I am trying to compute the hitting time of a linear Brownian motion on a two-sided boundary. More specifically, let $W_t$ be a (one-dimensional) Wiener process. Let $T = \inf \{t: |W_t| = a \}$ for some $ a > 0$. I want to find $\mathbb{P}\{ T > t\}$.

I know that probability distribution hitting time of a positive level, $\inf \, \{t: W_t = b\,, \ b > 0 \}$ can be computed quite easily, but I am not sure how to deal with it when dealing with the two-sided hitting time, i.e. with the absolute value. I am thinking of the minimum of hitting times of level $a$ and $-a$, but I can't get a promising conclusion.

  • $\begingroup$ Well, if $|W_t|=a$ then $W_t$ is equally likely to be $a$ or $-a$. I think that should let you use conditioning to finish, but when I did it I found that $P(T>t)=P(T_a>t)$, which is clearly incorrect. $\endgroup$ – Ian Mar 23 '15 at 21:06
  • $\begingroup$ This is equivalent to solving the heat equation on the space interval $[-a,a]$ with boundary conditions $P(x, 0)=1$ and $P(-a, t) = P(a, t) = 0$. I believe this gives a Fourier series solution with no nice closed form. Notes on the heat equation. $\endgroup$ – Ben Derrett Mar 23 '15 at 21:42
  • $\begingroup$ @BenDerrett Hi, could you please explain why this is true or give a hint? I can see that the density function of $W_t+x$ looks like the heat kernel, but I don't understand why this problem is equivalent to the heat equation with this boundary condition. $\endgroup$ – Syoung Apr 23 '17 at 15:29
  • $\begingroup$ @syoung see math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/… $\endgroup$ – Ben Derrett Apr 24 '17 at 19:08
  • $\begingroup$ @BenDerrett Thank you. $\endgroup$ – Syoung Apr 25 '17 at 0:34

I will give this a try.

For simplicity let $T_a=\inf \{t: |W_t| = a \}$

\begin{align} Pr(|W(t)|>a)&=P(|W(t)|>a|T_a<t)Pr(T_a<t)+P(|W(t)|>a|T_a>t)Pr(T_a>t)\\ \end{align}

$P(|W(t)|>a|T_a>t)=0$ since the time that $|W(t)|$ hits $a$ for the first time has not arrived, hence $|W(t)|$ can not be bigger than $a$.

Also note that $P(|W(t)|>a|T_a<t)=\frac12+Pr(W(t)<-2a)$ since we know that $|W(t)|$ has hit $a$ before $t$ (we have $T_a<t$). Therefore the event $\{|W(t)|>a|T_a<t\}$ is equivalent to $\{|a+W(t)|>a\}$.

Thus $$Pr(T_a<t)=\frac{2P(W(t)>a)}{\frac12+P(W(t)<-2a)}.$$

  • $\begingroup$ Interesting. $\{|W(t)|>a|T_a<t\}$ is not an event. $\endgroup$ – Ben Derrett Mar 24 '15 at 9:34
  • $\begingroup$ @BenDerrett Many thanks for taking a look. It is even more interesting to me :-) why is it not an event? $\endgroup$ – Math-fun Mar 24 '15 at 17:46
  • $\begingroup$ I will leave the comment to Ben. What I see strange in your reasoning is when you that two events in the end are equivalent. Who tells you have all the time t to cross 2a? As Ben said that probability is expressed by a series $\endgroup$ – Kolmo Mar 24 '15 at 18:31
  • $\begingroup$ Many thanks. I will think of this more. $\endgroup$ – Math-fun Mar 24 '15 at 18:43
  • $\begingroup$ @Math-fun An event is a set of outcomes to which a probability can be assigned. In particular, $\{|W(t)|>a\}$, $\{T_a\le t\}$ and $\{|W(t)|>a\text{ and } T_a\le t\}$ are events, but $\{|W(t)|>a| T_a\le t\}$ is not, since it doesn't refer to a particular set of outcomes. $\endgroup$ – Ben Derrett Mar 25 '15 at 8:53

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