# Finding steady-state probability of a Markov chain

Let $X_{n}$ be a Markov chain on state space $S = \{ 1,2 \dots, 23 \}$ with transition probability given by

$p_{i,i+1}= p_{i,i-1} = \frac {1}{2} \ \ \forall \ 2\le i \le 22 ,$

$p_{1,2}= p_{1,23} = \frac {1}{2}$

$p_{23,1}= p_{23,22} = \frac {1}{2}$

then we need to show that $P(X_n=i) = \frac {1}{23}$.

attempt :

( i thought of many results that i know but i could not figure it out )

i tried to solve equations

$\pi_1 = \frac {1}{2} \pi_2 + \frac {1}{2} \pi_{23}$

$\pi_2 = \frac {1}{2} \pi_1 + \frac {1}{2} \pi_3$

..

..

$\pi_{23} = \frac {1}{2} \pi_1 + \frac {1}{2} \pi_{22}$

but this seems confusing. Please suggest a proper method .

Note that $p(X_n=i)=\frac{1}{23}$ suggests that the distribution is independent of $n$ a result which You already used implicitly when you wrote $\pi P=\pi$ where $\pi$ is a vector and $P$ is the $23\times23$ transition matrix. This suggests that $\pi_n$ converge towards stationary distribution as $n\rightarrow \infty$ and that $\pi$ is the steady-state probability. Consider how You would compute $\pi$ as a result of infinite number of transitions. In particular, consider that $\pi_n=\pi_0 P^n$ and that $\lim_{n\rightarrow \infty} \pi_0 P^n= \lim_{n\rightarrow \infty} P^n= \pi$. You can then use the last equality to compute $\pi$ and to show that $\pi_i=\frac{1}{23}$.
• howcome $\lim_{n\rightarrow \infty} \pi_0 P^n$became $\lim_{n\rightarrow \infty} P^n$ why does $\pi_0$ vanish ? ...... – ANUJ NAIN Apr 19 '17 at 18:20