Functions of Processes with Markovian States--III
Martin Fox, Herman Rubin · The Annals of Mathematical Statistics · 1970
Abstract : Given a process (Yn), let epsilon be a state of finite rank. An example is given in which Yn = f(Xn), (Xn) is countable state, stationary, Markov, (Yn) is stationary with one state each of ranks 1 and 3, yet it is impossible to take (Xn) to be finite state Markov. In general it is proved that Yn = f(Xn) where epsilon = f(epsilon i) (i = 1,2,...), delta = f(delta) for delta not equal to epsilon and the epsilon i are Markovian states. If epsilon has rank 2 it is proved that two states suffice and that the rank of delta not equal to epsilon in (Xn) is the same as in (Yn). Finally, it is proved that if epsilon has rank 2 and (Yn) is stationary, (Xn) is stationary. (Author)