Kolmogorov $0-1$ Law and Borel-Cantelli Application

Question

Let $(X_{n})_{n\geq1}$ be a sequence of i.i.d. random variables such that $\mathbb{P}(X_{n} = -1) = \frac{1}{2}$ and $\mathbb{P}(X_{n} = 1) = \frac{1}{2}$. Let $S_{0} = 0$ and for $n\geq 1$ let $S_{n} = \sum_{k=1}^{n} X_{k}.$ We consider the following events:

$A_{x} = \{S_{n} = x$ for infinitely many $n\}$, where $x \in \mathbb{Z}$

$B_{+} = \{\limsup S_{n} = \infty\}, B_{\_} = \{\liminf S_{n} = -\infty\}$.

I want to show that $\mathbb{P}(B_{+}) = \mathbb{P}(B_{\_}) = 1$ and that $\mathbb{P}(A_{x}) = 1$ for all $x$. I have so far been able to show that $\mathbb{P}(B_{±})$ is either $1$ or $0$ using the Kolmogorov $0-1$ Law (since both events are in the tail $\sigma$-algebra of the $(X_{n})_{n\geq1}$ sequence).

I now think I have to firstly show that these two probabilities are equal, and then show that they are 1, but I am not sure how to approach it. For the $\mathbb{P}(A_{x})$ part, I think we need to use the Borel-Cantelli Lemma but I am not sure how to.

Answer 1

I don't think you need the 0-1 law. Try these steps

1. Show that for any $x\in \mathbb{Z}$

$$ \sum_n^\infty \mathbb{P}(S_n=x)=\infty. $$

2. Use reverse Borel-Cantelli to claim that $\mathbb{P}(A_x)=1$.

3. Use $\mathbb{P}(A_x)=1$ to infer $\mathbb{P}(B_{+})=1=\mathbb{P}(B_{-})$.