She gives an amazingly clear description of why, on a random walk on an undirected graphs, the stationary distribution puts probability on each node proportional to its degree.
Houston-Edwards also relies without proof on the following fact: The expected time to start and return to a vertex v is 1/p where p is the probability given to v in the stationary distribution. Why should that be true?
Didn't seem so obvious to me, so I looked it up. Here is an informal proof:
Let V1 be the random variable that is the expected length of time to get from v back to v. Let's take that walk again and let V2 be the expected length of time to get from v back to v the second time, and so on. Let An=V1+V2+..+Vn, the random variable representing the number of transitions taken to return to v n times. In a Markov chain each Vi is distributed the same so E(An) = n E(V1).
As n gets large the Markov chain approaches the stationary distribution and will be in state v about a p fraction of the time. After E(An) transitions we should return to v about p E(An) times, where we actually return n times. So we have p E(An) approaches n, or p n E(V1) approaches n and the only way this could happen is if E(V1)=1/p.