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Math Help - nonhomoegneous Poisson process

  1. #1
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    nonhomoegneous Poisson process

    Let {N(t): t≥0} be a nonhomoegneous Poisson process with continuous rate function λ(t)>0. Let T1<T2<... denote the times of the points. Show that
    E( |Tn-1| |N(1)=n)->1 as n->∞

    For nonhomoegneous Poisson process, I know that we have independent increments and that,
    Number of points in (t,t+s]
    =N(t,t+s] is Poisson distributed with mean
    t+s
    ∫ λ(u)du
    t
    But do I have to use this? I have no idea where to start and I would appreicate if someone can help.

    Thank you!

    Last edited by kingwinner; December 14th 2009 at 06:01 PM.
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    For nonhomoegneous Poisson process, I know that we have independent increments and that,
    Number of points in (t,t+s]
    =N(t,t+s] is equal to
    t+s
    ∫ λ(u)du
    t
    No, the number of points in (t,t+s] is a Poisson random variable whose parameter is the integral you wrote.

    About your question. You should try to use these facts (and similar ones):
    - T_n<1 when N(1)=n
    - ( 1-\varepsilon<T_n and N(1)=n) iff ( N(1-\varepsilon,1)\geq 1 and N(0,1)=n).
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  3. #3
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    Quote Originally Posted by Laurent View Post
    About your question. You should try to use these facts (and similar ones):
    - T_n<1 when N(1)=n
    - ( 1-\varepsilon<T_n and N(1)=n) iff ( N(1-\varepsilon,1)\geq 1 and N(0,1)=n).
    OK, but how should I begin? Do I need to find the conditional probability density function of |Tn-1| |N(1)=n?

    ( 1-\varepsilon<T_n and N(1)=n) iff ( N(1-\varepsilon,1)\geq 1 and N(0,1)=n)
    I can see why this is true, but with this we still cannot use the independent increments property becuase the intervals are overlapping, right?

    Thank you!
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