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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 .

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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 $\pi_i=\frac{1}{23}$.

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  • $\begingroup$ yes please provide some more hint . i still didn't understand $\endgroup$ – ANUJ NAIN Apr 19 '17 at 4:25
  • $\begingroup$ @ANUJNAIN added details $\endgroup$ – matus Apr 19 '17 at 13:43
  • $\begingroup$ howcome $\lim_{n\rightarrow \infty} \pi_0 P^n $became $ \lim_{n\rightarrow \infty} P^n $ why does $\pi_0$ vanish ? ...... $\endgroup$ – ANUJ NAIN Apr 19 '17 at 18:20
  • $\begingroup$ can you solve this question please . i am not able to understand your words $\endgroup$ – ANUJ NAIN Apr 19 '17 at 18:21
  • $\begingroup$ See the wikipedia section on steady state analysis for more information. $\endgroup$ – matus Apr 19 '17 at 18:55

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