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There are two bus lines (A & B) that stop at a bus station. The time until the next bus arrives is distributed as an exponential random variable with parameter λ buses/hour for both bus lines together. It is equally likely for a bus to arrive from line A and line B. What is the distribution for the waiting time for a bus from each line to arrive?

Attempt: I understand that after the first bus arrives, the number of arrivals it takes for a bus from the other line to come is distributed as a geometric RV (let's call it X). I think the distribution of the waiting time for a bus from each line to arrive would be given by the sum of the exponential RV over X. I'm not sure where to go from here to calculate the distribution.

user3869
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Let $A$ and $B$ be the waiting times.

Let's guess that these are exponentially distributed random variables with $P(A \geq t) = P(B \geq t) = e^{-\frac12 \lambda t}$.

By this answer, we have for the r.v. $C := \min\{A,B\}$ that $P(C \geq t) = e^{-\lambda t}$. This is consistent with the assumption (if I understand it correctly).

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user66081
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  • This is the intuitive answer: together they come in a memoryless process at an expected rate $\lambda$ buses per hour, and each is as likely as the other, so they each come at an expected rate of $\frac12\lambda$ buses per hour. Because of the memorylessness, you get exponential distributions. – Henry Nov 14 '24 at 21:07