Unequal Stage Sizes and Cumulative Enrollment

Why Unequal Stage Sizes Need Careful Accounting

A MAMS design need not enroll the same number of participants at every stage. Recruitment targets may rise or fall across stages, and different treatment arms may stop at different interim analyses. Consequently, the number enrolled in a later stage is not simply a fixed addition to every possible trial outcome: it is added only if the relevant arm reaches that stage. Correct calculations must track both the size of each stage’s enrollment increment and the probability that the increment occurs.

The central distinction is between stagewise enrollment, the new participants randomized at a particular stage, and cumulative enrollment, the total randomized up to and including that stage. In this tutorial, each participant is counted once in the allocation group to which they are randomized. Accounting for shared-control participants across comparisons requires a separate adjustment; the calculations below treat the enrollment increments as distinct allocation-group counts.

Definition: Let $i$ index an allocation group and $k=1,\ldots,K$ index stages. Write $n_{ik}$ for the number of new participants randomized to group $i$ at stage $k$, if that stage is reached. These are stagewise enrollment increments, not cumulative sample sizes. Let $R_{ik}=1$ if group $i$ reaches stage $k$, and $R_{ik}=0$ otherwise; define $\rho_{ik}=P(R_{ik}=1)$. The cumulative enrollment in group $i$ at the end of stage $m$ is $$ N_i(m)=\sum_{k=1}^{m}n_{ik}R_{ik}. $$

The first stage is entered by design, so $\rho_{i1}=1$. If $q_{ik}$ denotes the conditional probability of continuing from stage $k$ to stage $k+1$, given that group $i$ reached stage $k$, then its probability of reaching stage $k$ is $$ \rho_{ik}=\prod_{j=1}^{k-1}q_{ij},\qquad k\geq 2. $$ This product applies when the stated $q_{ij}$ are conditional probabilities along the path. If reach probabilities are supplied directly, use them directly instead. They must satisfy $1=\rho_{i1}\geq \rho_{i2}\geq\cdots\geq\rho_{iK}\geq0$: reaching a later stage requires reaching all earlier stages.

Expected and Maximum Cumulative Enrollment

Because $R_{ik}$ is an indicator, its expected value is its probability of being one: $E(R_{ik})=\rho_{ik}$. Linearity of expectation therefore gives the expected cumulative enrollment in group $i$ by stage $m$:

E\{N_i(m)\} =E\left\{\sum_{k=1}^{m}n_{ik}R_{ik}\right\} =\sum_{k=1}^{m}n_{ik}\rho_{ik}.

This result does not require stage-reach indicators to be independent. For the total enrollment across distinct allocation groups, define $N(m)=\sum_i N_i(m)$. Applying linearity again, $$ E\{N(m)\}=\sum_i\sum_{k=1}^{m}n_{ik}\rho_{ik}. $$ Arm-specific continuation patterns can therefore be handled separately and then added. The total expected enrollment is not obtained by assuming that all arms have one common continuation probability.

If all planned stages are reached, the maximum enrollment is the sum of every planned stagewise increment: $$ N_{\max}=\sum_i\sum_{k=1}^{K}n_{ik}. $$ For an individual group, its maximum is $N_{i,\max}=\sum_{k=1}^{K}n_{ik}$. These maxima presume the design permits every listed stage to occur and that the planned stage sizes are fixed. An arm with no enrollment planned at a stage has $n_{ik}=0$ there.

Worked Calculations

Worked Example: One Arm With Three Unequal Stages

Suppose one treatment group plans to randomize 80 participants at stage 1, 120 at stage 2, and 160 at stage 3. The conditional probabilities of continuing after stages 1 and 2 are 0.70 and 0.60, respectively. Find the maximum enrollment and the expected cumulative enrollment at the end of each stage.

The stagewise increments are $n_{11}=80$, $n_{12}=120$, and $n_{13}=160$. The reach probabilities are $\rho_{11}=1$, $\rho_{12}=0.70$, and $\rho_{13}=0.70(0.60)=0.42$. Thus the maximum is $$ N_{1,\max}=80+120+160=360\text{ participants}. $$

At the end of stage 1, expected cumulative enrollment is $80(1)=80$. At the end of stage 2, it is $$ E\{N_1(2)\}=80(1)+120(0.70)=80+84=164\text{ participants}. $$ At the end of stage 3, it is $$ E\{N_1(3)\}=80(1)+120(0.70)+160(0.42) =80+84+67.2=231.2\text{ participants}. $$

The expected number randomized at stage 3 is $160(0.42)=67.2$, not $160(0.60)$. The value 0.60 is conditional on reaching stage 2; the unconditional probability of reaching stage 3 is 0.42.

Worked Example: Adding Arms With Different Continuation Patterns

Consider two treatment groups with distinct enrollment counts. Group A plans 50, 70, and 90 participants at stages 1, 2, and 3. Its conditional continuation probabilities are 0.80 and 0.625. Group B plans 40, 60, and 80 participants, with conditional continuation probabilities 0.55 and 0.50. Calculate each group’s maximum and expected enrollment, then the totals.

For group A, the reach probabilities are $1$, $0.80$, and $0.80(0.625)=0.50$. Its maximum is $50+70+90=210$. Its expected enrollment is $$ E(N_A)=50(1)+70(0.80)+90(0.50) =50+56+45=151\text{ participants}. $$

For group B, the reach probabilities are $1$, $0.55$, and $0.55(0.50)=0.275$. Its maximum is $40+60+80=180$, and its expected enrollment is $$ E(N_B)=40(1)+60(0.55)+80(0.275) =40+33+22=95\text{ participants}. $$

Adding the distinct group counts gives a maximum of $210+180=390$ participants and expected enrollment of $151+95=246$ participants. No independence assumption between the groups’ stopping decisions is needed for this expectation calculation: the expected sum equals the sum of the expectations.

Worked Example: Unequal Stage Sizes and Arm-Specific Paths

A two-treatment design plans the following stagewise increments: group A has 40, 60, and 80 participants; group B has 40, 30, and 50 participants. The conditional probabilities of continuing after stages 1 and 2 are 0.70 and 0.60 for A, and 0.40 and 0.25 for B. Find the maximum total enrollment and the expected number enrolled at each stage, as well as the final expected total.

The reach probabilities are $1, 0.70, 0.70(0.60)=0.42$ for A and $1, 0.40, 0.40(0.25)=0.10$ for B. The maximum total enrollment, if both groups reach all stages, is $$ N_{\max}=(40+60+80)+(40+30+50)=300\text{ participants}. $$

The expected stagewise contributions are 80 participants at stage 1, $60(0.70)+30(0.40)=42+12=54$ at stage 2, and $80(0.42)+50(0.10)=33.6+5=38.6$ at stage 3. Therefore the expected final enrollment is $$ E\{N(3)\}=80+54+38.6=172.6\text{ participants}. $$

This stage-by-stage calculation and the group-by-group calculation agree: group A contributes $40+42+33.6=115.6$ in expectation, while group B contributes $40+12+5=57$; their sum is $115.6+57=172.6$. Group B’s lower reach probability at stage 3 does not change group A’s reach probability or planned enrollment.

Why the Calculation Works

The expectation formula follows from writing enrollment as a sum of stage contributions, each multiplied by an indicator for whether that contribution occurs. For one group, the $k$th contribution is $n_{ik}R_{ik}$. Since $n_{ik}$ is fixed and $P(R_{ik}=1)=\rho_{ik}$, $$ E(n_{ik}R_{ik})=n_{ik}E(R_{ik})=n_{ik}\rho_{ik}. $$ Summing over stages gives the stated formula. This argument also explains why unequal stage sizes pose no special mathematical difficulty: each increment is weighted by its own probability of occurring.

The calculation requires that planned stagewise increments and reach probabilities be defined for the same enrollment unit and stopping rule. In particular, $n_{ik}$ must mean newly randomized participants at stage $k$, not the cumulative number randomized by that stage. The reach probability must correspond to entry into that stage. If a design’s stated continuation probability is conditional on having reached the previous stage, it must be multiplied along the path before it is used to weight a later stage’s enrollment.

Diagnostics, Alternatives, and Limitations

Several checks can catch errors before expected enrollment is reported:

Check stage-size meaning. Stagewise increments should add to the maximum. If the inputs are cumulative sample sizes, first convert them into increments by taking differences.
Check the reach-probability scale. Reach probabilities are unconditional probabilities of reaching a stage. They cannot increase with stage number, and stage 1 has probability one.
Check bounds. With nonnegative stage sizes, expected total enrollment must be at least the enrollment planned for stage 1 and no greater than the maximum. A result above the maximum indicates an accounting or probability error.
Check arm-specific paths. Use each arm’s own probabilities when its stopping rule or continuation pattern differs. Do not transfer one arm’s probability to another without justification.
Check what is being counted. The sum over groups is appropriate when the increments represent distinct participants. If the same control participants contribute to multiple comparisons, counting a separate copy for each comparison overstates unique enrollment.

An equal-stage shortcut can be convenient when the stage increments really are equal, but it is not a substitute for summing the actual increments when sizes differ. A simulation can estimate expected enrollment by generating trial paths and averaging their enrollment totals; it is useful for checking a complex implementation. For fixed stage increments and known reach probabilities, however, the direct weighted sum is exact and usually easier to audit. Neither method supplies the continuation probabilities: those must come from the design’s interim decision behavior under the scenario of interest.

Expected enrollment is an average over possible stopping paths, not a promise that a particular trial will enroll that number. It also does not describe the variability of enrollment. Designs with the same expected enrollment can have different distributions of total enrollment, especially if their stopping events are dependent. The expectation calculation remains valid without independence, but further distributional or variability calculations require information about the joint stopping paths.

Common pitfall: Multiplying a later-stage size by the conditional probability of continuing only at the immediately preceding stage understates the path requirement. For example, if an arm has continuation probabilities 0.70 and 0.60, the probability of reaching stage 3 is $0.70(0.60)=0.42$, not 0.60. Use the unconditional reach probability for each stage, then multiply it by that stage’s enrollment increment.

Check Your Understanding

Work through the questions, then compare your results with the solutions below.

An arm plans 30, 50, and 70 participants at three stages. The conditional continuation probabilities are 0.80 after stage 1 and 0.50 after stage 2. What are its maximum and expected final enrollment?
For the same arm, how many participants are expected to be randomized at stage 3?
Two distinct groups have expected enrollments of 100 and 65. Must their stopping decisions be independent for total expected enrollment to be 165? Explain.
Stagewise increments are 40 at stage 1 and 100 at stage 2, with a stage-2 reach probability of 0.30. A calculation reports expected enrollment of 170. Identify the error and give the correct value.

Solutions. (1) The maximum is $30+50+70=150$. Reach probabilities are $1, 0.80, 0.80(0.50)=0.40$, so expected final enrollment is $30+50(0.80)+70(0.40)=30+40+28=98$ participants. (2) The expected stage-3 contribution is $70(0.40)=28$ participants. (3) No. Linearity of expectation gives a total expectation of $100+65=165$ regardless of dependence, provided the counts are defined as distinct enrollment contributions. (4) The reported value adds all planned increments as if stage 2 were certain. The correct expectation is $40(1)+100(0.30)=40+30=70$ participants.