2
$\begingroup$

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.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.