really appreciate some guidance on how to get the following equality:
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$\begingroup$ typo in the def of the running minimum mT := min Bs for s<=T $\endgroup$– ice_fox21Sep 5, 2021 at 15:36
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1 Answer
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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
$$ (*)= E\left[1_{\{2h-B_T<x\}}{\rm e}^{c(2h-B_T) - c^2T/2} \right] $$
