Central Limit Theorems for Bounded Arrays

Why Boundedness Can Simplify a Central Limit Theorem

For a triangular array, the summands may have different distributions within each row, and the distributions may change from row to row. A central limit theorem still applies when no single summand can make a substantial contribution to the row's total standard deviation. A useful sufficient condition is a deterministic bound on every centered summand that becomes negligible relative to that standard deviation.

The condition is useful because it converts a tail calculation into a simple comparison of two scales. We will express that comparison precisely, show how it verifies the Lindeberg condition, and contrast it with examples where one must examine the summands' tails directly.

The Bounded-Array Theorem

Definition: For each $n$, let $X_{n1},\ldots,X_{nk_n}$ be independent random variables with $E[X_{nk}]=0$ and finite variances. Define the row sum and its standard deviation by $$ S_n=\sum_{k=1}^{k_n}X_{nk},\qquad s_n^2=\sum_{k=1}^{k_n}\operatorname{Var}(X_{nk}). $$ A sequence $b_n\geq 0$ is a uniform row-wise bound if $P(|X_{nk}|\leq b_n)=1$ for every $k\leq k_n$.

The bound is on the centered summands themselves. If uncentered variables $W_{nk}$ satisfy $a\leq W_{nk}\leq b$ almost surely, for example, then $|W_{nk}-E[W_{nk}]|\leq b-a$. Thus a common bounded range gives a valid bound after centering.

Theorem (Central Limit Theorem for Bounded Arrays): Suppose the variables in each row are independent and centered, $0<s_n^2<\infty$, and there are deterministic bounds $b_n$ such that $|X_{nk}|\leq b_n$ almost surely for every $k\leq k_n$. If $$ \frac{b_n}{s_n}\longrightarrow 0, $$ then $$ \frac{S_n}{s_n}\xrightarrow{d}N(0,1). $$

The number of summands $k_n$ need not equal $n$. The theorem does not require identically distributed variables, nor does it require the same bound for every row. It requires that the largest possible centered summand, measured against the row's total standard deviation, vanishes.

Proof Through the Lindeberg Condition

For independent, mean-zero variables in row $n$, the Lindeberg condition is that, for every $\varepsilon>0$, $$ \frac{1}{s_n^2}\sum_{k=1}^{k_n}E\left[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon s_n\}\right]\longrightarrow 0. $$ The Lindeberg–Feller central limit theorem states that if $s_n^2$ is positive and finite, the summands in each row are independent and centered, and this condition holds, then $S_n/s_n\xrightarrow{d}N(0,1)$. We verify its condition directly.

Fix a threshold.
Choose any $\varepsilon>0$. Since $b_n/s_n\to 0$, there is an index $N$ such that $b_n<\varepsilon s_n$ whenever $n\geq N$.

Eliminate the tail event.
For $n\geq N$, every summand satisfies $|X_{nk}|\leq b_n<\varepsilon s_n$ almost surely. Therefore $P(|X_{nk}|>\varepsilon s_n)=0$, and the indicator in every Lindeberg term is zero almost surely.

Conclude the limit.
The entire Lindeberg sum is exactly zero for all $n\geq N$. The Lindeberg condition follows, so the Lindeberg–Feller theorem gives $S_n/s_n\xrightarrow{d}N(0,1)$. $\square$

The key is that the threshold $\varepsilon s_n$ eventually exceeds the bound for every summand, not merely for most of them. There is no interchange of a limit and an expectation: once $n\geq N$, each tail indicator is identically zero almost surely.

Consequences of the Uniform Bound

The same bound makes two other checks immediate. First, no individual summand can account for a nonnegligible share of the total variance. Second, the bound implies Lyapunov's moment condition for every positive order $\delta$, when that condition is useful.

Derived bounds: Under the bounded-array assumptions, $$ \max_{1\leq k\leq k_n}\frac{\operatorname{Var}(X_{nk})}{s_n^2} \leq \frac{b_n^2}{s_n^2}, \qquad \frac{\sum_{k=1}^{k_n}E|X_{nk}|^{2+\delta}}{s_n^{2+\delta}} \leq \left(\frac{b_n}{s_n}\right)^\delta \quad(\delta>0). $$ Thus $b_n/s_n\to0$ implies both that the largest variance fraction vanishes and that the Lyapunov ratio tends to zero.

To prove the first inequality, use $\operatorname{Var}(X_{nk})=E[X_{nk}^2]\leq b_n^2$, then divide by $s_n^2$ and take the maximum. For the second, the almost-sure bound gives $|X_{nk}|^{2+\delta}\leq b_n^\delta X_{nk}^2$. Taking expectations and summing yields $$ \sum_{k=1}^{k_n}E|X_{nk}|^{2+\delta} \leq b_n^\delta\sum_{k=1}^{k_n}E[X_{nk}^2] =b_n^\delta s_n^2. $$ Dividing by $s_n^{2+\delta}$ proves the claim. These are consequences, not extra assumptions needed for the bounded-array theorem.

Worked Examples

Worked Example: Increasing Weights on Uniform Variables

For each $n$, let $U_{n1},\ldots,U_{nn}$ be independent Uniform$(0,1)$ variables, and set $X_{nk}=k(U_{nk}-1/2)$. Determine the limiting distribution of the standardized row sum.

Each summand is centered because $E[U_{nk}]=1/2$. Since $|U_{nk}-1/2|\leq 1/2$, we have $|X_{nk}|\leq k/2\leq n/2$, so take $b_n=n/2$. A Uniform$(0,1)$ variable has variance $1/12$, hence $$ s_n^2=\sum_{k=1}^n\operatorname{Var}(X_{nk}) =\frac{1}{12}\sum_{k=1}^n k^2 =\frac{n(n+1)(2n+1)}{72}. $$

Consequently, $$ \frac{b_n}{s_n} =\frac{n/2}{\sqrt{n(n+1)(2n+1)/72}} \longrightarrow 0, $$ because the numerator has order $n$, whereas $s_n$ has order $n^{3/2}$. The bounded-array theorem applies: $$ \frac{\sum_{k=1}^n k(U_{nk}-1/2)} {\sqrt{n(n+1)(2n+1)/72}} \xrightarrow{d}N(0,1). $$ The summands are not identically distributed: their variances increase with $k$. The theorem accommodates this because the largest bound is still negligible relative to the combined standard deviation.

Worked Example: A Bound That Grows More Slowly Than Total Variation

Let $R_{nk}$ be independent variables taking values $-1$ and $1$, each with probability $1/2$. Set $X_{nk}=\sqrt{k}\,R_{nk}$ for $1\leq k\leq n$. Show that the normalized sum has a standard normal limit.

Symmetry gives $E[X_{nk}]=0$, and $X_{nk}^2=k$, so $\operatorname{Var}(X_{nk})=k$. Thus $$ s_n^2=\sum_{k=1}^n k=\frac{n(n+1)}{2}. $$ The largest absolute summand in row $n$ is $\sqrt n$, so choose $b_n=\sqrt n$. Then $$ \frac{b_n}{s_n} =\frac{\sqrt n}{\sqrt{n(n+1)/2}} =\sqrt{\frac{2}{n+1}} \longrightarrow 0. $$ The theorem therefore yields $$ \frac{\sum_{k=1}^n\sqrt{k}\,R_{nk}}{\sqrt{n(n+1)/2}} \xrightarrow{d}N(0,1). $$ Although the largest summand grows in absolute size, the total standard deviation grows faster. Bounded-array conditions are relative to $s_n$, not absolute conditions that require $b_n\to0$.

Worked Example: Direct Lindeberg Check and a Failed Bound Condition

Consider independent variables $X_{n1},\ldots,X_{nn}$. Let $X_{n1}$ take values $-1$ and $1$ with equal probability, and for $2\leq k\leq n$, let $X_{nk}$ take values $-1/n$ and $1/n$ with equal probability. Each variable is centered. Does the normalized sum converge to a standard normal distribution?

The total variance is $$ s_n^2=1+(n-1)\frac{1}{n^2}\longrightarrow 1. $$ A row-wise bound is $b_n=1$, so $b_n/s_n\to1$, not zero. In fact, the Lindeberg condition fails. Fix $\varepsilon=1/2$. For all sufficiently large $n$, $1>\varepsilon s_n$, while $1/n\leq\varepsilon s_n$. Thus only the first summand contributes, and $$ \frac{1}{s_n^2}\sum_{k=1}^n E\left[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon s_n\}\right] =\frac{1}{s_n^2}\longrightarrow 1, $$ rather than zero.

The failure is substantive. Write $T_n=\sum_{k=2}^nX_{nk}$. Then $E[T_n]=0$ and $\operatorname{Var}(T_n)=(n-1)/n^2\to0$, so Chebyshev's inequality gives $T_n\xrightarrow{p}0$. Since $s_n\to1$, the normalized sum $S_n/s_n$ converges in distribution to a variable taking $-1$ and $1$ with equal probability, not to $N(0,1)$. A summand retains nearly all the variance, so a normal limit should not be expected.

Uniform Bounds Versus Direct Lindeberg Calculations

A uniform bound is a particularly strong and convenient way to verify the Lindeberg condition: once $b_n<\varepsilon s_n$, the Lindeberg sum is exactly zero. But the bound is sufficient, not necessary. An array can have summands with large possible values and still satisfy Lindeberg if those values occur rarely enough that their total truncated second moment is negligible.

In a direct Lindeberg calculation, one fixes $\varepsilon>0$, identifies which summands can exceed $\varepsilon s_n$, and evaluates or bounds their contributions $E[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon s_n\}]$. The bounded-array shortcut avoids that tail analysis only when a deterministic bound works uniformly across the whole row. It is often useful to try the shortcut first, then use direct Lindeberg calculations if the bound is too crude or fails to shrink relative to $s_n$.

Common Mistakes

Bounding uncentered variables but forgetting to center. The theorem concerns $X_{nk}-E[X_{nk}]$. A valid bound for the original variables must be converted into a bound for their centered versions.
Checking $b_n\to0$ instead of $b_n/s_n\to0$. The bound may grow, as in the weighted Rademacher example, provided the total standard deviation grows faster.
Checking only the largest variance fraction. A vanishing largest variance fraction is not, by itself, the bounded-array assumption or a complete Lindeberg verification. The theorem's direct shortcut uses an almost-sure bound on every centered summand.
Assuming boundedness alone guarantees a normal limit. The failed example has summands bounded by one, but its total standard deviation stays near one and a single summand remains dominant.

Key takeaway: For independent centered rows, if every summand is bounded in absolute value by $b_n$ and $b_n/s_n\to0$, then the Lindeberg sum is eventually zero for each fixed threshold. The Lindeberg–Feller theorem then gives a standard normal limit.

Check Your Understanding

Use the row-wise standard deviation $s_n$ and the centered summands in each answer.

State the bounded-array condition that implies a standard normal limit.
Why does $b_n/s_n\to0$ make every Lindeberg tail event empty eventually for a fixed $\varepsilon>0$?
If $|X_{nk}|\leq b_n$, derive an upper bound for $\operatorname{Var}(X_{nk})/s_n^2$.
In the weighted Uniform$(0,1)$ example, why can the largest summand grow while the theorem still applies?
What feature of the two-point example prevents a normal limit, and how does it appear in the Lindeberg sum?