Reading Allocation Ratios as Sample-Size Shares

From Allocation Ratios to Shares

An allocation ratio describes sample sizes relative to one another; an allocation share describes each arm’s fraction of the total enrollment. The distinction matters in practice. A ratio such as 2:3 does not mean that 2 participants go to one arm and 3 to the other. It means that, among 5 total participants allocated according to that ratio, the corresponding shares are $2/5$ and $3/5$. For a planned total enrollment of 250, the same shares imply target sample sizes of 100 and 150.

For Neyman allocation, these shares are determined by the outcome variability in each arm and, for a general contrast, by the arm’s contrast coefficient. They describe how a fixed enrollment is divided to minimize the variance of the contrast under a specified model. In a response-adaptive trial, they can serve as target proportions for assignment, but the realized arm counts need not equal those targets exactly.

Definition: For $K$ arms with positive sample sizes $n_1,\ldots,n_K$ and total enrollment $N=\sum_{i=1}^K n_i$, the sample-size share for arm $i$ is $q_i=n_i/N$. Each share is between 0 and 1, and the shares sum to 1. An allocation ratio compares sample sizes, for example $n_1:n_2$, while a share states the fraction of the total assigned to an arm.

Assumptions Behind the Share Formula

The conversion from a Neyman ratio to shares relies on the same variance-minimization setup used to derive the ratio. Suppose arm $i$ has an outcome with variance $\sigma_i^2$, and the target is the linear contrast $\theta=\sum_{i=1}^K c_i\mu_i$, where $\mu_i$ is the mean outcome in arm $i$ and $c_i$ is its contrast coefficient. Assume independent observations across arms, independent observations within each arm with variance $\sigma_i^2$, and the usual sample-mean estimator $\hat{\theta}=\sum_{i=1}^K c_i\bar{Y}_i$. For fixed sample sizes, its variance is

\operatorname{Var}(\hat{\theta})=\sum_{i=1}^K\frac{c_i^2\sigma_i^2}{n_i}.

The allocation problem here uses a fixed total $N$, equal per-patient costs, known positive finite standard deviations, and no constraints that require particular arm sizes. The interior share formula also requires every contrast coefficient $c_i$ to be nonzero. These conditions matter: a ratio is not automatically a valid allocation rule if the target, costs, or constraints differ.

For a two-arm difference in means, the coefficients are $-1$ and $1$. The Neyman ratio is $n_1/n_2=\sigma_1/\sigma_2$, so converting it to shares gives

q_1=\frac{\sigma_1}{\sigma_1+\sigma_2}, \qquad q_2=\frac{\sigma_2}{\sigma_1+\sigma_2}.

Thus the arm with the larger outcome standard deviation receives the larger share. The ratio alone gives only a relative comparison; dividing each ratio component by their sum makes the shares explicit.

Worked Example: Turning a Two-Arm Ratio Into Shares

Suppose a hypothetical trial compares two treatments on an outcome measured in points. The planning standard deviations are $\sigma_1=8$ points and $\sigma_2=12$ points, and the total planned enrollment is $N=250$. Assume the two-arm difference in means is the target, observations are independent with the stated variances, and there are no cost or minimum-size constraints.

The Neyman ratio is

n_1:n_2=\sigma_1:\sigma_2=8:12=2:3.

The corresponding shares are found by dividing each ratio component by the sum $2+3=5$:

q_1=\frac{2}{5}=0.4000, \qquad q_2=\frac{3}{5}=0.6000.

The continuous target counts are $n_1=250(0.4000)=100$ and $n_2=250(0.6000)=150$. They sum to 250. The higher-variability arm receives 60% of the target enrollment, not because it is expected to have a better outcome, but because its larger standard deviation increases the variance contribution from each observation.

As a check, the estimated contrast variance under these target counts is $8^2/100+12^2/150=64/100+144/150=0.6400+0.9600=1.600$ points squared. Under equal allocation of 125 per arm, it would be $64/125+144/125=1.664$ points squared. Under the stated planning model, the Neyman split therefore has the smaller variance.

Multiarm Shares for a Weighted Contrast

For a multiarm contrast, the relevant allocation weight is not necessarily the standard deviation alone. It is the product of the absolute contrast coefficient and the arm’s standard deviation. A coefficient with larger absolute value means that uncertainty in that arm contributes more strongly to the contrast; a larger standard deviation also increases that contribution.

Result: If every $c_i\ne 0$, every $\sigma_i$ is positive and finite, and the fixed-total, equal-cost assumptions hold, the Neyman allocation shares are $$ q_i=\frac{|c_i|\sigma_i}{\sum_{j=1}^K |c_j|\sigma_j}, \qquad i=1,\ldots,K. $$ The corresponding allocation ratios are $n_i:n_j=|c_i|\sigma_i:|c_j|\sigma_j$.

The shares sum to one because their numerators add to the denominator. This gives a direct practical conversion: calculate each arm’s weight $|c_i|\sigma_i$, add the weights, and divide each individual weight by that total. Multiplying each share by $N$ gives the continuous target sample size for that arm.

Worked Example: Shares for a Three-Arm Contrast

Consider a hypothetical three-arm study estimating $\theta=\mu_1-\tfrac12\mu_2-\tfrac12\mu_3$. Suppose the planning standard deviations are 4, 6, and 10 outcome points, respectively, and the total enrollment is 240. Assume the independent-observation model and equal costs described above, with no minimum arm sizes.

The absolute coefficient-standard-deviation weights are

|c_1|\sigma_1=1(4)=4,\qquad |c_2|\sigma_2=\tfrac12(6)=3,\qquad |c_3|\sigma_3=\tfrac12(10)=5.

Their sum is $4+3+5=12$, so the shares are

q_1=\frac{4}{12}=\frac13,\qquad q_2=\frac{3}{12}=\frac14,\qquad q_3=\frac{5}{12}.

Multiplying by 240 gives $n_1=80$, $n_2=60$, and $n_3=100$. These are integer targets and sum to 240. Arm 3 receives the largest share because its standard deviation is high enough to outweigh its smaller contrast coefficient. Arm 2 has the smallest share because its coefficient is one-half and its standard deviation is lower than arm 3’s.

The variance under this allocation is

\frac{1^2(4^2)}{80} +\frac{(\tfrac12)^2(6^2)}{60} +\frac{(\tfrac12)^2(10^2)}{100} =0.2000+0.1500+0.2500 =0.6000

in squared outcome points. For comparison, equal allocation of 80 participants per arm gives $16/80+9/80+25/80=0.6250$ squared points. This comparison checks the calculation against the objective: under the assumed model, the Neyman shares give the smaller contrast variance.

Why the Neyman Shares Minimize Variance

A short inequality shows why the allocation weights have this form. With $N=\sum_i n_i$, apply the Cauchy–Schwarz inequality to the two vectors with components $\sqrt{n_i}$ and $|c_i|\sigma_i/\sqrt{n_i}$. For positive $n_i$, it gives

\left(\sum_{i=1}^K |c_i|\sigma_i\right)^2 \le \left(\sum_{i=1}^K n_i\right) \left(\sum_{i=1}^K\frac{c_i^2\sigma_i^2}{n_i}\right) = N\operatorname{Var}(\hat{\theta}).

The left side and $N$ are fixed, so this inequality supplies a lower bound on the contrast variance for any allocation with total $N$. Equality in Cauchy–Schwarz requires the two vectors to be proportional. When every coefficient is nonzero and every standard deviation is positive, this condition is equivalent to $n_i$ being proportional to $|c_i|\sigma_i$. Normalizing those proportional values to sum to $N$ gives the shares above. This justifies both the variance-minimizing interpretation and the conversion from relative weights to proportions.

Worked Example: Recovering Shares From a Multiarm Ratio

Suppose a hypothetical three-arm contrast has coefficients $1,-\tfrac12,-\tfrac12$, and the planning standard deviations are 2, 6, and 10 outcome points. Assume the same independent-observation, equal-cost model, and a total planned enrollment of 400.

The Neyman weights are $1(2)=2$, $\tfrac12(6)=3$, and $\tfrac12(10)=5$. The allocation ratio is therefore $2:3:5$. A ratio is relative, so its parts must be normalized: the total number of parts is $2+3+5=10$. The shares are

q_1=\frac{2}{10}=0.2000,\qquad q_2=\frac{3}{10}=0.3000,\qquad q_3=\frac{5}{10}=0.5000.

The corresponding target counts are $400(0.2000)=80$, $400(0.3000)=120$, and $400(0.5000)=200$. The counts sum to 400. In particular, the ratio $2:3:5$ does not mean to enroll only 10 participants; it specifies shares that can be scaled to the total enrollment.

Interpreting Targets and Recognizing Limits

The share formula is a planning result under a variance model, not a universal prescription for assigning patients. In a response-adaptive design, target shares may guide assignment probabilities as information accumulates. Realized counts can differ from $Nq_i$ because assignments are random, enrollment may stop at a fixed total, and the procedure may deliberately change probabilities over time. Furthermore, if assignments are adaptive, the variance of an estimator must be justified for that design; a target ratio alone does not establish that the fixed-allocation variance formula is exact.

The formula also depends on the objective. Unequal per-patient costs change the cost-constrained allocation weights. Minimum arm sizes or other restrictions can prevent the unconstrained shares from being attained. If standard deviations are estimated from pilot data rather than known, the shares are plug-in targets and may be sensitive to estimation error. These are reasons to check whether the assumptions and target match the design, rather than reasons to interpret a ratio differently.

There is an important exception when a contrast coefficient is zero. The positive-share formula and its interior equality case require all coefficients to be nonzero, as well as at least one nonzero coefficient. If $c_i=0$, arm $i$ contributes nothing to this contrast’s variance. With no requirement to enroll in that arm, the unconstrained variance-minimizing allocation assigns it zero participants and allocates the enrollment among the arms with nonzero coefficients. If that arm must receive positive enrollment, the resulting constrained problem must be handled separately; applying the positive-share formula to all arms does not solve it. If all coefficients are zero, the contrast is identically zero and the share formula has a zero denominator, so there is no variance-allocation problem of this form.

A practical diagnostic is to calculate the proposed shares explicitly and check that each is nonnegative and that they sum to one, then multiply by $N$ and verify that the counts sum to the planned total. Also check whether the contrast coefficients, variability assumptions, equal-cost condition, and constraints are consistent with the formula. If the calculated shares do not sum to one, a ratio has likely been mistaken for shares or the weights have been normalized incorrectly.

The main alternative for a simple benchmark is equal allocation, which uses $q_i=1/K$ regardless of the variances and contrast coefficients. It can be easier to implement and may be reasonable when variances are similar, but it is not generally the variance-minimizing allocation for the stated contrast. Cost-adjusted allocation is another alternative when patient costs differ, but it optimizes a different constrained problem. Neither alternative should be substituted without checking the study objective and assumptions.

Key takeaway: A Neyman ratio becomes a share by dividing each allocation weight by the sum of all weights. For a two-arm difference, the weights are the arm standard deviations; for a multiarm linear contrast, they are $|c_i|\sigma_i$. The resulting shares describe a fixed-enrollment target under the stated variance-minimization assumptions.

Check Your Understanding

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

For a two-arm difference with standard deviations 5 and 15, what are the Neyman shares? If $N=200$, what are the target counts?
A multiarm Neyman ratio is $1:4:5$. Convert it to shares and give the target counts when $N=300$.
For a contrast with coefficients $1,-1$ and standard deviations 7 and 9, which arm receives the larger Neyman share, and why?
Explain why the positive-share formula cannot be applied to every arm if one contrast coefficient is zero.
In a response-adaptive trial, why might realized sample-size proportions differ from the target shares?

Solutions. (1) The shares are $5/(5+15)=0.2500$ and $15/(5+15)=0.7500$, giving target counts $200(0.2500)=50$ and $200(0.7500)=150$. (2) The ratio parts sum to 10, so the shares are $0.1000,0.4000,0.5000$; the target counts are 30, 120, and 150, which sum to 300. (3) The second arm receives the larger share because $9>7$; for a two-arm difference, each absolute contrast coefficient is 1, so the weights are the standard deviations. (4) A zero coefficient means that arm contributes nothing to the contrast variance. The unconstrained optimum assigns it zero enrollment; a positive required enrollment introduces a separate constraint, and the all-arm positive-share formula no longer gives the solution. (5) Random assignment, changing assignment probabilities, and finite enrollment can all make realized counts differ from their target proportions; the target is not a guarantee of exact realized counts.