Verifying the Lindeberg Condition

What the Lindeberg Sum Measures

For a triangular array, the Lindeberg condition asks whether components that are large relative to the row’s total standard deviation contribute a negligible fraction of its total variance. It is a tail-truncation check: variance carried by small components is not enough by itself; we must also check how much variance lies beyond each fixed threshold on the normalized scale.

Let $\{X_{nk}:1\leq k\leq m_n,\ n\geq1\}$ be an array of independent random variables with finite second moments. Center each variable by writing $Y_{nk}=X_{nk}-E[X_{nk}]$, and let $B_n^2$ be the total variance in row $n$.

Definition: The array satisfies the Lindeberg condition if, for every fixed $\varepsilon>0$, $$ L_n(\varepsilon) = \frac{1}{B_n^2}\sum_{k=1}^{m_n} E\!\left[Y_{nk}^{\,2}\mathbf{1}\{|Y_{nk}|>\varepsilon B_n\}\right] \longrightarrow 0, \qquad B_n^2=\sum_{k=1}^{m_n}\operatorname{Var}(X_{nk}), $$ where $B_n^2>0$ for all sufficiently large $n$. The indicator $\mathbf{1}\{A\}$ is 1 when $A$ occurs and 0 otherwise. In the Lindeberg–Feller central limit theorem, the condition is used together with $B_n\to\infty$ to obtain a standard normal limit for the centered row sum divided by $B_n$.

The order of operations matters: center each variable, apply the threshold to its centered value, square that value, and then sum its expected contribution. The normalization is the total row variance $B_n^2$. In applications, writing this expression explicitly often reduces the check to a simple bound or to identifying which components can cross the threshold.

Bounded Arrays: A Direct Tail Check

A particularly useful sufficient condition is a deterministic bound on every centered variable. If that bound is negligible compared with the row standard deviation, then no variable reaches the Lindeberg threshold once $n$ is large enough.

Proposition: Suppose that, for each $n$, $|Y_{nk}|\leq c_n$ almost surely for every $1\leq k\leq m_n$, where $c_n$ is deterministic. If $c_n/B_n\to0$, then the array satisfies the Lindeberg condition.

Proof. Fix any $\varepsilon>0$. Since $c_n/B_n\to0$, there is an index $N$ such that $c_n<\varepsilon B_n$ for every $n\geq N$. For those $n$, each variable satisfies $|Y_{nk}|\leq c_n<\varepsilon B_n$ almost surely. Thus every indicator in the Lindeberg sum is zero almost surely, and $L_n(\varepsilon)=0$ for $n\geq N$. This holds for each fixed $\varepsilon>0$, so $L_n(\varepsilon)\to0$. $\square$

Worked Example: A Bounded Rademacher Array

In row $n$, let $Y_{n1},\ldots,Y_{nn}$ be independent, with each variable equal to $1/2$ or $-1/2$, each with probability $1/2$. Find $B_n^2$ and verify the Lindeberg condition.

Each variable is centered and has variance $$ \operatorname{Var}(Y_{nk}) = \frac12\left(\frac12\right)^2+\frac12\left(-\frac12\right)^2 = \frac14. $$ Therefore $B_n^2=n/4$, so $B_n=\sqrt{n}/2$. Every centered component has absolute value $1/2$. For a fixed $\varepsilon>0$, the threshold is $\varepsilon B_n=\varepsilon\sqrt{n}/2$, which exceeds $1/2$ whenever $n>1/\varepsilon^2$. For all such $n$, every tail indicator is zero, and hence $L_n(\varepsilon)=0$. Since this is true for every fixed positive $\varepsilon$, the condition holds.

This argument is stronger than merely bounding the tail sum: for sufficiently large rows, the tail is empty. The bound $c_n/B_n\to0$ is sufficient, not necessary; an array can satisfy the Lindeberg condition even when its largest possible component is not small in this uniform sense.

Bernoulli Variables with Changing Probabilities

For a Bernoulli variable, the centered value takes only two possible values. That makes its tail contribution explicit, even when the success probability varies with both $n$ and $k$.

Formula: If $X_{nk}\sim\operatorname{Bernoulli}(p_{nk})$, with $0\leq p_{nk}\leq1$, then $Y_{nk}=X_{nk}-p_{nk}$ equals $1-p_{nk}$ with probability $p_{nk}$ and $-p_{nk}$ with probability $1-p_{nk}$. For any $t>0$, $$ E\!\left[Y_{nk}^{\,2}\mathbf{1}\{|Y_{nk}|>t\}\right] = p_{nk}(1-p_{nk})^2\mathbf{1}\{1-p_{nk}>t\} + (1-p_{nk})p_{nk}^2\mathbf{1}\{p_{nk}>t\}. $$

The first term is the contribution from a success, whose centered value is $1-p_{nk}$; the second is the contribution from a failure, whose centered value is $-p_{nk}$. The row variance is $B_n^2=\sum_k p_{nk}(1-p_{nk})$. Since $|Y_{nk}|\leq1$, if $B_n\to\infty$, then for any fixed $\varepsilon>0$ the threshold $\varepsilon B_n$ eventually exceeds 1. At that point both indicators vanish for every component.

Worked Example: Heterogeneous Bernoulli Probabilities

For each row $n$, take independent variables $X_{nk}\sim\operatorname{Bernoulli}(k/(n+1))$, $1\leq k\leq n$. Verify the Lindeberg condition, including the row variance calculation.

The variance in row $n$ is $$ B_n^2 = \sum_{k=1}^{n}\frac{k}{n+1}\left(1-\frac{k}{n+1}\right) = \frac{1}{(n+1)^2}\sum_{k=1}^{n}k(n+1-k). $$ Using $\sum_{k=1}^{n}k=n(n+1)/2$ and $\sum_{k=1}^{n}k^2=n(n+1)(2n+1)/6$, the numerator of the sum on the right is $$ (n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(n+2)}{6}. $$ Thus $B_n^2=n(n+2)/6$, which tends to infinity. Each centered Bernoulli variable has absolute value at most 1. For any fixed $\varepsilon>0$, $B_n\to\infty$ implies $\varepsilon B_n>1$ for all sufficiently large $n$. Every tail indicator is then zero, so $L_n(\varepsilon)=0$ for those rows. The array satisfies the Lindeberg condition, despite the success probabilities changing across components and rows.

When the total variance does not grow, boundedness alone does not settle the question: the normalized threshold $\varepsilon B_n$ need not eventually exceed the maximum centered value. The exact Bernoulli formula shows how to check the contributions in that situation, but the central limit theorem requires its own full set of hypotheses.

Unequal Component Scales

A useful way to expose the effect of unequal scales is to multiply centered signs by deterministic weights. Let $R_{nk}$ be independent random variables taking values $1$ and $-1$ with equal probability, and set $Y_{nk}=a_{nk}R_{nk}$, where $a_{nk}\geq0$ is deterministic. Then $B_n^2=\sum_k a_{nk}^2$, and direct substitution gives $$ L_n(\varepsilon) = \frac{1}{B_n^2}\sum_{k=1}^{m_n}a_{nk}^2\mathbf{1}\{a_{nk}>\varepsilon B_n\}. $$ The tail sum is therefore the fraction of total variance carried by weights that exceed the normalized threshold. It is not enough to know the total variance; one must check where that variance is located.

Worked Example: Increasing but Diffuse Weights

In row $n$, let $Y_{nk}=kR_{nk}$, $1\leq k\leq n$, with independent Rademacher variables $R_{nk}$. Verify the condition even though the component scales are unequal.

The total variance is $$ B_n^2=\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}. $$ The largest component magnitude is $n$, and $$ \frac{n}{B_n} = \sqrt{\frac{6n}{(n+1)(2n+1)}} \longrightarrow0. $$ Fix $\varepsilon>0$. For sufficiently large $n$, $n<\varepsilon B_n$. Since $k\leq n$, no component exceeds $\varepsilon B_n$, and the displayed formula gives $L_n(\varepsilon)=0$. Thus the condition holds: although the scales increase with $k$, even the largest is eventually below each fixed normalized threshold.

Worked Example: One Component Carries a Persistent Tail

In row $n$, let $Y_{n1}=\sqrt{n}R_{n1}$, and let $Y_{nk}=R_{nk}$ for $2\leq k\leq n$, with all signs independent and Rademacher. Show that the Lindeberg condition fails.

The total variance is $B_n^2=n+(n-1)=2n-1$. Choose any fixed $\varepsilon$ with $0<\varepsilon<1/\sqrt{2}$. Then $$ \frac{\sqrt{n}}{B_n} = \sqrt{\frac{n}{2n-1}} \longrightarrow\frac{1}{\sqrt{2}}, $$ so the first component exceeds $\varepsilon B_n$ for all sufficiently large $n$. Its contribution to the normalized tail sum is $n/(2n-1)$. The remaining components have magnitude 1, while $\varepsilon B_n\to\infty$, so they eventually contribute nothing. Consequently, $$ L_n(\varepsilon)=\frac{n}{2n-1}\longrightarrow\frac12. $$ This positive limit shows that the condition fails: a non-negligible fraction of the total variance remains in the large component.

A Practical Verification Strategy

For a new array, start by calculating $B_n^2$, then write $L_n(\varepsilon)$ with the actual centered variables and threshold. The following sequence keeps the calculation focused on the tail condition rather than on unrelated properties of the array.

Find the row variance.
Use independence to add the component variances and identify the scale $B_n$.

Identify possible tail contributors.
Determine which centered components can exceed $\varepsilon B_n$, using their support or a direct tail calculation.

Normalize their squared contributions.
Bound or calculate the sum of their tail second moments divided by $B_n^2$, for an arbitrary fixed $\varepsilon>0$.

A common mistake is to say that the Lindeberg condition holds merely because every component has finite variance. That does not control how much variance lies beyond $\varepsilon B_n$. For instance, the unequal-scale example above has finite variance in every row, but one component contributes a limiting fraction $1/2$ to the tail sum. The correction is to evaluate the thresholded second moments themselves. Conversely, a bound such as $|Y_{nk}|\leq c_n$ is useful only after comparing $c_n$ with $B_n$; bounded components need not be below the normalized threshold if the row standard deviation fails to grow.

Key takeaway: The Lindeberg condition is a variance-tail condition. Verify it by locating the components that cross $\varepsilon B_n$ and showing that their combined squared contribution, as a fraction of $B_n^2$, vanishes for every fixed $\varepsilon>0$.

Check Your Understanding

Use the normalized tail sum, not just the total variance, to answer each question.

Suppose every centered component in row $n$ is bounded in absolute value by $c_n$, with $c_n/B_n\to0$. Explain why the Lindeberg sum is eventually zero for each fixed $\varepsilon>0$.
For $X\sim\operatorname{Bernoulli}(p)$, what are the two possible values of $X-E[X]$, and which probability weights belong to them?
For independent variables $Y_{nk}=a_{nk}R_{nk}$ with Rademacher signs, write the Lindeberg sum in terms of $a_{nk}$ and $B_n$.
In the one-dominant-component example, why does the first component prevent the Lindeberg sum from tending to zero?
Why does knowing that each component has finite variance, on its own, not verify the Lindeberg condition?