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Topic: From Littlewood's Book (Read 355 times) |
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Barukh
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From Littlewood's Book
« on: Sep 5th, 2007, 6:33am » |
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The following passage is taken from Littlewood’s book “Mathematician’s Miscellany”:
“Suppose busses on a given route average 10-minute intervals. If they run at exactly 10 minutes intervals, the average time a passenger arriving randomly at a stop will have to wait is 5 minutes. If the busses are irregular, the average time is greater; for one kind of random distribution it is 10 minutes; and what is more, the average time since the previous bus is also 10 minutes. For a certain other random distribution, both times become infinity.”
This may be easy, may be hard, so I put it in Medium. I don’t know the answer.
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Grimbal
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Re: From Littlewood's Book
« Reply #1 on: Sep 5th, 2007, 6:37am » |
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And what is the question? To prove the assertion? To find a distribution where both times go to infinity?
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Barukh
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Re: From Littlewood's Book
« Reply #2 on: Sep 5th, 2007, 6:44am » |
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Let's start by finding a distribution where both times equal 10 minutes.
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Grimbal
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Re: From Littlewood's Book
« Reply #3 on: Sep 5th, 2007, 6:47am » |
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That would be when 2 buses arrive simultaneously every 20 minutes.
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towr
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Re: From Littlewood's Book
« Reply #4 on: Sep 5th, 2007, 7:08am » |
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Isn't the average time to the next bus, and the time to the previous one always the same; it seems symmetric, since you can just reverse time.
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Barukh
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Re: From Littlewood's Book
« Reply #5 on: Sep 5th, 2007, 8:15am » |
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on Sep 5th, 2007, 6:47am, Grimbal wrote:| That would be when 2 buses arrive simultaneously every 20 minutes. |
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Yes, that does the trick (for both cases). Afterall, it was easy...
on Sep 5th, 2007, 7:08am, towr wrote:| Isn't the average time to the next bus, and the time to the previous one always the same; it seems symmetric, since you can just reverse time. |
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Grimbal
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Re: From Littlewood's Book
« Reply #6 on: Sep 5th, 2007, 9:01am » |
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on Sep 5th, 2007, 7:08am, towr wrote:| Isn't the average time to the next bus, and the time to the previous one always the same; it seems symmetric, since you can just reverse time. |
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I thought there might be non-symetric cases, especially if the waiting time raises steadily as the buses get more and more clustered, but I cannot think of a case where the two times converge to a different value when averaged over a larger and larger period.
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| « Last Edit: Sep 7th, 2007, 1:34am by Grimbal » |
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Eigenray
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Re: From Littlewood's Book
« Reply #7 on: Sep 6th, 2007, 3:47pm » |
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What if we have k buses arrive at time T(k)=k(k-1)/2, for each k? Then (# buses before time n)/n  1, while E(waiting time | arrive before time n) ~ n/3  .
Or are we supposed to consider a periodic solution, where the bus arrival times have a certain probability distribution? I don't think this is possible, because as long as there is a positive (constant) probability of a bus coming in each period, the expected waiting time is finite.
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Barukh
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Re: From Littlewood's Book
« Reply #8 on: Sep 6th, 2007, 10:48pm » |
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on Sep 6th, 2007, 3:47pm, Eigenray wrote:| Or are we supposed to consider a periodic solution, where the bus arrival times have a certain probability distribution? |
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Yes, that's what I thought at the beginning, feeling it may be hard to find such a distribution.
Quote:| I don't think this is possible, because as long as there is a positive (constant) probability of a bus coming in each period, the expected waiting time is finite. |
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But is it at least possible for 10-minutes waiting time case?
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