Working with Unequal Enrollment and Allocation Constraints

Why unequal enrollment makes constraints consequential

In a response-adaptive trial, the target allocation shares and the observed enrollment shares need not match at a particular step. A doubly adaptive biased coin design (DBCD) uses both: it favors arms whose current shares are below their target shares, while randomizing rather than assigning every participant deterministically. If an operational rule also limits how likely any arm may be to receive the next participant, the probabilities must be adjusted without confusing that rule with a change to the target itself.

The distinction matters because a restriction on next-patient probabilities acts on the immediate randomization, whereas a restriction on target shares changes the reference allocation that guides future randomizations. Neither is automatically a substitute for the other. We will write $N_k(n)$ for the number assigned to arm $k$ among the first $n$ participants, $s_k(n)=N_k(n)/n$ for its current enrollment share, $\rho_k$ for its target share, and $\gamma\geq 0$ for the DBCD bias parameter. When $n>0$, the $s_k(n)$ sum to one.

Definition: For positive target shares $\rho_k$, positive current shares $s_k(n)$, and $\gamma\geq0$, the nominal DBCD weight and probability for arm $k$ are $$ w_k(n)=\rho_k\left(\frac{\rho_k}{s_k(n)}\right)^\gamma, \qquad q_k(n)=\frac{w_k(n)}{\sum_{j=1}^K w_j(n)}. $$ The vector $q(n)$ is the nominal probability distribution for assigning participant $n+1$. It sums to one. An operationally constrained probability vector, denoted $p(n)$, is a feasible probability distribution chosen from a specified set of allowed next-participant probabilities.

This formula assumes the target shares are positive and sum to one, the current shares used in the calculation are positive, and the same $\gamma$ is applied across arms. If a current share is zero, this plug-in expression needs a separately specified rule; it is not made valid merely by imposing an operational cap. The earlier material on maintaining positive assignment chances and handling zero counts addresses such safeguards. The examples below use positive counts on every arm.

Probability constraints are not target-share constraints

A probability constraint limits $p_k(n)$, the chance that the next participant is assigned to arm $k$. For example, an operational requirement might impose $p_k(n)\leq 0.38$, or require every arm’s next-participant probability to be at least 0.05. These bounds are checked at each randomization step. They do not, by themselves, assert that the eventual fraction enrolled on that arm must be at most 0.38.

A target-share constraint instead restricts $\rho$, the reference allocation vector. If the target for an arm is limited to 0.50, that is a restriction on the desired share guiding the design. The resulting DBCD probabilities still depend on current shares and $\gamma$, and therefore need not equal the target shares. In particular, a bound on $\rho_k$ does not imply the same bound on $q_k(n)$ or $p_k(n)$.

To make an operational probability constraint precise, suppose the allowed set is described by lower and upper bounds $\ell_k\leq p_k\leq u_k$, together with $\sum_k p_k=1$. Feasibility requires $\ell_k\leq u_k$ for each arm and

\sum_{k=1}^K \ell_k \leq 1 \leq \sum_{k=1}^K u_k.

If these inequalities fail, no probability vector can satisfy the stated limits. A calculation cannot repair inconsistent operational requirements; the bounds or the design specification must be reconsidered.

Constraining nominal probabilities by proportional rescaling

One standard choice is to select the feasible vector closest to $q$ in Kullback–Leibler divergence:

\min_{p}\ \sum_{k=1}^K p_k\log\left(\frac{p_k}{q_k}\right) \quad\text{subject to}\quad \sum_{k=1}^Kp_k=1,\qquad \ell_k\leq p_k\leq u_k.

This is a design choice, not the only conceivable way to enforce bounds. It preserves the nominal relative weights among arms that are not at a bound, while adjusting arms that are constrained. The derivation explains why proportional rescaling appears.

Hold bound-active arms fixed.
Suppose a set of arms has been fixed at their applicable lower or upper bounds. The remaining probability mass $R$ must be assigned among the free arms $F$.

Differentiate on the free arms.
For the free coordinates, form the Lagrangian with multiplier $\lambda$ for their sum. Setting the derivative of $p_k\log(p_k/q_k)+\lambda p_k$ to zero gives $\log(p_k/q_k)+1+\lambda=0$.

Normalize the free probabilities.
The derivative equation implies $p_k=cq_k$ for every free arm, where $c=\exp(-1-\lambda)$. Since the free probabilities must sum to $R$, $c=R/\sum_{j\in F}q_j$. Thus $p_k=Rq_k/\sum_{j\in F}q_j$ for free arms, provided those values meet their bounds.

If a rescaled free-arm value violates another bound, that arm must also be fixed at the relevant bound and the remaining mass redistributed over the arms still free. This active-set procedure continues until the bounds are met. The objective is strictly convex for positive $q_k$, so on a nonempty feasible set this projection has a unique solution. Other projection rules, such as Euclidean distance, generally produce different probabilities and therefore define a different operational design.

Worked Example: Unequal enrollment before applying any constraint

Suppose a three-arm trial has target shares $\rho=(0.50,0.30,0.20)$. At a decision point, the current enrollment shares are $s=(0.60,0.30,0.10)$, corresponding, for example, to 60, 30, and 10 participants among 100. Let $\gamma=1$. Calculate the nominal probabilities.

The inputs satisfy the required conditions: both share vectors sum to one, each share is positive, and $\gamma\geq0$. The unnormalized DBCD weights are

w_1=0.50\left(\frac{0.50}{0.60}\right)=\frac{5}{12},\qquad w_2=0.30\left(\frac{0.30}{0.30}\right)=0.30,\qquad w_3=0.20\left(\frac{0.20}{0.10}\right)=0.40.

Their sum is $1.1166667$, or $67/60$. Dividing each unrounded weight by that sum gives

q=\left(\frac{25}{67},\frac{18}{67},\frac{24}{67}\right) \approx (0.3731,\,0.2687,\,0.3582).

The second arm is exactly at its target share, but its nominal probability is not necessarily 0.30: the probabilities jointly reflect all arms’ target-to-current ratios. Arm 3 is under-enrolled relative to its target and receives a higher probability than its target share.

Worked constraints at the probability level

Worked Example: An upper probability cap on one arm

Suppose the target shares are $(0.50,0.30,0.20)$, the current shares are $(0.60,0.25,0.15)$, and $\gamma=1$. The operational rule limits the next-participant probability for arm 1 to 0.38. Find the KL-projected probability vector.

All shares are positive and sum to one. The nominal weights are

w_1=0.50\left(\frac{0.50}{0.60}\right)=\frac{5}{12},\quad w_2=0.30\left(\frac{0.30}{0.25}\right)=0.36,\quad w_3=0.20\left(\frac{0.20}{0.15}\right)=\frac{4}{15}.

Using the exact weights, their sum is $1.0433333=313/300$, so the nominal probability vector is

q=\left(\frac{125}{313},\frac{108}{313},\frac{80}{313}\right) \approx(0.3994,\,0.3450,\,0.2556).

The nominal probability $q_1=125/313\approx0.3994$ exceeds the cap. Under the stated KL projection, arm 1 is fixed at 0.38. The remaining mass is $1-0.38=0.62$, and the two free arms retain their nominal ratio $108:80$. Therefore

p_2=0.62\left(\frac{108}{108+80}\right)\approx0.3562,\qquad p_3=0.62\left(\frac{80}{108+80}\right)\approx0.2638, \qquad p=(0.3800,\,0.3562,\,0.2638).

The probabilities sum to one, arm 1 meets its cap, and the relative probabilities of the two free arms are unchanged. This is a cap on the next assignment probability; it is not a claim that arm 1’s eventual enrollment share will be 0.38.

Worked Example: A target-share cap can leave a larger probability

Suppose an operational plan limits the target share for arm 2 to 0.50. Start from the proposed target $(0.20,0.60,0.20)$. For this example, specify that the excess 0.10 is redistributed between arms 1 and 3 in their original ratio $1:1$. The constrained target is then $\rho=(0.25,0.50,0.25)$. At the decision point, let current shares be $(0.50,0.30,0.20)$ and let $\gamma=1$. Calculate the nominal DBCD probabilities under the constrained target.

The target transformation is explicitly defined: subtract 0.10 from arm 2 and add 0.05 to each other arm. The resulting target and current share vectors are positive and sum to one. The weights are

w_1=0.25\left(\frac{0.25}{0.50}\right)=0.125,\quad w_2=0.50\left(\frac{0.50}{0.30}\right)=0.8333333,\quad w_3=0.25\left(\frac{0.25}{0.20}\right)=0.3125.

The sum is $1.2708333=61/48$, yielding

q=\left(\frac{6}{61},\frac{40}{61},\frac{15}{61}\right) \approx(0.09836,\,0.6557,\,0.2459).

Although the constrained target share for arm 2 is 0.50, its current share is only 0.30, so the DBCD probability for the next participant is about 0.6557. This is not a contradiction: the cap was imposed on the target, not on the per-step probability. If the operational requirement instead says $p_2\leq0.50$ at every decision, a probability-level constraint must also be applied.

Choosing and checking the constraint

Before calculating, translate the operational language into the object being constrained. “No more than 40% of the next participants should be assigned to arm 1” is ambiguous: it could refer to a per-step probability cap, a cumulative enrollment share, or a target share. These lead to different mathematical rules. A per-step probability cap is directly enforceable by constraining $p(n)$. A target-share limit modifies $\rho$. A cumulative enrollment requirement concerns the sequence of assignments and is not guaranteed solely by a one-step cap.

Restriction applies to	What is directly changed	What it does not guarantee
Next-participant probability	The feasible vector $p(n)$, obtained from $q(n)$ under specified per-step bounds	A fixed final enrollment share, especially under early stopping or variable enrollment
Target allocation share	The reference vector $\rho$, followed by recalculation of DBCD weights	The same numerical bound on $q_k(n)$ or on the next assignment probability
Final or cumulative enrollment share	A sequence-level requirement on realized counts	Exact attainment from randomization alone

Useful diagnostics are straightforward. At each decision, verify that nominal weights are positive and finite, that nominal probabilities sum to one within numerical tolerance, that the constrained probabilities sum to one, and that every probability bound is met. Also record which bounds are active. If the algorithm silently renormalizes after clipping a probability, check the final vector against all bounds: renormalization can push another arm outside its limit. If no feasible vector exists, or repeated active-set updates do not terminate, inspect the bounds and implementation rather than interpreting the output as a valid randomization distribution.

A probability projection preserves the nominal DBCD vector only when it is already feasible. Once constraints bind, the resulting probabilities generally differ from the unconstrained DBCD rule. A target transformation likewise changes the DBCD inputs and thus the subsequent allocation path. Which choice is appropriate depends on the operational objective, and any redistribution rule for target shares should be specified rather than inferred from the phrase “cap the target.”

Common pitfall

A common error is to observe that an arm’s target share is 0.40 and conclude that its next-participant probability cannot exceed 0.40. That reasoning treats a long-run reference share as if it were a one-step probability bound. In DBCD, current under-enrollment can make the nominal probability larger than the target, as the third worked example demonstrates. The correction is to identify the constrained quantity first: constrain the target vector if the requirement is about $\rho$, or constrain and renormalize the per-step probability vector if it is about the next assignment. Do not claim a cumulative enrollment guarantee unless the design explicitly enforces one.

Key takeaway: Compute the nominal DBCD probabilities from target shares and current enrollment shares first. Apply operational probability bounds to that probability vector using a stated feasible rule; apply target-share limits to the target vector and then recompute. The two constraints answer different design questions.

Check Your Understanding

For each problem, identify whether the restriction concerns a target share or a next-participant probability, then calculate or assess what is requested.

With target shares $(0.50,0.30,0.20)$, current shares $(0.60,0.30,0.10)$, and $\gamma=1$, calculate the nominal probability vector. Is a stated target of 0.50 for arm 1 itself a per-step cap?
A nominal vector is $q=(0.50,0.30,0.20)$, and a KL probability constraint requires $p_1\leq0.40$, with no other active bounds. Calculate $p$.
Can three arms have probability lower bounds $(0.40,0.40,0.30)$? Explain using the feasibility condition.
A design imposes $\rho_2\leq0.50$. Must the next-participant probability $p_2(n)$ be at most 0.50? Explain.

Solutions. (1) The weights are $5/12,0.30,0.40$, with sum $67/60$; hence $q=(25/67,18/67,24/67)\approx(0.3731,0.2687,0.3582)$. The target value is not itself a per-step cap. (2) Fix arm 1 at 0.40; the remaining mass is 0.60. The free-arm ratio is $0.30:0.20=3:2$, so $p=(0.40,0.36,0.24)$. (3) No. The lower bounds sum to $1.10>1$, so no probability vector can satisfy them. (4) No. The restriction concerns the target; current under-enrollment can make the next-participant probability exceed the target share unless a separate probability constraint is imposed.

Restriction applies to	What is directly changed	What it does not guarantee
Next-participant probability	The feasible vector \(p(n)\), obtained from \(q(n)\) under specified per-step bounds	A fixed final enrollment share, especially under early stopping or variable enrollment
Target allocation share	The reference vector \(\rho\), followed by recalculation of DBCD weights	The same numerical bound on \(q_k(n)\) or on the next assignment probability
Final or cumulative enrollment share	A sequence-level requirement on realized counts	Exact attainment from randomization alone