Probability related to min and max of Brownian motion

Question

Given the Brownian motion $X(t)$ and $s$ with value greater than $0$.

I want to ask about the probability:

\begin{align*} \Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) . \end{align*}

Intuitively, I believe $\Pr(\min_{0 \leq u \leq t} X(u) > 0)$ and $\Pr(\max_{0 \leq u \leq t} X(u) > s)$ are independent when $s > 0$. So \begin{align*} \Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) = \Pr(\min_{0 \leq u \leq t} X(u) > 0)\Pr(\max_{0 \leq u \leq t} X(u) > s). \end{align*} But I just could not prove it logically.

Can someone give me a hint? I really appreciate it. Thank you very much.

Answer 1

$$ \mathsf{P}\!\left(\min_{0\le u\le t}X_u>0,\max_{0\le u\le t}X_u>s\right)\le \mathsf{P}\!\left(\min_{0\le u\le t}X_u>0\right)=0. $$

Answer 2

If $B_t$ is Brownian motion, then with probability one, for all $\delta>0$, there will exists $0<t<\delta$ for which $B_t<0$. Therefore, $\min_{0\le t\le u}B_t$ is almost surely negative, so your probability is zero.

One way to prove this is to use the standard fact that $W_t:= tB_{1/t}$ is also a Brownian motion, and then apply the law of the iterated logarithm to $W_t$.