# Discrete Markov Transition Matrix

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}

• It is not periodic. – user940 Oct 14 '13 at 15:45
• "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. – leonbloy Oct 14 '13 at 15:46
• 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. – Cristian Eduardo Lehuede Lyon Oct 14 '13 at 15:50
• – leonbloy Oct 14 '13 at 16:01
• @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? – user940 Oct 14 '13 at 16:12

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.
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.