Functional Central Limit Theorem for Allocation Paths

Why Study Allocation Paths?

An allocation proportion at the end of a trial summarizes only one point in its history. A functional central limit theorem (FCLT) instead describes the fluctuations of the cumulative allocation count across the whole trial. This matters when asking whether random imbalances are concentrated early or late, or when a design changes its assignment probabilities as responses accrue.

The key distinction is between the realized allocation and its conditional expectation at each assignment. In an adaptive design, the assignment probability can change with the observed history. Centering each assignment by its own probability isolates the random allocation noise, even when those probabilities are not constant. The resulting theorem concerns a sequence of trials whose size $n$ tends to infinity; a single finite trial is an illustration, not an asymptotic limit.

Allocation Counts and the Functional Limit

Definition: Let $A_i=1$ if participant $i$ is assigned to arm 1 and $A_i=0$ otherwise. Let $\mathcal{F}_{i-1}$ denote the information available immediately before assignment $i$, and let $\pi_i=P(A_i=1\mid\mathcal{F}_{i-1})$ be the predictable assignment probability. The cumulative count is $N_1(m)=\sum_{i=1}^{m} A_i$. Its predictably centered, scaled path is $$ Z_n(t)=\frac{1}{\sqrt{n}}\sum_{i=1}^{\lfloor nt\rfloor}(A_i-\pi_i),\qquad 0\le t\le 1. $$

The summands have conditional mean zero, since $E(A_i-\pi_i\mid\mathcal{F}_{i-1})=0$. Their conditional variances are $\pi_i(1-\pi_i)$. Thus the accumulated conditional variance, on the same scale as $Z_n$, is

V_n(t)=\frac{1}{n}\sum_{i=1}^{\lfloor nt\rfloor}\pi_i(1-\pi_i).

A standard martingale functional central limit theorem gives a Gaussian limit when this variance process approaches a deterministic, continuous function and no single scaled increment can remain large. In this allocation setting the latter condition holds automatically: $|A_i-\pi_i|\le 1$, so each increment of $Z_n$ has absolute value at most $1/\sqrt n$.

Theorem: Suppose assignments are made sequentially with predictable probabilities $\pi_i$, and, as $n\to\infty$, $V_n(t)$ converges uniformly in probability on $0\le t\le1$ to a deterministic, continuous, nondecreasing function $v(t)$, with $v(0)=0$. Then the process $Z_n$ converges in distribution to $B(v(t))$, where $B$ is standard Brownian motion. The theorem describes a Gaussian fluctuation path with variance accumulated according to $v$, rather than necessarily at a constant rate.

Here “uniformly in probability” means that for every $\epsilon>0$, the probability that $\sup_{0\le t\le1}|V_n(t)-v(t)|>\epsilon$ tends to zero. In the usual process framework, the allocation paths are right-continuous step functions. This theorem is about their distribution as paths, not just the distribution of the final count.

Why the Martingale Conditions Give a Path Limit

For completeness, the central ingredients can be seen directly. Write $D_i=A_i-\pi_i$. The $D_i$ are martingale differences, and the conditional variance of the increment $D_i/\sqrt n$ is $\pi_i(1-\pi_i)/n$. Summing these conditional variances up to time $\lfloor nt\rfloor$ gives exactly $V_n(t)$. By assumption, that accumulated variance approaches $v(t)$.

The small-increment condition also holds: for any fixed $\epsilon>0$, $1/\sqrt n<\epsilon$ for all sufficiently large $n$. Consequently, no scaled assignment increment exceeds $\epsilon$, and the martingale Lindeberg condition is satisfied. The martingale functional central limit theorem then yields the stated Brownian time change. In particular, when $v(t)=ct$, the limit is $\sqrt c\,B(t)$. When $v$ grows at different rates over time, the path accumulates Gaussian variation at those different rates.

This reasoning depends on the variance-process condition, not on assignments being independent. Adaptive probabilities may depend on past outcomes, provided they are predictable and their accumulated conditional variances have the required limit. No response-distribution assumption is needed for the bounded assignment increments themselves; however, responses can affect the probabilities and hence whether the variance-process condition holds.

Worked Example: A Constant Allocation Probability

Suppose a hypothetical trial assigns each participant to arm 1 with probability $0.40$, independently of the assignment history. For each $i$, $\pi_i=0.40$, and the conditional variance is $0.40(0.60)=0.24$. With $n$ varying and tending to infinity,

V_n(t)=\frac{1}{n}\sum_{i=1}^{\lfloor nt\rfloor}0.40(0.60) =\frac{0.24\lfloor nt\rfloor}{n} \longrightarrow 0.24t.

The limiting centered allocation path is therefore $\sqrt{0.24}\,B(t)$, with standard deviation $\sqrt{0.24}\approx0.4899$ at $t=1$. This is a statement about the sequence of trials as $n\to\infty$.

For a separate finite-size illustration, at $n=1000$ and $t=0.3005$, the variance process equals $0.24(300)/1000=0.072$, because $\lfloor1000(0.3005)\rfloor=300$. The limiting function at that time is $0.24(0.3005)=0.07212$. The finite step-function value is close, but is not equal to the limit; one fixed $n=1000$ does not itself converge.

Worked Example: Allocation Probability Changes During the Trial

Consider a hypothetical design with deterministic predictable probabilities $\pi_i=0.30$ for assignments with $i/n\le0.5$, and $\pi_i=0.60$ thereafter. The conditional variances in the two portions are $0.30(0.70)=0.21$ and $0.60(0.40)=0.24$. As $n\to\infty$, the variance process approaches

v(t)=0.21\min(t,0.5)+0.24\max(t-0.5,0).

For example, at $t=0.75$, the first half of the trial contributes $0.21(0.5)=0.105$, and the next quarter contributes $0.24(0.25)=0.060$. Thus $v(0.75)=0.165$, and the limiting standard deviation of the centered, scaled count at that time is $\sqrt{0.165}\approx0.4062$. The path has a faster variance accumulation after the allocation probability changes, because $0.24>0.21$. The predictable centering remains essential: it subtracts the actual conditional assignment probability at each step.

Centering at a Fixed Target Is a Separate Question

A design may also be summarized relative to a fixed target proportion $p$. The exact decomposition is

\frac{N_1(\lfloor nt\rfloor)-\lfloor nt\rfloor p}{\sqrt n} =Z_n(t)+\frac{1}{\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}(\pi_i-p).

The first term is the allocation noise covered by the theorem. The second is the accumulated departure of the predictable probabilities from $p$. A fixed-target-centered path has the same limit only if that second term tends to zero (uniformly in $t$, for a path-level conclusion). If it tends to a nonzero deterministic function $h(t)$, the limit is shifted to $B(v(t))+h(t)$. If it diverges, the proposed $\sqrt n$ centering and scaling do not describe a finite limit.

Worked Example: A Small Systematic Departure From the Target

For each trial size $n$, suppose a hypothetical design uses the constant probability $\pi_{n,i}=0.40+0.20/\sqrt n$, which is between zero and one for all sufficiently large $n$. Center the count at the fixed target $p=0.40$. The accumulated predictable departure, divided by $\sqrt n$, is

\frac{1}{\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}(\pi_{n,i}-0.40) =\frac{\lfloor nt\rfloor}{n}(0.20) \longrightarrow 0.20t.

The conditional variance per assignment approaches $0.40(0.60)=0.24$, so the martingale part has limit $\sqrt{0.24}\,B(t)$. Consequently, the fixed-target-centered path has limit $\sqrt{0.24}\,B(t)+0.20t$, not a mean-zero Brownian path. The drift arises from the specified probability departure, not from the random assignment noise.

Extension to Several Arms

For $K$ arms, let $A_i$ be the $K$-component vector with a 1 in the assigned arm’s position and 0 elsewhere, and let $\boldsymbol{\pi}_i=E(A_i\mid\mathcal{F}_{i-1})$. The centered path is $n^{-1/2}\sum_{i\le\lfloor nt\rfloor}(A_i-\boldsymbol{\pi}_i)$. Its conditional covariance increment is $\operatorname{diag}(\boldsymbol{\pi}_i)-\boldsymbol{\pi}_i\boldsymbol{\pi}_i^{\mathsf T}$. If the accumulated matrices, divided by $n$, converge to $t\Sigma$ for a fixed probability vector, the multivariate martingale FCLT gives a Gaussian Brownian path with per-unit-time covariance matrix $\Sigma$. Since the arm counts sum to the total number assigned, their centered components sum to zero; the limiting covariance matrix is therefore singular in that direction.

Worked Example: Three Fixed Allocation Probabilities

Suppose a hypothetical three-arm trial assigns participants with fixed probabilities $(0.50,0.30,0.20)$. The conditional covariance matrix for one allocation is

\Sigma=\operatorname{diag}(0.50,0.30,0.20)- \begin{pmatrix}0.50\\0.30\\0.20\end{pmatrix} \begin{pmatrix}0.50&0.30&0.20\end{pmatrix} = \begin{pmatrix} 0.25&-0.15&-0.10\\ -0.15&0.21&-0.06\\ -0.10&-0.06&0.16 \end{pmatrix}.

At trial time $t$, the limiting covariance matrix of the centered, scaled allocation vector is $t\Sigma$. For example, the variance of the first component at $t=1$ is $0.25$, and its covariance with the second component is $-0.15$. The negative covariance is consistent with a fixed total: an assignment to one arm is not simultaneously an assignment to another. Each row of $\Sigma$ sums to zero, so the three limiting centered counts also sum to zero.

Diagnostics, Alternatives, and Limitations

Before applying the theorem, inspect the conditional variance process and the centering. A useful checklist is:

Predictability: Is each $\pi_i$ determined by information available before assignment $i$?
Variance stabilization: Does $V_n(t)$ approach a deterministic continuous function, rather than oscillating or retaining a random limit?
Scale: Are assignments occurring at order $n$ with nondegenerate accumulated variance on the $\sqrt n$ scale?
Centering: If the scientific target $p$ is used instead of $\pi_i$, does the accumulated probability departure vanish or have a specified limit?

The main alternative is an ordinary central limit theorem for the final allocation count at $t=1$. That can answer an endpoint question, but does not describe the joint fluctuation of the path at intermediate times. Under independent, fixed-probability randomization, the FCLT reduces to the familiar Brownian limit with variance rate $p(1-p)$. Under adaptive randomization, conditional-probability centering and the accumulated variance process are the corresponding tools.

A failure of deterministic variance stabilization can invalidate the stated Brownian time-change conclusion; a random limiting variance, for example, requires a different limit description. If an arm becomes so rare that its accumulated variance is not of order $n$, the usual $\sqrt n$ scaling may also be inappropriate. And a valid allocation-path limit alone does not establish a limit for response estimates or other trial statistics: those require additional arguments.

Common mistake. It is incorrect to replace the variance process by its limiting function at a fixed finite $n$, or to claim that one finite trial “converges” to a Brownian path. The theorem concerns a sequence as $n\to\infty$. For example, $0.24\lfloor nt\rfloor/n\to0.24t$ as $n\to\infty$; at $n=1000$, the left side is still a step function.

Check Your Understanding

Answer each question, then compare with the solutions below.

With fixed allocation probability $p=0.25$, what is the limiting variance function for the predictably centered allocation path?
Why does centering by $\pi_i$ differ from centering by a fixed target $p$ in an adaptive design?
For probabilities $0.20$ during the first half and $0.50$ during the second half, calculate $v(0.75)$.
In a three-arm trial, why should the limiting covariance matrix of centered allocation counts be singular?

Solutions. (1) The conditional variance per assignment is $0.25(0.75)=0.1875$, so $v(t)=0.1875t$. (2) The fixed-target-centered process includes the extra term $n^{-1/2}\sum_{i\le\lfloor nt\rfloor}(\pi_i-p)$, which may tend to a nonzero drift or fail to remain bounded. (3) The two variance rates are $0.20(0.80)=0.16$ and $0.50(0.50)=0.25$, so $v(0.75)=0.16(0.5)+0.25(0.25)=0.1425$. (4) The centered arm counts sum to zero because the total number assigned is fixed at each time. Thus the covariance matrix has a zero-variance direction along the all-ones vector and is singular.