Life Table Functions: Practice Problems and Solutions: Part 6

The life table function T_xrepresents "the total number of years lived beyond age x by the survivorship group with l₀ initial members" (Bowers, Gerber, et. al.). This function can be found as follows:

T_x=₀^∞∫l_x+tdt

Moreover, the following relationship holds:

T_x/l_x = ė_x

The life table function a(x) represents "the average number of years lived between ages x and x + 1 by those of the survivorship group who die between those ages." (Bowers, Gerber, et. al.).

The direct determination of a(x) is rather time-consuming:

a(x) = [₀¹∫t*l_x+t*μ_x+tdt]/[₀¹∫l_x+t*μ_x+tdt]

An approximation of a(x) which assumes that deaths are uniformly distributed in the year of age is

a(x) = ₀¹∫tdt = ½

A faster way to determine a(x) is through the following identity:

L_x = a(x)l_x + [1 - a(x)]l_x+1

Thus, a(x) = (L_x - l_x+1)/(l_x - l_x+1)

Meaning of variables:
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.).

ė_x is the complete-expectation-of-life for life (x).

l_x is the expected number of survivors at age x of an original group of l₀ members.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 66-67.

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

Problem S3L14-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Of a group of 7765 newborn triceratopses, how many total years will be lived beyond age 30 by the group's members?

Solution S3L14-1. We want to find T₃₀ = ₀^∞∫l_30+tdt. We know that l₀ = 7765, so l_x = l₀*s(x) = 7765e^-0.34x. Thus, T₃₀ = ₀^∞∫7765e^-0.34(30+t)dt = 7765e^-0.34(30) ₀^∞∫e^-0.34tdt =

7765e^-0.34(30)(-50/17)e^-0.34t│₀^∞ = 7765e^-0.34(30)(50/17) = 0.8489044841 years

Problem S3L14-2. For a certain group of burgundy crickets,

l_x = 354(1 - 0.0625x²) for 0 ≤ x ≤ 4 and 0 otherwise. Find ė_x for this group of burgundy crickets.

Solution S3L14-2.

We first find T_x by using the formula T_x = ₀^∞∫l_x+tdt. But the upper bound of our integral is not infinity, but 4-x, since at t = 4-x, x+t = x + (4 - x) = 4, and we know that l₄ = 0, so all members of the group of burgundy crickets will have died by age 4. Therefore,

₀^4-x∫354(1 - 0.0625(x+t)²)dt = ₀^4-x∫(354 - 22.125(x+t)²)dt = (354t - 7.375(x+t)³)│₀^4-x =

1416 - 354x - 7.375(4)³ + 7.375x³ = T_x = 944 - 354x+ 7.375x³ =

We use the formula T_x/l_x = ė_x to find that

ė_x = (944 - 354x+ 7.375x³)/[354(1 - 0.0625x²)].

We can simplify this expression as follows:

ė_x = [354(8/3 - x + (1/48)x³)]/[354(1 - 0.0625x²)].

ė_x = (8/3 - x + (1/48)x³)/(1 - 0.0625x²).

Problem S3L14-3. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. For a group of 7765 newborn triceratopses, find a(5).

Solution S3L14-3. We know that l₀ = 7765, so l_x = l₀*s(x) = 7765e^-0.34x.

Then we find L₅ using the formula L_x = ₀¹∫l_x+tdt.

L₅ = ₀¹∫7765e^-0.34(5+t)dt = 7765e^-0.34(5)₀¹∫e^-0.34tdt = 7765e^-0.34(5)(-50/17)e^-0.34t│₀¹ =

7765e^-0.34(5)(50/17)(1 - e^-0.34) = L₅ = 1202.543013 years.

We can also find l₅ = 7765e^-0.34*5 = 1418.537564 and l₆ = 7765e^-0.34*6 = 1009.67294.

Now we can use the formula a(x) = (L_x - l_x+1)/(l_x - l_x+1) = (L₅ - l₆)/(l₅ - l₆) =

(1202.543013 - 1009.67294)/(1418.537564 - 1009.67294) = a(5) = 0.4717211057

Problem S3L14-4. For a certain group of burgundy crickets,

l_x = 354(1 - 0.0625x²) for 0 ≤ x ≤ 4 and 0 otherwise. Find a(2) for this group of burgundy crickets.

Solution S3L14-4. We first find L₂ using the formula L_x = ₀¹∫l_x+tdt.

L₂ = ₀¹∫l_2+tdt = ₀¹∫354(1 - 0.0625(2+t)²)dt = ₀¹∫(354 - 22.125(2+t)²)dt =

(354t - 7.375(2+t)³)│₀¹ = 354 - 7.375*27 + 7.375*8 = 354 - 7.375*19 = L₂ = 213.875.

We can also find l₂ = 354(1 - 0.0625*2²) = 265.5 and l₃ = 354(1 - 0.0625*3²) = 154.875.

Now we can use the formula a(x) = (L_x - l_x+1)/(l_x - l_x+1) = (L₂ - l₃)/(l₂ - l₃) =

(213.875 - 154.875)/(265.5 - 154.875) = a(2) = 8/15 = about 0.5333333333333

Problem S3L14-5. In a certain group of ichthyosaurs, 6086 ichthyosaurs are alive at age 7, L₇ = 5440, and a(7) = 0.687. Find out how many of these ichthyosaurs will senselessly perish (i.e., die) before reaching their 8^th birthday. Round up to the nearest whole ichthyosaur.

Solution S3L14-5. We use the formula a(x) = (L_x - l_x+1)/(l_x - l_x+1). We want to find l₇ - l₈.

We first find l₈, for which we need to rearrange the formula thus:
a(x)(l_x - l_x+1) = (L_x - l_x+1)

a(x)l_x - a(x)l_x+1 = L_x - l_x+1

a(x)l_x - L_x = a(x)l_x+1 - l_x+1

a(x)l_x - L_x = (a(x) - 1)l_x+1

(a(x)l_x - L_x)/(a(x) - 1) = l_x+1

Thus, l₈ = (a(7)l₇ - L₇)/(a(7) - 1) = (0.687*6086 - 5440)/(0.687 - 1) = l₈ = 4022.102236

Thus, l₇ - l₈ = 6086 - 4022.102236 = 2063.897764 = 2064 ichthyosaurs.

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