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Let $(B_t)$ a Brownian motion. I denote $\tau_a=\inf\{t\geq 0\mid B_t=a\}$, $T_{a,b}=\tau a\wedge \tau b$ and $p_x\{A\}=p\{A\mid B_0=x\}$.

Why $$p_x\{B_{T_{a,b}}=b\}=p_x\{\tau_a<\tau_b\}\ \ \ ?$$

To me, $$p_x\{B_{T_{a,b}}=b\}=p_x\{T_{a,b}=\tau_b\}=p_x\{\tau_b<\tau_a\},$$ so what's wrong in my argument ?

idm
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  • Actually you are right; it should read $p_x(\tau_b<\tau_a)$ instead of $p_x(\tau_a<\tau_b)$. – saz Jun 10 '15 at 12:13
  • Thanks for your answer :-) Are you totally sure ? Because in my course my teacher does this mistakes ALL THE TIME... therefore, i'm a little bit suspicious. – idm Jun 10 '15 at 12:23
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    Yeah, pretty sure... $B_{T_{a,b}}= b$ means that the process hits $b$ before it hits $a$. Why not ask your teacher about it? – saz Jun 10 '15 at 12:25
  • Ok, I trust you :-) Because I don't have any more course with him (I prepare my exam which is the next week). I sent him some mail, but I don't have any answers. – idm Jun 10 '15 at 12:31

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