Central Limit Theorem for not identical distributed but independent centered random variables with variance one.

Question

so let's assume we have independent random variables $X_1,X_2, X_3, \ldots$ with $$\mathbb{E}[X_k]=0 \mbox{ and } \mathbb{Var}[X_k]=\sigma_k^2=1 \quad \forall k\in\mathbb{N}. $$

We define $$s_n^2:= \sum_{k=1}^n \mathbb{E}[(X_k-\mathbb{E}[X_k])^2]=\sum_{k=1}^n\sigma_k^2=n.$$

Now we check Lindenberg's condition: So for $\varepsilon>0$ we must check $$\lim\limits_{n\rightarrow \infty} \frac{1}{s_n^2}\sum_{k=1}^n \mathbb{E}[(X_k-\mathbb{E}[X_k])^2 \cdot \mathbf{1}_{|X_k| > \varepsilon s_n^2} ] =\lim\limits_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ].$$

We know, $$\mathbb{E}[X_k^2 ]=1,$$ so $$\mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]$$ converges quicker than $$\frac{1}{n},$$ whch means we find an $N\in\mathbb{N}$ such that for every $n\geq N$ we have
$$\mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]< \frac{1}{n}, $$ which means for every $\varepsilon>0$ we have $$\lim\limits_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ] \leq \lim\limits_{n\rightarrow \infty} \frac{N}{n} + \frac{1}{n} \sum_{k=N+1}^n \mathbb{E}[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} ]< \lim\limits_{n\rightarrow \infty} \frac{N}{n} + + \frac{1}{n^2 }=0,$$ so we know that the Central Limit Theorem holds.

Is that right?

Edit: What we really need is uniform integrability of ${ X_k^2: k\in\mathbb{N}} $$ and that gives us that Lindeberg holds.

Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!

Answer 1

It may be not true that $\mathbb{E}\left[X_k^2 \cdot \mathbf{1}_{|X_k| > \varepsilon n} \right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $\Pr(\left\lvert X_1\right\rvert >t)\sim t^{-2}(\log t)^{-2}$, then $\mathbb{E}[X_1^2 \cdot \mathbf{1}_{|X_1| > \varepsilon n} ]$ is of order $1/\log n$.

Note that it would suffices to have uniform integrability, that is, $$ \lim_{R\to +\infty}\sup_{k\geqslant 1}\mathbb{E}\left[X_k^2 \cdot \mathbf{1}_{|X_k| > R} \right]=0. $$

Otherwise, the are counter-examples.