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Transition Laws Between Adjacent Looks

Learn how the information added between adjacent looks determines the next statistic’s conditional mean and variance.

Advanced 10 min read

What You'll Learn

  • Express the next cumulative statistic as the earlier score plus an independent information increment.
  • Derive the conditional normal distribution of the next Z statistic given the current one.
  • Interpret how the information ratio controls conditional mean, variance, and correlation.
  • Calculate a conditional boundary-crossing probability from a stated prior statistic.
  • Recognize assumptions and diagnostics for using the canonical transition law.
  • Compare adjacent-look transitions with direct joint-normal conditioning.

Why an Adjacent-Look Transition Law Is Useful

A group sequential design accumulates evidence at planned looks. The cumulative statistic at the next look is related to the current statistic because the earlier observations are still part of the data. The transition law makes this dependence explicit: it describes the distribution of the next statistic conditional on the current one, with the amount of new information determining how much uncertainty is added.

This conditional description is useful for understanding how a boundary at a later look behaves after a particular result at the current look. It also gives a convenient way to factor a joint distribution across several looks. It does not make successive cumulative Z statistics independent; rather, it describes the new information as independent of the past under the canonical model.

Model and Notation

Consider two adjacent planned looks, \(k-1\) and \(k\), with deterministic information levels \(0<I_{k-1}<I_k\). Write \(S_j=\sqrt{I_j}Z_j\) for the unstandardized cumulative statistic at look \(j\). Under the canonical model under \(H_0\), \(S_j\) has mean zero and variance \(I_j\), and the increase in the cumulative statistic from one look to the next is an independent normal increment.

Definition: The information increment from look \(k-1\) to look \(k\) is \(\Delta I_k=I_k-I_{k-1}\). Under the canonical null model, \(S_k=S_{k-1}+\Delta S_k\), where \(\Delta S_k\sim N(0,\Delta I_k)\), and \(\Delta S_k\) is independent of the data accumulated through look \(k-1\). The transition law is the conditional distribution of \(Z_k\) given \(Z_{k-1}\).

The transition law is exact when the Gaussian independent-increment model is exact. In applications where the canonical model is an approximation, the transition law inherits that approximation. Its use here assumes the planned information levels are nonrandom, the information increases, the same cumulative statistic is carried forward, and the new increment has the stated normal distribution and independence. These conditions describe the model; they should not be assumed automatically for every statistic or monitoring scheme.

Deriving the Transition

Because the cumulative statistic at look \(k\) is the previous cumulative value plus the new increment, substitute \(S_{k-1}=\sqrt{I_{k-1}}Z_{k-1}\) and divide by \(\sqrt{I_k}\). Conditional on the earlier statistic, the only random term is the new increment.

$$ \begin{aligned} Z_k &=\frac{S_k}{\sqrt{I_k}}\\ &=\sqrt{\frac{I_{k-1}}{I_k}}Z_{k-1} +\frac{\Delta S_k}{\sqrt{I_k}}. \end{aligned} $$

The second term is independent of the earlier data, has mean zero, and has variance \(\Delta I_k/I_k\). Therefore, for any observed value \(z\) of \(Z_{k-1}\), the conditional distribution is

Theorem: Under the canonical null model,
$$ Z_k\mid Z_{k-1}=z \sim N\!\left( \sqrt{\frac{I_{k-1}}{I_k}}\,z,\, \frac{I_k-I_{k-1}}{I_k} \right). $$

The second argument of \(N(\mu,\sigma^2)\) is the variance. Thus the conditional standard deviation is \(\sqrt{(I_k-I_{k-1})/I_k}\). The transition’s conditional mean carries forward a fraction of the earlier Z statistic, while its variance reflects the fraction of total information that is new. Equivalently, the coefficient of \(z\) is the correlation between the two nested Z statistics under this model.

The conditional law given the entire history through look \(k-1\) has the same form, with \(z\) replaced by the observed \(Z_{k-1}\). This follows because the new increment is independent of that history. The current statistic summarizes the part of the accumulated score needed for this one-step transition.

Worked Example: Computing a Transition From the Information Ratio

Suppose two planned looks have \(I_{k-1}=100\) and \(I_k=160\), and the observed statistic at the earlier look is \(Z_{k-1}=1.20\). Assume the canonical null transition model applies. Find the conditional distribution at the next look.

The information increment is \(160-100=60\). The conditional mean and variance are

$$ \begin{aligned} E(Z_k\mid Z_{k-1}=1.20) &=\sqrt{\frac{100}{160}}(1.20) =0.9487,\\ \operatorname{Var}(Z_k\mid Z_{k-1}=1.20) &=\frac{60}{160} =0.3750. \end{aligned} $$

Thus \(Z_k\mid Z_{k-1}=1.20\sim N(0.9487,0.3750)\), with conditional standard deviation \(\sqrt{0.3750}=0.6124\). For example, the conditional probability that the next statistic exceeds 1.96 is

$$ P(Z_k>1.96\mid Z_{k-1}=1.20) = 1-\Phi\!\left(\frac{1.96-0.9487}{0.6124}\right) \approx 1-\Phi(1.6515) \approx 0.0493. $$

This is a probability under the specified transition model, conditional on the stated earlier statistic. It is not the unconditional probability of crossing a boundary at look \(k\), which would average over possible values at the earlier look.

What the Information Increment Changes

It is helpful to set \(r=I_{k-1}/I_k\), so \(0<r<1\). The transition mean is \(\sqrt{r}\,z\), and its variance is \(1-r\). These quantities move in opposite directions as the information added between looks changes.

  • If the new look adds little information relative to what was already accumulated, \(r\) is near 1. The next Z statistic tends to remain close to the current one, with relatively little conditional variance.
  • If the new look adds a large amount of information, \(r\) is smaller. The earlier value has less influence on the next standardized statistic, and the new increment contributes more conditional variance.
  • The conditional variance is positive for strictly increasing information and less than 1. If \(I_k=I_{k-1}\), there is no new information and the transition reduces to \(Z_k=Z_{k-1}\), not a fresh random draw.

These are statements about a conditional distribution under the model. They do not imply that a realized statistic must move toward zero or toward any particular boundary. The increment can be positive or negative.

Worked Example: A Negative Earlier Statistic

Suppose \(I_{k-1}=80\), \(I_k=200\), and \(Z_{k-1}=-0.400\). Under the canonical null model, the information ratio is \(80/200=0.400\), and the increment is \(120\).

The conditional mean is \(\sqrt{0.400}(-0.400)=-0.2530\), and the conditional variance is \(120/200=0.6000\). Hence

$$ Z_k\mid Z_{k-1}=-0.400 \sim N(-0.2530,0.6000), \qquad SD(Z_k\mid Z_{k-1})=\sqrt{0.6000}=0.7746. $$

The negative conditional mean reflects the part of the earlier cumulative statistic carried forward. The comparatively large conditional standard deviation reflects the substantial new information. It would be incorrect to treat the next Z statistic as having mean zero conditional on this observed earlier value.

Boundary Probabilities and a Common Error

For a one-sided upper boundary \(b\), the conditional crossing probability at the next look follows directly by standardizing the transition normal distribution:

$$ P(Z_k>b\mid Z_{k-1}=z) = 1-\Phi\!\left( \frac{b-\sqrt{I_{k-1}/I_k}\,z} {\sqrt{(I_k-I_{k-1})/I_k}} \right). $$

This calculation is useful as a conditional probability, but it is not by itself a calibrated group sequential error probability. Overall false-positive probability concerns crossing any boundary over the monitoring plan and requires accounting for the joint behavior at all looks.

Worked Example: Conditional Crossing From a Zero Earlier Statistic

Let \(I_{k-1}/I_k=0.2500\), let the earlier statistic be \(Z_{k-1}=0\), and consider an upper boundary \(b=1.645\). Under the canonical null transition, the conditional mean is zero and the variance is \(1-0.2500=0.7500\). Therefore the conditional standard deviation is \(\sqrt{0.7500}=0.8660254\).

Standardizing the boundary gives

$$ \frac{1.645-0}{\sqrt{0.7500}} = \frac{1.645}{0.8660254} \approx 1.8995. $$

Thus \(P(Z_k>1.645\mid Z_{k-1}=0)=1-\Phi(1.8995)\approx0.0288\). The calculation uses the conditional standard deviation, not the unconditional standard deviation of 1. Using the displayed rounded standard deviation also gives \(1.645/0.8660\approx1.8995\), which is consistent with the more precise value.

Relation to Joint-Normal Conditioning

The same transition can be obtained by conditioning a bivariate normal distribution. Under the canonical null model, \(Z_{k-1}\) and \(Z_k\) each have variance 1, and their correlation is \(\sqrt{I_{k-1}/I_k}\). For a bivariate normal pair with unit marginal variances and correlation \(\rho\), the conditional distribution of the second variable given the first value \(z\) has mean \(\rho z\) and variance \(1-\rho^2\). Substituting \(\rho=\sqrt{I_{k-1}/I_k}\) gives the transition law above.

The joint-normal route is convenient when only a pair of looks is being considered. The increment route makes the role of newly accumulated information explicit and extends naturally to a sequence: each step adds an independent increment, so the joint density can be represented as a product of one-step conditional densities. Both approaches give the same result when their assumptions hold.

1
Check the information schedule.
Confirm that the planned information levels are deterministic and satisfy \(0<I_{k-1}<I_k\).
2
Compute the ratio and increment.
Use \(r=I_{k-1}/I_k\) and \(\Delta I_k=I_k-I_{k-1}\); they imply the conditional mean coefficient \(\sqrt r\) and variance \(1-r\).
3
Condition on the observed earlier statistic.
Set the conditional mean to \(\sqrt r\,Z_{k-1}\), retain variance \(1-r\), and standardize any boundary using its square root.

Diagnostics, Limitations, and Pitfalls

A useful arithmetic check is that the conditional variance equals the fraction of total information added at the new look. It must lie strictly between zero and one when information increases. The squared conditional mean coefficient plus the conditional variance must equal one:

$$ \frac{I_{k-1}}{I_k} + \frac{I_k-I_{k-1}}{I_k} =1. $$

A result outside these constraints usually signals a reversed information ratio, an incorrect increment, or confusion between variance and standard deviation. In a simulation check under the null, consecutive increments should have approximately zero correlation, and the empirical conditional mean and variance across repetitions should be compatible with the stated formulas. Such checks can reveal coding errors or a mismatch between the simulated data mechanism and the assumed canonical model; they do not prove that the model is appropriate.

  • Wrong reasoning: “The next statistic is a new look, so it is independent of the earlier Z statistic and has conditional mean zero.”
  • Correction: The increment is independent of the earlier data. The cumulative statistic retains the earlier score, giving conditional mean \(\sqrt{I_{k-1}/I_k}\,Z_{k-1}\).
  • Wrong reasoning: “The conditional variance is the information ratio.”
  • Correction: The conditional variance is \(1-I_{k-1}/I_k\); the information ratio determines the squared carry-forward coefficient.
  • Limitation: If increments are dependent, information levels are random in a way not covered by the model, or the statistic is not adequately represented by the canonical normal process, this transition need not be exact.

Finally, if monitoring has already occurred at earlier looks, the probability of crossing at look \(k\) given that the trial has not stopped is not automatically the same as the probability conditional only on a particular \(Z_{k-1}=z\). When the current value is known, independent increments yield the stated one-step law even given the earlier history. When the current value is not fixed and one conditions only on survival, the distribution of that value must also be accounted for.

Check Your Understanding

Answer each question using the canonical null transition model. Solutions follow each question.

  1. If \(I_{k-1}=50\), \(I_k=125\), and \(Z_{k-1}=0.800\), find the conditional mean, variance, and standard deviation of \(Z_k\).
  2. For the same information levels as question 1, what fraction of the total information at look \(k\) is new?
  3. If \(I_{k-1}/I_k=0.640\), identify the conditional mean coefficient and conditional variance. What is the conditional standard deviation?
  4. Explain why the transition law does not imply that \(Z_k\) and \(Z_{k-1}\) are independent.

Solutions. (1) The ratio is \(50/125=0.400\), so the mean is \(\sqrt{0.400}(0.800)=0.5060\), the variance is \(1-0.400=0.6000\), and the standard deviation is \(\sqrt{0.6000}=0.7746\). (2) The new fraction is \((125-50)/125=75/125=0.6000\). (3) The mean coefficient is \(\sqrt{0.640}=0.8000\), the variance is \(1-0.640=0.3600\), and the standard deviation is \(0.6000\). (4) The independent quantity is the new score increment, not the next cumulative Z statistic. The next statistic includes the earlier cumulative score, and its conditional mean depends on \(Z_{k-1}\).

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