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?
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@saz I dont think it s $\frac34$. Could you pls give us a hint? – Tom Aug 06 '19 at 17:00
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@Tom Have a look at this question – saz Aug 06 '19 at 17:04
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@saz I would be able to compute $\mathbb{P}{T_1 < T_{-3}}$ and $\mathbb{P}{T_{-3} < T_{3}}$, which are $\frac34$ and $\frac12$ respectively.. This is quite easy and is worked out in the question you mentioned. But here the OP is talking about $\mathbb{P}{T_1 < T_{-3} < T_3}$. Sorry if I m missing something very obvious.. – Tom Aug 06 '19 at 20:12
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1@Tom Ah right, sorry, did some stupid mistake there. So, yes, it's a bit more complicated than the question which I linked and the answer is not $3/4$. Guess it's $3/4 * 1/3 = 1/4$, i.e. the OP was actually correct. First you need the probability hitting $1$ before $-3$ (which is $3/4$)... now, after arriving at $y=1$ you restart Brownian motion $B_t := W_{t+T_1}-W_{T_1}$ (using strong Markov property) and compute the probability that $W$ hits $y=-4$ before it hits $y=2$. Due to independence of $W$ and $B$ you can just multiply both probabilities to get the desired probability. – saz Aug 06 '19 at 20:30
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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 $$ \mathbb{P}^x \{ T_b < T_a < T_c \} = \frac{x-a}{b-a} \frac{c-b}{c-a} $$
Tom
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