The Lyapunov Central Limit Theorem

Why a Normal Limit Can Survive Unequal Distributions

The classical central limit theorem gives a normal approximation for sums of identically distributed observations under a finite-variance assumption. In applications, however, independent observations may have different distributions: patients can have different event probabilities, or measurements can have different variances. The Lyapunov central limit theorem gives a useful sufficient condition for a normal limit in this setting. It controls the contribution of large deviations through a moment of order strictly greater than two.

The key comparison is between the total higher-order moment and the total variance raised to the matching power. If the former becomes negligible, no single term or collection of unusually large terms prevents the standardized sum from behaving asymptotically like a normal variable.

The Lyapunov Condition and Theorem

Definition: Let $X_1,X_2,\ldots$ be independent random variables with finite means $\mu_i=E[X_i]$ and finite variances $\sigma_i^2=\operatorname{Var}(X_i)$. Define the centered partial sum and its variance by $$ S_n=\sum_{i=1}^n (X_i-\mu_i), \qquad s_n^2=\sum_{i=1}^n \sigma_i^2. $$ For a fixed $\delta>0$, the Lyapunov condition is $$ \frac{1}{s_n^{2+\delta}}\sum_{i=1}^n E\left[|X_i-\mu_i|^{2+\delta}\right]\longrightarrow 0, \qquad s_n^2>0. $$

Theorem (Lyapunov central limit theorem): Suppose $X_1,X_2,\ldots$ are independent, $E[|X_i|^{2+\delta}]<\infty$ for all $i$ for some fixed $\delta>0$, and $s_n^2\to\infty$. If the Lyapunov condition holds, then $$ \frac{S_n}{s_n}\xrightarrow{d}N(0,1). $$

The variables need not share a common distribution. The means and variances may change with $i$; centering each term and dividing by the square root of the total variance are essential. The condition is sufficient, not necessary: a normal limit can hold even when this particular higher-moment ratio does not tend to zero.

Worked Example: Identically Distributed Bernoulli Variables

Let $X_1,\ldots,X_n$ be independent Bernoulli variables with common success probability $p$, where $0<p<1$. Verify the Lyapunov condition with $\delta=1$.

Write $q=1-p$. Each $X_i$ has mean $p$ and variance $pq$, so $s_n^2=npq$. The centered variable $X_i-p$ equals $q$ with probability $p$, and equals $-p$ with probability $q$. Therefore $$ E[|X_i-p|^3]=pq^3+qp^3=pq(p^2+q^2). $$ The Lyapunov ratio is $$ \frac{n\,pq(p^2+q^2)}{(npq)^{3/2}} =\frac{p^2+q^2}{\sqrt{npq}} \longrightarrow 0. $$ Here $pq>0$ is fixed, so the denominator grows like $\sqrt n$. Also $s_n^2=npq\to\infty$. The theorem gives $$ \frac{\sum_{i=1}^n(X_i-p)}{\sqrt{npq}}\xrightarrow{d}N(0,1). $$

Worked Example: Independent Normals With Increasing Variances

Suppose the $X_i$ are independent with $X_i\sim N(\mu_i,i)$. Check the Lyapunov condition for $\delta=2$, without requiring the means $\mu_i$ to be equal.

For a normal variable with variance $i$, the centered fourth moment is $E[|X_i-\mu_i|^4]=3i^2$. The total variance and the numerator of the Lyapunov ratio are $$ s_n^2=\sum_{i=1}^n i=\frac{n(n+1)}{2}, \qquad \sum_{i=1}^n E[|X_i-\mu_i|^4] =3\sum_{i=1}^n i^2 =\frac{n(n+1)(2n+1)}{2}. $$ Thus the ratio is $$ \frac{n(n+1)(2n+1)/2}{[n(n+1)/2]^2} =\frac{2(2n+1)}{n(n+1)} \longrightarrow 0. $$ Since $s_n^2\to\infty$, Lyapunov’s theorem applies. In this example the centered sum is itself normal for every $n$, but the calculation illustrates how the theorem accommodates unequal variances.

Proof by Characteristic Functions

The proof reduces the stated condition to a Taylor expansion whose error is summable across the terms. Set $Y_i=X_i-\mu_i$, so $E[Y_i]=0$ and $\sum_{i=1}^n E[Y_i^2]=s_n^2$. Choose $\alpha=\min(\delta,1)$, which satisfies $0<\alpha\leq1$.

First, the Lyapunov condition at exponent $\delta$ implies the same type of condition at exponent $\alpha$: $$ \frac{1}{s_n^{2+\alpha}}\sum_{i=1}^n E[|Y_i|^{2+\alpha}] \leq \left( \frac{1}{s_n^{2+\delta}}\sum_{i=1}^n E[|Y_i|^{2+\delta}] \right)^{\alpha/\delta} \left(\frac{\sum_{i=1}^n E[Y_i^2]}{s_n^2}\right)^{1-\alpha/\delta} \longrightarrow 0. $$ For $0<\alpha<\delta$, this inequality follows by applying Hölder’s inequality to each term and then to the sum; when $\alpha=\delta$, it is equality. The final variance factor equals one.

For real $u$, Taylor’s formula for $e^{iu}$ and a bound for large $|u|$ give a constant $C_\alpha$ such that $$ \left|e^{iu}-1-iu+\frac{u^2}{2}\right| \leq C_\alpha |u|^{2+\alpha}. $$ Indeed, for $|u|\leq1$, the third-order remainder is bounded by a constant times $|u|^3$, which is at most a constant times $|u|^{2+\alpha}$. For $|u|>1$, bounding each displayed term separately gives the same inequality after increasing the constant.

Fix $t\in\mathbb R$. Let $u_{i,n}=E[\exp(itY_i/s_n)]$. Using $E[Y_i]=0$ in the Taylor bound yields $$ u_{i,n} =1-\frac{t^2 E[Y_i^2]}{2s_n^2}+r_{i,n}, \qquad \sum_{i=1}^n |r_{i,n}| \leq C_\alpha |t|^{2+\alpha} \frac{\sum_{i=1}^n E[|Y_i|^{2+\alpha}]}{s_n^{2+\alpha}} \longrightarrow 0. $$ To control the product of these characteristic functions, write $b_{i,n}=E[Y_i^2]/s_n^2$. Then $\sum_i b_{i,n}=1$. Jensen’s inequality gives \[ (E[Y_i^2])^{1+\delta/2}\leq E[|Y_i|^{2+\delta}], \] so the Lyapunov condition implies $\max_{i\leq n} b_{i,n}\to0$. Consequently $\sum_i b_{i,n}^2\leq \max_i b_{i,n}\to0$. If $a_{i,n}=u_{i,n}-1$, the expansion above now gives $$ \sum_{i=1}^n a_{i,n}\longrightarrow-\frac{t^2}{2}, \qquad \sum_{i=1}^n |a_{i,n}|^2\longrightarrow0. $$ For the second limit, the variance parts have squared sum tending to zero, while $\sum_i|r_{i,n}|\to0$ implies $\sum_i|r_{i,n}|^2\to0$.

In particular, $\max_i|a_{i,n}|\to0$. For all sufficiently large $n$, the power-series expansion of $\log(1+z)$ near zero gives $\log(1+a_{i,n})=a_{i,n}+O(|a_{i,n}|^2)$, with a uniform bound on the remainder. Independence lets us multiply the individual characteristic functions, and therefore $$ E\left[\exp\left(it\frac{S_n}{s_n}\right)\right] =\prod_{i=1}^n u_{i,n} =\exp\left\{\sum_{i=1}^n\log(1+a_{i,n})\right\} \longrightarrow e^{-t^2/2}. $$ The limit $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, continuous at zero. Lévy’s continuity theorem therefore implies convergence in distribution to $N(0,1)$. $\square$

A Nonidentical Bernoulli Application

Worked Example: Rare Events With Changing Probabilities

Let $X_i$ be independent Bernoulli variables with $P(X_i=1)=p_i=1/(i+1)$. Verify Lyapunov’s condition with $\delta=1$.

The variance of $X_i$ is \[ p_i(1-p_i)=\frac{i}{(i+1)^2}. \] As in the first example, the centered third absolute moment is \[ E[|X_i-p_i|^3] =p_i(1-p_i)\bigl(p_i^2+(1-p_i)^2\bigr) \leq p_i(1-p_i), \] because $p_i^2+(1-p_i)^2\leq1$. Hence the numerator of the Lyapunov ratio is at most $s_n^2$. Also \[ s_n^2=\sum_{i=1}^n\frac{i}{(i+1)^2} \geq \sum_{i=1}^n\frac{1}{4i}\longrightarrow\infty, \] where the inequality follows from $i^2/(i+1)^2\geq1/4$ for $i\geq1$. The ratio is therefore bounded above by \[ \frac{s_n^2}{s_n^3}=\frac{1}{s_n}\longrightarrow0. \] The theorem applies even though the success probabilities vary and decrease with $i$.

Interpretation and a Common Pitfall

The higher moment in the Lyapunov condition limits the aggregate impact of unusually large centered observations relative to the total standard deviation. It is not a requirement that each $X_i$ be identically distributed, nor does it say that every individual term has a normal distribution. The conclusion concerns the distribution of the centered, variance-standardized sum.

Wrong reasoning: “The variance of a sequence of sums diverges, so the sums cannot converge in distribution to a standard normal.” Diverging variances alone do not establish this: convergence in distribution does not, in general, imply convergence of variances.
Correction: A particular claim about a distributional limit needs a valid distributional argument. For example, let $\varepsilon_k$ be independent Rademacher variables, with probabilities $1/2$ each at $1$ and $-1$, and set $T_n=\sum_{k=1}^n(k/n)\varepsilon_k$. Its variance is $\sum_{k=1}^n(k/n)^2$, which diverges. But its characteristic function at $1$ is $\prod_{k=1}^n\cos(k/n)$. At least $\lfloor n/2\rfloor$ factors have $k/n\in[1/2,1]$, so their absolute values are at most $\cos(1/2)<1$; all other factors have absolute value at most one. Thus this characteristic function tends to zero, whereas the standard normal characteristic function at $1$ is $e^{-1/2}>0$. This directly rules out a standard normal limit for $T_n$. It does not rule one out for the properly standardized sum $T_n/\sqrt{\operatorname{Var}(T_n)}$.
Another common error: Checking only that every term has a finite variance. Finite variances do not by themselves verify Lyapunov’s condition; the higher moments must be finite and their normalized sum must tend to zero.

Key takeaway: Under independence, a finite $2+\delta$ moment and a vanishing Lyapunov ratio ensure that the centered sum, divided by its total standard deviation, converges in distribution to $N(0,1)$. Check the ratio for the sum as a whole, not just the moment of one term.

Check Your Understanding

Try these questions before moving on.

State the Lyapunov condition for a sequence of independent variables, including how the total variance is defined.
Why is a moment of order $2+\delta$, rather than just a finite variance, used in the proof?
For independent Bernoulli variables with a fixed success probability $p\in(0,1)$, what is the Lyapunov ratio for $\delta=1$, and why does it tend to zero?
In the normal example with $\operatorname{Var}(X_i)=i$, calculate the total variance $s_n^2$.
Why does divergence of variances alone not rule out convergence in distribution to a standard normal?