Markov chain: Obtaining transition matrix from recurrence probabilities - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:02:04Z http://mathoverflow.net/feeds/question/57945 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57945/markov-chain-obtaining-transition-matrix-from-recurrence-probabilities Markov chain: Obtaining transition matrix from recurrence probabilities Rahul Gupta 2011-03-09T13:19:52Z 2011-03-11T05:57:07Z <p>Consider a markov chain with finite space { 0,1,..n} with transition probability matrix whose entries are \$P_{ij}\$. Let</p> <p>\$f_{ij}^n\$ = probability that starting from state \$i \$ it goes to state \$j\$ first time. <br> </p> <p>Question 1. What are the necessary and sufficient condition arbitrary \$a_{ij}^{n}\$ needs to satisfy to be valid \$f_{ij}^{n}\$ of some markov chain ?<br></p> <p>Qustion 2. If existence of such a markov chain is shown, can we calculate \$P_{ij}\$ given valid \$f_{ij}^n\$ ? ( I mean, is there any algorithm to calculate?)</p> http://mathoverflow.net/questions/57945/markov-chain-obtaining-transition-matrix-from-recurrence-probabilities/57947#57947 Answer by Didier Piau for Markov chain: Obtaining transition matrix from recurrence probabilities Didier Piau 2011-03-09T13:34:56Z 2011-03-11T05:57:07Z <p>Re 2, for \$n=1\$, \$P_{ij}=f^1_{ij}\$ for every \$i\ne j\$ and \$1-P_{ii}\$ is the sum of \$f^1_{ij}\$ over \$j\ne i\$, hence one recovers trivially \$P\$ from \$f^1=(f^1_{ij})_{ij}\$. </p> <p>Re 2 again, on the contrary, there is no hope to recover \$P\$ from \$f^n=(f^n_{ij})_{ij}\$ in general, for any given \$n\ge2\$. For example, the two (deterministic) one-step rotations on the discrete circle with \$2n\$ vertices, clockwise and counterclockwide, yield the same matrix \$f^n\$ although their transition matrices \$P\$ are different.</p>