Gaussian Process Limits for Sequential Response Estimates

Why a Process Limit Is Needed

In an adaptive trial, the number of responses observed on an arm changes over time according to the allocation rule and the history of the trial. Consequently, an arm-specific estimate is itself a sequence of random quantities: its precision at an interim time can differ substantially from its precision at the end. A central limit theorem for the final estimate alone does not describe this whole sequence.

The process limit developed here separates two sources of time dependence. The accumulated response noise has a Gaussian limit whose time change is determined by the arm’s cumulative allocation. The estimate then divides that noise by the number allocated to the arm. This denominator produces an additional, essential time-dependent scaling.

Definition: Consider a trial with $K$ arms and $n$ total assignments. Let $A_{i,j}=1$ if participant $i$ is assigned to arm $j$, and $A_{i,j}=0$ otherwise. Let $Y_{i,j}$ denote the response that would be observed for participant $i$ on arm $j$, and define $N_j(m)=\sum_{i=1}^{m}A_{i,j}$. At trial time $t$, the sequential response estimate for arm $j$ is $$ \hat{\mu}_j(t)=\frac{\sum_{i=1}^{\lfloor nt\rfloor}A_{i,j}Y_{i,j}}{N_j(\lfloor nt\rfloor)} $$ whenever $N_j(\lfloor nt\rfloor)>0$. The rescaled cumulative allocation path is $q_{n,j}(t)=N_j(\lfloor nt\rfloor)/n$. A limiting allocation path $q_j(t)$ describes the asymptotic fraction of the full trial allocated to arm $j$ by time $t$.

Conditions for the Limit

We state conditions for one fixed arm $j$; the result can be applied separately to each arm. Assume responses on arm $j$ have a common mean $\mu_j$ and variance $\sigma_j^2$, with $0<\sigma_j^2<\infty$, and are independent across participants conditional on the trial history and assignment. The assignment decision for participant $i$ is made using information available before that participant’s response is observed. Thus assignment may depend on past responses, but not on the current response that has yet to be observed.

For a convenient sufficient moment condition, assume $E|Y_{i,j}-\mu_j|^{2+\delta}<\infty$ for some $\delta>0$, uniformly in $i$. Assume also that the allocation mechanism does not alter the conditional response mean or variance on this arm. In particular, after assignment to arm $j$, the centered response $Y_{i,j}-\mu_j$ has conditional mean zero and conditional variance $\sigma_j^2$, given the past and the assignment.

Finally, suppose the realized and predictable cumulative allocation paths, divided by $n$, converge in probability uniformly on $[0,1]$ to the same deterministic function $q_j(t)$. The function is continuous and nondecreasing, with $q_j(0)=0$. For the estimate process, work on a fixed interval $[\tau,1]$, where $\tau>0$ and $q_j(\tau)>0$. This ensures that the arm has a growing number of observations throughout the interval. These assumptions are stronger than necessary in some settings, but they provide a clear standard route to the limit.

Theorem: Under the conditions above, on any fixed interval $[\tau,1]$ with $q_j(\tau)>0$, $$ \left\{\sqrt{n}\bigl(\hat{\mu}_j(t)-\mu_j\bigr):\tau\leq t\leq 1\right\} \Rightarrow \left\{\frac{\sigma_j W(q_j(t))}{q_j(t)}:\tau\leq t\leq 1\right\} $$ in the space of right-continuous functions with left limits, with the Skorokhod $J_1$ topology. Here $W(u)$ is standard Brownian motion indexed by $u\geq 0$.

The theorem gives a process limit, not just a distribution at the end of the trial. At each time $t$, the limiting fluctuation is a Brownian motion evaluated at the cumulative allocation time $q_j(t)$, divided by that same allocation fraction. In particular, the limiting variance at a fixed $t$ is $\sigma_j^2/q_j(t)$. The factor $q_j(t)$ in the denominator is part of the result: omitting it gives a limit for the centered cumulative response sum, not for the sample mean.

Why the Denominator Changes the Scaling

Write the centered cumulative response noise on arm $j$ as $$ M_{n,j}(t)=\frac{1}{\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}A_{i,j}(Y_{i,j}-\mu_j). $$ Because the assignment is made before the current response is observed, the summands form martingale differences under the stated conditional-mean assumption. The predictable conditional variance accumulated up to time $t$, after division by $n$, converges to $\sigma_j^2 q_j(t)$. The moment condition ensures that large individual increments do not retain a non-negligible share of this variance. The martingale functional central limit theorem therefore gives $$ M_{n,j}(\cdot)\Rightarrow \sigma_j W(q_j(\cdot)). $$

Now use the identity for the sample mean and $q_{n,j}(t)=N_j(\lfloor nt\rfloor)/n$:

\sqrt n\bigl(\hat{\mu}_j(t)-\mu_j\bigr) = \frac{n^{-1/2}\sum_{i=1}^{\lfloor nt\rfloor}A_{i,j}(Y_{i,j}-\mu_j)} {N_j(\lfloor nt\rfloor)/n} = \frac{M_{n,j}(t)}{q_{n,j}(t)}.

On $[\tau,1]$, the limiting denominator is bounded below by $q_j(\tau)>0$. Uniform convergence of the allocation paths therefore makes division by $q_{n,j}(t)$ asymptotically well behaved. Applying the continuous mapping theorem to the numerator and denominator yields the stated limit. This argument also explains why the restriction away from time zero is substantive: $q_j(0)=0$, so the limiting ratio is generally undefined there.

Worked Example: Constant Allocation Probability

Suppose a two-arm trial assigns each participant to arm 1 with probability $\rho=0.40$, independently of the past. Responses on arm 1 have mean $\mu_1=10$ points and standard deviation $\sigma_1=2$ points, and satisfy the moment and response-independence assumptions above. The limiting cumulative allocation path is $q_1(t)=0.40t$. On $[0.2,1]$, it is positive, so the theorem applies.

At time $t=0.5$, the limiting variance of the $\sqrt n$-scaled estimate is $$ \frac{\sigma_1^2}{q_1(0.5)} = \frac{2^2}{0.40(0.5)} = \frac{4}{0.20} = 20\ \text{points}^2. $$ Thus the limiting standard deviation of $\sqrt n(\hat{\mu}_1(0.5)-10)$ is $\sqrt{20}=4.472$ points. For $n=1000$, the corresponding approximate standard error of the unscaled estimate at that time is $4.472/\sqrt{1000}=0.1414$ points.

The $1/\sqrt n$ factor here refers to the full trial size, whereas only about $0.20n$ participants have been assigned to arm 1 by time $0.5$. Equivalently, the estimate’s standard error is approximately $\sigma_1/\sqrt{nq_1(0.5)}$, which gives $2/\sqrt{200}=0.1414$ points.

Allocation That Changes Over Time

The allocation path need not be linear. If the arm’s allocation probability changes over trial time, $q_j(t)$ records the cumulative fraction allocated up to $t$, rather than the probability assigned at the instant $t$. The variance scaling uses this cumulative quantity.

Worked Example: A Changing Allocation Rate

Suppose arm 1’s allocation probability at trial time $t$ is approximately $0.30+0.40t$, so its limiting cumulative allocation path is $$ q_1(t)=\int_0^t(0.30+0.40s)\,ds=0.30t+0.20t^2. $$ The probability ranges from 0.30 to 0.70, so it is a valid allocation probability over $0\leq t\leq1$. Assume arm 1 responses have standard deviation $2$ points and meet the theorem’s moment and conditional response assumptions.

At $t=0.5$, $$ q_1(0.5)=0.30(0.5)+0.20(0.5)^2=0.15+0.05=0.20. $$ For a hypothetical trial with $n=1000$, this corresponds asymptotically to 200 arm 1 observations by that time. The limiting variance of the scaled estimate is $2^2/0.20=20$ points squared, and the approximate standard error of the unscaled estimate is $\sqrt{20}/\sqrt{1000}=0.1414$ points.

At $t=1$, $q_1(1)=0.50$, so the limiting variance is $4/0.50=8$ points squared and the approximate standard error is $\sqrt{8}/\sqrt{1000}=0.08944$ points. Although the instantaneous allocation probability at the end is 0.70, using 0.70 in the variance formula would be incorrect: the estimate at $t=1$ uses the cumulative allocation fraction 0.50.

When the Result Does Not Apply

The assumptions have practical consequences. First, estimates cannot be treated as a well-behaved process on an interval where an arm has no observations or only a bounded number of observations. For example, suppose a hypothetical design does not assign anyone to arm 1 until time $0.4$, and thereafter its allocation probability is 0.50. Then $q_1(t)=0$ for $t\leq0.4$, and $q_1(t)=0.50(t-0.4)$ for $t>0.4$. The theorem cannot describe $\hat{\mu}_1(t)$ on $[0.2,1]$, because the estimate is undefined before the first assignment to that arm. It can potentially describe it on $[0.6,1]$, where $q_1(0.6)=0.10>0$, provided the other assumptions hold.

Second, a central limit theorem for the numerator does not by itself justify a Gaussian limit for the estimate. The allocation path must converge and its limiting value must stay positive on the interval of interest. If an arm receives a vanishing fraction of assignments as $n$ grows, then $q_{n,j}(t)$ may not approach a positive function there. The normalization and limiting behavior may then differ from the theorem.

Third, if responses have infinite variance or sufficiently heavy tails that the martingale Lindeberg condition fails, Gaussian response-noise limits need not hold. The earlier material on heavy-tailed responses treats that distinct situation. Likewise, if current responses affect their own assignment decisions, the martingale argument given here is not justified. Adaptive assignment based on past responses is allowed; using information that is unavailable before the current response is observed is not.

Worked Example: Checking a Proposed Normalization

Suppose a hypothetical trial has $n=900$, and by time $t=0.6$ the limiting cumulative allocation fraction for arm 2 is $q_2(0.6)=0.30$. Assume arm 2 responses have standard deviation $3$ mg and the theorem’s conditions hold. A proposed calculation uses $q_2(0.6)$ as the variance divisor for the scaled estimate.

The correct limiting variance is $$ \frac{\sigma_2^2}{q_2(0.6)} = \frac{3^2}{0.30} = 30\ \text{mg}^2. $$ The limiting standard deviation of $\sqrt{900}(\hat{\mu}_2(0.6)-\mu_2)$ is $\sqrt{30}=5.477$ mg. The approximate standard error of the unscaled estimate is $5.477/\sqrt{900}=0.1826$ mg. This also follows from the approximate arm-specific sample size $900(0.30)=270$: $3/\sqrt{270}=0.1826$ mg.

A frequent incorrect calculation is to multiply the variance by $q_2(0.6)$, obtaining $9(0.30)=2.7$ mg squared. That is the wrong direction: allocating a smaller fraction gives fewer observations and a less precise estimate, so the variance must increase as the allocation fraction decreases.

How This Limit Relates to Other Approaches

A fixed-time central limit theorem describes $\hat{\mu}_j(t)$ at one chosen $t$. The process theorem gives a single approximation for the estimate trajectory over an interval, which is useful when studying multiple interim times or time-indexed summaries. It is not permission to treat estimates at different times as independent; they reuse earlier observations.

A deterministic-allocation calculation often replaces the arm’s sample size by its expected value and applies an ordinary central limit theorem. This can give the right leading variance at a fixed time under suitable conditions. The martingale process argument is more directly suited to adaptive assignment: it accounts for predictable allocation based on past data and establishes convergence of the entire numerator path before converting it to an estimate.

The limit is also not a universal claim that every adaptive design has the same Gaussian process. The response variance and the limiting allocation path both matter. The allocation path enters through $W(q_j(t))$ and through division by $q_j(t)$; changing the allocation trajectory changes the time evolution of the limiting estimate. The present result assumes a deterministic limiting path. If the limiting allocation path is random, a different treatment is required.

Key takeaway: For an arm with a positive limiting cumulative allocation fraction, the response-mean process has the limit $\sigma_j W(q_j(t))/q_j(t)$ after multiplication by $\sqrt n$. The denominator is cumulative allocation, and the result is valid only on time intervals where that denominator stays away from zero.

Check Your Understanding

Answer each question, then compare your reasoning with the solution given.

An arm has limiting allocation path $q(t)=0.25t$. What is the limiting variance of the $\sqrt n$-scaled estimate at $t=0.8$, in terms of its response variance $\sigma^2$?
Solution: $q(0.8)=0.25(0.8)=0.20$, so the variance is $\sigma^2/0.20=5\sigma^2$.
Why is it generally not enough to state a process limit for the centered cumulative response sum when the target is the sequential sample mean?
Solution: The sample mean divides the cumulative centered sum by $N_j(\lfloor nt\rfloor)$. After scaling, that denominator contributes the factor $1/q_j(t)$, which must be included in the process limit.
An arm’s limiting allocation path is zero through time $0.3$ and positive afterward. Can the stated theorem be applied to the estimate process on $[0.1,1]$?
Solution: No. The estimate is not available before assignments occur, and the limiting denominator is zero there. One may consider an interval $[\tau,1]$ with $\tau>0.3$, if the remaining assumptions hold and $q_j(\tau)>0$.
If the allocation fraction at time $t$ decreases while the response variance remains fixed, what happens to the limiting variance of the scaled estimate? Explain.
Solution: It increases, since the variance is $\sigma_j^2/q_j(t)$. A smaller allocation fraction means fewer responses and therefore less precision.
What role does the condition that assignment uses only past information play in the argument?
Solution: Together with the conditional response-mean assumption, it makes the centered assigned responses martingale differences. That supports the martingale functional central limit theorem for the cumulative response noise.