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====== The reflection principle ======

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

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$\bproof$
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. 
$\eproof$