and futility, the chosen effect assumption matters; here the emphasis is on carrying out and checking the normal calculation.

Definition: Let $Z_1=z_1$ be the observed interim Z statistic at information $I_1$, and let $Z_2$ be the final Z statistic at information $I_2$, where $0<I_1<I_2$. For an assumed treatment effect $\theta_A$ and final upper efficacy boundary $c$, the conditional power is $$ \operatorname{CP}(\theta_A)=P_{\theta_A}(Z_2\geq c\mid Z_1=z_1). $$ The subscript indicates that the probability for future data is calculated under the assumed effect $\theta_A$.

Model and Conditional Distribution

Suppose the endpoint is normal, the variance is known for this calculation, and observations are independent across participants. Let $\theta$ denote the treatment-minus-control mean difference, in endpoint units. Write $U(I)$ for the cumulative score at information $I$. Under the normal model, assume $U(I)\sim N(\theta I,I)$, with independent score increments over nonoverlapping information intervals. The standardized statistic is $Z(I)=U(I)/\sqrt{I}$; therefore its mean under effect $\theta$ is $\theta\sqrt{I}$.

Information is measured on the scale for which the variance of the estimated treatment difference is $1/I$. Thus $I$ has units inverse to the squared endpoint units, and $\theta\sqrt{I}$ is dimensionless. For example, with a normal endpoint and equal allocation, a known common variance $\sigma^2$, and total sample size $N$, the information for a difference in means is $I=N/(4\sigma^2)$.

At the interim, $U(I_1)=\sqrt{I_1}z_1$ is observed. The future increment $U(I_2)-U(I_1)$ is independent of the interim score and, under the assumed effect, has distribution $N(\theta_A(I_2-I_1),I_2-I_1)$. Since the final statistic is the total score divided by $\sqrt{I_2}$, its conditional distribution follows directly.

Z_2=\frac{\sqrt{I_1}z_1+\{U(I_2)-U(I_1)\}}{\sqrt{I_2}}, \qquad Z_2\mid Z_1=z_1\sim N\left( \frac{\sqrt{I_1}z_1+\theta_A(I_2-I_1)}{\sqrt{I_2}}, \frac{I_2-I_1}{I_2} \right).

The second argument of $N(\mu,\sigma^2)$ is the variance. Define the information fraction $t=I_1/I_2$. The conditional mean and standard deviation can then be written as $m=\sqrt{t}\,z_1+\theta_A\sqrt{I_2}(1-t)$ and $s=\sqrt{1-t}$. Standardizing the boundary gives the calculation used in each example.

\operatorname{CP}(\theta_A) =1-\Phi\left(\frac{c-m}{s}\right) =\Phi\left( \frac{\sqrt{t}\,z_1+\theta_A\sqrt{I_2}(1-t)-c} {\sqrt{1-t}} \right),

where $\Phi$ is the standard normal cumulative distribution function. The formula requires $0<I_1<I_2$. If the interim and final information are equal, there is no future increment: the final statistic is already determined by the interim data, so the probability is 1 or 0 according to whether $z_1$ crosses $c$. The formula also requires that the boundary $c$ match the test being planned; a one-sided upper-boundary formula is not a substitute for a two-sided or otherwise specified final decision rule.

Worked Calculations

Worked Example: Moderate Interim Evidence at the Planned Information

Suppose a hypothetical normal-endpoint trial has interim information $I_1=40$ and planned final information $I_2=100$, in inverse squared endpoint units. The interim statistic is $z_1=1.20$. The assumed treatment difference is $\theta_A=0.18$ endpoint units, and the one-sided final boundary is $c=1.96$. Calculate conditional power.

First, $t=40/100=0.40$. The conditional mean is $\sqrt{0.40}(1.20)+0.18\sqrt{100}(1-0.40)$. Substituting and calculating, $0.63246(1.20)+0.18(10)(0.60)=0.75895+1.0800=1.83895$. The conditional variance is $1-t=0.60$, so the conditional standard deviation is $\sqrt{0.60}=0.77460$.

Therefore, $(c-m)/s=(1.96-1.83895)/0.77460=0.1563$, and $\operatorname{CP}=\Phi(-0.1563)\approx 0.4379$.

Under the assumed treatment difference of 0.18 endpoint units, and conditional on the observed interim statistic of 1.20, the probability of crossing the final boundary is about 43.8%. This is not an unconditional probability over both interim and future data.

Worked Example: Recalculating at a Larger Final Information Level

Now consider a hypothetical trial with $I_1=40$, observed $z_1=0.60$, and an assumed treatment difference $\theta_A=0.18$ endpoint units. Suppose the final information level is set to $I_2=140$, and the upper boundary remains $c=1.96$. Calculate the conditional power for these inputs.

Here $t=40/140=2/7$. The conditional mean is $\sqrt{2/7}(0.60)+0.18\sqrt{140}(5/7)$. Equivalently, the second term is $0.18(140-40)/\sqrt{140}$. Substitution gives $0.53452(0.60)+0.18(100)/11.8322=0.32071+1.5213=1.8420$.

The conditional variance is $1-2/7=5/7=0.7143$, giving standard deviation $\sqrt{5/7}=0.84515$. Thus $(1.96-1.8420)/0.84515=0.1396$, and the conditional power is $\Phi(-0.1396)\approx 0.4445$.

For these specified inputs, conditional power is about 44.5%. The increase in final information does not guarantee a large conditional power: the observed interim statistic is modest, and the calculation still depends on the assumed effect and the final boundary.

Worked Example: A Favorable Interim Statistic Under a Zero-Effect Assumption

Suppose a hypothetical trial has $I_1=60$, $I_2=100$, an observed interim statistic $z_1=2.30$, and final upper boundary $c=1.96$. Calculate conditional power under the assumption $\theta_A=0$.

The information fraction is $t=0.60$. With $\theta_A=0$, the conditional mean is $\sqrt{0.60}(2.30)+0(10)(0.40)=0.77460(2.30)=1.7816$. The conditional standard deviation is $\sqrt{0.40}=0.63246$. Hence $(1.96-1.7816)/0.63246=0.2821$, and $\operatorname{CP}(0)=\Phi(-0.2821)\approx 0.3889$.

Even with a favorable interim statistic, the chance of crossing the final boundary is about 38.9% if the future increment has zero mean under this assumption. This is a conditional rejection probability under a particular effect assumption, not an estimate of the probability that the null hypothesis is true.

Justification and Interpretation

The key step is using the future score increment rather than treating the final statistic as independent of the interim statistic. The observed interim score remains part of the final cumulative score, while only the increment from $I_1$ to $I_2$ is random after conditioning. In particular, under $\theta_A$, the expected future score increment is $\theta_A(I_2-I_1)$, and its variance is $I_2-I_1$. Dividing the sum of interim and future scores by $\sqrt{I_2}$ yields the conditional mean and variance above.

This also explains why conditional power is not the same as ordinary power. Unconditional power averages over all possible interim results as well as future data under a specified effect. Conditional power holds the observed interim result fixed and averages only over future data. If the assumed effect is set equal to an estimate obtained from the interim data, that produces a plug-in conditional power; it is a different choice of effect assumption from using a prespecified planning effect.

Key takeaway: Conditional power is determined by four substantive inputs: the observed interim statistic, the assumed effect for future data, the information remaining, and the final efficacy boundary. Report the assumption and boundary alongside the probability so the number can be interpreted.

Checks, Alternatives, and Limitations

Before relying on a result, check that the information levels refer to the same endpoint and treatment contrast, that $I_2>I_1$, and that the assumed effect is expressed in compatible endpoint units. The information fraction must lie strictly between zero and one for the displayed formula. Check the boundary's sidedness and value against the final test: inserting a familiar critical value without confirming the design's boundary can give a numerically correct answer to the wrong question.

Dependence: The interim and final Z statistics are correlated. Treating them as independent discards the interim contribution to the final score and gives the wrong conditional mean and variance.
Effect assumption: A result under $\theta_A=0$ can differ substantially from one under a positive planning effect. Do not report conditional power without naming the assumed effect.
Information scale: A treatment difference in endpoint units must be combined with information on the reciprocal squared-unit scale. Mixing an effect measured in standardized units with information defined for an unstandardized contrast is inconsistent.
Model fit: The calculation relies on normal score increments with the stated variance and independent future data. Strong departures from the normal model, incorrect variance assumptions, or data dependence can make the numerical approximation unreliable.

A fixed-design power calculation is the main alternative when the question concerns success probability before any interim data are observed. Conditional power is useful after observing interim data, but answers a different question. A plug-in conditional calculation using the interim effect estimate is another alternative to using a fixed planning effect; it should be labelled as such and not treated as a neutral default. Conditional power alone also does not establish that a sample-size adaptation preserves the intended overall error rate; that requires the design's full testing procedure.

Common mistake: Calculating final power under the assumed effect as if the interim statistic had not been observed, then calling that conditional power. The correction is to condition on $Z_1=z_1$: retain the observed interim score in the final numerator, and model only the independent future increment under the assumed effect.

Check Your Understanding

For each question, identify the conditional-power inputs and then compare your reasoning with the solutions.

For $I_1=25$, $I_2=100$, $z_1=0.50$, $\theta_A=0.10$, and $c=1.96$, calculate the conditional mean and standard deviation.
Using the inputs in Question 1, calculate the conditional power to three significant figures.
At fixed $I_1$, $I_2$, $z_1$, and $c$, what happens to the conditional mean when the assumed effect increases? Explain the direction.
Why is conditional power not the probability that $H_0$ is false?
A calculation uses $I_1=80$ and $I_2=60$ in the displayed formula. What is wrong, and what must be checked?

Solutions. (1) $t=25/100=0.25$. The conditional mean is $\sqrt{0.25}(0.50)+0.10\sqrt{100}(0.75)=0.25+0.75=1.00$; the standard deviation is $\sqrt{0.75}=0.8660$. (2) The boundary standardization is $(1.96-1.00)/0.8660=1.109$, so the power is $\Phi(-1.109)\approx 0.134$. (3) The conditional mean increases by $\sqrt{I_2}(1-t)$ times the increase in the assumed effect, a positive amount when $I_2>I_1$. With a fixed boundary and variance, conditional power therefore increases. (4) It is a probability for future data under a specified effect, conditional on the observed interim statistic; it does not assign probabilities to hypotheses. (5) The final information cannot be below the interim information in this continuation calculation. Verify the information definitions and study plan; if information has already decreased or the design has changed, this formula's assumptions do not apply as written.