Asymptotics With Increasing Feedback Delays

Why Increasing Delays Change the Asymptotics

In response-adaptive randomization, an allocation decision at an enrollment update can use only the outcomes available by then. If delays grow with enrollment index, the data used at successive updates may become increasingly old relative to the total trial size. The key asymptotic question is not just whether fewer outcomes are available as a fraction of enrollment, but whether the number of usable outcomes still grows without bound.

This distinction matters because an estimate can become consistent from a growing, though increasingly small, fraction of the trial. Conversely, if the number of usable responses does not grow, the estimates driving adaptation need not settle near their population values. The results below concern delays that are exogenous to response and treatment assignment; informative or treatment-specific delays require additional assumptions.

Definition: Index enrollments by $i=1,2,\ldots$, with one enrollment per update. Let $D_i$ be the nonnegative integer delay for enrollment $i$, measured in updates, and let $i+D_i$ be its outcome’s return update. The number of usable outcomes by update $t$ is $$ m(t)=\#\{i:i+D_i\leq t\}. $$ The number newly returned at update $t$ is $m(t)-m(t-1)$. An outcome is usable only after it has returned.

A transparent benchmark is a deterministic delay schedule $D_i=\lceil c i^\gamma\rceil$, with constants $c>0$ and $\gamma>0$. Delays increase with enrollment index. Ignoring the rounding by the ceiling, the largest completed enrollment index $m$ approximately solves $m+c m^\gamma=t$. This yields three distinct growth regimes.

m(t)\asymp \begin{cases} t, & 0<\gamma<1,\\[2pt] t/(1+c), & \gamma=1,\\[2pt] (t/c)^{1/\gamma}, & \gamma>1. \end{cases}

When $0<\gamma<1$, almost all enrollments up to $t$ have returned: $m(t)/t\to1$, even though the backlog is of order $t^\gamma$. With linear delays, a positive fraction is available, but the fraction is only about $1/(1+c)$. With superlinear delays, $m(t)/t\to0$, while $m(t)$ still diverges. For $\gamma>1$, completion times become increasingly spaced, so an update may bring no new outcome; $m(t)$ describes cumulative availability, not a guarantee of one return at every update.

Worked Counts Under Three Delay Growth Rates

Worked Example: Sublinear Delays

Suppose, hypothetically, that $D_i=\lceil2\sqrt{i}\rceil$. Find the number of outcomes available by update $t=10{,}000$. The schedule is nondecreasing, so completed enrollments form a prefix: the answer is the largest $i$ satisfying $i+\lceil2\sqrt{i}\rceil\leq10{,}000$.

At $i=9{,}801$, $\sqrt{i}=99$, so the return update is $9{,}801+\lceil198\rceil=9{,}999$. At $i=9{,}802$, $2\sqrt{i}\approx198.010$, so the return update is $9{,}802+199=10{,}001$. Thus $m(10{,}000)=9{,}801$, and the available fraction is $9{,}801/10{,}000=0.9801$. The backlog is 199 outcomes. Although the delay increases, its square-root growth leaves nearly all outcomes available by this update.

Worked Example: Linear Delays

Suppose, hypothetically, $D_i=\lceil0.25i\rceil$. At update $t=1{,}000$, enrollment $i=800$ returns at $800+\lceil200\rceil=1{,}000$. Enrollment $i=801$ returns at $801+\lceil200.25\rceil=1{,}002$. Since return times increase with $i$, $m(1{,}000)=800$. The usable fraction is $800/1{,}000=0.8000$, consistent with the large-$t$ approximation $1/(1+0.25)=0.8000$.

This is not a bounded-delay situation: the most recently enrolled outcomes can remain pending for a delay proportional to enrollment index. Nevertheless, the number available grows linearly with trial size.

Worked Example: Superlinear Delays

Suppose, hypothetically, $D_i=\lceil0.01i^2\rceil$, and consider $t=1{,}000{,}000$. Ignoring the ceiling initially, solve $0.01i^2+i-1{,}000{,}000=0$. Multiplying by 100 gives $i^2+100i-100{,}000{,}000=0$, hence the positive root is

i=\frac{-100+\sqrt{100^2+4(100)(100{,}000{,}000)}}{2} =\frac{-100+\sqrt{400{,}010{,}000}}{2} \approx9{,}950.1.

Check the integer boundary with the ceiling. At $i=9{,}950$, $0.01i^2=990{,}025$, so the return update is $9{,}950+990{,}025=999{,}975$. At $i=9{,}951$, $0.01i^2=990{,}224.01$, so the rounded delay is 990,225 and the return update is $1{,}000{,}176$. Therefore $m(1{,}000{,}000)=9{,}950$, about $0.9950\%$ of the enrollments so far. The fraction is small, but the usable count is still substantial and increasing.

When Delayed Adaptation Still Converges

A growing $m(t)$ alone does not ensure convergence. The completed outcomes must be informative about the response distributions, and every treatment arm must continue to receive assignments. The following conditions give a useful sufficient result. There are finitely many treatment arms; each patient’s potential response on an assigned arm has that arm’s stable mean and finite variance; and response observations are independent across patients conditional on assignments. Delays are finite and determined independently of the patient’s response and assigned arm. At each assignment, the probability of assigning any arm is at least a fixed $\epsilon>0$. The adaptive probability rule is a continuous function $G$ of consistent estimates of the arm means, with any required initialization supplied before those estimates are available.

Theorem: Under these conditions, if the number of available outcomes tends to infinity, the completed-outcome estimates of each arm mean are consistent. Consequently, if the adaptive rule is $G$, the allocation probabilities converge to $G$ evaluated at the true arm means. The realized allocation proportions converge to those limiting probabilities as well, provided each assignment uses the stated probabilities conditional on the prior history.

The lower bound on assignment probabilities prevents an arm from being starved of data. Exogenous delays ensure that completion does not select patients according to their response or arm. With a common nondecreasing delay schedule, the completed observations at update $t$ are the first $m(t)$ enrollments. More generally, deterministic completion sets can be used if their size grows without bound and completion is independent of response and assignment.

Each arm continues to accrue observations.
Because its conditional assignment probability is at least $\epsilon$, the number assigned to each arm among a growing set of completed enrollment indices diverges with probability one. This follows from the law of large numbers for bounded martingale differences between each assignment indicator and its conditional probability.

The arm estimates converge.
For each arm, outcomes are independent with a common mean and finite variance, and the number observed on that arm diverges. The sample mean therefore converges to that arm’s true mean by the law of large numbers. Exogenous completion means the observed set has not been selected by response value.

The probabilities and realized proportions converge.
Continuity of $G$ implies that applying the allocation rule to estimates converging to the true means gives probabilities converging to $G$ at those means. For each arm, subtract its conditional assignment probability from its assignment indicator. The differences are bounded martingale differences, so their average converges to zero; the realized proportion therefore has the same limit as the average assignment probability.

The main rate consequence is important: superlinear delays can make the available fraction vanish without preventing consistency. If $m(t)$ grows like $t^{1/\gamma}$, estimates may stabilize much more slowly than under bounded delays, because they are based on far fewer responses at a given update. This result establishes convergence, not a particular convergence rate, and not that adaptive decisions are equally reliable at finite trial sizes.

Worked Example: A Vanishing Fraction Can Still Support Convergence

Continue with the hypothetical quadratic delay schedule above. At update $t=1{,}000{,}000$, suppose there are two arms and each assignment probability is at least $\epsilon=0.10$. With $m(t)=9{,}950$ completed outcomes, the lower bound on the expected completed count for either arm is $0.10(9{,}950)=995$. This expectation is not a guarantee of exactly 995 observations, but the probability floor ensures that each arm’s completed count grows without bound as $t$ increases.

If the response and delay assumptions hold, each arm’s completed-outcome mean is therefore consistent as the trial grows, even though the completed fraction under quadratic delays tends to zero. A continuous allocation rule can then converge to its target. At this finite update, however, the estimates may still be noisy; asymptotic consistency should not be mistaken for accurate decisions at every update.

Diagnostics, Alternatives, and Limitations

To assess whether the asymptotic argument is plausible in an operating trial, track both $m(t)$ and $m(t)/t$, not just the percentage pending. Also track completed counts by arm, the age of the most recently used outcomes, and the number of updates with no returned outcomes. Plotting cumulative returns against enrollment can reveal whether growth resembles a linear, sublinear, or slower pattern. A rising backlog is not by itself evidence of failure: under sublinear delays the backlog can grow while the usable fraction approaches one. The critical warning is that an arm’s completed count stops growing, or that delays depend on response or treatment in a way that changes which outcomes are observed.

Waiting for observed outcomes preserves direct use of measured responses but can leave the allocation rule operating on stale information. Predictive updating or imputation can use pending cases, as covered in the earlier material on pending responses, but those methods introduce model assumptions; they do not make informative delays harmless. The present convergence argument concerns what can be concluded from completed responses under exogenous delay, not the relative efficiency or patient-benefit consequences of alternative update strategies.

Failure of exploration: if an arm’s assignment probability approaches zero too quickly, its completed sample may fail to grow, so its response estimate and the adaptive rule need not converge as claimed.
Informative delay: if response values affect return times, completed outcomes can be a selected subset. A growing completed count does not by itself guarantee a consistent estimate.
Changing response distributions: the argument assumes stable arm-specific means. Calendar-time changes can make estimates of old completed outcomes unrepresentative of current responses.
Finite-sample instability: even when $m(t)\to\infty$, slow growth can leave large uncertainty for many updates. The asymptotic result does not certify a chosen finite-sample precision.

A common mistake is to say, “The available fraction tends to zero, so adaptive allocation cannot converge.” That conclusion confuses a relative rate with an absolute count. Under quadratic delays, for example, $m(t)$ grows approximately as $\sqrt{t/c}$: its fraction of $t$ vanishes, but its count diverges. The correction is to check whether each arm receives an unbounded number of usable outcomes and whether those outcomes remain representative. The opposite shortcut is also unsafe: finite delay for every individual guarantees eventual return of each fixed outcome, but without a delay-growth model it supplies no useful rate for how many outcomes are available at a given update.

Key takeaway: Increasing delays slow the information flow according to the growth of $m(t)$. A vanishing usable fraction does not rule out convergence; diverging, representative completed samples in every arm and a stable, continuous allocation rule are the essential conditions.

Check Your Understanding

Answer each question, then compare with the concise solutions below.

For $D_i=\lceil0.5i\rceil$, approximately what fraction of outcomes is available at a large update?
For $D_i=\lceil0.04i^2\rceil$, what is the approximate order of $m(t)$, and does the available fraction vanish?
Why is a positive lower bound on every arm’s assignment probability useful for convergence?
Give one reason a growing number of completed outcomes might still fail to yield consistent response estimates.

Solutions. (1) Here $\gamma=1$ and $c=0.5$, so $m(t)/t\to1/(1+0.5)=2/3$, approximately 0.667. (2) Here $\gamma=2$, so $m(t)$ is of order $(t/0.04)^{1/2}=5\sqrt{t}$; this diverges, while $m(t)/t$ is of order $5/\sqrt{t}\to0$. (3) The lower bound prevents an arm from being assigned only finitely often, allowing its completed sample count to grow and its response mean to be estimated consistently. (4) Delays may depend on response, causing the completed outcomes to be a selected subset; alternatively, the response distribution may change over time, so old completed outcomes do not estimate a stable current mean.