# Can Markov Chain state space be continuous?

I looked for a formal definition of Markov chain and was confused that all definitions I found restrict chain's state space to be countable. I don't understand purpose of such a restriction and I have feeling that it does not make any sense.

So my question is: can state space of a Markov chain be continuum? And if not then why?

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Since the state transitions themselves are countable, the state cannot enter more than a countable number of different states. So perhaps no generality is gained from having an uncountable state space. –  MJD May 30 '12 at 14:16
@MarkDominus As I understand this, if I'm walking along the street I'm not allowed to model this in terms of Markov chain because world coordinates are continuous rather than discrete. –  user32541 May 30 '12 at 14:24
I don't understand how you could model such a thing as a Markov process. What would the state transition be? You would need to have a continuous transition between states, and Markov processes are all about discrete transitions. –  MJD May 30 '12 at 14:29
If I make a step and one and another and each step I'm in this or this point according to some probability distribution. Yep, I need continuous transition in this case, but it's much like Markov chain and I'm wondering whether it can be called Markov chain. –  user32541 May 30 '12 at 14:41
"restrict chain's state space to be uncountable" -> sounds confusing. should "to be countable"? –  leonbloy May 30 '12 at 14:49

Yes, it can. In some quarters the "chain" in Markov chain refers to the discreteness of the time parameter. (A notable exception is the work of K.L. Chung.) The evolution in time of a Markov chain $(X_0,X_1,X_2,\ldots)$ taking values in a measurable state space $(E, {\mathcal E})$ is governed by a one-step transition kernel $P(x,A)$, $x\in E$, $A\in{\mathcal E}$: $${\bf P}[ X_{n+1}\in A|X_0,X_1,\ldots,X_n] = P(X_n,A).$$ Two fine references for the subject are Markov Chains by D. Revuz and Markov Chains and Stochastic Stability by S. Meyn and R. Tweedie.