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Well, I've been reading over the internet but I've been unable to find a straight answer. I've got a transition matrix for a Markov Discrete Chain. I've made the graph and according to my knowledge, this graph should be periodic. If the graph is periodic, then I shouldn't be able to find a stationary distribution by elevating the matrix times infinity right? Well, I added the matrix to MatLab, elevated it time 100, 1000,10000 and everytime I'm getting the same answer, even though I should be getting different ones. Can someone help me?

\begin{bmatrix} 0&0.4 &0 &0 &0 &0.6 \\ 0.3&0 &0.7 &0 &0 &0 \\ 0&0.5 &0 &0 &0.5 &0 \\ 0& 0 & 0.1 & 0 &0.9 & 0\\ 0.6&0.2 &0 &0 &0 &0.2 \\ 0&0 &0 &0.2 &0.8 &0 \end{bmatrix}

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  • $\begingroup$ It is not periodic. $\endgroup$
    – user940
    Oct 14, 2013 at 15:45
  • $\begingroup$ "this graph should be periodic" why? rather on the contrary (aperiodic). The bit about elevating the matrix to 100, 1000 ... and getting the same answer doesn't make sense to me neither. $\endgroup$
    – leonbloy
    Oct 14, 2013 at 15:46
  • $\begingroup$ Which is the definition of periodic? I found on wikipedia that a graph transition matrix is periodic if and only if there is a loop to the same node. $\endgroup$
    – Cristian Eduardo Lehuede Lyon
    Oct 14, 2013 at 15:50
  • $\begingroup$ No. en.wikipedia.org/wiki/Markov_chain#Periodicity $\endgroup$
    – leonbloy
    Oct 14, 2013 at 16:01
  • $\begingroup$ @CristianEduardoLehuedeLyon If there is a loop to the same node, then the graph is aperiodic. But not "if and only if". Could you link to this statement on Wikipedia? $\endgroup$
    – user940
    Oct 14, 2013 at 16:12

2 Answers 2

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You can return to state $1$ in two steps ($1\to2\to1$) or in three steps ($1\to6\to5\to1$) with positive probability. The period of the chain divides both 2 and 3, hence the period is 1. That is, the chain is aperiodic.

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  • $\begingroup$ Perfect answer, thanks $\endgroup$
    – Cristian Eduardo Lehuede Lyon
    Oct 14, 2013 at 16:02
  • $\begingroup$ Note, however, that the fact that the chain is aperiodic (instead of periodic) does not do away with the rest of your question. Rather, aperiodic chains are informally) the "nice chains", and it's for them the usual recipes for finding the stationary state apply. $\endgroup$
    – leonbloy
    Oct 14, 2013 at 16:36
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Compute the eigen-decomposition of your matrix. If you get one eigenvector that has eigenvalue $1$ and all other eigenvectors have eigenvalues of magnitude less than one, then your matrix is not periodic and any starting condition will converge to a single stationary point for the matrix.

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  • $\begingroup$ Perfect answer but Byron's one allows me to check it on a test rather than wasting time calculating the eigen values. $\endgroup$
    – Cristian Eduardo Lehuede Lyon
    Oct 14, 2013 at 16:01

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