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Math Help - Poisson Process + Erlang PDF Question

  1. #1
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    Poisson Process + Erlang PDF Question

    A claims process has one claim every ten hours. This is modeled as a Poisson process with \lambda = 0.1


    What is the probability that the 3rd claim will be processed in the 5th hour?


    How can I get the correct solution using the Erlang PDF function? I tried:


    \int_4^5{\frac{0.1^3 \cdot y^{2} \cdot e^{-0.1y}}{2!} \, dy} \approx 0.0065


    My simpler solution using the Poisson probability mass function is this: The probability of the third claim being processed during the fifth hour is the probability of exactly two claims in four hours combined with the independent probability of at least one in the next hour:


    p_X(2; 0.1 \cdot 4) \cdot (1 - p_X(0; 0.1 \cdot 1))


    = \frac{0.4^2 \cdot e^{-0.4}}{2!} \cdot \left(1 - \frac{0.1^0 \cdot e^{-0.1}}{0!}\right) \approx 0.0051


    Are both of these approaches correct? Why do these two approaches not match up?
    Last edited by VinceW; March 28th 2013 at 11:04 AM.
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  2. #2
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    Re: Poisson Process + Erlang PDF Question

    the approaches are not equivalent because it is not certain that claims 1&2 will both occur in the first four hours. you could have no claims in the first 4 hours and then 3 claims in hour 5.

    The poisson approach is therefore incomplete and gives a lower answer than your erlang method.
    Last edited by SpringFan25; March 28th 2013 at 04:14 PM.
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  3. #3
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    Re: Poisson Process + Erlang PDF Question

    Perfect! Thanks!

    I was just considering the scenario with 2 in the first 4 hours and 1+ in hour 5. When I also consider 0/3+ and 1/2+ cases and sum all the probabilities, the final probability value matches the Erlang integration approach.
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