Deriving Expected Sample Size Under Re-Estimation

Why Expected Sample Size Depends on the Interim Distribution

Under a re-estimation rule, the total number of participants is not fixed in advance: it depends on the interim result. A rule may assign a smaller total when the interim evidence is weak, a larger total for results in a prespecified promising range, and a capped total for stronger results. To describe the design’s average resource use, we need to average those possible assignments over the interim statistic’s sampling distribution.

This average is not a single property of the rule alone. The rule specifies what sample size follows each interim result; the sampling distribution specifies how often those results occur. Because that distribution generally depends on the true treatment effect, expected sample size can differ across possible true effects.

Definition: Let \(Z_I\) be the interim statistic and let \(N(z)\) be the total sample size assigned when \(Z_I=z\). The expected total sample size under a specified model and true parameter value is \(E[N(Z_I)]\), the average of the rule’s total sample-size assignment over the sampling distribution of \(Z_I\). Here, \(N\) counts all participants, including those enrolled before the interim analysis.

Model and Assumptions

For explicit calculations, use a two-arm normal model with equal allocation and a known, common standard deviation \(\sigma\). Assume independent observations, a fixed interim sample size of \(n_I\) participants per arm, and a true treatment-minus-control mean difference \(\Delta\). Define the interim statistic as the observed difference in sample means divided by its known standard error:

Z_I=\frac{\bar{x}_{T,I}-\bar{x}_{C,I}}{\sigma\sqrt{2/n_I}}.

Under these assumptions, the interim difference in sample means is normally distributed with mean \(\Delta\) and variance \(2\sigma^2/n_I\). Standardizing therefore gives:

Z_I\sim N(\mu_I,1), \qquad \mu_I=\frac{\Delta\sqrt{n_I}}{\sigma\sqrt{2}}.

This model is used to illustrate the integration. In an actual design, the probability model must match the endpoint, allocation, variance assumptions, interim timing, and analysis rule. The rule must also specify the total sample size for every possible interim result, including its lower and upper constraints. If a design permits early stopping, the total sample size for those interim results must be defined as the number enrolled at stopping.

Expected sample size is calculated under a particular value of \(\Delta\), or under a specified distribution of possible values of \(\Delta\). It is not obtained by averaging over interim results as if they were equally likely. Nor does this calculation by itself establish that the design controls the type I error rate or achieves a desired final power.

Deriving the Expectation

If \(Z_I\) has density \(f_{Z_I}(z)\), the expectation of the sample-size assignment is the integral of that assignment times the density. For a rule with regions \(A_1,\ldots,A_K\), assigning total \(N_k\) whenever \(Z_I\in A_k\), the integral becomes a weighted sum:

E[N(Z_I)] =\int_{-\infty}^{\infty}N(z)f_{Z_I}(z)\,dz =\sum_{k=1}^{K}N_kP(Z_I\in A_k).

The second equality follows by splitting the integral into the disjoint regions. Within region \(A_k\), the assigned total is constant, so its contribution is \(N_k\int_{A_k}f_{Z_I}(z)\,dz=N_kP(Z_I\in A_k)\). The regions must cover all possible interim results, apart from boundaries that have probability zero for a continuous statistic.

For the normal model above, \(P(Z_I\in[a,b))=\Phi(b-\mu_I)-\Phi(a-\mu_I)\), where \(\Phi\) is the standard normal cumulative distribution function. Thus the computation has two parts: find each region probability using the correct interim distribution, then multiply each probability by the total assigned in that region.

Worked Example: A Three-Region Rule

Suppose, hypothetically, the interim statistic follows \(N(1.2,1)\), and a prespecified rule assigns a total of 100 participants for \(Z_I<0\), 160 for \(0\le Z_I<1\), and 220 for \(Z_I\ge1\). The interim sample already enrolled is included in each total. Find the expected total sample size.

The region probabilities are:

\begin{aligned} P(Z_I<0)&=\Phi(-1.2)\approx0.115070,\\ P(0\le Z_I<1)&=\Phi(1-1.2)-\Phi(0-1.2)\\ &=\Phi(-0.2)-\Phi(-1.2)\approx0.305671,\\ P(Z_I\ge1)&=1-\Phi(-0.2)\approx0.579260. \end{aligned}

These probabilities sum to 1, up to rounding. Weighting each assigned total by its region probability gives:

\begin{aligned} E[N(Z_I)] &=100(0.11506967)+160(0.30567062)+220(0.57925971)\\ &\approx11.50697+48.90730+127.43714\\ &\approx187.8514. \end{aligned}

The expected total is therefore approximately 187.9 participants. It lies between the rule’s minimum of 100 and maximum of 220, as an average of values in that range must. The result is an average over repeated trials under the specified \(N(1.2,1)\) interim distribution; it does not mean that any individual trial enrolls 187.9 participants.

Rules That Vary Within a Region

Not every rule assigns a constant sample size throughout each region. If \(N(z)\) varies continuously with \(z\), the integral definition still applies, but a region-probability weighted sum alone is not sufficient. One useful normal identity is the first moment over a bounded region. For \(Z_I\sim N(\mu,1)\):

E[Z_I\,\mathbf{1}(a\le Z_I<b)] =\mu\{\Phi(b-\mu)-\Phi(a-\mu)\} +\phi(a-\mu)-\phi(b-\mu),

where \(\mathbf{1}(\cdot)\) is 1 when its condition holds and 0 otherwise, and \(\phi\) is the standard normal density. To verify the identity, set \(u=z-\mu\) in the defining integral. Then \(z=u+\mu\), and \(\int u\phi(u)\,du=-\phi(u)\), because \(\phi'(u)=-u\phi(u)\). Integrating from \(a-\mu\) to \(b-\mu\) yields the displayed expression.

Worked Example: Integrating a Linear Assignment

As a hypothetical mathematical rule, let \(Z_I\sim N(1.2,1)\), and assign \(N(z)=100\) when \(z<0\), \(N(z)=100+60z\) when \(0\le z<1\), and \(N(z)=160\) when \(z\ge1\). This rule is continuous at both boundaries. Find its expected assignment.

Separate the contributions from the three regions. The first and third contribute \(100P(Z_I<0)\) and \(160P(Z_I\ge1)\). On the middle region, \(N(Z_I)=100+60Z_I\), so its contribution is \(100P(0\le Z_I<1)+60E[Z_I\mathbf{1}(0\le Z_I<1)]\).

The middle-region probability is \(0.30567062\), as in the preceding calculation. The truncated first moment is:

\begin{aligned} E[Z_I\mathbf{1}(0\le Z_I<1)] &=1.2(0.30567062)+\phi(-1.2)-\phi(-0.2)\\ &\approx1.2(0.30567062)+0.194186-0.391043\\ &\approx0.169948. \end{aligned}

Therefore:

\begin{aligned} E[N(Z_I)] &=100(0.11506967)+100(0.30567062)\\ &\quad+60(0.169948)+160(0.57925971)\\ &\approx11.50697+30.56706+10.19688+92.68155\\ &\approx144.9525. \end{aligned}

The expected assignment is about 145.0 participants. The middle-region contribution uses the mean of \(Z_I\) restricted to that region, not the overall mean \(1.2\); substituting the overall mean there would give an incorrect result.

Changing the True Effect Changes the Average

The assignment rule may stay fixed while the expected sample size changes, because the interim statistic’s distribution shifts with \(\Delta\). The following calculation uses the same kind of three-region rule as before but a different hypothetical interim distribution.

Worked Example: Evaluating the Rule Under a Different Distribution

Suppose \(Z_I\sim N(0.1,1)\), and the rule assigns 120 participants for \(Z_I<0\), 180 for \(0\le Z_I<1\), and 240 for \(Z_I\ge1\). Calculate the expected total.

Standardizing the region boundaries gives:

\begin{aligned} P(Z_I<0)&=\Phi(-0.1)\approx0.460172,\\ P(0\le Z_I<1)&=\Phi(0.9)-\Phi(-0.1)\approx0.355768,\\ P(Z_I\ge1)&=1-\Phi(0.9)\approx0.184060. \end{aligned}

The weighted sum is:

\begin{aligned} E[N(Z_I)] &=120(0.46017216)+180(0.35576771)+240(0.18406013)\\ &\approx55.22066+64.03819+44.17443\\ &\approx163.4333. \end{aligned}

Under this specified distribution, the expected total is approximately 163.4 participants. It differs from the first example because both the assigned totals and the probabilities of entering each region matter. Even with the same rule, a different true treatment effect can alter the region probabilities and hence the average.

Checking the Calculation and Understanding Its Limits

A reliable calculation starts with the exact operational rule, not a rough verbal description. Write down the regions, the assigned total in each, and the distribution used for \(Z_I\). Then calculate region probabilities with the correct mean and variance, check that those probabilities sum to 1, and apply the weighted sum or integral. If the rule uses total sample size, do not accidentally add the interim sample a second time. If it instead specifies additional participants after the interim, total sample size equals the interim total plus that additional sample.

Several diagnostics can reveal problems. A calculated expectation outside the range of possible totals signals an error in the probabilities, assignments, or arithmetic. Region probabilities that do not sum to approximately 1 suggest a missing region or inconsistent boundaries. An average based on the null distribution may be inappropriate when the question concerns expected enrollment at a nonzero true effect. And if the actual procedure has a stopping rule, rounding, or additional operational constraints, those details must be included in \(N(z)\); integrating a simplified rule gives the expectation of that simplified rule, not necessarily of the implemented design.

For a piecewise-constant rule, the weighted-sum calculation is usually the clearest approach. For a continuous assignment, direct integration or truncated-normal moments may be useful. Simulation is an alternative numerical check: repeatedly draw interim statistics from the specified distribution, apply the rule to each draw, and average the resulting totals. Simulation does not replace defining the rule and model correctly, and small simulation runs can differ from the exact expectation due to Monte Carlo error.

Expected sample size describes average enrollment under a stated distribution. It is distinct from conditional power, which concerns the probability of final success conditional on interim information and assumptions about the effect. It is also distinct from final power, which averages final rejection over possible trial data under a specified effect. A low expected sample size alone is not evidence that a design has acceptable power or error rates; those properties require their own calculations.

Common mistake: averaging the assigned totals without weighting. The unweighted average gives equal importance to regions even when they have very different probabilities. Use \(\sum_k N_kP(Z_I\in A_k)\).
Common mistake: mixing rounded and unrounded calculations. Rounded probabilities can make the displayed sum differ slightly from the calculation. Keep sufficient precision for the weighted sum, then round the final answer.
Common mistake: treating the interim statistic’s distribution as fixed across effects. Under the normal model, its mean depends on \(\Delta\). State the effect value or distribution under which the expectation is being evaluated.
Common mistake: interpreting an expectation as a realized sample size. A single trial follows one branch of the rule. The expectation is the long-run average across repetitions under the specified model.

Key takeaway: Expected total sample size is the average of the rule’s total assignments over the interim statistic’s sampling distribution. For a region-based rule, multiply each assigned total by its region probability and sum; for a varying rule, integrate the assignment against the density.

Check Your Understanding

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

A rule assigns totals \(80\), \(140\), and \(200\) with respective region probabilities \(0.25\), \(0.50\), and \(0.25\). What is its expected total?
Why can the expected sample size for one fixed re-estimation rule differ under two true treatment effects?
If a rule assigns \(N(z)=100\) for \(z<0\) and \(N(z)=180\) for \(z\ge0\), write its expected total in terms of \(P(Z_I<0)\).
State one reason a simulated average might differ slightly from an exact integral.

Solutions. (1) \(80(0.25)+140(0.50)+200(0.25)=20+70+50=140\) participants. (2) The true effect changes the sampling distribution of the interim statistic, which changes the probabilities of entering the rule’s regions. (3) \(E[N(Z_I)]=100P(Z_I<0)+180P(Z_I\ge0)=100P(Z_I<0)+180\{1-P(Z_I<0)\}\). (4) A simulation uses a finite number of random draws, so its average has Monte Carlo variation around the exact expectation.