Expected Sample Size Under the Global Alternative

Why Calculate Expected Enrollment Under the Global Alternative?

A MAMS design has a maximum enrollment determined by its planned stage sizes, but the trial may stop before reaching that maximum. How often it does so depends on the treatment effects and on the design’s interim decision rules. Under the global alternative, every active treatment has its assumed target effect. Expected enrollment in this scenario describes the average enrollment across repeated trials when all treatments perform as planned.

This quantity is useful for understanding the enrollment typically required in a favorable design scenario. It is not a replacement for maximum enrollment: the maximum remains important for recruitment planning and resource commitments. Nor is the global-alternative expectation a guarantee about any one trial. It is calculated from the probabilities of reaching the design’s successive stages under a specified statistical model.

Definition and Calculation

Definition: The global alternative is the scenario in which every active treatment effect equals its assumed target effect. Let $n_j$ be the enrollment required in stage $j$, for $j=1,\ldots,J$, and let $R_j$ be the event that the trial reaches that stage. Define $p_j^A=P_A(R_j)$, the probability of reaching stage $j$ under the global alternative. The expected enrollment under the global alternative is $$ E_A(N)=\sum_{j=1}^{J} n_j p_j^A, \qquad p_1^A=1. $$

Here $N$ is the total enrollment in a realized trial. The maximum enrollment is $N_{\max}=\sum_{j=1}^{J}n_j$, attained if every stage is reached. The probability $p_j^A$ is an unconditional probability of reaching stage $j$, including all decisions at earlier stages. If instead a calculation supplies conditional continuation probabilities, they must first be combined to obtain $p_j^A$.

The formula assumes a prespecified stage schedule and a known enrollment cost $n_j$ whenever stage $j$ is reached. Enrollment is counted in participants, including all treatment and control participants specified by that stage’s design. The continuation probabilities must come from the joint model for the interim statistics and the actual decision rules. In particular, dependence between treatment comparisons must be included when relevant.

Why the Weighted Sum Gives Expected Enrollment

The calculation follows directly by expressing total enrollment in terms of stage-reach indicators. Let $I(R_j)$ equal 1 if stage $j$ is reached and 0 otherwise. Then the total enrollment is the sum of the stage sizes incurred:

N=\sum_{j=1}^{J} n_j I(R_j).

Taking expectations and using linearity of expectation gives

E_A(N) =E_A\left[\sum_{j=1}^{J} n_j I(R_j)\right] =\sum_{j=1}^{J}n_j E_A[I(R_j)] =\sum_{j=1}^{J}n_jP_A(R_j).

This derivation does not require the stage-reach events to be independent. They are nested: reaching a later stage requires reaching every earlier stage. If $q_j^A=P_A(R_{j+1}\mid R_j)$ is the conditional probability of proceeding from stage $j$ to stage $j+1$, then

p_1^A=1, \qquad p_j^A=\prod_{k=1}^{j-1}q_k^A \quad (j\ge 2).

The product is the chain rule for probability; it does not assume that interim statistics at different stages are independent. Each conditional probability must be appropriate to the history of the trial having reached that stage.

Worked Example: Two-Stage Design With Two Treatment Arms

Consider a hypothetical two-stage design with 60 participants in stage 1 and 90 additional participants in stage 2. Under the global alternative, suppose the two stage-1 treatment statistics are independent and each has distribution $N(1,1)$. The design proceeds to stage 2 only if neither statistic reaches the efficacy boundary $2$. There is no other early stopping rule in this illustration.

For either treatment statistic $Z_i$, the probability of not reaching the boundary is

P_A(Z_i<2)=\Phi\left(\frac{2-1}{1}\right)=\Phi(1).

By the stated independence assumption, the probability that both statistics remain below the boundary, and hence that the trial reaches stage 2, is

p_2^A=P_A(Z_1<2,\ Z_2<2)=\Phi(1)^2 \approx 0.70786098 \approx 0.7079.

Thus expected enrollment is

E_A(N)=60+90p_2^A =60+90(0.70786098) =123.707488 \approx 123.7\text{ participants}.

The maximum is $60+90=150$ participants. The expected enrollment is therefore about $26.29$ participants below the maximum, or $17.53\%$ lower as a proportion of the maximum. The independence assumption matters: if the statistics are correlated, multiplying their marginal probabilities is generally not correct, and the joint probability must be calculated under the appropriate joint model.

Worked Examples With Conditional and Simulated Probabilities

Worked Example: Three Stages From Conditional Continuation Probabilities

Suppose a hypothetical three-stage design enrolls 40 participants in each stage. Under the global alternative, the probability of proceeding from stage 1 to stage 2 is $q_1^A=0.80$, and the probability of proceeding from stage 2 to stage 3, conditional on having reached stage 2, is $q_2^A=0.70$. Assume these probabilities come from the design’s interim decision model.

The stage-reach probabilities are $p_1^A=1$, $p_2^A=q_1^A=0.80$, and

$$ p_3^A=q_1^Aq_2^A=(0.80)(0.70)=0.56. $$

Therefore,

E_A(N)=40(1)+40(0.80)+40(0.56) =40+32+22.4 =94.4\text{ participants}.

The maximum enrollment is $40+40+40=120$ participants. The result is less than the maximum because, under the assumed global alternative, some trials stop before stages 2 or 3. It would be incorrect to use $0.70$ alone as the probability of reaching stage 3: that is a conditional probability given that stage 2 was reached.

Worked Example: Estimating Expected Enrollment by Simulation

Suppose a hypothetical three-stage design requires 50 participants in stage 1, 40 additional participants in stage 2, and 60 additional participants in stage 3. A simulation generates 10,000 trials under a model in which every treatment has its assumed target effect. Of these trials, 7,240 reach stage 2 and 5,900 reach stage 3.

The estimated reach probabilities are $\hat p_2^A=7{,}240/10{,}000=0.724$ and $\hat p_3^A=5{,}900/10{,}000=0.590$. The latter cannot exceed the former because reaching stage 3 requires reaching stage 2; the reported counts satisfy this check. Substitution gives

\widehat{E_A(N)} =50+40(0.724)+60(0.590) =50+28.96+35.40 =114.36 \approx 114.4\text{ participants}.

The maximum is $50+40+60=150$ participants, so the simulated estimate is $35.64$ participants below the maximum. This is a Monte Carlo estimate, not an exact expectation: with a finite number of simulated trials, the estimated reach probabilities have simulation error. More simulations can reduce that error, but they cannot correct a misspecified model or incorrectly coded stopping rule.

Assumptions, Checks, and Common Pitfalls

Before interpreting an expected-enrollment calculation, check that the assumed scenario and the probability calculation match the design. The global alternative must specify the target effect for every treatment, along with the statistical model for the interim data. The stagewise rules must also be clear: for example, whether crossing an efficacy boundary stops the whole trial or only changes which treatments continue. The probabilities $p_j^A$ must reflect those rules and the dependence structure of the statistics.

Several checks help identify errors:

Probability ordering: $p_1^A=1$, and $1\ge p_2^A\ge\cdots\ge p_J^A\ge0$, because later-stage reach events are nested.
Enrollment bounds: $n_1\le E_A(N)\le N_{\max}$ when every stage size is nonnegative and stage 1 is always enrolled.
Conditional versus unconditional probabilities: multiply successive conditional continuation probabilities to obtain a later-stage reach probability; do not substitute a conditional probability directly for $p_j^A$.
Joint probabilities: do not multiply marginal probabilities across treatment arms unless independence is justified. Shared control data commonly induce dependence between comparisons.
Rounding: retain adequate precision in intermediate probabilities and round the final enrollment. For instance, $\Phi(1)^2$ is approximately $0.7079$; premature rounding of $\Phi(1)$ can change the final fourth decimal place.

A common mistake is to say that the trial will enroll the calculated expectation. An expectation is a long-run average over repeated trials under the specified scenario; a realized trial enrolls an amount determined by its observed interim decisions. Another mistake is to treat a global-alternative expectation as a conservative enrollment commitment. Early stopping can make the expectation substantially smaller than the maximum, but a trial can still reach that maximum, so the maximum remains relevant to planning.

Expected enrollment under the global alternative differs from two related summaries. Maximum enrollment is the fixed total required if all stages are reached. Expected enrollment under the global null uses continuation probabilities under the null scenario and can therefore be different. Neither scenario-specific expectation is an average over multiple possible treatment-effect patterns; such an average requires separately specified scenario probabilities.

Finally, the weighted-sum formula as written uses a fixed enrollment cost $n_j$ for each reached stage. If the number of participants required at a stage changes with the particular treatment arms still active, one fixed stage size may not represent the actual cost. The stage costs must then be defined to match the enrollment accounting being used. The present calculation should not be interpreted as resolving arm-specific cost changes by itself.

Key takeaway: Under a specified global-alternative model, expected enrollment is the sum of each stage’s enrollment cost multiplied by the probability of reaching that stage. The maximum enrollment is a separate design quantity, not the expected enrollment.

Check Your Understanding

Answer each question, then compare with the solutions below.

A two-stage design enrolls 45 participants in stage 1 and 75 in stage 2. Under the global alternative, the probability of reaching stage 2 is 0.60. Find expected and maximum enrollment.
In a three-stage design, the conditional probabilities of proceeding from stage 1 to 2 and from stage 2 to 3 are 0.90 and 0.50. What is the probability of reaching stage 3?
Why is it generally invalid to multiply marginal probabilities for two treatment statistics when they share control data?
A calculation gives stage-reach probabilities $1.00,\ 0.65,\ 0.72$ for stages 1, 2, and 3. What diagnostic check fails, and why?

Solutions. (1) $E_A(N)=45+75(0.60)=90$ participants; the maximum is $45+75=120$ participants. (2) The stage-3 reach probability is $0.90(0.50)=0.45$, by the chain rule. (3) Shared control data can make the treatment statistics dependent, so their joint probability is not generally the product of their marginal probabilities; use the joint model. (4) The ordering check fails because reaching stage 3 requires reaching stage 2, so $p_3^A$ cannot exceed $p_2^A$. The stated probabilities cannot describe nested stage-reach events.