This problem cropped up in some research I am doing. I imagine it is standard, but I cannot seem to find the answer.

Let $W_t$ be a standard Brownian motion. Suppose there are four values $a < 0 < b$ and $c<d$. For a given $t > 0$, I want to know the probability that $W_t$ reaches $b$ before reaching $a$ and then after reaching $b$ never falls below $c$ nor exceeds $d$ until time $t$.

Thanks all.

  • $\begingroup$ If it's a standard Brownian motion, then the probabilty that after some point it never falls below $c$ is $0$. $\endgroup$ – Michael Hardy Aug 12 '13 at 22:46
  • $\begingroup$ Sorry, I want to know the prob. it never falls below $c$ or exceeds $d$ until time $t$. $\endgroup$ – Brian Aug 12 '13 at 22:51

I can't add comments (not enough reputation) so I'm posting this as an answer :

Consider your first problem which is reaching $b$ before $a$. Denote $T_a$ and $T_b$ stopping times of $W_t$ reaching a and b and $T=\min(T_a,T_b)$. $W_t$ being a martingale (stopping theorems look wikipedia): $\mathbb{E}[W_{\min(t,T)}]=0=\mathbb{E}[a1_{T_a<T_b}]+\mathbb{E}[b1_{T_b\leq T_a}]$ And then : $$\mathbb{P}[T_a<T_b]=\frac{b}{b-a}$$

You can follow this method to get your results

  • $\begingroup$ That's not quite what I need. In addition to the probability that $T_b < T_a$, I also need to know the probability that $W_t$ remains within the interval $[c,d]$ (where $c < b < d$) during $[T_b,t]$. $\endgroup$ – Brian Aug 13 '13 at 0:38
  • $\begingroup$ Agreed, but the result you are looking for can be found in a similar manner. My answer is just the very basic case. What have you tried so far ? $\endgroup$ – Bertrand R Aug 13 '13 at 1:16
  • $\begingroup$ I can compute $P[T_b < T_a]$, $P[T_b < t]$, and $P[\{T_c > t\}\cap \{T_d > t\}]$ but since these events are not independent, I cannot seem to find their joint probability. $\endgroup$ – Brian Aug 13 '13 at 2:30
  • $\begingroup$ Hint : $W_{t+T_b}-W_{T_b}$ is a standard brownian motion and then reduce to the previous case. (edit : of course you need $c<b<d$ otherwise the result is trivial) $\endgroup$ – Bertrand R Aug 13 '13 at 2:32
  • $\begingroup$ Sorry for being slow, but I still don't see it. Define $f(t) = P(T_b \in dt,T_b<T_a)$ and $G_x(s) = P[\min_{0\leq j \leq s} W_j \geq c, \max_{0\leq j \leq s} W_s \leq d | W_0 = x]$. Then the way I would compute what I'm seeking if I knew $f(t)$ is $\int_0^t f(s)G_b(t-s)ds$. $\endgroup$ – Brian Aug 13 '13 at 2:48

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