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Poisson and uniform paradox

  1. Jun 23, 2009 #1
    Hi, all,

    Let's say we deploy some random points on a line of finite length according to a poisson distribution of density \lambda. Can I say that these points are also "uniformly" distributed on the same line?

    thks
     
  2. jcsd
  3. Jun 23, 2009 #2

    HallsofIvy

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    No, you can't. (And there is no "paradox".) With Poisson distribution, with parameter [itex]\lambda[/itex], points are as likely to be [itex]\le \lamba[/itex] as larger. With a "uniform" distribution, they are as likely to be less than or equal to the midpoint of the interval as to be above it. Further, points in a Poisson distribution are more likely to be close to \lambda than not while there is no number in an interval that points in a uniform distribution are more likely to be close to.
     
  4. Jun 23, 2009 #3
    The Poisson distribution gives the waiting time until the next event. I think he means, distribute points randomly along an interval such that their _waiting times_ are distributed as Poisson random variables.

    The resulting distribution of points is similar to what you would get if you sampled the same number of points from a uniform distribution along the same interval. There's at least one important difference, however: if you choose points with waiting times according to the Poisson distribution, you don't know when starting out how many points are going to fit in the interval.
     
  5. Jun 30, 2009 #4

    gel

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    That doesn't make sense, the Poisson distribution is unbounded, so not confined to a random line. Maybe you mean a random (and finite) set so that the number of points in any interval (a,b) is Poisson distributed with parameter [itex]\lambda(b-a)[/itex]? i.e., a http://books.google.co.uk/books?id=...bSNCw&sa=X&oi=book_result&ct=result&resnum=5".
    Then, yes, you get the same thing as choosing N independent and uniform random variables, where N itself has the Poisson distribution.

    (I don't see any paradox...)
     
    Last edited by a moderator: Apr 24, 2017
  6. Jul 2, 2009 #5
    To be more precise, for a Poisson process (where the interarrival times are exponential) given that there are N arrivals in [0,T] the arrival times T1,...,TN will be distributed according to the order statistics of N independent uniform variables.
     
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