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

Summary

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.

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.