# How to show that for any initial probability this Markov chain has the same distribution as $n \rightarrow \infty$?

I'm currently trying to solve this problem about Markov chains.

Let a Markov chain have the transition probability matrix:

$$P = \begin{bmatrix} 1 - \alpha & \alpha \\ \beta & 1 - \beta \end{bmatrix}$$

where $$|1 - \alpha - \beta| \lt 1$$.

1) Find the $$n$$-step transition matrix $$P^n$$ as $$n \rightarrow \infty$$.

2) Show that for any initial probability the Markov chain has the same probability distribution as $$n \rightarrow \infty$$.

For the first problem, I was able to find that the transition matrix is:

$$P^n = \begin{bmatrix} \dfrac{\alpha + \beta(1 - \alpha - \beta)^n}{\alpha + \beta} & \dfrac{\alpha - \alpha(1 - \alpha - \beta)^n}{\alpha + \beta} \\ \dfrac{\beta - \beta(1 - \alpha - \beta)^n}{\alpha + \beta} & \dfrac{\beta + \alpha(1 - \alpha - \beta)^n}{\alpha + \beta} \end{bmatrix}$$

and so for $$n \rightarrow \infty$$ we have:

$$P^{n \rightarrow \infty} = \begin{bmatrix} \dfrac{\alpha}{\alpha + \beta} & \dfrac{\alpha}{\alpha + \beta} \\ \dfrac{\beta}{\alpha + \beta} & \dfrac{\beta}{\alpha + \beta} \end{bmatrix}$$

If I'm incorrect, please let me know.

What's confusing me is the second problem. To be completely frank, I don't recall learning this in the course I'm taking and I'm having trouble finding resources online, which led me to ask a question here. How would I solve this problem? I'm having trouble from where to begin.

Any tips or feedback are appreciated. Thanks in advance!

• Let $\pi$ be the initial distribution, let $Q$ be your limit matrix. You have to show that $\pi Q$ does not depend on $\pi$. – kimchi lover Dec 14 '19 at 15:29

Let $$Q = \lim_{n\to\infty} P^n$$. For any initial distribution $$\alpha=(\alpha_0,\alpha_1)$$, we have $$\alpha Q=\begin{pmatrix}\alpha_0&\alpha_1\end{pmatrix}\left( \large\begin{array}{cc} \frac{\alpha }{\alpha +\beta } & \frac{\alpha }{\alpha +\beta } \\ \frac{\beta }{\alpha +\beta } & \frac{\beta }{\alpha +\beta } \\ \end{array} \right) = \left( \Large\begin{array}{cc} \frac{\alpha \alpha _0}{\alpha +\beta }+\frac{\beta \alpha _1}{\alpha +\beta } & \frac{\alpha \alpha _0}{\alpha +\beta }+\frac{\beta \alpha _1}{\alpha +\beta } \\ \end{array} \right)$$ which shows that $$\lim_{n\to\infty}\mathbb P(X_n=j\mid X_0=i)$$ is independent of $$i$$.