Markov Property: determined by just the law or also the realization? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:39:35Z http://mathoverflow.net/feeds/question/78397 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78397/markov-property-determined-by-just-the-law-or-also-the-realization Markov Property: determined by just the law or also the realization? Tom Alberts 2011-10-17T23:37:32Z 2011-10-18T14:17:48Z <p>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.</p> <p>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?</p> <p>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?"</p> http://mathoverflow.net/questions/78397/markov-property-determined-by-just-the-law-or-also-the-realization/78399#78399 Answer by psd for Markov Property: determined by just the law or also the realization? psd 2011-10-17T23:56:51Z 2011-10-17T23:56:51Z <blockquote> <p>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?</p> </blockquote> <p>It is a property solely of the law of the process.</p> <p>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.</p> <p>Edit: MathOverflow has highly variable responses to statistics questions, so maybe you should ask on the statistics stackexchange instead.</p> http://mathoverflow.net/questions/78397/markov-property-determined-by-just-the-law-or-also-the-realization/78403#78403 Answer by ShawnD for Markov Property: determined by just the law or also the realization? ShawnD 2011-10-18T00:43:26Z 2011-10-18T00:43:26Z <p>This sparked my curiosity. A little googling gives this: <a href="http://www.stat.cmu.edu/~cshalizi/754/notes/lecture-09.pdf" rel="nofollow">http://www.stat.cmu.edu/~cshalizi/754/notes/lecture-09.pdf</a>. 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.</p> <p>This should probably be a comment, but I'm not reputable enough . . .</p> http://mathoverflow.net/questions/78397/markov-property-determined-by-just-the-law-or-also-the-realization/78459#78459 Answer by Byron Schmuland for Markov Property: determined by just the law or also the realization? Byron Schmuland 2011-10-18T14:17:48Z 2011-10-18T14:17:48Z <p>I've posted a solution <a href="http://www.stat.ualberta.ca/people/schmu/preprints/markov_law.pdf" rel="nofollow">on my webpage</a>.</p>