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When one says that a stochastic process is Markovian, is this a property solely of the law of the process, or does the realization of the process also come in to play? I am asking even for the simplest examples, such as a process indexed by $\mathbb{N}$. Most abstract definitions are about being Markov with respect to some filtration, which indicates that it has to do with the realization also.

My guess is that by knowing only the law one can determine if there exists a version of the process that is Markovian, but that given a realization one cannot determine if it is Markovian solely by checking its law. If the latter case is true does anyone know of simple examples?

Edit: when I say ``realization'' I mean a collection of random variables with the given law. I do not mean the value of the random variables at a given point. So the question could be rephrased as: "Can one construct a collection of random variables that has the law of a Markov process, but such that the collection itself does not form a Markov process?"

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I've posted a solution on my webpage.

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This sparked my curiosity. A little googling gives this: http://www.stat.cmu.edu/~cshalizi/754/notes/lecture-09.pdf. On page 5 they discuss your question. If you make the filtration coarser, the Markov property can't go away, but you can add stuff to the filtration to kill the Markov property.

This should probably be a comment, but I'm not reputable enough . . .

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  • $\begingroup$ This is a helpful reference, thanks! In retrospect I guess I already knew that adding more info to the filtration could kill the Markov property, and the example on page 6 is a nice simple one. I guess in my edit about it should read: "but such that the collection itself does not form a Markov process with respect to its own filtration?" In this case the answer must be no. $\endgroup$
    – Tom Alberts
    Commented Oct 18, 2011 at 1:27
  • $\begingroup$ en.wikipedia.org/wiki/Adapted_process $\endgroup$
    – Steve Huntsman
    Commented Oct 18, 2011 at 3:41
  • $\begingroup$ In my mind this was initially confused by the fact that when you are dealing with the strong Markov property, you often have to ADD stuff to the filtration to make sure all of the things you want to be stopping times actually are stopping times (e.g. the first time you hit some Borel set) $\endgroup$
    – ShawnD
    Commented Oct 18, 2011 at 18:20
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When one says that a stochastic process is Markovian, is this a property solely of the law of the process, or does the realization of the process also come in to play?

It is a property solely of the law of the process.

As an imperfect analogy, you can think of two independent random variables versus a realization of some of their joint samples. Although for a particular realization the samples may have nonzero sample correlation, the random variables themselves are still independent.

Edit: MathOverflow has highly variable responses to statistics questions, so maybe you should ask on the statistics stackexchange instead.

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    $\begingroup$ Probability theory is alive and well as a mathematical research field. That MO has a variable response to such questions speaks more to who inhabits MO than the suitability of the topics. $\endgroup$
    – BSteinhurst
    Commented Oct 18, 2011 at 0:20
  • $\begingroup$ I should be a bit more precise in my questioning: by realization I mean a collection of random variables which has a given law, not the value of the random variables at a given point. I'll edit my question. $\endgroup$
    – Tom Alberts
    Commented Oct 18, 2011 at 0:32

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