# Average service time in a finite state Markov chain system

Consider a system with a large number of individuals that can belong to a finite number of states $$s_1,s_2,\dots,s_N$$. Transitions between states follow a Markov chain with stationary distribution $$\pi$$, so if an individual is randomly sampled from the system then his state is drawn from $$\pi$$.

Each individual needs to periodically update his or her personal information, and in state $$s_i$$ the time until the next update is given by $$f(s_i)$$, which is assumed to be deterministic and finite.

More precisely, if an individual starts in state $$s_i$$ at time 0, then the first updating of information takes place at time $$f(s_i)$$, at which time the individual new state, $$s$$, is drawn from the $$f(s_i)$$-step Markov chain distribution, and with time $$f(s)$$ until the next update.

If we look at the average time between updates for a single individual in a time interval $$[0,T]$$ with $$T\to\infty$$ (i.e., the long-run average time between updates), then it can be shown to be equal to \begin{align*} \mu = \sum_{i=1}^N\pi(s_i)\times f(s_i), \end{align*} irrespective of the starting state.

Is this quantity also the most natural measure of the average time between updates for all individuals in this system (i.e., the "overall" average time between updates), even if we consider a finite $$T$$? That is, if we consider a time interval $$[0,T]$$ where $$T$$ is large compared to the times between updates $$f(s_i)$$, $$i=1,\dots,N$$, but still finite.