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[Searched for this question and found something similar, but not identical. I hope this is not a duplicate. I don't know if this is a very easy question, I took my stochastic processes course too much time ago.]

Let $(B_t)_{0 \leq t \leq 1}$ be a standard Brownian motion and consider the process (for $0 \leq t \leq 1$, $\mu \in \mathbb{R}$, $\sigma > 0 \in \mathbb{R}$)

$$ X_t = \mu t + \sigma B_t $$

Consider the two stopping times

$$\tau_1 = \inf \left\{ 0 \leq t \leq 1: X_t =1 \right\}$$

and

$$\tau_{-1} = \inf \left\{ 0 \leq t \leq 1: X_t = -1 \right\} $$

How can I compute the following quantities?

$$ \mathbb{P}(\tau_1 \leq \tau_{-1} \leq 1) $$

$$ \mathbb{P}(\tau_{-1} \leq \tau_1 \leq 1) $$

$$ \mathbb{P}( \left\{ \tau_1 > 1 \right\} \cap \left\{\tau_{-1} > 1\right\}) $$

Did
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Gabriele Cacchioni
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    How well do you know martingales ? Can you solve this problem if t=0? If so, you should use Girsanov theorem (or find a transformation of $X_t$ that will be a martingale) in order to use stopping time theorems for martingales. An example of the stuff that you will need to use can be found in this question (that is not a duplicate but that handles the same concepts) : http://math.stackexchange.com/questions/1053294/density-of-first-hitting-time-of-brownian-motion-with-drift – WNG Sep 29 '16 at 12:31
  • The solution to just $P(\tau_1 \leq \tau_{-1})$ can be found using renewal theory: the solution is the solution to the BVP $(Lq)(x)=0$ inside the domain, $q(1)=1,q(-1)=0$, where $L$ is the generator. So in your case it is $\mu q'+\frac{1}{2} \sigma^2 q''=0$ so that $q(x)=c_1+c_2 e^{-\frac{2 \mu}{\sigma^2} x}$. Adjoin the boundary conditions to get the solution. – Ian Sep 30 '16 at 00:48
  • With the additional constraints involving $1$ the situation seems to be a bit more complicated. One approach, which is similar to the approach that WNG has suggested, is to compute the joint distribution of $(\tau_{-1},\tau_1)$ and integrate it. – Ian Sep 30 '16 at 00:48
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    @WNG I use Girsanov's theorem to obtain a probability $\mathbb{Q}$ absolutely continuous wrt $\mathbb{P}$ such that I get rid of the drift, namely I can write $dX_t = \sigma d \tilde{B}t$ where the new process $\tilde{B}_t = B_t + \frac{\mu}{\sigma}t$ is a $\mathbb{Q}$-bm (with this trasformation I know what is $\frac{d \mathbb{Q}}{d \mathbb{P}}$ so I can use it to compute expectations wrt the new measure). Moreover if I'm not wrong $X_t$ is also a $\mathbb{Q}$-martingale. I'm still puzzled about how to compute $\mathbb{P}(\tau_1 \leq \tau{-1} \leq 1)$ (even in the driftless case). – Gabriele Cacchioni Sep 30 '16 at 10:36
  • I'm not sure how to deal with the $1$ in that framework myself. One way to do it in the framework I suggested would be to consider a 2D Markov process where one of your components is time. Then you have $dX^1_t=\mu dt + \sigma dB$ and $dX^2_t=1 dt$, where $X^2$ now is just is an auxiliary variable for $t$. Now the probability to hit $1$ before $-1$ within time $1$ is just the probability to hit the set ${ (1,t) : t \in \mathbb{R} }$ before hitting the set ${ (-1,t) : t \in \mathbb{R} } \cup { (x,1) : x \in \mathbb{R} }$. – Ian Sep 30 '16 at 12:46
  • Keeping track of where the boundaries actually lie, that means your problem is on the domain $[-1,1] \times [0,1]$ and you have the boundary conditions $q(1,t)=1,q(-1,t)=0,q(x,1)=0$, and your PDE is $(Lq)=0$ where $L$ is the generator of the process $dX_t=\begin{bmatrix} \mu \ 1 \end{bmatrix} dt + \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} dB_t$, which I guess is $\mu \frac{\partial q}{\partial x} + \frac{\partial q}{\partial t} + \frac{1}{2} \sigma^2 \frac{\partial^2 q}{\partial x^2}$. Presumably this can be solved analytically by separation of variables... – Ian Sep 30 '16 at 12:53
  • One annoying thing is that this PDE has incompatible initial and boundary conditions. Today I read that a good way to approximate weak solutions to such problems is to make the boundary condition be time-dependent, initially being compatible with the initial data and then becoming aligned with the true boundary condition. – Ian Sep 30 '16 at 20:32
  • Specifically after reversing time you can look at it as $\frac{\partial q}{\partial t}=\mu \frac{\partial q}{\partial x} + \frac{1}{2} \sigma^2 \frac{\partial^2 q}{\partial x^2},q(−1,t)=0,q(1,t)=1,q(x,0)=0$ , where the goal is to find $q(x,1)$. You can instead look at the same ODE but with $q^\epsilon(1,t)=1−e^{-t/\epsilon}$ where $0<\epsilon \ll 1$ is a regularization parameter. Upon sending $\epsilon \to 0$ you get convergence to the true weak solution in a few standard senses (albeit not pointwise, of course). – Ian Sep 30 '16 at 20:41

1 Answers1

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Using renewal theory (please ask in the comments if you want some details), you can show that if $u(x,t)$ is the probability for $\sqrt{2}B_t$ to hit $1$ before $0$ given a time budget of $t$ starting at $x$, then $u$ satisfies the IBVP:

$$u_t= u_{xx} \\ u(x,0)=\begin{cases} 0 & x \in [0,1) \\ 1 & x=1 \end{cases} \\ u(0,t)=0 \\ u(1,t)=1.$$

The idea of this argument is to consider a Markov process $(t,\sqrt{2}B_t)$ and wait for it to hit the boundary of $[0,1]^2$ starting at some point $(0,x)$. Notably, the initial condition in this problem is not continuous, so the sense of any solution to it must be somewhat weak.

The solution to this IBVP is

$$u(x,t)=x+\sum_{k=1}^\infty \frac{(-1)^k}{\pi k} e^{-k^2 \pi^2 t} \sin \left ( k \pi x \right ).$$

You are welcome to translate this solution to $[-1,1]$. I don't know any simpler expression for this. Notably however, this solution actually happens to capture precisely the type of discontinuity in the initial data that we wanted in this case.

The answer to (a modification of) your question is $u(x,1)$. Notably, with these numbers, by $t=1$ only the first correction term is really significant anymore; $\frac{e^{-4 \pi^2}}{2 \pi}$ is about $10^{-18}$, and thereafter the series is converging faster than a geometric series with common ratio $e^{-4 \pi^2}$. So a good approximation of the solution here is just $x-\frac{e^{-\pi^2}}{\pi} \sin(\pi x)$. If you tweak the numbers you might need more terms; for example, with an interval of length $2$ and a standard Brownian motion, the eigenvalues are $-k^2 \pi^2/8$, so you'll want at least two correction terms, perhaps three.

You can imitate this argument to handle the general case with constant drift and diffusion; you will have $u_t=\mu u_x + \frac{1}{2} \sigma^2 u_{xx}$ with the same boundary conditions.

Ian
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