Conditional Lindeberg Conditions for Response Noise

Why Response Tails Matter Under Adaptive Assignment

A response-adaptive design changes treatment-assignment probabilities as information accumulates. Consequently, the contribution of a response to a normalized statistic can depend on the trial history before that response is observed. A martingale central limit theorem requires more than a well-behaved total conditional variance: it also requires that no small number of unusually large response-noise increments dominate the sum.

The conditional Lindeberg condition formalizes this “no dominating jumps” requirement. Its response-adaptive form must keep the timing of information straight. In particular, a coefficient that is taken outside a conditional expectation given the pre-assignment history must be known from that history. Being known before the response arrives is not enough if the coefficient can depend on the current assignment.

Sequential Setup and the Conditional Condition

Consider a triangular array of trials indexed by $n$, with participants $i=1,\ldots,n$ and treatment arms $a=1,\ldots,K$. Let $\mathcal{F}_{n,i-1}$ contain the information available immediately before participant $i$ is assigned. Write $A_{n,i}$ for the current assignment and $\pi_{n,i,a}=P(A_{n,i}=a\mid\mathcal{F}_{n,i-1})$ for its conditional probability. Thus the assignment probabilities may depend on the past, but are determined by it.

Let $Y_{n,i}$ be the observed response, and let $\mu_{n,i,a}$ be its conditional mean if arm $a$ is assigned. Define the response noise on arm $a$, denoted $\eta_{n,i,a}$, so that $Y_{n,i}-\mu_{n,i,a}=\eta_{n,i,a}$ when $A_{n,i}=a$. Assume the conditional mean-zero property $E(\eta_{n,i,a}\mid\mathcal{F}_{n,i-1},A_{n,i}=a)=0$. The conditional distribution may depend on the trial history and on the assigned arm; identical response distributions across participants are not required.

For each arm, let $b_{n,i,a}$ be a real-valued coefficient measurable with respect to $\mathcal{F}_{n,i-1}$. It is therefore fixed before the current assignment as well as before the response. The response-noise increment is $D_{n,i}=\sum_{a=1}^K I(A_{n,i}=a)b_{n,i,a}\eta_{n,i,a}$. Only the term for the assigned arm contributes. Conditional mean zero of the arm-specific noise implies $E(D_{n,i}\mid\mathcal{F}_{n,i-1})=0$, so these increments are martingale differences in the pre-assignment filtration, with the response revealed at each step.

Definition: The conditional Lindeberg condition for the array $D_{n,1},\ldots,D_{n,n}$ is that, for every fixed $\varepsilon>0$, $$ \sum_{i=1}^n E\!\left[D_{n,i}^2 I(|D_{n,i}|>\varepsilon)\mid\mathcal{F}_{n,i-1}\right]\ \xrightarrow{P}\ 0. $$ It requires the total conditional second-moment contribution from increments exceeding any fixed threshold to vanish in probability.

The threshold is applied to the weighted increment, not just to the raw response noise. A large response may be harmless if its coefficient is very small; a moderate response can matter if it receives a large coefficient. The convergence is in probability because the conditional tail sum itself can be random through the adaptive history.

Arm-Specific Tail Form

Since exactly one arm is assigned, condition first on the pre-assignment history and then sum over the possible assignments. The assumption that each $b_{n,i,a}$ is $\mathcal{F}_{n,i-1}$-measurable allows it to be treated as fixed in the conditional response expectation.

\begin{aligned} &E\!\left[D_{n,i}^2 I(|D_{n,i}|>\varepsilon)\mid\mathcal{F}_{n,i-1}\right]\\ &\quad=\sum_{a=1}^K \pi_{n,i,a}\,b_{n,i,a}^2 E\!\left[\eta_{n,i,a}^2 I\!\left(|b_{n,i,a}\eta_{n,i,a}|>\varepsilon\right) \mid\mathcal{F}_{n,i-1},A_{n,i}=a\right]. \end{aligned}

This is the arm-by-arm version of the condition. It shows how the design enters twice: through the probability of assigning arm $a$, and through the coefficient applied to the response noise on that arm. If $\pi_{n,i,a}=0$, that arm contributes zero for that participant. If $b_{n,i,a}=0$, its weighted increment is zero and contributes nothing to any positive-threshold tail.

The corresponding conditional variance is $$ E(D_{n,i}^2\mid\mathcal{F}_{n,i-1}) =\sum_{a=1}^K \pi_{n,i,a}b_{n,i,a}^2 E(\eta_{n,i,a}^2\mid\mathcal{F}_{n,i-1},A_{n,i}=a). $$ The Lindeberg condition is not a variance condition: it asks whether the part of this variance coming from large weighted increments disappears after summing over participants.

Worked Example: Bounded Noise and Shrinking Weights

Suppose there are two arms, each response noise has conditional mean zero, and $|\eta_{n,i,a}|\leq 2$ almost surely for both arms and every participant. Let the pre-assignment-measurable coefficients be $b_{n,i,a}=1/\sqrt{n}$, and allow the assignment probabilities to vary arbitrarily with the history.

For every assignment and every history, $|D_{n,i}|\leq 2/\sqrt{n}$, because only one arm is assigned. Fix $\varepsilon=0.1$. Whenever $n>400$, $2/\sqrt{n}<0.1$, so $I(|D_{n,i}|>0.1)=0$ almost surely for each $i$. Therefore the entire conditional Lindeberg sum is zero for $n>400$, and in particular converges to zero.

The assignment probabilities need not be constant for this argument: the deterministic bound holds for either arm. The assumptions are checked directly—noise is bounded as stated, the coefficients are known before assignment, and the threshold is eventually larger than the maximum weighted increment.

From Conditional Moments to a Tail Requirement

A convenient sufficient condition is a bound on conditional moments of order $2+\delta$, for some $\delta>0$. For any real $x$, the inequality $x^2 I(|x|>\varepsilon)\leq |x|^{2+\delta}/\varepsilon^\delta$ follows because $|x|^\delta>\varepsilon^\delta$ on the indicated event. Applying this inequality to each increment gives the justification below.

\begin{aligned} &\sum_{i=1}^n E[D_{n,i}^2 I(|D_{n,i}|>\varepsilon)\mid\mathcal{F}_{n,i-1}]\\ &\qquad\leq \varepsilon^{-\delta} \sum_{i=1}^n E[|D_{n,i}|^{2+\delta}\mid\mathcal{F}_{n,i-1}]\\ &\qquad=\varepsilon^{-\delta} \sum_{i=1}^n\sum_{a=1}^K \pi_{n,i,a}|b_{n,i,a}|^{2+\delta} E[|\eta_{n,i,a}|^{2+\delta}\mid\mathcal{F}_{n,i-1},A_{n,i}=a]. \end{aligned}

The last equality again uses pre-assignment measurability of $b_{n,i,a}$. Thus, if the arm-weighted conditional $2+\delta$ moment sum on the right converges to zero in probability, the conditional Lindeberg condition follows for every fixed $\varepsilon>0$. This is a sufficient criterion, not a necessary one: Lindeberg can hold even when no convenient higher-moment bound is available.

Worked Example: A Conditional Third-Moment Bound

Suppose there are two arms, the coefficients are $b_{n,i,a}=1/\sqrt{n}$, and the response distributions may depend on the past. Assume that for every arm, participant, and history, $E(|\eta_{n,i,a}|^3\mid\mathcal{F}_{n,i-1},A_{n,i}=a)\leq 4$. The assignment probabilities are history-dependent but sum to one across the two arms.

Use $\delta=1$. For each participant, the arm-weighted third-moment sum is bounded by \[ \sum_{a=1}^2\pi_{n,i,a}n^{-3/2}(4) =4n^{-3/2}\sum_{a=1}^2\pi_{n,i,a} =4n^{-3/2}. \] Summing over $i=1,\ldots,n$ gives at most $4/\sqrt{n}$. Consequently, for any fixed $\varepsilon>0$, the conditional Lindeberg sum is bounded by $\varepsilon^{-1}4/\sqrt{n}$, which converges to zero.

For example, at $n=10{,}000$ and $\varepsilon=0.5$, this upper bound is $2(4/100)=0.08$. It is an upper bound, not the exact tail sum. The conclusion follows from the stated conditional third-moment bound and the normalized coefficients; no independence or constant assignment probability is needed for this calculation.

Tail Conditions Can Be More Informative Than Moments

A direct tail check is often preferable when response distributions are skewed or heavy-tailed. For each arm and participant, inspect the conditional quantity $E[\eta_{n,i,a}^2 I(|b_{n,i,a}\eta_{n,i,a}|>\varepsilon)\mid\mathcal{F}_{n,i-1},A_{n,i}=a]$, multiply it by $\pi_{n,i,a}b_{n,i,a}^2$, and sum. This asks exactly how much conditional variance lies beyond the relevant weighted threshold. A uniform conditional moment bound may obscure that information or may fail even when the actual tail contribution is small.

For a rough diagnostic, the threshold in raw response units on arm $a$ is $\varepsilon/|b_{n,i,a}|$ when $b_{n,i,a}\ne0$. As weights shrink, this raw-response threshold grows. However, looking only at that threshold is not enough: the tail probability and the squared magnitude of the responses beyond it both matter. A rare event can still carry a non-vanishing share of the conditional variance.

Worked Example: Rare Large Responses Violate Lindeberg

Consider one arm with assignment probability one and $b_{n,i,1}=1/\sqrt{n}$. For every $i$, let the conditional response noise be $+\sqrt{n}$ with probability $1/(2n)$, $-\sqrt{n}$ with probability $1/(2n)$, and zero otherwise, independently of the past. Its conditional mean is zero and its conditional variance is \[ E(\eta_{n,i,1}^2\mid\mathcal{F}_{n,i-1}) =n\left(\frac{1}{2n}+\frac{1}{2n}\right)=1. \] Thus the sum of conditional variances of the weighted increments is $n(1/n)=1$.

But $D_{n,i}$ equals $+1$ or $-1$ with total probability $1/n$, and otherwise equals zero. For any $0<\varepsilon<1$, \[ E[D_{n,i}^2 I(|D_{n,i}|>\varepsilon)\mid\mathcal{F}_{n,i-1}] =\frac{1}{n}. \] Summing over all $n$ participants gives exactly $1$, not a quantity converging to zero. The conditional Lindeberg condition fails: the total variance does not disperse across small increments but remains in rare jumps of size one. This example has finite variance for every $n$, but its conditional third moment grows with $n$, so the preceding uniform moment argument cannot apply.

Checking the Information Timing

The response-adaptive setting makes measurability a substantive part of the calculation. If a coefficient $b_{n,i,a}$ is chosen using the current assignment, it is generally not measurable with respect to $\mathcal{F}_{n,i-1}$. One cannot then pull it outside a conditional expectation given $\mathcal{F}_{n,i-1}$ as if it were already known. The arm-specific expressions above require coefficients fixed from the pre-assignment history.

If an analysis genuinely uses a weight that depends on the current assignment, the conditioning must reflect that timing. One can first condition on the assignment using the enlarged information $\mathcal{F}_{n,i-1}\vee\sigma(A_{n,i})$, evaluate the response-noise tail for that assigned arm, and then average over assignments. The formula must retain any assignment-dependent weight inside the appropriate conditional expression; it cannot be replaced by the pre-assignment-measurable-weight formula without justification.

Common pitfall: “The weight is known before the response, so it can be taken outside the expectation conditional on the pre-assignment history.” This is incorrect if the weight can depend on the current assignment, because that assignment is not yet known given $\mathcal{F}_{n,i-1}$. For the arm-by-arm formula used here, require $b_{n,i,a}$ to be $\mathcal{F}_{n,i-1}$-measurable. Otherwise condition after assignment and keep assignment-dependent weights inside the relevant conditional calculation.

Alternatives, Diagnostics, and Limits

Three checks are useful, but they answer slightly different practical questions:

Check	What it establishes	Main limitation
Direct conditional tail sum	Verifies the Lindeberg condition itself at each fixed threshold.	May be difficult to calculate for complicated conditional response distributions.
Conditional $2+\delta$ moment bound	Gives a simple sufficient bound through the inequality derived above.	Can be too strong; a higher moment may not exist even when Lindeberg holds.
Truncation and tail assessment	Separates typical responses from rare extreme responses and can reveal which arm or time period drives the tail sum.	Truncating for a diagnostic does not itself prove the original, untruncated condition.

The condition is often assessed theoretically from a response model or empirically through arm-specific tail summaries and sensitivity checks. In finite data, observing no extreme response does not prove a conditional tail condition: an event can be rare yet contribute materially to the variance. In particular, check both the arm-specific assignment probabilities and the scale of the weights, and examine whether conditional moment or tail bounds are plausible in the regions of the history where the design assigns patients.

Conditional Lindeberg is a no-large-increments requirement used alongside the conditional-variance requirements of a martingale central limit theorem. It does not by itself establish a Gaussian limit, and it does not assert that responses are independent, identically distributed, or normally distributed. Its role is narrower: it rules out a persistent contribution to total variance from large normalized response-noise increments.

Fix the information set.
Identify $\mathcal{F}_{n,i-1}$ as the information available before the current assignment, and verify which coefficients are measurable with respect to it.

Write the assigned-arm increment.
Express the response noise as $D_{n,i}=\sum_a I(A_{n,i}=a)b_{n,i,a}\eta_{n,i,a}$, with a conditionally mean-zero response error on each arm.

Choose a tail or moment route.
Evaluate the arm-specific tail sum directly, or bound it using conditional $2+\delta$ moments and check that the resulting sum vanishes.

Look for concentration in rare jumps.
Check whether any arm, history, or small set of participants can retain a non-vanishing share of the weighted conditional variance above a fixed threshold.

Key takeaway: In adaptive trials, translate the Lindeberg condition into arm-specific weighted response tails. The assignment probabilities and response distributions may vary with history, but coefficients moved outside pre-assignment conditional expectations must be known before the current assignment.

Check Your Understanding

Answer the questions, then compare your reasoning with the solutions.

Why must $b_{n,i,a}$ be measurable with respect to $\mathcal{F}_{n,i-1}$ to use the arm-specific identity in this tutorial?
Suppose $|\eta_{n,i,a}|\leq 3$ and $b_{n,i,a}=1/\sqrt{n}$ for every arm. For a fixed $\varepsilon>0$, why does the conditional Lindeberg sum eventually equal zero?
If the arm-weighted conditional $2+\delta$ moment sum converges to zero in probability, what inequality implies the conditional Lindeberg condition?
In the rare-response example, what is the conditional variance sum, and why does Lindeberg nevertheless fail?

Solutions. (1) The coefficient must be fixed given the conditioning information; otherwise it can depend on the current assignment and cannot be pulled out of a pre-assignment conditional expectation. (2) Every increment has magnitude at most $3/\sqrt{n}$, which is eventually below any fixed positive $\varepsilon$, so every threshold indicator is zero. (3) Use $x^2I(|x|>\varepsilon)\leq |x|^{2+\delta}/\varepsilon^\delta$, then sum conditional expectations. (4) The variance sum is one, but for $0<\varepsilon<1$ the Lindeberg sum is also one, because each of $n$ increments contributes $1/n$ through rare jumps of magnitude one.