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When is a stochastic process a diffusion process?

Wikipedia says "A diffusion process is a Markov process with continuous sample paths for which the Kolmogorov forward equation is the Fokker–Planck equation." I don't know how to interpret this; doesn't this just mean it has the typical diffusion form \begin{equation}dX_t = \mu(X_t) dt + \sigma(X_t) dW_t, \;\; X_0 = x_0\end{equation} and how would one check this for an arbitrary process?

I understand this means there's an average drift and a variance around that drift. But I don't know if this form is valid for an arbitrary process $X_t$, since we can always define the "drift" as $ \mu(X_t) = \mathbb{E}[X_t]$ and variance $\sigma(X_t) = \text{Var}(X_t)$.

For example, define a process $X_t$ that increases by $dX$ with probability $p \in [0,1]$ in each time interval $dt$. This resembles a Poisson process with rate $p$ and infinitesimal jump size. Since the mean and variance of Poi$(p)$ are both $p$, then I guess $\mu(X_t)dt = p dt$ and $\sigma^2 = \sqrt{p}$. But I still don't know if the Kolmogorov forward equation is the Fokker-Planck.

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    $\begingroup$ I don't have an answer, but on Oksendal's book on SDEs, chapter 7 or 8 if I remember correctly there's a section entitled "when is an Ito process a diffusion?", You should check it out $\endgroup$
    – Chaos
    Commented Mar 3, 2021 at 21:44
  • $\begingroup$ For others interested: I looked up this chapter and it asks the question "if $X_t$ is an Ito diffusion, for what $C^2$ functions $\phi$ are $\phi(X_t)$ also an Ito diffusion?" I'm not sure whether/how this can answer my question. Does this mean I would have to look for a $\phi^{-1}$ that maps my given process back to a Brownian motion? $\endgroup$
    – 900edges
    Commented Mar 8, 2021 at 0:53

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I did some reading and here is a partial answer:

Definition: Diffusion process (see here)

A Markov process $X_t$ with transition probability $P(\Gamma, t|x,s)$ is a diffusion process if

  1. for every $x$ and every $\epsilon>0$, $$\int_{|x-y|>\epsilon} P(dy,t|x,s) = o(t-s).$$ I.e. the probability of moving more than $\epsilon$ in a small time frame is small.

  2. the drift is given by $a(x,s)$ such that for every $x$, and every $\epsilon >0$, $$ \int_{|y-x|\leq \epsilon} (y-x) P(dy,t|x,s) = a(x,s)(s-t) + o(s-t)$$

  3. Diffusion coefficient is given by $b(x,s)$ such that $$\int_{|y-x|\leq \epsilon} (y-x)^2 P(dy,t|x,s) = b(x,s)(s-t) + p(s-t).$$

I.e. the drift and diffusion coefficients specify the distribution of the movement, giving a necessary concentration of measure.

Characterization by Fokker-Planck equation In my question I said that a diffusion process is a Markov process whose forward equation is the Fokker-Planck equation. How can this be reconciled with the above definition?

This says that if we define a pdf for the transition density, $$ P(\Gamma,t|x,s) = \int_\Gamma p(s,x,t,y)dy$$ and that assume $a, b$ are differentiable, and assume 1., 2., 3. above, then the transition probability density satisfies the forward Kolmogorov equation.

So in addition note that the equation $dX_t = a(X,t) dt + b(X_t, t) dB_t, \;\; X_0 = x_0$, does not specify a diffusion process unless the coefficients are continuous.

Something I am still confused about: I saw somewhere else that the drift is given by $$\mathbb{E}[X_{t+dt}-X_t| X_t = x] = a(x,t)dt + O(dt^2)$$ I'm not sure how this is equivalent to the integral in 2) above, and why there is a $O(dt^2)$ term (since the expectation should be fixed).

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