Let $(X_n)$ be a sequence of i.i.d. r.v. such that $\mathbb{P}(X_n=1)=\mathbb{P}(X_n=-1)=\frac{1}{2}$
Also let $$S_n= \sum_{k=1}^{n} X_k$$
I am asked to show, using Borel–Cantelli lemma, that for every (integer) $k \geq 1$
$$\limsup_{n\rightarrow \infty}(S_{n+k}-S_n)=k\ \ \ \ \ \text{a.s.} $$
I am not sure how to interpret this and how to prove it.
My interpretation is that we are considering the events $A_n=\{(S_{n+k}-S_n)=k\}$
Then $$\sum \mathbb{P}(A_n) = \infty$$
but the events are not independent so we can not conlude with Borel–Cantelli lemma that
$$\mathbb{P} (A_n \ \ \text{i.o.}) = \mathbb{P}(\limsup_{n\rightarrow \infty}(S_{n+k}-S_n)=k) = 1 $$
My other idea was to show that the event "$A_n$ occours infinitely often" is a tail event, then by Fatou's lemma I could say that $$\mathbb{P}(\limsup_{n\rightarrow \infty}(S_{n+k}-S_n)=k) \geq \left(\frac{1}{2}\right)^k$$ and conclude by Kolmogorov 0-1 theorem…
I am not really sure and any help would be really appreciated!
Best Answer
Use Borel-Cantelli lemma with $$B_n:=\bigcap_{j=nk+1}^{(n+1)k}\{X_j=1\}.$$