Lindeberg's CLT v.s. Lyapunov's CLT

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Let $\{X_{nj}\}$, $n=1,2,...$, $j=1,2,...,k_n$, be a double array of random variables and for each $n$, $X_{n1},\ldots,X_{nk_n}$ are independent.  Define $S_n=\sum_{j=1}^{k_n}X_{nj}$ and $$\begin{array}{ll}
\mathscr{E}(X_{nj})=\alpha_{nj}, & \mathscr{E}(S_n)=\sum_{j=1}^{k_n}\alpha_{nj}=\alpha_n; \\
\sigma^2(X_{nj})=\sigma^2_{nj}, & \sigma^2(S_n)=\sum_{j=1}^{k_n}\sigma^2_{nj}=s^2_n; \\
\mathscr{E}\left(|X_{nj}-\alpha_{nj}|^{2+\delta}\right)=r^{2+\delta}_{nj}, \delta>0.&
\end{array}$$
$\diamondsuit$ Lyapunov's condition:

$\exists\,\delta>0$ such that $\gamma^{2+\delta}_{nj}$ exists for each $n$ and $j$, and $$\underset{n\rightarrow\infty}{\lim}\frac{1}{s^{2+\delta}_n}\sum_{j=1}^{k_n}r^{2+\delta}_{nj}=0.$$

$\diamondsuit$ Lindeberg's condition:
$$\underset{n\rightarrow\infty}{\lim}\frac{1}{s^2_n}\sum_{j=1}^{k_n}\mathscr{E}\left[(X_{nj}-\alpha_{nj})^2\,I\left(|x_{nj}-\alpha_{nj}|>\eta s_n\right)\right]=0, \forall\; \eta>0 $$

[Theorem] If the Lyapunov's condition holds, then so does the Lindeberg's condition.  The converse is NOT true.



$\bullet$ Proof. (Lyapunov $\implies$ Lindeberg)
WLOG, let $\alpha_{nj}=0$, otherwise we define new double array with $Y_{nj}=X_{nj}-\alpha_{nj}$.
For $\eta>0$, consider $$|X_{nj}|\geq\eta s_n\implies\left|\frac{X_{nj}}{\eta s_n}\right|\geq1\implies\left|\frac{X_{nj}}{\eta s_n}\right|^\delta\geq1\mbox{ for }\delta>0.$$We have $$\begin{array}{l}\frac{1}{s_n^2}\sum_{j=1}^{k_n}\mathscr{E}\left\{|X_{nj}|^2\cdot I(|X_{nj}|\geq\eta s_n)\right\}\\
\quad\leq\frac{1}{s_n^2}\sum_{j=1}^{k_n}\mathscr{E}\left\{\left|\frac{X_{nj}}{\eta s_n}\right|^\delta|X_{nj}|^2\cdot I(|X_{nj}|\geq\eta s_n)\right\}\\
\quad\leq\frac{1}{\eta^\delta s_n^{2+\delta}}\sum_{j=1}^{k_n}\mathscr{E}\left\{|X_{nj}|^{2+\delta}\right\}\\ \quad\rightarrow0\mbox{ as }n\rightarrow\infty\;\because\mbox{Lyapunov.}\end{array}$$Thus, Lyapunov's condition. $\implies$ Lindeberg's condition.

$\Box$

$\bullet$ Counterexample. (Lindeberg $\not\Rightarrow$ Lyapunov)
Let $X_1$, $X_2$, ..., $X_n$ be i.i.d. random variables with probability $$\begin{array}{ll}
\mathscr{P}\{X_i^2=nB_n^2\} &=\frac{1}{n^2\log{n}};\\
\mathscr{P}\{X_i^2=\frac{1}{n}B_n^2\} &=1-\frac{1}{\log{n}};\\
\mathscr{P}\{X_i^2=0\} &=\frac{1}{\log{n}}-\frac{1}{n^2\log{n}}. \end{array}$$Then $$s_n^2=\sum_{i=1}^n\mathscr{E}\{X_i^2\}=n\left[\frac{nB_n^2}{n^2\log{n}}+\frac{B_n^2}{n}\left(1-\frac{1}{\log{n}}\right)\right]=B_n^2.$$Choose $\eta<n$,$$\sum_{i=1}^{n}\mathscr{E}\left\{X_i^2\cdot I(|X_i|\geq\eta B_n)\right\}=n\cdot nB_n^2\frac{1}{n^2\log{n}}=\frac{B_n}{\log{n}}.$$Thus, $$\frac{1}{B_n^2}\sum_{i=1}^{n}\mathscr{E}\left\{X_i^2\cdot I(|X_i|\geq\eta B_n)\right\}=\frac{1}{\log{n}}\rightarrow0\mbox{ as }n\rightarrow\infty,$$ the Lindeberg condition is satisfied.

But, consider for $\delta>1$, $$\begin{array}{rl}\sum_{i=1}^{n}\mathscr{E}\left\{|X_i|^{2+\delta}\right\} &\geq\sum_{i=1}^{n}\mathscr{E}\left\{|X_i|^{2+\delta}\cdot I(|X_i|\geq\eta B_n)\right\}\\
&=\sum_{i=1}^{n}\left(nB_n^2\right)^{1+\frac{\delta}{2}}\frac{1}{n^2\log{n}}\\
&=B_n^{2+\delta}\frac{n^{\delta/2}}{\log{n}}. \end{array}$$Then we have $$\frac{1}{B_n^{2+\delta}}\sum_{i=1}^{n}\mathscr{E}\left\{|X_i|^{2+\delta}\right\}\geq\frac{1}{B_n^{2+\delta}}B_n^{2+\delta}\frac{n^{\delta/2}}{\log{n}}=\frac{n^{\delta/2}}{\log{n}}\rightarrow\infty\mbox{ as }n\rightarrow\infty.$$Thus, the Lyapunov's condition does not hold.

$\Box$