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.


1 Answer 1


(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 $t>0$.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.