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Optimal Allocation Targets · Tutorial 129 of 501

Ethical Allocation for Binary Response Outcomes

Learn to maximize expected binary-outcome successes while keeping the variance of the treatment-effect estimate within a stated limit.

Intermediate 9 min read

What You'll Learn

  • Express expected successes as a function of the treatment allocation.
  • Formulate a precision constraint using the variance of a difference in response proportions.
  • Find the feasible allocation range and check whether any integer allocation meets the constraint.
  • Choose the welfare-maximizing endpoint when treatment success probabilities differ.
  • Distinguish a patient-benefit target from a precision-minimizing allocation.
  • Identify the assumptions and limitations behind planning with binary response rates.

Binary Successes and the Allocation Decision

When an outcome is binary, each patient either has the defined success or does not. If success is the patient benefit of interest, the expected number of successes depends on both the response probabilities and the numbers assigned to each arm. A treatment-favored allocation may increase expected successes, but assigning too few patients to one arm can make the treatment-effect estimate too imprecise. The allocation decision can therefore be framed as maximizing expected successes subject to a precision requirement.

This tutorial considers two arms, a treatment arm and a control arm. The objective assigns equal value to each success and no additional value or penalty to either arm. That is a deliberately specific ethical objective: it is not a general claim that success is the only relevant patient outcome. The allocation is a planning target, not a rule for deciding which treatment an individual patient should receive.

Definition: Let \(n\) be the fixed total sample size and \(x\) the number assigned to treatment, so \(n-x\) are assigned to control. Let \(p_T\) and \(p_C\) be the respective probabilities of success. Under the objective that each success contributes one unit of patient benefit, expected total successes are \(W(x)=xp_T+(n-x)p_C\). If the inferential constraint is that the variance of the estimated treatment-minus-control response difference is no greater than \(V_{\max}\), a feasible target must also satisfy \(p_T(1-p_T)/x+p_C(1-p_C)/(n-x)\leq V_{\max}\), with \(1\leq x\leq n-1\).

The variance expression is the planning variance of the difference between two independent sample proportions, using the specified response probabilities. Its reciprocal square root is not used here: the standard error is the square root of the variance. A bound \(V_{\max}\) is therefore a variance bound; equivalently, a standard-error bound \(S_{\max}\) corresponds to \(V_{\max}=S_{\max}^2\).

Assumptions Behind the Planning Formula

The calculation assumes a fixed total sample size, two treatment groups, and binary outcomes defined consistently across arms. For planning, each patient's outcome is treated as a Bernoulli response with the relevant arm-specific probability: \(p_T\) under treatment and \(p_C\) under control. Outcomes are assumed independent across patients, and the response probabilities are treated as known planning inputs. In practice, they are estimates or assumptions and may be uncertain.

The variance formula also presumes that the treatment and control sample proportions are based on their respective assigned sample sizes and that all outcomes are observed. It does not account for missing outcomes, clustering, covariate adjustment, or dependence among patients. Those features can alter the variance and must be handled in the model used to set a target. In particular, if the probabilities are not credible planning values, a target that satisfies the calculated constraint may fail to satisfy the actual precision requirement.

Write \(a=p_T(1-p_T)\) and \(b=p_C(1-p_C)\). For a proposed allocation \(x\), the planning variance is \(V(x)=a/x+b/(n-x)\). The objective is linear:

$$ W(x)=np_C+x(p_T-p_C). $$

Thus, if \(p_T>p_C\), increasing \(x\) increases expected successes; if \(p_T<p_C\), increasing \(x\) decreases expected successes; and if the probabilities are equal, the expected-success objective is constant across allocations. The precision constraint determines which values of \(x\) are allowed. This simple direction-of-change argument is the key to selecting the ethical target once the feasible allocations are known.

Deriving the Feasible Allocation Range

Because \(x\) and \(n-x\) are positive for an allocation with both arms represented, multiplying the precision inequality by \(x(n-x)\) does not reverse its direction. The constraint becomes a quadratic inequality:

$$ \begin{aligned} \frac{a}{x}+\frac{b}{n-x} &\leq V_{\max}\\ a(n-x)+bx &\leq V_{\max}x(n-x)\\ V_{\max}x^2+(b-a-V_{\max}n)x+an &\leq 0. \end{aligned} $$

When \(V_{\max}>0\), the left side is an upward-opening quadratic. If it has real roots, the continuous values satisfying the inequality lie between the roots, inclusive. The actual target must be an integer in that interval and between 1 and \(n-1\). A real-valued interval alone is not enough: if it contains no integer, no allocation meets the stated constraint.

A useful feasibility check comes from minimizing \(V(x)\) over continuous \(x\). Differentiating gives \(V'(x)=-a/x^2+b/(n-x)^2\). Setting this to zero yields \(x/(n-x)=\sqrt{a/b}\) when \(a,b>0\). This is the precision-minimizing allocation, not generally the patient-benefit-maximizing allocation. If even the minimum achievable variance exceeds \(V_{\max}\), the constraint is infeasible. If the constraint is feasible, the benefit objective chooses an endpoint of the feasible range, not necessarily the precision-minimizing point.

Target-selection result: Find all integer allocations satisfying the precision constraint. If \(p_T>p_C\), choose the largest feasible \(x\); if \(p_T<p_C\), choose the smallest feasible \(x\). If \(p_T=p_C\), every feasible allocation has the same expected number of successes under this objective, so another prespecified criterion is needed to choose among them.

Worked Allocations

Worked Example: Expected Successes at a Proposed Allocation

Suppose a hypothetical trial plans to enroll \(n=60\) patients. The assumed success probabilities are \(p_T=0.60\) and \(p_C=0.45\). Consider assigning \(x=36\) patients to treatment and 24 to control. These planning probabilities are assumed to apply to all patients in their respective arms, and the calculation concerns expected successes, not a guarantee about the realized count.

The expected number of treatment successes is \(36(0.60)=21.6\), and the expected number of control successes is \(24(0.45)=10.8\). Therefore,

$$ W(36)=36(0.60)+24(0.45)=21.6+10.8=32.4. $$

For comparison, an equal allocation of 30 patients per arm has expected successes \(30(0.60)+30(0.45)=18+13.5=31.5\). Under the stated assumptions, the 36-to-24 allocation has 0.9 more expected successes. This comparison alone does not establish that the allocation meets any precision requirement; that must be checked separately.

Worked Example: Maximizing Expected Successes Under a Precision Bound

Now consider a hypothetical trial with \(n=80\), \(p_T=0.65\), and \(p_C=0.50\). Require the planning variance of the difference in response proportions to be at most \(V_{\max}=0.0125\). The corresponding Bernoulli variances are \(a=0.65(0.35)=0.2275\) and \(b=0.50(0.50)=0.25\). The assumptions are fixed sample size, independent binary responses, and complete outcome measurement.

Substitution into the quadratic form of the constraint gives

$$ \begin{aligned} 0.0125x^2+(0.25-0.2275-0.0125(80))x+0.2275(80)&\leq 0\\ 0.0125x^2-0.9775x+18.2&\leq 0. \end{aligned} $$

The discriminant is \((-0.9775)^2-4(0.0125)(18.2)=0.04550625\), whose square root is approximately \(0.21332\). The roots are

$$ x=\frac{0.9775\pm\sqrt{0.04550625}}{0.025}, \qquad x\approx 30.57,\ 47.63. $$

The continuous feasible interval is approximately \(30.57\leq x\leq47.63\), so the feasible integer treatment allocations are \(31,32,\ldots,47\). Since \(p_T-p_C=0.15>0\), expected successes increase with \(x\); the largest feasible integer, \(x=47\), maximizes the stated benefit objective. Check the boundary directly:

$$ V(47)=\frac{0.2275}{47}+\frac{0.25}{33} \approx 0.004840+0.007576=0.01242\leq0.0125. $$

At \(x=48\), \(V(48)=0.2275/48+0.25/32\approx0.004740+0.007813=0.01255\), which exceeds the bound. The chosen allocation is therefore the most treatment-heavy feasible integer allocation. Its expected successes are \(47(0.65)+33(0.50)=30.55+16.5=47.05\). This is an expectation under the planning model, not a prediction that the observed number of successes will equal 47.05.

Worked Example: When the Control Response Probability Is Higher

Keep \(n=80\) and \(V_{\max}=0.0125\), but suppose the hypothetical planning probabilities are \(p_T=0.35\) and \(p_C=0.50\). Then \(a=0.35(0.65)=0.2275\) and \(b=0.25\). The same quadratic and roots as in the preceding calculation apply, so the feasible integer treatment allocations are again 31 through 47. This setup also illustrates why the precision-minimizing allocation should not be confused with the ethical target.

First, the continuous precision minimum solves \(x/(80-x)=\sqrt{0.2275/0.25}=\sqrt{0.91}\approx0.9539\). Solving for \(x\),

$$ x=\frac{80(0.9539)}{1+0.9539}\approx39.06. $$

The precision-minimizing integer is near 39, but \(p_T-p_C=-0.15\), so expected successes decrease as treatment allocation \(x\) increases. Among feasible integers, the ethical target under the stated success-only objective is therefore the smallest one, \(x=31\), not the precision minimum. Its expected successes are \(31(0.35)+49(0.50)=10.85+24.5=35.35\). At the other feasible endpoint, \(x=47\), expected successes are \(47(0.35)+33(0.50)=16.45+16.5=32.95\). Both meet the variance constraint, but the smaller treatment allocation has the higher expected success count under these assumed probabilities.

Diagnostics, Alternatives, and Limitations

Several checks help detect a faulty target calculation. Confirm that \(x\) and \(n-x\) are both positive; verify that the variance terms use the correct arm's probability and sample size; and substitute the chosen integer back into the original variance formula. Check the adjacent integer in the direction of greater patient benefit as well. If that adjacent value violates the constraint, the boundary choice is supported. If it does not, the proposed allocation may not be the best feasible integer.

The continuous quadratic interval can be empty, or it can contain no allowable integer after rounding and enforcing arm-size restrictions. In either case, there is no target satisfying the stated requirement. The correct response is to revise a design input transparently—for example, increase the total sample size or reconsider the precision requirement—not to report an infeasible allocation as if it met the constraint.

The principal alternative targets answer different questions. Equal allocation is simple and may be preferred for reasons outside this objective, but it need not maximize expected successes or minimize variance. A precision-minimizing allocation selects the point that minimizes the variance of the treatment-effect estimator; it may sacrifice expected successes relative to the feasible welfare-maximizing endpoint. A purely welfare-maximizing allocation would assign everyone to the arm with the higher assumed response probability, but it cannot estimate a between-arm difference when an arm has no participants. The precision constraint rules out such unsupported extremes when they fail the bound.

These calculations depend on the response probabilities. If they are misspecified, both the expected-success comparison and the variance constraint can be misleading. Further, equal value for each success omits possible differences in the importance of outcomes, treatment-related harm, costs, and patient preferences. Those considerations require a different utility specification rather than an informal adjustment to the response-probability calculation. Finally, a target allocation is not itself a complete randomization procedure: it specifies a desired overall allocation, while the mechanism for assigning each arriving participant is a separate design choice.

A common mistake is to select the Neyman, or precision-minimizing, ratio and call it the ethical target because it gives the smallest variance. That reasoning optimizes precision alone. Under a precision constraint, first determine the feasible allocations, then use the sign of \(p_T-p_C\) to select the feasible endpoint that maximizes expected successes. Another mistake is to round a continuous endpoint without checking it. Since the constraint can fail at the next integer, always evaluate the original variance at the final allocation.

Key takeaway: For binary successes and a fixed total sample size, expected successes are linear in the treatment count, while the variance constraint defines the allowed range. The welfare-maximizing target is the largest feasible treatment count when treatment has the higher assumed response probability, and the smallest feasible count when control does.

Check Your Understanding

Answer each question using the stated success-only objective and, where relevant, the planning variance constraint.

  1. For \(n=50\), \(p_T=0.60\), and \(p_C=0.40\), write the expected-success function \(W(x)\). Does it increase or decrease with \(x\)?
  2. Why must a proposed allocation with \(x=0\) be excluded from the two-arm variance formula?
  3. If a precision constraint permits integer allocations from 18 through 27, and \(p_T<p_C\), which feasible allocation maximizes expected successes?
  4. What does it mean if the continuous feasible interval contains no integer allocation?
  5. Does a precision-minimizing allocation necessarily maximize expected successes subject to a precision bound? Briefly explain.

Solutions. 1. \(W(x)=0.60x+0.40(50-x)=20+0.20x\), so it increases with \(x\). 2. With \(x=0\), there are no treatment observations, so the treatment response proportion and its variance term are undefined; there is no two-arm comparison. 3. Choose \(x=18\), the smallest feasible treatment count, because \(p_T<p_C\) makes expected successes decrease with \(x\). 4. No integer allocation satisfies the stated precision constraint, so the design as specified is infeasible. 5. No. Precision minimization selects the allocation with the lowest variance; the constrained benefit objective selects the feasible endpoint favored by the larger response probability.

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