Unequal Response Variances and Arm-Specific Limits

Why arm-specific limits matter

In an adaptive trial, the number assigned to each treatment is part of the evolving design. If responses also have different variances across treatments, uncertainty in an arm mean depends on both the variability of that arm’s responses and how many participants it receives. A high-variance arm can have a less precise mean even when its allocation fraction matches that of a lower-variance arm.

The central calculation is simple to state but important to get right: if arm \(k\) receives a fraction \(\pi_k\) of \(n\) participants in the limit, then its sample mean has asymptotic variance \(\sigma_k^2/(n\pi_k)\). Equivalently, scaling its estimation error by \(\sqrt n\) gives limiting variance \(\sigma_k^2/\pi_k\). The fraction divides the response variance; it does not multiply it.

Setup and assumptions

Consider a trial with \(K\) treatment arms and \(n\) sequentially enrolled participants. Let \(A_t\) denote the arm assigned to participant \(t\), and let \(N_k(n)=\sum_{t=1}^n I(A_t=k)\) be the number assigned to arm \(k\). The observed response from a participant assigned to arm \(k\) is denoted \(Y_{k}\) when discussing that arm’s response distribution; its mean and variance are \(\mu_k\) and \(\sigma_k^2\). The observed sample mean in arm \(k\) is \(\bar Y_k\).

For the basic joint limit, use these conditions:

Assignments are made using information available before the response of the participant being assigned. Thus the assignment rule does not use that participant’s as-yet-unobserved response.
Within each arm, responses follow a stable distribution with mean \(\mu_k\) and finite variance \(\sigma_k^2\). The response innovation for a participant assigned to arm \(k\) has that mean and variance conditional on the history and assignment.
Arm-specific response innovations have no contemporaneous cross-arm contribution: at a given participant only one arm is assigned. The conditional variance of the arm-\(k\) centered contribution is \(\sigma_k^2\) when that arm is assigned.
Allocation fractions converge in probability to fixed constants \(\pi_k>0\), with \(\sum_{k=1}^K\pi_k=1\).
The centered response contributions satisfy the conditional Lindeberg condition required by the martingale central limit theorem. Informally, no single unusually large response contributes a non-negligible share of the normalized sum asymptotically.

The positivity condition ensures each arm’s sample size grows proportionally to \(n\), so every arm mean is eventually defined and has a root-\(n\) limit. The finite-variance and Lindeberg conditions justify a Gaussian approximation; allocation fractions alone do not establish normality.

Definition: The limiting allocation fraction for arm \(k\) is \(\pi_k\) when \(N_k(n)/n\) converges in probability to \(\pi_k\). It describes the asymptotic share of participants assigned to that arm. Under the assumptions above, the arm-specific root-\(n\) error is \(\sqrt n(\bar Y_k-\mu_k)\), whose limiting variance is \(\sigma_k^2/\pi_k\).

Deriving the joint limit

Define the centered arm-\(k\) response sum through participant \(n\) by \(S_k(n)=\sum_{t=1}^n I(A_t=k)(Y_t-\mu_k)\), where \(Y_t\) is the response observed at time \(t\). Since \(S_k(n)=N_k(n)(\bar Y_k-\mu_k)\) whenever \(N_k(n)>0\), we can first find the joint limit of the centered sums and then divide by the arm counts.

The predictable assignment rule makes the centered contributions martingale differences: before observing participant \(t\)’s response, the conditional expectation of \(I(A_t=k)(Y_t-\mu_k)\) is zero. Their accumulated conditional variances for arm \(k\), divided by \(n\), converge to \(\pi_k\sigma_k^2\). Contributions for distinct arms at the same participant have product zero because only one arm is assigned. The martingale central limit theorem, with the stated Lindeberg condition, therefore gives the centered-sum limit. Slutsky’s theorem then converts it to the limit for the means.

\left(\frac{S_1(n)}{\sqrt n},\ldots,\frac{S_K(n)}{\sqrt n}\right) \ \xrightarrow{d}\ N_K\!\left(0,\operatorname{diag}(\pi_1\sigma_1^2,\ldots,\pi_K\sigma_K^2)\right).

For each arm, \(\sqrt n(\bar Y_k-\mu_k)=\{n/N_k(n)\}\{S_k(n)/\sqrt n\}\), and \(n/N_k(n)\) converges in probability to \(1/\pi_k\). Multiplying each coordinate by its limiting factor gives the joint arm-mean result:

Theorem: Under the assumptions above,

\sqrt n \begin{pmatrix} \bar Y_1-\mu_1\\ \vdots\\ \bar Y_K-\mu_K \end{pmatrix} \ \xrightarrow{d}\ N_K\!\left( 0,\operatorname{diag}\!\left(\frac{\sigma_1^2}{\pi_1},\ldots,\frac{\sigma_K^2}{\pi_K}\right) \right).

The diagonal form describes the limiting covariance under these conditions. In finite samples, adaptive allocation can make arm means dependent because assignments respond to previous outcomes. The zero off-diagonal entries in this particular limit are not a claim that finite-sample dependence is impossible; they follow from the asymptotic martingale argument and its assumptions.

Worked calculations

Worked Example: Three Arms With Unequal Variances

Suppose a hypothetical three-arm trial has fixed limiting allocation fractions \((\pi_1,\pi_2,\pi_3)=(0.50,0.30,0.20)\). Suppose the response variances are \((\sigma_1^2,\sigma_2^2,\sigma_3^2)=(4,9,16)\), in squared response units. Assume predictable allocation, stable arm-specific response distributions with finite variances, and the conditional Lindeberg condition.

The theorem gives each component variance by dividing the arm’s response variance by its fraction:

\frac{\sigma_1^2}{\pi_1}=\frac{4}{0.50}=8,\qquad \frac{\sigma_2^2}{\pi_2}=\frac{9}{0.30}=30,\qquad \frac{\sigma_3^2}{\pi_3}=\frac{16}{0.20}=80.

Thus the root-\(n\) vector of mean errors has limiting covariance matrix \(\operatorname{diag}(8,30,80)\). For large \(n\), the approximate variance of \(\bar Y_3\) is \(80/n\), whereas that of \(\bar Y_1\) is \(8/n\). Arm 3’s mean is less precise both because its responses are more variable and because it receives a smaller fraction. The units of each variance are squared response units.

Worked Example: A Two-Arm Contrast

Consider a hypothetical trial with \(n=200\), 140 participants assigned to arm 1 and 60 to arm 2. Suppose the allocation fractions are treated as tending to \(\pi_1=0.70\) and \(\pi_2=0.30\), and the arm response variances are estimated as \(s_1^2=4\) and \(s_2^2=25\) response-units squared. The observed arm means are \(\bar Y_1=12.4\) and \(\bar Y_2=8.1\). For this large-sample calculation, assume the same regularity conditions as above and use the sample variances as estimates of the population variances.

The estimated contrast is \(\bar Y_1-\bar Y_2=12.4-8.1=4.3\) response units. The asymptotic variance of the root-\(n\) contrast is the sum of the two arm-specific variances:

\frac{s_1^2}{\pi_1}+\frac{s_2^2}{\pi_2} =\frac{4}{0.70}+\frac{25}{0.30} =5.7142857\ldots+83.3333333\ldots =89.047619\ldots\approx89.048.

Therefore the estimated standard error of the contrast is \(\sqrt{89.048/200}=0.667\) response units. A normal-approximation 95% interval is \(4.3\pm1.96(0.667)=[2.993,5.607]\) response units. This interval uses estimated variances and an asymptotic approximation; it is not an exact finite-sample interval. Its interpretation is that the procedure, under repeated trials satisfying the model and allocation conditions, has approximate 95% coverage for the mean difference.

The calculation shows why the less frequently assigned, higher-variance arm can dominate uncertainty: arm 2 contributes \(25/0.30=83.333\) of the \(89.048\) units of root-\(n\) contrast variance.

Worked Example: A Random Limiting Allocation Fraction

Suppose, hypothetically, an adaptive design has a random limiting allocation vector: with probability one-half it is \((\Pi_1,\Pi_2)=(0.25,0.75)\), and with probability one-half it is \((0.75,0.25)\). Let the arm response variances be \(\sigma_1^2=1\) and \(\sigma_2^2=9\). Assume in addition that a stable martingale limit applies, with Gaussian innovations conditionally independent of the limiting allocation vector. This additional condition is made explicit because convergence of allocation fractions by itself does not guarantee a conditional Gaussian limit.

In the first allocation scenario, the conditional limiting covariance matrix for the root-\(n\) arm-mean errors is \(\operatorname{diag}(1/0.25,9/0.75)=\operatorname{diag}(4,12)\). In the second scenario it is \(\operatorname{diag}(1/0.75,9/0.25)=\operatorname{diag}(1.333,36)\).

Consequently, the unconditional limit is a mixture of two centered bivariate normal distributions, not generally a single bivariate normal distribution with one fixed covariance matrix. For example, its first-coordinate variance is the average \(0.5(4)+0.5(1.333)=2.667\), and its second-coordinate variance is \(0.5(12)+0.5(36)=24\). Those averages summarize unconditional variances, but replacing the mixture by a single normal distribution with those variances can give an inaccurate description of its tails.

Random limits, checks, and alternatives

When \(N_k(n)/n\) converges to a random \(\Pi_k>0\), the variance scaling suggests the conditional covariance \(\operatorname{diag}(\sigma_k^2/\Pi_k)\). A mixed-normal conclusion of that form requires an appropriate stable martingale central limit result, including conditions on the relation between the limiting allocation and the Gaussian innovations. Random convergence of the fractions alone is not enough. This distinction matters when allocation feedback makes the limiting fractions depend on the response history.

A practical diagnostic is to check the assumptions arm by arm and on the allocation path. Examine whether each \(N_k(n)\) grows proportionally to \(n\), whether any fraction approaches zero, and whether response variances are plausibly stable over enrollment. Investigate extreme responses and influential observations because the Lindeberg condition rules out a single response dominating the normalized sum. If the allocation fractions remain unstable, an interior-fraction limit may not describe the data; if an arm’s fraction vanishes, its root-\(n\) scaling need not apply.

For a treatment contrast \(c^\mathsf{T}\bar{\mathbf Y}\), where \(\bar{\mathbf Y}\) is the vector of arm means and \(c\) is a fixed contrast vector, the limiting variance of \(\sqrt n(c^\mathsf{T}\bar{\mathbf Y}-c^\mathsf{T}\boldsymbol\mu)\) is \(\sum_{k=1}^K c_k^2\sigma_k^2/\pi_k\) under the diagonal-covariance result. In a two-arm difference, \(c=(1,-1)\), so both arm variances enter positively. This contrasts with an equal-variance pooled calculation, which imposes a common variance and can misstate uncertainty when arm variances differ. A fixed-allocation analysis uses the same arm-specific formula if its fixed fractions are positive; adaptivity changes how the fractions arise, not the variance scaling once the stated limit holds.

Common mistake: Multiplying an arm’s response variance by its allocation fraction gives \(\pi_k\sigma_k^2\), which is the variance contribution for the centered sum divided by \(\sqrt n\), not for the arm mean scaled by \(\sqrt n\). Since the arm mean averages about \(n\pi_k\) responses, its root-\(n\) variance is \(\sigma_k^2/\pi_k\). A smaller allocation fraction therefore increases the mean’s asymptotic variance.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

An arm has response variance \(16\) and limiting allocation fraction \(0.40\). What is the limiting variance of its root-\(n\) mean error?
Two arms have variances \(4\) and \(9\), with limiting fractions \(0.50\) and \(0.50\). What is the limiting variance of the root-\(n\) difference of their sample means?
Why is a positive limiting allocation fraction needed for the stated arm-specific root-\(n\) result?
If the allocation fraction has a random limit, is that alone sufficient to conclude a mixed-normal limit? Explain.

Solutions. (1) Divide by the fraction: \(16/0.40=40\). (2) Add the arm contributions: \(4/0.50+9/0.50=8+18=26\). (3) A positive fraction ensures the arm count grows proportionally to \(n\), so the arm mean has the stated root-\(n\) scaling; if the fraction vanishes, that scaling may fail. (4) No. Random fraction convergence alone does not establish a conditional Gaussian limit; an appropriate stable martingale central limit result and its additional conditions are required.