Annually 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).

An annually increasing whole life insurance policy will pay 1 unit in benefits if death occurs during the first year, 2 units in benefits if death occurs during the second year, and n units in benefits if death occurs during the nth year. The following functions characterize annually increasing whole life insurance.

b_t = └t + 1┘ for t ≥ 0;

v_t = v^t for t ≥ 0;

Z = └T+ 1┘v^Tfor T ≥ 0.

The function└x┘is the greatest integer function, where└x┘is the greatest integer that is less than or equal to x.

The actuarial present value of annually increasing whole life insurance is denoted as (IĀ)_x and can be found via the following formula:
E[Z] = (IĀ)_x =₀^∞∫└t + 1┘v^t*_tp_x*μ_x(t)dt

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 105.

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

Problem S3L30-1. The lives of magenta hippopotami follow the survival function s(x) = e^-0.09x. The present-value discount factor among magenta hippopotami is computed in an unusual manner such that v^t = 1/└t + 1┘.Sumatopoppih the Magenta Hippopotamus is 3 years old and has an annually increasing whole life insurance policy that pays a benefit of n Golden Hexagons (GH) if death occurs in the nth year of the policy. Find the actuarial present value of such a policy.

Solution S3L30-1. We use the formula (IĀ)_x = ₀^∞∫└t + 1┘v^t*_tp_x*μ_x(t)dt. Here, x = 3.

We find _tp₃ = s(3 + t)/s(3) = e^-0.09t.

We find μ₃(t) = -s'(x)/s(x) = 0.09e^-0.09t/e^-0.09t = 0.09.

Moreover, we note that └t + 1┘v^t =└t + 1┘(1/└t + 1┘) = 1.

Thus, (IĀ)₃ = ₀^∞∫0.09e^-0.09tdt = -e^-0.09t│₀^∞ = e^-0.09*0 - e^-∞ = 1 - 0 = (IĀ)₃ = 1 GH.

Problem S3L30-2. 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. Spartacus the Triceratops is currently 2 years old has a 2-year annually increasing life insurance policy, which will pay 1 Triceratops Currency Unit (TCU) if he dies within 1 year from now and 2 Triceratops Currency Units (TCU) if he dies between 1 and 2 years from now. Find the actuarial present value of this policy.

Solution S3L30-2. We use the formula (IĀ)_x = ₀^∞∫└t + 1┘v^t*_tp_x*μ_x(t)dt. Here, however, the upper bound of the integral is not infinity but 2, since the policy only applies to the next two years.

We are given that x = 2 and v = e^-0.09.

We find _tp_x = s(2 + t)/s(2) = e^-0.34t.

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

Thus, (IĀ)_2:2 = ₀²∫└t + 1┘ e^-0.09t*e^-0.34t*0.34dt = ₀²∫0.34└t + 1┘e^-0.43tdt

We note that from t = 0 to t = 1, └t + 1┘= 1 and from t = 1 to t = 2, └t + 1┘= 2.

Thus, we can decompose the integral above into two integrals that are more convenient to work with.

₀²∫0.34└t + 1┘e^-0.43tdt = ₀¹∫0.34e^-0.43tdt + ₁²∫0.68e^-0.43tdt = (-34/43)e^-0.43t│₀¹ + (-68/43)e^-0.43t│₁² =

(34/43)(1 - e^-0.43) + (68/43)(e^-0.43 - e^-0.86) = (IĀ)₂ = about 0.635867154 TCU.

Problem S3L30-3. From his current vantage point, Morgan the Mortal has a probability of 0.05 of dying in every year starting now, where 0 ≤ t ≤ 20. He takes out annually increasing whole life insurance policy that pays n Golden Hexagons (GH) if he dies in the nth year. The annual force of interest is 0. Find the actuarial present value of Morgan's policy.

Solution S3L30-3. For every nth year between n = 1 and n = 20, inclusive, Morgan's policy will pay

n if he dies in that year. The probability of Morgan dying in the nth year is 0.05. So Morgan's expected value with regard to each year's potential payment is 0.05*n, which is also the expected present value of that year's payments, since the force of interest is zero.

Thus, the actuarial present value of Morgan's policy is _n=0ⁿ⁼²⁰Σ0.05*n = 0.05(20)(20+1)/2 = 10.5 GH.

Problem S3L30-4. After getting away with massive life insurance fraud once, Imhotep the Immortal decides to engage in it on a repeated basis. He takes out an annually increasing whole life insurance policy that pays n Golden Hexagons (GH) if he dies in the nth year. He then fakes his own death on an annual basis, collects the payments, and then misrepresents his identity as that of some living policyholder who had a life insurance policy identical to his own. Thus, Imhotep expects to be able to collect n GH in every nth year. However, insurance companies have become wary of Imhotep's machinations, so they thwart his attempts with a probability of 0.32. The annual force of interest is 0.2, and Imhotep only wants to engage in this scheme during every year for which the current present value of his expected revenues is increasing. What is the actuarial present value of Imhotep's entire scheme to him?

Solution S3L30-4. During the nth year, Imhotep has 0.68 probability of collecting a payment of n, which will have a present value of e^-0.2n to him now. We want to find the value of n that maximizes 0.68ne^-0.2n, the present value of Imhotep's revenues from the nth year. Once that value of n is exceeded, Imhotep will move on to a different life insurance fraud scheme.

To find the maximum value of f(n) = 0.68ne^-0.2n, we first find f'(n).

f'(n) = 0.68e^-0.2n - 0.136ne^-0.2n.

At the maximum value for f(n), f'(n) = 0.

Thus, 0.68e^-0.2n - 0.136ne^-0.2n = 0

0.68e^-0.2n = 0.136ne^-0.2n

0.68 = 0.136n and n = 5.

Thus, Imhotep will only engage in this scheme for 5 years. The actuarial present value of this scheme to him then is 0.68e^-0.2 + 0.68*2e^-0.2*2 + 0.68*3e^-0.2*3 + 0.68*4e^-0.2*4 + 0.68*5e^-0.2*5 = 5.060912795 GH.

Problem S3L30-5. 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. Tullius the Giant Pin-Striped Cockroach is 77 years old and has a 3-year annually increasing life insurance policy that will pay n Golden Hexagons (GH) if he dies in the nth year of the policy, up to the policy's expiration. Find the actuarial present value of Tullius's policy.

Solution S3L30-5. We use the formula (IĀ)_x = ₀^∞∫└t + 1┘v^t*_tp_x*μ_x(t)dt. Here, however, the upper bound of the integral is not infinity but 3, since the policy only applies to the next 3 years.

Here, x = 77, v = e^-0.02.

We find _tp_x = (1 - (77+t)/94)/(1-77/94) = (17- t)/17

We find μ_x(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = 1/(94 - x) = 1/(94 - (77+t)) = 1/(17 - t).

Conveniently enough, _tp_xμ_x(t) = ((17 - t)/17)/(17 - t) = (1/17).

Thus, we have (IĀ)_77:3 = ₀³∫└t + 1┘e^-0.02t*(1/17)dt

We note that from t = 0 to t = 1, └t + 1┘= 1, from t = 1 to t = 2, └t + 1┘= 2, and from t = 2 to t = 3,

└t + 1┘= 3.

Thus, we can decompose the integral above into three integrals that are more convenient to work with.

₀³∫└t + 1┘e^-0.02t*(1/17)dt = ₀¹∫e^-0.02t*(1/17)dt + ₁²∫e^-0.02t*(2/17)dt + ₂³∫e^-0.02t*(3/17)dt =

(-50/17)e^-0.02t│₀¹ + (-100/17)e^-0.02t│₁² + (-150/17)e^-0.02t│₂³ =

(50/17)(1 - e^-0.02) + (100/17)(e^-0.02 - e^-0.04) + (150/17)(e^-0.04 - e^-0.06) = about 0.3402779756 GH.

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