Preserving Control Allocation

Why Reserve Allocation for Concurrent Control?

Response-adaptive allocation can favor experimental arms when current information makes their outcomes look promising. That may increase assignment to those arms, but it can also leave relatively few participants assigned to concurrent control. A design may therefore specify a minimum probability of assigning each new participant to control, regardless of how the adaptive rule would otherwise allocate.

The key is to impose that reserve without needlessly changing the adaptive rule’s comparisons among experimental arms. If the control reserve binds, the experimental probabilities can be reduced by a common factor. Their relative allocation ratios then remain the same, while the control probability meets the prespecified minimum.

Definition: Suppose an adaptive rule gives a raw allocation vector $q=(q_C,q_1,\ldots,q_K)$, where $q_C$ is the probability of assignment to concurrent control and $q_j$ is the probability of assignment to experimental arm $j$. The probabilities are nonnegative and sum to one. For a specified control reserve $\rho$, with $0\leq \rho<1$, the control-preserving rule sets $$ p_C=\max(q_C,\rho). $$ If $q_C\geq \rho$, set $p_j=q_j$ for every experimental arm. If $q_C<\rho$, set $$ p_j=(1-\rho)\frac{q_j}{1-q_C},\qquad j=1,\ldots,K. $$ Here $p_C,p_1,\ldots,p_K$ are the probabilities actually used for randomization.

The raw probabilities may come from probability matching, expected utility, a softmax rule, or another adaptive procedure. The reserve is a constraint applied to those probabilities; it is not itself a method for estimating treatment effects or choosing the raw allocation. The rule assumes that there is one concurrent-control category and at least one experimental arm. It also requires $0\leq\rho<1$, so a positive amount of probability remains available to experimental arms.

Why the Rescaling Works

When $q_C<\rho$, the experimental arms together have raw probability $1-q_C$. The reserve gives them total probability $1-\rho$. Multiplying every experimental probability by the same factor, $(1-\rho)/(1-q_C)$, changes their total to $1-\rho$ but keeps their proportions unchanged.

\sum_{j=1}^{K}p_j = \sum_{j=1}^{K}(1-\rho)\frac{q_j}{1-q_C} = \frac{1-\rho}{1-q_C}\sum_{j=1}^{K}q_j = \frac{1-\rho}{1-q_C}(1-q_C) = 1-\rho.

Thus $p_C+\sum_{j=1}^{K}p_j=\rho+(1-\rho)=1$ when the reserve binds. For any two experimental arms $j$ and $k$ with positive raw probabilities, their ratio is unchanged:

\frac{p_j}{p_k} = \frac{(1-\rho)q_j/(1-q_C)}{(1-\rho)q_k/(1-q_C)} = \frac{q_j}{q_k}.

If $q_C\geq\rho$, no adjustment is needed: the raw vector already satisfies the minimum. In either case, the implemented control probability is at least $\rho$. When the reserve binds, it equals $\rho$; otherwise, it remains $q_C$. This rule therefore preserves the raw adaptive allocation unless the control constraint requires a change.

Worked Example: Rescaling Three Experimental Arms

Suppose a hypothetical adaptive rule gives raw probabilities $q_C=0.12$, $q_1=0.48$, $q_2=0.27$, and $q_3=0.13$. The probabilities sum to one. A protocol specifies a concurrent-control reserve of $\rho=0.25$. Since $0.12<0.25$, the reserve binds.

The experimental arms have total raw probability $1-0.12=0.88$, and their required total after reserving control is $1-0.25=0.75$. The common rescaling factor is $0.75/0.88$. Therefore,

p_C=0.25,\qquad p_1=0.75\frac{0.48}{0.88}=0.4091,\qquad p_2=0.75\frac{0.27}{0.88}=0.2301,\qquad p_3=0.75\frac{0.13}{0.88}=0.1108.

The total is $0.25+0.4091+0.2301+0.1108=1.0000$, up to rounding. Before adjustment, the ratio of the first two experimental probabilities was $0.48/0.27=1.778$. After adjustment, it is $0.4091/0.2301\approx1.778$. The reserve reduces experimental allocation overall but does not change their relative allocation.

Worked Example: When the Reserve Does Not Bind

In a hypothetical update, suppose the raw probabilities are $q_C=0.34$, $q_1=0.40$, and $q_2=0.26$, with reserve $\rho=0.25$. The probabilities sum to one, and $q_C\geq\rho$. The implemented probabilities are therefore unchanged: $p_C=0.34$, $p_1=0.40$, and $p_2=0.26$.

The control reserve is a minimum, not a target that control must equal. Reducing control from $0.34$ to $0.25$ would change the raw adaptive allocation even though no constraint requires it. Under the stated rule, the experimental arms retain their raw probabilities and control remains above its minimum.

Worked Example: Translating a Probability Reserve Into Expected Assignments

Consider a hypothetical trial with 100 sequential assignments and a control reserve of $\rho=0.30$. Suppose the reserve is applied at every assignment, using the information available at that point. Let $I_t$ equal 1 if participant $t$ is assigned to control and 0 otherwise. At each assignment, conditional on the information available immediately before randomization, $P(I_t=1\mid\text{current information})\geq0.30$.

By adding the expected indicators,

E\left[\sum_{t=1}^{100}I_t\right] = \sum_{t=1}^{100}E(I_t) \geq 100(0.30) = 30.

Thus the expected number assigned to control is at least 30. This does not guarantee that at least 30 of the 100 participants will actually receive control: random assignment can produce a realized count below its expectation. The guarantee is about each participant’s conditional assignment probability, not a deterministic final count.

Choosing and Implementing the Reserve

The value of $\rho$ should be specified in the design, along with the rule for applying it. It represents a design constraint, not a value that should be changed informally in response to interim results. The choice can reflect the need to continue collecting concurrent-control data while retaining adaptive allocation among experimental arms. There is no universally appropriate reserve: a larger value gives control more assignment probability whenever the constraint binds, and leaves less probability for all experimental arms combined.

At each update, the allocation calculation can be implemented as a short sequence:

Obtain the raw probabilities.
Compute $q_C,q_1,\ldots,q_K$ using the prespecified adaptive rule and check that they are nonnegative and sum to one.

Compare raw control allocation with the reserve.
If $q_C\geq\rho$, use the raw vector. If $q_C<\rho$, set $p_C=\rho$.

Rescale experimental allocation if needed.
When $q_C<\rho$, multiply each $q_j$ by $(1-\rho)/(1-q_C)$.

Check the result before randomizing.
Confirm that the probabilities sum to one, $p_C\geq\rho$, and that experimental-arm ratios are preserved when the reserve binds.

A useful diagnostic is to inspect both raw and implemented probabilities across updates. If the reserve binds frequently, the constrained allocation may differ substantially from the unconstrained adaptive rule. That can be appropriate, but it should be visible in design review and monitoring. Also check boundary cases, rounding, and any additional allocation constraints. For example, this control reserve alone does not ensure a minimum allocation probability for any experimental arm.

What the Reserve Does and Does Not Guarantee

A per-assignment reserve is stronger than a requirement on the average raw probability: it applies at each randomization, conditional on the information used to make that assignment. Consequently, if it is applied at all $N$ assignments, the expected control count is at least $N\rho$. It does not fix the realized control count, guarantee balance within a particular subgroup, or ensure a specified amount of information for every comparison.

Nor does a control reserve make the adaptive rule’s conclusions more reliable by itself. It changes the allocation probabilities, not the likelihood, prior, estimand, or analysis model. Its practical effect on inference depends on the outcomes, the allocation history, the number of participants, and the analysis plan. Those features should be considered when evaluating a design; the probability constraint should not be mistaken for a guarantee about statistical precision.

The rule also presumes that the raw allocation assigns some probability to experimental treatment when the reserve binds: $q_C<\rho<1$ implies $1-q_C>0$, so the rescaling denominator is positive. If raw probabilities for particular experimental arms are zero, rescaling leaves them zero. A control reserve cannot revive an arm that the underlying allocation rule has already assigned probability zero.

Comparison With Other Allocation Choices

Approach	What it does	How it differs from a control reserve
Control-preserving floor	Raises control to $\rho$ only when raw control probability is below $\rho$, and proportionally rescales experimental arms.	Leaves the raw allocation unchanged when it already satisfies the minimum.
Fixed control ratio	Assigns control a specified probability at every update.	Can reduce control even when the adaptive rule would assign more than the minimum; a floor does not require control to equal $\rho$.
Unconstrained adaptive allocation	Uses the raw probabilities without a control constraint.	May assign control less than the design’s required minimum.
Separate experimental-arm constraints	Imposes lower or upper limits on probabilities for experimental arms.	Addresses arm-specific allocation restrictions; it is not implied by reserving control.

Common Mistakes

Setting control to the reserve and leaving the experimental probabilities untouched. If the raw probabilities sum to one, this makes the new total exceed one. When the reserve binds, the experimental probabilities must be rescaled to sum to $1-\rho$.
Taking the maximum and then normalizing every entry. This does not generally preserve the intended control floor. For example, with $q_C=0.12$ and $\rho=0.25$, replacing $q_C$ by $0.25$ while leaving the experimental entries unchanged gives a total of $1.13$. Normalizing that vector would make control $0.25/1.13\approx0.2212$, below the required $0.25$. Set control to the reserve and rescale the experimental allocation to the remaining probability instead.
Calling the reserve a guaranteed control count. The constraint is on randomization probabilities; the realized number assigned to control remains random.
Treating the reserve as an equality. If the raw control probability exceeds $\rho$, the rule leaves it above the reserve.

Key takeaway: A control reserve is imposed by setting $p_C=\max(q_C,\rho)$. When raw control allocation falls below the reserve, assign the remaining probability $1-\rho$ to experimental arms in proportion to their raw probabilities. This guarantees the per-randomization control minimum while preserving relative experimental allocation.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Raw probabilities are $q_C=0.20$, $q_1=0.50$, and $q_2=0.30$, with $\rho=0.25$. What are the implemented probabilities?
Raw probabilities are $q_C=0.35$, $q_1=0.45$, and $q_2=0.20$, with $\rho=0.25$. Does the rule change them?
If the reserve binds, what happens to the ratio $p_1/p_2$ compared with $q_1/q_2$, assuming both raw probabilities are positive?
A trial applies a reserve of $0.20$ at each of 80 assignments. What lower bound does this give on the expected control count, and what does it not guarantee?
Why is $\rho=1$ not a suitable reserve when experimental arms must remain available for randomization?

Solutions. (1) The raw control probability is below the reserve. The factor is $(1-0.25)/(1-0.20)=0.75/0.80=0.9375$, so $p_C=0.25$, $p_1=0.50(0.9375)=0.46875$, and $p_2=0.30(0.9375)=0.28125$. These sum to one. (2) No. Since $q_C=0.35\geq0.25$, use $(0.35,0.45,0.20)$ unchanged. (3) The ratio is unchanged: $p_1/p_2=q_1/q_2$, because both experimental probabilities are multiplied by the same factor. (4) The expected control count is at least $80(0.20)=16$. The reserve does not guarantee at least 16 realized control assignments. (5) A reserve of one leaves zero probability for all experimental arms, since the probability remaining for them is $1-\rho=0$.

Approach	What it does	How it differs from a control reserve
Control-preserving floor	Raises control to \(\rho\) only when raw control probability is below \(\rho\), and proportionally rescales experimental arms.	Leaves the raw allocation unchanged when it already satisfies the minimum.
Fixed control ratio	Assigns control a specified probability at every update.	Can reduce control even when the adaptive rule would assign more than the minimum; a floor does not require control to equal \(\rho\).
Unconstrained adaptive allocation	Uses the raw probabilities without a control constraint.	May assign control less than the design’s required minimum.
Separate experimental-arm constraints	Imposes lower or upper limits on probabilities for experimental arms.	Addresses arm-specific allocation restrictions; it is not implied by reserving control.

What You'll Learn