Occupation time of a continuous-time Markov chain (${\int }_0^{t}1\{Z(u)=s\}du$)

The occupation time records the duration spent in a state of a continuous-time Markov chain. Conditional on $Z(0)=r$, its expected value is ${\int }_0^t{p}_{rs}(u)du$, where $p_{rs}$ is the transition probability.

For a continuous-time Markov chain with absorbing endpoints $0,N$ and rates $q_{j,j\pm 1}={\lambda }_j/2$, the jump chain is a simple symmetric random walk. On leaving $i$, absorption before returning has probability $p_i=\frac{1}{2}(1/i+1/(N-i))$, by gambler's ruin on each side. The geometric return count before absorption and independent rate-${\lambda }_i$ holding times give ${\mathbb{E}}_{i}I=1/(p_i{\lambda }_i)$. The rates at other states affect elapsed excursion times but not their return probabilities.

For a finite-state homogeneous continuous-time Markov chain with generator matrix $Q$, integrate the indicator of occupying state $j$ over $[0,T]$. Interchanging the bounded time integral and expected value gives the displayed integral of transition probabilities. It counts all visits, not only the first holding time. Summing over states gives $T$. A zero eigenvalue of $Q$ does not cause a problem for this integral, although it prevents casually writing it with an inverse matrix $Q^{-1}$.