Continuously Increasing Whole Life Insurance: Practice Problems and Solutions

As in Section 21, the following is defined to be the present-value function.

z_t = Z = b_tv_t

z_t = Z is the present value, at policy issue, of the benefit payment.

b_tis the benefit function.

v_tis 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).

A countinuously increasing whole life insurance policy will pay n unit in benefits if death occurs at time n, irrespective of whether n is a whole number or not. The following functions characterize continuously increasing whole life insurance.

b_t = t for t ≥ 0;

v_t = v^t for t ≥ 0;

Z = Tv^Tfor T ≥ 0.

The actuarial present value of a continuously increasing whole life insurance policy is denoted as (ĪĀ)_x and can be found using the following formula:

(ĪĀ)_x = ₀^∞∫t*v^t*_tp_x*μ_x(t)dt = ₀^∞∫ _s│Ā_x ds

Meaning of variables:

_tp_x = probability that life (x) will survive for t more years.

μ_x(t) = force of mortality that life (x) will experience at age (x + t).

_s│Ā_x= the actuarial present value of an s-year deferred life insurance policy that pays one unit in benefits.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 106-107.

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

Problem S3L31-1. An s-year deferred life insurance policy on the life of a 3-year-old rabbitskinned rabbit has the following actuarial present value: e^-0.02s. Find the actuarial present value of a continuously increasing whole life insurance policy on the life of a 3-year-old rabbitskinned rabbit.

Solution S3L31-1. We use the formula (ĪĀ)_x = ₀^∞∫ _s│Ā_x ds. We are given _s│Ā₃ = e^-0.02s.

Thus, (ĪĀ)₃ = ₀^∞∫e^-0.02sds = (-50)e^-0.02s│₀^∞ = (ĪĀ)₃ =50.

Problem S3L31-2. 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 0.09. Anaxagoras the Triceratops is currently 10 years old has a continuously increasing whole life insurance policy, which will pay n Triceratops Currency Unit (TCU) if he dies at time n after the policy's inception. Find the actuarial present value of this policy.

Solution S3L31-2. We use the formula (ĪĀ)_x = ₀^∞∫t*v^t*_tp_x*μ_x(t)dt.

We are given that x = 10 and v = e^-0.09.

We find _tp_x = s(10 + t)/s(10) = e^-0.34t.

We find μ_x(t) = -s'(x)/s(x) = 0.34e^-0.34t/e^-0.34t = 0.34.

Thus, (ĪĀ)₁₀ = ₀^∞∫t*v^t*_tp_x*μ_x(t)dt = ₀^∞∫t* e^-0.09t* e^-0.34t *0.34dt = ₀^∞∫0.34t*e^-0.43tdt

We use the Tabular Method of Integration by Parts:

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

+......0.34t....e^-0.43t

-........0.34...(-100/43)e^-0.43t

+......0.......(10000/1849)e^-0.43t

Thus, (ĪĀ)₁₀ = [0.34t(-100/43)e^-0.43t - 0.34(10000/1849)e^-0.43t ]│₀^∞

= 0.34(10000/1849) = (ĪĀ)₁₀ =about 1.8388318 TCU.

Problem S3L31-3. From his current vantage point, Miltiades the Mortal has a probability of 0.04 of dying in every year starting now, where 0 ≤ t ≤ 25. He takes out a continuously increasing whole life insurance policy that pays n Golden Hexagons (GH) if he dies at time n. The annual force of interest is 0.08. Find the actuarial present value of Miltiades's policy.

Solution S3L31-3. We use the formula (ĪĀ)_x = ₀^∞∫t*v^t*_tp_x*μ_x(t)dt, noting that _tp_x*μ_x(t) = f_T(t) = 0.04 for 0 ≤ t ≤ 25. Also, v^t = e^-0.08t. Thus,

(ĪĀ)_x = ₀²⁵∫0.04te^-0.08t dt, since the longest Miltiades can survive is 25 years.

We use the Tabular Method of Integration by Parts:

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

+.......0.04t.....e^-0.08t

-........0.04....(-100/8)e^-0.08t

+.........0......(10000/64)e^-0.08t

Thus, (ĪĀ)_x = [0.04t(-100/8)e^-0.08t - 0.04(10000/64)e^-0.08t] │₀²⁵ = [ (-t/2)e^-0.08t - 6.25e^-0.08t] │₀²⁵

= (-12.5)e^-0.08*25 - 6.25e^-0.08*25 + 6.25 = (-12.5)e^-2 - 6.25e^-2 + 6.25 = 6.25 - 18.75e^-2 =

6.25 - 18.75(0.135335283) = 6.25 - 2.53753655625 = about 3.7124634475 GH.

Problem S3L31-4. The life of a giant pin-striped cockroach has the following survival function associated with it: s(x) = 1 - x/94, for 0 ≤ x ≤ 94 and 0 otherwise. The annual force of interest is 0.02. Depirtsnip the Giant Pin-Striped Cockroach is 60 years old and has a continuously increasing whole life insurance policy that will pay n Golden Hexagons (GH) if he dies in year n. Find the actuarial present value of Depirtsnip's policy.

Solution S3L31-4. We use the formula (ĪĀ)_x = ₀^∞∫t*v^t*_tp_x*μ_x(t)dt, noting that _tp_x*μ_x(t) = f_T(t).

Since Depirtsnip has not died for the first 60 years of life, and deaths for giant pin-striped cockroaches are uniformly distributed, Depirtsnip has 1/34 probability of dying in each of the next 34 years. So _tp_x*μ_x(t) = 1/34. The upper bound of our integral is not infinity, but 34, since Depirtsnip can live at most 34 more years.

We also know that v^t = e^-0.02t. Thus, (ĪĀ)₆₀ = ₀³⁴∫(1/34)te^-0.02t.

We use the Tabular Method of Integration by Parts:

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

+........(1/34)t...e^-0.02t

-........(1/34)...-50e^-0.02t

+.........0.......2500e^-0.02t

Thus, (ĪĀ)₆₀ = [(-50/34)te^-0.02t - (2500/34)e^-0.02t] │₀³⁴ =

(2500/34) - 50e^-0.68 - (2500/34)e^-0.68 =

73.592412 - 123.592412e^-0.68 = 73.592412 - 123.592412*0.506616992 = 73.592412 -62.582099 = about 11.010313 GH.

Problem S3L31-5. For irate rats, you are given that (ĪĀ)₆ = 5 and (ĪĀ)₈ = 2. Moreover,

_s│Ā_x = e^-(k_x)s for all values of x and s, where k_x is some value dependent on the value of x.

Find _2│Ā₆ - _2│Ā₈ for irate rats.

Solution S3L31-5. We use the formula (ĪĀ)_x = ₀^∞∫ _s│Ā_x ds.

We find k₆ for x = 6. (ĪĀ)₆ = 5 = ₀^∞∫e^-(k_6)sds

5 = (-1/k₆)e^-(k_6)s│₀^∞

5 = (1/k₆), so k₆ = 0.2

Thus, _s│Ā₆ = e^-0.2s and _2│Ā₆ = e^-0.4.

We find k₈ for x = 8. (ĪĀ)₈ = 2 = ₀^∞∫e^-(k_8)sds

2 = (-1/k₈)e^-(k_8)s│₀^∞

2 = (1/k₈), so k₈ = 0.5.

Thus, _s│Ā₈ = e^-0.5s and _2│Ā₈ = e^-1.

Hence,_2│Ā₆ - _2│Ā₈ = e^-0.4 - e^-1 = 0.302440605.

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