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Let $(X_t)_{t\geq 0}$ a non-homogeneous Markov process. I have read several times, that the associated space-time process $(t, X_t)_{t\geq 0}$ is then a homogeneous Markov process.

I tried to come up with a proof, but it seems not very intuitive to me. Grateful for any hints, that push me in the right direction.

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up vote 2 down vote accepted

I have not heard the term space-time process before, given the rest of your question I think I understand though (if not, I apologise beforehand).

Let $\{Y_t\}$ be the what you call the space-time process. Now consider the probability a transition $\operatorname P \left (Y_{t + d} = (\tau, s') \mid Y_t = (\tau_0, s) \right)$. The transition probability can be non-zero only if $d = \tau - \tau_0$ (otherwise the transition is not possible since the time parameter is "deterministic").

If $d = \tau - \tau_0$ then the probability is given by the transition probability $\operatorname P \left ( X_\tau = s' \mid X_{\tau_0} = s \right )$, independently of $t$.

Hope that answers your question.

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Yes, that clarifies the problem. Thank you. – user13655 Jul 23 '11 at 8:38

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