I want to ask for the following problem. Let $(W_t)_{t\geq 0}$ be the standard Brownian motion. For each $t>0$, we call $$m_t =\inf_{0 \leq s \leq t} W_s, \qquad M_t = \sup_{0 \leq s \leq t} W_s.$$ Let $0< \alpha_1 < \alpha_2 \leq 1$ and $0<a_1<a_2$ and $b>0$. Can we prove that \begin{equation} \label{goal1} \mathbb{P}(m_1\geq -b \mid M_{\alpha_1} \leq a_1, M_{\alpha_2} \leq a_2) \geq \mathbb{P}(m_1 \geq -b \mid M_1 \leq a_1)? \end{equation} In fact, I want to prove this for general sequences $0< \alpha_1 < \alpha_2<\ldots<\alpha_k \leq 1$ and $0<a_1<a_2<\ldots<a_k$, i.e. \begin{equation} \label{goal2} \mathbb{P}(m_1\geq -b \mid M_{\alpha_1} \leq a_1, \ldots, M_{\alpha_k} \leq a_k) \geq \mathbb{P}(m_1 \geq -b \mid M_1 \leq a_1)? \end{equation} Heuristically, conditioning on $M_1 \leq a_1$ would make the Brownian motion go deeper and thus make $m_1$ be smaller than conditioning on $M_{\alpha_1} \leq a_1, \ldots, M_{\alpha_k} \leq a_k$ I think.
Do you have any idea/reference to turn this prediction be a rigorous proof, at least for the case $k=2$?
Thanks.