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$$M_n \le r\sqrt{\log n}?$$

## 2 Answers

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.

**A necessary condition is that $r> \sqrt 2$.**

Indeed, if $M_n$ has this property then so does $B_n-B_{n-1}$. Let $Y_n = B_n - B_{n-1}$. The variance of $B_n-B_{n-1}$ is 1.

The probability that $Y_n>r\sqrt{\log n}$ is $$1-\Phi(r\sqrt{\log n}) = \Phi(-r\sqrt{\log n})$$ where with the standard normal CDF, $\Phi$, we have for $x>0$, $$ \sqrt{2\pi}\Phi(-x)\ge e^{-x^2/2}/x - e^{-x^2/}/x^3. $$ Thus $$ \sum_n \Phi(-r\sqrt{\log n}) \ge $$ $$ \sum_n\sqrt{2\pi}^{-1}{n^{-r^2/2}}\left(\frac{1}{r\sqrt{\log n}} - \frac{1}{r^3(\log n)^{3/2}}\right)$$ This diverges if $r^2/2\le 1$, i.e., if $r\le\sqrt{2}$.

By the second Borel-Cantelli lemma, infinity many of the events $Y_n>\sqrt{\log n}$ occur.