Prove that Brownian motion $(X_t)$ is such that $P(|X_{t+h} − X_t|> ε)\ll h$ when $h\to 0$

Question

I am facing this problem, but don't have the knowledge to solve it. I need to prove that given a simple Brownian motion $X_t$, I have that

$$\frac{P(|X_{t+h} − X_t|> ε)}{h}→ 0 \mbox{ when }h\to 0.$$

Thanks!

Answer 1

Hint: Use the stationarity of the increments and the Markov inequality:

$$\mathbb{P}(|X|>\epsilon) \leq \frac{\mathbb{E}(|X|^4)}{\epsilon^4}.$$

Answer 2

Here is another approach:

$X_{t+h}-X_{t}$ is just a normal random variable with mean 0, variance $h$. So $$\mathbb P(|X_{t+h}-X_{t}|>\epsilon) = \mathbb P(|\frac{X_{t+h}-X_{t}}{\sqrt h}|>\epsilon/\sqrt h)$$

Now, $\frac{X_{t+h}-X_{t}}{\sqrt h}$ is a standard normal random variable, with density $\sim e^{-x^2/2}$, which is bounded above by $xe^{-x^2/2}$ for $x>1$. Note that, $xe^{-x^2/2}$ has close form integral, which gives you a close form upper bound for what you want to estimate.