Equilibrium distribution of Ehrenfest's urn

Question

(I'll post my own answer to this, but others may be of interest, so post your own if you have one.) (PS: In reply to comments posted below: Stackexchange encourages posting an answer to one's own question.)

Paul and Tatiana Ehrenfest posed the problem of two urns containing some marbles. At each step in a random process, one marble is chosen randomly, all having the same chance of being chosen. The chosen marble is moved from whichever urn it's in to the other one. The number of marbles in the first urn is the state of a Markov chain. The question is: what is the equilibrium distribution of this Markov chain? The answer is: it is the binomial distribution of the number of successes in $n$ trials with probability $1/2$ of success on each trial, where $n$ is the number of marbles in the two urns. So now the question is: how do you prove that?

Answer 1

My preferred way to answer this is to consider the related Markov chain, whose state at any time is not the number, but the set of marbles in the first urn. By symmetry, one shows that in equilibrium, all subsets of the whole set of marbles are equally probable.

Thus the probability that the set has $k$ of the $n$ marbles is the number of subsets of size $k$ divided by the whole number of subsets of the set of $n$ marbles.

Answer 2

Consider the Markov chain in continuous time on the discrete cube $\{0,1\}^n$ such that each coordinate changes independently of the others at rate $1$. The coordinates are independent Markov chains on $\{0,1\}$, and each converges in distribution to the uniform distribution on $\{0,1\}$, hence the Markov chain on $\{0,1\}^n$ converges in distribution to the uniform distribution on $\{0,1\}^n$, in particular, the process of the number of ones, known as the Ehrenfest chain, converges in distribution to the sum of $n$ independent uniform Bernoulli random variables, which follows the binomial distribution $\left(n,\frac12\right)$, QED.

The argument shows that the Ehrenfest chain requires at most $n\log n$ steps to converge to the uniform distribution, in total variation, starting from any initial distribution.