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For $B_t$ Brownian Motion with drift $\mu<0$, I have the max value, $X = \max_{0<t<\infty}B_t$ .

I need to prove with the Strong Markov Property that, $P(X>c+d)=P(X>c)P(X>d)$

a. It seems weird to me since one of the right hand terms is redundant...

b. I'll appreciate any hints, especially on how to use the markov property on X. I know I have it on the BM process itself, but not sure how they relate.

Thanks.

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1 Answer 1

up vote 2 down vote accepted

Let $T=\inf\{t\geqslant0\mid B_t\geqslant c\}$. Then $[T\lt\infty]=[X\geqslant c]$ and, conditionally on the event $[T\lt\infty]$, $(B_{t+T})_{t\geqslant0}$ is distributed like $(c+B_t)_{t\geqslant0}$ and is independent on $(B_t)_{t\leqslant T}$.

Hence, conditionally on the event $[T\lt\infty]$, the maximum of $(B_{t+T})_{t\geqslant0}$ is distributed like $c+X$. Thus, $\mathrm P(X\geqslant c+d\mid T\lt\infty)=\mathrm P(c+X\geqslant c+d)=\mathrm P(X\geqslant d)$.

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