I am nearly done with a question:
Let $(B_t)$ be a Brownian motion on $\mathbb{R}$. For a fixed $x >0$, let $\tau$ be a stopping time defined by $$ \tau = \inf \{t \geq 0 : B_t \not \in (-x,x) \}.$$
I wish to compute, for $\sigma>0$, $$ \mathbb{E} \bigg(\exp \bigg\{ -\frac{{\sigma}^2}{2} \tau \bigg\} \bigg).$$
What I have done so far:
Noting that $\bigg\{\exp \bigg[ \sigma B_t -\frac{{\sigma}^2}{2} t \bigg] \bigg\}_{t \geq 0} $ is a martingale w.r.t the natural filtration, I applied the optional stopping theorem and the dominated convergence theorem to conclude that $$ \mathbb{E} \bigg\{\exp \bigg[ \sigma B_{\tau} -\frac{{\sigma}^2}{2} \tau \bigg] \bigg\} =1. $$ Then, I get $$ e^{\sigma x} \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =x\}} \bigg) + e^{- \sigma x} \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =-x\}} \bigg) =1.$$
Then I can't proceed from here. Is it true that $$ \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =x\}} \bigg) = \mathbb{E} \bigg( e^{-\frac{\sigma^2}{2} \tau} \mathbf{1}_{\{B_{\tau} =-x\}} \bigg) \quad ? $$ (I am not sure whether $B_{\tau}$ has the same distribution as $-B_{\tau}$.)