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Alice is quite popular. She gets called on her cell-phone in a Poisson$(\lambda)$ manner. She answers her calls when possible, and ignores them when in the middle of conversation. Since you know her personally, you know that the length of her conversation is distributed by r.v. $X$, independently of the discussion/person.

Now, the first phone-call she receives will surely get through to her w.p. 1. However after some time both distributions (exponential and $X$) will "mix", and also the probability that some call gets through to her should start converging towards some value $p$. I'm most interested in $p$ and in the so-called "mixing time".


The answer should be in terms of moments (mean, variance, etc.) of $X$, and of course $\lambda$. But if the question is too general, I'd also be happy with some specific answer, say when also $X$ is $\lambda'$-exponentially distributed.

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    $\begingroup$ I think you could either a. use the renewal theorem, for which a good source is Feller vol 2. He does a similar example with a Geiger counter which is locked for a fixed amount of time after an event, or b. make a long run argument which gives me $ p = \frac 1 {1 + \lambda E(X)}$ $\endgroup$
    – user83457
    Dec 15, 2017 at 11:50
  • $\begingroup$ Thank you @michael, long run argument indeed easily shows the mentioned value, which corresponds to the expected fraction of time the line is free. Also, I'm guessing that the "mixing time" is not relevant in this case, by the fact that the calls arrive in Poisson manner, is this correct? $\endgroup$ Dec 16, 2017 at 10:36

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