Limit superior of $Y_1 + \ldots, Y_n$ with $Y_i$ bounded, i.i.d. and $\Pr(Y_1 \neq 0)>0$. 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.
 A: Since $P(Y_=0)\neq 1$ we can normalize $Y_i$'s so that the variance is $1$. In that case the Law of Iterated Logarithms (en.wikipedia.org/wiki/Law_of_the_iterated_logarithm ) shows that $\lim \sup \frac {X_n} {\sqrt {2nlog(log\, n))}} =1$ and $\lim \inf \frac {X_n} {\sqrt {2nlog(log\, n))}} =-1$. This implies that $\lim \sup X_n =\infty$ and $\lim \inf X_n =-\infty$.
A: My question already has a good answer but I was wondering if the following approach is also valid. To prove $\limsup X_n = \infty $ we need to produce a subsequence that diverges to $\infty$ with probability one. Define the stopping time
$$N_M = \inf\{n \geq 1: X_n \geq M\}$$
If we could show that $\Pr(N_M<\infty) = 1$ for all $M>0$ then it would follow that $\limsup X_n = \infty$ almost surely, as the subsequence
$$\left\{X_{N_M} \right\}_{M \in \mathbb{N}}$$
diverges to $\infty$ almost surely. Set $Z_n = X_{N_M \wedge n}$. Then $Z_n$ is a martingale that is bounded on one side since $Z_n < M + \sup_n |Y_n| < \infty$. Therefore, it converges a.s. by the martingale convergence theorem. Now suppose that $N_M = \infty$. Then
$$N_m = \infty \implies |Z_{n+1} - Z_n| = |X_{n+1}-X_n| = |Y_{n+1}|,$$
and $\Pr(|Y_{n+1}|>0) = \Pr(|Y_{1}|>0) >0$, so the sequence $Z_n$ cannot converge almost surely, a contradiction. Therefore, $N_M < \infty$ almost surely.
A: 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.
