It is a standard result that if $W_t$ is a Brownian Motion and $S$ is a stopping time of the standard filtration $F_t$ then we have that $B_t = W_{S+t} - W_S$ is a Brownian Motion.

I quote the theorem above because I think it might be useful for showing the following:

I am trying to show that $W_{t-T_b}$ is symmetric in the sense that $P^{b}[W_{t-T_b} \leq a] = P^b[W_{t-T_b} \geq 2b -a]$ on $\{T_b \leq t\}$, where $T_b$ is the first hitting time of $W$ to $b$.

Any help would be appreciated. Thanks.

  • $\begingroup$ Is $P^b(\cdot) = P(\cdot \mid T_b \leq t)$? $\endgroup$ – Tom Jun 25 '14 at 19:50
  • $\begingroup$ Sorry, I just realized this question is trivial because $P^b[W_{t-T_b}\leq a] = P^b[W_{t-0}\leq a] = P^b[W_{t-0}\geq 2b-a] = P^b[W_{t-T_b}\geq 2b-a]$. $\endgroup$ – user30201 Jun 25 '14 at 23:14
  • $\begingroup$ Then, close it? $\endgroup$ – Did Jul 5 '14 at 12:43

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