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Let $X_1,X_2,...$ be independent and for any n $\ge 1$ and $\alpha>0$ $$X_n = \left\{ \begin{array}{rl} n^\alpha & \text{with } Pr(X_n= n^\alpha) = \frac{1}{2n^{2\alpha}},\\ -n^\alpha & \text{with }Pr(X_n= -n^\alpha) = \frac{1}{2n^{2\alpha}},\\ 0 & \text{with } Pr(X_n= 0) = 1- \frac{1}{n^{2\alpha}}. \end{array} \right.$$ Let $S_n = X_1+ \dots +X_n$ and $B_n^2 = \sigma_1^2+\dots+\sigma_n^2.$ Does $\frac{S_n}{B_n}\rightarrow Z \sim N(0,1)$ in distribution.

Solving this question is an example of using the Lindeberg-Feller CLT. I found that,

$E[X_n]= n^\alpha(\frac{1}{2n^{2\alpha}})-n^\alpha(\frac{1}{2n^{2\alpha}})+0(1-\frac{1}{n^{2\alpha}}) = 0$

and

$E[X_n^2]=(n^{\alpha})^2(\frac{1}{2n^{2\alpha}}) + (-n^{\alpha})^2(\frac{1}{2n^{2\alpha}})+0^2(1-\frac{1}{n^{2\alpha}})=1.$

Therefore $\sigma_n^2 = 1$ and $B_n = \sqrt{n}$.

If the Lindeberg condition holds, i.e., for any $\epsilon > 0$ $$\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} E[X_k^2 I_{\{|X_k|>\epsilon B_k\}}]}{B_n^2} = 0.$$

Then $\frac{S_n}{B_n}\rightarrow Z \sim N(0,1)$ in distribution.

In our case, since $\sigma_k=1$ We have to deal with for any $\epsilon> 0,$ $\sum_{k=1}^{n} E[X_k^2 I_{\{\frac{|X_k|}{\sqrt{k}}>\epsilon\}}]$ for the numerator. I am stuck now because I dont know how to represent $E[X_k^2 I_{\{\frac{|X_k|}{\sqrt{k}}>\epsilon\}}]$.

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    $\begingroup$ You messed up the most important part of the condition: $I_{\{|X_k|>\epsilon B_k\}}$ with $B_k$ instead of $\sigma_k$ and $B_n^2$ in the denominator instead of $B_n$. This (as well as simpler Lyapunov's condition) will be enough to conclude convergence only if $\alpha<1$. $\endgroup$
    – A.S.
    Commented Mar 9, 2016 at 4:08
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    $\begingroup$ Correction. Condition is satisfied iff $|X_k|\le \epsilon \sqrt k$ eventually, hence we have convergence iff $\alpha<\frac 1 2$. $\endgroup$
    – A.S.
    Commented Mar 9, 2016 at 4:28

1 Answer 1

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The condition we have to check for Lindeberg's condition is thtat for all positive $\varepsilon$, $$ \lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon B_n\}}\right]}{B_n^2} = 0, $$ and since $B_n=\sqrt n$, this is equivalent to $$ \lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]}{n} = 0. $$ If $\alpha\lt 1/2$, then for each fixed $\varepsilon$, there exists a $n_0$ such that $n^\alpha\leqslant \varepsilon n^{1/2}$ for all $n\geqslant n_0$ hence for such $n$, $\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]=0$.

If $\alpha\geqslant 1/2$ and $\varepsilon\in (0,1)$, the term $\mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]$ is one if $k^\alpha\gt \varepsilon \sqrt n$, that is , if $k\gt \varepsilon^{\alpha}n^{1/(2\alpha)}$ and $0$ otherwise hence $$ \sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]\geqslant \sum_{k=\lfloor \varepsilon^{\alpha}n^{1/(2\alpha)}\rfloor}^n \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]= \left(n-\lfloor \varepsilon^{\alpha}n^{1/(2\alpha)}\rfloor\right) $$ hence Lindeberg's condition is not satistified.

Since $\max_{1 \le i \le n} \sigma_i/ s_n \to 0$, Lindeberg's condition is equivalent to the convergence of $\left(S_n/b_n\right)_n$ to a standard normal distribution. We thus conclude that the convergence of $S_n/B_n$ to a standard normal distribution holds if and only if $\alpha<1/2$.

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  • $\begingroup$ Hello! I am a little confused by this proof. I think that $E[X_k^2 I_{|X_k| > \epsilon \sqrt{n}}] =1$ if $k^\alpha > \epsilon \sqrt{n}$. $\endgroup$ Commented Nov 23, 2020 at 2:59
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    $\begingroup$ @mrsergazinov I have edited the answer; thanks for pointing this out. $\endgroup$ Commented Nov 23, 2020 at 10:51

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