Why a Triangular Array Is Needed
In a fixed-distribution model, each response contribution is often treated as having the same law as the sample grows. Adaptive trials and other evolving designs need not satisfy that simplification. The response distribution may depend on the sample size, the participant’s position in the sequence, or the observed history that determines the next allocation. A central limit theorem must therefore allow the distribution of the \(i\)th contribution in a sample of size \(n\) to differ from the distribution of the \(i\)th contribution in a sample of size \(n+1\).
A triangular array records this possibility: row \(n\) contains the \(n\) contributions used at sample size \(n\). The key is not that distributions stay fixed, but that no contribution dominates the total fluctuation and that the total variance has a stable, nonzero scale. In a sequential trial, the corresponding conditional requirements matter because the response distribution can depend on information available before the response is observed.
Independent Triangular-Array Central Limit Theorem
For independent terms within each row, the Lindeberg condition gives a general criterion for a Gaussian limit. It requires that the fraction of total variance contributed by unusually large centered terms becomes negligible. The threshold is measured relative to the row’s standard deviation \(s_n\), not relative to a fixed response scale.
The condition allows considerable variation: terms need not be identically distributed, and their variances need not be equal. It rules out a different problem—one or a few rare, very large contributions carrying a non-negligible share of the total variance. If \(s_n^2/n\to v\) for some \(0<v<\infty\), then the unstandardized sum has fluctuations of order \(\sqrt{n}\), and its standardized version has the stated standard normal limit.
Worked Example: Normal Responses With Changing Variances
Suppose that, in row \(n\), the independent centered response contributions satisfy \(Z_{n,i}\sim N(0,1+i/n)\), for \(i=1,\ldots,n\). The variance changes across positions and rows. Find the total variance and determine the limit of the standardized sum.
The row variance is $$ s_n^2=\sum_{i=1}^n\left(1+\frac{i}{n}\right) =n+\frac{1}{n}\frac{n(n+1)}{2} =\frac{3n+1}{2}. $$ Thus \(s_n^2/n\to 3/2\), so the standard deviation grows at order \(\sqrt n\).
Every centered normal term has fourth moment \(E(Z_{n,i}^4)=3(1+i/n)^2\leq 12\). For any \(\varepsilon>0\), the inequality \(Z^2\mathbf{1}\{|Z|>\varepsilon s_n\}\leq Z^4/(\varepsilon^2s_n^2)\) gives $$ \frac{1}{s_n^2}\sum_{i=1}^n E\!\left[Z_{n,i}^2\mathbf{1}\{|Z_{n,i}|>\varepsilon s_n\}\right] \leq \frac{12n}{\varepsilon^2s_n^4}\longrightarrow 0. $$ The Lindeberg condition holds. In this particular example the sum is itself normal, so the standardized sum is exactly \(N(0,1)\) for each \(n\); the calculation also verifies the general array condition rather than relying on a common response distribution.
Uniform Moment Bounds as a Practical Check
Checking the Lindeberg tail expression directly can be inconvenient. A uniform moment bound provides a simple sufficient condition. Suppose there are constants \(\delta>0\) and \(C<\infty\), not depending on \(n\) or \(i\), such that \(E|Z_{n,i}|^{2+\delta}\leq C\). If \(s_n^2\geq c n\) for some \(c>0\) and all sufficiently large \(n\), then $$ Z_{n,i}^2\mathbf{1}\{|Z_{n,i}|>\varepsilon s_n\} \leq \frac{|Z_{n,i}|^{2+\delta}}{(\varepsilon s_n)^\delta}. $$ Taking expectations, summing, and dividing by \(s_n^2\) yields $$ \frac{1}{s_n^2}\sum_{i=1}^n E\!\left[Z_{n,i}^2\mathbf{1}\{|Z_{n,i}|>\varepsilon s_n\}\right] \leq \frac{Cn}{\varepsilon^\delta s_n^{2+\delta}} \leq \frac{C}{\varepsilon^\delta c^{1+\delta/2}n^{\delta/2}} \longrightarrow 0. $$ This proves the Lindeberg condition. The variance lower bound matters: a moment bound alone does not ensure that the total variance grows on the required scale.
Worked Example: Bernoulli Responses With Two Distributions
For each even \(n\), suppose the first \(n/2\) independent responses are Bernoulli with success probability \(0.20\), and the remaining \(n/2\) are Bernoulli with success probability \(0.60\). Let \(Z_{n,i}=X_{n,i}-E(X_{n,i})\). Verify a Gaussian limit for their sum.
A Bernoulli variable with success probability \(p\) has variance \(p(1-p)\). The first group therefore contributes variance \(0.20(0.80)=0.16\) per response, and the second contributes \(0.60(0.40)=0.24\). Hence $$ s_n^2=\frac n2(0.16)+\frac n2(0.24)=0.20n. $$ In particular, \(s_n^2/n=0.20>0\).
Each centered response satisfies \(|Z_{n,i}|\leq1\), so \(E|Z_{n,i}|^3\leq1\) uniformly. The moment argument with \(\delta=1\) applies: for any \(\varepsilon>0\), the Lindeberg expression is at most \(n/(\varepsilon s_n^3)=1/(\varepsilon(0.20)^{3/2}\sqrt n)\), which tends to zero. Therefore \(\sum_i Z_{n,i}/\sqrt{0.20n}\) converges in distribution to \(N(0,1)\). Different success probabilities do not invalidate the limit; the variance and tail conditions are what matter.
Conditional Requirements for Sequential Responses
In an adaptive trial, independence may fail after conditioning is handled, because the next assignment and response can depend on the observed history. A useful formulation is in terms of martingale differences. Let \(\mathcal F_{n,i}\) represent the information available through step \(i\) in row \(n\). A contribution \(D_{n,i}\) is a martingale difference if it is measurable using the information through step \(i\) and \(E(D_{n,i}\mid\mathcal F_{n,i-1})=0\). Its conditional variance is \(v_{n,i}=E(D_{n,i}^2\mid\mathcal F_{n,i-1})\).
The conditional variance requirement says that the accumulated predictable variance, after normalization, approaches one in probability. Conditional Lindeberg requires that the accumulated conditional second-moment contribution from terms larger than the normalized threshold vanishes in probability. Together, these are standard sufficient conditions for a martingale-array central limit theorem.
A conditional uniform moment bound can verify the conditional tail requirement. If \(E(|D_{n,i}|^{2+\delta}\mid\mathcal F_{n,i-1})\leq C\) almost surely for all \(n,i\), and \(s_n^2\geq cn\), then the conditional Lindeberg expression is bounded by \(Cn/(\varepsilon^\delta s_n^{2+\delta})\), which tends to zero. This bound is uniform over the observed histories, not merely an average over them. The conditional variance condition remains a separate requirement; a moment bound cannot replace it.
Worked Example: Adaptive Conditional Success Probabilities
Consider a sequential binary response \(Y_{n,i}\) whose success probability \(q_{n,i}\) is determined by the history before step \(i\). Suppose the adaptive rule always sets \(q_{n,i}\) to either \(0.30\) or \(0.70\), possibly depending on that history. Define \(D_{n,i}=Y_{n,i}-q_{n,i}\). Check the conditional requirements.
By construction, \(E(D_{n,i}\mid\mathcal F_{n,i-1})=0\). Also, \(|D_{n,i}|\leq1\), and $$ E(D_{n,i}^2\mid\mathcal F_{n,i-1})=q_{n,i}(1-q_{n,i})=0.21, $$ because both allowed probabilities give the same variance. Thus for \(s_n^2=0.21n\), the normalized conditional variance sum is exactly one for every history: $$ \frac{1}{s_n^2}\sum_{i=1}^n E(D_{n,i}^2\mid\mathcal F_{n,i-1}) =\frac{0.21n}{0.21n}=1. $$ For any fixed \(\varepsilon>0\), once \(\varepsilon\sqrt{0.21n}>1\), no \(|D_{n,i}|\) can exceed \(\varepsilon s_n\). The conditional Lindeberg sum is then zero. The normalized sum \(\sum_iD_{n,i}/\sqrt{0.21n}\) consequently has a standard normal limit. The calculation allows the success probabilities to respond to past data, but it relies on their bounded outcomes and the stable conditional variance.
How to Choose and Diagnose a Condition
For independent rows, Lindeberg is the direct condition, while Lyapunov’s condition is often easier to check: for some \(\delta>0\), \(\sum_i E|Z_{n,i}|^{2+\delta}/s_n^{2+\delta}\to0\). The moment-bound derivation above shows why Lyapunov implies Lindeberg. Lyapunov is convenient but stronger, so failure to verify it does not prove that a Gaussian limit fails. For sequential data, use conditional moments and conditional variances: unconditional bounds alone can conceal histories under which large increments or unstable variance are likely.
| Setting | Variance requirement | Tail requirement | Useful interpretation |
|---|---|---|---|
| Independent triangular array | Total variance \(s_n^2\) is positive and supplies the normalization | Unconditional Lindeberg condition | No large term carries a non-negligible share of total variance |
| Independent array using Lyapunov | Same total-variance normalization | A higher-moment ratio tends to zero | Convenient sufficient check, generally stronger than Lindeberg |
| Sequential martingale array | Normalized sum of conditional variances tends to one in probability | Conditional Lindeberg condition in probability | Variance and tail control hold relative to the observed history |
A practical diagnostic sequence is to identify the centered contribution being summed, calculate its total or conditional variance, establish that this variance is of the proposed order, and then bound the contribution from terms exceeding \(\varepsilon s_n\). For adaptive responses, also check that the conditional variance sum stabilizes across histories. A variance that collapses toward zero, or remains highly random after normalization, cannot be repaired by asserting a uniform moment bound.
Failure Mode: A Rare Large Contribution
Uniformly bounded responses are not essential, but without some tail control a rare extreme contribution can survive normalization. For example, let \(B_n\) be Bernoulli with probability \(1/n\), and set \(Z_{n,1}=n(B_n-1/n)\). Its variance is \(n(1-1/n)\), of order \(n\), even though the large value occurs with probability only \(1/n\). If the other \(n-1\) terms are independent \(N(0,1)\), the row total variance is also of order \(n\). For a fixed \(\varepsilon>0\), the exceptional value of \(Z_{n,1}\) is larger than \(\varepsilon s_n\) for all sufficiently large \(n\), and its contribution to the Lindeberg numerator is approximately \(n\). Dividing by a total variance of order \(2n\) does not make this contribution vanish. Thus the Lindeberg condition fails. The example shows why a large total sample size and finite variance in each row do not by themselves guarantee a normal limit.
Check Your Understanding
Answer each question using the variance and tail conditions, not an assumption that all responses share one distribution.
- An independent array has \(s_n^2/n\to 0\). Does this fact alone verify the usual nondegenerate \(\sqrt n\)-scale condition? Explain.
- For an independent array, suppose \(E|Z_{n,i}|^3\leq 8\) uniformly and \(s_n^2\geq 2n\). What does the moment bound imply about Lindeberg?
- In a sequential array, does a bounded response by itself establish the martingale-array normal limit? Name the other key requirement.
- Why can Lyapunov be useful even though it is stronger than Lindeberg?
Solutions. (1) No. A positive finite limiting variance per observation is needed for a nondegenerate \(\sqrt n\)-scale limit under that normalization; \(s_n^2/n\to0\) indicates a smaller variance scale. (2) Use \(\delta=1\), \(C=8\), and \(c=2\) in the moment bound. The Lindeberg expression is at most \(8n/(\varepsilon s_n^3)\leq 8/(\varepsilon 2^{3/2}\sqrt n)\), which tends to zero. (3) No. One must also verify that the normalized sum of conditional variances converges to one in probability. (4) Lyapunov replaces a tail calculation with a higher-moment calculation; satisfying it guarantees Lindeberg, although some arrays satisfy Lindeberg without satisfying Lyapunov.