Probability Density Functions and Cumulative Distribution Functions of Present-Value Random Variables: Practice Problems and Solutions

Z is the random variable representing the present value of a life insurance policy with a benefit of one unit. How might the probability density function (p.d.f.) of Z and the cumulative distribution function (c.d.f.) of Z be expressed?

We assume we know the p.d.f. and c.d.f. of T, the future-lifetime random variable for life (x). These will be denoted as f_T(t) and F_T(t).

Then

F_Z(z) = 0 for z ≤ 0;

F_Z(z) = 1 - F_T(ln(z)/ln(v)) for 0 < z < 1;

F_Z(z) = 1 for 1 ≤ z.

Moreover,

f_Z(z) = f_T(ln(z)/ln(v))(1/(δz)) for 0 < z < 1 and 0 elsewhere.

Meaning of variables:

δ = annual force of interest = ln(1+r), where r is the annual effective interest rate.

v = the discount factor for one year ((1/(1+r)), where r is the annual effective interest rate.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 97-98.

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

Problem S3L26-1. The future lifetime of 2-year-old centaurs can be modeled by the following

p. d. f.: f_T(t) = 0.05e^-0.05t. The annual force of interest is 0.07. Find f_Z(z) for a whole life insurance policy entered into by 2-year-old centaurs that pays a one-unit benefit upon death.

Solution S3L26-1. We use the formula f_Z(z) = f_T(ln(z)/ln(v))(1/(δz)) for 0 < z < 1. Here, δ = 0.07 and v = e^-0.07. So ln(v) = -0.07. Thus, f_Z(z) = 0.05e^{-0.05(ln(z)/ln(v))}(1/(0.07z))

f_Z(z) = (5/7z)e^(5/7)ln(z)

f_Z(z) = (5/7z)e^{ln(z^(5/7))}

f_Z(z) = (5/7z)z^5/7

f_Z(z) = (5/7)z^-2/7 for 0 < z < 1 and 0 elsewhere.

Problem S3L26-2. The future lifetime of 2-year-old centaurs can be modeled by the following

p. d. f.: f_T(t) = 0.05e^-0.05t. The annual force of interest is 0.07. Find F_Z(z) for a whole life insurance policy entered into by 2-year-old centaurs that pays a one-unit benefit upon death.

Solution S3L26-2. We use the formula F_Z(z) = 1 - F_T(ln(z)/ln(v)) for 0 < z < 1.

f_T(t) = 0.05e^-0.05t implies that s_T(t) = e^-0.05t and F_T(t) = 1 - s_T(t) = 1 - e^-0.05t. (The future lifetime of centaurs follows an exponential distribution.)

Then F_Z(z) = 1 - (1 - e^{-0.05(ln(z)/ln(v))})

F_Z(z) = e^{-0.05(ln(z)/ln(v))}

We know from Solution S3L26-1 that ln(v) = -0.07, so

F_Z(z) = e^{-0.05(ln(z)/-0.07)}

F_Z(z) = e^(5/7)ln(z)

F_Z(z) = e^{ln(z^(5/7))} for 0 < z < 1.

Then

F_Z(z) = 0 for z ≤ 0;

F_Z(z) = z^5/7 for 0 < z < 1;

F_Z(z) = 1 for 1 ≤ z.

Problem S3L26-3. The future lifetime of 11-year-old microdragons can be modeled by the following p. d. f.: f_T(t) = 1/66 for 0 ≤ t ≤ 66 and 0 otherwise. The annual force of interest is 0.1. Find f_Z(z) for a whole life insurance policy entered into by 11-year-old microdragons that pays a one-unit benefit upon death.

Solution S3L26-3. We use the formula f_Z(z) = f_T(ln(z)/ln(v))(1/(δz)) for 0 < z < 1. Here, δ = 0.1

and f_T(ln(z)/ln(v)) = 1/66, since the value of f_T(□) does not depend on the value of □.

Thus, f_Z(z) = (1/66)(1/0.1z) = f_Z(z) = 10/(66z), but for which bounds?

We know that Pr{T > 66} = 0, so Pr{0 < Z < v⁶⁶} = 0 (There is no way that the present value of one unit of benefit will be greater than v⁶⁶.)

So our bounds are v⁶⁶ < z < 1. We find v⁶⁶ = e^-0.1*66 = about 0.001360368

Thus, f_Z(z) = 10/(66z) for 0.001360368 < z < 1 and 0 elsewhere.

Problem S3L26-4. The future lifetime of 11-year-old microdragons can be modeled by the following p. d. f.: f_T(t) = 1/66 for 0 ≤ t ≤ 66 and 0 otherwise. The annual force of interest is 0.1. Find F_Z(z) for a whole life insurance policy entered into by 11-year-old microdragons that pays a one-unit benefit upon death.

Solution S3L26-4. We know from Solution S3L26-3 that f_Z(z) = 10/(66z) for 0.001360368 < z < 1 and 0 elsewhere.We also know that F_Z(z) = ∫f_Z(z)dz = (10/66)ln(z) for 0.001360368 < z < 1.

Then

F_Z(z) = 0 for z ≤ 0.001360368;

F_Z(z) = (10/66)ln(z) for 0.001360368 < z < 1;

F_Z(z) = 1 for 1 ≤ z.

Problem S3L26-5. The cumulative distribution function for the future lifetime of 4-year-old wolf-rabbits is F_T(t) =1 - exp(-0.2x³). The annual effective interest rate is 0.04. Find F_Z(z) for a life insurance policy for 4-year-old wolf-rabbits that pays a one-unit benefit upon death.

Solution S3L26-5. We use the formula F_Z(z) = 1 - F_T(ln(z)/ln(v)) for 0 < z < 1.

1 - F_T(ln(z)/ln(v)) = 1 - [1 - exp(-0.2(ln(z)/ln(v))³)]

Since r = 0.04, it follows that v = 1/1.04 = 0.9615384615 and thus ln(v) = -0.0392207132.

Thus, F_Z(z) = exp(-0.2(ln(z)/-0.0392207132)³)

F_Z(z) = exp(3315(ln(z))³) for 0 < z < 1.

Then

F_Z(z) = 0 for z ≤ 0;

F_Z(z) = exp(3315(ln(z))³)for 0 < z < 1;

F_Z(z) = 1 for 1 ≤ z.

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