Chapter 7
Multiple Decrement Models
7.1 Multiple Decrement Models
Individual (x) is subject to several causes of decrement which are mutually exclusive. e.g. failure due to death, disability or lapse; analysis of mortality by cause of death, etc.
Model
time until failure of (x)
cause of failure of (x)
,
joint density of T and J
i.e. satisfies
and cause of death
Recall for a single decrement
density function of
Similarly, joint density of T(x) and J(x)
P{(x)dies at t and by cause j}
=P{(x) survives to t}X P{(x) dies at (t,t+ ) and caused by j survived to t}
where
Thus
i.e.
Now
But
i.e.
Summary
1.
3. ,
4.
5.
6.
7.
8.
Example 7.1.1 In a multiple decrement table associated with a certain body of lives, there
are three forces of decrement and where . What is the probability that a life age 10 will remain in the body of lives until age 60?
Solution
We want
i.e.
Example 7.1.2 For a double decrement table, you are given:
(i) ,
(ii) ,
Which of the following are true?
I. ,
II. ,
III.
Solution
( I) is true.
II.
( II) is false
III.
( III) is true
7.1.1 Survivorship Group
initial number of lives age (x)
expected number of survivors to age x+t
expected total decrements between x and x+t
expected decrements between x and x+t from cause j
In particular, expected decrements between x and x+1 from cause j
Also,
where expected number of lives who will terminate eventually from cause j.
7.1.2 Multiple Decrement Table
Life table for multiple decrements which for an initial radix of lives and a given set of
probabilities of decrement , produces values of and
Example 7.1.3 Given the following double decrement probabilities
x
63 0.1 0.1
64 0.2 0.4
65 0.5 0.5
for a radix of 1000 lives age 63, we get the resulting double decrement table:
63 0.1 0.1 0.8 1000 100 100
64 0.2 0.4 0.4 800 160 320
65 0.5 0.5 0 320 160 160
66 0
Example 7.1.3 A two year junior college expects that its students will leave in accordance with the following probabilities:
Class Withdrawal for Academic Failure Withdrawal for Other Causes
First year 0.10 0.30
Second year 0.05 0.20
a. How many new students should be admitted at the beginning of each school year if 180 graduates are desired at the end of each year?
Solution
Let initial number of students
We want
Solving,
b. How many new students should be admitted at the beginning of each school year if the total enrollment is to be 800 students?
Solution
2nd year students
Want
i.e.
7.1.3 Relationship between Multiple Decrement Probabilities
and Life Table Functions
In Particular,
Since
i.e.
In particular,
and
Similarly,
In particular, and
Since
and
Similarly,
Example 7.1.4 For a double decrement table, you are given:
(i)
(ii)
Calculate for
Solution
i.e.
X
30 --- 0.75 --- --- 130
31 0.2 0.5 1850 ---
---32 --- --- --- --- 54
---Calculate
Solution
Solving,
Since
i.e.
Example 7.1.6 Given
Solution
i.e.
7.1.4 Central Rates of Multiple Decrement
Recall for a single decrement
central death rate at age x
central rate of decrement from all causes
central rate of decrement from cause j
Example 7.1.7 For a multiple decrement table, you are given:
(i) ,
(ii) ,
Calculate
Solution
i.e.
7.1.5 Associated Single Decrement Tables
Definition:
and
Interpretation:
P{(x) dies in the coming year by cause j assuming no other cause of death exists}
P{(x) survives the coming year by not dying and the only cause of death is j}
absolute rate of decrement
Fact
1. In the context of a single decrement, is a legitimate probability. When several
decrements are operating simultaneously, is not necessarily a legitimate probability.
2. Since is a legitimate force of mortality,
and
i.e.
for at least one j
i.e. for all j necessarily
3. Since
and
i.e.
But and
i.e.
4. Since
i.e.
5. Since
Since
i.e.
i.e.
7.1.6 Central Rate Decrement for Associated Single Document
Recall
Corresponding central rate for associated single decrement is defined by
Fact
1.
3. If is an increasing function of t, then
If is an decreasing function of t,
7.1.7 Constant Force Assumption
Then
i.e.
But and
i.e. ………..(1)
Given ( and hence ), we can get and
We can hence solve for using (1)
i.e.
i.e.
Example 7.1.8 For a triple-decrement table, you are given:
(i)
(ii)
(iii) ,
Assume a Constant Force of decrement for each decrement over each year of age.
Calculate
Solution
i.e.
i.e.
7.1.8 Uniform Distribution of Deaths Assumption
Then
and
i.e. as in the Constant Force of mortality assumption
i.e.
i.e.
Example 7.1.9 For a double decrement table, you are given:
(i)
(ii)
(iii) Each decrement is uniformly distributed over each year of age in the double decrement table.
Calculate
Solution
i.e.
i.e.
Example 7.1.9 For a double decrement table, you are given:
(i)
(ii)
Solution
i.e.
But
i.e.
i.e.
i.e.
and
7.1.9 Construction of a multiple decrement table
Given the absolute rates of decrement , we can construct a multiple decrement table
from an initial radix of lives by assuming either a Constant Force of Mortality assumption or a Uniform Distribution of Deaths assumption.
Example 7.1.11 Given the following absolute rates of decrement table for a double decrement model:
x
63 0.1 0.2
64 0.2 0.4
65 0.5 0.5
Construct the multiple decrement table for an initial radix of 1000 lives.
Solution
where ,
and
We get the following:
x
63 0.1 0.2 0.9 0.8 0.72 0.28 0.09 0.19
64 0.2 0.4 0.8 0.6 0.48 0.52 0.16 0.36
65 0.5 0.5 0.5 0.5 0.25 0.75 0.38 0.38
The multiple decrement table is then:
x
63 0.09 0.19 0.72 1000 90 190
64 0.16 0.36 0.48 720 115 259
65 0.38 0.38 0.24 346 131 131
7.1.10 Other approximations
A. Assumption of UDD on associated single decrement:
We assume ,
Then
and
It follows that
For 2 decrements
Similarly,
and
B. Central Rate Bridge assumption : Assume (i) UDD on associated single decrement
(ii) UDD on multiple decrement probabilities
i.e.
(iii) for each cause j
i.e.
Solving for , we get
i.e.
where
Example 7.1.12 Suppose ,
Calculate , for each of the following:
a. Assuming Constant Force or UDD on each multiple decrement probability b. Assuming UDD on associated single decrement.
c. Under Central Rate Bridge assumption.
Solution
a.
i.e.
b. ………(1)
i.e.
i.e.
i.e.
i.e.
i.e.
i.e.
Solving
Since (2)-(1) gives
c.
Similarly,
Example 7.1.13 Suppose ,
Calculate for each of the following:
b. UDD on
c. Central Rate Bridge assumption
Solution
a.
;
i.e.
b.
c.
Example 7.1.14 Given ,
Suppose satisfies the UDD assumption and all occurs at mid-year.
a. for
b. and
Solution
a.
,
,
Thus for
for
b.
i.e.
7.2 Mixed Distribution Assumption for
Multiple Decrements
Example 7.2.1 For a double decrement table, you are given the double decrement
probabilities which satisfy
i.e. Uniform Distribution of Death Assumption
i.e. second decrement occurs at the beginning of the year.
Find expressions for in terms of
i.e.
In terms of a survivorship group
To get using the relationship
i.e.
Thus
Example 7.2.2 A multiple decrement table has two decrements, deaths (d) and
withdrawals (w). Withdrawals occur once a year three-fourths of the way through the year of age. Deaths in the associated single decrement are uniformly distributed over each year of age. You are given:
(i)
(ii)
(iii)
Calculate
Solution
Given
for
for
i.e.
i.e.
i.e.
Thus
and
Example 7.2.3 In a double decrement table, determine the following:
A. in terms of assuming is UDD and all occurs at ,
B. in terms of assuming is UDD and all occurs at ,
Solution
A.
But for
for
i.e.
:
:
B.
for
for
i.e.
k=0:
k=1:
7.2.1 Net Single Premiums in a Multiple Decrement Context
Assume m decrements. Then the net single premium for insurance benefit of one payable at the moment of exit of an insured (x) is
Under the assumption of UDD on each multiple decrement,
Note
n.s.p. for an insurance benefit of one payable at the moment of exit of an insured (x) by decrement j.
General: Typically, the insurance benefit varies depending on the cause of decrement.
Let insurance benefit at x+t if cause of decrement is j.
Then
Example 7.2.4 A multiple decrement table has two causes of decrement:
1. death by accident
2. death other than by accident
You are given:
(i) A fully continuous whole life insurance issued to (x) pays $2 if death results by accident and $1 if death results other than by accident.
(ii) , the force of interest.
Calculate the net single premium for this insurance.
Solution
n.s.p.
