Expected Sample Size Near an Interim Decision Threshold

Why Near-Threshold Means Matter

A MAMS design can enroll substantially different numbers of participants under two plausible assumptions about treatment activity. The difference can be especially noticeable when an interim statistic has a mean near the boundary for continuing its arm: a small change in the assumed mean can then change the probability of continuing by a meaningful amount. This tutorial quantifies that local sensitivity for a fixed design and a specified set of assumed effects.

The calculations concern expected enrollment under the design’s continuation rules. They do not change the design or its error-rate properties. The design’s thresholds and stage sizes are treated as fixed; the question is how the expected enrollment responds when the interim-statistic means are near those thresholds.

Definition: For arm \(j\), let \(Z_j\) be its interim statistic and let \(c_j\) be its arm-continuation threshold, with the rule that the arm continues when \(Z_j>c_j\). Its continuation probability is \(q_j=P(Z_j>c_j)\). A mean is near the threshold when the assumed mean of \(Z_j\) is close to \(c_j\), on the scale of the statistic’s standard deviation.

Probability and Enrollment Sensitivity

Suppose the interim statistic for one arm is approximately normal with mean \(\mu\) and standard deviation \(\tau\), and the continuation rule is \(Z>c\). The continuation probability is then

q(\mu)=P(Z>c)=1-\Phi\!\left(\frac{c-\mu}{\tau}\right) =\Phi\!\left(\frac{\mu-c}{\tau}\right),

where \(\Phi\) is the standard normal cumulative distribution function. In the common case of a standardized statistic with variance 1, \(\tau=1\). Differentiating with respect to the assumed mean gives

\frac{dq}{d\mu} =\frac{1}{\tau}\phi\!\left(\frac{\mu-c}{\tau}\right),

where \(\phi\) is the standard normal density. This derivative is largest at \(\mu=c\), since \(\phi(x)\) is largest at \(x=0\). Thus a given small change in the mean changes continuation probability most when the mean is at the threshold. At the threshold \(q=0.5\); moving the mean far above or below it makes the continuation probability closer to 1 or 0 and less locally sensitive.

Key result: For \(Z\sim N(\mu,\tau^2)\) and continuation rule \(Z>c\), the local change in continuation probability per unit change in mean is \(\phi((\mu-c)/\tau)/\tau\). It is greatest at the boundary, where it equals \(1/(\tau\sqrt{2\pi})\).

For a two-stage design with a fixed first-stage enrollment \(n_1\) and a fixed additional enrollment \(n_2\) if the arm continues, the expected total enrollment is

E(N)=n_1+n_2q(\mu).

Consequently, \(dE(N)/d\mu=n_2\,dq/d\mu\). The expected enrollment is most sensitive near the threshold when a large second-stage enrollment is at stake. In a multistage design, the same principle applies at each decision: a continuation probability near 0.5 can make the expected enrollment sensitive to the assumed mean, with the size of the effect also depending on the enrollment that the decision makes possible.

Worked Enrollment Calculations

Worked Example: One Arm Near Its Boundary

Consider a hypothetical two-stage design. It enrolls 120 participants in stage 1 and, if the arm continues, 80 additional participants in stage 2. Suppose its interim statistic has standard deviation 1 and continuation threshold \(c=2.0\). Calculate expected enrollment when the assumed interim means are 1.5, 2.0, and 2.5.

The model is \(Z\sim N(\mu,1)\), and continuation requires \(Z>2.0\). For \(\mu=1.5\), \(q=\Phi((1.5-2.0)/1)=\Phi(-0.5)=0.3085\), so \(E(N)=120+80(0.3085)=144.68\), or about 145 participants.

For \(\mu=2.0\), \(q=\Phi(0)=0.5000\), and \(E(N)=120+80(0.5000)=160\) participants. For \(\mu=2.5\), \(q=\Phi(0.5)=0.6915\), and \(E(N)=120+80(0.6915)=175.32\), or about 175 participants.

The means 0.5 units below and above the threshold produce continuation probabilities 0.3085 and 0.6915. The resulting expected enrollments differ by about 30.6 participants. These are expectations under the stated model, not guarantees about the enrollment in any one trial.

In a MAMS design, stage costs may depend on how many arms remain. A shared control cohort may be enrolled if at least one arm continues, while each continuing arm has its own additional treatment cohort. In that setting, arm-specific probabilities alone are not always enough: the probability that at least one arm continues depends on the joint distribution of the interim statistics.

Enrollment accounting: Let \(S\) denote the set of arms that continue at an interim analysis, and let \(C(S)\) be the next-stage enrollment required for that set. The expected next-stage enrollment is \(E[C(S)]\). If costs are fixed by the active set, this can be written as \(\sum_S P(S)C(S)\). Probabilities for joint arm-continuation patterns must reflect the dependence between interim statistics.

Worked Example: Two Arms and a Shared Control Cohort

Consider a hypothetical two-stage MAMS design with 160 participants enrolled in stage 1. In stage 2, 30 shared-control participants are enrolled if at least one arm continues, and 40 participants are enrolled for each arm that continues. Suppose the standardized interim statistics are independent. Arm 1 has mean equal to its threshold, so \(q_1=0.5000\); arm 2 has mean 0.5 standard deviations below its threshold, so \(q_2=\Phi(-0.5)=0.3085\).

Under independence, the probability that neither arm continues is \((1-q_1)(1-q_2)=(0.5000)(0.6915)=0.34575\). Thus the probability that the shared control cohort is enrolled is \(1-0.34575=0.65425\). The expected number of continuing arms is \(q_1+q_2=0.8085\), by linearity of expectation.

The expected stage 2 enrollment is therefore \(30(0.65425)+40(0.8085)=19.6275+32.34=51.9675\) participants. Total expected enrollment is \(160+51.9675=211.9675\), or about 212 participants.

Now shift arm 1’s mean to 0.5 standard deviations below its threshold, leaving arm 2 unchanged. Both continuation probabilities are 0.3085. The probability that at least one continues is \(1-(0.6915)^2=0.5218\), and the expected number of continuing arms is \(0.3085+0.3085=0.6170\). Expected stage 2 enrollment is \(30(0.5218)+40(0.6170)=15.65+24.68=40.33\), giving total expected enrollment of about 200 participants. This comparison isolates how moving one mean away from its threshold can change expected enrollment under the stated independence assumption.

Dependence Can Change the Shared-Cost Calculation

Even when each arm’s marginal continuation probability is known, it does not determine the probability that at least one arm continues. This matters when enrollment includes a shared cost triggered by any continuing arm. The following calculation uses a specified bivariate normal model to make the dependence assumption explicit.

Worked Example: Correlated Statistics at Both Thresholds

Suppose two hypothetical standardized interim statistics are jointly normal, each with variance 1 and mean exactly at its continuation threshold. Their correlation is \(\rho=0.6\). Each arm therefore continues with probability 0.5. Use the same stage 1 enrollment and stage 2 costs as in the preceding example: 160 participants in stage 1, 30 shared-control participants if either arm continues, and 40 participants per continuing arm.

For two standard normal variables with zero means and correlation \(\rho\), the probability that both are below zero is \(1/4+\arcsin(\rho)/(2\pi)\). Here the threshold-standardized statistics have zero means, so \(P(\text{neither continues})=0.25+\arcsin(0.6)/(2\pi)=0.3524\). Therefore, \(P(\text{at least one continues})=1-0.3524=0.6476\).

The expected number of continuing arms remains \(0.5+0.5=1\), whether or not the statistics are correlated. Expected stage 2 enrollment is \(30(0.6476)+40(1)=19.43+40=59.43\). Total expected enrollment is \(160+59.43=219.43\) participants.

If independence had instead been assumed, the probability that at least one arm continues would be \(1-(0.5)(0.5)=0.75\), producing expected total enrollment of \(160+30(0.75)+40=222.5\). The marginal continuation probabilities are identical in the two calculations; the shared-control expectation differs because the joint continuation pattern differs.

Derivation, Checks, and Limitations

The single-arm enrollment result follows by writing total enrollment as a random quantity. Stage 1 always enrolls \(n_1\) participants. Stage 2 enrolls \(n_2\) participants exactly when the continuation event \(A=\{Z>c\}\) occurs. Using the indicator \(I(A)\), which is 1 when \(A\) occurs and 0 otherwise,

N=n_1+n_2 I(A), \qquad E(N)=n_1+n_2 E[I(A)] =n_1+n_2P(A) =n_1+n_2q(\mu).

This makes the interpretation of the sensitivity direct: increasing \(q\) by 0.01 increases expected enrollment by \(0.01n_2\) when the stage 2 cost is fixed. Differentiating the normal probability gives the local rate of change derived above. Near the boundary, a first-order approximation for a small mean shift \(\Delta\) is \(q(\mu+\Delta)\approx q(\mu)+\phi((\mu-c)/\tau)\Delta/\tau\). This approximation is local; for larger shifts, calculate the normal probabilities directly.

Before using these calculations, check that the modeled statistic and its scale match the design’s continuation rule, that the threshold is applied in the correct direction, and that the assumed stage sizes and shared costs reflect the enrollment plan. The normal approximation should be credible for the interim data and the statistic being used. If the interim statistic has a different variance, use its standard deviation \(\tau\), rather than silently treating it as 1. For multiple arms, check whether the assumed joint distribution is defensible; shared-control comparisons commonly induce dependence.

A further check is to calculate several nearby means, not just one. The continuation probability must increase as \(\mu\) increases for a rule \(Z>c\), and expected enrollment must not decrease when greater continuation probability can only add participants. If either check fails, revisit the boundary direction, probability calculation, or enrollment accounting.

Approach	What it tells you	Main caution
Expected enrollment at the assumed mean	A direct planning summary under one specified model.	By itself, it can conceal sensitivity to the mean near a boundary.
Nearby-mean calculations or a local derivative	Shows how expected enrollment responds to small changes in the assumed interim mean.	A local derivative is not a reliable substitute for recalculation after a large change.
Maximum planned enrollment	Gives the design’s enrollment ceiling under the specified stage plan.	It does not describe the expected enrollment or its sensitivity.
Joint-pattern calculation for multiple arms	Accounts for dependence when stage costs depend on which arms continue.	Results rely on the specified joint distribution, not only marginal continuation probabilities.

Near-threshold sensitivity is a planning diagnostic, not evidence that an arm will or will not continue in a future trial. It describes how expected enrollment changes under a probability model. It also has limits: it does not represent uncertainty about the model itself, guarantee realized enrollment, or determine whether the design is desirable. When the assumed effect is uncertain near a boundary, report the expected enrollment at the stated assumption and examine plausible nearby means rather than treating one calculation as definitive.

Common Pitfall

Wrong reasoning: “Each arm has a 0.5 probability of continuing, so the probability that at least one of two arms continues must be 1.” This adds marginal probabilities as if they were probabilities of disjoint events.
Correction: The union probability depends on the joint pattern. Under independence it is \(1-(1-0.5)^2=0.75\); under positive dependence it can be smaller. Use the joint distribution when a shared stage cost is triggered by any arm continuing. The expected number of continuing arms, in contrast, is the sum of marginal probabilities, even when the arms are dependent.
Another warning: “The mean is just above the boundary, so the arm will continue.” A mean describes a distribution, not a realized statistic. Continuation is probabilistic unless the interim value itself is known.

Key takeaway: With a normal interim model and a fixed continuation boundary, continuation probability changes most rapidly when the mean is near the boundary. Expected enrollment inherits that sensitivity in proportion to the enrollment unlocked by continuation; shared stage costs additionally require appropriate joint continuation probabilities.

Check Your Understanding

Try each question, then compare your work with the solutions below.

A hypothetical design enrolls 90 participants in stage 1 and 60 more if an arm continues. If \(Z\sim N(1.6,1)\) and the threshold is 1.8, calculate the continuation probability and expected enrollment.
For a standard deviation of 1, what is the derivative of continuation probability with respect to the mean at the threshold? If continuation unlocks 120 participants, what is the derivative of expected enrollment there?
Two independent arms each have continuation probability 0.4. What is the probability that at least one continues?
Why does a correlation assumption matter for shared-control enrollment but not for the expected number of continuing arms?

Solutions. 1. \(q=\Phi((1.6-1.8)/1)=\Phi(-0.2)=0.4207\). Thus \(E(N)=90+60(0.4207)=115.242\), or about 115 participants. 2. At the threshold, \(dq/d\mu=\phi(0)=1/\sqrt{2\pi}=0.3989\) per unit of mean. The expected-enrollment derivative is \(120(0.3989)=47.87\) participants per unit of mean. 3. Under independence, the probability neither continues is \((1-0.4)^2=0.36\), so the probability at least one continues is \(1-0.36=0.64\). 4. The expected number continuing is the sum of the two marginal probabilities by linearity of expectation, regardless of dependence. The probability at least one continues depends on the joint distribution, so dependence changes the chance that the shared cohort is needed.