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Consider the line $a+bt$ where $a,b>0$. Let $B(t)$ be Brownian motion and let $\tau=\inf\{t>0:B(t)=a+bt\}$ be the first hitting time of that line, with the understanding that $\tau=\infty$ if the line is never hit. I want to compute the probability that Brownian motion hits that line, i.e. $P(\tau<\infty)$.

There are two steps in my work that I am not confident in.

Define the process $X(t)=B(t)-bt$ and let $T_a=\inf\{t>0:X(t)=a\}$, again with $T_a=\infty$ if $X(t)$ does not hit $a$. Then $\tau$ and $T_a$ have the same distribution (unsure about this) so we may instead calculate $P(T_a<\infty)$.

Let $\tilde{a}>0$ and define $T_{a,-\tilde{a}}=\inf\{t>0:X(t)=a \text{ or } X(t)=-\tilde{a}\}$ with $T_{a,-\tilde{a}}=\infty$ if $\{a,-\tilde{a}\}$ is never hit. The process $e^{2bX(t)}$ is a martingale and Doob's Optional Stopping theorem gives

$ \begin{align*} E(e^{2bX(T_{a,-\tilde{a}})})&=1, \\ e^{2ba}P(X(T_{a,-\tilde{a}})=a)+e^{-2b\tilde{a}}P(X(T_{a,b})=-\tilde{a})&=1. \end{align*} $

Letting $\tilde{a}\to\infty$,

$\begin{align*} e^{2ba}\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)&=1, \\ \lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)&=e^{-2ba}, \\ P(T_a<\infty)&=e^{-2ba}. \end{align*} $

The last step where I say $\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)=P(T_a<\infty)$ I am also unsure about.


marked as duplicate by Behrouz Maleki, Daniel W. Farlow, Raskolnikov, Shailesh, Leucippus Aug 25 '16 at 2:52

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  • $\begingroup$ I looked at that but I decided to post my question since it looked like my method was different and I wanted to know if my method was correct. Thanks. $\endgroup$ – user363512 Aug 25 '16 at 0:20
  • $\begingroup$ Believe in yourself! This is great work $\endgroup$ – BronchoX May 10 '18 at 23:56
  • $\begingroup$ So we found another solution on the internet and they did the same thing, so we're pretty confident that you're doing it right $\endgroup$ – BronchoX May 10 '18 at 23:58

To answer the question why $\tau$ and $T_a$ have the same distribution: Since the process $X(t)$ is defined as $X(t) = B(t) - bt$, the equations $B(t) = a +bt$ and $X(t) = a$ are equivalent. Therefore, $\tau$ and $T_a$ are the infimum of the same set, hence they are equal as random variables and in particular identically distributed.

As for the second question why $$\lim_{\tilde a \to \infty} P(X(T_{a,-\tilde{a}}) = a) = P(T_a < \infty)$$ holds: The random variable $T_{a,-\tilde a}$ is the first time that $X(t)$ hits one of the boundaries of the region $-\tilde a < x < a$. The event $X(T_{a,-\tilde a}) = a$ occurs iff $X(t)$ hits the upper bound $a$ before ever hitting the lower bound $-\tilde a$. We can assume that $X(t)$ is continuous since the Brownian motion $B(t)$ is continuous almost surely. Then it is clear that the events $\{\omega : X(T_{a,-\tilde a}) = a\}$ are increasing as $\tilde a \to \infty$ (if $\tilde b > \tilde a$ and $X(t)$ never goes below $-\tilde a$ before hitting $a$, it doesn't go below $-\tilde b$ either) and that their union is $\{\omega: T_a < \infty\}$ (if $X(t)$ hits $a$ at some point, then there exists a lower bound $-\tilde a$ such that $X(t)$ doesn't go below $-\tilde a$ before hitting $a$). This implies $$\lim_{\tilde a \to \infty} P(X(T_{a,-\tilde{a}}) = a) = \lim_{n \to \infty} P(X(T_{a,-n}) = a)$$ and the claim follows from the general fact that for a countable increasing sequence of events $A_1 \subseteq A_2 \subseteq \ldots$ one has $P(\bigcup_n A_n) = \sup_n P(A_n)$.

  • $\begingroup$ When you write $\lim_{\tilde{a}\to\infty}P(X(T_{a,-\tilde{a}})=a)=\lim_{n\to\infty}P(X(T_{a,-n})=a)$, that's possible because $\{\omega:X(T_{a,-\tilde{a}})=a\}$ is an increasing set in $\tilde{a}$ so the limit of the probability on the left hand side exists, allowing us to instead look at the particular sequence (n). Would that be right? $\endgroup$ – user363512 Aug 25 '16 at 0:17
  • $\begingroup$ Yes, exactly. - $\endgroup$ – marlu Aug 25 '16 at 12:25

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