A problem about Central Limit Theorem without identically distributed

Question

${X_n}$ is an independent random variable sequence, $P(X_{n}=\pm n^{\alpha})=\frac{n^{1-2\alpha}}{2},\quad P(X_{n}=0)=1-n^{1-2\alpha},\quad n\geq1,\quad\alpha\geq\frac{1}{2}.$

Define $S_{n}=\sum_{i=1}^{n}X_{i}$ and $\sigma_{n}^{2}=\operatorname{Var}\left(S_{n}\right).$

Prove that when $\alpha\in\left[1/2,1\right)$, $S_{n}/\sigma_{n}\xrightarrow{d}N(0,1)$, but when $\alpha>1$, this conclusion does not hold.

My question is: How to prove when $\alpha>1$, $S_{n}/\sigma_{n}\xrightarrow{d}N(0,1)$ does not hold?

Answer 1

Too long for a comment. I would be happy if someone have a better idea.

I would use Levy's theorem stating that a sequence of random variables $(X_n)_n$ converges in distribution to a random variable $X$ if, and only if, the sequence $(\phi_{X_n})_n$ of characteric functions of $(X_n)_n$ converges pointwise to the characteristic function $\phi_X$ of $X$.

To apply this strategy to your problem, first, notice that $\mathrm{E}(X_n)=0$ and $\mathrm{Var}(X_n)=\mathrm{E}({X_n}^2)=n$, so that the independence of the $X_n$ yields $$\mathrm{E}(S_n)=0,\quad {\sigma_n}^2=\frac{n(n+1)}{2},\quad \phi_{S_n/\sigma_n}(t)=\mathrm{E}(e^{itS_n/n})=\prod_{k=1}^n\phi_{X_n}(t/\sigma_n),$$ but, besides, Euler's formula shows that: $$\phi_{X_n}(t)=\mathrm{E}(e^{itX_n})=1-n^{1-2\alpha}+n^{ 1-2\alpha}\cos(tn^\alpha).$$ I think from there the result is a matter of real analysis (series expansion).

Answer 2

Define $A_j:=\{X_j\neq 0\}$. Then $\mathbb P(A_j)=j^{1-2\alpha}$ hence by the Borel-Cantelli lemma, $\limsup_{n\to\infty}A_n$ has probability zero. This implies that $(S_n)$ converges almost surely. Indeed, for each $\omega$ in the complement of $\limsup_{n\to\infty}A_n$, there exists $N(\omega)$ such that for each $j\geqslant N(\omega)$, $\omega\notin A_j$.

Since $\sigma_n\to\infty$, we have $S_n/\sigma_n\to 0$ almost surely if $\alpha>1$.