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Test2-version1-Solutions

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

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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 0 , p M 0 , of 0 : 2 . 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 is a zero-modi°ed POIS ( ± = 2 : 7) with probability mass at 0 equal to 0 : 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 ± 0 : 2 1 ± e ° 3 ° e 3( t ° 1) ± e ° 3 ± . It follows that P N p ( t ) = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ° e 3(0 : 1+0 : 9 t ° 1) ± e ° 3 ± = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ° e 3(0 : 9)( t ° 1) ± e ° 3 ± = 0 : 2 + 1 ± 0 : 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 0 equal to 0 : 2147 , one shall prove that P N p ( t ) = 0 : 2147 + 1 ± 0 : 2147 1 ± e ° 2 : 7 ° e 2 : 7( t ° 1) ± e ° 2 : 7 ± . Simple modi°cations of (1) lead to P N p ( t ) = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ° e 2 : 7( t ° 1) ± e ° 2 : 7 + e ° 2 : 7 ± e ° 3 ± = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ² e ° 2 : 7 ± e ° 3 ³ + 1 ± 0 : 2 1 ± e ° 3 ° e 2 : 7( t ° 1) ± e ° 2 : 7 ± = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ² e ° 2 : 7 ± e ° 3 ³ + 1 ° 0 : 2 1 ° e ° 3 ² 1 ± e ° 2 : 7 ³ 1 ± e ° 2 : 7 ° e 2 : 7( t ° 1) ± e ° 2 : 7 ± = 0 : 2 + 1 ± 0 : 2 1 ± e ° 3 ² 1 ± e ° 3 ± ² 1 ± e ° 2 : 7 ³³ + 1 ° 0 : 2 1 ° e ° 3 ² 1 ± e ° 2 : 7 ³ 1 ± e ° 2 : 7 ° e 2 : 7( t ° 1) ± e ° 2 : 7 ± = 1 ± 1 ± 0 : 2 1 ± e ° 3 ² 1 ± e ° 2 : 7 ³ + 1 ° 0 : 2 1 ° e ° 3 ² 1 ± e ° 2 : 7 ³ 1 ± e ° 2 : 7 ° e 2
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