Life Table Probability Functions: Practice Problems and Solutions: Part 5

The function L_x is "the total expected number of years lived between ages x and x+1 by survivors of the initial group of l₀ lives" (Bowers, Gerber, et. al.). We can find it as follows.
L_x=₀¹∫l_x+tdt

The function m_x is the central-death-rate at age x. We can find it as follows.

m_x= (l_x - l_x+1)/L_x

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. p. 65.

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

Problem S3L13-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Of a group of 13503 newborn triceratopses, how many years will members of this group in aggregate live between the ages of 5 and 6?

Solution S3L13-1. We use the formula L_x = ₀¹∫l_x+tdt.Here,l₀ = 13503. We recall from Section 6 that l_x = l₀*s(x), so l_x+t = l₀*s(x+t) = 13503e^-0.34(x+t) and therefore l₀*s(5+t) = 13503e^-0.34(5+t).

We want to find L₅ = ₀¹∫13503e^-0.34(5+t)dt = 13503e^-0.34(5)₀¹∫e^-0.34tdt = 13503e^-0.34(5)(-50/17)e^-0.34t│₀¹ = 13503e^-0.34(5)(50/17)(1 - e^-0.34) = L₅ = 2091.170419 years.

Problem S3L13-2. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. For a group of 13503 newborn triceratopses, find the central-death-rate at age 5.

Solution S3L13-2. We want to find m₅, which we will do using the formulam_x = (l_x - l_x+1)/L_x.

We know from Solution S3L13-1 that L₅ = 2091.170419 years. Moreover, we recall from Section 6 that l_x = l₀*s(x), so l₅ = 13503e^-0.34(5) and l₆ = 13503e^-0.34(6). Thus,

m₅ = (l₅ - l₆)/L₅ = (13503e^-0.34(5) -13503e^-0.34(6))/2091.170419 = m₅ = 0.34.

Problem S3L13-3. For a particular group of polka-dotted zebras, you are given l_x = 95/x⁴, for all x > 1. Find L₄.

Solution S3L13-3. We use the formula L_x = ₀¹∫l_x+tdt. Here, x = 4, so l_x+t = 95/(4+t)⁴ and

L₄ = ₀¹∫[95/(4+t)⁴]dt = -95/[3(4+t)³]│₀¹ = 95/(3(4)³) - 95/(3(5)³) = L₄ = 1159/4800 = about 0.2414583333.

Problem S3L13-4. For a particular group of polka-dotted zebras, you are given l_x = 95/x⁴, for all x > 1. Find m₄.

Solution S3L13-4. We use the formulam_x = (l_x - l_x+1)/L_x.

We know from Solution S3L13-3 that L₄ = 0.2414583333years.

Moreover, l₄ = 95/4⁴ and l₅ = 95/5⁴.

Thus, m₄ = (95/4⁴ - 95/5⁴)/0.2414583333 = m₄ = about 0.9073770493.

Problem S3L13-5. Black swans always survive until age 16. After age 16, the lifetime of a black swan can be modeled by the cumulative distribution function F(x) = 1 - 4x^-1/2, x > 16. There is a cohort of 3511 newborn black swans. How many years will members of this group in aggregate live between the ages of 31 and 32?

Solution S3L13-5. We use the formula L_x = ₀¹∫l_x+tdt. We recall from Section 6 that l_x = l₀*s(x). Here, l₀ = 3511 and s(x) = 1 - F(x) = 4x^-1/2, x > 16. Thus, for x > 16, l_x+t = 3511*4(x+t)^-1/2. We seek to find L₃₁, so here x = 31. Thus, L₃₁ = ₀¹∫3511*4(31+t)^-1/2dt = 3511*8(31+t)^1/2│₀¹ = 3511*8[(32)^1/2 - (31)^1/2] = L₃₁ = 2502.356737 years.

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