Suppose that we have two independent alternating renewal processes such that both alternate between states "0" and "1" independently. The amount of time each of them is in state "1" and state "0" are independent exponential random variables with rates $\lambda_1$ and $\lambda_0$, respectively.
Given that one of them is currently in state "1", what is the waiting time until both are in state "1" (for the first time).
I think we should start like this:
suppose the two processes are $U(t)$ and $V(t)$. At time $0$, $U(t)$ is in state "1" and $V(t)$ in an arbitrary state. Suppose that the time until $U(t)$ leaves state "1" is $\tau$ and the time until $V(t)$ enters state "1" is $s$; the expected waiting time is, $$E[W]=\int_{\tau=0}^{\infty}\int_{s=0}^{\tau}E[W|s<\tau]p(s)p(\tau)dsd\tau+\int_{\tau=0}^{\infty}\int_{s=\tau}^{\infty}E[W|s>\tau]p(s)p(\tau)dsd\tau\\ \int_{\tau=0}^{\infty}\int_{s=0}^{\tau}sp(s)p(\tau)dsd\tau+\int_{\tau=0}^{\infty}\int_{s=\tau}^{\infty}E[W|s>\tau]p(s)p(\tau)dsd\tau$$ I do not know how to continue.
Any help is appreciated!
Let $t_{xy}$ denote the mean time before U and V are both in state 1 knowing that, at time $0$, they are in states $x$ and $y$ respectively. Starting from U and V in any pair of states, the mean time before a transition occurs is $s=1/(\lambda_1+\lambda_2)$, then the state which changes is U with probability $p=\lambda_1/(\lambda_1+\lambda_2)$ and is V with probability $1-p$. Thus, $$ t_{10}=s+pt_{00},\quad t_{00}=s+pt_{10}+(1-p)t_{01},\quad t_{01}=s+(1-p)t_{00}, $$ hence the desired mean time $t_{10}$ is $$ t_{10}=\frac{2-p}{1-p}s=\frac1{\lambda_2}+\frac1{\lambda_1+\lambda_2}. $$