Open Access
June, 1968 Splitting a Single State of a Stationary Process into Markovian States
S. W. Dharmadhikari
Ann. Math. Statist. 39(3): 1069-1077 (June, 1968). DOI: 10.1214/aoms/1177698339

Abstract

Let $\{Y_n, n \geqq 1\}$ be a stationary process with a finite state-space $J$. Let $\delta$ denote a state of $J$ and let $s, t$ denote finite sequences of states of $J$. If $s = (\delta_1, \cdots, \delta_n)$, let $p(s) = P\lbrack(Y_1, \cdots, Y_n) = s\rbrack$. The rank $n(\delta)$ of a state $\delta$ is defined to be the largest integer $n$ such that we can find $2n$ sequences $s_1, \cdots, s_n, t_1, \cdots, t_n$ such that the $n \times n$ matrix $|p(s_i\delta t_j)|$ is nonsingular. The number $n(\delta)$ was first defined by Gilbert [5] and the term rank was first used by Fox and Rubin [4]. A state $\delta$ is called Markovian if $n(\delta) = 1$. It is easy to check that $\delta$ is Markovian if, and only if, $p(s\delta t) = p(s\delta)p(\delta t)/p(\delta)$ for all $s$ and $t$. Suppose that $\mu$ is a fixed state of $J$. Let $J' = J - \{\mu\}$. Assume that $n(\mu) < \infty$. Fox and Rubin have shown that there exists a stationary process $\{X_n\}$ with a countable state-space $I = J' \cup J''$ and a function $f$ on $I$ onto $J$ such that (a) $f(i) = \mu$ if $i \varepsilon J''$ and $f(\delta) = \delta$ if $\delta \varepsilon J'$; (b) states of $J''$ are Markovian states of $\{X_n\}$; and (c) $\{Y_n\}$ and $\{f(X_n)\}$ have the same distribution. Gilbert [5] has shown that $J''$ must have at least $n(\mu)$ elements whereas Fox and Rubin [4] have given an example to show that $J''$ cannot always be chosen to be finite. For $\delta \varepsilon J'$ let $\nu(\delta)$ denote the rank of $\delta$ in $\{X_n\}$. In general $\nu(\delta) \geqq n(\delta)$. But Fox and Rubin have shown that $\{ X_n\}$ can be constructed in such a way that $\nu(\delta) = 1$ whenever $n(\delta) = 1$. Finally they have shown that, if $n(\mu) = 2$, then $\{X_n\}$ can be chosen in such a way that $J''$ has 2 elements and $\nu(\delta) = n(\delta)$ for all $\delta \varepsilon J'$. In this paper we give some conditions under which $J''$ can be chosen to be finite. These conditions are similar to those imposed in [2]. It is shown that $\{X_n\}$ can be constructed in such a way that, for $\delta \varepsilon J', \nu(\delta) = 1$ whenever $n(\delta) = 1$. Finally it is proved that if $N(\mu) = n(\mu)$, then $\nu(\delta) = n(\delta)$ for all $\delta \varepsilon J'$. This generalizes the result proved by Fox and Rubin for the case $n(\mu) = 2$. However, they have given results for the non-stationary case also. The results of this paper were partially reported in [3].

Citation

Download Citation

S. W. Dharmadhikari. "Splitting a Single State of a Stationary Process into Markovian States." Ann. Math. Statist. 39 (3) 1069 - 1077, June, 1968. https://doi.org/10.1214/aoms/1177698339

Information

Published: June, 1968
First available in Project Euclid: 27 April 2007

zbMATH: 0159.46301
MathSciNet: MR224154
Digital Object Identifier: 10.1214/aoms/1177698339

Rights: Copyright © 1968 Institute of Mathematical Statistics

Vol.39 • No. 3 • June, 1968
Back to Top