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Users arrive according to Poisson process with rate $\lambda$. If every third user is removed, then do the remaining users form a Poisson process with rate $2\lambda/3$? If every other user is removed, then do the remaining users form a Poisson process with rate $\lambda/2$? and so on..?

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It depends on how the removal occurs: are you saying each user is removed independently with probability $1/3$ or that literally every third user is systematically removed? If the latter is the case, then consider what that does to the patterns of inter-arrival times and recall that in a Poisson process the inter-arrival times must be independent. – whuber May 2 '14 at 21:46
I am saying that every other user is removed. – user100503 May 2 '14 at 22:11
I believe the resulting process will not be Poisson after doing some research. Can anyone please verify? – user100503 May 2 '14 at 22:59
up vote 2 down vote accepted

No, deterministic deletion of arrivals (such as "delete every third arrival") does not leave a Poisson process, but random deletion does, e.g. for every arrival, make a decision (independent of all past and future decisions) to delete with probability $\frac{1}{3}$ and to not delete with probability $\frac 23$. Then what is left after such random deletions is a Poisson process with rate $\frac{2\lambda}{3}$. Read, for example, the last paragraph of this answer.

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whuber pretty much covered it already in his comment, but I'll give a slightly different way of thinking about it. I'll try to come back with a slightly longer one soon

Short answer:

The interarrival time is now exponential($\lambda$) for some users and Gamma(2,$\lambda$) for others (time to removed user, plus time from removed user to next user). It's therefore not a Poisson process.

[If the removals were random rather than every third user, however, it would be a Poisson process.]

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