Math 472 HW 2 Solutions

# Math 472 HW 2 Solutions - annual beneﬁt premiums are...

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MATH 472/567: Actuarial Theory II/Topics in Actuarial Theory I Homework #2: Spring 2011 Assigned January 26, due February 2 1. Re-do Problem 3(b) in the “Chapter 8: Lecture Examples” handout, using a retrospective approach to calculate 2 ¯ V . (27.76) 2. For a special fully continuous life insurance on (40): (i) b t = 0 for 0 t < 15; b t = 1000 e 0 . 04 t for t 15. (ii) π t = π for 0 t < 15; π t = 0 for t 15. (iii) Mortality follows de Moivre’s Law with a limiting age of 90. (iv) δ = 0.06 (a) Calculate π , using the equivalence principle. (43.23) (b) Calculate 20 ¯ V , the benefit reserve at the end of the twentieth year. (1673.56) 3. For a fully discrete 10-payment whole life insurance of 100,000 on (x): (i) i = 0.05 and q x +9 = 0.011 (ii) The level benefit premium is 2078. (iii) The terminal benefit reserve for the ninth year is 32,535. Calculate the initial benefit reserve for year 11. (35,635.64) ————THERE ARE MORE PROBLEMS ON THE BACK ———— 1

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4. For a fully discrete whole life insurance of 1000 on (45), you are given that

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Unformatted text preview: annual beneﬁt premiums are level and: k 1000 k V 45 q 45+ k 22 235 0.015 23 255 0.020 24 272 0.025 Calculate the 25th year terminal beneﬁt reserve. (286.04) 5. For a fully discrete whole life insurance with non-level beneﬁts on (70): (i) The annual level beneﬁt premium for this insurance is equal to P 50 . (ii) q 70+ k = q 50+ k + 0.01 for k = 0, 1, . .., 19 (iii) q 60 = 0.01368 (iv) k V = k V 50 for k = 0, 1, . .., 19 (v) 11 V 50 = 0.16637 Calculate the net amount at risk for year 11. (0.4816) 6. For a special fully discrete 3-year endowment insurance on (65): (i) The death beneﬁt is 5000 plus the beneﬁt reserve at the end of the year of death. (ii) The pure endowment beneﬁt is 5000. (iii) The annual level beneﬁt premium is π . (iv) p 65+ h = 0.97 for h = 0, 1, 2. (v) i = 0.06 Calculate: 1 V . (1570.55) 2...
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