 A proof of the Markov chain tree theoremV. Anantharam and P. Tsoucas Statistics & Probability Letters, 1989, vol. 8, issue 2, 189-192 Abstract: Let X be a finite set, P be a stochastic matrix on X, and = limn --> [infinity] (1/n)[summation operator]n-1k=0Pk. Let G = (X, E) be the weighted directed graph on X associated to P, with weights pij. An arborescence is a subset a [subset, double equals] E which has at most one edge out of every node, contains no cycles, and has maximum possible cardinality. The weight of an arborescence is the product of its edge weights. Let denote the set of all arborescences. Let ij denote the set of all arborescences which have j as a root and in which there is a directed path from i to j. Let [short parallel][short parallel], resp. [short parallel]ij[short parallel], be the sum of the weights of the arborescences in , resp. ij. The Markov chain tree theorem states that ij = [short parallel]ij[short parallel]/[short parallel][short parallel]. We give a proof of this theorem which is probabilistic in nature. Keywords: arborescence; Markov; chain; stationary; distribution; time; reversal; tree (search for similar items in EconPapers) Downloads: (external link) Related works: Export reference: BibTeX RIS (EndNote, ProCite, RefMan) HTML/Text Persistent link: https://EconPapers.repec.org/RePEc:eee:stapro:v:8:y:1989:i:2:p:189-192 Ordering information: This journal article can be ordered from Access Statistics for this article Statistics & Probability Letters is currently edited by Somnath Datta and Hira L. Koul More articles in Statistics & Probability Letters from Elsevier |
Is your work missing from RePEc? Here is how to contribute.
Questions or problems? Check the EconPapers FAQ or send mail to .
EconPapers is hosted by the
School of Business at Örebro University.