FUNCTIONS OF PROCESSES WITH MARKOVIAN STATES.
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)
Document Details
- Document Type
- Technical Report
- Publication Date
- Aug 21, 1967
- Accession Number
- AD0657176
Entities
People
- Herman Rubin
- Martin Fox
Organizations
- Michigan State University