Complete-Expectation-of-Life: Practice Problems and Solutions

The function ė_x is the complete-expectation-of-life and is equivalent to E[T(x)], the expected value of the future lifetime of life (x) - a life that has already attained the age of x. The following expressions are equal to ė_x.

ė_x = E[T(x)] = ₀^∞∫t*_tp_x*μ_x+t*dt = ₀^∞∫_tp_xdt

It is also useful to be able to find E[T(x)²] and Var[T(x)].

E[T(x)²] = ₀^∞∫t²*_tp_x*μ_x+t*dt = 2₀^∞∫t*_tp_xdt

Var[T(x)] = E[T(x)²] - E[T(x)]² = 2₀^∞∫t*_tp_xdt - ė_x²

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 62-63.

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

Problem S3L10-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find ė₉.

Solution S3L10-1. First, we want to find _tp₉ = s(9 +t)/s(9) = e^-0.34(9+t)/e^-0.34(9) = e^-0.34t. Thus, we can use the formula ė_x = ₀^∞∫_tp_xdt = ₀^∞∫_tp₉dt = ₀^∞∫e^-0.34tdt = (-50/17)e^-0.34t│₀^∞ = ė₉ = 50/17 = about 2.941176471

Problem S3L10-2. 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. Find E[T(34)²].

Solution S3L10-2. First, we want to find _tp₃₄ = s(34 +t)/s(34) = [1 - (34 + t)/94]/[1 - 34/94] =

(60 - t)/60. Thus, we can use the formula E[T(x)²] = 2₀^∞∫t*_tp_xdt = 2₀⁶⁰∫t(60 - t)/60dt = 2₀⁶⁰∫(t - t²/60)dt =

2(t²/2 - t³/180)│₀⁶⁰ = 2(60²/2 - 60³/180) = E[T(34)²] =1200. (Note: The upper bound of the integral was changed to 60, because no giant pin-striped cockroach can live more than 60 years past the age of 34.)

Problem S3L10-3. Zigzag-striped plankton never survive past the age of 2. The cumulative distribution function for the lives of zigzag-striped plankton is as follows: G(t) = t³/8, 0 ≤ t ≤ 2, 0 otherwise. Find ė₁.

Solution S3L10-3. First, we want to find _tp₁ = s(1 + t)/s(1). Since G(t) = t³/8, s(t) = 1 - G(t) = 1 - t³/8. Thus, s(1 + t)/s(1) = (1 - (1+t)³/8)/(1 - 1³/8) = (1 - (1+t)³/8)/(7/8) = _tp₁ = (8 - (1+t)³)/7. Hence, we can use the formula ė_x = ₀^∞∫_tp_xdt = ₀¹∫[(8 - (1+t)³)/7] dt = [8t/7 - (1+t)⁴/28]│₀¹ = [8/7 - (2)⁴/28] - [0 - 1/28] =

ė₁ = 17/28 = about 0.6071428571. (Note: The upper bound of the integral was changed to 1, because no zigzag-striped plankton can live more than 1 year past the age of 1.)

Problem S3L10-4. Zigzag-striped plankton never survive past the age of 2. The cumulative distribution function for the lives of zigzag-striped plankton is as follows: G(t) = t³/8, 0 ≤ t ≤ 2, 0 otherwise. Find E[T(1)²].

Solution S3L10-4. We use the formula E[T(x)²] = 2₀^∞∫t*_tp_xdt. From Solution S3L10-3, we know that _tp₁ = (8 - (1+t)³)/7. Thus, E[T(1)²] = 2₀¹∫[t(8 - (1+t)³)/7]dt = = 2₀¹∫[(8t - t(1+t)³)/7]dt =

2₀¹∫[(8t - t(1+ 3t + 3t² + t³))/7]dt = 2₀¹∫[(8t - t(1+ 3t + 3t² + t³))/7]dt = 2₀¹∫[(8t - t - 3t² - 3t³ - t⁴)/7]dt =

2₀¹∫[(7t - 3t² - 3t³ - t⁴)/7]dt = 2[t²/2 - t³/7 - 3t⁴/28 - t⁵/35] │₀¹ = 2(½ - 1/7 - 3/28 - 1/35) = E[T(1)²] = 31/70 = about 0.4428571429. (Note: The upper bound of the integral was changed to 1, because no zigzag-striped plankton can live more than 1 year past the age of 1.)

Problem S3L10-5. Zigzag-striped plankton never survive past the age of 2. The cumulative distribution function for the lives of zigzag-striped plankton is as follows: G(t) = t³/8, 0 ≤ t ≤ 2, 0 otherwise. Find Var[T(1)].

Solution S3L10-5. We use the formula Var[T(x)] = E[T(x)²] - E[T(x)]². From Solutions S3L10-3 and S3L10-4, we know that E[T(1)²] = 31/70 and E[T(1)] = ė₁ = 17/28. Thus,

Var[T(1)] = (31/70) - (17/28)² = Var[T(1)] = 291/3920 = about 0.0742346939

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