Question: I wish to show the following claim.

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> Associate a graph with a Markov process${}^1$ by letting $(j,k)$ be an edge if $q(j,k) > 0$ or $q(k,j) > 0$. Show that if the graph associated with a stationary Markov process is a tree, then the process is reversible.
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A proof of this is given [here (Proposition 1.6)](https://books.google.co.uk/books?id=2BnTBwAAQBAJ&pg=PA6&lpg=PA6&dq=associated%20graph%20of%20continuous%20markov%20chain%20tree%20reversible&source=bl&ots=5Bvv5KZZqU&sig=7c2BgTFtB5Xooe5Jv3HcRNwuvHk&hl=en&sa=X&ved=0ahUKEwiphr_Qr53XAhWCoRoKHQYBBHcQ6AEIXDAJ#v=onepage&q=associated%20graph%20of%20continuous%20markov%20chain%20tree%20reversible&f=false) or [here (Lemma 1.5)](http://www.statslab.cam.ac.uk/~frank/BOOKS/book/ch1.pdf). Unfortunately, I don't completely follow the proofs. I understand the basics, eg that removing an edge from a tree disconnects it. However, I don't understand why this then reduces full balance for $j$ (or $k$) down to detailed balance for the pair $(j,k)$. There's no mention of stationarity in the proof -- I image that this is the understanding part that I'm missing...

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${}^1$ by 'Markov process' I mean a continuous-time Markov chain on a countable state space

Answer: 
Let us consider discrete-time -- the continuous-case follows similarly. Define
$$ Q(A,B) = \sum\_{x \in A} \sum\_{y \in B} \pi(x) p(x,y). $$
One can show using only stationarity (see, eg, [Lemma 1.4](http://www.statslab.cam.ac.uk/~frank/BOOKS/book/ch1.pdf)), for any set $S$, that
$$ Q(S,S^c) = Q(S^c,S). $$
This *does not* require reversiblity.

Now let us consider our problem. Take $i$ and $j$ with $(i,j)$ an edge. Let $S$ be the set of all vertices 'on one side of $j$', along with $j$, ie the set of all vertices that are connected to $j$ not using the edge $(i,j)$. Then the equality
$$ Q(S,S^c) = Q(S^c,S) $$
*does* reduce to the detailed balance equation for $i$ and $j$.

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This is fairly straightfoward, but I do believe that it isn't adequately described in the linked answers -- indeed, one says to take $S$ to be $j$, which *isn't* what one wants to do (or, at least, I can't see how to do it using that). Hopefully this answer can be a reference for those with a similar issue to what I had!
Let $\{X\_t:t\geqslant0\}$ be the CTMC on state space $E=\{1,2,\ldots,n\}$. It is clear that $\{X\_t\}$ is irreducible as trees are connected graphs, and so there exists a unique stationary distribution $\pi$ (as the state space is finite). Let $Q$ be the generator of $\{X\_t\}$ and let $i,j\in E$. Assume that $\mathbb P(X\_t=j)=\pi\_j$, so that the process is stationary.

If $i=j$ then trivially $\pi\_iQ\_{ii}=\pi\_iQ\_{ii}$. If $Q\_{ij}=0$ then $\pi\_iQ\_{ij}=\pi\_jQ\_{ji}=0$ (since $Q\_{ij}>0\iff Q\_{ji}>0$). If $Q\_{ij}>0$, consider the cut formed by removing edge $\{i,j\}$ from the graph - let $i\in A$ and $j\in A^c$. Because the graph is a tree, the detailed balance equation
$$
\sum\_{k\in A}\sum\_{\ell\in A^c}\pi\_kQ\_{k\ell} = \sum\_{k\in A^c}\sum\_{\ell\in A}\pi\_kQ\_{k\ell}
$$
reduces to 
$$
\pi\_i Q\_{ij} = \pi\_jQ\_{ji},
$$
and so it follows that $\{X\_t\}$ is reversible.