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