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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 $>0$?

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    $\begingroup$ Are you familiar with Blumenthal zero one law? $\endgroup$
    – clark
    Nov 25, 2019 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, 2019 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, 2019 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, 2019 at 5:12
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    $\begingroup$ 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, 2019 at 0:20

1 Answer 1

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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 $\mathbb{P}(B)=1$.

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