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Meeting a friend at a train stop

  1. Feb 21, 2012 #1


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    Consider Problem #1 here. The author seems to assume that the train comes immediately when one of the friends gives up on waiting for the other. What if the train arrival times follow a Poison process with rate λ?

    I have some ideas but I won't post anything in case anyone else wants a stab at it.
  2. jcsd
  3. Feb 22, 2012 #2
    I think that he's only asking "what is the probability that the two friends arrive within 5 minutes of each other?" I think that you are asking "what is the probability that a Poisson train arrives before the friend given that 5 minutes has passed since the first person's arrival?" Given that the first arrival has waited 5 minutes without meeting his friend he can restart the Poisson process and ask for the probability that the first train arrival time is less than the random arrival time of his friend. I might start with something simple to figure out what's going on and just find the expected arrival times of his friend and the first train given the conditions above, then go from there.
  4. Feb 22, 2012 #3


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    Yup that's exactly what I'm asking.

    Yup, the nice thing about Poisson processes is that it's memory less, so you don't have to worry about whether the one friend had seen a train when he was waiting for the other friend. Waiting time is always exponential with rate λ.

    Well the expected arrival time isn't really going to give you enough information, you could find that the expected arrival time of the second friend is less than the expected arrival time of the train or vice versa depending on λ but that's not enough to tell you about the probability of one arriving before the other. It is a start though.
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