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Consider two users who arrive to a system with exponential arrival times with parameters $\lambda_a$ and $\lambda_b$. Once they arrive, the users stay in the system for an exponentially distributed time (parameters $\mu_a$ $\mu_b$) before leaving. The four exponential RVs are independent. What is the probability that A arrives before B AND leaves after B.

Letting $T_a$, $T_b$ denote the arrival times of each, and $S_a$, $S_b$ denote the service times. This probability is then $Pr(T_a < T_b \cap T_a + S_a > T_b + S_b)$ We can solve this by conditioning but it gets quite messy, is there a simpler way to tackle this? Perhaps something involving the memoryless property of the exponential distribution?

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There are three exponential time 'contests', and each is independent of the others.

First, A must arrive before B. Second (no memory), B must arrive before A finishes being served. Third (no memory), B must finish service before A finishes service.

Thus, the desired probability is the product of three ratios: $$\frac{\lambda_a}{\lambda_a + \lambda_b} \frac{\lambda_b}{\mu_a + \lambda_b} \frac{\mu_b}{\mu_a + \mu_b}.$$

For example, if both arrival rates are 1 and both service rates are 2, then the result is $(1/2)(1/3)(1/2) = 1/12.$

The following brief simulation in R illustrates to about three-place accuracy:

 ta = rexp(10^6, 1);  tb = rexp(10^6, 1)
 sa = rexp(10^6, 2);  sb = rexp(10^6, 2)
 mean(ta < tb & ta+sa > tb+sb)
 ## 0.082959
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  • $\begingroup$ Thanks I think that makes sense. Can a similar approach be used to find the expected time of the final departure? I know if I define Xa = Ta+Sa, Xb = Tb+Sb, then this will be E[Xa | Xa > Xb]P(Xa>Xb) + E[Xb | Xa < Xb]P(Xa < Xb), but this will get messy quick it seems. $\endgroup$ – user67081 May 2 '15 at 21:34
  • $\begingroup$ Unclear. Time until departure of whom and starting from when? $E(X_a) = E(T_a)+E(S_a) = 1/\lambda_a + 1/\mu_a.$ The sum of two exponentials is not gamma unless the rates are equal. PDF at arxiv.org/pdf/1307.3945 $\endgroup$ – BruceET May 2 '15 at 22:15
  • $\begingroup$ Essentially the quantity i'm looking for would be E[max(Sa+Ta, Sb+Tb)] $\endgroup$ – user67081 May 2 '15 at 23:27
  • $\begingroup$ As is the case for very many queueing problems, simulation is a lot easier than analysis. (The original problem really clever precisely because an analytic solution is not difficult.) For simulation, append 'mean(pmax(ta+sa,tb+sb))' to the first two lines in my Answer. I got 2.08. The expectation of each element of the max is 1.5, so at least that seems reasonable for the expectation of the max. $\endgroup$ – BruceET May 2 '15 at 23:41
  • $\begingroup$ I agree, simulation is far simpler, but this is a textbook problem so I'm expecting there is another trick for solving it. Unfortunately a sum of exponentials no longer has the memoryless property which would have made finding E[max(Sa+Ta,Sb+Tb)] easy $\endgroup$ – user67081 May 3 '15 at 19:46

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