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Poisson Process

  1. Jan 21, 2014 #1
    I need some help on the following question: Let N() be a poisson process with parameter [tex] \lambda [/tex].

    I need to find that probability that

    [tex] N((1,2]) = 3 [/tex] given [tex] N((1,3]) > 3 [/tex]

    I know that this is equal to the probability that

    [tex] P(A \cap B) / P(B) [/tex] where A = N((1,2]) and B = N((1,3]) > 3, but I'm not sure where to go from there.
     
  2. jcsd
  3. Jan 21, 2014 #2

    haruspex

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    Yes, that's the right start. Can you write down the value of P(B)?
    For P(A&B), you have "N((1,2])=3 and N((1,3]) > 3". Can you translate that into a combination of the event A and some fact concerning N((2,3])?
     
  4. Jan 22, 2014 #3

    Ray Vickson

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    Are you sure you have copied the problem correctly? Getting P{N(1,2]=3|N(1,3]>3} is not too difficult (just use the definition and known expressions), but the answer is not particularly enlightening. However, the alternative problem P{N(1,3]>3|N(1,2]=3} gives a much nicer answer, and one that reveals an important property of Poisson processes.
     
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