Allocation Floors and Caps

Why Put Bounds on Adaptive Allocation?

An adaptive randomization rule can increasingly favor an arm as evidence accumulates. That responsiveness may be useful, but an allocation probability that becomes very small can leave an arm with little opportunity to accrue participants; one that becomes very large can sharply reduce allocation to the alternatives. A protocol can limit these extremes by specifying a minimum assignment probability, or floor, and a maximum assignment probability, or cap, for each arm.

The task is not simply to replace any probability below a floor or above a cap. Assignment probabilities must still sum to one. A constrained rule therefore needs to redistribute the probability mass released or required by the bounds. The method here does that by scaling the starting probabilities for arms that are not held at a floor or cap.

Definition: Suppose there are $K$ arms and an adaptive rule has produced base probabilities $q_1,\ldots,q_K$, where $q_i>0$ and $\sum_{i=1}^K q_i=1$. For arm $i$, let $\ell_i$ be its floor and $u_i$ its cap, with $0\leq \ell_i\leq u_i\leq 1$. The constrained probabilities are defined by $$ \pi_i=\min\{u_i,\max\{\ell_i,cq_i\}\},\qquad i=1,\ldots,K, $$ where $c>0$ is chosen so that $\sum_{i=1}^K\pi_i=1$. Thus, each scaled base probability $cq_i$ is raised to its floor if it is too small and lowered to its cap if it is too large.

The base probabilities may come from a response-adaptive rule such as probability matching, an expected-utility rule, or a softmax rule. This tutorial concerns the additional constraint step, not how the base probabilities are calculated. At an allocation update, the trial uses $\pi_i$ as the probability of assigning the next eligible participant to arm $i$.

Feasibility and the Scaling Constant

Before calculating $c$, check that the bounds permit any probability vector at all. The necessary and sufficient feasibility condition is $$ \sum_{i=1}^K \ell_i\leq 1\leq \sum_{i=1}^K u_i. $$ The floors cannot collectively demand more than the available total probability, and the caps must collectively allow at least that total. If either inequality fails, no choice of probabilities can satisfy all bounds and sum to one; the design specifications must be changed.

For feasible bounds and strictly positive base probabilities, define the total after scaling and clipping as $$ S(c)=\sum_{i=1}^K \min\{u_i,\max\{\ell_i,cq_i\}\}. $$ Each term is continuous and nondecreasing in $c$, so $S(c)$ is continuous and nondecreasing. At $c=0$, $S(c)=\sum_i\ell_i\leq1$; as $c$ becomes sufficiently large, every arm reaches its cap and $S(c)=\sum_i u_i\geq1$. Therefore a value of $c$ exists for which $S(c)=1$. In ordinary cases the solution can be found by checking which arms are at bounds and solving a linear equation. If all arms are fixed at bounds and their probabilities sum to one, the resulting vector is determined even if more than one value of $c$ yields it.

Key calculation: Let $L$ be the set of arms at their floors, $U$ the set at their caps, and $I$ the set strictly between their bounds. For a proposed classification of the arms, normalization requires $$ \sum_{i\in L}\ell_i+\sum_{i\in U}u_i+c\sum_{i\in I}q_i=1. $$ If $\sum_{i\in I}q_i>0$, solve this equation for $c$, then verify that every arm really satisfies its proposed floor, cap, or interior condition.

Worked Calculations

Worked Example: Bounds That Do Not Change the Allocation

Suppose a hypothetical three-arm trial has base probabilities $q=(0.50,0.30,0.20)$, floors $\ell=(0.10,0.25,0.10)$, and caps $u=(0.70,0.60,0.60)$. Find the constrained probabilities.

The bounds are feasible: the floors sum to $0.45$, and the caps sum to $1.90$. Test $c=1$. The scaled probabilities are $(0.50,0.30,0.20)$, and each is within its arm's bounds. In particular, the second arm's probability $0.30$ is above its floor $0.25$, so it is not at the floor.

The total is already one: $$ S(1)=0.50+0.30+0.20=1. $$ Therefore $c=1$, and the constrained allocation is $\pi=(0.50,0.30,0.20)$. Bounds need not alter an allocation when the base probabilities already satisfy them.

Worked Example: Raising an Arm to Its Floor

For a hypothetical trial, let $q=(0.60,0.25,0.15)$, $\ell=(0.10,0.30,0.10)$, and $u=(0.80,0.70,0.60)$. The second arm's base probability is below its floor. Suppose arms 1 and 3 remain interior while arm 2 is at its floor. Normalization then requires $$ 0.30+c(0.60+0.15)=1. $$

Solving gives $c=0.70/0.75=0.9333$. The scaled values are $cq=(0.5600,0.2333,0.1400)$. The second value is below its floor and is replaced by $0.30$; the other two values are between their respective floors and caps. Thus $$ \pi=(0.5600,0.3000,0.1400),\qquad 0.5600+0.3000+0.1400=1.000. $$

This verifies the proposed classification as well as the total. The probability taken from the scaled value for arm 2 is redistributed proportionally through the common multiplier $c$ among the interior arms.

Worked Example: A Cap and a Floor at the Same Time

Consider four hypothetical arms with $q=(0.55,0.25,0.15,0.05)$, floors $\ell=(0.05,0.10,0.10,0.10)$, and caps $u=(0.40,0.60,0.50,0.30)$. Suppose arm 1 is capped and arm 4 is at its floor, while arms 2 and 3 are interior. Normalization gives $$ 0.40+0.10+c(0.25+0.15)=1. $$

Thus $c=0.50/0.40=1.25$. The scaled values are $(0.6875,0.3125,0.1875,0.0625)$. The first exceeds its cap $0.40$, the fourth is below its floor $0.10$, and the second and third are within their bounds. Clipping accordingly gives $$ \pi=(0.4000,0.3125,0.1875,0.1000). $$ The total is $0.4000+0.3125+0.1875+0.1000=1.0000$, and the assumed active bounds are consistent with the calculated multiplier.

Why the Active-Set Check Matters

A convenient calculation can still be wrong if the assumed set of arms at floors and caps is inconsistent with the value of $c$. For a proposed floor arm, check that $cq_i\leq\ell_i$; for a proposed cap arm, check that $cq_i\geq u_i$; for a proposed interior arm, check that $\ell_i<cq_i<u_i$. At equality, either adjacent classification gives the same probability, but the total must still equal one.

The consistency check is essential because the equation for $c$ depends on which arms are treated as fixed. If an arm classified as being at its floor actually has $cq_i>\ell_i$, the equation wrongly removes its contribution from the scaled group. Recalculate with the correct active set rather than accepting a plausible-looking total.

Check feasibility.
Confirm that the sum of floors is at most one and the sum of caps is at least one.

Propose the active bounds.
Identify arms expected to be held at a floor or cap; treat the others as interior.

Solve normalization.
Use the fixed bound probabilities and scale the base probabilities for the interior arms so that the total is one.

Verify and report.
Check every arm against its proposed bound status, then check that the constrained probabilities sum to one.

Common Mistake, Diagnostics, and Alternatives

A frequent error is to clip the base probabilities at the bounds and then renormalize the clipped values. Renormalization changes every value, including those that were supposed to remain fixed at a cap or floor. For example, take base probabilities $(0.70,0.20,0.10)$ with a cap of $0.60$ on arm 1. Clipping alone gives $(0.60,0.20,0.10)$, whose sum is $0.90$. Dividing by $0.90$ produces approximately $(0.6667,0.2222,0.1111)$, which violates the cap. The correction is to find $c$ while keeping the capped arm at $0.60$; the other arms must receive the remaining $0.40$ in proportion to their base weights.

Useful checks in implementation and protocol review include the following:

Verify feasibility of every specified set of bounds before trial opening.
At every update, check that each $\pi_i$ lies between $\ell_i$ and $u_i$ and that $\sum_i\pi_i=1$, up to a stated numerical tolerance.
Record which arms are at bounds and check that their status agrees with the multiplier and scaled probabilities.
Check that the software handles equality at a bound consistently and does not introduce rounding errors large enough to change the total materially.

One alternative is fixed randomization, which uses pre-specified probabilities rather than adapting base weights to accumulating information. Another is simple truncation followed by renormalization, but that does not generally preserve the specified caps and floors. Proportional clipping instead retains the relative ratios $q_i/q_j$ for any two interior arms, because $\pi_i=cq_i$ and $\pi_j=cq_j$. Once an arm reaches a bound, its probability no longer follows that proportional scaling.

Floors and caps constrain the probability of an assignment at a given update; they do not guarantee exact realized arm counts. Random allocation can produce counts above or below their expectations, especially in small samples. Bounds also do not by themselves ensure adequate power, balance on prognostic variables, or protection against bias from other features of a trial. Their values should therefore be justified in the design and implemented as part of a fully specified allocation rule.

Key takeaway: Constrained adaptive allocation is a normalization problem with bounds. Scale the base probabilities by one common multiplier, clip at arm-specific floors and caps, and verify feasibility, active-set consistency, and the unit total.

Check Your Understanding

For each problem, first identify what the constraints require. Solutions follow the questions.

A three-arm rule has floors $(0.20,0.20,0.20)$ and caps $(0.50,0.40,0.40)$. Are these bounds feasible?
With base probabilities $(0.50,0.30,0.20)$, floors $(0.10,0.25,0.10)$, and caps $(0.70,0.60,0.60)$, is the second arm at its floor when $c=1$? What is the constrained allocation?
For base probabilities $(0.50,0.30,0.20)$, suppose arm 1 is at cap $0.40$, arms 2 and 3 are interior, and neither is at a floor. Find $c$ and the allocation for this active-set assumption.
Why is it incorrect to clip probabilities at their bounds and then divide all the clipped values by their sum?

Solutions. (1) The floors sum to $0.60$, and the caps sum to $1.30$, so $0.60\leq1\leq1.30$; the bounds are feasible. (2) At $c=1$, the second arm has probability $0.30$, which is greater than its floor $0.25$. All base probabilities already satisfy their bounds and sum to one, so $\pi=(0.50,0.30,0.20)$. (3) Normalization gives $0.40+c(0.30+0.20)=1$, so $c=0.60/0.50=1.20$. The interior values are $0.36$ and $0.24$, yielding $\pi=(0.40,0.36,0.24)$; both are interior if their floors and caps contain those values. (4) Dividing by the clipped total changes the values that were clipped, so a cap can be exceeded or a floor can be violated. Instead, solve for the common multiplier with the bound arms held fixed, then verify the resulting probabilities.

What You'll Learn