Neyman Allocation for Binary Outcomes

Why Binary Outcomes Need Their Own Allocation Calculation

When the outcome is binary, each participant either has the defined response or does not. The variability of that outcome depends on the response probability: a response rate near one-half produces more variability than a rate near zero or one. Therefore, for a fixed total sample size, the variance-minimizing allocation between two arms depends on their response rates, not just on the difference between them.

This tutorial derives that target for estimating a difference in response proportions. It treats the two-arm allocation as a planning target based on specified response rates. A target is not, by itself, a rule for how to randomize each participant during a trial; in a response-adaptive trial, assignments and observed outcomes can be dependent, and the analysis must account for the actual design.

Definition: Let arm 1 be the treatment arm and arm 0 the comparator. For a binary response, let \(p_1\) and \(p_0\) be the response probabilities in the two arms. The estimand is the difference in response proportions, \(p_1-p_0\). For arm sample sizes \(n_1\) and \(n_0\), the unadjusted estimator is \(\hat{p}_1-\hat{p}_0\), where \(\hat{p}_a\) is the observed response proportion in arm \(a\). Under independent Bernoulli outcomes and fixed arm sizes, its variance is \(p_1(1-p_1)/n_1+p_0(1-p_0)/n_0\).

The allocation question is: with a fixed total of \(N=n_1+n_0\) participants and the same per-participant cost in both arms, how should \(N\) be divided to minimize the variance of \(\hat{p}_1-\hat{p}_0\)? The response probabilities are treated as specified planning values for this calculation. The resulting targets are continuous; they need not be integers.

Deriving the Binary Neyman Ratio

For a Bernoulli response with probability \(p_a\), the outcome variance in arm \(a\) is \(p_a(1-p_a)\). Write \(v_1=p_1(1-p_1)\) and \(v_0=p_0(1-p_0)\). The variance to minimize is \(v_1/n_1+v_0/n_0\), subject to \(n_1+n_0=N\). Assuming both variances are positive, substitute \(n_0=N-n_1\) and differentiate with respect to \(n_1\).

V(n_1)=\frac{v_1}{n_1}+\frac{v_0}{N-n_1}, \qquad \frac{dV}{dn_1}=-\frac{v_1}{n_1^2}+\frac{v_0}{(N-n_1)^2}.

At an interior minimum, the derivative is zero. Since sample sizes are positive, taking positive square roots gives the ratio below. The second derivative, \(2v_1/n_1^3+2v_0/(N-n_1)^3\), is positive, so this stationary point minimizes the variance rather than maximizing it.

\frac{v_1}{n_1^2}=\frac{v_0}{n_0^2} \quad\Longrightarrow\quad \frac{n_1}{n_0}=\sqrt{\frac{v_1}{v_0}} =\sqrt{\frac{p_1(1-p_1)}{p_0(1-p_0)}}.

Result: For two arms with equal per-participant costs, the binary-outcome Neyman allocation assigns participants in proportion to the within-arm standard deviations of the binary outcomes. With \(N=n_1+n_0\), its continuous targets are \(n_1^*=N\sqrt{p_1(1-p_1)}/[\sqrt{p_1(1-p_1)}+\sqrt{p_0(1-p_0)}]\) and \(n_0^*=N-n_1^*\).

The formula is the two-arm Neyman allocation expressed in terms of Bernoulli variances. It allocates more participants to the arm with the larger outcome standard deviation, because additional observations there reduce the variance contribution more. At the optimum, the minimized variance can also be written as the squared sum of the two arm standard deviations divided by the total sample size:

V_{\min} = \frac{\left(\sqrt{p_1(1-p_1)}+\sqrt{p_0(1-p_0)}\right)^2}{N}.

Worked Targets From Specified Response Rates

Worked Example: A Moderately Higher Treatment Response Rate

Suppose a hypothetical planning calculation uses a treatment response rate of \(p_1=0.70\), a comparator response rate of \(p_0=0.50\), and a total of \(N=400\) participants. Assume the response is measured for everyone, outcomes are independent Bernoulli variables within each arm, arm response probabilities are constant, and costs are equal.

The arm variances are \(v_1=0.70(1-0.70)=0.21\) and \(v_0=0.50(1-0.50)=0.25\). Their standard deviations are \(\sqrt{0.21}=0.4583\) and \(\sqrt{0.25}=0.5000\). Thus:

\frac{n_1}{n_0} = \sqrt{\frac{0.21}{0.25}} = 0.9165, \qquad \frac{n_1^*}{400} = \frac{0.4583}{0.4583+0.5000} = 0.4782.

The continuous targets are \(n_1^*=400(0.4782)=191.3\) participants in the treatment arm and \(n_0^*=400-191.3=208.7\) in the comparator arm. The target puts slightly more participants in the comparator arm because its assumed response variance is larger. These decimal values express the theoretical target; an implemented trial needs integer assignments.

Worked Example: A Large Difference Does Not Imply Unequal Allocation

Consider a second hypothetical plan with \(p_1=0.80\), \(p_0=0.20\), and \(N=300\). Suppose the same independence, constant-probability, complete-outcome, and equal-cost assumptions hold. The specified response-rate difference is \(0.60\), but the allocation follows the two variances:

v_1=0.80(0.20)=0.16, \qquad v_0=0.20(0.80)=0.16, \qquad \frac{n_1}{n_0}=\sqrt{\frac{0.16}{0.16}}=1.

Therefore \(n_1^*=300/2=150\) and \(n_0^*=150\). This is an important distinction: Neyman allocation is driven by outcome variability, not by the size of the treatment effect. Here the response rates are far apart but their Bernoulli variances are identical.

Worked Example: More Allocation to the More Variable Arm

For a third hypothetical plan, use \(p_1=0.90\), \(p_0=0.60\), and \(N=600\). Assume independent Bernoulli responses, complete outcomes, constant response probabilities within arms, and equal costs. The variances are \(v_1=0.90(0.10)=0.09\) and \(v_0=0.60(0.40)=0.24\), so the standard deviations are \(0.3000\) and \(0.4899\).

\frac{n_1}{n_0} = \sqrt{\frac{0.09}{0.24}} = 0.6124, \qquad \frac{n_1^*}{600} = \frac{0.3000}{0.3000+0.4899} = 0.3798.

The targets are \(n_1^*=600(0.3798)=227.9\) and \(n_0^*=600-227.9=372.1\). The comparator receives more participants because its response probability of 0.60 gives it the larger Bernoulli variance. As a check, the minimized variance is approximately \((0.3000+0.4899)^2/600=0.001040\), in squared proportion units.

Assumptions, Diagnostics, and Alternatives

The calculation is appropriate when the target is the unadjusted difference in two response proportions, the binary outcome is observed for each participant, the arm-specific response probabilities adequately describe the planning scenario, and participants’ outcomes are independent under the design. It also assumes equal per-participant costs. If the trial uses a different estimand, substantial clustering, missing outcomes, covariate adjustment, or unequal costs, this formula may not be the relevant variance calculation. The cost-adjusted allocation result addresses unequal costs; it should not be replaced by the equal-cost ratio without checking the cost assumption.

The probabilities in a planning calculation are assumptions, not facts established by the formula. A useful diagnostic is to calculate each arm’s \(p_a(1-p_a)\) explicitly before forming the ratio. Check that each rate lies between zero and one, that the resulting standard deviations are nonnegative, and that the two target shares sum to one. Also check the direction: the arm with the higher Bernoulli variance should receive the larger target. If both response rates are equal, their variances match and the target is equal allocation. If rates are complementary, such as 0.80 and 0.20, their variances also match.

Equal allocation is a simple alternative and may be preferred for practical or scientific reasons, but it is not generally the variance-minimizing allocation under unequal variances. The Neyman target instead minimizes the variance of the stated difference in proportions for a fixed total sample size and equal costs. It does not maximize expected patient benefit, guarantee ethical superiority, or automatically maximize power under every analysis or design.

A further distinction matters in response-adaptive randomization. The derivation assumes fixed arm sizes when evaluating the variance. An adaptive procedure can make future assignments depend on earlier outcomes, so its realized arm sizes are random and the simple fixed-size variance expression need not describe its full sampling variance. A Neyman target can inform a desired allocation, but it does not by itself justify an adaptive algorithm or an inferential procedure.

Common mistake: “The treatment response rate is higher, so Neyman allocation must assign more participants to treatment.” This reasoning confuses the treatment effect with outcome variance. The correct comparison is between \(\sqrt{p_1(1-p_1)}\) and \(\sqrt{p_0(1-p_0)}\); the larger standard deviation gets the larger target under equal costs. A high response rate can correspond to low variance when it is near one.

Check Your Understanding

Use the two-arm binary Neyman formula. Treat all response rates as specified hypothetical planning values, and assume independent Bernoulli outcomes, complete outcomes, constant response probabilities within arms, and equal costs unless a question says otherwise.

For \(p_1=0.40\) and \(p_0=0.50\), which arm receives the larger Neyman target, and what is the sample-size ratio \(n_1/n_0\)?
For \(p_1=0.90\) and \(p_0=0.10\), explain why the Neyman target is equal allocation despite the large response-rate difference.
If the two arm variances are \(0.09\) and \(0.16\), respectively, calculate the ratio of the first arm’s target to the second arm’s target.
State one reason why a fixed-size Neyman target does not, by itself, establish the variance of an adaptive trial estimator.

Solutions. (1) The variances are \(0.40(0.60)=0.24\) and \(0.50(0.50)=0.25\). Thus \(n_1/n_0=\sqrt{0.24/0.25}=0.9798\), so arm 0 receives the slightly larger target. (2) Both variances equal \(0.90(0.10)=0.10\), so the ratio is one and the target is equal. (3) The ratio is \(\sqrt{0.09/0.16}=0.7500\). (4) In an adaptive trial, assignments and realized arm sizes may depend on earlier outcomes; therefore arm sizes are random and the fixed-size variance expression need not capture the full sampling variance.

Neyman Allocation for Binary Outcomes

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