Equal Spacing Means Equal Increments of Information
A group-sequential design schedules interim analyses before the trial begins. When those looks are equally spaced, the relevant spacing is in information: each look adds the same increment of information toward the final analysis. Equal intervals in calendar time or equal numbers of newly enrolled participants do not, by themselves, establish equal information increments.
Let \(I_k\) be the cumulative information at look \(k\), and let \(I_K\) be the planned maximum information at the final look \(K\). The information fraction is \(t_k=I_k/I_K\). For \(K\) equally spaced looks, including the final analysis, the planned fractions are \(1/K,2/K,\ldots,1\). Thus four equally spaced looks occur at fractions \(0.25,0.50,0.75,1.00\). The first look is at one quarter of the planned total information, not at one quarter of the number of interim analyses before the final look.
The standard canonical model for planned efficacy analyses assumes that, under \(H_0\), the successive standardized test statistics have a joint multivariate normal distribution, each with mean 0 and variance 1, and \(\operatorname{Corr}(Z_i,Z_j)=\sqrt{t_i/t_j}\) for \(i<j\). This dependence reflects the fact that later analyses reuse information already present in earlier analyses. It is not appropriate to treat the \(K\) statistics as independent tests.
From Information Fractions to Boundary Locations
A one-sided efficacy boundary is a sequence of critical values \(b_1,\ldots,b_K\). The trial stops for efficacy at the first look \(k\) where \(Z_k\ge b_k\). A common classical family is the Wang–Tsiatis family, in which the boundary shape is
Here \(c\) is a positive constant selected to attain the target overall one-sided type I error \(\alpha\), while \(\Delta\) controls how the boundary changes with information. The shape alone does not determine \(\alpha\): \(c\) must be calibrated under the joint distribution of the planned statistics. In this family, \(\Delta=0.5\) gives Pocock’s constant boundary, and \(\Delta=0\) gives the O’Brien–Fleming information-scaled shape \(b_k=c/\sqrt{t_k}\).
The calibration is a first-crossing calculation. Under the null, the probability of at least one efficacy crossing is
For a specified shape and schedule, numerical integration of the canonical joint normal distribution can evaluate this probability as a function of \(c\). Calibration chooses \(c\) so that it equals the target \(\alpha\). Because the statistics are correlated, multiplying a single-look tail probability by \(K\) is generally not this calculation.
Why the Canonical Dependence Matters
The dependence structure can be seen by writing the statistic at look \(k\) as a combination of the previous statistic and new information. Let \(t_0=0\). Under the canonical null model, take independent standard normal increments \(\epsilon_k\), so that
For \(k=1\), the first term is zero, so \(Z_1=\epsilon_1\). For later looks, the first term carries forward the earlier information and the second represents new information. The conditional variance is \(1-t_{k-1}/t_k\), which is positive when information increases between looks.
This representation also justifies the correlation formula. For \(i<j\), the contribution of \(Z_i\) to \(Z_j\) is multiplied by \(\sqrt{t_i/t_j}\); the remaining increments after look \(i\) are independent of \(Z_i\). Therefore \(\operatorname{Cov}(Z_i,Z_j)=\sqrt{t_i/t_j}\operatorname{Var}(Z_i)=\sqrt{t_i/t_j}\), since \(\operatorname{Var}(Z_i)=1\). In equal spacing, \(t_i/t_j=i/j\), so the joint dependence is fixed by the number and order of the planned looks.
Worked Example: Locating a Four-Look Wang–Tsiatis Boundary
Suppose a hypothetical trial plans four equally spaced information looks and uses the Wang–Tsiatis shape with \(\Delta=0.25\). For illustration, take \(c=2.10\). This value is chosen only to demonstrate the boundary arithmetic; it is not claimed to be calibrated to any particular \(\alpha\).
The planned fractions are \(t=(0.25,0.50,0.75,1.00)\). Substituting each fraction into \(b_k=2.10t_k^{-0.25}\) gives:
Suppose the hypothetical observed statistics are \(Z=(2.80,2.55,2.30,2.15)\). At look 1, \(2.80<2.970\), so the boundary is not crossed. At look 2, \(2.55\ge2.497\), so efficacy is declared at the second look and monitoring stops. The later values are shown only to make the hypothetical sequence explicit; they would not be used after stopping.
These calculations locate the cutoffs for the chosen shape and \(c\), but they do not establish the design’s type I error. That requires calibrating \(c\) for the target \(\alpha\) using all four correlated statistics.
Reading the Shapes at Equally Spaced Looks
At equal information fractions, the classical shapes have especially direct interpretations. Pocock’s \(\Delta=0.5\) boundary has \(b_k=c\) at every look. Its flatness means the standardized evidence threshold does not change over time. The O’Brien–Fleming shape has \(b_k=c/\sqrt{t_k}\), so its threshold is largest early and declines toward the final look. This makes early efficacy crossing more demanding relative to the later looks.
The following calculations illustrate shape, not matched-alpha performance. The constants are deliberately specified rather than presented as calibrated design values.
Worked Example: Comparing Pocock and O’Brien–Fleming Shapes
For four equally spaced looks, use the illustrative Pocock constant \(c=2.40\). Its boundary is \(2.40\) at all four fractions. For an illustrative O’Brien–Fleming shape, take \(c=1.96\). The boundary values are:
The Pocock cutoffs are flat; the O’Brien–Fleming-shaped cutoffs decrease as information accumulates. For example, a hypothetical \(Z_1=3.00\) crosses the illustrative Pocock boundary because \(3.00\ge2.40\), but does not cross the illustrative O’Brien–Fleming boundary because \(3.00<3.920\). These constants were not calibrated to the same \(\alpha\), so this comparison demonstrates geometry only; it does not establish that one design has a higher or lower type I error than the other.
The boundary is interpreted at the first crossing, not by checking only the final statistic or by counting how many statistics exceed their cutoffs. If an earlier crossing has already stopped the trial, later hypothetical values do not change the decision. The calibrated overall type I error describes the probability, under \(H_0\), of at least one crossing over the complete planned monitoring rule.
Information Spacing Is Not Automatically Sample-Size Spacing
In some settings, information is proportional to sample size, so equal sample-size increments produce equal information increments. In other settings, information can depend on event accrual, variability, allocation, or other features of the data. The schedule must therefore be expressed in information fractions before assigning equally spaced cutoffs. When information fractions are unequal, use the actual \(t_k\) values in the boundary formula and recalibrate for that schedule rather than substituting \(k/K\).
Worked Example: Checking a Non-Equally Spaced Information Schedule
Consider a hypothetical study with cumulative sample sizes \(80,170,270,390\) at its four looks. Assume for this calculation that information is proportional to cumulative sample size, so the information fractions are the sample-size fractions:
These are not the equally spaced fractions \(0.25,0.50,0.75,1.00\). For instance, the second fraction is \(0.4359\), not \(0.50\), and the third is \(0.6923\), not \(0.75\). The increments are \(0.2051\), \(0.2308\), \(0.2564\), and \(0.3077\), which also are not equal.
For an illustrative O’Brien–Fleming-shaped boundary with \(c=1.96\), the first cutoff would be \(1.96/\sqrt{0.2051}=4.328\), rather than \(1.96/\sqrt{0.25}=3.920\). The later cutoffs must likewise use \(0.4359\), \(0.6923\), and \(1.000\). As above, this \(c\) is for arithmetic illustration only and is not asserted to control a specified \(\alpha\). A real design would calibrate its constant using these actual fractions.
The sample-size ratios in this example are valid information fractions only because the example explicitly assumes proportionality. If, for example, information accrues through observed events rather than enrollment alone, the event-based information may be the appropriate quantity. Applying equal-look cutoffs to unequal information fractions can distort the intended boundary and invalidate the planned type I error calibration.
Assumptions, Diagnostics, and Limitations
The calculations above rely on planned analyses, a correctly specified information schedule, and the canonical joint normal null model. The test statistics must be standardized consistently across looks, and the stopping rule must be the one used for calibration. The one-sided boundary here concerns efficacy; a two-sided design needs its own specified rule and calibration rather than an automatic reuse of the one-sided cutoffs.
Before applying the boundary, check that the observed analysis timing corresponds reasonably to planned information fractions, that information has increased at each look, and that the final planned information is correctly identified. If enrollment, event accrual, or data availability causes a look to occur at a materially different information fraction, the planned \(t_k\) values may no longer describe the analysis. A boundary calibrated for one schedule does not automatically retain its type I error under a changed schedule.
A common mistake is to call four looks “equally spaced” because there are four scheduled meetings, or because each meeting is a similar number of weeks after the last one. The correction is to compute cumulative information fractions and verify \(t_k=k/K\). Another mistake is to use the shape formula alone and assume it guarantees the desired \(\alpha\). The formula specifies the pattern of cutoffs; only calibration under the joint null distribution sets the overall error probability.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For five equally spaced information looks, what are the planned information fractions?
- For a four-look O’Brien–Fleming-shaped boundary with illustrative \(c=2.00\), calculate the cutoff at \(t=0.25\).
- Why is it generally incorrect to calculate the overall type I error by multiplying a single-look tail probability by the number of looks?
- For cumulative sample sizes \(60,150,240\) and a final total of \(300\), assuming information is proportional to sample size, are three looks equally spaced in information?
Solutions. (1) The fractions are \(0.20,0.40,0.60,0.80,1.00\), since \(t_k=k/5\). (2) \(b=2.00/\sqrt{0.25}=2.00/0.50=4.00\). (3) The statistics are correlated because each later statistic incorporates earlier information; the overall error is the probability of the union of crossing events under their joint distribution, not generally the sum of marginal probabilities. (4) The three listed sample sizes are before the final analysis. Including the final analysis gives four looks with fractions \(0.20,0.50,0.80,1.00\), rather than the equally spaced four-look fractions \(0.25,0.50,0.75,1.00\); therefore, the schedule is not equally spaced in information.