Test2-version1-Solutions

Test2-version1-Solutions - ACTSC 431 - Loss Models 1 FALL...

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Unformatted text preview: ACTSC 431 - Loss Models 1 FALL 2007 TEST #2 Name : ID Number : 1. (12 marks) Suppose that the per payment r.v. Y p is EXP ( & = 50) distributed. In addition, the number of payments N p has a zero-modi&ed POIS( = 3 ) distribution with a probability mass at , p M , of : 2 . The actuary responsible of reviewing the previous actuarys work decides to use a slightly di/erent model for the next period based on the claim experience : & the per payment r.v. Y p has the same distribution & the distribution of N p (see above) will be used instead for the number of losses N L We assume that a loss results in a (non-zero) payment with probability : 9 (independently of N L ). (a) (4 marks) Prove that the new distribution for N p is a zero-modi&ed POIS ( = 2 : 7) with probability mass at equal to : 2147 . Solution: We know that P N p ( t ) = P N L (0 : 1 + 0 : 9 t ) , where P N L ( t ) = 0 : 2 + 1 : 2 1 e & 3 & e 3( t & 1) e & 3 . It follows that P N p ( t ) = : 2 + 1 : 2 1 e & 3 & e 3(0 : 1+0 : 9 t & 1) e & 3 = : 2 + 1 : 2 1 e & 3 & e 3(0 : 9)( t & 1) e & 3 = : 2 + 1 : 2 1 e & 3 & e 2 : 7( t & 1) e & 3 . (1) To prove that N p is a zero-modi&ed POIS ( = 2 : 7) with probability mass at equal to : 2147 , one shall prove that P N p ( t ) = 0 : 2147 + 1 : 2147 1 e & 2 : 7 & e 2 : 7( t & 1) e & 2 : 7 . Simple modi&cations of (1) lead to P N p ( t ) = : 2 + 1 : 2 1 e & 3 & e 2 : 7( t & 1) e & 2 : 7 + e & 2 : 7 e & 3 = : 2 + 1 : 2 1 e & 3 e & 2 : 7 e & 3 + 1 : 2 1 e & 3 & e 2 : 7( t & 1) e & 2 : 7 = : 2 + 1 : 2 1 e & 3 e & 2 : 7 e & 3 + 1 & : 2 1 & e & 3 1 e & 2 : 7 1 e & 2 : 7 & e 2 : 7( t & 1) e & 2 : 7 = : 2 + 1 : 2 1 e & 3 1 e & 3 1 e & 2 : 7 + 1 & : 2 1 & e & 3 1 e & 2 : 7 1 e & 2...
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This note was uploaded on 06/28/2009 for the course ACTSC 431 taught by Professor Laundriualt during the Spring '09 term at Waterloo.

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Test2-version1-Solutions - ACTSC 431 - Loss Models 1 FALL...

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