Life Table Probability Functions: Practice Problems and Solutions: Part 1

Before we begin to explore various life table functions, some actuarial notation is in order.

(x) refers to a life of age x.

T(x) refers to the future lifetime of a life currently at age x.

X refers to the age of death of a particular life.

F(x) is the cumulative distribution function of X.

s(x) is the survival function of X.

The following are true of F(x) and s(x):

F(x) = Pr(X ≤ x), x ≥ 0.

s(x) = 1 - F(x) = Pr(X > x), x ≥ 0.

It is always assumed in these models that F(0) = 0 and s(0) = 1.

The following two expressions are particularly important in studying life tables.

_tq_x is the probability that life (x) will die within t years. It can also be expressed as

_tq_x = Pr(T(x) ≤ t), t ≥ 0.

_tp_x is the probability that life (x) will survive during the next t years and reach the age x + t. It can also be expressed as

_tp_x = 1 - _tq_x = Pr(T(x) > t), t ≥ 0.

When a life's current age is zero, _xp₀ = s(x), x ≥ 0.

When t = 1, we can omit the "t" subscript to the left of the "p" or the "q." Thus,

q_x is the probability that life (x) will die within 1 year.

p_x is the probability that life (x) will survive during the next 1 year and reach the age x + 1.

Here are some convenient ways of calculating _tp_x and _tq_x.

_tp_x = s(x +t)/s(x)

_tq_x = 1 - s(x +t)/s(x)

The above calculations will be the focus of this section, as it is vital to be able to do them in order to move on to understanding other concepts relevant to Exam 3L.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1986. First Edition. Society of Actuaries: Itasca, Illinois. pp. 45-48.

Problem S3L1-1. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. Find ₃p₅.

Solution S3L1-1. We use the formula _tp_x = s(x +t)/s(x). Thus, ₃p₅ = s(5 +3)/s(5) = s(8)/s(5) =

e^-0.34*8/e^-0.34*5 = e^-0.34*3 = ₃p₅ =e^-1.02 = 0.3605949402

Problem S3L1-2. 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 ₅₅q₃₄.

Solution S3L1-2. We use the formula _tq_x = 1 - s(x +t)/s(x). Thus, ₅₅q₃₄ = 1 - s(55 + 34)/s(34) =

1 - s(89)/s(34) = 1 - (1 - 89/94)/(1 - 34/94) = ₅₅q₃₄ = 11/12 = about 0.916666666667

Problem S3L1-3. 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 65 will survive to age 68?

Solution S3L1-3. We wish to find ₃p₆₅ = s(65 +3)/s(65) = (1/68)/(1/65) = ₃p₆₅ = 65/68 = about 0.9558823529.

Problem S3L1-4. 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. What is the probability that a black swan that has survived to age 33 will survive to age 34?

Solution S3L1-4. We seek to find p₃₃ = s(33+1)/s(33) = s(34)/s(33). We know that s(x) = 1 - F(x) =

1- (1 - 4x^-1/2) = 4x^-1/2. Thus, s(34)/s(33) = (4*34^-1/2)/(4*33^-1/2)= p₃₃ = √(33/34) = 0.9851843661.

Problem S3L1-5. The lives of Unicorn-Pegasi can be modeled by the following cumulative distribution function: F(x) = x/36 - e^-0.45x. If any Unicorn-Pegasus reaches the age of 35, it will get to live forever. Find the probability that a Unicorn-Pegasus currently aged 34 will not get to live forever.

Solution S3L1-5. We wish to find q₃₄, the probability that the Unicorn-Pegasus aged 34 will die within 1 year. We use the formula _tq_x = 1 - s(x +t)/s(x). Here, s(x) = 1 - F(x) = 1 - x/36 + e^-0.45x.

q₃₄ = 1 - s(35)/s(34) = 1 - (1 - 35/36 + e^-0.45*35)/(1 - 34/36 + e^-0.45*34) = about 0.4999994386.

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