Lindeberg's condition vs Lyapunov's condition

Question

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.

Answer 1

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.