I have been told that one of the property of the continuous-time random walk in two dimensions is that:

$$\int_{Z} \, G(z, t | p_1) \, G(z, t | p_2) \,dz = \,G(p_1,p_2,2t)$$

where Z defines the coordinates of a two-dimensional infinite space, t is time, $p_1$ and $p_2$ are the geographical position of two particles, and $G(z, t | p)$ is the probability density function that a given particle at position $p$ came from position $z$ at time $t$. $G(z, t | p)$ is expressed as a free diffusion process:

$$G(z, t | p) = \frac{1}{4\pi Dt} \exp[-(p-z)^2/4Dt]$$

In few words, it seems that one of the property of a random walk is that the probability density function that two particles were at nearly the same position at some point in Z at time t is given by $G(p_1,p_2,2t)$.

I have two questions related to this:

  1. Can someone point out a source of the proof for such statement?
  2. Is this equivalence true also in the case of a random-walk in a finite bounded space? For instance using PDF in a 1-Dimensional box as Eq. 13 in https://www.frontiersin.org/articles/10.3389/fphy.2020.583202/full?

Concerning question 2, the PDF in 1 dimension , equivalent to $G(x, t | x_0, 0)$ is:


where $\tau = 2(\frac{L}{2})^2\frac{1}{\sigma^2}$ and $n\to\infty$

In two-dimension, the PDF is the product of the PDF in 1 dimension for x- and y-directions.

I did a numerical test with the PDF in a confined box, by comparing the results obtained from $$\int_{Z} \, G(z, t | p_1) \, G(z, t | p_2) \,dz$$ with results obtained from $$G(p_1,p_2,2t)$$.

For instance, assuming that there are two particles A and B at position $A = {xA0, yA0}$ and $B = {xB0, yB0}$, respectively, $\sigma^2 = 0.003$ and $L = 11$ in both x- and y-directions, and setting n = 10000, I do get:

enter image description here


1 Answer 1


In general, for this kind of stochastic processes, and $t_f > t > t_i$, we have : $$G(x_f,t_f|x_i,t_i) = \int G(x_f,t_f|x,t)G(x,t|x_i,t_i)\text dx$$ Representing the fact that the particle must be somewhere at the intermediary time $t$.

Since the process in OP is time-independent, we can write $G(p_2,t|z)= G(p_2,2t|z,t)$. It is also translation invariant and isotropic : $G(p_1,t|z)$ only depends on $|p_1-z|$ and therefore $G(p_1,t|z) = G(z,t|p_1)$. Using these properties, we have : \begin{align} \int_{Z} \, G(z, t | p_1) \, G(z, t | p_2) \text dz &= \int_Z G(p_2,2t|z,t)G(z,t|p_1,0) \\ &= G(p_2,2t|p_1,0) \\ &= G(p_1,p_2,2t) \end{align}

For a random walk in a finite bounded space, the relation $G(p_1,t|z) = G(z,t|p_1)$ will no longer hold and the result fails to be true.

  • $\begingroup$ thanks for the answer. However, I have tried to compare the integral and the PDF numerically, and even in the case of a 2D BOX the relation seem still to hold. I am updating my question with the figure, so that we can discuss on that. $\endgroup$ Jun 27, 2022 at 19:25

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