From Crossing to No Crossing
At an interim analysis, a one-sided efficacy rule rejects \(H_0\) if the standardized test statistic reaches or exceeds its upper boundary. Across several analyses, the rejection event is the union of the individual crossing events. Those events are dependent, so adding their probabilities generally does not give the cumulative Type I error. A useful alternative is to calculate the probability of the complementary event: remaining below every boundary.
This tutorial derives that calculation for the canonical group sequential model. The result is a recursive equation for the next boundary, not a rule for adding a nominal amount of error at each look. Its role is to choose the cutoff whose cumulative crossing probability equals the specified cumulative spending.
Let \(A(t_i)\) be the desired cumulative Type I error spending through look \(i\), and abbreviate it by \(A_i\). The cumulative rejection event through look \(i\) is \(\{\text{there is some }j\le i\text{ with }Z_j\ge b_j\}\). Its complement is the event that every statistic through look \(i\) remains below its boundary.
Because the joint distribution is continuous, replacing a strict inequality by a non-strict one does not change these probabilities. The no-crossing formulation is especially convenient because it lets us update the surviving probability from one analysis to the next.
Conditional Transition Between Looks
For adjacent looks, write \(\rho_i=\sqrt{t_{i-1}/t_i}\), for \(i\ge2\). Under the canonical model, the next statistic conditional on the current one is normal:
The variance is positive because the information times increase strictly. Equivalently, for an independent \(N(0,1)\) variable \(\varepsilon_i\), the transition can be written \(Z_i=\rho_iZ_{i-1}+\sqrt{1-\rho_i^2}\,\varepsilon_i\). This follows by standardizing the independent score increment between looks: the earlier score contributes the fraction \(t_{i-1}/t_i\) of the variance at look \(i\), and the new increment contributes the remaining fraction.
This transition is the step that makes a recursion possible. Given the statistic at look \(i-1\), the probability of staying below the next boundary is a normal cumulative probability. The earlier boundaries still matter: only paths that survived all of them are included in the update.
Deriving the No-Crossing Recursion
Let \(\phi\) and \(\Phi\) denote the standard normal density and cumulative distribution function. Define \(q_i(y)\) as the joint density at \(Z_i=y\) together with survival below every boundary through look \(i\). It is an unnormalized density: its integral is the probability of surviving, not 1.
Only values below \(b_1\) survive, so \(q_1(y)=\phi(y)\mathbf{1}(y<b_1)\), where \(\mathbf{1}\) is 1 when its condition holds and 0 otherwise.
For \(i\ge2\), integrate over surviving values \(x\) at look \(i-1\), multiply by the conditional density of \(Z_i=y\), and retain only \(y<b_i\).
Integrate the updated density over its surviving range and choose \(b_i\) so that the result is \(1-A_i\).
In equations, the propagation step and the resulting cutoff equation are:
The second equality integrates out \(y\), using the conditional probability \(P(Z_i<b_i\mid Z_{i-1}=x)\). The recursion therefore carries forward exactly the probability mass associated with paths that have not crossed. For the first look, \(S_1=\Phi(b_1)\), so \(b_1=\Phi^{-1}(1-A_1)\) when \(0<A_1<1\).
When Does a Finite Cutoff Exist?
Assume the cumulative spending is nondecreasing and \(A_{i-1}<1\), so the previous survivor probability \(S_{i-1}=1-A_{i-1}\) is positive. For a candidate cutoff \(b\), define \(G_i(b)=P(Z_1<b_1,\ldots,Z_{i-1}<b_{i-1},Z_i<b\mid H_0)\). As \(b\) increases, \(G_i(b)\) increases continuously from 0 to \(S_{i-1}\). The upper limit is the probability of having survived the earlier looks, since a cutoff tending to \(+\infty\) imposes no restriction at look \(i\).
Consequently, a unique finite cutoff solves \(G_i(b_i)=1-A_i\) precisely when the target is strictly between those limits. With nondecreasing spending, this is the condition \(A_{i-1}<A_i<1\). A positive spending increment alone is not sufficient: if \(A_i=1\), the target survivor probability is zero, which no finite cutoff attains. If \(A_i=A_{i-1}\), the target equals \(S_{i-1}\) and the limiting cutoff is \(+\infty\). If \(A_i=1\), the target is zero and the limiting cutoff is \(-\infty\). These are limiting cases, not finite boundary solutions.
Worked Example: Calibrating the First Look
Suppose a hypothetical one-sided design targets \(A_1=0.025\) at its first look. Under the stated null model, \(Z_1\sim N(0,1)\), so the no-crossing equation is \(\Phi(b_1)=1-0.025=0.975\).
Thus \(b_1=\Phi^{-1}(0.975)=1.960\), to three significant figures. The null probability of crossing this boundary at the first look is \(P(Z_1\ge1.960\mid H_0)\approx0.0250\), as required. The cutoff is in standardized \(Z\)-score units; it is not a treatment-effect estimate in the trial’s original measurement units.
Worked Example: Setting a Second-Look Cutoff
Consider a hypothetical two-look design with \(t_1=0.50\), \(t_2=1.00\), first-look spending \(A_1=0.010\), and target cumulative spending \(A_2=0.025\). The information times are strictly increasing, and the canonical null model gives \(\rho_2=\sqrt{0.50/1.00}=0.7071\). At look 1, \(b_1=\Phi^{-1}(0.990)=2.326\), and \(q_1(x)=\phi(x)\) for \(x<2.326\).
The finite-cutoff condition holds: \(0.010<0.025<1\). The equation for the second boundary is \( \int_{-\infty}^{2.326}\phi(x)\Phi((b_2-0.7071x)/\sqrt{0.50})\,dx=1-0.025=0.975.\) Numerical quadrature followed by one-dimensional root-finding gives \(b_2\) approximately \(2.08\) (the reported value is rounded). Substituting the unrounded numerical root into the integral gives a survivor probability of approximately \(0.9750\); hence the cumulative null rejection probability is \(1-0.9750=0.0250\), to the displayed precision.
The second boundary is calibrated conditional on the full surviving distribution at look 1, not by treating \(Z_2\) as an independent new test. In particular, the integral includes only paths with \(Z_1<2.326\) and accounts for the correlation between the two statistics.
Numerical Implementation and Checks
A practical implementation can propagate \(q_i\) by numerical quadrature, evaluate \(G_i(b)\), and use a bracketing root-finder to solve \(G_i(b)=1-A_i\). A bracket is available whenever \(A_{i-1}<A_i<1\): choose a sufficiently low cutoff with \(G_i(b)<1-A_i\) and a sufficiently high one with \(G_i(b)>1-A_i\). Continuity then guarantees a root between them, and strict increase gives uniqueness. A bracketed method is preferable to an unguarded Newton iteration when the target is near a limiting case.
Check the computation at each look by integrating the propagated density and verifying \(\int q_i(y)\,dy\approx1-A_i\). Also verify that the numerical cumulative crossing probabilities are nondecreasing and that the computed cutoff moves in the expected direction: increasing the target \(A_i\), with earlier boundaries held fixed, lowers the cutoff. Integration limits, grid resolution, and root tolerance should be tightened until the reported spending is stable at the desired precision.
A common failure is to use the marginal probability \(P(Z_i<b_i)=\Phi(b_i)\) as though it were the probability of surviving through look \(i\). It ignores earlier crossings and generally gives the wrong cumulative error. Another failure is to add nominal increments as if all look-specific crossing events were disjoint. They are not; paths can cross more than one boundary, although the trial would have stopped at the first crossing. The recursion avoids both errors by retaining only the surviving paths.
Alternative Calculations and Limitations
Direct multivariate-normal integration can evaluate \(P(Z_1<b_1,\ldots,Z_i<b_i)\) from the joint distribution. It is mathematically equivalent to the recursion and can be useful as an independent check, especially for a small number of looks. The recursive formulation exposes the conditional structure and allows the calculation to proceed one look at a time. Simulation under \(H_0\) is another check: simulate complete statistic paths, count paths that cross by each look, and compare the empirical cumulative rejection proportions with the targets. Simulation is not a substitute for accurate calibration when precise Type I error control is required, because its estimates have Monte Carlo error.
The recursion depends on the assumed joint null distribution, the information times, and the specified deterministic boundaries. Incorrect information times, a poor normal approximation, or a statistic whose increments do not follow the assumed model can make the numerical result fail to control the actual Type I error. The method described here addresses upper efficacy boundaries under \(H_0\); additional stopping rules or a different null model require calibrating the relevant joint stopping event rather than silently reusing this equation.
Check Your Understanding
Answer each question, then compare with the solutions below.
- At a first look, the target cumulative spending is \(0.05\). Write the no-crossing equation for \(b_1\) and give its approximate value.
- Why does the second-look cutoff equation integrate only over \(x<b_1\)?
- Suppose \(A_{i-1}=0.02\) and \(A_i=0.02\). Is there a finite cutoff that attains the target survivor probability? State the limiting cutoff.
- Suppose \(A_{i-1}=0.90\) and \(A_i=1\). Explain why a positive spending increment does not imply a finite cutoff.
Solutions. (1) The equation is \(\Phi(b_1)=1-0.05=0.95\), so \(b_1=\Phi^{-1}(0.95)\approx1.645\). (2) Values \(x\ge b_1\) have already crossed at the first look and are not part of the event of surviving through look 2; including them would overstate the survivor probability. (3) No finite cutoff attains the target: it equals the previous survivor probability, so the limiting cutoff is \(+\infty\), which adds no rejection probability at this look. (4) The target survivor probability is \(1-A_i=0\). For every finite cutoff there is positive probability of \(Z_i\) falling below it while the earlier looks survived, so the target-zero limit requires \(b_i\to-\infty\), not a finite boundary.