Overdispersed Poisson Distribribution I'm interested in finding the probability distribution of the following process:
Let $X \sim Poisson(\lambda_1)$, and $Y \sim Poisson(\lambda_2)$.
If $X$ is observed, the system 'waits' for time $t_x$ before being able to observe $X$ or $Y$ again. That is, time continues to advance, but during the waiting time $t_x$, neither $X$ nor $Y$ can be observed. Similarly for $Y$ with time $t_y$. 
What is the probability of observing $X$ in the following t seconds, starting in an un-waiting state?
What is the probability that the next observation is $X$, and what is the expected time to observe it? 
I'm finding this problem difficult to solve from an analytical perspective. I've already simulated it but I'd like to solve it analytically.
Any help is greatly appreciated. I really don't know where to start with incorporating the waiting times :(
 A: I think the system may work like this:  We have two independent Poisson arrival processes $X_1(t)$ and $X_2(t)$, with rates $\lambda_1, \lambda_2$, respectively (in units of arrivals/sec).  The timeline is segmented into renewal frames $i \in \{1, 2, 3, ...\}$ that work like this: 


*

*We start each renewal frame $i$ in observation mode, which lasts until the next arrival. 

*If the next arrival is type 1, we go to sleep for time $T_1$ seconds. On the other hand, if the next arrival is type $2$, we go to sleep for time $T_2$ seconds. Here, $T_1$ and $T_2$ are fixed nonnegative constants.  

*We ignore all arrivals that occur while we are asleep.  

*The next renewal frame $i+1$ occurs when we wake up from sleep. 

The basic theory is that if $X_1(t)$ and $X_2(t)$ are independent Poisson processes of rates $\lambda_1, \lambda_2$, then the combined process is also a Poisson process, has rate $\lambda_1 + \lambda_2$, and each new arrival of the combined process is independently type 1 with prob $\lambda_1/(\lambda_1+\lambda_2)$ (independent of the history of interarrival times and types before this new arrival).  
So if we start a new renewal frame $i \in \{1, 2, 3, ...\}$, the time $Y_i$ to the next arrival is exponentially distributed with rate $\lambda_1+\lambda_2$, so it has average time $1/(\lambda_1+\lambda_2)$.  Regardless of the value of $Y_i$, the new arrival is type 1 with prob $\lambda_1/(\lambda_1+\lambda_2)$, and it is  type 2 with prob $\lambda_2/(\lambda_1+ \lambda_2)$.
Renewal theory allows us to compute the steady state fraction of time we are in the observation mode, the sleep-type-1 mode, and the sleep-type-2 mode: Let $H(t)$ be the total time we are in the observation mode during $[0,t]$, and let $S_1(t), S_2(t)$ be the total time we are in the sleep-type-1 mode and sleep-type-2 mode, respectively, during $[0,t]$.  Then:
\begin{align}
\lim_{t\rightarrow\infty} \frac{H(t)}{t} &= \frac{\frac{1}{\lambda_1+\lambda_2}}{\frac{1}{\lambda_1+\lambda_2}+ \frac{\lambda_1T_1}{\lambda_1+\lambda_2} + \frac{\lambda_2T_2}{\lambda_1+\lambda_2}} \quad \mbox{ with prob 1} \\
\lim_{t\rightarrow\infty} \frac{S_1(t)}{t} &= \frac{\frac{\lambda_1T_1}{\lambda_1+\lambda_2}}{\frac{1}{\lambda_1+\lambda_2}+ \frac{\lambda_1T_1}{\lambda_1+\lambda_2} + \frac{\lambda_2T_2}{\lambda_1+\lambda_2}} \quad \mbox{ with prob 1} \\
\lim_{t\rightarrow\infty} \frac{S_2(t)}{t} &= \frac{\frac{\lambda_2T_2}{\lambda_1+\lambda_2}}{\frac{1}{\lambda_1+\lambda_2}+ \frac{\lambda_1T_1}{\lambda_1+\lambda_2} + \frac{\lambda_2T_2}{\lambda_1+\lambda_2}} \quad \mbox{ with prob 1} 
\end{align}
Notice that these fractions of time sum to 1.
