1
There are three sets of new material:
1. Section 8.5 is to be added at the end of lesson 8. No exercises are provided for this
material.
2. Lesson 28 replaces the corresponding lesson in the 15th edition.
3. Lesson 61 is a new lesson and comes between lesson 60 and lesson 61 of the 15th
edition.
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8.5. EXTREME VALUE DISTRIBUTIONS
Using this distribution, it proves that if e (x ) ≥ e (0) for all x , which would be true for a decreasing
hazard rate, then the coefficient of variation is at least 1, and if e (x ) ≤ e (0) for all x , then the coefficient
of variation is at most 1.
I doubt the equilibrium distribution will be tested on.
EXAMPLE 8I X follows an exponential distribution with mean θ .
Determine the distribution function of the equilibrium distribution for X , or X e .
ANSWER: S (x ) = e −x /θ . So
e −x /θ
θ
which is the density function of an exponential with mean θ . So X e has the same distribution as X . ƒ
f e (x ) =
EXAMPLE 8J X follows a two-parameter Pareto distribution with parameters α and θ .
Determine the distribution function of the equilibrium distribution for X , or X e .
α
ANSWER: S (x ) = θ /(θ + x ) . So
f e (x ) =
α
θ /(θ + x )
θ /(α − 1)
=
(α − 1)θ α−1
(θ + x )α
which is the density function of a Pareto distribution with parameters α − 1 and θ .
8.5
ƒ
Extreme value distributions
The following material was added to the syllabus for October 2013. It is purely descriptive (no calculations),
so exam coverage of it is bound to be light.
We will mention two heavy-tailed distributions used for risk management.
The first distribution arises as the limit of the maximum of a set of observations as the size of the set goes
to infinity. The Fisher-Tippett Theorem states that this limit has only one of three possible distributions. Of
the three, we will mention only one of them, the Fréchet distribution. This is the same as the inverse Weibull
distribution.
The second distribution arises as the limit of the excess loss random variable as the truncation point (the
deductible) goes to infinity. The Balkema-de Haan-Pickands Theorem states that this limit has only one of
three possible distributions. We will mention only two of them. You could guess what these two are. We’ve
learned that the excess loss of an exponential is exponential and the excess loss of a Pareto is Pareto, so these
two are certainly limiting distributions. And in fact, an exponential distribution is the limiting distribution
for excess loss for lighter-tailed distributions, while the Pareto distribution is a limiting distribution for excess
loss for other distributions. Thus a Pareto distribution may be a good model for high-deductible insurances.
(The exponential distribution is not heavy-tailed, so we didn’t count it when we said that we’ll mention two
heavy-tailed distributions.)
Extreme value theory calls the set of limiting distributions for mean excess loss “generalized Pareto“, but
this is not the same distribution as the generalized Pareto in the Loss Models appendix.
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Lesson 28
Mortality Table Construction
Reading: Loss Models Fourth Edition 12.4
While the techniques in this lesson can be used to estimate any random variable, they are usually used for
mortality table construction, or for modeling other random variables that are a function of time.
Mortality tables are constructed by specifying the probability that a person age x will die in the next year,
denoted by q x , for every integral age x . These tables do not specify probabilities of surviving fractional years,
or survival functions for a person at a fractional age. They are built off large data sets with almost all data
censored and truncated. Parametric functions are usually inadequate for fitting mortality data.
We will consider two categories of techniques for estimating q x . The first category uses individual data.
The second category uses data grouped into intervals.
In the methods we discuss in this lesson, the mortality rate q x or the hazard rate is estimated by dividing
number of deaths by number of exposures. The definition of exposure depends on the method.
We will use the following notation from Exam MLC/3L:
nq x
is the probability of death within n years for a person age x .
np x
is the probability of survival for n years for a person age x .
You are not responsible for this notation for Exam C/4, but it will be convenient for some formulas.
28.1
Individual data based methods
In the methods of this section, exposure is measured policy by policy. We discuss two methods for estimating
q x . The two methods differ in the definition of exposure and in what is estimated (mortality rate or hazard
rate).
The first method uses exact exposure. The exact exposure for each individual is the amount of time from
entry into the study until leaving the study. Entry into the study may be at the start of the study or may be
late. Leaving the study may be by death, withdrawal, or termination of the study. Exposure is tabulated by
age. Here’s an example.
EXAMPLE 28A A 3-year mortality study uses records starting on Jan. 1, 2009 and ending on Dec. 31, 2011. For
simplicity, assume all events including births, policy issue, withdrawals, and deaths occur on the first day of
the month and that a month is 1/12 of a year. The ending day of the study will be treated as if it is 1/1/2012.
The following data is recorded for 6 individuals in the study:
Number
Birth Date
Policy Issue
Date of Withdrawal
Date of Death
1
2
3
4
5
6
3/1944
10/1944
12/1944
1/1945
4/1945
7/1945
5/2004
6/2010
2/2009
12/2008
1/2010
11/2010
5/2011
—
—
4/2010
—
—
—
—
4/2011
—
8/2011
—
Calculate the total exact exposure for every age that has any exposure.
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28. MORTALITY TABLE CONSTRUCTION
at that time. He leaves
ANSWER: The first person enters the study when it starts on 1/2009 and is age 64 10
12
2
at age 67 12
. Therefore there are 2 months of exposure for age 64, 12 months apiece for ages 65 and 66, and
2 months for age 67.
8
and leaves at the end of the study on
The second person enters the study on 6/2010 at age 65 12
3
12/31/2011 at age 67 12 . Therefore there are 4 months of exposure for age 65, 12 months for age 66, and
3 months for age 67.
2
4
The third person enters the study at age 64 12
and leaves at age 66 12
. There are 10 months of exposure at
age 64, 12 months at age 65, and 4 months at age 66.
3
. There are
The fourth person enters the study when it starts on 1/2009 at age 64 and leaves at age 65 12
12 months of exposure for age 64 and 3 months for age 65.
4
9
and leaves at age 66 12
. There are 3 months of exposure for
The fifth person enters the study at age 64 12
age 64, 12 for age 65, and 4 for age 66.
4
6
The sixth person enters the study at age 65 12
and stays to the end at 12/31/2011 and is age 66 12
at that
time. There are 8 months of exposure at age 65 and 6 months at age 66.
Total exposure in months is:
Number
Age 64
Age 65
Age 66
Age 67
1
2
3
4
5
6
2
0
10
12
3
0
12
4
12
3
12
8
12
12
4
0
4
6
2
3
0
0
0
0
Total
27
51
38
5
We assume the hazard rate is constant for each age. If e j is exact exposure for age j and d j is the number
of deaths, the estimate of the hazard rate at that age is h j = d j /e j . As we will learn in Subsection 33.1.1, this is
the maximum likelihood estimate of the hazard rate. If we are estimating a mortality rate for one year, then
the integral of the hazard rate over the year is H j = d j /e j , and q j = 1 − e −d j /e j . More generally, to estimate a
mortality rate over n years, if we assume the hazard rate is constant over the entire period, H j = nd j /e j and
q j = 1 − e −n d j /e j
(28.1)
where d j and e j are the number of deaths and the exposure over the entire period.
EXAMPLE 28B In the previous example, calculate the mortality rates using the exact exposure method.
ANSWER: There are two deaths at age 66 (#3,#5) and no other deaths, so q̂64 = q̂65 = q̂67 = 0 and q̂66 = 1 −
e −2/(38/12) = 0.468248 .
ƒ
The second method uses actuarial exposure. Actuarial exposure is the same as exact exposure except
for those who die. For those who die, exposure is counted until the end of the year of death, even if that is
after the termination of the study. Then setting e j equal to actuarial exposure, the estimate of the one-year
mortality rate is q̂ j = d j /e j . If we wanted to estimate a mortality rate over n years, the estimate would be
q̂ j =
nd j
ej
(28.2)
where d j and e j are the number of deaths and the exposure rate over the entire period.
The actuarial estimator is an inconsistent estimator, but it is not much different from the exact exposure
estimator for a large data set with a low mortality rate. On the other hand, it replicates the empirical estimator
if data is complete. (The exact exposure method does not replicate the empirical estimator.)
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28.1. INDIVIDUAL DATA BASED METHODS
EXAMPLE 28C In Example 28A, calculate actuarial exposure and the estimated mortality rates.
2
ANSWER: Actuarial exposure is the same as exact exposure except for #3 and #5. For #3, entry is at age 64 12
and leaving is on 12/2011 at age 67, so there are 8 additional months of exposure at age 66. For #5, leaving
is on 4/2012, so there are 8 additional months of exposure at age 66. Thus there are 38 + 16 = 54 months of
ƒ
exposure at age 66 and q̂66 = 2/(54/12) = 0.444444 . Mortality at the other ages is still 0.
?
Quiz 28-1 For a mortality study from 1/1/2011 to 12/31/2012, there are 3 individuals born on 2/1/1981.
The first purchases a policy on 2/1/2010 and neither dies nor withdraws. The second purchases a policy on
2/1/2011 and withdraws on 8/1/2011. The third purchases a policy on 7/1/2011 and dies on 1/1/2012.
Calculate the total actuarial exposure for q30 of these three individuals.
Notice that actuarial exposure for those who die is not necessarily an integer, because entry may occur at
a fractional age. However, in practice, insureds are assigned an integral insuring age. When a person buys a
policy, his birthday is set equal to the date the policy is issued. The age assigned to the person may be the age
last birthday or the age nearest birthday. Since this age controls the premiums and policy values, mortality
is estimated based on the insuring age rather than the real age.
EXAMPLE 28D Redo Example 28A using age last birthday as the insuring age. For convenience, the data are
repeated here:
Number
Birth Date
Policy Issue
Date of Withdrawal
Date of Death
1
2
3
4
5
6
3/1944
10/1944
12/1944
1/1945
4/1945
7/1945
5/2004
6/2010
2/2009
12/2008
1/2010
11/2010
5/2011
—
—
4/2010
—
—
—
—
4/2011
—
8/2011
—
Compute mortality rates using the exact exposure method and the actuarial exposure method.
ANSWER: First we use exact exposure. In each case, the birth date is advanced to the next policy anniversary.
The first person gets a birthday of 5/1944. The second person gets a birthday of 6/1945. The third person
gets a birthday of 2/1945. The fourth person gets a birthday of 12/1945. The fifth person gets a birthday of
1/1946. The sixth person gets a birthday of 11/1945. The resulting exposure in months is:
Number
Age 63
Age 64
Age 65
Age 66
1
2
3
4
5
6
0
0
0
11
0
0
4
0
12
4
12
0
12
12
12
0
7
12
12
7
2
0
0
2
Total
11
32
55
23
#3 dies at age 66 but #5 dies at age 65 based on the insuring age. So the mortality estimates are q̂63 = q̂64 =
0, q̂65 = 1 − e −1/(55/12) = 0.196021 , q̂66 = 1 − e −1/(23/12) = 0.406513 .
Now let’s use actuarial exposure. For #3, actuarial exposure terminates at the next birthday in February,
or at 2/2012, so actuarial exposure adds 10 months of exposure at age 66 for #3. For #5, actuarial exposure
terminates at the next birthday in January, or 1/2012, so actuarial exposure adds 5 months of exposure at
ƒ
age 65 for #5. Thus q̂65 = 1/(60/12) = 0.2 and q̂66 = 1/(33/12) = 0.363636 .
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28. MORTALITY TABLE CONSTRUCTION
Examples like the previous one can be confusing. If you’re trying to calculate exposure for just one age,
it may be helpful to create a table specifying exactly when the exposure for that age starts and when it ends.
For example, if you wanted the actuarial exposure at age 64 in the previous example, your table would look
like this:
Number
Insuring
Birth Date
Age 64 Exposure
Begins
Age 64 Exposure
Ends
Months of
Exposure
1
2
3
4
5
6
5/1944
6/1945
2/1945
12/1945
1/1946
11/1945
1/2009
6/2010
2/2009
12/2009
1/2010
11/2010
5/2009
6/2010
2/2010
4/2010
1/2011
11/2010
4
0
12
4
12
0
and for age 65, it would look like this:
Number
Insuring
Birth Date
Age 64 Exposure
Begins
Age 64 Exposure
Ends
Months of
Exposure
1
2
3
4
5
6
5/1944
6/1945
2/1945
12/1945
1/1946
11/1945
5/2009
6/2010
2/2010
—
1/2011
11/2010
5/2010
6/2011
2/2011
—
1/2012
11/2011
12
12
12
0
12
12
The above study is calendar based, or date-to-date. To avoid fractional years of exposure due to the study
ending, some studies are anniversary based, or anniversary-to-anniversary. This means that the study starts
and ends on the policy anniversary. This method, however, throws away data before the anniversary in the
first calendar year and after the anniversary in the last calendar year.
EXAMPLE 28E Redo Example 28A using age nearest birthday as the insuring age as an anniversary-toanniversary study, starting with anniversaries in 2009 and ending at anniversaries in 2011. Calculate exposure using the actuarial exposure method.
ANSWER: The first person gets a birthday of 5/1944. The second person gets a birthday of 6/1944, since that
is nearer to the actual birthday than 6/1945. The third person gets a birthday of 2/1945. The fourth person
gets a birthday of 12/1944. The fifth person gets a birthday of 1/1945. The sixth person gets a birthday of
11/1945.
For the first person, exposure begins on 5/2009 and ends on 5/2011, for 12 months apiece at age 65 and 66.
For the second person, exposure begins on 6/2010 and ends on 6/2011, for 12 months at age 66.
For the third person, exposure begins on 2/2009 and ends on 2/2011. Therefore, the death is ignored.
There are 12 months apiece of exposure at ages 64 and 65.
For the fourth person, exposure begins on 12/2009 and ends on 4/2010. Therefore, there are 4 months of
exposure at age 65.
For the fifth person, exposure begins on 1/2010 and ends on 1/2011. Once again the death is ignored,
and there is 12 months of exposure at age 65.
For the sixth person, exposure begins on 11/2010 and ends at 11/2011, for 12 months of exposure at
age 65.
Exposure is summed in the following table:
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28.2. INTERVAL-BASED METHODS
28.1.1
Number
Age 64
Age 65
Age 66
1
2
3
4
5
6
0
0
12
0
0
0
12
0
12
4
12
12
12
12
0
0
0
0
Total
12
52
24
Variance of estimators
The estimate of the variance of the exact exposure estimator for mortality over n years is
d
V
ar(q̂ j ) = (1 − q̂ j )2 n 2
dj
e j2
(28.3)
We will derive this estimate in Section 34.4. To remember it, it’s like a combination of the Greenwood approximation (the first factor, S (x )2 ) with the variance of the Nelson-Åalen estimator (the last factor, d j /e j2 ).
To estimate the variance of the actuarial estimator for mortality over n years, treat the estimate as the
empirical estimate with complete data. The number of members of the population is set equal to exposure
divided by the length of the period, e j /n, since each member would contribute n to exposure if they were
present for the entire n-year period. Then use a version of equation (23.1). Note that q x is the complement
of the probability of survival for one year, so Var(q̂ x ) is the same as the variance of the probability of survival
for one year. The resulting formula is
q̂ j (1 − q̂ j )
d
(28.4)
V
ar(q̂ j ) =
e j /n
Since q̂ j = d j /e j , this can also be written as
d j (e j − d j )
e j3 /n
.
EXAMPLE 28F In Example 28A, estimate the standard deviation of the estimates of q̂66 using exact exposure
and using actuarial exposure.
ANSWER: With exact exposure, e66 =
38
12
v
t
and q̂66 = 0.468248. The standard deviation of q̂66 is
(1 − 0.468248)2
2
= 0.23748
(38/12)2
With actuarial exposure, e66 = 54/12, as derived in Example 28C, and in that exercise we obtain q̂66 = 4/9,
so the standard deviation of q̂66 is
v
t (4/9)(5/9)
= 0.24103
ƒ
51/12
28.2
Interval-based methods
In an interval based method, data is summarized by age or age group.
Assume data are grouped into intervals, each one starting at c j and ending at c j +1 , where c0 < c1 < · · · < ck .
The following notation is taken from the textbook. On past exams, they have occasionally used textbook
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482
28. MORTALITY TABLE CONSTRUCTION
notation from this topic with no explanation, expecting you to know what it means. Note that previous
editions of the textbook used different notation.
Pj is the population at time c j . This is the number of members of the population after all deaths and withdrawals from the previous interval and after any new entries at the beginning of the current interval.
n j is the number of new entrants in the interval [c j , c j +1 ). n stands for new! Initially, the textbook uses this
notation only for new entrants during the interval, but not at the beginning of the interval. Later on,
the textbook introduces the notation n jb , the number of entrants at the beginning of the interval at
exact time c j , and n m
j , the number of entrants at any time other than the very beginning of the interval
(m for middle).
We include the left endpoint of the interval [c j , c j +1 ) but not the right endpoint. If someone enters the
study at exact age 45, for example, he’s included when calculating the mortality rate for 45-year olds.
This is consistent with the rules for ties mentioned in the paragraph after equation (24.1).
w j is the number of withdrawals in the interval (c j , c j +1 ]. Initially, the textbook uses this notation only for
withdrawals during the interval, but not at the end of the interval. Later on, the textbook introduces
the notation w je , the number of withdrawals at the end of the interval at exact time c j +1 , and w jm , the
number of withdrawals not at the very end of the interval.
We include the right endpoint of the interval (c j , c j +1 ], but not the left one. If someone leaves the study
at exact age 46, for example, he’s included when calculating the mortality rate for 45-year olds and not
considered as having withdrawn at age 45; withdrawals occurring at the same time as a death count in
the risk set for those deaths. This is consistent with the rules for ties mentioned in the paragraph after
equation (24.1).
d j is the number of deaths in the interval (c j , c j +1 ]. Do not confuse this with d j defined on page 389 for
individual data. There you are given d j for each individual and it means entry time. Here in intervalbased estimation, d j is the count of number of deaths in an interval.
The populations are computed recursively starting with P0 = n0b . Then
m
e
b
Pj = Pj −1 + n m
j −1 − d j −1 − w j −1 − w j −1 + n j
(28.5)
Using interval data, we can estimate exposure by assuming that entries and withdrawals during the interval (not at the beginning or end of the interval) contribute 1/2 of a year (or whatever the interval length
is) of exposure. Thus the exact exposure method would use an exposure of
m
e j = Pj + 0.5(n m
j − wj − dj )
in its calculation of q̂ j = 1 − e −d j /e j . The actuarial exposure method would use
m
e j = Pj + 0.5(n m
j − wj )
(28.6)
(28.7)
when calculating q̂ j = d j /e j .
Another method is Kaplan-Meier style. Assuming entries and withdrawals during the interval occur in
the middle of the interval and deaths are uniform, entries and withdrawals contribute half a period to the
m
risk set, so the risk set is Pj + 0.5(n m
j − w j ). This results in the actuarial estimator of the previous paragraph.
Notice that unlike in individual data methods, when we talk about exposure in grouped data methods,
it refers to the number of individuals in the study, rather than the number of individuals times the size of
the interval. Thus q̂ j is the estimated proportion of deaths throughout the interval, not the annual mortality
rate. If the interval length is 5 years, for example, q̂ j is the estimated probability of dying in five years. So
“exposure” and “risk set” are one and the same in this context.
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28.2. INTERVAL-BASED METHODS
EXAMPLE 28G (Data from Example 28A) A 3-year mortality study reviews the company’s records starting on
Jan. 1, 2009 and ending on Dec. 31, 2011. For simplicity, assume all events occur on the first of each month
and that a month is 1/12 of a year; however, the ending day of the study will be treated as if it is 1/1/2012.
The following data is recorded for 6 individuals in the study:
Number
Birth Date
Policy Issue
Date of Withdrawal
Date of Death
1
2
3
4
5
6
3/1944
10/1944
12/1944
1/1945
4/1945
7/1945
5/2004
6/2010
2/2009
12/2008
1/2010
11/2010
5/2011
—
—
4/2010
—
—
—
—
4/2011
—
8/2011
—
Insuring age is age last birthday. Data are grouped by age.
Calculate the exposure both on an exact basis and on an actuarial basis. Then calculate the mortality
rates on an actuarial basis.
ANSWER: Let’s calculate each individual’s entry age and withdrawal/death age.
Number
Entry age
Withdrawal age
Death age
1
2
3
4
5
6
8
64 12
67
7
66 12
—
—
2
66 12
—
7
65 12
—
65
64
1
63 12
64
65
4
64 12
—
2
66 12
m
b
We’ll use ages as subscripts of our symbols. Then looking down the entry age column, n63
= 1, n64
= 2,
m
b
m
m
e
n64 = 1, n65 = 2, and all other n j ’s are 0. Looking down the withdrawal age column, w64 = 1, w66 = 2, w66 = 1,
e
Notice that #1 leaves at exact age 67 and so is counted as w66
. Looking down the death age column, d 65 = 1
and d 66 = 1. The populations are
P63 = 0
m
b
P64 = P63 + n63
+ n64
=0+1+2=3
m
b
m
+ n65
− w64
=3+1+2−1=5
P65 = P64 + n64
P66 = P65 − d 65 = 5 − 1 = 4
Exact exposure is
m
e
P67 = P66 − d 66 − w66
− w66
=4−1−2−1=0
m
e63 = P63 + 0.5n63
= 0 + 0.5 = 0.5
m
m
e64 = P64 + 0.5(n64
− w64
) = 3 + 0.5(1 − 1) = 3
e65 = P65 − 0.5(d 65 ) = 5 − 0.5 = 4.5
m
e66 = P66 − 0.5(w66
+ d 66 ) = 4 − 0.5(2 + 1) = 2.5
Actuarial exposure doesn’t subtract 0.5d j at j = 65 and 66, so e65 = 5 and e66 = 3. The resulting mortality
rates are q̂63 = q̂64 = 0, q̂65 = 1/5 = 0.2 , q̂66 = 1/3 = 0.333333 .
The next example is one of the few examples in this lesson not involving mortality tables.
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484
28. MORTALITY TABLE CONSTRUCTION
EXAMPLE 28H You offer two types of automobile comprehensive policies: one with no deductible and a
policy limit of 5000, and another with a deductible of 500 and a maximum covered loss of 10,000. You have
the loss experience shown in the following table.
Number of Losses in Range
0 deductible, 5000 limit 500 deductible, 10,000 limit
Loss Range
(0, 500]
(500, 1000]
(1000, 2000]
(2000, 5000]
(5000, 7500]
(7500, 10000]
At limit
Total
20
18
16
19
32
24
21
18
10
5
110
27
100
The ground-up loss distribution for both types of policy is assumed to be the same.
Estimate the probability that a loss will be no more than 7500 using the actuarial method.
ANSWER: Since the deductibles and limits are at interval endpoints, the only reasonable assumption is that
all of entries and withdrawals occur at integer endpoints. Exposure in each interval (or the risk set) equals
the population at the beginning of each interval. The policies with 500 deductible enter the study at 500, so
we have the following table:
j
cj
Pj
n jb
w je
dj
ej
Ŝ (c j +1 )
0
1
2
3
4
5
0
500
1000
2000
5000
7500
100
190
140
100
33
100
110
0
0
0
0
0
0
27
0
20
50
40
40
18
100
190
140
100
33
0.8
0.8(14/19)
0.8(10/19)
0.8(6/19)
0.8(6/19)(15/33)
As hinted in the last column of the above table, the intervals (500, 1000], (1000, 2000], and (2000, 5000] may
be combined into a single interval with risk set 190 and events 50+40+40 = 130, since there is no truncation
or censoring between 500 and 5000.
The probability of survival to 7500 is 0.8(6/19)(15/33) = 0.114833, so the probability that a loss is no more
than 7500 is 1 − 0.114833 = 0.885167 .
ƒ
EXAMPLE 28I You perform a mortality study on 950 lives age 45. Additional lives enter the study at ages 45
and later. You are given the following information from the mortality study:
j
Age
cj
Number of lives
entering at
this age
nj
0
1
2
3
4
45
46
47
48
49
50
30
40
30
10
Number of lives
withdrawing
before next age
wj
Number of deaths
dj
100
90
80
70
60
3
5
6
4
8
All entries, other than the original 950 lives, and all withdrawals are assumed to occur uniformly throughout each year.
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28.2. INTERVAL-BASED METHODS
Estimate the probability of survival for an individual age 45 to each of ages 46, 47, 48, 49, 50 using the
actuarial method.
m
ANSWER: Calculate the population recursively using Pj +1 = Pj + n m
j − w j − d j . The number of terminations
at the end of the study, w5e , is the number of lives remaining after death and withdrawal, or 684, and wi = 0
for i 6= 5.
m
Calculate the exposure as e j = Pj + 0.5(n m
j − w j ). Note that the original 950 lives are part of the starting
population and count fully, unlike the late entrants (including the 50 entrants at age 45), which count as
0.5 lives in the exposure. Similarly, the 684 lives surviving at the end count fully, unlike withdrawals, which
count as 0.5 lives apiece.
j
cj
Pj
nm
j
w jm
w je
dj
ej
p̂ j0
0
j +1 p̂45
0
1
2
3
4
5
45
46
47
48
49
50
950
897
832
786
742
0
50
30
40
30
10
100
90
80
70
60
0
0
0
0
684
3
5
6
4
8
925
867
812
766
717
922/925
862/867
806/812
762/766
709/717
922/925 = 0.996757
0.996757(862/867) = 0.991008
0.991008(806/812) = 0.983686
0.983686(762/766) = 0.978549
0.978549(709/717) = 0.967631
Comment
In all of the above examples involving mortality studies, we assumed data by age, but the methods we have
discussed apply equally well to data arranged into larger intervals, such as quinquennial ages.
?
Quiz 28-2 You are given the following information regarding five individuals in a study:
dj
uj
xj
0
0
1
2
4
2
—
6
—
6
—
3
—
7
—
The data are grouped into intervals of 5 time units. Entries and withdrawals are assumed to be uniformly
distributed within each group.
Calculate the probability of death in the period (0, 5] using this approximation and the actuarial estimator.
Multiple Decrement
The estimators using the Kaplan-Meier method are all single-decrement rate estimators. You may multiply
survival probabilities for several decrements to obtain an estimate of the probability of survival from all
decrements combined.
EXAMPLE 28J In a study of 100 life insurance policies, you are given that there are 2 deaths and 10 withdrawals in the first year, and 3 deaths and 10 withdrawals in the second year.
Use the actuarial estimator to estimate the death and withdrawal rates, assuming that deaths and withdrawals are uniformly distributed throughout each year. Then determine the probability that a policy on a
newly issued life will not terminate by death or withdrawal within two years.
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28. MORTALITY TABLE CONSTRUCTION
(τ)
ANSWER: We will use MLC/3L notation p j to indicate the probability of surviving all decrements.
0(d )
0(w )
There are no late entrants. The populations are P0 = 100 and P1 = 100 − 2 − 10 = 88. Let q̂ j and q̂ j be
the estimated rates of decrement from death and withdrawal respectively for j = 0, 1.
From the perspective of the mortality study, there are 10 withdrawals apiece in years 1 and 2. The actuarial exposures are e0 = 100 − 0.5(10) = 95 and e1 = 88 − 0.5(10) = 83, so
93
= 0.978947
95
80
0(d )
1 − q̂1 =
= 0.963855
83
0(d )
1 − q̂0
=
using the actuarial estimator.
From the perspective of the withdrawal study, there are 2 censored observations in the first year and
3 censored observations in the second year. The actuarial exposures are e0 = 100 − 0.5(2) = 99 and e1 =
88 − 0.5(3) = 86.5, so
89
= 0.898990
99
76.5
0(w )
= 0.884393
1 − q̂1 =
86.5
0(w )
1 − q̂0
=
using the actuarial estimator. Then
(τ)
p̂0 = (0.978947)(0.898990) = 0.880064
(τ)
p̂1 = (0.963855)(0.884393) = 0.852427
(τ)
and the probability of survival for 2 years is 2 p̂0
= (0.880064)(0.852427) = 0.750190 . Notice that this is
(τ)
different from naively estimating that since 25 lives either died or withdrew, which would lead to 2 p̂0
(100 − 25)/100 = 0.75.
=
ƒ
Exercises
28.1. For a calendar year mortality study for 1/1/2009–12/31/2011, a person born on 3/1/1984 buys an
insurance policy on 9/1/2006 and withdraws on 8/1/2011.
Insuring age is age last birthday.
Calculate the exact exposure for age 27.
28.2. For a calendar year mortality study for 2012, you are given the following data:
Person
Date of Birth
Policy Issue Date
Date of Termination
Cause of Termination
1
2
3
4
6/1/1971
1/1/1972
4/1/1972
6/1/1972
5/1/2009
5/1/2009
6/1/2012
3/1/2000
—
10/1/2012
10/1/2012
—
—
surrender
death
—
Calculate the absolute difference between the Kaplan-Meier product limit estimator of q40 and the actuarial estimator of q40 .
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EXERCISES FOR LESSON 28
Table 28.1: Summary of Mortality Table Construction Formulas
Individual data
Exact exposure: For each insured, time from entry to leaving, whether leaving is by death, withdrawal, or
termination of study.
Estimate of q j (mortality rate over n-year period) using exact exposure:
Variance of estimate:
q̂ j = 1 − e −n d j /e j
d
V
ar(q̂ j ) = (1 − q̂ j )2 n 2
(28.1)
dj
(28.3)
e j2
Actuarial exposure: For each insured, time from entry to leaving if leaving is by withdrawal or termination
by study. If leaving is by death, time from entry to end of year (of age) of death.
Estimate of q j (mortality rate over n-year period) using actuarial exposure:
q̂ j =
Variance of estimate:
d
V
ar(q̂ j ) =
dj
(28.2)
ej
q̂ j (1 − q̂ j )
(28.4)
e j /n
Insuring age: Age based on setting age to integer at policy issue. The integer may be age nearest birthday
or age last birthday.
Interval-based data
Population:
Exact exposure:
m
e
b
Pj +1 = Pj + n m
j − w j − w j − d j + n j +1
(28.5)
m
e j = Pj + 0.5(n m
j − wj − dj )
(28.6)
m
e j = Pj + 0.5(n m
j − wj )
(28.7)
Actuarial exposure, or risk set for Kaplan-Meier approximation for large data sets:
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28. MORTALITY TABLE CONSTRUCTION
28.3. For an anniversary-to-anniversary study for years 2007–2012, you are given the following data:
Person
Date of Birth
Policy Issue Date
Date of Termination
Cause of Termination
1
2
3
4
8/1/1962
11/1/1962
2/1/1963
6/1/1963
12/1/2004
6/1/2007
10/1/2008
1/1/2000
6/1/2011
10/1/2011
—
—
death
surrender
—
—
Insuring age is age last birthday.
Calculate q̂48 using the exact exposure method.
28.4.
For an anniversary-to-anniversary study for years 2007–2010, you are given the following data:
Person
Date of Birth
Policy Issue Date
Date of Termination
Cause of Termination
1
2
3
4
5
8/1/1967
11/1/1967
1/1/1968
2/1/1968
6/1/1970
1/1/2004
6/1/2007
6/1/2005
10/1/2009
2/1/2005
2/1/2008
5/1/2008
1/1/2011
—
—
death
surrender
surrender
—
—
Insuring age is age nearest birthday.
Calculate q̂40 using the actuarial estimator.
28.5. In a mortality study, you have the following individual data. For each triplet, the first number is the
age in months of entry to the study, the second number is the age in months of exit from the study, and the
letter is s for withdrawal or termination of study and x for death.
(32-0,34-0,s )
(33-0,35-0,s )
(33-2,33-9,x )
(33-5,33-11,s )
(33-8,35-5,s )
You estimate q33 in two ways:
1.
The actuarial estimator based on individual data.
2.
The actuarial estimator based on data grouped by age. Entries and withdrawals are assumed to occur
uniformly between integral ages, unless they occur at an exact integer.
Calculate the absolute difference between the two estimates.
28.6.
For a calendar year mortality study from 1/1/2009 to 12/31/2011:
(i) 25 individuals born on 7/1/1949 were issued policies on 3/1/2010. One of them withdrew on
5/1/2010.
(ii) 5 individuals born on 4/1/1950 were issued policies before 1/1/2009. One of them withdrew on
10/1/2010 and one died on 5/1/2011.
(iii) 1 individual born on 10/1/1950 was issued a policy on 2/1/2009 and died on 1/1/2011.
Calculate the variance of the exact exposure estimate of q̂60 ,
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EXERCISES FOR LESSON 28
28.7. For a calendar year mortality study from 1/1/2007 to 12/31/2011:
(i) 25 individuals born on 5/1/1967 were issued policies on 3/1/2004. One of them died on 11/1/2009,
and two withdrew on 1/1/2010.
(ii) 20 individuals born on 7/1/1967 were issued policies on 8/1/2009. One of them died on 6/1/2010,
and one withdrew on 2/1/2010.
Estimate the variance of the actuarial estimator for q̂42 .
28.8.
In a calendar year mortality study from 1/1/2006 to 12/31/2008:
(i) Insuring age is used, and is equal to age nearest birthday.
(ii) On Jan. 1, 2006, the following policies with insuring age 25 at their anniversaries in 2006 are in force:
Policy anniversary
Number
1/1
3/1
8/1
12/1
24
42
15
22
(iii) One of the insureds with policy anniversary 3/1 dies on 9/20/2007.
(iv) Three of the policies with anniversary 8/1 withdraw on 2/1/2006, 4/1/2007, and 11/1/2008.
Actuarial exposure is used to estimate 3 q̂25 , the probability that a person age 25 dies within 3 years.
Estimate the standard deviation of this estimate.
28.9. In a mortality study, you have the following individual data. For each triplet, the first number is the
age in months of entry to the study, the second number is the age in months of exit from the study, and the
letter is s for withdrawal or termination of study and x for death.
(32-9,37-9,s )
(33-2,36-7,s )
(33-2,35-3,x )
(33-5,38-5,s )
(33-7,35-1,s )
You estimate the probability that a person exact age 33 dies within 3 years using the exact exposure
method.
Estimate the variance of the estimator.
28.10.
In a study of population mortality, you have the following data:
Age
Lives at this age
Lives leaving the study
at the next age
Deaths before
the next age
5
10
20
50
65
10,000
9,450
8,645
6,827
525
780
1768
1400
25
25
50
70
All lives leaving the study are assumed to leave the study at the end of the age interval.
Mortality is estimated using the actuarial estimator. Linear interpolation is used to estimate mortality
between interval endpoints.
Determine the estimate of the probability that a life age 5 survives to age 45, or 40 p̂5 .
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28.11.
28. MORTALITY TABLE CONSTRUCTION
You are given the following information about time to remission for a cohort of 70 cancer patients:
Weeks
(lower–upper)
Number Who
Experienced Remission
Number Who
Withdrew from Study
0–1
1–2
2–3
3–4
3
6
12
8
2
10
2
6
To estimate time to remission, you use the actuarial estimator with an assumption of uniform withdrawals within each week.
Determine the resulting estimate for the proportion experiencing remission within 3 weeks.
28.12.
You are given the following data from a mortality study on ages 40–41:
Age
Entrants
Beginning
of Year
Entrants
During
Year
Withdrawals
During
Year
Withdrawals
End
of Year
Deaths
40
41
125
150
10
8
5
10
25
20
1
2
Estimate q41 using the exact exposure method.
28.13.
From a study on lives ages 35–37, you are given:
Age
Lives
Deaths
Withdrawals
35
36
37
1000
1100
1000
3
4
5
100
99
100
You use the actuarial estimator with the assumption of uniform withdrawals between integral ages.
0
Estimate the 3-year single-decrement mortality rate for a person age 35, 3 q̂35
.
28.14.
A three-year mortality study starts out with 800 lives age 40. You are given the following statistics:
Year
New entries
Withdrawals
Deaths
1
2
3
40
38
35
80
90
85
2
3
4
You use the actuarial estimator with the assumption of uniform entries and withdrawals within each year.
0
Estimate the 3-year mortality rate for a person age 40, 3 q̂40
.
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EXERCISES FOR LESSON 28
28.15. A study of agent persistency for their first four years under contract starts with 60 agents. Statistics
for this population are:
(i)
(ii)
(iii)
(iv)
15 agents persist for four years.
2 agents die at times 1.5 and 3.5.
5 agents become disabled at times 0.3, 1.3, 2.5, 3.3, and 3.9.
The other 38 agents withdraw from their contract, with 22 withdrawing in the first year, 10 in the
second year, and 3 apiece in the third and fourth years.
The actuarial estimator is used. All decrements are assumed to be uniformly distributed within each
contract year.
Determine the single-decrement probability of an agent persisting for four years in a population of agents
not subject to death or disability.
28.16. An insurance policy is available at deductibles of 0 and 500, and at maximum covered losses of 5000
and 10,000. The underlying loss distribution is assumed to be independent of deductible and maximum
covered losses.
You have the following data on loss sizes. Losses at 5000 and 10000 shown in the table are censored
observations at the limits.
Deductible
Loss Range
0
500
(0, 500]
(500, 1000]
(1000, 5000]
(5000, 10000]
At 5000
At 10,000
35
20
25
10
8
2
15
18
9
7
1
Total
100
50
Using the actuarial estimator, estimate the probability that loss size will be no more than 10,000.
28.17. An insurance policy is available with deductibles of 250 and 500, and may or may not have a maximum covered losses of 1000. The underlying loss distribution is assumed to be independent of deductible
and maximum covered losses.
You have the following data on loss sizes. Losses at 1000 are censored observations at the limit.
Deductible
Loss Range
250
500
(250, 500]
(500, 1000]
(1000, 5000]
(5000, 10000]
At 1000
8
12
14
10
6
15
18
17
0
Total
50
50
Using the actuarial estimator, estimate the probability that a loss upon which a positive payment is made
from a policy with a 500 deductible will exceed 5000.
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28.18.
28. MORTALITY TABLE CONSTRUCTION
You have the following information from a study starting with 100 policies on insureds age 20:
j
(c j , c j +1 ]
dj
uj
xj
0
1
2
(20, 40]
(40, 60]
(60, 80]
4
10
5
24
20
8
3
5
6
Superscripts (d ) and (w ) indicate decrements due to death and withdrawal respectively.
0(d )
60 q20
is estimated from this study using the actuarial estimator, with all new entries assumed to occur
at the beginning of each interval and withdrawals assumed to occur at the end of each interval.
0(w )
For another population assumed to have the same mortality, 60 q20 = 0.6.
(τ)
Estimate 60 p20 for that population.
28.19.
[4-F01:27] You are given the following information about a group of 10 claims:
Claim Size
Interval
(0, 15000]
(15000, 30000]
(30000, 45000]
Number of Claims
in Interval
1
1
4
Number of Claims
Censored in Interval
2
2
0
Assume that claim sizes and censorship points are uniformly distributed within each interval.
Estimate, using the actuarial estimator, the probability that a claim exceeds 30,000.
(A) 0.67
(B) 0.70
(C) 0.74
(D) 0.77
(E) 0.80
28.20. [C-S05:7] Loss data for 925 policies with deductibles of 300 and 500 and maximum covered losses
of 5,000 and 10,000 were collected. The results are given below:
Range
(300, 500]
(500, 1,000]
(1,000, 5,000)
(5,000, 10,000)
At 5,000
At 10,000
Total
Deductible
300
500
50
—
50
75
150
150
100
200
40
80
10
20
400
525
Total
50
125
300
300
120
30
925
Let X be the random variable for the ground-up loss size.
Using the actuarial estimator, estimate F (5000 | X > 300).
(A) 0.25
(B) 0.32
(C) 0.40
(D) 0.51
(E) 0.55
Additional released exam questions: C-F06:17, C-S07:26
Solutions
28.1. The insuring birthday is 9/1/1984 and the person withdraws on 8/1/2011, so there is 0 exposure at
age 27.
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EXERCISE SOLUTIONS FOR LESSON 28
28.2. For the Kaplan-Meier product limit estimator, the risk set on 10/1/2012 is 3 since all 3 policies are in
force and age 40, and ties for surrender count. The first policy, however is age 41. Thus q̂40 = 1/3.
For the actuarial estimator, age 40 exposure in 2012 is 5 months for the first policy, 9 months for the
second, 10 months for the third since deaths count up to the next birthday, and 7 months for the fourth.
Total exposure is 31 months, and q̂40 = 12/31.
The absolute difference is 12/31 − 1/3 = 0.053763 .
28.3. For the first life, insured birthday is 12/1/1962 and age 48 is attained at 12/1/2010, so there are 6
months of exposure.
For the second life, insured birthday is 6/1/1963 and age 48 is attained at 6/1/2011, so there are 4 months
of exposure.
For the third life, insured birthday is 10/1/1963 and age 48 is attained at 10/1/2011, so there are 12
months of exposure.
For the fourth life, insured birthday is 1/1/1964 and age 48 is attained at 1/1/2012. Since it is an
anniversary-to-anniversary study, the study ends at the anniversary in 2012 and there is no exposure.
Total exposure is 6 + 4 + 12 = 22 months. The exact exposure estimator is 1 − e −12/22 = 0.420422 .
28.4. Insuring birthdays are 1/1/1968, 6/1/1967, 6/1/1968, 10/1/1967, and 2/1/1970 for the five people
respectively. Actuarial exposure at age 40 is 12 months, 11 months, 12 months, 0 months, and 0 months
respectively. The actuarial estimator is q̂40 = 1/(35/12) = 0.3429 .
28.5. For individual data, actuarial exposure at age 33 is 12 months, 12 months, 10 months, 6 months, and
4 months respectively, for a toal of 44 months. The actuarial estimate is q̂33 = 1/(44/12) = 3/11.
For grouped data, the population P33 is 2, since the first two people are in the study at exact age 33. There
are 3 new entrants (the other 3 people). The fourth person withdraws in the middle of the year. Exposure is
2 + 0.5(3 − 1) = 3. The actuarial estimate is q̂33 = 1/3.
The absolute difference is |3/11 − 1/3| = 2/33 .
28.6. Exact exposure at age 60 for the first group of 25 is 4 months apiece, except for the one who withdrew
for whom it is 2 months. For the group of 5 individuals, there are 12 months of exposure, including from the
one who died, except for the one who withdrew who provides 6 months of exposure. For the third group of 1
individual, exact exposure is 3 months. The total exposure in months is 24(4)+2+4(12)+6+3 = 155. There is
one death at age 60, namely the individual from the third group. The death from the first group is at age 61.
The estimate of q60 is q̂60 = 1 − e −12/155 = 0.074498. The variance is
1
2
Var(q̂60 ) = (1 − 0.074498)
= 0.00513397
(155/12)2
28.7. Exposure for age 42 is 23 years from those who persisted or died from the first group, plus 2(2/3)
for the two who withdrew from the first group. Exposure for age 42 is 19(11/12) for those who persisted or
died from the second group, plus 1/2 from the one who withdrew from the second group. Total exposure is
23 + 4/3 + 19(11/12) + 1/2 = 42.25. Then q̂42 = 2/42.25. The estimated variance is
d
V
ar(q̂ j ) =
q̂ j (1 − q̂ j )
ej
=
(2/42.25)(1 − 2/42.25)
= 0.00106737
42.25
28.8. Actuarial exposure for the 1/1 policies is 36 months apiece.
Actuarial exposure for the 3/1 policies is 34 months apiece, except for the one who dies who contributes
24 months of exposure.
Actuarial exposure for the 8/1 policies is 29 months apiece, except for those who withdraw who contribute 0 months, 8 months, and 27 months respectively.
Actuarial exposure for the 12/1 policies is 25 months apiece.
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28. MORTALITY TABLE CONSTRUCTION
Total exposure in months is 24(36) + 41(34) + 24 + 12(29) + 8 + 27 + 22(25) = 3215. The estimate of 3q 25 is
1/(3215/12) = 0.0037325. The standard deviation is
v
q
t (0.0037325)(1 − 0.0037325)
d
V
ar(3 q̂25 ) =
= 0.0064528
3215/36
28.9. Exposure is:
(32-9,37-9,s )
(33-2,36-7,s )
(33-2,35-3,x )
(33-5,38-5,x )
(33-7,35-1,s )
36 months
34 months
25 months
31 months
18 months
When computing exposure, treat the ages as exact, so that someone entering who is exactly 33 years and
2 months old will have 10 more months until age 34 and then another 24 months, for a total of 34 months.
The other lines can be computed in a similar way. Total exposure is 36 + 34 + 25 + 31 + 18 = 144 months or 12
years. Then 3 q̂33 = 1 − e −3/12) = 0.221199. The variance of the estimator is
d
V
ar(3 q̂33 ) = (1 − 0.221199)2 (32 )
1
= 0.037908
122
28.10. Since there are no new entrants (the question does not mention them and the lives at each age are
the previous age’s lives minus deaths and withdrawals), e j = Pj , so
10000 − 25
= 0.9975
10000
9450 − 25
= 0.994861
15 p̂5 = 0.9975
9450
8645 − 50
= 0.989107
45 p̂5 = 0.994861
8645
5 p̂5
=
Using linear interpolation,
5
1
5
1
(0.989107) +
(0.994861) = 0.990066
40 p̂5 =
45 p̂5 +
15 p̂5 =
6
6
6
6
28.11. There are no new entrants. We are told to assume uniform withdrawals, or e j = Pj − 0.5u j in the
absence of new entrants.
Week
j
Lives
Pj
Remissions
dj
Withdrawals
wj
Exposure
ej
0
j +1 p̂0
0
1
2
70
65
49
3
6
12
2
10
2
69
60
48
66/69 = 0.956522
0.956522(54/60) = 0.860870
0.860870(36/48) = 0.645652
Since we want the proportion experiencing remission, the answer is the complement of 3 p̂00 , or 3 q̂00 =
1 − 0.645652 = 0.354348 .
28.12. P41 = 125 + 10 − 5 − 25 − 1 + 150 = 254. We assume uniform entry and withdrawal, so exact exposure
is 254 + 0.5(8 − 10 − 20 − 2) = 242. The mortality estimate is 1 − e −2/242 = 0.0082304 .
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EXERCISE SOLUTIONS FOR LESSON 28
28.13.
The exposures for the three ages are then 950, 1050.5, and 950. The survival rate for three years is

‹
‹
‹
3
4
5
0
p̂
=
1
−
1
−
1
−
= (0.9968)(0.9962)(0.9947) = 0.9878
3 35
950
1050.5
950
0
The three-year mortality rate is 3 q̂35
= 1 − 0.9878 = 0.0122 .
28.14. Note that all 800 lives begin at age 40, not half of them. Since we assume uniform entries and withdrawals, e j = Pj + 0.5(n j − w j ). Since there are no entries at the beginning of ages after 40, population is
calculated recursively as Pj +1 = Pj + n j − w j − d j . The resulting calculation is shown in the following table:
j
Pj
nj
wj
dj
ej
0
1
2
800
758
703
40
38
35
80
90
85
2
3
4
780
732
678
0
0
Therefore 3 p̂40
= (778/780)(729/732)(674/678) = 0.9875 and 3 q̂40
= 1 − 0.9875 = 0.0125 .
28.15. Withdrawal is the decrement of interest, while death and disability are censored observations. The
risk sets are e j = Pj +0.5(n j −w j ). There are no new entrants. There is 1 censored observation in the first year
(the disability at time 0.30), 2 in the second year (death at 1.5, disability at 1.3), 1 in the third year (disability
at 2.5), and 3 in the fourth year (death at 3.5, disability at 3.3 and 3.9). The following table computes the risk
sets.
j
Pj
nj
wj
dj
ej
0
1
2
3
60
37
25
21
0
0
0
0
1
2
1
3
22
10
3
3
59.5
36.0
24.5
19.5
0
j +1 p0
37.5/59.5 = 0.630252
0.630252(26/36) = 0.455182
0.455182(21.5/24.5) = 0.399445
0.399445(16.5/19.5) = 0.337992
The probability of persisting four years in the absence of other decrements is 0.337992 .
28.16. We estimate S (10,000). For the interval (0, 500], the risk set is the 100 policies with deductible 0 and
there are 35 losses, so Ŝ (500) = 0.65. We can group together the two intervals (500, 1000] and (1000, 5000],
since there is no censoring or new entrants between 500 and 5000. In the interval (500, 5000], the risk set is
115 (all losses except the 35 observations in (0, 500]) and
the number of losses in this combined interval is
20 + 25 + 15 + 18 = 78, so Ŝ (5000) = 0.65 (115 − 78)/115 = 0.2091. In the interval (5000, 10000], the risk set is,
adding up the count of observed losses in the interval plus the 2+1 observations censored at the 10,000 limit,
10 + 9 + 2 + 1 = 22 (or if you prefer, start with the population of 37 at 5000 and remove the 15 observations
censored at 5000). There are 19 losses in the interval (5000, 10000], so Ŝ (10,000) = 0.2091 (22 − 19)/22 =
0.02852. The following table summarizes our calculations:
cj
0
500
5000
10000
Pj − n j
0
65
22
nj
wj
dj
ej
Ŝ (c j )
100
50
0
0
15
3
35
78
19
100
115
22
0.65
0.2091
0.02852
The probability that loss size will be no more than 10,000 is therefore 1 − 0.02852 = 0.97148 .
28.17. Even though we are estimating the probability for a 500 deductible, all information from the study
for both deductibles must be used. We are assuming that the underlying loss distribution is independent of
deductible, so the information from policies with a 250 deductible is relevant to our estimate.
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28. MORTALITY TABLE CONSTRUCTION
Since we are assuming a 500 deductible, we just have to calculate a conditional survival function, the
condition being that the loss is over 500. So we begin the estimation procedure with the interval (500, 1000].
The only logical assumption for the risk set is that all entries and withdrawals occur at endpoints.
j
cj
Pj
nj
wj
dj
ej
0
1
500
1000
92
59
0
0
6
0
27
32
92
59
In the above table, the population at 500 is all losses except for the 8 in the range (250, 500]. The population at 1000 is 92 − 6 − 27 = 59. Based on this table, the probability that a loss is greater than 5000 given
that it is greater than 500 is (92 − 27)/92 (59 − 32)/59 = 0.3233 .
28.18.
The mortality estimate is obtained by setting e j = Pj + n j . The following table shows the results.
j
(c j , c j +1 ]
0
1
2
(20, 40]
(40, 60]
(60, 80]
Pj − n j
100
77
62
nj
wj
dj
4
10
5
24
20
8
3
5
6
ej
0(d )
20( j +1) p̂20
104
87
67
101/104 = 0.971154
0.971154(82/87) = 0.915340
0.915340(61/67) = 0.833370
0(d )
So we have 60 p̂20 = 0.833370. In the other population,
(τ)
60 p20
= (0.833370)(1 − 0.6) = 0.3333
28.19. In each of the first two intervals, because of uniform distribution, half the censored claims are censored below the claim in the interval. The risk set in the first interval is
10 − 0.5(2) = 9, and in the
therefore
second interval, 10 − 1 − 2 − 0.5(2) = 6. Then the actuarial estimate is 89 65 = 0.7407 . (C)
28.20. The risk set at 500 is the 400 policies with deductible 300, so Ŝ (500) = (400 − 50)/400 = 0.875. The
risk set between 500 and 5000 are the 875 remaining policies (925 minus the 50 with claims below 500), and
125 + 300 = 425 claims are between 500 and 5000 (no censoring occurs below 5000, so we can combine the
2 intervals), so Ŝ (5000) = 0.875 (875 − 425)/875 = 0.45. Thus Fˆ (5000 | X > 300) = 1 − 0.45 = 0.55 . (E)
Quiz Solutions
28-1. The first one has a full 12 months in the study at age 30. the second one turns 30 on 2/1/2011 and
stays in the study for 6 months, providing 6 months of exposure. The third one purchases a policy at age 30
and 5 months and dies during the year, so exposure is until the next birthday, or 7 months. Total exposure
for all three in months is 12 + 6 + 7 = 25 .
28-2. Notice that d j is used in its individual-data sense to mean the time of a new entry.
There are 5 entering lives and 1 leaving life, so the risk set is 0.5(5− 1) = 2, the number of deaths is 1, and
the mortality probability is 0.5 .
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Lesson 61
Simulation—Special Techniques
Reading: Loss Models Fourth Edition 20.2
In this lesson we discuss some situations in which the inversion method is not applied directly to the simulated variable. We also discuss another simulation method: the polar method.
61.1
Mixtures
It is usually difficult or impossible to invert a mixed distribution. It is easier to instead first select the component and then select the value of the component using a pair of uniform random variables to perform
the simulation. The first component of the pair is used to select the component of the mixture; the second
component is used to simulate the distribution of that component.
EXAMPLE 61A A random variable is a mixture of an inverse Pareto distribution, with weight 0.7, and a Pareto
distribution, with weight 0.3. The parameters of the inverse Pareto distribution are θ = 1000, τ = 2. The
parameters of the Pareto distribution are α = 3, θ = 800. You are to simulate three observations of this
mixture using the following three pairs of uniform random numbers on [0, 1):
(0.324,0.871)
(0.555,0.666)
(0.932,0.488)
Determine the mean of the three resulting numbers.
ANSWER: This mixture is a discrete mixture. Therefore, the selection of the component is simulated as a
discrete random variable with probability 0.7 of having the first distribution and 0.3 of having the second
distribution. You need to know which distribution is first and which is second, and an exam will have some
way of telling you. Here, we’ll use the order in which the two distributions appear in the question (and you
should generally use this order unless told otherwise). So if the first random number of a pair is less than 0.7
we select an inverse Pareto distribution and otherwise we select a Pareto distribution. We select an inverse
Pareto distribution for the first two pairs and a Pareto distribution for the third pair.
The tables state that for an inverse Pareto distribution, VaRp (X ) = θ (p −1/τ − 1)−1 . Using the second component of the first two pairs,
0.871 → 1000(0.871−1/2 − 1)−1 = 13,986.62
0.666 → 1000(0.666−1/2 − 1)−1 = 4,437.39
−1
The tables tell us that for a Pareto distribution, VaRp (Z ) = θ (1 − p )−1/α − 1 , so
0.488 → 800(0.512−1/3 − 1)−1 = 3200
The average is (13,986.62 + 4,437.39 + 3200)/3 = 7208 .
ƒ
Although the textbook discusses this method only for discrete mixtures, it may be used for continuous
mixtures as well.
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61.2
Multiple decrements
If you are simulating the number of deaths on a group of insurance policies, you are simulating a binomial
random variable with m being the size of the group and q the probability of death. However, if other decrements1 are possible, you are simulating a multinomial random variable. If there are k possibilities, you will
need k uniform random numbers. You must specify the order in which the decrements are to be simulated.
If there are n members in the group and the probabilities of the k decrements are p1 , p2 ,. . . pk , then the first
decrement is simulated as a binomial random variable with parameters n and p1 . Each successive decrement is simulated as a binomial random variable. The first parameter of the binomial is equal to the first
parameter for the previous decrement reduced by the outcome of the previous decrement. The second parameter of the binomial is the probability p j divided by the complement of the sum of the probabilities of

P j −1 Š
the previous decrements, or p j 1 − i =1 pi .
EXAMPLE 61B For a group of 500 policyholders, the following may occur in the next year:
1. Accidental death with probability 0.005.
2. Death by other causes with probability 0.008.
3. Surrender with probability 0.03.
Use the following uniform random numbers to simulate the number of policyholders who leave the
group in each way in the next year:
0.94
0.52
0.01
Determine the simulated number who leave the group in each way.
ANSWER: For accidental death, the binomial parameters are m = 500 and q = 0.005. Then
p0 = 0.995500 = 0.081572
p1 = 500(0.995499 )(0.005) = 0.204954
500
p2 =
(0.995498 )(0.0052 ) = 0.256965
2
500
p3 =
(0.995497 )(0.0053 ) = 0.214353
3
500
p4 =
(0.995496 )(0.0054 ) = 0.133836
4
500
p5 =
(0.995495 )(0.0055 ) = 0.066716
5
The sum of the first five probabilities is 0.891681 and the sum of the first six probabilities is 0.958397, so 5
accidental deaths occur.
1A
decrement is a cause for leaving the group, such as death, surrender, disability.
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For death by other causes, the binomial parameters are m = 500−5 = 495 and q = 0.008/0.995 = 0.008040.
p0 = (1 − q )495 = 0.018389
p1 = 495(1 − q )494 q = 0.073781
495
p2 =
(1 − q )493 q 2 = 0.147711
2
495
p3 =
(1 − q )492 q 3 = 0.196749
3
495
p4 =
(1 − q )491 q 4 = 0.196151
4
The sum of the first four probabilities is 0.4366 and the sum of the first five probabilities is 0.6328, so 4
deaths by other causes occur.
For surrenders, the binomial parameters are 495−4 = 491 and q = 0.03/(1−0.005−0.008) = 0.030395. We
P6
P7
find that j =0 p j = 0.007335 and j =0 p j = 0.017520, so there are 7 surrenders.
ƒ
The computation of the binomial random variable when m is large can be laborious. In Section 61.3, we
will discuss an alternative method which, however, also requires many calculations when generating larger
random numbers. For now, let’s discuss using the normal approximation to simulate a binomial random
variable.
To use the normal approximation, calculate the mean µ = mq and the variance σ2 = mq (1 − q ) of the
binomial random variable being simulated, simulate a standard normal random variable z u , and then set
xi = µ + z u σ. Round the result xi to the nearest integer.
Let’s redo Example 61B with the normal approximation. We will use the inversion method to generate
normal random numbers.
1. For accidental deaths, mean is 500(0.005) = 2.5 and variance is 2.5(0.995) = 2.4875. Invert 0.94, which
is halfway in between 0.9394 and 0.9406, so using the SOA rounding method for the normal table, we
−1
get z u = Φp
(0.94) = 1.55 or 1.56, but it doesn’t matter whether we use 1.55 or 1.56. Using xi = µ + z u σ,
2.5 + 1.56 2.4875 = 4.96 which rounds to 5, and using 1.55 would also get a number rounding to 5, so
there are 5 accidental deaths.
2. For other deaths, the mean is 495(0.008/0.995) = 3.97990 and the variance is 3.97990(1−0.008/0.995) =
−1
3.94790.
p Since Φ(0.05) = 0.5199, it follows that Φ (0.52) = 0.05. The normal approximation is 3.97990+
0.05 3.94790 = 4.079, which rounds to 4.
3. For surrenders, the mean is 491(0.03/0.987) = 14.92401 and the variance is 1.92401(1
p − 0.03/0.987) =
4.47039. The inverse of 0.01, by SOA rounding rules, is −2.33. Then 14.92401 − 2.33 14.47039 = 6.06
which rounds to 6. This is different from the exact binomial random number which is 7.
Sometimes in a multiple-decrement situation you wish to simulate both the year in which the decrement
happened and the decrement that happened. This can be done with one random number, rather than one
random number for the year and a second one for the decrement, by ordering all the outcomes and treating
the outcome as a single discrete random variable.
EXAMPLE 61C A person buys a 3-year insurance policy. During the 3 years, the policyholder may die, surrender, or convert the policies. The probabilities of these outcomes are
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Probability that newly
issued policy leaves due to
Year
Death
Surrender
Conversion
1
2
3
0.01
0.02
0.03
0.15
0.08
0.07
0.04
0.05
0.06
Set up a simulation rule so that both the year of decrement and the type of decrement are simulated with
one random number.
ANSWER: We order the 9 decrement outcomes, plus the probability of no decrement, as follows:
Number
Decrement
Probability
Cumulative Probability
1
2
3
4
5
6
7
8
9
10
Maintain policy for 3 years
Die first year
Surrender first year
Convert first year
Die second year
Surrender second year
Convert second year
Die third year
Surrender third year
Convert third year
0.49
0.01
0.15
0.04
0.02
0.08
0.05
0.03
0.07
0.06
0.49
0.50
0.65
0.69
0.71
0.79
0.84
0.87
0.94
1.00
Then if a uniform random number is at least equal to a number in the last column of this table and less
than the following one, the simulated outcome is the one on the line of the number it is at least equal to. For
example, 0.66 would give surrender in the first year.
ƒ
On an exam, they would have to tell you which order to put the outcomes in. As an alternative, the
outcomes could be given monetary value, and then you would order the outcomes in the order of increasing
monetary value.
EXAMPLE 61D For a 2-year insurance policy, if the policyholder dies in the first two years, the company pays
1000 at the end of the year. If the policyholder surrenders in the first year, the company pays 100 at the end
of the year. If the policyholder surrenders at the end of the second year, the company pays 200 at the end of
the year.
The present value of the company’s payment is computed using i = 0.05.
The probability of death in the first year is 0.01 and the probability of surrender is 0.09. If the policyholder
survives one year, the probability of death in the second year is 0.02 and the probability of surrender is 0.08.
Use the following four uniform random numbers on [0, 1) to simulate the company’s payment:
0.08
0.82
0.99
0.77
Determine the average simulated present value of the company’s payment.
ANSWER: The present values for death are 1000/1.05 = 952.38 for death in the first year and 1000/1.052 =
907.03 for death in the second year. The present values for surrender are 100/1.05 = 95.24 for surrender
in the first year and 200/1.052 = 181.41 for surrender in the second year. Thus the order of the outcomes,
from lowest to highest, is persisting (not dying or surrendering) for 2 years, surrendering in the first year,
surrendering in the second year, dying in the second year, dying in the first year. The probability of persisting
for 1 year is 1 − 0.09 − 0.01 = 0.9, and the probability of persisting for two years is (0.9)(1 − 0.8 − 0.2) = 0.81.
The following table lists the probabilities of all outcomes:
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61.3. SIMULATING (a , b , 0) DISTRIBUTIONS
Decrement
Probability
Cumulative Probability
Persisting for 2 years
Surrendering in first year
Surrendering in second year
Dying in second year
Dying in first year
(0.9)(0.9) = 0.81
0.09
(0.9)(0.08) = 0.072
(0.9)(0.02) = 0.018
0.01
0.81
0.90
0.972
0.990
1.00
There is no payment for the uniform random numbers 0.08 and 0.77. The uniform random number 0.82
goes to surrender in the first year. The uniform random number 0.99 goes to dying in the first year, since
when a number is on the border of an interval it goes to the higher outcome. The average simulated value
ƒ
is (0 + 95.24 + 952.38 + 0)/4 = 261.90 .
61.3
Simulating (a , b , 0) distributions
A random variable from an (a , b , 0) distribution may be simulated like any other discrete random variable by
setting up a table of the cumulative distribution function. For Poisson and negative binomial distributions,
which have an infinite number of possible values, one would compute the cumulative distribution function
up to the highest uniform random number. One could then store the table and look up values as needed.
An alternative is to model the random variable as a stochastic process. A stochastic process generates
events. Depending on how the stochastic process is defined, the number of events in a unit of time may
be an (a , b , 0) random variable. The amount of time between events is exponential. Exponential random
variables can be simulated using the inversion method.
This is the procedure for generating a random variable from the (a , b , 0) class using stochastic processes:
1. Start at time t = 0 with k = 0.
2. Determine the rate at which the next event occurs, λk . λk will vary based on the distribution. We will
discuss how to determine λk later.
3. The mean amount of time until the next event occurs is 1/λk . Generate an exponential random variable with mean 1/λk . This variable is sk = − ln(1 − u k ) λk .
4. Add sk to t ; t = t + sk . This is the amount of time that has passed so far.
5. If t ≤ 1, set k = k + 1 and generate another event.
6. If t > 1, the generated random number is k .
Now we discuss the calculation of λk for each distribution in the (a , b , 0) class.
Poisson
For a Poisson distribution λk = λ for all k . This means that you sum up − ln(1 − u k ) λ until it exceeds 1,
and then the Poisson random number is 1 less than the number of numbers that were summed up. For
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61. SIMULATION—SPECIAL TECHNIQUES
computational purposes, this can be simplified as follows:
n
X
ln(1 − u k )
−
>1
λ
k =0
−
n
X
k =0
− ln
ln(1 − u k ) > λ
n
Y
k =0
n
Y
k =0
(1 − u k ) > λ
(1 − u k ) < e −λ
In other words, keep multiplying 1− u k until the product is less than e −λ , and then the result is one less than
the number of numbers in the product.
EXAMPLE 61E You are to generate a Poisson random variable with mean 2.5 using a stochastic process. Use
the following uniform random numbers on [0, 1):
0.82
0.28
0.52
0.12
0.97
Determine the resulting random number.
ANSWER: e −2.5 = 0.082085. Multiplying the complements of the uniform random numbers:
(0.18)(0.72) = 0.1296
(0.1296)(0.48) = 0.0622 < 0.082085
It took 3 numbers to get below 0.082085, so the Poisson random number is 2 .
ƒ
The shortcut of turning the sum into a product is not available for non-Poisson distributions.
Binomial
For a binomial distribution, let d = ln(1 − q ) and c = −md . Then λk = c + d k .
EXAMPLE 61F You are to generate a binomial random variable with m = 250, q = 0.01 using a stochastic
process. Use the following uniform random numbers on [0, 1):
0.82
0.28
0.52
0.12
0.97
Determine the resulting random number.
ANSWER: d = ln(1 − 0.01) = −0.010050 and c = 2.51258.
λ0 = 2.51258
λ1 = 2.51258 − 0.010050 = 2.50253
λ2 = 2.50253 − 0.010050 = 2.49248
ln 0.18
= 0.682484
2.51258
ln 0.72
s1 = −
= 0.131269
2.50253
ln 0.48
s2 = −
= 0.294473
2.49248
s0 = −
t = 0.682484
t = 0.682484 + 0.131269 = 0.813753
t = 0.813753 + 0.294473 = 1.108226
Exactly two events occur before time 1, so the binomial random variable is 2 .
ƒ
If the stochastic process generates m events before time 1, then the binomial random number is m . No more
events can be generated, since λm = 0.
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61.4. NORMAL RANDOM VARIABLES: THE POLAR METHOD
Negative binomial
For a negative binomial distribution, let d = ln(1 + β ) and c = r d . Then λk = c + d k .
EXAMPLE 61G You are to generate a negative binomial random variable with r = 5, β = 0.5 using a stochastic
process. Use the following uniform random numbers on [0, 1):
0.82
0.28
0.52
0.12
0.97
Determine the resulting random number.
ANSWER: d = ln(1 + 0.5) = 0.405465 and c = 5(0.405465) = 2.027325
λ0 = 2.027325
λ1 = 2.027325 + 0.405465 = 2.432790
λ2 = 2.432790 + 0.405465 = 2.838255
ln 0.18
= 0.845843
2.027325
ln 0.72
= 0.135032
s1 = −
2.432790
ln 0.48
s2 = −
= 0.258599
2.838255
s0 = −
t = 0.845843
t = 0.845843 + 0.135032 = 0.980874
t = 0.980874 + 0.258599 = 1.239473
Exactly two events occur before time 1, so the negative binomial random variable is 2 .
ƒ
In all the (a , b , 0) cases, the stochastic process is only used to generate a random number. There is no
implication that these stochastic processes are actually occurring. Even if the actual situation you are simulating is a stochastic process, it is not necessarily true that the times of the events generated by the simulation
are appropriate for the actual stochastic process. The actual stochastic process may be a different process,
yet generate a binomial/negative binomial/Poisson number of events in a given amount of time.
?
Quiz 61-1 Simulate a Bernoulli random variable with q = 0.2 using a stochastic process. Use the following
uniform random numbers on [0, 1), one for each simulation:
0.34
0.14
Is the result different from the straightforward use of the inversion method?
61.4
Normal random variables: the polar method
To simulate a normal random variable X with mean µ and variance σ2 , we simulate a standard normal random variable ni and then set xi = µ + σxi . To simulate a lognormal random variable with parameters µ and
σ, we simulate a normal random variable with parameters µ and σ and then exponentiate it. So simulating
normal and lognormal random variables boils down to simulating standard normal random variables.
The inversion method may be used to simulate standard normal random variables. This is because there
are polynomial approximations to the inverse of the cumulative distribution function of the standard normal distribution. However, these functions do not work as well at the tails of the distribution as at central values, so extreme values are underrepresented. An alternative method for generating standard normal
random variables is the Box-Muller transformation. This method generates two standard normal random
variables
from two uniform
p random variables on [0, 1), u 1 and u 2 . The two standard normal numbers are
p
−2 ln u 1 cos 2πu 2 and −2 ln u 1 sin 2πu 2 .
EXAMPLE 61H You are simulating a random variable having a normal distribution with µ = 5 and σ2 = 16.
Use the following two uniform random numbers on [0, 1):
0.73
0.18
and the Box-Muller transformation to generate two normal random numbers.
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61. SIMULATION—SPECIAL TECHNIQUES
ANSWER: First,
0.71785 .
p
−2 ln 0.73 = 0.79336. Then 0.79336 cos 2π(0.18) = 0.33780 and 0.79336 sin 2π(0.18) =
ƒ
p
Notice that −2 ln u i is like a radius and 2πu 2 is like an angle, so you are representing a point in polar
coordinates. However, computing sine and cosine requires a polynomial approximation. A more efficient
way to carry out this method is the polar method. In this method, you randomly select a point from the circle
of radius 1 centered at the origin. The way you do this is you select a random point from the square of side 2
centered at the origin. If the random point is not in the circle of radius 1 centered at the origin, you discard
the point and try again. In other words:
1. Generate two uniform random numbers on [0, 1].
2. Transform them into uniform random numbers on [−1, 1]. If u 1 and u 2 are the original uniform random
numbers, then the transformed random numbers are v1 = 2u 1 − 1 and v2 = 2u 2 − 1.
3. Calculate S = v12 + v22 .
4. If S > 1, discard the pair of uniform random numbers and start again from the first step.
p
After you get a usable pair, then compute T = −(2 lnS )/S . The two standard normal random numbers are
v1 T and v2 T .
Don’t forget to transform the uniform random numbers you start off with into uniform random numbers on [−1, 1], by multiplying each one by 2 and subtracting 1.
EXAMPLE 61I You are simulating a random variable having a normal distribution with µ = 5 and σ2 = 16.
Use the following two uniform random numbers on [0, 1):
0.73
0.18
and the polar method to generate two normal random numbers.
ANSWER: v1 = 2(0.73) − 1 = 0.46 and v2 = 2(0.18) − 1 = −0.64. Then S = 0.462 + 0.642 = 0.6212 ≤ 1, so we
can use this pair of numbers. (Otherwise we’d need to be provided with more uniform numbers.) Then T =
p
−(2 ln 0.6212)/0.6212 = 1.23808. The two standard normal random numbers are 0.46(1.23808) = 0.56952
ƒ
and −0.64(1.23808) = −0.79237 .
Exercises
61.1. A random variable is a mixture of two 2-parameter Pareto distributions. The first component has a
weight of 0.6, α = 2, and θ = 100. The second component has a weight of 0.4, α = 3, and θ = 1000. You use
the first uniform random number to determine the component and the second uniform random number to
determine the result. Use the following three pairs of uniform random numbers:
{0.3,0.7}
{0.7,0.3}
{0.5,0.9}
Calculate the mean of the resulting random numbers.
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EXERCISES FOR LESSON 61
Table 61.1: Summary of Special Simulation Techniques
For mixture distributions, use two uniform random numbers; one to select the component of the mixture,
the other to generate a random number from that component.
To generate multinomial random variables with population size n and k categories having probabilities
p1 , p2 , . . . , pk :
1. Generate k uniform random numbers on [0, 1): u 1 , . . . u k .
2. Generate x1 as binomial m = n, q = p1 .

Pi −1 Š
Pi −1
3. Generate xi as binomial with m = n − j =1 x j , q = pi 1 − j =1 p j . Do this for i = 2, . . . , k
A typical situation with a multinomial random variable is an insured population of size n subject to k
decrements.
The binomial distribution may be approximated with a normal distribution if m is large.
To generate both the year of a decrement and the type of a decrement with one uniform random variable,
order all the outcomes.
The (a , b , 0) class may be simulated as a stochastic process. Interevent times are exponential with mean
1/λi , i = 0, 1, . . ., where
• For Poisson, λi = λ
• For binomial, λi = −m ln(1 − q ) + i ln(1 − q )
• For negative binomial, λi = r ln(1 + β ) + i ln(1 + β )
The Box-Muller transformation of p
two uniform random variables
on [0, 1], u 1 and u 2 , results in two
p
standard normal random variables −2 ln u 1 cos 2πu 2 and −2 ln u 1 sin 2πu 2 .
To generate standard normal random variables using the polar method,
1. Generate two uniform random variables on [0, 1], u 1 and u 2 .
2. v1 = 2u 1 − 1 and v2 = 2u 2 − 1.
3. If S = v12 + v22 > 1, discard the pair and start again.
p
4. T = −(2 lnS )/S
5. Random numbers are T v1 and T v2 .
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61. SIMULATION—SPECIAL TECHNIQUES
61.2. Number of claims follows a Poisson distribution with mean λ. λ is uniformly distributed on (0, 0.2).
You simulate number of claims by simulating λ with the first uniform random number and then simulating the Poisson distribution with the second uniform number using the inversion method.
Use the following pair of uniform random numbers on [0, 1): {0.438,0.805}.
Determine the resulting number of claims.
61.3. A portfolio of 100 insurance policies is subject to two decrements, death and surrender. The probability of death for each individual is 0.015 and the probability of surrender is 0.05.
You are to simulate the number of deaths and surrenders, in that order. Use the following uniform random numbers on [0, 1): 0.15 for the number of deaths and 0.25 for the number of surrenders. Use the inversion method to generate binomial random variables.
Determine the resulting number of deaths and surrenders.
61.4. A group of 200 employees is subject to three decrements, death, disability, and termination. The
probabilities of these decrements are 0.01, 0.08, and 0.20.
You are to simulate the three decrements in the order death, disability, and termination. Generate random numbers using the normal approximation and the inversion method. Use the following three uniform
random numbers on [0, 1) in the order specified:
0.8212
0.7088
0.1357
Determine the number of terminations.
61.5. For an insurance policy, the probability of death in the first year is 0.01 and the probability of surrender is 0.10. For policies that are in force at the end of one year, the probability of death in the second year is
0.02 and the probability of surrender is 0.05.
The outcome (year of termination, and whether termination is by death or surrender) is simulated. The
outcomes are ordered as follows:
1.
Policy in force for two years
2.
Death in first year
3.
Death in second year
4.
Surrender in first year
5.
Surrender in second year
Five policies are simulated with the following five uniform random numbers on [0, 1):
0.08
0.92
0.84
0.32
0.65
Determine the number of policies for each of the five outcomes.
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EXERCISES FOR LESSON 61
61.6. [4B-F98:15] (1 point) You wish to generate a single random number from a Poisson distribution with
mean 3. A random number generator produces the following uniform numbers in the unit interval [0, 1):
Position in
Generation Sequence
Random Number
1
2
3
4
5
0.30
0.70
0.30
0.50
0.90
Representing the Poisson random number as the result of a stochastic process, determine the simulated
number.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
61.7. [4B-F94:21] (2 points) You wish to generate single random number from a Poisson distribution with
mean 3. A random number generator produces the following uniform numbers in the unit interval [0, 1):
Position in
Generation Sequence
Random Number
1
2
3
4
5
0.909
0.017
0.373
0.561
0.890
Representing the Poisson random number as the result of a stochastic process, determine the simulated
number.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
61.8. [130-S87:9] The number of flies In a cup of soup has a Poisson distribution with mean 2.5. You are
to simulate z , the number of flies in two cups of soup by treating the Poisson distribution as a stochastic
process. You are to use the following uniform random numbers on [0, 1) in the order given:
First Cup: 0.20 0.90 0.90 0.90
Second Cup: 0.60 0.60 0.90 0.80
Determine z .
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
61.9. [130-S88:1] You are to generate a random observation from the Poisson distribution with mean λ = 3
treating the Poisson distribution as a stochastic process. Use the following sequence of random numbers
from the uniform distribution over [0, 1):
0.36
0.50
0.75
0.25
0.50
0.33 0.95 0.85
What is your random observation?
(A) 0
(B) 2
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(C) 4
(D) 6
(E) 7
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61. SIMULATION—SPECIAL TECHNIQUES
61.10. For a population of 250, you are simulating the number of deaths in a year. The probability of death
is 0.01. You will model the random variable for the number of deaths as a stochastic process. Use the following uniform random numbers on [0, 1) in the order given:
0.67
0.35
0.57
0.22
0.40
Determine the resulting number of deaths.
61.11. You are to generate a random observation from a binomial distribution with m = 4, q = 0.4, using a
stochastic process. Use the following uniform random numbers on [0, 1) in the order given:
0.12
0.67
0.28
0.55
Determine the resulting number.
61.12. You are to generate a random observation from a negative binomial distribution with r = 2, β = 1
using a stochastic process. Use the following uniform random numbers on [0, 1) in the order given:
0.12
0.67
0.28
0.55
0.80
Determine the resulting number.
61.13. You are to generate a random observation from a negative binomial distribution with r = 20, β =
1.5 using a stochastic process. After generating 10 uniform numbers, the cumulative sum of the generated
numbers of the process is 0.989027.
Determine the greatest lower bound for the eleventh uniform number such that the negative binomial
random number that is generated is 10.
61.14. [4B-S90:51] (2 points) Simulate two observations, x1 and x2 from a normal distribution with mean 0
and standard deviation 1 using the polar method. Use the two random numbers, 0.30 and 0.65, taken from
the uniform distribution on [0, 1).
Determine x1 and x2 .
(A) −1.33 and 1.00
(B) −0.67 and 0.50
(E) Cannot be determined
(C) −2.67 and 2.00
(D) −0.65 and 0.48
61.15. [4B-S92:12] (2 points) You are given the following ordered pairs of points from the uniform distribution on [0, 1):
(0.111, 0.888);
(0.652, 0.689);
(0.194, 0.923)
Using the polar method, determine the first two simulated standard normal values, X (1) and X (2).
(A) (−0.243, 0.242)
(B) (0.083, 0.661)
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(C) (0.374, 0.313)
(D) (1.066, 1.326)
(E) (1.599, −1.357)
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EXERCISES FOR LESSON 61
61.16. [4B-F95:26, 4B-F98:22] (3 points) Y and Z are a pair of independent, identically distributed lognormal random variables with parameters µ = 5 and σ = 3.
Use the polar method to determine the first simulated pair of observations of Y and Z , denoted (y1 , z 1 )
from the following list of pairs of random numbers generated from the uniform distribution over [0, 1):
0.20, 0.95
0.60, 0.85
0.30, 0.15
Calculate y1 + z 1 .
(A)
(B)
(C)
(D)
(E)
Less than 10
At least 10, but less than 100
At least 100, but less than 1,000
At least 1,000, but less than 10,000
At least 10,000
61.17. [130-F87:3] You are to use the polar method to generate a pair of independent normal random variables with mean 0 and standard deviation 1. You are to use the following uniform random numbers on [0, 1)
in the order given:
0.82
p
p
Let Q = − ln 0.36 and R = − ln 1.082.
0.09
0.50
0.20
Calculate the mean of the two generated normal variables.
(A) −0.7Q
(B) −0.1R
(C) 0.1R
(D) 1.0R
(E) 0.7Q
61.18. [130-F88:7] You are to generate a pair of independent normal variables, X (l ) and X (2), with mean 0
and standard deviation 1, using the polar method.
You are to first generate a random point, (R , θ ), inside the unit circle as follows:
R 2 = U1
θ = 2πU2
where R and θ represent the random point in polar coordinates, and each Ui represents a uniform random
variable on [0, 1).
You are given
1
e
U2 = 0.500
U1 =
Determine X (1) and X (2).
(A)
(B)
(C)
(D)
(E)
−1.4, 0.0
−0.8, 0.0
−0.6, 0.0
−0.4, 0.0
0.0, 0.6
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61. SIMULATION—SPECIAL TECHNIQUES
61.19. [130-S89:4] You are to generate an ordered pair (X 1 , X 2 ) of independent standard normal random
variables using the polar method. Let (V1 , V2 ) be the pair of independent random variables, each uniformly
distributed on [−1, 1], used to generate (X 1 , X 2 ). If V1 = e −1 cos 30◦ and V2 = e −1 sin 30◦ , determine the value
of X 2 .
p
(E) e
(A) e −2
(B) e −1
(C) 1
(D) 2
Solutions
61.1. According to the tables, the inversion formula for a Pareto is θ (1 − u)−1/α − 1 . For the first number
and third numbers, we use α = 2 and θ = 100, and for the second, α = 3 and θ = 1000.
0.7 → 100 0.3−1/2 − 1 = 82.57419
0.3 → 1000 0.7−1/3 − 1 = 126.24788
0.9 → 100 0.1−1/2 − 1 = 216.22777
The average of the three numbers is 141.68 .
61.2. We generate λ = 0.2(0.438) = 0.0876. Then p0 = e −0.0876 = 0.91613, and 0.805 < 0.91613, so there are
0 claims.
61.3. For deaths, the probability of no deaths is 0.985100 = 0.2206, and 0.15 < 0.2206, so the number of
deaths is 0 .
For surrenders, we use the probability q = 0.05/(1 − 0.015) = 0.050761
p0 = (1 − q )100 = 0.005464
100
p1 =
(1 − q )99 q = 0.029221
1
100
p2 =
(1 − q )98 q 2 = 0.07735
2
100
p3 =
(1 − q )97 q 3 = 0.13512
3
100
p4 =
(1 − q )96 q 4 = 0.17522
4
The sum up to p3 is 0.2472 and the sum up to p4 is 0.4224, so 3 surrenders occur.
61.4. For deaths, the mean is 2 and the variance is 2(0.99)p= 1.98. The uniform number goes to the standard
normal random number 0.92, which then goes to 2 + 0.92 1.98 = 3.29, which rounds to 3.
For disability, the binomial parameters are m = 200 − 3 = 197 and q = 0.08/(1 − 0.01). The mean is
m q = 15.9192 and the variance is mq (1 − q ) = 14.6328. The uniform number
p goes to the standard normal
random number 0.55. The resulting number of disabilities is 15.9192 + 0.55 14.6328 = 18.02, which rounds
to 18.
For terminations, the binomial parameters are m = 197−18 = 179 and q = 0.2/(1−0.01−0.08) = 0.219780.
The mean is mq = 39.34066 and the variance is mq (1 − q ) = 30.69436. The uniform number goes to −1.1.
p
The resulting number of terminations is 39.34066 − 1.1 30.69436 = 33.25, which rounds to 33 .
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EXERCISE SOLUTIONS FOR LESSON 61
61.5. The probabilities of the five outcomes are (0.89)(0.93) = 0.8277, 0.01, (0.89)(0.02) = 0.0178, 0.10, and
(0.89)(0.05) = 0.0445. Accumulating, we get 0.8277, 0.8377, 0.8555, 0.9555, 1 respectively. Thus 0.08, 0.32, and
0.65 go to policy in force for two years, 0.92 goes to surrender in first year, and 0.84 goes to death in second
year. Thus the numbers for each outcome are 3,0,1,1,0 respectively.
61.6. We generate exponential random numbers with mean 1/3, and sum them up until they’re greater
than 1. So we want
X
1
ln(1 − u i ) > 1
−
3
X
−
ln(1 − u i ) > 3
Y
(1 − u i ) < e −3 = 0.049787
The product of the first two numbers is (0.7)(0.3) = 0.21. The product of the first three numbers is
(0.21)(0.7) = 0.147. The product of the first four numbers is 0.147(0.5) = 0.0735. The product of the first
five numbers is (0.0735)(0.1) = 0.00735 < 0.049787. It took 5 events to go past time 1, so the random number
is 4 . (E)
61.7. We multiply (1 − u i ) until the product is below e −3 = 0.049787.
(1 − 0.909)(1 − 0.017) = 0.089453
(0.089453)(1 − 0.373) = 0.056087
(0.056087)(1 − 0.561) = 0.024622
It took 4 numbers, so the answer is 3 . (C)
61.8. e −2.5 = 0.082. Multiplying the first cup’s numbers, (0.8)(0.1) = 0.08 < 0.082, so we have 1 fly. For the
second cup, 0.4(0.4) = 0.16 and (0.16)(0.1) = 0.016, so we have 2 flies. The total is 3 . (C)
61.9.
e −3 = 0.0498.
(0.36)(0.5) = 0.18
(0.18)(0.25) = 0.045
It took 3 numbers to get below e −3 , so we generate 2 . (B)
61.10. ln(1−q ) = ln 0.99 = −0.010050. The first λ is −250(−0.010050) = 2.512584, with resulting exponential
random number −(ln 0.33)/2.512584 = 0.441244. The next numbers are
ln 0.65
= 0.172139
2.502534
ln 0.43
−
= 0.338606
2.492484
ln 0.78
−
= 0.100088
2.842434
−
These 4 numbers add up to 1.052. It took 4 numbers to get the sum above 1, so there are 3 deaths.
61.11. ln(1 − q ) = ln 0.6 = −0.51083. The first λ is −4(−0.51083) = 2.04330. The resulting exponential random number is −(ln 0.88)/2.04330 = 0.062562. The next numbers are
ln 0.33
= 0.72345
1.53248
ln 0.72
−
= 0.32154
1.02165
−
These 3 numbers add up to 1.108. Thus the binomial random number is 2 .
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61.12. ln(1 + β ) = ln 2 = 0.6931472. The first λ is 2(0.6931472) = 1.38630. The resulting exponential random
number is −(ln 0.88)/1.38630 = 0.092212. The next numbers are
ln 0.33
2.07944
ln 0.72
2.77259
ln 0.45
3.46574
ln 0.20
4.15888
= 0.533154
= 0.118483
= 0.230401
= 0.386988
These 5 numbers add up to 1.361238. Since 5 numbers are required to exceed 1, the negative binomial
number is 4 .
61.13. ln(1+β ) = ln 2.5 = 0.9162907. The 11th hazard rate is λ10 = 20(0.9162907)+10(0.9162907) = 27.48872.
We want
ln(1 − u)
−
= 1 − 0.989027 = 0.010973
27.48872
Therefore, u > 1 − e −27.48872(0.010973) = 0.260391 .
61.14.
Translate the numbers to uniform on [−1, 1):
0.3 → 2(0.3) − 1 = −0.4
0.65 → 2(0.65) − 1 = 0.3
p
p
Then S = 0.42 + 0.32 = 0.25 and −(2 lnS )/S = −(2 ln 0.25)/0.25 = 3.3302. The two standard normal numbers are −0.4(3.3302) = −1.3321 and 0.3(3.3302) = 0.9991 . (A)
61.15.
The first pair gives
0.111 → 2(0.111) − 1 = −0.778
0.888 → 2(0.888) − 1 = 0.776
(−0.778)2 + (0.776)2 = 1.207 > 1
so that pair is discarded. The second pair gives
0.652 → 2(0.652) − 1 = 0.304
0.689 → 2(0.689) − 1 = 0.378
0.3042 + 0.3782 = 0.2353
v
t −2 ln 0.2353
= 3.5069
0.2353
The resulting standard normal numbers are 0.304(3.5069) = 1.066 and 0.378(3.5069) = 1.326 . (D)
61.16. For the first pair, 0.20 → 2(0.20) − 1 = −0.6 and 0.95 → 2(0.95) − 1 = 0.9, and 0.62 + 0.92 > 1, so we
2
2
discard
pthis pair. For the second pair, 0.60 → 2(0.60)−1 = 0.2 and 0.85 → 2(0.85)−1 = 0.7, and 0.2 +0.7 = 0.53.
Then −(2 ln 0.53)/0.53 = 1.547827. The standard normal random numbers generated are
0.2(1.547827) = 0.309565
0.7(1.547827) = 1.083479
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Then we multiply by σ = 3, add µ = 5, and exponentiate to obtain the desired lognormal random variables.
y1 = e 3(0.309565)+5 = 376
z 1 = e 3(1.083479)+5 = 3829
y1 + z 1 = 376 + 3829 = 4205
(D)
61.17. 0.82 → 2(0.82) − 1 = 0.64 and 0.09 → 2(0.09) − 1 = −0.82. Then 0.642 + 0.822 = 1.082 > 1, so we discard
that pair.
0.50 → 2(0.50) − 1 = 0 and 0.20 → 2(0.20) − 1 = −0.6. Then
S=
p
02 + 0.62 = 0.36
p
−(2 ln 0.36)/0.36 = 2/0.36Q
The mean of the two generated normal numbers is
v
0 − 0.6 t 2
Q = = −0.707Q
2
0.36
(A)
61.18.
Usually we’re given the point in Cartesian coordinates. Here,
p
p
x1 = R cos θ = 1/e cos π = − 1/e
p
x2 = R sin θ = 1/e sin π = 0
p
p
p
p
p
So S = 1/e and −(2 lnS )/S = 2e . The numbers are X (1) = − 1/e 2e = − 2 = −1.414 and X (2) = 0 .
(A)
p
p
p
p
You could also use the Box-Muller transformation directly.
−2 lnU1 = −2 ln 1/e = (−2)(−1) = 2,
p
and cos 2πU2 = cos π = −1, sin 2πU2 = sin π = 0. So we get {− 2, 0}, or answer (A).
p
p
61.19. S = V12 + V22 = e −2 . So −(2 lnS )/S = 4e 2 = 2e . Then
p
−1
◦
−1 1
X 2 = e sin 30 2e = e
(2e ) = 1
(C)
2
Quiz Solutions
61-1. d = ln(1 − q ) = ln 0.8; c = − ln 0.8; λ0 = − ln 0.8. Then
ln(1 − 0.34)
>1
− ln 0.8
ln(1 − 0.14)
−
<1
− ln 0.8
−
So the first simulation generates 0 since no events occur before time 1, and the second simulation generates 1 since one event occurs before time 1. We see that if the random number is 0.2 or higher, then the
numerator of the exponential random variable will be the logarithm of a number less than or equal to 0.8,
whose absolute value will be greater than or equal to that of the denominator, and if the random number
is less than 0.2, then the numerator of the exponential random variable will be the logarithm of a number
greater than 0.8, whose absolute value will be less than that of the denominator. In the first case, the ratio
is greater than 1 so 0 is generated, and in the second case one event occurs before time 1 so 1 is generated.
This is the opposite of the inversion method, which generates 0 for uniform numbers less than 0.8 and 1 for
uniform numbers greater than or equal to 0.8.
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61. SIMULATION—SPECIAL TECHNIQUES