Lyapunov's Theorem for Nonidentical Sums

Why a Normal Limit Can Hold Without Identical Distributions

A central limit theorem does not require every summand to have the same distribution. In many applications, independent contributions have different variances or different shapes, and their distributions may change as the number of terms grows. The central question is whether the total variation in the sum is spread across enough contributions, rather than being dominated by a few unusually large ones.

Lyapunov’s theorem gives a convenient sufficient condition for a normal limit in this setting. It compares the sum of higher absolute moments with the total variance raised to the corresponding power. To apply it well, first center each summand, then calculate the row’s variance scale, and finally bound the aggregate moment at that scale.

Lyapunov’s Theorem for Independent Sums

The theorem is naturally stated for a triangular array: for each $n$, there are $n$ independent variables in row $n$, and their distributions may vary both within and across rows. Write the variables as $X_{n,1},\ldots,X_{n,n}$, and let $Y_{n,k}=X_{n,k}-E[X_{n,k}]$ be their centered versions. Define the total variance $s_n^2=\sum_{k=1}^n \operatorname{Var}(X_{n,k})$.

Theorem (Lyapunov): Suppose that, within each row $n$, $X_{n,1},\ldots,X_{n,n}$ are independent, have finite means and variances, and $0<s_n^2<\infty$. If there is a $\delta>0$ such that $$ \frac{1}{s_n^{2+\delta}}\sum_{k=1}^n E\!\left[|X_{n,k}-E[X_{n,k}]|^{2+\delta}\right]\longrightarrow 0, $$ then $$ \frac{\sum_{k=1}^n(X_{n,k}-E[X_{n,k}])}{s_n}\xrightarrow{d}N(0,1). $$

The ratio in the condition is dimensionless: the numerator has the scale of a $2+\delta$ moment, and $s_n^{2+\delta}$ is the matching power of the total standard deviation. The condition must hold for one fixed $\delta>0$; it is not enough for a different choice of $\delta$ to work in each row. The result is sufficient, not necessary: failure of this moment condition does not by itself rule out a normal limit.

Why the Moment Condition Works

The connection to the Lindeberg–Feller central limit theorem makes Lyapunov’s condition useful. In a centered row, the Lindeberg condition asks that, after division by $s_n^2$, the total variance contributed by summands larger than any fixed fraction of $s_n$ tends to zero. Lyapunov’s moment bound forces this tail contribution to vanish.

Result: Under the hypotheses of Lyapunov’s theorem, for every $\varepsilon>0$, $$ \frac{1}{s_n^2}\sum_{k=1}^n E\!\left[Y_{n,k}^2\mathbf{1}\{|Y_{n,k}|>\varepsilon s_n\}\right]\longrightarrow 0. $$

To prove this, on the event $\{|Y_{n,k}|>\varepsilon s_n\}$, the inequality $|Y_{n,k}|^2\leq |Y_{n,k}|^{2+\delta}/(\varepsilon s_n)^\delta$ holds. Therefore

\begin{aligned} \frac{1}{s_n^2}\sum_{k=1}^n E\!\left[Y_{n,k}^2\mathbf{1}\{|Y_{n,k}|>\varepsilon s_n\}\right] &\leq \frac{1}{\varepsilon^\delta s_n^{2+\delta}} \sum_{k=1}^n E[|Y_{n,k}|^{2+\delta}]\\ &=\varepsilon^{-\delta} \left(\frac{1}{s_n^{2+\delta}}\sum_{k=1}^n E[|Y_{n,k}|^{2+\delta}]\right) \longrightarrow 0. \end{aligned}

This is the Lindeberg condition for the centered array. Since the variables are independent within each row and their total variance is $s_n^2$, the Lindeberg–Feller central limit theorem gives the claimed normal limit. Thus Lyapunov’s condition is a convenient moment-based way to check a condition that controls the large summands. $\square$

A Practical Verification Strategy

A reliable calculation separates the work into three parts. First write the centered summands explicitly; the moment condition applies to $X_{n,k}-E[X_{n,k}]$, not automatically to $X_{n,k}$. Next compute $s_n^2$ by adding the variances, using independence within the row. Finally choose a convenient $\delta$ and bound or calculate the sum of the centered $2+\delta$ moments. Often $\delta=1$ or $\delta=2$ gives manageable expressions.

Center each term.
Set $Y_{n,k}=X_{n,k}-E[X_{n,k}]$, so the sum being tested has mean zero.

Find the variance scale.
Use independence to calculate $s_n^2=\sum_k\operatorname{Var}(X_{n,k})$, then normalize by $s_n$.

Compare aggregate moments.
For some fixed $\delta>0$, show that $\sum_k E[|Y_{n,k}|^{2+\delta}]/s_n^{2+\delta}$ tends to zero.

Worked Example: Alternating Bernoulli Probabilities

In row $n$, let $X_{n,k}$ be independent Bernoulli variables with success probability $1/4$ for odd $k$ and $3/4$ for even $k$. Show that the centered sum, divided by its standard deviation, converges in distribution to $N(0,1)$.

For either success probability, the variance is $p(1-p)=3/16$. Hence $s_n^2=3n/16$, which is positive for every $n$. For a Bernoulli variable $X$, $$ E[|X-p|^3]=p(1-p)^3+(1-p)p^3 =p(1-p)\big((1-p)^2+p^2\big). $$ At $p=1/4$ or $p=3/4$, this equals $(3/16)(10/16)=15/128$. Thus, using $\delta=1$, the Lyapunov ratio is $$ \frac{\sum_{k=1}^n E[|X_{n,k}-E[X_{n,k}]|^3]}{s_n^3} =\frac{15n/128}{(3n/16)^{3/2}} =\frac{5}{2\sqrt{3n}}\longrightarrow 0. $$ Lyapunov’s theorem applies, so the centered sum divided by $\sqrt{3n/16}$ converges in distribution to $N(0,1)$. The success probabilities differ within each row, but the theorem only requires independence and the stated moment condition.

Worked Example: Normal Summands with Increasing Variances

Let $X_{n,k}\sim N(0,k)$, independently for $k=1,\ldots,n$. Verify Lyapunov’s condition using $\delta=2$.

The variables are already centered, and their total variance is $$ s_n^2=\sum_{k=1}^n k=\frac{n(n+1)}{2}. $$ A centered normal variable with variance $k$ has fourth moment $3k^2$. Therefore $$ \frac{\sum_{k=1}^n E[|X_{n,k}|^4]}{s_n^4} =\frac{3\sum_{k=1}^n k^2}{\left(\sum_{k=1}^n k\right)^2} =\frac{3n(n+1)(2n+1)/6}{n^2(n+1)^2/4} =\frac{2(2n+1)}{n(n+1)} \longrightarrow 0. $$ The variance of an individual summand grows with $k$, so equal variances are not part of the argument. The aggregate fourth moment is nevertheless small relative to the square of the total variance. The standardized sum consequently converges in distribution to $N(0,1)$.

Worked Example: Poisson Variables with Increasing Means

Let $Z_k$ be independent Poisson variables with means $k$, and consider the sum $\sum_{k=1}^n Z_k$. Check Lyapunov’s condition for the centered variables using $\delta=1$.

Write $Y_k=Z_k-k$. Then $\operatorname{Var}(Y_k)=k$, so $s_n^2=\sum_{k=1}^n k=n(n+1)/2$. The centered fourth moment of a Poisson variable with mean $k$ is $k+3k^2$. By Cauchy–Schwarz, $$ E[|Y_k|^3] =E[|Y_k|\,Y_k^2] \leq \sqrt{E[Y_k^2]E[Y_k^4]} =\sqrt{k(k+3k^2)} \leq 2k^{3/2}, $$ where the last inequality uses $1+3k\leq 4k$ for $k\geq1$. Consequently, $$ \frac{\sum_{k=1}^n E[|Y_k|^3]}{s_n^3} \leq \frac{2\sum_{k=1}^n k^{3/2}}{(n(n+1)/2)^{3/2}} \leq \frac{2n^{5/2}}{(n^2/2)^{3/2}} =\frac{4\sqrt{2}}{\sqrt{n}} \longrightarrow 0. $$ The variables have different means and variances, but after centering, their aggregate third absolute moment is negligible relative to $s_n^3$. The centered sum divided by $s_n$ therefore converges in distribution to $N(0,1)$.

Common Pitfall: Using the Wrong Scale

A frequent error is to calculate the sum of higher moments and argue that it is “small compared with $n$.” The Lyapunov condition does not compare that sum with $n$ in general. It compares it with the matching power $s_n^{2+\delta}$ of the actual total standard deviation. When variances differ across summands, $s_n^2$ need not be proportional to $n$.

For example, if one uses $\delta=1$, the denominator is $s_n^3$, not $s_n^2$. Another error is to insert uncentered moments without justification: the condition concerns $E[|X_{n,k}-E[X_{n,k}]|^{2+\delta}]$. Center first, calculate the variance scale from the centered variables, and then compare like powers.

Key takeaway: Lyapunov’s theorem handles independent, nonidentically distributed sums by comparing the aggregate centered $2+\delta$ moments with the total variance scale $s_n^{2+\delta}$. A vanishing ratio guarantees that no large summands contribute too much variance, and yields a standard normal limit after centering and scaling.

Check Your Understanding

For each question, identify the scale or bound needed to apply Lyapunov’s theorem.

For independent variables in row $n$, define $s_n^2=\sum_k\operatorname{Var}(X_{n,k})$. What expression must tend to zero for the Lyapunov condition with $\delta=1$?
Why does a bound on $\sum_k E[|X_{n,k}|^3]$ alone not necessarily verify the condition for a centered sum?
If $s_n^2$ is of order $n^2$ and $\delta=2$, what order of aggregate fourth moment would be sufficient for the Lyapunov ratio to vanish?
In the alternating Bernoulli example, which feature of the calculation shows that identical distributions are not required?
Does failure of a chosen Lyapunov condition prove that the standardized sum has no normal limit? Explain.