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

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 ?

• This does not contradict the Centeral Limit Theorem because the $X_{i}$'s are not identically distributed (support of each $X_{i}$ is dependent on $i$). That is common mistake many make with central limit theorem, you must be summing over IID random variables Aug 27, 2014 at 22:31
• @user159813 of course for iid, thanks. Aug 27, 2014 at 22:46
• I am bit confused with your goal to prove. If $Var(X_n)=1$ then $Var(S_n)=n$ (they are independent,so we can sum the variances). But then $Var(S_n/\sqrt(n))=1$, so how it can converge to $0$ almost surely? Did I miss something? Aug 27, 2014 at 23:24
• Of of what Alexander Vigodner pointed out are you sure you didn't mean $S_{n}/n=\bar{X}$ because then the $Var(\bar{X})=0$ as $n\rightarrow\infty$ thus I could see $\bar{X}\rightarrow 0$ a.s. possibly Aug 28, 2014 at 6:34
• Variances are not that useful to study almost sure convergence.
– Did
Aug 28, 2014 at 7:48

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.