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Thread: Problem with recursion

  1. #1
    Actuary.com - Level I Poster
    Join Date
    Oct 2006

    Problem with recursion

    Problem is taken from exams for Polish Actuaries from 10th October 2005, Exercice 1, General Insurance Mathematics. Here is the problem:

    Damage which occurred in year t is liquidated:
    - in the same year with probability 0.3
    - in the next year with probabilitu 0.3
    - in the year t+k (k>1) with probability 0.8 * (0.5)^k
    We don't care in which period of the year the damage occurred - e.g. if one damage occurred on the 1st January and any other damage occurred on 30th December the probability that they are liquidated by the end of the k-th year is the same.
    Let R(t) denotes the number of damages which were NOT liquidated at the end of the year t (R(t) consists both of the unliquidated damages which occurred in year t as well as of the unliquidated damages which occurred before the year t and weren't liquidated at the end of year t-1).
    Let n(t) denotes the number of damages in year t.
    Assuming that
    R(t-1) = 1100
    n(t) = 900
    n(t-1) = 800
    find the expected value of R(t) under the three conditions given above.

    Answer is R(t) =1220.

    The most sensible way of solving the problem seems to be finding the recursion formula for R(t) as the function of (R(t-1), n(t) and n(t-1).
    As I was trying to solve the problem I met with the problem that n(t-2) which value is not given couldn't be reduced. I suppose I do it in the wrong way and don't know what to do with it.
    If anyone could help me with this exercice I'll be extremely grateful.

    Last edited by Nobody1111; May 10th 2007 at 01:16 PM.

  2. #2
    Actuary.com - Level II Poster
    Join Date
    Mar 2007
    I don't see how it can be solved recursively. I can reduce it to R(t) = 1760 - .5 R(t-2) - .05 n(t-2), but I don't think there is enough information for a recursive approach. I thought I was on to something when I managed to make the infinite sums go away...The only thing I can think of is that you could use a distribution type to determine E[X], but that is probably a dead end, too. If you find the answer, please share.

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