MATH 4513/RMBI 4220 Solution-7

MATH 4513/RMBI 4220 Solution-7 Problem 1: Find the 20th year terminal reserve for a fully discrete whole life insurance of 1 issued at age 50 for each of the following cases. (a) Exam LTAM Life Table at 5%. (b) Survival follows ℓx = 110 − x and i = .06. (c) Constant force of mortality of . 035 at all ages, and force of interest of 6%. Solution: Prospective form 20V50 = A70−P50 ·a¨70 = A70− A50 a¨50 ·a¨70, insurance form tVx = Ax+t−Ax 1−Ax , annuity form tVx = 1 − a¨x+t a¨x . (a) Prospective form: 20V50 = A70 − P50 · a¨70 = A70 − A50 a¨50 · a¨70 = .42818 − .18931 17.0245
· (12.0083) = .295. Insurance form: 20V50 = A70−A50 1−A50 = .42818−.18931 1−.18931 = .295. Annuity form: 20V50 = 1 − a¨70 a¨50 = 1 − 12.0083 17.0245 = .295. (b) A50 = 1 110−50 · a110¯−50| = .26936, A70 = 1 110−70 · a 110−70| = .37616, Insurance form: 20V50 = A70−A50 1−A50 = .37616−.26936 1−.26936 = .146. (c) For the constant force model, Ax = q q+i for all x. Therefore, tVx = Ax+t−Ax 1−Ax = 0 for all x and t. Problem 2: A whole life insurance policy of face amount $1, 000 is issued to (40). On the basis of some mortality table and a 4% interest rate, the following values are determined relative to the first year of the policy: (i) net premium = $20 (due at the start of the year), and (ii) terminal net reserve at the end of the year = $18. What is q40? A) .00274 B) .00280 C) .00285 D) .00290 E) .00300 Solution: 20 × (1, 04) − q40 × (1000) = p40 × (18) → q40 = 20(1.04)−18 1000−18 = .00285. Answer: C. 1 Problem 3: Solution: P¯ (10a¯40) = 10¯a40 a¯40::0| . With µ = .03 and δ = .05, we get ¯a40:10| = R 10 0 e −δt · tp40dt = R 10 0 e −.05t · e −.03tdt = 1−e−.8 .08 = 6.8834, and a¯40 = 1 µ+δ = 12.5, so that 10a¯40 = ¯a40 − a¯40:10 = 5.6166 (alternatively, 10    a¯40 = e −10δ · 10p40 · a¯50 = e −10(δ+µ) · 1 µ+δ = 5.61