I was trying to prove that some specific stopping time is finite almost surely (which was given), but I got stuck and was hoping someone could help me out. The setting is like this: $X_i \sim N(0,1)$ for all $i \in \mathbb{N}$, $S_0 = 0$, $S_n = \sum_{i=1}^n X_i$ for $n \geq 1$ and $\tau = \inf\{n \in \mathbb{N}: S_n > 1\}$.
Now we want to show that $P(\tau = \infty) = 0$. I have the feeling as if we want to use the Kolmogorov Doob inequality here (note $S_n$ is just an additive martingale), which is: $$ P(\max_{1\leq k \leq n}S_k \geq \epsilon n) \leq \frac{\mathbb{E}(|S_n|)}{n\epsilon}. $$ For $\epsilon > 0$ I thought we could use this in the following way: $$ P(\tau=\infty) \leq P(\tau>n) \leq P(\max_{1\leq k \leq n}S_k \leq 1). $$ But note that here we actually want the inequality inside the probability to be reversed, since then we can just take $\epsilon=\frac{1}{n}$ and we can use the inequality above. But it really seems that we need this here, since the event that the stopping time $\tau$ is larger than $n$ is a subset of the event that the maximum over all $k$ from $1$ to $n$ of $S_k$ is smaller or equal to one, as otherwise we would already have that $\tau \leq n$.
Thoughts
Am I using the Kolmogorov Doob inequality wrong? Can we transform it such that it makes our argument work? Is it correct that $\tau > n$ is a subset of $\{\max_{1\leq k \leq n}S_k \leq 1\}$? Maybe this can be proven in another way, but I am really not sure. We have used nowhere that our $X_n$ are normally distributed, which also could be of great use.