Expected Sample Size Under the Global Null

Why Expected Enrollment Depends on Interim Decisions

A MAMS design may enroll fewer participants than its maximum because treatment arms can stop at interim analyses. Under the global null, every active treatment has zero effect relative to control. The continuation probabilities evaluated under this scenario determine how often the design reaches later stages. Expected enrollment therefore combines the planned stage sizes with the probability that each stage is actually entered.

A shared control group creates an important distinction. At a later stage, the number of treatment groups enrolled depends on how many arms continue. The common control group, however, is enrolled once if at least one arm continues, not once for every continuing arm. Its expected contribution depends on the probability that any arm remains active.

Definition: The global null is the configuration in which every treatment effect is zero. The expected enrollment under the global null is the mean total number of participants enrolled when the design is run under that configuration, averaging over the interim data and resulting continuation decisions.

Enrollment Formula With a Shared Control

Suppose a design has $K$ treatment arms and $J$ stages. At stage $j$, the planned enrollment is $n_{Tj}$ participants per active treatment arm and $n_{Cj}$ participants in one shared control group. Stage sizes are fixed in advance; a stage is entered only if at least one treatment arm remains active. Participants enrolled in different stages are distinct, and all randomized participants count toward enrollment.

Let $C_{ij}$ be the event that treatment arm $i$ remains active after the interim decision at the end of stage $j$. Let $M_{j-1}$ be the number of arms active when stage $j$ is about to begin. Before stage 1, all $K$ arms are active, so $M_0=K$. For $j>1$, $M_{j-1}=\sum_{i=1}^{K} I(C_{i,j-1})$, where $I(\cdot)$ is 1 when its event is true and 0 otherwise.

Enrollment result: If each stage uses $n_{Tj}$ participants for each active treatment arm and enrolls one shared control group of $n_{Cj}$ participants whenever at least one arm is active, then $$ E(N)=\sum_{j=1}^{J}\left\{n_{Tj}E(M_{j-1})+n_{Cj}P(M_{j-1}>0)\right\}. $$ The probabilities and expectations are evaluated under the global null.

The first term counts treatment participants arm by arm. The second counts a shared control group only when the stage occurs. For stage 1, $E(M_0)=K$ and $P(M_0>0)=1$. For later stages, the design reaches stage $j$ precisely when at least one arm continued past stage $j-1$.

Write $r_{i,j}=P(C_{ij})$ for the probability that arm $i$ remains active after stage $j$, and $g_j=P(\bigcup_{i=1}^{K}C_{ij})$ for the probability that at least one arm remains. Since $M_j$ is the sum of the arm-specific indicators, linearity of expectation gives $E(M_j)=\sum_i r_{i,j}$. This identity does not require independence among arms.

E(N)=K n_{T1}+n_{C1} +\sum_{j=2}^{J}\left\{n_{Tj}\sum_{i=1}^{K}r_{i,j-1}+n_{Cj}g_{j-1}\right\}.

Why the Formula Works

For each stage $j$, let $N_{Tj}$ be the number of treatment participants enrolled there and $N_{Cj}$ the number of shared-control participants. Under the stated design, $N_{Tj}=n_{Tj}M_{j-1}$, while $N_{Cj}=n_{Cj}I(M_{j-1}>0)$. Total enrollment is the sum of these quantities across stages. Taking expectations and using linearity gives the result:

E(N)=\sum_{j=1}^{J}\{E(N_{Tj})+E(N_{Cj})\} =\sum_{j=1}^{J}\{n_{Tj}E(M_{j-1})+n_{Cj}P(M_{j-1}>0)\}.

This derivation also shows why the two continuation summaries are not interchangeable. $E(M_{j-1})$ determines expected treatment enrollment, whereas $P(M_{j-1}>0)$ determines expected shared-control enrollment. In general, neither quantity can be recovered from the other without information about the joint continuation events.

Worked Example: A Two-Stage Design With Three Arms

Consider a hypothetical design with three treatment arms. Stage 1 enrolls 40 participants per treatment arm and 40 in a shared control group. At the stage-1 interim, each arm has probability 0.60 of continuing under the global null, and the probability that at least one arm continues is 0.85. If stage 2 is reached, it enrolls 30 participants per continuing arm and 30 in one shared control group. Calculate expected total enrollment.

Stage 1. All three arms enter, so enrollment is $3(40)+40=160$ participants.

Stage 2 treatment enrollment. The expected number of continuing arms is $3(0.60)=1.80$. Thus expected stage-2 treatment enrollment is $30(1.80)=54$ participants.

Stage 2 control enrollment. The shared control group is enrolled with probability 0.85, so its expected contribution is $30(0.85)=25.5$ participants.

Therefore, $$ E(N)=160+54+25.5=239.5 $$ participants. The result is an average over repeated runs of this hypothetical design; it need not be an integer, even though each realized trial enrolls an integer number of people.

Calculating Continuation Probabilities From the Null Model

The probabilities $r_{i,j}$ and $g_j$ must describe the design’s actual interim rules under the global null. They can be obtained from the interim data model and boundaries, or from a simulation that generates data under zero effects and applies the specified decisions. The calculation must preserve the dependence induced by shared control data and any other common sources of variation.

If continuation is described by conditional probabilities, the chain rule can be used for each arm. For example, if $c_{i\ell}=P(C_{i\ell}\mid C_{i,\ell-1})$, then $r_{i,j}=\prod_{\ell=1}^{j}c_{i\ell}$, with $r_{i,0}=1$. This is valid when each conditional probability refers to the stated event of that arm having survived the preceding stages. It does not imply that the arms are independent. To obtain $g_j$, joint continuation information is still needed.

In a symmetric design under the global null, the arm-specific probabilities may be equal, so $r_{i,j}=r_j$ and $E(M_j)=Kr_j$. Symmetry alone does not determine $g_j$: arms may continue together more or less often depending on their dependence. The next examples show how these pieces enter an enrollment calculation.

Worked Example: A Three-Stage Calculation Under the Global Null

Consider a hypothetical three-arm design with these stage sizes: stage 1 has 30 participants per treatment arm and 30 in shared control; stage 2 has 20 per active arm and 25 in shared control; stage 3 has 15 per active arm and 20 in shared control. Suppose the null-model continuation calculations are symmetric across arms, with $r_1=0.55$ and $g_1=0.82$ after stage 1, and $r_2=0.30$ and $g_2=0.58$ after stage 2.

Stage 1 enrollment is $3(30)+30=120$. For stage 2, the expected number of active arms is $3(0.55)=1.65$, giving expected treatment enrollment $20(1.65)=33$. The shared-control contribution is $25(0.82)=20.5$.

For stage 3, the expected number of active arms is $3(0.30)=0.90$, giving expected treatment enrollment $15(0.90)=13.5$. Stage 3 is entered with probability $g_2=0.58$, so expected control enrollment is $20(0.58)=11.6$.

Combining the stage contributions, $$ E(N)=120+33+20.5+13.5+11.6=198.6 $$ participants. The probabilities are internally consistent with stopping: $r_2\le r_1$ and $g_2\le g_1$. They must also be calculated from a joint null model compatible with the design, rather than assumed to be independent across arms.

Worked Example: Why the Shared-Control Probability Is Not the Mean Number of Arms

In a hypothetical three-arm design, suppose each arm has probability 0.50 of continuing after stage 1. The pairwise intersection probabilities are 0.20 for each of the three arm pairs, and the probability all three continue is 0.10. Stage 2 enrolls 20 participants per continuing arm and 30 in a shared control group.

The expected number of continuing arms is $3(0.50)=1.50$. To calculate the chance that at least one continues, use inclusion-exclusion: $$ g_1=P(C_{11}\cup C_{21}\cup C_{31}) =3(0.50)-3(0.20)+0.10=1.00. $$ Thus stage 2 is reached with probability 1.00. Expected treatment enrollment is $20(1.50)=30$, and expected shared-control enrollment is $30(1.00)=30$, for total expected stage-2 enrollment of 60 participants.

Multiplying the control size by the expected number of arms would give $30(1.50)=45$, which is incorrect: it would count more than one shared control group on average. The expected number of arms and the probability of any arm continuing answer different questions.

Assumptions, Diagnostics, and Alternatives

The calculation relies on the enrollment rules matching the formula. In particular, stage sizes must be fixed as specified, the control group must be shared and enrolled once per reached stage, and a stage with no active treatment arms must not enroll further participants. If control enrollment instead changes with the number of active arms, replace $n_{Cj}I(M_{j-1}>0)$ by the actual control-enrollment function of $M_{j-1}$ and take its expectation. Similarly, planned sample sizes should be adjusted if the target is actual enrollment including attrition or unusable outcomes; the formula as written counts the stated randomized participants.

A useful check is to compare expected enrollment with its possible range. It must be at least the stage-1 enrollment and no greater than the maximum enrollment obtained if every arm reaches every stage. Also, $0\le r_{i,j}\le1$, continuation probabilities cannot increase for an arm that can only be dropped, and $g_j$ cannot increase as stages progress. Under symmetry, $E(M_j)=Kr_j$, and $r_j\le g_j\le \min(1,Kr_j)$. Violations usually indicate inconsistent probabilities or a mismatch in event definitions.

The maximum sample size is a useful planning bound, but it assumes every planned stage is reached and therefore does not represent average use under the null. A shortcut based only on the expected number of active arms correctly captures treatment enrollment but generally miscounts a shared control group. Conversely, multiplying by the probability that any arm continues captures shared-control enrollment but not the number of treatment groups. Both summaries are required.

A further failure mode is using continuation probabilities from a model that omits dependence created by common controls. Marginal arm probabilities may still suffice for expected treatment enrollment, but the probability of at least one arm continuing can be wrong if the joint distribution is wrong. Check that the interim model, decision rule, and continuation probabilities all refer to the same global-null scenario and stopping rule.

List stage costs.
Record treatment enrollment per active arm and shared-control enrollment for each stage.

Obtain null continuation probabilities.
For each interim, calculate arm-specific survival probabilities and the probability that at least one arm survives.

Weight each stage by its chance of being reached.
Multiply treatment cost by the expected number of active arms and control cost by the probability any arm is active.

Check the result.
Verify stagewise probability ordering and compare the expected total with stage-1 and maximum enrollment.

Key takeaway: Under the global null, expected treatment enrollment at a stage is based on the expected number of continuing arms. Expected shared-control enrollment is based on the probability that at least one arm continues. Preserve both quantities when dependence is present.

Check Your Understanding

Answer each question, then compare with the solutions below.

A two-arm design enrolls 25 participants per arm and 30 in shared control at stage 1. At stage 2, it enrolls 20 per continuing arm and 24 in shared control. Under the global null, each arm continues with probability 0.40 and at least one continues with probability 0.65. Find expected total enrollment.
Why is the expected number of active arms not generally an appropriate multiplier for shared-control sample size?
For four symmetric arms, suppose each arm has probability 0.25 of continuing after an interim. What is the expected number of continuing arms? Can the probability that at least one continues be determined from this marginal probability alone?
For a design with three stages, what probability determines whether the shared control group is enrolled in stage 3?

Solutions. (1) Stage 1 enrollment is $2(25)+30=80$. The stage-2 expected number of active arms is $2(0.40)=0.80$, so expected treatment enrollment is $20(0.80)=16$. Expected control enrollment is $24(0.65)=15.6$. Total expected enrollment is $80+16+15.6=111.6$ participants. (2) A single shared group is enrolled once whenever at least one arm continues; its expected enrollment is the group size times the probability of that union event, not the group size times the number of continuing arms. (3) The expected number is $4(0.25)=1.00$. The probability at least one continues cannot generally be determined from the marginals alone because it depends on the joint continuation pattern. (4) It is the probability that at least one arm remains active after stage 2, $P(M_2>0)$, evaluated under the global null.