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

A continuous n-year deferred life annuity begins to make continuous payments n years after the present time. The actuarial present value of an n-year deferred life annuity is denoted by _n│ā_x and can be found as follows:

_n│ā_x = _n^∞∫v^t*_tp_x*dt = _n^∞∫_tE_x*dt

If Y is the present value of such an n-year deferred life annuity, then

Var(Y) = (2/δ)v²ⁿ_np_x(ā_x+n - ²ā_x+n) - (_n│ā_x)²

A continuous n-year certain and life annuity makes guaranteed payments for the first n years of its existence, no matter what happens to the annuitant. After the first n-years, this annuity behaves the same as a life annuity. The actuarial present value on a continuous n-year certain and life annuity is denoted by the symbol ā^---_x:n¬ (with the line drawn directly over the "x:n¬" whenever one's writing and publishing interface permits) and can be found as follows.

ā^---_x:n¬ = ā_n¬+ _n^∞∫v^t*_tp_x*dt, where ā_n¬is the symbol representing the present value of a continuous n-year annuity-certain. We recall that ā_n¬= (1 - e^-nδ)/δ.

Moreover, the following relationships hold.

ā^---_x:n¬ = ā_n¬+ (ā_x - ā_x:n¬ )

ā^---_x:n¬ = ā_n¬+ _n│ā_x

Meaning of Symbols:

Ā_x = the actuarial present value of a whole life insurance policy for life (x).

²Ā_x = the second moment of the actuarial present value of a whole life insurance policy for life (x).

ā_x = the actuarial present value of a continuous whole life annuity for life (x).

ā_T¬= the present-value random variable for a continuous whole life annuity for an annuitant with a future-lifetime random variable of T.

δ = the annual force of interest.

ā_x:n¬= the actuarial present value of a continuous n-year term life annuity for life (x).

_tE_x= the actuarial present value of a t-year pure endowment for life (x).

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 136-140.

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

Problem S3L39-1. 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. Diocletian the Triceratops has a continuous 3-year deferred life annuity. What is the actuarial present value _3│ā_x of Diocletian's annuity?

Solution S3L39-1. We use the formula _n│ā_x = _n^∞∫v^t*_tp_x*dt.

Since the lifetimes of triceratopses are exponentially distributed, _tp_x = e^-0.34t for all x. Moreover, v^t = e^-0.06t. Thus, _3│ā_x = ₃^∞∫e^-0.06t* e^-0.34t *dt = ₃^∞∫e^-0.4t *dt = (-5/2)e^-0.4t│₃^∞ = (5/2)(e^-1.2) =

_3│ā_x = about 0.7529855298.

Problem S3L39-2. 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. Diocletian the Triceratops has a continuous 3-year deferred life annuity. What is the variance of Diocletian's annuity?

Solution S3L39-2. We use the formula Var(Y) = (2/δ)v²ⁿ_np_x(ā_x+n - ²ā_x+n) - (_n│ā_x)². Since the lifetimes of triceratopses are exponentially distributed, triceratopses exhibit a constant force of mortality. Conveniently enough, for constant forces of mortality, the actuarial present value of a life annuity does not depend on the initial age of the annuitant! Thus, ā_x+n= ā_x. We use the formula ā_x = 1/(δ + μ) = 1/(0.06 + 0.34) = 1/0.4 = 2.5 for triceratopses. Thus, ā_x+3 = 2.5 as well.

Moreover, from Solution S3L39-1, we know that _3│ā_x = 0.7529855298 and that ₃p_x = e^-0.34*3 =

e^-1.02. Here, 2n = 6, so v²ⁿ = e^-0.06*6 = e^-0.36.

We have only to find ²ā_x+3, which is the same as ²ā_x for triceratopses.

By the Rule of Moments, if : ā_x = 1/(δ + μ), then ²ā_x = 1/(2δ + μ) = 1/(0.12 + 0.34) = about 2.173913043.

Thus, Var(Y) = (2/0.06)e^-0.36*e^-1.02(2.5 - 2.173913043) - (0.7529855298)² =

Var(Y) = about 2.167562282.

Problem S3L39-3. 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. Jorgax the Triceratops has a continuous 3-year certain and life annuity. What is the actuarial present value

ā^---_x:3¬ of Jorgax's annuity?

Solution S3L39-3. We use the formula ā^---_x:n¬= ā_n¬ + _n│ā_x.
We know from Solution S3L39-1 that _3│ā_x = 0.7529855298.

We need to find ā_3¬ using the formula ā_n¬= (1 - e^-nδ)/δ.

ā_3¬= (1 - e^-0.18)/0.06 = ā_3¬ = about 2.745496476.

Thus, ā^---_x:3¬= 2.745496476 + 0.7529855298 = ā^---_x:3¬= about 3.498482006.

Problem S3L39-4. For 5-year-old jumping giraffes, the following assets giving the same continuous payoff have the following actuarial present values:

- A continuous whole life annuity: 45 Golden Hexagons (GH)

- A continuous 6-year temporary life annuity: 4.9 GH

- A continuous 6-year certain and life annuity: 45.7 GH.

Find the actuarial present value of a continuous 6-year annuity-certain for 5-year-old jumping giraffes.

Solution S3L39-4. We use the formula ā^---_x:n¬= ā_n¬+(ā_x - ā_x:n¬ ), which we rearrange as follows:

ā_n¬= ā^---_x:n¬ - ā_x + ā_x:n¬ .

We are given that ā₅ = 45, ā_5:6¬ ---_5:6¬ = 45.7.

Thus, ā_6¬ = 45.7 - 45 + 4.9 = ā_6¬ = 5.6 GH.

Problem S3L39-5. The future lifetimes of 11-year-old pygmy whales are uniformly distributed with probability of death 1/55 in every subsequent year until age 66. The annual force of interest is 0.1. Ymgyp the Pygmy Whale has a continuous 20-year certain and life annuity whose actuarial present value is ā^---_11:20¬ . Find ā^---_11:20¬ .

Solution S3L39-5. We use the formula ā^---_x:n¬ = ā_n¬ + _n^∞∫v^t*_tp_x*dt.

The probability that Ymgyp will survive to year t is _tp_x = (55 - t)/55.

The discount factor is v^t = e^-0.1t.

Here, n = 20 and ā_20¬ = (1 - e^-20δ)/δ = (1 - e^-2)/0.1 = 8.646647168.

Thus, ā^---_11:20¬ = 8.646647168 + ₂₀^∞∫e^-0.1t*((55 - t)/55)*dt.

ā^---_11:20¬ 20^∞∫e^-0.1t dt - ₂₀^∞∫(1/55)te^-0.1t dt.

₂₀^∞∫e^-0.1t dt = 10e ^-2 = about 1.353352832.

To find ₂₀^∞∫(1/55)te^-0.1t dt, we use the Tabular Method of integration by parts:

Sign.......u........dv

+......(1/55)t......e^-0.1t

-.......(1/55).....-10e^-0.1t

+..........0.......100e^-0.1t

Thus, ₂₀^∞∫(1/55)te^-0.1t dt = [(-10/55)te^-0.1t - (100/55)e^-0.1t ]│₂₀^∞ =

(200/55)e^-2+(100/55)e^-2 = about 0.738192454 .

Thus, ā^---_11:20¬= 8.646647168 + 1.353352832 - 0.738192454.

So ā^---_11:20¬= about 9.261807546.

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