# supremum of brownian motion almost surely > 0

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

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

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