Deferred 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 m-year deferred insurance policy pays a benefit to the insured only if the insured person dies at least m years from the time the policy was issued. The following functions are associated with an m-year deferred insurance policy that pays one unit in benefits.

b_t = 1 for t > m;

b_t = 0 for t ≤ m;

v_t = v^t for t > 0;

Z = v^Tif T > m;

Z = 0 if T ≤ m.

The actuarial present value of an m-year deferred insurance policy that pays one unit in benefits is denoted by _m│Ā_x and can be found via the following formula:

_m│Ā_x= _m^∞∫v^t*_tp_x*μ_x(t)dt

The rule of moments applies to deferred life insurance policies as well:

E[Z^j] @ δ_t = E[Z] @ jδ_t

E[Z^j] = _m^∞∫(v^t)^j*_tp_x*μ_x(t)dt

E[Z^j] = _m^∞∫e^-δjt*_tp_x*μ_x(t)dt

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. p. 103.

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

Problem S3L29-1. 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. Spotarecirt the Triceratops is currently 11 years old has a 5-year deferred insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) if he dies more than 5 years from now. Find the actuarial present value of this policy.

Solution S3L29-1. We use the formula _m│Ā_x = _m^∞∫v^t*_tp_x*μ_x(t)dt.

Here, x = 11, v = e^-0.09, and m = 5.

We find _tp_x = s(11 + t)/s(11) = e^-0.34t.

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

Thus, _5│Ā₁₁ = ₅^∞∫ e^-0.09t*e^-0.34t*0.34dt = ₅^∞∫0.34e^-0.43tdt = (-34/43)e^-0.43t│₅^∞ = (34/43)e^-0.43*5 =

_5│Ā₁₁ = about 0.0921037527 TCU.

Problem S3L29-2. 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. Spotarecirt the Triceratops is currently 11 years old has a 5-year deferred insurance policy, which will pay him 1 Triceratops Currency Unit (TCU) if he dies more than 5 years from now. Find the variance of this policy.

Solution S3L29-2. If Z is the present-value random variable for this policy, then

Var(Z) = E(Z²) - E(Z)². We know from Solution S3L29-1 that E(Z) = 0.0921037527.

We find E(Z²) using the formula E[Z^j] = _m^∞∫e^-δjt*_tp_x*μ_x(t)dt.

Here, x = 11, m = 5, j = 2, _tp_x = e^-0.34t, μ_x(t) = 0.34, and δ = 0.09. Thus,

E[Z²] = ₅^∞∫e^-0.18t*e^-0.34t*0.34dt = ₅^∞∫0.34e^-0.52tdt = (-34/52)e^-0.52t│₅^∞ = (34/52)e^-0.52*5 =

E[Z²] = 0.0485634934. Thus, Var(Z) = E(Z²) - E(Z)² = 0.0485634934 - 0.0921037527² =

Var(Z) =about 0.0400803922.

Problem S3L29-3. 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. Dwarvar the Giant Pin-Striped Cockroach is currently 30 years old and has a 22-year deferred life insurance policy which will pay 1 Golden Hexagon (GH) upon death, if he dies more than 22 years from now. Find the actuarial present value of Dwarvar's policy.

Solution S3L29-3. We use the formula _m│Ā_x = _m^∞∫v^t*_tp_x*μ_x(t)dt.

Here, x = 30, v = e^-0.02, and m = 22.

We note that the upper bound of our integral is not infinity, since the most Dwarvar can live is

94 - 30 = 64 more years. Thus, 64 is the upper bound of the integral.

We find _tp_x = (1 - (30+t)/94)/(1-30/94) = (64 - t)/64

We find μ_x(t) = -s'(x)/s(x) = (1/94)/(1 - x/94) = 1/(94 - x) = 1/(94 - (30+t)) = 1/(64 - t).

Conveniently enough, _tp_xμ_x(t) = ((64 - t)/64)/(64 - t) = (1/64).

Thus, _22│Ā₃₀ = ₂₂⁶⁴∫(1/64)e^-0.02tdt = (-50/64)e^-0.02t│₂₂⁶⁴ = (50/64)(e^-0.02*22 - e^-0.02*64) = _22│Ā₃₀ = about 0.285936813 GH.

Problem S3L29-4. 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. Dwarvar the Giant Pin-Striped Cockroach is currently 30 years old and has a 22-year deferred life insurance policy which will pay 1 Golden Hexagon (GH) upon death, if he dies more than 22 years from now. Find the variance of Dwarvar's policy.

Solution S3L29-4. If Z is the present-value random variable for this policy, then

Var(Z) = E(Z²) - E(Z)². We know from Solution S3L29-3 that E(Z) = 0.285936813.

We find E(Z²) using the formula E[Z^j] = _m^∞∫e^-δjt*_tp_x*μ_x(t)dt.

Here, x = 30, m = 22, j = 2, _tp_xμ_x(t) = 1/64 (from Solution S3L29-3), and δ = 0.02. Thus,

E[Z²] = ₂₂⁶⁴∫(1/64)e^-0.04tdt = (-25/64)e^-0.04t│₂₂⁶⁴ = (25/64)(e^-0.04*22 - e^-0.04*64) = E[Z²] = about 0.1318274106.

Var(Z) = E(Z²) - E(Z)² = 0.1318274106 - 0.285936813² = Var(Z) = about 0.0500675496.

Problem S3L29-5. For burgundy crickets, the survival function is s(x) = (1 - 0.0625x²) for 0 ≤ x ≤ 4 and 0 otherwise. Among burgundy crickets, interest is determined in an unusual manner, and the discount factor v^t is equal to (1/3)/(1/3 + (1/6)t). Tekcirc the Burgundy Cricket is 0 years old and has a 1-year deferred life insurance policy paying 1 Golden Hexagon (GH) upon death. Find the actuarial present value of this policy.

Solution S3L29-5. We use the formula _m│Ā_x = _m^∞∫v^t*_tp_x*μ_x(t)dt.

Here, x = 0, v = (1/3)/(1/3 + (1/6)t), and m = 1.

_tp₀ = s(t) = (1 - 0.0625t²)

μ₀(t) = -s'(x)/s(x) = 0.125x/(1 - 0.0625x²) = 0.125t/(1 - 0.0625t²)

Thus, _tp₀μ₀(t) = (1 - 0.0625t²)*0.125t/(1 - 0.0625t²) = 0.125t.

Thus, v^t*_tp₀μ₀(t) = ((1/3)/(1/3 + (1/6)t))0.125t = 2*0.125t/(2 + t) = 0.25t/(2 + t).

The upper bound of our integral will be 4, since the largest age a newborn burgundy cricket can attain is 4.

Thus, _1│Ā₀ = ₁⁴∫(0.25t/(2 + t))dt

We use the Tabular Method of Integration by Parts:

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

+......0.25t.....1/(2+t)

- ......0.25......ln(2+t)

+........0........(2+t)ln(2+t) - (2+t)

Thus, ₁⁴∫(0.25t/(2 + t))dt = [0.25tln(2+t) - 0.25(2+t)ln(2+t) + 0.25(2+t)] │₁⁴ =

[0.25*4*ln(6) - 0.25(6)ln(6) + 0.25(6)] - [0.25ln(3) - 0.25(3)ln(3) + 0.25(3)] = about 0.403264097 GH.

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