Bias in Adaptive Arm Means

Why an Adaptive Arm Mean Needs Care

An arm-specific sample mean is familiar: add the observed outcomes in an arm and divide by the number assigned to that arm. Under fixed sampling, or when inclusion is independent of outcomes, it is often a natural estimator of the arm’s response probability. Under response-adaptive randomization, however, a previous outcome can change the probability that a later patient is assigned to the same arm. The final sample size and the observed outcomes can then be related.

This tutorial isolates that mechanism in a two-stage binary-outcome example. The calculation concerns the mean in one treatment arm, not a comparison between treatment arms. Its purpose is to show exactly how an apparently ordinary sample mean can have a nonzero expectation bias under a specified adaptive rule.

Definition: For treatment arm A, let \(N_A\) be the number of patients assigned to A, and let \(Y_i\) be patient \(i\)'s observed binary outcome when assigned to A. The adaptive arm mean is \( \bar{Y}_A=\sum_{i:\,A_i=A}Y_i/N_A \), provided \(N_A>0\). Its expectation bias for the arm response probability \(p\) is \(E(\bar{Y}_A)-p\). A negative bias means the expectation of the estimator is below \(p\); it does not mean every realized estimate is below \(p\).

A Two-Stage Bernoulli Model

Suppose the first patient is assigned to A as an initialization step. The first outcome is \(Y_1\sim\mathrm{Bernoulli}(p)\), where \(p\) is the response probability for patients assigned to A. At stage two, the patient is assigned to A with probability \(q(Y_1)\), where \(q(0)=q_0\) and \(q(1)=q_1\). Otherwise the patient is assigned to another arm, B. Let \(D=1\) if the second patient is assigned to A and \(D=0\) otherwise.

For the calculation, assume the second patient's potential outcome under A, \(Y_2\), is also Bernoulli with response probability \(p\), and is independent of \(Y_1\) and the randomization used to generate \(D\). Outcomes are binary, patients are exchangeable for the A response probability, and one patient's assignment does not change another patient's outcome. The assignment mechanism uses \(Y_1\), but it does not use \(Y_2\), which is not yet observed when the decision is made. These assumptions isolate the effect of response-dependent assignment; they are not a claim that every clinical trial follows this simple model.

The arm mean is always defined here because the first patient is assigned to A. If the second patient also goes to A, it averages two outcomes; otherwise it consists of \(Y_1\) alone. In either case,

\bar{Y}_A=\frac{Y_1+D Y_2}{1+D} =Y_1+\frac{D}{2}(Y_2-Y_1).

The second expression makes the key feature visible: the first outcome remains in the estimate, while the second outcome changes the estimate only when \(D=1\). But the probability of that change depends on the first outcome.

Worked Example: A Rule That Favors A After a Success

Assume the response probability in A is \(p=0.60\). If the first patient responds, assign the second patient to A with probability \(q_1=0.75\); if the first patient does not respond, use \(q_0=0.25\). Find the expectation and bias of the final A arm mean.

From the displayed expression and independence of \(Y_2\) from the stage-two assignment decision conditional on \(Y_1\),

E(\bar{Y}_A) =p+\frac{1}{2}E\{q(Y_1)(p-Y_1)\} =p+\frac{1}{2}p(1-p)(q_0-q_1).

Substituting the values gives

E(\bar{Y}_A) =0.60+\frac{1}{2}(0.60)(0.40)(0.25-0.75) =0.60-0.0600 =0.540.

Thus the bias is \(0.540-0.600=-0.0600\), or −6.00 percentage points. The adaptive rule more often adds a second A outcome after an initial success. That new outcome has expected value 0.60, so averaging it with an initial success tends to pull the arm mean down from 1. The calculation is an expectation over both patient outcomes and the random assignment decision, not a prediction that any particular trial will yield 0.540.

Deriving the Expectation

The expectation formula follows by conditioning on the first outcome. If \(Y_1=0\), the second patient joins A with probability \(q_0\), and the expected change in the mean when that happens is \(p/2\). If \(Y_1=1\), joining A changes the mean by an expected amount \((p-1)/2\), and this occurs with probability \(q_1\). Weighting these cases by their probabilities gives

\begin{aligned} E(\bar{Y}_A) &=(1-p)\left(0+q_0\frac{p}{2}\right) +p\left(1+q_1\frac{p-1}{2}\right)\\ &=\frac{1}{2}p(1-p)q_0+p -\frac{1}{2}p(1-p)q_1\\ &=p+\frac{1}{2}p(1-p)(q_0-q_1). \end{aligned}

Equivalently, subtracting \(p\) shows the bias directly:

E(\bar{Y}_A)-p =\frac{1}{2}p(1-p)(q_0-q_1).

For \(0<p<1\), the sign is determined by \(q_0-q_1\). If A is more likely to receive the second patient after a success (\(q_1>q_0\)), this arm mean has negative bias in this two-stage model. If A is more likely after a failure (\(q_0>q_1\)), the bias is positive. If the probabilities are equal, the bias is zero. The bias is also zero at \(p=0\) or \(p=1\), because outcomes then have no variability.

This derivation is specific to the stated design: one initial A observation, one possible additional A observation, and the sample mean as estimator. The sign and size should not be carried over mechanically to other adaptive rules, longer trials, or estimators.

Worked Example: Adaptation After a Failure

Let \(p=0.40\), and suppose the second patient is assigned to A with probability \(q_0=0.75\) after a failure and \(q_1=0.25\) after a success. Calculate the expectation and interpret the result.

Using the derived expression,

E(\bar{Y}_A) =0.40+\frac{1}{2}(0.40)(0.60)(0.75-0.25) =0.40+0.0600 =0.460.

The bias is \(0.460-0.400=0.0600\), or +6.00 percentage points. In this rule, an initial failure is more likely to be followed by a second A observation. When the first outcome is 0, adding a second outcome with expected value 0.40 tends to raise the average. This is the opposite sign from the previous example, consistent with \(q_0>q_1\).

What Creates the Bias?

The mechanism is not that a Bernoulli outcome is intrinsically a poor summary, nor that the assignment probabilities are necessarily calculated incorrectly. The issue is that the estimator averages a random set of observations whose membership depends on an earlier outcome. The sample size \(N_A=1+D\) and the contribution of \(Y_1\) are linked through the rule for \(D\). Consequently, averaging over trials does not generally reproduce the response probability \(p\).

A useful contrast is an outcome-independent random decision at stage two. If \(q_0=q_1=q\), then \(D\) is independent of \(Y_1\), and the formula reduces to \(E(\bar{Y}_A)=p\). More directly, conditional on either value of \(D\), the observations included in A are independent Bernoulli outcomes with response probability \(p\), so their average has expectation \(p\). This comparison shows why random sample size alone is not enough to establish bias: the relationship between inclusion and outcomes matters.

Worked Example: A Random Sample Size Without Outcome Dependence

Suppose \(p=0.30\), and the second patient is assigned to A with probability \(0.50\) regardless of \(Y_1\). Then \(q_0=q_1=0.50\), so

E(\bar{Y}_A) =0.30+\frac{1}{2}(0.30)(0.70)(0.50-0.50) =0.300.

The arm contains either one or two patients, but the decision to include the second outcome does not respond to the first outcome. In this model the random denominator does not create bias. This is a narrow conclusion under the assumptions given, not a general guarantee for all ways of stopping or sampling.

Diagnostics, Alternatives, and Limits

In an actual design, inspect the assignment rule and ask whether an observed outcome can change later probabilities of assignment to the arm whose mean is being reported. Then check whether the estimator’s denominator is random and whether the set of outcomes in that denominator depends on the same history. For a small design, enumerate all possible outcome and randomization paths, as above. For a larger design, simulation under clearly stated response probabilities can estimate the operating characteristics, including the expectation of the reported arm mean.

A useful simulation check is to repeat the full assignment-and-outcome process many times, calculate \(\bar{Y}_A\) in each replicate, and compare the average of those means with the specified \(p\). The simulated average should approach the formula as the number of replicates grows, subject to Monte Carlo error. A discrepancy may signal a coding error, an incorrect conditioning assumption, or a different estimator from the one analyzed.

The main alternative for this particular question is a design with allocation independent of previous outcomes, such as fixed allocation or outcome-independent random allocation. Under the exchangeable Bernoulli assumptions above, that removes the selection mechanism in this example. Adaptive allocation may still be chosen for other design goals, but the simple observed arm mean should not automatically be treated as unbiased merely because every patient’s outcome has response probability \(p\).

The calculation has important limits. It concerns the expectation over repeated trials, not the conditional expectation given one observed history. It assumes a common response probability, independent outcomes, no interference, and a specific two-stage allocation rule. Different rules, covariate-dependent assignment, delayed outcomes, or departures from exchangeability can change the result. The formula is therefore best used to understand the mechanism and check this design, not as a universal correction for adaptive arm means.

Common mistake: “Each outcome assigned to A is Bernoulli with response probability \(p\), so the sample mean must be unbiased.” That reasoning overlooks that inclusion in the final arm sample can depend on a previous outcome. In the example, the first outcome helps determine whether a second one is averaged with it, and the expectation of the resulting ratio can differ from \(p\).

Check Your Understanding

Use the two-stage model and the formula \(E(\bar{Y}_A)-p=\frac{1}{2}p(1-p)(q_0-q_1)\) to answer each question.

For \(p=0.50\), \(q_0=0.20\), and \(q_1=0.60\), calculate the expectation and bias of the arm mean.
In this model, what happens to the bias when \(q_0=q_1\), and why?
For \(0<p<1\), which direction of adaptation gives positive bias: \(q_0>q_1\) or \(q_1>q_0\)?
Why can a sample mean be biased here even though the potential outcomes for A are independent Bernoulli variables with a common response probability?
What happens to the bias when \(p=0\) or \(p=1\), regardless of \(q_0\) and \(q_1\)?

Solutions. 1. The bias is \(\frac{1}{2}(0.50)(0.50)(0.20-0.60)=-0.0500\), so the expectation is \(0.50-0.0500=0.450\). 2. The bias is zero because \(q_0-q_1=0\); assignment of the second patient does not depend on the first outcome. 3. \(q_0>q_1\) gives positive bias because every other factor in the expression is positive. 4. The inclusion and denominator depend on the first outcome, so the final set being averaged is not selected independently of the observed response history. 5. The bias is zero, since \(p(1-p)=0\); outcomes are constant when \(p\) is 0 or 1.

What You'll Learn