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Conditional and Predictive Power · Tutorial 333 of 699

Conditional Power for a Single Final Look

Learn how to calculate and interpret the probability of crossing a single final one-sided boundary, conditional on the interim Z-statistic and a specified treatment effect.

Advanced 12 min read

What You'll Learn

  • State the assumptions for a canonical normal test with independent information increments.
  • Derive the final Z-statistic distribution conditional on the observed interim Z-statistic.
  • Calculate single-final-look conditional power under a specified effect and critical value.
  • Distinguish conditional power from unconditional design power and predictive probability.
  • Diagnose arithmetic, boundary, and modeling errors in conditional-power calculations.

Why Condition on the Interim Result?

An interim analysis provides information about the evidence already observed and the data still to come. Conditional power asks a focused question: given the interim Z-statistic, how likely is the final test to cross its success boundary if a specified treatment effect is the truth? It is therefore a probability about future evidence under an assumed effect, conditional on the evidence already observed.

This tutorial considers one one-sided normal test and exactly one remaining final look. The final critical value is fixed, and the assumed effect is specified rather than estimated from the interim result. The key calculation comes from the conditional distribution of the final Z-statistic given the interim Z-statistic.

Model, Assumptions, and Definition

Let \(I_1\) be the information at the interim analysis and \(I_2\) the total information at the final analysis, with \(0<I_1<I_2\). Write \(Z_1\) and \(Z_2\) for the interim and final Z-statistics. The direction of benefit is positive, so the final test rejects \(H_0\) when \(Z_2\geq c\), where \(c\) is the prespecified final critical value. For a one-sided test of size \(\alpha\), a common choice is the upper standard-normal quantile satisfying \(P\{N(0,1)\geq c\}=\alpha\).

Use the canonical normal model under an assumed effect \(\theta\): the information-scaled statistic \(U(I)=\sqrt{I}Z(I)\) has mean \(\theta I\), variance \(I\), and independent increments over disjoint information intervals. Equivalently, \(U(I_1)\sim N(\theta I_1,I_1)\), and \(U(I_2)-U(I_1)\sim N(\theta(I_2-I_1),I_2-I_1)\), independently of \(U(I_1)\). Here \(\theta\) is the effect parameter in this canonical model, so the mean of \(Z(I)\) is \(\theta\sqrt I\).

This model is appropriate when the normal approximation is adequate, the information values are treated as fixed, the final analysis uses the stated Z-statistic, and the future increment is independent of the interim data. For example, it applies to a normal endpoint with known variance under the usual fixed-design normal model. It does not automatically account for an estimated variance, changing information, an additional future look, or an unmodeled change to the analysis. The independent-increments result is the conditional-distribution foundation used here.

Definition: For an observed interim value \(Z_1=z_1\), assumed effect \(\theta\), and fixed final boundary \(c\), the single-final-look conditional power is \(P_{\theta}(Z_2\geq c\mid Z_1=z_1)\). The subscript emphasizes that the probability is calculated under the specified effect \(\theta\), not averaged over uncertainty about that effect.

Deriving the Conditional Distribution and Power

Once \(Z_1=z_1\) is observed, the interim information-scaled statistic is known: \(U(I_1)=\sqrt{I_1}z_1\). The final statistic adds the independent increment from the remaining information \(I_2-I_1\). Dividing that sum by \(\sqrt{I_2}\) converts it back to the final Z-statistic.

1
Write the final statistic as interim plus future increment.
\(Z_2=\{U(I_1)+[U(I_2)-U(I_1)]\}/\sqrt{I_2}\).
2
Condition on the observed interim value.
Given \(Z_1=z_1\), the first term in the numerator is the fixed value \(\sqrt{I_1}z_1\). The independent future increment is normal with mean \(\theta(I_2-I_1)\) and variance \(I_2-I_1\).
3
Divide the conditional mean and variance by the appropriate powers of \(\sqrt{I_2}\).
The resulting conditional distribution has mean \(\sqrt{I_1/I_2}\,z_1+\theta(I_2-I_1)/\sqrt{I_2}\) and variance \((I_2-I_1)/I_2\).

In full, the conditional distribution and its upper-tail probability are

$$ Z_2\mid Z_1=z_1 \sim N\left( \sqrt{\frac{I_1}{I_2}}\,z_1+ \frac{\theta(I_2-I_1)}{\sqrt{I_2}}, \frac{I_2-I_1}{I_2} \right). $$
$$ \operatorname{CP}_{\theta}(z_1) = 1-\Phi\left[ \frac{ c-\sqrt{I_1/I_2}\,z_1-\theta(I_2-I_1)/\sqrt{I_2} }{ \sqrt{(I_2-I_1)/I_2} } \right], $$

where \(\Phi\) is the standard normal cumulative distribution function. This follows by subtracting the conditional mean from both sides of \(Z_2\geq c\) and dividing by the positive conditional standard deviation. The numerator of the standardized boundary is the distance from the conditional mean to the final boundary; a larger conditional mean therefore increases conditional power.

Worked Example: A Favorable Interim Result Under a Specified Effect

Suppose a hypothetical trial has \(I_1=50\) units of information at interim and \(I_2=100\) at the final analysis. The observed interim statistic is \(z_1=1.00\). Use the assumed effect \(\theta=0.20\) and the one-sided final boundary \(c=1.645\). Calculate conditional power.

The information values satisfy \(0<I_1<I_2\), and the calculation assumes a fixed information plan, the stated normal model with independent increments, and no further look before the final analysis. The remaining information is \(100-50=50\). The conditional mean and standard deviation are

$$ \begin{aligned} \mu_{2\mid 1} &=\sqrt{\frac{50}{100}}(1.00) +\frac{0.20(100-50)}{\sqrt{100}}\\ &=0.7071068+1.0000\\ &=1.7071068,\\[4pt] \sigma_{2\mid 1} &=\sqrt{\frac{100-50}{100}} =\sqrt{0.5} =0.7071068. \end{aligned} $$

Standardizing the final boundary gives

$$ \frac{1.645-1.7071068}{0.7071068}\approx -0.08783. $$

Thus \(\operatorname{CP}_{0.20}(1.00)=1-\Phi(-0.08783)=\Phi(0.08783)\approx 0.5350\). Under the assumed effect of \(0.20\), and conditional on this interim result, the probability of exceeding the final boundary is about 53.50%. This is not a 53.50% probability that the treatment works; it is a probability of a future test outcome under the specified model.

How the Assumed Effect Changes the Answer

The interim statistic alone does not determine conditional power. The assumed treatment effect matters because it determines the mean drift of the future increment. Holding \(z_1\), \(I_1\), \(I_2\), and \(c\) fixed, a larger positive \(\theta\) moves the conditional final mean toward and beyond the upper boundary. This makes conditional power useful for asking how the same interim evidence would translate into prospects under different effect assumptions.

Worked Example: A Smaller Assumed Effect

Keep the hypothetical trial inputs \(I_1=50\), \(I_2=100\), \(z_1=1.00\), and \(c=1.645\), but assume \(\theta=0.10\). As before, the information is fixed, the normal independent-increments model applies, and there is one final look.

The conditional mean changes because the expected contribution from the remaining 50 information units is smaller:

$$ \begin{aligned} \mu_{2\mid 1} &=\sqrt{\frac{50}{100}}(1.00) +\frac{0.10(50)}{\sqrt{100}}\\ &=0.7071068+0.5000\\ &=1.2071068,\\ \sigma_{2\mid 1}&=\sqrt{0.5}=0.7071068. \end{aligned} $$

The standardized boundary is

$$ \frac{1.645-1.2071068}{0.7071068}\approx 0.619274. $$

Therefore \(\operatorname{CP}_{0.10}(1.00)=1-\Phi(0.619274)\approx 0.2679\). Under this smaller assumed effect, the chance of crossing the final boundary, conditional on the same interim Z-statistic, is about 26.79%. The decrease from the previous example is due to the changed effect assumption, not a change in the observed interim evidence or in the critical value.

A Different Interim Result and a Numerical Check

The observed interim statistic also affects the conditional mean. A less favorable \(z_1\) generally leaves the final statistic farther from an upper boundary. A useful diagnostic is to calculate the conditional mean and standard deviation separately before standardizing; this helps reveal sign errors or an incorrect use of the remaining information.

Worked Example: A Weak Interim Result

For a third hypothetical calculation, let \(I_1=25\), \(I_2=100\), \(z_1=0\), \(\theta=0.20\), and \(c=1.645\). Assume the same fixed-design normal model, independent increments, and one final look.

The future increment has mean \(0.20(100-25)=15\) on the information-scaled statistic, which becomes \(15/\sqrt{100}=1.5000\) on the final Z scale. Its standard deviation on that scale is

$$ \sqrt{\frac{100-25}{100}}=\sqrt{0.75}=0.8660254. $$

Since \(z_1=0\), the interim contribution to the final conditional mean is zero. Hence \(\mu_{2\mid 1}=1.5000\), and the standardized boundary is

$$ \frac{1.645-1.5000}{0.8660254}\approx 0.167432. $$

The conditional power is \(1-\Phi(0.167432)\approx 0.4335\), or about 43.35%. The result is below one half because the conditional mean is below the final boundary, although the future increment still has substantial variability.

Interpretation, Alternatives, and Failure Modes

Conditional power is conditional on the interim Z-statistic and on a chosen effect value. It is not the same as unconditional design power, which averages over all possible interim and final data under a specified effect. For comparison, if there were no interim conditioning and the final statistic followed \(N(\theta\sqrt{I_2},1)\), its one-sided crossing probability would be \(1-\Phi(c-\theta\sqrt{I_2})\). A design with interim stopping rules requires accounting for those rules when calculating its overall rejection probability; the single conditional calculation above is not a substitute for that whole-design calculation.

It is also distinct from a predictive probability that averages over uncertainty about the effect. Here \(\theta\) is fixed for the calculation. Choosing a different plausible \(\theta\) can materially change the result, as the examples show. Accordingly, a conditional-power value should always be reported together with the assumed effect, observed interim statistic, information values, and final boundary.

The formula depends on the model matching the analysis. If the future information is not fixed, the increments are not independent, the final test is not the stated one-sided Z-test, or the variance model is inadequate, the calculated value may not describe the actual chance of success. In practice, check that \(I_2>I_1\), that \(c\) is the final boundary being used, and that the intended effect is expressed in the same canonical parameterization as the model. A computed probability outside \([0,1]\) signals an arithmetic or implementation error.

A common mistake is to treat \(Z_2\) as independent of the observed \(Z_1\) and calculate power from the final marginal distribution alone. That ignores the information already observed. The correction is to condition on \(Z_1=z_1\): the known interim contribution is \(\sqrt{I_1}z_1\), and only the remaining increment is random. Another frequent error is to use \(\theta I_2/\sqrt{I_2}\) as the future mean contribution. That is the mean contribution for all information, not just what remains; the correct future contribution is \(\theta(I_2-I_1)/\sqrt{I_2}\).

Finally, a high conditional power is not evidence that the assumed effect is true, and a low value does not prove that the null hypothesis is true. The calculation describes a probability under an assumption, conditional on the current result. Its value is limited by the plausibility of that assumption and by whether the model represents the planned final analysis.

Key takeaway: For one fixed final upper boundary, first find the conditional mean and variance of the final Z-statistic using the observed interim value and the remaining independent increment. Then standardize the boundary and take the upper normal tail.

Check Your Understanding

Use the single-final-look normal model in each question. Solutions follow so you can check both the setup and the interpretation.

  1. With \(I_1=40\), \(I_2=100\), \(z_1=0.50\), \(\theta=0.10\), and \(c=1.645\), write the conditional mean and standard deviation of \(Z_2\).
  2. For the inputs in Question 1, express conditional power using \(\Phi\). Is it above or below 50%?
  3. If \(I_2=I_1\), why can the displayed conditional-power formula not be used as written?
  4. State one difference between conditional power and unconditional design power.

Solutions. 1. The assumptions are \(I_2>I_1\), a fixed final boundary, and normal independent increments under the specified \(\theta\). Thus \(\mu_{2\mid 1}=\sqrt{40/100}(0.50)+0.10(60)/\sqrt{100}=0.316228+0.6000=0.916228\), and \(\sigma_{2\mid 1}=\sqrt{60/100}=0.774597\). 2. The standardized boundary is \((1.645-0.916228)/0.774597\approx0.9409\), so conditional power is \(1-\Phi(0.9409)\), approximately \(0.1734\), below 50%. 3. There is no remaining information, so the conditional standard deviation is zero; the future statistic is not a nondegenerate normal increment, and the formula’s standardization divides by zero. The final result is already determined by the interim data in this idealized model. 4. Conditional power fixes the observed interim statistic and evaluates future success under a specified effect; unconditional design power averages over possible data under the design, including the variability of interim results and, when applicable, the stopping rule.

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