I am considering a biased random walk:
$X_1,X_2,\dots$ iid with $\mathbb{P}(X_1=1)=p$ and $\mathbb{P}(X_1=-1)=1-p$ with $p\in[0,1]\backslash\{1/2\}$, $S_n=X_1+\dots+X_n$.
In this setting I want to examine the event $A=\{S_n=0$ infinitely often$\}$. I suppose that $\mathbb{P}(A)=1$ or at least $\mathbb{P}(A)\in\{0,1\}$ but I can't prove it.
My attempts are towards Borel-Cantelli or Kolmogorov 0-1. Using Kolmogorov I should try to prove that this is a tail event and that's my problem. One can rewrite $A$ as the $\text{limsup }A_n$ with $A_n=\{S_n=0\}$ but now I am a bit lost. I would appreciate any hints (Is this a tail event at all?, Is it correct that $\mathbb{P}(A)\in\{0,1\}$?, $\dots$). Thanks a lot in advance!
We use Borel-Cantelli. First note that $\mathbb{P}(S_{2m+1}=0)=0~\forall m$. $$\sum_{k=1}^{\infty}\mathbb{P}(S_{2k}=0)=\sum_{n=1}^{\infty}{2n \choose n}p^n(1-p)^n.$$ We get (in order to use the ratio test) $$\text{limsup}_{n\to\infty}\sqrt[n]{{2n \choose n}p^n(1-p)^n}=\text{limsup }_{n\to\infty}p(1-p)\sqrt[n]{\frac{(2n)!}{n!^2}}=4p(1-p).$$ This is less than $1$ iff $p(1-p)<\frac{1}{4}$ which is the case by definition of $p$. Hence by the ratio test, the sum converges and Borel-Cantelli leads to $$\mathbb{P}(\text{limsup}_{n\to\infty}\{S_{2k=0}\})=\mathbb{P}(A)=0.$$