Suppose $(x_t)_{t=0}^\infty$ is a random starting from $x_0=0$ with transition probability $P(x_{t+1}-x_t)=p\mathbf 1(x_{t+1}-x_t=1)+q\mathbf 1(x_{t+1}-x_t=-1)$ where $p\ge0,q\ge0,p+q=1$. Given $a>0$ and $T\in\mathbf N$. How do we prove the following assertion? $$P\Big(\bigcap_{n=0}^\infty\bigcup_{T=n}^\infty \Big\{\sum_{t=0}^T \mathbf 1(x_{t+1}-x_t=1)>p(T+1)+a\Big\}\Big)=1.$$
I am thinking of taking this yet-to-be-resolved approach. Maybe there is a more direct method.
The approached I attempted linked to in the question is a bit too strong to resolve this question. The assertion can be inferred from either the law of the iterated logarithm or the central limit theorem and Kolmogorov's zero-one law.
(to be continued)