MATH 3286
Mathematics of Finance
Alex Karassev
COURSE OUTLINE
•
Theory of Interest
1.
2.
3.
4.
5.
•
Interest: the basic theory
Interest: basic applications
Annuities
Amortization and sinking funds
Bonds
Life Insurance
6.
7.
8.
9.
Preparation for life contingencies
Life tables and population problems
Life annuities
Life insurance
Chapter 1
INTEREST: THE BASIC THEORY
•
•
•
•
•
•
Accumulation Function
Simple Interest
Compound Interest
Present Value and Discount
Nominal Rate of Interest
Force of Interest
1.1 ACCUMULATION FUNCTION
Definitions
• The amount of money initially invested is
called the principal.
• The amount of money principal has grown to
after the time period is called the
accumulated value and is denoted by
A(t) – amount function.
 t ≥0 is measured in years (for the moment)
• Define Accumulation function a(t)=A(t)/A(0)
• A(0)=principal
• a(0)=1
• A(t)=A(0)∙a(t)
Natural assumptions on a(t)
• increasing
• (piece-wise) continuous
a(t)
a(t)
(0,1)
t
(0,1)
Note: a(0)=1
a(t)
t
(0,1)
t
Definition of Interest
and
Rate of Interest
• Interest = Accumulated Value – Principal:
Interest = A(t) – A(0)
• Effective rate of interest i (per year):
a(1)  a(0) A(1)  A(0)

a(0)
A(0)
(since A(t)  A(0)  a(t))
i  a(1)  1 
• Effective rate of interest in nth year in:
in 
A(n)  A(n - 1) a(n)  a(n - 1)

A(n - 1)
a(n - 1)
Example (p. 5)
a(t)=t2+t+1
•
•
•
•
•
Verify that a(0)=1
Show that a(t) is increasing for all t ≥ 0
Is a(t) continuous?
Find the effective rate of interest i for a(t)
Find in
Two Types of Interest
( ≡ Two Types of Accumulation Functions)
• Simple interest:
– only principal earns interest
– beneficial for short term (1 year)
– easy to describe
• Compound interest:
– interest earns interest
– beneficial for long term
– the most important type of accumulation
function
1.2 SIMPLE INTEREST
a(t)=1+it, t ≥0
•Amount function:
A(t)=A(0) ∙a(t)=A(0)(1+it)
•Effective rate is i
•Effective rate in nth year:
i
in 
1  i (n  1)
a(t) =1+it
1+i
(0,1)
t
1
a(t)=1+it
Example (p. 5)
Jack borrows 1000 from
the bank on January 1,
1996 at a rate of 15%
simple interest per year.
How much does he owe
on January 17, 1996?
t=?
Solution
A(0)=1000
i=0.15
A(t)=A(0)(1+it)
=1000(1+0.15t)
How to calculate t in practice?
• Exact simple interest
t = number of days
365
• Ordinary simple interest (Banker’s Rule)
number
of
days
t=
360
Number of days: count the last day but not the first
A(t)=1000(1+0.15t)
Number of days (from Jan 1 to Jan 17) = 16
• Exact simple interest
 t=16/365
 A(t)=1000(1+0.15 ∙ 16/365) = 1006.58
• Ordinary simple interest (Banker’s Rule)
 t=16/360
 A(t)=1000(1+0.15 ∙ 16/360) = 1006.67
1.3 COMPOUND INTEREST
Interest earns interest
• After one year:
a(1) = 1+i
• After two years:
a(2) = 1+i+i(1+i) = (1+i)(1+i)=(1+i)2
• Similarly after n years:
a(n) = (1+i)n
COMPOUND INTEREST
Accumulation Function
a(t)=(1+i)t
•Amount function:
A(t)=A(0) ∙a(t)=A(0) (1+i)t
•Effective rate is i
•Moreover effective rate in
nth year is i (effective rate
is constant):
(1  i) n  (1  i) n1
in 
 1  i 1  i
n 1
(1  i)
a(t)=(1+i)t
1+it
1+i
(0,1)
t
1
How to evaluate a(t)?
• If t is not an integer,
first find the value for
the integral values
immediately before and
after
• Use linear interpolation
• Thus, compound
interest is used for
integral values of t and
simple interest is used
between integral values
a(t)=(1+i)t
(1+i)2
1+i
1
1
t 2
Example (p. 8)
Jack borrows 1000 at 15%
compound interest.
a) How much does he owe after 2
years?
b) How much does he owe after 57
days, assuming compound interest
between integral durations?
c) How much does he owe after 1 year
and 57 days, under the same
assumptions as
in (b)?
d) How much does he owe after 1 year
and 57 days, assuming linear
interpolation between integral
durations
e) In how many years will his principal
have accumulated to 2000?
a(t)=(1+i)t
A(t)=A(0)(1+i)t
A(0)=1000, i=0.15
A(t)=1000(1+0.15)t
1.4 PRESENT VALUE AND DISCOUNT
The amount of money that will accumulate
to the principal over t years is called the
present value t years in the past
t
-t
PRESENT
VALUE
PRINCIPAL
ACCUMULATED
VALUE
Calculation of present value
•
•
•
•
t=1, principal = 1
Let v denote the present value
v (1+i)=1
v=1/(1+i)
v=1/(1+i)
In general:
• t is arbitrary
• a(t)=(1+i)t
• [the present value of 1 (t years in the past)]∙ (1+i)t = 1
• the present value of 1 (t years in the past) = 1/ (1+i)t = vt
t
 1 
t
v 

(
1

i
)

1 i 

t
t
a(t)=(1+i)
gives the value
of one unit
(at time 0)
at any time t,
past or future
a(t)=(1+i)t
(0,1)
t
If principal is not equal to 1…
present value = A(0) (1+i)t
t<0
t=0
t>0
PRESENT
VALUE
PRINCIPAL
A (0)
ACCUMULATED
VALUE
A(0) (1+i)t
A(0) (1+i)t
Example (p. 11)
Solution
The Kelly family buys a new
house for 93,500 on
May 1, 1996.
How much was this house
worth on May 1, 1992 if real
estate prices have risen at a
compound rate for 8 % per
year during that period?
a(t)=(1+i)t
• Find the present value of
A(0) = 93,500
• 996 - 1992 = 4 years
in the past
• t = - 4, i = 0.08
• Present value =
A(0) (1+i)t =
93,500 (1+0.8) -4 =
68,725.29
If simple interest is assumed…
• a (t) = 1 + it
• Let x denote the
present value of one
unit t years in the
past
• x ∙a (t) = x (1 + it) =1
• x = 1 / (1 + it)
NOTE:
In the last formula,
t>0
a(t) =1+it
Thus, unlikely to the
case of compound
interest, we cannot
use the same formula
for present value and
accumulated value in
the case of simple
interest
1
1 / (1 + it)
a(t) =1+it
1 / (1 - it)
1
t
Discount
Alternatively:
• We invest 100
• After one year it
accumulates to 112
• The interest 12 was
added at the end of
the term
• Look at 112 as a
basic amount
• Imagine that 12
were deducted from
112 at the beginning
of the year
• Then 12 is amount
of discount
Rate of Discount
Definition
d=
=
Effective rate of discount
d
accumulated value after 1 year – principal = A(1) – A(0)
A(1)
accumulated value after 1 year
A(0) ∙a(1)– A(0)
A(0) ∙a(1)
=
a(1) – 1
a(1)
Recall:
i=
accumulated value after 1 year – principal
principal
=
a(1) – 1
a(0)
In nth year…
a(n)  a(n  1)
dn 
a ( n)
Identities relating d to i and v
a(1) ia(0) (1  i)  1
i
d d 
 Note: d <i
a1(
1)i
1 i
1 i
1  d iv (1  i )  i
1
1 d  1


v
1 i
1 i
1 i
d
i
1 d
Present and accumulated
values in terms of d:
1
1 d  v 
1 i
• Present value = principal * (1-d)t
• Accumulated value = principal * [1/(1-d)t]
If we consider positive and negative values of t then:
a(t) = (1 - d)-t
Examples (p. 13)
1. 1000 is to be accumulated by January 1,
1995 at a compound rate of discount of 9%
per year.
a) Find the present value on January 1, 1992
b) Find the value of i corresponding to d
2. Jane deposits 1000 in a bank account on
August 1, 1996. If the rate of compound
interest is 7% per year, find the value of
this deposit on August 1, 1994.
1.5
NOMINAL RATE OF INTEREST
Example (p. 13)
A man borrows 1000 at an effective rate of
interest of 2% per month. How much does
he owe after 3 years?
Note: t is the number of
effective interest periods
in any particular problem
More examples… (p. 14)
• You want to take out a mortgage on a
house and discover that a rate of interest
is 12%
per year.
However,
you find out
In both
examples
the
given rates
that this(12%
rate isand
“convertible
semiof interest
18%) were
annually”.
the effective rate of
nominal
rates Is
of 12%
interest
interest per year?
• Credit card charges 18% per year
convertible monthly. Is 18% the effective
rate of interest per year?
…yet another example
• You have two credit card offers:
– 17% convertible semi-annually
– 16% convertible monthly
• Which is better?
Definition
• Suppose we have interest convertible
m times per year
• The nominal rate of interest i(m) is
defined so that i(m) / m is an effective
rate of interest in 1/m part of a year
Note:
If i is the effective
rate of interest per
year, it follows that
 i 
1  i  1 

m 

(m)
m
Equivalently:
i
(m)
 [1  i ]
1/ m
m
1
In other words,
i is the effective rate
of interest
convertible annually
which is equivalent
to the effective rate
of interest i(m) /m
convertible mthly
Examples (p. 15)
1. Find the accumulated value of 1000 after
three years at a rate of interest of 24 %
per year convertible monthly
2. If i(6)=15% find the equivalent nominal
rate of interest convertible semi-annually
Formula that relates
nominal rates of interest
m
 i 
 i 
1 
  1  i  1 

m 
n 


(m)
(n)
n
Nominal rate of discount
• The nominal rate of discount d(m) is
defined so that d(m) / m is an effective
rate of discount in 1/m part of a year
• Formula:
 d 
1  d  1 

m 

(m)
m
Formula relating nominal rates of
interest and discount
 d 
1  d  1 

n 

(n)
n
1
1 d 
1 i

1  i  (1  d )
 i 
1  i  1 

m 

(m)
 i 
 d 
1 
  1 

m 
n 


(m)
1
m
m
(n)
n
Example
• Find the nominal rate of discount
convertible semiannualy which is
equivalent to a nominal rate of interest
of 12% convertible monthly
m
 i 
 d 
1 
  1 

m 
n 


(m)
(n)
n
1.6 FORCE OF INTEREST
• What happens if the number m of
periods is very large?
• One can consider mathematical model
of interest which is convertible
continuously
• Then the force of interest is the nominal
rate of interest, convertible continuously
Definition
Nominal rate of interest
equivalent to i:
Let m approach infinity:
i
( m)
lim i
 m[(1  i)
1/ m
( m)
m
 1]
1/ m
 lim m[(1  i)
 1]
1/ m
 lim m[(1  i)
 1]
m
We define the
δ
force of interest
equal to this limit:
  lim i
m 
( m)
m 
Formula
• Force of interest
• Therefore
• and
δ = ln (1+i)
eδ = 1+i
a (t) =
t
δt
(1+i) =e
• Practical use of δ: the previous
formula gives good approximation
to a(t) when m is very large
Example
• A loan of 3000 is taken out on June 23,
1997. If the force of interest is 14%, find
each of the following:
– The value of the loan
on June 23, 2002
– The value of i
– The value of i(12)
Remark
t
t

[(1  i ) ]  (1  i )  ln( 1  i )  (1  i )  
t

[(1  i )t ] a(t )


t
(1  i )
a(t )
The last formula shows that it is reasonable to define force
of interest for arbitrary accumulation function a(t)
Definition
The force of interest corresponding to a(t):
a(t )
t 
a(t )
Note:
1) in general case,
force of interest depends on t
2) it does not depend on t ↔ a(t)= (1+i)t !
Example (p. 19)
• Find in δt the case of simple interest
• Solution
a(t ) (1  it )
i
t 


a(t )
1  it
1  it
How to find a(t)
if we are given by δt ?
We have:
Consider differential equation
in which a = a(t)
is unknown function:
a(t )
t 
a(t )
a    a
t
t
Since a(0) = 1
its solution is given by

a(t )  e 0
 r dr
Applications
• Prove that if δt = δ is a constant then
a(t) = (1+i)t for some i
• Prove that for any amount function A(t) we have:
n
 A(t ) dt  A(n)  A(0)
t
0
• Note: δt dt represents the effective rate of
interest over the infinitesimal “period of time” dt .
Hence A(t)δt dt is the amount of interest earned
in this period and the integral is the total amount
Remarks
• Do we need to define the force of discount?
• It turns out that the force of discount coincides
with the force of interest!
(Exercise: PROVE IT)
• Moreover, we have the following inequalities:
d    d ( m )  d ( m 1)        i ( m1)  i ( m )   i
• and formulas:
1 1
  1 and
d i
1
d
(m)

1
i (m)
1

m
Chapter 2
INTEREST: BASIC APPLICATIONS
• Equation of Value
• Unknown Rate of Interest
• Time-Weighted Rate of Return
2.1 Equation of Value
• Four numbers:
• principal A(0)
• accumulated value A(t) = A(0) ∙ a(t)
• period of investment t (determine
effective period to find t)
• rate of interest i
• Time diagram
• Bring all entries of the diagram to the
same point in time and write
the equation of value
Examples
1. Find the accumulated value of 500 after 173 months
at a rate of compound interest of 14% convertible
quarterly (p. 30)
2. Alice borrows 5000 from FF Company at a rate of
interest 18% per year convertible semi-annually. Two
years later she pays the company 3000. Three years
after that she pays the company 2000. How much
does she owe seven years after the loan is taken out?
(p. 31)
3. Eric deposits 8000 on Jan 1, 1995 and 6000 on
Jan 1, 1997 and withdraws 12000 on Jan 1, 2001.
Find the amount in Eric’s account on Jan 1, 2004 if the
rate of compound interest is 15% per year (p. 31)
More Examples…
Unknown time
John borrows 3000 from FFC. Two years
later he borrows another 4000. Two years
after that he borrows an additional 5000.
At what point in time would a single loan
of 12000 be equivalent if i = 0.18 ? (p. 32)
Unknown rate of interest
Find the rate of interest such that an
amount of money will triple itself over 15
years (p. 32)
2.2 UNKNOWN RATE OF INTEREST
• We need to find the rate of interest i
• Set up equation of value and solve it for i
• Very often the resulting equation is a polynomial
equation in i of degree higher than 2
• In general, there is no formula for solutions of
equation of degree ≥ 5 (and the formulas for
degrees 3 or 4 are very complicated)
• Use approximations (numerical methods)
Examples
1. Joan deposits 2000 in her bank
account on January 1, 1995, and then
deposits 3000 on January 1, 1998. If
there are no other deposits or
withdrawals and the amount of money
in the account on January 1, 2000 is
7100, find the effective rate of interest.
2. Obtain a more exact answer to the
previous question
2.3 TIME-WEIGHTED RATE OF RETURN
• Let B0, B1, … , Bn-1, Bn denote balances in a fund
such that precisely one deposit or withdrawal
(denoted by Wt) is made immediately after Bt
starting from t=1
• Let W1, … , Wn-1 denote the amounts of deposits
(Wt > 0) or withdrawals (Wt > 0) and let W0 = 0
• Determine rate of interest earned in the time period
between balances:
Bt
it 
Bt 1  Wt 1
1
The time-weighted rate of return is defined by
i = (1+i1 ) (1+i2 ) … (1+in ) - 1
Example (p. 35)
On January 1, 1999, Graham’s stock portfolio is worth
500,000. On April 30, 1999, the value has increased to
525,000. At that point, Graham adds 50,000 worth of stock
to this portfolio. Six month later, the value has dropped to
560,000, and Graham sells 100,000 worth of stock. On
December 31, 1999, the portfolio is again worth 500,000.
Find the time-weighted rate of return for Graham’s portfolio
during 1999.
Chapter 3
ANNUITIES
• Basic Results
• Perpetuities
• Unknown Time and
Unknown Rate of Interest
• Continuous Annuities
• Varying Annuities
3.2 Basic Results
• Definition: Annuity is a series of
payments made at regular intervals
• Practical applications: loans, mortgages,
periodic investments
• Basic model: consider an annuity under
which payments of 1 are made at the end
of each period for n equal periods
Formulas
Series of n payments
of 1 unit
present value
of annuity an|
0
1
1
1
1
2
3
accumulated value
of annuity sn|
1
…..
n
The present (accumulated) value of the series of payments is equal
to the sum of present (accumulated) values of each payment
nn
2
n nn 1
aan vvvv2 2vv3 3

v

v
(
1

v

v



v v )  n
1



v
n 1
n
sn  an (1 ai )n  
(1  i ) 

i i 
n
n
n
n

1 v
1 v
1 v
1 v
v

 n n
nn
n
n

1
1
(
1

i
)

v
(
1

i
)
(
1

i
)
1
1 svn  ( i1) an1  i 
i
sn 


i
v
i
i
Examples (p. 46)
1.
(Loan) John borrows 1500 and wishes to pay it
back with equal annual payments at the end of
each of the next ten years. If i = 17% determine the
size of annual payment
2.
(Mortgage) Jacinta takes out 50,000 mortgage. If
the mortgage rate is 13% convertible semiannually,
what should her monthly payment be to pay off the
mortgage in 20 years?
3.
(Investments) Eileen deposits 2000 in a bank
account every year for 11 years. If i = 6 % how
much has she accumulated at the time of the last
deposit?
One more example… (p. 47)
•
Elroy takes out a loan of 5000 to buy a
car. No payments are due for the first 8
months, but beginning with the end of
9th month, he must make 60 equal
monthly payments. If i = 18%, find:
a) the amount of each payment
b) the amount of each payment if there
is no payment-free period
Annuity-immediate and Annuity-due
an|
Annuity-immediate
(payments made at the end)
1
1
1
0
än|
1
2
3
…..
sn|
1
n
..
sn|
Annuity-due
(payments made at the beginning)
1
1
1
1
2
3
an  an (1  i )
1
…..
n
n+1
s1n vsnn (1  i ) 1  v n 1  v n
an  an (1  i ) 
(1  i )  i 
i
di ) n  1
n
1

i
(
1

1 v
sn 
an 
d
d
Example (p. 50)
•
Henry takes out a loan of 1000 and
repays it with 10 equal yearly payments,
the first one due at the time of the loan.
Find the amount of each payment if
i = 16%
3.3 Perpetuities
• Definition: Perpetuity is an annuity
whose payments continue forever
• Practical applications: perpetual bonds
• Basic model: consider perpetuity under
which payments of 1 are made forever
Formulas
present value of
perpetuity a∞|
0
1
1
1
1
2
3
…..

a  v  v  v    v  ...   v
2
3
n
k 1
k
1
a 
i
1 v
1
a  lim an  lim

n 
n 
i
i
n
1
1 i 1
a  a (1  i )  (1  i ) 

i
i
d
1
a 
d
1.
2.
– Unknown
Time
3.4Examples
Unknown
Time and
Unknown Rate
of Interest
A fund of 5000 is used to award scholarships of amount
500, one per year, at the end of each year for as long as
possible. If i=9% find the number of scholarships which
can be awarded, and the amount left in the fund one
year after the last scholarship has been awarded
A trust fund is to be built by means of deposits of amount
5000 at the end of each year, with a terminal deposit, as
small as possible, at the end of the final year. The
purpose of this fund is to establish monthly payments of
amount 300 into perpetuity, the first payment coming
one month after the final deposit. If the rate of interest is
12% per year convertible quarterly, find the number of
deposits required and the size of the final deposit
Example – Unknown Rate of Interest
3. At what effective yearly rate of interest is
the present value of 300 paid at the end
of every month, for the next 5 years,
equal to 15,000?
• 1st method: linear interpolation
• 2nd method: successive approximations
t
mn
mn1


1
i
1
1
t
t
(m)
3.5
Continuous
Annuities
   vm  vm  vm 
a   1 
n|
m
m
m
m
t

1
t

1
0


Let effective period be 1/m part of the year and i(m)t /m
(m)
mn
be the effective
mn rate of interest: (1+
m in(m)/m)m = 1 + i
1 1  vm
1 1  ( vm )
1 1 v
 vm
 vm
 vm

1  vm Series
m of n∙m1 payments
vm
m ( m1)  vm
(m
m)
a
s
of 1/m
n
n n|
n
n|
1/m n1/m 1/m
1/m
1 1 v
1 1 v
1 1 v
1 v



 (m)
(m)
(m)
m0 11/m 12/m m3/mi
m i
i
…..
 1mn
1
n = nm(1/m)
(
m
)
nm
vm
m
 i 
1 v
n
1 
  ( m ) (1  i ) 
(m)
n
|
n
|
m
i
a
s


n
n
n|
1/m
1/m
1 1/m
v
(1 1/m
i ) n1|
(nm )
( mn)
 1 ( m)
1 avn|  i ( mn) (1 sin)| 
i
 0( m )1/m(12/m
 )3/m …..
(m)
n = nm(1/m)
i
i
s ( m)  a
( m)
( m)
Formula:
n
Let m approach infinity…
• Present
1 i
(m)
1 
a  limofacontinuous
 lim annuity
m 
m 
mn
value
n|
m
n|
t 1

t
a  limmna (1m ) t / m
 n|  lim
v 

m


n
|
m 
m 
t 1 m
(m)
n
n
n
(m)
v

1
v

1
v

1
• Accumulated
value
t
s

lim
s
of
dt 

n| m 
annuity
n|
0 vcontinuous
1
ln v

  ln 1  i 
ln 

1 i 
Formulas:
1 vn
1 vn
n

n


ln
1

i

(1  i )  1
1 v
n
a 
n|

sn| 

3.6 Varying Annuities
• Arithmetic annuities
– increasing
– decreasing
• Arithmetic increasing perpetuities
Arithmetic annuity
• Definition: Arithmetic annuity is a series
of payments made at regular intervals
such that
– the first payment equals P
– payments increase by Q every year
• Thus payments form
arithmetic progression:
P, P+Q, P+2Q, …, P+(n-1)Q
Formulas
Series of n payments
k-th payment = P+ (k-1)Q
accumulated value
k=1,2,…, n
present value
S
A
P P+Q P+2Q
P+(n-1)Q
0
1
2
3
 an  nv n 

A  Pan  Q 
i


…..
n
 sn  n 
S  Psn  Q 

 i 
1) Increasing annuity with P = 1, Q = 1
Series of n payments
k-th payment = k
k=1,2,…, n
present value
(Ia)n|
1
We have:
0
2
3
 an  nv n  n

A  Pan  Q  n
i


P=1
 an  nv
Q = 1 ( Ia ) n  an  
i

n
n
 ssn n n
S  Psn  Q  n 
( Is ) n  i 
a  nv
i
(Is)n|
n
Two
Special
Cases
…..
1
2
3
( Ia ) n 
accumulated value



i
 sn  n 
( Is ) n  sn  

 i 
2) Decreasing annuity with P = n, Q = -1
present value
(Da)n|
Series of n payments
k-th payment = n – k + 1
k=1,2,…, n
n
0
n-1 n-2
1
2
3
We have:
 an  nv n 

A  Pan  Q 
 n i

( Da ) 
na
P=n
Q = -1 n
 an  nv n 

( Da ) n  nan  
i


i
accumulated value
(Ds)n|
1
…..
n
 sn  n 
S  Psn  Q n(1 i ) n
 i 
( Ds ) n   s  n 
n
i
( Ds ) n  nsn  
 i 
 sn
Increasing perpetuity
k-th payment = k
continues forever
present value
(Ia)∞|
1
0
1
2
3
2
3
…..
n
1

v
n
n

nv
an  nv
( Ia )   lim ( Ia ) n  lim
 lim d
n 
n 
n 
i
i
1 1
( Ia )    2
i i
Examples (p. 62 - 65)
1. Find the value, one year before the first payment, of a
series of payments 200, 500, 800,… if i = 8% and the
payments continue for 19 years
2. Find the present value of an increasing perpetuity
which pays 1 at the end of the 4th year, 2 at the end of
the 8th year, 3 at the end of the 12th year, and so on, if i
= 0.06
3. Find the value one year before the first payment of an
annuity where payments start at 1, increase by annual
amounts of 1 to a payment of n, and then decrease by
annual amounts of 1 to a final payment of 1
4. Show both algebraically and verbally that
(Da)n| = (n+1)an| - (Ia)n|
More examples: Geometric annuities
1. Geometric annuity
An annuity provides for 15 annual payments. The first is
200, each subsequent is 5% less than the one
preceding it. Find the accumulated value of annuity at
the time of the final payment if i = 9%
2. Inflation and real rate of interest
In settlement of a lawsuit, the provincial court ordered
Frank to make 8 annual payments to Fred. The first
payment of 10,000 is made immediately, and future
payments are to increase according to an assumed rate
of inflation of 0.04 per year. Find the present value of
these payments assuming i = .07
Chapter 4
AMORTIZATION AND
SINKING FUNDS
• Amortization
• Amortization Schedule
• Sinking Funds
• Yield Rates
4.1 Amortization
• Amortization method: repay a loan by means of
instalment payments at periodic intervals
• This is an example of annuity
• We already know how to calculate the amount
of each payment
• Our goal: find the outstanding principal
• Two methods to compute it:
– prospective
– retrospective
Two Methods
• Prospective method:
outstanding principal at any point in time is equal
to the present value at that date of all remaining
payments
• Retrospective method:
outstanding principal is equal to the original
principal accumulated to that point in time minus
the accumulated value of all payments previously
made
• Note: of course, this two methods are equivalent.
However, sometimes one is more convenient than
the other
Examples (p. 75-76)
• (prospective) A loan is being paid off with
payments of 500 at the end of each year for the
next 10 years. If i = .14, find the outstanding
principal, P, immediately after the payment at the
end of year 6.
• (retrospective) A 7000 loan is being paid of with
payments of 1000 at the end of each year for as
long as necessary, plus a smaller payment one
year after the last regular payment. If i = 0.11
and the first payment is due one year after the
loan is taken out, find the outstanding principal,
P, immediately after the 9th payment.
One more example… (p. 77)
• (Renegotiation) John takes out 50,000 mortgage at
12.5 % convertible semi-annually. He pays off the
mortgage with monthly payments for 20 years, the
first one is due one month after the mortgage is
taken out. Immediately after his 60th payment, John
renegotiates the loan. He agrees to repay the
remainder of the mortgage by making an immediate
cash payment of 10,000 and repaying the balance
by means of monthly payments for ten years at 11%
convertible semi-annually. Find the amount of his
new payment.
4.2 Amortization Schedule
• Goal: divide each payment (of annuity) into
two parts – interest and principal
• Amortization schedule – table, containing
the following columns:
– payments
– interest part of a payment
– principal part of a payment
– outstanding principal
Example
5,000 at 12 % per year repaid by 5
annual payments
Amortization schedule:
Duration
Payment
Interest
Principal
Repaid
0
Outstanding
Principal
5000.00
1
1387.05
600.00
787.05
4212.95
2
1387.05
505.55
881.50
3331.45
3
1387.05
399.77
987.28
2344.17
4
1387.05
281.30
1105.75
1238.42
5
1387.05
148.61
1238.44
0
Outstanding
principal
P
Payment
X
t-1
• Interest earned during
interval (t-1,t) is iP
• Therefore interest portion
of payment X is iP
and principal portion is
X - iP
t
Recall: in practical problems, the outstanding principal P
can be found by prospective or retrospective methods
Example
A 1000 loan is repaid by annual payments of 150, plus a
smaller final payment. If i = .11, and the first payment is
made one year after the time of the loan, find the amount
of principal and interest contained in the third payment
General method
outstanding principal at t
present value =
outst. principal at 0
an-t|
an|
1
1
1
1
…..
0
1
2
1
…..
t
t+1
n
interest portion of (t+1)-st payment = i a n-t| = 1 – vn-t
principal portion of (t+1)-st payment = 1 – (1 – vn-t ) = vn-t
If each payment is X then
interest part of kth payment = X (1 – vn-k+1 )
principal part of kth payment = X∙vn-k+1
Example (p. 79)
• A loan of 5000 at 12% per year is to be
repaid by 5 annual payments, the first due
one year hence. Construct an amortization
schedule
General rules to obtain an amortization schedule
Duration
Payment
Interest
Principal
Repaid
0
Outstanding
Principal
i = 12 %
5000.00
1
1387.05
600.00
787.05
4212.95
2
1387.05
505.55
881.50
3331.45
3
1387.05
399.77
987.28
2344.17
4
1387.05
281.30
1105.75
1238.42
5
1387.05
148.61
1238.44
0
Take the entry from “Outs. Principal” of the previous row, multiply it by
i, and enter the result in “Interest”
II. “Payment” – “Interest” = “Principal Repaid”
III. “Outs. Principal” of prev. row - “Principal Repaid” = “Outs. Principal”
IV. Continue
I.
Example (p. 80)
A 1000 loan is repaid by annual payments of 150, plus a
smaller final payment. The first payment is made one
year after the time of the loan and i = .11. Construct an
amortization schedule
Duration
0
1
2
3
4
5
6
7
8
9
10
11
12
Payment Interest
150
110.00
Principal Repaid
Outstanding Principal
1000
40.00
960.00
4.3 Sinking Funds
• Alternative way to repay a loan – sinking
fund method:
– Pay interest as it comes due keeping
the amount of the loan (i.e.
outstanding principal) constant
– Repay the principal by a single
lump-sum payment at some point in
the future
lump-sum payment L
interest
iL
iL
iL
…..
0
1
2
n
Loan L
• Lump-sum payment L is accumulated by periodic
deposits into a separate fund, called the sinking fund
• Sinking fund has rate of interest j usually different from
(and usually smaller than) i
• If (and only if) j is greater than i then sinking fund method
is better (for borrower) than amortization method
Examples (p. 82)
• John borrows 15,000 at 17% effective annually. He agrees to
pay the interest annually, and to build up a sinking fund which
will repay the loan at the end of 15 years. If the sinking fund
accumulates at 12% annually, find
– the annual interest payment
– the annual sinking fund payment
– his total annual outlay
– the annual amortization payment which would pay off this
loan in 15 years
• Helen wishes to borrow 7000. One lender offers a loan in
which the principal is to be repaid at the end of 5 years. In the
mean-time, interest at 11% effective is to be paid on the loan,
and the borrower is to accumulate her principal by means of
annual payments into a sinking fund earning 8% effective.
Another lender offers a loan for 5 years in which the
amortization method will be used to repay the loan, with the
first of the annual payments due in one year. Find the rate of
interest, i, that this second lender can charge in order that
Helen finds the two offers equally attractive.
4.4
Yield Rates
• Investor:
– makes a number of payments at various points in
time
– receives other payments in return
• There is (at least) one rate of interest for which the
value of his expenditures will equal the value of the
payments he received (at the same point in time)
• This rate is called the yield rate he earns on his
investment
• In other words, yield rate is the rate of interest which
makes two sequences of payments equivalent
• Note: to determine yield rate of a certain investor, we
should consider only payments made directly to, or
directly by, this investor
Examples
• Herman borrows 5000 from George and
agrees to repay it in 10 equal annual
instalments at 11%, with first payment due in
one year. After 4 years, George sells his right
to future payments to Ruth, at a price which will
yield Ruth 12% effective
– Find the price Ruth pays.
– Find George’s overall yield rate.
• At what yield rate are payments of 500 now
and 600 at the end of 2 years equivalent to a
payment of 1098 at the end of 1 year?
Chapter 5
BONDS
• Price of a Bond
• Book Value
• Amortization
Schedule
• Other Topics
5.1 Price of a Bond
Bonds:
• certificates issued by a corporation or government
• are sold to investors
• in return, the borrower (i.e. corporation or government) agrees:
– to pay interest at a specified rate (the coupon rate) until a specified
date (the maturity date)
– and, at that time, to pay a fixed sum (the redemption value)
• Usually:
– the coupon rate is a nominal rate convertible semiannually and is
applied to the face (or par) value stated on the front of the bond
– the face and redemption values are equal (not always)
• Thus we have:
– regular interest payments
– lump-sum payment at the end
Example of a bond
• Face amount = 500
• Redemption value = 500
• Redeemable in 10 years with semiannual coupons
at rate 11%, compounded semiannually
• Then in return investor receives:
• 20 half-yearly payments of (.055)(500) = 27.50 interest
• a lump-sum payment of 500 at the end of the 10 years
Notations
• F = the face value (or the par value)
• r = the coupon rate per interest period (we assume
that the quoted rate will be a nominal rate 2r
convertible semiannually)
• Note: the amount of each interest payment (coupon)
is Fr
• C = the redemption value (often C = F, i.e. bond
“redeemable at par”)
• i = the yield rate per interest period
• n = the number of interest periods until the
redemption date (maturity date)
• P = the purchase price of a bond to obtain yield rate i
coupon
(interest)
Fr
redemption value C
Fr
Fr
Note: time is
measured in
half-years
…..
0
1
2
n
purchase price P
Equation to find yield rate i
P  ( Fr )an|i  C (1  i)
Note: often C = F
n
Examples (p. 94 – p. 96)
• A bond of 500, redeemable at par after 5
years, pays interest at 13% per year
convertible semiannually. Find the price to
yield an investor
– 8% effective per half-year
– 16% effective per year
• Remarks
– P < C since the yield rate is higher than the
coupon rate, i > r
– therefore the investor is buying the bond at a
discount
– otherwise (if i < r) we would have P > C and
then the investor would have to buy the bond
at a premium of P - C
• A corporation decides to issue 15-years
bonds, redeemable at par, with face
amount of 1000 each. If interest payments
are to be made at a rate of 10% convertible
semiannually, and if George is happy with
a yield of 8% convertible semiannually,
what should he pay for one of these
bonds?
• A 100 par-value 15-year bond with coupon
rate 9% convertible semiannually is selling
for 94. Find the yield rate.
5.2 Book Value
• The book value of a bond at a time t is an analog of an
outstanding balance of a loan
• The book value Bt is the present value of all future
payments
• At time t where the tth coupon has just been paid we have:
Bt  ( Fr )ant|  Cv
n t
• Remarks
– Usually C = F
– In the last formula, an-t and v are computed using the
yield rate i
– P = B0 < Bt < Bt+1< Bn = C or P = B0 > Bt > Bt+1 > Bn = C
Examples (p. 96 - 97)
1. Find the book value immediately after
the payment of 14th coupon of a 10-year
1,000 par-value bond with semiannual
coupons, if r=.05 and the yield rate is
12% convertible semiannually.
2. Let Bt and Bt+1 be the book values just
after the tth and (t+1)th coupons are paid.
Show that Bt+1 = Bt (1+i) – Fr
3. Find the book value in 1) exactly 2
months after the 14th coupon is paid.
How do we find the book value
between coupon payment dates?
Assume
simple interest
at rate i per period
between adjacent
coupon payments
Example
Find the book value
exactly 2 months after the
14th coupon is paid of a
10-year 1,000 par-value
bond with semiannual
coupons, if r=.05 and the
yield rate is 12%
convertible semiannually.
Alternative approach
•
•
•
•
Since Bt+1 = Bt (1+i) – Fr we can view Bt (1+i) as the book value just
before next (i.e. (t+1)th) coupon is paid
Book value calculated using simple interest between coupon dates
is called the flat price of a bond
Using linear interpolation between Bt+1 and Bt we obtain the market
price (or the amortized value) of the bond
Clearly market price ≤ flat price at any given moment
(1+i) Bt
Fr
Bt
Bt+1
t
t+1
Example (p. 98)
Find the market price exactly 2 months
after the 14th coupon is paid of a
10-year 1,000 par-value bond with
semiannual coupons, if r =.05 and the
yield rate is 12% convertible
semiannually.
5.3 Bond Amortization Schedule
• Goal: trace changes of the book value
• Bond amortization schedule – table, containing the following
columns:
– time
Time Coupon Interest
Principal
Book Value
adjustment
– coupon
0
1037.17
– interest
1
40
31.12
8.88
1028.29
– principal adjustment
2
40
30.85
9.15
1019.14
– book value
3
40
30.57
9.43
1009.71
4
40
30.29
9.71
1000.00
Example
1000 par value two-year bond which pays interest at 8%
convertible semiannually; yield rate is 6% convertible
semiannually
Algorithm
•
•
•
•
Book value at time t is Bt
Amount of coupon at time t+1 is Fr
The amount of interest contained in this coupon is iBt
Fr – iBt represents the change in the book value
between these dates
Time Coupon Interest
Principal Book Value
adjustment
0
1037.17
1
40
31.12
8.88
1028.29
2
40
30.85
9.15
1019.14
3
40
30.57
9.43
1009.71
4
40
30.29
9.71
1000.00
Example (p. 99)
• Consider 1000 par-value 10-year bond
with semiannual coupons, r = .05 and the
yield rate i = 0.06 effective semiannually.
Find the amount of interest and change in
book value contained in the 15th coupon
of the bond.
Example (p. 99)
• Construct a bond amortization schedule
for a 1000 par-value two-year bond which
pays interest at 8% convertible
semiannually, and has a yield rate of 6%
convertible semiannually
Time
0
1
2
3
4
Coupon
Interest
Principal Adjustment
40
31.12
8.88
Book Value
1037.17
1028.29
5.4 Other Topics
• Different frequency of coupon payments
• Increasing or decreasing coupon
payments
• Different yield rates
• Callable bonds
Examples (p. 101 – p. 102)
• (Different frequency) Find the price of a 1000 par-value
10-year bond which has quarterly 2% coupons and is
bought to yield 9% per year convertible semiannually
• (Increasing coupon payments) Find the price of a 1000
par-value 10-year bond which has semiannual coupons of
10 the first half-year, 20 the second half-year,…, 200 the
last half-year, bought to yield 9% effective per year
•
(Different yield rates) Find the price of a 1000 par-value
10-year bond with coupons at 11% convertible
semiannually, and for which the yield rate is 5% per halfyear for the first 5 years and 6% per half-year for the last 5
years
Callable bonds
• A borrower (i.e. corporation, government etc.) has the right
to redeem the bond at any of several time points
• The earliest possible date is the call date and the latest is
the usual maturity date
• Once the bond is redeemed, no more coupons will be paid
possible
redemption
Purchase Date
Call Date
Maturity Date
Examples (p. 103 – p. 105)
• Consider a 1000 par-value 10-year bond with semiannual
5% coupons. Assume this bond can be redeemed at par at
any of the last 4 coupon dates. Find the price which will
guarantee an investor a yield rate of
– 6% per half-year
– 4% per half-year
• Consider a 1000 par-value 10-year bond with semiannual
5% coupons. This bond can be redeemed for 1100 at the
time of the 18th coupon, for 1050 at the time of the 19th
coupon, or for 1000 at the time of 20th coupon. What price
should an investor pay to be guaranteed a yield rate of
– 6% per half-year
– 4% per half-year
Chapter 6
PREPARATION FOR LIFE
CONTINGENCIES
• Introduction
• Contingent
Payments
6.1 Introduction
• Ideal situation: all payments are made
• Real-life situations:
– failure to make a payment
– default on a loan
– bad credit ratings
– life contingencies and life insurance
• Contingent payments (we need to combine
the theory of interest and elementary
probability)
6.3 Contingent Payments
• Assume that for each payment of the loan (annuity,
bond etc.) there is a probability that this payment is
made
• Finding the present value of such sequence of
payments we need to take into account these
probabilities
• Example Henry borrows 1000 from Amicable Trust and
agrees to repay the loan in one year. If payment were
certain, the company would charge 13% interest. From
prior experience, however, it is determined that there is
a 5% chance that Henry will not repay any money at all.
What should Amicable Trust ask Henry to repay?
Examples (p. 119 – p. 122)
• (Loan) The All-Mighty Bank lends 50,000,000 to a small Central
American country, with the loan to be repaid in one year. It is felt that
there is a 20% chance that a revolution will occur and that no money
will be repaid, a 30% chance that due to inflation only half the loan will
be repaid, and a 50% chance that the entire loan will be repaid. If
payments were certain, the bank would charge 9%. What rate of
interest should the bank charge?
• (Payments contingent upon survival) Mrs. Rogers receives 1000 at
the end of each year as long as she is alive. The probability is 80%
she will survive one year, 50% she will survive 2 years, 30% she will
survive 3 years, and negligible that she will survive longer than 3
years. If the yield rate is 15%, what should Mrs. Rogers place on
these payments now?
• (Life insurance) An insurance company issues a policy which pays
50,000 at the end of the year of death, if death should occur during
the next two years. The probability that a 25-year-old will live for one
year is .99936, and the probability he will live for two years is .99858.
What should the company charge such a policyholder to earn 11% on
its investments?
• (Annuity) Alphonse wishes to borrow some money form
Friendly Trust. He promises to repay 500 at the end of
each year for the next 10 years, but there is a 5% chance
of default in any year. Assume that once default occurs,
no further payments will be received. How much can
Friendly Trust lend Alphonse if it wishes to earn 9% on its
investments?
• Redo the last example, without the restriction that once
default occurs, no further payments will be received
• (Bond) A 20-year 1000 face value bond has coupons at
14% convertible semiannually and is redeemable at par.
Assume a 2% chance that, in any given half-year, the
coupon is not issued, and that once default occurs, no
further payments are made. Assume as well that a bond
can be redeemed only if all coupons have been paid. Find
the purchase price to yield on investor 16% convertible
semiannulllay.
Chapter 7
LIFE TABLES AND POPULATION
PROBLEMS
• Introduction
• Life Tables
• The Stationary
Population
• Expectation of Life
7.1 Introduction
• How do we find probabilities?
• Data obtained from practice
• Data required to find
probabilities of surviving
to certain ages (or, equivalently, of
dying before certain ages) are
contained in life tables
7.2 Life Tables
Age
lx
dx
0
1,000,000
1580
1.58
1
998,420
680
68
2
997,740
485
.49
3
997,255
435
.44
1000
qx
• lx – number of lives survived to age x
• Thus l0 = 1,000,000 is a starting population
• Survival function S(x) = lx / l0 is the probability of surviving
to age x
• dx = lx – lx+1 – number of lives who died between (x, x+1)
• qx = dx /lx – probability that x – year-old will not survive to
age x + 1
Examples (p. 129 – p. 130)
Age
lx
dx
0
1,000,000
1580
1.58
1
998,420
680
68
2
997,740
485
.49
3
997,255
435
.44
1000
qx
• Find
– the probability that a newborn will live to age 3
– the probability that a newborn will die between
age 1 and age 3
• Find an expression for each of the following:
– the probability that an 18-year-old lives to age 65
– the probability that a 25-year-old dies between ages
40 and 45
– the probability that a 25-year-old does not die
between ages 40 and 45
– the probability that a 30-year-old dies before age 60
• There are four persons, now aged 40, 50, 60 and 70.
Find an expression for the probability that both the 40year-old and the 50-year-old will die within the five-year
period starting ten years from now, but neither the 60year-old nor the 70-year-old will die during that five-year
period
Note:
• If we are already given by probabilities qx and
starting population l0 we can construct the
whole life table step-by-step since dx = qx lx
and lx+1 = lx - dx
Example
• Given the following probabilities of deaths
q0 = .40, q1 = .20, q2 = .30, q3 = .70, q4 = 1
and starting with l0 = 100 construct a life table
More notations…
• qx = dx / lx – probability that x – year-old will not
survive to age x + 1
• px = 1- qx – probability that x – year-old will survive to
age x + 1
• Note: qx = (lx – lx+1) / lx and px = lx+1 / lx
•
nqx –
probability that x – year-old will not survive to age
x+n
•
np x
= 1- nqx – probability that x – year-old will survive to
age x + n
• Note: nqx = (lx – lx+n) / lx and n px = lx+n / lx
• Formulas:
lx – lx+n = dx + dx+1 + …+ dx+n
n+mpx = mpx ∙ npx+m
What is mPx when m is not integer?
• tPx where 0 < t <1
• Assuming that deaths are distributed
uniformly during any given year we can
use linear interpolation to find tpx :
l x  td x l x  t (l x  l x 1 ) (1  t )l x  tlx 1


t px 
lx
lx
lx
Examples (p. 132 – p. 133)
• 30% of those who die between ages 25 and 75 die
before age 50. The probability of a person aged 25 dying
before age 50 is 20%. Find 25P50
• Using the following life table and assuming a uniform
distribution of deaths over each year, find:
– 4/3P1
– The probability that a newborn will survive the first year but die in
the first two months thereafter
Age
lx
dx
0
1,000,000
1580
1.58
1
998,420
680
68
2
997,740
485
.49
3
997,255
435
.44
1000
qx
7.4 The Stationary Population
• Assume that in every given year (or, more precisely, in any
given 12-months period) the number of births and deaths is the
same and is equal to l0
• Then after a period of time the total population will remain
stationary and the age distribution will remain constant
• px and qx are defined as before
• lx denote the number of people who reach heir xth birthday
during any given year
• dx = qx lx represent the number of people who die before
reaching age x+1
• Also, dx represent the number of people who die during any
given year between ages x and x + 1
• Similarly lx – lx + n represent the number of people who die
during any given year between ages x and x + n
Number of people aged x
• Let Lx denote the number of people aged x
(last birthday) at any given moment
• Note: Lx ≠ lx
• Assuming uniform distribution of deaths we
1
1
1
obtain:
L  l  d  l  (l  l )  (l  l )
x
x
• More precisely:
1
Lx   l x t dt
0
2
x
x
2
x
x 1
2
x
x 1
Number of people aged x and over
• Let Tx denote the number of people
aged x and over at any given moment

• Then Tx   Lx i
i 0
• Assuming uniform distribution we obtain:


1
1
Tx   (l x i  l x i 1 )  l x   l x i
2
i 0 2
i 1
• More precisely:

Tx   l x t dt
0
Example (p. 139)
• An organization has a constant total membership. Each
year 500 new members join at exact age 20.
20% leave after 10 years, 10% of those remaining
leave after 20 years, and the rest retire at age 65.
Express each of the following in terms of life table
functions:
– The number who leave at age 40 each year
– The size of the membership
– The number of retired people alive at any given time
– The number of members who die each year
7.5 Expectation of Life
• What is the average future life time ex of
a person aged x now?
• The answer is given by expected value
(or mathematical expectation) and is
called the curtate expectation:

ex   t ( t p x  t 1 p x ) 1(1 p x  2 p x )  2( 2 p x  3 p x )  3( 3 p x  4 p x )   
t 1

 px  2 px 3 px     t px
t 1
Complete Expectation

(1)
Complete expectation: (4)
ex   t p x
t 1
e 

(2)
x
Tx   l x t dt
px dt
(2)&(3)&(4) imply:
l x t
t px 
lx

e
x
Tx

lx
Approximation:
Since

1
Tx  l x   l x i
2
i 1
t
0
0
(3)


we get

e
x
1
 ex 
2
Remarks

e
x
Tx

lx
Thus Tx can be
viewed as the total
number of years of
future life of those
who form group lx

Tx  l x e x
Note: this interpretation
makes sense in any
population (stationary or
not)
Average age at death
• Average age at death of a person currently
aged x is given by

x  ex
Tx
 x
lx
• Letting x = 0 we obtain the average age at
death for all death among l0 individuals

e
0
T0

l0
Examples (p. 142 – 143)
• If tp35 = (.98)t for all t, find e35 and eo35
without approximation. Compare the
value for eo35 with its approximate value
• Interpret verbally the expression
Tx-Tx+n – nlx+n
• Fin the average age at death of those
who die between age x and age x+n
Chapter 8
LIFE ANNUITIES
• Basic Concepts
• Commutation Functions
• Annuities Payable mthly
• Varying Life Annuities
• Annual Premiums and Premium Reserves
8.1 Basic Concepts
• We know how to compute present value of
contingent payments
• Life tables are sources of probabilities of
surviving
• We can use data from life tables to compute
present values of payments which are
contingent on either survival or death
Example (pure endowment), p. 155
• Yuanlin is 38 years old. If he reaches
age 65, he will receive a single payment
of 50,000. If i = .12, find an expression
for the value of this payment to Yuanlin
today. Use the following entries in the
life table: l38 = 8327, l65 = 5411
Pure Endowment
• Pure endowment: 1 is paid t years from
now to an individual currently aged x if
the individual survives
• Probability of surviving is t px
• Therefore the present value of this
payment is the net single premium for the
pure endowment, which is:
–t = v t p
E
=
(
p
)
(1
+
t)
t x
t x
t x
Example (life annuity), p. 156
• Aretha is 27 years old. Beginning one year
from today, she will receive 10,000 annually
for as long as she is alive. Find an expression
for the present value of this series of
payments assuming i = .09
• Find numerical value of this expression if
px = .95 for each x
Life annuity
Series of payments of 1 unit
as long as individual is alive
present value
(net single premium)
of annuity
ax
1
age
probability
x
1
x+1 x+2
px
2px
1
…..
x+n
…..
npx

a x  vpx  v 2 2 p x  v 3 3 p x    v n n p x    v t t p x
t 1
Temporary life annuity
Series of n payments of 1 unit
(contingent on survival)
present value
last payment
ax:n|
1
age
probability
x
1
x+1 x+2
px
2px
1
…..
x+n
npx
n
a x:n|  vpx  v 2 2 p x  v 3 3 p x    v n n p x  v t t p x
t 1
n - years deferred life annuity
Series of payments of 1 unit as long as individual is alive
in which the first payment is at x + n + 1
present value
first payment
n|ax
1
age
x
x+1 x+2
…
1
x + n x + n +1 x + n + 2
probability
n+1px
…
n+2px
n2
n 3
n t
n1
|
a

v
p

v
p

v
p



v
n
x
n 1 x
n2 x
n 3 x
n t p x   

  v n t n t p x 
t 1
Note:
n
| ax  ax  ax:n|

s
v
 s px
s  n 1
Life annuities-due
äx
1
x
äx:n|
1
x
1
…..
x+1 x+2
px
1
1
1
px
…
x+n
2px
ax  1  a x  1   v t t p x
t 1
npx
1
x+1 x+2

1
n 1
….. x + n-1
ax:n|  1  a x:n 1|  1   v t t p x
t 1
x+n
n-1px
2px
n|äx
1
x
x+1 x+2
…
1
1
x + n x + n +1 x + n + 2
npx
n+1px
n+2px
…
n
| ax  n1 | ax
Note
ax:n|  1  ax:n1|
but
ax:n|  (1  i)ax:n|
ax1:n|  vpx1ax:n|
8.2 Commutation Functions
Recall: present value of a pure endowment
of 1 to be paid n years hence to a life
currently aged x
xn


v lxn
n
n lxn
n Ex  v n px  v 
 x
v lx
 lx 
= v xlx
Then nEx = Dx+n / Dx
Denote Dx
Life annuity and commutation functions


a x  v t p x   t E x
t
t 1
t 1
Since nEx
we have
= Dx+n / Dx

Dxt
1
Dx1  Dx2  Dx3  
ax  

Dx
t 1 Dx
Define commutation
Nx 
function Nx as follows:
Then:


t 0
t 0
x t
D

v
 x t  l x t
N x 1
ax 
Dx
Identities for other types of life annuities
temporary life annuity
Dx t N x 1  N x  n1


Dx
t 1 Dx
n
ax:n|
n-years delayed l. a.
N x  n 1
n | ax 
Dx
temporary l. a.-due
N x  N xn
ax:n| 
Dx
Accumulated values of life annuities
temporary life annuity
since a  N x 1  N x  n 1
x:n|
Dx
we have
s x:n|
ax:n| n Ex  sx:n|
Dx  n
and n E x 
Dx
N x 1  N x  n 1

Dx  n
similarly for temporary life annuity-due:
ax:n| n Ex  sx:n|
and
sx:n|
N x  N xn

Dx  n
Examples (p. 162 – p. 164)
• (life annuities and commutation functions) Marvin, aged
38, purchases a life annuity of 1000 per year. From
tables, we learn that N38 = 5600 and N39 = 5350. Find
the net single premium Marvin should pay for this
annuity
– if the first 1000 payment occurs in one year
– if the first 1000 payment occurs now
• Stay verbally the meaning of (N35 – N55) / D20
• (unknown rate of interest) Given Nx = 5000, Nx+1=4900,
Nx+2 = 4810 and qx = .005, find i
Select group
• Select group of population is a group with the
probability of survival different from the
probability given in the standard life tables
• Such groups can have higher than average
probability of survival (e.g. due to excellent
health) or, conversely, higher mortality rate
(e.g. due to dangerous working conditions)
Notations
• Suppose that a person aged x is
in the first year of being in the select group
• Then p[x] denotes the probability of survival for 1 year
and q[x] = 1 – p[x] denotes the probability of dying during
1 year for such a person
• If the person stays within this group for subsequent
years, the corresponding probabilities of survival for 1
more year are denoted by p[x]+1, p[x]+2, and so on
• Similar notations are used for life annuities:
a[x] denotes the net single premium for a life annuity of 1
(with the first payment in one year) to a person aged x in
his first year as a member of the select group
• A life table which involves a select group is called a
select-and-ultimate table
Examples (p. 165 – p. 166)
• (select group) Margaret, aged 65, purchases a life annuity
which will provide annual payments of 1000 commencing at
age 66. For the next year only, Margaret’s probability of
survival is higher than that predicted by the life tables and, in
fact, is equal to p65 + .05, where p65 is taken from the
standard life table. Based on that standard life table, we have
the values D65 = 300, D66 = 260 and N67 = 1450. If i = .09,
find the net single premium for this annuity
• (select-and-ultimate table) A select-and-ultimate table has a
select period of two years. Select probabilities are related to
ultimate probabilities by the relationships p[x] = (11/10) px and
p[x]+1 = (21/20) px+1. An ultimate table shows D60 = 1900,
D61 = 1500, and ä 60:20| = 11, when i = .08. Find the select
temporary life annuity ä[60]:20|
• The following values are based on a unisex life table:
N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865,
N42 = 4625.
It is assumed that this table needs to be set forward
one year for males and set back two years for
females. If Michael and Brenda are both age 40, find
the net single premium that each should pay for a life
annuity of 1000 per year, if the first payment occurs
immediately.
Chapter 8
LIFE ANNUITIES
• Basic Concepts
• Commutation Functions
• Annuities Payable mthly
• Varying Life Annuities
• Annual Premiums and Premium Reserves
8.1 Basic Concepts
• We know how to compute present value of
contingent payments
• Life tables are sources of probabilities of
surviving
• We can use data from life tables to compute
present values of payments which are
contingent on either survival or death
Example (pure endowment), p. 155
• Yuanlin is 38 years old. If he reaches
age 65, he will receive a single payment
of 50,000. If i = .12, find an expression
for the value of this payment to Yuanlin
today. Use the following entries in the
life table: l38 = 8327, l65 = 5411
Pure Endowment
• Pure endowment: 1 is paid t years from
now to an individual currently aged x if
the individual survives
• Probability of surviving is t px
• Therefore the present value of this
payment is the net single premium for the
pure endowment, which is:
–t = v t p
E
=
(
p
)
(1
+
t)
t x
t x
t x
Example (life annuity), p. 156
• Aretha is 27 years old. Beginning one year
from today, she will receive 10,000 annually
for as long as she is alive. Find an expression
for the present value of this series of
payments assuming i = .09
• Find numerical value of this expression if
px = .95 for each x
Life annuity
Series of payments of 1 unit
as long as individual is alive
present value
(net single premium)
of annuity
ax
1
age
probability
x
1
x+1 x+2
px
2px
1
…..
x+n
…..
npx

a x  vpx  v 2 2 p x  v 3 3 p x    v n n p x    v t t p x
t 1
Temporary life annuity
Series of n payments of 1 unit
(contingent on survival)
present value
last payment
ax:n|
1
age
probability
x
1
x+1 x+2
px
2px
1
…..
x+n
npx
n
a x:n|  vpx  v 2 2 p x  v 3 3 p x    v n n p x  v t t p x
t 1
n - years deferred life annuity
Series of payments of 1 unit as long as individual is alive
in which the first payment is at x + n + 1
present value
first payment
n|ax
1
age
x
x+1 x+2
…
1
x + n x + n +1 x + n + 2
probability
n+1px
…
n+2px
n2
n 3
n t
n1
|
a

v
p

v
p

v
p



v
n
x
n 1 x
n2 x
n 3 x
n t p x   

  v n t n t p x 
t 1
Note:
n
| ax  ax  ax:n|

s
v
 s px
s  n 1
Life annuities-due
äx
1
x
äx:n|
1
x
1
…..
x+1 x+2
px
1
1
1
px
…
x+n
2px
ax  1  a x  1   v t t p x
t 1
npx
1
x+1 x+2

1
n 1
….. x + n-1
ax:n|  1  a x:n 1|  1   v t t p x
t 1
x+n
n-1px
2px
n|äx
1
x
x+1 x+2
…
1
1
x + n x + n +1 x + n + 2
npx
n+1px
n+2px
…
n
| ax  n1 | ax
Note
ax:n|  1  ax:n1|
but
ax:n|  (1  i)ax:n|
ax1:n|  vpx1ax:n|
8.2 Commutation Functions
Recall: present value of a pure endowment
of 1 to be paid n years hence to a life
currently aged x
xn


v lxn
n
n lxn
n Ex  v n px  v 
 x
v lx
 lx 
= v xlx
Then nEx = Dx+n / Dx
Denote Dx
Life annuity and commutation functions


a x  v t p x   t E x
t
t 1
t 1
Since nEx
we have
= Dx+n / Dx

Dxt
1
Dx1  Dx2  Dx3  
ax  

Dx
t 1 Dx
Define commutation
Nx 
function Nx as follows:
Then:


t 0
t 0
x t
D

v
 x t  l x t
N x 1
ax 
Dx
Identities for other types of life annuities
temporary life annuity
Dx t N x 1  N x  n1


Dx
t 1 Dx
n
ax:n|
n-years delayed l. a.
N x  n 1
n | ax 
Dx
temporary l. a.-due
N x  N xn
ax:n| 
Dx
Accumulated values of life annuities
temporary life annuity
since a  N x 1  N x  n 1
x:n|
Dx
we have
s x:n|
ax:n| n Ex  sx:n|
Dx  n
and n E x 
Dx
N x 1  N x  n 1

Dx  n
similarly for temporary life annuity-due:
ax:n| n Ex  sx:n|
and
sx:n|
N x  N xn

Dx  n
Examples (p. 162 – p. 164)
• (life annuities and commutation functions) Marvin, aged
38, purchases a life annuity of 1000 per year. From
tables, we learn that N38 = 5600 and N39 = 5350. Find
the net single premium Marvin should pay for this
annuity
– if the first 1000 payment occurs in one year
– if the first 1000 payment occurs now
• Stay verbally the meaning of (N35 – N55) / D20
• (unknown rate of interest) Given Nx = 5000, Nx+1=4900,
Nx+2 = 4810 and qx = .005, find i
Select group
• Select group of population is a group with the
probability of survival different from the
probability given in the standard life tables
• Such groups can have higher than average
probability of survival (e.g. due to excellent
health) or, conversely, higher mortality rate
(e.g. due to dangerous working conditions)
Notations
• Suppose that a person aged x is
in the first year of being in the select group
• Then p[x] denotes the probability of survival for 1 year
and q[x] = 1 – p[x] denotes the probability of dying during
1 year for such a person
• If the person stays within this group for subsequent
years, the corresponding probabilities of survival for 1
more year are denoted by p[x]+1, p[x]+2, and so on
• Similar notations are used for life annuities:
a[x] denotes the net single premium for a life annuity of 1
(with the first payment in one year) to a person aged x in
his first year as a member of the select group
• A life table which involves a select group is called a
select-and-ultimate table
Examples (p. 165 – p. 166)
• (select group) Margaret, aged 65, purchases a life annuity
which will provide annual payments of 1000 commencing at
age 66. For the next year only, Margaret’s probability of
survival is higher than that predicted by the life tables and, in
fact, is equal to p65 + .05, where p65 is taken from the
standard life table. Based on that standard life table, we have
the values D65 = 300, D66 = 260 and N67 = 1450. If i = .09,
find the net single premium for this annuity
• (select-and-ultimate table) A select-and-ultimate table has a
select period of two years. Select probabilities are related to
ultimate probabilities by the relationships p[x] = (11/10) px and
p[x]+1 = (21/20) px+1. An ultimate table shows D60 = 1900,
D61 = 1500, and ä 60:20| = 11, when i = .08. Find the select
temporary life annuity ä[60]:20|
• The following values are based on a unisex life table:
N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865,
N42 = 4625.
It is assumed that this table needs to be set forward
one year for males and set back two years for
females. If Michael and Brenda are both age 40, find
the net single premium that each should pay for a life
annuity of 1000 per year, if the first payment occurs
immediately.
8.3 Annuities Payable mthly
• Payments every mth part of the year
• Problem: commutation functions reflect
annual probabilities of survival
• First, we obtain an approximate formula
for present value
• Assume for a moment that the values Dy
are also given for non-integer values of y
Usual life annuity
ax
1
age
x
1
x+1 x+2
1
…..
x+n
…..
Annuity payable every 1/m part of the year
a(m)x
age
x
1/m
1/m
x+
1/m
x+
2/m
1/m
…..
x+
(m-1)/m
1/m
x+1
…..
a(m)x
age
Annuity payable every 1/m part of the year
1/m
1/m
x+
1/m
x+
2/m
x

1/m
…..
x+
(m-1)/m
1/m
x+1
…..

1 1/ m
a  v 1/ m p x  v 2 / m 2 / m p x    v i  j / m i  j / m p x   
m
l

1  1/ m l x 1/ m
2 / m lx2 / m
i  j / m x i  j / m
 v
v
 v
  
m
lx
lx
lx

(m)
x
x i  j / m
x 1 / m
x2 / m


v
l x i  j / m
l x 1/ m v
lx2 / m
1 v
 


  
x
x
x
m  v l x
v lx
v lx

Dx  i  j / m

1  Dx 1/ m Dx  2 / m
 


 
m  Dx
Dx
Dx

Annuity payable every 1/m part of the year
a(m)x
age
a
(m)
x
x
1/m
1/m
x+
1/m
x+
2/m
1/m
…..
x+
(m-1)/m
1/m
x+1
Dx i  j / m

1  Dx 1/ m Dx  2 / m
 


  
m  Dx
Dx
Dx

…..
Dx 1/ m  Dx  2 / m    Dx  ( m 1) / m  Dx 1  



Dx 11/ m  Dx 1 2 / m    Dx 1 ( m 1) / m  Dx  2   
m

1 m
1  m

Dx  i  j / m
 Dx 1 j / m   Dx 1 j / m   

mDx  j 1
j 1
 mDx i 0 j 1
1

mDx
Using linear interpolation for Dx+i+j/m
Dx i  j / m  Dx i  mj Dx i 1  Dx i 
m
D
j 1
m
x i  j / m
  Dx i  mj Dx i 1  Dx i  
j 1
m
 mDx i  Dx i 1  Dx i  mj  mDx i  Dx i 1  Dx i 
j 1
 mDx i  Dx i 1  Dx i 

m
 D
i  0 j 1
x i  j / m
m(m  1)

2m
(m  1)
2

(m  1)
(m  1)
(m  1)
 mDx 1  Dx  2  Dx 1 
 mDx  2  Dx 3  Dx  2 

2
2
2
(m  1)
Dx1  Dx   Dx 2  Dx1   Dx 3  Dx 2   
 mDx 1  Dx  2    mDx 
2
m 1
 mN x 1 
Dx
2
mDx  Dx 1  Dx 
Using linear interpolation for Dx+i+j/m

m
 D
i 0 j 1
a
( m)
x
x i  j / m
1

mDx
m 1
 mN x 1 
Dx
2
m  1  N x 1 m  1
m 1

mN x1  2 Dx   D  2m  ax  2m
x
a
(m)
x
m 1
 ax 
2m
Continuous life annuity
a x  lim a
m 
(m)
x
m 1 
1

 lim  a x 
 ax 

m 
2m 
2

a x  lim ax( m ) 
m 



1 1/ m
 lim
v 1/ m px  v 2 / m 2 / m px    v i  j / m i  j / m px     v t t p x dt
m  m
0
1
a x  ax 
2

a x   v t p x dt
t
0
Annuity payable m-thly, deferred
a(m)x+n
n|a(m)x
1/m
age
x
…
x+n
na

(m)
x
v
Dx  n
Dx
na
1/m
x+
x+
n+1/m n+2/m
n
n
px a
(m)
xn
1/m
…..
x +n+
(m-1)/m
1/m
x + n+1
…..
Dx  n ( m )

axn 
Dx
Dx  n  m  1 
m 1 

a


n
a

 xn



x
2m 
Dx  2 m 

( m)
x
Dx  n  m  1 
 n ax 


Dx  2m 
Annuity payable m-thly, temporary
a(m)x:n|
age
a
(m)
x:n |
x
a
1/m
1/m
x+
1/m
x+
2/m
(m)
x
na
(m)
x
1/m
…..
x +n+
(m-1)/m
1/m
x+n
Dx  n  m  1  
m 1  

  ax 
   n a x 

  
2m  
Dx  2 m  

 Dx  n  m  1
 Dx  n  m  1


 a x  n a x  1 
 a x:n |  1 
Dx  2 m
Dx  2 m




a
( m)
x:n |
 Dx  n  m  1

 ax:n |  1 
Dx  2m

Examples
• Page 168, 8.10
8.4 Varying Life Annuities
• Arithmetic increasing annuities
• It is sufficient to look at the sequence 1,2,3,….
• Temporary decreasing annuities
Example
• Ernest, aged 50, purchases a life
annuity, which pays 5,000 for 5 years,
3,000 for 5 subsequent years, and
8,000 each year after. If the first
payment occurs in exactly 1 year, find
the price in terms of commutation
functions.
Arithmetic increasing annuity
(Ia)x
1
age
x
n
…..
x+1 x+2
px
probability
2
2px
x+n
…..
npx
Ia x  vpx  2v 2 2 px  3v 3 3 px    nv n n px   



t 1
t 2
t 3
  v t t p x   v t t p x   v t t p x    a x 1 | a x  2 | a x  


N x t 1
  t | ax  
Dx
t 0
t 0

S x   N x t
t 0
Ia x
S x 1

Dx
Arithmetic increasing annuity, temporary
(Ia)x:n|
1
age
probability
x
2
x+1 x+2
px
Ia x:n |
2px
n
…..
x+n
npx
S x 1  S x  n 1  nN x  n 1

Dx
x + n+1
Arithmetic decreasing annuity, temporary
(Da)x:n|
n
x
n-1
x+1 x+2
1
…..
x+n
x + n+1
Arithmetic decreasing annuity, temporary
(Da)x:n|
n
(Ia)x:n|x
x+1 x+2
1
x
n-1
1
…..
2
x+1 x+2
x+n
x + n+1
n
…..
x+n
x + n+1
Arithmetic decreasing annuity, temporary
(Da)x:n|
n
(Ia)x:n|x
x+1 x+2
1
(n+1)ax:n|
x
n-1
1
…..
2
x+1 x+2
x+1 x+2
x + n+1
n
…..
n+1 n+1
x
x+n
x+n
n+1
…..
x+n
Ia x:n |  Da x:n |  (n  1)ax:n |
x + n+1
Arithmetic decreasing annuity, temporary
(Da)x:n|
n
age
x
n-1
x+1 x+2
px
probability
1
…..
2px
x+n
x + n+1
npx
Ia x:n |  Da x:n |  (n  1)ax:n |
Ia x:n |  S x 1  S x  n1  nN x  n1
Dx
a x:n|
N x 1  N x  n 1

Dx
Da x:n |
nN x 1  ( S x  2  S x  n  2 )

Dx
Examples
• Georgina, aged 50, purchases a life annuity which will pay
her 5000 in one year, 5500 in two years, continuing to
increase by 500 per year thereafter. Find the price if S51 =
5000, N51 = 450, and D50 = 60
• Redo the previous example if the payments reach a
maximum level of 8000, and then remain constant for life.
Assume S58 = 2100
• Two annuities are of equal value to Jim, aged 25. The first is
guaranteed and pays him 4000 per year for 10 years, with
the first payment in 6 years. The second is a life annuity with
the first payment of X in one year. Subsequent payments are
annual, increasing by .0187 each year.
If i = .09, and from the 7% -interest table, N26=930 and D25=
30, find X.
8.5 Annual Premiums and
Premium Reserves
• Paying for deferred life annuity with a series of payments
instead of a single payment
• Premium reserve is an analog of outstanding principal
• Premiums often include additional expenses and
administrative costs
• In such cases, the total payment is called
gross premium
• Loading = gross premium – net premium
• General approach: actuarial present values of two
sequences of payments must be the same (equation of
value)
Annual premiums P = tP(n|äx)
P
age
x
P
x+1
P
…
x + t-1 x +t
1
… x+n
1
x + n +1
1
x+n+2
…
• t is the number of premium payments
• Present value of premiums is P äx:t|
• Present value of benefits is n|äx
• Therefore P äx:t| = n|äx
N xn

Dx
N xn
n ax
P


N x  N x t N x  N x t
ax:t |
Dx
N xn

t P( n a x ) 
N x  N x t
Example
• Arabella, aged 25, purchases a deferred life annuity of 500
per month, with the first benefit coming in exactly 20 years.
She intends to pay for this annuity with a series of annual
payments at the beginning of each year for the next 20
years. Find her net annual premium if D25 = 9000, D 45 =
5000, ä25 = 15 and ä45 = 11.5
Reserves
P
age
x
P
x+1
P
P
… x + t -1 … x +n-1
1
x+n
1
x + n +1
1
x+n+2
…
n V ( |ä )
t
n x
• Analog of outstanding principal immediately after premium t has
been paid
• Assume that the number of premium payments is n
• Reserve
n V ( |ä ) = PV of all future benefits – PV of all future premiums
t
n x
N x t

ax t 
, tn

Dx t
n

tV ( n a x )  
N x  n  P( N x t  N x  n )
 n t | ax t  Pax t:n t| 
, tn

Dx t
Loading and Gross premiums
• Arabella, aged 25, purchases a deferred life annuity of 500
per month, with the first benefit coming in exactly 20 years.
She intends to pay for this annuity with a series of annual
payments at the beginning of each year for the next 20
years. Assume that 50% of her first premium is required for
initial underwriting expenses, and 10% of all subsequent
premiums are needed for administration costs. In addition,
100 must be paid for issue expenses. Find Arabella’s annual
gross premium, if D25 = 9000, D 45 = 5000, ä25 = 15,
and ä45 = 11.5
Chapter 9
LIFE INSURANCE
• Basic Concepts
• Commutation Functions and Basic Identities
• Insurance Payable at The Moment of Death
• Varying Insurance
• Annual Premiums and Premium Reserves
9.1 Basic Concepts
• Benefits are paid upon the death of the insured
• Types of insurance
– Whole life policy
– Term insurance
– Deferred insurance
– Endowment insurance
Whole life policy
• Benefit (the face value) is paid to the beneficiary
at the end of the year of death of inured person
• If the face value is 1 and insurance is sold to a
person aged x, the premium is denoted by Ax
Ax
1
age
probability
x
x+1 x+2
px
2px
…..
x+t
x + t+1
tqx
Whole life policy

Ax   t p x  q x t  v
t 1
t 0
Ax
1
age
probability
x
x+1 x+2
px
2px
…..
x+t
x + t+1
qx+t
Term insurance
• Benefit (the face value) is paid to the beneficiary
at the end of the year of death of inured person,
only if the death occurs within n years
• If the face value is 1 and insurance is sold to a
person aged x, the premium is denoted by A1x
n 1
1
x:n |
A
  t p x  q x t  v
t 0
t 1
Deferred insurance
• Does not come into force until age x+n
• If the face value is 1 and insurance is sold to a
person aged x, the premium is denoted by A1x:n|
Ax  A  n | Ax
1
x:n |
n
| Ax  Ax  A
1
x:n |
n-year endowment insurance
• Benefit (the face value) is paid to the beneficiary at the end
of the year of death of inured person, if the death occurs
within n years
• If the insured is still alive at the age x+n, the face value is
paid at that time
• If the face value is 1 and insurance is sold to a person
aged x, the premium is denoted by Ax:n|
Exercise:
Ax:n|  A  n Ex
1
x:n |
1
x:n |
A
 Ax  Ax:n |
Examples
• Rose is 38 years old. She wishes to purchase a life
insurance policy which will pay her estate 50,000 at the end
of the year of her death. If i=.12, find an expression for the
actuarial present value of this benefit and compute it,
assuming px = .94 for all x.
• Michael is 50 years old and purchases a whole life policy
with face value 100,000. If lx= 1000(1-x/105) and i=.08, find
the price of this policy.
• Calculate the price of Rose’s and Michael’s policies if both
policies are in force for a term of only 30 years.
• Calculate the price of Rose’s and Michael’s policies if both
policies are to be 30 years endowment insurance.
9.2 Commutation Functions
• Recall:
Dx  v l x
x

N x   Dx  t
t 0
Dx  n
n Ex 
Dx
Commutation Functions

Ax   t p x  q x t  v
• Recall:
t 1
t 0
• So we need:
t
p x  q x t  v
t 1
Cx  d xv
x  t 1
l x t d x t t 1 d x t t 1 d x t v


v 
v 
x
l x l x t
lx
lxv
x 1
t
p x  q x t  v
t 1
C x t

Dx
Whole life insurance


C x t
1
t 1
Ax   t px  qxt  v  

Dx
t 0
t 0 Dx

M x   C x t
t 0
Mx
Ax 
Dx

C
t 0
x t
Term insurance
n 1
1
x:n |
A
  t p x  q x t  v
t 0
n 1
A1x:n |
t 1
C x t  C x t  C x t M x M x  n





Dx
Dx
t 0 Dx
t 0 Dx
t  n Dx
1
x:n |
A
M x  M xn

Dx
n-year endowment insurance
Ax:n|  A  n Ex
1
x:n |
Ax:n |
M x  M x  n Dx  n M x  M x  n  Dx  n



Dx
Dx
Dx
Note
• We can represent insurance premiums
in terms of actuarial present values of
annuities, e.g. Ax = 1 – d äx
• Hence they also can be found using
“old” commutation functions
Examples
• Juan, aged 40, purchases an insurance policy paying
50,000 if death occurs within the next 20 years, 100,000 if
death occurs between ages 60 and 70, and 30,000 if death
occurs after that. Find the net single premium for this policy
in terms of commutation functions.
• Phyllis, aged 40, purchases a whole life policy of 50,000. If
N40 = 5000, N41 = 4500, and i = .08, find the price.
9.3 Insurance Payable
at the Moment of Death
• We consider scenario when the benefit is paid at
the end of the year of death
• Alternatively, the benefit can be paid at the
moment of death
Divide each year in m parts
Ax( m ) 
v1/ m 1/ m q x  v 2 / m 1/ m p x 1/ m q x 1/ m    v i  j / m i  ( j 1) / m p x 1/ m q x  ( j 1) / m   
1/ m
v
l x  l x 1/ m
2 / m l x 1 / m l x 1 / m  l x  2 / m
v


lx
lx
l x 1/ m
 v
i j / m
l x i  ( j 1) / m l x i  ( j 1) / m  l x i  j / m

 
lx
l x i  ( j 1) / m
1/ m
v
v
l x  l x 1/ m
2 / m l x 1 / m  l x  2 / m
v

lx
lx
i j / m
l x i  ( j 1) / m  l x i  j / m
lx
 
 1/ m l x  l x 1/ m

2 / m l x 1 / m  l x  2 / m
v
v
 

1
1/ m
1/ m


mlx  i  j / m l x i  ( j 1) / m  l x i  j / m


v




1/ m
1/ m
'
2/ m
'


v

l

v

l
1
x  s ( 0 ,1) / m
x  s ( 0, 2) / m  

 i j / m '

mlx  v
 l x  s (i , j ) / m  

1  m i j / m '

v
 l x  s (i , j ) / m

mlx i 0 j 1
Taking the limit as m→∞ we get:
Ax  lim A
m 

( m)
x
1
 lim 
m  ml
x


m
i j / m
'
v

l

x  s (i , j ) / m 
i  0 j 1
'

l
l
l
t
t
x t
x t


v

dt

v



0
0 lx  lxt
lx
'
x t


dt   v t t p x  x t dt

0
Premium for insurance payable at the
moment of death
• Whole life policy:

Ax   v t  t px   x t dt
0
• Term policy:
n
A : n |  v t  t px   x t dt
0
Examples
• Find the net single premium for a 100,000 life insurance
policy, payable at the moment of death, purchased by a
person aged 30 if i = .06 and tp30 = (.98)t for all t
• Solve the previous example if it is 20 years endowment
insurance, force of interest is .06 and
lx = 105 – x, 0 ≤ x ≤ 105.
Remarks
• Using integration by parts, we can get
Āx = 1 – δ ā x
• Approximate formula:
Āx ≈ (i/δ) Ax
• To obtain it, use linear interpolation in the following
expression:
A


1 1/ m
v (l x  l x 1/ m )  v 2 / m (l x 1/ m  l x  2 / m )    v i  j / m (l x i  ( j 1) / m  l x i  j / m )  
lx