Why Require a Minimum Number of Arms to Continue?
At an interim analysis, an experimental arm may continue only if its interim score meets a prespecified cutoff. This threshold rule is transparent, but it can leave very few arms for the remaining stages. If the design requires a minimum number of experimental options to remain under evaluation, the continuation policy must say what happens when too few arms meet the cutoff.
A minimum-continuation constraint answers that operational question. It does not change the evidence from the interim data, and it does not make a below-cutoff arm satisfy the cutoff. Instead, when the usual rule selects fewer than the required minimum, a prespecified ranking identifies additional arms to carry forward. The distinction between meeting the evidence criterion and being retained to satisfy the quota must remain clear.
This definition assumes that a larger score consistently represents a more favorable interim result. The score might be a standardized selection score or another measure established by the design; the minimum rule does not prescribe how to construct it. The cutoff, score direction, minimum \(m\), and ranking convention must be specified before using the rule. The scores must be available for all arms being ranked and comparable for the intended ranking.
The rule also presumes that \(1 \leq m \leq q\). If \(m>q\), the minimum is impossible to meet because fewer than \(m\) experimental arms are available. That is a design-feasibility problem, not something the interim data can resolve. If two arms have identical scores at the boundary of selection, the design needs a prespecified tie-handling convention; without one, the selected set may not be uniquely determined.
Writing the Rule as a Selection Set
Let \(E=\{i:S_i\geq c\}\) be the set of qualifying arms, and let \(r=|E|\) be the number of qualifying arms. The retained set \(R\) can be described in two cases. When \(r\geq m\), the threshold rule alone selects all qualifying arms. When \(r<m\), the retained set contains \(E\) and the \(m-r\) highest-ranked arms among those not in \(E\).
Here \(B_{m-|E|}\) denotes the set of the \(m-|E|\) highest-ranked nonqualifying arms. In either case, \(R\) contains at least \(m\) arms. It may contain more than \(m\): if several arms meet the cutoff, the rule retains all of them rather than discarding qualifying arms to impose an upper limit. Thus, this is a minimum, not an exact continuation count.
Calculate each arm’s prespecified score and identify the arms with \(S_i \geq c\).
Compare the number \(r\) of qualifying arms with the minimum \(m\).
If \(r\geq m\), retain all qualifiers. If \(r<m\), add the highest-ranked \(m-r\) arms from those below the cutoff.
Label arms as cutoff-qualifying or quota-retained, so the operational decision is not mistaken for evidence that every retained arm met the cutoff.
Worked Example: Filling a Minimum of Three Arms
At a hypothetical interim analysis, five experimental arms have standardized selection scores. Larger scores favor continuation. The cutoff is \(c=1.00\), and the minimum is \(m=3\).
| Arm | Score \(S_i\) | Meets cutoff? |
|---|---|---|
| A | 1.42 | Yes |
| B | 0.91 | No |
| C | 0.73 | No |
| D | 0.22 | No |
| E | -0.15 | No |
Only arm A qualifies because \(1.42 \geq 1.00\), while each of the other four scores is below 1.00. Thus \(r=1\), and the number of additional arms needed is \(m-r=3-1=2\). Ranking the nonqualifying arms from highest score to lowest gives B, C, D, E. The two highest-ranked nonqualifiers are B and C.
The retained set is therefore \(\{A,B,C\}\). Arm A qualifies by the cutoff; B and C are retained only to satisfy the minimum. In particular, their retention does not mean their scores reached 1.00.
Why the Fill Rule Has This Form
When fewer than \(m\) arms qualify, any retained set that includes all qualifiers needs at least \(m-r\) additional arms. Choosing exactly that many additional arms makes the final set contain exactly \(m\) arms in this case. To favor the strongest available interim results among the nonqualifiers, the rule chooses the \(m-r\) largest scores below the cutoff.
This gives a simple justification for the ranking step. Sort all \(q\) scores from largest to smallest, using the prespecified tie convention if needed. When \(r<m\), all \(r\) scores at or above the cutoff are among the leading scores, and the next \(m-r\) largest scores complete the first \(m\) positions. Any alternative set of \(m\) arms that keeps all qualifiers but substitutes a lower-scoring nonqualifier for a higher-scoring one has not followed the ranking principle. The fill operation therefore honors both parts of the policy: keep every qualifying arm, then select the strongest available arms needed to meet the minimum.
This justification concerns the deterministic selection rule, not the probability that an arm is truly effective. The ranking depends on observed scores, which can be noisy. Assessing how often the rule retains a promising arm or advances a less promising one requires evaluating the rule under plausible joint distributions of the interim scores, including their dependence. A shared control can induce dependence between arm-specific scores; that dependence does not prevent the ranking rule from being applied, but it matters when evaluating its operating characteristics.
Worked Example: More Arms Qualify Than the Minimum
In a second hypothetical analysis, four experimental arms have scores \(1.31\), \(1.08\), \(0.84\), and \(0.61\). The cutoff is \(c=0.80\), and the minimum is \(m=2\).
The first three arms qualify because \(1.31\geq0.80\), \(1.08\geq0.80\), and \(0.84\geq0.80\). The fourth does not, because \(0.61<0.80\). Therefore \(r=3\), and \(3\geq2\). The rule retains all three qualifying arms.
The retained count is three, not two. Discarding the third qualifying arm merely to make the count equal to \(m\) would convert a minimum into an exact top-two rule, which is a different policy.
Worked Example: No Arm Reaches the Cutoff
In a third hypothetical analysis, the scores are \(0.76\), \(0.53\), \(0.19\), and \(-0.08\). The cutoff is \(c=1.00\), and the prespecified minimum is \(m=2\).
Every score is below 1.00, so \(r=0\). The number of arms to add is \(m-r=2-0=2\). Ranking the four nonqualifying arms gives scores \(0.76\), \(0.53\), \(0.19\), and \(-0.08\). The two highest-ranked arms are the first two, so those are retained to meet the minimum.
Neither retained arm met the cutoff. The decision follows the minimum-continuation policy, and its interpretation must not be upgraded to “two arms showed sufficient interim evidence.” The design should distinguish that operational outcome from cutoff qualification in its records and reporting.
Assumptions, Diagnostics, and Limitations
The rule is mechanical, but its meaning depends on several design choices. Before an interim decision, check that:
- The minimum \(m\) is feasible for the number of experimental arms available at that look.
- The score is defined for every arm, points in the intended direction, and can support the planned ranking.
- The cutoff and ranking rule are prespecified, including how exact ties are handled.
- The decision procedure states whether all cutoff qualifiers continue when their number exceeds \(m\). The rule defined here retains all qualifiers.
- Records distinguish qualification under the cutoff from retention solely to satisfy the minimum.
Potential warning signs include an undocumented change in score direction, a missing score for an arm, or a selection record that retains fewer than \(m\) arms even though \(q\geq m\). Another diagnostic is to reconstruct the decision from the recorded scores, cutoff, and tie convention. If the resulting set differs from the recorded set, either the rule was not applied as written or some design specification is missing.
A minimum constraint necessarily weakens the threshold-only rule’s ability to exclude all low-scoring arms: if too few qualify, at least some nonqualifiers continue. That is the intended trade-off, not a statistical contradiction. It can preserve options for later evaluation, but it can also use later-stage resources on arms with weaker interim scores. The minimum should therefore be chosen as part of the design, with its resource and selection consequences assessed in advance.
The main alternatives answer different questions. A threshold-only rule retains exactly the qualifying arms and may retain fewer than \(m\). A fixed top-\(k\) rule retains a specified number based solely on rank, whether or not those arms meet a cutoff. The minimum rule combines these ideas: it retains all qualifiers and uses ranking only when the qualifying set is too small. It is not equivalent to lowering the cutoff until \(m\) arms qualify, because lowering the cutoff changes the qualification criterion itself.
Common Mistake
Incorrect reasoning: “The minimum is three, so the three retained arms all passed the interim cutoff.” This confuses the operational requirement with the evidence criterion. If only one arm qualifies, two additional arms may be retained despite falling below the cutoff.
Correction: Report the two statuses separately. State which arms met \(S_i\geq c\), which additional arms were retained by rank to satisfy \(m\), and how many arms continue. The minimum rule determines continuation; it does not change observed scores or establish that a nonqualifying arm met the cutoff.
Check Your Understanding
Answer each question using the rule that all cutoff qualifiers continue and that higher scores rank above lower scores. Solutions are included for checking.
- There are five experimental arms, \(c=0.50\), and \(m=2\). Their scores are \(0.91\), \(0.48\), \(0.44\), \(0.20\), and \(-0.10\). Which arms continue, and why?
- There are four arms, \(c=0.30\), and \(m=2\). Their scores are \(0.75\), \(0.42\), \(0.29\), and \(0.12\). How many continue?
- There are three available experimental arms but the protocol specifies \(m=4\). Can the minimum rule be implemented? Explain.
- Why is “lower the cutoff until \(m\) arms pass” not necessarily the same policy as filling the minimum by rank?
Solutions. (1) Only the arm with score \(0.91\) qualifies. Since \(r=1<m=2\), add the highest-scoring nonqualifier, \(0.48\). Those two arms continue; only the first qualifies. (2) The scores \(0.75\) and \(0.42\) both meet \(0.30\), so \(r=2=m\). Both continue, and neither below-cutoff arm is added. (3) No. With \(q=3\), at most three experimental arms are available, so a requirement of four is infeasible and must be resolved in the design specification. (4) Lowering the cutoff changes which scores count as qualifying. Filling the minimum keeps the original cutoff fixed and labels added arms as nonqualifiers selected by rank; the operational count may match, but the evidence classification does not.