Why Moment Assumptions Matter
A central limit theorem for an adaptively sampled response sum needs more than a rule for assigning participants to treatment arms. It also needs control over the response fluctuations that accumulate under that rule. Bounded responses make this control straightforward, but boundedness is stronger than necessary. Under a fixed response distribution with finite variance, a Gaussian limit can still hold; higher moments mainly provide stronger bounds on large observations and can support sharper approximations.
The key distinction is between a condition that guarantees the first-order limit and a condition that makes the argument easier or more quantitative. For adaptive designs, this distinction must be made alongside a check of the allocation process: if the limiting allocation fraction is random, the limiting variance can be random as well.
Setup and Assumptions
Consider \(K\) treatment arms and \(n\) sequential assignments. Let \(A_t\) be the arm assigned at step \(t\), and let \(Y_t\) be the observed response. Write \(\mathcal{F}_{t-1}\) for the information available before the assignment at step \(t\), and suppose the design assigns arm \(k\) with conditional probability \(q_{t,k}=P(A_t=k\mid\mathcal{F}_{t-1})\). The allocation may depend on all earlier assignments and responses.
Assume that, conditional on the past and on assignment to arm \(k\), the response has a fixed arm-specific distribution with mean \(\mu_k\) and variance \(\sigma_k^2\). Responses are generated freshly at each assignment: their conditional distribution does not otherwise depend on earlier responses. For this tutorial, assume \(0<\sigma_k^2<\infty\) for every arm. These conditions make the centered response \(X_t=Y_t-\mu_{A_t}\) have conditional mean zero and finite arm-specific variance.
The sequential increments in \(M_k(n)\) have conditional mean zero given the observed past: the assignment probabilities are chosen from that past, and each arm-specific centered response has mean zero. Thus these sums are martingales. Their conditional variance accumulates according to the allocation probabilities. For example, the predictable variance for arm \(k\)'s sum is \(\sigma_k^2\sum_{t=1}^n q_{t,k}\). The realized count \(N_k(n)\) differs from \(\sum_t q_{t,k}\) by a sum of bounded, mean-zero allocation fluctuations; divided by \(n\), this difference converges to zero in probability.
A basic sufficient allocation condition is \(N_k(n)/n\to\pi_k\) in probability for each arm, where \(\pi_k\) is a constant. Under this condition, the predictable variance divided by \(n\) converges to \(\pi_k\sigma_k^2\). A martingale central limit theorem then gives a Gaussian limit provided the large-increment contribution, called the conditional Lindeberg term, vanishes. The moment assumptions determine how that last condition is checked.
Three Levels of Moment Control
| Assumption on centered responses | What it provides | What it does not automatically provide |
|---|---|---|
| Uniform boundedness: \(|Y_t-\mu_k|\le C\) | Rules out large individual increments and implies all finite moments | A fixed limiting variance or a nonrandom allocation limit |
| Finite arm-specific variance: \(E[(Y_t-\mu_k)^2\mid A_t=k]=\sigma_k^2<\infty\) | For fixed arm distributions, gives the Lindeberg condition and is enough for the first-order Gaussian limit | A quantitative rate of convergence without further conditions |
| A higher moment: \(E[|Y_t-\mu_k|^{2+\delta}\mid A_t=k]\le C\), for \(\delta>0\) | Bounds the Lindeberg term at a rate and can support sharper error bounds | Elimination of random allocation variance or bias in an estimator |
The boundedness condition is convenient, not essential. The finite-variance statement above relies on fixed arm-specific response distributions (or, more generally, conditions that ensure a uniform conditional Lindeberg property). If response distributions change with time, merely having finite variance at each step need not control the accumulation of rare large observations uniformly. A higher-moment bound that holds uniformly over steps is one way to provide that control.
Why Finite Variance Gives the Lindeberg Condition
For a scalar contrast, choose fixed weights \(c_1,\ldots,c_K\), and define \(M_c(n)=\sum_k c_kM_k(n)\). Its increment at time \(t\) is \(D_t=c_{A_t}(Y_t-\mu_{A_t})\). Suppose \(v_n=n^{-1}\sum_t E(D_t^2\mid\mathcal{F}_{t-1})\) converges in probability to a constant \(v>0\). The martingale central limit theorem says that \(M_c(n)/\sqrt n\) converges in distribution to \(N(0,v)\) if, for every \(\varepsilon>0\), $$ \frac{1}{n}\sum_{t=1}^n E\!\left[D_t^2 I(|D_t|>\varepsilon\sqrt n)\mid\mathcal{F}_{t-1}\right] \ \xrightarrow{P}\ 0. $$ This is the conditional Lindeberg condition: after scaling by \(\sqrt n\), no small set of exceptionally large increments contributes a non-negligible share of the variance.
For fixed arm distributions with finite variances, the condition follows by truncating each arm's response tail. In expectation, the left side is bounded by \(\sum_k c_k^2 E_k[X_k^2 I(|c_kX_k|>\varepsilon\sqrt n)]\), where \(X_k=Y-\mu_k\) under arm \(k\). Each term tends to zero because \(E_k[X_k^2]<\infty\); there are only finitely many arms. Markov's inequality then gives convergence in probability of the nonnegative conditional Lindeberg term.
For the scalar contrast above, the limiting variance is \(v=\sum_k c_k^2\pi_k\sigma_k^2\). If allocation fractions converge instead to random limits, the conditional variance may also be random. In that case, a Gaussian approximation with one fixed variance is not justified just by finite response variances: the unconditional limit can be a mixture of normal distributions. The deterministic-fraction theorem should not be used without checking this distinction.
Worked Examples
Worked Example: Bounded Responses Under Adaptive Allocation
Suppose a hypothetical two-arm trial has Bernoulli responses: arm 1 has response probability \(0.30\), and arm 2 has response probability \(0.50\). The allocation rule is adaptive, but assume its observed fractions converge in probability to \(\pi_1=0.60\) and \(\pi_2=0.40\). Find the limiting variance of \(M_1(n)/\sqrt n\).
A Bernoulli response is bounded between 0 and 1. For arm 1, \(\mu_1=0.30\) and \(\sigma_1^2=0.30(1-0.30)=0.21\). The limiting variance is therefore $$ \pi_1\sigma_1^2=0.60(0.21)=0.1260. $$ Under the stated allocation and response assumptions, \(M_1(n)/\sqrt n\) converges in distribution to \(N(0,0.1260)\). This calculation concerns the centered response sum, not the uncentered total response, which also includes its arm-specific mean.
Worked Example: Finite Variance Without a Higher Moment
Consider a hypothetical response \(Y=\mu+X\) on one arm, where \(X\) has a symmetric distribution with mean 0 and variance 1, but no finite \(2+\delta\) moment for any \(\delta>0\). Such a distribution can be constructed with a sufficiently heavy tail whose survival probability is proportional to \(x^{-2}(\log x)^{-2}\) for large \(x\), followed by scaling to variance 1. Suppose adaptive assignments to this arm satisfy \(N_1(n)/n\to0.70\) in probability. What Gaussian limit follows for its centered sum?
The conditional variance per assignment to arm 1 is 1, so the limiting variance of its sum divided by \(\sqrt n\) is $$ \pi_1\sigma_1^2=0.70(1)=0.70. $$ The response is not bounded and does not have a finite moment above order 2. Nevertheless, because its arm-specific distribution is fixed and its variance is finite, the tail-truncation argument establishes the Lindeberg condition. Thus \(M_1(n)/\sqrt n\) converges in distribution to \(N(0,0.70)\). The example shows that a \(2+\delta\) moment is not required for this first-order limit.
Worked Example: What a Higher Moment Adds
Suppose arm 1 responses are normal with mean \(\mu_1\) and variance 1, and its allocation fraction tends to \(0.60\). For a standard normal centered response \(X\), \(E|X|^3=2\sqrt{2/\pi}\approx1.596\). Use this third absolute moment to bound the Lindeberg term at \(n=10{,}000\) and \(\varepsilon=0.50\).
Here \(\delta=1\). Since \(X^2 I(|X|>\varepsilon\sqrt n)\le |X|^3/(\varepsilon\sqrt n)\), the normalized Lindeberg term is bounded above by $$ \frac{1.596}{0.50\sqrt{10{,}000}} =\frac{1.596}{50} =0.03192. $$ The bound illustrates how a higher moment supplies an explicit finite-sample control that finite variance alone does not quantify. The limiting variance remains \(0.60(1)=0.60\), so the centered arm sum divided by \(\sqrt n\) has limit \(N(0,0.60)\). The bound is an upper bound, not the exact Lindeberg term.
Using and Diagnosing a Gaussian Limit
The following checks separate the response-moment argument from the allocation argument:
Use \(Y_t-\mu_{A_t}\), so the response increments have conditional mean zero under the stated model.
Establish convergence of the predictable variance divided by \(n\). A limiting allocation fraction and finite arm variances identify its limit.
Bounded responses imply Lindeberg directly; fixed finite-variance distributions imply it by tail truncation; a uniform \(2+\delta\) moment gives an explicit bound.
If the allocation fractions have random limits, determine whether the limiting variance is random before reporting a single normal distribution.
In practice, response histograms and observed extremes can reveal gross departures from a proposed bounded or light-tailed model, while arm-specific sample variances help assess the scale of the Gaussian approximation. They cannot prove that a population has finite variance, nor do they establish the martingale assumptions. Also inspect allocation fractions: an arm receiving very few assignments may not support a useful normal approximation even if the response distribution has finite variance. If the response law depends on time or past outcomes beyond arm assignment, the fixed-distribution tail argument needs replacement by an appropriate uniform conditional condition.
The main alternatives differ in what they assume and what they buy. Boundedness is the easiest sufficient condition but may be implausible for continuous outcomes. Finite variance is weaker and, for fixed arm distributions, sufficient for the first-order martingale limit. A higher-moment bound is stronger than finite variance but yields a direct rate for the Lindeberg bound. None of these response assumptions alone guarantees a deterministic variance: allocation behavior must also be checked.
Common Pitfall
- Wrong reasoning: “Every arm has a finite variance, so the response sum must converge to a normal distribution with a fixed variance.”
- Correction: Finite variance controls the response tails under the fixed-distribution assumptions, but a Gaussian limit also requires variance accumulation to converge. If adaptive allocation fractions have a random limit, the limiting variance may be random, so a single fixed-variance normal law need not describe the unconditional limit.
- Another mistake: Requiring bounded responses or a finite third moment as if either were necessary for every Gaussian limit. Both are useful sufficient conditions; finite variance can suffice under the conditions stated here.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- For an arm with response variance \(4\) and limiting allocation fraction \(0.25\), what is the limiting variance of its centered response sum divided by \(\sqrt n\)?
- Why does boundedness imply the conditional Lindeberg condition for a fixed response bound?
- Is a finite \(2+\delta\) moment necessary for a first-order Gaussian limit under fixed arm-specific response distributions with finite variance?
- What issue arises if the allocation fraction converges to a random variable rather than a constant?
Solutions. (1) The limiting variance is \(0.25(4)=1\). (2) If \(|Y_t-\mu_k|\le C\), then for \(n>(C/\varepsilon)^2\), no increment can exceed \(\varepsilon\sqrt n\), so the Lindeberg sum is zero. (3) No. Finite variance suffices here because the fixed arm-specific tails make the Lindeberg term vanish. (4) The limiting variance may be random, producing a mixture of normal distributions rather than a single normal distribution with fixed variance.