Asymptotic Independence of Allocation and Outcome Noise

Why Independence Is a Limit-Theory Question

In an adaptive trial, the mechanism assigning participants and the responses observed after assignment are connected over time: past responses may influence later allocation probabilities. That connection does not, by itself, settle whether an allocation fluctuation and a response-noise fluctuation are dependent in the limit. The relevant question is whether their suitably normalized sums converge jointly to a Gaussian vector with zero cross-covariance.

This tutorial gives a precise sufficient condition. It concerns two components of a common sequence of martingale increments: one recording allocation innovations and one recording response noise. Conditional orthogonality means that their one-step conditional cross-covariance is zero, or that the accumulated cross-covariance vanishes at the normalization scale. Under a joint martingale central limit theorem with a deterministic limiting covariance matrix, that condition gives independent Gaussian limits. It does not generally give independence at finite sample sizes.

Setup and the Joint Limit Result

Definition: Let \(\mathcal F_{n,i}\) describe the information available through step \(i\) in a sequence of trials with \(n\) steps. An allocation increment \(A_{n,i}\) and a response-noise increment \(Y_{n,i}\) are martingale differences if they are measurable with respect to \(\mathcal F_{n,i}\) and \(E[A_{n,i}\mid\mathcal F_{n,i-1}]=0\) and \(E[Y_{n,i}\mid\mathcal F_{n,i-1}]=0\). Their dimensions may be greater than one. Write \(D_{n,i}=(A_{n,i}^{\mathsf T},Y_{n,i}^{\mathsf T})^{\mathsf T}\).

For example, an allocation difference may be the current treatment indicator minus its probability conditional on past information. A response-noise increment may be a centered response residual or score contribution. The exact definitions depend on the statistic under study; both components must be centered conditional on the same past-information filtration for the martingale argument below.

Here is a finite-dimensional sufficient result for normalized terminal sums. Assume the increment dimensions are fixed as \(n\) grows, the increments have finite conditional second moments, and the following conditions hold. First, for fixed nonrandom matrices \(\Sigma_A\) and \(\Sigma_Y\), the accumulated conditional covariance blocks satisfy

\frac{1}{n}\sum_{i=1}^{n}E[A_{n,i}A_{n,i}^{\mathsf T}\mid\mathcal F_{n,i-1}] \xrightarrow{P}\Sigma_A,\qquad \frac{1}{n}\sum_{i=1}^{n}E[Y_{n,i}Y_{n,i}^{\mathsf T}\mid\mathcal F_{n,i-1}] \xrightarrow{P}\Sigma_Y,

and the accumulated conditional cross-covariance vanishes:

\frac{1}{n}\sum_{i=1}^{n}E[A_{n,i}Y_{n,i}^{\mathsf T}\mid\mathcal F_{n,i-1}] \xrightarrow{P}0.

Second, with \(\|\cdot\|\) denoting the Euclidean norm, the joint conditional Lindeberg condition holds for every \(\epsilon>0\):

\frac{1}{n}\sum_{i=1}^{n} E\!\left[ \|D_{n,i}\|^2 \mathbf 1\{\|D_{n,i}\|>\epsilon\sqrt n\} \mid\mathcal F_{n,i-1} \right] \xrightarrow{P}0.

Theorem: Under these conditions, the joint martingale central limit theorem gives \(\left(n^{-1/2}\sum_{i=1}^n A_{n,i},\,n^{-1/2}\sum_{i=1}^n Y_{n,i}\right)\xrightarrow{d}N(0,\operatorname{blockdiag}(\Sigma_A,\Sigma_Y))\). Consequently, the two limiting Gaussian vectors are independent.

The factors \(1/n\) in the covariance and Lindeberg conditions correspond to the \(n^{-1/2}\) normalization of the sums. The covariance of \(n^{-1/2}\sum_iD_{n,i}\) is built from \(n^{-1}\sum_i E[D_{n,i}D_{n,i}^{\mathsf T}\mid\mathcal F_{n,i-1}]\), not from the unscaled sum of conditional covariances.

Why Block Diagonality Gives Independence

Combine the two increment types into \(D_{n,i}\). The conditional covariance matrix has allocation, response, and cross-covariance blocks. The first two blocks converge to \(\Sigma_A\) and \(\Sigma_Y\); the cross block and its transpose converge to zero. Thus the limiting covariance is block diagonal. The joint martingale central limit theorem gives a jointly Gaussian limit with this covariance. For a jointly Gaussian vector, blocks with zero cross-covariance are independent; this remains true if a block is degenerate.

This last step matters. Zero covariance alone does not imply independence for arbitrary distributions. It implies independence here because the joint limit is Gaussian. Also, the theorem assumes fixed, nonrandom limiting covariance matrices. A random limiting covariance can lead to a mixture of Gaussian distributions, for which blockwise zero conditional covariance alone does not generally establish unconditional independence.

Center the increments.
Check that allocation and response components each have conditional mean zero given the same preceding information.

Check the normalized covariance blocks.
Verify convergence of each \(1/n\)-scaled sum and that the accumulated cross block tends to zero.

Control large increments.
Check the joint conditional Lindeberg condition at threshold \(\epsilon\sqrt n\).

Use the joint Gaussian limit.
A block-diagonal covariance gives independent limiting allocation and response vectors.

This is a finite-dimensional statement about terminal sums. A corresponding assertion for whole allocation and response paths requires a joint process-level limit and appropriate conditions for that process; the terminal-vector result alone does not establish it.

Worked Examples

Worked Example: Independent Allocation and Response Innovations

Consider a hypothetical two-arm trial with \(n=1000\) participants. At each step, conditional on the past, assign to arm 1 with fixed probability \(p=0.4\). Let \(Z_i\) be the arm 1 indicator and define the allocation increment \(A_i=Z_i-p\). Suppose the response-noise increment is \(Y_i=\varepsilon_i\), where the \(\varepsilon_i\) are independent \(N(0,9)\) variables, independent of assignments and past information.

Both increments have conditional mean zero. Their conditional variances and cross-covariance are

E[A_i^2\mid\mathcal F_{i-1}]=p(1-p)=0.4(0.6)=0.24,\quad E[Y_i^2\mid\mathcal F_{i-1}]=9,\quad E[A_iY_i\mid\mathcal F_{i-1}]=0.

Therefore, the normalized accumulated covariance is

\frac{1}{1000}\sum_{i=1}^{1000} \begin{pmatrix}0.24&0\\0&9\end{pmatrix} = \begin{pmatrix}0.24&0\\0&9\end{pmatrix}.

The allocation increments are bounded, and the Gaussian response increments have finite second moments, so the joint Lindeberg condition holds. The theorem gives independent limiting normal variables with variances \(0.24\) and \(9\). In particular, the standard deviations of the two normalized sums are \(\sqrt{0.24}\approx0.4899\) and \(3\). This is a statement about their joint limiting distribution, not a claim that every allocation-based statistic in a real adaptive trial is independent of every response-based statistic.

Worked Example: Orthogonality Without Finite-Sample Independence

For a hypothetical illustration, let the pairs \((A_i,Y_i)\) be independent across \(i\), with \(A_i\) taking values \(-1,0,1\) with probabilities \(1/4,1/2,1/4\), respectively. Set \(Y_i=\mathbf 1\{A_i=0\}-1/2\). Treat \(A_i\) and \(Y_i\) as allocation and response-noise innovations in this abstract example.

Their means are zero. The variances and covariance are

E[A_i^2]=\frac14(1)+\frac12(0)+\frac14(1)=\frac12,\qquad E[Y_i^2]=\frac12\left(\frac12\right)^2+\frac12\left(-\frac12\right)^2=\frac14,

E[A_iY_i] =\frac14(-1)\left(-\frac12\right)+\frac12(0)\left(\frac12\right) +\frac14(1)\left(-\frac12\right)=0.

Yet \(Y_i\) is determined by \(A_i\), so these increments are not independent. They are bounded, and the covariance matrix for the normalized sums is \(\left(\begin{smallmatrix}1/2&0\\0&1/4\end{smallmatrix}\right)\). The multivariate central limit theorem therefore gives independent Gaussian limits with variances \(1/2\) and \(1/4\). The example shows exactly what the result does and does not say: orthogonality need not remove dependence at a single step, but it can yield independent Gaussian limits for the accumulated sums.

Worked Example: Shared Noise Produces Dependent Limits

Now suppose a hypothetical trial’s allocation and response innovations share a random input. Let \(G_i\) and \(H_i\) be independent standard normal variables, independent across steps and of the past. Set \(A_i=G_i\) and \(Y_i=0.6G_i+0.8H_i\). Both increments have mean zero and variance one, but

$$ E[A_iY_i]=E[G_i(0.6G_i+0.8H_i)]=0.6. $$

Thus the normalized accumulated covariance is

\frac{1}{n}\sum_{i=1}^{n} E\!\left[ \begin{pmatrix}A_i\\Y_i\end{pmatrix} \begin{pmatrix}A_i&Y_i\end{pmatrix} \right] = \begin{pmatrix}1&0.6\\0.6&1\end{pmatrix}.

The sums are jointly Gaussian, and their correlation is \(0.6\) for every \(n\); their limits are not independent. The failure is not that the central limit theorem fails, but that the limiting cross-covariance does not vanish. Shared random inputs or shared response shocks can create this problem in applications.

Diagnostics, Alternatives, and Limitations

The useful diagnostic is to derive the conditional cross-covariance from the actual increments, rather than infer it from the names “allocation” and “response.” For each step, condition on \(\mathcal F_{n,i-1}\), calculate \(E[A_{n,i}Y_{n,i}^{\mathsf T}\mid\mathcal F_{n,i-1}]\), and then examine its normalized accumulated sum. Check separately that the increments are conditionally centered and that the Lindeberg condition is plausible. A nonzero cross-covariance limit warns against claiming independent limits; failure of a Gaussian limit condition also means this theorem cannot justify independence.

Approach	What it establishes	Key qualification
Independent random streams	Can provide a design-based route to independence of the underlying random inputs.	Statistics computed from the streams may still be coupled through adaptation or shared data.
Zero covariance at one step	Shows conditional orthogonality for that increment, if the conditional cross-covariance is zero.	Does not by itself prove finite-sample independence.
Joint martingale limit with zero accumulated cross block	Yields independent Gaussian limits under the stated covariance and Lindeberg conditions.	Requires a suitable deterministic Gaussian limit; it is not a general independence result for non-Gaussian limits.

The main limitation is that this is a sufficient asymptotic argument for specified normalized sums. It does not assert that allocation and outcomes are unrelated, nor does it remove dependence caused by the adaptation rule. Earlier material on covariance kernels and variance decomposition helps identify which increments contribute to a statistic, but independence must still be justified for the particular joint limit being claimed.

Common mistake: “The conditional covariance is zero, so allocation and response noise are independent.” Zero covariance does not generally imply independence, and conditional orthogonality does not imply finite-sample independence. The correction is to verify a joint Gaussian limit with vanishing normalized cross-covariance; only then does zero cross-covariance imply independence of the limiting blocks.

Key takeaway: Asymptotic independence follows when the allocation and response-noise sums have a joint Gaussian limit and their normalized accumulated conditional cross-covariance vanishes. The \(1/n\) scaling is essential for sums normalized by \(1/\sqrt n\).

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

For sums normalized by \(1/\sqrt n\), what scaling belongs in front of the accumulated conditional covariance?
Does zero conditional covariance of two increments imply that the increments are independent?
In the shared-noise example, what is the limiting cross-covariance?
Why does a block-diagonal covariance establish independence in the stated theorem?

Solutions. (1) Use \(1/n\), since the covariance of \(n^{-1/2}\sum_iD_{n,i}\) has the corresponding \(1/n\) factor. (2) No; zero covariance is not enough for arbitrary distributions, and the discrete example has dependent increments despite zero covariance. (3) It is \(0.6\), so the limiting Gaussian variables are dependent. (4) The joint limit is Gaussian, and jointly Gaussian blocks with zero cross-covariance are independent.