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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.

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  • $\begingroup$ 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. $\endgroup$ – saz Dec 7 '18 at 7:51
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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 $P(\lim_{t \to \infty} M_t > a) \ge \lim_{u \to \infty} P(M_u > a) = 1$.

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  • $\begingroup$ What is $\lim_{t\to\infty} M_t$? $\endgroup$ – thomasb Sep 20 at 16:37
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Sep 20 at 17:45
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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 $$ \mathbb{P}\left(\lim_{t\to+\infty}\sup_{s\in[0,t]}B_s>a\right)=1. $$

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