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Classical Boundaries · Tutorial 95 of 699

A Two-Look Pocock Calibration Example

Learn how a two-look Pocock boundary is calibrated from the information schedule and the joint null distribution of the interim statistics.

Intermediate 10 min read

What You'll Learn

  • Calculate information fractions from planned information at each look.
  • Derive the null correlation between two cumulative Z statistics.
  • Set up the familywise type I error equation for a two-look, two-sided Pocock design.
  • Interpret the numerically calibrated common cutoff in a sequential decision.
  • Recognize when the planned calibration may not apply and how it differs from simpler alternatives.

Why Work Through a Two-Look Calibration?

A Pocock design uses the same efficacy cutoff at each planned analysis. That common cutoff is not simply the ordinary final-analysis critical value: looking more than once creates more than one opportunity to cross a boundary. Calibration must account for the chance of crossing at either look under the null hypothesis. For two looks, the calculation is small enough to inspect directly, making it a useful way to connect the information schedule, the dependence between interim statistics, and the overall type I error.

This example concerns a planned two-sided design with familywise type I error \(\alpha=0.05\), two analyses at information fractions \(0.5\) and \(1\), and no other stopping rules. The Pocock rule rejects \(H_0\) if the absolute value of either cumulative Z statistic reaches a common cutoff \(c\). The calculation uses the null joint distribution of those statistics, not an assumption that the two looks are independent.

Definition: In a two-look Pocock design, the efficacy boundary is the same positive cutoff \(c\) at both analyses. For a two-sided test, stop and reject \(H_0\) at the first look \(k\) for which \(|Z_k|\ge c\). Choose \(c\) so that, under \(H_0\), the probability of crossing at either planned look is the target familywise type I error \(\alpha\).

Information Fractions Determine the Correlation

Let \(I_1\) and \(I_2\) be the planned information at the first and second looks, with \(0<I_1<I_2\). Their information fractions are \(t_1=I_1/I_2\) and \(t_2=1\). In a canonical Z-statistic design under \(H_0\), the cumulative score at look 2 contains the score accumulated by look 1 plus the new information accumulated between looks. With independent increments, the score variances are \(I_1\) and \(I_2\), and the covariance of the two cumulative scores is \(I_1\). Standardizing each score gives

$$ \operatorname{Corr}(Z_1,Z_2) =\frac{I_1}{\sqrt{I_1I_2}} =\sqrt{\frac{I_1}{I_2}} =\sqrt{t_1}. $$

Thus \((Z_1,Z_2)\) has a bivariate standard normal null distribution with correlation \(\rho=\sqrt{t_1}\). This result assumes the planned information is correctly specified, the cumulative statistics have the canonical normal form under \(H_0\), and information is accumulated through independent increments. It does not say the statistics are identical: the later statistic includes new data. Nor does it say they are independent; for an interim look before the final analysis, their correlation is positive.

Worked Example: Finding the Information Fraction and Correlation

Suppose a hypothetical trial plans information \(I_1=400\) at the interim analysis and \(I_2=1000\) at the final analysis. Assume a canonical Z-statistic design with independent information increments under \(H_0\). Find the information fraction at the interim look and the null correlation between the two statistics.

The interim information fraction is

$$ t_1=\frac{I_1}{I_2}=\frac{400}{1000}=0.400. $$

The null correlation is therefore

$$ \rho=\sqrt{t_1}=\sqrt{0.400}=0.6325. $$

The interim analysis occurs at 40.0% of the planned final information, and the two cumulative Z statistics have correlation about 0.6325 under \(H_0\). The assumed independent increments and canonical null distribution justify using this correlation formula. Counting participants alone would give this same fraction only if participant count were proportional to information for this design.

Calibrating the Common Cutoff

For a two-sided design, the null event of no rejection is that both statistics lie strictly between \(-c\) and \(c\). Consequently, the familywise type I error is the probability of the complementary event: at least one of the two statistics has absolute value at least \(c\). Because the statistics are correlated, this probability must be calculated from their joint distribution.

For standard normal variables with correlation \(\rho\), the conditional distribution of \(Z_2\) given \(Z_1=z\) is normal with mean \(\rho z\) and variance \(1-\rho^2\). Integrating the conditional probability that \(Z_2\) is inside the cutoff over values of \(Z_1\) inside the cutoff gives the probability of no crossing:

$$ P(|Z_1|<c,\ |Z_2|<c) = \int_{-c}^{c}\phi(z) \left[ \Phi\left(\frac{c-\rho z}{\sqrt{1-\rho^2}}\right) - \Phi\left(\frac{-c-\rho z}{\sqrt{1-\rho^2}}\right) \right]\,dz. $$

Here \(\phi\) and \(\Phi\) are the standard normal density and cumulative distribution functions. The calibration sets one minus this integral equal to the target \(\alpha\). In general, the integral is evaluated numerically and the cutoff is found by numerical root-finding. The dependence matters: replacing the joint probability with a calculation that treats the looks as independent will not give the correct calibration.

Worked Example: Calibrating a Two-Look, Two-Sided Design

Consider a hypothetical design with \(I_1=500\) and \(I_2=1000\), a two-sided target \(\alpha=0.05\), and the canonical normal assumptions stated above. There are no other stopping rules. First, the information fraction and correlation are

$$ t_1=\frac{500}{1000}=0.500, \qquad \rho=\sqrt{0.500}=0.7071. $$

The cutoff solves

$$ 1- \int_{-c}^{c}\phi(z) \left[ \Phi\left(\frac{c-0.7071z}{\sqrt{1-0.7071^2}}\right) - \Phi\left(\frac{-c-0.7071z}{\sqrt{1-0.7071^2}}\right) \right]\,dz =0.05. $$

Numerically evaluating the integral and solving for \(c\) gives \(c\approx2.178\). Substituting this cutoff into the familywise error expression gives approximately \(0.0500\), to the displayed precision. Thus the design rejects \(H_0\) at either look if \(|Z_k|\ge2.178\), and otherwise continues to the next planned analysis.

At either look considered by itself, the null probability of exceeding this two-sided cutoff is \(2\Phi(-2.178)\approx0.0294\). That is a marginal probability for one statistic, not the familywise probability across both looks. The calibrated familywise error is 0.05 because it counts the union of the two crossing events, with their dependence included.

Applying the Calibrated Rule

Calibration determines the cutoff before outcomes are examined. At analysis time, the decision is made by comparing each observed statistic with that cutoff, in the direction or directions specified by the design. For a two-sided design, the sign of the statistic indicates the direction of the observed effect, while the absolute value determines whether the boundary has been crossed.

Worked Example: Interpreting Statistics at Both Looks

Suppose a hypothetical trial uses the calibrated design above. Its observed interim statistic is \(Z_1=2.10\), and, if the trial continues, its final statistic is \(Z_2=2.24\). Assume the planned information schedule and null model remain applicable.

At the interim look,

$$ |Z_1|=|2.10|=2.10<2.178. $$

The interim efficacy boundary is not crossed, so this rule does not reject \(H_0\) at that look. At the final look,

$$ |Z_2|=|2.24|=2.24\ge2.178. $$

The final boundary is crossed, so the design rejects \(H_0\). The conclusion is based on the pre-calibrated sequential rule, not on comparing the final statistic with the ordinary single-analysis cutoff. The result is evidence against \(H_0\) at the design’s familywise type I error level; it does not establish that \(H_0\) has a particular probability of being true or that the treatment caused an effect.

Worked Example: Early Crossing Is Not Undone by a Later Statistic

In another hypothetical trial using the same cutoff, suppose \(Z_1=-2.20\) at the interim look. The statistic is negative, but the design is two-sided:

$$ |Z_1|=|-2.20|=2.20\ge2.178. $$

The boundary is crossed at the interim analysis, so the rule stops for efficacy and rejects \(H_0\). A later statistic is not used to reverse that decision because the design stops at the first crossing. In context, the observed statistic points in the negative direction; the two-sided rule allows a crossing in either direction. This example also shows why specifying one-sided versus two-sided testing before calibration is essential.

Derivation, Diagnostics, and Alternatives

The key justification for the correlation calculation can be seen by writing the cumulative scores explicitly. Let \(S_1\) be the score accumulated by the first look and let \(S_{\mathrm{new}}\) be the additional score between looks. Under the canonical null model, assume these increments are independent, with variances \(I_1\) and \(I_2-I_1\). Then \(S_2=S_1+S_{\mathrm{new}}\), so

$$ \operatorname{Var}(S_2)=I_1+(I_2-I_1)=I_2, \qquad \operatorname{Cov}(S_1,S_2) =\operatorname{Cov}(S_1,S_1+S_{\mathrm{new}}) =I_1. $$

Since \(Z_1=S_1/\sqrt{I_1}\) and \(Z_2=S_2/\sqrt{I_2}\), standardization yields \(\operatorname{Corr}(Z_1,Z_2)=I_1/\sqrt{I_1I_2}=\sqrt{I_1/I_2}\). The common cutoff then follows by applying the probability of the union of the two crossing events, rather than adding two marginal probabilities as if the events were independent.

In practice, check that the planned information fractions correspond to the actual analysis schedule and that the analysis statistics are compatible with the assumed canonical joint null distribution. If the interim information differs materially from plan, the correlation implied by the planned fraction may no longer describe the data. Likewise, changes in the statistic, variance estimation, or dependence structure can require an appropriate recalculation. Do not assume that the square-root correlation formula applies merely because analyses are labelled “interim” and “final.”

A common mistake is to use the ordinary two-sided final-analysis cutoff \(1.96\) at both looks because it corresponds to a single-analysis \(\alpha=0.05\). This ignores the chance of crossing at the interim look as well as at the final look, so it does not provide the stated familywise error control. A different shortcut is Bonferroni allocation: assign a two-sided error probability \(0.025\) to each look, giving a per-look cutoff of approximately \(2.241\). This controls the familywise error by the union bound but is generally conservative because it does not use the correlation. Pocock calibration instead finds a common cutoff from the joint null distribution. Other sequential designs can use different cutoffs across looks; their calibration and boundary shapes are distinct from the common-cutoff calculation here.

The result is specific to its assumptions and plan. It is not a universal cutoff for all two-look designs: the cutoff depends on \(\alpha\), the information fraction, whether the test is one-sided or two-sided, the null distribution, and the stopping rule. An additional rule that stops for futility or changes the analysis schedule must be described when calculating the operating characteristics; the simple equation above represents the stated two-look efficacy rule.

Key takeaway: For two canonical cumulative Z statistics, the null correlation is the square root of the first information fraction. A two-sided Pocock cutoff is calibrated by setting the probability of crossing either common boundary equal to the target familywise \(\alpha\), using the joint rather than independent null distribution.

Check Your Understanding

Answer each question using the two-look canonical normal framework described above. Solutions are included for checking.

  1. A design plans \(I_1=360\) and \(I_2=900\). What is the interim information fraction, and what is the null correlation between \(Z_1\) and \(Z_2\)?
  2. For a two-sided design with cutoff \(c=2.178\), the interim statistic is \(Z_1=2.18\). Does the design cross the boundary at the interim look? Explain.
  3. Why is it incorrect to treat the two cumulative statistics as independent when calculating the familywise type I error?
  4. What is the main distinction between the Pocock calibration and applying a Bonferroni correction at two looks?

Solutions. (1) \(t_1=360/900=0.400\), so \(\rho=\sqrt{0.400}=0.6325\). (2) Yes: \(|2.18|=2.18\ge2.178\), so the two-sided boundary is crossed. (3) The second cumulative score contains the first-look score, giving \(\operatorname{Cov}(S_1,S_2)=I_1\) and \(\operatorname{Corr}(Z_1,Z_2)=\sqrt{I_1/I_2}>0\); independence would give the wrong joint crossing probability. (4) Bonferroni assigns an error allowance to each look and uses a union bound, generally conservatively. Pocock calibration uses the joint null distribution to choose one common cutoff for the target familywise error.

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