The Progress of Liberty

The Actuary’s Free Study Guide for Exam 3L - Section 41

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 41 of the Study Guide. See an index of all sections by following the link in this paragraph.

A discrete whole-life annuity-due makes payments every year at the beginning of each year, so long as the annuitant survives. For life (x), a discrete whole-life annuity-due paying a benefit of 1 per year has actuarial present value ä_x, which can be determined as follows:

ä_x = _k=0^∞Σv^k*_kp_x

If the discrete whole life annuity is an annuity-immediate, making payments at the beginning of each year, then the actuarial present value of such an annuity is denoted by a_x and can be found as follows.

a_x = _k=1^∞Σv^k*_kp_x

The following recursive relationship exists between ä_x and ä_x+1.

ä_x = 1 + v*p_x*ä_x+1

Moreover, if we know A_x, the actuarial present value of a whole life insurance policy with benefits payable at the end of the year of death (Reminder: A_x = _k=0^∞∑v^k+1*_kp_x*q_x+k), we can figure out ä_x as follows.

ä_x = (1 - A_x)/d

Here, d is the annual rate of discount, which compares to the annual effective interest rate i as follows: d = i/(1+i). Thus, ä_x = (1+i)(1 - A_x)/i.

The variance of the present value of a discrete whole-life annuity-due is denoted as Var(ä_K+1¬) and can be found as follows:

Var(ä_K+1¬) = (²A_x – (A_x)²)/d²

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 143-144, 146.

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

Problem S3L41-1. 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 currently 0.06. Triton the Triceratops is currently 7 years old and takes out a discrete whole life annuity-due paying a benefit of 1 Golden Hexagon (GH) at the beginning of each year. Find the actuarial present value of this annuity.

Solution S3L41-1. We use the formula ä_x = _k=0^∞Σv^k*_kp_x. Since the lifetimes of triceratopses are exponentially distributed, _kp_x = e^-0.34k for all x. Moreover, v^k = e^-0.06k. Thus,

ä₇ = _k=0^∞Σe^-0.06k*e^-0.34k = _k=0^∞Σe^-0.4k= 1/ (1-e^-0.4) = ä₇ = about 3.033244782 GH.

Problem S3L41-2. The life of a yellow whale has the following survival function associated with it: s(x) = e^-0.07x. The annual force of interest is currently 0.02. Horatius the Yellow Whale is currently 54 years old and takes out a discrete whole life annuity-immediate paying a benefit of 1 Golden Hexagon (GH) at the end of each year. Find the actuarial present value of this annuity.

Solution S3L41-2. We use the formula a_x = _k=1^∞Σv^k*_kp_x. Since the lifetimes of yellow are exponentially distributed, _kp_x = e^-0.07k for all x. Moreover, v^k = e^-0.02k. Thus,

a₅₄ = _k=1^∞Σe^-0.02k*e^-0.07k = _k=1^∞Σe^-0.09k = e^-0.09/(1 - e^-0.09) = a₅₄ = about 10.6186101 GH.

Problem S3L41-3. A 10-year-old chicken-cricket can obtain a discrete whole life annuity-due paying a benefit of 1 Golden Hexagon (GH) at the beginning of each year for 45 GH. The annual force of interest is 0.07, and only 89% of 10-year-old chicken-crickets survive to age 11. Find the actuarial present value of a discrete whole life annuity-due offering the same conditions to an 11-year-old chicken-cricket.

Solution S3L41-3. We use the formula ä_x = 1 + v*p_x*ä_x+1, which we rearrange as follows:

v*p_x*ä_x+1 = ä_x – 1

ä_x+1 = (ä_x – 1)/(v*p_x)

For x = 10, we are given that ä₁₀ = 45, v = e^-0.07, and p₁₀ = 0.89.

Thus, ä₁₁ = (45 – 1)/(e^-0.07*0.89) = ä₁₁ = about 53.02287638 GH. (This means that a chicken-cricket who survives to age 11 is a fortunate chicken-cricket indeed, having gotten over the mortality spike during the tenth year of life.)

Problem S3L41-4. A 15-year-old heptapus (think octopus with seven tentacles) can get a whole life insurance policy with a benefit of 1 Golden Hexagon (GH) payable at the end of the year of death for 0.67 GH. The annual effective rate of interest is 0.04. Find the actuarial present value of a discrete whole life annuity-due paying annual benefits of 1 GH and available to a 15-year-old heptapus.

Solution S3L41-4. We use the formula ä_x = (1+i)(1 - A_x)/i for x = 15, i = 0.04, and A₁₅ = 0.67.

Thus, ä₁₅ = (1.04)(1 – 0.67)/0.04 = ä₁₅ = 8.58 GH.

Problem S3L41-5. Under an annual force of interest of 0.05, the actuarial present value of a whole life insurance policy with a benefit of 1 Golden Hexagon (GH) payable at the end of the year of death for a 20-year-old squawking fish is 0.44 GH. The second moment of the present value of this policy is 0.31 GH. Find the variance of the present value of a discrete whole life annuity-due paying annual benefits of 1 GH and available to a 20-year-old squawking fish.

Solution S3L41-5. We use the formula Var(ä_K+1¬) = (²A_x – (A_x)²)/d².

For x = 20, we are given that ²A₂₀ = 0.31 and A₂₀ = 0.44.

We find d = r/(1+r) = (e^δ – 1)/e^δ = (e^0.05 – 1)/e^0.05 = d = 0.0487705755

Thus, Var(ä_K+1¬) = (0.31 – (0.44)²)/0.0487705755² = about 48.93698619.

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

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