# Why a positive recurrent Markov chain implies positive limiting probability?

Let $X=\{X_0,X_1,\ldots\}$ be an irreducible, positive recurrent, and aperiodic Markov chain with the state space $S=\{0,1,2,\ldots\}$ then how do we show that the probability $$\lim_{n\rightarrow\infty}P(X_n=k\mid X_0=j)$$ is independent of $j$ and strictly positive? That is, why $$\lim_{n\rightarrow\infty}P(X_n=k\mid X_0=j)=g(k)>0?$$

• You need the Markov chain to be irreducible and aperiodic, not just positive recurrent. – Robert Israel Sep 28 '14 at 22:24
• @ Robert: You are absolutely right. I forgot to mention that in my question but now I add this assumption too. Thanks! – user176667 Sep 28 '14 at 22:25
• @Inj You really need it to be aperiodic as well, otherwise the limit may not exist. – Sasha Sep 28 '14 at 22:33
• @ Sasha: Yes, that's true. – user176667 Sep 28 '14 at 22:34