Brownian Motion, $\tau = \inf \{ t > 0; B_t \notin (−a, a) \}$ and $\sigma = \inf \{ t > 0; B_t \notin (−b, b) \}$

Question

Let $B_t$ be a Brownian motion with $B_0 = 0$, and for $0 < a < b$, let

$\tau = \inf \{ t > 0; B_t \notin (−a, a) \}$ and $\sigma = \inf \{ t > 0; B_t \notin (−b, b) \}$

Find $P_0(B_\tau = −a, B_\sigma = b)$.

Some thoughts of mine:

Using the following theorem,

"Let $T = \inf \{ t : B_t \notin (a, b) \}$, where $a < 0 < b$. $E_0T = −ab,$"

we can say that $E\tau = a^2$ and $E\sigma=b^2$.

Is this in the right direction? Or is there some other theorems to make use of?

Answer 1

You can rewrite that probability by the strong Markov property as $P_0(B_\tau = -a) P_{-a}(B_\sigma = b) = \frac{1}{2} P_{-a}(B_\sigma = b)$. Then there's a formula for $P_{x}(B_\sigma = b)$ as a function of $x$, do you know it? (Hint: it's a harmonic function, which means in $1$-d that its second derivative is $0$, and it's equal to $1$ at $b$ and $0$ at $-b$)