Example of CLT when moments do not exist

Consider $X_n = \begin{cases} 1 & \text{w.p. } (1 – 2^{-n})/2\\ -1 & \text{w.p. } (1 – 2^{-n})/2\\ 2^k & \text{w.p. } 2^{-k} \text{ for } k > n\\ \end{cases}$

I need to show that even though this has infinite moments, $$\sqrt{n}(\bar{X}_n) \overset{d}{\to} N(0,1) $$

I have tried showing this by using Levy’s Continuity Theorem, i.e, tried showing that the characteristic function of the left side converges to the characteristic function of the standard normal. However, this seemed impossible to show.

A hint provided for this problem was to truncate each $X_i$, i.e. letting $Y_{ni} = X_i I\{X_i \leq n\}$ and using Lindeberg condition to show that $\sqrt{n} \bar{Y}_n \overset{d}{\to} N(0,1)$.

However, I have not been able to show that the Lyapunov condition is satisfied. This is mainly because $Y_{ni}$ does not behave like I would want it to. I would want $Y_{ni}$ to only take values -1 and 1, however, the way it is constructed, it can take values $-1, 1, 2^{i+1}, 2^{i+2}, \dots, 2^{\lfloor\log_2 n\rfloor}$


Here’s an answer based on @cardinal’s comment:

Let the sample space be that of paths of the stochastic processes $(X_i)_{i=0}^{\infty}$ and $(Y_i)_{i=0}^{\infty}$, where we let $Y_i=X_i \mathbb{1}_{\{X_i\leq 1\}}$. The Lindeberg condition (conforming with Wikipedia’s notation) is satisfied, for:
$$\frac{1}{s_n^2} \sum_{i=0}^n \mathbb E (Y_i^2 \mathbb{1}_{\{|Y_i|> \epsilon s_n^2\}})\leq \frac{1}{s_n^2} \sum_{i=0}^n P(|Y_i|> \epsilon s_n^2)\to0,$$
for any $\epsilon$ as $s_n^2\to \infty$ whenever $n\to \infty.$

We also have that $P(X_i\neq Y_i, i.o.) = 0$ by Borel-Cantelli since $P(X_i \neq Y_i)=2^{-i}$ so that $\sum_{i=0}^{\infty} P(X_i \neq Y_i) = 2<\infty$. Stated differently, $X_i$ and $Y_i$ differ only finitely often almost surely.

Define $S_{X,n}=\sum_{i=0}^{n} X_i$ and equivalently for $S_{Y,n}$. Pick a sample path of $(X_i)_{i=1}^{\infty}$ such that $X_i > 1$ only for finitely many $i$. Index these terms by $\mathcal{J}$. Require also from this path that the $X_j,j\in \mathcal{J}$ are finite. For such a path, $$\frac{S_{\mathcal{J}}}{\sqrt{n}} \to 0,\text{ as }n\to \infty$$ where $S_{\mathcal{J}}:=\sum_{j\in \mathcal{J}}X_j$. Moreover, for large enough $n$,

Using the Borel-Cantelli result together with the fact that $X_i$ is almost surely finite, we see that the probability of a sample path obeying our requirements is one. In other words, the differing terms go to zero almost surely. We thus have by Slutsky’s theorem that for large enough $n$, $$\frac{1}{\sqrt{n}}S_{X,n}=\frac{S_{Y,n}+S_{\mathcal{J}}}{\sqrt{n}}\overset{d}{\to}\xi+0,$$ where $\xi\sim N(0,1)$.

Source : Link , Question Author : Greenparker , Answer Author : ekvall

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