Let $B\_t$ be a standard Brownian motion. Let$$M\_n = \max\{|B\_t - B\_{n-1}| : n - 1 \le t \le n\}.$$For which $r > 0$ is it the case with probability one, for all $n$ sufficiently large$$M\_n \le r\sqrt{\log n}?$$



For which $r > 0$ is it the case with probability one, for all $n$ sufficiently large $M_n \le r\sqrt{\log n}$?
The condition $r>\sqrt 2$ is also sufficient. The random variables $(M\_n)\_{n\geq1}$ are iid (in fact, we need that they are identically distributed), due to the independence of the increments and the translation invariance of Brownian motion. We shall use the first Borel-Cantelli lemma to prove that for $r>\sqrt 2$,
$$
\mathbb{P}[M\_n\geq r\sqrt{\log n}\text{ infinitely often}]=0.
$$
To this end, it is sufficient to prove that
$$
\sum\_n \mathbb{P}[M\_n\geq r\sqrt{\log n}]<\infty.
$$
Indeed,
$$
\mathbb{P}[M\_n\geq r\sqrt{\log n}]=\mathbb{P}[M\_1\geq r\sqrt{\log n}]\leq 2\mathbb{P}[U\_1\geq r\sqrt{\log n}],
$$
where $U\_t=\max\{B\_s:s\in[0,t]\}$ is the running maximum (the inequality above holds since $M\_1$ is the largest between $U\_1$ and $-L\_1$, and $U\_t$ and $-L\_t$ have the same distribution, where $L\_t$ is the running minimum). By the reflection principle,
$$
\mathbb{P}[U\_1\geq r\sqrt{\log n}]=2\mathbb{P}[B\_1\geq r\sqrt{\log n}]
=2(1-\Phi(r\sqrt{\log n})),
$$
and the series converges for $r>\sqrt 2$.


Notice that since $\mathbb{P}[M\_1\geq r\sqrt{\log n}]\geq\mathbb{P}[U\_1\geq r\sqrt{\log n}]$ and since $(M\_n)\_{n\geq1}$ are independent, a similar argument and the second Borel-Cantelli lemma (as in the answer of @Bjørn Kjos-Hanssen) provide the necessity of the condition.


