Nobody1111
May 10th 2007, 02:13 PM
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.
Nobody1111
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.
Nobody1111