Math 630 Problem Set 2

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1 Math 63 Problem Set 2 1. AN n-year term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of.572. Given µ x+t =.7 and δ =.5, find n. (Answer: 11) 2. Given: Ā 1 x:n =.4275, δ =.55, µ x+t =.45. Calculate Āx:n. (Answer:.4775) 3. For a whole { life insurance of 1 on (x) with{ benefits payable at the moment of death,.4, < t 1.6, < t 1 given δ = and µ.5, t > 1 x+t =. Calculate the single.7, t > 1 benefit premium (i.e. EPV) for this insurance. (Answer: ) 4. Given: i = 5%, CFM, e x = 16.. Find 2 Ā x. (Answer:.664) 5. For a 5-year deferred whole life insurance of 1 payable at the moment of death, you are given that µ =.4 and δ =.1. Calculate the variance of the present value of the benefit. (Answer:.369) 6. A whole life insurance provides a death benefit of 1 at the moment of death plus a return of the net single premium with interest at δ =.8. The net single premium is calculated using µ =.4 and δ =.16. Calculate the net single premium. (Answer: 3/1) 7. Given that mortality follows DML with ω = 1 and δ =.6. Calculate Ā1 4:1. (Answer:.12533) 8. Mortality for (4) follows DML with ω = 1. Z represents the present value of a whole life insurence payable at the moment of death. δ =.6. Calculate V(Z). (Answer:.6578) 9. Given: A whole life insurance whose benefit is t for t ; µ =.4; d = ; benefits are payable at the moment of death. Calculate E(Z). (Answer: 4) 1. CFM, δ =.6, 2 Ā x =.25. Calculate (ĪĀ) x. (Answer: 4)

2 11. The purchase price of a washing machine is 6. Company ABC provides a 1-year warranty. In the event of failure for t, ABC will pay 6(1.1t)at time t of failure. There is a constant force of failure of.2, δ =.8. Calculate the APV (i.e. EPV) of the warranty given it is paid for at the time of purchase. (Answer: ) 12. Given: µ =.5, δ =.6, Z is the prsent value random variable of a 5-year deferred whole life insurance of 1 on (x) payable at the moment of death. Calculate the 9th percentile of Z. (Answer: ) 13. Z is the present value for a whole life insurance of 1 on (x) payable at the moment of death. Given that µ x+t =.1t and δ =.4, calculate the 8th percentile of Z. (Answer:.7655) 14. A maintenance contract on a hotel promises to replace burned out light bulbs at the end of each year for 3 years. The hotel has 1, light bulbs. The light bulbs are all new. If a replacement bulb burns out it will be replaced with a new bulb. Given that q =.1, q 1 =.3, q 2 =.5. Each light bulb costs 1. i =.5. Calculate the APV of this contract. (Answer: ) 15. Gary, age 3, is subject to µ =.12 and wants to buy a 3-year $1 endowment insurance. δ =.9. Determine the net single premium for this insurance. (Answer: ) 16. For a special 3-year term insurance on (x) you are given q x+k =.2(k + 1), k =, 1, 2. The following are death benefits payable at the end of year of death. k 1 2 b k+1 3, 35, 4, i =.6. Calculate the EPV. (Answer: ) 17. Given: DML with k q x = 1/75 for all x, δ =.5, Z is the present value of a whole life insurance of $1 payable at the end of the year of death issued to (x). Determine V(Z). (Answer:.6222) 18. For a 2-year pure endowment of 1 on (x), you are given 2 p x =.65 and V(Z) =.5E(Z) where Z is the present value. Calculate i. 19. Given: A 4 =.3, A 4:2 =.45, 2 p 4 =.9, i =.4. Calculate A 6. (Answer:.12186) (Answer: )

3 2. Given: A x+1 A x =.15, i =.6, q x =.5. Calculate A x + A x Given: 1(IA) 5 = , 1A 1 5:1 = 5.58, 1A 51 = 249.5, i = 6%. Calculate 1(IA) Deaths are UDD over each year of age. i =.1, q x =.5, q x+1 =.8. Calculate Ā1 x:2 (Answer: ) (Answer: 573.7) (Answer: ) 23. Assuming UDD and A (4) x = 1.248A (2) x, find i. (Answer:.2) 24. Problems from Chapter 5 of the text: Exercises Solutions or Hints to Selected Problems 2. Ā x:n = Ā1 x:n + ne x. However, Ā x = Ā1 x:n + ne x Ā x+n µ δ + µ = µ ne x δ + µ.45 = n E x.45 n E x =.5. So, Ā x:n = = Ā x = Ā1 x:1 + v1 1p x Āx+1. We have v 1 1p x = e.4 e.6 = e 1 = , 1 Ā 1 x:1 = v t tp x µ x+t dt = 1 e.4t e.6t (.6) dt = Ā x+1 = µ δ + µ = = Therefore, Ā x = ( ) ( ) = The EPV of 1 is 1Āx = Under CFM, e x = 1/µ µ = 1/16. i = 5% δ = ln(1.5). 2 Āx = 2 v t tp x µ x+t dt = 2 e δt e µt µ dt = = µ δ + µ e 2(δ+µ) = Let E be the single net premium. The present value of the benefit is then Z = v Tx (1 + Ee.8Tx ) = v Tx + Ee.8Tx v Tx where v = e.16 E = E(Z) = E(v Tx ) + EE(e.8Tx v Tx ) =.4 + E E = E E = 3/ Note that E(e.8Tx v Tx ) = E(e.8Tx e.16tx ) = E(e.8Tx ) is the EPV of a continuous.4 whole life insurance of 1 with µ =.4 and δ =.8, which is equal to..8+.4

4 8. First, recall that for DML, t p x µ x+t = 1 ω x. So, tp 4 µ 4+t = 1 E(Z) = 6 v t tp 4 µ 4+t dt = E(Z 2 ) = 6 6 e.6t 1 6 dt = 1 6 v 2t tp 4 µ 4+t dt = V(Z) = = (1 e.6 6 ) = e.12t 1 dt = First v = 1 d = = δ = ln v =.6 E(Z) = ( ĪĀ) = tv t x tp x µ x+t dt = te δt e µt µ dt =.4 Use integration by parts to finish. 1. First, 2 Ā x = µ.25 = µ µ =.4. 2δ+µ.12+µ (ĪĀ ) = tv t x tp x µ x+t dt = (Same as the previous problem.) te.6t e.4t.4 dt =, 4 te.1t dt te.1t dt 11. The is a 1-year term insurance with a variable warranty (i.e. benefit) payment of b t = 6(1.1t) at time t. The APV is E(Z) = 1 b t v t tp µ +t dt = First note that tp x = e t µ x+s ds = e.5t2 6(1.1t)e.8t e.2t.2 dt The present value is Z = v Tx = e.4tx. Let z = e.4t be the 8th percentile. Using a graph of z = e.4t, we have.8 = P (Z z ) = P (T x t ) = t p x = e.5t2. Solving the above equation for t, we get t = So, the 8th percentile is z = e.4t = Consider one bulb first. The possible replacement scenarios (outcomes) are listed in the following chart, where each is the duration of a year and a dot means that a replacement occurs at that time. Scenario PV of the Cost Probability p p 1 p 2 =.315 v 3 p p 1 q 2 =.315 v 2 p q 1 p =.243 v q p p 1 =.63 v 2 + v 3 p q 1 q =.27 v + v 2 q q p =.9 v + v 3 q p q 1 =.27 v + v 2 + v 3 q q q =.1

5 The average (mean) cost for one bulb is the sum of the products of the present value and its probability. Therefore, for 1 bulbs the value is 1 [.315v v (v + v 3 ) +.1(v + v 2 + v 3 ) ] = = 1v + 28v v 3 = Another solution is to use a chart to keep track of the replacements according to the given mortality rates (probabilities). p p 1 q p q 1 p q p 1 q 9 63 p q p q 1 9 q The numbers in red are the numbers of bulbs replaced at that times. The present value is then 1v + (27 + 1)v 2 + ( )v 3 = First, A 1 3:3 = vq 3 + v 2 1 q 3 + v 3 2 q 3, and 3 E 3 = v 3 3p 3. Then A 3:3 = A 1 3:3 + 3E 3. The answer is 1A 3: First note that k q x = k p x q x+k = 1 life time. Also note that v = e.5. = 1 ω x 75 ω x = 75, the maximal remaining 74 E(Z) = v k+1 k q x = 1 75 (v + v2 + + v 75 ) = k=

6 74 E(Z 2 ) = v 2(k+1) k q x = 1 75 (v2 + v v 15 ) = k= V(Z) = = Use A 4 = A 1 + 4:2 v2 2p 4 A 6 = (A 4:2 v 2 2p 4 ) + v 2 2p 4 A First, A 1 = vq 5:1 5 q 5 = A 1 (1 + i) p 5:1 5 = 1 q 5. Then use (IA) 5 = A 5 + vp 5 (IA) 51 = (A 1 + vp 5:1 5A 51 ) + vp 5 (IA) Ā 1 = i x:2 δ A1 where x:2 A1 = vq x:2 x + v 2 1 q x = vq x + v 2 p x q x A (4) x = i(2) A (2) i (4) x i(2) = ( 1+i 1) i (4) 4( 4 = i+1 = i 1+i 1 2

Insurance Benefits. Lecture: Weeks 6-8. Lecture: Weeks 6-8 (STT 455) Insurance Benefits Fall 2014 - Valdez 1 / 36

Insurance Benefits. Lecture: Weeks 6-8. Lecture: Weeks 6-8 (STT 455) Insurance Benefits Fall 2014 - Valdez 1 / 36 Insurance Benefits Lecture: Weeks 6-8 Lecture: Weeks 6-8 (STT 455) Insurance Benefits Fall 2014 - Valdez 1 / 36 An introduction An introduction Central theme: to quantify the value today of a (random)