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Say we have some non-Markovian process, $X_{t}$ which depends on some identifiable information contained in the sigma algebra generated by the collection of random variables up to time $s<t$, $\mathcal{F}_{s}^{X}$, such that $P(X_{t}|\mathcal{F}_{s}^{X}) \not = P(X_{t}|X_{s})$. Should one be able to extend the state space to support another process, say $V_{t}$, such that the all information for which $X_{t}$ was dependent on in ($\mathcal{F}_{s}^{X}- \sigma(X_{s})$) is encoded in it and such that $V_{t}$ is markov, is the process $(X_{t},V_{t})$ Markov in the product space? I have seen this alluded to and it makes intuitive sense, though I have not seen this property stated formally.

An example I've seen in the wild would be a M/G/1 queue which does not have an exponential service time but $(X_{t},V_{t})$ being the double which records the current queue count ($X_{t}$) and the elapsed service time of the current job ($V_{t}$) being Markovian.

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Yes, you can take $V_t=(1[s\leq t]\cdot X_s)_{s}$, so that the state space of $V_t$ is now a space of functions (although I will purposely avoid saying what $\sigma$-algebra to put on this huge space to avoid getting into the weeds of measurability technicalities...) and the entire process $(V_t)_t$ is a path in a function space, or equivalently $V$ is a random field indexed by $V_{s,t}$.

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