Sunday, 25 April 2010

pr.probability - When is a 1-block factor of a non-Markovian process Markov?

Some necessary conditions for Z to be Markov are easy to understand and to write down.



For every y, y and y in the state space S of Y, write p3(yyy) for the probability that [Yt=y,Yt+1=y,Yt+2=y], which is independent of time t. Assume that Z=phi(Y). For every z, z and z in phi(S), write q3(zzz) for the sum of p3(yyy) over every y, y and y such that z=phi(y), z=phi(y) and z=phi(y). Then a necessary condition is that q3 can be factorized, in the sense that there exist functions r and s such that q3(zzz)=r(zz)s(zz), for every z, z and z in phi(S).



Of course, this condition is far from sufficient. In fact, to be able to say anything even moderately interesting about this problem, one should probably specify the kind of processes Y and Z one has in mind.

No comments:

Post a Comment