6
$\begingroup$

Is it true that \begin{equation} M_t = \sup_{s \leq t} B_s > 0 \ \ \text{a.s.} \end{equation} for all $t>0$? I remember reading this somewhere, but intuitively, can't the Brownian motion B stay below 0 for some time with probability $>0$?

$\endgroup$
5
  • 2
    $\begingroup$ Are you familiar with Blumenthal zero one law? $\endgroup$ – clark Nov 25 '19 at 2:20
  • $\begingroup$ I have heard of it and I just checked on Wikipedia. Does that really explain the above? $\endgroup$ – Felix P. Nov 25 '19 at 2:35
  • $\begingroup$ Yes, you can argue using that. Another way to see it is if you know that $tB_{1/t}$ is a Brownian motion. $\endgroup$ – clark Nov 25 '19 at 3:39
  • $\begingroup$ @clark Could you elaborate how I can use the fact that $tB_{1/t}$ is a BM? $\endgroup$ – Felix P. Nov 25 '19 at 5:12
  • 1
    $\begingroup$ Since $B_t$ as $t\to \infty$ cannot stay always positive, it means that $tB_{1/t}$ must be alternating sign as $t\to 0$. $\endgroup$ – clark Nov 26 '19 at 0:20
2
$\begingroup$

One way to argue is using Blumenthal's zero one law. Define $A_n=\{B_{1/n}>\frac{1}{\sqrt{n}}\}$, and set $B=\{B_{1/n}>\frac{1}{\sqrt{n}} ~\text{i.o.}\} $

Then, \begin{align*} \mathbb{P}(B) &=\mathbb{P}(\limsup_n A_n)\\ &\geq \limsup_n\mathbb{P}( A_n)\\ &= \limsup_n\mathbb{P}(B_{1/n}>\frac{1}{\sqrt{n}})\\ &=\limsup_n\mathbb{P}(N(0,1)>1)\\ &=\mathbb{P}(N(0,1)>1)=M>0 \end{align*}

By Blumenthal's zero one law $\mathbb{P}(B)= 1 ~\text{or}~0$. So $\mathbb{P}(B)=1$.

$\endgroup$

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.