BMA3105 Actuarial Mathematics II
Lecturer 1 : Actuarial Survival Models
Notes Prepared
Gikonyo Kiguta
Mount Kenya University
Contents
1
Actuarial Survival Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.1
Introduction
1.2
Age-At-Death Random Variable
1.2.1
1.2.2
1.2.3
1.2.4
1.2.5
The Cumulative Distribution Function of X . . .
The Survival Distribution Function of X . . . . . .
The Probability Density Function of X . . . . . . .
Force of Mortality of X (hazard rate function)
The Mean and Variance of X . . . . . . . . . . . . .
1.3
Selected Parametric Survival Models
1.3.1
1.3.2
1.3.3
1.3.4
1.3.5
The Uniform or De Moivre’s Model . . . . . . . . . . . . . . . . .
The Exponential Model . . . . . . . . . . . . . . . . . . . . . . . . .
The Gompertz Model . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Modified Gompertz Model: The Makeham’s Model
The Weibull Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4
Time-Until-Death Random Variable
1.4.1
1.4.2
1.4.3
1.4.4
1.4.5
1.4.6
The Survival Function of T (x) . . . . . . . . . . . .
The Cumulative Distribution Function of T (x)
Probability Density Function of T (x) . . . . . . .
Force of Mortality of T (x) . . . . . . . . . . . . . . .
Mean and Variance of T (x) . . . . . . . . . . . . .
Central Death Rates . . . . . . . . . . . . . . . . . .
1.5
The Life Table Format
1.5.1
1.5.2
Constructing a life table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Life table functions at non-integer ages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
5
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28
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31
32
33
36
37
1. Actuarial Survival Models
1.1
Introduction
There are insurance policies that provide a benefit on the death of the policyholder. Since the death
date of the policyholder is unknown, the insurer when issuing the policy does not know exactly
when the death benefit will be payable. Thus, an estimate of the time of death is needed. For
that, a model of human mortality is needed so that the probability of death at a certain age can be
calculated. Survival models provide such a framework.
A survival model is a special kind of a probability distribution. In the actuarial context, a survival
model can be the random variable that represents the future lifetime of an entity that existed at time
0.
1.2
Age-At-Death Random Variable
The central difficulty in issuing life insurance is that of determining the time of death of the insured.
In this section, we introduce the first survival model: The age-at-death distribution. Let time 0
denote the time of birth of an individual. We will always assume that everyone is alive at birth .
The age-at-death of the individual can be modeled by a positive continuous random variable X:
1.2.1
The Cumulative Distribution Function of X
The cumulative distribution function (CDF) of X is given by
F(x) = Pr(X ≤ x) = Pr(X < x)
That is, F(x) is the probability that death will occur prior to (or at) age x. Since X is positive, the
event {s : X(s) ≤ 0} is impossible so that F(0) = 0. Also, we know that F(∞) = 1, F(x) is non
decreasing, F(X) is right-continuous, and that F(x) = 0 for x < 0
Chapter 1. Actuarial Survival Models
6
Recall
Relationship between PDF and CDF for a Continuous Random Variable
Let X be a continuous random variable with pdf f and codf F. - By definition, the cdf is found by
integrating the pdf:
Z
x
F(x) =
f (t)dt
−∞
- By the Fundamental Theorem of Calculus, the pdf can be found by differentiating the cdf:
f (x) =
d
[F(x)]
dx
Example
Can the function F(x) =
death random variable?
1
x+1 , x
≥ 0 be a legitimate cumulative distribution function of an age-at-
Solution
since F(0) = 1 6= 0, the given function can not be a CDF of an age-at-death random variable
Example
Express the following probability statement in terms of F(x) : Pr(a < X ≤ b | X > a)
Solution.
By Bayes’ formula, we have
Pr[(a < X ≤ b) ∩ (X > a)]
Pr(X > a)
Pr(a < X ≤ b) F(b) − F(a)
=
=
Pr(X > a)
1 − F(a)
Pr(a < X ≤ b | X > a) =
Example
Let
x 61
, 0 ≤ x ≤ 120
F(x) = 1 − 1 −
120
Determine the probability that a life aged 35 dies before the age of 55.
Solution
The probability is given by
Pr(35 < X ≤ 55 | X > 35) =
F(55) − F(35)
= 0.0437
1 − F(35)
An alternative notation for F(x) used by actuaries is
x q0
= F(x) = Pr(X ≤ x)
1.2 Age-At-Death Random Variable
7
Example
Given that x q0 = 1 − e−0.008x , x ≥ 0. Find the probability that a newborn baby dies between age 60
and age 70.
Solution.
We have
Pr(60 < X ≤ 70) = 70 q0 − 60 q0 = e−0.48 − e−0.56 = 0.04757
Example
Let X be a continuous random variable with PDF
3
4x 0 < x ≤ 1
fX (x) =
0
otherwise
Find P X ≤
2
3
|X >
1
3
Solution.
We have
P 13 < X ≤ 32
2
1
P X ≤ |X >
=
3
3
P X > 13
R
2
3
1
3
= R1
1
3
=
4x3 dx
4x3 dx
3
16
Problem
The CDF of a continuous random variable is given by
F(x) = 1 − e−0.34x , x ≥ 0
Find Pr(10 < X ≤ 23)
Problem
Let
x 61
, 0 ≤ x ≤ 120
F(x) = 1 − 1 −
120
Determine the probability that a newborn survives beyond age 25.
Problem
Chapter 1. Actuarial Survival Models
8
The CDF of an age-at-death random variable is given by F(x) = 1− e−0.008x , x ≥ 0. Find the
probability that a newborn baby dies between age 60 and age 70 .
1.2.2
The Survival Distribution Function of X
The survival distribution function is given by
s(x) = 1 − F(x) = Pr(X > x), x ≥ 0.
Thus, s(x) is the probability that a newborn will survive to age x. From the properties of F we see
that:
• s(0) = 1, (everyone is alive at birth)
• s(∞) = 0 (everyone dies eventually)
• s(x) is right continuous
• s(x) is non-increasing
These four conditions are necessary and sufficient so that any non-negative function s(x) that
satisfies these conditions serves as a survival function.
In actuarial notation, the survival function is denoted by x p0 = s(x) = Pr(X > x) = 1 − x q0 .
Example
Consider an age-at-death random variable X with survival distribution defined by
1
1
(100 − x) 2 , 0 ≤ x ≤ 100.
10
(a) Explain why this is a suitable survival function.
(b) Find the corresponding expression for the cumulative probability function.
(c) Compute the probability that a newborn with survival function defined above will die between
the ages of 65 and 75 .
s(x) =
Solution.
1
1
(a) We have that s(0) = 1, s0 (x) = − 20
(100 − x)− 2 ≤ 0, s(x) is right continuous, and s(100) = 0.
Thus, s satisfies the properties of a survival function.
1
1
(b) We have F(x) = 1 − s(x) = 1 − 10
(100 − x) 2 .
(c) We have
Pr(65 < X ≤ 75) = s(65) − s(75) =
1
1
1
1
(100 − 65) 2 − (100 − 75) 2 ≈ 0.092
10
10
Example
The survival distribution function for an individual is determined to be
s(x) =
75 − x
,
75
0 ≤ x ≤ 75.
(a) Find the probability that the person dies before reaching the age of 18.
(b) Find the probability that the person lives more than 55 years.
1.2 Age-At-Death Random Variable
9
(c) Find the probability that the person dies between the ages of 25 and 70 .
Solution.
(a) We have
Pr(X < 18) = Pr(X ≤ 18) = F(18) = 1 − s(18) = 0.24
(b) We have
Pr(X > 55) = s(55) = 0.267
(c) We have
Pr(25 < X < 70) = F(70) − F(25) = s(25) − s(70) = 0.60
Problem
Consider an age-at-death random variable X with survival distribution defined by
s(x) = e−0.34x , x ≥ 0
Compute Pr(5 < X < 10).
Problem
Consider an age-at-death random variable X with survival distribution
x p0
= e−0.34x , x ≥ 0.
Find x q0 .
Problem
2
x
Find the cumulative distribution function corresponding to the survival function s(x) = 1 − 100
for
x ≥ 0 and 0 otherwise.
Problem
x
The survival distribution is given by s(x) = 1 − 100
for 0 ≤ x ≤ 100 and 0 otherwise.
(a) Find the probability that a person dies before reaching the age of 30 .
(b) Find the probability that a person lives more than 70 years.
1.2.3
The Probability Density Function of X
The probability density function of an age-at-death random variable X measures the relative
likelihood for death to occur for a given age.
The PDF of X relates to a point and is considered as an instantaneous measure of death at age x
compared to F(x) and s(x) which are probabilities over time intervals so they are considered as
time-interval measures.
Mathematically, the probability density function, denoted by f (x), is given by
f (x) =
d
d
F(x) = − s(x), x ≥ 0
dx
dx
Chapter 1. Actuarial Survival Models
10
whenever the derivative exists. Note that f (x) ≥ 0 for x ≥ 0 since F is non-decreasing for x ≥ 0.
When the density function is defined over the relevant interval, we have
Pr(a < X ≤ b) =
Z b
f (x)dx,
a
Z x
Z x
F(x) =
f (t)dt =
−∞
f (t)dt
0
s(x) = 1 − F(x) =
Z ∞
f (t)dt −
Z x
Z ∞
f (t)dt =
0
0
f (t)dt.
x
Using laws of probability, we can use either F(x) or s(x) to answer probability statements.
Example
1
Explain why the function f (x) =
5
, x ≥ 0 and 0 otherwise can not be a PDF.
(x+1) 2
Solution.
If f (x) is a PDF then we must have
Z ∞
0
R∞
0
f (x)dx = 1. But
dx
(x + 1)
5
2
=−
∞
2
3(x + 1)
3
2
=
0
2
6= 1
3
Example
Suppose that the survival function of a person is given by
s(x) =
80 − x
, 0 ≤ x ≤ 80.
80
Calculate
(a) F(x).
(b) f (x).
(c) Pr(20 < X < 50).
Solution.
(a) We have
F(x) = 1 − s(x) =
x
,x ≥ 0
80
(b) We have
f (x) = −s0 (x) =
1
, x > 0.
80
(c) We have
Pr(20 < X < 50) = s(20) − s(50) = 0.375
1.2 Age-At-Death Random Variable
11
Example
Find a formula both in terms of F(x) and s(x) for the conditional probability that an entity will die
between the ages of x and z, given that the entity is survived to age x.
Solution.
We have
Pr[(x < X ≤ z) ∩ (X > x)]
Pr(X > x)
Pr(x < X ≤ z)
=
Pr(X > x)
F(z) − F(x)
=
1 − F(x)
s(x) − s(z)
=
s(x)
Pr(x < X ≤ z | X > x) =
Example
An age-at-death random variable is modeled by an exponential random variable with PDF f (x) =
0.34e−0.34x , x ≥ 0. Use the given PDF to estimate Pr(1 < X < 1.02).
Solution.
We have
Pr(1 < X < 1.02) ≈ 0.02 f (1) = 0.005
Problem
The density function of a random variable X is given by f (x) = xe−x for x ≥ 0. Find the survival distribution function of X.
Problem
Consider an age-at-death random variable X with survival distribution defined by
s(x) = e−0.34x , x ≥ 0
Find the PDF and CDF of X.
Problem
Find the probability density function of a continuous random variable X with survival function
s(x) = e−λ x , λ > 0, x ≥ 0.
Chapter 1. Actuarial Survival Models
12
1.2.4
Force of Mortality of X (hazard rate function)
This may be defined as the "instantaneous rate of mortality", i.e.
µx = lim
h qx
h
h→0+
Theorem
Suppose that µx is continuous on [α, w). We have
µx = −
Proof.
we have
s0 (x)
l0
=− x
s(x)
lx
−[s(x + h) − s(x)] −s0+ (x)
=
h→0+
hs(x)
s(x)
µx = lim
where s0+ (x) denotes the R.H. derivative of s(x). But s(x) and s0+ (x) = −s(x)µx are continuous on
[α, w), so s0 (x) exists and equals −s(x)µx
Hence
µx = −
s0 (x)
l0
=− x
s(x)
lx
We remark that
µx = −
Hence
−
Z x
Z x
d
(log ly ) dy
dy
= [log ly ]xα
µy dy =
α
d
log lx
dx
α
= log lx − log lα
lx
= log
lα
from which we obtain the important formulae:
Zx
lx = lα exp −
µy dy
α
and
Zx
s(x) = exp −
µy dy
α
It follows that
Z x+t
Zt
lx+t
= exp −
µy dy = exp − µx+r dr
t px =
lx
x
0
Note
We can also express the hazard rate in terms of s(s)and f (x):
µ(x) = lim
h→0
Pr(x < X ≤ x + h | X > x)
f (x)
f (t)
=
=
= hazard rate at age t years .
h
s(x)
1 − F(t)
1.2 Age-At-Death Random Variable
13
Example
A life aged 50 has a force of mortality at age 50 equal to 0.0044. Estimate the probability
that the person dies on his birthday.
Solution.
From the above definition with x = 50, µ(50) = 0.0044, and h =
we can write
1
365
= 0.00274
Pr(50 < X < 50 + 0.00274 | X > 50) ≈ µ(50) × 0.0044
= (0.00274)(0.0044) = 1.2 × 10−5
We can relate the mortality function to the survival function from birth as shown in the next example.
Example
Show that
µ(x) = −
d
s0 (x)
= − [ln s(x)]
s(x)
dx
Solution.
The equation follows from f (x) = −s0 (x) and
d
dx [ln s(x)]
=
s0 (x)
s(x)
Example
Find the hazard rate function of an exponential random variable with parameter µ.
Solution.
We have
µ(x) =
f (x) µe−µx
= −µx = µ.
s(x)
e
The exponential random variable is an example of a constant force model
Cumulative hazard function (integrated hazard function
The function Λ(x) is called the cumulative hazard function or the integrated hazard function
(CHF). The CHF can be thought of as the accumulation of hazard up to time x.
Λ(x) = − ln(s(x))
Summarizing, if we know any one of the functions µ(x), Λ(x), s(x) we can derive the other two
functions:
d
ln s(x)
dx
Λ(x) = − ln(s(x))
µ(x) = −
Chapter 1. Actuarial Survival Models
14
−s(x) = e−Λ(x)
Now, for a function µ(x) to be an acceptable force of mortality, the function s(x) = e−Λ(x) must be
an acceptable survival function.
Example
Show that
s(x) = e−Λ(x)
where
Z x
Λ(x) =
µ(s)ds.
0
Solution.
Z x
µ(s)ds = −
0
Z x
d
0
ds
[ln s(s)]ds = ln s(0) − ln s(x) = ln 1 − ln s(x) = − ln s(x).
Example
Find the cumulative hazard function of the exponential random variable with parameter µ.
Solution.
We have
Z x
Λ(x) =
Z x
µ(s)ds =
0
µdx = µx
0
Example
The force of mortality is
µx =
1
100 − x
Calculate 10 p50 .
Solution.
Z 10
µ50+s ds
10 p50 = exp −
0
Z 10
ds
= exp −
0 50 − s
= exp(ln(50 − 10) − ln(50))
40
=
= 0.8
50
˜
Example
1.2 Age-At-Death Random Variable
15
A person age 70 is subject to the following force of mortality:
(
0.01 t ≤ 5
µ70 (t) =
0.02 t > 5
Calculate 20 p70 for this person.
Solution.
20
Z
20 p70 = exp −
µ70 (s)ds
0
Break the integral up into two parts:
Z 5
0
Z 5
µ70 (t)dt =
0.01 dt = 0.05
0
Z 20
5
Z 20
µ70 (t)dt =
0.02 dt = 0.3
5
It follows that 20 p70 is the product of the negative exponentiated integrals,
20 p70
= e−0.05−0.3 = 0.70469
.
Example
Which of the following functions can serve as a force of mortality?
(I) µ(x) = BCx , B > 0, 0 < C < 1, x ≥ 0.
1
(II) µ(x) = B(x + 1)− 2 , B > 0, x ≥ 0.
(III) µ(x) = k(x + 1)n , n > 0, k > 0, x ≥ 0.
Solution.
(I) Finding s(x), we have
s(x) = e−
Rx
0
µ(t)dt
B
x
= e lnC (1−C ) .
B
We have s(x) ≥ 0, s(0) = 1, s(∞) = e lnC 6= 0. Thus, the given function cannot be a force of mortality.
(II) Finding s(x), we have
s(x) = e−
Rx
0
µ(t)dt
1
−2B (x+1) 2 −1
=e
.
We have, s(x) ≥ 0, s(0) = 1, s(∞) = 0, s0 (x) < 0, and s(x) is rightcontinuous. Thus, µ(x) is a
legitimate force of mortality.
(III) Finding s(x), we have
s(x) = e−
We have, s(x) ≥ 0, s(0) = 1, s(∞) =
legitimate force of mortality
0, s0 (x)
Rx
0
µ(t)dt
= e−
kxn+1
n+1
.
< 0, and s(x) is right-continuous. Thus, µ(x) is a
Chapter 1. Actuarial Survival Models
16
Problem
An age-at-death random variable has a survival function
s(x) =
1
1
(100 − x) 2 , 0 ≤ x ≤ 100.
10
Find the force of mortality of this random variable.
Problem
Consider an age-at-death random variable X with survival distribution defined by
s(x) = e−0.34x , x ≥ 0
Find µ(x).
Problem
Let
x 61
F(x) = 1 − 1 −
, 0 ≤ x ≤ 120
120
Find µ(40).
Problem
The force of mortality of a survival model is given by µ(x) =
f (x).
1.2.5
3
x+2 , x ≥ 0.
Find Λ(x), s(x), F(x), and
The Mean and Variance of X
For a continuous random variable X, the mean of X, also known as the (unconditional) first moment
of X, is given by
Z
∞
E(X) =
x f (x)dx
0
provided that the integral is convergent.
Z ∞
E(X) =
x f (x)dx = −
0
=
Z ∞
xs0 (x)dx
0
−xs(x)]∞
0 +
Z ∞
Z ∞
s(x)dx =
0
s(x)dx.
0
In the actuarial context, E(X) is known as the life expectancy or the complete expectation of life
◦
at birth and is denoted by ex .
◦
Z ∞
ex =
0
Z ∞
t px dt
=
s(x)dx
0
Likewise, the mean of X 2 or the (unconditional) second moment of X is given by
Z
E X2 =
0
∞
x2 f (x)dx
1.2 Age-At-Death Random Variable
17
provided that the integral is convergent.
Using integration by parts
E X
2
Z ∞
=2
xs(x)dx
0
The variance of X is given by
Var(X) = E X 2 − [E(X)]2 .
Example
An actuary models the lifetime in years of a random selected person as a random variable X
1
x 2
with survival function s(x) = 1 − 100
, 0 ≤ x ≤ 100. Find e◦0 and Var(X).
Solution.
The mean is
◦
e0 = E(X) =
Z 100 1−
0
x 23
2
x 12
1−
dx = −
100
3
100
100
=
0
200
3
The second moment of X is
Z
E X2 =
100
x2 f (x)dx =
0
1
200
Z 100
0
x − 12
16000
x2 1 −
dx =
100
3
Thus,
16000 40000 80000
Var(X) = E X 2 − [E(X)]2 =
−
=
3
9
9
Two more important concepts about X are the median and the mode.
The median of a continuous random variable X is defined to be the value m satisfying the equation
Pr(X ≤ m) = Pr(X ≥ m) =
1
2
Thus, the median age-at-death m is the solution to
F(m) = 1 − s(m) =
1
2
We define the mode of X to be the value of x that maximizes the PDF f (x).
Example
Consider an age-at-death random variable X with cumulative distribution defined by
x 12
F(x) = 1 − 1 −
, 0 ≤ x ≤ 100.
100
Find the median and the mode of X.
Chapter 1. Actuarial Survival Models
18
Solution.
m
To find the median, we have to solve the equation 1 − 100
12
1
x
To find the mode, we first find f (x) = F 0 (x) = − 200
1 − 100
= 0.5 which gives m = 75.
− 12
.
The maximum of this function on the interval [0, 100] occurs when x = 0 (you can check this
either analytically or graphically) so that the mode of X is 0
Problem
Consider an age-at-death random variable X with survival function s(x) = (1 + x)e−x , x ≥ 0.
Calculate E(X).
Problem
The age-at-death random variable has the PDF f (x) = 1k , 0 ≤ x ≤ k. Suppose that the expected
age-at-death is 4 . Find the median age-at-death.
Problem
An actuary models the lifetime in years of a random selected person as a random variable X
x2
with survival function s(x) = 1 − 8100
, 0 ≤ x ≤ 90. Calculate ė0 and Var(X).
Problem
A survival distribution has a force of mortality given by µ(x) =
◦
1
720−6x , 0
≤ x < 120.
(a) Calculate e0 .
(b) Find Pr(30 ≤ X ≤ 60).
Problem
The PDF of a survival model is given by f (x) =
24
2+x .
Find the median and the mode of X.
Curtate expectation of life
We now consider the discrete random variable K (or K(x) if the age x is not clear) defined by
K = the integer part of T
= the number of complete years to be lived
in the future by (x)
Now it follows by formula 1.3.16 that
Pr{K = k} = k | qx
(k = 0, 1, 2, . . .)
1.2 Age-At-Death Random Variable
19
This variable is used in many actuarial calculations. In particular, the curtate expectation of life at
age x, written ex , is the mean of K; that is
∞
ex =
Theorem
∞
∑ k · k qx =
∑ k·k
k=0
k=1
∞
ex =
∑ k px =
k=1
Proof.
qx
lx+1 + lx+2 + . . .
lx
dx+1
dx+2
dx+3
+2
+3
+...
lx
lx
lx
1
= [(dx+1 + dx+2 + dx+3 + . . .)
lx
+ (dx+2 + dx+3 + . . .)
ex =
+ (dx+3 + . . .)
+ . . .]
lx+1 + lx+2 + lx+3 + . . .
=
lx
ly = (ly − ly+1 ) + (ly+1 − ly+2 ) + . . .
= dy + dy+1 + . . .
[since We may also evaluate Var(K) by the formula
Var(K) = E K 2 − [E(K)]2
∞
=
∑ k2
k=1
dx+k
− (ex )2
lx
As is clear by general reasoning,
◦
ex + ex +
1
2
Example
In a certain population, the force of mortality equals 0.025 at all ages. (We are assuming here that
there is no upper limit to age.)
Calculate:
(i) the probability that a new-born baby will survive to age 5
(ii) the probability that a life aged exactly 10 will die before age 12
(iii) the probability that a life aged exactly 5 will die between ages 10 and 12
(iv) the complete expectation of life of a new-born baby
(v) the curtate expectation of life of a new-born baby.
Solution.
(i)
Z5
p
=
exp
−
0.025dt
= e−0.125 = 0.88250
5 0
0
Chapter 1. Actuarial Survival Models
20
(ii)
Z2
0.025dt = 1 − e−0.05 = 0.04877
2 q10 = 1 − 2 p10 = 1 − exp −
0
(iii)
5
| 2 q5 = 5 p5 · 2q10 = e−0.125 × 0.04877 = 0.88250 × 0.04877 = 0.04304
(iv)
◦
e0 =
Z ∞
0
Z ∞
t p0 dt =
e−0.025t dt = −
0
1 −0.025t ∞
1
e
=
= 40
0
0.025
0.025
(v)
∞
e0 =
1.3
∞
∑ kp0 =
∑ e−0.025k =
k=1
k=1
e−0.025
= 39.5
1 − e−0.025
Selected Parametric Survival Models
Parametric survival models are models for which the survival function s(x) is given by a mathematical formula. In this section, we explore some important parametric survival models.
1.3.1
The Uniform or De Moivre’s Model
Let X be a uniform random variable on the interval [a, b]. It is easy to see 3 that the PDF of this
1
random variable is f (x) = b−a
for a ≤ x ≤ b and zero elsewhere. This model is an example of a
two-parameter model with parameters a and b.
If X is the age-at-death random variable, we take a = 0 and b = ω where ω is the maximum
or terminal age by which all people have died then the PDF is f (x) = ω1 for 0 ≤ x ≤ ω and 0
otherwise. In the actuarial context, this survival model is known as De Moivre’s Law.
Example
Consider the uniform distribution model as defined above. Find F(x), s(x), and µ(x).
Solution.
We have
Z x
x
ω
0
ω −x
s(x) = 1 − F(x) =
ω
f (x)
1
µ(x) =
=
s(x)
ω −x
F(x) =
f (s)ds =
Example
◦
Consider the uniform distribution model as defined above. Find e0 and Var(X).
Solution.
We have
◦
e0 = E(X) =
Z ω
x f (x)dx =
0
ω
2
1.3 Selected Parametric Survival Models
21
and
Z
E X2 =
ω
x2 f (x)dx =
0
ω2
.
3
Thus,
Var(X) =
ω2 ω2 ω2
−
=
3
4
12
Problem
◦
Suppose that the age-at-death random variable X is uniform in [0, ω]. Find Var(X) if e0 = 45.
Problem
Suppose that the age-at-death random variable X is uniform in [0, ω] with Var(X) =
625
3 .
Find ω.
Problem
x
, 0 ≤ x ≤ 90. Calculate
The survival function in De Moivre’s Law is given by s(x) = 1 − 90
(a) µ(x)
(b) F(x)
(c) f (x)
(d) Pr(20 < X < 50).
1.3.2
The Exponential Model
In this survival model, the age-at-death random variable follows an exponential distribution with
survival function given by
s(x) = e−µx , x ≥ 0, µ > 0.
This model is also known as the constant force model since the force of mortality is constant.
FT (x) (t) = 1 − e−µt
sT (x) (t) = e−µt
fT (x) (t) = µe−µt
µx (t) = µ
= s(x) = e−µt
1
E[T (x)] =
µ
1
Var(T (x)) = 2
µ
t px
Example
Consider the exponential model. Find F(x), f (x), and µ(x).
Solution.
Chapter 1. Actuarial Survival Models
22
We have
F(x) = 1 − s(x) = 1 − e−µx
f (x) = F 0 (x) = µe−µx
f (x)
µ(x) =
=µ
s(x)
Example
◦
Consider the exponential model. Find e0 and Var(X).
Solution.
We have
Z ∞
◦
e0 = E(X) =
Z ∞
x f (x)dx =
0
µxe−µx dx =
0
1
µ
and
Z
E X2 =
∞
x2 f (x)dx =
0
Z ∞
µx2 e−µx dx =
0
2
.
µ2
Thus,
Var(X) =
2
1
1
−
= 2
µ2 µ2
µ
Example
Consider an exponential model with density function f (x) = 0.01e−0.01x , x > 0 . Calculate x q0 and
x p0 .
Solution.
We have
x q0
= Pr(X ≤ x) = F(x) = 1 − e−0.01x
x p0
= 1 − x q0 = e−0.01x
Example
Let the age-at-death be exponential with density function f (x) = 0.01e−0.01x . Find Pr(1 < x < 2).
Solution.
We have
Pr(1 < X < 2) = s(1) − s(2) = 1 p0 − 2 p0 = e−0.01 − e−0.02 = 0.00985
Example
1.3 Selected Parametric Survival Models
23
Given the survival function


1,
s(x) = 1 − 0.01ex


0.
0≤x≤1
1 < x < 4.5
x ≥ 4.5
Find µ(4).
Solution.
We have
µ(4) = −
0.01e4
s0 (4)
=
= 1.203
s(4)
1 − 0.01e4
Problem
Consider an exponential survival model with s(x) = e−0.04x where x in years. Find the probability that a newborn survives beyond 20 years.
Problem
An exponential model has a force of mortality equals to 0.04. Find the probability that a newborn
dies between the age of 20 and 30 .
Problem
x
An exponential model has a survival function s(x) = e− 2 , x ≥ 0. Calculate µ(40).
Problem
Find the value of µ and the median survival time for an exponential survival function if s(3) = 0.4.
Problem
The age-at-death random variable is described by the PDF f (x) =
◦
x
1 − 60
.
60 e
Find e0 and Var(X).
1.3.3
The Gompertz Model
Gompertz proposed the force of mortality
µx = Bcx
with appropriate parameters B and c > 1. 1 He based this law on studies in which the logarithm of
the force of mortality over a wide range of ages above 20 appeared to fit well to a straight line.
Example
Find the survival function of Gompertz model.
Chapter 1. Actuarial Survival Models
24
Solution.
We have
s(x) = e−
Rx
0
µ(s)ds
= e−
Rx
0
Bcs ds
B
x
= e ln c (1−c )
Example
Under Gompertz’ Law, you are given that µ(20) = 0.0102 and µ(50) = 0.025. Find µ(x).
Solution.
The given hypotheses lead to the two equations
Bc20 = 0.0102
and
Bc50 = 0.025
.
Thus, taking ratios we find c30 =
0.025
0.0102 .
Solving for c we find c = 1.03. Thus,
B=
0.0102
= 0.0056
1.0320
. The force of mortality is given by
µ(x) = 0.0056(1.03)x
Example
You are given that mortality follows Gompertz with B = 0.01 and c = 1.1. Calculate
(i) µ(10)
(ii) the probability of a life aged 20 to attain age 30
(iii) the probability of a life aged 20 to die within 10 years.
Solution.
(i) We have
µ(10) = BC10 = 0.01(1.1)10 = 0.025937
(ii) We have
Pr(X > 30) s(30)
=
Pr(X > 20) s(20)
0.01
30
e ln 1.1 (1−1.1 )
= 0.01
= 0.32467
20
e ln 1.1 (1−1.1 )
Pr(X > 30 | X > 20) =
(iii) We have
Pr(X < 30 | X > 20) = 1 − Pr(X > 30 | X > 20) = 1 − 0.32467 = 0.67533
1.3 Selected Parametric Survival Models
25
Problem
Suppose that the lives of a certain species follow Gompertz’s Law. It is given that µ(0) = 0.43 and
µ(1) = 0.86. Determine µ(4).
Problem
Suppose that Gompertz’ Law applies with µ(30) = 0.00013 and µ(50) = 0.000344. Find
s0 (x)
s(x) .
Problem
A survival model follows Gompertz’ Law with parameters B = 0.0004 and c = 1.07. Find the
cumulative distribution function.
1.3.4
The Modified Gompertz Model: The Makeham’s Model
Makeham improved Gompertz’s law by adding a third parameter A :
µx = A + Bcx
A represents the constant part of the force of mortality, mortality that is independent of age and
is due to accidental causes. Bcx , with c > 1, represents mortality resulting from deterioration and
degeneration, which increases exponentially by age.
Example
Find the survival function of Makeham’s model.
Solution.
We have
s(x) = e−
Rx
0
µ(s)ds
= e−
Rx
s
0 (A+Bc )ds
B
x
= e−Ax− ln c (c −1)
Example
A survival model follows Makeham’s Law. You are given the following information: - 5 p70 = 0.70.
- 5 p80 = 0.40. - 5 p90 = 0.15. Determine the parameters A, B, and c.
Solution.
B
x
t
Let α = e−A and β = e ln c in the previous problem so that t px = α t β c (1−c ) . From the given
hypotheses we can write
70
5
α 5 β c (1−c ) = 0.70
80
5
α 5 β c (1−c ) = 0.40
90
5
α 5 β c (1−c ) = 0.15.
Chapter 1. Actuarial Survival Models
26
Hence,
0.40
0.70
0.15
ln
0.40
ln
= c70 1 − c5
c10 − 1 ln β
= c80 1 − c5
c10 − 1 ln β
and
From these two last equations, we find
0.15
0.40
c10 = ln
÷ ln
=⇒ c = 1.057719.
0.40
0.70
Substituting this above we find
0.40
÷ c70 1 − c5 c10 − 1 =⇒ β = 1.04622
ln β = ln
0.70
Solving the equation
B
e ln 1.057719 = 1.04622
we find B = 0.002535. Also,
h
i
70 5
A = −0.2 ln 0.70β c (c −1) = −0.07737.
Finally, the force of mortality is given by
µ(x) = −0.07737 + 1.04622(1.057719)x , x ≥ 0
Problem
The force of mortality of Makeham’s model is given by µ(x) = 0.31+ 0.43 (2x ) , x ≥ 0. Find
f (x) and F(x)
Problem
The force of mortality of Makeham’s model is given by µ(x) = 0.31+ 0.43 (2x ) , x ≥ 0. Calculate s(3)
Problem
The following information are given about a Makeham’s model:
- The accident hazard is 0.31.
- The hazard of aging is 1.72 for x = 2 and 3.44 for x = 3.
Problem
The force of mortality of a Makeham’s Law is given by µ(x) = A + 0.1(1.003)x , x ≥ 0 . Find
µ(5) if s(35) = 0.02.
1.3 Selected Parametric Survival Models
1.3.5
27
The Weibull Model
The Weibull law of mortality is defined by the hazard function
µ(x) = kxn ,
where k > 0, n > 0, x ≥ 0. That is, the death rate is proportional to a power of age. Notice that the
exponential model is a special case of Weibull model where n = 0.
Example
Find the survival function corresponding to Weibull model.
Solution.
We have
s(x) = e−
Rx
0
µ(s)ds
= e−
Rx
0
ksn ds
=e
kxn+1
n+1
Example
Suppose that X follows an exponential model with µ = 1. Define the random variable Y =
1
h(X) = X 3 . Show that Y follows a Weibull distribution. Determine the values of k and n.
Solution.
(y)
We need to find the force of mortality of Y using the formula µY (y) = sfYY (y)
. We first find the
CDF of Y . We have
1
3
FY (y) = Pr(Y ≤ y) = Pr X 3 ≤ y = Pr X ≤ y3 = 1 − e−y .
Thus,
sY (y) = 1 FY (y) = e−y
and
3
3
fY (y) = FY0 (y) = 3y2 e−y .
The force of mortality of Y is
µY (y) =
fY (y)
= 3y2 .
sY (y)
It follows that Y follows a Weibull distribution with n = 2 and k = 3
Example
You are given that mortality follows Weibull with k = 0.00375 and n = 1.5. Calculate
(i) µ(10)
R 30
(ii) 10 p20 = e− 20 µ(x)dx .
(iii) 10 q20 = 1 − 10 p20 .
Solution.
Chapter 1. Actuarial Survival Models
28
(i)
µ(10) = kxn = 0.00375(10)1.5 = 0.11859
(ii)
10 p20
= e−
R 30
20
0.00375x1.5 dx
= 0.089960
(iii)
10 q20
= 1 − 10 p20 = 1 − 0.089960 = 0.99100
Problem
Consider a Weibull model with k = 2 and n = 1. Calculate s(15) and µ(15).
Problem
Consider a Weibull model with k = 3. It is given that µ(2) = 12. Calculate s(4).
Problem
Consider a Weibull model with n = 1. It is given that s(20) = 0.14. Determine the value of
k.
Problem
A survival model follows a Weibull Law with mortality function µ(x) = kxn . It is given that
µ(40) = 0.0025 and µ(60) = 0.02. Determine the parameters k and n
1.4
Time-Until-Death Random Variable
Let (x) indicate that a newborn is known to be alive at age x or simply a life aged x. We let T (x) be
the continuous random variable that represents the additional time (x) might survive beyond the
age of x.
We define T (x) as the time-until-death or the future-lifetime random variable.
From this definition, we have a relationship between the age-at-death random variable X and
the time-until-death T (x) given by
X = x + T (x)
1.4.1
The Survival Function of T (x)
For t ≥ 0, the actuarial notation of the survival function of T (x) is t px . Thus, t px is the probability
of a life aged x to attain age x + t. That is, t px is a conditional probability. In symbol,
t px
= sT (x) (t) = Pr(X > x + t | X > x) = Pr[T (x) > t]
In the particular case of a life aged 0 , we have T (0) = X and x p0 = s(x), x ≥ 0 .
Using the fact that the event {X > x + t} is a subset of the event {X > x} and the conditional
1.4 Time-Until-Death Random Variable
probability formula Pr(E | F) =
t px
Pr(E∩F)
Pr(F)
29
we find
Pr[(X > x + t) ∩ (X > x)]
Pr(X > x)
Pr(X > x + t) s(x + t)
=
=
Pr(X > x)
s(x)
= Pr(X > x + t | X > x) =
Example
Find t px in the case X is exponentially distributed with parameter µ.
Solution.
We have
t px
=
s(x + t) e−µ(x+t)
= −µx = e−µt
s(x)
e
Example
Suppose that t px =
75−x−t
75−x , 0
≤ t ≤ 75 − x. Find the probability that a 35 -year-old reaches age 70 .
Solution.
We have
70 − 35p35 =
75 − 35 − 35
5
=
= 0.143
75 − 35
35
Problem
x
You are given the survival function s(x) = 1 − 75
, 0 ≤ x ≤ 75. Find the survival function and
the probability density function of T (x).
Problem
x
The survival function is given by s(x) = 1 − 100
, x ≥ 0.
(a) Express the probability that a person aged 35 will die between the ages of 52 and 73 using the p
notation.
(b) Calculate the probability in (a).
Problem
The PDF of X is given by f (x) =
1
,x
(x+1)2
≥ 0. Find t p3 .
Chapter 1. Actuarial Survival Models
30
1.4.2
The Cumulative Distribution Function of T (x)
From the survival function we can find the cumulative distribution function of T (x) which we
denote by t qx = FT (x) (t). Thus,
t qx
= Pr[T (x) ≤ t] = 1 − t px .
Note that t qx is the probability of (x) does not survive beyond age x +t or the conditional probability
that death occurs not later than age x + t, given survival to age x. It also follows that
t qx
= 1−
s(x + t)
s(x)
Example
Find 3 p5 and 4 q7 if s(x) = e−0.12x .
Solution.
We have
3 p5 =
s(3 + 5) e−0.12(8)
= −0.12(5) = 0.69768.
s(5)
e
Likewise,
4 q7 = 1 −
e−0.12(11)
s(4 + 7)
= 1 − −0.12(7) = 0.38122
s(7)
e
Problem
100
Given the survival function s(x) = 100+x
age 25 to survive an additional year?
1.1
. What is the probability of a newborn who lived to
Problem
The survival function is given by
s(x) = (1 − 0.01x)0.5 ,
0≤x≤1
What is the probability that a life aged 40 to die between the ages of 60 and 80 ?
Problem
You are given the hazard rate function
µ(x) =
Find a formula for t q20 .
1.1
, x ≥ 0.
100 + x
1.4 Time-Until-Death Random Variable
1.4.3
31
Probability Density Function of T (x)
Now, from the CDF of T (x) we can find an expression for the probability density function of T (x).
Indeed, we have
d
d s(x + t)
f (x + t)
fT (x) (t) = [t qx ] = −
=
.
dt
dt
s(x)
s(x)
In words, fT (x) (t) is the conditional density of death at time t given survival to age x or the conditional density at age x + t, given survival to age x.
Example
2
x
for 0 ≤ x < 100.
An age-at-death random variable has the survival function s(x) = 1 − 10000
(a) Find the survival function of T (x).
(b) Find the PDF of T (x).
Solution.
(a) The survival function of T (x) is
2
(x+t)
s(x + t) 1 − 10000
10000 − (x + t)2
=
, 0 ≤ t ≤ 100 − x.
=
t px =
2
x
s(x)
10000 − x2
1 − 10000
(b) The PDF of T (x) is
d
fT (x) (t) = −
dt
s(x + t)
s(x)
=
2(x + t)
, 0 ≤ t ≤ 100 − x
10000 − x2
Problem
x
Consider the survival function s(x) = 1 − 90
, 0 ≤ x ≤ 90. Find the survival function and the
probability density function of T (x).
Problem
t
Suppose that t px = 1 − 90−x
, 0 ≤ t ≤ 90 − x. Find the density function of T (x).
Problem
1
An age-at-death random variable has the CDF F(x) = 1 − 0.10(100 − x) 2 for 0 ≤ x ≤ 100. Find
fT (36) (t).
Problem
Let age-at-death random variable X have density function:
f (x) =
Find fT (2) (t).
x
, 0 ≤ x ≤ 10
50
Chapter 1. Actuarial Survival Models
32
1.4.4
Force of Mortality of T (x)
Recall that the hazard rate function µ(x) is the death rate at age x given survival to age x. That is,
µ(x) = lim
h→0
Pr(x < X ≤ x + h | X > x)
h
Likewise, we have
Pr(t < T (x) ≤ t + h | T (x) > t)
h
Pr(x + t < X ≤ x + t + h | X > x + t)
= lim
h→0
h
= µ(x + t)
µT (x) (t) = lim
h→0
That is, µT (x) (t) is the death rate at age x + t given survival to age x + t. The PDF of T (x) can be
expressed in terms of the distribution of X as shown in the next example.
Example
Show that fT (x) (t) = t px µ(x + t), x ≥ 0,t ≥ 0.
Solution.
We have
d
d
FT (x) (t) = t qx
dt dt s(x + t)
d
1−
=
dt
s(x)
0
s(x + t)
s (x + t)
=
−
= t px µ(x + t)
s(x)
s(x + t)
fT (x) (t) =
Example
For a group of lives aged 30 , containing an equal number of smokers and non-smokers, you
are given:
(i) For non-smokers, µ N (x) = 0.08, x ≥ 30
(ii) For smokers, µ S (x) = 0.16, x ≥ 30
Calculate q80 for a life randomly selected from those surviving to age 80 .
Solution.
We have
ST (30) (t) = Pr(T (30) > t | S) Pr(S) + Pr(T (30) > t | N) Pr(N) = 0.5e−0.16t + 0.5e−0.08t .
Hence,
q80 = 1 − P80 = 1 −
ST (30) (51)
ST (30) (50)
0.5e−0.16(51) + 0.5e−0.08(51)
0.5e−0.16(50) + 0.5e−0.08(50)
= 1 − 0.922 = 0.078
= 1−
1.4 Time-Until-Death Random Variable
33
Problem
The CDF of T (x) is given by
(
FT (x) (t) =
t
100−x ,
0 ≤ t < 100 − x
t ≥ 100 − x
1,
Calculate µ(x + t) for 0 ≤ t < 100 − x.
Problem
Find the hazard rate function of T (x) if X is exponentially distributed with parameter µ.
Problem
You are given the following information: fT (x) (t) = 0.015e−0.015t and t px = e−0.015t . Find µ(x + t)
1.4.5
Mean and Variance of T (x)
Like any random variable, the expected value is defined (using actuarial notation) by
Z ∞
◦
ex = E[T (x)] =
t fT (x) (t)dt
0
◦
◦
ex is called the complete expectation of life at age x. We next derive an expression for ex in terms
of the survival function of X. Indeed, we have
◦
Z ∞
ex =
0
t fT (x) (t)dt =
Z ∞
f (x + t)
t
s(x)
0
dt
∞
1
t f (x + t)dt
=
s(x) 0
Z ∞
1
∞
=
− ts(x + t)|0 +
s(x + t)dt
s(x)
0
Z ∞
Z ∞
s(x + t)
=
dt =
t px dt
s(x)
0
0
Z
Next, the variance of T (x) is given by
Var(T (x)) = E
where
E[T (x)] =
Z
E (T (x))2 =
0
∞
T (x)2 − E(T (x)]2
Z ∞
s(x + t)
0
and
s(x)
Z ∞
dt =
t 2 fT (x) (t)dt =
0
t px dt
Z ∞
f (x + t)
0
s(x)
dt
∞
1
=
t 2 f (x + t)dt
s(x) 0
Z ∞
1
∞
2
=
2ts(x + t)dt
− t s(x + t) 0 +
s(x)
0
Z ∞
Z ∞
2ts(x + t)
=
dt = 2
t · t px dt
s(x)
0
0
Z
Chapter 1. Actuarial Survival Models
34
Example
◦
The age-at-death random variable is uniformly distributed in [0, 90]. Find e30 .
Solution.
Since X is uniform on [0, 90], T (30) is uniform on [0, 60] and
0 + 60
= 30
2
◦
e30 = E(T (30)) =
Example
Let the age-at-death X be exponential with density function f (x) = 0.05e−0.05x , x ≥ 0 . Calculate the variance of T (x).
Solution.
The CDF is given by
F(x) = 1 − e−0.05x
and the SDF
s(x) = e−0.05x .
Thus,
Z ∞
E(T (x)) =
e−0.05t dt = 20
0
Z
E (T (x))2 =
∞
2te−0.05t dt = 80
0
Hence,
Var(T (x)) = 80 − 202 = 40
Problem
The SDF of an age-at-death random variable X is given by
1
s(x) = 0.1(100 − x) 2 , 0 ≤ x ≤ 100.
Find the expected value of T (25).
Problem
The CDF of an age-at-death random variable X is given by
1
F(x) = 1 − 0.1(100 − x) 2 , 0 ≤ x ≤ 100.
Find the variance of T (25).
1.4 Time-Until-Death Random Variable
35
Curtate-future-lifetime
The curtate-future-lifetime of a life aged x, denoted by K(x), is the random variable representing
the number of full years lived after age x. That is, K(x) is the integer part of T (x).
∞
ex = E[K(x)] =
∑ k Pr(K(x) = k)
k=0
= (1 px − 2 px ) + 2 (2 px − 3 px ) + 3 (3 px − 4 px ) + · · ·
∞
= 1 px + 2 px + 3 px + · · · =
∑ kpx
k=1
Example
◦
Suppose that s(x) = e−µx , x ≥ 0. Find ex and ex .
Solution.
We have
t px
=
s(x + t)
= e−µt
s(x)
Z ∞
◦
ex =
ex =
0
∞
Z ∞
t px dt =
eµt dt =
0
1
µ
∞
∑ kpx = ∑ e−µk
k=1
k=1
e−µ
=
1 − e−µ
Example
You are given px =
99−x
100 , x
= 90, 91, · · · , 99. Calculate Var[K(96)].
Solution.
First note that q99 = 1 so that the limiting age is 99 . We have
99−96
E[K(96)] = e96 =
∑
kp96 = p96 + 2 p96 + 3 p96
k=1
= p96 + p96 p97 + p96 p97 p98
= 0.3 + 0.3 × 0.2 + 0.3 × 0.20 × 0.10 = 0.366
99−96
=
∑
(2k − 1)k p96 = 1 (1 p96 ) + 3 (2 p96 ) + 5 (3 p96 )
k=1
E K(96)2 = 0.3 + 3(0.3 × 0.2) + 5(0.3 × 0.20 × 0.10) = 0.51
Var[K(96)] = E K(96)2 − (E[K(96)])2
= 0.51 − 0.3662 = 0.376044
Chapter 1. Actuarial Survival Models
36
Problem
A survival model has the survival function s(x) = 0.1(100 − x)0.5 , 0 ≤ x ≤ 100 . Find the probability
mass function of K(x).
Problem
◦
x
, 0 ≤ x ≤ 100. Find ex and ex .
A survival model has the survival function s(x) = 1 − 100
Problem
Consider a mortality model with the property pk =
1.4.6
1
2
for all k = 1, 2, · · · . Find ex and Var(K(x)).
Central Death Rates
The central rate of death or the average hazard on the interval [x, x + 1] is the continuous weighted
average of the force of mortality µ(y) with weight function
s(y)
w(y) = R x+1
.
s(t)dt
x
We denote the weighted average by
R x+1
mx =
x
s(y)µ(y)dy
R x+1
x
s(y)dy
.
Equivalently we can express mx in terms of f (x) and s(x) by using the fact that f (x) = µ(x)s(x) :
R x+1
mx = Rxx+1
x
f (y)dy
s(y)dy
Example
Find mx if X is exponentially distributed with parameter µ.
Solution.
s(x) = e−µx and µ(x) = µ.
Thus,
R x+1
µe−µt dt
=µ
mx = Rx x+1
e−µt dt
x
Example
Assume that the force of mortality follows the DeMoivre’s Law, where µ(x) =
Calculate m20 .
1
80−x
for 0 ≤ x < 80.
1.5 The Life Table Format
37
Solution.
We have
R 21
m20 =
20
s(y)µ(y)dy
R 21
20
=
s(y)dy
1
80y − 0.5y2 |21
20
R 21 dy
= R 21
20
20 80
y dy
1 − 80
= 0.01681
Example
You are given that µ = 0.02. Calculate 10 m75 .
Solution.
We have
10 m75
= µ = 0.02
Example
You are given x q0 =
x2
10,000 , 0
< x < 100. Calculate n mx .
Solution.
We have
s(x) = 1 −
x2
2x
and µ(x) =
10, 000
10, 000 − x2
Thus,
n mx
R n x+t
0 5,000 dt
= R n (10,000−(x+t)2 )
0
10,000
dt
Rn
= Rn
0
=
2(x + t)dt
[10, 000 − (x + t)2 ] dt
0
(x + n)2 − x2
3
10, 000n − 13 (x + n)3 + x3
2x + n
=
1
10, 000 − 3 (3x2 + 3nx + n2 )
1.5
The Life Table Format
Non-Select Life Table
The calculation of the relevant probabilities can be simplified by use of life tables.
The life table is a device for calculating probabilities such as t px and t qx using a one-dimensional
array. The key to the definition of a life table is the relationship:
t+s px
= t px × s px+t = s px × t px+s
Chapter 1. Actuarial Survival Models
38
This result is called the principle of consistency.
1.5.1
Constructing a life table
To construct a life table, we choose a starting age, which will be the lowest age in the table. We
denote this lowest age α.
We also need to choose the highest age in the table, ω, which is the age beyond which survival is assumed to be impossible. ω is referred to as the limiting age of the table.
We also start with a group of newborns known in the actuarial terminology as the cohort. The
original number of individuals `0 in the cohort is called the radix.
Define
`x as the number of survivors at exact age.
n dx is the expected number of deaths between the ages of x and x + n
n dx
= `x − `x+n
We introduce the important notation:
t px
t qx
= Pr{T (x) > t} = Probability that x survives to age x + t
= Pr{T (x) ≤ t} = Probability that x dies between ages x and x+t
When t = 1, it may be omitted, so that
px = Pr{(x) survives for at least a year }
qx = Pr{(x) dies within a year}
It is clear that, for all t ≥ 0
t px +t
qx = 1
and,
s(x + t) lx+t
=
s(x)
lx
t px
=
t qx
= 1−
and
lx+t
lx
That is, the (cumulative) distribution function of the variable T is
(
1 − lx+t
, if t ≥ 0
lx
F(t) = Pr{T ≤ t} =t qx =
0
, if t < 0
Deferred probabilities
The symbol m | indicates deferment for m years; for example,
1.5 The Life Table Format
m |n qx
39
= P{(x) will die between ages x + m and x + m + n}
By elementary probability, this equals Pr{(x) will die before age x+m+n}−Pr{(x) will die before age x+
m}
=m+n qx −m qx
=m px −m+n px
= lx+m −llxx+m+n
That is,
m|n qx
=
lx+m − lx+m+n
lx
This may be remembered by the following rule: of lx lives aged x, lx+m − lx+m+n is the (expected)
number of deaths aged between ages x + m and x + m + n.
If n = 1 it may be omitted, so we have
m|qx = Pr{(x) will die between the ages x + m and x + m + 1}
dx+m
lx
=
We remark that there is no such thing as
m|n px
We may also use the result that
m |n qx
= Pr{(x) survives to age x + m} · Pr{(x + m) dies within n years }
=m px·n qx+m
Example
Consider the life table
Age
0
1
2
3
4
5
`x
100,000
99,499
98,995
98,489
97,980
97,468
dx
501
504
506
509
512
514
(a) Find the number of individuals who die between ages 2 and 5.
(b) Find the probability of a life aged 2 to survive to age 4 .
Solution.
Chapter 1. Actuarial Survival Models
40
(a) 2 d3 = `2 − `5 = 98, 995 − 97, 468 = 1527.
(b) 2 p2 =
`4
`2
=
97,980
98,995
= 0.98975
Example
You are given:
`x = 10000(100 − x)2 , 0 ≤ x ≤ 100
Calculate the probability that a person now aged 20 will reach retirement age of 65 .
Solution.
We have
45 p20
=
`65 10000(100 − 65)2
=
= 0.1914
`20 10000(100 − 20)2
Example
Using the extract from ELT15 (Males) given above, calculate the values of:
(i) p2
(ii) 2 p3
(iii) 4 q1
(iv) l6
Solution.
(i) p2 = ll32 = 99,086
99,124 = 0.99962 or
p2 = 1 − q2 = 1 − 0.00038 = 0.99962
(ii)
2 p3
=
I5
I3
=
99,032
99,086
= 0.99946
(iii) 4 q1 = 1 − 4
p1 = 1 − ll51 = 1 − 99,032
99,186 = 0.00155
(iv)
I6 = I5 − d5 = 99, 032 − 22 = 99, 010
Example
x
Given `x = 1000 1 − 105
, determine each of the following:
(a) `0
(b) `35
(c) q20
(d)15 p35
(e)15 q25
(f) The probability that a 30 -year-old dies between ages 55 and 60 .
1.5 The Life Table Format
41
(g) The probability that a 30 -year-old dies after age 70 .
Solution.
0
(a) `0 = 1000 1 − 105
= 1000.
35
(b) `35 = 1000 1 − 105 = 667 (note the answer must be an integer).
21
(c) q20 = `20`−`
= 1 − .98824 = .01176.
20
(d) 15 p35 =
`50
`35
= .78571.
40
(e) 15 q25 = 1 − 15 p25 = 1 − ``25
= .1875.
60
(f) `55`−`
= .0667.
30
(g) This is equal to the probability that a 30 -year-old reaches age 70 , which is 40 p30 =
`70
`30
= .4667.
Example
Suppose that in a particular life table:
Ix = 100 − x
for 0 ≤ x ≤ 100
Calculate the complete expected future lifetime of a life currently aged 50.
Solution.
Using the given expression for Ix :
t p50
=
50 − t
I50+t
=
I50
50
for 0 ≤ t ≤ 50
We only consider values of t up to 50 , as there is a zero probability of surviving to age 100 in this
life table. So the complete expected future lifetime of a life currently aged 50 is:
◦
e50 −
Z 50
0
p50 dt −
1
1 2 50 1
1 2
2
dt −
50t − t
−
50 − 50 − 25
50
50
2 0
50
2
Z 50
50 − t
0
Example
Using ELT15 (Males) mortality, calculate the probability of a 37-year old dying between age
65 and age 75.
Solution.
The required probability is:
28 | 10q37 =
I65 − I75
I37
Looking up the values in the Tables gives:
28 | 10q37 =
79, 293 − 53, 266
= 0.26851
96, 933
Chapter 1. Actuarial Survival Models
42
Example
Use the extract from life table to calculate:
(a) l40
(b)the probability that a life currently aged exactly 30 dies between ages 35 and 36
(c) 10 p30 and
(d) 5 q30
x
30
31
32
33
34
35
36
37
38
39
lx
10000.00
9965.22
9927.12
9885.35
9839.55
9789.29
9734.12
9673.56
9607.07
9534.08
dx
34.78
38.10
41.76
45.81
50.26
55.17
60.56
66.49
72.99
80.11
Solution.
(a)
l40 = l39 − d39 = 9453.97
(b) This probability is 5| q30 so.
5| q30
=
l35 − l36 d33
=
= 0.00552
l30
l30
10 p30
=
L40 9453.97
=
= 0.94540
I30
10000
(c)
(d)
5 q30
=
l30 − l33
= 0.02107
l30
Problem
1. Write the symbol for the probability that (52) lives to at least age 77
2. Write the symbol for the probability that a person age 74 dies before age 91
1.5 The Life Table Format
43
3. Write the symbol for probability that (33) dies before age 34.
4. Write the symbol for probability that a person age 43 lives to age 50, but doesn’t survive to age
67
5. Write 5|6 qx
Problem
x
Suppose that s(x) = 1 − 10
, 0 ≤ x ≤ 10.
(a) Find `x .
(b) Using life table terminology, find p2 , q3 , 3 p7 , and 2 q7 .
Problem
Complete the entries in the following table:
x
0
1
2
3
4
5
1.5.2
`x
100,000
99,499
98,995
98,489
97,980
97,468
dx
px
qx
Life table functions at non-integer ages
Life table functions such as Ix , qx or µx are usually tabulated at integer ages only, but sometimes we
need to compute probabilities involving non-integer ages or durations, such as 2.5p37.5 . We can do
so using approximate methods. We will show two methods.
Uniform distribution of deaths (UDD)
The first method is based on the assumption that, for integer x and 0 ≤ t ≤ 1, the function t px µx+t
is a constant. Since this is the density (PDF) of the time to death from age x, it is seen that this
assumption is equivalent to a uniform distribution of the time to death, conditional on death falling
between these two ages. Hence it is called the Uniform Distribution of Deaths (or UDD) assumption.
Let x be fixed. We may say there is a uniform distribution of deaths (U.D.D.) between ages
x and x + 1 if, for 0 ≤ t ≤ 1,
lx+t = (1 − t)lx + tlx+1
(i.e. ly is linear for x ≤ y ≤ x + 1 )
This equation may be written in the form
lx+t = lx − tdx
(0 ≤ t ≤ 1)
Theorem
The following conditions are each equivalent to the assumption of U.D.D. between ages x and x + 1
:
t qx = t · qx (0 ≤ t ≤ 1)
t px µx+t
= qx
(0 ≤ t < 1)
Chapter 1. Actuarial Survival Models
44
Also, using t px = s px × t−s px+s , it can be shown that for integer age x and 0 ≤ s < t ≤ 1 :
t−s qx+s
=
(t − s)qx
1 − sqx
Example
In a certain non-select mortality table, there is a uniform distribution of deaths between any
two consecutive integer ages. Find formulae in terms of l30 , l31 and l32 for
(i) 1.5p30.5
(ii) µ30.5
Solution.
(i)
l32
l30.5
=
l32
1
2 (l30 +l31 )
since l30+t is linear for 0 < t < 1
(ii) t p30 µ30+t = q30 for 0 ≤ t ≤ 1, so µ30.5 =
q30
1
2 p30
=
l30 −l31
l30 12
=
l30 −l31
1
2 (l30 +l31 )
Example
Calculate the value of 0.5p55.5 using ELT15 (Females) mortality, assuming a uniform distribution
of deaths between integer ages.
Solution.
Using the UDD assumption, we have:
0.5p55.5 = 1 − 0.5 q55.5 = 1 −
0.5q55
0.5 × 0.00475
= 1−
= 0.99762
1 − 0.5q55
1 − 0.5 × 0.00475
Example
Calculate the value of 0.5 p55.5 using ELT15 (Females) mortality and linear interpolation, based on
the assumption of a uniform distribution of deaths between integer ages.
Solution.
Using the UDD assumption, we have:
0.5 p55.5
=
I56
I56
94, 082
94, 082
=
=
=
= 0.99761
I55.5 0.5I55 + 0.5I56 0.5 × 94, 532 + 0.5 × 94, 082 94, 307
This answer is slightly different from the answer obtained previously for this probability, due to the
fact that the values shown in the Tables are rounded.
Example
Given that q80 = 0.02, calculate 0.6 p80.3 under the assumption of a uniform distribution of deaths.
1.5 The Life Table Format
45
Solution.
We have
`80.9
`80.3
0.9p80
=
0.3p80
1 − 0.9q80
=
1 − 0.3q80
1 − 0.9(0.02)
=
= 0.9879
1 − 0.3(0.02)
0.6p80.3 =
Example
Consider the following life table.
Age
0
1
2
3
4
5
`x
100,000
99,499
98,995
98,489
97,980
97,468
dx
501
504
506
509
512
514
Under the uniform distribution of deaths assumption, calculate
(a) 1.4 q3 and
(b) 1.4 q3.5 .
Solution.
(a) We have
1.4q3 = 1 − 1.4 p3 = 1 − (p3 ) (0.4p4 )
`4
= 1 − · (1 − 0.4q4 )
`3
`4
d4
= 1 − · 1 − 0.4
`3
`4
97980
512
= 1−
· 1 − 0.4 ×
= 0.0072
98489
97980
(b) We have
1.4q3.5 = 1 − 1.4 p3.5
`4.9
= 1−
`3.5
`4 − 0.9d4
= 1−
`3 − 0.5d3
= 0.007281
Constant Force of Mortality
Chapter 1. Actuarial Survival Models
46
The second fractional age assumption says that the force of mortality is constant between integer ages. That is, for integer x ≥ 0 we have µ(x + t) = µx for all 0 ≤ t < 1. The constant µx can
be expressed in terms of px as follows:
px = e−
R1
µ(x+t)dt
0
= e−µx =⇒ µx = − ln px
Further, under the constant force of mortality assumption we can write
t px
= sT (x) (t) = e−
Rt
0
µ(x+s)ds
t
= e−µxt = eln px = ptx , 0 ≤ t < 1.
= FT (x) (t) = 1 − t px = 1 − ptx = 1 − (1 − qx )t
d
fT (x) (t) = FT (x) (t) = − ln px ptx = µx ptx = t px µ(x + t).
dt
t qx
Example
Consider the following extract from a life table.
Age
0
1
2
3
4
5
`x
100,000
99,499
98,995
98,489
97,980
97,468
dx
501
504
506
509
512
514
Under the constant force of mortality assumption, find (a) 00.75 p2 and (b) 1.25 p2 .
Solution.
(a) We have
0.75 p2
=
p0.75
2
=
`2
`1
0.75
=
98, 995
99, 499
0.75
= 0.9962
(b) We have
`4 0.25
1.25 p2 = p2 · 0.25 p3 =
`3
0.25
98, 489
97, 980
=
= 0.9936
98, 995
98, 489
`3
`2
Example
You are given
(i) qx = 0.02
(ii) The force of mortality is constant between integer ages. Calculate 0.5 qx+0.25 .
Solution.
1.5 The Life Table Format
47
In general, for s,t > 0 and s + t < 1 we have
−
s px+t = e
Rs
0
µx dy
= psx .
Thus,
0.5 qx+0.25
0.5
= 1 − 0.5 px+0.25 = 1 − p0.5
= 0.01
x = 1 − (1 − 0.02)
Example
The life table is given as follows
x
26
27
28
qx
0.0213
0.0232
0.0254
Find the probability that (26.5) will survive to age 28.25 under
(i)) the UDD assumption
(ii) the constant-force assumption
Solution.
We have
28.25−26.5 p26.5
= 1.75 p26.5 = 0.5 p26.5 p270.25 p28 .
(i) Under the UDD assumption, we have
0.5q26
0.5(0.0213)
= 1−
= 0.989235
1 − 0.5q26
1 − 0.5(0.0213)
p27 = 1 − q27 = 1 − 0.0232 = 0.9768
0.5 p26.5
= 1 − 0.5 q26.5 = 1 −
0.25 p28
= 1 − 0.25 q28 = 1 − 0.25q28 = 1 − 0.25(0.0254) = 0.99365
1.75 p26.5
= (0.989235)(0.9768)(0.99365) = 0.960149.
(ii) Under the constant force of mortality, we have
0.5 p26.5
0.5
0.5
= 1 − 0.5 q26.5 = 1 − 1 − p0.5
= 0.989293
26 = p26 = (1 − q26 )
p27 = 1 − q27 = 1 − 0.0232 = 0.9768
0.25 p28
1.75 p26.5
= (1 − q28 )0.25 = (1 − 0.0254)0.25 = 0.993589
= (0.989293)(0.9768)(0.993589) = 0.960145.
Example
Calculate 3 p62.5 based on PFA92C20 mortality using:
(i) the CFM assumption
(ii) the UDD assumption.
Chapter 1. Actuarial Survival Models
48
Solution.
First of all, we can split up the probability at integer ages as follows:
3 p62.5
= 0.5 p62.5 × 2 p63 × 0.5p65
Looking up values from PFA92C20 in the Tables:
2 p63
=
l65 9, 703.708
= 0.992617
=
l63 9, 775.888
(i) Under the constant force of mortality assumption:
0.5 p62.5
= (p62 )0.5 = (1 − q62 )0.5 = (1 − 0.002885)0.5 = 0.998556
Also:
0.5 p65
= (p65 )0.5 = (1 − q65 )0.5 = (1 − 0.004681)0.5 = 0.997657
So, overall:
3 p62.5
= 0.998556 × 0.992617 × 0.997657 = 0.988861
(ii) Under the UDD assumption:
0.5 p62.5
and:
0.5 p65
= 1 − 0.5 q62.5 = 1 −
0.5q62
0.5 × 0.002885
= 1−
= 0.998555
1 − 0.5q62
1 − 0.5 × 0.002885
= 1 − 0.5 q65 = 1 − 0.5q65 = 1 − 0.5 × 0.004681 = 0.997660 So overall:
3 p62.5
= 0.998555 × 0.992617 × 0.997660 = 0.988863
Alternatively, assuming UDD, we could take the following approach:
3 p62.5
l65.5 0.5l65 + 0.5l66
=
l62.5 0.5l62 + 0.5l63
0.5(9, 703.708 + 9, 658.285) 9, 680.9965
=
=
= 0.988863
0.5(9, 804.173 + 9, 775.888) 9, 790.0305
=
Example
In a certain population, the force of mortality equals 0.025 at all ages. (We are assuming here that
there is no upper limit to age.)
Calculate:
(i) the probability that a new-born baby will survive to age 5
(ii) the probability that a life aged exactly 10 will die before age 12
(iii) the probability that a life aged exactly 5 will die between ages 10 and 12
Solution.
(i)
Z5
p
=
exp
−
0.025dt
= e−0.125 = 0.88250
5 0
0
1.5 The Life Table Format
49
(ii)
Z2
0.025dt = 1 − e−0.05 = 0.04877
2 q10 = 1 − 2 p10 = 1 − exp −
0
(iii)
5|2 q5
= 5 p5 × 2 q10 = e−0.125 × 0.04877 = 0.88250 × 0.04877 = 0.04304
Problem
You are given:
(i) p90 = 0.75
(ii) a constant force of mortality between integer ages.
Calculate 1 q90 and 1 q90+ 11 .
12
12
12
Problem
You are given:
(i) qx = 0.1
(ii) a constant force of mortality between integer ages.
Calculate 0.5 qx and 0.5 qx+0.5 .
Select mortality Table
We now consider the situation when mortality rates (or the force of mortality ) depend on two
factors: (i) age, and (ii) the time (duration) since a certain event, known as “selection”.
Select rates are usually studied by modelling the force of mortality µ as a function of the age at
joining the population and the duration since joining the population. The usual notation is:
[x] + r age at date of transition
[x]
age at date of joining population
r
duration from date of joining the population until date of transition
µ{x]+r the transition rate (force of mortality) at exact duration r having joined the population at
age [x]
l[x]+r expected number of lives alive at duration r having joined the population at age [x], based
on some assumed radix