Propagating Alpha After Early Endpoint Rejection

Why an Interim Rejection Changes the Alpha Ledger

In a multiple-endpoint group sequential design, each endpoint may be examined at several interim looks. The multiplicity procedure determines both which endpoint hypotheses can be rejected and how much type I error is allocated to their sequential tests. If an endpoint is formally rejected early, a prespecified graphical procedure may let some of its allocation flow to other endpoints that remain active. That opportunity is useful only when the graph and sequential testing rules permit it.

The key distinction is between an endpoint’s allocated alpha and the cutoff used at one particular look. An allocation is a budget for the endpoint’s test over its scheduled looks, not a fresh nominal level to spend at every look. A transferred allocation therefore changes the testing rule for eligible remaining hypotheses; it does not authorize an unadjusted extra test.

Definition: Let \(H_1,\ldots,H_m\) be endpoint hypotheses in a multiplicity graph. Each active hypothesis \(H_i\) has an allocation \(\alpha_i\), and the graph specifies transition weights \(g_{ij}\) from endpoint \(i\) to endpoint \(j\). When \(H_i\) is formally rejected under the prespecified procedure, its allocation may be transferred to eligible active endpoints according to those weights. A hypothesis is active if it has not been formally rejected and the procedure still permits it to be tested.

For clarity, first consider a graph with no self-transitions, in which each row of transition weights sums to one. If \(H_r\) is formally rejected, the direct allocation update for each still-active endpoint \(j\) is

\alpha_j^{\text{new}}=\alpha_j^{\text{old}}+\alpha_r^{\text{old}}g_{rj}.

The update uses the allocation of the rejected endpoint immediately before the transfer. A positive \(g_{rj}\) means that the graph directs some of that allocation to \(H_j\); a zero weight means there is no direct transfer to \(H_j\). Graphs with self-transitions or unassigned portions require their prespecified update rule rather than this simplified formula.

What Counts as a Rejection

A transfer is triggered by a formal rejection under the current multiplicity-controlled group sequential procedure. For endpoint \(i\), this means that its statistic crossed the efficacy boundary applicable to that endpoint, look, and current allocation, with the required multiplicity and sequential calibration. A small-looking statistic p-value or a nominal threshold considered on its own is not enough.

The allocation \(\alpha_i\) concerns the endpoint’s sequential test across its looks. Its boundaries must control the probability of a false rejection over that testing process. If an endpoint receives more alpha after another endpoint is rejected, a valid sequential procedure must specify how the updated allocation affects future testing. With no earlier testing of the recipient, a future single-look test can use its updated level directly. If the recipient has already had looks, simply applying the full updated level at a later look can spend error twice. In that situation, the updated rule must retain valid sequential calibration, for example through an appropriate conditional-error implementation.

Key condition: Transfer only after a formal rejection, only to recipients allowed by the prespecified graph update, and only through sequential boundaries valid for the updated procedure. Alpha is not transferable merely because a statistic is promising or a nominal p-value is below a familiar threshold.

Worked Example: Updating Three Endpoint Allocations

Suppose a hypothetical three-endpoint design has total one-sided familywise level \(0.025\). Its initial allocations are \(\alpha_A=0.0125\), \(\alpha_B=0.0075\), and \(\alpha_C=0.0050\). The graph directs 60% of \(A\)’s allocation to \(B\) and 40% to \(C\). At an interim look, \(A\) crosses its formally specified efficacy boundary. Assume \(B\) and \(C\) remain active and the sequential implementation permits the update.

The amounts transferred are

0.0125(0.60)=0.0075 \quad\text{to } B, \qquad 0.0125(0.40)=0.0050 \quad\text{to } C.

Thus the updated allocations are

\alpha_B^{\text{new}}=0.0075+0.0075=0.0150, \qquad \alpha_C^{\text{new}}=0.0050+0.0050=0.0100.

The total remains \(0.0150+0.0100=0.0250\). Endpoint \(A\)’s allocation has been redistributed, not added to the family’s original error budget. The numbers describe endpoint-wide allocations; they do not say that \(B\) and \(C\) may each use those amounts independently at every remaining look.

Eligibility Depends on the Graph and the Testing History

The graph identifies where released alpha can go. Being an endpoint in the same trial is not, by itself, enough to make a hypothesis a recipient. A hypothesis can use transferred allocation only if it remains active, is reachable under the current graph update, and has a valid test rule for the resulting allocation. Its clinical role, whether it is primary or secondary, and any gatekeeping restrictions matter only insofar as they are encoded in that prespecified procedure.

After an endpoint is rejected, the graph may also update transition weights among the remaining nodes. That updated graph determines possible later transfers. Consequently, receiving some alpha now does not necessarily make a hypothesis a direct recipient of every future transfer. Record both the allocation update and any transition-weight update required by the graph; do not infer new edges from the observed statistics.

Worked Example: An Active Hypothesis Without an Incoming Edge

Suppose endpoint \(D\) has allocation \(0.010\) and has outgoing weights \(g_{DE}=1\) and \(g_{DF}=0\). Endpoints \(E\) and \(F\) are both active when \(D\) is formally rejected. The graph update gives

\alpha_E^{\text{new}}=\alpha_E^{\text{old}}+0.010, \qquad \alpha_F^{\text{new}}=\alpha_F^{\text{old}}.

Although \(F\) remains active, it receives none of \(D\)’s allocation because its incoming transition weight from \(D\) is zero. It may receive alpha later if a subsequent graph update directs allocation to it, but that possibility must follow from the graph—not from a discretionary decision made after seeing the interim results. If the procedure specifies that a fraction can remain unallocated, that fraction also stays unavailable unless the update rule says otherwise.

A useful way to check eligibility is to ask three questions in order:

Was the source formally rejected?
Confirm that it crossed the applicable efficacy boundary under the current sequential multiplicity procedure.

Does the graph send allocation to this recipient?
Apply the prespecified outgoing weights and any required graph update, restricting attention to eligible active endpoints.

Can the recipient use the updated level validly?
Use its calibrated sequential test. If earlier looks have occurred, do not treat the entire updated endpoint allocation as fresh alpha at the next look.

Boundary Updates and a Numerical Decision

When a recipient has not yet been tested, the updated allocation can determine its level for its first test, subject to the design’s stated assumptions. For example, for a one-sided standard normal statistic \(Z\) at a single scheduled look, a test at level \(\alpha\) rejects for \(Z\geq z_{1-\alpha}\), where \(z_{1-\alpha}\) is the corresponding standard normal quantile. With multiple looks, the relevant boundaries must instead control the sequential type I error across those looks.

Worked Example: Testing a Recipient at Its First Look

Continue the hypothetical allocations from the first example. Suppose \(B\) has not yet been examined, and its first and only scheduled test is a one-sided standard normal test. After the valid rejection of \(A\), \(B\)’s updated level is \(0.0150\). Assume its statistic under \(H_B\) has a standard normal null distribution and the prespecified graph-sequential procedure permits this single-look use of the updated allocation. The critical value is

z_{1-0.0150}=z_{0.9850}\approx 2.170.

Suppose the hypothetical observed statistic for \(B\) is \(Z_B=2.24\). Since \(2.24\geq2.170\), \(H_B\) is rejected at that look under the updated rule. Its one-sided p-value is approximately \(1-\Phi(2.24)=0.0125\), which is below \(0.0150\). This conclusion relies on the stated single-look assumptions. If \(B\) had already been tested, the critical value \(2.170\) could not simply be applied as a new full-level boundary without checking the sequential error calibration.

Why the Update Can Preserve Error Control

The rationale for recycling is that the testing procedure controls the familywise error rate while allocating its error budget through a prespecified graph. In a graphical closed-testing construction, local tests for the relevant hypothesis intersections must themselves be valid. When a hypothesis is rejected, the graph reallocates its local level among remaining hypotheses; it does not create new error budget. With group sequential testing, those local tests must also account for repeated looks.

The sequential requirement is essential. Let the information observed by an interim look be denoted by \(\mathcal{F}_k\). A testing rule for a remaining hypothesis that has already been examined must account for the conditional probability of a future false rejection given \(\mathcal{F}_k\). Replacing its boundary with an ordinary single-look cutoff at the newly enlarged allocation can ignore the probability already spent at earlier looks. A valid implementation uses boundaries or conditional-error rules that preserve the required error control after the graph update.

This gives a practical distinction between allocation arithmetic and boundary validity. The graph determines how much alpha is assigned to each eligible node. The sequential test determines how that node’s assigned amount is used over time. Correct arithmetic alone does not establish that the updated boundaries are valid.

Worked Example: Why a Later Cutoff Is Not Automatically the Updated Level

Suppose endpoint \(C\) originally has total sequential allocation \(0.010\), and it has already been examined at one interim look without crossing its boundary. After another endpoint is rejected, the graph raises \(C\)’s total allocation to \(0.016\). It would be incorrect to conclude automatically that \(C\) may now use a new one-sided boundary of \(z_{0.984}\approx2.144\) at the next look.

That cutoff is the usual single-look normal cutoff for level \(0.016\), but \(C\) has already had a look. The probability of a false rejection at the first look must be included in the sequential design’s error accounting. The correct next boundary depends on the earlier look, the original sequential rule, the update rule, and the information available at the new look. Therefore the graph update supplies \(0.016\) as the endpoint’s updated allocation, while the validated sequential procedure supplies the boundary. Without that procedure, the stated cutoff is not justified.

Common Mistakes and Practical Checks

Recycling after nominal significance. A nominal p-value below a customary cutoff is not the trigger. The source endpoint must be formally rejected under the current multiplicity-controlled sequential boundary.
Sending alpha to every remaining endpoint. Only recipients specified by the outgoing weights and current graph update can receive the source allocation.
Using the transfer repeatedly. The same source allocation cannot be transferred twice. Maintain a ledger showing the pre-transfer allocation, the transfer fractions, and the new allocations.
Applying the updated total at every look. An endpoint-wide allocation is not a per-look allowance. Earlier looks must be included in the sequential calibration.
Changing the graph after seeing results. A data-driven change to recipients or transfer weights is not the prespecified recycling rule and can invalidate error control.

A sound implementation therefore records the source rejection, the active recipient set, each applied transition weight, the updated allocations, and the boundary rule used for every remaining look. When an endpoint is not yet available or cannot be tested at that time, its allocation should not be treated as immediately usable by another endpoint unless the prespecified procedure explicitly allows that transfer.

Key takeaway: A formally rejected endpoint can release alpha only through the prespecified multiplicity graph. An active hypothesis can use the transferred allocation only if it is an eligible recipient and its sequential testing rule remains valid after the update.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Endpoint \(A\) has allocation \(0.008\), and its graph sends 25% to \(B\) and 75% to \(C\). If \(A\) is formally rejected and both recipients are active, how much alpha does each receive?
Endpoint \(D\) has no outgoing edge to active endpoint \(F\). Can \(F\) use some of \(D\)’s allocation because it is the only endpoint with results available?
Why is crossing a nominal p-value threshold insufficient to trigger alpha transfer?
A recipient has already been tested at an earlier look. Why might it be invalid to use its entire updated allocation as a new single-look level?

Solutions. (1) \(B\) receives \(0.008(0.25)=0.002\), and \(C\) receives \(0.008(0.75)=0.006\). These amounts sum to the released \(0.008\). (2) No. Availability of results does not create an edge; \(F\) can receive alpha only if the prespecified graph update directs some to it. (3) Transfer requires formal rejection under the applicable sequential multiplicity-controlled boundary, not a comparison with a generic nominal cutoff. (4) The earlier look may already have used part of the sequential type I error allowance. A new single-look cutoff can ignore that prior spending, so the updated boundaries must be calibrated for the full testing history.