First passage time in Brownian motion

Question

Let $X_t,\, t\geqslant 0,$ be a Brownian motion and consider the stopping times $T_a := \inf \{t \mid X_t = a\}$. Find the probability $\mathbb P\{T_{2}< T_{-1} < T_{3}\}$, for instance.

So we have two events $\{T_2 < T_{-1}\}$ and $\{T_{-1}< T_3\}$. Separately, the probabilities are clear. My intuition says that we can multiply the probabilites for the initial problem, but I'm not really satisfied with this intuitive mumbo-jumbo. What if there are some funny cases, when it doesn't work..

Yet, I have no idea how to formally explain this.

Answer 1

They are definitely not independent.

You didn't specify the starting value for the Brownian motion, but let's say it's 0. On the event $\{T_3 < T_{-1}\}$, where the Brownian motion reaches 3 before -1, it must have passed through the value 2 even earlier, by continuity. So we have $\{T_2 < T_{-1}\} \subset \{T_3 < T_{-1}\}$. As such, $\{T_2 < T_{-1}\}$ cannot be independent of $\{T_3 < T_{-1}\}$, unless they both had probability 0 or 1 which is not the case. So $\{T_2 < T_{-1}\}$ is also not independent of $\{T_3 < T_{-1}\}^c = \{T_{-1} < T_3\}$.

(This same argument works if the starting point is anything less than 2. If it's greater or equal to 2, then $P(T_2 < T_{-1}) = 1$ and the events are trivially independent.)

You do end up multiplying probabilities, but not because of independence per se. The strong Markov property yields a certain conditional independence statement, if you like, but you have to be very careful to frame it the right way.

Answer 2

Ok, so we restart the process at a fixed time. $$ \mathbb P\{T_2<T_{-1}<T_3\} = \mathbb P \{T_{2}<T_{-1}\} \mathbb P\{T_{-3}<T_1\} = \frac{1}{3} \cdot \frac{1}{4}, $$ where the second probability is centered around $2$.