Test2-S08-Solutions

Test2-S08-Solutions - ACTSC 431 - Loss Models 1 Spring 2008...

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Unformatted text preview: ACTSC 431 - Loss Models 1 Spring 2008 TEST #2 1. (22 marks) Suppose that the ground-up loss X is a discrete random variable (r.v.) with probability mass function (p.m.f.) p X ( x ) = 8 < : : 5 ; x = 2 : 3 ; x = 4 : 2 ; x = 5 . and that the number of losses N L has a zero-modi&ed NB( r = 5 ;& = 0 : 4 ) distribution with a probability mass at , p M , of : 1 . We consider a policy with an ordinary deductible of 2 . (a) (7 marks) Prove that the distribution of the number of (non-zero) payments N p is a zero-modi&ed NB ( r = 5 ;& = 0 : 2) with probability mass at equal to : 3387 . Solution: Here, the probability that a loss results in a zero-payment is the probability that the ground-up loss is 2 since when X = 2 , Y L = max ( X & 2 ; 0) = 0 . It follows that Pr ( Y L = 0) = Pr ( X = 2) = 0 : 5 . We also know that P N p ( t ) = P N L (0 : 5 + 0 : 5 t ) , where P N L ( t ) = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 & : 4 ( t & 1) 5 & 1 1 + 0 : 4 5 ! . One arrives at P N p ( t ) = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 & : 4 (0 : 5 + 0 : 5 t & 1) 5 & 1 1 + 0 : 4 5 ! = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 & : 2 ( t & 1) 5 & 1 1 + 0 : 4 5 ! . (1) To prove that N p is a zero-modi&ed NB ( r = 5 ;& = 0 : 2) with probability mass at equal to : 3387 , one shall prove that P N p ( t ) = 0 : 3387 + 1 & : 3387 1 & 1 1 : 2 5 1 1 & : 2 ( t & 1) 5 & 1 1 : 2 5 ! . 1 Simple modi&cations of (1) lead to P N p ( t ) = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 & : 2 ( t & 1) 5 & 1 1 + 0 : 2 5 + 1 1 + 0 : 2 5 & 1 1 + 0 : 4 5 ! = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 + 0 : 2 5 & 1 1 + 0 : 4 5 ! + 1 & : 1 1 & & 1 1+0 : 4 5 1 1 & : 2 ( t & 1) 5 & 1 1 + 0 : 2 = 0 : 1 + 1 & : 1 1 & & 1 1+0 : 4 5 1 & 1 1 + 0 : 4 5 & 1 & 1 1 + 0 : 2 5 !!...
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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-S08-Solutions - ACTSC 431 - Loss Models 1 Spring 2008...

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