Stat 475
Life Contingencies I
Chapter 2: Survival models
The future lifetime random variable — Notation
We are interested in analyzing and describing the future
lifetime of an individual.
We use (x) to denote a life age x for x ≥ 0.
Then the random variable describing the future lifetime, or
time until death, for (x) is denoted by Tx for Tx ≥ 0.
We also have notation for the density, distribution and
survival functions of Tx :
Density: fx (t) =
d
dt Fx (t)
Distribution: Fx (t) = Pr [Tx ≤ t]
Survival: Sx (t) = Pr [Tx > t] = 1 − Fx (t)
2
The future lifetime random variable — Relationships
We now consider the set of random variables {Tx }x≥0 and the
relationships between them:
Important Assumption
Pr [Tx ≤ t] = Pr [T0 ≤ x + t|T0 > x]
Important Consequence
S0 (x + t) = S0 (x) · Sx (t)
This idea can be generalized:
Important General Relationship
Sx (t + u) = Sx (t) · Sx+t (u)
3
Survival function conditions and assumptions
We will require any survival function corresponding to a future
lifetime random variable to satisfy the following conditions:
Necessary and Sufficient Conditions
1
2
Sx (0) = 1
lim Sx (t) = 0
t→∞
3
Sx (t) must be a non-increasing function of t
In addition, we will make the following assumptions for the survival
functions used in this course:
Assumptions
1
2
Sx (t) is differentiable for all t > 0.
lim t · Sx (t) = 0
t→∞
3
lim t 2 · Sx (t) = 0
t→∞
4
Future lifetime random variable example
Assume that the survival function for a newborn is given by
S0 (t) = 8(t + 2)−3
1
Verify that this is a valid survival function.
2
Find the density function associated with the future lifetime
random variable T0 , that is, find f0 (t).
3
Find the probability that a newborn dies between the ages of
1 and 2.
4
Find the survival function corresponding to the random
variable T10 .
5
Force of Mortality
We will define the force of mortality as
µx = lim +
dx→0
Pr [T0 ≤ x + dx|T0 > x]
dx
Then for small values of dx, we can use the approximation
µx dx ≈ Pr [T0 ≤ x + dx|T0 > x]
We can relate the force of mortality to the survival function:
− d S0 (x)
µx = dx
S0 (x)
Z t
Sx (t) = exp −
µx+s ds
and
0
6
Some commonly used forms for the force of mortality
It is sometimes convenient to describe the future lifetime random
variable by explicitly modeling the force of mortality. Some of the
parametric forms that have been used are:
Gompertz: µx = Bc x , 0 < B < 1, c > 0
Makeham: µx = A + Bc x , 0 < B < 1, c > 0
de Moivre1 : µx =
1
, 0≤x <ω
ω−x
Exponential1 : µx = k, 0 ≤ x ≤ ∞
If there is a limiting age, it is typically denoted by ω.
1
When these models are used to model human mortality, it is primarily for
mathematical convenience rather than realism.
7
Force of mortality example
Again, assume that the survival function for a newborn is given by
S0 (x) = 8(x + 2)−3
1
Find the force of mortality, µx .
2
Consider the shape of µx as a function of x.
Does it seem to be a realistic survival model for describing
human lifetimes?
What would you expect the force of mortality corresponding to
humans to look like?
8
Actuarial Notation
Thus far we have described the properties of the future lifetime
random variable using statistical notation. Actuaries have also
developed their own (different) notation to describe many of the
same concepts:
t px
= Pr [Tx > t] = Sx (t)
t qx
= Pr [Tx ≤ t] = Fx (t)
u |t qx
= Pr [u < Tx ≤ u + t] = Fx (u + t) − Fx (u)
In any of the above expressions, the subscript t may be omitted
when its value is 1.
The density of the random variable Tx can be written as
fX (t) = t px µx+t .
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Some Basic Relationships
We can express some relationships in this actuarial notation:
t px
= 1 − t qx
u |t qx
= u px − u+t px
u |t qx
= u px t qx+u
t+u px
= t px u px+t
Z t
µx+s ds
t px = exp −
0
Z
t qx
=
t
s px
µx+s ds
0
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Mean of the Future Lifetime Random Variable
One quantity that is often of interest is the mean of the future
lifetime random variable, E [Tx ], which is called the complete
expectation of life.
Z ∞
◦
E [Tx ] = e x =
t fx (t) dt
0
Z ∞
t t px µx+t dt
=
0
Z ∞
=
t px dt
0
We can also find the term expectation of life.
Z n
Z ∞
◦
E [min(Tx , n)] = e x:n =
t t px µx+t dt +
nt px µx+t dt
n
Z0 n
=
t px dt
0
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Variance of the Future Lifetime Random Variable
In order to calculate the variance of Tx , we need its second
moment:
Z ∞
2
E Tx =
t 2 fx (t) dt
0
Z ∞
t 2 t px µx+t dt
=
0
Z ∞
=2
t t px dt
0
Then we can calculate the variance in the usual way:
V [Tx ] = E Tx2 − (E [Tx ])2
◦ 2
= E Tx2 − e x
12
Summary of notation for the future lifetime RV
We can summarize the statistical and actuarial notation for the
properties of Tx covered thus far:
Concept
Statistical Notation
Actuarial Notation
Expected Value
E [Tx ]
ex
Distribution Function
Fx (t)
t qx
Survival Function
Sx (t)
t px
Force of Mortality
λ(x) or λx
µx
Density
fx (t)
Deferred Mortality Prob.
Pr [u < Tx ≤ u + t]
13
◦
t px
µx+t
u |t qx
Curtate Future Lifetime Random Variable
We are often interested in the integral number of years lived in the
future by an individual. This discrete random variable is called the
curtate future lifetime and is denoted by Kx .
We can consider the pmf (probability mass function) of Kx :
Pr [Kx = k] = Pr [k ≤ Tx < k + 1]
= k |qx
= k px qx+k
We also have the relation Kx = bTx c.
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Moments of the Curtate Future Lifetime Random Variable
We can find the mean (called the curtate expectation of life and
denoted by ex ) and variance of Kx :
E [Kx ] = ex =
∞
X
k px
k=1
∞
X
E Kx2 = 2
k k px − ex
k=1
V [Kx ] = E Kx2 − (ex )2
Using the trapezoidal rule, we can also find an approximate
relation between the means of Tx and Kx :
1
◦
e x ≈ ex +
2
15
Exercise 2.2
S0 (x) =
18000 − 110x − x 2
18000
(a) What is the implied limiting age?
[90]
(b) Is it a valid survival function?
(c) Calculate
20 p0 .
h
(d) S20 (t)
(e)
10|10 q20
[0.8556]
i
2
15400−150t−t
15400
[0.1169]
(f) µ50
[0.021]
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Exercise 2.1/2.6
(
1− 1−
F0 (t) =
1
t 1/6
120
for 0 ≤ t ≤ 120
for t > 120
(a)
30 p0
[0.9532]
(b)
20 p30
[0.041]
(c)
25 p40
[0.9395]
h
i
1
720−6x
(d) Derive µx
◦
(e) Calculate e x and Var [Tx ] for (30)
◦
(f) Calculate e x and Var [Tx ] for (80)
17
77.143, 21.3962
34.286, 9.5092
Exercise 2.5
F0 (t) = 1 − e −λt
e −λt
(a) S0 (t)
(b) µx
λ
(c) ex
1
e λ −1
(d) Is this reasonable for human mortality?
18
Exercise 2.6
px = 0.99
px+1 = 0.985
3 px+1
(a) px+3
= 0.95
qx+3 = 0.02
0.98
(b) 2 px
0.97515
(c) 2 px+1
0.96939
(d) 3 px
0.95969
(e)
0.03031
1|2 qx
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SOA #32


1
S0 (t) = 1 −


0
et
100
Calculate µ4 . [1.20]
20
for 0 ≤ t < 1
for 1 ≤ t < 4.5
for 4.5 ≤ t
SOA #200
The graph of a piecewise linear survival function, S0 (t), consists of
3 line segments with endpoints (0, 1), (25, 0.50), (75, 0.40), (100,
0).
Calculate
20|55 q15
55 q35
. [0.6856]
21