# Limiting distribution of a Markov Chain

The problem can be found in Markov Chains - dice problem.

You start with five dice. Roll all the dice and put aside those dice that come up 6. Then roll the remaining dice, putting aside those dice that come up 6. And so on. Let $$X_n$$ be the number of dice that are sixes after n rolls.

Question: Prove that $$X_n$$ has a limiting distribution

In using the notations and results of Markov Chains - dice problem, we have that the transition matrix, denoted, $$P$$ is diagonalizable. So $$P=A D A^{-1}$$ with
$$D = \begin{pmatrix} \left(\frac{5}{6}\right)^{5} & 0 & 0 & 0 & 0 & 0 \\ 0 & \left(\frac{5}{6}\right)^{4} & 0 & 0 & 0 & 0 \\ 0 & 0 & \left(\frac{5}{6}\right)^{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & \left(\frac{5}{6}\right)^{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \left(\frac{5}{6}\right) & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

It implies that $$P^{n}$$ converge to $$A D^{*} A^{-1}$$ with

$$D^{*} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

However this is not enough to prove that the Markov Chain has a limiting distribution. We also need to prove that $$A D^{*} A^{-1}$$ has identical rows to conclude. And I am not sure how to show that.
Another way to prove that the Markov Chain has a limiting distribution would be to use the fact that it is a Doeblin Chain. But I am looking for more elegant solutions.

• What exactly does the product $AD^{*}A^{-1}$ look like? You can determine this without doing any calculation, because you can use either the definition of matrix multiplication or reasoning about its eigenvalues to establish that it contains a great many zeros. – whuber Nov 23 '19 at 22:43
• For a finite state Markov chain, a limiting distribution exists iff it is irreducible and aperiodic. This is the most "elegant" solution I can think of. – Math1000 Dec 19 '19 at 2:21