Two-state excursion visit law
= Two-state excursion visit law
Let $a=\mathbb P_i(T_j<T_i)$ and $b=\mathbb P_j(T_i<T_j)$. Before returning to $i$, the chain visits $j$ zero times with probability $1-a$ and, for $k\geq1$, exactly $k$ times with probability
$$
ab(1-b)^{k-1}.
$$
When $b>0$, the expected number of visits is $a/b$.