Characteristic-Function Proof for Triangular Arrays

From a Tail Condition to a Normal Limit

For a triangular array, the summands can have different distributions in each row, so the familiar identical-summand characteristic-function calculation does not apply directly. The Lindeberg condition supplies the control that calculation needs: after normalization, the total variance coming from summands larger than any fixed threshold must vanish. We will use it to expand each component characteristic function, bound the sum of the remainders, and then take the product across a row.

Let \(X_{n,1},\ldots,X_{n,k_n}\) be independent random variables within row \(n\), with \(E[X_{n,k}]=0\) and finite variances. Independence between different rows is not required. Define the total row variance and its positive square root by

s_n^2=\sum_{k=1}^{k_n}\operatorname{Var}(X_{n,k}) =\sum_{k=1}^{k_n}E[X_{n,k}^2], \qquad s_n>0.

The quantity of interest is the normalized row sum \(S_n/s_n\), where \(S_n=\sum_{k=1}^{k_n}X_{n,k}\). The normalization gives this sum mean zero and variance one.

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

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

Here \(\mathbf{1}\{A\}\) is 1 when event \(A\) occurs and 0 otherwise. Equivalently, for \(Y_{n,k}=X_{n,k}/s_n\), the total second moment of the normalized summands beyond any fixed threshold tends to zero.

The Component Expansion and Its Remainder

Write \(Y_{n,k}=X_{n,k}/s_n\). Then \(E[Y_{n,k}]=0\) and \(\sum_k E[Y_{n,k}^2]=1\) for every \(n\). For a fixed real \(t\), the characteristic function of one component is

\phi_{n,k}(t)=E[e^{itY_{n,k}}] =1-\frac{t^2}{2}E[Y_{n,k}^2]+r_{n,k}(t),

where the linear term vanishes because \(E[Y_{n,k}]=0\), and

r_{n,k}(t)=E\!\left[e^{itY_{n,k}}-1-itY_{n,k} +\frac{t^2Y_{n,k}^2}{2}\right].

The key is not merely that each individual remainder is small; we need the sum of the absolute remainders across the entire row to vanish. For real \(u\), Taylor’s theorem gives \(|e^{iu}-1-iu+u^2/2|\le |u|^3/6\). Also, integrating the second derivative of \(e^{iu}\) gives \(|e^{iu}-1-iu|\le u^2/2\), and therefore \(|e^{iu}-1-iu+u^2/2|\le u^2\) for all real \(u\).

Remainder bound: If the array satisfies the Lindeberg condition, then for every fixed \(t\),

\sum_{k=1}^{k_n}|r_{n,k}(t)|\longrightarrow 0.

To prove the bound, fix \(\delta>0\) and split each expectation according to whether \(|Y_{n,k}|\le\delta\). On this event use the cubic Taylor bound; on its complement use the global quadratic bound. Summing gives

\begin{aligned} \sum_k |r_{n,k}(t)| &\le \frac{|t|^3\delta}{6}\sum_k E[Y_{n,k}^2] +t^2\sum_k E\!\left[Y_{n,k}^2 \mathbf{1}\{|Y_{n,k}|>\delta\}\right]\\ &=\frac{|t|^3\delta}{6} +t^2\sum_k E\!\left[Y_{n,k}^2 \mathbf{1}\{|Y_{n,k}|>\delta\}\right]. \end{aligned}

For fixed \(\delta\), the last sum tends to zero by the Lindeberg condition. Hence the limit superior of the left side is at most \(|t|^3\delta/6\). Since this holds for every \(\delta>0\), letting \(\delta\) decrease to zero proves the bound. \(\square\)

Taking the Product Across a Row

Independence within row \(n\) makes the characteristic function of the normalized sum the product of its component characteristic functions. To control this product, set \(a_{n,k}=E[Y_{n,k}^2]\) and \(z_{n,k}=\phi_{n,k}(t)-1\). The expansion just proved gives

z_{n,k}=-\frac{t^2}{2}a_{n,k}+r_{n,k}(t), \qquad \sum_k z_{n,k}\longrightarrow-\frac{t^2}{2},

because \(\sum_k a_{n,k}=1\) and the sum of the remainders tends to zero.

We also need the largest individual variance contribution to vanish. For any \(\delta>0\),

a_{n,k}\le \delta^2+ E\!\left[Y_{n,k}^2\mathbf{1}\{|Y_{n,k}|>\delta\}\right].

Indeed, on \(|Y_{n,k}|\le\delta\), the second moment is at most \(\delta^2\), and the probability of that event is at most one. Taking the maximum over \(k\), the tail term is bounded by the sum of all row tail terms. That sum tends to zero by Lindeberg. Thus \(\limsup_n\max_k a_{n,k}\le\delta^2\), and letting \(\delta\) decrease to zero yields \(\max_k a_{n,k}\to0\). From the formula for \(z_{n,k}\) and the remainder bound, \(\max_k|z_{n,k}|\to0\) as well.

Moreover, \(\sum_k|z_{n,k}|\le t^2/2+\sum_k|r_{n,k}(t)|\), so these sums are bounded for all sufficiently large \(n\). Consequently,

\sum_k|z_{n,k}|^2 \le \left(\max_k|z_{n,k}|\right)\sum_k|z_{n,k}| \longrightarrow 0.

For complex \(z\) with \(|z|\le1/2\), the logarithm defined by its power series satisfies \(\log(1+z)=z+O(|z|^2)\), with a uniform constant on that disk. Since \(\max_k|z_{n,k}|\to0\), this expansion applies to every component for all sufficiently large \(n\). Independence now gives

\begin{aligned} E[e^{itS_n/s_n}] &=\prod_k\phi_{n,k}(t) =\prod_k(1+z_{n,k})\\ &=\exp\!\left\{\sum_k\log(1+z_{n,k})\right\} \longrightarrow \exp(-t^2/2). \end{aligned}

The last step follows because \(\sum_k z_{n,k}\to-t^2/2\) and the total logarithm error is bounded by a constant times \(\sum_k|z_{n,k}|^2\), which tends to zero.

Theorem (Lindeberg–Feller central limit theorem, characteristic-function proof): Suppose each row consists of independent, mean-zero random variables with finite variances, \(s_n^2=\sum_k\operatorname{Var}(X_{n,k})>0\), and the Lindeberg condition holds. Then

\frac{\sum_{k=1}^{k_n}X_{n,k}}{s_n} \xrightarrow{d}N(0,1).

Proof. The preceding argument shows that for each real \(t\), the characteristic function of the normalized sum converges to \(e^{-t^2/2}\), the characteristic function of \(N(0,1)\). The Lévy continuity theorem states that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution to the law with that limiting characteristic function. Since \(e^{-t^2/2}\) is continuous at zero, the claimed convergence follows. \(\square\)

Worked Applications

Worked Example: A Rademacher Array

For each \(n\), let \(X_{n,1},\ldots,X_{n,n}\) be independent, with each variable equal to \(1\) or \(-1\) with probability \(1/2\). Here \(E[X_{n,k}]=0\), each variance is 1, and \(s_n^2=n\). For any fixed \(\varepsilon>0\), once \(n>\varepsilon^{-2}\), every summand satisfies \(|X_{n,k}|=1\le\varepsilon\sqrt n\). Thus all the Lindeberg indicators are zero for sufficiently large \(n\), and the condition holds.

The theorem gives \(\sum_kX_{n,k}/\sqrt n\xrightarrow{d}N(0,1)\). The characteristic function also illustrates the product argument directly: it is \(\{\cos(t/\sqrt n)\}^n\). Using \(\cos u=1-u^2/2+O(u^4)\) as \(u\to0\), this equals \(\{1-t^2/(2n)+O(n^{-2})\}^n\), which tends to \(e^{-t^2/2}\).

Worked Example: Nonidentical Centered Bernoulli Variables

Let the variables in row \(n\) be independent Bernoulli variables centered by their means. Set \(p_{n,k}=1/4\) when \(k\) is odd and \(p_{n,k}=3/4\) when \(k\) is even, and let \(X_{n,k}=B_{n,k}-p_{n,k}\), for \(1\le k\le n\). Every variable has variance \(p_{n,k}(1-p_{n,k})=3/16\), so \(s_n^2=3n/16\).

The largest possible absolute value of \(X_{n,k}\) is \(3/4\). Hence

\frac{|X_{n,k}|}{s_n} \le\frac{3/4}{\sqrt{3n}/4} =\frac{\sqrt3}{\sqrt n}.

For any fixed \(\varepsilon>0\), this bound is at most \(\varepsilon\) for all sufficiently large \(n\). The Lindeberg tail terms are then all zero, so the normalized sum converges in distribution to \(N(0,1)\), despite the different Bernoulli distributions within each row.

Worked Example: A Rare-Jump Array That Does Not Have a Normal Limit

For \(n\ge2\), let \(B_{n,1},\ldots,B_{n,n}\) be independent Bernoulli variables with success probability \(1/n\), and put \(X_{n,k}=B_{n,k}-1/n\). Each variable has mean zero and variance \((1/n)(1-1/n)\), so \(s_n^2=1-1/n\).

Take \(\varepsilon=1/2\). For sufficiently large \(n\), \(1-1/n>\tfrac12\sqrt{1-1/n}\), while \(1/n\le\tfrac12\sqrt{1-1/n}\). Thus only the success value \(X_{n,k}=1-1/n\) contributes to the Lindeberg tail, and

\frac{1}{s_n^2}\sum_{k=1}^n E\!\left[X_{n,k}^2\mathbf{1}\{|X_{n,k}|>s_n/2\}\right] = \frac{n(1/n)(1-1/n)^2}{1-1/n} =1-\frac1n\longrightarrow1.

So Lindeberg fails. In fact, the normalized row sum is \((\sum_k B_{n,k}-1)/\sqrt{1-1/n}\). Its characteristic function converges to \(\exp(e^{it}-1-it)\), not to \(e^{-t^2/2}\). At \(t=2\pi\), the first expression equals 1, whereas the second is strictly less than 1. This rare-jump example shows why a fixed total variance does not by itself ensure a normal limit.

Common Pitfall and Why the Proof Matters

A common mistake is to argue that a sum must be approximately normal because it is a sum of many independent variables and has been scaled to variance one. That reasoning overlooks how the variance is distributed across the summands. In the rare-jump example, the normalized variance is one, but a nonvanishing amount is carried by jumps that remain large; the row sum therefore does not have a normal limit.

The characteristic-function proof makes the role of Lindeberg precise. The condition controls the total quadratic remainder contributed by large normalized summands. For the small summands, the Taylor remainder is cubic and therefore bounded by a small threshold times their total variance. Together these facts make the cumulative expansion error vanish. The product of the component characteristic functions then has the same limit as the exponential of the sum of their quadratic terms, which is the normal characteristic function.

This proof technique is useful beyond recognizing the theorem’s conclusion: it identifies exactly where independence, centering, variance normalization, and tail control enter the argument. In particular, the product step depends on independence within each row, while the remainder estimate depends on the Lindeberg condition.

Check Your Understanding

Use the normalized summands and component characteristic functions to answer these questions.

Why does the linear term in the expansion of \(\phi_{n,k}(t)\) vanish?
In the remainder bound, which estimate is used when \(|Y_{n,k}|\le\delta\), and which is used when \(|Y_{n,k}|>\delta\)?
Explain why the Lindeberg condition implies \(\max_k E[Y_{n,k}^2]\to0\).
Where in the proof is independence within each row needed?
Why does convergence of the normalized sums’ characteristic functions to \(e^{-t^2/2}\) imply convergence in distribution?