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

The actuarial present value of a whole life insurance policy for life (x) with one unit in benefits payable at the end of the year of death is denoted as A_x. The following formulas pertain to A_x.

A_x = _k=0^∞∑v^k+1*_kp_x*q_x+k

l_x*A_x = _k=0^∞∑v^k+1*d_x+k

Recursion formula: A_x = vq_x + vp_xA_x+1

l_x*(1 + i)A_x = l_xA_x+1 + d_x(1 - A_x+1)

A_x+1 - A_x = iA_x - q_x(1 - A_x+1)

In the last of these formulas, q_x(1 - A_x+1) is the annual cost of insurance, and i is the annual effective interest rate.

Meaning of variables:

_kp_x = probability that life (x) will survive for k more years.

q_x+k = probability that life (x+k) will die within 1 year.

v = 1/(1+i) = one-year present value discount factor.

l_x = number of survivors in a deterministic survivorship group to age x. l_x = l₀*_xp₀, where the original number of members in a deterministic survivorship group is l₀.

d_x = number of deaths aged x in a deterministic survivorship group.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 111-115.

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

Problem S3L34-1. The actuarial present value of a whole life insurance policy for three-year-old ordinary green gremlins that pays one Golden Hexagon (GH) in benefits at the end of the year of death is 0.44. The actuarial present value of a whole life insurance policy for four-year-old ordinary green gremlins that pays one Golden Hexagon (GH) in benefits upon death is 0.39. A three-year-old ordinary green gremlin has only a 0.77 probability of surviving to age 4. Find the annual effective interest rate.

Solution S3L34-1. We use the formula A_x+1 - A_x = iA_x - q_x(1 - A_x+1), where x = 3. We rearrange the formula thus:

iA_x = A_x+1 - A_x + q_x(1 - A_x+1)

i = x+1 - A_x + q_x(1 - A_x+1)]/A_x

We are given that q₃ = 1 - 0.77 = 0.23, A₃ = 0.44 and A₄ = 0.39.

Thus, i = [0.39- 0.44+ 0.23(1 - 0.39)]/0.44 = i = about 0.2052272727.

Problem S3L34-2. 10-year-old white elephants have a 0.46 probability of reaching age 11. The annual force of interest is 0.05. The actuarial present value of a whole life insurance policy for an 11-year-old white elephant that pays one Golden Hexagon (GH) in benefits at the end of the year of death is 0.77. Find the actuarial present value of a whole life insurance policy for a 10-year-old white elephant that pays one Golden Hexagon (GH) in benefits at the end of the year of death.

Solution S3L34-2. We use the formula A_x = vq_x + vp_xA_x+1, where x = 10. We are given A₁₁ = 0.77 and p₁₀ = 0.46, so q₁₀ = 1 - 0.46 = 0.54. Moreover, v = e^-0.05. Thus,

A₁₀ = e^-0.05(0.54 + 0.46*0.77) = A₁₀ = about 0.8505893514 GH.

Problem S3L34-3. 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. Jefferson the Triceratops is currently 19 years old and takes out a whole insurance policy paying a benefit of 1 Golden Hexagon (GH) at the end of the year of death. Find the actuarial present value of this policy.

Solution S3L34-3. We use the formula A_x = _k=0^∞∑v^k+1*_kp_x*q_x+k.

We know that δ = 0.06, so v = e^-0.06. We are given that x = 19.

We find _kp₁₉ = s(x + k)/s(x) = s(19 + k)/s(19) = e^-0.34(19+k)/e^-0.34(19) = e^-0.34k

We find q_19+k = 1 - s(20 + k)/s(19 + k) = 1 - e^-0.34.

Thus, A₁₉ = _k=0^∞∑e^-0.06(k+1)*e^-0.34k*(1 - e^-0.34)

A₁₉ = e^-0.06(1 - e^-0.34)_k=0^∞∑e^-0.06k*e^-0.34k

A₁₉ = e^-0.06(1 - e^-0.34)_k=0^∞∑e^-0.4k

A₁₉ = e^-0.06(1 - e^-0.34)/(1 - e^-0.4) = A₁₉ = 0.8233575754.

Problem S3L34-4. 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. Madison the Triceratops is currently 18 years old and takes out a whole insurance policy paying a benefit of 1 Golden Hexagon (GH) at the end of the year of death. Use a recursion formula and a previous solution to find the actuarial present value of this policy.

Solution S3L34-4. We already know from Solution S3L34-3 that for triceratopses, A₁₉ = 0.8233575754. We seek to find A₁₈. We use the formula A_x = vq_x + vp_xA_x+1, where x = 18 and v = e^-0.06.

Since the deaths of triceratopses follow an exponential distribution, it follows that for every x,

p_x = e^-0.34 and q_x = 1 - e^-0.34.

Thus, A₁₈ = (e^-0.06)(1 - e^-0.34 + 0.8233575754e^-0.34) = A₁₈ = 0.8233575754.

(This is as can be expected from an exponential distribution, where the values of p_x and q_x are each independent of the value of x. Thus, the value of the whole life insurance policy should not be expected to change on the basis of the age of the insured.)

Problem S3L34-5. There are 5470 7-year-old yellow bluebirds, of whom only 5012 will reach their 8^th birthday. The annual effective interest rate is 0.03, and the actuarial present value for a whole insurance policy for 8-year-old yellow bluebirds paying a benefit of 1 Golden Hexagon (GH) at the end of the year of death is 0.83. Find the actuarial present value for a whole insurance policy for 7-year-old yellow bluebirds paying a benefit of 1 Golden Hexagon (GH) at the end of the year of death.

Solution S3L34-5. We use the formula l_x*(1 + i)A_x = l_xA_x+1 + d_x(1 - A_x+1), which we rearrange thus:

A_x = [l_xA_x+1 + d_x(1 - A_x+1)]/[l_x*(1 + i)]. We are given that l₇ = 5470 and l₈ = 5012. Thus, l₈ - l₇ = d₇ = 458.

Moreover, i = 0.03 and A₈ = 0.83. Thus,

A₇ = [5470*0.83 + 458(1 - 0.83)]/[5470*1.03] = A₇ = about 0.8196446637.

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