# Stationary distribution of an infinite Markov chain

The exercise is asking for the stationary distribution, the estimated time to get from state $$0$$ to state $$4,$$ and to conclude if the chain is time-reversible.

So I have the following transition matrix:

$$P=\left[\begin{matrix} 1/2&1/2&0&0&0&\cdots\\ 1/2&0&1/2&0&0&\cdots\\ 1/2&0&0&1/2&0&\cdots\\ 1/2&0&0&0&1/2&\cdots\\ 1/2&0&0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{matrix}\right]$$

This means $$P_{i,0} =1/2,$$ $$P_{i,i+1} = 1/2,$$ and $$P_{ij}=0$$ in every other case.

I have progressed up to this point:

\begin{align*} n_1&=\frac12\,n_1+\frac12\,n_2+\frac12\,n_3+\dots\\ n_2&=\frac12\,n_1\\ n_3&=\frac12\,n_2\\ &\vdots\\ n_k&=\frac12\,n_{k-1}\\ \sum_{k=1}^\infty n_k&=1. \end{align*}

I'm not sure how to solve the problem, though.

Any help would be very much appreciated.

We can rewrite manipulate the first equation by noticing that $$\pi_1+\pi_2+\pi_3+\cdots=1$$. This gives us the following:
\begin{align} \pi_1&=\frac12\pi_2+\frac12\pi_3+\frac12\pi_4+\cdots\\ &=\frac12\left(\pi_2+\pi_3+\pi_4+\cdots\right)\\ &=\frac12(1-\pi_1) \end{align}
Solving $$\pi_1=\frac12(1-\pi_1)$$ gives us $$\pi_1=\frac13$$. Then, each term is half of the previous one ($$\pi_n=\frac12\pi_{n-1}$$), so we have $$\pi_n=\frac13\left(\frac12\right)^{n-1}$$.