Revision as of 03:35, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> In the course of a walk with Snell along Minnehaha Avenue in Minneapolis in the fall of 1983, Peter Doyle<ref group="Notes" >Private communication.</ref> suggested the following explanation for the constancy of ''Kemeny's constant'' (see Exercise...")
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BBy Bot
Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

In the course of a walk with Snell along Minnehaha Avenue in Minneapolis

in the fall of 1983, Peter Doyle[Notes 1] suggested the following explanation for the constancy of Kemeny's constant (see Exercise Exercise). Choose a target state according to the fixed vector [math]\mat{w}[/math]. Start from state [math]i[/math] and wait until the time [math]T[/math] that the target state occurs for the first time. Let [math]K_i[/math] be the expected value of [math]T[/math]. Observe that

[[math]] K_i + w_i \cdot 1/w_i= \sum_j P_{ij} K_j + 1\ , [[/math]]

and hence

[[math]] K_i = \sum_j P_{ij} K_j\ . [[/math]]

By the maximum principle, [math]K_i[/math] is a constant. Should Peter have been given the prize?

Notes

  1. Private communication.