TITLE: Limit superior of $Y_1 + \ldots, Y_n$ with $Y_i$ bounded, i.i.d. and $\Pr(Y_1 \neq 0)>0$.
QUESTION [5 upvotes]: I am looking to establish that if $Y_1, Y_2, \ldots$ are bounded i.i.d. random variables satisfying $E(Y_1) = 0$ and $\Pr(Y_1 \neq 0) >0$ then for $X_n = Y_1 + \ldots, Y_n$, with probability one 

$$ \limsup_n X_n= \infty \quad \text{and} \quad \liminf_n X_n = -\infty  $$.

Now, obviously these are tail events and as a result they have probability either zero or one. Hence it suffices to show that the probability is positive. For this,
$$\Pr(\limsup_n X_n \geq M) \geq \limsup_n  \Pr(X_n \geq M) $$ so that it suffices to show that the right hand side is positive and this is where I am stuck since I cannot translate this into a statement for the $Y_i$. Could you please give me a hint? 
I know that this may be done with the help of the Central Limit Theorem but I am looking for a solution that avoids this.

REPLY [3 votes]: Here is another approach. For a fixed $M$, the event $A_M:=\{\limsup_{n\ge 1}X_n\ge M\}$ belongs to the exchangeable sigma algebra. Thus, by the Hewitt–Savage 0-1 law, $\mathsf{P}(A_M)\in \{0,1\}$. However,
$$
\mathsf{P}(A_M)\ge \limsup_{n\ge 1}\mathsf{P}(X_n\ge M)= \limsup_{n\ge 1}\mathsf{P}\!\left(X_n/\sqrt{n}\ge M/\sqrt{n}\right)>0
$$
by the CLT, and, therefore, $\mathsf{P}(A_M)=1$. Consequently, $
\mathsf{P}(A_{\infty})=\lim_{M\to\infty}\mathsf{P}(A_M)=1$.

Note that when $\mathsf{P}(Y_1=0)=1$, $X_n=0$ for all $n\ge 1$ a.s.