Let $\pi$ be a stationary distribution of a Markov Chain. Show that if $\pi (x) >0$ and x leads to y, then $\pi (y) >0$.
I understand that since x leads to y, $P^{n}(x,y)>0$ for some $n < \infty$. By definition, the stationary distribution satisfies that $\pi(y) = \sum_{i} \pi(i)P^{n}(i,y)$. I think I can break this up as $\pi(y) = \pi (x) P^{n}(x,y) +\sum_{i \neq x} \pi(i)P^{n}(i,y)$. I understand the first part of this sum is greater than 0 by assumptions and definitions stated earlier, but how can I prove or how do I know the second part of this sum is greater than 0?
Please let me know if you know! Thanks!
$\pi(y) = \sum_{i} \pi(i)P^{n}(i,y) \geq \pi (x)P^{n} (x,y)$ since $\pi(x)P^{n}(x,y)$ is one of the terms in the sum and all the terms are non-negative. Hence, $\pi(y)>0$.