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Conditional and Predictive Power · Tutorial 337 of 699

Conditional Power for Two-Sided Success Criteria

Learn to calculate and interpret the conditional probability of crossing either final success boundary in a two-sided normal test, given the interim result and a specified design effect.

Advanced 10 min read

What You'll Learn

  • State the normal independent-increments assumptions needed for a one-remaining-look calculation
  • Derive the conditional distribution of the final test statistic given the interim statistic
  • Add the conditional probabilities of crossing the upper and lower final boundaries
  • Distinguish a two-sided success event from an upper-tail-only calculation
  • Diagnose errors caused by using the wrong boundary, effect assumption, or information fraction
  • Interpret conditional power as a probability conditional on the observed interim data and specified effect

Why Two-Sided Conditional Power Has Two Tails

At an interim analysis, conditional power asks how likely the trial is to meet its final success criterion, given the information observed so far and an assumption about the effect from this point onward. For a two-sided test, success can occur in either direction: the final statistic may cross an upper boundary or a lower boundary. Conditional power must include both possibilities.

The two tail probabilities are not generally equal. The observed interim statistic and the assumed treatment effect determine the conditional distribution at the final analysis. If that distribution is shifted toward the upper boundary, the upper-tail contribution will usually be larger; if it is shifted toward the lower boundary, the reverse may hold. The calculation therefore requires the actual final success boundaries and a clearly stated effect assumption.

Definition: For a trial with one remaining final analysis, conditional power is the probability, conditional on the observed interim data and under a specified assumption about the treatment effect, that the final test statistic will meet the trial's final success criterion. For a two-sided criterion, success is crossing either the upper or lower final critical value.

Model and Conditional Distribution

Suppose there is one interim analysis at information \(I_1\) and one final analysis at information \(I_2\), with \(0<I_1<I_2\). Write \(t=I_1/I_2\) for the interim information fraction, and let \(Z_1=z_1\) be the observed interim statistic. Assume a canonical normal model with known variance or a valid normal approximation: the standardized score accumulated over each information increment is normally distributed, and increments from disjoint information intervals are independent. Under a fixed effect \(\theta\), an increment over information \(\Delta I\) has mean \(\theta\Delta I\) on the score scale and variance \(\Delta I\).

For the conditional calculation, take the assumed effect to be fixed, the final information \(I_2\) to be fixed, and the observed interim statistic to contain the relevant interim information. In particular, there is no unaccounted adaptation that changes the final test's reference distribution or success boundaries. Define the assumed standardized final drift by \(\delta=\theta\sqrt{I_2}\). The statistic at the final analysis is the accumulated score divided by \(\sqrt{I_2}\).

$$ Z_2\mid Z_1=z_1 \sim N\!\left(\sqrt{t}\,z_1+(1-t)\delta,\;1-t\right). $$

Here the second argument of \(N\) is the variance. Thus the conditional mean is \(\mu_c=\sqrt{t}\,z_1+(1-t)\delta\), and the conditional standard deviation is \(s_c=\sqrt{1-t}\). The observed interim value and the assumed design effect both contribute to the conditional mean; the remaining information fraction determines the conditional variance.

Why This Conditional Distribution Holds

Let \(W_1=\sqrt{I_1}Z_1\) be the interim score. The independent score increment from \(I_1\) to \(I_2\), denoted \(W_2-W_1\), has distribution \(N(\theta(I_2-I_1), I_2-I_1)\) under the model. Conditional on the observed interim statistic, the final score is \(W_2=W_1+(W_2-W_1)\), where only the increment remains random.

$$ Z_2 =\frac{\sqrt{I_1}z_1+(W_2-W_1)}{\sqrt{I_2}}. $$

Dividing the conditional mean and variance of this expression by the appropriate powers of \(\sqrt{I_2}\) gives

$$ E(Z_2\mid Z_1=z_1) =\sqrt{t}\,z_1+\frac{\theta(I_2-I_1)}{\sqrt{I_2}} =\sqrt{t}\,z_1+(1-t)\delta, \qquad \operatorname{Var}(Z_2\mid Z_1=z_1) =\frac{I_2-I_1}{I_2} =1-t. $$

This derivation explains why the conditional variance is not 1: the interim score is already observed, so only the remaining information contributes uncertainty. It also shows why the assumed effect enters as \((1-t)\delta\), not as the full final drift \(\delta\).

Two-Sided Conditional Power Formula

Suppose the final test declares success if \(Z_2\geq c_U\) or \(Z_2\leq -c_L\), where \(c_U>0\) and \(c_L>0\) are the actual upper and lower critical-value magnitudes. The events are disjoint, so their conditional probabilities add. With \(\Phi\) denoting the standard normal cumulative distribution function, conditional power is

$$ \begin{aligned} CP(z_1;\delta) &=P(Z_2\geq c_U\mid Z_1=z_1) +P(Z_2\leq -c_L\mid Z_1=z_1)\\ &=1-\Phi\!\left(\frac{c_U-\mu_c}{s_c}\right) +\Phi\!\left(\frac{-c_L-\mu_c}{s_c}\right). \end{aligned} $$

For symmetric final boundaries, \(c_U=c_L=c\). A fixed-sample two-sided test at level \(\alpha\) commonly uses \(c=z_{1-\alpha/2}\), but a group sequential design may use a final boundary different from that value to account for its monitoring plan. Use the boundary specified by the design, not an automatic substitution of 1.96. The formula applies to the final analysis after the observed interim analysis; it does not calculate the probability of success at several future looks.

Worked Example: Symmetric Boundaries With a Specified Design Effect

Suppose \(t=0.25\), the observed interim statistic is \(z_1=0.40\), and the prespecified assumed standardized final drift is \(\delta=0.40\). Let the final success boundaries be symmetric at \(\pm1.96\). Calculate conditional power.

The conditional mean and standard deviation are

$$ \mu_c=\sqrt{0.25}(0.40)+(1-0.25)(0.40) =0.20+0.30=0.50, \qquad s_c=\sqrt{0.75}=0.8660. $$

The upper-boundary contribution is \(P(Z_2\geq1.96\mid Z_1=0.40)\), and the lower-boundary contribution is \(P(Z_2\leq-1.96\mid Z_1=0.40)\). Substituting into the two-tail formula gives

$$ \begin{aligned} CP &=1-\Phi\!\left(\frac{1.96-0.50}{\sqrt{0.75}}\right) +\Phi\!\left(\frac{-1.96-0.50}{\sqrt{0.75}}\right)\\ &\approx 0.04816. \end{aligned} $$

Thus the conditional probability of crossing either final boundary is about \(4.816\%\) under the specified effect assumption and model. The small probability reflects the interim information and the assumed drift; it is not the unconditional power calculated before the interim result was observed.

Worked Example: A Negative Interim Result and Symmetric Boundaries

Suppose \(t=0.50\), \(z_1=-1.20\), and the assumed design effect is \(\delta=0\). The final boundaries are \(\pm1.96\). This example asks for conditional power under the null-effect assumption, not the probability that the null hypothesis is true.

First calculate the conditional distribution:

$$ \mu_c=\sqrt{0.50}(-1.20)+(1-0.50)(0) =-0.8485, \qquad s_c=\sqrt{0.50}=0.7071. $$

The upper standardized value is \((1.96-(-0.8485))/0.7071\approx3.971\), giving an upper-tail probability of about \(0.000036\). The lower standardized value is \((-1.96-(-0.8485))/0.7071\approx-1.572\), giving a lower-tail probability of about \(0.0580\). Therefore,

$$ CP\approx0.000036+0.0580=0.0580. $$

Almost all of this conditional success probability comes from crossing the lower boundary. Calculating only the upper tail would miss the direction toward which the conditional distribution is shifted.

Worked Example: Unequal Two-Sided Boundaries

Suppose a design uses final success boundaries \(Z_2\geq2.10\) or \(Z_2\leq-1.80\). At the interim analysis, \(t=0.40\) and \(z_1=0.50\); the assumed standardized final drift is \(\delta=0.60\). Calculate the two-sided conditional power using the stated boundaries.

The conditional mean and standard deviation are

$$ \mu_c=\sqrt{0.40}(0.50)+(0.60)(0.60) =0.3162+0.3600=0.6762, \qquad s_c=\sqrt{0.60}=0.7746. $$

The upper standardized value is \((2.10-0.6762)/0.7746\approx1.838\), so the upper-tail probability is about \(0.0330\). The lower standardized value is \((-1.80-0.6762)/0.7746\approx-3.197\), so the lower-tail probability is about \(0.00069\). Adding the disjoint crossing probabilities gives

$$ CP\approx0.0330+0.00069=0.0337. $$

The asymmetry of the boundaries matters: the lower critical value is not \(-2.10\), so using a symmetric-boundary formula here would answer a different question.

Checking the Calculation and Interpreting It

A reliable calculation keeps the assumptions, the conditional distribution, and the success event aligned. Before evaluating normal probabilities, check that \(0<t<1\), that \(s_c=\sqrt{1-t}\) is positive, and that the value of \(\delta\) corresponds to the effect assumption being used. Then verify that the upper and lower thresholds are the design's final boundaries. If the trial's boundary is adjusted for sequential monitoring, replacing it with a conventional fixed-sample critical value can materially change the answer.

The two crossing events cannot occur together for positive upper and lower boundaries, so their probabilities add without a correction for overlap. Under a continuous normal model, equality at a boundary has probability zero; using \(>\) or \(\geq\) therefore gives the same probability. In a discrete test, equality may have positive probability and the success rule must specify how boundary equality is treated.

Conditional power is conditional on the interim data and on the chosen effect assumption. It is not the probability that \(H_a\) is true, nor is it the probability of success averaged over all possible interim results. A different assumed effect produces a different conditional power even with the same observed \(z_1\). The value can also be low while unconditional design power was adequate, because the interim result has updated the chance of eventual success.

Common Mistakes, Alternatives, and Limits

  • Using only one tail. The reasoning “the alternative is positive, so only the upper boundary matters” is wrong for a two-sided success rule. Add the conditional lower-tail probability as well, even if it is small.
  • Using the full final drift in the conditional mean. The term from the remaining information is \((1-t)\delta\), not \(\delta\). The interim statistic already reflects information accumulated through \(I_1\).
  • Using variance instead of standard deviation in a z-score. The conditional variance is \(1-t\), but the denominator that standardizes a difference is \(\sqrt{1-t}\).
  • Substituting a familiar critical value without checking the design. A two-sided fixed-sample critical value need not be the final boundary in a group sequential design. Use the stated final boundary on each side.

A one-sided conditional-power calculation is appropriate only when the trial's success criterion is genuinely one-sided; it is not a shortcut for a two-sided test. An unconditional power calculation averages over possible interim outcomes instead of conditioning on the observed one. Conditional power is also distinct from predictive probability, which averages success probability over uncertainty about the effect rather than fixing the effect at a specified value.

The normal calculation is limited by its assumptions. It may be inaccurate when the normal approximation is poor, the variance is estimated in a way not represented by the model, or the future information and test statistic are altered by an adaptation that changes the conditional distribution. With several remaining analyses, success is a union of crossing events across multiple looks and requires a joint sequential calculation; the one-final-look formula above is not sufficient.

Key takeaway: Find the conditional mean and standard deviation of the final statistic, standardize each final boundary separately, and add the upper- and lower-crossing probabilities. Report the interim result, effect assumption, and actual final boundaries alongside the conditional-power value.

Check Your Understanding

Use the one-remaining-final-look model. For each question, state the relevant formula or reasoning before giving the result.

  1. With \(t=0.36\), \(z_1=0\), and \(\delta=0.50\), what are the conditional mean and standard deviation of \(Z_2\)?
  2. For symmetric boundaries \(\pm c\), write the conditional-power formula in terms of \(\mu_c\), \(s_c\), and \(\Phi\).
  3. Why is it incorrect to calculate two-sided conditional power using only \(P(Z_2\geq c\mid Z_1=z_1)\)?
  4. If \(t=0.80\), what happens to the conditional standard deviation compared with the case \(t=0.20\), and why?

Solutions. (1) \(\mu_c=\sqrt{0.36}(0)+(1-0.36)(0.50)=0.32\), and \(s_c=\sqrt{0.64}=0.80\). (2) \(CP=1-\Phi((c-\mu_c)/s_c)+\Phi((-c-\mu_c)/s_c)\). (3) A two-sided success rule includes both \(Z_2\geq c\) and \(Z_2\leq-c\); omitting the lower-tail probability omits one route to success. (4) At \(t=0.80\), \(s_c=\sqrt{0.20}\approx0.4472\); at \(t=0.20\), \(s_c=\sqrt{0.80}\approx0.8944\). More information has already been observed at the interim analysis in the first case, leaving less information—and therefore less conditional variance—remaining.

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