Counting of brownian particles: Point Process

Question

Imagine a point process defined by the passage time of purely brownian particles through a given point (in 1D), line (2D) or plane (3D). I'm interested in the variance of the counts (number of particles passing the points) as a function of sampling time.

Unlike brownian motion, I expect my signal to show non-vanishing correlation, because a particle that just crossed the point, has a high probability to cross it again. Thus the correlation should look like a dirac delta function (classic for discrete point process) plus a decreasing time function.

However, I'm a bit confuse with the way to calculate it analytically. I search in areas of photon counting, geiger counter, and gaz dynamics, but I didn't find any explicit calculation of the variance.

Two idea I had:

1/ Studying the Spatio-Temporal correlation function of a field of Brownian particles (see Gardiner, Stochastic Methods, eq. 13.3.22 for instance):

$G(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\left(-x^2/(4Dt)\right)$

in 1d. We should have: $Var(T)=\lim_{x\rightarrow 0} \int_T\int_T G(x,t) dt dt'$. However, the last expression is hard to compute.

2/ Using the probability of return time to origin of Brownian motion (the correlation arise from successive passage of the same particle)

Any idea or suggestions would be welcomed !

Answer 1

The quantity you are dealing with is the local time of the Brownian motion at one point. It is rather difficult to deal with local times usely.

What you want to compute is the distribution of the quantity $$\lambda(\vec r_0)=\int_0^t\delta^{(d)}\left(\vec r_0-\vec B_u\right)\mathrm du.\tag{1}$$ Note that $\lambda$ is a spatial density of time, it has dimension of $[L^{-d}T]$, where $d$ is the dimension of space.

First compute the characteristic function of the Brownian motion starting at $\vec r=0$, at time $t>0$ $$\phi(\vec q,t)=\left\langle\exp\left[\mathrm i\vec q\cdot\vec B_t\right]\right\rangle= \int\mathrm e^{\mathrm i\vec q\cdot \vec r}\frac{\mathrm e^{-\vec r^2/4Dt}}{\left(4\pi Dt\right)^{d/2}}\mathrm d\vec r=\frac{\mathrm e^{-Dt\vec q^2}}{\left(4\pi Dt\right)^{d/2}}.\tag{2}$$

Now, we transform the delta-function in (1) into a Fourier integral thanks to the equation (2) and compute the average of $\lambda(\vec r_0)$ $$\left\langle\lambda(\vec r_0)\right\rangle=\left\langle\int_0^t\int\frac{\mathrm d\vec q}{(2\pi)^d}\mathrm e^{-\mathrm i\vec q\cdot(\vec r_0-\vec B_u)}\mathrm du\right\rangle =\int_0^t\int\frac{\mathrm d\vec q}{(2\pi)^d}\mathrm e^{-\mathrm i\vec q\cdot \vec r_0}\phi(\vec q,u)\mathrm du.$$ The average is thus $$\left\langle\lambda(\vec r_0)\right\rangle =\int_0^t\frac{\mathrm e^{-\vec r_0^2/4Du}}{\left(4\pi Du\right)^{d/2}}\mathrm du.$$

There are exact results in one two and three dimensions for the average.

In one dimension, we have $$\left\langle\lambda(\vec r_0)\right\rangle=\sqrt{\frac t{\pi D}}\mathrm e^{-\vec r_0^2/4Dt}-\frac{|\vec r_0|}{2D}\mathrm{erfc}\left(\frac{|\vec r_0|}{2\sqrt{Dt}}\right);$$

in two dimensions $$\left\langle\lambda(\vec r_0)\right\rangle=\frac1{4\pi D}\Gamma\left(0,\frac{\vec r_0^2}{4Dt}\right)$$ ($\Gamma$ is the incomplete gamma function) and in three dimensions $$\left\langle\lambda(\vec r_0)\right\rangle=\frac{1}{4\pi D|\vec r_0|}\mathrm{erfc}\left(\frac{|\vec r_0|}{2\sqrt{Dt}}\right)$$

I have used the same method in a recent paper for a Brownian bridge.

To compute $\left\langle\lambda(\vec r_0)^2\right\rangle$ use the same computation technique (also described in the paper). You need to evaluate $$\int_0^t\mathrm du\int_0^{t-u}\mathrm du'\int\frac{\mathrm d\vec q}{(2\pi)^d}\frac{\mathrm d\vec q'}{(2\pi)^d} \mathrm e^{-\mathrm i\vec q\cdot \vec r_0}\phi(\vec q,u)\phi(\vec q',u').$$

But the distribution you are looking for is not defined in dimensions larger than one: a regularization length is needed. This comes from the properties of the Brownian motion.

In the case of the Browian (the case studied in the paper), once it is regularized, the distribution is as you said: there is a $\delta$-function plus a decreasing function with the distance. In three dimensions, the decreasing function is a simple exponential.

Answer 2

A possibility is maybe to consider the probability flow. At the origin of the modelisation of Brownian motion or heat equation, you have a conservation equation $\frac{\partial \rho}{\partial t} + div \vec j = 0$

Here $\rho(x,t)$ must be considered as a probability density and $\vec j(x,t)$ as a probability flow.

Then, supposing a simple relation $\vec j = - D\vec \nabla \rho$, we get the usual "heat" equation : $\frac{\partial \rho}{\partial t} - D \nabla^2 \rho = 0$, with the Kernel $G(x,t)$.

So $\vec j(x,t)$ could be the quantity in which you are interested. If so, the above relations allow you to have it simply from the probability density $\rho(x,t)$