Why Two-Sided Crossing Probabilities Need Care
A two-sided sequential rule can reject at a look because the statistic is either sufficiently small or sufficiently large. These are separate rejection regions, but the probabilities at later looks must be calculated only among paths that have not already crossed a boundary. Otherwise, a path that crossed early can be counted again at a later look, even though the trial would already have stopped.
The calculation uses the survivor-density recursion and Gaussian transition kernels from the earlier material on those subjects. The key distinction is between a marginal probability of being beyond a boundary at a particular look and a first-crossing probability at that look. Only first-crossing probabilities can be added across looks to obtain the probability of ever rejecting.
For the calculations below, assume a continuous canonical Gaussian sequence under the null hypothesis \(H_0\). Write the information at look \(k\) as \(I_k\), with \(0<I_1<\cdots<I_K\). Under \(H_0\), \(Z_k\) has a standard normal marginal distribution, and for successive looks the transition is
The second argument of \(N(\cdot,\cdot)\) is the variance. This model assumes the canonical Gaussian information structure and the stated information ordering. The boundary rule is also assumed to stop the trial at the first crossing. With continuous distributions, the probability of landing exactly on a boundary is zero, so the choice of strict or non-strict inequality at a boundary does not change the probabilities.
First Crossings From Survivor Mass
Let \(g_k(z)\) denote the unnormalized survivor density at look \(k\): its integral is the probability of surviving through look \(k\), not one. At the first look, this density is the standard normal density restricted to the continuation interval. At subsequent looks, it is obtained by transitioning only the mass that survived the preceding look, then retaining the new continuation interval.
Here \(\phi\) is the standard normal density, \(\mathbf{1}\{\cdot\}\) is an indicator, and \(K_k(z\mid x)\) is the Gaussian transition density specified above. The lower and upper first-crossing probabilities at look \(k\geq2\) are obtained by integrating the transition tails over the previous survivor density:
At look 1, use the marginal distribution directly: \(q_{1,L}=P(Z_1\leq a_1)\) and \(q_{1,U}=P(Z_1\geq b_1)\). The cumulative probability of a first crossing through look \(m\) is the sum of the two tails at every look up to \(m\):
The lower and upper events at one look are disjoint because \(a_k<b_k\). First-crossing events at different looks are also disjoint: a path assigned to look \(k\) survived all earlier looks and stops at \(k\). These disjointness facts justify the sum.
A Short Probability-Conservation Justification
For a path that survived through look \(k-1\), exactly one of three outcomes occurs at look \(k\): it crosses the lower boundary, crosses the upper boundary, or survives again. Integrating the conditional probabilities of those outcomes against \(g_{k-1}\) gives
Thus the surviving mass decreases by exactly the amount of first-crossing mass added at that look. At look 1 the corresponding identity is \(q_{1,L}+q_{1,U}+\int g_1(z)\,dz=1\). Adding these identities across looks yields the useful check
This identity assumes there are no additional stopping outcomes, such as a separate futility rule. If the specified design includes other stopping regions, their probabilities must also be included in the partition.
Worked Example: Why Marginal Look Probabilities Cannot Be Added
Suppose a hypothetical two-look calculation gives \(P(R_1)=0.0400\), where \(R_k\) means that the statistic is beyond either boundary at look \(k\) in the underlying Gaussian sequence. Suppose also \(P(R_2)=0.0600\) and \(P(R_1\cap R_2)=0.0150\). These are illustrative probabilities, not results from a real study.
Adding the two marginal probabilities gives \(0.0400+0.0600=0.1000\). This double-counts the \(0.0150\) probability of paths outside a boundary at both looks. The probability of a first crossing at look 2 is instead
The total probability of rejection by look 2 is therefore \(0.0400+0.0450=0.0850\). In a recursive calculation, the survivor-density integral directly supplies the \(0.0450\) first-crossing contribution, without needing to calculate the overlap separately.
Worked Gaussian Calculation
For later looks, a conditional tail probability is straightforward to evaluate. If \(r=I_{k-1}/I_k\), the transition mean at the previous state \(x\) is \(\sqrt{r}x\), and the transition standard deviation is \(\sqrt{1-r}\). Consequently, the conditional lower-tail probability is the standard normal CDF evaluated at \((a_k-\sqrt{r}x)/\sqrt{1-r}\); the upper-tail probability is one minus the CDF evaluated at \((b_k-\sqrt{r}x)/\sqrt{1-r}\).
Worked Example: Integrating Both Tails at the Second Look
Consider a hypothetical two-look design under \(H_0\), with \(I_1/I_2=0.25\), boundaries \((-2.5,2.5)\) at look 1 and \((-1.8,1.8)\) at look 2. The look-1 crossing probability is \(2\Phi(-2.5)=0.01242\), so the survivor mass is approximately \(0.9876\). For illustration, suppose a coarse quadrature of the look-1 survivor density uses nodes \(x=-1.5,0,1.5\) with weights \(0.1646,0.6584,0.1646\), respectively. The weights sum to \(0.9876\), as required by the survivor mass. The following calculations use this quadrature approximation.
The transition standard deviation is \(\sqrt{1-0.25}=\sqrt{0.75}=0.8660\), and the conditional mean at node \(x\) is \(0.5x\). At \(x=1.5\), the mean is \(0.75\), giving
At the same node, the lower-tail probability is \(\Phi((-1.8-0.75)/0.8660)=\Phi(-2.945)\approx0.00162\). At \(x=-1.5\), symmetry swaps these two probabilities. At \(x=0\), each tail is \(\Phi(-1.8/0.8660)=\Phi(-2.078)\approx0.01885\). The quadrature approximation to the upper first-crossing probability at look 2 is therefore
By symmetry, \(q_{2,L}\approx0.03123\). The cumulative crossing probability through look 2 is approximately \(0.01242+0.03123+0.03123=0.07488\). The total survivor mass at look 2 is consequently about \(1-0.07488=0.9251\). This example illustrates the tail integration and mass accounting; the accuracy of the numerical answer depends on refining the quadrature.
Worked Example: Keeping Asymmetric Tails Separate
Consider another hypothetical two-look design under the same canonical null model. At look 1, take \(a_1=-2.0\) and \(b_1=2.4\). The lower crossing probability is \(\Phi(-2.0)=0.02275\), and the upper crossing probability is \(1-\Phi(2.4)=0.00820\). Thus the look-1 survivor mass is \(1-0.02275-0.00820=0.9691\).
Suppose recursive integration at look 2, using the specified boundaries and transition kernel, returns \(q_{2,L}=0.0216\) and \(q_{2,U}=0.0304\). These are hypothetical quadrature outputs. The lower and upper contributions differ, so they should be reported separately before combining them. The probability of a first crossing at look 2 is \(0.0216+0.0304=0.0520\), and the cumulative crossing probability through look 2 is
The implied survivor mass is \(1-0.08295=0.9171\), which is less than the look-1 survivor mass, as it must be. Combining lower and upper contributions is appropriate for the probability of any rejection, but keeping them separate is important when checking an asymmetric boundary calculation.
Numerical Checks, Alternatives, and Limitations
A probability calculation is only as trustworthy as its treatment of survivor mass. The following checks help identify common failures:
- Check the mass balance. At each look, the lower crossing mass, upper crossing mass, and new survivor mass should sum to the previous survivor mass, up to numerical integration error.
- Check that cumulative crossing probability is nondecreasing. Each added first-crossing contribution is nonnegative, and the cumulative value cannot exceed one.
- Check both tails and the interval endpoints. Reversing a tail inequality, using the wrong boundary, or truncating the continuation interval incorrectly can distort one tail while leaving the other apparently plausible.
- Check convergence. Refine the integration grid or quadrature rule and compare the resulting lower-tail, upper-tail, and cumulative probabilities. A stable total can conceal errors that cancel between the two tails.
A common mistake is to calculate \(P(Z_k\leq a_k)+P(Z_k\geq b_k)\) from the marginal distribution at each look and then add those values across looks. That counts paths that would already have stopped, and generally overstates the probability of ever rejecting. The correction is to integrate each conditional tail against \(g_{k-1}\), which contains only paths still at risk of crossing.
Recursive integration is natural when the boundary rule is applied at each look: it explicitly carries forward only surviving paths. A direct multivariate normal calculation can instead express the probability of surviving all looks as a probability over a multivariate continuation region, with crossing probability obtained by subtraction from one. That approach can be useful as an independent check, but it may become demanding as the number of looks grows. Either approach depends on the assumed joint Gaussian model, correct boundaries, and adequate numerical accuracy.
The recursion also depends on accurately representing density near each boundary. Coarse cells, inaccurate interpolation, or a truncated state space can lose or misallocate probability mass. Probability conservation is a diagnostic, not proof that every tail is accurate: complementary errors may still leave the total near one. Reported values should therefore be checked for grid or quadrature stability, especially when one tail is small.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- Why is the survivor density at look \(k-1\) not normalized to integrate to one?
- If the lower and upper first-crossing probabilities at a look are \(0.012\) and \(0.018\), what is the total first-crossing probability at that look?
- A calculation adds marginal boundary probabilities across three looks. What is the central error, and what should be summed instead?
- What probability-mass identity should hold at a look after including both crossing tails and continued survival?
Solutions. (1) Its integral is the probability of having survived through look \(k-1\), which is generally less than one; renormalizing it would discard the probability that has already crossed. (2) The two tails are disjoint, so the total is \(0.012+0.018=0.030\). (3) Marginal probabilities include paths that may already have crossed and stopped; sum lower and upper first-crossing probabilities at each look instead. (4) The lower crossing mass plus the upper crossing mass plus the new survivor mass must equal the previous survivor mass.