Exercises in Life Insurance Mathematics
Edited by
Bjarne Mess
Jakob Christensen
University of Copenhagen
Laboratory of Actuarial Mathematics
Introduction
This collection of exercises in life insurance mathematics replaces the collection of
Steen Pedersen and all other exercises and problems in any text or article in the
FM0L curriculum.
The following abbreviations are being used for the contributors of exercises:
AM
BM
BS
FW
JC
JH
MSC
MS
RN
SH
SK
SP
SW
Bowers et al. “Actuarial Mathematics”,
Society of Actuaries, Itasca, Il 1986
Bjarne Mess
Bo Søndergaard
Flemming Windfeld
Jakob Christensen
Jan Hoem “Elementær rentelære”,
Universitetsforlaget, Oslo, 1971.
Michael Schou Christensen
Mogens Steffensen
Ragnar Norberg
Svend Haastrup
Stephen G. Kellison “The theory of Interest”,
Richard D. Erwing, Inc., Homewood, Il 1970
Steen Pedersen “Opgaver i livsforsikringsmatematik”
Schwartz “Numerical Analysis”, Wiley, 1989
Copenhagen, August 2, 1997
Bjarne Mess
Jakob Christensen
4
Exercises in Life Insurance Mathematics
Exercises
1.
FM0 S91, 1
FM0 S93, 2
Interest
FM0 S92, 1
FM0 S94, 1
FM0 S93, 1
Exercise 1.1 Show that an | < an| < än| when i > 0.
(JH(1), 1971)
Exercise 1.2 Show that an | /n decreases as n increases and i > 0.
(JH(2), 1971)
Exercise 1.3 Show that an | (i1 )/an| (i2 ) decreases as n increases if i1 > i2 .
(JH(3), 1971)
Exercise 1.4 A man needs approximately $2.500 and raises in that connection a
loan in a bank. The principal, which is to be fully repaid after 6 months, is $2.600.
From this amount the bank deducts the future interest $84,50 and other fees of $5,70,
so that the bank pays out the man $2.509,80 cash. The interest rate of the bank is
6.5%.
(a) What is the effective interest rate p. a. for the bank?
(b) What is the effective interest rate p. a. for the borrower?
(JH(4 rev.), 1971)
Exercise 1.5 One day a company receives an american loan offer: Principal of
$5.000.000, rate of course 99% and nominal interest rate 6.5%. The loan is free of
installments for 5 years and is after that to be amortized over 15 years: Interests
and installments are due annually. Assume that the dollar-rate of exchange decreases
exponentially from DKK 6,75 at the initial time to DKK 4,75 at the end of the loan.
How can one determine the effective interest rate of the loan?
Interest
5
(SP(19))
Exercise 1.6 By calculation of the interest rate for a fraction of a year, a bank will
usually calculate with linear payment of interest instead of exponential payment of
interest. If the interest rate is i p. a. and we have to calculate interest for a period
of time α (0 < α < 1), the bank will calculate interest as αi per kr. 1. – in capital,
instead of calculating the interest as
(1 + i)α − 1
per. kr. 1. – in capital. Is this for the benefit of the borrower?
(JH(7), 1971)
Exercise 1.7 A man is going to buy new furniture on an installment plan. In the
hire-purchase agreement he finds the following account:
Cash payment for the furniture
− In advance to the salesman
= Net balance
+ Installment fee for 18 mths.
= Net balance for installment
The monthly installments are
$6.306,00
$2.217, 00
$4.089,00
$501, 00
$4.590, 00
$255, 00
What effective interest rate p. a.
(JH(10), 1971)
Exercise 1.8 A man has been promised some money. He can choose from two
alternatives for the payment.
Under alternative (i) A5 = $4.495 and A9 = $5.548 are paid out after 5 and 9 years
respectively.
Under alternative (ii) B7 = $10.000 is paid out after 10 years.
Denote the market interest rate by i.
For which value (values) of i is (i) just as good as (ii), and when is (i) more profitable
for the man? What if A9 = $5.562, i. e. $14 more?
(JH(11), 1971)
6
Exercises in Life Insurance Mathematics
Exercise 1.9 A says to B: “I would like to borrow $208 in one year from today.
In return for your kindness I will pay $100 cash now, and $108,15 in two years from
today by the end of the loan.”
What is the effective interest rate p. a. for B if he accepts this?
(JH(12), 1971)
Exercise 1.10 Consider a usual annuity loan with principal H, interest rate i and
n installments. Show that the installment which falls due in period t is
Ft =
i(1 + i)t−1
H,
(1 + i)n − 1
and find an expression for the remaining debt immediately after this period.
(SP(6))
Exercise 1.11 Consider a linearly increasing annuity. At time t = 1, 2, . . . the
amount t is being paid. The present value of this cash flow is denoted by Ian | .
(a) Show that
Ian| =
n−1
X
t| an−t| ,
t=0
and interpret this equation intuitively.
(b) Give an explicit expression for Ian | .
(c) What does the symbol Ia n| mean? Give expressions corresponding to the ones
from (a) and (b).
(SP(10))
Exercise 1.12 A person has a table of annual annutities with different interest rates
and durations to his disposal. However, he needs some present values of half-year
annuities. These annuities do all have the same interest rate and the corresponding
whole-year annuities can be found in the table.
What is the easiest way to find the desired annuities?
(SP(11))
Interest
7
Exercise 1.13 A debtor is going to pay an amount of 1 some time in the future.
He does not know this point of time in advance; he only knows that it is a stochastic variable T with a known distribution. Now consider the expected present value
denoted by
Aδ(T ) = E(e−δT ).
(a) Show that the variance of the present value is given by
δ
2
Var(e−δT ) = A2δ
(T ) − (A(T ) ) .
Now assume that creditor has to pay a continuous T -year annuity. The present value
is aδT | . Let the expected present value be denoted by
aδ(T ) = E(aδT | ).
(b) Show that
Aδ(T ) = 1 − δaδ(T )
and interpret the equation.
(c) Give an expression for the variance Var(a δT | ) corresponding to the one from (a).
(d) Show that
Aδ(T ) > v ET ,
and find a similar inequality for aδ(T ) . (Hint: Use Jensen’s inequality.)
(e) Find at least two situations where these considerations are relevant.
(SP(16))
Exercise 1.14 Consider a loan, principal H, nominal interest rate i 1 , rate of course
k and installment Ft in period t, where t = 1, . . . , N . Show that the effective interest
rate ie satisfies
1−S
i1
ie =
k−S
where
N
1 X
S=
Ft vet .
H t=1
(SP(20))
Exercise 1.15 A loan with principal H and fixed interest rate i 1 has to be amortized
annually over a period of N years. The borrower can each year deduct half the interest
8
Exercises in Life Insurance Mathematics
expenses on his tax declaration. Construct the installment plan in a way so that the
amount of amortisation minus deductible (actual net payment) will be the same in
all periods (assume tax is payable by the end of each year). Find the annual net
installment.
(RN “Opgaver til FM0 (rentelære)” 18.05.93)
Exercise 1.16 Consider a general loan. Show that for t ≥ 1 we have
At = (1 + i1 )Rt−1 − Rt
Rt =
n−t
X
At+j v1j
(1.1)
j=1
v1t Rt = H −
t
X
Aj v1j
(1.2)
j=1
with the use of standard notation. Formula (1.1) is called the prospective formula for
the remaining debt. Formula (1.2) is called the retrospective formula for the remaining
debt.
(JH(20), 1971)
Exercise 1.17 A loan is being repaid by 15 annual payments. The first five installments are $400 each, the next five $300 each, and the final five are $200 each. Find
expressions for the remaining debt immediately after the second $300 installment –
(a) prospectively,
(b) retrospectively.
(SK(1) p. 122, 1970)
Exercise 1.18 A loan of $1.000 is being repaid with annual installments for 20 years
at effective interest of 5% . Show that the amount of interest in the 11th installment
is
50
.
1 + v 10
(SK(10) p. 123, 1970)
Exercise 1.19 A borrower has mortgage which calls for level annual payments of 1 at
the end of each year for 20 years. At the time of the seventh regular payment he also
Interest
9
makes an additional payment equal to the amount of principal that according to the
original amortisation schedule would have been repaid by the eighth regular payment.
If payments of 1 continue to be made at the end of the eighth and succeding years
until the mortgage is fully repaid, show that the amount saved in interest payments
over the full term of the mortgage is
1 − v 13 .
(SK(16) p. 124, 1970)
Exercise 1.20 A man has some money invested at an effective interest rate i. At
the end of the first year he withdraws 162.5% of the interest earned, at the end of
the second year he withdraws 325% of the interest earned, and so forth with the
withdrawal factor increasing in arithmetic progression. At the end of 16 years the
fund exhausted. Find i.
(SK(40) p. 127, 1970)
Exercise 1.21 A loan of a25 | is being repaid with continuous payments at the annual
rate of 1 p. a. for 25 years. If the interest rate i is 0.05, find the total amount of interest
paid during the 6th through the 10th years inclusive.
(SK(42) p. 127, 1970)
Exercise 1.22 After having made six payments of $100 each on a $1.000 loan at
4% effective, the borrower decides to repay the balance of the loan over the next five
years by equal annual principal payments in addition to the annual interest due on
the unpaid balance. If the lender insists on a yield rate of 5% over this five-year
period, find the total payment, principal plus interest, for the ninth year.
(SK(45) p. 127, 1970)
Exercise 1.23 A student has heard of a bank that offers a study loan of L = 10.000
kr. The rate of interest is 3% p. a. and the student applicates for the loan on the
following conditions:
(i) The first m = 5 years he will only pay an interest of 300 kr. per year.
(ii) After that period of time he will pay interests and installments of 900 kr. per year
until the loan is fully amortized (the last installment may be reduced).
(a) For how long N does he have to pay installments and how big is the last amount
of amortisation?
10
Exercises in Life Insurance Mathematics
(b) Which amount αt has to be paid at time t if the loan (with interest earned) is to
be fully paid back at time t (t = 1, 2, . . . , N )?
(Aktuarembetseksamen, Oslo 1960)
Aggregate Mortality
11
2.
Aggregate Mortality
Exercise 2.1 Let T be a stochastic variable with distribution function F . Assume F
is concentrated on the interval [a, b] and that F is continuous with continuous density
f . Assume F (t) < 1 for t ∈ [a, b). Define
µ(t) =
f (t)
, t ∈ [a, b).
1 − F (t)
We say that µ is the intensity of F .
(a) Show that
Rb
a
µ(t)dt = ∞.
(b) Can we conclude that µ(t) → ∞ for t → b − ?
Let T be the life length of a newly born. Let a = 0 and b = ω where ω is the maximum
life length.
(c) Show that µ is the force of mortality.
(HRH(1))
Exercise 2.2 Use the decrement tables of G82M to find the following probabilities:
(a) The probability that a 1 year old person dies after his 50th year, but before his
60th year.
(b) The probability that a 30 year old dies within the next 37 years.
(c) The probability that two persons now 26 and 31 years old, and whose remaining
life times are assumed to be stochastically independent, both are alive in 12 years.
(SP(28))
Exercise 2.3 Explain why each of the following functions cannot serve in the role
indicated by the symbol:
µx = (1 + x)−3 , x ≥ 0
F (x) = 1 −
22x 11x2 7x3
+
−
, 0≤x≤3
12
8
24
12
Exercises in Life Insurance Mathematics
f (x) = xn−1 e−x/2 , x ≥ 0, n ≥ 1.
(AM(3.4) p. 77, 1986)
Exercise 2.4 Consider a population, where the distribution functions for a man’s
and a woman’s total life lengths are x q0M and x q0K respectively. Assume that these
K
probabilities are continuous so that the forces of mortality µ M
x and µx are defined.
Let s0 denote the probability that a newly born is a female. Assume moreover that s 0
and the forces of mortality are not being altered during the period of time considered
in this exercise.
(a) Find the distribution function x q0 for the total life time for a person of unknown
sex and find t qx . Find the force of mortality µx for a person of unknown sex.
(b) What is the probability sx that a person aged x is a woman?
K
Using decrement series `M
x and `x for men and women respectively, work out a decrement serie `x for the total population:
K
(c) How should one appropriately choose ` M
0 and `0 ?
K
(d) Express ax in terms of aM
x and ax .
(SP(32))
Exercise 2.5 Consider a random survivorship group consisting of two subgroups:
(1) The survivors of 1.600 births.
(2) The survivors of 540 persons joining 10 years later at age 10.
An excerpt from the appropriate mortality table for both subgroups follows:
x
0
10
70
`x
40
39
26
If γ1 and γ2 are the numbers of survivors under the age of 70 out of subgroups (1)
and (2) respectively, estimate a number c such that P (γ 1 + γ2 > c) = 0.05. Assume
the lives are independent.
Aggregate Mortality
13
(AM(3.13) p. 78, 1986)
Exercise 2.6 When considering aggregate mortality the probability that an x-year
old person is going to die between x + s and x + s + t, is denoted by the symbol s|t qx .
(a) Express this probability by the distribution function of the person’s remaining life
time.
(b) Is there a connection between
s|t qx
and t qx ?
(c) Show that
s|t qx =
Z
s+t
u px µx+u du,
s
and interpret this expression.
When t = 1 we write s| qx = s|1 qx .
(d) Show that for integer x and n we have
n| qx
(e) Show that
s|t qx
=
dx+n
.
`x
can be expressed similarly (use the function ` instead of d).
(f) Prove the following identities:
n|m qx
=
n px
n| qx
=
n px
n+m px
=
n px
− n+m px
· qx+n
· m px+n .
(SP(24))
Exercise 2.7 Let e◦x:n| denote the expected future lifetime of (x) between the ages
of x and x + n. Show that
e◦x:n|
=
=
n
Z
tt px µx+t dt + nn px
Z0 n
0
This is called the partial life expectancy.
t px dt.
14
Exercises in Life Insurance Mathematics
(AM(3.14) p. 78, 1986)
Exercise 2.8 The force of mortality µ x is assumed to be
µx = βcx .
Three persons are x, y and z years old respectively. What is the probability of dying
in the order x, y, z?
(Tentamen i försikringsmatematik, Stockholms Högskola 1954)
Exercise 2.9 If F (x) = 1 − x/100, 0 ≤ x ≤ 100, find µ x , F (x), f (x) and P (10 <
X < 40).
(AM(3.5) p. 77, 1986)
Exercise 2.10 If µx = 0.0001 for 20 ≤ x ≤ 25, evaluate
2|2 q20 .
(AM(3.7) p. 77, 1986)
Exercise 2.11 Assume that the force of mortality µ x is Gompertz-Makeham, i. e.
µx = α+βcx . For at certain cause of death, the force of mortality is given by α 1 +β1 cx .
Show that the probability of dying from the above disease for an x-year old is
β1 α1 β − αβ1
ex .
+
β
β
(Tentamen i försikringsmatematik, Stockholms Högskola 1954)
Exercise 2.12 Show that constants a and b can be determined so that
µx = a log(1 − qx ) + b log(1 − qx+1 ).
when µx can be put as a linear function for x < t < x + 2.
(Tentamen i försikringsmatematik, Stockholms Högskola 1954)
Exercise 2.13 Assuming the force of mortality to be Gompertz-Makeham, i. e.
µx = α + βcx , show that for each age x we have
−
log c
log(1 − qx ) < µx < − log(1 − qx ).
c−1
Aggregate Mortality
15
(Tentamen i försikringsmatematik, Stockholms Högskola, 1954)
Exercise 2.14 Given that `x+t is strictly decreasing for t ∈ [0, 1] show that
(a) if `x+t is concave down, then qx > µx ,
(b) if `x+t is concave up, then qx < µx .
(AM, 1986)
Exercise 2.15 Prove the following expressions:
d
d
`x µx < 0, when
µx < µ2x
dx
dx
d
d
`x µx = 0, when
µx = µ2x
dx
dx
d
d
`x µx > 0, when
µx > µ2x .
dx
dx
(AM(3.12) p. 77, 1986)
Exercise 2.16 Show the following identities:
∂ t px
∂t
∂ t px
∂x
= −µx+t · t px
= (µx − µx+t ) · t px
1 =
`x =
Z
Z
ω−x
t px µx+t dt
0
ω−x
`x+t µx+t dt.
0
(SP(26))
Exercise 2.17 If the force of mortality µ x+t , 0 ≤ t ≤ 1, changes to µx+t − c where c
is a positive constant, find the value of c for which the probability of (x) dying within
a year will be halved. Express the answer in terms of q x .
(AM(3.34) p. 80, 1986)
Exercise 2.18 From a standard mortality table, a second table is prepared by
doubling the force of mortality of the standard table. Is the rate of mortality, q x , at
16
Exercises in Life Insurance Mathematics
any given age under the new table, more than double, exactly double or less than
double the mortality rate, qx , of the standard table?
(AM(3.35) p. 80, 1986)
Exercise 2.19 If µx = Bcx , show that the function `x µx has a maximum at age x0 ,
where µx0 = log c. (Hint: Exercise 2.15).
(AM(3.36) p. 80, 1986)
Exercise 2.20 Assume
µx =
Acx
1 + Bcx
for x > 0.
(a) Find the survival function F (x).
(b) Verify that the mode of the distribution of X, the age of death, is given by
x0 =
log(log c) − log A
.
log c
(AM(3.37) p. 80, 1986)
Exercise 2.21. (Interpolation in Life Annuity Tariffs) Consider a table with the
present value ax:u−x| for an integer expiration age u with age at issue x = 0, 1, . . . , u
and futhermore a table of the one year survival probabilities p x for the same ages. Let
t be a real number, 0 ≤ t < 1. We are trying to find a way to determine a x+t:u−x−t|
from the data of the table; this method will, of course, depend on how the mortality
varies with the age.
Assume that the force of mortality is constant on one year age intervals, i. e. µ x+t = µx
for all t with 0 ≤ t < 1.
(a) Find
t−s px+s
in terms of px for 0 ≤ s < t ≤ 1.
(b) Find an expression for ax+t:1−t| .
(c) Show that for every t there exists a λ so that
ax+t:u−x−t| = λax:u−x| + (1 − λ)ax+1:u−x−1| ,
and express λ in terms of the discount rate v, t and p x .
(d) How should one interpolate in a corresponding table for A x:u−x| .
Aggregate Mortality
17
(FM1 exam, summer 1983)
Exercise 2.22 For insurances where the policies are issued on aggravated circumstances, one operates with excess mortality. Let µ x be the force of mortality corresponding to the normal mortality. A person is said to have an excess mortality if his
force of mortality is given by
µ0x = (1 + k)µx .
(a) Show that for all positive k, x and t we have
0
t qx
< (1 + k)t qx .
(b) Show that if there exists a constant ∆ so that µ 0x = µx+∆ is valid for all x then
a0x = ax+∆
for all x; in this case the insurance is issued with an increase of age ∆.
(c) Show that the condition in (b) is fulfilled if the mortality satisfies Gompertz’s law,
i. e. there exist constants β and γ so that µ x = β exp(γx) for all x.
(d) Show, oppositely, that if µx is strictly increasing in x and if there for any k ≥ 0
exists a constant ∆k so that
µx+∆k = (1 + k)µx ,
then the mortality satisfies Gompertz’s law.
(SP(43))
18
Exercises in Life Insurance Mathematics
3.
Insurance of a Single Life
FM0 S92, 3
Exercise 3.1 Prove the identities:
äx = 1 + vpx äx+1
1 − n Ex = äx:n| − ax:n|
dax:n|
= (µx + δ)ax:n| + n Ex − 1.
dx
(SP(29))
Exercise 3.2 Let µx be a weakly increasing function of x and assume that µ x → ∞
as x → ∞.
(a) Show that ax → 0 as x → ∞.
(b) Examine if Ax has a finite limit as x tends to infinity.
(SP(31))
Exercise 3.3 Consider an insurance contract issued to an x-year old. At death
within the first n years, the level continuous premium is paid back with interest and
compound interest earned. Rewrite the integral expression for the present value after
t years (t < h) of the future return of premium per unit of the premium in order to
show that this value is
Dx+h
(st| + ax+t:n−t| ) −
s ,
Dx+t h|
where st| is defined by JH. Interpret the expression.
(Eksamen i Forsikringsvidenskab og Statistik, KU, winter 1943-44)
Exercise 3.4 Show that
n Ex
= 1 − iax:n| − (1 + i)A1x:n| ,
and interpret this formula (i is the interest rate).
(Eksamen i Forsikringsvidenskab og Statistik (rev.), winter 1946-47)
Insurance of a Single Life
19
Exercise 3.5 Assume there exist positive constants k and α, so that
x
`x = k(1 − )α
ω
for all x ∈ [0, ω].
(a) Find an expression for µx .
(b) Find an expression for e◦x (see exercise 2.7).
Now assume that α = 1.
(c) Show that
n| qx
is independent of n.
(d) Show that for n = ω − x we have
ax =
n − an|
.
nδ
(SP(30))
Exercise 3.6 Ax:n| denotes the expected present value of a life insurance contract,
where the amount of 1 is to be paid out by the end of the year in which the insured
dies, not later than n years after the time of issue, or if he survives until the age of
x + n. x is the age at entry.
(a) Give an expression for Ax:n| and show that
Ax:n| = 1 − däx:n| ,
where d is the discount rate.
A1x:n| denotes the expected present value of a life insurance where the amount of 1
is paid out by the end of the year, during which he dies if he dies before the age of
x + n. x is the age at entry.
(b) Give an expression for A1x:n| and show that
A1x:n| = 1 − n Ex − däx:n| .
(c) Try to interpret the formulas in (a) and (b).
(SP(33))
Exercise 3.7 Assume that active persons have force of mortality µ ax as a function
of age and force of disability νx as a function of age. Assume moreover that disabled
20
Exercises in Life Insurance Mathematics
persons have force of mortality µix . There is no recovery. The force of interest is
denoted by δ.
The four quantities defined below are the single net premiums an insured with age
at entry x has to pay for a level continuous annuity with sum 1 p. a. The insurance
cancels n years after issue.
aix:n | The single net premium for an insured who is disabled at the time of issue. The
annuity is payable from issue until the time of death of the insured.
aax:n | Single net premium for active persons. The annuity is payable from the time of
issue until death of the insured.
aaa
x:n | As above except that the premium is due for a contract that cancels by death
or by disability of the insured.
aai
x:n | Single net premium for an active. The annuity is payable if the insured is being
disabled within n years from time of issue. Expires if he dies.
a
i
(a) Express aai
x:n| in terms of µx , νx , µx and δ.
aa
i
(b) Assume that µax = µix for all x. Express aai
x:n| in terms of ax:n| and ax:n| .
(c) Assume that µax = µix + ε and νx = ν where ε and ν are independent of x and
aa
i
ε 6= ν. Express aai
x:n| by ax:n| and ax:n| (and ε and ν).
(Aktuarembetseksamen i Oslo (rev.), fall 1953)
Exercise 3.8 Assume the force of mortality is a strictly increasing function of the
age, when this is greater than or equal to a certain x 0 .
Show that for x ≥ x0 and 0 < n ≤ ∞ the following inequalities hold:
ax:n|
∂ax:n|
∂x
<
1 − v n n px
,
µx + δ
< 0,
ax <
1
.
δ
(SP(37))
Insurance of a Single Life
21
Exercise 3.9 The quantity
e◦x:n| =
Z
n
t px dt
0
is the expected period of insurance for a term insurance or an n-year temporary
annuity, age of entry x and age of expiration x + n.
Define
=
ex:n|
ëx:n|
=
n
X
t=1
n−1
X
t px
t px .
t=0
Give a similar interpretation of these identities. Define
at|
=
ät|
=
1 − vt
i
1 − vt
d
and show that for integer values of t the following inequalities are valid:
ax:n|
< aex:n ||
äx:n|
< äëx:n |.
|
(SP(39))
Exercise 3.10 Show that
ax =
Z
∞
t px Ax+t dt.
0
(SP(41))
Exercise 3.11 Assume that µx is a weakly increasing function of x and consider
for given x two insurance contracts with initial age x: First consider a whole-life life
insurance with sum insured 1 and secondly a whole-life continuous annuity with level
payment intensity determined so that the expected present value of the two insurance
contracts are equal. The one with the biggest variance of the present value is naturally
the one with the biggest risk for the company.
Show that there exists an x0 ≥ 0 so the annuity is more risky than the life insurance
iff x > x0 .
22
Exercises in Life Insurance Mathematics
(SP(42))
Exercise 3.12. (Multiplicative Hazard Model) The mortality in a population varies
from person to person; some has greater or lesser mortality than the average. This
can be modelled as follows:
There exists a underlying force of mortality µ x and for each person a constant θ
independent of age exists so that the force of mortality for a person aged x is θµ x .
The value of θ for a randomly chosen person is assumed to be a realisation of a
stochastic variable Θ, and we assume moreover that EΘ = 1.
Show that in this model the expected present value of a continuous temporary n-year
annuity with payment intensity 1 is greater than or equal to
ax:n| =
Z
n
e−δt t px dt,
0
and examine under which conditions the two present values are equal.
(SP(44))
Exercise 3.13 Consider an n-year endowment insurance, sum insured S, age at
entry x and premium paid continuously during the entire period with level intensity
p. Expenses are disregarded.
(a) What is the surplus of this contract for the company in terms of the remaining
life time of the insured?
(b) Find the mean and variance of the surplus.
(c) Explain how p should be determined so that the probalility of getting a negative
surplus is lesser than a certain ε. (Hint: Tchebychev’s inequality.)
(d) Show (by applying the central limit theorem) how it is possible to obtain a probability of a negative surplus for the entire portfolio lesser than ε by using a smaller p
than the one found in (c).
(SP(50))
We have so far worked with continuous insurance benefits - annuities that are due
continuously and life insurances that are due upon death. The pure endowment seems
to be of another origin, because the time of possible single payment is determined in
advance. In the next exercise we will consider more general kinds of non-continuous
or discrete benefits. For at start consider an x-year old whose remaining life time T
Insurance of a Single Life
23
is determined by the survival function
F (t | x) = e−
R∞
0
µx+τ dτ
,
where µx+t is the force of mortality at the age of x + t, t > 0. As usual let v denote
the annual discount rate.
The results of the following exercise will show that continuous benefits can be concidered as limits for discrete benefits. We will also see that both continuous and discrete
annuities and life insurances are closely related to pure endowment benefits.
Exercise 3.14 The present value of a t-year pure endowment with sum 1 is
Cte = v t 1{T >t} .
(a) Find the expectation, t Ex , of Cne and find Cov(Cse , Cte ) for s 6= t.
A brute-forth generalisation of the pure endowment is produced by summing more
benefits like this. A simple variant is the n-year temporary deferred annuity payable
annually with fixed amounts as long as the insured is alive. This is the sum of n pure
endowments with deferment times 1, . . . , n. The present value at time t = 0 is
Cna(1)
=
n
X
Cte .
t=1
If the annuity is payable h times a year with fixed amounts
be
Cna(h) =
hn
X
1
,
h
the present value will
e
Ct/h
.
t=1
a(h)
(b) Find the expectation, a(h)
.
x:n| , and the variance of the present value C n
An n-year temporary life insurance with sum insured 1, payable at the end of the
year of death, has present value
Cnti(1) =
n
X
v t 1{t−1<T ≤t} ,
t=1
and the corresponding insurance payable at the end of the
occurs, has present value
Cnti(h) =
hn
X
t=1
t
v n 1{ t−1
t .
h <T ≤ h }
1
th
h
year, in which death
24
Exercises in Life Insurance Mathematics
(c) Express the present value Cnti(h) in terms of present values of annuities given by
Cna(h) . Compare with similar expressions for continuous benefits.
(h)
(d) Find the expectation, Ax:n| , and the variance of the present value C nti(h) .
(e) Use the results from (b) and (d) to prove the well known formulas
ax:n|
1
Ax:n|
=
=
n
Z
Z0 n
0
t
v F (t | x)dt =
Z
n
t Ex dt
0
v t F (t | x)µx+t dt
= 1 − δax:n| − n Ex ,
for the expectations of Cna and Cnti and also to find their variances. (Hint: By monom
tone convergence we have Cnα(2 ) % Cnα as m → ∞, α ∈ {a, ti}. Then use the
monotone convergence for the expectation).
(f) Use the technique in (e) to find formulas for the expectations and variances of
continuous benefits in the usual Markov model. Consider an annuity, payable with
level intensity of 1 by staying in state j, and an insurance where a sum of 1 is paid
upon every transition j → k.
(RN “Opgave E7” 29.01.90)
Exercise 3.15 The functions that occur in insurance mathematics often depend on
several variables, e. g. m| ax:n| , and are often hard to tabulate. In order to solve this
problem, we introduce the so-called commutation functions. In connection with life
insurances of one life we consider the following:
Cx
Dx
Mx
Nx
Rx
Sx
=
=
=
=
=
=
v x dx
v x `x
Cx
Dx
Mx
Nx
Rx
Sx
Pω
v ξ dξ
Pξ=x
ω
ξ
ξ=x v `ξ
Pω
(ξ − x)v ξ dξ
Pξ=x
ω
ξ
ξ=x (ξ
− x)v `ξ
=
=
=
=
=
=
R x+1 ξ
v ` µ dξ
Rxx+1 ξ ξ ξ
v ` dξ
Rxω ξ ξ
v
` µ dξ
Rxω ξ ξ ξ
v `ξ dξ
Rxω
(ξ
− x)v ξ `ξ µξ dξ
Rxω
ξ
x
(ξ − x)v `ξ dξ.
(a) Show that
m| ax:n|
=
N x+m − N x+m+n
.
Dx
1
(b) Find corresponding expressions for a x:n| , Ax:n| , Ax , Ax:n| , n Ex , ax:n| and äx:n| .
(c) What can we possibly use R x for?
Insurance of a Single Life
25
(SP(35))
26
Exercises in Life Insurance Mathematics
4.
The Net Premium Reserve and Thiele’s Differential Equation
FM0 S91, 1
FM0 S92, 2
FM0 S93, 2
FM0 S94, 2
FM0 S95, 1
Exercise 4.1 Consider an n-year pure endowment, sum insured S, premium payable
continuously during the insurance period with level intensity π. Upon death two
thirds of the premium reserve is being paid out.
(a) Put up Thiele’s differential equation for V t . What are the boundary conditions?
(b) Find an expression for the premium reserve at time t, t ∈ [0, n)
(c) Determine the premium intensity π by adopting the equivalence principle.
(SH and MSC, 1995)
Exercise 4.2 We have the choice of two different premium payment schemes.
• For an insurance of a single life a level continuous premium is due with force
p as long as the insured is alive at the most for n years from the issue of the
contract.
• Every year an annual premium of the size
p(1) = p · a1|
is paid in advance. If the insured dies during the insurance period the return of
premium is
a |
R = p(1) · θ = p · aθ| ,
a1|
where θ denotes the remaining part of the year at time of death.
(a) Show that these two premium payment schemes are equivalent in the manner
that regardless of when the insured is going to die, the present values of the premium
payments under the two schemes are equal.
(b) Find the expected present value of the return of premium at the time of issue.
Let the prospective reserves at time t from the time of issue of the two premium
schemes be denoted by V t and Vt .
The Net Premium Reserve and Thiele’s Differential Equation
(c) Show that for t ≤ n
27
Vt = V t + p · a[t]−t |
where [t] is the integer part of t.
(SP(53))
Exercise 4.3 Consider a linearly increasing n-year term insurance. If the insured
dies at time t after the time of issue where t < n, the amount tS is paid out. If the
insured is alive at time n, the amount nS is paid out at this time. The age of the
insured at entry is x. The net premium determined by the equivalence principle, is
due continuously with level intensity p.
(a) Find an expression for p.
(b) Find a prospective and a retrospective expression for the reserve V t at any time
t, 0 < t < n.
(c) Show that the two expressions found in (b) are equal for all t, 0 ≤ t < n.
(d) Derive Thieles differential equation.
(e) Find the savings premium and the risk premium.
(SP(54))
Exercise 4.4 An n-year insurance contract has been issued to a person (x). The
premium is composed of a single premium π 0 at the beginning of the contract and
by a continuous intensity (πt )t∈(0,n) as long as (x) is alive, at the most for n years.
The benefits are a pure endowment, sum insured S n at time n, a term insurance, sum
insured St at time t ∈ (0, n), and a continuous flow with intensity (s t )t∈(0,n) as long
as (x) is alive during the insurance period.
(a) Put up Thiele’s differential equation.
(b) Find a boundary condition without assuming the equivalence principle.
(c) Find a prospective expression for the premium reserve by solving the differential
equation.
(d) Adopt the equivalence principle and find an alternative boundary condition.
(e) Find, by applying the new boundary condition a retrospective expression for the
premium reserve.
28
Exercises in Life Insurance Mathematics
Now assume that the benefits moreover consist of a pure endowment, sum insured S
at time m (0 < m < n).
(f) Has this altered Thiele’s differential equation?
(g) Which extra boundary condition are now to be added in order to solve the differential equation?
Assume that π0 = 0, πt = π for t ∈ (0, m) and that π is determined by the equivalence
principle.
(h) Find the net premium.
(i) Find the premium reserve at any time.
(SH “Opgave til 14/10-94” (rev.))
Exercise 4.5. (Prospective Widow Pension in Discrete Time) Consider a policy
with widow pension, insurance period n years, issued to a man aged x and his wife
aged y. During the insurance period the premium Π falls due annually in advance
as long as both are alive. If the man dies, the benefit is a widow pension of sum 1
paid out on every following anniversary of the policy during the insurance period if
the widow is still alive. All expenses are disregarded.
(a) Put up an expression for the premium reserve for this policy at its tth anniversary.
(b) Show how the premium reserve at any time t can be expressed in terms of the
reserve at time t + 1 for t = 0, 1, . . . , n − 1 so that the premium reserves can be
calculated recursively.
(c) Define the savings premium and the risk premium and find an interpretable expression for the latter.
(FM1 exam, summer 1977)
Exercise 4.6 Consider an n-year term insurance, sum insured S, age at entry x.
Continuous premium during the entire period with level intensity p determined by
the equivalence principle.
(a) Find Thieles differential equation for the premium reserve V t .
(b) Which initial conditions would be natural to use for t = 0 and t = n − respectively?
(c) Solve the differential equation with each of the initial conditions and compare the
solutions.
The Net Premium Reserve and Thiele’s Differential Equation
29
(SP(55))
Exercise 4.7 If the insured dies before time n (from the time of issue) the benefit is
a continuous annuity with force 1, duration m from the time of death. If he is alive
at time n, the benefit is a similar annuity from this time and if he is still alive at time
n + m he receives a whole-life annuity with intensity 1. There is a single net premium
at the time of issue and the equivalence principle is adopted.
(a) Find the single net premium.
(b) Find the prospective premium reserve at any time.
(c) Derive Thieles differential equation.
Let Rc denote the risk sum at time t.
(d) Prove that
Rt =

m

 (1 − v )ax+t:n−t| − n+m−t| ax+t
−n+m−t| ax+t

 −a
x+t
(t < n)
(n ≤ t < n + m)
(n + m ≤ t).
In particular we have
and since
R0 = (1 − v m )ax:n| − n+m| ax ,
lim R0 = (1 − v m ax ) > 0,
n→∞
we have R0 > 0 if only n is big enough. Assume that R 0 > 0. Because Rt is a strictly
increasing continuous function on [0, n] and R n < 0, there exists a unique τ ∈ (0, n)
so that Rτ = 0.
(e) Show that this τ is determined by
N x+τ = N x+n +
N x+n+m
.
1 − vm
(SP(56))
Exercise 4.8 Consider a pension insurance contract, where the benefit is an n-year
annuity of 1 deferred m years (expected present value m|n ax ). Premium is paid with
level intensity c during the deferment period (expected present value ca x:m| ).
(a) What is the equivalence premium c and the development of the reserve when
x = 30, m = 30, n = 20 and the technical basis is G82M, i. e. i = 0.045 and µ x =
0.0005 + 10−4.12+0.038x .
30
Exercises in Life Insurance Mathematics
(b) Do similar calculations as in (a) for an extended contract where k times the
premium reserve is being paid out by possible death during the deferment period,
k = 0.5, k = 1.
(RN “Opgave til FM0” 15.10.93)
Exercise 4.9 Consider a single-life status (x) with force of mortality µ x . Define the
premium reserve by Vt = E(U[t,∞) | T > t) as usual.
(a) Show that the premium reserve always is right continuous.
(b) Discuss under which conditions it is left continuous.
Consider the following benefits at time t
• st dt1{T >t} , annuities
• St 1{T ∈dt} , life insurances
• Bt 1{T >t} , pure endowment,
and show that if the premium is being paid with level intensity π and there isno lump
sum at time t then Vt is left continuous.
(c) If there is a lump sum at time t, what does V t − Vt− look like?
NB: Assume that µx+t , st , St are continuous, and assume that there only exist a finite
number of t’s where Bt 6= 0.
(SH “Opgave til 13/10-95”)
Exercise 4.10 Consider an insurance of a single life, age at entry x. At time t
(t = 0, 1, 2, . . .) the premium Pt is being paid if the insured is still alive, and if he dies
during [t − 1, t) then St is the benefit. The equivalence principle is adopted for the
insurance and all expenses are disregarded. Let the premium reserve at time t be V t
and let the stochastic variable Gt be given by
Gt =


 0
V + P − vS
t
t
t+1

 V + P − vV
t
t
t+1
(the insured is dead at time t− )
(the insured dies during the interval [t, t + 1))
(the insured is alive at time t + 1).
Let the stochastic variable Y be the present value at time 0 of the company’s surplus
of the insurance.
The Net Premium Reserve and Thiele’s Differential Equation
(a) Interpret Gt and show that
Y =
∞
X
31
v t Gt .
t=0
(b) Show that
VarY =
∞
X
Var(v t Gt ).
t=0
G1 , G2 , . . . are not necessarily stochastically independent, but note that (b) is valid
regardless of whether G0 , G1 , . . . are stochastically independent or not.
(c) Show that Hattendorf’s Formula
VarY =
∞
X
t=0
t px v
2t+2
px+t qx+t (St+1 − Vt+1 )2
is valid.
(SP(58))
Exercise 4.11 In this exercise we are to study an endowment insurance with return
of premium paid at death before expiration. Consider a person of age x who wishes
to buy an insurance with age of expiration x + n, and where the premium is paid
continuously with level intensity p as long as he is alive during the insurance period.
The lump sum S = 1 is paid if he is alive at age x + n and if he dies before that
the premium paid so far will be returned with interest (basic interest i) earned. We
disregard expenses.
(a) Show that the variable payment at death is given by
Bt = pat| (1 + i)t .
(b) Determine the continuous premium intensity p.
(c) Find the net premium reserve Vt at time t, t ∈ [0, n).
(d) Put up Thiele’s differential equation and determine the risk sum.
(e) Comment on the results and evaluate whether or not you will recommend the
insurance company to issue this kind of insurance.
(FM1 exam, summer 1985)
32
Exercises in Life Insurance Mathematics
Exercise 4.12 An n-year deferred whole-life annuity on the longest lasting life has
been issued to two persons (x) and (y). Annual amount of 1 and level continuous
premium on the longest lasting life with intensity π during the deferment period. The
forces of mortality are denoted by µ x and νy
Put up a retrospective expression for the net reserve at time t assuming that both
are alive.
(Eksamen i Forsikringsvidenskab og Statistik (rev.), winter 1944-45)
Exercise 4.13 A family annuity is an insurance contract of one life that assures
payment of a continuous annual annuity from the possible death of the insured during
the insurance period until the expiration of the contract after n years. Force of
mortality µx , interest rate i, age at entry x.
(a) Put up the formulas for the net premium reserves, prospectively and retrospectively, with level continuous premium payment π. Show that if the insured does not
die during the insurance period the reserve will at least once become negative.
(b) Discuss how the total premium reserve of this contract for a portfolio of identical
contract issued at the same time will develop during the insurance period. Show that
this reserve never becomes negative.
(The students of 1946 had 10 hours to complete this exercise!)
(Aktuarembetseksamen (rev.), Oslo fall 1946)
Exercise 4.14 During construction of a technical basis with mortality of death and
mortality of survival it is a problem that the premium for an insurance can depend
on whether the insurance stands alone or it is combined with other insurances.
This exercise describes the attempts made under construction of G82 in order to solve
this problem of additivity.
Consider an insurance
Ax:n| + s · n| ax
issued against a single premium.
Thiele’s differential equation for the net premium reserve is
∂Vx (t)
=
∂t
(
δVx (t) − µ̃x+t (1 − Vx (t)) (0 < t < n)
δVx (t) − s + µ̃x+t Vx (t)
(n < t)
(4.1)
The Net Premium Reserve and Thiele’s Differential Equation
33
where µ̃x+t is the actual expected force of mortality. Introduce the first order forces
of mortality µ and µ̂ which satisfy
µx+t < µ̃x+t < µ̂x+t ,
and replace (4.1) by
∂Vx (t)
=
∂t
(
(δ + µx+t )Vx (t) − µ̂x+t (0 < t < n)
(δ + µx+t )Vx (t) − s
(n < t).
(4.2)
Thus we get a smaller increase of the reserve and we get a technical basis “on the safe
side”.
(a) Determine the single premium on the first order technical basis by solving (4.2).
Consider the special case
µ̂x = (µx + g2 )(1 + g1 ),
and let
µ∗x = µx + g2
and
δ ∗ = δ − g2 .
(b) What will the single premium be in this case? Comment on the result.
Now consider an educational endowment a x|n| .
(c) Put up the differential equations (4.1) and (4.2) and solve (4.2) for the special
case above. What is the problem in this case?
Finally consider a survival annuity a x|y .
(d) Answer the same question as in (c).
(SP(99))
34
Exercises in Life Insurance Mathematics
5.
Expenses
Exercise 5.1 Work out the details in RN in the case where α 00 = β 00 = γ 00 = 0.
(RN(1) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93)
Exercise 5.2 Express Vtg and cg in terms of Vtn and cn and α0 in the case where
α00 = β 00 = γ 00 = γ 000 = 0.
(RN(2) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93)
Exercise 5.3 Treat the case of a level annuity payable upon death in (m, n) against
level premiums in (0, m).
(RN(3) “Expenses, gross premiums and reserves” 12.10.90 rev. 13.03.93)
Exercise 5.4 Consider an n-year deferred whole-life annuity, age at entry x, payable
with level continuous intensity S. Level gross continuous premium intensity p payable
during the deferment period. The premium is determined by the equivalence principle. For now assume that the expenses are initial expenses αS, loading for collection
fees due continuously with level intensity βp and administration costs also due continuously with level intensity γS.
(a) Find p and the prospective gross premium reserve V tg .
Because of inflation, loading for collection fees and administration expenses are paid
with intensities βf (t)p and γf (t)S at time t.
(b) Put up exspressions for p and Vtg .
(b) Find p and Vtg when f (t) = 1 + kt and where f (t) = exp(ct).
(SP(62))
Exercise 5.5 Consider a whole-life life insurance, sum insured 1, age at entry x,
single net premium B.
(a) Show that the expected effective interest rate for the insured is
Z
∞
0
1
B − t t px µx+t dt − 1.
Expenses
35
Assume that the company has some initial expenses α, but no other expenses.
(b) What is the expected effective interest rate?
(SP(60))
Exercise 5.6 A simple capital insurance, sum insured S, duration n, pays out S at
time n from the time of issue no matter if the insured is alive or not.
(a) Put up an expression for the net payment for this insurance and explain why it is
independent of the age at entry.
A simple capital insurance only makes sense if it is not paid by a single payment (when
dealing with insurance). Assume that the premium is paid continuously during the
entire insurance period with level intensity p, but only if the insured is alive. The
premium is determined by the equivalence principle.
(b) What will the net premium be?
In the gross premium p, initial expenses αS are included as well as loading for collection fees βp and administration expenses γS paid continuously.
(c) Determine p.
(d) Put up an expression for the prospective gross premium reserve, both when the
insured is alive as well as when he is dead.
(SP(61))
The following exercise examines what happens to the insurance technical quantities
when we bring surrender into consideration.
Exercise 5.7 Consider an n-year endowment insurance, age of entry x, benefits are
S1 if one dies during the insurance period and S 2 if one obtains the age of x + n.
Life conditioned equivalence premium is paid continuously until time n (from the age
of entry). Moreover assume that surrender can take place at any time during the
premium payment period, and that the present value of the conventionally calculated
gross premium reserve, liquidated by surrender, is positive. By surrender at time t
the company pays out G(t).
(a) Now disregard all expenses and assume that G(t) is lesser than or equal to the
conventionally calculated (net) premium reserve at time t. Instead of using a conventional technique, the company could itself bring surrender into consideration and
36
Exercises in Life Insurance Mathematics
into its own technical basis. Show that the equivalence principle then would lead to a
premium P 0 ≤ P . Discuss conditions for P 0 = P and give a lower limit for how small
P 0 can get when G(·) varies. Here and in the following it might be useful to study
Thiele’s differential equation.
Now assume that some administration costs are not neglectible. The expenses consist
of the amount α in initial expenses, of γ in administration costs per time unit and
of a fraction β of the actual annual gross premium P in loading for collection fees, P
calculated conventionally. Upon surrender 100θ% of the gross premium reserve is paid
out if the reserve is positive, 0 ≤ θ ≤ 1. By surrender where the gross premium reserve
is positive, a fixed percentage of the reserve is deducted to cover the loss experienced
by surrender where the gross premium reserve is negative. For now disregard expenses
that fall upon surrender. This gross premium reserve is calculated without respect to
surrender.
(b) Show that you will get a lesser gross premium reserve if you bring surrender
into consideration. Assume that θ is chosen so that the equivalence principle can be
applied anyway.
(c) Now assume that in the above situation a constant expense ξ is associated with
the actual payment of G(t). If G(t) calculated in (b) is smaller than ξ, nothing is paid
out by surrender. When the value upon surrender mentioned exceeds ξ the difference
is paid out. Show that the actual gross premium reserve still will be lesser than the
conventional when surrender is brought into consideration. Can θ still be determined
so that the equivalence premium still can be applied?
(FM1 exam (rev.), summer 1979)
Exercise 5.8 It has been proposed that the administration expenses should be calculated as being proportional to the gross premium reserve instead of being proportional
to the sum insured. Now consider an n-year endowment insurance, level continuous
gross premium intensity p, sum insured S, age at entry x. Acquisition expenses αS.
Loading for collection fees βp and γV t at time t, respectively. Vt denotes the gross
premium reserve. Administration costs fall due continuously with level intensity γV t
at time t.
(a) Put up Thiele’s differential equation for V t .
(b) Solve the differential equation with initial conditions for t = 0 and t = n − and
show that the solutions can be expressed by expected present values for annuities
with another interest rate than the interest rate of the technical basis.
(c) Determine the equivalence premium.
Expenses
37
In G82 the interest rate is i = 5% p. a. but gross premiums and gross reserves are
calculated with an interest rate of 4.5% p. a.
(d) What is the corresponding value of γ?
(SP(64))
Exercise 5.9. (Equipment Insurance) By an equipment insurance, sum insured S,
insurance period n, the sum S is paid out at time n if the insured is still alive; if he
dies during the insurance period, the company returns the up till now paid premiums
without interest earned. The level gross premium intensity p is payable during the
entire insurance period. The expenses are initial expenses αS, loading for collection
fees due with continuous level intensity βp and continuous administration costs due
with level intensity γS.
Put up an expression for p applying the equivalence principle.
(SP(66))
Exercise 5.10. (Child’s Insurance) A person aged x has been issued a child’s
insurance: If the insured dies during [x, y) the gross premium is paid back with
earned interest according to the technical basis. If he dies during [y, u) the sum S is
immediately paid out and if he is alive at age u, then S is paid out. The level gross
premium p, the administration costs γS and loading for collection fees βp fall due
continuously during the insurance period. Acquisition expenses are αS.
(a) Put up Thiele’s differential equation for this insurance.
(b) Find the prospective gross premium reserve at any time during the insurance
period.
(c) What is the gross (equivalence) premium intensity, and what is the risk sum at
any time with this premium.
(SP(68))
Exercise 5.11 Consider an n-year endowment insurance, sum insured S, age at
issue x, premium payable until time m, initial expenses αS, administration costs and
loading for collection fees due during the entire insurance period continuously with
level intensities γS and βpg respsctively, where pg is the level gross premium intensity.
Assume that m ≤ n and γ < δ.
(a) Give an expression for pg applying the equivalence principle and prove that it can
38
Exercises in Life Insurance Mathematics
be cast as
pg =
Ax:n| + γ̃ax:n|
(1 − β̃)ax:m|
S.
(b) Show that β̃ > β and γ̃ > γ.
The numerator of the expression is the so-called passive with added sum, because it
is produced from the net passive Ax:n| increased by the present value of γ̃ during the
S
entire insurance period. This passive is denoted by A x:n| .
Let Vt be the net premium reserve at time t (calculated from the time of issue) and
let Vt1 be Vt increased by the reserve of the future administration costs.
(c) Give expressions for Vt and Vt1 .
The company ought to set aside the reserve V t1 but normally the reserve
S
Vt2 = SAx+t:n−1| − (1 − β̃)pg ax+t:m−t|
is set aside.
(d) Compare Vt , Vt1 and Vt2 and try to explain why one prefers to set aside V t2 instead
of Vt1 .
If the insured wishes to surrender his contract at time t, the company pays him the
surrender value of the contract Gt , which is the reserve Vt1 less the part of the initial
expenses that have not yet been amortized.
(e) Show that the surrender value can be cast as
Gt = S(Ax+t:n−t| + γax+t:n−t| ) − (1 − β)pg ax+t:m−t| .
g
The coefficient for S is the surrender value passive and is denoted by A x:n| .
(f) Find an expression for the difference between the net premium reserve and the
surrender value and prove that for m = n it is α(S − V t ).
If the insured wishes to cancel the payment of premiums without entirely to surrender
the contract, it is called a premium free policy. The size of this policy is determined by
letting the surrender value of the new policy equal the surrender value of the original
policy at the time of change.
(g) Give an expression for the sum of the premium free policy and show that its
reserve at the time of change is lesser than the reserve of the original policy.
Expenses
39
Assume that the annual gross premium is given by
p̈g = εpg ,
where ε is called the continuity factor.
(h) Explain why the surrender value and the reserve can be cast as
g
SAx+t:n−t| − ζ p̈g ax+t:m−t|
and
S
SAx+t:n−t| − η p̈g ax+t:m−t| ,
respectively.
(SP(70 rev.))
Exercise 5.12 A married man considers a life insurance on the following conditions:
(i) If he dies before time r from time of issue of the contract, the company has to pay
a continuous pension with level intensity s in a period of time m.
He furthermore considers a supplementary pension, also with level intensity s which is
due to initiate right after the expiration of the pension (i). He considers two options:
(ii) The supplementary pension is due as long as his wife lives.
(iii) The supplementary pension is due as long as his wife lives, at the most until time
n from the time of issue, n > m + r.
As a second alternative he considers a survival annuity, also with payment intensity
s. Here he considers two options:
(iv) The annuity initiates if the man dies before time r from the time of issue and is
due as long as his wife lives.
(v) The annuity initiates if the man dies before time r from the time of issue and is
due as long as his wife lives, at the most until time n from the time of issue, n > r.
(a) Put up an expression for the single net premium for these five contracts.
Assume that the above contracts are issued against an annual premium payment paid
in advance as long as the man and his wife are alive, at the most until time r from
the time of issue. If one of the two dies during the insurance period, the amount
(aθ| /a1| )P is being returned (in Danish: Ristorno), where θ is the remaining part of
the last premium payment period and P is the term premium.
40
Exercises in Life Insurance Mathematics
(b) How would you calculate the annual net premiums?
(c) Put up an expression for the net premium reserve at any time during the insurance
period for the last insurance (v) applying the recently described premium payment
principles.
When calculating the gross premiums, the company uses the following expense rates:
Initial expenses αS, loading for collection fees β times the gross premium, administration costs due continuously with intensityand γ times the gross premium reserve
at any time, and finally payment costs of ε times the amount paid out.
(d) Find the continuous gross premium intensity, applying the equivalence principle.
(e) Find the gross premium when the premium payment takes place as described
before (b).
(SP(76) rev.)
Select Mortality
41
6.
Select Mortality
Exercise 6.1
(a) What could be the meaning of the symbol
(b) Put up an expression for
period of selection is 5 years.
2|6 q[30]+2
s|t q[x]+u ?
in terms of ` under the assumption that the
(c) Express the following three quantities with one symbol:
• The probability that a person now 50 years old who got insured 3 years ago dies
between the ages of 58 and 59, presuming the period of selection is 5 years,
• the probability that a new born dies between 67 and 72 years of age,
• the number of deaths between the ages of 29 and 30 in the third year of an
insurance portfolio, presuming the period af selection now is 3 years.
(SP(25))
Exercise 6.2 In this exercise we will try to explain the presence of select mortality
for a portfolio of insured and study its properties.
For two functions f and g we shall use the obvious notation
f g(t) = f (t)g(t), (f + g)(t) = f (t) + g(t).
The portfolio is assumed to be divided between the two states active and disabled
according to the figure below where the course of events is modelled by a Markov
process {Xt }t≥0 , and t is the age of the insured.
σ(t)
1. Active
K
KKK o
KK
/
ρ(t)
µ(t)K
KKK
KK%
2. Disabled
sss
sss
ν(t)
ss
s
s
ys
s
3. Dead
The transition probabilities of the model are denoted by
pij (s, t) = P (Xt = j | Xs = i), s ≤ t, i, j = 1, 2, 3
42
Exercises in Life Insurance Mathematics
and the intensities µij (t) are assumed to exist and are given by
µij (t) = lim
h&0
pij (t, t + h)
, i 6= j.
h
We assume that the intensities are continuous functions. Define
µ(t) = µ13 (t), ν(t) = µ23 (t), σ(t) = µ12 (t), ρ(t) = µ21 (t)
and
µ1 (t) = µ(t) + σ(t), µ2 (t) = ρ(t) + ν(t).
We take it that
µ(t) < ν(t), ∀t ≥ 0,
i. e. the mortality for a disabled is always greater than for an active person.
When we cannot observe whether an insured is active or disabled at any time after
entry (as an active), one gets a filtration (of the above Markov model) which is
determined by the force of mortality for a random insured. Let µ̃(τ, x) denote this
intensity for an insured of age x with age of entry τ, τ ≤ x.
(a) Explain that µ̃ is given by
p12
p11
(τ, x) + ν(x)
(τ, x)
p11 + p12
p11 + p12
p12
= µ(x) + {ν(x) − µ(x)}
(τ, x)
p11 + p12
µ̃(τ, x) = µ(x)
and thus give the grounds for the presence of select mortality.
We obviously want τ → µ̃(τ, x) to be decreasing for fixed x which will be shown by
differentiation in the following.
(b) Explain that τ → µ̃(τ, x) is decreasing iff the fraction τ → (p 12 /p11 )(τ, x) is
decreasing and give an interpretation of this.
(c) Show that
d
dτ
σ(τ )(p12 p21 − p11 p22 )(τ, x)
p12
(τ, x) =
p11
p211 (τ, x)
and
d
(p12 p21 − p11 p22 )(τ, x) = {µ1 (τ ) + µ2 (τ )}(p12 p21 − p11 p22 )(τ, x),
dτ
and explain why τ → µ̃(τ, x) is decreasing.
Let p̂ij (s, t) denote the transition probabilities corresponding to the model without
recovery, i. e. ρ(t) = 0, ∀t ≥ 0.
Select Mortality
43
(d) Find expressions for p̂11 (τ, x) and p̂12 (τ, x) as a function of the intensities and
show that
p12
p̂12
(τ, x) <
(τ, x).
p11
p̂11
Give an interpretation of this and explain how µ̃ is affected by changing to the model
without recovery.
It is a common opinion that the selection the insured goes through at entry disappears
after a period of time, called the period of selection. We will try to explain this
phenomenon mathematically. Lad τ0 be the age of entry for an insured.
(e) Show that
x → exp
Z
x
τ0
µ1 (s)ds p11 (τ0 , x), x ≥ τ0
is an increasing function. Assume that µ 2 (t) ≥ µ1 (t), ∀t ≥ 0 and that there exists a
w > τ0 so that
Z
w
τ0
Show that
(µ2 (t) − µ1 (t))dt = ∞.
d
lim
x%w dτ
monotonically with
p12
p11
(τ0 , x) = 0
d p12
(τ0 , x) = 0, ∀x ≥ w,
dτ p11
and explain why this verifies the presence of a period of selection of w.
(FM1 exam, 1989-ordning, opgave 1, summer 1995)
Exercise 6.3 Mortality in a portfolio of insured lives will usually be different from
the mortality of the general population because the insured lives are a selected part
of the population. We will in this exercise study one relationship that is assumed to
contribute a great deal to the effect of selection, i. e. the fact that people with illnesses that cause severe excess mortality are not allowed to underwrite life insurances
(under the usual terms). In the following such persons will be called “ill”. Thus the
population can be divided according to the figure below.
ρx
/ 1. Insured, not ill
QQQ
QQQ κ
nnn
κx nnnn
QQQx
n
QQQ
n
QQQ
nnn
(
vnnn
σx
4. DeadhP
PPP
mm6
m
P
m
PPPλx
λx mmm
PPP
mmm
m
PPP
m
m
P m
m
0. Not insured, not ill
σx
3. Not insured, ill
2. Insured, ill
44
Exercises in Life Insurance Mathematics
Now assume that every person enters state “0” at birth and that the transitions
between the states afterwards go on as a time continuous Markov chain with transition
intensities only dependent on the age x as indicated in the figure. Excess mortality
for the ill persons means that
λx ≥ κx , x > 0,
(6.1)
with “>” for some values of x.
With the usual notation for the transition probabilities the following are satisfied
p11 (x − t, x) = e
p12 (x − t, x) =
−
Z
e
p02 (0, x) =
p03 (0, x) =
Z
Rx
0
Rx
0
x
e
0
0
x
(σu +κu )du
−
x−t
−
Z
x−t
x
p00 (0, x) = e−
p01 (0, x) = e
Rx
Rz
x−t
,
(σu +κu )du
(σu +κu +ρu )du
(σu +κu )du
−
e−
Rz
0
Rz
0
(6.2)
σz e−
Rx
z
λu du
dz, 0 < t < x;
,
(1 − e
(σu +κu )du
(6.3)
(6.4)
−
Rx
0
(1 − e
(σu +κu +ρu )du
ρu du
−
Rz
σz e−
0
ρu du
Rx
z
),
(6.5)
)σz e
λu du
−
Rx
z
λu du
dz, 0 < x.
dz,
(6.6)
(6.7)
(a) Prove the formulas (6.2) and (6.3) by putting up and solving differential equations.
(b) Assume that (6.4) is given. Give direct, informal grounds for the expressions (6.5)
– (6.7).
The insured lives are either in state “1” or in state “2” (those in state “2” received
the insurance contract before they were struck by illness). The insurance company
does not observe in which of the two states the insured is. All the company knows is
the time of entry and age. Let µ[x−t]+t denote the force of mortality for an insured of
age x who received the insurance t years ago.
(c) Derive an expression for µ[x−t]+t . Show that under the condition (6.1), µ [x−t]+t is
a non-decreasing function of t for constant x (it might be desirable to express µ [x−t]+t
as a weighted average of κx and λx ). How will you explain this result to a person who
has no knowledge of actuarial science?
(d) Discuss the formula for µ[x−t]+t to find theoretical explanations as to why insurance
companies operate with a period of selection s so that the mortality is considered to
be aggregate for t > s.
Let µx denote the force of mortality for a randomly chosen person of age x in the
population (that is, we do not observe in which of the states “0” – “3” the person is).
Select Mortality
45
(e) Find an expression for µx . Show that under the condition (6.1) the inequality
µx ≥ µ[x−t]+t , 0 < t < x
is satisfied. The result clarifies the preliminary remarks of this exercise. Try to give an
explanation that is comprehensible for a person without any knowledge of actuarial
mathematics.
(FM1 exam (1), winter 1985/86)
46
Exercises in Life Insurance Mathematics
7.
Markov Chains in Life Insurance
FM0 S94, 1
Exercise 7.1 Consider a model for competing risks with k + 1 states, 0: “alive” and
1, . . . , k denoting death from k different reasons; denote the partial probabilities of
death by
Z
(j)
t qx
t
= 1 − exp −
0
µ0j
,
x+τ dτ
and define
(j)
t px
= 1 − t qx(j) .
(a) Prove that
00
t px
=
k
Y
(j)
t px .
j=1
(b) Prove that
k
X
0j
t px
=
j=1
k
X
(j)
t qx
j=1
+
−
X
X
(i) (j)
t qx t qx
1≤i<j≤k
(h) (i) (j)
t qx t qx t qx
1≤h<i<j≤k
− · · · + (−1)k+1 t qx(1) · · · t qx(k) .
Now assume that for given j there exist constants c and t 0 > 0 so that
(j) 0j
t px µx+t
= c,
for all t ∈ [0, t0 ].
(c) Prove that for all t ∈ [0, t0 ]
(j)
t qx
=
0j
t px
=
t
· t q (j)
t0 0 x
Z t
1
πh6=j τ px (h) dτ.
· t qx(j)
t
0
(SP(83))
Exercise 7.2. (Life Insurance with Exemption from Payment of Premium by Disability) In this exercise we will consider a policy that can attend one of N states
Markov Chains in Life Insurance
47
enumerated 1, . . . , N . The development of the policy is being described by a Markov
process with transition intensities µ jk
t for transition from state j to state k at time t
from issue.
In the contract it is stated that the amount B tjk is payable upon transition from state
j to state k at time t. As long as the policy stays in state j, a continuous payment
with intensity Btj is due, i. e. in the time interval [t, t + ∆t) the amount B tj ∆t + o(∆t)
is paid out. Assume that all amounts B tjk are non-negative and that the force of
interest δ is independent of time. For now we disregard administration expenses.
(a) There are no assumptions regarding the sign of B tj . How should negative values
of Btj be interpreted?
Let Vtj denote the premium reserve in state j at time t. It is defined as the expected
present value of the out payments in the time interval [t, ∞) discounted back until
time t, given that the policy is in state j at time t.
(b) Show that the premium reserve satisfies the differential equation system
−Btj =
X jk jk
d j
µt (Bt + Vtk − Vtj ), j = 1, . . . , N.
Vt − δVtj +
dt
k6=j
(c) Interpret this differential equation system intuitively in terms of savings premium
and risk premium.
A special case of the this general Markov model is the disability model. This model
has three states, 1, 2 and 3, corresponding to active, disabled and dead. The insured’s
age at entry is x and the intensities are denoted by
µ12
= σx+t (from active to disabled),
t
µ21
= ρx+t (recovery),
y
µ13
= µx+t (dead as active),
y
µ23
= νx+t (dead as disabled).
t
All other transition intensities are 0.
Apply this model for an investigation of an endowment insurance with exemption
from payment of premiums by disability. Assume the insured is active at entry. The
insurance period is n, so the insurance cancels when the insured has reached the age
of x + n or if he dies before that.
In question (d) it is assumed that the equivalence premium is paid continuously with
level intensity π as long as the insured is active and that a constant sum insured S
48
Exercises in Life Insurance Mathematics
is payable upon the expiration of the policy; hence, with the notation from above we
have
Bt1 = −π, Bt2 = 0, Bt13 = Bt23 = S.
(d) Come up with formulas for π, Vt1 and Vt2 for 0 ≤ t < n in terms of the transition
probabilities and transition intensities in the model and the discounting rate v, by
direct prospective reasoning.
If the premium and the benefits depend on the reserves, the premium and the premium
reserve cannot be determined directly as in (d); instead the differential equation
system from (b) must be solved with appropriate boundary conditions. Now it is
assumed that the premium and the payment of benefit at age x + n are due as above,
but upon death of the insured before time x + n, the premium reserve of the policy is
paid out as a supplement to the sum insured; B tj3 = S +Vtj for j = 1, 2 and 0 < t < n.
(e) Show that the differential equation system from (b) gives the grounds for a differential equation of first order in V t1 − Vt2 . What is the initial condition? Solve the
differential equation.
(f) What are π, Vt1 and Vt2 for 0 ≤ t < n.
(g) Show that if νx+t ≥ µx+t for all t < n then Vt2 > Vt1 for all t < n. Interpret this
result.
Assume that we have the following expenses: Initial expenses due at time 0 with
an amount of αS, loading for collection fees βp, where p is the level gross premium
intensity and administration costs due with an intensity at time t equal to γV tj if the
policy is in state j.
(h) Explain, without performing any detailed calculations, what changes would follow
from these assumptions in the theory discussed in (e) – (g).
(FM1 exam , summer 1984)
Exercise 7.3 An active person aged x considers a disability annuity, which falls due
continuously with level intensity b upon disability before the age of x + n. Premium is
payable at rate π as long as the person is active during the contract period. Assume
that the state of the policy is S(t) at time t after the time of issue where {S(t)} t≥0 is
a time continuous Markov model with state space and transitions as follows
Markov Chains in Life Insurance
0. Active
K
49
σ(x+t)
KKK
KK
µ(x+t)
K
KKK
KK%
/ 1. Disabled
ttt
ttt
ν(x+t)
tt
t
t
yt
t
2. Dead
(a) Give, without any proof, expressions for the transition probabilities
pjk (s, t) = P (S(t) = k | S(s) = j), 0 ≤ s ≤ t, j, k ∈ {0, 1, 2}.
(7.1)
The present value at time s of benefits less premiums in [s, n] can be cast as
C(s) =
Z
n
s
v t−s (b1{S(t)=1} − π1{S(t)=0} )dt,
where 1A is the indicator function for the event A.
(b) Find, for 0 ≤ s ≤ n and j = 0, 1, 2, the conditional expection
Vj (s) = E(C(s) | S(s) = j)
(7.2)
Zj = Var(C(s) | S(s) = j)
(7.3)
and the conditional variance
as expressions of integrals of functions of the transition probabilities from (7.1).
(c) Find EC(s) and VarC(s) in terms of the transition probabilities (7.1) above and
the functions (7.2) and (7.3).
(d) Now assume that the equivalence principle is being adopted, i. e. V 0 (0) = 0, where
π is the net premium intensity and Vj (s) in (7.2) above is the net premium reserve in
state j at time s. Does the net reserve ever become negative?
(FM1 exam opgave 1, summer 1989)
Exercise 7.4 The figure below illustrates an expansion of the model discussed in
exercise 7.3. There are two states of disability i 1 and i2 representing two degrees of
disability. Assume that ν2 (x+t) ≥ ν1 (x+t). Let {S̃}t≥0 be the corresponding Markov
chain and define the stochastic process {S(t)} t≥0 by S(t) = S̃(t) for S̃(t) ∈ {a, d} and
S(t) = i for S̃(t) ∈ {i1 , i2 }. If one does not know the degree of disability, {S(t)} is
the observable process.
50
Exercises in Life Insurance Mathematics
i1 .
rr8
r
r
rr
σ1 (x+t)
r
rrr
r
r
r
/
σ2 (x+t)
a. LL
i2 .
LLL
rr
r
ν1 (x+t)
LL
r
rr
µ(x+t)
ν2 (x+t)
LLL
r
r
LLL
r
& xrrr
d.
The transition intensities for the S(t)-process are
µjk (t, {S(s)}s<t ) = lim
dt&0
P (S(t + dt) = k | S(t) = j, {S(s)}s<t )
dt
for j 6= k and any specification of {S(s)} s<t .
(a) Prove that
µai (t, {S(s)}s<t ) = σ1 (x + t) + σ2 (x + t)
and
µad (t, {S(s)}s<t ) = µ(x + t)
(both independent of {S(s)}s<t ) and
µid (t; S(s) = a for 0 ≤ s < τ, S(s) = i for τ ≤ s < t)
=
P2
j=1
σj (x + τ ) exp(−
P2
j=1
R t−τ
0
σj (x + τ ) exp(−
νj (x + τ + u)du)νj (x + t)
R t−τ
0
νj (x + τ + u)du)
(7.4)
(a function of x + τ and t − τ ). Give a sufficient condition for {S(t)} t≥0 being Markov.
(b) Interpret the expression in (7.4). Give sufficient conditions for finding an approximating function to (7.4) depensing only on x + t for sufficiently great t − τ , say
t − τ ≥ s0 (the time of selection).
(FM1 exam opgave 2, summer 1989)
Exercise 7.5. (Markov Chains in Life Insurance) Give a detailed proof of the
Kolmogorov backward differential equations (2.15) and the corresponding integral
equations (2.21).
(RN “Problems in Life Insurance” 14.10.93, problem 7)
Exercise 7.6. (Markov Chains in Life Insurance) consider the model for disabilities,
recoveries, and deaths in Paragraph H in RN “Markov Chains in Life Insurance”
04.03.94.
Markov Chains in Life Insurance
51
(a) Find the second order differential equation for p 00 (s, ·) as outlined in the text,
making appropriate assumptions about differentiability of the intensities.
From now on consider the special case with constant intensities:
(b) Find explicit solutions for p00 (s, t) and p01 (s, t). Note that the solution depends
on s and t only through t − s, hence put s = 0. Discuss how the probabilities depend
on t and the intensities.
(c) Calculate for t = 0, 10, 20, . . . , 100 and draw graphs of the probabilities for some
different choices of the intensities, including as key references
1. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.005,
2. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.01,
3. σ = 0.01, ρ = 0.01, µ = 0.005, ν = 0.1.
(You can program the formulas and compute directly or you can employ the program
’retres’.)
(d) Explain how the result 3. in item (c) above can be used to find the probabilites
for the case σ = 0.1, ρ = 0.1, µ = 0.05, ν = 1. (The ratios between the intensities are
the essential feature.)
(e) The probability p02 (s, t) gives the mortality law for a person who is known to be
active at age s. Discuss how it depends on the intensities, with special attention to
the case where µ = ν.
(f) Consider the special case with no mortality (µ = ν = 0), whereby the number of
states essentially becomes 2. Find p 01 (0, t), and discuss how the expression depends
on t and the intensities. Find the limit as t → ∞ and discuss the expression.
(RN “Problems in Life Insurance” 14.10.93, problem 8)
Exercise 7.7. (Markov Chains in Life Insurance) Formulate suitable continuous
time Markov chain models for solution of the following problems, where N is assumed
to be a Poisson variate with parameter 1:
(a) Find the probability that N ∈ {0, 2, 4, . . .}.
(b) Find the probability that N ∈ {0, 3, 6, . . .}. Think of other variations of the
problem.
52
Exercises in Life Insurance Mathematics
(RN “Problems in Life Insurance” 14.10.93, problem 9)
Exercise 7.8. (Markov Chains in Life Insurance) Prove that the four definitions
(2.1), (2.2), (2.3), and (2.5) of the Markov property are equivalent (assuming that the
sample paths of the process are as stated in Paragraph 1A).
(RN “Problems in Life Insurance” 14.10.93, problem 10)
Exercise 7.9. (Reserves) At time 0 a person (x) aged x buys a standard pension
insurance policy specifying that, conditional on survival, premiums are payable with
level intensity c from time 0 to time m and pensions are payable continuously with
level intensity b from time m to time n, m < n. There are two states, 0: “alive”
and 1: “dead”.
Let µx+t be the force of mortality at age x + t, and denote by
Rt
p
=
exp(−
µ
t x
0 x+s ds) the probability that (x) survives to age x + t. Assume that
interest is earned at a constant rate δ so that v(t) = v t , with v = e−δ the annual
discount rate. Throughout b is taken as fixed and c is to be determined by the
equivalence principle.
(a) Put up prospective and retrospective expressions for the reserves in both states
at any time t ∈ [0, n) after issue of the policy. As an exercise, find the reserves also
by solving the appropriate differential equations. Determine c.
(b) Find the conditional variances of the individual reserves (the present values of
future and past payments) at time t, given the state of the policy at time t.
Henceforth the standard policy is referred to as policy ’S’. Consider a modified policy
’P’, by which the prospective reserve in state 0 is to be repaid upon death of (x)
during the period [0, n).
(c) Find the statewise reserves for ’P’. (Differential equations must now be used.)
Determine c.
(d) Find the conditional variances of the individual reserves for ’P’, corresponding to
those in (b).
Consider another modified policy ’R’, by which the retrospective reserve in state 0 is
to be paid upon the death of (x) during [0, n).
(e) Find the statewise reserves for ’R’ and determine c.
(f) Find the conditional variances of the individual reserves by policy ’R’, corresponding to those in (b).
(g) Compare the results obtained for ’S’, ’P’, and ’R’.
Markov Chains in Life Insurance
53
Finally, consider a policy ’M’ with a mixed rule for repayment of the reserve, by which
the retrospective reserve in state 0 is to be repaid upon the death of (x) during the
premium period [0, m], whilst the prospective reserve in state 0 is to be repaid upon
death during the pension period [m, n].
(h) Find the statewise reserves for the policy ’M’, and determine c.
(RN “Problems in Life Insurance” 14.10.93, problem 15)
Exercise 7.10. (Reserves) At time 0 a person buys a life insurance policy specifying
that an amount b (the sum insured) is provided immediately upon death before time
n and premiums are payable with level intensity c as long as the person is alive
and active up to time n (premium waiver by disability). Assuming that recovery
is impossible, the relevant Markov model can be sketced below. Assume that the
discount function is v(t) = v t = e−δt . The premium c is to be determined by the
equivalence principle.
σ(x)
0. Active
J
JJJ
JJ
µ(x)J
JJJ
JJ
%
/ 1. Disabled
ttt
ttt
ν(x)
tt
t
t
yt
t
2. Dead
(a) Put up integral expressions for the transition probabilities.
(b) Put up expressions for the prospective and retrosprective reserves in all states
at any time t ∈ [0, n). Find the reserves also by solving the appropriate differential
equations. Determine the premium intensity c.
Suppose that instead of full premium waiver, the premium during disability is made
dependent on the past savings on the contract. More specifically, assume that premiums during disability fall due with intensity c − c 0 V1− (t) at time t if the policy then
is in state 1.
(c) Which relation must c and c0 satisfy?
(RN “Problems in Life Insurance” 14.10.93, problem 16)
54
Exercises in Life Insurance Mathematics
Exercise 7.11 Consider the usual disability model without recovery:
σ(x)
0. Active
J
JJJ
JJ
µ(x)J
/ 1. Disabled
ttt
ttt
ν(x)
ttt
t
t
yt
JJJ
JJ%
2. Dead
An insurance is issued to a person (x). Premium is payed continuously with level
intensity π as long as (x) is alive, at the most for n years. As long as (x) is disabled,
he receives a payment with level intensity c until his death. If (x) dies before the age
of x + n the sum D is paid out. If (x) is active at the time m ∈ (0, n) the sum S is
paid out.
(a) Put up the statewise reserves.
(b) Put up Thiele’s differential equation and determine the risk sums.
(BS “Opgave til FM0” 1995)
Exercise 7.12 A person considers a life insurance policy specifying that an amount
S is paid immediately upon death before time of expiration n. If the insured is alive
at time n he is also provided the sum insured S. Premiums are payable with level
intensity p as long as the insured is active up to time n. That is, the insured has the
right to exemption from payment of premium if he is disabled. Interest is earned at
a constant rate δ. The Markov model used is illustrated below.
σ(x)
0. Active
J
JJJ o
JJ
/
ρ(x)
µ(x)J
JJJ
JJ%
1. Disabled
ttt
ttt
ν(x)
ttt
t
t
yt
2. Dead
(a) Put up the following:
1. Thiele’s differential equation for the statewise reserves V 0 (t) and V1 (t).
2. The statewise reserve V0 (t) as an integral function of p00 (t, u) and V1 (t).
3. The statewise reserve V0 (t) as an integral function of p00 (t, u) and p01 (t, u).
4. The statewise reserve V1 (t) as an integral function of p00 (t, u) and V0 (t).
5. The statewise reserve V1 (t) as an integral function of p11 (t, u) and p10 (t, u).
Markov Chains in Life Insurance
55
(b) Differentiate both expressions for V 0 (t) in order to verify Thiele’s differential
equation.
(c) How will you in practice find the equivalence premium for this kind of insurance?
(d) Let µt = νt . Put up a differential equation for V 1 (t) − V0 (t). Solve it and use the
result to find out how σt and ρt ought to be chosen in relation to σt0 and ρ0t so that
the effect will be the desired increase of the premium. Give an interpretation of all
results.
(e) Find expressions for the safety loadings for the two states and consider the difference. Which sign does it have with a reasonable choice of parameters?
(f) Expand the model with the state “surrender”. What would you consider a reasonable payment in connection with surrender from the state of active and disabled
respectively. How should the first order intensity of surrender be put in relation to
the intensity on the second order basis with your choice of surrender value?
(MS “Opgave til FM1” 19.09.95)
Exercise 7.13 Consider the usual disability model with four level transition intensities µ, ν, σ and ρ. Let α = µ + σ, κ = ν + ρ and assume that α 6= κ.
(a) Show that the probability for an active person being active in t years, after having
been disabled once and only once is
σρ
e−αt − e−κt
te−αt +
κ−α
α−κ
.
(b) Find the probability for an active person being disabled for the second time in t
years.
(c) Explain how one relying, on information about death, disability and recovery in
some population, can estimate the probabilities in (a) and (b) in two different ways.
(HRH “Opgaver til FM1”)
56
Exercises in Life Insurance Mathematics
8.
Bonus Schemes
FM0 S95, 1
Exercise 8.1 An insurance company uses the following technical basis: Force of
mortality µx = 0.0005+105.88+0.038x−10 (G82M), interest rate i = 5% p. a., acquisition
costs 3% of the sum insured, loading for collecting fees 5% of the premium and
administration expenses due continuously with intensity γV tg where Vtg is the gross
premium reserve and
1, 05
γ = log
.
1, 045
The technical basis of second order has interest rate ĩ = 5.5% p. a. and force of
mortality µ̃x = µx − (δ̃ − δ), where δ and δ̃ are the forces of interest corresponding to
i and ĩ respectively. Same expenses as above.
A 30-year old person considers an endowment insurance, insurance period n years.
The premium, calculated according to the equivalence principle, is due continuously
with level intensity during the entire period of insurance.
(a) Calculate the gross premium intensity and the gross premium reserve after 20 and
40 years, respectively.
(b) Put up a differential equation for the safety loading S t
Assume for the time being that the bonus fund is entirely paid out after 40 years.
(c) Confirm that this gives the greatest possible bonus fund during the entire insurance
period.
(d) Calculate this bonus fund after 20 and 40 years.
Now assume that the bonus scheme consists of two discrete payments: One after 20
years and one again after 40 years.
(e) Calculate these two payments.
Finally assume that the bonus scheme is as previously, but that the payment after 20
years is used as a deposit for a life conditioned capital insurance, duration n years,
calculated on the technical basis of second order without expense contributions.
(f) Show that the amount, payable at the 40th year, is the same as the payment according to the first bonus scheme and explain the difference between the two schemes.
Bonus Schemes
57
(g) Critisize the considered bonus schemes especially related to the insured that passes
away during the insurance period and try to suggest a more reasonable bonus scheme.
(SP(74))
Exercise 8.2 Consider two lives (x) and (y) ages x and y respectively with remaining
life times Tx and Ty . Assume Tx and Ty to be stochastically independent.
For premium calculation the company has established a technical basis of first order;
(x)
force of interest δ, force of mortality µ x+t for (x) at age x + t and force of mortality
(y)
µy+t for (y) at age y + t. (x) and (y) consider a widow insurance (whole-life annuity).
The possible whole-life annuity is due continuously with intensity 1 as long as (y) is
alive and (x) is dead.
(a) What is the present value of the possible whole-life annuity?
(b) Derive an expression for the expected present value of the possible whole-life annuity in terms of one-life and both-life annuity expressions. Give a direct interpretation
of this expression.
For the possible annuity a level continuous premium is due as long as they both are
alive.
(c) What is the equivalence premium intensity?
(d) Give an expression for the premium reserve by prospective reasoning.
(e) Derive Thiele’s differential equation.
Introducing the technical basis of second order, the interest rate is δ̃ and the forces
(y)
(x)
of mortality are µ̃x+t and µ̃y+t .
(f) Which principles in general should be the basis for choosing the technical basis of
second order? How would you determine the elements of the technical basis of second
order compared to the ones of the first order for the possible whole-life annuity?
(g) Derive an expression for the safety loading.
We now introduce the following bonus scheme: As long as (x) and (y) both are alive,
no return of safety margin is due to payment. If (y) dies before (x) no return is due
either. If (x) dies before (y), a continuous amount B is added to the annuity payment
of one. The amount B is paid out as a continuous bonus.
(h) Put up an expression for determination of B.
58
Exercises in Life Insurance Mathematics
Instead of the bonus scheme above, we would like a return of an amount equal to the
present value of the added amount B at the time of (x)’s death if (x) dies before (y).
The payment of bonus consists of a continuous payment of an amount B as long as
the whole-life annuity is due.
(i) What is the difference between the two bonus schemes? Which Bonus scheme
would be preferable from a safety margin view?
(FM100, Oslo 1987 05.12.87 (ex. 1))
Exercise 8.3 By an endowment insurance an amount of S is paid either upon death
before year n after the time of issue or at the latest n years after the time of issue.
The premium is due with level intensity B during the insurance period, the force of
interest is δ, the force of mortality µ x , expenses αS by issue and continuous expenses
with intensity γS + βB per year during the insurance period.
(a) Determine the premium B by applying the equivalence principle. Determine the
net and gross premium reserves at any time during the insurance period. Find a
connection between the two reserves and explain the reason for this difference.
The expenses α, β, γ do not contain any safety loadings. On the contrary there are
assumed to be safety loadings κ and λ included in the force of interest and in the
force of mortality respectively. Hence, a realistic force of interest of second order is
δ̃ = δ + κ and a realistic force of mortality is µ̃ x = µx − λ.
(b) Put up the expression for the safety loading, the policy contributes at any time
during the insurance period.
The return of premium scheme is as follows: (i) An amount (κ − λ)V t is paid out
during (t, t + dt) during the insurance period, where V t is the premium reserve for
covering the outstanding claims and return of premiums (i. e. the insured earns an
interest (κ − λ)dt of his part of the total fund); (ii) He receives an additional amount
K to the sum insured upon death during the insurance period.
(c) Determine K so the total safety loadings are being allotted to the insured.
(FM1, winter 1985-86 (ex. 2))
Exercise 8.4 Consider an endowment insurance, sum insured S, duration n, age of
entry x. The company adopts the following technical basis for calculation of premiums
and premium reserves: Force of interest δ, force of mortality µ x and expenses βπ paid
continuously where π is the level net premium, paid continuously during the entire
period. In the technical basis of first order we assume the force of surrender to be
zero.
Bonus Schemes
59
(a) Put up Thiele’s differential equation and an expression for the equivalence premium intensity.
The actual development with respect to the technical basis of second order for the
company is as follows: The interest rate has dropped to δ 0 and the administration
expenses have dropped to β 0 . Now assume that the force of surrender exists and is
denoted by νx+t t years after the time of issue where ν x is independent of the time
the person has been insured.
(b) Discuss if this assumption of independence i reasonable.
Until now the company has used the safety loadings to bring down the future premiums every year, or bonus has been paid out in connection with death or surrender,
where the total reserve is allotted. A new bonus scheme is allotting bonus cash every
whole year or by death or surrender.
(c) Put up Thiele’s differential equation for the premium reserve on the technical
basis of second order. What are the boundary conditions? Intuitively, why does ν x
vanish?
(d) Put up the differential equations for the bonus funds. What are the boundary
conditions for the two bonus schemes?
(e) Try to figure out the variance of the bonus funds of the 2 bonus schemes.
(FW (rev.), 1994)
60
Exercises in Life Insurance Mathematics
9.
Moments of Present Values
Exercise 9.1 Consider the standard setup, where the development of the life insurance is described by a continuous Markov Chain on a state space J = {0, . . . , J},
and the contract specifies that a0g (t)dt is payable if the policy stays in state g in the
time interval (t, t + dt) and a0gh (t) is payable upon transition from state g to state h at
time t. (More general annuity payments can easily be dealt with, but the expression
becomes more messy.) Assume for the time being that the discount function v is
deterministic.
To calculate the variance of
V =
Z
∞
v(τ )A(dτ ),
0
the present value at time 0 of future benefits less premiums, one needs to find
EV
2
= E
= E
∞
Z
v(τ )A(dτ )
Z0 ∞
2
2
2
v (τ )(A(dτ )) + 2
0
Z
∞
v(τ )A(dτ )
0
Z
v(ϑ)A(dϑ) .
ϑ>τ
(9.1)
(a) Prove that (9.1) can be recast as
EV 2 =
Z
∞
v 2 (τ )
0
+
X
p0g (0, τ ){2a0g Vg+ (τ )
g
X
µgh (τ )a0gh (τ )(a0gh (τ ) + 2Vh+ (τ ))}dτ,
(9.2)
h6=g
where Vg+ denotes the prospective reserve in state g at time t.
The variance is obtained by subtracting the square of the mean present value from
the mean of the square. Formula (9.2) appears to offer an escape from the double
integration that has to be performed in (9.1). It requires that the prospective reserve
in different states be computed (they are essentially the inner integral, of course) and
stored in memory beforehand. However, we have standard programs for that.
(b) Use (9.2) to calculate the variance for a simple term insurance and for a simple
life annuity, for which the results are well-known.
Another simple formula for the variance, which shall not be proved here, is
VarV =
Z
∞
0
v 2 (τ )
X
g6=h
p0g (0, τ )µgh (τ )(a0gh (τ ) + Vh+ (τ ) − Vg+ (τ ))2 dτ
(9.3)
(c) Prove that (9.2) essentially remains valid by stochastic discount function v(t) =
exp(−∆(t)) if the process {∆(t)}t≥0 has independent increments.
Moments of Present Values
61
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 1))
Exercise 9.2 Consider a temporary life insurance issued to a person at age x. The
policy specifies that the sum insured S is payable immedieately upon (possible) death
of the insured before time n and that premium is due with fixed amount c at times
0, 1, . . . , n − 1 as long as the insured is alive. Assume that administration costs incur
continuously with constant intensity γS throughout the duration of the policy and
that the force of interest δ is fixed.
(a) Determine the premium c as a function of S and γ by the equivalence principle.
Find an expression for the prospective reserve of the policy at time t ∈ [0, n).
(b) Find the variances and the covariances of the present values at time 0 of the
insurance payment, the premiums and the administration costs. Find the variance of
the present value at time 0 of the total cash flow of payments generated by the policy.
To be continued as exercise 11.3, page 71.
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 2))
Exercise 9.3 Consider an n-year term insurance with equivalence premium payable
continuously with level intensity π throughout the duration of the policy.
(a) Put up the prospective reserve and the variance of the present value at issue at
time 0 of benefits less premiums.
(b) Suppose that in case of surrender the insured immedieately gets the current net
value of the policy defined as the prospective reserve at the time of surrender. Assume
that surrender takes place with intensity γ(t) at time t < n. (Thus we consider an
extended model with three states “insured”, “withdrawn” and “dead”.) Find the
prospective reserve in state “insured” and the variance of the present value at issue
of benefits less premiums. Compare with the results in (a).
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 4))
It is possible to find differential equations which determines any n-order moment
(n)
Vk (t) recursively in any state k for a generalized Markov-model. In the next exercise
we consider the 2nd order moment for an endowment insurance.
Exercise 9.4 Assume that a person at age x buys an n-year endownment insurance
with level premium intensity π payable as long as the insured is alive. Upon death
the amount S is paid out immediately. The force of mortality is given by µ x at age x.
Let as usual Ut denote the present value at time t of benefits less premiums in the
62
Exercises in Life Insurance Mathematics
future.
(a) Find an expression for V (2) (t) = E(Ut2 | T > t).
(b) Derive the expression from (a) with respect to t to find that you will get a differential equation which determines V (2) (t) recursively from V (t).
(BM and JC, 1995)
Inference in the Markov Model
10.
63
Inference in the Markov Model
Exercise 10.1 The time continuous Markov Model shown below
1
kk5
kkk
0 E
E
µ (x)
kk 1
kkk
µ2 (x)
EE
EE
EE
EE
/
2
..
EE
.
EE
EE
EE
E"
µh (x)
E
h
is called the model for competing risks with h reasons for resignation. The function
P
µ = hi=1 µi is the total intensity of resignation.
Define t px = p00 (x, x + t) and t qx(k) = p0k (x, x + t).
(a) Prove that
t px
= p00 (x, x + t) = e−
and
(k)
t qx = p0k (x, x + t) =
Z
Rt
o
µ(x+s)ds
x+t
p00 (x, s)µk (s)ds, for k = 1, 2, . . . , h.
x
Assume that L independent persons of the same age are being observed during one
year. Let the state “0” in the model above represent the state “alive” and let the
h reasons of resignation be reasons of deaths. It is possible to assume the forces of
mortality, µ1 , . . . , µh , to be constant and the same for all L persons. The number of
deaths from the h reasons are denoted D 1 , . . . , Dh and the sum of life time is T .
(b) Determine the maximum likelihood estimators µ̂ 1 , . . . , µ̂h and the asymptotic distributions of the estimators.
(HRH “Opgaver til FM3” (ex. 2))
Exercise 10.2 Consider the time continuous Markov model as follows
64
Exercises in Life Insurance Mathematics
σ(x)
/
0. Working
KKK o
KKK
µ(x)K
KKK
KK
%
1. Not working
rrr
rrr
ρ(x)
µ(x)
rrr
r
r
yr
2. Dead
(a) Find explicit expressions for the transition probabilities.
(b) Assume that all members in a portfolio of N independent lives are being observed
during the period of time [0, 1] and assume that all transition intensities are constant
during the time of observation. How would you estimate σ, ρ and µ? Determine the
asymptotic distributions of the estimators.
(HRH “Opgaver til FM3” (ex. 12))
Exercise 10.3 An insurance company is to perform a mortality study based on
complete records for n life insurance policies with unlimited term period. Policy
number i was issued zi years ago to a person who was then aged x i . The actuary sets
out to maximize the likelihood
n
Y
µ(xi + Ti , θ)
Di
exp
i=1
Z
xi +Ti
!
µ(s, θ)ds ,
xi
where the notation is obvious.
One employee in the department objects that the method represents a neglect of
information; it is known that the insured have survived, not only the period they
were insured, but also the period from birth until entry into the scheme. Thus, he
claims, the appropriate likelihood is rather
n
Y
i=1
µ(xi + Ti , θ)
Di
exp
Z
xi +Ti
!
µ(s, θ)ds .
0
Settle this apparent paradox. (A suitible framework for discussing the problem is an
enriched model with three states, “uninsured”, “insured” and “dead”.)
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 5))
Exercise 10.4 Reference is to Paragraph 2C in the paper RN “Inference in the
Markov Model”, 09.02.93, the Gompertz-Makeham mortality study.
Inference in the Markov Model
65
(a) Modify the formulas to the situation where person number i entered the study z i
years ago at age xi .
(b) Find explicit expressions for the entries of the asymptotic covariance matrix of
the MLE.
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 6))
Exercise 10.5. (Transformation of Data and Analytical Smoothening) Consider the
mortality model where n independent persons are being observed during the interval
of age [x, x + z), where x and z are integers. Exact times of death are observed and
we disregard censoring during the period except by age z. The force of mortality µ(t)
at the age of t is assumed to be piecewise constant over one-year intervals of age so
that
µ(t) = µk , for t ∈ [k, k + 1), k = x, . . . , x + z + 1.
We introduce µ = (µx , . . . , µx+z−1 )t . Let further more Dk and Tk denote the number
of deaths occured and the observed time of risk in [k, k + 1) respectively and let
µ̂ = (µˆx , . . . , µ̂x+z−1 )t , where
µ̂k =
Dk
, k = x, . . . , x + z − 1.
Tk
(a) Assume that µk can be cast as
µk = gk (θ), k = x, . . . , x + z − 1,
where gk (θ) = g(ξk ; θ) and ξk ∈ [k, k + 1). Here g(t; θ) denotes a function of the age
t and of an unknown parameter θ = (θ1 , . . . , θp )t , p < z. Assume moreover that the
exists and has full rank p for g(θ) = (g x (θ), . . . , gx+z−1 (θ))t . Let
Jacobian Dg = Dg(θ)
dθ
Rz+ = {η = (ηx , . . . , ηx+z−1 ) ∈ Rz | ηk > 0, k = x, . . . , x + z − 1},
and let L : Rz+ → Rz be a differentiable mapping, so that the z × z Jacobian DL =
DL(η)
has full rank z. Define at last α̂ = (α̂ x , . . . , α̂x+z−1 )t and g̃(θ) by
dη
α̂ = L(µ̂)
and
g̃(θ) = L ◦ g(θ) = L(g(θ)).
The parameter θ can be determined in two ways by analytical smoothening. You can
either by modified χ2 -minimizing determine the value of θ that brings g(θ) “closest”
to µ̂, or you can by modified χ2 -minimizing determine the value of θ that brings g̃(θ)
“closest” to α̂. Denote these two estimators by θ̂ and θ̈ respectively and show that θ̂
and θ̈ has the same asymptotic variance. Is θ̂ = θ̈?
66
Exercises in Life Insurance Mathematics
(b) Now assume that
µk = βcξk , k = x, . . . , x + z − 1.
Show how it is possible to use the theory in (a) to construct an estimator for (β, c)
that is easy to calculate and give the asymptotic variance of the estimators.
(FM3 exam, winter 1985-86)
Exercise 10.6 Verify (1.13) – (1.16) in the paper RN “Inference in the Markov
Model”, 09.02.93.
(RN “Inference in the Markov Model” 09.02.93 (Problem 1))
Exercise 10.7 In the situation of paragraph 1E in the paper RN “Inference in the
Markov Model”, 09.02.93, consider the problem of estimating µ from the D i alone,
the interpretation being that it is only observed whether survival to z takes place or
not. Show that the likelihood based on D i , i = 1, . . . , n, is
q N (1 − q)n−N ,
with q = 1 − e−µz , the probability of death before z. (Trivial: It is the binomial
situation.)
Note that N is now sufficient, and that the class of distributions is a regular exponential class. The MLE of q is
N
q∗ =
n
with the first two moments
Eq ∗ = q, Varq ∗ =
q(1 − q)
.
n
The MLE of µ = − log(1 − q)/z is µ∗ = − log(1 − q ∗ )/z. Apply (6.6) in the Appendix
of the paper to show that
µ∗ ∼as N µ,
q
.
2
nz (1 − q)
The asymptotic efficiency of µ̂ relative to µ ∗ is
asVarµ∗
=
asVarµ̂
e
µz
2
− e−
µz
µz
2
!2
=
sinh(µz/2)
µz/2
2
(sinh is the hyperbolic sine function defined by sinh(x) = (e x − e−x )/2). This function
measures the loss of information suffered by observing only death/survival by age z
Inference in the Markov Model
67
as compared to inference based on complete obervation throughout the time interval
(0, z). It is ≥ 1 and increases from 1 to ∞ as µz increses from 0 to ∞. Thus, for small
µz, the number of deaths is all that matters, whereas for large µz, the life lengths are
all that matters. Reflect over these findings.
(RN “Inference in the Markov Model” 09.02.93 (Problem 2))
Exercise 10.8 Use the general theory of Section 2 of the paper RN “Inference in
the Markov Model”, 09.02.93, to prove the special results in Section 1.
(RN “Inference in the Markov Model” 09.02.93 (Problem 3))
Exercise 10.9 Work out the details leading to (2.9) – (2.11) in the paper RN
“Inference in the Markov Model”, 09.02.93.
(RN “Inference in the Markov Model” 09.02.93 (Problem 5))
68
Exercises in Life Insurance Mathematics
11.
Numerical Methods
In this chapter we will use some numerical methods on the theory of life inurance that
we have already seen. One does not always have a program that can calculate the
premiums, the development of the reserves etc. and it can be necessary to develop
your own programs.
Inspired by SW let us at first deal with numerical integration. We wish to calculate
the definite integral
I=
Z
b
f (x)dx.
(11.1)
a
A commonly used numerical method for evaluating (11.1) is the summed Simpson’s
rule. In all it’s simplicity is states that
I=
Z
b
a
N
−1
X
h
f (x)dx '
f (a) + 4f (x1 ) + f (b) + 2
(f (x2k ) + 2f (x2k+1 ))
3
k=1
"
#
with h = (b − a)/2N, xj = a + jh, j = 1, 2, . . . , 2N − 1. The accuracy is good for small
values of h and the implementation of this method is straightforward.
Exercise 11.1 A man aged 25 years considers a pure endowment of 1.000.000, age
of expiration 60. The technical basis of the company is G82M, i. e. the interest rate
is i = 4.5% and the force of mortality is
µx = 0.0005 + 105.88+0.038x−10 .
The premium is a net continuous premium with level intensity π. Disregard expenses.
(a) Put up Thiele’s differential equation and the expression for the equivalence premium π.
(b) Apply the summed Simpson’s rule in order to calculate π numerically.
(c) Solve Thiele’s differential equation and apply the same algorithm as in (b) to
evaluate the premium reserve at times t = 10, 20, 30.
Some expenses are, however, not neglectible and the insured has to pay some expenses during the insurance period in order to cover the administration expenses β,
some fraction of the net premium intensity π. Administration costs are due with a
continuous intensity γVt . Assume that γ is lesser than the force of interest δ.
(d) How is Thiele’s differential equation and the equivalence premium (which is now
the gross premium) modified?
(e) Use your program in (c) to calculate the gross premium and the gross premium
reserve at times t = 10, 20, 30.
Numerical Methods
69
(BM og JC, 1995)
The following is inspired by SW. Now consider the function f . When evaluating the
net premium reserve in the above we solved the differential equation theoretically and
then applied Simpson for evaluation. This is not always possible. Another way is to
solve Thiele’s differential equation numerically; methods for this are plentyful and the
d
Runge-Kutta method of fourth degree is highly recommended. Let dx
y = f (x, y(x)),
xk = x0 + kh, k integer and define
k1 = hf (xk , yk ),
k2 = hf (xk + 0.5h, yk + 0.5k1 ),
k3 = hf (xk + 0.5h, yk + 0.5k2 ),
k4 = hf (xk + h, yk + k3 )
then
(k1 + k2 + 2k3 + k4 )
.
6
The initial condition is y0 = x0 . It turns out that a surprisingly big h can be chosen
when the slope is not too big. A sixth order Runge-Kutta can also be applied, but
the difference from the fourth order R-K is really not that great. It is possible to use
the difference between the fourth and the sixth order Runge-Kutta in order to find
appropriate and varying h’s. The sixth order Runge-Kutta looks like this:
y(xk+1 ) ' y(xk ) +
k1 = hf (xk , yk ),
k2 = hf (xk + 0.5h, yk + 0.5k1 ),
k3 = hf (xk + 0.5h, yk + 0.5k2 ),
k4 = hf (xk + h, yk + k3 ),
7
10
1
2
k5 = hf (xk + h, yk + k1 + k2 + k4 ),
3
27
27
27
1
28
1
546
54
378
k6 = hf (x + h, yk +
k1 − k2 +
k3 +
k4 −
k5 ),
5
625
5
625
625
625
and hence
5
27
125
1
k1 + k4 + k5 +
k6 .
24
48
56
336
In both the fourth and the sixth order R-K, we have initial condition y 0 = x0 .
y(xk+1 ) ' y(xk ) +
It is easy to generalisize the R-K to a system of differential equations. For such a
system of differential equations



y(x) = 


y1 (x)
y2 (x)
..
.
yn (x)









f(x, y) = 


f1 (x, y1 (x), . . . , yn (x))
f2 (x, y1 (x), . . . , yn (x))
..
.
fn (x, y1 (x), . . . , yn (x))






 y =


0


(1)
y0
(2)
y0
..
.
(n)
y0






70
Exercises in Life Insurance Mathematics
we have, similar to the fourth order R-K,
k 1 = hf (xk , y k ),
k 2 = hf (xk + 0.5h, y k + 0.5k1 ),
k 3 = hf (xk + 0.5h, y k + 0.5k2 ),
k 4 = hf (xk + h, y k + k3 ),
y(xk+1 ) ' y(xk ) +
(k 1 + k 2 + 2k 3 + k 4 )
,
6
d
y(x) = f(x, y(x)) and y(x0 ) = y 0 . Remember this method when evaluating
where dx
a system of simultaneous differential equations. It is obvious that this method for
solving differential equation systems simultaneously has a great applicability when
studying the simultaneous development of the statewise reserves in a general Markov
model for a policy. The functions are allowed to depend on each other just as the
statewise reserves depend on each other, compare with the expression for the statewise
reserves.
Exercise 11.2 Consider an endowment insurance, duration 20 years, age at entry
x = 30, interest rate i = 4.5%, force of mortality
µx = 0.0005 + 105.88+0.038x−10 .
The sum insured is 500.000. The only premium is a single net premium Π upon issue
of the contract. The equivalence principle is adopted.
(a) Use the fourth order Runge-Kutta to find this single net premium Π. Use the
same program to study the development of the net premium reserve.
Instead the man does not want to compose any initial capital, but a level continuous
premium with intensity π during the insurance period.
(b) Use Runge-Kutta to calculate this premium.
(c) Our man is having a hard time deciding, but he chooses to compose an initial
capital of 5.000 and then a level continuous premium with intensity π during the
insurance period. What will π be, again applying Runge-Kutta?
(BM and JC, 1995)
Exercise 11.3. (Continued from exercise 9.2) Compute quantities in items (a) and
(b) numerically in the case with the G82M mortality, δ = log(1.045), x = 30, n = 10
and b = 1. (A numerical integration must be performed to find the second order moments. Recall formulas (9.1)-(9.2) from exercise 9.1, page 61. Formula (9.1) requires
Numerical Methods
71
integration in two dimensions. Formulas (9.2) and (9.3) require integration in one
dimension when a table of reserves has been generated.)
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 2))
Exercise 11.4. (Moments of Present Values) Consider the Markov model for disabilities, recoveries and deaths with intensities as in G82M technical basis; no recoveries,
non-differential mortality intensity µ(x) = 0.0005 + 10 −4.12+0.038x at age x and disability intensity σ(x) = 0.0004 + 10 −5.46+0.06x at age x. As annual interest rate use
4.5%.
(a) Compute the expected value and standard deviation of the present value at time 0
of benefits less premiums for a disability pension insurance issued to an active person
at age x = 30, with insurance period n = 20 years and specifying that pensions are
payable continuously with intensity 1 during disability and premiums determined by
the equivalence principle. Perform the calculations also for x = 30, n = 40 and for
x = 50, n = 20.
(b) Now consider a portfolio of I insurance contracts and let V denote the present
value √
of future benefits less premiums for the entire insurance portfolio. Suppose
V + 2 V is to be provided as a reserve. Assume all I contracts are identical pension
insurance policies as described above, with x = 30 and n = 20 and that the individual
life histories
√ are stochastically independent. Study e. g. the “fluctuation loading per
policy”, 2 V /I, as function of I.
(c) Perform calculations parallel to those in (a) for a modified contract where the
benefit, instead of pensions during disability, consists of a lump sum payment of
Z
n
v u−t e−
t
Ru
t
µ(x+s)ds
du
upon onset of disability at time t < n. (That is, the sum paid is the value of the
pension described in (a), capitalised upon onset of disability.) Take x = 30 and
n = 20. Compare with the corresponding result in (a) and comment.
(RN “Problems in Life Insurance Mathematics” 14.10.93 (Problem 3))
Exercise 11.5 Consider a 30-year term insurance, issued on G82M, sum insured
DKK 100.000, age at entry 40. The equivalence principle is adopted. Assume at first
that we have a single net premium.
(a) What is the single net premium?
(b) Put up an expression for the premium reserve V t at time t and calculate it for
t = 0, 5, 10, 15, 20, 25 and 30.
72
Exercises in Life Insurance Mathematics
(c) Now assume that the premium is paid continuously during the entire insurance
period with level intensity π. What is π?
(d) Put up an expression for the premium reserve and evaluate it using the same
values of t as in (b).
(SP(57))
Exercise 11.6 A man aged 40 years has been issued a 20-year term insurance with
sum insured 200.000 and level continuous premium intensity during the insurance
period.
Find the premium intensity and the net premium reserve 10 years after the time of
issue of the contract using the technical basis G82, 4.5% net, assuming the equivalence
princple is adopted.
(SP(52))
Exercise 11.7 Consider the disability model outline below with recovery and excemption from payment of premium by disability. The insurance contract is a 40-year
term insurance, sum insured S = 800.000, age of entry x = 25, level premium intensity π as long as the person is active at the most for 40 years. The interest rate is
4.5%.
σ(x)
0. Active
J
JJJ o
JJ
/
ρ(x)
µ(x)J
JJJ
JJ%
1. Disabled
ttt
ttt
ν(x)
tt
t
t
yt
t
2. Dead
Assume that the intensities are
µx = 0.0005 + 105.88+0.038x−10 ,
σx = 0.0004 + 104.54+0.06x−10 ,
ρx = 0.15,
νx = 10 · µx ,
where µx and σx correspond to the technical basis G82M (including GA82M).
(a) Put up differential equations for the statwise reserves and for the transition probabilities. What are the boundary conditions?
Numerical Methods
73
(b) Find the equivalence premium intensity π applying Runge-Kutta to finde the
transition probabilities and some numerical integration method for evaluating the
integrals.
(c) Study the development of the reserves simultaneously, again applying RungeKutta. What are the statewise reserves at the times t = 10, 20, 30?
(BM and JC, 1995)
Exercise 11.8
(a) Construct graphs for the reserves for the following four life insurances
1. Pure endowment, sum 1 against single net premium.
2. Pure endowment, sum 1 against level continuous premium during the entire
period.
3. Term insurance, sum 1 upon death before time n against level continuous premium.
4. Endowment insurance, sum 1 upon death or at time x + n, if the insured is still
alive, against level continuous premiums.
All insurances are issued on G82, i. e. µ x = 0.0005 + 10−4.12+0.038x and i = 0.045.
(b) Find the premiums above.
(c) What is the expected insurance period for product 3?
(d) Consider product 1. Assume that 50% of the reserve is paid out upon death
before the age of x + n. Construct graphs of the reserve and calculate the single net
premium.
(MSC “Tillæg til opgave E3” FM0 12.10.94)
Exercise 11.9 We consider a force of mortality which is Gompertz-Makeham, i. e.
µx = α + βcx .
As usual the survival function is denoted by F . Consider an x-year old person with
remaining life time T . The survival function for the person is
F (t | x) =
F (x + t)
.
F (x)
The person can sign different kinds of insurance contracts:
74
Exercises in Life Insurance Mathematics
• A pure endowment, where an amount of 1 is payable at time x + n if the person
then is alive. Expected present value for this benefit is
n Ex
= v n F (n | x),
where v = 1/(1 + i) is the one-year discount rate at interest rate i.
• An n-year temporary annuity, payable continuously with force 1 per year until
death, at the most for n years. Expected present value for this contract is
ax:n| =
Z
n
0
t
v F (t | x)dt =
Z
n
t Ex dt.
0
• An n-year term insurance, where an amount of 1 is payable upon death before
age x + n. The expected present value of this contract is
1
Ax:n|
=
Z
n
0
v t F (t | x)µx+t dt
= 1 − δax:n| − n Ex ,
where δ = log(1 + i) is the force of interest.
• An endowment insurance which is the sum of a pure endowment and a term
insurance. The expected present value of this benefit is
Ax:n| = 1 − δax:n| .
(a) Work out tables for µx , F (x) and the density f (x) = F (x)µx for x = 0, 1, . . . , 100.
Use the values in the Danish technical basis G82, i. e. α = 0.0005, β = 10 −4.12 ,
c = 100.038 and interest rate i = 0.045.
(b) Calculate the expected present values for the contracts above for x = 30, i = 0.045.
(c) Redo the calculations in (b) using other values of x and n. In particular, let n
vary for fixed x = 30 and let x vary for fixed n = 30.
(d) Construct tables that show how the expected present values above depend on
i, α, β and c.
(RN FM1 89/90 opg. E3 05.10.89)
Numerical Methods
75
76
Exercises in Life Insurance Mathematics
Insurance Terms
Aggravated circumstance
Allot
Amount allotted
Annuity
Bonus scheme
Capital insurance
Child’s insurance
Collection costs
Deduct
Discount rate
Down payment
Duration
Endowment insurance
Entry
Exemption from payment of premium
Excess mortality
Force of interest
Force of mortality
Hire-purchase agreement
Installment
Insurance period
Interest rate
Issue
Life annuity
Loading for collection costs
Portfolio
Premium free policy
Principal
Pure endowment
Rate of course
Return of premium
Second order technical basis
Single net premium
Sum insured
Surrender
Surrender value
Technical basis
Term insurance
Time of expiry
Underwriting
Skærpede vilkår
Tilbageføre
Tilbageføringsbeløb
Annuitet, rente
Bonusplan
Kapitalforsikring
Børneforsikring
Inkassoomkostninger
At fratrække (skat)
Diskonteringsfaktor
Kontant udbetaling (på lån)
Varighed
Livsforsikring med udbetaling
evt. sammensat livsforsikring
Indtrædelse
Præmiefritagelse
Overdødelighed
Renteintensitet
Dødelighedsintensitet
Købskontrakt
Afdrag
Forsikringstid
Rentefod
Udstedelse
Livrente
Inkassotillæg
Portefølje, bestand
Fripolice
Hovedstol
Ren oplevelsesforsikring
Kurs
Ristorno
Teknisk grundlag af anden orden
Nettoengangspræmie
Forsikringssum
Tilbagekøb, genkøb
Tilbagekøbsværdi, genkøbsværdi
Teknisk rundlag, beregningsgrundlag
Ophørende livsforsikring
Udløbsdato
Processen at tegne en forsikring