Worked Multiarm Allocation Example

From Multiarm Targets to a Numerical Allocation

With several treatment arms, allocation is not determined by a single ratio. Each arm has its own outcome variability, so a target must balance the precision of all the arm means. This tutorial works through that calculation and checks what it does—and does not—optimize. The allocations are design targets: they do not, by themselves, specify a response-adaptive randomization rule.

Suppose a trial has \(K\) arms and a fixed total sample size \(N\). Let \(n_i\) be the number assigned to arm \(i\), and let \(\sigma_i\) be the within-arm outcome standard deviation for that arm. In practice, the \(\sigma_i\) values used to plan may be estimated from earlier information; examples below label those inputs \(s_i\). The objective considered here is to minimize the sum of the variances of the \(K\) arm-mean estimates, subject to \(\sum_{i=1}^K n_i=N\).

Definition: Under the assumptions that observations within each arm are independent and have a common variance \(\sigma_i^2\), the variance of the sample mean in arm \(i\) is \(\operatorname{Var}(\bar Y_i)=\sigma_i^2/n_i\). If \(s_i\) is an estimate of \(\sigma_i\), the corresponding estimated variance of the arm mean is \(s_i^2/n_i\). The multiarm Neyman target for minimizing the sum of arm-mean variances is \(n_i=N\sigma_i/\sum_{j=1}^K\sigma_j\); replacing \(\sigma_i\) by planning estimates gives estimated targets.

The common-variance condition is within each arm, not across all arms: the arms may have different variances. Independence alone is not enough for the displayed variance formula. If observations within an arm are independent but have unequal variances, the variance of their mean is the sum of their individual variances divided by \(n_i^2\), rather than necessarily \(\sigma_i^2/n_i\). For estimated variance calculations, \(s_i^2\) is the usual sample variance within arm \(i\), calculated from that arm’s outcomes.

Why the Standard-Deviation Targets Arise

The calculation uses the sum \(\sum_i \sigma_i^2/n_i\) as its precision criterion. This is a specific, transparent objective: reducing the total marginal variance across the arm means. It is not a claim that each arm mean, or every possible contrast, is individually as precise as possible.

\begin{aligned} \text{Minimize}\quad & \sum_{i=1}^K \frac{\sigma_i^2}{n_i} \quad\text{subject to}\quad \sum_{i=1}^K n_i=N,\\ \mathcal{L} &= \sum_{i=1}^K \frac{\sigma_i^2}{n_i} +\lambda\left(\sum_{i=1}^K n_i-N\right),\\ \frac{\partial \mathcal{L}}{\partial n_i} &= -\frac{\sigma_i^2}{n_i^2}+\lambda=0 \quad\Longrightarrow\quad n_i=\frac{\sigma_i}{\sqrt{\lambda}},\\ \sum_{i=1}^K n_i=N &\quad\Longrightarrow\quad \sqrt{\lambda}=\frac{\sum_{j=1}^K \sigma_j}{N},\\ n_i&=N\frac{\sigma_i}{\sum_{j=1}^K\sigma_j}. \end{aligned}

For positive standard deviations, each term \(\sigma_i^2/n_i\) is strictly convex in \(n_i>0\), since its second derivative is \(2\sigma_i^2/n_i^3>0\). Thus the stationary allocation is the unique minimum among positive continuous allocations satisfying the fixed-total constraint. At this target, writing \(S=\sum_j\sigma_j\), the variance for arm \(i\) is \(\sigma_i S/N\), and the minimum sum is \(S^2/N\). If the standard deviations are estimated rather than known, these are plug-in calculations based on the estimates, not a guarantee that the true variances will attain those values.

The formula assigns a larger share to an arm with greater outcome standard deviation. This can seem counterintuitive if the goal is simply to estimate that arm’s mean: more observations are indeed needed to estimate a high-variance mean precisely. But the allocation is balancing the sum of all arm-mean variances under a fixed total, not equalizing those variances. In fact, at the optimum, arm \(i\)’s variance is proportional to \(\sigma_i\), not identical across arms.

Worked Example: Three Arms With Different Variability

Suppose a hypothetical three-arm trial has \(N=180\) participants. Planning estimates of the outcome standard deviations are \(s_1=4\), \(s_2=6\), and \(s_3=8\) outcome units. Assume independent observations within and across arms, a common variance within each arm, and no allocation constraints beyond the fixed total. These values are planning inputs, not observations from a real study.

Their sum is \(4+6+8=18\). The estimated Neyman targets are \(n_1=180(4/18)=40\), \(n_2=180(6/18)=60\), and \(n_3=180(8/18)=80\). They sum to 180, so no integer adjustment is needed.

The estimated arm-mean variances are \(s_i^2/n_i\). Substitution gives \(4^2/40=16/40=0.4000\) squared outcome units for arm 1, \(6^2/60=36/60=0.6000\) squared outcome units for arm 2, and \(8^2/80=64/80=0.8000\) squared outcome units for arm 3. Their sum is \(0.4000+0.6000+0.8000=1.800\) squared outcome units.

For comparison, equal allocation gives 60 participants per arm. Its estimated variances are \(16/60=0.2667\), \(36/60=0.6000\), and \(64/60=1.0667\), with sum \(1.933\) squared outcome units. The Neyman target therefore lowers this specified sum by \(1.9333-1.8000=0.1333\), or \(0.1333/1.9333\approx 6.897\%\) relative to equal allocation. It does not lower every arm’s variance: arm 1 has fewer participants under the target and a larger estimated variance than under equal allocation.

Comparing Targets and Variance Estimates

The next example shows the same calculation with four arms, including two with the same estimated standard deviation. Equal standard deviations produce equal target sample sizes, while arms with larger standard deviations receive proportionally more participants.

Worked Example: Four Arms and an Equal-Allocation Benchmark

In a hypothetical trial, let \(N=200\), with planning standard-deviation estimates \(s_1=3\), \(s_2=5\), \(s_3=5\), and \(s_4=7\) outcome units. Assume the same within-arm independence and common-variance conditions as above. The total of the estimates is \(3+5+5+7=20\).

The targets are \(n_1=200(3/20)=30\), \(n_2=200(5/20)=50\), \(n_3=200(5/20)=50\), and \(n_4=200(7/20)=70\). These are integers and add to 200.

The estimated variances are \(3^2/30=9/30=0.3000\), \(5^2/50=25/50=0.5000\), \(5^2/50=0.5000\), and \(7^2/70=49/70=0.7000\), all in squared outcome units. The total is \(0.3000+0.5000+0.5000+0.7000=2.000\).

Under equal allocation, each arm has 50 participants. The estimated variances are \(9/50=0.1800\), \(25/50=0.5000\), \(25/50=0.5000\), and \(49/50=0.9800\). Their sum is \(2.160\) squared outcome units. The Neyman target reduces the sum by \(0.160/2.160\approx 7.407\%\). Notice that the low-variability first arm has a larger variance estimate under Neyman allocation than under equal allocation; the high-variability fourth arm has a smaller one. The comparison is about the total specified by the objective.

A useful check is whether the stated objective matches the scientific question. When the aim is the sum of arm-mean variances, the calculation above applies. Equal allocation is a simple alternative and is optimal for this objective when all arm standard deviations are equal. If the objective instead emphasizes a particular contrast or unequal patient costs, the allocation calculation changes; the targets here should not be presented as universally optimal.

Checking a Candidate Allocation

Once targets have been calculated, the same variance formula can evaluate another proposed allocation. Comparing the candidate’s sum with the target’s minimum is a direct diagnostic of the cost, under this objective, of departing from the target. It does not diagnose whether the planning standard deviations accurately describe future outcomes.

Worked Example: Quantifying a Departure From the Target

Consider three hypothetical arms with planning estimates \(s_1=2\), \(s_2=4\), and \(s_3=6\) outcome units, and \(N=120\). Assume independent observations within and across arms and a common variance within each arm. The sum of the standard deviations is 12, so the targets are \(120(2/12)=20\), \(120(4/12)=40\), and \(120(6/12)=60\).

At the target, the estimated variances are \(2^2/20=0.2000\), \(4^2/40=0.4000\), and \(6^2/60=0.6000\) squared outcome units. Their sum is \(1.200\). Now consider the feasible candidate allocation \((30,40,50)\). Its estimated variances are \(2^2/30=0.1333\), \(4^2/40=0.4000\), and \(6^2/50=0.7200\), summing to \(1.2533\).

The candidate’s sum exceeds the target sum by \(1.2533-1.2000=0.0533\) squared outcome units, or \(0.0533/1.2000\approx 4.444\%\). The candidate improves precision for arm 1 but worsens it for arm 3; the net result is a larger total variance. The calculation does not establish that the candidate is inferior for every scientific purpose, only that it is less efficient for the stated sum-of-variances criterion under these inputs.

Assumptions, Alternatives, and Common Pitfalls

Before using a multiarm allocation calculation, check the assumptions behind both its variance estimates and its optimization target. The formula \(s_i^2/n_i\) is appropriate when the arm’s observations are independent and share a common variance, with \(s_i^2\) estimating that variance. Dependence—for example, clustered outcomes—or unequal variances among individuals within an arm requires a variance calculation suited to that structure. The fixed-\(N\) optimization also presumes that participants can be assigned across arms without additional cost or minimum-size constraints.

Unequal allocation is the intended output. When variability differs, equal allocation generally does not minimize the sum of arm-mean variances. When the variability estimates are equal, the target reduces to equal allocation.
Estimates are uncertain. The calculation treats the supplied standard deviations as inputs. If they poorly represent the trial outcomes, the target can be inefficient. Compare plausible inputs and inspect whether the resulting targets are sensitive to them.
Check every target and variance. Targets must sum to \(N\), each variance estimate must use the matching arm’s \(s_i^2\) and \(n_i\), and the variance units must be squared outcome units. A denominator accidentally taken from another arm can conceal a large error.
Do not confuse a target with an adaptive rule. A target specifies desired sample-size proportions. It does not describe how assignments are updated over time or account for all inferential consequences of an adaptive assignment process.

Common mistake: “The arm with the largest standard deviation should receive fewer participants because its mean is hard to estimate.” That reverses the allocation logic for the stated objective. Since \(\operatorname{Var}(\bar Y_i)=\sigma_i^2/n_i\), the Neyman target gives more participants to a higher-variability arm, in proportion to its standard deviation. Another mistake is to say that this makes every arm equally precise; the examples show why that is false. The allocation minimizes the sum, not each component separately.

Key takeaway: For a fixed total and the objective of minimizing the sum of arm-mean variances, allocate in proportion to the within-arm standard deviations. Compute \(s_i^2/n_i\) arm by arm, verify the total allocation, and interpret the comparison only in relation to that objective and its assumptions.

Check Your Understanding

Use the stated objective of minimizing the sum of arm-mean variances. Assume the within-arm independence and common-variance conditions hold unless a question asks you to assess them.

Three arms have standard-deviation estimates 2, 2, and 4, with \(N=80\). Find the Neyman targets and the estimated variance sum.
For three arms with standard-deviation estimates 3, 4, and 5, and \(N=120\), find the target sample sizes.
Why does independence alone not justify using \(\sigma_i^2/n_i\) for an arm mean?
If all arm standard-deviation estimates are equal, what target does the formula give, and how does it compare with equal allocation?

Solutions. (1) The standard-deviation sum is \(2+2+4=8\), so the targets are \(80(2/8)=20\), \(20\), and \(40\). The estimated variance sum is \(2^2/20+2^2/20+4^2/40=0.2000+0.2000+0.4000=0.8000\) squared outcome units. (2) The sum is 12, giving \(120(3/12)=30\), \(120(4/12)=40\), and \(120(5/12)=50\). (3) Independent observations can have different variances; then the variance of their mean is the sum of their variances divided by \(n_i^2\). The formula \(\sigma_i^2/n_i\) requires a common variance within the arm, in addition to independence. (4) Equal standard deviations yield equal target sample sizes, so this target agrees with equal allocation.