Showing $\limsup \frac{|S_n|}{n}=\infty$

Question

$X_n$'s are i.i.d symmetric with $E|X_1|=\infty$. Then $\limsup \frac{|S_n|}{n}=\infty$. How do I show $\limsup \frac{S_n}{n}=\infty$ and $\liminf \frac{S_n}{n}=-\infty$?

My attempt: Let $c=\limsup \frac{S_n}{n}=\limsup \frac{-S_n}{n}=-\liminf \frac{S_n}{n}$. (Since $X_n\overset{d}{=}-X_n$)

As $\limsup \geq \liminf$, we've $c\geq -c \Rightarrow c\geq 0$

Now $\infty=\limsup \frac{|S_n|}{n} \geq \limsup \frac{S_n}{n}=c\hspace{5pt}$ i.e. $0\leq c \leq \infty$ which is trivially true. How do I show $c=\infty$? I appreciate any kind of hint/help. Thank you,

Answer 1

Set

$$\begin{align*} A &:= \left\{ \limsup_{n \to \infty} \frac{S_n}{n} = \infty \right\} \\ B &:= \left\{ \liminf_{n \to \infty} \frac{S_n}{n} = - \infty \right\}. \end{align*}$$

Hints:

1. By symmetry, we have $\mathbb{P}(A) = \mathbb{P}(B)$.
2. By Hewitt-Savage's 0-1-law, $\mathbb{P}(A)=\mathbb{P}(B) \in \{0,1\}$.
3. Deduce from the first two steps and the fact that $$\mathbb{P}\left(\limsup_n \frac{|S_n|}{n} = \infty \right)=1$$ that $\mathbb{P}(A) = \mathbb{P}(B)=1$.