A time reversible Markov chain problem on urns Question: (Ross Probability Models, Ch. 4, Ex. 70) A total of $m$ white balls and $m$ black balls are distributed into two urns such that each urn contains $m$ balls. At each stage, a ball is selected from either urn, and the selected balls are switched. Let $X_n$ denote the number of black balls in the first urn after the $n$th switch. Give (a) the transition probabilities, (b) find/guess the limiting probabilities of the chain without computation, and (c) actually compute the limiting probabilities and show that the chain is time reversible.
Problem: I don't know how to reason through (b), and I can't find a simple form solution for (c).
Attempt: I will first do part (a). Let $P$ be the transition matrix and $\pi$ the limiting probabilities vector. Then
$$P_{ii}=2\left(\frac{i}{m}\right)\left(\frac{m-i}{m}\right)$$
Since we choose $i$ blacks from the first urn and $m-i$ blacks from the second, or vice versa regarding whites.
$$P_{i,i+1}=\left(\frac{m-i}{m}\right)^2$$
Since we choose from the $m-i$ whites from the first urn and $m-i$ blacks from the second. Similarly, then,
$$P_{i,i-1}=\left(\frac{i}{m}\right)^2$$
Noting that $P_{i,i-1}+P_{ii}+P_{i,i+1}=1$, all other entries in $P$ for each $i$ is $0$.
Now I will solve (c). The chain is time reversible because for any states $i,j$, it's not possible to witness $i$ followed by $j$ twice without witnessing $j$ followed by $i$ someplace in between. Thus, we have
$$\pi_iP_{i,i-1}=\pi_{i-1}P_{i-1,i}\Longrightarrow\pi_i=\pi_{i-1}\frac{P_{i-1,i}}{P_{i,i-1}}=\pi_{i-1}\left(\frac{m-i+1}{i}\right)^2$$
Then
$$\pi_i=\pi_0\prod_{j=1}^i\left(\frac{m-j+1}{j}\right)^2$$
And
$$\sum_{k=0}^m\pi_k=\pi_0\sum_{k=0}^m\prod_{j=1}^k\left(\frac{m-j+1}{j}\right)^2=1$$
Which gives me a very messy solution for $\pi_i$ for all $i$. Unfortunately, I can't think of any way to simplify this with the squared term, but for various reasons I suspect that the solution has a nice representation.
Update: We have that
$$\pi_0\sum_{k=0}^m\prod_{j=1}^k\left(\frac{m-j+1}{j}\right)^2=\pi_0\sum_{k=0}^m{m\choose k}^2=\pi_0\sum_{k=0}^m{m\choose k}{m\choose{m-k}}=\pi_0{{2m}\choose m}=1$$
Or
$$\pi_0=\frac1{{2m}\choose m}$$
And
$$\pi_i=\pi_0\prod_{j=1}^i\left(\frac{m-j+1}{j}\right)^2=\pi_0{m\choose i}^2=\pi_0{m\choose i}{m\choose{m-i}}=\frac{{m\choose i}{m\choose{m-i}}}{{2m}\choose m}$$
 A: I think you already have a decent answer to the problem. I would love to share another perspective of view.
For (b), since the balls are uniformly chosen each time, it is natural to conjecture that in the long run, the distribution of the balls in two urns is as if we uniformly choose $m$ balls from $m$ white balls and $m$ black balls, and put these $m$ chosen balls in urn 1 and the rest of $m$ balls in urn 2.
Therefore, if the conjecture is true, we then have
$$
\pi_i = \frac{ {{m}\choose{i}} {{m}\choose{m-i}} }{ {{2m}\choose{m}} },
$$
where there are in total ${{2m}\choose{m}}$ ways of choosing $m$ balls from $2m$ balls (ignoring order) and to have $i$ black balls in urn 1, we have to choose $i$ black balls (from $m$ black balls in total) and $m-i$ white balls (from $m$ balls in total).
So far, it is merely an intuitive argument.
For (c), to verify that the $\{\pi_i\}$ are indeed the stationary probabilities, since the Markov chain is inreducible and positive recurrent, the stationary equations have only a unique solution, therefore it suffices to verify that $\{\pi_i\}$ satisfy the stationary equations.
In particular, the Markov chain can only transition to neighbouring states, (also as indicated in the problem that the MC is time reversible) so we are encouraged to verify
$$
\pi_{i} P_{i j}=\pi_{j} P_{j i} \quad \text { for all } i, j, 
$$
and
$$
\sum_{i=0}^{m} \pi_{i} = 1 .
$$
If this is true, we know that (1) the MC is indeed time reversible and (2) $\{\pi_i\}$ are indeed the stationary probs. (See, for example, Section 4.8 of Ross Probability Models 12th Edition.)
The former eqn can be verified after a little algebra, and the latter eqn follows from our previous probabilistic argument (for (b)).
