exercise:Ef04d1ba60: Difference between revisions
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(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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\newcommand{\secstoprocess}{\all} | \newcommand{\secstoprocess}{\all} | ||
\newcommand{\NA}{{\rm NA}} | \newcommand{\NA}{{\rm NA}} | ||
\newcommand{\mathds}{\mathbb}</math></div> Complete the following alternate proof of | \newcommand{\mathds}{\mathbb}</math></div> Complete the following alternate proof of [[guide:87cf36f969#thm 11.2.3|Theorem]]. 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 | ||
11.2.3 | |||
we | |||
compute <math>b_{ij}</math> in terms of the possibilities on the outcome of the first | |||
step, then | |||
we have the equation | |||
<math display="block"> | <math display="block"> | ||
b_{ij} = p_{ij} + \sum_k p_{ik} b_{kj}\ , | b_{ij} = p_{ij} + \sum_k p_{ik} b_{kj}\ , | ||
</math> | </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 | 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 display="block"> | <math display="block"> | ||
\mat{B} = \mat{N}\mat{R}\ . | \mat{B} = \mat{N}\mat{R}\ . | ||
</math> | </math> | ||
Latest revision as of 00:58, 16 June 2024
[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. 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]]