This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: MATH 472/567: Actuarial Theory II/Topics in Actuarial Theory I Final Exam Additional Problems: Spring 2011 Note: This problem set only includes problems for material that was NOT covered on the two midterms. The final exam is cumulative: refer to the previous two additional problem sets for the rest of the material. 1. Consider two independent lives, (25) and (40). Each life has mortality that follows the Illustrative Life Table. Calculate: (a) The probability that both (25) and (40) will survive the next 20 years. (0.8423) (b) The probability that exactly one of (25) and (40) will survive the next 20 years. (0.1526) (c) The probability that at least one of (25) and (40) will die within the next 20 years. (0.1577) (d) The probability that both (25) and (40) will die within the next 20 years. (0.0051) 2. You are given: (i) For males: s ( x ) = 1 - x 75 , 0 x 75. (ii) For females: mortality follows de Moivres Law with limiting age . (iii) At age 60, the force of mortality for females is 60% of the force of mortality for males. For two independent lives, a male age 65 and a female age 60, calculate the expected time until the first death. (4.33) 3. For independent lives (35) and (45): (i) 5 p 35 = 0.90 (ii) 5 p 45 = 0.80 (iii) q 40 = 0.03 (iv) q 50 = 0.05 Calculate: 5 | q 35:45 . (0.0105) 1 4. For two independent lives ages 30 and 34: x 30 31 32 33 34 35 36 37 q x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Calculate the probability that the last death of these two lives will occur during the fourth year. (0.2230) 5. Suppose both (x) and (y) have independent future lifetimes and mortality that is subject to = 0.05. Calculate: (a) e xy . (30) (b) var [ T ( xy )]. (500) (c) cov [ T ( xy ) , T ( xy )]. (100) 6. For a special fully continuous last-survivor insurance of 1 on independent lives (x) and (y): (i) The death benefit is payable at the moment of the second death. (ii) Level annual benefit premiums of are payable only while exactly one of (x) and (y) is alive. (iii) = 0.05, and x ( t ) = y ( t ) = 0.04 for t > 0. Calculate: . (0.04) 7. For a last-survivor insurance of 10,000 on independent lives (70) and (80): (i) The benefit, payable at the end of the year of death, is paid only if the second death occurs during year 5....
View Full Document
- Spring '08