91Ó°ÊÓ

We have a Markov chain whose state at any given time is represented by a vector \(\left(n_1, \ldots, n_m\right)\), where \(n_i\) denotes the number of balls in urn \(i\). The state space is given by \(S = \{(n_1,\ldots, n_m) \mid \sum_{i=1}^m n_i = M\text{ and }0 \le n_i \le M \text{ for all }i\}\).

A natural guess for the limiting probabilities would be the uniform distribution. We hypothesize that all states in the Markov chain are equally likely, so the limiting probabilities would be given by \[\pi\left(n_{1}, \ldots, n_{m}\right) = \frac{1}{\binom{M + m - 1}{m - 1}}.\] The reason for this choice is that there are a total of \(\binom{M + m - 1}{m - 1}\) ways to distribute the \(M\) balls and \(m\) urns, and we are assuming that each configuration is equally likely.

To confirm our guess, let's show that the detailed balance equations hold for the proposed limiting probabilities: \[\pi\left(n_{1}, \ldots, n_{m}\right)Q\left(\left(n_{1}, \ldots, n_{m}\right),\left(n_{1}', \ldots, n_{m}'\right)\right) = \pi\left(n_{1}', \ldots, n_{m}'\right)Q\left(\left(n_{1}', \ldots, n_{m}'\right),\left(n_{1}, \ldots, n_{m}\right)\right).\] Let's analyze the transition probabilities between two adjacent states, \((n_1, \ldots, n_m)\) and \((n_1',\ldots,n_m')\). There are two ways to make a transition between these two states: 1. Move a ball from urn \(i\) to urn \(j\) (where \(i \neq j\)): In this case, \(n_i' = n_i - 1\), \(n_j' = n_j + 1\), and \(n_k' = n_k\) for \(k \neq i,j\). The transition probability is given by \[Q\left(\left(n_{1}, \ldots, n_{m}\right),\left(n_{1}', \ldots, n_{m}'\right)\right) = \frac{n_i}{M}\frac{M-n_j}{M-1} = \frac{n_i(M-n_j)}{M(M-1)}.\] 2. Move a ball from urn \(j\) to urn \(i\) (where \(i \neq j\)): In this case, \(n_i' = n_i + 1\), \(n_j' = n_j - 1\), and \(n_k' = n_k\) for \(k \neq i,j\). The transition probability is given by \[Q\left(\left(n_{1}', \ldots, n_{m}'\right),\left(n_{1}, \ldots, n_{m}\right)\right) = \frac{n_j}{M}\frac{M-n_i}{M-1} = \frac{n_j(M-n_i)}{M(M-1)}.\] Now, let's plug in the values for the proposed limiting probabilities: \[\begin{aligned} &\pi\left(n_{1}, \ldots, n_{m}\right)Q\left(\left(n_{1}, \ldots, n_{m}\right),\left(n_{1}', \ldots, n_{m}'\right)\right)\\ =&\frac{1}{\binom{M + m - 1}{m - 1}}\frac{n_i(M-n_j)}{M(M-1)}\end{aligned}\] and \[\begin{aligned} &\pi\left(n_{1}' , \ldots, n_{m}'\right)Q\left(\left(n_{1}', \ldots, n_{m}'\right),\left(n_{1}, \ldots, n_{m}\right)\right)\\ =&\frac{1}{\binom{M + m - 1}{m - 1}}\frac{n_j(M-n_i)}{M(M-1)}, \end{aligned}\] which indeed satisfy the detailed balance equations: \[\frac{1}{\binom{M + m - 1}{m - 1}}\frac{n_i (M - n_j)}{M (M - 1)} = \frac{1}{\binom{M + m - 1}{m - 1}}\frac{n_j (M - n_i)}{M(M - 1)}.\]

Since we have shown that the detailed balance equations hold for the limiting probabilities, we can now conclude that the Markov chain is time-reversible. In other words, the probability of moving from a given state to another is the same as the probability of moving back to the original state when considering the limiting probabilities. As a result, we have shown that the limiting probabilities for this Markov chain are given by the uniform distribution, and the Markov chain is indeed time-reversible.