Checking Lyapunov Conditions in Examples

What a Lyapunov Check Requires

The Lyapunov central limit theorem is useful when summands are independent but need not have the same distribution. In applications, the main work is often not invoking the theorem but showing that a particular sum has a sufficiently small contribution from large deviations. Lyapunov’s condition measures that contribution through moments of order greater than two, relative to the total variance.

For each \(n\), consider independent random variables \(X_{n,1},\ldots,X_{n,m_n}\) with finite means and variances. The distributions may vary between summands and may also depend on \(n\). Define the centered variables \(Y_{n,k}=X_{n,k}-E[X_{n,k}]\), and let \(s_n^2\) be the variance of their sum.

Definition: For a fixed \(\delta>0\), the Lyapunov ratio is

L_n(\delta)= \frac{\displaystyle\sum_{k=1}^{m_n}E\!\left[|Y_{n,k}|^{2+\delta}\right]} {s_n^{2+\delta}}, \qquad s_n^2=\sum_{k=1}^{m_n}\operatorname{Var}(X_{n,k}). $$ The Lyapunov condition holds if \(L_n(\delta)\to 0\) for some fixed \(\delta>0\). The Lyapunov central limit theorem states that, if the variables in each sum are independent, \(0<s_n^2<\infty\), the displayed moments are finite, and the condition holds, then $$ \frac{\sum_{k=1}^{m_n}(X_{n,k}-E[X_{n,k}])}{s_n} \xrightarrow{d}N(0,1).

The denominator is the total standard deviation raised to the same power as the moments in the numerator. A useful check therefore compares each summand’s higher moment with the scale of the whole sum. Choosing \(\delta=1\) uses third absolute moments; choosing \(\delta=2\) uses fourth moments. The theorem requires one fixed choice of \(\delta>0\) for which the ratio tends to zero.

A Bound for Centered Bounded Variables

A simple moment bound often makes Lyapunov’s condition immediate. Suppose the centered summands in the \(n\)-th sum satisfy \(|Y_{n,k}|\le b_n\) almost surely, with the same bound \(b_n\) applying to every summand in that sum. Since \(2+\delta>2\), multiplying \(|Y_{n,k}|^2\) by \(|Y_{n,k}|^\delta\) gives a direct bound.

Theorem: If \(|Y_{n,k}|\le b_n\) almost surely for every \(k\), then, for every \(\delta>0\),

\sum_{k=1}^{m_n}E[|Y_{n,k}|^{2+\delta}] \le b_n^\delta s_n^2, \qquad L_n(\delta)\le \left(\frac{b_n}{s_n}\right)^\delta.

In particular, if \(b_n/s_n\to0\), Lyapunov’s condition holds for every fixed \(\delta>0\).

Proof. The almost-sure bound implies \(|Y_{n,k}|^\delta\le b_n^\delta\), and hence

|Y_{n,k}|^{2+\delta} =|Y_{n,k}|^2|Y_{n,k}|^\delta \le b_n^\delta |Y_{n,k}|^2.

Taking expectations gives \(E[|Y_{n,k}|^{2+\delta}]\le b_n^\delta E[|Y_{n,k}|^2]=b_n^\delta\operatorname{Var}(X_{n,k})\), because \(Y_{n,k}\) is centered. Summing over \(k\) yields the first inequality. Dividing by \(s_n^{2+\delta}\), and using \(\sum_k\operatorname{Var}(X_{n,k})=s_n^2\), gives the ratio bound. If \(b_n/s_n\to0\), its fixed positive power also tends to zero. \(\square\)

Worked Example: Bounded Bernoulli Summands with a Fixed Success Probability

Let \(X_1,\ldots,X_n\) be independent Bernoulli random variables with \(P(X_i=1)=p\), where \(0<p<1\) is fixed. Check Lyapunov’s condition for their centered sum.

Write \(Y_i=X_i-p\) and \(q=1-p\). The centered variable equals \(q\) with probability \(p\), and \(-p\) with probability \(q\). Thus

\operatorname{Var}(X_i)=pq,\qquad E[|Y_i|^3]=pq^3+qp^3 =pq(p^2+q^2).

Using \(\delta=1\), the total variance is \(s_n^2=npq\), so the Lyapunov ratio is

L_n(1) =\frac{npq(p^2+q^2)}{(npq)^{3/2}} =\frac{p^2+q^2}{\sqrt{npq}}.

Since \(p\) and \(q\) are fixed and positive, the numerator is constant and the denominator grows proportionally to \(\sqrt n\). Therefore \(L_n(1)\to0\). For instance, if \(p=0.2\) and \(q=0.8\), then \(p^2+q^2=0.68\) and \(pq=0.16\), giving \(L_n(1)=1.7/\sqrt n\). The Lyapunov theorem consequently gives a standard normal limit for the centered sum divided by \(\sqrt{npq}\).

Unequal Variances: Compare the Largest Scale with the Total

For variables with unequal variances, a common strategy is to bound each third moment in terms of its own standard deviation. Write \(\sigma_{n,k}^2=\operatorname{Var}(X_{n,k})\). If \(E[|Y_{n,k}|^3]\le C\sigma_{n,k}^3\) for a constant \(C\) not depending on \(n\) or \(k\), then the third-moment ratio has a convenient upper bound.

Theorem: Under the moment bound \(E[|Y_{n,k}|^3]\le C\sigma_{n,k}^3\),

L_n(1) \le C\frac{\sum_{k=1}^{m_n}\sigma_{n,k}^3} {\left(\sum_{k=1}^{m_n}\sigma_{n,k}^2\right)^{3/2}} \le C\frac{\max_k\sigma_{n,k}} {\sqrt{\sum_{k=1}^{m_n}\sigma_{n,k}^2}}.

Thus, if the largest standard deviation divided by the total standard deviation tends to zero, the third-moment Lyapunov condition holds.

Proof. Summing the assumed moment bounds gives the first inequality, since \(s_n^2=\sum_k\sigma_{n,k}^2\). For the second, each \(\sigma_{n,k}\le\max_j\sigma_{n,j}\), so

\sum_k\sigma_{n,k}^3 \le \left(\max_k\sigma_{n,k}\right)\sum_k\sigma_{n,k}^2.

Dividing by \((\sum_k\sigma_{n,k}^2)^{3/2}\) proves the claimed bound. If its final fraction tends to zero, then \(L_n(1)\to0\). \(\square\)

Worked Example: Independent Normal Variables with Increasing Variances

For each \(n\), let \(X_1,\ldots,X_n\) be independent with \(X_i\sim N(0,i)\). The distributions do not have a common variance. Check the Lyapunov condition with \(\delta=1\).

If \(Z\sim N(0,1)\), direct integration of the normal density gives \(E[|Z|^3]=2\sqrt{2/\pi}\). Since \(X_i\) has the same distribution as \(\sqrt{i}Z\),

E[|X_i|^3]=2\sqrt{\frac{2}{\pi}}\,i^{3/2}, \qquad s_n^2=\sum_{i=1}^{n}i=\frac{n(n+1)}{2}.

Consequently,

L_n(1) = 2\sqrt{\frac{2}{\pi}}\, \frac{\sum_{i=1}^n i^{3/2}}{\left(n(n+1)/2\right)^{3/2}}.

For every \(i\le n\), \(i^{3/2}=\sqrt{i}\,i\le\sqrt{n}\,i\). Therefore

L_n(1) \le 2\sqrt{\frac{2}{\pi}}\, \frac{\sqrt n\,n(n+1)/2}{\left(n(n+1)/2\right)^{3/2}} = 2\sqrt{\frac{2}{\pi}}\sqrt{\frac{2}{n+1}}.

The right-hand side tends to zero, so Lyapunov’s condition holds. This bound shows why a single summand with a larger variance does not automatically prevent a normal limit: what matters here is its scale relative to the variance accumulated by the full sum.

Changing Distributions and Moment Bounds

When distributions change with \(n\), do not assume a moment is constant just because the variables have the same form. Calculate or bound the moment at the current value of \(n\), sum those bounds, and compare the result with \(s_n^{2+\delta}\). Uniform bounds are helpful, but the relevant question is how their total compares with the denominator.

Worked Example: Bernoulli Probabilities Decreasing with Sample Size

For each \(n\ge2\), let \(X_{n,1},\ldots,X_{n,n}\) be independent Bernoulli variables with success probability \(p_n=n^{-1/2}\). Check Lyapunov’s condition with \(\delta=1\).

Set \(q_n=1-p_n\). As in the fixed-probability calculation, each variable has variance \(p_nq_n\) and centered third absolute moment \(p_nq_n(p_n^2+q_n^2)\). Thus

s_n^2=np_nq_n =\sqrt n\left(1-\frac{1}{\sqrt n}\right), \qquad L_n(1)= \frac{p_n^2+q_n^2}{\sqrt{np_nq_n}}.

The numerator satisfies \(p_n^2+q_n^2\to1\). The denominator squared is \(np_nq_n=\sqrt n(1-1/\sqrt n)\), which tends to infinity. More explicitly, its square root is of order \(n^{1/4}\), so \(L_n(1)\to0\). Lyapunov’s condition holds even though the success probability changes with \(n\).

Contrast this with \(p_n=1/n\). Then \(np_nq_n=1-1/n\to1\), while \(p_n^2+q_n^2\to1\), and therefore

L_n(1)=\frac{p_n^2+q_n^2}{\sqrt{np_nq_n}}\longrightarrow 1,

not zero. This calculation shows that the same boundedness of the centered summands does not ensure Lyapunov’s condition when the total variance fails to grow. It shows only that this particular sufficient condition cannot be used; failure of Lyapunov’s condition alone does not prove that a normal limit is impossible.

Choosing a Moment Bound and Avoiding a Pitfall

A good Lyapunov check is usually short once the right bound has been identified. For bounded centered variables, the bound \(E[|Y|^{2+\delta}]\le b_n^\delta\operatorname{Var}(X)\) is effective when the common bound \(b_n\) is small relative to \(s_n\). For variables whose moments scale with their standard deviations, a bound such as \(E[|Y|^3]\le C\sigma^3\) reduces the check to comparing the largest \(\sigma\) with the total standard deviation.

Higher moments can also help, but they must be normalized consistently. For example, using \(\delta=2\) requires summing fourth absolute moments and dividing by \(s_n^4\); a fourth-moment bound is useful only if that resulting ratio tends to zero. The theorem needs a fixed \(\delta>0\), not an exponent that changes with \(n\).

Common pitfall: “Every summand is bounded, so Lyapunov’s condition holds.” Boundedness alone is not enough. The relevant bound is \(L_n(\delta)\le(b_n/s_n)^\delta\), and it tends to zero only if the bound on an individual centered summand is negligible relative to the total standard deviation. In the Bernoulli example with \(p_n=1/n\), summands are bounded but the Lyapunov ratio tends to one. The correction is to calculate the denominator as well as the individual moment bound.

The practical value of Lyapunov checks is that they turn a central limit theorem’s hypothesis into concrete moment arithmetic. They allow different variances and changing distributions, while making explicit which control is doing the work: bounded summands relative to total variance, or higher moments that grow no faster than an appropriate power of each summand’s standard deviation. Keep the conclusion appropriately limited: a vanishing Lyapunov ratio establishes the theorem’s condition; a nonvanishing ratio merely leaves that route unavailable.

Check Your Understanding

For each question, identify the moment or scale comparison that decides the Lyapunov check.

If centered variables satisfy \(|Y_{n,k}|\le b_n\), what condition on \(b_n\) and \(s_n\) guarantees Lyapunov’s condition?
For independent Bernoulli summands with a fixed success probability \(p\), derive the third absolute centered moment in terms of \(p\) and \(q=1-p\).
If \(E[|Y_{n,k}|^3]\le C\sigma_{n,k}^3\), which comparison involving the largest \(\sigma_{n,k}\) suffices to make \(L_n(1)\to0\)?
For Bernoulli variables with \(p_n=1/n\) and \(n\) summands, what happens to the third-moment Lyapunov ratio, and what does that alone allow you to conclude?