Exit time of a stochastic process defined by a SDE

Solution 1:

Here is a common method used for calculating exit time/stopping times of stochastic processes.

We need to find a non-random function $u(h)$ which is a function of the stochastic process on which the stopping criteria is applied. We will use Itô's lemma to find a differential equation satisfied by $u$ and the stopping criteria will be the boundary condition for this function.

For the stochastic process specified in the question, we can apply Itô's lemma to get,

$du(h_t) = u_h(h_t)dh_t + \frac{1}{2}u_{hh}(h_t)d[h,h]_t$

we can now write the dynamic of $h_t$ as following,

\begin{equation} h_t = \Psi(x_t) + \frac{1}{2}mv_t^2 \\ dh_t = \nabla \Psi(x_t)dx_t + mv_tdv_t + \frac{1}{2}md[v,v]_t \\ dh_t= v_t\nabla \Psi(x_t)dt + v_t(-\nabla \Psi(x_t)dt - \gamma v_tdt+\sigma dW_t) + \frac{\sigma^2}{2m}dt\\ d[h,h]_t = \sigma^2v_t^2dt \end{equation}

Plugging the above into the differential of $u$, we get

\begin{equation} du = u_h(-\gamma v_t^2+\frac{\sigma^2}{2m})dt + u_hv_t\sigma dW_t+ \frac{1}2 u_{hh}\sigma^2v_t^2dt \end{equation}

Integrating the above between $t=0$ and $t=\tau$, we get,

\begin{equation} u(h(\tau)) = u(h(0)) + \int_0^\tau{\left(u_h(-\gamma v_t^2+\frac{\sigma^2}{2m})+u_{hh}\sigma^2v_t^2\right)dt} + \int_0^\tau{u_hv_t\sigma dW_t} \end{equation}

Now $h(\tau) = r$ and if we can find $u(h)$ which satisfies the differential equation,

$u_h(-\gamma v^2+\frac{\sigma^2}{2m})+u_{hh}\sigma^2v^2=-1$,

and boundary condition, $u(r)=0$. Then we can take expectation of the Stochastic Integral equation and use the fact that expectation of a well-behaved Itô integral is $0$, and we will get

$\mathbb{E}[\tau] = u(h(x_0,v_0))$.

We can use the relation between $h$ and $v$ to eliminate $v$ from the ODE. Alternate formulation of the function $u$ could be as a function of $(x,v)$ which will lead to a PDE based on similar application of Itô's lemma.

Special Case: If we can assume some relation between $\gamma, \sigma$ and $m$, then the established ODE could be simplified to get nice results. For example if $\gamma = 0$, then $u(h) = \frac{2m}{\sigma^2}(r-h)$ satisfies the ODE and boundary condition. $\mathbb{E}[\tau]$ would be then equal to $\frac{2m}{\sigma^2}(r-h(x_0,v_0))$.