Why Unequal Information Increments Matter
Sequential calculations often describe each analysis by its cumulative information, rather than by calendar time or the number of participants alone. The gap between successive analyses can vary: one interval may add relatively little information, while another adds much more. That variation changes the conditional distribution used to move the test statistic from one look to the next. It also affects the probability mass assigned to stopping and surviving at each look.
The calculation uses the canonical Gaussian model: the score accumulates with information, and its increments over disjoint information intervals are independent and normal. The information times are treated as fixed design quantities. Under this model, unequal increments require a look-specific transition mean and variance; using a single transition kernel throughout can give incorrect crossing probabilities.
Here \(\theta=0\) represents the null model, while a nonzero \(\theta\) represents a specified alternative drift per square root of information. The transition formula below assumes that the score process has the stated independent-increment normal distribution and that the information times are fixed. It is not automatically justified when information is estimated adaptively in a way that changes the distribution of the statistic.
Deriving the Transition for an Unequal Increment
Write the score added between looks \(i-1\) and \(i\) as \(S(t_i)-S(t_{i-1})\). Under the model, this increment is independent of the score through look \(i-1\), with mean \(\theta\Delta_i\) and variance \(\Delta_i\). Dividing the accumulated score at look \(i\) by \(\sqrt{t_i}\) gives
Conditional on \(Z_{i-1}=z\), the second term is normal, independent of \(Z_{i-1}\), with mean \(\theta\Delta_i/\sqrt{t_i}\) and variance \(\Delta_i/t_i\). Therefore,
The transition coefficient, drift contribution, and variance are all determined by the information times and the drift. The variance is positive because \(t_i>t_{i-1}\). Under the null, set \(\theta=0\); the conditional mean then contains only the rescaled previous statistic. At the first look, there is no previous statistic to condition on: \(Z_1\sim N(\theta\sqrt{t_1},1)\).
Worked Example: Computing Two Unequal Transitions
Suppose three hypothetical analyses occur at \(t_1=0.4\), \(t_2=1.1\), and \(t_3=2.5\). Calculate the transitions under the null and under the hypothetical drift \(\theta=0.3\).
For the second look, \(\Delta_2=1.1-0.4=0.7\). The coefficient on \(Z_1\) is \(\sqrt{0.4/1.1}=0.6030\), and the transition variance is \(0.7/1.1=0.6364\). Under \(\theta=0.3\), the drift contribution is \(0.3(0.7)/\sqrt{1.1}=0.2002\). Thus
Under the null the same transition has mean \(0.6030z\), with the same variance \(0.6364\). At the first look, \(Z_1\sim N(0,1)\) under the null, while under \(\theta=0.3\) its mean is \(0.3\sqrt{0.4}=0.1897\).
For the third look, \(\Delta_3=2.5-1.1=1.4\). The coefficient is \(\sqrt{1.1/2.5}=0.6633\), the variance is \(1.4/2.5=0.5600\), and the drift contribution is \(0.3(1.4)/\sqrt{2.5}=0.2660\). Hence \(Z_3\mid Z_2=z\sim N(0.6633z+0.2660,\ 0.5600)\) under the stated alternative. These calculations show why the two transitions cannot be represented by one common kernel.
Using the Transitions in Boundary Integrals
Suppose the stopping boundaries at look \(i\) are \(L_i\) and \(U_i\), with \(L_i<U_i\). A statistic below \(L_i\) or above \(U_i\) stops the trial; a value strictly between them survives to the next look. Since the Gaussian distribution is continuous, whether endpoints are included does not change the probabilities.
Let \(g_{i-1}(z)\) be the density of \(Z_{i-1}\) among paths that have survived all earlier looks, without renormalizing it to integrate to one. The surviving density at look \(i\) is
Here \(\phi(x;m,v)\) is the normal density at \(x\) with mean \(m\) and variance \(v\). The previous-look integral is restricted to its survivor interval; the indicator at the current look retains only current survivor states. To calculate a crossing probability at look \(i\), integrate the same transition density over the relevant stopping region instead. For example, the upper-bound crossing mass is
For a two-sided stopping rule, the lower-bound crossing contribution uses \(\Phi((L_i-\mu_i(z))/\sqrt{v_i})\); the two contributions are added if the stopping regions do not overlap. The survivor and crossing integrals must use the same transition parameters. In numerical quadrature, the limits for integrating over the previous state are the previous look’s survivor limits, while limits for integrating over the current state are determined by the current survivor or stopping region.
Worked Example: A Conditional Upper-Bound Crossing Probability
At two hypothetical looks, let \(t_1=0.5\) and \(t_2=1.5\). Under the null, suppose the upper stopping boundary at the second look is \(U_2=2.0\). Find the probability of crossing that boundary conditional on \(Z_1=0.8\).
The increment is \(\Delta_2=1.0\), so the conditional mean is \(\sqrt{0.5/1.5}(0.8)=0.4619\), and the variance is \(1.0/1.5=0.6667\). The standard deviation is \(\sqrt{0.6667}=0.8165\). Therefore,
Thus, conditional on the stated previous statistic and under the null model, the probability that the next statistic exceeds 2.0 is about 2.98%. This is a conditional transition probability, not the unconditional probability of crossing at the second look; the latter also averages over the distribution of states that survived the first look.
Worked Example: Setting Quadrature Limits for Unequal Times
Consider hypothetical times \(t_1=0.6\) and \(t_2=2.0\), drift \(\theta=0.2\), first-look survivor interval \((-1.4,1.6)\), and second-look upper boundary \(U_2=2.2\). Set up and approximate the probability of an upper crossing at look 2 among paths that survived look 1.
Here \(\Delta_2=1.4\), the coefficient is \(\sqrt{0.6/2.0}=0.5477\), the drift contribution is \(0.2(1.4)/\sqrt{2.0}=0.1980\), and the variance is \(1.4/2.0=0.7000\), with standard deviation \(0.8367\). At look 1, \(Z_1\sim N(0.2\sqrt{0.6},1)=N(0.1549,1)\), so its density is \(\phi(z;0.1549,1)\). The required crossing mass is
The integration limits \(-1.4\) and \(1.6\) come from survival at the first look, not from the second-look boundary. For a two-node Gauss-Legendre approximation, the nodes on this interval are \(z=0.1\pm1.5/\sqrt{3}\), or approximately \(-0.7660\) and \(0.9660\), and each weight is \(1.5\). At these nodes, the conditional upper-tail probabilities are approximately \(0.0019\) and \(0.0392\), while the first-look densities are approximately \(0.2611\) and \(0.2872\). The resulting approximation is
This two-node calculation is an illustrative coarse approximation, not a guarantee of numerical accuracy. It estimates the unconditional probability of reaching the upper stopping region at look 2 along paths that survive look 1. Refinement and convergence checks are needed before relying on a numerical result.
Diagnostics, Alternatives, and Limitations
A useful diagnostic is probability-mass accounting. At a given look, the mass among previous survivors should divide into lower crossing mass, current survivor mass, and upper crossing mass. These nonnegative quantities should sum, up to numerical error, to the previous survivor mass. A substantial discrepancy can indicate incorrect limits, a missing tail, an incorrectly normalized kernel, or inadequate quadrature.
Check the transition parameters independently before integrating: verify that \(\Delta_i=t_i-t_{i-1}>0\), that \(0<\Delta_i/t_i<1\) when \(t_{i-1}>0\), and that the conditional variance is \(\Delta_i/t_i\), not \(\Delta_i\). Under the null, the drift term must be zero. Under an alternative, the drift contribution must be \(\theta\Delta_i/\sqrt{t_i}\), not \(\theta\sqrt{\Delta_i}\). If an interval has very small information increment, its transition variance is small; if it has a large increment relative to \(t_i\), its transition variance is correspondingly larger.
Recursive integration is convenient because it carries forward only the surviving density. Direct integration of the joint multivariate normal distribution of all \(Z_i\) is an alternative: for \(i<j\), the canonical model gives \(\operatorname{Cov}(Z_i,Z_j)=\sqrt{t_i/t_j}\). Both approaches describe the same model when implemented correctly. The recursive method makes stage-specific survivor limits explicit, while direct joint integration can become cumbersome as the number of looks increases.
| Approach | What varies with unequal information | Practical consideration |
|---|---|---|
| Recursive survivor-density integration | Each transition has its own coefficient, drift, and variance; each stage has its own integration limits. | Tracks stopping and surviving mass look by look. |
| Direct joint-normal integration | The covariance matrix depends on the cumulative information times. | Provides a joint formulation but may require high-dimensional integration. |
| One repeated transition kernel | It does not account for changes in the information ratios. | Generally wrong unless the transitions truly have matching parameters. |
A common mistake is to reason that equally spaced information times produce the same transition coefficient and variance at every look. For example, at times \(1,2,3\), the raw increments are both 1, but the second-look fraction is \(1/2\) and the third-look fraction is \(1/3\); likewise, the corresponding ratios \(t_{i-1}/t_i\) are \(1/2\) and \(2/3\). The correction is to calculate each ratio from its own pair of cumulative times. Equal spacing makes the raw increments equal, not the standardized transitions identical.
These formulas rely on the canonical normal independent-increment model and fixed information times. They do not by themselves address numerical truncation, interpolation, grid refinement, or how to choose boundaries to achieve a target error rate. Nor should information increments be inferred from calendar spacing unless the design justifies that identification. When the assumptions or numerical checks fail, the resulting crossing probabilities may not describe the intended design.
Check Your Understanding
Answer each question using the transition model and integration rules above. Short solutions are included after the questions.
- For \(t_{i-1}=2\) and \(t_i=5\), find \(\Delta_i\), the coefficient on \(Z_{i-1}\), and the transition variance.
- Under the null, what is the conditional mean of \(Z_i\) when \(Z_{i-1}=z\)?
- If the previous survivor interval is \((-1,1.5)\), which limits are used to integrate over the previous state when calculating upper-bound crossings at the next look?
- Why does equal spacing of information times not generally produce an identical standardized transition at every look?
Solutions. (1) \(\Delta_i=5-2=3\), the coefficient is \(\sqrt{2/5}=0.6325\), and the variance is \(3/5=0.6000\). (2) The conditional mean is \(\sqrt{t_{i-1}/t_i}\,z\), because \(\theta=0\). (3) Integrate over the previous survivor interval, from \(-1\) to \(1.5\); the current upper boundary determines the conditional tail probability, not these previous-state limits. (4) Although equal spacing keeps \(\Delta_i\) fixed, the cumulative denominator \(t_i\) changes, so \(t_{i-1}/t_i\) and \(\Delta_i/t_i\) generally change with the look.