Balancing Prognosis While Adapting to Responses

Why Combine Prognostic Balance and Response Adaptation?

A response-adaptive design uses outcomes observed so far to alter the probability of assigning the next participant to each treatment. A covariate-adaptive design instead uses the incoming participant’s baseline characteristics and the accrued allocation counts to encourage prognostic balance. These aims can point in different directions: the response component may favor one treatment, while the balance component may favor the other. A combined rule makes that trade-off explicit while retaining randomized assignment.

The design below combines two probabilities. The first favors the treatment that would reduce a prespecified covariate imbalance score. The second is the response-driven probability produced by a prespecified outcome-model allocation procedure. A fixed mixing weight determines how much influence the balance component has. The example uses two treatments, A and B; the same principle can be extended, but this tutorial focuses on the two-treatment rule.

Definition: Let $S_A$ and $S_B$ be the covariate imbalance scores that would result if the incoming participant were assigned to A or B. Let $b_A$ be the balance-component probability of assigning A: set $b_A=1-\rho$ if $S_A>S_B$, $b_A=\rho$ if $S_A<S_B$, and $b_A=0.5$ if the scores tie, where $0.5<\rho<1$. Let $q_A$ be the response-component probability of assigning A. For a prespecified balance weight $0\leq\lambda\leq1$, the combined probability is $p_A=\lambda b_A+(1-\lambda)q_A$. Assign B with probability $1-p_A$.

The value $\rho$ controls how strongly the balance component favors the lower-score assignment. The value $\lambda$ controls the relative influence of balance and response adaptation. Both should be specified before enrollment. For example, with $\rho=0.8$, the balance component assigns probability 0.8 to the lower-score treatment rather than making that component deterministic. If $\lambda=0.5$, the combined rule gives the balance and response components equal weight.

Defining the Prognostic Balance Component

For a categorical prognostic factor with levels $k=1,\ldots,K$, let $n_{Ak}$ and $n_{Bk}$ be the numbers already assigned to A and B in level $k$. One simple imbalance score is the sum of the absolute within-level differences:

S=\sum_{k=1}^{K} w_k\lvert n_{Ak}-n_{Bk}\rvert,

where each prespecified weight $w_k$ is nonnegative. To calculate $S_A$, update the count for the incoming participant’s level as if that participant received A, then calculate the score. Calculate $S_B$ in the same way under assignment to B. The lower score is the assignment that better improves—or less worsens—the defined balance criterion. More elaborate scores can include multiple factors or levels; the choice of score and weights must be settled before the rule is used.

This score is a design device, not a measure of treatment effect. It compares possible allocation histories for the next participant. It does not guarantee that the final groups will be exactly balanced, and it only reflects covariates and weights included in the score.

Worked Example: Calculating the Balance Preference

Suppose a trial uses one binary prognostic factor, risk group, with high- and low-risk levels. Each level has weight 1. Before the next assignment, the counts are high risk: A = 2 and B = 1; low risk: A = 1 and B = 1. The incoming participant is high risk.

If this participant receives A, the high-risk counts become 3 and 1, and the low-risk counts remain 1 and 1. Thus $S_A=\lvert3-1\rvert+\lvert1-1\rvert=2+0=2$.

If the participant receives B, the high-risk counts become 2 and 2, and the low-risk counts remain 1 and 1. Thus $S_B=\lvert2-2\rvert+\lvert1-1\rvert=0+0=0$. B has the lower score.

With $\rho=0.8$, the balance component therefore sets $b_A=1-\rho=0.2$ and $b_B=0.8$. It favors B, but it still leaves a 0.2 probability of assigning A through this component.

Combining the Components

The response component must be defined by the trial’s response-adaptive procedure. In this tutorial, $q_A$ denotes the probability that procedure would assign A, based on outcomes observed and processed before the next randomization. For covariate-adjusted adaptation, the outcome model can account for baseline covariates and produce allocation targets using the prespecified rule. The combined design uses that response probability as an input; it does not replace the outcome model or determine how that model is fitted.

The mixing rule can be implemented in either of two equivalent ways. One can calculate $p_A$ directly and draw a random assignment from it. Alternatively, one can first select the balance component with probability $\lambda$, otherwise select the response component, and then draw from the selected component’s assignment probability. The second implementation helps explain why the combined probability is a weighted average.

Justification: Let $C$ indicate which component is selected, with $P(C=\text{balance})=\lambda$. The probability of assigning A is the sum of the probabilities of assigning A through either component: $$ P(A)=P(A\mid C=\text{balance})P(C=\text{balance})+P(A\mid C=\text{response})P(C=\text{response}) =\lambda b_A+(1-\lambda)q_A. $$ Because $b_A$ and $q_A$ are probabilities in $[0,1]$, and $0\leq\lambda\leq1$, their weighted average is also in $[0,1]$. The probability of B is $1-p_A$, so the rule defines a valid randomized assignment.

This justification establishes the probability calculation, not the performance of a particular design. How well the rule balances covariates or adapts to outcomes depends on the score, the response procedure, the weight $\lambda$, and the trial’s enrollment and outcome process.

Worked Enrollment Sequence

Consider a hypothetical trial with the same high- and low-risk factor, equal score weights, $\rho=0.8$, and $\lambda=0.5$. Before the sequence begins, five participants have enrolled: three on A and two on B. The allocation counts are high risk: A = 2, B = 1; low risk: A = 1, B = 1. The response procedure fits its prespecified covariate-adjusted model using outcomes available so far. For illustration, the model’s standardized predicted success probabilities are $\theta_A=0.60$ and $\theta_B=0.50$.

Suppose the response rule for this illustration maps the predicted difference to $q_A=\min(0.8,\max(0.2,0.5+2(\theta_A-\theta_B)))$. This deliberately bounded mapping is part of the hypothetical protocol, not a universal response-adaptive rule. Initially, $q_A=0.5+2(0.60-0.50)=0.70$. The balance score for the next high-risk participant was calculated above, giving $b_A=0.20$.

Worked Example: Tracing Three Consecutive Assignments

Assignment 1: high-risk participant. The response component gives $q_A=0.70$; the balance component gives $b_A=0.20$. Therefore,

$$ p_A=0.5(0.20)+0.5(0.70)=0.10+0.35=0.45. $$

A random number $U=0.30$, drawn uniformly between 0 and 1, is below 0.45, so the participant is assigned to A. Suppose the participant’s outcome is a failure. The high-risk allocation counts are now A = 3 and B = 1. The model is refitted under the protocol; for this hypothetical sequence, it reports $\theta_A=0.60$ and $\theta_B=0.35$.

Assignment 2: low-risk participant. Before this assignment, the low-risk counts are A = 1 and B = 1. Assigning this participant to either arm creates a low-risk difference of 1, while the high-risk difference is already 2. Thus $S_A=S_B=3$, so the balance scores tie and $b_A=0.50$. The response probability is $q_A=0.5+2(0.60-0.35)=1.00$, truncated to the protocol’s upper bound of 0.80. Hence,

$$ p_A=0.5(0.50)+0.5(0.80)=0.25+0.40=0.65. $$

If $U=0.40$, the assignment is A because $0.40<0.65$. Suppose this participant has a successful outcome. The low-risk allocation counts become A = 2 and B = 1. Under the stipulated model update for this illustration, the next standardized predictions are $\theta_A=0.65$ and $\theta_B=0.35$.

Assignment 3: low-risk participant. Assigning the incoming participant to A would make the low-risk counts 3 versus 1, while assigning to B would make them 2 versus 2. The high-risk difference remains 2 either way. Thus $S_A=2+2=4$ and $S_B=2+0=2$; B is favored, giving $b_A=0.20$. The response rule gives $q_A=0.5+2(0.65-0.35)=1.10$, truncated to 0.80. The combined probability is

$$ p_A=0.5(0.20)+0.5(0.80)=0.10+0.40=0.50. $$

If $U=0.25$, this participant is assigned to A. The probability is 0.50 even though the balance component favors B: the two components have equal weight, and their probabilities lie equally far on either side of 0.50. The random draw, rather than either component alone, determines the realized assignment.

What the Weight Changes

For a given participant and trial history, the combined probability lies between the component probabilities. If balance favors B with $b_A=0.20$ and the response component favors A with $q_A=0.80$, then $p_A=0.20+0.60(1-\lambda)$. At $\lambda=0$, the rule is purely response-driven and $p_A=0.80$; at $\lambda=1$, it is purely balance-driven and $p_A=0.20$; at $\lambda=0.5$, it is 0.50. Intermediate values compromise between the two components without silently changing either component’s definition.

Worked Example: Comparing Balance Weights

Suppose the balance component gives $b_A=0.20$ and the response component gives $q_A=0.70$. With $\lambda=0.25$, the combined probability is $p_A=0.25(0.20)+0.75(0.70)=0.05+0.525=0.575$. With $\lambda=0.75$, it is $p_A=0.75(0.20)+0.25(0.70)=0.15+0.175=0.325$.

Increasing the balance weight moves the probability toward the balance component, which favors B in this example. It does not make balance certain: with $\lambda=0.75$, A still has probability 0.325. Nor does a larger balance weight guarantee better final balance in every realized trial; it changes the assignment probabilities, while the randomization still permits different allocation sequences.

Checks, Alternatives, and Limitations

Before enrollment, verify that the score is computed consistently under both candidate assignments, that the response probability is based only on outcomes available at the decision time, and that $\rho$, $\lambda$, and any bounds on $q_A$ are documented. During enrollment, retain the component probabilities, the combined probability, the covariate counts used for the score, the response-model output, and the randomization result. These records make it possible to check whether the implemented probability matches the protocol.

A common mistake is to calculate a balance preference and a response preference, then choose whichever one seems more compelling for the current participant. That is not the specified combined rule: it replaces the prespecified probability with an ad hoc decision. The correction is to calculate $b_A$, obtain $q_A$, and apply the same mixing formula at every eligible randomization.

The main alternatives answer different design priorities. Pure minimization uses the balance component alone; it does not adapt allocation in response to outcomes. Pure response adaptation uses $q_A$ alone; it does not incorporate the incoming participant’s effect on the chosen balance score. Stratified randomization uses prespecified strata and randomization within them, rather than continually updating a global score after each assignment. The combined rule explicitly blends two probabilities, but does not remove the design choices or limitations of either component.

In particular, a combined rule cannot repair an unsuitable balance score, unreliable or delayed outcome information, or an inadequately specified response procedure. A score that omits an important prognostic factor cannot balance that factor, and an aggressive response component can still make allocation probabilities highly unequal. Conversely, giving balance a large weight can substantially dilute the intended response adaptation. The two components also do not establish that the treatment with a higher modeled response probability causes better outcomes; the probability is an allocation decision based on the information and assumptions of the design.

Key takeaway: A transparent combined design specifies a covariate score, a randomized balance preference, a response-driven probability, and a fixed mixing weight. For each participant, calculate both component probabilities first, then combine them before making the random assignment.

Check Your Understanding

Answer each question using the definitions and rules in this tutorial. The short solutions follow the questions.

If $S_A=4$, $S_B=2$, and $\rho=0.8$, what is $b_A$, and why?
For $b_A=0.20$, $q_A=0.70$, and $\lambda=0.25$, calculate $p_A$.
If the balance and response components both give $b_A=q_A=0.60$, what is the combined probability for any allowed $\lambda$? Explain.
Name two quantities that should be recorded for checking whether the implemented rule follows the protocol.

Solutions. (1) $b_A=0.20$, because $S_B<S_A$, so the balance component favors B with probability $\rho=0.80$. (2) $p_A=0.25(0.20)+0.75(0.70)=0.05+0.525=0.575$. (3) $p_A=\lambda(0.60)+(1-\lambda)(0.60)=0.60$; averaging identical component probabilities leaves them unchanged. (4) Suitable answers include the two component probabilities, the combined probability, the covariate counts used in the score, and the response-model output.