In the usual derivation of the Fokker-Planck, e.g. here, we start with a generic continuous-time process $X_t$ and look at the integral $\int_{-\infty}^\infty h(Y) \frac{\partial P(Y,t|X)}{\partial t}dY$.
In equation (9), the Taylor expansion of $h(Y)$ for $Y\approx Z$ brings about the term $$ D^{(n)}(Z) = \frac{1}{n!} \lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{-\infty}^\infty (Y-Z)^n P(Y,\Delta t|Z) dY.$$
To derive the Fokker-Planck, we assume that $D^{(n)}(Z) = 0$ for $n>2$. This seems to be the condition that describes a Markov process $X_t$ as a diffusion process (since a diffusion process is a continuous Markov process whose distribution over time satisfies the Fokker-Planck).
What does it mean for $D^{(n)}(Z) = 0$ for $n>2$? I'm guessing it's some kind of regularity condition that prevents a large jump in a small time interval, but I don't understand how to interpret the equation.