The theorem is "If a transition matrix for an irreducible Markov chain with a finite state space S is doubly
stochastic, its (unique) invariant measure is uniform over S."
If a Markov Chain has a doubly-stochastic transition matrix, I read that its limiting probabilities make up the uniform distribution, but I do not quite understand why.
I have been trying to come up with, and locate, an understandable proof for this. But the proofs I find all gloss over details I don't understand, like proposition 15.5 here (why does it work to just use the [1,…1] vectors?) Could someone point me to (or write) a more simple/detailed proof?
(Though not part of anything I will hand in at school, it is part of a course I'm taking so I guess I'll tag it with homework in either case.)
Best Answer
Suppose we have an $M+1$-state irreducible and aperiodic Markov chain, with states $m_j$, $j=0,1,\ldots, M$, with a doubly stochastic transition matrix (i.e., $\sum_{i=0}^M P_{i,j}=1$ for all $j$). Then the limiting distribution is $\pi_j=\frac{1}{M+1}$.
Proof
First note that the $\pi_j$ is the unique solution to $\pi_j=\sum_{i=0}^M \pi_iP_{i,j}$ and $\sum_{i=0}^M\pi_i=1$.
Try $\pi_i=1$. This gives $\pi_j=\sum_{i=0}^M \pi_iP_{i,j}=\sum_{i=0}^M P_{i,j}=1$ (because the matrix is doubly stochastic). Thus $\pi_i=1$ is a solution to the first set of equations, and to make it a solution to the second normalize by dividing by $M+1$.
By uniqueness, $\pi_j=\frac{1}{M+1}$.