Endowment Insurance Payable at the End of the Year of Death: Practice Problems and Solutions

As in Section 33, the present value of the benefit payment for an insurance policy with the payment made at the end of the year of death is z_k+1 and is expressed as follows.

z_k+1 = b_k+1v_k+1.

z_k+1 = Z is the present value, at policy issue, of the benefit payment.

b_k+1is the benefit function.

V_k+1is 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).

Then for a policy of n-year endowment insurance with one unit in benefits payable at the end of the year of death, the following functions apply.

b_k+1 = 1 for all positive integer values k;

v_k+1 = v^k+1 for all integer values of k such that 0 ≤ k ≤ n-1;

v_k+1 = vⁿ for all integer values of k such that n ≤ k;

Z = v^K+1 for all integer values of K such that 0 ≤ K ≤ n-1;

Z = vⁿ for all integer values of K such that n ≤ K.

The actuarial present value of an n-year endowment insurance policy paying one unit in benefits at the end of the year of death is A_x:n¬ and can be found as follows.

A_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k + vⁿ*_np_x

A_x:n¬ = A¹_x:n¬ + A_x:n¹_¬,

Where A¹_x:n¬ is the actuarial present value of an n-year term life insurance policy paying one unit in benefits at the end of the year of death and A_x:n¹_¬ is the actuarial present value of an n-year pure endowment. (Please, please forgive me for this notation! I did not invent it, and I recognize the rather confusing nature of notation for two different concepts where the only difference in notation is in the horizontal placement of the superscript 1. I henceforth take the liberty of calling the actuarial present value of an n-year pure endowment Ǻ_x:n¹_¬ , which means that A_x:n¬ = A¹_x:n¬ + Ǻ_x:n¹_¬.)

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 115.

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

Problem S3L35-1. The actuarial present value of a 10-year term life insurance policy for newborn red bluefish paying one unit in benefits at the end of the year of death is 0.56. The actuarial present value of a 10-year endowment insurance policy for newborn red bluefish paying one unit in benefits at the end of the year of death is 0.88. Find the actuarial present value of a 10-year pure endowment for newborn red bluefish paying one unit in benefits.

Solution S3L35-1. We use the formulaA_x:n¬ = A¹_x:n¬ + Ǻ_x:n¹_¬, which we rearrange thus:

Ǻ_x:n¹_¬ = A_x:n¬ - A¹_x:n¬
We are given that A_0:10¬ = 0.88 and A¹_0:10¬ = 0.56.

Thus, Ǻ_0:10¹_¬ = 0.88 - 0.56 = Ǻ_0:10¹_¬ = 0.32.

Problem S3L35-2. For 2-year-old white-whiskered wallabies, the actuarial present value of a 13-year term life insurance policy paying one unit in benefits at the end of the year of death is twice the actuarial present value of a 13-year pure endowment paying one unit in benefits, which is three times the present value of one dollar payable in 13 years at an annual force of interest of 0.23. Find the actuarial present value of a 13-year endowment insurance policy for 2-year-old white-whiskered wallabies, paying one unit in benefits at the end of the year of death.

Solution S3L35-2. The present value of one dollar payable in 13 years at an annual force of interest of 0.23 is e^-0.23*13 = 0.0502874367.

Thus, Ǻ_2:13¹_¬ = 3*0.0502874367 = 0.1508623102.

We are also given that A¹_2:13¬ = 2*Ǻ_2:13¹_¬ = 0.3017246203.

Thus, A_2:13¬ = A¹_2:13¬ + Ǻ_2:13¹_¬ = 0.1508623102 + 0.3017246203 = about 0.4525869305.

Problem S3L35-3. 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 currently 0.06. Alaric the Triceratops is currently 12 years old and takes out a 3-year endowment insurance policy paying a benefit of 1 at the end of the year of death. Find the actuarial present value of this policy.

Solution S3L35-3. We use the formula A_x:n¬ = _k=0^n-1∑v^k+1*_kp_x*q_x+k + vⁿ*_np_x.

Here, x = 12 and n = 3.

For an exponential distribution with survival function s(x) = e^-0.34x, _kp_x = e^-0.34k

and _np_x = e^-0.34n. Here, ₃p_x = e^-0.34n = e^-1.02.

We know that v = e^-0.06 and so vⁿ = e^-0.06*3 = e^-0.18.

So vⁿ*_np_x = e^-0.18e^-1.02 = e^-1.2.

We find q_12+k = 1 - s(13 + k)/s(12 + k) = 1 - e^-0.34.

So _k=0²∑ e^-0.06(k+1)*e^-0.34k*(1 - e^-0.34) =

e^-0.06*(1 - e^-0.34) + e^-0.12*e^-0.34*(1 - e^-0.34) + e^-0.18*e^-0.68*(1 - e^-0.34) =

(1 - e^-0.34)(e^-0.06 + e^-0.46 + e^-0.86) = about 0.5753670393.

To get A_12:3¬ , we add 0.5753670393 + e^-1.2 = A_12:3¬ = about 0.8765612512.

Problem S3L35-4. Mauricius the Mortal has a 0.36 probability of dying in 34.5 years, a 0.2 probability of dying in 44.2 years, a 0.14 probability of dying in 54.6 years, and a 0.3 probability of dying in 1004.7 years. The annual force of interest hereafter and forevermore is 0.04. Mauricius has a 50-year endowment insurance policy that pays 100,000 Golden Hexagons (GH) at the end of the year of death. Find the actuarial present value of this policy.

Solution S3L35-4. The values of the curate-future-lifetime random variable K for Mauricius are either 34, 44, 54, or 1004. Because this is a 50-year endowment insurance policy, the following present value factors correspond to these values of K:

34 → v³⁵

44 → v⁴⁵

54 → v⁵⁰

1004 → v⁵⁰

Thus, the probability of the present-value factor being v⁵⁰ is 0.14 + 0.3 = 0.44.

Hence, the actuarial present value of Mauricius's policy is

100000(0.36*e^-0.04*35+ 0.2*e^-0.04*45 + 0.44*e^-0.04*50) = about $18,138.22

Problem S3L35-5. 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 currently 0.06. For 12-year old triceratopses, find the value of n for which an n-year pure endowment with one unit in benefits half the actuarial present value of a 3-year pure endowment with one unit in benefits.

Solution S3L35-5. The value of an n-year pure endowment is vⁿ*_np_x, which for triceratopses is

e^-0.06n*e^-0.34n= e^-0.4n. The value of a 3-year pure endowment, then, is e^-0.4*3 = e^-1.2.

Then we want to find n such that e^-0.4n = (1/2)e^-1.2.

This means that n = ln((1/2)e^-1.2)/-0.4 = n = about 4.732867951 years.

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