I am wondering whether there is a one-to-one correspondence between transition functions and homogeneous Markov processes?

We say that $(X_t,\mathcal{F}_t)_{t\geq 0}$ is a Markov process if $\mathbb{P}(A\vert \mathcal{F}_t)=\mathbb{P}(A\vert X_t)$ for all $A\in\sigma(X_s, s\geq t)$.

On the other hand a transition function $[0,\infty)\times E\times \mathcal{E}\ni(s,x,A)\mapsto P_s(x,A)$ satisfies

(i) $x\mapsto P_s(x,A)$ is measurable for all fixed $s,A$;

(ii) $P_s(x,\cdot)$ is a probability measure (or it satisfies $\leq 1$, in which case we extend the state space $E$ by a point $\Delta$ so it becomes a probability measure) and

(iii) the Chapman-Kolmogorov property.

It is clear that a transition function that satisfying (i)-(iii) induces a Markov process.

If $(X_t,\mathcal{F}_t)$ is a Markov process, the following is well defined for any Markov process

$P_s(x,A):=\mathbb{P}[X_s\in A\vert X_0=x]$.

Now, this should satisfy (i) which can be showed using the Radon-Nikodym theorem. (ii) and (iii) should also be satisfied.

However, I am confused as there are Markov processes $X$ which have branching points, i.e., $P_0(x,{x})\neq 1$. But if $X$ is a Markov process, isn't $\mathbb{P}(X_0\in A\vert X_0=x)=\delta_x(A)$?

  • $\begingroup$ You are probably referring to cases when $$\lim_{t\to0+}P_t(x,x)\ne1,$$ rather than $P_0(x,x)\ne1$. $\endgroup$ – Did Jul 23 '16 at 17:39
  • $\begingroup$ @Did No, I mean what I wrote. In Vol 1 of Diffusions and Markov processes, Rogers&Williams refer to such points as branch points and call transitions functions without them "normal" (see III.3, III.37 of the reference just mentioned). It seems to me that the machinery of Ray processes is needed to properly understand a relaxation of the "normality" condition of transition functions. I have still not come up with a satisfying answer to my last question above. Any ideas? $\endgroup$ – mathsquestion88 Jul 24 '16 at 12:44

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