# Symmetric random walk does not converge a.s. using CLT

Let $(X_n)_n$ be a sequence of iid random variables with $P(X_1 = 1) = \frac{1}{2}$ and $P(X_1 = -1) = \frac{1}{2}$. Define $S_n := \sum_{i = 1}^n X_i$. Prove that $S_n$ does not converge almost surely by contradiction using the Central Limit Theorem.

The CLT states that: $\frac{S_n-nE(X_1)}{\sqrt{Var(X_1)n}} = \frac{S_n}{\sqrt{n}} \overset{\mathcal{L}}{\longrightarrow} \mathcal{N}(0,1)$.

I don't really have an idea how to deduce that $S_n$ is not almost surely convergent using the CLT.

## 1 Answer

Suppose $S_n\to S$ almost surely for some random variable $S$. Then $$\frac{S_n}{\sqrt{n}}\to 0\quad a.s.$$

One the other hand, almost surely convergence implies convergence in distribution. What contradiction can you have?

• Thank you for the answer but why is $\frac{S_n}{\sqrt{n}} \to 0$ almost surely? The contradiction would be as you mentioned with the convergence in distrubution. According to the CLT $\frac{S_n}{\sqrt{n}} \overset{\mathcal{L}}{\longrightarrow} \mathcal{N}(0,1)$ whereas a constant random variable $c$, especially 0, has the distribution function $\mathbb{1}_{[c,\infty)}$. Hence, the necessary contradiction. Dec 18, 2016 at 22:50
• Think about the case for real sequence. If $a_n\to a$ for some real number $a$, then what can you say about $\lim_{n\to\infty}\dfrac{a_n}{\sqrt{n}}$?
– user9464
Dec 18, 2016 at 22:53
• Oh well, of course you are right. $\lim_{n \to \infty} a_n/ \sqrt{n} = 0$ and this also applies for the almost surely convergence since you assumed that $S_n \to S$ a.s. Dec 18, 2016 at 22:57