Convergence of discrete random variables, show $\frac{S_n}{\sqrt{n}}\to0$ a.s.

Question

Let $X_n$ be a sequence of independent discrete real random variables, with discrete density $$p_{X_n}(x):=\Pr(X_n=x)= \cases{ 1-\frac1{n^2} & if $\;x= 0$\cr \frac1{2n^2} & if $\;x=n,-n$\cr 0 & \text{else} }$$

If $S_n=\sum_{i=1}^{n}X_i$, show that $\displaystyle\frac{S_n}{\sqrt{n}}\to0$ a.s.

I computed, $\mathbf E(X_n)=0$ and $\mathrm{Var}(X_n)=1$, $\forall n$

• Does it not contradict the Central Limit Theorem ? $$\left(\frac{S_n-n\mu}{\sigma\sqrt{n}}\to Z,\text{where } Z \text{ is standard normal}\right)$$

• With Borel-Cantelli I obtained; $$\sum_n\Pr(X_n\neq0)<\infty\Longrightarrow\Pr(\limsup\limits_n\{X_n\neq0\})=0$$ What does it mean now? The set $\limsup\limits_n\{X_n=0\}$ has probability $1$, using the set definition of $\limsup$; $$\limsup\limits_n\{X_n=0\}=\bigcap_{n\ge 1}\bigcup_{k\ge n}\{X_k=0\}=\lim_{n\to\infty}\bigcup_{k=n}^{\infty}\{X_k=0\}$$ How can I conclude from that $X_n\neq0$ only for a finite number of $n$, otherwise it is difficult to prove the claim.

• and the last question, If we had $\{\limsup\limits_n X_n=0\}$ (limsup in the curly brackets), then we couldn't use the set definition, is that correct ?

Answer 1

CLT does not apply to the present setting because (the simplest version of) CLT assumes that $(X_n)$ is i.i.d. while here the distribution of $X_n$ depends on $n$.

With Borel-Cantelli I obtained; $$\sum_n\Pr(X_n\neq0)<\infty\Longrightarrow\Pr(\limsup\limits_n\{X_n\neq0\})=0$$ What does it mean now? The set $\limsup\limits_n\{X_n=0\}$ has probability $1$ (...)

Actually, $\Pr(\limsup\limits_n\{X_n\neq0\})=0$ shows more that what you write, that is, that $\liminf\limits_n\{X_n=0\}$ has probability $1$. Thus, there exists some integer valued random variable $N$, almost surely finite, such that $X_n=0$ for every $n\geqslant N$.

Now, this is a fact of mere logic that every deterministic sequence $(x_n)$ which is zero for every $n$ large enough is such that $(s_n)$ is bounded, where $s_n=\displaystyle\sum\limits_{i=1}^nx_i$ for every $n$. A fortiori, $\dfrac{s_n}{\sqrt{n}}\to0$.

Applying this to your setting, one gets the desired result that $\dfrac{S_n}{\sqrt{n}}\to0$ almost surely, and even the stronger result that, for every positive $\alpha$, $\dfrac{S_n}{n^\alpha}\to0$ almost surely, since $(S_n)$ is almost surely bounded.