I found two diffent ways to define Markov processes. It can be defined with a transition function and a collection of probability measures or just by the Markov property. I want to understand the connection between these two possibilities. To be precise I have the following problem:

Let $(\Omega ,{\mathcal {F}},\mathbb {P} )$ be a probability space with a filtration $(\mathcal{F}_t,\ t \in T)$, for some (totally ordered) index set $T$; and let $(E,\mathcal{E})$ be a measure space. Let an $E$-valued stochastic process $X=(X_t,\ t\in T)$ be adapted to the filtration, such that for each $s,t\in T$ with $s < t$ and every $f:E\to E$ measurable and bounded, $$ \mathbb{E}(f(X_t)|\mathcal{F}_s)=\mathbb{E}(f(X_t)|\sigma(X_s)). $$ Then I want to prove the following:

There exists a transition function $(P_t)_t$ ($P_t:E\times\mathcal{E}\to[0,1]$), such that for every distribution $\nu$ on $(E,\mathcal{E})$ there exists a probability distribution $\mathbb{P}_\nu$ on $(\Omega,\mathcal{F})$ with $$ (i)\quad \mathbb{P}_\nu(X_0\in B)=\nu(B)\text{ for every }B\in\mathcal{E} $$ $$ (ii)\quad \mathbb{E}_\nu(f(X_{t+s})|\mathcal{F}_s)=P_tf(X_s)=\mathbb{E}_{X_s}f(X_t),\quad \mathbb{P}_\nu-\text{a.s.} $$ where $\mathbb{E}_\nu$ is the expectation induced by $\mathbb{P}_\nu$ and $$ \mathbb{E}_{X_s}f(X_t):\Omega\to\mathbb{R},\quad\omega\mapsto \mathbb{E}_{X_s(\omega)}f(X_t) $$ and furthermore $\mathbb{P}=\mathbb{P}_{\delta_0}$.

Can someone tell me how to find/construct suitable transition kernels $P_t$ and probability distributions $\mathbb{P}_\nu$? Is there some kind of uniqunes in these choices?

  • $\begingroup$ By definition of conditional expectations, $E(f(X_t)\mid X_s)=g(X_s)$ and the mapping $f\mapsto g$ is linear. Your transition semi-group is then $P_{s,t}f=g$. $\endgroup$ – Did Mar 9 '17 at 7:27
  • $\begingroup$ Yes, that makes sense. Do you also know how we get the $\mathbb{P}_\nu$? $\endgroup$ – drogan Mar 9 '17 at 11:57
  • $\begingroup$ This is just a notation to specify that the initial distribution is $\nu$. $\endgroup$ – Did Mar 9 '17 at 14:34
  • $\begingroup$ Do you think the claim above is not correct? Is it not true that for every $\nu$ exists a probability measure $\mathbb{P}_\nu$ with the above properties? $\endgroup$ – drogan Mar 9 '17 at 14:57
  • $\begingroup$ ?? Where would I have said / hinted / whatever, that the claim does not hold? $\endgroup$ – Did Mar 9 '17 at 15:01

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