Expectation and limit of a stop-and-go traveler markov chain The velocity $V(t)$ of a stop and go traveler is a two-state Markov chain whose generator is given by $$
\begin{array}{cc}
&\begin{matrix}0&1\end{matrix}\\
\
\begin{matrix}0\\ 1\end{matrix}
&\pmatrix{-\alpha&\alpha\\ \beta&-\beta}
\end{array}
$$  
The distance traveled in time $t$ is the integral of the velocity:  $S(t) = \int_0^t V(u) \, du$.  
If I assume that the distribution of the velocity at time $t=0$ is given by $P\{V(0)=0\}=a=1-P\{V(0)=1\}$, how would I determine the mean of $S(t)$ and find the limit as t goes to $\infty$ of $\frac{1}{t} E[S(t)]$?  Wouldn't $S(t)=V(t)$ by the fundamental theorem of calculus?
 A: Since $V(\tau) \in \{0,1\}$, we have: 
$$ S(t) = \int_0^t V(\tau)d\tau = \int_0^t 1\{V(\tau)=1\}d\tau $$
where $1\{V(\tau)=1\}$ is an indicator function that is 1 if $V(\tau)=1$, and 0 else. Then $\lim_{t\rightarrow\infty} S(t)/t$ is the limiting fraction of time being in state $1$, which you can work out via basic steady state theory (and the answer is independent of the probability distribution at time 0).  If you like, you can argue (via the bounded convergence theorem, noting that $0 \leq S(t)/t \leq 1$ for all $t$) that (with prob 1) the answer is the same as $\lim_{t\rightarrow\infty} E[S(t)]/t$. 
As for the mean of $S(t)$ at any finite time $t$:  You can take expectations of $S(t)$ to get: 
$$ E[S(t)] = \int_0^t Pr[V(\tau)=1]d\tau $$
where we have used the fact that the expectation of an indicator function is just the probability that the function is 1. 
Computation of the above finite-horizon integral seems to require a transient analysis of the probability $V(\tau)=1$, not a steady state analysis. You can find $Pr[V(\tau)=1]$ via standard continuous time Markov chain theory using whatever differential equations you are most comfortable with. For example, define $p(t) = Pr[V(t)=1]$ and, for small $\delta>0$, write an equation for $p(t+\delta)$ in terms of $p(t)$. You will eventually get a linear ODE that you can exactly solve.
