Brownian motion does not vary more than $\sqrt{n}$ on each interval $[n,n+1]$

Question

Let $(B_t)_{t\geq 0}$ be a Brownian motion in $\mathbb{R}$ defined on a measure space $(\Omega,\mathscr{F},\mathbb{P})$.

Can you give me some hints on how to prove the following: for a.e. $\omega\in\Omega$ there is $N(\omega)$ such that for all $n\geq N(\omega)$, $ sup_{t\in[n,n+1]}|B_t(\omega)-B_n(\omega)|\leq \sqrt{n}. $

I thought of using the fact that the BM is Holder continuous with exponent $\alpha<1/2$ on each interval $[n,n+1]$, which implies that there is $c_n$ s.t. $sup_{t\in[n,n+1]}|B_t(\omega)-B_n(\omega)|\leq c_n$ (actually, continuity is enough for this), but then I don't know how to glue intervals together (since the $c_n$ might go to infinity faster than $\sqrt{n}$).

Answer 1

You need to use the Borel-Cantelli lemma and show that $\sum \limits_{n=0}^{\infty} \mathbb{P}\big(\sup \limits_{n \le t \le n+1} |B_t - B_n| \ge \sqrt{n}\big) < \infty$.

Assuming this result holds, then the lemma ensures that almost surely, $\sup \limits_{n \le t \le n+1} |B_t - B_n| < \sqrt{n}$ for all but finitely many $n$.

Proof of the summability: it is equivalent to showing that $\sum \limits_{n=0}^{\infty} \mathbb{P}\big(S \ge \sqrt{n}\big) < \infty$, where $S = \sup \limits_{0 \le t \le 1} |B_t|$, since $\sup \limits_{n \le s \le n+1} |B_s - B_n|$ has the same distribution as $S$ for all $n$.

Less is known about $S$ than about $M_1 = \sup \limits_{0 \le t \le 1} B_t$, so we will first find a bound for $\mathbb{P}(S \ge x)$. Without using the exact formula, we can write:

\begin{align*} \mathbb{P}(S \ge x) & = \mathbb{P}\Big(\big(\sup \limits_{0 \le t \le 1} S_t \ge x\big) \cup \big(\inf \limits_{0 \le t \le 1} S_t \le -x\big)\Big) \\ & \le \mathbb{P}\Big(\sup \limits_{0 \le t \le 1} S_t \ge x\Big) + \mathbb{P}\Big(\inf \limits_{0 \le t \le 1} S_t \le -x\Big) \\ & \le 2 \mathbb{P}\Big(\sup \limits_{0 \le t \le 1} S_t \ge x\Big) \\ & \le 2 \mathbb{P}\big(|B_1| \ge x\big) \end{align*}

since it is well-known the running maximum is a half-normal. Using a standard bound on the upper tail of a normal distribution, we get $\mathbb{P}\big(S \ge x\big) \le \frac{4\exp(-x^2/2)}{x\sqrt{2\pi}}$. Hence $\sum \limits_{n=1}^{\infty} \mathbb{P}\big(S \ge \sqrt{n}\big) \le \sum \limits_{n=1}^{\infty} \frac{4\exp(-n/2)}{\sqrt{2\pi n}} < \infty$.

Answer 2

Attempted solution after reading the comments:

Applying the reflection principle, get $$ \mathbb{P}(\sup_{t\in[n,n+1]}|B_t-B_n|\geq \sqrt{n})\leq\mathbb{P}(\max_{t\in[n,n+1]}(B_t-B_n)\geq\sqrt{n})+\mathbb{P}(\min_{t\in[n,n+1]}(B_t-B_n)\leq-\sqrt{n})=4\mathbb{P}(B_1\geq \sqrt{n}). $$ Now check that $$ \mathbb{P}(B_1\geq \sqrt{n})\leq\frac{E(B^4)}{n^2}=\frac{c}{n^2}. $$ Since the series $\sum_{n\geq1}\frac{c}{n^2}$ converges, by Borel-Cantelli get that $$ \limsup(\sup_{t\in[n,n+1]}|B_t-B_n|\geq \sqrt{n})=0, $$ which implies the conclusion.

Is this right? It seems that I have not used the fact that the sets $(\sup_{t\in[n,n+1]}|B_t-B_n|\geq \sqrt{n})$ are independent, and that the same proof also shows that in the statement of the theorem one can take any sequence $a_n$ such that there is a $k>0$ such that $\sum_n \frac{1}{a_n^k}$ converges.