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Let $B(t)$ be a Brownian Motion and $$M(t) = max_{s:s \leq t} B(s)$$ and $$\tau_a = min_t{B(t) = a}$$

Then, $P(\tau_a < t) = P(B(t) - B(\tau_a) > 0 \: |\: \tau_a < t) + P(B(t) - B(\tau_a) < 0 \: |\: \tau_a < t)$

So this says that the probability that the first time we hit 'a' has already occurred, is equal to the probability that the distance we have moved since we hit 'a' is greater than 0 plus the probability that the distance we have moved since we hit 'a' is less than 0? I don't understand why this is true, could someone please help me with the intuition?

jj238sdfaaa
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    Your defintion of $M(t)$ doesn't make sense to me. Did you mean to write $B(s)$ instead of $B(t)$? So, $M$ is the biggest $B$ has become so far? – MPW May 11 '15 at 02:25
  • Yes, that's correct – jj238sdfaaa May 11 '15 at 03:01
  • Okay. Your notation is poor, having a hard time understanding your question. Do I also understand correctly that $$\tau_a = \min{s:B(s)=a}?$$ So $\tau_a$ is the first hit time for $a$? – MPW May 11 '15 at 03:58
  • Your proposition says something like P(bar)=P(foo|bar)+P(not foo|bar). Also, note that $B(\tau_a)\equiv a$. – MPW May 11 '15 at 04:03
  • Yes, $\tau_a = $min ${s: B(s) = a}$ – jj238sdfaaa May 11 '15 at 12:35

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