# Showing that lim sup of sum of iid binary variables $X_i$ with $P[X_i = 1] = P[X_i = -1] = 1/2$ is a.s. infinite

Let $(X_i)_{i\in\mathbb{N}}$ be an i.i.d. sequence of binary random variables with $$P[X_i = 1]=P[X_i = -1] = \frac{1}{2}$$ and let $$S_n = \sum_{i=1}^{n} X_i.$$

I'd like to show that $$P[\lim \sup_{n \rightarrow \infty} S_n = \infty] = 1$$ with the means of basic probability theory and the Borel–Cantelli lemma or Kolmogorov's 0-1 law. Could somebody give me a hint?

• If you just want a hint, try to show the stronger claim that $P(\limsup \frac{S_n}{\sqrt{n}} = \infty) = 1$ using the 0-1 law and the CLT. If you want a full solution using this hint, I wrote it up here: math.stackexchange.com/questions/210131/… – Chris Janjigian Jan 12 '15 at 1:29
• I managed to sign up to math.stackexchange such that I don't own my original question above. Since I also can't comment yet, I'll write here and then later delete this reply. Chris, thanks for your comment, but I was looking for a proof that doesn't require the CLT. – Markus Jan 12 '15 at 1:49

By the Hewitt-Savage's 0-1 law, the random variable $\limsup_{n\to \infty}S_n$ is almost surely constant (the constant may be $+\infty$ or $-\infty$ and is denoted by $c$). Defining $S'_n:=S_{n+1}-X_1$, the sequence $(S'_n)_{n\geqslant 1}$ has the same distribution as the sequence $(S_n)_{n\geqslant 1}$, hence $c=c-X_1$. Since $X_1$ is not degenerated, then $c\in\{+\infty,-\infty\}$.
By symmetry of $X_1$ hence that of $S_n$, we actually have $c= +\infty$.
• Any reasons that $\lim\sup S_n$ is a constant a.s. ? It seems that there is no clear reason that $\{\lim\sup S_n <a\}$ is a tale event. – Brian Ding Feb 17 '15 at 22:38
• Hewitt-Savage requires that $(S_n)_{n \in \mathbb N}$ are independent random variables. Why should this be true? – D Ford Nov 17 '17 at 18:25
• Actually, $\limsup S_n$ is a tail event. See Achim Klenke's "Probability Theory: A Comprehensive Introduction", 2nd ed, Example 2.36(ii). – D Ford Nov 14 '18 at 15:18