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So I got following problem:

Let $B_t$ be a $\{H_t\}_{t \in \mathbb{R}_+}$ Brownian motion where $\{H_t\}_{t \in \mathbb{R}_+}$ is a right-continuous complete filtration and consider $$S_t := \inf \Bigl\{s \geq 0, B_s = \frac{t}{\sqrt{2}} \Bigr\}$$ $$T_t := \lim_{s \to t^+} S_t$$ Now we have to prove, that $T_t$ is a stopping time and almost surely finite.

We got the hint, that $P(\sup B_t = \infty, \inf B_t = - \infty) = 1$ for a standard Brownian motion and that a $H_t$-Brownian motion is also a martingale with respect to $H_t$.

I am totally lost. I tried (unsuccessfully) to use Borel-Cantelli. In addition, if a Brownian motion oscillates between $-\infty$ and $\infty$, I guess, it "has to catch the equality", but I wasn't able to formulate that thought further. So I would be thankful for every input.

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1 Answer 1

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(I presume that $t$ is to be $>0$.)

1. Convince yourself that $$ \{T_t<u\}=\cup_{r\in(0,u)\cap\Bbb Q}\{B_r>t/\sqrt{2}\}, $$ for each $u>0$.

2. What does the oscillation property that you mention tell you about the range $\{B_t(\omega): t\ge 0\}$ for a typical $\omega$? (And then notice that $T_t\le S_{2t}$, at least if $t>0$.)

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  • $\begingroup$ To 1.- "$\supset$": If it exists such $r \in (0,u)$, then by the continuity of $t \to B_t$, one can assume, that it exists $c \in (0,r) \subset (0,u)$ such that $B_t = t/\sqrt{2}$. So the Inf is smaller than u and therefore also the limit, right? "$\subset$": If the limit is $L < u$, then $B_L = t/\sqrt{2}$ (or a sequence ...) and then it exists a $L < r < u$, such that $B_r > t/\sqrt{2}$, cause otherwise the Brownian motion would have a local maximum ... which is not possible? I would say, the range of $\{B_t(\omega) \} = (-\infty, \infty)$, but I still can't see how that helps me :/ $\endgroup$
    – AHoppla
    Nov 8, 2021 at 23:34
  • $\begingroup$ It tells you that $S_t<\infty$ a.s. $\endgroup$ Nov 10, 2021 at 0:08
  • $\begingroup$ Ah thanks, I will try to finish the prove from now on! $\endgroup$
    – AHoppla
    Nov 10, 2021 at 14:55

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