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world:exobrownian


2017/10/07 23:39 ·

# The reflection principle

Exercise Let $B$ be a standard Brownian motion and denote $M(t)=\max_{s \in [0,t]} B(s)$. Then $M(t)\eqlaw |B(t)|$

$\bproof$ Let $T_a=\inf\set{t\geq 0}{B_t \geq a}$ be the first time that the Brownian motion crosses the threshold $a$. Note that \begin{align*} \PP(B(t)\geq a)&=\PP(B(t)\geq a, M(t)\geq a)+ \underbrace{\PP(B(t)\geq a, M(t)< a)}_{=0}\\ &=\PP(B(T_a+t-T_a)-a\geq 0| \underbrace{M(t)\geq a}_{T_a \leq t}) \PP(M(t)\geq a)=\PP(M(t)\geq a)/2 \end{align*} since $(B(s+t)-B(s))_{t\geq 0}$ is a standard brownian motion and from this, we can show that for the almost surely finite stopping time $T_a$, then $(B(T_a+t)-B(T_a))_{t\geq 0}=(B(T_a+t)-a)_{t\geq 0}$ is a standard brownian motion. Finally, $$\PP(M(t)\geq a)=2 \PP(B(t)\geq a)=\PP(B(t)\geq a)+\PP(-B(t)\geq a)=\PP(|B(t)|\geq a)$$ which concludes the proof. $\eproof$