I have the following problem:
Problem. Let $ (X_{n})_{n \in \mathbb{N}} $ be a sequence of independent random variables such that $$ \mathbf{Pr} \! \left( X_{n} = \sqrt{n} + 1 \right) = \mathbf{Pr} \! \left( X_{n} = - \sqrt{n} - 1 \right) = \frac{1}{2 (\sqrt{n} + 1)^{2}} $$ and $$ \mathbf{Pr} \! \left( X_{n} = 0 \right) = 1 - \frac{1}{(\sqrt{n} + 1)^{2}}. $$ Then prove that the random variable $ \dfrac{\sum_{i = 1}^{n} X_{i}}{\sqrt{n}} $ converges in distribution to one with a standard normal distribution.
I have to use Lindeberg’s Condition together with the Central Limit Theorem in order to solve this, but I’m stuck. Any assistance would be greatly appreciated. Thanks!
We want to check that for a fixed positive $\varepsilon$, the convergence $$\lim_{n\to\infty}\frac 1{s_n^2}\sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon s_n \}] =0 $$ takes place, with $s_n^2=\sum_{j=1}^n \operatorname{Var}(X_j)=n$ in order to apply Lindeberg-Lévy-Feller theorem. We thus have to prove that $$\tag{*} \lim_{n\to\infty}\frac 1n \sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=0.$$ If $j$ is such that $\sqrt j +1\leqslant \varepsilon\sqrt n$, then $\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=0$ and if $j$ is such that $\sqrt j +1\gt \varepsilon\sqrt n$, then $\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=1$, hence $$\frac 1n \sum_{j=1}^n\mathbb E[X_j^2\mathbf 1\{ |X_j|\gt \varepsilon \sqrt n \}]=\frac 1n\sum_{j\gt(\varepsilon\sqrt n-1)^2}^n1, $$ which does not converge to $0$.
By Theorem 2.1. page 331 of the book Probability: A Graduate Course by Allan Gut, since $\max_{1\leqslant j\leqslant n}\frac{\operatorname{Var}(X_j)}{s_n^2}\to 0$, the sequence $\left(n^{-1/2}\sum_{j=1}^nX_j \right)_{n\geqslant 1} $ cannot converge to a standard normal distribution.