Test2-version2-Solutions

Test2-version2-Solutions - ACTSC 431 - Loss Models 1 FALL...

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ACTSC 431 - Loss Models 1 FALL 2007 TEST #2 Name : ID Number : 1. (8 marks) Suppose that the per payment r.v. Y p is PAR ( = 5 = 200) distributed. In addition, the number of payments N p n = 4 ;q = 0 : 4 ) distribution with a probability mass at 0 , p M 0 , of 0 : 1 . The actuary responsible of reviewing the previous actuary±s 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 0 : 9 (independently of N L ). (a) (4 marks) Prove that the new distribution for N p ( n = 4 ;q = 0 : 36) with proba- bility mass at 0 equal to 0 : 1395 . Solution: We know that P N p ( t ) = P N L (0 : 1 + 0 : 9 t ) , where P N L ( t ) = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 6 + 0 : 4 t ) 4 ± (0 : 6) 4 ± . It follows that P N p ( t ) = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 6 + 0 : 4 (0 : 1 + 0 : 9 t )) 4 ± (0 : 6) 4 ± = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 64 + 0 : 36 t ) 4 ± (0 : 6) 4 ± . (1) To prove that N p ( n = 4 ;q = 0 : 36) with probability mass at 0 equal to 0 : 1395 , one shall prove that P N p ( t ) = 0 : 1395 + 1 ± 0 : 1395 1 ± (0 : 64) 4 (0 : 64 + 0 : 36 t ) 4 ± (0 : 64) 4 ± . P N p ( t ) = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 64 + 0 : 36 t ) 4 ± (0 : 64) 4 + (0 : 64) 4 ± (0 : 6) 4 ± = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 64) 4 ± (0 : 6) 4 ± + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 64 + 0 : 36 t ) 4 ± (0 : 64) 4 ± = 0 : 1 + 1 ± 0 : 1 1 ± (0 : 6) 4 1 ± (0 : 6) 4 ± 1 ± (0 : 64) 4 ±± + 1 ± 0 : 1 1 ± (0 : 6) 4 (0 : 64 + 0 : 36 t ) 4 ± (0 : 64)
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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-version2-Solutions - ACTSC 431 - Loss Models 1 FALL...

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