Let $S_n = X_1 + \ldots + X_n$, $X_i \sim \mathbf{N}(0, 1)$. The claim here is that
$\mathbb{P}\Big(\max_{1 \le k \le n} \dfrac{|S_k|}{\sqrt{k}} > c \Big) \approx 2 \cdot \mathbb{P}\Big(X_1 > c \Big)$ as $n \rightarrow \infty$.
why is it so?
A few observations (don't know how relevant they are) -
- Since $X_1$ is normal, it is symmetric. Therefore, $2 \cdot \mathbb{P}\Big(X_1 > c \Big) = \mathbb{P}\Big(|X_1| > c \Big)$.
- Clearly this is reminiscent of Levy's Inequality (whose proof I can't find anywhere), but only in one direction, whence in view of observation (1), looks like the intention here is to prove (again, at least in one direction) $\mathbb{P}(\max_{1 ≤ k ≤ n} |S_k| > c)≤ 2 \cdot \mathbb{P}(|S_n| > c )$, and normalizing will yield the result by the Central Limit Theorem.