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Thread: Poisson Mixture problem

  1. #1
    Actuary.com - Level I Poster
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    Poisson Mixture problem

    Nov 2000 #13
    A claim count distribution can be expressed as a mixed Poisson distribution. The mean of the Poisson distribution is uniformly distributed over the interval [0,5].
    Calculate the probability that there are 2 or more claims.

    I know the way i solved this question is wrong. Can you help me identify why my logic is flawed

    I let X represent the mixed poisson distribution

    E(X) = E(E(X|lamba )) = E(lamba) = mean of uniform [0, 5] = 2.5

    P(X>= 2) = 1- P(x =0) - P(x =1) = 1- e(-2.5) -2.5e(-2.5) = 0.71

    The answer 0.61 and is solved by this method (number 137) http://math.illinoisstate.edu/actuary/examm/m-09-05.pdf

  2. #2
    Actuary.com - Level III Poster
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    Quote Originally Posted by liltich View Post
    Nov 2000 #13
    A claim count distribution can be expressed as a mixed Poisson distribution. The mean of the Poisson distribution is uniformly distributed over the interval [0,5].
    Calculate the probability that there are 2 or more claims.

    I know the way i solved this question is wrong. Can you help me identify why my logic is flawed

    I let X represent the mixed poisson distribution

    E(X) = E(E(X|lamba )) = E(lamba) = mean of uniform [0, 5] = 2.5

    P(X>= 2) = 1- P(x =0) - P(x =1) = 1- e(-2.5) -2.5e(-2.5) = 0.71

    The answer 0.61 and is solved by this method (number 137) http://math.illinoisstate.edu/actuary/examm/m-09-05.pdf
    You cannot simply find the mean of lambda by (5/2).

    P(0)= 1/5 * integral from 0 to 5 (e^(-lambda)) d(lambda)
    This is the right way. you should be able to read it even it is messy.

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