# Lindeberg's condition vs Lyapunov's condition

Can someone construct an example where Lindeberg's condition holds but Lyapunov's condition does not? This is a problem from Billingsley's/Chung's book.

## 1 Answer

Consider a random variable $$X$$ taking values in $$\mathbb Z$$ and $$P(X=i) = \frac{c}{|i|^3\log^2|i|}, i\in \mathbb Z \setminus \{0,1,-1\}$$ where $$c$$ is the correct normalizing constant. Notice $$E(X) = 0$$ from symmetry, $$\sigma^2 :=EX^2 < \infty$$ and for any $$\delta>0$$, $$E(|X|^{2+\delta}) = \infty$$. Let $$X,X_1,X_2, \ldots,$$ be i.i.d. Clearly Lyapunov condition does not hold.

Now lets check the Lindeberg condition. $$s_n^2 = \sum_{i=1}^n \sigma_i^2 = n\sigma^2$$. Also for all $$\epsilon >0$$, $$\frac 1 {s_n^2}\sum_{k=1}^nE(|X_k|^2 1_{|X_k| > \epsilon s_n}) = \frac{n}{\sigma^2 n} \sum_{|i|>\epsilon \sigma \sqrt {n}}i^2 \frac{c}{|i|^3\log^2 |i|} < c' \int_{\epsilon \sigma \sqrt{n}}^\infty \frac{dz}{z\log^2z} < c'' \int_{\frac{1}{2}\log n}^{\infty}\frac{du}{u^2} =O\left(\frac {1}{\log n}\right),$$ where $$c', c''$$ are some absolute constants. Hence Lindeberg condition is satisfied.

• Thank you, that is a really nice example, also explained very clearly. Commented Nov 16, 2014 at 16:16