Math 630 Problem Set 2
1. AN n-year term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of 0.0572. Given µx+t= 0.007 and δ = 0.05, find n.
(Answer: 11) 2. Given: ¯A1
x:n|= 0.4275, δ = 0.055, µx+t= 0.045. Calculate ¯Ax:n|.
(Answer: 0.4775) 3. For a whole life insurance of 1000 on (x) with benefits payable at the moment of death,
given δ = 0.04, 0 < t ≤ 10
0.05, t > 10 and µx+t=
0.06, 0 < t ≤ 10
0.07, t > 10 . Calculate the single benefit premium (i.e. EPV) for this insurance.
(Answer: 593.868) 4. Given: i = 5%, CFM, e◦x= 16.0. Find 20|A¯x.
(Answer: 0.06064) 5. For a 5-year deferred whole life insurance of 1 payable at the moment of death, you are given that µ = 0.04 and δ = 0.1. Calculate the variance of the present value of the benefit.
(Answer: 0.030069) 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 = 0.08. The net single premium is calculated using µ = 0.04 and δ = 0.16. Calculate the net single premium.
(Answer: 3/10) 7. Given that mortality follows DML with ω = 100 and δ = 0.06. Calculate ¯A140:10|.
(Answer: 0.12533) 8. Mortality for (40) follows DML with ω = 100. Z represents the present value of a
whole life insurence payable at the moment of death. δ = 0.06. Calculate V(Z). (Answer: 0.06578) 9. Given: A whole life insurance whose benefit is t for t ≥ 0; µ = 0.04; d = 0.0582355;
benefits are payable at the moment of death. Calculate E(Z).
(Answer: 4) 10. CFM, δ = 0.06, 2A¯
x = 0.25. Calculate ( ¯I ¯A)x.
11. The purchase price of a washing machine is 600. Company ABC provides a 10-year warranty. In the event of failure for t ≤ 0, ABC will pay 600(1 − 0.1t)at time t of failure. There is a constant force of failure of 0.02, δ = 0.08. Calculate the APV (i.e. EPV) of the warranty given it is paid for at the time of purchase.
(Answer: 44.14533) 12. Given: µ = 0.05, δ = 0.06, Z is the prsent value random variable of a 5-year deferred whole life insurance of 1000 on (x) payable at the moment of death. Calculate the 90th percentile of Z.
(Answer: 628.19) 13. Z is the present value for a whole life insurance of 1 on (x) payable at the moment of
death. Given that µx+t= 0.01t and δ = 0.04, calculate the 80th percentile of Z.
(Answer: 0.7655) 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 10,000 light bulbs. The light bulbs are all new. If a replacement bulb burns out it will be replaced with a new bulb. Given that q0 = 0.1, q1 = 0.3, q2 = 0.5. Each light bulb costs 1. i = 0.05. Calculate the APV of
this contract.
(Answer: 6688.26) 15. Gary, age 30, is subject to µ = 0.12 and wants to buy a 3-year $1000 endowment
insurance. δ = 0.09. Determine the net single premium for this insurance.
(Answer: 787.61) 16. For a special 3-year term insurance on (x) you are given qx+k = 0.02(k + 1), k = 0, 1, 2.
The following are death benefits payable at the end of year of death.
k 0 1 2
bk+1 300,000 350,000 400,000
i = 0.06. Calculate the EPV.
(Answer: 36829.06) 17. Given: DML withk|qx = 1/75 for all x, δ = 0.05, 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: 0.06222) 18. For a 20-year pure endowment of 1 on (x), you are given 20px = 0.65 and V(Z) =
0.05E(Z) where Z is the present value. Calculate i.
(Answer: 0.102186) 19. Given: A40= 0.3, A40:20| = 0.45, 20p40 = 0.9, i = 0.04. Calculate A60.
20. Given: Ax+1− Ax = 0.015, i = 0.06, qx= 0.05. Calculate Ax+ Ax+1.
(Answer: 1.18318) 21. Given: 1000(IA)50= 4996.75, 1000A150:1| = 5.58, 1000A51= 249.05, i = 6%.
Calculate 1000(IA)51.
(Answer: 5073.07) 22. Deaths are UDD over each year of age. i = 0.10, qx = 0.05, qx+1 = 0.08.
Calculate ¯A1 x:2|
(Answer: 0.113592) 23. Assuming UDD and A(4)x = 1.00248A(2)x , find i.
(Answer: 0.02) 24. Problems from Chapter 5 of the text: Exercises 5.1–5.9.
Solutions or Hints to Selected Problems
2. ¯Ax:n| = ¯A1x:n|+nEx. However, ¯Ax = ¯A1x:n|+nExA¯x+n ⇒ µ δ + µ = 0.4275 +nEx µ δ + µ ⇒ 0.45 = 0.4275 +nEx· 0.45 ⇒ nEx = 0.05. So, ¯Ax:n| = 0.4275 + 0.05 = 0.4775. 3. A¯x= ¯A1x:10|+ v1010px· ¯Ax+10. We have v10 10px = e−0.4e−0.6 = e−1 = 0.3678794412, ¯ A1x:10|= Z 10 0 vttpxµx+tdt = Z 10 0 e−0.04te−0.06t(0.06) dt = 0.3792723353 ¯ Ax+10 = µ δ + µ = 0.07 0.05 + 0.07 = 0.5833333333 Therefore, ¯Ax = 0.3792723353 + (0.3678794412) · (0.5833333333) = 0.593868676 ⇒ The EPV of 1000 is 1000 ¯Ax = 593.87. 4. Under CFM, e◦x= 1/µ ⇒ µ = 1/16. i = 5% ⇒ δ = ln(1.05). 20|A¯x = Z ∞ 20 vttpxµx+tdt = Z ∞ 20 e−δte−µtµ dt = · · · = µ δ + µe −20(δ+µ) = 0.0606413904
6. Let E be the single net premium. The present value of the benefit is then Z = vTx(1 + Ee0.08Tx) = vTx + Ee0.08TxvTx
where v = e−0.16 ⇒ E = E(Z) = E(vTx) + EE(e0.08TxvTx) = 0.04
0.16+0.04 + E 0.04 0.08+0.04 ⇒
E = 204 +124 E ⇒ E = 3/10.
Note that E(e0.08TxvTx) = E(e0.08Txe−0.16Tx) = E(e−0.08Tx) is the EPV of a continuous
8. First, recall that for DML, tpxµx+t= ω−x1 . So, tp40µ40+t = 601. E(Z) = Z 60 0 vttp40µ40+tdt = Z 60 0 e−0.06t 1 60dt = 1 60· 1 0.06(1 − e −0.06·60 ) = 0.270188 E(Z2) = Z 60 0 v2ttp40µ40+tdt = Z 60 0 e−0.12t 1 60dt = 0.138785 ⇒ V(Z) = 0.138785 − 0.2701882 = 0.06578. 9. First v = 1 − d = 1 − 0.0582355 = 0.9417645 ⇒ δ = − ln v = 0.06 E(Z) = I ¯¯Ax = Z ∞ 0 tvttpxµx+tdt = Z ∞ 0 te−δte−µtµ dt = 0.04 Z ∞ 0 te−0.1tdt Use integration by parts to finish.
10. First, 2A¯ x = 2δ+µµ ⇒ 0.25 = 0.12+µµ ⇒ µ = 0.04. ¯ I ¯Ax= Z ∞ 0 tvttpxµx+tdt = Z ∞ 0 te−0.06te−0.04t0.04 dt = 0, 04 Z ∞ 0 te−0.1tdt (Same as the previous problem.)
11. The is a 10-year term insurance with a variable warranty (i.e. benefit) payment of bt = 600(1 − 0.1t) at time t. The APV is
E(Z) = Z 10 0 btvttp0µ0+tdt = Z 10 0 600(1 − 0.1t)e−0.08te−0.02t0.02 dt
13. First note that tpx= e− Rt
0µx+sds = e−0.005t2
The present value is Z = vTx = e−0.04Tx. Let z
0 = e−0.04t0 be the 80th percentile. Using
a graph of z = e−0.04t, we have
0.8 = P (Z ≤ z0) = P (Tx ≥ t0) =t0px = e
−0.005t2 0.
Solving the above equation for t0, we get t0 = 6.68047. So, the 80th percentile is
z0 = e−0.04t0 = 0.7655.
14. 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
— — — 0 p0p1p2 = 0.315 — — —· v3 p0p1q2 = 0.315 — —· — v2 p 0q1p0 = 0.243 —· — — v q0p0p1 = 0.063 — —· —· v2+ v3 p0q1q0 = 0.027 —· —· — v + v2 q 0q0p0 = 0.009 —· — —· v + v3 q 0p0q1 = 0.027 —· —· —· v + v2+ v3 q 0q0q0 = 0.001
The average (mean) cost for one bulb is the sum of the products of the present value and its probability. Therefore, for 10000 bulbs the value is
100000.315v3+ 0.243v2+ · · · + 0.027(v + v3) + 0.001(v + v2+ v3) = = 1000v + 2800v2+ 3700v3 = 6688.262607
Another solution is to use a chart to keep track of the replacements according to the given mortality rates (probabilities).
0 1 2 3 10000 @ @ @ @ @ @ R 9000 1000 p0 q0 * H H H H H H j * H H H H H H j 6300 2700 900 100 p1 q1 p0 q0 : XX XX XXz : XX XX XXz : XX XX XXz : XX XX XXz 3150 3150 2430 270 630 270 90 10 p2 q2 p0 q0 p1 q1 p0 q0
The numbers in red are the numbers of bulbs replaced at that times. The present value is then 1000v + (2700 + 100)v2+ (3150 + 270 + 270 + 10)v3 = 6688.262607 15. First, A1 30:3| = vq30+ v 2 1|q30+ v32|q30, and 3E30 = v33p30. Then A30:3| = A130:3|+3E30.
The answer is 1000A30:3|.
17. First note that k|qx = kpx· qx+k = ω−x1 = 751 ⇒ ω − x = 75, the maximal remaining
life time. Also note that v = e−0.05.
E(Z) = 74 X k=0 vk+1k|qx = 1 75(v + v 2 + · · · + v75) = 0.25394
E(Z2) = 74 X k=0 v2(k+1)k|qx= 1 75(v 2+ v4+ · · · + v150) = 0.126708 ⇒ V(Z) = 0.126708 − 0.253942 = 0.06222. 19. Use A40= A140:20|+ v2020p40A60 = (A40:20| − v2020p40) + v2020p40A60 21. First, A1 50:1| = vq50 ⇒ q50= A 1 50:1|(1 + i) ⇒ p50= 1 − q50. Then use
(IA)50= A50+ vp50(IA)51= (A150:1|+ vp50A51) + vp50(IA)51
22. ¯A1 x:2|= i δA 1 x:2| where A 1 x:2|= vqx+ v 2 1|qx = vqx+ v2pxqx+1. 23. A(4)x = i (2) i(4)A (2) x ⇒ i (2) i(4) = 1.00248 ⇒ 2(√1+i−1) 4(√41+i−1 = 1.00248 ⇒ 4 √ 1+i+1 2 = 1.00248 ⇒ i
