# First passage time of Brownian motion

I am currently trying to understand why $P(\tau_b < \infty)$ for $b \in \mathbb{R}$, where $\tau_b := \inf\{t \geq 0: B_t = b\}$. There was a little step in the argument below which I did not understand.

Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Let us write $$\sup_{t \geq 0}B_t \geq \sup_{n \in \mathbb{N}}B_n = \sup_{n \in \mathbb{N}} (\xi_1 +...+\xi_n)$$ where $\xi_j=B_j-B_{j-1}$ are i.i.d. standard normal random variables. If we further write $$S:=\sup_{n\in \mathbb{N}}(\xi_1 +...+\xi_n) = \xi_1 + \sup_{n \in \mathbb{N},n \geq 2}(\xi_2 +...+\xi_n) =: \xi_1 + S'$$ then $S$ and $S'$ have the same distribution and $\xi_1$ and $S'$ are independent. I need to somehow conclude, since $\xi_1$ is standard normal distributed, that then $S=\infty$ a.s. How do I see this?

Thanks a lot in advance!

By 0-1 law for exchangeable random variables $P\{|S|<\infty\}$ is $0$ or $1$. If it is $1$ then $Ee^{itS}Ee^{it\xi}=Ee^{itS}$ for all $t$ and $Ee^{itS}=Ee^{itS}$ because $S$ and $S'$ have the same distribution. This gives $Ee^{it\xi}=1$ at least for $|t|$ sufficiently small, which is a contradiction because$Ee^{it\xi}=e^{-t^{2}/2}$.Hence $P\{|S|=\infty\}=1$. By definition $S>-\infty$ almost surely so we get $S=\infty$ almost surely.
• Doesn't it hold for arbitrary $t$ instead of sufficiently small $|t|$? – vaoy Aug 30 '18 at 9:26
• I cannot cancel $Ee^{itS}$ unless I know it is not zero. For $|t|$ small it is close to $1$ and hence non-zero. – Kavi Rama Murthy Aug 30 '18 at 9:29