The Lindeberg–Feller Central Limit Theorem

Why a Central Limit Theorem for Arrays?

The classical central limit theorem describes sums of identically distributed variables, but many statistical sums have independent terms whose distributions change with the sample size. A triangular array represents this situation: each row is a separate finite collection, and the number and distributions of its terms may vary from row to row. The Lindeberg–Feller theorem gives a precise condition under which such a sum, centered and divided by its own standard deviation, approaches a standard normal distribution.

The key is to control the variance contribution from terms that are large relative to the row’s total standard deviation. The variables need not have the same distribution, and the variances need not be equal. What matters is that no appreciable fraction of the total variance is carried by unusually large terms.

Triangular Arrays and Total-Variance Scaling

Write the array as \(\{X_{nk}:1\leq k\leq m_n,\ n\geq1\}\), where \(m_n\) is the number of variables in row \(n\). Assume that variables within each row are independent, have finite second moments, and are centered. Independence between different rows is not required. Define the row sum and its total variance by

S_n=\sum_{k=1}^{m_n}X_{nk}, \qquad s_n^2=\operatorname{Var}(S_n) =\sum_{k=1}^{m_n}\operatorname{Var}(X_{nk}).

The last equality follows from independence within the row. Assume \(s_n^2>0\) for each \(n\). The natural normalization is \(S_n/s_n\): it has mean zero and variance one for every row. In particular, \(s_n\) is the square root of the sum of the row variances, not the square root of the number of terms unless the variances make those quantities proportional.

Definition: The array satisfies the Lindeberg condition if, for every fixed \(\varepsilon>0\),

\frac{1}{s_n^2}\sum_{k=1}^{m_n} E\!\left[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon s_n\}\right] \longrightarrow 0.

Here \(\mathbf{1}\{A\}\) is 1 when the event \(A\) occurs and 0 otherwise. Equivalently, set \(Y_{nk}=X_{nk}/s_n\). Then \(\sum_k E[Y_{nk}^2]=1\), and the condition is \(\sum_k E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|>\varepsilon\}]\to0\) for every \(\varepsilon>0\).

The condition concerns the aggregate second-moment contribution from terms exceeding a fixed fraction of the total standard deviation. It is not enough that each term has mean zero, or that the total variance is finite in each row; large terms must not retain a nonnegligible share of the normalized variance as \(n\) grows.

The Lindeberg–Feller Theorem

Theorem (Lindeberg–Feller): For each \(n\), let \(X_{n1},\ldots,X_{nm_n}\) be independent, mean-zero random variables with finite variances. Let \(s_n^2=\sum_{k=1}^{m_n}\operatorname{Var}(X_{nk})>0\). If the array satisfies the Lindeberg condition, then

\frac{S_n}{s_n} = \frac{\sum_{k=1}^{m_n}X_{nk}}{s_n} \xrightarrow{d}N(0,1).

This is a sufficient-condition theorem: the stated hypotheses guarantee the normal limit. The limit is standard normal because the row sums have been centered and divided by their total standard deviation. For variables \(W_{nk}\) that are not centered, apply the theorem to \(X_{nk}=W_{nk}-E[W_{nk}]\), provided the centered variables meet its assumptions.

Two Consequences of the Lindeberg Condition

The condition places restrictions on individual terms as well as on the row as a whole. The following consequences help explain why the condition rules out a single term carrying substantial variance or remaining large after normalization.

Consequence: If the Lindeberg condition holds, then the largest normalized individual variance tends to zero:

\max_{1\leq k\leq m_n}\frac{\operatorname{Var}(X_{nk})}{s_n^2} \longrightarrow 0.

Proof. Let \(Y_{nk}=X_{nk}/s_n\), so \(E[Y_{nk}]=0\). For any fixed \(\varepsilon>0\) and every \(k\),

E[Y_{nk}^2] = E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|\leq\varepsilon\}] + E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|>\varepsilon\}] \leq \varepsilon^2+ E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|>\varepsilon\}].

The second term for any one \(k\) is no greater than its sum over the row. Taking the maximum and then the limit superior gives

\limsup_{n\to\infty}\max_k E[Y_{nk}^2] \leq \varepsilon^2+ \lim_{n\to\infty}\sum_k E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|>\varepsilon\}] =\varepsilon^2.

This holds for every \(\varepsilon>0\), so the limit superior is zero. Since \(E[Y_{nk}^2]=\operatorname{Var}(X_{nk})/s_n^2\), the claimed result follows. \(\square\)

There is also a probability version. For fixed \(\varepsilon>0\), the union bound and the inequality \(\mathbf{1}\{|y|>\varepsilon\}\leq y^2\mathbf{1}\{|y|>\varepsilon\}/\varepsilon^2\) give

P\!\left(\max_{1\leq k\leq m_n}|Y_{nk}|>\varepsilon\right) \leq \sum_k P(|Y_{nk}|>\varepsilon) \leq \frac{1}{\varepsilon^2}\sum_k E[Y_{nk}^2\mathbf{1}\{|Y_{nk}|>\varepsilon\}] \longrightarrow 0.

Thus, with probability tending to one, no normalized term exceeds any fixed positive threshold. Both consequences follow from the Lindeberg condition; neither replaces the condition in the theorem.

Worked Applications

Worked Example: Independent Normal Terms with Unequal Variances

In row \(n\), let \(X_{n1},\ldots,X_{nn}\) be independent with \(X_{nk}\sim N(0,p_{nk})\), where \(p_{nk}=2k/[n(n+1)]\). Since \(\sum_{k=1}^{n}k=n(n+1)/2\), the row variances sum to 1, so \(s_n^2=1\). Also \(\max_k p_{nk}=2/(n+1)\).

For \(Z\sim N(0,1)\), each \(X_{nk}\) has the same distribution as \(\sqrt{p_{nk}}Z\). Therefore, for fixed \(\varepsilon>0\), the Lindeberg sum is

\sum_{k=1}^{n}E[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon\}] = \sum_{k=1}^{n}p_{nk} E\!\left[Z^2\mathbf{1}\!\left\{|Z|>\frac{\varepsilon}{\sqrt{p_{nk}}}\right\}\right].

Since \(p_{nk}\leq 2/(n+1)\), the threshold in each expectation is at least \(\varepsilon\sqrt{(n+1)/2}\). The sum is consequently bounded above by

\left(\sum_{k=1}^{n}p_{nk}\right) E\!\left[Z^2\mathbf{1}\!\left\{|Z|>\varepsilon\sqrt{\frac{n+1}{2}}\right\}\right] = E\!\left[Z^2\mathbf{1}\!\left\{|Z|>\varepsilon\sqrt{\frac{n+1}{2}}\right\}\right].

Here \(Z\) is one fixed standard normal variable, not a sequence of changing variables. For this fixed \(Z\), the indicator tends pointwise to zero, and the integrand is bounded by the integrable random variable \(Z^2\), with \(E[Z^2]=1\). Dominated convergence shows that the bound tends to zero. The Lindeberg condition holds, so \(\sum_{k=1}^n X_{nk}\xrightarrow{d}N(0,1)\).

Worked Example: Repeating a Fixed Finite-Variance Distribution

Let \(W_1,W_2,\ldots\) be independent copies of a variable \(W\) with \(E[W]=\mu\) and \(0<\operatorname{Var}(W)=\sigma^2<\infty\). Take \(X_{nk}=W_k-\mu\), \(1\leq k\leq n\). Then \(s_n^2=n\sigma^2\). For fixed \(\varepsilon>0\), the Lindeberg expression is

\frac{1}{n\sigma^2}\sum_{k=1}^{n} E\!\left[(W_k-\mu)^2 \mathbf{1}\{|W_k-\mu|>\varepsilon\sigma\sqrt n\}\right] = E\!\left[\frac{(W-\mu)^2}{\sigma^2} \mathbf{1}\{|W-\mu|>\varepsilon\sigma\sqrt n\}\right].

The equality uses that every \(W_k\) has the same distribution as the single variable \(W\). For this fixed \(W\), the indicator tends pointwise to zero, while the integrand is bounded by \((W-\mu)^2/\sigma^2\), whose expectation is 1. Dominated convergence gives a limit of zero. Thus the Lindeberg condition holds and the theorem yields

\frac{\sum_{k=1}^{n}(W_k-\mu)}{\sigma\sqrt n} \xrightarrow{d}N(0,1).

The finite second moment is essential to this verification: it supplies the integrable dominating variable.

Worked Example: A Row of Weighted Random Signs

For each \(n\), let \(X_{nk}=kR_{nk}\), \(1\leq k\leq n\), where the \(R_{nk}\) are independent and take \(1\) and \(-1\) with probability \(1/2\) each. Every term has mean zero and variance \(k^2\). Hence

s_n^2=\sum_{k=1}^{n}k^2 =\frac{n(n+1)(2n+1)}{6}.

The absolute value of every \(X_{nk}\) is at most \(n\). For fixed \(\varepsilon>0\), the smallest possible value of \(\varepsilon s_n\) relative to \(n\) is measured by

\frac{\varepsilon s_n}{n} = \varepsilon\sqrt{\frac{(n+1)(2n+1)}{6n}} \longrightarrow\infty.

Thus, for all sufficiently large \(n\), \(\varepsilon s_n>n\), and every indicator \(\mathbf{1}\{|X_{nk}|>\varepsilon s_n\}\) is zero. The Lindeberg sum is then exactly zero for those rows. The theorem gives \(\sum_{k=1}^n kR_{nk}/s_n\xrightarrow{d}N(0,1)\).

Interpreting the Condition and Avoiding a Pitfall

Lindeberg’s condition is a variance-tail condition, not a requirement that every variable have a particular distribution or that all terms be identically distributed. Its threshold is \(\varepsilon s_n\), the same total-standard-deviation scale used to normalize the sum. The theorem therefore accommodates different row sizes, unequal variances, and changing distributions, provided large terms contribute vanishing total normalized variance.

Common mistake: using a fixed raw threshold. It is incorrect to check whether \(E[X_{nk}^2\mathbf{1}\{|X_{nk}|>\varepsilon\}]\) is small and conclude that the Lindeberg condition holds. The cutoff must grow or shrink with the row’s total standard deviation: it is \(\varepsilon s_n\), and the sum must be divided by \(s_n^2\). For example, when \(s_n\) grows, a fixed raw cutoff may count moderate values as “large” even though they are negligible on the scale of the normalized sum. Conversely, examining only the largest term without summing its variance contribution can miss many individually small tail contributions.

The practical check is to calculate the total variance first, standardize the array, and then examine the aggregate second moment beyond each fixed normalized threshold. The theorem’s conclusion is a distributional limit for the centered row sum; it does not assert that finite-row sums are exactly normal.

Key takeaway: For independent centered variables in row \(n\), normalize by \(s_n=(\sum_k\operatorname{Var}(X_{nk}))^{1/2}\). If the total normalized second-moment contribution from terms larger than \(\varepsilon s_n\) vanishes for every \(\varepsilon>0\), the normalized row sum converges in distribution to \(N(0,1)\).

Check Your Understanding

Use the theorem’s hypotheses and scaling in each answer.

For independent centered variables in one row, how is \(s_n^2\) calculated, and why does independence matter?
Write the Lindeberg condition using the normalized variables \(Y_{nk}=X_{nk}/s_n\).
In the unequal-variance normal example, why can dominated convergence be applied to a single fixed variable \(Z\)?
What conclusion about the largest normalized individual variance follows from the Lindeberg condition?
Why is a fixed raw cutoff \(\varepsilon\), rather than \(\varepsilon s_n\), not the cutoff in the Lindeberg condition?