# T5-Answer-232-2010W - P ¨ a x 5 = 1000 ¯ A 1 x 10 P ¨ s...

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ANSWERS TO TUTORIAL 5 – ACTSC 232, WINTER 2010 1. (a) By the equivalence principle, P ¯ a 60 = 1000 ¯ A 60 . Thus, P = 41 . 83. (b) We have L = 1000 v T x - P ¯ a T x = (1000 + P δ ) v T x - P δ . Thus, 0 . 3 = Pr { L > 0 } = Pr ( T x < 1 δ log P + 1000 δ P !) . Hence, 0 . 3 = 1 δ log P + 1000 δ P ! × 1 40 and P = 69 . 23. 2. By the equivalence principle, P ¨ a 60: 10 = 5000 A 1 60: 10 + 10 P A 1 60: 10 . Thus, P = 250 . 20. 3. Note that the annual premium is 12 P . (a) By the equivalence principle, 2000 ¯ A 50 = 12 P ¨ a (12) 50 . Thus, P UDD = 2000 i δ A 50 12 ¨ a (12) 50 = i δ × 2 × 249 . 0475 12 × 12 . 8024 = 3 . 3386 . (b) By the equivalence principle, 2000 A (12) 50 = 12 P ¨ a (12) 50 . Thus, P UDD = 2000 i i (12) A 50 12 ¨ a (12) 50 = i i (12) × 2 × 249 . 0475 12 × 12 . 8024 = 3 . 3304 . 4. By the equivalence principle, we have
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Unformatted text preview: P ¨ a x : 5 = 1000 ¯ A 1 x : 10 + P ¨ s 5 (1 + i ) 5 10 E x . Hence, P = 223 . 26. 5. Let P 1 be the premium rate for L 1 . Then, P 1 = 1-δ ¯ a x ¯ a x and V ar ( L 1 ) = V ar ( v T x ) ( δ ¯ a x ) 2 . Let P 2 be the premium rate for L 2 . Then, V ar ( L 2 ) = ± 1 + P 2 δ ² 2 V ar ( v T x ) = 1 . 85 . Furthermore, E ( L 2 ) = ¯ A x-P 2 ¯ a x =-. 8. Thus, 1 . 85 = 1 . 8 2 V ar ( v T x ) ( δ ¯ a x ) 2 = 1 . 8 2 V ar ( L 1 ) . Hence, V ar ( L 1 ) = 0 . 571. 1...
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