\epsilon n)$, $B(t)$ is a Brownian motion

Asked Mar 29 '17 at 23:32

Active Mar 29 '17 at 23:32

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When I try to read the proof of Law of Large Numbers, there is a step as the tittle of the question, I would like to know why it works. Thanks beforehand.

asked Mar 29 '17 at 23:32

Fly_back

This is called the reflection principle. It tells you the distribution of the maximum of Brownian motion over an interval: https://en.wikipedia.org/wiki/Reflection_principle_(Wiener_process). – kccu Mar 29 '17 at 23:37
@kccu thanks so much. – Fly_back Mar 29 '17 at 23:40

Why $\mathbb{P}(\max\limits_{0 \le t \le 1} B(t) > \epsilon n) = 2\mathbb{P}(B(1) > \epsilon n)$, $B(t)$ is a Brownian motion

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