Conditional Probability about Brownian motion

Question

Let $W_t$ be standard Brownian motion, $T_a$ be the hitting time, I would like to compute the probability $$P(T_{1} < T_{-3} < T_3)$$ My computation is ${1\over 4}$, can anyone help check?

Answer 1

I tried to make Saz's argument formal (plus a bit of generalization) to check my understanding and I am posting what I have done since it may benefit others (e.g. OP).

Let $X$ be a strong Markov process on $\mathbb{R}$ with continuous sample paths and let $T_x = \inf \{t: X_t = x\}$.

Given three real numbers $a < b < c$, we look for the probability of hitting $b$ before $a$ and $a$ before $c$, having started the process from $x$. Of course, this problem is equivalent to the (easier) problem of computing $\mathbb{P}^x \{ T_b < T_a\}$ unless $x \in (a,b)$.

In general, we have

$$\begin{eqnarray} \mathbb{P}^x \{ T_b < T_a < T_c \} &=& \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_a < T_c\}} \\ &=& \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_b < T_c\}} (\mathbb{I}_{\{ T_a < T_c\}} \circ \theta_{T_b}) \\ &=& \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_b < T_c\}} \mathbb{E}^x_{T_b}(\mathbb{I}_{\{ T_a < T_c\}} \circ \theta_{T_b}) \\ &=& \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_b < T_c\}} \mathbb{E}^{X_{T_b}}\mathbb{I}_{\{ T_a < T_c\}} \\ &=& \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_b < T_c\}} \mathbb{E}^b\mathbb{I}_{\{ T_a < T_c\}} \\ \end{eqnarray}$$

With respect to the first term, we have that $$ \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} \mathbb{I}_{\{ T_b < T_c\}} = \begin{cases} 0 & \text{if } x \in (-\infty, a] \\ \mathbb{E}^x \mathbb{I}_{\{ T_b < T_a\}} & \text{if } x \in (a, b) \\ \mathbb{E}^x \mathbb{I}_{\{ T_b < T_c\}} & \text{if } x \in [b, c) \\ 0 & \text{if } x \in [c, \infty) \\ \end{cases}$$

So let's take from now on $x \in (a,b)$.

If $X$ has a scale function $s$ (I think this is always the case if $X$ is regular), then we can conclude from the reasoning above that $$ \mathbb{P}^x \{ T_b < T_a < T_c \} = \frac{s(x)-s(a)}{s(b)-s(a)} \frac{s(c)-s(b)}{s(c)-s(a)} $$

If, as it is the case for a standard Brownian Motion, the process is in natural scale (i.e. $s(x) = x$ is a scale function), then we can further simplify our expression $$ \mathbb{P}^x \{ T_b < T_a < T_c \} = \frac{x-a}{b-a} \frac{c-b}{c-a} $$