# inordinatum

Physics and Mathematics of Disordered Systems

## How does an absorbing boundary modify the propagator of a stochastic process?

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A recurring theme throughout my blog are stochastic processes, widely used to model quantities under the influence of both deterministic evolution laws and random fluctuations. A complete solution for a stochastic process is its propagator: The probability density for the final state at some time $t_f$ given an initial condition at time $t_i$. The propagator depends on the boundary conditions imposed on the stochastic process. For example, an absorbing boundary is useful for describing extinction in a stochastic model of population dynamics: When the population of a given species reaches zero, it cannot be re-introduced.
Typically, however, the “free” propagator for a stochastic process — i.e. the propagator without any boundaries — is much easier to determine than the propagator with any given boundary condition.

However, common sense tells us that the free propagator for a stochastic process contains already all there is to know about it. So I have wondered for quite a while if and how one can recover e.g. the propagator with an absorbing boundary, knowing only the free propagator. I found the solution only quite recently in this preprint on arXiv, and think that it is interesting enough to explain it here briefly.

## The result

Consider a one-dimensional, stationary, Markovian stochastic process. Let us denote its free propagator by

$\displaystyle G(x_i,x_f|t_f-t_i).$

This is the density for the final value $x_f$ at time $t_f$, given an initial value $x_i$ at time $t_i$. Stationarity of the process implies that $G_0$ is a function of the time difference $\tau := t_f-t_i$ only. Define the Laplace transform of the free propagator by

$\displaystyle \hat{G}(x_i,x_f|\lambda) := \int_{0}^{\infty} \mathrm{d}\tau\, e^{\lambda \tau}G(x_i,x_f|\tau).$

Now let us add an absorbing boundary at $x=a$. Then, the propagator

$\displaystyle G_a(x_i,x_f|t_f-t_i)$

is the density for the final value $x_f$ at time $t_f$, given an initial value $x_i$ at time $t_i$, without touching the line $x=a$ between $t_i$ and $t_f$. The upshot of this post is the following simple relationship between the Laplace transforms of the free propagator $G$ and of the propagator with absorbing boundary $G_a$ :

$\displaystyle \hat{G}_a(x_i,x_f|\lambda) = \hat{G}(x_i,x_f|\lambda)-\frac{\hat{G}(x_i,a|\lambda)\hat{G}(a,x_f|\lambda)}{\hat{G}(a,a|\lambda)}.$      (1)

In case the denominator on the right-hand side is zero, one may have to perform some limiting procedure in order to make sense of this formula. We can now check how nicely it works for simple cases, like Brownian motion with drift, where the result is known by other means.

## Example: Brownian motion with drift

Let us consider an example, the Brownian motion with drift $\mu$. Its stochastic differential equation is

$\displaystyle \partial_t X(t) = \mu + \xi(t),$

where $\xi$ is white noise with $\overline{\xi(t)\xi(t')} =2\sigma\delta(t-t')$.

The free propagator is a simple Gaussian

$\displaystyle G(x_f,x_i|\tau) = \frac{1}{\sqrt{4\pi\sigma\tau}}\exp\left[-\frac{(x_f-x_i-\mu \tau)^2}{4\sigma \tau}\right].$

Its Laplace transform is given by:

$\displaystyle \begin{array}{rl} \hat{G}(x_f,x_i|\lambda) =& \int_{0}^{\infty} \mathrm{d}\tau\, e^{\lambda \tau}G(x_i,x_f|\tau) \\ =&\frac{1}{\sqrt{\mu^2-4\lambda\sigma}}\exp\left[\frac{\mu}{2\sigma}(x_f-x_i) -\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}|x_f-x_i|\right]. \end{array}$

To compute $G_a$, let us assume $x_i > a$, $x_f > a$. Note this is without loss of generality, since if $x_i$ and $x_f$ are on opposite sides of the absorbing boundary the propagator $G_a$ is zero. Applying the result (1), we have

$\displaystyle \begin{array}{rl} \hat{G}_a(x_i,x_f|\lambda) =& \hat{G}(x_i,x_f|\lambda)-\frac{\hat{G}(x_i,a|\lambda)\hat{G}(a,x_f|\lambda)}{\hat{G}(a,a|\lambda)} \\ \quad=& \frac{1}{\sqrt{\mu^2-4\lambda\sigma}}\exp\left[\frac{\mu}{2\sigma}(x_f-x_i) -\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}|x_f-x_i|\right] - \\ & \frac{1}{\sqrt{\mu^2-4\lambda\sigma}}\exp\left[\frac{\mu}{2\sigma}(a-x_i) -\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}|a-x_i| \right. \\ &\quad \left.+\frac{\mu}{2\sigma}(x_f-a) -\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}|x_f-a|\right] \\ \quad=& \frac{1}{\sqrt{\mu^2-4\lambda\sigma}}e^{\frac{\mu}{2\sigma}(x_f-x_i)}\left[e^{-\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}|x_f-x_i|} - e^{-\frac{1}{2\sigma}\sqrt{\mu^2-4\lambda\sigma}(x_i+x_f-2a)}\right]. \end{array}$

The inverse Laplace transform of this expression is:

$\displaystyle \begin{array}{rl} G_a(x_i,x_f|\tau) =& G(x_i,x_f|\tau)-e^{\frac{\mu}{\sigma}(a-x_i)}G(2a-x_i,x_f|\tau) \\ =& \frac{1}{\sqrt{4\pi\sigma\tau}}e^{-\frac{\mu^2 \tau}{4\sigma} + \mu\frac{x_f-x_i}{2\sigma}}\left[e^{-\frac{(x_f-x_i)^2}{4\sigma\tau}}-e^{-\frac{(2a-x_f+x_i)^2}{4\sigma\tau}}\right]. \end{array}$

This is, of course, exactly the same result (the so-called Bachelier-Levy formula) which I had already discussed earlier in a guest blog post at Ricardo’s blog, using slightly different methods (a variant of the Cameron-Martin-Girsanov theorem). Note that some signs which were wrong in the original post are corrected here.

## The proof

Now let us go back to the case of a general stochastic process, and discuss how formula (1) can be proven. Let us assume that the stochastic process in question has a path-integral representation, i.e. that

$\displaystyle G(x_i,x_f|\tau) = \int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D}[x(t)]\,e^{-S[x(t)]}$

for some action $S$.

Now, to add an absorbing boundary at $x=a$ we weigh each crossing of $x=a$ by a factor $e^{-\gamma}$ in the path integral, and then let $\gamma \to \infty$. This effectively retains only those trajectories in the path integral, which never touch $x=a$. We thus have:

$\displaystyle G_a(x_i,x_f|\tau) = \lim_{\gamma \to \infty}\int_{x(t_i)=x_i}^{x(t_f)=x_f} \mathcal{D}[x(t)]\,e^{-S[x(t)]-\gamma\delta(x(t)-a)}$.

Now, to evaluate the path integral, we expand it in a series in $\gamma$ (or, in other words, in the number $k$ of crossings of $x=a$). We get:

$\displaystyle \begin{array}{rl} \lefteqn{G_a^{(\gamma)}(x_i,x_f|\tau) = G(x_i,x_f|\tau) + }&\\ & + \sum_{k=1}^{\infty} (-\gamma)^k \int_{t_i}^{t_f}\mathrm{d}t_1\, G(x_i,x_1|t_1-t_i) \int_{t_1}^{t_f}\mathrm{d}t_2\, G(x_1,x_2|t_2-t_1) \cdots \\ & \cdots \int_{t_{k-1}}^{t_f}\mathrm{d}t_k\,G(x_{k-1},x_k|t_k-t_{k-1})\,G(x_k,x_f|t_f-t_k) \end{array}$.

Note that here I cancelled the factor $\frac{1}{k!}$, coming from the Taylor expansion of $e^{-\gamma \delta(x(t)-a)}$, against the $k!$ possibilities of ordering the times $t_1\cdots t_k$. This is why I can assume them to be in ascending order above. Also, the $\delta$ functions actually set all $x_1=x_2=\cdots=x_k=a$.
Taking Laplace transforms, the time integrals (which are a $k$-fold convolution) simplify to a simple product:

$\displaystyle \begin{array}{rl} \hat{G}_a^{(\gamma)}(x_i,x_f|\lambda) =& \hat{G}(x_i,x_f|\lambda) + \\ & + \sum_{k=1}^{\infty} (-\gamma)^k \hat{G}(x_i,a|\lambda)\hat{G}(a,a|\lambda)^{k-1} \hat{G}(a,x_f|\lambda) \\ =& \hat{G}(x_i,x_f|\lambda) - \frac{\gamma \hat{G}(x_i,a|\lambda)\hat{G}(a,x_f|\lambda)}{1+\gamma \hat{G}(a,a|\lambda)}\\ =& \hat{G}(x_i,x_f|\lambda) - \frac{\hat{G}(x_i,a|\lambda)\hat{G}(a,x_f|\lambda)}{\gamma^{-1}+ \hat{G}(a,a|\lambda)} \end{array}$.

Now, one easily obtains the claimed result:

$\displaystyle \begin{array}{rl} \hat{G}_a(x_i,x_f|\lambda) =& \lim_{\gamma \to \infty}\hat{G}_a^{(\gamma)}(x_i,x_f|\lambda) \\ =& \hat{G}(x_i,x_f|\lambda) - \frac{\hat{G}(x_i,a|\lambda)\hat{G}(a,x_f|\lambda)}{\hat{G}(a,a|\lambda)} \end{array}$.

This computation can, of course, be generalized to more complicated situations like several boundaries, higher dimensions, etc. For details see again the original preprint by C. Grosche. Have fun and let me know if you have questions!