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 $E[X_k^2 I_{\{\frac{|X_k|}{\sqrt{k}}>\epsilon\}}]$.
Best Answer
$E\left[X_k^2 I_{\left\{\frac{|X_k|}{\sqrt{k}}>\epsilon\right\}}\right]$ is either $1$ or $0$ depending on $k$ and $\epsilon$.
You are interested in the case of fixed $\epsilon$ while $k$ increases, and in particular whether $\frac{k^\alpha}{\sqrt{k}}>\epsilon$ is true, i.e. whether $k^{\alpha-1/2} >\epsilon$ is true, in which case the expectation is $1$.
When $\alpha < \frac12$, for any $k > \epsilon^{-1/(1/2-\alpha)}$ you will have $k^{\alpha-1/2} \not > \epsilon$, so the Lindeberg condition will hold for that $\alpha$
When $\alpha > \frac12$, for all $k > \epsilon^{1/(\alpha-1/2)}$ you will have $k^{\alpha-1/2} > \epsilon$, so the Lindeberg condition will not hold for that $\alpha$
When $\alpha = \frac12$ and $0 < \epsilon \lt 1$, for all $k$ you will have $k^{\alpha-1/2} > \epsilon$, so the Lindeberg condition will not hold for that $\alpha$
So for the condition to hold, you need $\alpha < \frac12$.
As a check, this is clearly true when $\alpha = 0$ as you would then have the simple Central Limit Theorem