Is there a boundary in probability for Brownian motion? For a standard Brownian motion $W_t$ and a given crossing probability $\alpha < 1$, I want to have a boundary function $f(t) > 0$, such that the probability that $W_t$ ever crosses the boundary is bounded by the given probability, i.e.
$$ P(|W_t| > f(t), \exists t > 0) \le \alpha $$
By Law of the iterated logarithm, any $f(t) = O(\sqrt{t\log\log t})$ does not meet this condition because
$$ \lim\sup_{n\rightarrow \infty} \frac{W_t}{\sqrt{t\log\log t}} = \sqrt{2} $$ if I get it right. 
For my case, I would like an $f(t) = o(t)$. In other words, I want to have a function such that 
$$ \lim_{t\rightarrow \infty}\frac{f(t)}{t} = 0 $$. 
 A: Fix $\alpha \in (0,1)$. Let $M_t := \sup_{s \leq t} W_s$ and $m_t := \inf_{s \leq t} W_s$ be the running maximum and mimimum, respectively. Then
$$\mathbb{P}(\exists t \in [0,n]: |W_t|>c_n) = \mathbb{P}(\{M_n>c_n\} \cup \{m_n < -c_n\})$$
for any constant $c_n>0$ and $n \in \mathbb{N}$. Since both $M_n$ and $m_n$ are real-valued, we can choose $c_n>0$ sufficiently large such that
$$\mathbb{P}(\exists t \in [0,n]: |W_t|>c_n) = \mathbb{P}(\{M_n>c_n\} \cup \{m_n < -c_n\}) \leq (1-\alpha) \alpha^n.$$
If we define
$$f(t) := \sum_{n=1}^{\infty} c_n 1_{(n-1,n]}(t)$$
then
$$\mathbb{P}(\exists t>0: |W_t|>f(t)) \leq \sum_{n \in \mathbb{N}} \mathbb{P}(\exists t \in [0,n]: |W_t|>c_n) \leq  (1-\alpha) \sum_{n \in \mathbb{N}} \alpha^n = \alpha$$
which shows that $f$ does the job.
Remarks:


*

*The distribution of $M_n$ and $m_n$ is known; in fact, $M_n \sim -m_n \sim |W_n|$ (se e.g. Brownian Motion - An Introduction to Stochastic Processes by Schilling & Partzsch, Chapter 6, for a proof). This allows us to compute bounds for the constants $c_n$.

*This construction of $f$ works, more generally, for any stochastic process $(X_t)_{t \geq 0}$ with continuous sample paths.
A: If $f(t)=b+t$, where $b>0$, then $ P(W_t > f(t)$ for some $t > 0)=e^{-2b}$, so if you choose $b=-\log\sqrt{\alpha/2}$, then
$$
P(|W_t| > f(t)\hbox{ for some }t > 0)\le 2P(W_t > f(t)\hbox{ for some }t > 0)=2e^{-2b}=\alpha.
$$
