Life Table Probability Functions: Practice Problems and Solutions: Part 2

Here, we introduce another life table probability function.

_t│uq_x is the probability that life (x) will survive for the next t years and die within the subsequent u years. That is, _t│uq_x is the probability that life (x) will die between ages x + t and x + t + u.

When u = 1, this function can simply be written as _t│q_x.

Other expressions for _t│uq_x are as follows:

_t│uq_x= Pr[t < T(x) ≤ t + u]

_t│uq_x= _{t + u}q_x - _tq_x

_t│uq_x= _tp_x - _{t + u}p_x

We can calculate _t│uq_x as follows:

_t│uq_x = (s(x + t) - s(x + t + u))/s(x) = _tp_x* _uq_x+t

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 47-48.

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

Problem S3L2-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find _6│4 q₂.

Solution S3L2-1. We use the following formula: _t│uq_x = (s(x + t) - s(x + t + u))/s(x).Thus,

_6│4 q₂= (s(2 + 6) - s(2 + 6 + 4))/s(2) = (s(8) - s(12))/s(2) = (e^-0.34*8 - e^-0.34*12)/(e^-0.34*2) = _6│4 q₂ = about 0.096655409

Problem S3L2-2. 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 _5│5q₂₂.

Solution S3L2-2. We use the following formula: _t│uq_x = (s(x + t) - s(x + t + u))/s(x).Thus,

_5│5q₂₂= (s(22 + 5) - s(22 + 5 + 5))/s(22) = (s(27) - s(32))/s(22) = ((1 - 27/94) - (1 - 32/94))/(1 - 22/94) = (5/94)/(72/94) = _5│5q₂₂ = 5/72 = about 0.0694444444444

Problem S3L2-3. Three-headed donkeys always survive until age 1. Thereafter, the survival function for the life of a three-headed donkey is s(x) = 1/x for all x > 1. What is the probability that a three-headed donkey that has survived to age 23 will senselessly perish (i.e., die) between the ages of 45 and 69?

Solution S3L2-3. We wish to find _22│24q₂₃ = (s(45) - s(69))/s(23) = (1/45 - 1/69)/(1/23) = _22│24q₂₃ = 8/45 = about 0.17777777778

Problem S3L1-4. You know the following about the lives of jumping slugs. For a jumping slug that has survived to age 39, the probability of surviving to age 45 is 0.835. For a jumping slug that has survived to age 45, the probability of dying within the subsequent 8 years is 0.23. What is the probability that a jumping slug that has survived to age 39 will senselessly perish (i.e., die) between the ages of 45 and 53?

Solution S3L1-4. We use the following formula: _t│uq_x = _tp_x * _uq_x+t. We wish to find _6│8q₃₉ = ₆p₃₉ * ₈q₄₅, where we are given that ₆p₃₉ = 0.835 and ₈q₄₅ = 0.23. Thus, 0.835*0.23 = _6│8q₃₉ = 0.19205

Problem S3L1-5. There is a 0.05 probability that an ancient Greek god who has survived to age 1245 will die between the ages of 1298 and 1403. Zeus is currently 1245 years old, and his probability of surviving to age 1298 is 0.95. What is the probability that, if Zeus survives to age 1298, he will also survive past age 1403?

Solution S3L1-5. We use the following formula: _t│uq_x = _tp_x * _uq_x+t. We wish to find ₁₀₅p₁₂₉₈. We are given that _53│105q₁₂₄₅ = 0.05 and ₁₂₉₈p₁₂₄₅ = 0.95. We thus know that 0.05 = 0.95*₁₀₅q₁₂₉₈, and so ₁₀₅q₁₂₉₈ = 1/19. 1- ₁₀₅q₁₂₉₈ = ₁₀₅p₁₂₉₈ = 18/19 = 0.0526315789

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