As for the puzzle I like your explanation better, I was trying to warm up my confusing math speak since its a bit rusty. The switching of certain edges with directed paths reminds me of time-ordering operators in QFT, except there you just switch the order of terms rather than replacing them with possibly more terms.

We’ve got a connected component of a weakly reversible graph, and we’re trying to show it’s strongly connected. So, given two vertices v and w, we know there’s an undirected path of edges from v to w, and we’re trying to show there’s a directed one. Look at each edge in this path—say the edge between some vertex x and the next vertex on the path, y. Either it’s pointing the right way—from x to y—or it’s not. If it’s pointing the right way, don’t mess with it. If it’s pointing the wrong way—from y to x—we know by weak reversibility that there’s a directed path going back from x to y. So, replace this edge by that directed path.

We thus get a directed path from v to w, built from the edges in the original path that we pointing the right direction, and the directed paths we used to replace the edges that were pointing the wrong direction.

(Look, ma—no subscripts!)

By the way, maybe it’s time to announce that Blake Pollard is now starting grad school at U.C. Riverside! When are you actually going there, Blake?

A general directed multigraph looks a bit like this:

Actually this is just a directed graph: it doesn’t have multiple edges going from one vertex to another. It also doesn’t have edges from a vertex to itself. But as I’ve explained, these features are irrelevant for the Markov process! When studying Markov processes, it’s enough to consider directed graphs with positive numbers labelling their edges.

The strongly connected components are shaded in blue. If we collapse each strongly connected component to a point, and combine all edges from one component to another into a single edge, we get a directed acyclic graph, meaning a directed graph without any directed cycles.

What can the equilibria of the Markov process look like? In the terminology of this post, an equilibrium is a probability distribution with

A directed acyclic graph gives a partial order on the set of vertices, where v ≤ w exactly when there exists a directed path from v to w. Let’s say a vertex v is maximal if there’s no vertex w with v < w.

So, there's a partial order on the strongly connected components of a directed graph, and we can talk about a 'maximal' strongly connected component. In this picture:

the strongly connected component containing f and g is maximal.
If you imagine what happens with a probability distribution as it evolves in time according to a Markov process with the graph, you’ll see that probability will flow into this component, but never leave it.

In general there could be a bunch of maximal strongly connected components. Probability will flow into these, but never leave.

For this reason, I think it’s obvious that an equilibrium can only be nonzero on the strongly connected components that are maximal.

So what are these like? We can write the equilibrium as a sum of pieces, each supported on a different maximal strongly connected component. (By ‘supported on’ I mean that it’s zero outside this component.)

Each of these pieces will itself be an equilibrium—since no probability can flow out, and none will be flowing in from other maximal strongly connected components, either.

So, what’s an equilibrium like if it’s supported on just a single maximal strongly connected component?

Since each strongly connected component is connected and weakly reversible, the theorem I described in this post says there’s exactly one equilibrium probability distribution supported on this component. It’s positive everywhere, and every equilibrium is a scalar multiple of this.

Conclusion: for a general Markov process on a finite set, we get one equilibrium probability distribution supported on each maximal strongly connected component. Every equilibrium is a linear combination of these. They form a basis.

By the way, stuff like \usepackage, or macros don’t work here. You get what you get and that’s all you get. According to WordPress the blog comes pre-equipped with amsmath, amsfonts and amssymb. But they don’t list all the stuff that doesn’t work. All sorts of fancy formatting commands don’t work. So don’t push your luck.

Thanks—fixed!

is a directed path from to and since the graph is weakly reversible we have a directed path path back from to for every pair of vertices in the connected component, hence it is strongly connected. What you really run into is exactly what you see in your example of an undirected path, namely vertices in the undirected path that are only the source or only the target for two different transitions, using weak reversibility on the one of these that goes with your ordering you end up with an ordered path. It’s interesting you can either choose to go along and use weak reversibility at each vertex to initially create a directed path back from to or you can do what I did and use weak reversibility to direct your undirected path and then use weak reversibility again to show that component is strongly connected. Of course you need both for strongly connected.