Revision as of 02:33, 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> Complete the following alternate proof of Theorem \ref{thm 11.2.3}. Let <math>s_i</math> be a transient state and <math>s_j</math> be an absorbing state. If we compute <math>b_{ij}</math> in terms of the possibilities on the outcome of the fi...")
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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]

Complete the following alternate proof of

Theorem \ref{thm 11.2.3}. Let [math]s_i[/math] be a transient state and [math]s_j[/math] be an absorbing state. If we compute [math]b_{ij}[/math] in terms of the possibilities on the outcome of the first step, then we have the equation

[[math]] b_{ij} = p_{ij} + \sum_k p_{ik} b_{kj}\ , [[/math]]

where the summation is carried out over all transient states [math]s_k[/math]. Write this in matrix form, and derive from this equation the statement

[[math]] \mat{B} = \mat{N}\mat{R}\ . [[/math]]