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Unblinded Re-Estimation and the Promising Zone · Tutorial 422 of 699

Diagnostic: Variance Estimation and Re-Estimation

Learn to quantify how noisy interim variance estimates can produce unstable sample size revisions and misleading power expectations.

Advanced 14 min read

What You'll Learn

  • Relate the pooled interim variance estimate to a chi-square distribution under normality and equal variances.
  • Calculate how interim degrees of freedom determine the variability of a revised sample size.
  • Translate a selected sample size into approximate power under a specified true variance.
  • Diagnose instability using variance-ratio distributions, quantiles, and sensitivity checks.
  • Distinguish plug-in power calculations from the operating power of the complete adaptive design.

Why Variance-Estimation Noise Matters

For a continuous outcome, a sample size calculation depends on the variance as well as the effect size, significance level, and target power. In unblinded re-estimation, an interim variance estimate can replace the planning variance in a sample size rule. That substitution is useful when the planning variance is uncertain, but it also makes the revised sample size random: an unusually small estimate can reduce the sample size, while an unusually large estimate can increase it. The diagnostic question is not merely whether the estimate is unbiased. It is how much noise it carries and how that noise propagates into the design.

The calculations below use a normal approximation to a final two-sided test. For a balanced two-arm design, \(n\) denotes the number of participants per arm, \(\delta\) is the treatment-effect difference in outcome units, and \(\sigma\) is the common within-arm standard deviation. These calculations diagnose variance-driven changes; they are not a substitute for evaluating the complete group sequential and re-estimation procedure.

Assumptions for the basic variance diagnostic: Outcomes are independent and normally distributed within each arm, with a common variance \(\sigma^2\). At an interim look with \(n_I\) observations per arm, the pooled within-arm variance estimate \(S_p^2\) has \(df=2n_I-2\) degrees of freedom and satisfies \(df\,S_p^2/\sigma^2\sim\chi^2_{df}\). The treatment effect and variance are expressed in the same outcome units throughout. Departures from normality, independence, or equal variances can invalidate this reference distribution or make the pooled estimate a poor summary of the variance relevant to the final analysis.

A convenient normal-approximation planning formula for a fixed final test is

$$ n_0=\frac{2\sigma_{\mathrm{plan}}^2\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{\delta_{\mathrm{plan}}^2}, \qquad n^*=n_0\frac{S_p^2}{\sigma_{\mathrm{plan}}^2}. $$

Here \(n_0\) is the planned number per arm using planning standard deviation \(\sigma_{\mathrm{plan}}\); \(n^*\) is the unrounded variance-adjusted target, before any minimum, maximum, or rounding rule is applied. This proportional rule follows because the required sample size is proportional to the assumed variance when the other planning inputs are held fixed. Its diagnostic value is that relative variance-estimation error transfers directly to relative sample-size error.

How Interim Degrees of Freedom Control Instability

If the true variance equals the planning variance, define the variance ratio \(R=S_p^2/\sigma_{\mathrm{plan}}^2\). Then \(n^*=n_0R\) and \(R\sim\chi^2_{df}/df\). Since the mean and variance of a chi-square variable with \(df\) degrees of freedom are \(df\) and \(2df\), respectively,

$$ E(R)=1,\qquad \operatorname{SD}(R)=\sqrt{\frac{2}{df}},\qquad \operatorname{SD}(n^*)=n_0\sqrt{\frac{2}{df}}. $$

Thus the plug-in target is unbiased before caps and rounding, but it can still be highly variable. The coefficient of variation of \(n^*\) is \(\sqrt{2/df}\). This is a measure of relative spread, not a guarantee that any single estimate will lie within that distance of \(n_0\). With small \(df\), the ratio distribution is also right-skewed: occasional large variance estimates can produce especially large targets.

Worked Example: Translating Interim Degrees of Freedom Into Sample-Size Noise

Suppose a hypothetical design has a continuous, unrounded planning target of \(n_0=168.169\) participants per arm. Compare an interim pooled variance estimate based on 10 observations per arm with one based on 40 per arm. Assume the normal, independent, equal-variance model and that the true variance equals the planning variance.

With 10 per arm, \(df=2(10)-2=18\). The coefficient of variation of the target is

$$ \sqrt{\frac{2}{18}}=0.333333\ldots $$

Therefore its standard deviation is

$$ 168.169\sqrt{\frac{2}{18}}=56.0563\ldots\approx56.1 $$

participants per arm. With 40 per arm, \(df=78\), giving

$$ \sqrt{\frac{2}{78}}=0.160128\ldots,\qquad 168.169\sqrt{\frac{2}{78}}=26.9286\ldots\approx26.9 $$

participants per arm. These are standard deviations of the unrounded target under the stated model, not confidence limits for a particular target. The larger interim sample reduces the target’s standard deviation because it supplies more information about the variance. In practice, integer rounding and sample-size bounds alter the distribution of the implemented target.

From a Variance Estimate to Approximate Power

A revised target is not itself a power calculation. To see the consequence of a particular variance estimate, evaluate power at the selected sample size using a stated true variance and effect. Under the balanced two-arm normal approximation, the standardized treatment-effect estimator has noncentrality \(\lambda=\delta\sqrt{n/(2\sigma^2)}\). If the final test uses the fixed two-sided normal critical value \(z_{1-\alpha/2}\), its approximate power is

$$ \Phi\left(\lambda-z_{1-\alpha/2}\right)+ \Phi\left(-\lambda-z_{1-\alpha/2}\right). $$

This expression is a fixed-sample approximation evaluated at a selected \(n\). In a group sequential design, the final boundary need not equal the fixed-test critical value. Moreover, a sample size chosen from interim data makes the full procedure adaptive. Therefore plug-in power at \(n^*\) is a diagnostic, not a proof of the operating power or type I error of the complete design.

Worked Example: How a Low Variance Estimate Can Reduce Approximate Power

Consider a hypothetical balanced two-arm trial with \(\delta=4\) outcome units, true \(\sigma=10\) outcome units, and two-sided \(\alpha=0.05\). Use a baseline target \(n_0=100\) per arm and a plug-in rule \(n^*=100R\). Assume normal independent outcomes with equal variances; the values below illustrate consequences of different variance estimates, rather than claims about a real study.

If the interim estimate gives \(R=0.5\), then \(n^*=50\) per arm. The noncentrality under the true variance is

$$ \lambda=4\sqrt{\frac{50}{2(10^2)}}=2.000. $$

Using \(z_{0.975}=1.960\), approximate power is

$$ \Phi(2.000-1.960)+\Phi(-2.000-1.960) =\Phi(0.040)+\Phi(-3.960)\approx0.516. $$

If instead \(R=1.5\), then \(n^*=150\) per arm and

$$ \lambda=4\sqrt{\frac{150}{200}}=3.464, \qquad \Phi(3.464-1.960)+\Phi(-3.464-1.960)\approx0.934. $$

The same planning effect and true variance therefore yield very different approximate powers at these two selected sample sizes. The calculation isolates the variance-driven sample-size pathway; it does not account for a sequential final boundary, interim-effect-based decisions, or dependence introduced by the exact adaptive testing procedure.

A related one-sample or paired-difference calculation helps illustrate a common arithmetic check. If a hypothetical final sample has \(n=118\), \(\delta=5\) outcome units, and true standard deviation \(\sigma=20\) outcome units, then \(\lambda=\delta\sqrt{n}/\sigma=5\sqrt{118}/20=2.715695\). The fixed two-sided normal approximation at \(\alpha=0.05\) is

$$ \Phi(2.715695-1.960)+\Phi(-1.960-2.715695) \approx0.775085\approx0.775. $$

The one-sample formula is not interchangeable with the balanced two-arm formula: their standard errors differ. In either setting, retain unrounded quantities in intermediate calculations. For example, multiplying \(168.169\) by the rounded coefficient \(0.333\) gives a slightly different result than using \(\sqrt{2/18}\); the latter gives 56.1 participants, not 56.0 when rounded to one decimal place.

Quantiles, Caps, and Practical Diagnostics

A standard deviation summarizes spread but does not show the asymmetry of the variance ratio or the fraction of targets that may be extreme. Quantiles of \(\chi^2_{df}/df\) provide a direct diagnostic range. If a target is capped, the probability of reaching the cap can also be assessed under the variance model. A cap limits the implemented sample size; it does not make the underlying estimate less noisy.

Worked Example: Comparing the Central Spread at Two Interim Sizes

Suppose the baseline target is \(n_0=100\) per arm, with true variance equal to the planning variance. Compare interim pooled estimates from 5 per arm (\(df=8\)) and 20 per arm (\(df=38\)). Approximate 5th and 95th percentiles of \(\chi^2_8\) are 2.733 and 15.507; for \(\chi^2_{38}\), they are 24.884 and 53.384.

For \(df=8\), the corresponding variance-ratio quantiles are \(2.733/8=0.3416\) and \(15.507/8=1.9384\). Multiplying by the baseline target gives an approximate central 90% range of

$$ 100(0.3416)=34.16 \quad\text{to}\quad 100(1.9384)=193.84 $$

participants per arm, before bounds or rounding. For \(df=38\), the ratio quantiles are \(24.884/38=0.6548\) and \(53.384/38=1.4048\), giving

$$ 100(0.6548)=65.48 \quad\text{to}\quad 100(1.4048)=140.48 $$

participants per arm. These wide ranges show why a very early variance estimate can produce substantial sample-size variability even when its expected value is correct. The ranges are model-based quantiles, not a prediction interval that accounts for misspecification, adaptive testing, or operational constraints.

For a real design, check the variance re-estimation rule against the assumptions and implementation, not just its formula. Useful diagnostics include:

  • Compute the interim degrees of freedom and the implied coefficient of variation. A large value signals that target variability should be expected.
  • Examine quantiles of the variance ratio and the frequency with which the minimum or maximum sample size would bind.
  • Repeat the calculation over plausible true variances and interim sample sizes. Report the range of selected targets and their approximate power, rather than only the target at the planning variance.
  • Inspect whether outliers, skewness, unequal variances, clustering, or missing outcomes make the pooled interim estimate unrepresentative of the final analysis variance.
  • Check whether the estimate is computed from independent within-arm residual information as intended. Under the stated normal model, pooled variance and treatment-effect estimates are independent; that fact does not automatically hold under model departures or every analysis procedure.

Alternatives and Limits of the Diagnostic

The main alternative is to keep the sample size fixed at its planning value. That avoids variance-estimate-driven changes but leaves the design exposed to a planning variance that may be wrong. A second approach is to use a conservative planning variance or a prespecified upper bound, trading expected enrollment for protection against underestimation. A third is to use a stabilized estimate, such as a prespecified combination of interim information and planning information. Stabilization can reduce extreme revisions, but changes the sampling distribution and must be specified and evaluated rather than treated as an automatic correction. Blinded variance re-estimation may reduce access to treatment-effect information, but it does not eliminate variance-estimation noise or guarantee unchanged operating characteristics.

The chi-square diagnostic is exact only for the stated normal, independent, common-variance model. Small interim samples, heavy tails, outliers, variance differences between arms, and changes in outcome measurement can all undermine its usefulness. Rounding, a minimum sample size, and a maximum sample size can also make the implemented target biased relative to the unbounded plug-in target. Finally, fixed-sample power evaluated at each selected \(n\) is not generally the unconditional power of a sequential adaptive design. The latter depends on the complete decision and testing rule and should be assessed with a full design evaluation.

A Short Derivation and a Common Mistake

The sample-size variability result follows directly from the chi-square model. Let \(U=df\,S_p^2/\sigma^2\), so that \(U\sim\chi^2_{df}\). When \(\sigma^2=\sigma_{\mathrm{plan}}^2\), the rule gives \(n^*=n_0U/df\). Because \(E(U)=df\) and \(\operatorname{Var}(U)=2df\),

$$ E(n^*)=n_0\frac{E(U)}{df}=n_0,\qquad \operatorname{Var}(n^*)= \left(\frac{n_0}{df}\right)^2 2df =\frac{2n_0^2}{df}. $$

Taking the square root yields \(\operatorname{SD}(n^*)=n_0\sqrt{2/df}\). This derivation concerns the unbounded, unrounded target. If the true variance differs from the planning variance, then \(E(S_p^2)=\sigma^2\), so \(E(n^*)=n_0\sigma^2/\sigma_{\mathrm{plan}}^2\): the target shifts as well as remaining noisy.

Common mistake: “The interim variance estimate is unbiased, so re-estimation preserves the planned sample size and target power.” Unbiasedness says only that the average variance estimate equals the true variance under the model. It does not make each estimate precise, prevent a low estimate from reducing the target, or ensure that a capped adaptive design achieves its nominal power. Quantify the ratio’s spread and evaluate the full rule under plausible conditions.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. In a balanced two-arm trial with 12 interim observations per arm, what are the pooled variance degrees of freedom and the coefficient of variation of the unrounded target when the true and planning variances agree?
  2. If \(n_0=80\) per arm and the interim variance estimate is \(1.25\) times the planning variance, what is the unrounded target before caps and rounding?
  3. Why does a maximum sample-size cap not remove the variability of the interim variance estimate?
  4. For a two-sided normal test, explain why power at a selected sample size is not automatically the operating power of the full adaptive group sequential design.

Solutions. (1) \(df=2(12)-2=22\), and the target coefficient of variation is \(\sqrt{2/22}=0.3015\), or about 30.2%. (2) \(n^*=80(1.25)=100\) participants per arm. This is the unrounded target before any design constraints. (3) The cap changes large realized targets to the maximum, but the variance estimate and its uncapped target remain random; the cap can also change the distribution of implemented sample sizes. (4) A fixed-sample calculation uses a specified critical value and selected \(n\). The adaptive design’s final rejection probability also depends on interim decisions, the group sequential boundary, and how sample size depends on interim data, so its operating power must be evaluated for the whole procedure.

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