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If $B(t)$ is a standard Brownian motion, let $T_x= \min \{ t>0: B(t)=x\}$.

I have already proved that if $a,b>0$ then $P[T_a<T_{-b}]=b/(a+b)$, using this, and its complement I was trying to compute $P[T_1<T_{-1}<T_2]$ but I could write this probability in terms of the mentioned property. Any suggestions?

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Once BM hits $1$, you use the strong Markov property to view this as a new BM $\leadsto\mathbb P[T_1<T_{-1}<T_2]=\mathbb P[T_1<T_{-1}]\cdot \mathbb P[T_{-2}<T_1]$.

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