Application of Lyapunov's Central Limit Theorem (4)

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Let $\{X_n,\,n\geq1\}$ be a sequence of uniformly bounded independent random variables, and $S_n=\sum_{i=1}^nX_i$.  Suppose $\sigma^2_n=\mbox{Var}(S_n)\rightarrow\infty$ as $n\rightarrow\infty$, then $$\frac{S_n-\mathscr{E}(S_n)}{\sigma_n}\overset{d}{\rightarrow}\Phi.$$
$\bullet$ Proof.

$X_i$'s are uniformly bounded $\implies|X_i|\leq M<\infty,\,\forall\,i=1,2,\ldots.$  Let $\alpha_i=\mathscr{E}(X_i)$, we have $|X_i-\alpha_i|\leq|X_i|+|\alpha_i|\leq2M$.  Define $\gamma_i=\mathscr{E}\left(|X_i-\alpha_i|^3\right)$, then the Lyapunov's condition  $$\begin{array}{rl}\frac{1}{\sigma^3_n}\sum_{i=1}^n\gamma_i
&=\frac{1}{\sigma^3_n}\sum_{i=1}^n\mathscr{E}\left(|X_i-\alpha_i|^3\right) \\
&=\frac{1}{\sigma^3_n}\sum_{i=1}^n\mathscr{E}\left(|X_i-\alpha_i|(X_i-\alpha_i)^2\right) \\
&\leq\frac{2M}{\sigma^3_n}\sum_{i=1}^n\mathscr{E}(X_i-\alpha_i)^2 \\
&=\frac{2M}{\sigma^3_n}\sigma^2_n =\frac{2M}{\sigma_n}\rightarrow0\mbox{ as }n\rightarrow\infty, \\ \end{array}$$since $\sigma^2_n\rightarrow\infty.$  Thus the Lyapunov's condition holds and then we have $$\frac{S_n-\mathscr{E}(S_n)}{\sigma_n}\overset{d}{\rightarrow}\Phi.$$

$\Box$