# Brownian motion reflection principle result

I'm studying about the reflection principle of the brownian motion, and I found that this result is a direct consequence of this principle:

Let $$B_t$$ a brownian motion, then for every $$a \in \mathbb{R} \$$,

$$\mathbb{P}(\lim_{t \to \infty} \sup_{s\in [0,t]} B_s > a) = 1$$

I'm trying to prove this statement using the reflection principle but I'm totally lost. I can't see how are those results related.

• Which formulation of the reflection principle do you know/use? It is direct consequence of the reflection principle that $$M_t := \sup_{s \leq t} B_s$$ equals in distribution $|B_t$|. Knowing this, it shouldn't be difficult to prove the assertion. – saz Dec 7 '18 at 7:51

Let $$M_t = \sup_{s \in [0,t]} B_s$$. For $$a < 0$$ the statement is trivial, so take $$a \ge 0$$.
The reflection principle says that $$P(M_t > a) = 2 P(B_t > a)$$. Now since you know $$B_t \sim N(0,t)$$, you can compute $$\lim_{t \to \infty} 2 P(B_t > a)$$ and determine that it equals 1. Thus $$\lim_{t \to \infty} P(M_t > a) = 1$$.
This is not the same as the desired statement, but can be used to prove it in this case. Note that $$M_t$$ is increasing in $$t$$, so necessarily $$\lim_{t \to \infty} M_t$$ exists (as a random variable) and $$\lim_{t \to \infty} M_t \ge M_u$$ for any fixed $$u$$. Hence $$P(\lim_{t \to \infty} M_t > a) \ge P(M_u > a)$$. Passing to the limit as $$u \to \infty$$ and using what we previously proved, we have $$P(\lim_{t \to \infty} M_t > a) \ge \lim_{u \to \infty} P(M_u > a) = 1$$.
• What is $\lim_{t\to\infty} M_t$? – thomasb Sep 20 at 16:37
• It's the usual thing. It's the random variable $M$ defined as $M(\omega) = \lim_{t \to \infty} M_t(\omega)$. We know by definition of $M_t$ that for each $\omega$, $M_t(\omega)$ is an increasing function of $t$, so the limit exists in $[0,+\infty]$ for each $\omega$ and $M$ is well defined. It is an exercise to verify that it is measurable. Of course, we end up proving that $P(M > a) = 1$ for every $a$, from which it follows that in fact $M = +\infty$ almost surely. – Nate Eldredge Sep 20 at 17:45
From the Law of Iterated Logarithm: $$\limsup_{t\to+\infty}\frac{B_t}{\sqrt{2t\ln\ln t}} \overset{\mathbb{P}\rm -a.s.}{=} 1,$$ so $$\mathbb{P}\left(\lim_{t\to+\infty}\sup_{s\in[0,t]}B_s>a\right)=1.$$