Summary
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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?




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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.



Derive the solution of the diffusion equation from the solution of a random walk
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.


![](https://i.stack.imgur.com/tHEwT.png)
