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I'm trying to prove an increment property for the Brownian motion, but I'm unable to figure it out, maybe someone can help me out.

Consider two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ of positive real numbers and let $\mathbb{B}$ be a standard Brownian motion on a probability space $(\Omega, \mathcal{A}, \mathbb{P})$.

The statement I'm trying to prove is the following.

Assume that $a_n - b_n = o(n)$ as $(n\to\infty)$. Or in other words $\lim_{n \to \infty} \frac{\vert a_n - b_n\vert}{n} = 0$, then already $\max_{1 \leq k \leq n} \vert \mathbb{B}(a_k) - \mathbb{B}(b_k) \vert = o\left(\sqrt{n \log(n)}\right)$. Or equivalently

$\lim_{n \to \infty} \frac{\max_{1 \leq k \leq n} \vert \mathbb{B}(a_k) - \mathbb{B}(b_k) \vert}{\sqrt{n \log(n)}} = 0$

I initially thought that I might have to use a result like Levy's modulus of continuity for large increments (Something like this), but I still wasn't able to figure it out. I'm thankful for any hints.

Davide Giraudo
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student7481
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1 Answers1

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Let $t_k=|b_k-a_k|$ and $\epsilon_k=\sqrt{2t_k/k}$. We are given that $\epsilon_k \to 0$ as $k \to \infty$. Also, denote $\psi(k)=\sqrt{2k \log k}$. Then by the tail bound for a standard Normal variable, $P(|Z|>r) \le e^{-r^2/2}$, applied to $Z_k=[B(a_k)-B(b_k)]t_k^{-1/2}$, we have $$P\bigl(|B(a_k)-B(b_k)|>\epsilon_k \psi(k)\bigr) \le \exp\Bigl(\frac{-\epsilon_k^2 \psi^2(k)}{2t_k}\Bigr)=k^{-2}\,.$$

By Borel Cantelli, with probability 1 the inequality $$|B(a_k)-B(b_k)| \le \epsilon_k \psi(k)$$ holds for all sufficiently large $k$, so $$P\Bigl(\lim_{k \to\infty} \frac{|B(a_k)-B(b_k)|}{\psi(k)} =0\Bigr)=1 \,, $$ whence $$P\Bigl(\lim_{n \to\infty} \max_{k \in [1,n]}\frac{|B(a_k)-B(b_k)|}{\psi(n)} =0\Bigr)=1 \,. $$

Yuval Peres
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