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I Poisson distribution with conditional probability

  1. Oct 19, 2016 #1
    Hi guys,
    I have a question about computing conditional probabilities of a Poisson distribution.
    Say we have a Poisson distribution P(X = x) = e^(−λ)(λx)/(x!) where X is some event.
    My question is how would we compute P(X > x1 | X > x2), or more specifically P(X> x1 ∩ X > x2) with x1 > x2?
    I originally thought that P(X > x1 ∩ X > x2) = P(X > x1) but recently read about the memorylessness property of exponential distributions and I'm not sure if it applies to Poisson distributions.
     
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  3. Oct 19, 2016 #2

    Stephen Tashi

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    If x1 > x2 then ##\{X: X > x1, X > x2\}## is the same event as ##\{X:X>x1\}## , isn't it?
     
  4. Oct 20, 2016 #3
    Yes. Also I know that the events X = x of a Poisson distribution are independent of one another but surely P(X >= 70) and P(X >= 80) for example can't be, because given at least 70 events happen, the probability that at least 80 events happen would be 10 no?
     
  5. Oct 20, 2016 #4

    Stephen Tashi

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    But my remark wasn't about the independence of events. If ##A \subset B ## then ##Pr(A \cap B) = Pr(A)##.

    As far as independence goes, in most cases if ##A \subset B## then ##A## and ##B## are not independent events. Exceptions would be cases like ##Pr(A) = Pr(B) = 0 ## or ##Pr(A) = Pr(B) = 1 ##.

    To find ##Pr(X > 80 | X > 70)##, what does Bayes theorem tell you ?

    In the current Wikipedia article on "Memorylessness" https://en.wikipedia.org/wiki/Memorylessness there is the interesting claim:

    - surprising (to me), if true.
     
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