# Whole Life Insurance: Practice Problems and Solutions

**This section of sample problems and solutions is a part of** **The Actuary's Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 23 of the Study Guide. See an index of all sections by following the link in this paragraph.**

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

**z _{t} = Z = b_{t}v_{t}**

**z _{t} = Z** is the present value, at policy issue, of the benefit payment.

**b _{t}**is the

**benefit function.**

**v _{t}**is 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).

**Whole life insurance** makes a payment at the time of death of the insured person, no matter when that time might be. A policy for which a sum of 1 is paid at the death of the insured person has the following functions associated with it, where t is the time from the present moment until death.

**b _{t}** =

**1 for t ≥ 0;**

**v _{t}** =

**v**

^{t}for t ≥ 0;**Z** = **v ^{T} for t ≥ 0.**

The **actuarial present value** for life (x) of a whole life insurance policy for which the benefit is 1 is denoted as **E[Z] = ? _{x} = _{0}^{∞}∫v^{t}*_{t}p_{x}*μ_{x}(t)dt.**

**Source:** Bowers, Gerber, et. al. *Actuarial Mathematics.* 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp.96-97.

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

**Problem S3L23-1.** Maximus the Mortal takes out a whole life insurance policy for $100,000. 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. Maximus 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 S3L23-1.** For the first 40 years from now, the annual discount factor will be 1/1.04. For all subsequent years, the annual discount factor will be 1/1.01. Maximus's estate will collect $100,000 at the time of his death irrespective of when he dies. Thus, the actuarial present value of Maximus's policy is

100000*(0.3/1.04^{15} + 0.3/(1.04^{40}*1.01^{3})+ 0.2/(1.04^{40}*1.01^{25})+ 0.2/(1.04^{40}*1.01^{80})) =

**about $27850.44**.

**Problem S3L23-2.** 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 0.07. François the Triceratops is currently 3 years old has a whole life insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) upon death. Find the actuarial present value of this policy.

**Solution S3L23-2.** We use the formula ?_{x} = _{0}^{∞}∫v^{t}*_{t}p_{x}*μ_{x}(t)dt.

We know that x = 3 and δ = 0.07.

We find _{t}p_{3} = s(x + t)/s(x) = s(3 + t)/s(3) = e^{-0.34(3+t)}/e^{-0.34(3)} = e^{-0.34t}

We find μ_{3}(t) = -s'(x)/s(x) = 0.34e^{-0.34t}/e^{-0.34t} = 0.34

We find v^{t} = e^{-0.07t}

Thus, ?_{3} = _{0}^{∞}∫ e^{-0.07t}*e^{-0.34t}*0.34dt.

?_{3} = _{0}^{∞}∫0.34e^{-0.41t}dt = (-34/41)e^{-0.41t}?_{0}^{∞} = **34/41 = about 0.8292682927 TCU**.

**Problem S3L23-3.** Lysander the Spiky Tarantula is 4 years old and has a whole life insurance policy, which will pay 5 Golden Hexagons (GH) upon death. The probability density function for the future lifetime of 4-year-old spiky tarantulas is f_{T}(t) =0.4462871026*0.64^{t}. The annual effective interest rate is 0.04. Find the actuarial present value of Lysander's policy.

**Solution S3L23-3.** We use the formula ?_{x} = _{0}^{∞}∫v^{t}*_{t}p_{x}*μ_{x}(t)dt = _{0}^{∞}∫v^{t}*f_{T}(t)dt.

We want to find 5?_{4}. We know that f_{T}(t) =0.4462871026*0.64^{t} and v^{t} = (1/1.04)^{t}.

Thus,

5?_{4} = _{0}^{∞}∫(5/1.04^{t})*0.4462871026*0.64^{t}dt

5?_{4} = _{0}^{∞}∫2.231435513*0.6153846154^{t}dt

5?_{4} = [2.231435513/ln(0.6153846154)]0.6153846154^{t} ?_{0}^{∞}

5?_{4} = -[2.231435513/ln(0.6153846154)]

**5? _{4} =about 4.596085667 GH.**

**Problem S3L23-4.** 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. Hcaorkcoc the Giant Pin-Striped Cockroach is currently 56 years old and has a whole life insurance policy which will pay 10 Golden Hexagons (GH) upon death. The annual force of interest is 0.02. Find the actuarial present value of Hcaorkcoc's policy.

**Solution S3L23-4.** We use the formula ?_{x} = _{0}^{∞}∫v^{t}*_{t}p_{x}*μ_{x}(t)dt.

We want to find 10?_{56}. Since no giant pin-striped cockroach lives past the age of 94, our integral's upper bound will be 94-56 = 38, because Hcaorkcoc will not live for more than 38 additional years.

We find _{t}p_{56} = s(x + t)/s(x) = s(56 + t)/s(56) = (1 – (56+t)/94)/(1 – 56/94) = (38 – t)/38

We find μ_{56}(t) = -s'(x)/s(x) = (1/94)/(1 – x/94) = (-1/94)/[(94 – x)/94] = 1/(94 – x) =

1/(94 – (56+t)) = 1/(38 – t). Conveniently enough, _{t}p_{56}* μ_{56}(t) = ((38 – t)/38)(1/(38 – t)) = 1/38. We find v^{t} = e^{-0.02t}.

Thus, 10?_{56} = 10*_{0}^{38}∫e^{-0.02t}*(1/38)dt

10?_{56} = 10*(-50/38)e^{-0.02t}?_{0}^{38} = 10(50/38)(1 – e^{-0.76}) = **10? _{56} = about 7.004389118 GH**.

**Problem S3L23-5.** For burgundy crickets, the survival function is s(x) = (1 – 0.0625x^{2}) 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 (1/3)/(1/3 + (1/6)t). Hatshepsut the Burgundy Cricket is 2 years old and has a whole life insurance policy paying 1 Golden Hexagon (GH) upon death. Find the actuarial present value of this policy.

**Solution S3L23-5.**

We use the formula ?_{x} = _{0}^{∞}∫v^{t}*_{t}p_{x}*μ_{x}(t)dt.

We want to find ?_{2}. Since no giant pin-striped cockroach lives past the age of 4, our integral's upper bound will be 4-2 = 2, because Hatshepsut will not live for more than 2 additional years.

_{t}p_{2} = s(x + t)/s(x) = s(2 + t)/s(2) = (1 – 0.0625(2+t)^{2})/(1 – 0.0625(2)^{2}) =

(1 – 0.0625(2+t)* ^{2}*)/(0.75) = (0.75 – 0.25t – 0.0625t

^{2})/(0.75) =

_{t}p

_{2}= 1 – (1/3)t – (1/12)t

^{2}.

μ_{2}(t) = -s'(x)/s(x) = 0.125x/(1 – 0.0625x^{2}) = 0.125*(2+t)/(1 – 0.0625*(2+t)^{2}) =

(0.25 + 0.125t)/(0.75 – 0.25t – 0.0625t^{2}) = μ_{2}(t) = (1/3 + (1/6)t)/(1 – (1/3)t – (1/12)t^{2}).

It is given that v^{t} =(1/3)/(1/3 + (1/6)t).

Thus, ?_{2} = _{0}^{2}∫v^{t}*_{t}p_{2}*μ_{2}(t)dt =

_{0}^{2}∫[(1/3)/(1/3 + (1/6)t)](1 – (1/3)t – (1/12)t^{2})((1/3 + (1/6)t)/(1 – (1/3)t – (1/12)t^{2}))dt =

_{0}^{2}**∫(1/3)dt =? _{2} = 2/3 = about 0.6666666666**

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