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Suppose that $B_t$ is a standard Brownian motion. And $T_a$, $T_b$ are the hitting time whereas $a<0$, $b>0$. Then are these two random variables independent?

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No, hitting times for the same Brownian Motion are not independent. Think of it this way: given $T_a = s$, $B_s = a$ so if $a \ne b$, $B_t$ is much less likely to be close to $b$ when $t$ is near $s$, and therefore $T_b$ is much less likely to be close to $s$.

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I see. Thanks very much! –  Julie Sep 28 '11 at 22:53
    
May I know what's the probability of T_b < T_a –  user31694 May 18 '12 at 17:35
    
if $c=\mathbb{P}(T_b<T_a)$, then, by OST, $0=\mathbb{E}[B_{T_a\wedge T_b}]=a(1-c)+bc$. Solve for $c$. –  Tom Artiom Fiodorov May 18 '12 at 17:39

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