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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 $-1, 1, 2^{i+1}, 2^{i+2}, \dots, 2^{\lfloor\log_2 n\rfloor}$

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    $\begingroup$ If you're truncating at $n$, check that last paragraph carefully for the values the truncated variable can take on. At any rate, try truncating at $1$ instead, use Borel-Cantelli and then Slutsky to get the result. You should be able to use Lindeberg or Lyapunov on the truncated piece (though I didn't actually check that). $\endgroup$ – cardinal Mar 9 '14 at 1:36
  • $\begingroup$ Sorry about that. Changed it to "infinite" moments $\endgroup$ – Greenparker Mar 9 '14 at 1:53
  • $\begingroup$ @cardinal I went over the possible values $Y_{ni}$ can take again, and added a floor to the log term. Otherwise, the values seem right. If I truncate at 1, I will get the values I want for $Y_{ni}$ and will be able to apply the Lindeberg condition to get convergence to the normal. However, I do not see how this will imply convergence to normal for the $\sqrt{n} \bar{X}_n$ $\endgroup$ – Greenparker Mar 9 '14 at 2:18
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    $\begingroup$ What is "$\bar{X}_n$"? You have not described a context in which there are samples or multiple instances of each $X_n$, whence -- given what is stated in the question -- about the only possible reading of this notation is that it refers to the mean of $X_n$, which is always infinite and is a number, not a distribution. We therefore have to imagine you are contemplating iid samples of $X_n$, but you need to tell us this and you especially need to stipulate what the sample sizes are. $\endgroup$ – whuber Aug 3 '14 at 15:22
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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 $n$, $$S_{X,n}-S_{Y,n}=S_{\mathcal{J}}.$$

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 $\xi\sim N(0,1)$.

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