MATH 3286 Mathematics of Finance

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MATH 3286Mathematics of Finance

• Alex Karassev

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COURSE OUTLINE

• Theory of Interest
• Interest the basic theory
• Interest basic applications
• Annuities
• Amortization and sinking funds
• Bonds
• Life Insurance
• Preparation for life contingencies
• Life tables and population problems
• Life annuities
• Life insurance

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Chapter 1INTEREST THE BASIC THEORY

• Accumulation Function
• Simple Interest
• Compound Interest
• Present Value and Discount
• Nominal Rate of Interest
• Force of Interest

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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 theaccumulated value
  and is denoted byA(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)

5
Natural assumptions on a(t)

• increasing
• (piece-wise) continuous

a(t)
a(t)
a(t)
(0,1)
(0,1)
(0,1)
t
t
t
Note a(0)1
6
Definition of Interest andRate of Interest

• Interest Accumulated Value
  PrincipalInterest A(t) A(0)
• Effective rate of interest i (per year)

• Effective rate of interest in nth year in

7
Example (p. 5)
a(t)t2t1

• 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

8
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 accumulationfunction

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1.2 SIMPLE INTEREST

• a(t)1it, t 0

• Amount functionA(t)A(0) a(t)A(0)(1it)
• Effective rate is i
• Effective rate in nth year

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Example (p. 5)
a(t)1it
Solution

• 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?

A(0)1000
i0.15
A(t)A(0)(1it)1000(10.15t)
t?
11
How to calculate t in practice?

• Exact simple interest number of days
  365
• Ordinary simple interest (Bankers Rule)
  number of days 360

t
t
Number of days count the last day but not the
first
12
A(t)1000(10.15t)
Number of days (from Jan 1 to Jan 17) 16

• Exact simple interest
• t16/365
• A(t)1000(10.15 16/365) 1006.58
• Ordinary simple interest (Bankers Rule)
• t16/360
• A(t)1000(10.15 16/360) 1006.67

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1.3 COMPOUND INTEREST
Interest earns interest

• After one yeara(1) 1i
• After two yearsa(2) 1ii(1i)
  (1i)(1i)(1i)2
• Similarly after n yearsa(n) (1i)n

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COMPOUND INTEREST Accumulation Function
a(t)(1i)t

• Amount functionA(t)A(0) a(t)A(0) (1i)t
• Effective rate is i
• Moreover effective rate in nth year is i
  (effective rateis constant)

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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

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Example (p. 8)

• Jack borrows 1000 at 15 compound interest.
• How much does he owe after 2 years?
• How much does he owe after 57 days, assuming
  compound interest between integral durations?
• How much does he owe after 1 year and 57 days,
  under the same assumptions asin (b)?
• How much does he owe after 1 year and 57 days,
  assuming linear interpolation between integral
  durations
• In how many years will his principal have
  accumulated to 2000?

a(t)(1i)t
A(t)A(0)(1i)t
A(0)1000, i0.15
A(t)1000(10.15)t
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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
18
Calculation of present value

• t1, principal 1
• Let v denote the present value
• v (1i)1
• v1/(1i)

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In general
v1/(1i)

• t is arbitrary
• a(t)(1i)t
• the present value of 1 (t years in the past)
  (1i)t 1
• the present value of 1 (t years in the past) 1/
  (1i)t vt

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a(t)(1i)t
gives the valueof one unit(at time 0)at any
time t,past or future
21
If principal is not equal to 1

• present value A(0) (1i)t

t lt 0
t 0
t gt 0
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Example (p. 11)
Solution
a(t)(1i)t
The Kelly family buys a new house for 93,500
onMay 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?

• Find the present value ofA(0) 93,500
• 996 - 1992 4 yearsin the past
• t - 4, i 0.08
• Present value A(0) (1i)t 93,500
  (10.8) -4 68,725.29

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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 gt 0
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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
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Discount
Alternatively

• 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

• We invest 100
• After one year it accumulates to 112
• The interest 12 was added at the end of the term

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Rate of Discount
Definition Effective rate of discount d
accumulated value after 1 year principal
accumulated value after 1 year
A(1) A(0) A(1)
d

A(0) a(1) A(0) A(0) a(1)
a(1) 1 a(1)


Recall
accumulated value after 1 year principal
principal
i
a(1) 1 a(0)

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In nth year
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Identities relating d to i and v
Note d lt i
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Present and accumulated values in terms of d

• 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
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Examples (p. 13)

• 1000 is to be accumulated by January 1, 1995 at a
  compound rate of discount of 9 per year.
• Find the present value on January 1, 1992
• Find the value of i corresponding to d
• 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.

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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 ofeffective interest
periodsin any particular problem
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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 that this rate is
  convertible semi-annually. Is 12 the
  effective rate of interest per year?
• Credit card charges 18 per year convertible
  monthly. Is 18 the effective rate of interest
  per year?

In both examples the given ratesof interest (12
and 18) werenominal rates of interest
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yet another example

• You have two credit card offers
• 17 convertible semi-annually
• 16 convertible monthly
• Which is better?

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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

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Note
If i is the effective rate of interest per year,
it follows that
In other words,i is the effective rate of
interestconvertible annually which is equivalent
to the effective rate of interest i(m) /m
convertible mthly
Equivalently
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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

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Formula that relatesnominal rates of interest
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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

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Formula relating nominal rates of interest and
discount
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Example

• Find the nominal rate of discount convertible
  semiannualy which is equivalent to a nominal rate
  of interest of 12 convertible monthly

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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

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Definition
Nominal rate of interest equivalent to i
Let m approach infinity
We define theforce of interest dequal to this
limit
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Formula

• Force of interest d ln (1i)
• Therefore ed 1i
• and a (t) (1i)t edt
• Practical use of d the previous formula gives
  good approximation to a(t) when m is very large

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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 loanon June 23, 2002
• The value of i
• The value of i(12)

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Remark
The last formula shows that it is reasonable to
define forceof interest for arbitrary
accumulation function a(t)
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Definition
The force of interest corresponding to a(t)

• Note
• in general case,force of interest depends on t
• it does not depend on t ? a(t) (1i)t !

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Example (p. 19)

• Find in dt the case of simple interest
• Solution

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How to find a(t) if we are given by dt ?
We have
Consider differential equation in which a a(t)
is unknown function
Since a(0) 1 its solution is given by
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Applications

• Prove that if dt d is a constant thena(t)
  (1i)t for some i
• Prove that for any amount function A(t) we have
• Note dt dt represents the effective rate of
  interest over the infinitesimal period of time
  dt . Hence A(t)dt dt is the amount of interest
  earned in this period and the integral is the
  total amount

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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
• and formulas

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Chapter 2INTEREST BASIC APPLICATIONS

• Equation of Value
• Unknown Rate of Interest
• Time-Weighted Rate of Return

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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 writethe equation of value

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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 Erics account on Jan 1, 2004
   if the rate of compound interest is 15 per year
   (p. 31)

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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)

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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)

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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

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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 t1
• Let W1, , Wn-1 denote the amounts of deposits
  (Wt gt 0) or withdrawals (Wt gt 0) and let W0 0
• Determine rate of interest earned in the time
  period between balances

The time-weighted rate of return is defined byi
(1i1 ) (1i2 ) (1in ) - 1
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Example (p. 35)
On January 1, 1999, Grahams 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 Grahams portfolio during 1999.
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Chapter 3ANNUITIES

• Basic Results
• Perpetuities
• Unknown Time andUnknown Rate of Interest
• Continuous Annuities
• Varying Annuities

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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

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Formulas
Series of n paymentsof 1 unit
accumulated valueof annuity
present valueof annuity
1
1
1
1
..
n
0
1
3
2
The present (accumulated) value of the series of
payments is equalto the sum of present
(accumulated) values of each payment
62
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?

63
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
• the amount of each payment
• the amount of each payment if there is no
  payment-free period

64
Annuity-immediate and Annuity-due
Annuity-immediate(payments made at the end)
1
1
1
1
..
n
0
1
3
2
Annuity-due (payments made at the beginning)
1
1
1
1
..
n
n1
1
3
2
65
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 ifi 16

66
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

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Formulas
present value ofperpetuity
1
1
1
..
0
1
3
2
68
3.4 Unknown Time and Unknown Rate of Interest
Examples Unknown Time

1. 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 i9 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
2. 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

69
Example Unknown Rate of Interest

• 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

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3.5 Continuous Annuities
Let effective period be 1/m part of the year and
i(m)/m be the effective rate of interest (1
i(m)/m)m 1 i
Series of nm paymentsof 1/m
1/m
1/m
1/m
1/m
..
0
1/m
3/m
2/m
n nm(1/m)
Formula
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Let m approach infinity

• Present valueof continuous annuity
• Accumulated valueof continuous annuity

Formulas
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3.6 Varying Annuities

• Arithmetic annuities
• increasing
• decreasing
• Arithmetic increasing perpetuities

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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 formarithmetic progression
• P, PQ, P2Q, , P(n-1)Q

74
Formulas
Series of n paymentsk-th payment P (k-1)Q
k1,2,, n
accumulated value S
present value A
P
PQ
P2Q
P(n-1)Q
..
n
0
1
3
2
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1) Increasing annuity with P 1, Q 1
Series of n paymentsk-th payment k k1,2,, n
accumulated value (Is)n
present value (Ia)n
1
2
3
n
Two Special Cases
..
n
0
1
3
2
We have
P 1 Q 1
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2) Decreasing annuity with P n, Q -1
Series of n paymentsk-th payment n k 1
k1,2,, n
accumulated value (Ds)n
present value (Da)n
n
n-1
n-2
1
..
n
0
1
3
2
We have
P n Q -1
77
Increasing perpetuity
k-th payment kcontinues forever
present value (Ia)8
1
2
3
..
0
1
3
2
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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
   (n1)an - (Ia)n

79
More examples Geometric annuities

1. Geometric annuityAn 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 interestIn 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

80
Chapter 4AMORTIZATION AND SINKING FUNDS

• Amortization
• Amortization Schedule
• Sinking Funds
• Yield Rates

81
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

82
Two Methods

• Prospective methodoutstanding principal at any
  point in time is equal to the present value at
  that date of all remaining payments
• Retrospective methodoutstanding 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

83
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.

84
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.

85
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

86
Example
5,000 at 12 per year repaid by 5 annual payments
Amortization schedule
Duration Payment Interest PrincipalRepaid Outstanding Principal
0 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
87
Example

• Interest earned during interval (t-1,t) is iP
• Therefore interest portion of payment X
  is iP and principal portion is X - iP

Outstanding principalP
PaymentX
t - 1
t
Recall in practical problems, the outstanding
principal P can be found by prospective or
retrospective methods
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
88
General method
outstanding principal at t
present value outst. principal at 0
1
1
1
1
1
..
..
n
0
1
t
2
t1
interest portion of (t1)-st payment i a n-t
1 vn-t
principal portion of (t1)-st payment 1 (1
vn-t ) vn-t
If each payment is X then interest part of kth
payment X (1 vn-k1 ) principal part of kth
payment Xvn-k1
89
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

90
General rules to obtain an amortization schedule
Duration Payment Interest PrincipalRepaid Outstanding Principal
0 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
i 12

1. Take the entry from Outs. Principal of the
   previous row, multiply it by i, and enter the
   result in Interest
2. Payment Interest Principal Repaid
3. Outs. Principal of prev. row - Principal
   Repaid Outs. Principal
4. Continue

91
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

92
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 singlelump-sum payment
  at some point in the future

93
lump-sum payment L
interestiL
iL
iL
..
0
1
2
n

• 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

94
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.

95
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

96
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 Georges 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?

97
Chapter 5BONDS

• Price of a Bond
• Book Value
• AmortizationSchedule
• Other Topics

98
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

99
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

100
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

101
redemption value C
Note time is measured in half-years
coupon (interest)Fr
Fr
Fr
..
0
1
2
n
Equation to find yield rate i
Note often C F
102
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 lt C since the yield rate is higher than the
  coupon rate, i gt r
• therefore the investor is buying the bond at a
  discount
• otherwise (if i lt r) we would have P gt C and then
  the investor would have to buy the bond at a
  premium of P - C

103

• 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.

104
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

• Remarks
• Usually C F
• In the last formula, an-t and v are computed
  using the yield rate i
• P B0 lt Bt lt Bt1lt Bn C or P B0 gt Bt gt
  Bt1 gt Bn C

105
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 Bt1 be the book values just after the
   tth and (t1)th coupons are paid. Show that Bt1
   Bt (1i) Fr
3. Find the book value in 1) exactly 2 months after
   the 14th coupon is paid.

106
How do we find the book value between coupon
payment dates?
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.

• Assumesimple interestat rate i per period
  between adjacent coupon payments

107
Alternative approach

• Since Bt1 Bt (1i) Fr we can view Bt (1i)
  as the book value just before next (i.e. (t1)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 Bt1 and Bt we
  obtain the market price (or the amortized value)
  of the bond
• Clearly market price flat price at any given
  moment

(1i) Bt
Fr
flat price
Bt
Bt1
market price
t1
t
108
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.

109
5.3 Bond Amortization Schedule

• Goal trace changes of the book value
• Bond amortization schedule table, containing
  the following columns
• time
• coupon
• interest
• principal adjustment
• book value

Time Coupon Interest Principaladjustment Book Value
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 1000 par value two-year bond which pays
interest at 8 convertible semiannually yield
rate is 6 convertible semiannually
110
Algorithm

• Book value at time t is Bt
• Amount of coupon at time t1 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 Principaladjustment Book Value
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
111
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.

112
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

113
5.4 Other Topics

• Different frequency of coupon payments
• Increasing or decreasing coupon payments
• Different yield rates
• Callable bonds

114
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 half-year for the first 5 years
  and 6 per half-year for the last 5 years

115
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
Maturity Date
Purchase Date
Call Date
116
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

117
Chapter 6PREPARATION FOR LIFE CONTINGENCIES

• Introduction
• ContingentPayments

118
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)

119
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?

120
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?

121

• (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.

122
Chapter 7LIFE TABLES AND POPULATION PROBLEMS

• Introduction
• Life Tables
• The StationaryPopulation
• Expectation of Life

123
7.1 Introduction

• How do we find probabilities?
• Data obtained from practice
• Data required to findprobabilities of
  survivingto certain ages (or, equivalently, of
  dying before certain ages) are contained in life
  tables

124
7.2 Life Tables
Age lx dx 1000 qx
0 1,000,000 1580 1.58
1 998,420 680 68
2 997,740 485 .49
3 997,255 435 .44

• 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 lx1 number of lives who died between
  (x, x1)
• qx dx /lx probability that x year-old will
  not survive to age x 1

125
Examples (p. 129 p. 130)
Age lx dx 1000 qx
0 1,000,000 1580 1.58
1 998,420 680 68
2 997,740 485 .49
3 997,255 435 .44

• Find
• the probability that a newborn will live to age 3
• the probability that a newborn will die between
  age 1 and age 3

126

• 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 40-year-old and the 50-year-old will die
  within the five-year period starting ten years
  from now, but neither the 60-year-old nor the
  70-year-old will die during that five-year period

127
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 lx1
  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

128
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 lx1) / lx and px lx1 / lx
• nqx probability that x year-old will not
  survive to age x n
• npx 1- nqx probability that x year-old will
  survive to age x n
• Note nqx (lx lxn) / lx and n px lxn /
  lx
• Formulaslx lxn dx dx1 dxn
  nmpx mpx npxm

129
What is mPx when m is not integer?

• tPx where 0 lt t lt1
• Assuming that deaths are distributed uniformly
  during any given year we can use linear
  interpolation to find tpx

130
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 1000 qx
0 1,000,000 1580 1.58
1 998,420 680 68
2 997,740 485 .49
3 997,255 435 .44
131
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 x1
• 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

132
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
  obtain
• More precisely

133
Number of people aged x and over

• Let Tx denote the number of peopleaged x and
  over at any given moment
• Then
• Assuming uniform distribution we obtain
• More precisely

134
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

135
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

136
Complete Expectation
Complete expectation (4)
(1)
(2)
(2)(3)(4) imply
(3)
Approximation
Since
we get
137
Remarks
Thus Tx can be viewed as the total number of
years of future life of those who form group lx
Note this interpretation makes sense in any
population (stationary or not)
138
Average age at death

• Average age at death of a person currently aged x
  is given by
• Letting x 0 we obtain the average age at death
  for all death among l0 individuals

139
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 expressionTx-Txn nlxn
• Fin the average age at death of those who die
  between age x and age xn

140
Chapter 8LIFE ANNUITIES

• Basic Concepts
• Commutation Functions
• Annuities Payable mthly
• Varying Life Annuities
• Annual Premiums and Premium Reserves

141
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

142
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

143
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 Ex (t px ) (1 t) t v t t px
144
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 ifpx
  .95 for each x

145
Life annuity
Series of payments of 1 unitas long as
individual is alive
present value(net single premium)of annuity ax
1
1
1
..
..
x
x 1
x 2
x n
age
px
2px
npx
probability
146
Temporary life annuity
Series of n payments of 1 unit(contingent on
survival)
present valueaxn
last payment
1
1
1
..
x
x 1
x 2
x n
age
px
2px
npx
probability
147
n - years deferred life annuity
Series of payments of 1 unit as long as
individual is alivein which the first payment is
at x n 1
present valuenax
first payment
1
1


x n 1
x
x 1
x 2
age
x n 2
x n
n1px
n2px
probability
Note
148
Life annuities-due
äx
1
1
1
1
..

x
x 1
x 2
x n
äxn
px
2px
npx
1
1
1
1
..
x n
x n-1
x
x 1
x 2
n-1px
px
2px
näx
1
1
1


x
x 1
x 2
x n 1
x n 2
x n
n2px
n1px
npx
149
Note
but
150
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
• Denote Dx vxlx
• Then nEx Dxn / Dx

151
Life annuity and commutation functions
Since nEx Dxn / Dxwe have
Define commutation function Nx as follows
Then
152
Identities for other types of life annuities
temporary life annuity
n-years delayed l. a.
temporary l. a.-due
153
Accumulated values of life annuities
temporary life annuity
since
and
we have
similarly for temporary life annuity-due
and
154
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,
  Nx14900, Nx2 4810 and qx .005, find i

155
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)

156
Notations

• Suppose that a person aged x isin the first year
  of being in the select group
• Then px denotes the probability of survival for
  1 year and qx 1 px 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
  px1, px2, and so on
• Similar notations are used for life
  annuitiesax 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

157
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, Margarets 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 px (11/10)
  px and px1 (21/20) px1. An ultimate table
  shows D60 1900, D61 1500, and ä 6020 11,
  when i .08. Find the select temporary life
  annuity ä6020

158

• 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.

159
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

160
Usual life annuity
ax
1
1
1
..
..
x
x 1
x 2
x n
age
Annuity payable every 1/m part of the year
1
1
1
1
ax
..
..
x
x 1/m
x 2/m
x (m-1)/m
age
x n