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ACTSC 232 Homework Problem

# ACTSC 232 Homework Problem - ACTSC 232 Spring 2011...

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ACTSC 232 – Spring 2011 Solutions to Chapter 4 Book Problems Exercise 4.1 (a) 5 E 35 = v 5 5 p 35 = 1 1 . 06 5 98 , 485 . 68 100 , 000 . 00 = 0 . 73594 (b) A 1 35: 5 = A 35 - 5 | A 35 = A 35 - 5 E 35 A 40 = 0 . 151375 - 0 . 13872 = 0 . 01266 (c) 5 | A 35 = 5 E 35 A 40 = 0 . 13872 (d) ¯ A 35: 5 = ¯ A 1 35: 5 + A 1 35: 5 UDD = i δ A 1 35: 5 + 5 E 35 UDD = (1 . 02971) (0 . 01266) + 0 . 73594 UDD = 0 . 74897 1

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Exercise 4.2 First we note that under UDD, for 0 < t, 1 m < 1 and 0 < t + 1 m < 1 t p x = 1 - t q x and 1 m q x + t = 1 m q x 1 - t q x so that t p x 1 m q x + t = t | 1 m q x = 1 m q x Then we have A ( m ) x = X k =0 m - 1 X y =0 v k + y m + 1 m k + y m | 1 m q x = X k =0 m - 1 X y =0 v k + y m + 1 m k p x y m | 1 m q x + k = X k =0 m - 1 X y =0 v k + y +1 m k p x 1 m q x + k = X k =0 k p x q x + k v k m - 1 X y =0 1 m v y +1 m = X k =0 k p x q x + k v k a ( m ) 1 = i i ( m ) i ( m ) i X k =0 k p x q x + k v k a ( m ) 1 = i i ( m ) X k =0 k p x q x + k v k i ( m ) i a ( m ) 1 = i i ( m ) X k =0 k p x q x + k v k a 1 = i i ( m ) X k =0 k p x q x + k v k v = i i ( m ) X k =0 k | q x + k v k +1 = i i ( m ) A x 2
Exercise 4.6 (a) ( IA ) 1 x : n = n - 1 X k =0 v k +1 ( k + 1) k | q x = v q x + n - 1 X k =1 v k +1 ( k + 1) k | q x = v q x + n - 1 X k =1 v k +1 ( k + 1) k p x q x + k = v q x + n - 1 X k =1 v k +1 ( k + 1) p x k - 1 p x +1 q x + k = v q x + p x n - 1 X k =1 v k +1 ( k + 1) k - 1 p x +1 q x + k = v q x + v p x n - 1 X k =1 v k ( k + 1) k - 1 | q x +1 = v q x + v p x n - 1 X k =1 v k ( k ) k - 1 | q x +1 + n - 1 X k =1 v k k - 1 | q x +1 !

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ACTSC 232 Homework Problem - ACTSC 232 Spring 2011...

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