Neyman Targets With Minimum Arm Sizes

Why Minimum Arm Sizes Change Neyman Targets

A variance-minimizing allocation gives more participants to arms whose outcomes contribute more uncertainty to the estimator. But a purely variance-based target can assign very few participants to an arm. A trial may therefore require a prespecified minimum sample size in every arm, for example to ensure that each treatment is represented by enough participants for the planned analysis. The resulting target is still chosen to minimize variance, but it must respect those lower bounds.

The key consequence is that the usual Neyman proportions cannot simply be used unchanged. If an unconstrained target falls below a required minimum, that arm is fixed at its minimum, and the sample remaining after that assignment is allocated among the other arms according to their relative Neyman weights. It may take more than one pass to identify all the binding minimums.

Definition: Consider $K$ treatment arms, a fixed total sample size $N$, and a planned contrast $\theta=\sum_{k=1}^{K}c_k\mu_k$. Let $\sigma_k^2$ be the outcome variance in arm $k$, and let $m_k$ be its required minimum sample size. A constrained Neyman target is a set of arm sizes $n_1,\ldots,n_K$ that minimizes the planned variance of the contrast subject to $n_k\geq m_k$ and $\sum_{k=1}^{K}n_k=N$.

For this tutorial, assume independent observations across participants and arms, finite arm-specific variances, and a sample-mean estimator in each arm. Treat the variances and contrast coefficients as fixed planning inputs. The continuous targets $n_k$ are initially allowed to be non-integers; converting them to integer sample sizes is a separate step. We assume $b_k=|c_k|\sigma_k>0$ for each arm with a nonzero contrast coefficient. If an arm has zero contrast coefficient, its size does not contribute to this contrast's variance, and the positive-weight formula below does not apply to that arm without additional design requirements.

The Constrained Variance-Minimization Problem

The estimator of the contrast is $\hat{\theta}=\sum_{k=1}^{K}c_k\bar{x}_k$, where $\bar{x}_k$ is the sample mean in arm $k$. Under the stated independence assumptions, its variance is the sum of the arm-specific contributions:

\operatorname{Var}(\hat{\theta})=\sum_{k=1}^{K}\frac{c_k^2\sigma_k^2}{n_k} =\sum_{k=1}^{K}\frac{b_k^2}{n_k}, \qquad b_k=|c_k|\sigma_k.

The optimization problem is to minimize this variance subject to the total-sample and minimum-size constraints. Before solving it, check feasibility: the minimums must not require more participants than are available.

\sum_{k=1}^{K}m_k\leq N.

If this condition fails, no allocation can satisfy all the minimums with the stated total. If it holds, the objective is strictly convex in positive $n_k$, because the second derivative of $b_k^2/n_k$ is $2b_k^2/n_k^3>0$. Thus, with a feasible set, the constrained minimum is unique.

Deriving the Bound-Adjusted Neyman Rule

For arms not held at their minimum, minimize the variance while keeping the total sample fixed. Introduce a multiplier $\lambda$ for the total-size constraint and differentiate the objective with respect to a free $n_k$:

-\frac{b_k^2}{n_k^2}+\lambda=0 \quad\Longrightarrow\quad n_k=\frac{b_k}{\sqrt{\lambda}}.

Therefore, all arms that are not constrained at their minimum have sizes proportional to $b_k$. For an arm at its lower bound, optimality requires that assigning it more participants would not improve the objective enough to justify taking those participants from the free arms. The first-order condition at the bound is $\lambda\geq b_k^2/m_k^2$, or equivalently $b_k/\sqrt{\lambda}\leq m_k$. These conditions give the full rule:

Theorem: If the minimum-size constraints are feasible and all $b_k>0$, the unique continuous constrained Neyman target has the form $$ n_k=\max\left(m_k,\frac{b_k}{\sqrt{\lambda}}\right), $$ where $\lambda>0$ is chosen so that $\sum_{k=1}^{K}n_k=N$. An arm is active at its lower bound when $n_k=m_k$; otherwise it receives its unconstrained Neyman share of the remaining total.

This expression is sometimes applied by an active-set calculation. Start with the unconstrained Neyman allocation, $n_k=N b_k/\sum_j b_j$. If any proposed size is below its minimum, fix such an arm at its minimum, remove it from the proportional allocation, and distribute the remaining sample among the unfixed arms in proportion to their $b_k$ values. Recheck the remaining arms: another proposed size may now fall below its minimum. Continue until every free arm meets its bound.

Check feasibility.
Confirm that the sum of the prescribed minimums is no greater than $N$.

Allocate proportionally among free arms.
For a current free set $F$, allocate the residual sample in proportion to $b_k$, using $n_k=(N-\sum_{j\notin F}m_j)b_k/\sum_{j\in F}b_j$.

Clamp violations and repeat.
If a proposed free-arm size is below its minimum, set it to that minimum, remove it from the free set, and recalculate the residual allocation.

Verify the result.
Check that all targets meet their minimums, sum to $N$, and that the free-arm ratios $n_k/b_k$ are equal.

Worked Allocations

Worked Example: One Minimum Becomes Active

Suppose three arms have planning weights $b_1=2$, $b_2=4$, and $b_3=6$, with total $N=120$ and minimums $m_1=30$, $m_2=20$, and $m_3=10$. These weights could, for example, be the arm standard deviations for a contrast with coefficients of magnitude one. Find the continuous constrained targets.

The minimums are feasible because $30+20+10=60\leq120$. Without constraints, the Neyman targets are proportional to $2:4:6$:

(n_1,n_2,n_3) =120\left(\frac{2}{12},\frac{4}{12},\frac{6}{12}\right) =(20,40,60).

Arm 1's proposed target of 20 is below its minimum of 30, so set $n_1=30$. The remaining $120-30=90$ participants are allocated between arms 2 and 3 in the ratio $4:6$:

n_2=90\left(\frac{4}{10}\right)=36, \qquad n_3=90\left(\frac{6}{10}\right)=54.

Both remaining targets exceed their minimums, and $30+36+54=120$. Thus arm 1 is active at its lower bound, while arms 2 and 3 are free. As a check, their ratios are $36/4=54/6=9$. The planned variance objective is $\frac{2^2}{30}+\frac{4^2}{36}+\frac{6^2}{54}=0.1333+0.4444+0.6667=1.244$, in the squared units of the contrast.

Worked Example: Two Minimums Become Active in Sequence

Take four arms with $b=(1,2,3,4)$, total $N=100$, and minimums $m=(35,30,10,5)$. The sum of minimums is 80, so a feasible allocation may exist. Identify all active bounds and calculate the targets.

The unconstrained allocation is $(10,20,30,40)$. Arm 1 is below its minimum of 35. Fix it at 35, leaving 65 participants for arms 2 through 4. Proportional allocation over those arms gives:

(n_2,n_3,n_4) =65\left(\frac{2}{9},\frac{3}{9},\frac{4}{9}\right) =(14.44,21.67,28.89).

The proposed size for arm 2, 14.44, is below its minimum of 30. Fix arm 2 at 30. Now 35 participants remain for arms 3 and 4, whose weights are 3 and 4:

n_3=35\left(\frac{3}{7}\right)=15, \qquad n_4=35\left(\frac{4}{7}\right)=20.

The final targets are $(35,30,15,20)$. They sum to 100; arms 3 and 4 meet their minimums and retain the free-arm ratio $15/3=20/4=5$. Arms 1 and 2 are active. The objective is $\frac{1}{35}+\frac{4}{30}+\frac{9}{15}+\frac{16}{20}=1.562$ approximately. Sequential rechecking matters: after arm 1 was fixed, arm 2 also fell below its bound, so a single adjustment would not have produced a feasible optimum.

Worked Example: Bounds for a Weighted Contrast

Suppose the target is the contrast $\theta=\mu_1-\frac{1}{2}\mu_2-\frac{1}{2}\mu_3$. Planning standard deviations are $\sigma_1=2$, $\sigma_2=4$, and $\sigma_3=2$; the total is $N=60$, and minimums are $(15,28,8)$. Calculate the constrained targets.

The contrast coefficients are $c=(1,-1/2,-1/2)$. The weights are therefore not just the standard deviations: $b_k=|c_k|\sigma_k$, giving $b=(2,2,1)$. The unconstrained targets are $60(2/5,2/5,1/5)=(24,24,12)$. Arm 2 would receive 24, less than its minimum of 28, so set $n_2=28$. The remaining 32 are allocated to arms 1 and 3 in their weight ratio $2:1$:

n_1=32\left(\frac{2}{3}\right)=21.33, \qquad n_3=32\left(\frac{1}{3}\right)=10.67.

Thus the continuous targets are approximately $(21.33,28,10.67)$. They sum to 60, and both free arms exceed their minimums. The contrast variance is $\frac{2^2}{21.33}+\frac{2^2}{28}+\frac{1^2}{10.67}\approx0.4241$, in squared outcome units. Using $\sigma_k$ alone would miss the contrast coefficients and could give the wrong allocation.

Checks, Alternatives, and Limitations

The constrained solution is the usual Neyman allocation when no minimum is restrictive. When minimums do bind, it preserves proportional Neyman allocation only among the free arms; it does not generally preserve the original unconstrained proportions across all arms. Equal allocation is simpler and may be chosen for other design reasons, but it is not generally variance-minimizing for the stated contrast. In contrast, the lower-bound method directly minimizes that contrast's variance under the specified constraints.

Several checks help reveal errors. First, verify feasibility before allocating. Second, confirm that the targets sum to $N$ and that no target is below its minimum. Third, among free arms, confirm that $n_k/b_k$ is constant. Finally, check each bound: an arm left at its minimum should have an unconstrained share at the final multiplier no greater than that minimum. A target that fails any of these checks is not the constrained Neyman solution.

A common mistake is to calculate unconstrained shares and then merely raise every undersized arm to its minimum without reducing the other allocations. That changes the total sample size. The correction is to account for the sample consumed by each binding minimum and redistribute only the residual among the free arms, rechecking for further violations. Another mistake is to treat a feasible continuous target as an implementable integer allocation without adjustment. Continuous targets describe the variance-minimizing proportions at the planning stage; integer sample-size conversion requires its own total-preserving step.

These targets depend on the assumed variances and on the contrast being optimized. They do not guarantee that the allocation is best for a different estimand, and they do not account for uncertainty in estimated planning variances. The result is also a target allocation, not a rule for how assignments are generated during enrollment. It addresses one specific design question: how to minimize the planned variance of a specified contrast, given a fixed total and prespecified minimum arm sizes.

Check Your Understanding

Use the constrained Neyman rule to answer the questions. Solutions are provided below each problem.

For $b=(1,3)$, $N=40$, and minimums $m=(15,5)$, find the continuous targets and identify any active bound.
Solution: The unconstrained targets are $(10,30)$, so arm 1 is below its minimum. Set $n_1=15$; the 25 remaining participants go to arm 2. The targets are $(15,25)$, which sum to 40 and meet both minimums. Arm 1 is active.
For $b=(2,2,4)$, $N=80$, and minimums $(10,10,10)$, is any minimum active?
Solution: The unconstrained targets are $80(2/8,2/8,4/8)=(20,20,40)$. Each exceeds its minimum, so none is active and the unconstrained Neyman allocation remains optimal.
For $b=(1,2)$, $N=20$, and minimums $(12,9)$, can a feasible allocation be made?
Solution: No. The minimums sum to $12+9=21$, which is greater than 20. No allocation can satisfy both bounds and the fixed total.
For a contrast with coefficients $c=(1,-1)$ and standard deviations $(3,5)$, what are the Neyman weights? What changes if the coefficients are $c=(1,-1/2)$?
Solution: With coefficients $(1,-1)$, the weights are $b=(|1|3,|-1|5)=(3,5)$. With coefficients $(1,-1/2)$, they are $b=(3,2.5)$. The second contrast gives arm 2 a smaller contribution to variance, so its proportional allocation weight is smaller.