Endowment 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 policy of n-year endowment insurance makes a payment either upon the beneficiary's death or upon the beneficiary's survival to the end of a term of n years. The earliest of these two times is the payment of death. For n-year endowment insurance that pays one unit in benefits, we have the following functions:

b_t = 1 for t ≥ 0;

v_t = v^t for t ≤ n;

v_t = vⁿ for t > n;

Z = v^Tif T ≤ n;

Z = vⁿ if T > n.

The actuarial present value of n-year endowment insurance is denoted by the symbol Ā_x:n¬.

n-year endowment insurance can be expressed as a sum of an n-year pure endowment and an n-year term life insurance policy. The actuarial present value of an n-year endowment insurance policy is the sum of the actuarial present values of the corresponding n-year pure endowment and n-year term life insurance policy. Thus,

Ā_x:n¬ = A¹_x:n¬ + Ā¹_x:n¬

We let Z₁ = the present-value random variable for an n-year term life insurance policy.

We let Z₂ = the present-value random variable for an n-year pure endowment.

We let Z₃ = the present-value random variable for an n-year endowment insurance policy.

Thus, Z₃ = Z₁ + Z₂ and Var(Z₃) = Var(Z₁) + Var(Z₂) + 2Cov(Z₁ , Z₂).

The covariance of Cov(Z₁ , Z₂) can be expressed as follows:

Cov(Z₁ , Z₂) = -Ā¹_x:n¬*A¹_x:n¬

From Section 22, it is known that Var(Z₁) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)².

From Section 27, it is known that Var(Z₂) = ²A¹_x:n¬ - (A¹_x:n¬)².

Thus, Var(Z₃) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)² + ²A¹_x:n¬ - (A¹_x:n¬)² - 2Ā¹_x:n¬*A¹_x:n¬

Var(Z₃) = ²Ā¹_x:n¬ + ²A¹_x:n¬ - (Ā¹_x:n¬ + A¹_x:n¬)²

Reminders:

For n-year term life insurance, Ā¹_x:n¬ = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt and E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt.

For n-year pure endowments, A¹_x:n¬ = vⁿ*_np_x and Var(Z) = v²ⁿ*_np_x*_nq_x.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 101-102.

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

Problem S3L28-1. 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 0.09. Hasdrubal the Triceratops is currently 7 years old has a 2-year endowment insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) upon death. Find the actuarial present value of this policy.

Solution S3L28-1. We use the formula Ā_x:n¬ = A¹_x:n¬ + Ā¹_x:n¬.

We first need to find

Ā¹_x:n¬ =₀ⁿ∫v^t*_tp_x*μ_x(t)dt and

A¹_x:n¬ = vⁿ*_np_x

We know that v = e^-0.09, x = 7, and n = 2.

We find _tp_x = s(x+t)/s(x) = s(7+t)/s(7) = e^-0.34t. So _np_x = ₂p₇ = e^-0.34*2.

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

Thus, Ā¹_7:2¬ =₀²∫e^-0.09t*e^-0.34t*0.34dt = ₀²∫0.34e^-0.43tdt = (-34/43)e^-0.43t│₀² = (34/43)(1 - e^-0.43*2) =

Ā¹_7:2¬ = 0.4561044

Also, A¹_7:2¬ = e^-0.09*2e^-0.34*2 = e^-0.86 = A¹_7:2¬ = 0.4231620823.

Thus, Ā_7:2¬ = A¹_7:2¬ + Ā¹_7:2¬ = 0.4231620823 + 0.4561044 = Ā_7:2¬ = about 0.8792664823.

Problem S3L28-2. 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. Hasdrubal the Triceratops is currently 7 years old has a 2-year endowment insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) upon death. If Z₁ is the present-value random variable for a 2-year term life insurance policy for Hasdrubal, and Z₂ is the present-value random variable for an n-year pure endowment for Hasdrubal, find the covariance of Z₁ and Z₂.

Solution S3L28-2. We use the formula Cov(Z₁ , Z₂) = -Ā¹_x:n¬*A¹_x:n¬.

We know from Solution S3L28-1 that A¹_7:2¬ = 0.4231620823 and Ā¹_7:2¬ = 0.4561044.

Thus, Cov(Z₁ , Z₂) = -0.4561044*0.4231620823 = about -0.1930060877.

Problem S3L28-3. 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 0.09. Hasdrubal the Triceratops is currently 7 years old has a 2-year endowment insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) upon death. Find the variance of the present-value random variable for this policy.

Solution S3L28-3. We use the formula Var(Z₃) = Var(Z₁) + Var(Z₂) + 2Cov(Z₁ , Z₂), with Z₃ being the present-value random variable for this policy and Z₁ and Z₂ as defined in Problem S3L28-2.

We know from Solution S3L28-2 that Cov(Z₁ , Z₂) = -0.1930060877.

We can find Var(Z₂) using the formula Var(Z₂) = v²ⁿ*_np_x*_nq_x.

We know that v = e^-0.09, x = 7, and n = 2. We also know from Solution S3L28-1 that

_np_x = ₂p₇ = e^-0.34*2 = e^-0.68. Thus, _nq_x = 1 - _np_x = 1 - e^-0.68.

Thus, Var(Z₂) = e^-0.09*4*e^-0.68(1 - e^-0.68) = Var(Z₂) = 0.174388534.

We need to find Var(Z₁) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)².

We know from Solution S3L28-1 that Ā¹_7:2¬ = 0.4561044.

We can find E[Z₁²] = ²Ā¹_7:2¬ using the formula E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt from Section 22.

We find _tp_x = s(x+t)/s(x) = s(7+t)/s(7) = e^-0.34t.

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

Thus, E[Z₁²] = ₀²∫e^-2*0.09t*e^-0.34t*0.34dt = ₀²∫0.34*e^-0.52tdt = (-34/52)e^-0.52t│₀² = (34/52)(1 - e^-0.52*2) = ²Ā¹_7:2¬ = 0.4227411695.

Thus, Var(Z₁) = 0.4227411695 - (0.4561044)² = Var(Z₁) = 0.2147099458.

Now we can apply the formula Var(Z₃) = Var(Z₁) + Var(Z₂) + 2Cov(Z₁ , Z₂).

Var(Z₃) = 0.2147099458 + 0.174388534 + 2(-0.1930060877)

Var(Z₃) = about 0.0030863045. (That is an outstandingly predictable insurance policy!)

Problem S3L28-4. 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. Odoacer the Giant Pin-Striped Cockroach is currently 44 years old and has a 13-year endowment which will pay 1 Golden Hexagon (GH) if he reaches age 57. The annual force of interest is 0.02, and the actuarial present value of Odoacer's policy iscurrently 0.5705781735. Odoacer is not satisfied with this present value and seeks to turn his pure endowment into a 13-year endowment insurance policy. What will the present value of his new policy be?

Solution S3L28-4. We use the formula Ā_x:n¬ = A¹_x:n¬ + Ā¹_x:n¬. We know that

A¹_44:13¬ = 0.5705781735.

We find Ā¹_44:13¬ using the formula Ā¹_x:n¬= ₀ⁿ∫v^t*_tp_x*μ_x(t)dt.

We know that x = 44, n = 13, and v = e^-0.02.

We find _tp_x = (1 - (44+t)/94)/(1-44/94) = (50 - t)/50

We find μ_x(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = 1/(94 - x) = 1/(94 - (44+t)) = 1/(50 - t).

Interestingly enough, _tp_xμ_x(t) = ((50 - t)/50)/(50 - t) = (1/50).
Thus, Ā¹_44:13¬= ₀¹³∫e^-0.02t(1/50)dt = (-50/50)e^-0.02t│₀¹³ = (1 - e^-0.02*13) = Ā¹_44:13¬= 0.2289484142.

Thus, Ā_44:13¬ = A¹_44:13¬ + Ā¹_44:13¬ = 0.5705781735 + 0.2289484142 = about 0.7995265877.

Problem S3L28-5. 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. Odoacer the Giant Pin-Striped Cockroach is currently 44 years old and has a 13-year endowment which will pay 1 Golden Hexagon (GH) if he reaches age 57. The annual force of interest is 0.02, and the variance of Odoacer's policy is currently 0.1143857534. Odoacer is not satisfied with this variance and seeks to turn his pure endowment into a 13-year endowment insurance policy. What will the variance of his new policy be?

Solution S3L28-5. We know that x = 44, n = 13, and v = e^-0.02. We use the formula

Var(Z₃) = Var(Z₁) + Var(Z₂) + 2Cov(Z₁ , Z₂), where Z₁ is the present value of a 13-year term life insurance policy and Z₂ is the present value of a 13-year pure endowment.

We are given Var(Z₂) = 0.1143857534.

We use the formula Cov(Z₁ , Z₂) = -Ā¹_x:n¬*A¹_x:n¬, knowing from Solution S3L28-4 that

A¹_44:13¬ = 0.5705781735 and Ā¹_44:13¬ = 0.2289484142.

Thus, Cov(Z₁ , Z₂) = -0.2289484142*0.5705781735 = Cov(Z₁ , Z₂) = -0.130632968.

We need to find Var(Z₁) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)².

We can find E[Z₁²] = ²Ā¹_7:2¬ using the formula E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt from Section 22.

We find _tp_x = (1 - (44+t)/94)/(1-44/94) = (50 - t)/50

We find μ_x(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = 1/(94 - x) = 1/(94 - (44+t)) = 1/(50 - t).

Interestingly enough, _tp_xμ_x(t) = ((50 - t)/50)/(50 - t) = (1/50).
Thus, E[Z₁²] = ₀¹³∫e^-0.02*2*t*(1/50)dt = ₀¹³∫(1/50)e^-0.04tdt = (-1/2)e^-0.04t│₀¹³ = (1/2)(1 - e^-0.52) = ²Ā¹_44:13¬ = 0.202739726.

Thus, Var(Z₁) = 0.202739726 - 0.2289484142² = Var(Z₁) = 0.1503223497

Now we can apply the formula Var(Z₃) = Var(Z₁) + Var(Z₂) + 2Cov(Z₁ , Z₂).

Var(Z₃) = 0.1503223497 + 0.1143857534 + 2*-0.130632968.

Var(Z₃) = about 0.0034421671.

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