Asymptotic Variance Decomposition for Adaptive Estimators

Why Decompose the Variance?

In a two-arm adaptive trial, the number assigned to each arm is random and can depend on earlier outcomes. That randomness matters when the estimator weights observed responses according to the realized arm sizes. Its uncertainty is therefore not always just the variability of individual responses: fluctuation in the arm counts, and dependence between those fluctuations and response noise, can also affect the limiting variance.

This tutorial studies a specific estimator: the overall average of the observed responses, interpreted relative to a target mixture of the two arm means. It is not the usual treatment-effect contrast. Keeping the estimand explicit makes the allocation contribution visible and gives an exact decomposition to analyze.

Model, Estimator, and Assumptions

Definition: At trial step $t$, let $I_t\in\{0,1\}$ denote the assigned arm and let $Y_t$ be the response. Write $N_1(n)=\sum_{t=1}^n I_t$ and $N_0(n)=n-N_1(n)$. Assume the response model $Y_t=\mu_{I_t}+\varepsilon_t$, where $\mu_j$ is the mean response in arm $j$, and $\varepsilon_t$ has conditional mean zero given the trial history and current assignment. Let $\rho\in(0,1)$ be the limiting target proportion assigned to arm 1, and define $\Delta=\mu_1-\mu_0$, $\theta=\rho\mu_1+(1-\rho)\mu_0$, and $\hat\theta_n=n^{-1}\sum_{t=1}^nY_t$.

The result below uses these regularity conditions. First, the allocation proportion converges to its target: $N_1(n)/n\to\rho$ in probability. Second, responses have stable conditional variances: $E(\varepsilon_t^2\mid\text{history before response }t,\ I_t=j)=\sigma_j^2$, with finite $\sigma_j^2$. Third, the scaled response-noise sum satisfies a central limit theorem, with a Lindeberg condition ruling out a non-negligible contribution from a single unusually large response. Finally, assume the allocation fluctuation $A_n=(N_1(n)-n\rho)/\sqrt n$ and the response-noise sum $R_n=n^{-1/2}\sum_{t=1}^n\varepsilon_t$ converge jointly to $(A,Z_R)$, and that their second moments and covariance converge. These moment conditions matter when passing from a distributional limit to a variance limit.

Under the stated conditional-variance and allocation conditions, the response-noise limit has variance $v_R=\rho\sigma_1^2+(1-\rho)\sigma_0^2$. Joint convergence does not require $A$ and $Z_R$ to be independent. In an adaptive design, their covariance can be nonzero because earlier response noise may influence later assignment decisions.

The Exact Decomposition and Its Limit

The decomposition follows by adding and subtracting the target arm means. Since $N_0(n)=n-N_1(n)$, the estimator satisfies

\begin{aligned} \sqrt n(\hat\theta_n-\theta) &=\frac{1}{\sqrt n}\sum_{t=1}^n\varepsilon_t +\frac{N_1(n)\mu_1+N_0(n)\mu_0-n\theta}{\sqrt n}\\ &=R_n+\Delta A_n. \end{aligned}

For the second equality, the mean part in the numerator is $N_1(n)\mu_1+(n-N_1(n))\mu_0-n[\rho\mu_1+(1-\rho)\mu_0]$. Collecting the terms multiplying $\mu_1-\mu_0$ gives $(N_1(n)-n\rho)\Delta$, which proves the identity.

Theorem: Under the assumptions above, if $(R_n,A_n)$ converges jointly to $(Z_R,A)$, then $\sqrt n(\hat\theta_n-\theta)$ converges in distribution to $Z_R+\Delta A$. If second moments also converge, its limiting variance is $$ V=v_R+\Delta^2\operatorname{Var}(A)+2\Delta\operatorname{Cov}(Z_R,A). $$

The first term is response noise. The second is the variance contribution from random sample-size fluctuation: a fluctuation in arm 1’s count changes the estimator’s mean part by $\Delta$ times that fluctuation. The third term is the allocation-feedback covariance. It can be positive or negative, so it is not itself a variance. The sum of all terms must be nonnegative.

A short justification of the variance formula is to center the limiting sum and expand its square: $\operatorname{Var}(Z_R+\Delta A)=E[((Z_R-EZ_R)+\Delta(A-EA))^2]$. The square has two individual variance terms and twice the cross-product expectation, which is $2\Delta\operatorname{Cov}(Z_R,A)$. This also shows why dropping the covariance requires a justification, such as independence or zero covariance; adaptivity alone does not provide one.

Worked Examples

Worked Example: Three Contributions to the Variance

Suppose a hypothetical two-arm design has target $\rho=0.60$, arm means $\mu_1=5$ and $\mu_0=3$, response variances $\sigma_1^2=4$ and $\sigma_0^2=9$, and limiting allocation fluctuation with $\operatorname{Var}(A)=0.25$ and $\operatorname{Cov}(Z_R,A)=0.30$. Assume $E(A)=E(Z_R)=0$ and the joint limit is Gaussian. Find the limiting variance.

Here $\Delta=5-3=2$. The response-noise variance is $v_R=0.60(4)+0.40(9)=2.4+3.6=6.0$. The random-count contribution is $\Delta^2\operatorname{Var}(A)=2^2(0.25)=1.0$. The covariance contribution is $2\Delta\operatorname{Cov}(Z_R,A)=2(2)(0.30)=1.2$. Therefore

$$ V=6.0+1.0+1.2=8.2. $$

The scaled estimation error has a centered Gaussian limit with variance $8.2$ squared-response units, under the stipulated centered joint Gaussian limit. The positive covariance increases the variance relative to the sum of the first two terms alone.

Worked Example: A Gaussian Limit Need Not Be Centered

Suppose a hypothetical allocation schedule has $N_1(n)=0.60n+0.50\sqrt n+o(\sqrt n)$, with integer rounding that does not change the stated limit. Then $A_n\to 0.50$, so $A=0.50$ deterministically. Let $\Delta=2$, and suppose the response-noise limit is $Z_R\sim N(0,6)$.

The decomposition gives $\sqrt n(\hat\theta_n-\theta)\Rightarrow Z_R+2(0.50)=Z_R+1$. Consequently the limit is $N(1,6)$: it is Gaussian but not centered. Its variance is $6$, because a deterministic shift has zero variance. Its squared error about zero has limiting expectation $6+1^2=7$, so the nonzero mean affects mean squared error even though it does not add to variance.

This distinction is important: centering requires $E[Z_R+\Delta A]=0$. In particular, centered $Z_R$ and $E(A)=0$ suffice, but joint Gaussianity by itself does not.

Worked Example: Negative Feedback Covariance

For a further hypothetical design, suppose $v_R=6$, $\Delta=2$, $\operatorname{Var}(A)=0.25$, and $\operatorname{Cov}(Z_R,A)=-0.60$. The covariance is feasible because its squared value, $0.36$, is no greater than $v_R\operatorname{Var}(A)=1.5$.

The variance is $6+2^2(0.25)+2(2)(-0.60)=6+1-2.4=4.6$. The negative covariance reduces the total variance: allocation fluctuations tend to occur alongside response-noise fluctuations that offset their effect on the estimator. This does not mean that either source of variation disappears; it means their sum is less variable than it would be without that negative dependence.

Interpretation, Diagnostics, and Alternatives

The size of $\Delta$ determines how strongly arm-count fluctuations affect this estimator’s mean component. If $\mu_1=\mu_0$, then $\Delta=0$, and the allocation terms vanish from the exact identity, even if the allocation count itself fluctuates substantially. When the arm means differ, the random-count term can matter, and feedback may increase or decrease the total variance.

Useful diagnostics should examine the parts of the decomposition rather than only a single reported standard error. In repeated hypothetical trial runs, one can track $A_n$, $R_n$, their empirical variances, and their covariance. Check that allocation proportions approach the assumed interior target, that neither arm’s count remains too small, and that unusually large responses do not dominate the noise sum. A covariance estimate that changes markedly with trial length, or an arm proportion that does not settle near $\rho$, is evidence that the asymptotic approximation may be unreliable.

A fixed allocation with $N_1(n)-n\rho=o(\sqrt n)$ has $A=0$, so this estimator’s limiting variance reduces to $v_R$. That is a useful benchmark, but using it for an adaptive design is justified only if the allocation fluctuation and its covariance have no first-order effect. For other estimands, such as an arm-specific mean or the treatment-effect contrast, the estimator has a different decomposition; the formula here should not be transferred without deriving the relevant identity.

The result is also limited to a stable two-arm response model with an interior limiting allocation proportion and the stated moment and joint-limit conditions. If response distributions change over time, an arm probability approaches zero, or a required moment fails, the displayed variance need not apply. The decomposition identifies what must be assessed; it does not establish those assumptions automatically.

A Common Mistake

Common mistake: Treating the response-noise term and the random allocation term as independent, then reporting $v_R+\Delta^2\operatorname{Var}(A)$ as the full variance. Correction: Include $2\Delta\operatorname{Cov}(Z_R,A)$ unless zero covariance has been established. Also distinguish variance from centering: a Gaussian limit can have nonzero mean $\Delta E(A)+E(Z_R)$, and a nonzero mean affects mean squared error even when it does not alter variance.

Check Your Understanding

Answer each question using the decomposition, then compare with the concise solutions.

If $\Delta=0$, which terms involving allocation fluctuation remain in the limiting variance?
Suppose $v_R=5$, $\Delta=1.5$, $\operatorname{Var}(A)=0.16$, and $\operatorname{Cov}(Z_R,A)=0.20$. Calculate the limiting variance.
If $A$ is the constant $0.4$, what is $\operatorname{Var}(A)$, and what does it contribute to the limiting mean?
Why is the variance formula incomplete if it omits the covariance term without explanation?

Solutions. (1) Neither the $\Delta^2\operatorname{Var}(A)$ term nor the covariance term remains, since both contain $\Delta$; the variance is $v_R$. (2) $5+(1.5)^2(0.16)+2(1.5)(0.20)=5+0.36+0.60=5.96$. (3) A constant has variance zero, so its variance contribution is zero; it shifts the limiting mean by $\Delta(0.4)$. (4) Response noise and allocation fluctuation may be dependent under adaptive assignment, so the covariance can be nonzero and changes the variance of their sum.