Making sense of . . .
Traditional survival models:
aggregate, select, or select & ultimate
Excerpt #2b from Jim Daniel’s LTAM seminars
James W. Daniel
Jim Daniel’s Actuarial Seminars
www.actuarialseminars.com
August 7, 2020
c Copyright 2020 by James W. Daniel; reproduction in whole or in part without the
express permission of the author is forbidden.
Foreword
This document briefly describes aggregate, select, or select & ultimate survival models. Essentially
it’s part of one of the lessons I gave in my face-to-face LTAM exam-prep seminars.
NOTE: The numbered items—Examples, Equations, Definitions, Sections, et cetera—can be
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Chapter 1
Traditional survival models:
aggregate, select, or
select & ultimate
1.1
Aggregate models versus select models
As you’ll see in detail below, an aggregate model for future-lifetime random variables allows you to
deduce the probability distribution of Tx for any age x from knowledge of the probability distribution
of T0 , the future-lifetime random variable of a newborn. On the other hand, with a select model
the probability distribution of the future-lifetime random variable at one age cannot in general be
deduced from the probability distribution of the future-lifetime random variable at a different age.
A select & ultimate model is a special type of select model that is a sort of blend of a select and an
aggregate model.
In an aggregate model, a life aged x is usually denoted by (x) and its future-lifetime random
variable by Tx ; in a select model or select & ultimate model, a life aged x is usually denoted by [x]
and its future-lifetime random variable by T[x] .
1.1.1
Aggregate models
Suppose that the probability distribution of the future-lifetime random variable T0 of a newborn life
(0) in an aggregate model is known. In particular, suppose that its survival function S0 , for which
S0 (t) = Pr[T0 > t], is known. I want to deduce the probability distribution of Tx for any x > 0.
The idea in an aggregate model is that when you consider an (x), a life aged x, absolutely the
only thing you know is that (x) was once a typical ordinary newborn (0) who has survived to age
x—you have no additional information, such as that (x) has just been diagnosed with a serious
disease or that (x) just passed a rigorous physical with flying colors.
At first glance it might then seem as though saying that (x) survives at least t years is equivalent
to saying that (0) survives x + t years—so it might seem as though t px , the probability that (x)
survives at least t years, should equal S0 (x + t), the probability that a newborn survives at least
x + t years. But this thought ignores the fact that you know that the newborn has already fought
off Mother Nature and managed to survive the first x years. In fact, the number of years (x) lives is
the number of years the newborn lives minus the age x, conditioned on knowing that the newborn
has survived x years:
Tx = T0 − x T0 > x.
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From this, you can find t px :
t px
= Pr[Tx > t] = Pr[T0 − x > t T0 > x] = Pr[T0 > x + t T0 > x]
Pr[T0 > x + t and T0 > x]
Pr[T0 > x + t]
=
Pr[T0 > x]
Pr[T0 > x]
S0 (x + t)
=
.
S0 (x)
=
This shows that the probability distribution of Tx is determined by that of T0 in an aggregate model.
It’s also easy to find the force of mortality µx (t). Intuitively, this measures how hard Mother
Nature is trying to kill off the x-year-old after t years, which seems as though that should be the
same as how hard she is trying to kill off a newborn after x + t years—which is just µ0 (x + t), also
denoted by µ(x + t) or µx+t . This is in fact correct:
d S0 (x+t)
d
− dt
− dt
t px
S0 (x)
= S (x+t)
µx (t) =
0
p
t x
S0 (x)
S 0 (x + t)
=− 0
= µ0 (x + t) = µ(x + t) = µx+t .
S0 (x + t)
Since in general
t px
= e−
Rt
0
µx (z) dz
,
it follows that for an aggregate model
t px
= e−
Rt
0
µ(x+z) dz
.
In summary:
KEY ⇒
Fact 1.1 (Aggregate models) Suppose that µ0 (t) = µ(t) = µt is the force of mortality for a
newborn after t years, and that S0 is the survival function for a newborn. Then, in an aggregate
model,
Rt
S0 (x + t)
= e− 0 µ(x+z) dz
t px =
S0 (x)
and
µx (t) = µ0 (x + t) = µ(x + t) = µx+t .
KEY ⇒
Exams! ⇒
Example 1.2 A commonly tested aggregate survival model is the Exponential model or forever
constant force model. In this model, the force of mortality for a newborn is some positive constant
µ, which by Key Fact 1.1 makes t px = e−µt for all ages x. This means that Tx is an Exponential
random variable with mean µ1 , for all ages x. This is an unrealistic model for human lives, but an
easy model for computations on exams.
End of Example 1.2
1.1.2
Select models
In a select model, a life aged x is usually denoted by [x] to distinguish it from a life aged x in
an aggregate model, where the notation is (x). The intuitive idea is that you have some special
information about [x], such as that [x] was just diagnosed with some chronic illness that makes the
survival probabilities for [x] quite different from those of a normal x-year-old in an aggregate model.
Thus the formulas in Key Fact 1.1 do not hold. Note that the force of mortality µ[x] (t) is sometimes
denoted by µ[x]+t .
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Example 1.3 Suppose that for each age x ≥ 0 the force of mortality for a select life aged x is
constant and is given by
µ[x] (t) = 0.01(x + 1).
Then µ[9] (t) = 0.1 and µ[0] (9 + t) = 0.01 are quite different from one another, unlike for an aggregate
model in which µ9 (t) = µ0 (9 + t).
End of Example 1.3
1.2
Select & ultimate models
In the previous Subsection on select models, Subsection 1.1.2, I mentioned that with select models
you have some special information about the select life [x] that makes the survival probabilities for
[x] quite different from those for a normal x-year-old in an aggregate model. In some cases—such as
when what makes [x] special is that [x] just passed a demanding physical exam including a thorough
medical history—the impact of the special information may only last a few years; just because you
were in great shape at 25 and had lower mortality probabilities than the average 25-year-old for
the next few years doesn’t mean that that will still be true in 10 or 15 years. In such a case,
a select & ultimate model might be appropriate, with special one-year survival probabilities (the
select probabilities) for a few years, after which the one-year survival probabilities (the ultimate
probabilities) are the same as for a typical person of that age. This means, in particular, that the
force of mortality will be special for [x] for the first few years during the so-called select period. The
length of that select period—the number of years with special survival and death probabilities—is
usually denoted by d.
One way to define a select & ultimate model—an approach not commonly seen on Exam LTAM—
is to provide one formula for the force of mortality during the d-year select period and another after
that period. That is, µ[x] (t) would be given by a special formula for 0 ≤ t ≤ d but then would
equal µ(x + t) for t > d, where µ(z) is the force of mortality for a typical newborn. Then the future
lifetime of the life [x] + d would have the same distribution as that of an ordinary (x + d)-year-old.
Example 1.4 Suppose that the force of mortality for a normal life is forever constant at µ = 0.006,
so that µ(z) = 0.006 at all ages z. Suppose that a select & ultimate model has a two-year select
period, and that µ[x] (t) = 0.006(3 − t) for t ≤ 2 while µ[x] (t) = µ(x + t) = 0.006 for t > 2. Then
t p[x]
= e−
Rt
0
µ[x] (z) dz
= e−
Rt
0
0.006(3−z) dz
2
= e−0.018t+0.003t for 0 ≤ t ≤ 2,
while
t p[x]
= 2 p[x] t−2 p[x]+2 = 2 p[x] t−2 px+2 = e−0.024 e−0.006(t−2) for t > 2.
End of Example 1.4
On Exam LTAM, it’s more common for a problem to describe a select & ultimate model by using
either a mortality table or a life table.
1.2.1
Select & ultimate mortality tables
One of the common ways questions on Exam LTAM describe select & ultimate models is to provide
a mortality table—a table of one-year death probabilities for a select life [x] at each age x. With a
d-year select period, the death probabilities q[x]+j are special for the d years with 0 ≤ j ≤ d − 1 and
then revert to ordinary probabilities after d years, so that q[x]+j = qx+j for j ≥ d.
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Example 1.5 Here’s an example of a part of a select & ultimate mortality table with a two-year
select period.
x
90
91
92
q[x]
0.20
0.30
0.40
q[x]+1
0.25
0.35
0.45
qx+2
0.21
0.32
0.45
x+2
92
93
94
Notice that to find the sequence of mortality or survival probabilities that apply to a select life
[x], you move across the row corresponding to the age x and then down the column of ultimate (or
ordinary) probabilities. For instance,
4 p[90]
= (1 − 0.2) × (1 − 0.25) × (1 − 0.21) × (1 − 0.32) = 0.32232.
For more complicated probabilities such as 1|3 q[91] , it’s usually simpler to work with the life table
corresponding to a mortality table, as indicated in the following Subsection 1.2.2.
End of Example 1.5
1.2.2
Select & ultimate life tables
It’s often easier to compute probabilities intuitively using the life table that corresponds to the mortality table provided in an Exam LTAM question; if you’re unfamiliar with computing probabilities
intuitively using life tables, I suggest you take a look at my LTAM Seminar Excerpt Traditional
survival models: intuitive calculations using life tables before continuing here. I’ll illustrate constructing a corresponding select & ultimate life table by using the select & ultimate mortality table
above in Example 1.5.
As usual in a life table, I can choose the value of one ` arbitrarily; I’ll choose `[90] = 10000.
Using the survival probabilities starting with [90] deduced from the mortality table in Example 1.5
produces part of the corresponding life table:
x
90
91
92
93
`[x]
10000
—
—
—
`[x]+1
8000
—
—
—
`x+2
6000
4740
3223.2
1772.76
x+2
92
93
94
95
Now I can’t choose an arbitrary value for `[91] , because after two years it has to lead to `[91]+2 =
`91+2 = `93 that has already been computed to be `93 = 4740. To correspond to the mortality table
in Example 1.5, I first need to choose `[91]+1 so that
p[91]+1 = 1 − q[91]+1 = 1 − 0.35 =
`[91]+2
4740
=
,
`[91]+1
`[91]+1
which gives `[91]+1 = 7292.31 approximately. Then I need to choose `[91] so that
p[91] = 1 − q[91] = 1 − 0.30 =
`[91]+1
7292.31
=
,
`[91]
`[91]
which gives `[91] = 10417.58 approximately. Inserting these values into the life table gives
x
90
91
92
93
`[x]
10000
10417.58
—
—
`[x]+1
8000
7292.31
—
—
`x+2
6000
4740
3223.2
1772.76
x+2
92
93
94
95
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Similarly, I can’t choose an arbitrary value for `[92] , because after two years it has to lead to
`[92]+2 = `92+2 = `94 that has already been computed to be `94 = 3223.2. Instead, I have to make my
way backwards from right to left just as I did for the line starting at age 91. That same process first
produces `[92]+1 = 5860.36 approximately and then `[92] = 9767.27 approximately. The complete
select & ultimate life table corresponding to the select & ultimate mortality table in Example 1.5 is
thus
x
90
91
92
93
`[x]
10000
10417.58
9767.27
—
`[x]+1
8000
7292.31
5860.36
—
`x+2
6000
4740
3223.2
1772.76
x+2
92
93 .
94
95
Complicated probabilities are easy to compute intuitively in the usual manner with life tables.
Example 1.6 Suppose that an Exam LTAM problem gave me the select & ultimate mortality
table given in Example 1.5 above and asked me to compute 1|3 q[91] . My approach would be to first
create the part of the select & ultimate life table applying to [91] as found above and then compute
intuitively. Although 1|3 q[91] is of course the probability that [91] survives one year and then dies
in the following three years, I think of it intuitively as the fraction of select 91-year-olds that die
between ages 92 and 95. The number of select 91-year-olds that die between ages 92 and 95 is
(expected to be) `[91]+1 − `[91]+4 = `[91]+1 − `95 = 7292.31 − 1772.76 = 5519.55. The number of
select 91-year-olds is `[91] = 10417.58, so the fraction of select 91-year-olds that die between ages 92
5519.55
and 95 is 10417.58
= 0.52983. That is, 1|3 q[91] = 0.52983.
End of Example 1.6
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