Recovering components of a Poisson process Consider a scenario with two independent input Poisson processes with rates $P_1 \sim Pois(\lambda_1)$ and $P_2 \sim Pois(\lambda_2)$. These enter a queueing system to form a combined process of rate $\lambda_1 + \lambda_2$ and then incur some additional exponentially-distributed service delay (say with mean $1/\mu$). If the queue is M/M/1 and $\mu > \lambda_1 + \lambda_2$, the output process should also be Poisson with rate $\lambda_1 + \lambda_2$. If an observer of the output process observes just the samples corresponding to $P_1$ (resp. $P_2$), does she see a Poisson process with rate $\lambda_1$ (resp. $\lambda_2$)? If not, what does this process look like?

Yes, she sees a Poisson process. This is because the customers leave in the same order they entered. The input process is a process with rate $\lambda_1+\lambda_2$ in which each customer is independently marked as type $i$ with probability $\frac{\lambda_i}{\sum_i\lambda_i}$. The customers leave in the same order, so the output process is again a process with rate $\lambda_1+\lambda_2$ in which the customers are marked with the same distribution, and thus it can be considered as composed by two Poisson sub-processes, just like the input process. The fact that the marking is coupled to the marking of the input process doesn't change this (just like the fact that the combined output process is coupled to the combined input process doesn't change the fact that it looks like a Poisson process to the observer).