From master equation to Fokker-Planck equation For the continuous master equation in real space and time, we have for the distribution $f(x,t)$:
$$\frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\mathrm{d}x'$$
In the statistical mechanics book by P.K. Pathria, the right hand side is Taylor expanded to obtain (keeping up to second order)
$$\frac{\partial f(x,t)}{\partial t}=-\frac{\partial}{\partial x}[f(x,t)\int_{-\infty}^{\infty}\xi W(x;\xi)\mathrm{d}\xi]+\frac{1}{2}\frac{{\partial}^2}{\partial x^2}[f(x,t)\int_{-\infty}^{\infty}{\xi}^2 W(x;\xi)\mathrm{d}\xi],$$
where $\xi=x'-x$, and $W(x;\xi)=W(x,x')$, which is the transition probability density from $x$ to $x'$.
This is pretty much confusing to me and I have the following questions:


*

*How is the expansion carried out? It seems that we should expand around $x'$, after which a change of integration variable is done. However, it doesn't seem to lead to the above expression.

*The difference $\xi$ between $x$ and $x'$ is not necessarily small and in fact as the integration variable it goes all the way to $\infty$. Then in this case, how is the keeping up to second order and neglect all the higher orders justified?
Any help is much appreciated.
 A: For the first question, we start by rewriting
$$
\frac{\partial f(x,t)}{\partial t}=\int_{-\infty}^{\infty}[f(x',t)W(x',x)-f(x,t)W(x,x')]\, dx'
$$
as
$$
\begin{aligned}
\frac{\partial f(x,t)}{\partial t}
&=\int_{-\infty}^{\infty}[f(x',t)W(x';x-x')-f(x,t)W(x;x'-x)] \, d (x' - x)
\\
&=
\int_{-\infty}^{\infty} f(x-\xi,t) W(x-\xi;\xi) \, d \xi
-f(x, t) \int_{-\infty}^{\infty} W(x; \xi) \, d \xi,
\qquad (1)
\end{aligned}
$$
where $\xi \equiv x - x'$.
For the first term, the change of the minus sign in $d(-\xi) \to d\xi$ is compensated by swapping the lower and upper limits of the integral:
$$
\int_{-\infty}^\infty g(\xi) d(-\xi)
=
-\int_{\infty}^{-\infty} g(\xi) d\xi
=
\int_{-\infty}^{\infty} g(\xi) d\xi
$$
Next we can expand the first term using the Kramers-Moyal expansion
$$
f(x-\xi,t) W(x-\xi;\xi)
=
f(x,t) W(x;\xi)
-\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right]
+\frac{ \xi^2 } {2!} \frac{\partial^2 }{ \partial x^2} \left[ f(x,t) W(x;\xi) \right]
+ \dots,
\qquad (2)
$$
in which the first term of right-hand side will cancel the second term of (1).
So
$$
\begin{aligned}
\frac{\partial f(x,t)}{\partial t}
&=
\int_{-\infty}^\infty
\left(
-\xi \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right]
+\frac{ \xi^2 } {2!} \frac{\partial }{ \partial x} \left[ f(x,t) W(x;\xi) \right]
\right) \, d\xi,
\\
&=
-\frac{\partial }{ \partial x}
\left[f(x,t)
\int_{-\infty}^\infty
\xi  W(x;\xi)
\, d\xi\right]
+
\frac{ 1 } {2!}
\frac{\partial^2 }{ \partial x^2}\left[
 f(x,t)
\int_{-\infty}^\infty
\xi^2 W(x;\xi) \, d\xi
\right].
\end{aligned}
$$

For the second question, you are absolutely correct to say that we cannot always justify the truncation. Van Kampen has a entire chapter devoted to this (chapter X) among complaints in other places. The formally correct expansion, Kramers-Moyal expansion, yields
$$
\begin{aligned}
\frac{\partial f(x,t)}{\partial t}
&=
\sum_{k = 1}^\infty
\frac{(-1)^k}{k!}
\frac{\partial^k }{ \partial x^k}
\left[f(x,t)
\int_{-\infty}^\infty
\xi^k  W(x;\xi)
\, d\xi\right].
\end{aligned}
$$
The problem is that this expansion is equivalent to the master equation, and it is generally too difficult to solve.
