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 ?

  • 2
    $\begingroup$ 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 $\endgroup$
    – Kamster
    Aug 27, 2014 at 22:31
  • $\begingroup$ @user159813 of course for iid, thanks. $\endgroup$
    – inequal
    Aug 27, 2014 at 22:46
  • 2
    $\begingroup$ 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? $\endgroup$ Aug 27, 2014 at 23:24
  • $\begingroup$ 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 $\endgroup$
    – Kamster
    Aug 28, 2014 at 6:34
  • $\begingroup$ Variances are not that useful to study almost sure convergence. $\endgroup$
    – Did
    Aug 28, 2014 at 7:48

1 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .