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A bus will depart every 10 minutes from the origin, and the time it takes to travel to station $A$ follows a Poisson distribution with expectation of 10 minutes.

  1. Alice arrives at station $A$ around 9:00 AM, and her arrival time follows a uniform distribution with $a = -10, b = 10$. What is Alice's expected wait time station $A$?

  2. Bob arrives at station $A$ at 10:00 AM on the dot, what is Bob's expected wait time?

After reading this post: Please explain the waiting paradox

I think the answer to the first question is 5, since we can swap the role of the bus and Alice in the problem in Glen_b's answer, and arrive at the midpoint of the 10-minute intervals.

But for the second problem, what we are calculating is how long after 10:00 AM does the bus arrive, can we say it's 10 minutes by somehow shift the distribution, since 10:00 AM is a constant?

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    $\begingroup$ There's something seriously weird about this question: "the time it takes to travel to station A follows a Poisson distribution with expectation of 10 minutes" ... they're using a Poisson distribution for elapsed time. Times are continuous, the Poisson is discrete. In a Poisson process, it's event counts that are Poisson and the inter-event times are exponential. If the question really means what it says, be very careful about applying the information from the link, which does not use this strange formulation. ... $\endgroup$
    – Glen_b
    Commented May 14, 2022 at 3:58
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    $\begingroup$ The properties of the Poisson don't make sense for times because the units don't match up. Standard deviation of times should be in minutes but the Poisson has variance = mean. Change the units to seconds or hours and you lose variance=mean, so it can't be Poisson. $\:$ Indeed I am very concerned that whoever wrote the question potentially misunderstands the models being used so badly that they might expect almost any answer at all, depending on what else they misunderstand at the same time. $\endgroup$
    – Glen_b
    Commented May 14, 2022 at 3:59
  • $\begingroup$ @Glen_b Thanks; I didn't expect to see you here! For the objection regarding a discrepancy of units, wouldn't it apply to all instances of Poisson distribution? For example, if the number of policies an insurance agent sells per week follows a Poisson distribution with mean of 3, the standard deviation would be $\sqrt{3}$ policies, and variance would be 3 policy squared? If this is indeed correct, would it help if we discretize time into minutes? Although it doesn't make sense to talk about a fractional policy from the insurance agent, we indeed got a non-integral for stand deviation. $\endgroup$
    – user101998
    Commented May 14, 2022 at 17:49
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    $\begingroup$ Counts are unitless (yes, they're a count of something, but the count itself does not have units). There's no problem there making variance=mean. Times are not unitless. There's definitely a problem making variance=mean there -- certainly enough that if you decided to use a Poisson for times, you'd need to explain how that worked (e.g. an underlying quasi-Poisson model that just happened for some reason to have $\phi=1$ in this instance for these units; but that sort of explanation makes no sense here, since the time is not a scaled count either). $\endgroup$
    – Glen_b
    Commented May 14, 2022 at 23:31
  • $\begingroup$ It doesn't matter whether you round/truncate times to integers, the unit problem persists $\endgroup$
    – Glen_b
    Commented May 14, 2022 at 23:34

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