Moments of the Present Value of a Term Life Insurance Policy: Practice Problems and Solutions

In Section 21, the formula for Ā¹_x:n¬, the actuarial present value of an n-year term life insurance policy paying a benefit of 1 upon the death of life (x), was introduced:
E[Z] = Ā¹_x:n¬ = ₀^∞∫z_t*f_T(t)dt = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt

Here, we introduce formulas for the j-th moment of the distribution of Z, denoted by E[Z^j]. In one of these formulas, δ, the force of interest, will be used. We recall from Course 2/FM that v = e^-δ.

E[Z^j] = ₀ⁿ∫(v^t)^j*_tp_x*μ_x(t)dt

E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt

We note that E[Z^j] when the force of interest is δ is equal to E[Z] when the force of interest is jδ. More generally, for any deterministic force of interest, expressed as δ_t, the following property, known as the rule of moments, holds.

E[Z^j] @ δ_t = E[Z] @ jδ_t

The rule of moments enables us to calculate the variance of Z:
Var(Z) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)²

Meaning of variables:

Ā¹_x:n¬= the actuarial present value of n-year term life insurance which pays a benefit of 1 upon the death of life (x). The force of interest is assumed to be δ.

²Ā¹_x:n¬ = E[Z²] = the actuarial present value of n-year term life insurance which pays a benefit of 1 upon the death of life (x). Important: The force of interest for ²Ā¹_x:n¬is assumed to be 2δ.

f_T(t) = the probability density function of T, the time to death of life (x).

μ_x(t) = the force of mortality for life (x) at time t.

_tp_x = the probability that life (x) will survive to time x+t.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 95-96.

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

Problem S3L22-1. An n-year term life insurance policy has an actuarial present value of ψ for a constant force of interest 0.07. At what force of interest will the 3rd moment of the distribution of the present-value random variable for this policy be equal to ψ?

Solution S3L22-1. By the rule of moments, E[Z^j] @ δ_t = E[Z] @ jδ_t.

Here, j = 3. Moreover, E[Z] = ψ.

Thus, E[Z³] @ δ_t = ψ @ 3*δ_t.

We are given that 3*δ_t = 0.07. Thus, δ_t = 0.07/3 = about 0.02333333333.

Problem S3L22-2. 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. Theodosius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1. Find the actuarial present value of this policy.

Solution S3L22-2. We use the formula Ā¹_x:n¬ = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt. We are asked to find Ā¹_2:3¬ .

We find _tp₂ = s(x + t)/s(x) = s(2 + t)/s(2) = e^-0.34(2+t)/e^-0.34(2) = e^-0.34t

We find μ₂(t) = -s'(x)/s(x) = 0.34e^-0.34x/e^-0.34x = 0.34

We find v^t = e^-0.06t, since δ = 0.06

Thus, Ā¹_2:3¬ = ₀³∫e^-0.06t*e^-0.34t*0.34dt = ₀³∫0.34*e^-0.40tdt = (-34/40)e^-0.40t│₀³ = (34/40)(1 - e^-0.40*3)

Ā¹_2:3¬ = about 0.5939849199.

Problem S3L22-3. 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. Theodosius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1. Find the second moment of the distribution of the present-value random variable for this policy.

Solution S3L22-3. We want to find E[Z²], which we can do using the formula

E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt, using j = 2.

As in Solution S3L22-1, we have _tp₂ = e^-0.34t, μ₂(t) = 0.34, and n = 3.

However, v^t is changed from e^-0.06t to e^-0.06*2t = e^-0.12t.

Thus, E[Z²] = ₀³∫e^-0.12t*e^-0.34t*0.34dt = ₀³∫0.34*e^-0.46tdt = (-34/46)e^-0.46t│₀³ = (34/46)(1 - e^-0.46*3)

²Ā¹_x:n¬ = E[Z²] = about 0.5531810695.

Problem S3L22-4. 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. Theodosius the Triceratops is currently 2 years old and takes out a 3-year term life insurance policy paying a benefit of 1. Find the variance of the distribution of the present-value random variable for this policy.

Solution S3L22-4. We use the formula Var(Z) = ²Ā¹_x:n¬ - (Ā¹_x:n¬)².

From Solution S3L22-1, we know that Ā¹_2:3¬ = 0.5939849199.

From Solution S3L22-2, we know that ²Ā¹_x:n¬ = 0.5531810695.

Thus, Var(Z) = 0.5531810695 - 0.5939849199² = Var(Z) = about 0.2003629844.

Problem S3L22-5. 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. Sparkles the Giant Pin-Striped Cockroach is currently 11 years old and has a 50-year term life insurance policy which will pay 10 Golden Hexagons (GH) upon death. The annual force of interest is 0.3. Find the fifth moment of the distribution of the present-value random variable for this policy.

Solution S3L22-5.
The random variable for the present value of the insurance policy is 10Z.
We seek to find E[(10Z)⁵]. We use the formula E[Z^j] = ₀ⁿ∫e^-δjt*_tp_x*μ_x(t)dt.

We know that j = 5, δ = 0.3, x = 11, and n = 50.

We find _tp₁₁ = s(x + t)/s(x) = s(11 + t)/s(11) = (1 - (11+t)/94)/(1 - 11/94) = (83 - t)/83

We find μ₁₁(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = (-1/94)/[(94 - x)/94] = 1/(94 - x) = 1/(94 - (11-t)) = 1/(83 - t).

Conveniently enough, _tp₁₁*μ₁₁(t) = ((83 - t)/83)(1/(83 - t)) = (1/83) We find e^-δjt = e^-5*0.3t = e^-1.5t

Thus, E[(10Z)⁵] = 10⁵E[Z⁵] = 10⁵*₀⁵⁰∫e^-1.5t(1/83)dt
10⁵E[Z⁵] = 10⁵*(-2/249)e^-1.5t│₀⁵⁰ = 10⁵*(2/249)(1 - e^-75) = 10⁵*(2/249) (for all practical purposes).
10⁵E[Z⁵] = 200000/249 = about 803.2128514.

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