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I have the following question:

Why do some books work with the canonical Markov process instead of just the abstract one, as from my point of view they both share exactly the same properties in terms of Markov property and strong Markov property etc. Some authors mention the shift operator but I don't really see any advantage of using the shift operator.

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    $\begingroup$ Could you add definitions or references? What's a canonical Markov process? $\endgroup$
    – Kostya_I
    Commented Sep 27, 2021 at 14:43
  • $\begingroup$ Sure ,let a probability space be given and a Markov process $X$ on it (using the filtration generated by X) with state space $(E,\mathcal{E}).$ Then we can consider $X$ as a random map $X: \Omega \rightarrow , \omega \mapsto (t \mapsto X_t(\omega)) \in E^{[0,\infty)} $ where $E^{[0,\infty)} := \{f:[ 0,\infty) \mapsto E \}.$ Further, let us define the following process $Y$ on $E^{[0,\infty)}:$ \begin{align*} Y_t: &E^{[0,\infty)} \rightarrow E \\ &f \mapsto Y_t(f)= f(t) \end{align*} Then $Y$ is a Markov on $E^{[0,\infty)}$ in regard to distribution of $X.$ $Y$ is the canonical version $\endgroup$
    – Oli Bernet
    Commented Sep 27, 2021 at 16:00
  • $\begingroup$ It is actually similar to a Brownian motion. If a Brownian motion is given we can always find its canonical version, same process as above, but usually we just work with the abstract version, so what is the advantage of having a canonical version? $\endgroup$
    – Oli Bernet
    Commented Sep 27, 2021 at 16:05
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    $\begingroup$ First of all, the canonical probability space is not really canonical, as it can refer to any of the following: the class of all paths, the class of càdlàg paths, or the class of continuous paths. It is convenient to work with the canonical realisation of a Markov process (that is, the one defined on the canonical probability space): shift operators, time-reversal and other path transformations are then defined with no difficulties. On the other hand, it is often necessary to work in a general setting — in order to have more then one adapted process or variables independent of the process. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 27, 2021 at 19:43
  • $\begingroup$ I see, I thought there might be some bigger advantage than just an ease for defining things, but that clears thing up thank you very much Mateusz $\endgroup$
    – Oli Bernet
    Commented Sep 27, 2021 at 20:17

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