Let $X_1,X_2,...$ be a discrete time process, which takes values in $S$ that is finite or countable. I want to prove that there exists a Markov chain $Y_1,Y_2...$ taking values in countable $\hat S$, and a function $f:\hat S \rightarrow S$ such that $f(Y_n)=X_n$ for any $n$.

How to prove this? It seems untrue as the discrete time process can depend on the past values, while the Markov process only depends on the last value.

  • $\begingroup$ What is $\hat{S}$? $\endgroup$ – user10354138 Oct 30 '18 at 3:35
  • $\begingroup$ @user10354138 it's the set that the markov chain takes values in. $\endgroup$ – Dole Oct 30 '18 at 4:31
  • $\begingroup$ In case I'm allowed to be pretend to be a lawyer and use the fact that you have not explicitly put any assumptions on $\hat S$, I'd take $\hat S = S^{\Bbb N}$, i.e. the space of all infinite histories of $X$. Otherwise indeed, there may not exist such a Markov Chain. $\endgroup$ – Ilya Oct 30 '18 at 16:36
  • $\begingroup$ @Ilya It has to be countable unfortunately. Doesn't work when $S$ is infinite. $\endgroup$ – Dole Nov 2 '18 at 6:49
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    $\begingroup$ Well, maybe then one could work around with $\hat S = \bigcup_{n\in \Bbb N}S^n$, which is countable $\endgroup$ – Ilya Nov 5 '18 at 10:11

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