really appreciate some guidance on how to get the following equality: enter image description here

  • $\begingroup$ typo in the def of the running minimum mT := min Bs for s<=T $\endgroup$
    – ice_fox21
    Sep 5 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:

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

Further noting that

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

as $h>x$, we get

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

  • $\begingroup$ it's clear, thank you $\endgroup$
    – ice_fox21
    Sep 5 at 19:42

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