Term Life Insurance: Practice Problems and Solutions

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).

n-year term life insurance makes a payment if and only if the insured person dies within n years of the policy's issue. 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 if t ≤ n;

b_t = 0 if t > n.

v_t = v^t for t ≥ 0;

Z = v^T if T ≤ n;

Z = 0 if T > n.

The expectation of Z, E[Z], is the actuarial present value of the insurance. E[Z] is also denoted as

Ā¹_x:n¬ for life (x) and can be found using the following formulas:

Ā¹_x:n¬ = ₀^∞∫z_t*f_T(t)dt

Ā¹_x:n¬ = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt

In the symbol Ā¹_x:n¬, the superscript "1" means that 1 unit of money is paid for a life currently aged x if that life dies prior to reaching the age of x + n. It is possible to have some other amount than 1, say, k, paid instead, in which case the symbol for the actuarial present value of the insurance would become Ā^k_x:n¬.

Meaning of variables:

f_T(t) = the probability density function of T, the time to death of life (x).

μ_x(t) = the force of mortality for life (x) at time t.

_tp_x = the probability that life (x) will survive to time x+t.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 94-95.

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

Problem S3L21-1. Imhotep the Immortal lives forever and has decided to commit life insurance fraud. He plans to take out a 56-year term life insurance policy for $100,000 and fake his own death 55 years from now, thereafter pretending to be his own non-existent son (Imhotep II) and claiming the benefits from his policy. Imhotep can foresee that the annual effective rate of interest will be 0.03 for the next 35 years and 0.05 for every year thereafter. What is the present value of his policy to him, given that he is sure of his ability to collect on it? Round your answer to the nearest cent.

Solution S3L21-1. Imhotep will collect $100,000 after 55 years. For the first 35 of those years, the annual discount factor will be 1/1.03. For the remaining 20 years, the annual discount factor will be 1/1.05. Thus, the present value of Imhotep's policy is 100000(1/1.03)³⁵(1/1.05)²⁰= about $13,394.03

Problem S3L21-2. Mandy the Mortal takes out a 56-year term life insurance policy for $100,000. She has a 0.02 probability of dying 5 years from now, a 0.1 probability of dying 23 years from now, a 0.6 probability of dying 35 years from now, a 0.2 probability of dying 53 years from now, and a 0.08 probability of dying 999 years from now. Mandy can foresee that the annual effective rate of interest will be 0.03 for the next 35 years and 0.05 for every year thereafter. Find the actuarial present value of Mandy's policy to her. Round your answer to the nearest cent.

Solution S3L21-2. For the first 35 years from now, the annual discount factor will be 1/1.03. For all subsequent years, the annual discount factor will be 1/1.05. Mandy will only collect if she dies within the next 56 years. That is, if she dies 999 years from now, she will not collect. Thus, the actuarial present value of Mandy's policy is 100000*(0.02/1.03⁵ + 0.1/1.03²³ + 0.6/1.03³⁵ + 0.2/(1.03³⁵*1.05¹⁸)) =

about $31068.52

Problem S3L21-3. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual effective interest rate in Triceratopsland is 0.07. Jerry the Triceratops is currently 3 years old has a 6-year term life insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) upon death. Find the actuarial present value of this policy. Set up the integral and then use any calculator to evaluate it.

Solution S3L21-3. We use the formula Ā¹_x:n¬ = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt.

We know that x = 3 and n = 6.

We find _tp₃ = s(x + t)/s(x) = s(3 + t)/s(3) = e^-0.34(3+t)/e^-0.34(3) = e^-0.34t

We find μ₃(t) = -s'(x)/s(x) = 0.34e^-0.34t/e^-0.34t = 0.34

We find v^t = (1/1.07)^t

Thus, Ā¹_3:6¬ = ₀⁶∫(1/1.07)^t*e^-0.34t*0.34dt.

Ā¹_3:6¬ = ₀⁶∫(0.34e^-0.34t/1.07^t)dt = about 0.7617676461 TCU.

Problem S3L21-4. Xerxes the Spiky Tarantula is 4 years old and has a 4-year term 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 Xerxes's policy. Set up the integral and evaluate by hand, using a calculator only for the numerical computations.

Solution S3L21-4. We use the formula Ā^k_x:n¬ = ₀^∞∫z_t*f_T(t)dt.

Here, k = b_t= 5, and v_t = (1/1.04)^t .

So z_t = b_tv_t = 5/1.04^t for t ≤ 4 and 0 otherwise.

It is given that f_T(t) =0.4462871026*0.64^t.

Thus, Ā⁵_4:4¬ = ₀⁴∫(5/1.04^t)*0.4462871026*0.64^tdt

Ā⁵_4:4¬ = ₀⁴∫2.231435513*0.6153846154^tdt

Ā⁵_4:4¬ = [2.231435513/ln(0.6153846154)]0.6153846154^t │₀⁴

Ā⁵_4:4¬ = [2.231435513/ln(0.6153846154)]0.6153846154⁴ -[2.231435513/ln(0.6153846154)]

Ā⁵_4:4¬ = about 3.936950241GH.

Problem S3L21-5. 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. Orgox the Giant Pin-Striped Cockroach is currently 23 years old and has a 40-year term life insurance policy which will pay 10 Golden Hexagons (GH) upon death. The annual effective interest rate is 0.11. Find the actuarial present value of Orgox's policy.

Solution S3L21-5. We use the formula Ā^k_x:n¬ = k* ₀ⁿ∫v^t*_tp_x*μ_x(t)dt.

We know that x = 23 and n = 40.

We find _tp₂₃ = s(x + t)/s(x) = s(23 + t)/s(23) = (1 - (23+t)/94)/(1 - 23/94) = (71 - t)/71

We find μ₂₃(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = (-1/94)/[(94 - x)/94] = 1/(94 - x) = 1/(94 - (23-t)) = 1/(71 - t). Conveniently enough, _tp₂₃* μ₂₃(t) = ((71 - t)/71)(1/(71 - t)) = 1/71.

We find v^t = (1/1.11)^t

Thus, Ā¹⁰_23:40¬ = 10*₀⁴⁰∫(0.9009009009)^t(1/71)dt = (10/71)*(1/ln(0.9009009009))(0.9009009009)^t │₀⁴⁰

= -1.349607606(0.9009009009)^t │₀⁴⁰ = 1.349607606(1 - 0.9009009009⁴⁰) = about 1.328844689 GH.

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