Markov Property: determined by just the law or also the realization? - MathOverflow

Markov Property: determined by just the law or also the realization?

2011-10-17T23:37:32Z

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?"

Answer by psd for Markov Property: determined by just the law or also the realization?

2011-10-17T23:56:51Z

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.

Answer by ShawnD for Markov Property: determined by just the law or also the realization?

2011-10-18T00:43:26Z

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 . . .

Answer by Byron Schmuland for Markov Property: determined by just the law or also the realization?

2011-10-18T14:17:48Z

I've posted a solution on my webpage.