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,

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


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

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


2 Answers 2


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


I believe that the key here is to show that the indicator function $I$ will be zero for all k, greater than some number. Since the expected value of $X^2$ is just 1, an all 1s indicator function would yield a limit of $\frac{n}{n}$, technically $\frac{2}{3}$ since we know a third of the $X_ks$ are 0.

But since $I$ will be zero for all k, $$k > \left(\frac{1}{2\epsilon}\right)^{\left(\frac{1}{2\alpha-\frac{1}{2}}\right)}$$ that is if my algebra is right ;)

Then the limit should be some finite number over n, which goes to infinity.


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