Derivation of Differential Chapman-Kolmogorov equation, Kramers-Moyal expansion I'm stuck with the derivation of the differential Chapman-Kolmogorov equation provided in Gardiner 1985, section 3.4. This is supposed to be some middle ground between the master equation and the Fokker-Planck equation since it allows for jumps to be present in addition to diffusion, while it has the virtue of jump and diffusion to be neatly separated. It is however slightly less general than the Chapman-Kolmogorov equation, since it assumes that the transition probability evolves in time in a differentiable way.
Gardiner starts by defining the jump term $W(x|z,t)$ and the drift and diffusion terms in eqs.(3.4.1,...,3.4.3). I will repeat them here for convenience (I drop the big-Os and replace by limits):
\begin{align}
W(x|z,t)&=\lim_{\Delta t\rightarrow 0} \frac{p(x,t+\Delta t|z,t)}{\Delta t},\\
A_i(z,t)&=\lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\,(x_i-z_i)p(x,t+\Delta t|z,t),\\
B_{ij}(z,t)&=\lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\,(x_i-z_i)(x_j-z_j)p(x,t+\Delta t|z,t),
\end{align}
assuming that the limits exist uniformly in $\epsilon,z,t$.
He then goes on to argue that all higher order terms must vanish. In eq. (3.4.7), this argument is done for the specific case of the third order moment. However, he does not comment on the various lines of the calculation. Let me print it here for you. The term he want to prove vanishes reads
\begin{equation}
C_{ijk}(z,t)=\lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx(x_i-z_i)(x_j-z_j)(x_k-z_k)p(x,t+\Delta t|z,t),
\end{equation}
and he defines
\begin{equation}
\bar C(\alpha,z,t)=\sum_{i,j,k}\alpha_i\alpha_j\alpha_kC_{ijk}(z,t).
\end{equation}
Then, he writes
\begin{equation}
\begin{split}
 \left |\bar C(\alpha,z,t)\right | &\leq \lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t}\int_{|x-z|<\epsilon} |\alpha\cdot(x-z)|[\alpha\cdot(x-z)]^2p(x,t+\Delta t|z,t) dx+\mathcal{O}(\epsilon)\\ 
&\leq |\alpha|\epsilon \lim_{\Delta t\rightarrow 0} \int [\alpha\cdot(x-z)]^2p(x,t+\Delta t|z,t) dx+\mathcal{O}(\epsilon) \\
&=\epsilon |\alpha| [\alpha_i \alpha_j B_{ij}(z,t)+\mathcal{O}(\epsilon)]+\mathcal{O}(\epsilon)\\
&=\mathcal{O}(\epsilon),
\end{split}
\end{equation}
such that the left-hand side must vanish for $\epsilon\rightarrow 0$.
I have a bit of trouble with this derivation. First of all, I think that the integration limits and a factor $1/\Delta t$ are missing in the second line. I think that I'm fine with the first inequality, which I spell out just in case
\begin{equation}
\begin{split}
 \left |\bar C(\alpha,z,t)\right | &= \left |   \sum_{i,j,k}\alpha_i\alpha_j\alpha_k \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx(x_i-z_i)(x_j-z_j)(x_k-z_k)p(x,t+\Delta t|z,t) \right | \\
&=  \left |   \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\sum_{i,j,k}\alpha_i\alpha_j\alpha_k (x_i-z_i)(x_j-z_j)(x_k-z_k)p(x,t+\Delta t|z,t) \right | \\
&=   \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t}  \left | \int_{|x-z|<\epsilon}dx[\alpha\cdot(x-z)]^3p(x,t+\Delta t|z,t) \right | \\
&\leq   \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\left | [\alpha\cdot(x-z)]^3\right |p(x,t+\Delta t|z,t)  \\
&=   \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\left | \alpha\cdot(x-z)\right |[\alpha\cdot(x-z)]^2p(x,t+\Delta t|z,t)  
\end{split}
\end{equation}
The second line holds because of the linearity of limits and integrals. The third is due to the continuity of the absolute value and because $\Delta t$ is positive (an unstated assumption). The fourth line is due to monotonicity of the integral and the last line is trivial.
Now I suppose the reasoning is to say that the $\left | \alpha\cdot(x-z)\right |$ in the integrand is smaller than $\epsilon |\alpha|$ within the integration region, which can be taken outside the integral and limits. The remaining integrand is just the right term which gives $\alpha_i \alpha_j B_{ij}(z,t)$, when integrated and combined with the $1/\Delta t$ and in the limit $\Delta t \rightarrow 0$. Ok, it seems that I understand the whole thing? No!
Here's the thing. Suppose I did the same reasoning for
\begin{equation}
\bar B(\alpha,z,t)=\sum_{i,j}\alpha_i\alpha_jB_{ij}(z,t).
\end{equation}
I would get, by the same manipulations as above,
\begin{equation}
\begin{split}
 \left |\bar B(\alpha,z,t)\right | &\leq ... \\
&=   \lim_{\epsilon\rightarrow 0}\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\left | \alpha\cdot(x-z)\right |\left | \alpha\cdot(x-z)\right |p(x,t+\Delta t|z,t) \\
&\leq  \lim_{\epsilon\rightarrow 0}\epsilon |\alpha|\lim_{\Delta t\rightarrow 0} \frac{1}{\Delta t} \int_{|x-z|<\epsilon}dx\left | \alpha\cdot(x-z)\right |p(x,t+\Delta t|z,t).
\end{split}
\end{equation}
Now, while the remaining term in the integral is not exactly the one which would give rise to $\alpha_i A_i$, I don't see what would prevent the integral to be at least finite such that the limit vanishes. This would imply that the diffusion matrix is zero in general, which is certainly false. Clearly, I'm misunderstanding Gardiner's argument and missing a crucial point which makes the argument for $C$ work and the one for $B$ break down.
On a related note, other derivations normally include all the jump moments, which lead to the Kramers-Moyal expansion, which reads (in 1d)
\begin{equation}
\partial_t p(x,t|x',t')=\sum_{l=1}^{\infty} \frac{(-1)^l}{l!}\frac{\partial^l}{\partial x^l}\left [ D^l(x,t)p(x,t|x',t')\right ],
\end{equation}
where the corrspondence to Gardiner is $D^1=A$ and $D^2=B$, but all higher terms are retained. Usually this is followed by some argument along the lines 'most of the time, third- and higher-order terms can be dropped, but I cannot tell you why'. How is this reconciled with Gardiner's view?
 A: Regarding your first question:
Let's just consider a simple example; namely, the (one-dimensional) Brownian motion. Then it follows from the homogenity of Brownian motion (in time and space) that
$$\begin{align*} A(z,t) &:= \lim_{\varepsilon \to 0} \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x-z|<\varepsilon} (x-z) p(x,t+ \Delta t \mid z,t) \, dx \\ &= \lim_{\varepsilon \to 0} \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x|<\varepsilon} x q_{\Delta t}(x) \, dx =0 \end{align*}$$
where $q_{\Delta t}$ denotes the density of the Normal distribution with mean $0$ and variance $\Delta t$. That's not surprising - obviously, the Brownian motion has no drift. In contrast, if we consider the corresponding integral with the absolute value inside the integral (and that's exactly the integral which pops up in your question), then we see that
$$\begin{align*} I(\Delta t,\varepsilon):= \frac{1}{\Delta t} \int_{|x|<\varepsilon} |x| q_{\Delta t}(x) \, dx &= \frac{2}{\Delta t} \int_{0}^{\varepsilon} |x| \frac{1}{\sqrt{2\pi \Delta t}} \exp \left(- \frac{x^2}{2\Delta t} \right) \, dx \\ &= \sqrt{\frac{2}{\pi}} \frac{1}{\sqrt{\Delta t}} \int_0^{\varepsilon/\sqrt{\Delta t}} |z| \exp \left(- \frac{z^2}{2} \right) \, dz. \end{align*}$$
Consequently,
$$I(\Delta t,\varepsilon) \to \infty \qquad \text{as} \, \Delta t \to 0.$$
Back to your question: The fact that the remaining term (in your estimate for $\bar{B}(\alpha,z,t)$) is not exactly the one from the definition of $A$, means that Gardiner's argument does not apply in this case since the limit is not necessarily finite.
(And sorry, I don't know the answer to your second question.)
