Brownian Reflection Principle for Joint Terminal Value and Maximum

Summary

The document uses the Brownian reflection principle to relate the terminal value of Brownian motion to its running maximum. On the specified region, where the barrier height exceeds the terminal-value threshold, reflecting paths at the barrier maps an event involving a terminal value below the threshold and a maximum above the barrier to an event involving the reflected terminal value. This also gives equality of the corresponding joint densities.

The argument applies that equality inside an expectation weighted by an exponential function of the terminal value. It then observes that, under the stated domain restriction, the reflected-terminal-value event already implies that the path crossed the barrier, allowing the maximum condition to be removed. This is a focused proof step, not a general treatment of Brownian maxima or a trading strategy. Its event simplification depends on the stated inequalities and should not be used outside that domain without further justification.

Key ideas

• Reflection at a barrier relates Brownian paths with different terminal values but the same maximum event.
• On the specified domain, the joint densities of the terminal value and maximum obey a reflection symmetry.
• That symmetry can be applied to expectations weighted by an exponential of the terminal value.
• The maximum condition can be dropped from the reflected event only under the stated domain restriction.

Tags

# Reflection principle of the Brownian motion


# Reflection principle of the Brownian motion












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

## Answer by ir7 (score 5)

https://quant.stackexchange.com/a/67778

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

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.