# Properties of random-walk in infinite and finite two-dimensional space: probability of two particles being in the same location at time t

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:

$$P(x,t|x_0,0)=\frac{1}{L}+\frac{2}{L}\sum_{i=1}^{n}\exp[-(\frac{i\pi}{2})^2\frac{t}{\tau}]\cos(\frac{ix\pi}{L})\cos(\frac{ix_0\pi}{L})$$

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: 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.