There is an exercise that I'm trying to do:
$\pi$ is a stationary distribution if and only if $\pi \textbf{1} = \textbf{1} $, and $F(A,A^c)=F(A^c,A)$, where the “probability flow” from a set $A$ of states to its complement under a distribution $\pi$ is $$F(A,A^c) = \sum_{i \in A } \sum_{j \in A^c } \pi(i)p_{i,j}$$
My attempt 1:
If we suppose that $\pi$ is stationary then the balanced equations hold, and fixing a set $A \subset S$, we have $$\pi = \sum_{j \in S} \pi(j) p_{j,i} = \sum_{j \in A} \pi(j) p_{j,i} + \sum_{j \in A^c} \pi(j) p_{j,i} $$
then I tried to multiply $\pi(j)$ by $\sum_{j \in S} p_{j,i}$, then $$\pi \sum_{j \in S} p_{j,i}= \sum_{j \in S} \pi(j) p_{j,i} = \sum_{j \in A} \pi(j) p_{j,i} (\sum_{j \in S} p_{j,i})+ \sum_{j \in A^c} \pi(j) p_{j,i}(\sum_{j \in S} p_{j,i}) $$
and then I got stuck, How can I simplify this? I think $\sum_{j \in S} p_{j,i}$ can be removed but I do not how to merge this term in the sum.
My attempt 2:
If I suppose that the probability flow equation hold, then if I fix $A = \{i\}$, then
$$F(A,A^c) = \sum_{i \in A } \sum_{j \in A^c } \pi(i)p_{i,j} = \sum_{i \in A } \sum_{j \in A^c } \pi(j)p_{j,i}= F(A^c,A)$$
$$ \Rightarrow \sum_{i \in A } \sum_{j \in A^c } \pi(i)p_{i,j} - \sum_{i \in A } \sum_{j \in A^c } \pi(j)p_{j,i} = 0$$
Here, can I group the transition matrices?
Could someone help me to solve this, please. Thanks for your time and hints, everyone.