I am trying to understand in the original derivation of Brownain diffusion, where does the assumption of Markovian and Gaussian nature factor in.
In Albert Einstein's original work on Brownian motion in 1905, he derives the probability distribution density of particles using a random walk scheme as follows.
If the particle is at a position $x$ at at a time $t$ with the the probability density $p(x,t)$, then it makes a tiny jump of length $\Delta$ in a span of time $\tau$, where $\Delta$ is a random number with PDF $\phi(\Delta)$. The probabilty of finding the particle at any $x$ at a time $t+\tau$ can be given as,
$$ p(x,t+\tau)=\int_{-\infty}^{\infty} p(x-\Delta,t)\phi(\Delta)d\Delta $$ Question 1.
Does the above equation correspond to a Markovian approximation? I would assume so because the position of the particle at the next step depends only the previous step.
To proceed further, Einstein considers a Taylor expansion of $p(x,t)$ in both $x$ and $t$ about $\Delta$ and $\tau$ respectively. With this he obtains,
$$ \tau \frac{\partial p}{\partial t} + ... = \frac{1}{2}\frac{\partial^2 p}{\partial x^2}\left[ \int_0^\infty \Delta^2 \phi(\Delta) d\Delta \right] + ... $$
In these above steps, the first order term in \Delta vanishes for a symmetric distribution of jumps (i.e. $\phi(\Delta)$ is even).
Question 2.
Now, can we say that the truncation of the series to only upto the second term in $\Delta$ account to Gaussian approximation?
Lastly, it also seems like if we consider the complete expansion in $\Delta$, one can reproduce the Kramers-Moyal expansion. Is this essentially true?