The probability distribution (pdf) of a random walk in 1 dimension is represented by a Bessel function. On the other hand, the pdf of a Brownian motion in free space is represented by a Gaussian distribution. While it is possible to derive the diffusion equation from the master equation of a random walk by taking the limit of small jumps, is it possible to directly derive the Gaussian distribution as some asymptotic form of the Bessel function? In other words, can I derive the pdf of a Brownian motion directly from the pdf of a random walk, rather than from the equation of the random walk?

Consider a continuous time random walk on a neareast-neighbour 1d lattice, where each jump is of size $d$. The probability of being at site $n$, i.e. being at a distance $nd$ of the origin is described by: $$\dot{P}_{n}(t)=\frac{1}{2}\left(P_{n-1}(t)+P_{n+1}(t)\right)-P_{n}(t)$$

The solution is $P_n(t)=I_n(t)e^{-t}$, where $I_n(t)$ is the Bessel function.

When we extend the random walk on the real line, i.e $d\to 0$, the probability is described by the diffusion equation:

$$\frac{\partial p(x,t)}{\partial t}=D\frac{\partial^2p(x,t)}{\partial x^2}$$

The solution is $p(x,t)=\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}$.

Is it possible to derive the expression of $p(x,t)$ from our expression of $P_n(t)=I_n(t)e^{-t}$, instead of starting from the master equation? I.e. how does the Bessel function relate to the Gaussian distribution in the limit $d\to0$?

If the question is not clear, or you would like more details, please let me know and I will edit the question.


1 Answer 1


To carry out the limit, it helps to start from an integral representation of the Bessel function, $$P_n(T)=e^{-T}I_n(T)=\frac{1}{2\pi}\int_{-\pi}^\pi \exp [i k n+T \cos k-T]\,dk.$$ For $T\gg 1$ this may be approximated by expansion of the exponent to second order in $k$, $$P_n(T)\approx\frac{1}{2\pi}\int_{-\infty}^\infty \exp[ikn-\tfrac{1}{2}Tk^2]\,dk=(2\pi T)^{-1/2}e^{-n^2/2T}.$$

Define $x=nd$, $t=\tau T$, $D=\tfrac{1}{2}d^2/\tau$ to arrive at $$d^{-1}P_n(T)\approx \frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}\equiv p(x,t). $$

Here is a plot for $d=0.1$, $D=1$, $t=1$; blue is $p(x,t)$, red is $d^{-1}P_n(T)$, the two curves are indistinguishable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.