# Reflection principle of the Brownian motion

really appreciate some guidance on how to get the following equality:

• typo in the def of the running minimum mT := min Bs for s<=T Sep 5, 2021 at 15:36

I'll only show it for $$M_T = \max_{u\leq T} B_u$$ and $$(x,h)$$-domain

$$\{ h> 0, h > x \}.$$

By the reflection principle we have:

$$P\left( B_T < x, M_T > h \right) = P\left( 2h - B_T < x, M_T > h \right),$$ on the above domain, and hence we also have the following equality of the joint densities of $$(B_T,M_T)$$ and $$(2h-B_T, M_T)$$: $$P\left(B_T \in dx, M_T \in dh \right) = P\left(2h-B_T \in dx, M_T \in dh \right),$$ on the same domain.

By using it, we get:

$$E\left[1_{\{B_Th\}}{\rm e}^{cB_T - c^2T/2} \right] = E\left[1_{\{2h-B_T h\}}{\rm e}^{c(2h-B_T) - c^2T/2} \right] = (*)$$

Further noting that

$$\{2h-B_Th\} = \{2h-B_T < x \},$$

as $$h>x$$, we get

$$(*)= E\left[1_{\{2h-B_T

• it's clear, thank you Sep 5, 2021 at 19:42