Chapter_8a_-_Question_Set_-_Solution

Chapter_8a_-_Question_Set_-_Solution - ACTSCI 331 - Life...

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Unformatted text preview: ACTSCI 331 - Life Contingencies I Exercises (Chapter 8a) 1. Let T ( x ) and T ( y ) be two independent rv&s. Given that 20 p x = 0 : 9 and 20 p y = 0 : 8 , nd 20 p xy and 20 p xy . (Answers : 0.72 and 0.98). Solution: We have 20 p xy = Pr ( T ( xy ) > 20) = Pr (min( T ( x ) ;T ( y )) > 20) = Pr ( T ( x ) > 20 ;T ( y ) > 20) = IND Pr ( T ( x ) > 20) Pr ( T ( y ) > 20) = 20 p x 20 p y = (0 : 9) (0 : 8) = 0 : 72 , and 20 p xy = 1 & 20 q xy = 1 & Pr (max( T ( x ) ;T ( y )) 20) = 1 & Pr ( T ( x ) 20 ;T ( y ) 20) = IND 1 & Pr ( T ( x ) 20) Pr ( T ( y ) 20) = 1 & 20 q x 20 q y = 1 & (1 & : 9) (1 & : 8) = 0 : 98 . 1 2. Let the joint p.d.f. of ( T ( x ) ;T ( y )) be given by f T ( x ) ;T ( y ) ( t;s ) = 1 4 & 2 e & & ( t + s ) + 3 4 2 e & ( t + s ) ; t;s > ; with & = 0 : 01 and = 0 : 02 . Calculate Pr(20 < T ( xy ) < 50) and Pr(20 < T ( xy ) < 50) . (Answers: : 311 and : 249 ). Solution: We have Pr (20 < T ( xy ) < 50) = Pr ( T ( xy ) > 20) & Pr ( T ( xy ) > 50) = 20 p xy & 50 p xy , where n p xy = Pr ( T ( x ) > n;T ( y ) > n ) = Z 1 n Z 1 n & 1 4 & 2 e & & ( t + s ) + 3 4 2 e & ( t + s ) dsdt = 1 4 e & 2 &n + 3 4 e & 2 n . Therefore, Pr (20 < T ( xy ) < 50) = & 1 4 e & 2(20)(0 : 01) + 3 4 e & 2(20)(0 : 02) & & 1 4 e & 2(50)(0 : 01) + 3 4 e & 2(50)(0 : 02) = 0 : 311 . Also, Pr (20 < T ( xy ) < 50) = Pr ( T ( xy ) < 50) & Pr ( T ( xy ) < 20) = 50 q xy & 20 q xy , where n q xy = Pr ( T ( x ) < n;T ( y ) < n ) = Z n Z n & 1 4 & 2 e & & ( t + s ) + 3 4 2 e & ( t + s ) dsdt = 1 4 1 & e & &n 2 + 3 4 1 & e & n 2 . Thus, Pr (20 < T ( xy ) < 50) = 1 4 1 & e & 50(0 : 01) 2 + 3 4 1 & e & 50(0 : 02) 2 & & 1 4 1 & e & 20(0 : 01) 2 + 3 4 1 & e & 20(0 : 02) 2 = 0 : 249 . 2 3. Suppose & 10 p x = 0 : 9 and 30 p x = 0 : 5 & 10 q y = 0 : 15 and 30 q y = 0 : 45 Under the independent assumption, &nd (a) Pr (the &rst death amongst ( x ) and ( y ) occurs between times 10 and 30 ). (Answer: 0.49) Solution: We have Pr (10 < min ( T ( x ) ;T ( y )) < 30) = Pr ( T ( xy ) > 10) Pr ( T ( xy ) > 30) = 10 p xy 30 p xy = IND 10 p x 10 p y 30 p x 30 p y = 0 : 9 (1 : 15) : 5 (1 : 45) = 0 : 49 . (b) Pr (at most one death amongst ( x ) and ( y ) occurs between times 10 and 30 ). (Answer: 0.88); Solution: We have 1 Pr ( 2 deaths between 10 and 30 ) = 1 Pr (10 < T ( x ) < 30 ; 10 < T ( y ) < 30) = IND 1 Pr (10 < T ( x ) < 30) Pr (10 < T ( y ) < 30) = 1 ( 10 p x 30 p x ) ( 10 p y 30 p y ) = 1 (0 : 9 : 5) ((1 : 15) (1 : 45)) = 0 : 88 . (c) Pr (the second death amongst ( x ) and ( y ) occurs between times 10 and 30 ). (Answer: 0.21); Solution : By de&nition, Pr (10 < T ( xy ) < 30) = Pr ( T ( xy ) < 30) Pr ( T ( xy ) < 10) = 30 q xy 10 q xy = IND 30 q x 30 q y 10 q x 10 q y = IND (1 : 5) (0 : 45) (1 : 9) (0 : 15) = 0 : 21 ....
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This note was uploaded on 01/04/2012 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.

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Chapter_8a_-_Question_Set_-_Solution - ACTSCI 331 - Life...

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