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Sample Size and Expected Sample Size · Tutorial 212 of 499

Expected Sample Size Under Partial Treatment Activity

Specify an effect for each active arm, then use arm-specific continuation and joint continuation probabilities to calculate expected enrollment under partial treatment activity.

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What You'll Learn

  • Represent partial activity with a separate assumed effect for each treatment arm.
  • Translate arm-specific effects into probabilities of passing interim continuation boundaries.
  • Account for dependence between treatment-arm statistics when finding the chance that any arm continues.
  • Calculate expected enrollment from stage-specific costs and probabilities of active-arm sets.
  • Compare partial-activity enrollment with global-null and global-alternative scenarios.
  • Identify assumptions and diagnostics that can make an expected-enrollment estimate unreliable.

Why Partial Treatment Activity Changes Expected Enrollment

A MAMS design may begin with several experimental treatments, but the assumed benefit need not be the same for every arm. One treatment might have the effect used in the design’s power calculation, another might have no effect, and a third might have a smaller effect. These differences change the chances that individual arms pass interim continuation rules—and therefore change how many participants are enrolled at later stages.

The global null and global alternative describe useful limiting scenarios: all experimental arms have no effect, or all have their assumed beneficial effects. Partial activity asks a different planning question: what expected enrollment results from a specified pattern of effects across arms? The answer depends on both each arm’s chance of continuing and the dependence among arms’ interim statistics.

Definition: A partial-activity scenario specifies an assumed treatment effect separately for each experimental arm. Under that scenario, the expected sample size is the probability-weighted average total enrollment over the possible interim decisions and resulting paths through the design. It is not the maximum sample size, nor does it describe a guaranteed enrollment.

Modeling Arm-Specific Continuation

Suppose there are \(J\) experimental arms, indexed by \(j=1,\ldots,J\), and a common control. Let \(Z_{jk}\) be the interim test statistic for arm \(j\) at stage \(k\). For illustration, assume that under a specified effect scenario the vector of stage-\(k\) statistics follows a multivariate normal model:

$$ \mathbf{Z}_k \sim N_J(\boldsymbol{\eta}_k,R_k). $$

Here \(\boldsymbol{\eta}_k\) contains the arm-specific mean shifts on the \(Z\)-statistic scale, and \(R_k\) is the correlation matrix, which can reflect the shared control. If arm \(j\) continues when \(Z_{jk}>c_{jk}\), its marginal continuation probability is

$$ q_{jk}=P(Z_{jk}>c_{jk})=1-\Phi(c_{jk}-\eta_{jk}), $$

where \(\Phi\) is the standard normal cumulative distribution function. The mean shift \(\eta_{jk}\) is determined by the assumed effect for that arm and the stage’s information. A beneficial effect generally raises the continuation probability relative to a no-effect assumption, but does not guarantee that the arm continues.

These marginal probabilities do not, by themselves, determine whether any arm continues. If statistics share a control, they are typically correlated. For two arms, for example,

$$ P(\text{at least one continues}) =q_{1k}+q_{2k}-P(Z_{1k}>c_{1k},Z_{2k}>c_{2k}). $$

The joint probability must be calculated using the joint model, such as a bivariate normal probability with the appropriate correlation. Multiplying marginal probabilities as if the arms were independent is justified only when independence is a reasonable assumption for the decision statistics.

Worked Example: Translating a Partial-Activity Scenario Into Continuation Probabilities

Consider three hypothetical treatment arms at an interim analysis. Each continues if its statistic exceeds \(c=0.5\). Assume the statistics have unit variance and arm-specific mean shifts \(\eta_1=1.2\), \(\eta_2=0\), and \(\eta_3=0.8\). These values represent, respectively, a stronger assumed effect, no assumed effect, and a smaller positive assumed effect. The marginal calculations use only the stated normal distributions; they do not require the arms to be independent.

For arm 1,

$$ q_1=1-\Phi(0.5-1.2)=1-\Phi(-0.7)=\Phi(0.7)=0.7580. $$

For arm 2, whose assumed mean shift is zero,

$$ q_2=1-\Phi(0.5-0)=1-\Phi(0.5)=0.3085. $$

For arm 3,

$$ q_3=1-\Phi(0.5-0.8)=1-\Phi(-0.3)=\Phi(0.3)=0.6179. $$

Thus the assumed effects yield different chances of continuing: 0.7580, 0.3085, and 0.6179. These are marginal probabilities for each arm. To calculate a later-stage shared-control cost, we would also need the joint probabilities, or equivalently the probability that the set of continuing arms is nonempty.

From Continuation Decisions to Expected Enrollment

Let \(A_k\) denote the set of experimental arms still active when stage \(k\) begins. The first-stage set contains all \(J\) arms; later sets depend on earlier decisions. Let \(g_k(A_k)\) be the number of participants enrolled at stage \(k\) when the active set is \(A_k\). If no arms remain, the design stops and \(g_k(\varnothing)=0\). Total enrollment is the sum of stage enrollments along the realized path.

Key result: Expected total enrollment under a specified effect scenario is the sum, over stages, of expected stage enrollment:
$$ E(N)=\sum_{k=1}^{K} E\{g_k(A_k)\} =\sum_{k=1}^{K}\sum_{A\subseteq\{1,\ldots,J\}} P(A_k=A)\,g_k(A). $$

The second expression makes the scenario dependence explicit: the probabilities of the possible active sets are determined by the arm-specific effects, the interim rules, and the dependence structure. The formula does not require equal effects or equal continuation probabilities across arms.

A common allocation rule makes the calculation especially transparent. Suppose a stage enrolls \(n_C\) control participants if at least one arm remains and \(n_T\) participants per active treatment arm. Then

$$ E\{g_k(A_k)\} =n_C P(A_k\ne\varnothing) +n_T\sum_{j=1}^{J}P(j\in A_k). $$

The treatment-arm term follows by writing the number of active arms as a sum of indicators: \(|A_k|=\sum_j I(j\in A_k)\). Taking expectations gives \(E(|A_k|)=\sum_jP(j\in A_k)\). The control term is different: one control group is enrolled if any arm remains, so its expected contribution uses \(P(A_k\ne\varnothing)\), not the sum of marginal continuation probabilities. This distinction matters whenever control enrollment is shared.

Worked Example: Two-Stage Enrollment With a Shared Control Group

Consider a hypothetical two-arm, two-stage design. Stage 1 enrolls 40 control participants and 40 participants per treatment arm, for \(40+2(40)=120\) participants. At stage 2, the design enrolls 30 control participants if at least one arm continues, plus 30 participants for each arm that continues. Suppose the stage-1 continuation probabilities are \(q_1=0.7580\) and \(q_2=0.3085\), as in the first two calculations above. Assume the stage statistics have correlation 0.5 and that bivariate normal integration gives \(P(\text{both continue})=0.286\). This joint probability is stated for the hypothetical example; the calculation accounts for dependence rather than assuming independence.

The expected number of continuing treatment arms is the sum of the marginal probabilities:

$$ E(|A_2|)=0.7580+0.3085=1.0665. $$

The probability that at least one arm continues is

$$ P(A_2\ne\varnothing)=0.7580+0.3085-0.286=0.7805. $$

Therefore expected stage-2 enrollment is

$$ E\{g_2(A_2)\} =30(0.7805)+30(1.0665) =23.415+31.995 =55.41. $$

Expected total enrollment is \(120+55.41=175.41\), or about 175 participants when reported to the nearest whole participant. The maximum is \(120+30+2(30)=210\), but the expected value is lower because stage 2 is not always reached with both arms active. The calculation uses the assumed arm-specific effects through the continuation probabilities; changing either assumed effect can change the expected enrollment.

More Than One Interim Stage

With additional stages, an arm must remain active through all earlier decisions to contribute treatment participants at a later stage. For instance, if an arm’s stagewise continuation decisions are independent across stages and its continuation probabilities are \(q_{j1}\) and \(q_{j2}\), its probability of reaching stage 3 is \(q_{j1}q_{j2}\). Independence of decisions across stages is a modeling assumption, often motivated by independent increments; it should not be imposed if the design’s statistics or decision rules make it inappropriate.

Across arms, the probability that at least one reaches a later stage still depends on their joint behavior. One useful general calculation is to enumerate possible active sets, determine each set’s probability under the multivariate interim model, and multiply by the enrollment cost for that set. For many arms, direct enumeration may be inefficient; numerical integration or simulation from the specified joint model can estimate the same expectation.

Worked Example: Three Stages With Arm-Specific Activity

Suppose a hypothetical design has two arms and three stages. Stage 1 enrolls 10 control participants and 20 per treatment arm, so its enrollment is \(10+2(20)=50\). At each later stage, enrollment is 10 control participants if any arm remains and 20 participants per remaining arm. For a deliberately simplified calculation, assume independence between arms and between stagewise continuation decisions. Let arm A have continuation probability 0.8 at each interim and arm B have probability 0.3 at each interim. These probabilities represent different assumed effects and are held constant across the two decisions.

The probability of reaching stage 2 with at least one arm is

$$ 1-(1-0.8)(1-0.3)=1-(0.2)(0.7)=0.86. $$

The expected number of arms at stage 2 is \(0.8+0.3=1.1\). Thus expected stage-2 enrollment is

$$ 10(0.86)+20(1.1)=8.6+22=30.6. $$

To reach stage 3, each arm must continue at both interim decisions. The respective probabilities are \(0.8^2=0.64\) and \(0.3^2=0.09\). Under the stated independence assumption, the probability that at least one reaches stage 3 is

$$ 1-(1-0.64)(1-0.09)=1-(0.36)(0.91)=0.6724. $$

The expected number of arms at stage 3 is \(0.64+0.09=0.73\), so expected stage-3 enrollment is

$$ 10(0.6724)+20(0.73)=6.724+14.6=21.324. $$

Adding stage costs gives \(50+30.6+21.324=101.924\), or about 102 participants. The independence assumptions make this arithmetic possible, but shared-control statistics would generally invalidate independence between arms. In that setting, joint probabilities from the appropriate correlated model must replace the products used here.

Assumptions, Diagnostics, and Comparisons

An expected-enrollment calculation is conditional on a complete set of design and scenario assumptions. Specify the effect for every arm, the information and continuation rule at each interim, the joint distribution of the test statistics, and the enrollment cost for every possible active set. If the design uses a shared control, the model should reflect the resulting correlation. If stage sizes or allocation ratios change when arms are dropped, use the actual stage-specific enrollment function \(g_k(A_k)\), not a constant cost for each stage.

Check the calculation in several ways. First, the expected enrollment must lie between the minimum enrollment and the maximum permitted by the design. Second, the probability of any arm continuing must be at least as large as each individual continuation probability and no larger than their sum (or 1, if the sum exceeds 1). Third, when one arm’s assumed effect is increased while all else is held fixed in a monotone continuation design, its marginal continuation probability should not decrease. Finally, compare a direct sum over active sets with simulation from the same joint model when practical; material disagreement can reveal an error in indexing, dependence, or stage costs.

The global-null calculation sets all arm effects to zero; the global-alternative calculation assigns the assumed beneficial effect to every arm. Partial activity lies between these cases only in an informal sense: enrollment need not be a simple numerical interpolation, because continuation probabilities and the chance of any arm remaining are nonlinear functions of the effects and their dependence. Averaging expected enrollment across several scenarios is a separate planning choice that requires scenario weights; a single partial-activity calculation does not supply those weights.

A common mistake is to calculate each arm’s expected treatment enrollment correctly, then use \(q_1+q_2\) as the probability of enrolling the shared control group. That sum is the expected number of continuing arms, not the probability that at least one continues. The correction is to use the joint probability of the active set; for two arms, \(P(A\ne\varnothing)=q_1+q_2-P(\text{both continue})\). Another failure is to assign every arm the same effect simply because the design has one common alternative hypothesis. Under partial activity, the scenario must explicitly state the effect assumed for each arm.

Expected enrollment is a planning summary, not a prediction of the realized sample size in a particular trial. It is also sensitive to uncertain effects, covariance assumptions, and operational features such as delayed outcomes or missed assessments. When these inputs are uncertain, report the scenario-specific expectations and the assumptions behind them rather than presenting one value as universally applicable.

Check Your Understanding

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

  1. Two arms have marginal continuation probabilities 0.60 and 0.40, with probability 0.25 that both continue. What is the probability that at least one continues?
  2. A later stage enrolls 25 participants per active treatment arm and 20 control participants if any arm is active. If two arms have continuation probabilities 0.70 and 0.50 and their joint continuation probability is 0.35, find expected enrollment at that stage.
  3. With two arms and independent interim decisions across stages, an arm has continuation probabilities 0.8 and 0.5 at the first and second interim. What is its probability of reaching the third stage?
  4. Why is it generally incorrect to assume that two treatment-versus-shared-control test statistics are independent?

Solutions. (1) By inclusion–exclusion, \(0.60+0.40-0.25=0.75\). (2) The chance of any arm continuing is \(0.70+0.50-0.35=0.85\), and the expected number of active arms is \(0.70+0.50=1.20\). Expected enrollment is \(20(0.85)+25(1.20)=17+30=47\) participants. (3) The arm must continue at both decisions, so the probability is \(0.8(0.5)=0.40\), under the stated independence assumption. (4) The two statistics use the same control data, so their sampling errors are generally correlated; the joint continuation probability must reflect that dependence.

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