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Let $B$ be a Brownian motion with $B(0)=0$, $x,y>0$ and $B^*$ the Brownian motion reflected at $-x$.

I came across the following:

$$ \mathbb P_0(\inf_{s\in [0,t]}B(s)<-x, B(t)\geq y-x) = \mathbb P_0(B^*(t)\leq -x-y) $$

How does the use of the reflected Brownian motion get rid of the infimum condition?

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up vote 0 down vote accepted

If $B^*(t)$ is less than $-x-y$, in particular it is less than $-x$, so by continuity it had to cross the level $-x$ at some time. At this time, $B$ also reached the level $-x$, so $\inf B(s) \le -x$. It is possible that $\inf B(s) = -x$, i.e. that $B$ reached the level $-x$ but never crossed it, but this event has probability 0.

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