Jtdylktuy

Let $X_1,X_2,...$ be independent and for any n $ge 1$ and $alpha>0$
$$X_n = left{
begin{array}{rl}
n^alpha & text{with } Pr(X_n= n^alpha) = frac{1}{2n^{2alpha}},\
-n^alpha & text{with }Pr(X_n= -n^alpha) = frac{1}{2n^{2alpha}},\
0 & text{with } Pr(X_n= 0) = 1- frac{1}{n^{2alpha}}.
end{array} right.$$
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,

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

and

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

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 $E[X_k^2 I_{{frac{|X_k|}{sqrt{k}}>epsilon}}]$.

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

If $alphageqslant 1/2$ and $varepsilonin (0,1)$, the term $mathbb Eleft[X_k^2 I_{{|X_k|>epsilon sqrt n}}right]$ is zero if $k^alphagt varepsilon sqrt n$, that is , if $kgt varepsilon^{alpha}n^{1/(2alpha)}$ hence
$$
sum_{k=1}^{n} mathbb Eleft[X_k^2 I_{{|X_k|>epsilon sqrt n}}right]=
sum_{k=1}^{lfloor varepsilon^{alpha}n^{1/(2alpha)}rfloor} mathbb Eleft[X_k^2 I_{{|X_k|>epsilon sqrt n}}right]
$$
and for $1leqslant kleqslant lfloor varepsilon^{alpha}n^{1/(2alpha)}rfloor$, the term $mathbb Eleft[X_k^2 I_{{|X_k|>epsilon sqrt n}}right]$ is one hence
$$frac{sum_{k=1}^{n} mathbb Eleft[X_k^2 I_{{|X_k|>epsilon sqrt n}}right]}{n} =frac 1nlfloor varepsilon^{alpha}n^{1/(2alpha)}rfloor $$
hence Lindeberg's condition holds if and only if $alpha>1/2$.

Since $max_{1 le i le n} sigma_i/ s_n to 0$, Lindeberg's condition is equivalent to the convergence of $left(S_n/b_nright)_n$ to a standard normal distribution.