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If one talks about homogeneous Markov diffusion $$ \mathrm dX_t = \mu(X_t)\mathrm dt+\sigma(X_t)\mathrm dw_t $$ with $\mu,\sigma$ sufficiently differentiable and of appropriate dimensions, there is nice equation for a function $m_f(x,t) = \mathsf E_x f(X_t)$ for $f\in C^2(\mathbb R)$: $$ \begin{cases} \frac{\partial m_f}{\partial t} &= \mu\frac{\partial m_f}{\partial x}+\frac12\sigma^2\frac{\partial^2 m_f}{\partial x^2}, \\ m(x,0)&=f(x). \end{cases} $$

On the other hand, while answering on the question Stochastic model I advised to use Fokker-Plank equation for the density of the process rather then the very same equation for $m_f$.

The problem I had is the following. Since $\mu$ and $\sigma$ are time-dependent there, one can construct a process $Z_t = (X_t,Y_t)$ with $\mathrm dY_t = \mathrm dt$ and obtain everything for this process just using theory of homogeneous/time-independent Markov processes (how it is usually written in the books). Unfortunately, if you derive an infinitesimal generator for this process then you obtain $$ \mathcal A_Zg(x,y) = \frac{\partial g}{\partial y}+\mu\frac{\partial g}{\partial x}+\frac12\sigma^2\frac{\partial^2 g}{\partial x^2}. $$

For sure, now one shoud define $m_f(\tau|x,t) = \mathsf E_{x,t}f(X_{t+\tau})$ where $\tau\in \mathbb R_{\geq 0}$. If I am not wrong then it follows $$ \frac{\partial m_f}{\partial \tau} = \frac{\partial m_f}{\partial t}+\mu\frac{\partial m_f}{\partial x}+\frac12\sigma^2\frac{\partial^2 m_f}{\partial x^2}\quad(*) $$ which is kind of strange equation.

My question has three parts:

  1. is an equation $(*)$ correct? if yes, are there developed methods for its solutions?

  2. if this equation is not correct, what is the right equation?

  3. I usually have problems when deal with non-homogeneous Markov processes since trick $Z_t = (t,X_t)$ does not help me. Could you refer me to literature where authors consider non-homogeneous Markov processes in details (rather then saying that this trick will help to use provided theory of homogeneous processes)?

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Also asked it here:… – Ilya Jul 15 '11 at 16:17

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