Below problem is excerpted from Stochastic Processes (2e, Ross). The solution for 5.12(b) can be found here.
5.12 Suppose that the “state” of the system can be modeled as a two-state continuous-time Markov chain with transition rates $\nu_0=\lambda$, $\nu_1=\mu$ . When the state of the system is $i$, “events” occur in accordance with a Poisson process with rate $\alpha_i$, $i=0, 1$. Let $N(t)$ denote the number of events in $(0, t)$.
(b) If the initial state is state $0$, find $E\big(N(t)\big)$.
I am not sure about why $E(N(t))=\alpha_0 E(T_0(t))+\alpha_1 E(T_1(t))$, where $T_i(t)$ stands for the time spent in state $i$ before time $t$. It is quite obvious for Poisson process. But I fail to understand it in this case. (I know little about renewal theory, so is it possible to circumvent it?) Also, I think it is a Doubly stochastic Poisson Process (Cox process) instead of the typical conditional Poisson process. Is it true?
Sketch of the solution: Let $M(t) = E(N_t|X_0=0)$. Conditioning on the state at $t$, we have $M^{'}(t)=\alpha_0P_{00}(t)+\alpha_1 P_{01}(t).$
The transition probability $P_{0i}$ can be solved via the Kolmogorov forward equation.