Q1-S-2010W - R 1 t p 50 dt + p 50 R 1 s p 51 ds = R 1 ( p...

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ANSWERS TO QUIZ 1 – ACTSC 232, WINTER 2010 1. (a) The probability is 9 12 q 0 = 1 - S 0 ( 3 4 ) = 0 . 0133 . (b) μ x = - S 0 0 ( x ) S 0 ( x ) = 225 - 2 x (110 - x )(115 - x ) , 0 x < 110. (c) 5 | 12 q 95 = 5 p 95 - 17 p 95 = S 0 (100) S 0 (95) - S 0 (112) S 0 (95) = 10 × 15 15 × 20 - 0 = 0 . 5 . 2. (a) For 0 t < 100, we have t p 20 = e - R 20+ t 20 μ x dx = ± 100 - t 100 ² 2 . For t 100, we have t p 20 = 0. (b) The probability is Pr { K 20 = 9 } = 9 p 20 - 10 p 20 = 0 . 0181 . (c) e 20 = k =1 k p 20 = 99 k =1 ( 100 - k 100 ) 2 = 1 100 2 99 n =1 n 2 = 32 . 835. 3. (a) We have n d x = l x - l x + n = l 0 S 0 ( x ) - l 0 S 0 ( x + n ) = l 0 Z x + n x f 0 ( t ) dt = l 0 Z x + n x t p 0 μ t dt = Z x + n x l t μ t dt = Z n 0 l x + s μ x + s ds. or Z n 0 l x + s μ x + s ds = l x Z n 0 s p x μ x + s ds = l x Z n 0 f x ( s ) ds = l x n q x = n dx. (b) We have l x + t = t p x l x = (1 - t q x ) l x . Under the UDD assumption, we have for 0 t 1, t q x = 1 - tq x . Thus, under the UDD assumption, l x + t = (1 - tq x ) l x = (1 - t (1 - l x +1 l x )) l x = (1 - t ) l x + tl x +1 . 4. (a) We have e 50: 2 = R 1 0 t p 50 dt + R 2 1 t p 50 dt = R 1 0 t p 50 dt + p 50 R 2 1 t - 1 p 51 dt = R 1 0 t p 50 dt + p 50 R 1 0 s p 51 ds. Under the constant force assumption, we have
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Unformatted text preview: R 1 t p 50 dt + p 50 R 1 s p 51 ds = R 1 ( p 50 ) t dt + p 50 R 1 ( p 51 ) s ds = 1 . 896 . (b) The probability is 18 12 q 50 . 3 = 1-1 . 5 p 50 . 3 = 1-l 51 . 8 l 50 . 3 . Under the UDD assumption, 1-l 51 . 8 l 50 . 3 = 1-. 2 l 51 + 0 . 8 l 52 . 7 l 50 + 0 . 3 l 51 = 1-. 2 p 50 + 0 . 8 2 p 50 . 7 + 0 . 3 p 50 = 0 . 08183 . 1...
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This note was uploaded on 03/20/2011 for the course ACTSC 232 taught by Professor Matthewtill during the Summer '08 term at Waterloo.

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