- (ii The probability that two 80-year olds are both alive in 20 years is 0.01(iii There is an 8 chance of a 70-year old surviving the next 30

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MATH 471: Actuarial Theory I Homework #1: Fall 2009 Assigned August 26, due September 2 1. If F ( x ) = x 110 for 0 x 110 (de Moivre’s Law), find expressions for the following; where 0 x 110 for (a) – (b), and 0 t 70 for (c) – (e): (a) f ( x ) (b) s ( x ) (c) t p 40 (d) t q 40 (e) the p.d.f. of T(40) 2. Let s ( x ) = ( 100 100+ x ) 2 for x 0. Calculate: (a) The probability that (25) will survive at least 40 more years. (0.5739) (b) The probability that (30) will die within the next 20 years. (0.2489) (c) The probability that (35) will die between ages 45 and 60. (0.1549) 3. You are given the following information: (i) The probability that two 70-year olds are both alive in 20 years is 0.16.
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Unformatted text preview: (ii) The probability that two 80-year olds are both alive in 20 years is 0.01. (iii) There is an 8% chance of a 70-year old surviving the next 30 years. (iv) All lives are independent and have the same expected mortality. Calculate the probability that an 80-year old will survive to age 90. (0.5) 4. Suppose mortality follows modified (or generalized) de Moivre’s Law , where: s ( x ) = ( ω-x ω ) α for 0 ≤ x ≤ ω , α > 0. Note that regular de Moivre’s Law is a special case of this modified law with α = 1. Show that: t p x = ( ω-x-t ω-x ) α for 0 ≤ t ≤ ( ω-x ), α > 0....
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This note was uploaded on 02/14/2011 for the course MATH 471 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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- (ii The probability that two 80-year olds are both alive in 20 years is 0.01(iii There is an 8 chance of a 70-year old surviving the next 30

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