Annually Decreasing Term Life Insurance: Practice Problems and Solutions

As in Section 21, the following is defined to be the present-value function.

z_t = Z = b_tv_t

z_t = Z is the present value, at policy issue, of the benefit payment.

b_tis the benefit function.

v_tis the discount function. v is the one-year discount factor by which a sum of money payable one year from now is multiplied to get its present value today. If the annual effective interest rate is r, then v = 1/(1+r).

An annually decreasing term life insurance policy will pay n unit in benefits if death occurs during year 1, n-1 units in benefits if death occurs during year 2, and n-k+1 in benefits if death occurs in year k. The following functions characterize annually decreasing term life insurance.

b_t = n -└t┘ for t ≤ n;

b_t = 0, t > n;

v_t = v^t for t ≥ 0;

Z = v^T(n -└T┘) for T ≤ n;

0 for T > n.

The actuarial present value of an annually decreasing term life insurance is denoted as (DĀ)¹_x:n¬ and can be found via the following formula.

(DĀ)¹_x:n¬ = ₀ⁿ∫(n -└t┘)v^t*_tp_x*μ_x(t)dt

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 108.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L32-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is 0.09. Cato the Triceratops is currently 5 years old has a 2-year annually decreasing life insurance policy, which will pay 2 Triceratops Currency Unit (TCU) if he dies within 1 year from now and 1 Triceratops Currency Units (TCU) if he dies between 1 and 2 years from now. Find the actuarial present value of this policy.

Solution S3L32-1. We use the formula (DĀ)¹_x:n¬ = ₀ⁿ∫(n -└t┘)v^t*_tp_x*μ_x(t)dt.

We are given that x = 5, n = 2, and v = e^-0.09.

We find _tp_x = s(5 + t)/s(5) = e^-0.34t.

We find μ_x(t) = -s'(x)/s(x) = 0.34e^-0.34t/e^-0.34t = 0.34.

We note that from t = 0 to t = 1, └t ┘= 0 → n -└t┘= 2 and from t = 1 to t = 2,

└t┘= 1 → n -└t┘= 1.

Thus, (DĀ)¹_5:2¬ = ₀²∫(2 -└t┘)e^-0.09t*e^-0.34t*0.34dt.

We can decompose this integral as follows:
(DĀ)¹_5:2¬ = ₀¹∫(2*0.34)e^-0.43tdt + ₁²∫0.34e^-0.43tdt

(DĀ)¹_5:2¬ = (-68/43)e^-0.43t│₀¹ + (-34/43)e^-0.43t│₁²

(DĀ)¹_5:2¬ = (68/43)(1 - e^-0.43) + (34/43)(e^-0.43 - e^-0.86)

(DĀ)¹_5:2¬ = about 0.7324460461

Problem S3L32-2. The life of a giant pin-striped cockroach has the following survival function associated with it: s(x) = 1 - x/94, for 0 ≤ x ≤ 94 and 0 otherwise. The annual force of interest is 0.02. Lothar the Giant Pin-Striped Cockroach is 91 years old and has a 3-year annually decreasing life insurance policy that will pay 3-k+1 Golden Hexagons (GH) if he dies in the kth year of the policy, up to the policy's expiration. Find the actuarial present value of Tullius's policy.

Solution S3L32-2. We use the formula (DĀ)¹_x:n¬ = ₀ⁿ∫(n -└t┘)v^t*_tp_x*μ_x(t)dt, noting that _tp_x*μ_x(t) = f_T(t). Since Lothar has not died for the first 91 years of life, and deaths for giant pin-striped cockroaches are uniformly distributed, Lothar has 1/3 probability of dying in each of the next 3 years. So _tp_x*μ_x(t) = 1/3.

We also know that v^t = e^-0.02t. Thus,

(DĀ)¹_91:3¬ = ₀³∫(3 -└t┘)(1/3)e^-0.02t= ₀¹∫(3)(1/3)e^-0.02t + ₁²∫(2)(1/3)e^-0.02t + ₂³∫(1/3)e^-0.02t

(DĀ)¹_91:3¬ = (-50)e^-0.02t │₀¹ + (-100/3)e^-0.02t │₁² + (-50/3)e^-0.02t │₂³

(DĀ)¹_91:3¬ = 50(1 - e^-0.02) + (100/3)(e^-0.02 - e^-0.04) + (50/3)(e^-0.04 - e^-0.06)

(DĀ)¹_91:3¬ =about 1.954122566

Problem S3L32-3. For burgundy crickets, the survival function is s(x) = (1 - 0.0625x²) for 0 ≤ x ≤ 4 and 0 otherwise. Among burgundy crickets, interest is determined in an unusual manner, and the discount factor v^t is equal to 2/(2 + t). Lysistrata the Burgundy Cricket is 1year old and has a 3-year annually decreasing term life insurance policy whose benefit in the last year of its existence is 1 Golden Hexagon (GH). Find the actuarial present value of this policy.

Solution S3L32-3. We use the formula (DĀ)¹_x:n¬ = ₀ⁿ∫(n -└t┘)v^t*_tp_x*μ_x(t)dt.

Here, x = 1 and n = 3.

_tp₁ = s(x + t)/s(x) = s(1 + t)/s(1) = (1 - 0.0625(1+t)²)/(1 - 0.0625(1)²) =

(1 - 0.0625(1+t)²)/(0.9375).

μ₁(t) = -s'(x)/s(x) = 0.125x/(1 - 0.0625x²) = 0.125*(1+t)/(1 - 0.0625*(1+t)²)

Thus, _tp₁*μ₁(t) = [(1 - 0.0625(1+t)²)/(0.9375)][0.125*(1+t)/(1 - 0.0625*(1+t)²)] =

0.125*(1+t)/(0.9375) = (2/15)(1+t)

Thus, v^t*_tp₁*μ₁(t) = [2/(2 + t)] (2/15)(1+t) = (4/15)(1+t)/(2 + t) = (4/15)(1 - 1/(2 + t))

So (DĀ)¹_1:3¬ = ₀³∫(3 -└t┘)(4/15)(1 - 1/(2 + t))dt.

(DĀ)¹_1:3¬ = ₀¹∫(3)(4/15)(1 - 1/(2 + t))dt + ₁²∫(2)(4/15)(1 - 1/(2 + t))dt + ₂³∫(4/15)(1 - 1/(2 + t))dt

(DĀ)¹_1:3¬ = [0.8t - 0.8ln(2+t)] │₀¹ + [(8/15)t - (8/15)ln(2+t)] │₁² + [(4/15)t - (4/15)ln(2+t)] │₂³

(DĀ)¹_1:3¬ = 0.8 - 0.8ln(3) + 0.8ln(2) + (16/15) - (8/15)ln(4) - (8/15) + (8/15)ln(3) + 0.8 - (4/15)ln(5) - (8/15) + (4/15)ln(4)

(DĀ)¹_1:3¬ = about 1.062692528.

Problem S3L32-4. Milton the Mortal takes out a 55-year annually decreasing term life insurance policy that would pay him $100,000 if he died 55 years from now. He has a 0.3 probability of dying 15 years from now, a 0.3 probability of dying 43 years from now, a 0.2 probability of dying 65 years from now, and a 0.2 probability of dying 120 years from now. Milton can foresee that the annual effective rate of interest will be 0.04 for the next 40 years and 0.01 for every year thereafter. Find the actuarial present value of Maximus's policy to him. Round your answer to the nearest cent.

Solution S3L32-4. The policy will only pay Milton if he dies 15 years from now or 43 years from now. If he dies 15 years from now, he will collect 100000(55-15+1) = $4,100,000.

If he dies 43 years from now, he will collect 100000(55-43+1) = $1,300,000.

Thus, the actuarial present value of Milton's policy is

0.3*4100000/1.04¹⁵ + 0.3*1300000/(1.04⁴⁰*1.01³) = about $761,819.02

Problem S3L32-5. Imhotep the Immortal is scamming life insurance companies again. This time, he takes out a 6-year decreasing term life insurance policy which pays $1,000,000 in the last year of its existence. He plans to fake his death at the end of each year, collect the life insurance payment, and then misrepresent his identity as that of another living policyholder with an identical life insurance policy to his own. Insurance companies have become more sophisticated at preventing Imhotep's attempts at fraud, and they especially focus on guarding against the largest losses. Thus, every year for which Imhotep expects to get $n million by fraud, his probability of getting caught is 0.15n. The annual force of interest is 0.1. Find the expected present value of the fraudulent life insurance scheme to Imhotep. Round your answer to the nearest cent.

Solution S3L32-5. Duringevery year k, where 1 ≤ k ≤ 6, Imhotep tries to collect $(7 - k) million. His probability of success is (1 - 0.15(7 - k)), and the discount factor applied to the kth year's amount is e^-0.1k.

Thus, the expected present value of Imhotep's scheme (in millions of dollars) is

6*0.1*e^-0.1 + 5*0.25*e^-0.2 + 4*0.4*e^-0.3 + 3*0.55*e^-0.4 + 2*0.7*e^-0.5 + 1*0.85*e^-0.6 = about $5,173,285.94.

See other sections of The Actuary's Free Study Guide for Exam 3L.