I'm reading Einstein article : See section 4 page 5.
Suppose we have a total of $n$ particles in a liquid. In a time $\tau$, the $x-$coordinates of the individual particle will increase by $\Delta $. A certain probability distribution law will hold for $\Delta $ : the number $dn$ of particles experiencing a displacement that lies between $\Delta $ and $\Delta +d\Delta $ in time interval $\tau$ is expressed by $$dn=n\varphi (\Delta )d\Delta ,\tag{*}$$ where $\int_{\mathbb R}\varphi (\Delta )d\Delta =1$, $\varphi $ differ from $0$ only for very small values of $\Delta $ and satisfy $$\varphi (\Delta )=\varphi (-\Delta ).$$
Q1) I really don't understand why $(*)$ hold neither why $\varphi (\Delta )\neq 0$ only for small value of $\Delta $ nor why $\varphi (\Delta )=\varphi (-\Delta )$.
Suppose that the number $\nu$ of particle per unit of volume depend only on $x$ and $t$, i.e. $\nu=f(x,t)$. From the definition of $\varphi $, we can easely obtain the number of particles found at time $t+\tau$ between $x$ and $x+dx$. One obtain $$f(x,t+\tau)dx=dx\int_{-\infty }^\infty f(x+\Delta )\varphi (\Delta )d\Delta .\tag{**}$$
Q2) What is $f=f(x+\Delta )$ in the previous formula ? Should it be $f(x+\Delta ,t)$ or $f(x+\Delta , t+\tau)$ ? Also, how did he gets $(**)$ ? It look to appear a bit from no where.
The important thing you need to keep in mind is that $\Delta$ is the displacement of the particles. That is, a particle with $\Delta = 0$ will end up at the same $x$-location after time $\tau$.
$\Delta$ is an intrinsically stochastic variable, and in the paper it has a distribution function labeled $\varphi(\Delta)$. In other words $\phi(\Delta){\rm d}\Delta$ represents the probability a particle moves (along $x$) a distance between $\Delta$ and $\Delta + {\rm d}\Delta$. Two important things to realize here
Since $\Delta$ is a displacement its sign can be both positive and negative. Meaning that particles can move to the right or to the left, respectively. Physically, both motions must occur with the same probability, therefore, $\color{blue}{\varphi(\Delta) = \varphi(-\Delta)}$
Note that $\tau$ is a (small) finite number, that means that a particle can only move so far from its original position in that period of time. That is for arbitrarily large $\Delta$ the probability distribution of $\phi$ must vanish, otherwise you'd be assuming particles could move infinitely fast, which is clearly not the case. Hence $\varphi$ must have finite support.
Since $\phi(\Delta){\rm d}\Delta$ represents the probability a particle moves (along $x$) a distance between $\Delta$ and $\Delta + {\rm d}\Delta$, the amount of particles ${\rm d}n$ that you will find in this interval after a time $\tau$ it is just the fraction that moves times the total number of particles $n$, or
$$ \color{blue}{{\rm d}n = n \phi(\Delta){\rm d}\Delta} $$
The idea is that at time $t + \tau$ you can get the new particle density, by counting where the particles were at $t$ and displacing them over all posible values, weighted of course by the probability of such displacement
$$ \color{blue}{f(x, t + \tau)~{\rm d}x = {\rm d}x ~\int_{-\infty}^{+\infty}f(x + \Delta, t)\varphi(\Delta)~{\rm d}\Delta} $$