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We have two urns (blue and red) that are connected, and two particles, $p_1$ and $p_2$, are traveling between these urns independently. The amount of time $Z_1$ that $p_1$ spends in blue urn is iid with distribution $\exp{(\lambda_1)}$, and the amount of time it spends in red urn $Y_1$ is iid with distribution $F$. The amount of time $Z_2$ that $p_2$ spends in blue urn is iid with distribution $\exp(\lambda_2)$, and the amount of time it spends in red urn $Y_2$ is iid with distribution $G$. Both $F$ and $G$ are nonlattice.

I look at the urns at time $t$ and see $p_1$ is in the blue urn and $p_2$ is in the red urn. What is the expected time $E[T]$ until for the first time, both particles are in the blue urn ($t\to\infty$).

If we had just $p_2$, the mean time until for the first time it is in the blue urn is: $$E[T']=\frac{E[Y_2^2]}{2E[Y_2]}$$ but here we have two renewal processes.

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  • $\begingroup$ Excellent question--but I guess you will have to explain in more details your ideas about it... $\endgroup$ – Did Jan 6 '14 at 18:20
  • $\begingroup$ I am thinking of modeling it as a four state semi-Marcov process, with states: \{br,rb,bb,rr\}, where bb is the absorbing state. Beginning from state br, how long does it take until the semi-Markov process is absorbed. $\endgroup$ – Mah Jan 6 '14 at 20:47
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    $\begingroup$ When F and G are not exponential, I am not sure this would be even semi-Markov... $\endgroup$ – Did Jan 6 '14 at 21:28
  • $\begingroup$ You are right, it is not semi-Markov. $\endgroup$ – Mah Jan 9 '14 at 19:21

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