The <law of the iterated logarithm for a simple symmetric random walk> states that almost surely
$$
\limsup_{n\to\infty}\frac{|S_n|}{\sqrt{2n\log\log n}}=1.
$$
Since $\sqrt{2\log\log n}$ is unbounded, $|S_n|/\sqrt n$ exceeds every fixed $c>0$ at some finite time. Hence
$$
T_c=\inf\{n\geq0:|S_n|>c\sqrt n\}<\infty
$$
almost surely.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 / 1 / b / Solution

For a <simple symmetric random walk>, almost surely
$$
\limsup_{n\to\infty}\frac{S_n}{\sqrt{2n\log\log n}}=1,
\qquad
\liminf_{n\to\infty}\frac{S_n}{\sqrt{2n\log\log n}}=-1.
$$

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