Let $B = (B_t)_t$ be a Standard Brownian motion and $\tau$ is defined as: $$ \tau = \inf\{t\geq 0: B_t = -1 \text{ or } B_t = 5\} $$
I want to show that $\mathbb E B_\tau = 0.$
What I have done:
I first try to do this exercise with $\tau$ defined when the hitting was $-1$ or $1$, that is, $\tau = \{t\geq 0: B_1 = 1 \text{ or } B_t = -1\}$. In this case, $\tau$ can be written as $\min(\tau_1, \tau_{-1})$ where $\tau_i$ are the hitting times with respect to $1$ and $-1$ respectively. Hence, I thought it would be useful to write $B_\tau = B_\tau\mathbb{1}_{\{\tau_1 < \tau_{-1}\}} + B_\tau\mathbb{1}_{\{\tau_1 > \tau_{-1}\}} = \mathbb{1}_{\{\tau_1 <\tau_{-1}\}} - \mathbb{1}_{\{\tau_1>\tau_{-1}\}}$. Then, when calculating the expectation: $$ \mathbb E B_\tau = \mathbb P\{\tau_1 <\tau_{-1}\} - \mathbb P\{\tau_1 >\tau_{-1}\} $$
I have some intuition that why this thing is equal to $0$, because this probabilities should (I think) be equal, but I don't see how to prove it easily.
However, in the first situation, there's a $5$ that turns the expectation into:
$$ \mathbb E B_\tau = \mathbb P\{\tau_1 <\tau_{-1}\} - 5\mathbb P\{\tau_1 >\tau_{5}\} $$ and here I completely get lost. Can anyone help me, please?
