Choosing Treatment-Control Allocation for Assurance

Why allocation matters

At a fixed total sample size, assigning more participants to one arm leaves fewer for the other. The best split depends on how much outcome variability each arm has and, when resources are constrained by cost, how expensive it is to enroll participants in each arm. The allocation can therefore affect the probability that a trial meets its prespecified Bayesian success rule, even when the total number enrolled is unchanged.

A useful starting point is the variance of the estimated treatment-control difference. Minimizing this variance is a standard design principle, but it is not automatically equivalent to maximizing assurance for every possible prior and decision rule. The connection holds under identifiable conditions; outside them, assurance should be evaluated directly for candidate allocations.

Model and assurance criterion

Consider a two-arm trial with $n_t$ participants assigned to treatment and $n_c$ assigned to control. Let the outcome in arm $j$ have mean $\mu_j$ and known variance $\sigma_j^2$, where $j$ is $t$ or $c$. Assume outcomes are independent between participants and arms, and that the sampling model is normal. The estimator of the treatment effect is the difference in sample means, $\hat{\theta}=\bar{x}_t-\bar{x}_c$, for the estimand $\theta=\mu_t-\mu_c$.

Definition: Under this model, the sampling variance of the estimated treatment effect is $V(n_t,n_c)=\sigma_t^2/n_t+\sigma_c^2/n_c$. Assurance is the probability, averaged over the design distribution for unknown treatment effects and any other uncertain parameters, that the planned trial's data satisfy its specified success rule.

For a transparent derivation, suppose the analysis uses a flat prior for $\theta$ and the known sampling variance $V$. The posterior distribution is then normal with mean $\hat{\theta}$ and variance $V$. Define success as a posterior probability greater than $q$ that the treatment effect exceeds a clinically relevant threshold $\delta$, with $q>0.5$. This criterion can be written as

P(\theta>\delta\mid \text{data})>q \quad\Longleftrightarrow\quad \hat{\theta}>\delta+z_q\sqrt{V},

where $z_q$ is the $q$-quantile of the standard normal distribution. To calculate assurance, suppose the design prior is $\theta\sim N(m_0,\tau^2)$, independently of the sampling error. Conditional on $\theta$, $\hat{\theta}\sim N(\theta,V)$. Averaging over this design prior gives $\hat{\theta}\sim N(m_0,\tau^2+V)$, so the assurance for this rule is

A(V)=1-\Phi\left( \frac{\delta+z_q\sqrt{V}-m_0}{\sqrt{\tau^2+V}} \right),

where $\Phi$ is the standard normal cumulative distribution function. This expression shows exactly how allocation affects assurance in this particular model: allocation changes $V$, which changes both the decision threshold and the distribution of the estimated effect.

Fixed total sample size: the variance-based allocation

Suppose the total sample size $N=n_t+n_c$ is fixed, with both arm sizes positive. Substituting $n_c=N-n_t$ into $V$ gives a function of one allocation choice. Differentiating and setting the derivative to zero yields the variance-minimizing split.

V(n_t)=\frac{\sigma_t^2}{n_t}+\frac{\sigma_c^2}{N-n_t}, \qquad \frac{dV}{dn_t}=-\frac{\sigma_t^2}{n_t^2} +\frac{\sigma_c^2}{(N-n_t)^2}=0.

At the stationary point, $\sigma_t/n_t=\sigma_c/n_c$, and therefore $n_t/n_c=\sigma_t/\sigma_c$. The second derivative is positive for $0<n_t<N$, so this is the minimum. Thus more participants are assigned to the arm with the larger standard deviation, because an additional participant there reduces the variance more.

Result: With a fixed total sample size and the stated independent, known-variance normal model, the variance-minimizing allocation is $n_t:n_c=\sigma_t:\sigma_c$. It is equal allocation when the two arm variances are equal. If assurance decreases as $V$ increases over the candidate designs, this allocation also maximizes assurance.

That final condition matters. In the normal example above, if $m_0\geq\delta$ and $q>0.5$, assurance decreases as $V$ increases. To see this, set $s=\sqrt{V}$, $a=\delta-m_0$, and $x=(a+z_qs)/\sqrt{\tau^2+s^2}$, so $A=1-\Phi(x)$. Differentiation gives

\frac{dx}{ds} = \frac{z_q\tau^2-a s}{(\tau^2+s^2)^{3/2}}.

Here $a\leq0$ and $z_q>0$, so $dx/ds\geq0$. Since $\Phi$ is increasing, $A$ cannot increase as $V$ increases. If the design prior mean is below the target, however, the derivative need not have this sign throughout the design range. In that case, the variance-minimizing allocation need not maximize assurance, and the assurance formula or the relevant assurance computation should be evaluated directly.

Worked Example: Fixed Sample Size With Unequal Variances

Suppose a hypothetical trial has $N=100$ participants, with known outcome variances $\sigma_t^2=4$ and $\sigma_c^2=1$. Use the fixed-size variance formula to compare equal allocation with the variance-minimizing allocation.

The standard deviations are $\sigma_t=2$ and $\sigma_c=1$, so the optimal continuous ratio is $n_t/n_c=2/1$. With $n_t+n_c=100$, the continuous allocation is $n_t=66.67$, $n_c=33.33$. A nearby integer allocation is 67 treatment and 33 control participants. Its variance is

V(67,33)=\frac{4}{67}+\frac{1}{33} =0.05970+0.03030=0.09000.

Equal allocation gives

V(50,50)=\frac{4}{50}+\frac{1}{50} =0.08000+0.02000=0.1000.

Thus the 67:33 split has lower sampling variance. To illustrate the assurance connection, take a hypothetical design prior $N(0.5,0.3^2)$, target $\delta=0.2$, and success threshold $q=0.95$. Using $z_{0.95}=1.645$, the 67:33 posterior-mean cutoff is $0.2+1.645\sqrt{0.09}=0.6935$. Its assurance is approximately $1-\Phi((0.6935-0.5)/\sqrt{0.09+0.09})=0.324$. For equal allocation, the cutoff is $0.2+1.645\sqrt{0.10}=0.7201$, giving assurance approximately $1-\Phi((0.7201-0.5)/\sqrt{0.09+0.10})=0.307$. Under these assumptions, the variance reduction corresponds to an increase in assurance.

A continuous optimum is a guide, not an implementable design when sample sizes must be integers. For a fixed $N$, calculate $V$, or the actual assurance, at nearby integer splits that sum to $N$. If the endpoint, success rule, or model is more complex than the example, compare candidate allocations using that full assurance calculation rather than assuming that the variance formula settles the question.

Costs and a fixed budget

Costs affect the statistical optimum when the constraint is a budget rather than a fixed total sample size. Let $c_t$ and $c_c$ be the per-participant costs in the treatment and control arms. Under a fixed budget $B$, the constraint is $c_tn_t+c_cn_c=B$. To minimize $V$ subject to this constraint, form the Lagrangian $L=V+\lambda(c_tn_t+c_cn_c-B)$. Setting its two partial derivatives to zero gives $\sigma_t^2/n_t^2=\lambda c_t$ and $\sigma_c^2/n_c^2=\lambda c_c$. Dividing these equations and taking positive square roots gives the cost-adjusted allocation ratio.

\frac{n_t}{n_c} = \sqrt{\frac{\sigma_t^2c_c}{\sigma_c^2c_t}} = \frac{\sigma_t}{\sigma_c}\sqrt{\frac{c_c}{c_t}}.

The ratio increases with treatment-arm variability and decreases with treatment-arm cost. Under the same monotonicity condition that links smaller $V$ to greater assurance, this is the assurance-maximizing ratio for a fixed budget in the stated model. As before, a different assurance function can change the answer.

Worked Example: Cost-Adjusted Allocation

Consider a hypothetical fixed budget of $B=\$120{,}000$, treatment-arm variance $144$, control-arm variance $100$, and per-participant costs $c_t=\$900$ and $c_c=\$400$. Assume the continuous allocation approximation and the known-variance model.

The cost-adjusted ratio is

\frac{n_t}{n_c} = \sqrt{\frac{144(400)}{100(900)}} = \sqrt{0.64}=0.8000.

Set $n_t=0.8n_c$ in the budget constraint. Then $900(0.8n_c)+400n_c=120{,}000$, so $1{,}120n_c=120{,}000$, giving $n_c=107.14$ and $n_t=85.71$. The corresponding variance is approximately

V=\frac{144}{85.71}+\frac{100}{107.14} =1.680+0.9333=2.613.

A real design requires integer sample sizes and may also require minimum or maximum arm sizes. Candidate integer allocations should respect the budget and be checked for their variance or, preferably when the decision rule is not variance-monotone, their assurance.

If the total sample size is fixed and both allocations have the same number of participants, different per-participant costs do not alter the statistical variance objective $V$. Costs may make one split financially infeasible, or may matter to a separate economic objective, but they do not enter the fixed-$N$ variance formula. If the budget is fixed instead, the total sample size generally changes with the split, and costs do enter the optimum. These are different design constraints and should not be conflated.

Comparing allocation approaches and checking a design

Approach	Constraint or objective	When it is useful
Equal allocation	Same number in each arm	Simple and variance-optimal when arm variances are equal and costs are equal.
Variance-based allocation	Minimize $V$ at fixed $N$	Appropriate when variances differ and assurance improves as $V$ falls.
Cost-adjusted allocation	Minimize $V$ at fixed budget	Appropriate when arm costs differ and the same assurance-monotonicity condition holds.
Direct assurance comparison	Maximize the specified assurance over feasible allocations	Needed when the assurance function is not monotone in $V$, or the model or success rule is more complex.

A practical design check is to state the constraint first, derive a continuous candidate, round to feasible integer allocations, and compare the objective across nearby candidates. For direct assurance optimization, use the same design prior, analysis model, and success rule for every allocation. If assurance is estimated by simulation, its numerical uncertainty should be considered before treating small differences between allocations as meaningful.

The formulas here rely on independent outcomes, a correctly specified normal sampling model, known arm-specific variances, and positive sample sizes. If variances are uncertain, outcomes are clustered, or the endpoint is binary or time-to-event, the simple expression for $V$ may not apply. Likewise, a prior or decision criterion that makes assurance nonmonotone in $V$ can produce an optimum that differs from the variance-minimizing ratio. These are reasons to calculate the assurance under the planned model, not to apply a familiar ratio without checking its assumptions.

Common pitfalls

Using the variance ratio instead of the standard-deviation ratio. At fixed $N$, the optimal ratio is $n_t/n_c=\sigma_t/\sigma_c$, not $\sigma_t^2/\sigma_c^2$. The squared variances appear in $V$, but solving the derivative equation takes their square roots.
Claiming that treatment cost changes the fixed-$N$ statistical optimum. With total sample size fixed, costs do not appear in $V$. Costs alter the optimum under a budget constraint, or restrict which fixed-size allocations are feasible.
Assuming lower variance always means higher assurance. That link depends on the design distribution and success rule. Check monotonicity or evaluate assurance directly when the condition is not established.
Reporting a fractional allocation as a final design. Continuous formulas identify a target ratio; a trial enrolls integer numbers of participants. Compare feasible rounded alternatives and account for any operational constraints.

Key takeaway: At fixed total sample size, the variance-based ratio is the ratio of arm standard deviations. At fixed budget, the ratio also includes the square root of the inverse relative costs. Either ratio maximizes assurance only when the planned success probability is maximized by reducing the treatment-effect sampling variance.

Check Your Understanding

Try each question, then check the brief solutions below.

At fixed total sample size, suppose $\sigma_t=3$ and $\sigma_c=2$. What is the variance-minimizing treatment-to-control ratio?
If the total sample size is fixed, do unequal per-participant costs by themselves change the variance-minimizing ratio? Explain.
For a fixed budget, suppose $\sigma_t^2=9$, $\sigma_c^2=4$, $c_t=4$, and $c_c=1$. Find the cost-adjusted ratio $n_t/n_c$.
Why should a designer calculate assurance directly when the design prior mean is below the clinical target?

Solutions. (1) $n_t/n_c=\sigma_t/\sigma_c=3/2=1.5$. (2) No: with fixed $N$, costs do not enter $V=\sigma_t^2/n_t+\sigma_c^2/n_c$; they may affect feasibility but not this variance objective. (3) The ratio is $\sqrt{9(1)/(4(4))}=\sqrt{9/16}=0.7500$. (4) In that case the derivative governing the assurance's dependence on $V$ need not keep one sign, so the variance-minimizing allocation may not maximize the specified assurance.