Users arrive according to Poisson process with rate λ. If every third user is removed, then do the remaining users form a Poisson process with rate 2λ/3? If every other user is removed, then do the remaining users form a Poisson process with rate λ/2? and so on..?
2026-03-26 00:58:51.1774486731
Poisson arrival and selective removal
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No. If they are removed with probability $p$ independently of each other and independently of the Poisson process, then one does indeed get another Poisson process, this time with rate $\lambda(1-p)$.
But suppose one removes alternate "users". In a Poisson process, the waiting time from one user to the next has a memoryless exponential distribution. If the average waiting time is one hour then the probability the next user arrives in the next minute, given the time one has waited so far, does not depend on the time one has waited so far. But with this alternate-removal process, one would have one non-removed arrival every two hours on average. So consider the probability of a non-removed arrival in the next minute, given that one has only waited ten minutes. It is improbable that the first removed arrival has occurred because ten minutes is a short time, hence improbable the next non-removed arrival will be in the next minute. Now suppose one has waited $2.5$ hours so far. Then it is probable that the first removed arrival has already arrived, hence more probable that the next non-removed arrival is in the next minute, than it would be if one had just waited ten minutes.
Since the waiting time to the next non-removed arrival is thus not memoryless, one does not have a Poisson process.