Test1-Solutions-S09

Test1-Solutions-S09 - ACTSC 432 - Loss Models 2 TEST #1 1....

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Unformatted text preview: ACTSC 432 - Loss Models 2 TEST #1 1. (14 marks) Assume that Y j & = & has density function f Y j & ( y j & ) = p & &e & &y + (1 & p ) & & 2 ye & &y , y > , with < p < 1 . Also, let & be a gamma distributed random variable with density & ( & ) = & & 1 ( ) e & & , & > , for > 3 . (a) (4 marks) Find the (unconditional) density function of Y . Solution: By de&nition, the unconditional density function of Y is given by f Y ( y ) = Z 1 f Y j & ( y j & ) & ( & ) d& = Z 1 p & &e & &y + (1 & p ) & & 2 ye & &y & & 1 ( ) e & & d& = p Z 1 & &e & &y & & 1 ( ) e & & d& + (1 & p ) Z 1 & & 2 ye & &y & & 1 ( ) e & & d& = p Z 1 & ( ) e & ( 1 + y ) & d& + (1 & p ) Z 1 y & +1 ( ) e & ( 1 + y ) & d& = p 1 ( ) ( + 1) 1 + y +1 Z 1 1 + y +1 ( + 1) & e & ( 1 + y ) & d& + (1 & p ) y ( ) ( + 2) 1 + y +2 Z 1 1 + y +2 ( + 2) & +1 e & ( 1 + y ) & d& = p (1 + y ) +1 + (1 & p ) y ( + 1) 2 (1 + y ) +2 ....
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Test1-Solutions-S09 - ACTSC 432 - Loss Models 2 TEST #1 1....

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