Random walk with zero reflection derivations
Problem image
Does anyone know how to derive the proof that if p < 1/2, then all states are positive recurrent? I've tried using this formula: $q*1 + p*(T_1*q + p*(2T_1*q + ...$
where $T_1$ is the expected time to revisit the first state. This doesn't seem to work and reduces to a finite value for any $p < 1$, so I'm assuming the logic is flawed. I've attached a picture of the problem
markov-chains random-walk
asked Nov 27 at 5:08
Brian Sun
11
add a comment |
Problem image
Does anyone know how to derive the proof that if p < 1/2, then all states are positive recurrent? I've tried using this formula: $q*1 + p*(T_1*q + p*(2T_1*q + ...$
where $T_1$ is the expected time to revisit the first state. This doesn't seem to work and reduces to a finite value for any $p < 1$, so I'm assuming the logic is flawed. I've attached a picture of the problem
markov-chains random-walk
asked Nov 27 at 5:08
Brian Sun
11
add a comment |
Problem image
Does anyone know how to derive the proof that if p < 1/2, then all states are positive recurrent? I've tried using this formula: $q*1 + p*(T_1*q + p*(2T_1*q + ...$
where $T_1$ is the expected time to revisit the first state. This doesn't seem to work and reduces to a finite value for any $p < 1$, so I'm assuming the logic is flawed. I've attached a picture of the problem
markov-chains random-walk
asked Nov 27 at 5:08
Brian Sun
11
Problem image
Does anyone know how to derive the proof that if p < 1/2, then all states are positive recurrent? I've tried using this formula: $q*1 + p*(T_1*q + p*(2T_1*q + ...$
where $T_1$ is the expected time to revisit the first state. This doesn't seem to work and reduces to a finite value for any $p < 1$, so I'm assuming the logic is flawed. I've attached a picture of the problem
markov-chains random-walk
markov-chains random-walk
asked Nov 27 at 5:08
Brian Sun
11
asked Nov 27 at 5:08
Brian Sun
11
asked Nov 27 at 5:08
Brian Sun
11
asked Nov 27 at 5:08
Brian Sun
11
asked Nov 27 at 5:08
Brian Sun
11
11
add a comment |
add a comment |
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