Discrete Deferred Life Annuities and Certain and Life Annuities: Practice Problems and Solutions

The actuarial present value of a discrete n-year deferred whole life annuity-due is denoted as

_n│ä_xand can be found via the following formula.

_n│ä_x= _k=n^∞Σv^k*_kp_x

The actuarial present value of a discrete n-year deferred whole life annuity-immediate is denoted as _n│a_xand can be found via the following formula.

_n│a_x= _k=n+1^∞Σv^k*_kp_x

The actuarial present value of a discrete n-year certain and whole life annuity-due is denoted as ä^---_x:n¬(with a line drawn directly over the subscripts whenever possible) and can be found via the following formula.

ä^---_x:n¬= ä_n¬ + _k=n^∞Σv^k*_kp_x = ä_n¬ + _n│ä_x

(We recall from annuity theory that ä_n¬= (1+i)(1 - (1+i)^-n)/i, where i is the annual effective interest rate.)

The actuarial present value of a discrete n-year certain and whole life annuity-immediate is denoted as a^---_x:n¬(with a line drawn directly over the subscripts whenever possible) and can be found via the following formula.

a^---_x:n¬= a_n¬ + _k=n+1^∞Σv^k*_kp_x = a_n¬ + _n│a_x

(We recall from annuity theory that a_n¬= (1 - (1+i)^-n)/i.)

The actuarial accumulated value of an n-year temporary annuity-due paying 1 every year while life (x) survives is denoted by ﯖ_x:n¬("s" with two dots directly over it. The proxy symbol used here is the closest I could find.). It can be found as follows:

ﯖ_x:n¬ = ä_x:n¬/_nE_x = _k=0ⁿ⁺¹Σ(1/_n-kE_x+k)

We recall that _nE_x = vⁿ*_np_x.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 145-148.

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

Problem S3L43-1. 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. Montezuma the Triceratops is currently 5 years old and takes out a discrete 2-year deferred 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 S3L43-1. We use the formula _n│ä_x = _k=n^∞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, for n = 2,

_2│ä₅ = _k=2^∞Σe^-0.06k*e^-0.34k = _k=2^∞Σe^-0.4k = e^-0.8/(1 - e^-0.4) = about 1.362924736 GH.

Problem S3L43-2. 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 currently 0.06. Montezuma II the Triceratops is currently 5 years old and takes out a discrete 2-year certain and 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 S3L43-2. We use the formula ä^---_x:n¬ = ä_n¬ + _k=n^∞Σv^k*_kp_x = ä_n¬ + _n│ä_x. We know from Solution S3L43-2 that _2│ä₅ = 1.362924736.We now find ä_2¬ = (1+i)(1 - (1+i)^-2)/i.

We find i = e^0.06 - 1. Thus, ä_2¬ = (e^0.06)(1 - (e^-0.12))/(e^0.06 - 1) = ä_2¬ = about 1.941764535.

Thus, ä^---_5:2¬ = 1.941764535 + 1.362924736 = about 3.30468927 GH.

Problem S3L43-3. 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. Oywell the Yellow Whale is currently 15 years old and takes out a discrete 7-year deferred 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 S3L43-3. We use the formula _n│a_x = _k=n+1^∞Σv^k*_kp_xfor n = 7. Since the lifetimes of yellow whales are exponentially distributed, _kp_x = e^-0.07k for all x. Moreover, v^k = e^-0.02k. Thus,

_7│a₁₅ = _k=8^∞Σe^-0.02k*e^-0.07k = _k=8^∞Σe^-0.09k = e^-0.72/(1 - e^-0.09) = _7│a₁₅ = about 5.655384677 GH.

Problem S3L43-4. 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. Oywell II the Yellow Whale is currently 15 years old and takes out a discrete 7-year certain and 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 S3L43-4. We use the formula a^---_x:n¬ = a_n¬ + _k=n+1^∞Σv^k*_kp_x = a_n¬ + _n│a_x.

From Solution S3L43-3, it is known that _7│a₁₅ = 5.655384677.

We find i = e^0.02 - 1. Thus, a_7¬= (1 - e^-0.14)/(e^0.02 - 1) = about 6.466985083 GH.

Hence, a^---_15:7¬ = 5.655384677 + 6.466985083 = about 12.12236976 GH.

Problem S3L43-5. 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. Eduardo the Triceratops is currently 14 years old and takes out a discrete 5-year life annuity-due paying a benefit of 1 Golden Hexagon (GH) at the beginning of each year. Assuming that Eduardo survives to the end of the 5-year period, what will be the actuarial accumulated value of his annuity then?

Solution S3L43-5. We use the formula ﯖ_x:n¬ = ä_x:n¬/_nE_x.

To find ä_14:5¬, we use the formula ä_x:n¬ = _k=0^n-1Σv^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, for n = 5,

ä_14:5¬ = _k=0⁴Σe^-0.06k*e^-0.34k = _k=0⁴Σe^-0.4k = 1 + e^-0.4 + e^-0.8 + e^-1.2 + e^-1.6 = about 2.62273974 GH.

We find ₅E₁₄ = v⁵*₅p_x = e^-0.06*5e^-0.34*5 = e^-0.4*5 = e^-2.

Thus, our desired answer is 2.62273974e² = about 19.37957107 GH.

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