About function V in geometric drift condition for Markov Chain
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When I read the geometric ergodicity of Markov chain in Meyn and Tweedie, I note that in the drift condition $PV(x)<lambda V(x)+bI_{xin C}$, where $V(x)ge 1$ is required. Why $V(x)ge 1$ rather than $V(x)ge 0$ in this equation?
dynamical-systems markov-chains ergodic-theory monte-carlo
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
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$begingroup$
When I read the geometric ergodicity of Markov chain in Meyn and Tweedie, I note that in the drift condition $PV(x)<lambda V(x)+bI_{xin C}$, where $V(x)ge 1$ is required. Why $V(x)ge 1$ rather than $V(x)ge 0$ in this equation?
dynamical-systems markov-chains ergodic-theory monte-carlo
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
$endgroup$
add a comment |
$begingroup$
When I read the geometric ergodicity of Markov chain in Meyn and Tweedie, I note that in the drift condition $PV(x)<lambda V(x)+bI_{xin C}$, where $V(x)ge 1$ is required. Why $V(x)ge 1$ rather than $V(x)ge 0$ in this equation?
dynamical-systems markov-chains ergodic-theory monte-carlo
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
$endgroup$
When I read the geometric ergodicity of Markov chain in Meyn and Tweedie, I note that in the drift condition $PV(x)<lambda V(x)+bI_{xin C}$, where $V(x)ge 1$ is required. Why $V(x)ge 1$ rather than $V(x)ge 0$ in this equation?
dynamical-systems markov-chains ergodic-theory monte-carlo
dynamical-systems markov-chains ergodic-theory monte-carlo
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
edited Jan 9 at 12:18
Xiaol.Song
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
asked Dec 24 '18 at 10:06
Xiaol.SongXiaol.Song
33
33
add a comment |
add a comment |
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