Given the Brownian motion $X(t)$.
I want to ask about the conditional probability:
\begin{align*} \Pr \left \{\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s, X(t) < s)|X(0) = a \right \}, \end{align*} where $s,a >0$, and $s > a$.
I tried to use the reflection principle to solve this problem but it seemed inoperative due to the intersection of both $min$ and $max$ events.
Can someone give me a hint? I really appreciate it. Thank you very much.
PS: Or else, please give me some hint about the probability \begin{align*} \Pr \left \{\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right \}. \end{align*}
Thank you all again!
The law of $\left(\min_{0 \leq u \leq t} X(u), \max_{0 \leq u \leq t} X(u), X_t \right)$ is called Lévy’s triple law. According to René L. Schilling, Lothar Partzsch, Brownian Motion: An Introduction to Stochastic Processes, 6.5 Lévy’s triple law $(6.16)$
$$P \left( m_t > a, M_t <b, X(t) \in dx)|X(0) = 0 \right) = \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}} \right)$$ where $a<0<b$, $m_t = \min_{0 \leq u \leq t} X(u)$ and $M_t = \max_{0 \leq u \leq t} X(u)$
Hence, \begin{align*} P \left( m_t > a, M_t <b\right) &= \int_{\mathbb{R}}P \left( m_t > a, M_t <b, X(t) \in dx)|X(0) = 0 \right) \\ &= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\int_{\mathbb{R}}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}} \right)e^{-\frac{x^2}{2t}}dx\\ &= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{(b-a)^2n^2}{2t}}-e^{-\frac{((b-a)n+a)^2}{2t}} \right)\\ \end{align*}
Return back to your question, it suffices to put $(a,b) \leftarrow (-a,s-a)$
$$P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )=1-\frac{1}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{s^2n^2}{2t}}-e^{-\frac{(sn-a)^2}{2t}} \right)$$
For $P\left( \min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$, just notice that this probability is equal to $$P\left( \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )-P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$$
and $\left(\max_{0 \leq u \leq t} X(u)|X(0) = a\right)$ follows the same distribution as $\left(|X(t)-a|+a\right)$ according to the law of maximum of brownian motion.