Timeline for Vorticity Matrix for Markov chain

6 events

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Jun 12, 2020 at 22:16	comment	added	user121747		That means your Q matrix represents a reversible Markov chain. In this case, $\Gamma$ is indeed zero. Actually, one checks whether a Markov chain is reversible by checking if $[\pi]Q=Q^\top[\pi]$. I am not aware of how to construct a non-zero $\Gamma$ matrix in this case.
Jun 6, 2020 at 10:19	comment	added	user3322017		Since I can't add image of graph in comments, I am giving vertices and edges for graph. Vertex set V = {1,2,3,4,5}. Edge set E = { (1,2), (1,4), (2,5), (4,5), (3,4) }. All edges are undirected
Jun 6, 2020 at 10:16	comment	added	user3322017		$\Gamma = [\pi]Q - Q^{\top}[\pi]$ , turns out to be zero matrix for simple random walk in which neighbour node is chosen uniformly $$ Q = \left( \begin{matrix} 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \end{matrix} \right) $$ $$ \pi = \left( \begin{matrix} \frac{2}{10} & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{10} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{10} & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{10} & 0 \\ 0 & 0 & 0 & 0 & \frac{2}{10} \end{matrix} \right) $$
May 27, 2020 at 19:12	review	Late answers
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May 27, 2020 at 18:54	history	answered	user121747	CC BY-SA 4.0