Let $S_n$ be the symmetric random walk on $\mathbb{Z}$. How do i calculate $P(\limsup_{n\rightarrow\infty} S_n=\infty)$? I already know that the probability is 1 but I don't really know how to start? Anyone got some advice? The more basic the better!
Thanks a lot!
Let $A_M=[\limsup\limits_{n\to\infty}S_n\leqslant M]$ for some fixed integer $M$, then $A_M\subseteq\liminf [S_n\leqslant M]$. Assuming that you know that $\liminf\limits_{n\to\infty}P[S_n\leqslant M]=\ell$ with $\ell\lt1$ ${}^{(1)}$, this yields $P[A_M]\leqslant\ell$. According to a comment, you also know that $A_M$ is an asymptotic event, hence $P[A_M]=0$. This holds for every $M$ hence $\limsup\limits_{n\to\infty}S_n=+\infty$ almost surely.
${}^{(1)}$ To show this, proceed by contradiction, then $P[S_n\leqslant M]\to1$ hence $P[S_n\geqslant-M]\to1$ by symmetry, and $P[|S_n|\leqslant M]\to1$. The distribution of $S_n$ is binomial $(\frac12,n)$ hence there exists some absolute finite constant $c$ such that $P[S_n=k]\leqslant c/\sqrt{n}$ for every $k$ and $n$. In particular, $P[|S_n|\leqslant M]\leqslant(2M+1)c/\sqrt{n}\to0$, a contradiction.