I am applying the Feynman-Kac theory for solving a PDE with boundary conditions.

For the SDE simulation I use the Euler-approximation, which introduces a time-step $h$ for the Brownian Motion, and since the PDE has some boundary conditions, we need to compute its hitting time $\tau$.

Numerically speaking, I start with $\tau = 0$ and increase it progressively.

So at each step:

  • the process position is increased by a normal depending (also) on $h$ (the SDE);
  • the variable $\tau$ is increased by some $k$;
  • the hitting condition is checked.

My question is: is there a specific connection between $h$ and $k$? Below my reasoning.

EDIT: I am adding more details with the hope of expressing better my question.

Consider the 1-dimensional heat on $[0,1]$, $u_t = u_{xx}$ with zero boundary conditions. So $u_0$ a known initial state, and $v(t,0)=v(t,1)=0$ for each time. Suppose we want to compute the solution at time $T=0.05$.

The stopping time to insert into Feynman-Kac expectation is:

$ \tau = \inf \{ k \geq 0 : T-k \leq 0,$ or $X_s\geq 1$, or $X_s \leq 0 \} $

where with $X_t$ solves $dX_t = \sqrt{2} {dB}_t$.

I approximate the SDE say with a time-step $h=0.01$, so if we start from $x \in [0,1]$ (point on which we are going to compute the PDE solution) and set $k=0$ (hitting time approximation), at every step we:

  1. move by adding a normal $N(0, 2h)$
  2. increase $k$ by $\sqrt{h}$

Again: when (1) goes to 0 or 1, or $T - k$ goes to 0, the hit is considered done.

Now, (1) works great: the process take its time for hitting the boundary. Conversely, since $\sqrt(0.01)=0.1$, $T - \sqrt{h} = 0.05-0.1 = -0.05$ so the boundary is immediately hit!

As a result, in the simulation the boundary is hit always after one step and I obtain just a small shift of my initial condition $u_0$.

If, conversely, in (2) I subtract something like $h^{1.8}$, rather than $\sqrt{h}$, the results are pretty nice...


Computing Exit Times with the Euler Scheme by F. Buchmann addresses this problem in detail.

For a random walk with position $x_{k+1}=x_k+\delta x_k$, incremented by a normally distributed stochastic variable $\delta x_k$ with variance $h$, the simplest hitting time estimate is $\tau=Kh$ with $K$ the smallest $k$ such that $x_k$ is outside of the boundary. This is an overestimate, because it is possible that both $x_k$ and $x_{k+1}$ are inside of the boundary, and yet the Brownian motion crossed the boundary between times $kh$ and $(k+1)h$. Buchmann constructs interpolants that test for this intermediate excursion.

  • $\begingroup$ Thank you so much, Carlo. Your suggestion has been very helpful, although at the end it didn't really solved the problem. I accepted your answer, and added more details on mine allowing it to be fully reproducible, in case someone else wanted to have a look. $\endgroup$ – user133725 Jan 22 at 8:45

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