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Write $q_{ij}=c{m}_{ij}$ for $i\ne j$, with a common positive constant $c$, and put $d_i={\sum }_{j\ne i}{m}_{ij}$. The infinitesimal generator matrix is symmetric because the graph is undirected. Thus the uniform distribution ${\pi }_i=1/|I|$ satisfies detailed balance, and the continuous-time random walk is reversible when started uniformly.

Its jump chain chooses a link uniformly from those incident to its current vertex:The flux ${\hat{\pi }}_i{\hat{P}}_{ij}=m_{ij}/{\sum }_j{d}_j$ is symmetric, so this random walk on a multigraph is also reversible. The continuous-time equilibrium is uniform because higher-degree vertices have shorter holding times; at jump times the equilibrium is degree-weighted. If the intended proportionality constant depends on the starting vertex, $q_{ij}=c_i{m}_{ij}$, reversibility still holds with ${\pi }_i\propto 1/c_i$, and the same jump chain results.

First realize the chain as a random walk on a multigraph: put three parallel edges between $a,b$, two between $a,c$, and one between $b,c$. There are six edges and vertex degrees $5,4,3$. Choosing an incident edge uniformly gives exactly the stated transition probabilities. The stationary distribution of a graph random walk is the degree divided by twelve, $\pi =(5,4,3)/12$, and detailed balance follows because each edge multiplicity contributes the same flux in both directions.

For the first positive return time, Kac's lemma givesFor passage to $b$, put $h_a=k(a,b)$ and $h_c=k(c,b)$. The first-step equations are $h_a=1+(2/5)h_c$ and $h_c=1+(2/3)h_a$, since reaching $b$ stops the clock. ThusA multigraph is essential: a simple graph on three vertices cannot have six edges.