Markov chain: Obtaining transition matrix from recurrence probabilities - MathOverflow most recent 30 from http://mathoverflow.net2013-05-18T06:02:04Zhttp://mathoverflow.net/feeds/question/57945http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/57945/markov-chain-obtaining-transition-matrix-from-recurrence-probabilitiesMarkov chain: Obtaining transition matrix from recurrence probabilitiesRahul Gupta2011-03-09T13:19:52Z2011-03-11T05:57:07Z
<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#57947Answer by Didier Piau for Markov chain: Obtaining transition matrix from recurrence probabilitiesDidier Piau2011-03-09T13:34:56Z2011-03-11T05:57:07Z<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>