Exchap10-11-students-solutions

# Exchap10-11-students-solutions - ACTSCI 331 Life...

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Unformatted text preview: ACTSCI 331 - Life Contingencies I Exercises (Chapters 10-11) 1. a. We have f T (50) ;J (15 ; 1) = 15 p ( & ) 50 & (1) (65) = 15 p (1) 50 15 p (2) 50 & (1) (65) = e & : 01(15) 50 & 15 50 : 01 , f T (50) ;J (15 ; 2) = 15 p ( & ) 50 & (2) (65) = 15 p (1) 50 15 p (2) 50 & (2) (65) = e & : 01(15) 50 & 15 50 1 100 & 65 , and & ( & ) (65) = & (1) (65) + & (2) (65) = 0 : 01 + 1 100 & 65 . b. One deduces that 10 p ( & ) 40 = 10 p (1) 40 10 p (2) 40 = e & : 01(10) 60 & 10 60 , 10 q (2) 40 = Z 10 t p ( & ) 40 & (2) (40 + t ) dt = Z 10 t p (1) 40 t p (2) 40 & (2) (40 + t ) dt = Z 10 e & : 01 t 1 60 dt = 1 60 1 & e & : 01(10) : 01 , 1 and 10 q (1) 40 = Z 10 t p ( & ) 40 & (1) (40 + t ) dt = Z 10 t p (1) 40 t p (2) 40 & (1) (40 + t ) dt = Z 10 e & : 01 t 60 & t 60 (0 : 01) dt = : 01 60 " (60 & t ) e & : 01 t & : 01 & & & & t =10 t =0 & Z 10 e & : 01 t : 01 dt # = : 01 60 " 60 : 01 & 50 e & : 01(10) : 01 & 1 & e & : 01(10) (0 : 01) 2 # = 1 & 50 60 e & : 01(10) & 1 60 1 & e & : 01(10) : 01 . c. It follows that 15 j 10 q (2) 40 = 15 p ( & ) 40 10 q (2) 55 = e & : 01(15) 60 & 15 60 ¡Z 10 t p ( & ) 55 & (2) (55 + t ) dt ¢ = e & : 01(15) 60 & 15 60 ¡Z 10 e & : 01 t 1 45 dt ¢ = e & : 01(15) 1 60 1 & e & : 01(10) : 01 , and 1 q (2) 40 = Z 60 t p ( & ) 40 & (2) (40 + t ) dt = Z 60 t p (1) 40 t p (2) 40 & (2) (40 + t ) dt = Z 60 e & : 01 t 1 60 dt = 1 60 1 & e & : 01(60) : 01 . 2. a. expected number of survivors at age 31 = 1000 p ( & ) 30 = 1000 £ 1 & q ( & ) 30 ¤ = 1000 (1 & (0 : 01 + 0 : 2 + 0 : 005)) expected number of survivors at age 32 = 1000 2 p ( & ) 30 = 1000 p ( & ) 30 p ( & ) 31 = 1000 (1 & (0 : 01 + 0 : 2 + 0 : 005)) (1 & (0 : 01 + 0 : 2 + 0 : 004)) b. expected number ... = 1000 £ 3 q (2) 30 + 3 q (3) 30 ¤ = 1000 £ 3 q (2) 30 + 3 q (3) 30 ¤ , where 3 q (2) 30 = q (2) 30 + p ( & ) 30 q (2) 31 + 2 p ( & ) 30 q (2) 32 = : 2 + (1 & : 01 & : 2 & : 05) (0 : 2) + (1 & (0 : 01 + 0 : 2 + 0 : 005)) (1 & (0 : 01 + 0 : 2 + 0 : 004)) (0 : 2) 2 and 3 q (3) 30 = q (3) 30 + p ( & ) 30 q (3) 31 + 2 p ( & ) 30 q (3) 32 = : 005 + (1 & (0 : 01 + 0 : 2 + 0 : 05)) (0 : 004) + (1 & (0 : 01 + 0 : 2 + 0 : 005)) (1 & (0 : 01 + 0 : 2 + 0 : 004)) (0 : 002) c. expected number ... = 1000 2 p ( & ) 30 q (1) 32 = 1000 (1 & (0 : 01 + 0 : 2 + 0 : 005)) (1 & (0 : 01 + 0 : 2 + 0 : 004)) (0 : 02) d. expected number .... = 1000 2 p ( & ) 30 & q (1) 32 + q (3) 32 ¡ = 1000 (1 & (0 : 01 + 0 : 2 + 0 : 005)) (1 & (0 : 01 + 0 : 2 + 0 : 004)) (0 : 02 + 0 : 002) 3. We know E [ T (50) j J = 1] = Z 1...
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Exchap10-11-students-solutions - ACTSCI 331 Life...

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