Transcript MathFinLec3

LOGO
MATH 2040
Introduction to
Mathematical Finance
Instructor: Miss Liu Youmei
1
Chapter 3 Basic Annuities (年金）
Definition of an annuity
Annuity-immediate
Annuity-due
Annuity values on any date
Perpetuity-immediate &perpetuity-due
Problem with unknown time
Problem with unknown rate of interest
2
Definition of an annuity
• An annuity is a series of payments made at equal
intervals of time (annually or otherwise).
• Payments made for certain for a fixed period of
time are called an annuity-certain.
• In this chapter, we deal with annuities with:
– The payment frequency and the interest conversion
period are equal.
– The payments are leveled (all equal to each other).
3
Annuity-immediate
• Payments of 1 are made at the end of every year for n
years.
• The present value (at t = 0) of an annuity-immediate, where
the annual effective rate of interest is i, shall be denoted as
and is calculated as follows:
4
Present value of annuity-immediate
5
Calculation of annuity-immediate
6
Accumulated value of annuity-immediate
• The accumulated value (at t = n) of an
annuity-immediate, where the annual effective
rate of interest is i, shall be denoted by
,
and is calculated as follows:
7
Accumulated value of annuity-immediate
8
First basic relationship
• The first basic relationship is:
• Consider an n-year investment where 1 is
invested at time 0.
• Also consider the stream of payments of i at the
end of each year for n years and then 1 is
refunded at time t = n.
1
9
First basic relationship – cont’d
• The present value of these multiple payments
(at time t = 0) is
.
• Hence we have
• Note that
10
Second basic relationship
• The second basic relationship is:
PV = FV v n and PV(1 + i )n = FV
• If the FV (future value) at time n,
, is
discounted back to time 0, then we will have the PV
(present value),
.
• So
11
Second basic relationship – cont’d
• If the PV at time 0,
, is accumulated to time n,
then we will have the FV,
.
• So
12
Second basic relationship
13
Third basic relationship
• The third basic relationship is:
• Suppose a loan of 1 is to be paid back over n
years with equal annual payments of P made at
the end of each year, then
or
• Alternatively, suppose the same loan of 1 is to
be repaid by interest payments of i made at the
end of each year, and the loan amount 1 is
paid back at time n.
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Third basic relationship – cont’d
• In order to produce the loan amount at time n, annual
payments of D, at the end of each year for n years,
will be made into an account that credits interest at an
annual effective rate of interest i.
• The future value of all these payment D must be equal
to 1, so we have
or
• The total payment will therefore be
15
Third basic relationship – cont’d
• Note that
• Therefore, a level annual payment on a loan is
the same as
– making an annual interest payment each year plus
– making annual deposits that will accumulate to the loan
amount.
16
Third basic relationship
1
1
17
Interest repayment option one
•
•
•
Given a loan of 1, there are 3 options in
repaying the loan over the next n years.
Pay back the loan and all interest due at time n.
Total interest paid = A(n) – A(0)
= Loan  (1 + i)n  Loan
= Loan  [(1 + i)n  1]
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Interest repayment option two
•
•
Pay at the end of each year, the interest that
comes due on the loan and then pay back the
loan at time n.
Annual interest payment = i  Loan
Total interest paid
= i  Loan  n
= Loan  (i  n)
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Interest repayment option three
•
•
Pay a level annual amount at the end of each year
for the next n years.
Annual payment
•
Total payments
•
Total interest paid
20
Comparing interest repayment options
•
•
•
Option 1 and 2 is a comparison between compound
v.s. simple interest.
Therefore, less interest is paid under option 2. This
would make sense because if you pay off interest as it
becomes due, the loan cannot grow as it does under
option 1.
Option 2 and 3 is a mathematical comparison that
shows less interest being paid under option 3. This can
be proved by making use of the basic relationship
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Example 3.1
•
•
•
Find the present value of an annuity which pays
$500 at the end of each half-year for 20 years if the
rate of interest is 9% convertible semi-annually.
We can view the annuity as one that pays $500 at
the end of each period for 40 periods with interest
rate 4.5% per period.
So the answer is:
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Example 3.2
•
•
•
If a person invests $1000 at 8% p.a. convertible quarterly,
how much can be withdrawn at the end of every quarter to
use up the fund exactly at the end of 10 years?
Since the interest rate is 8% p.a. convertible quarterly, and
withdrawals are made quarterly, we can view the problem
as one with interest at 2% per period, and the fund is to be
withdrawn in 40 periods.
So the answer is
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Example 3.3
•
Compare the total amount of interest that would be
paid on a $1000 loan over a 10-year period, if the
effective rate of interest is 9% per annum, under
the following three repayment methods:
– (1) The entire loan plus interest is paid in one lumpsum at the end of 10 years.
– (2) Interest is paid each year and the principal is
repaid at the end of 10 years.
– (3) The loan was repaid by level payments over the
10 year period.
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Example 3.3 (1)
• (1) The entire loan plus interest is paid in
one lump-sum at the end of 10 years.
• The accumulated value of the loan at the end
of 10 years is 1000(1.09)10 = 2367.36
• The total amount of interest paid is therefore:
$2367.36 - 1000 = $1367.36
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Example 3.3 (2)
– (2) Interest is paid each year and the
principal is repaid at the end of 10
years.
– Interest paid each year is
$1000  0.09 = $ 90
– Total interest for 10 years is
$90  10 = $ 900.
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Example 3.3 (3)
– (3) The loan was repaid by level payments over the
10 year period.
Let the level payment be R. Then
Which gives
So the total interest paid is
10  $155.82  $1000 = $588.20.
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Example 3.3 - remark
•
The later the principal was repaid, the more is
interest paid.
– (1) The entire principal plus accumulated interest was
paid at the end of 10 years, so the most interest was
paid for this option.
– (2) Interest is paid each year, less interest was paid on
this option.
– (3) The loan was repaid by level payments. In each
payment, some interest and some principal were paid.
So the least interest amount was paid for this option.
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Annuity due(期初年金）
• For annuity immediate, payments of 1 are made at
the end of every year for n years.
• In another type of annuity, called annuity due,
payments of 1 are made at the beginning of every
year for n years.
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Annuity due
• The present value (at t = 0) of an annuity-due, where
the annual effective rate of interest is i, shall be
denoted as
and is calculated as follows:
30
Annuity due – cont’d
• The accumulated value at t = n of an annuity-due,
where the annual effective rate of interest is i, shall
be denoted as
and is calculated as follows:
31
Basic relationships -Annuity due
• Similar to annuity immediate, there are also
three basic relationships on annuity-due.
• (1)
• (2) PV(1 + i)n = FV and PV = FV  v n
• (3)
• There are similar interpretation of all these
relationships as those for annuity-immediate.
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Basic relationships - Annuity due (1)
• Consider an n-year investment where 1 is
invested at time 0.
• Suppose this investment produces
– annual interest payments of d at the beginning of
each year
– the amount of 1 at t = n.
• The present value of this is
• It follows that
• Note that
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Basic relationships - Annuity due (2)
• If the future value at time n,
, is discounted back
to time 0, then we obtained its present value,
.
• Similarly, if the present value at time 0,
, is
accumulated forward to time n, then we get
So
.
34
Basic relationships - Annuity due (3)
• Consider a loan of 1, to be repaid back over n
years with equal annual payments of P made at
the beginning of each year.
• Suppose an effective rate of interest, i, is used.
• The present value of the single payment loan must
be equal to the present value of the multiple
payment stream, i.e.
35
Basic relationships - Annuity due (3)
• Consider an alternative method of repaying the same loan of 1,
where the annual interest due on the loan d, is paid at the
beginning of each year for n years, and the loan amount 1 is
paid at time n.
• To produce the loan amount 1 at time n, annual payments, D,
are deposited at the beginning of each year for n years so that
it will grow with the same effective rate of interest as the loan.
• The future value of the multiple deposits should equal the
future value of 1 at time n. i.e.
• The total annual payment is then
36
More basic relationships
• In addition to the above three, there are also
three additional basic relationships on
annuities.
• (4)
and
• (5)
• (6)
37
Basic relationship (4)
• An annuity-due starts one period ahead of
an annuity-immediate, and as a result it
earns interest for one more period.
• That applies to both present value and
accumulated value.
• So we have
•
and
38
Basic relationship (5)
• An annuity-due starts immediately. That
results in an additional payment at time 0.
• The remaining (n – 1) payments can be
viewed as an annuity-immediate for (n – 1)
years
• Hence we get
39
Basic relationship (6)
• The annuity due for (n – 1) years may be
viewed as one that starts at the end of year
one and finishes at the end of the (n – 1)-th
year.
• An additional payment of 1 at time n
results in
becoming n payments
that now starts at the end of each year.
• So we have
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Example 3.4
• An investor wishes to accumulate $1000 in a fund at the end
of 12 years. He plans to make deposits at the end of each year,
the final payment to be made at the end of the 11-th year. What
should the annual deposits be if the fund earns 7% effective?
• Suppose the annual deposit is $ D. Number of payments made
is 11, and we want to get the accumulated value of these 11
payments at the end of 12 years, so
• Solving for D, we get D = 59.21
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Annuity values on any date
• There are several other dates to valuing annuities
rather than at the beginning of the term (t = 0) or at
the end of the term (t = n).
– Present values more than one period before the first
payment date
– Accumulated values more than one period after the last
payment date
– Current value between the first and the last payment
dates.
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Values before first payment (1)
• Consider a series of payments of 1 that are made at time t
= 3, 4, 5, 6, 7, 8 and 9.
• At t = 2, there exists 7 future payments whose present
value is represented by
. If this value is discounted back
to time t = 0, then the value of this series of payments is (2
periods before the first end-of-the-year payment) is
.
• Alternatively, at t = 3, there exists 7 future payments
whose present value is represented by
. If this is
discounted back to time t = 0, then the value of this series
of payments (3 periods before the first beginning-of-theyear payment) is
43
Values before first payment (2)
• Another way to examine this situation is to pretend there
are 9 end of year payments. This can be done by adding 2
more payments to the existing 7. In this case, let the 2
additional payments be made at t = 1 and 2.
• At time t = 0, the value of the 9 payments and the two
added-on payments are
and
respectively.
• Therefore the present value at t = 0 is
• Note that this results in the formula:
• A more general formula is: For positive integers m and n
44
Values before first payment (3)
• With the annuity due version, suppose 3 more payments
are added to the existing 7 payments. In this case, let the
3 additional payments be made at t = 0, 1 and 2.
• At time t = 0, the value of the 10 payments and the three
added-on payments are
and
respectively.
• Therefore the present value at t = 0 is
• Note that this results in the formula:
• A more general formula is: For positive integers m and n
45
Values before first payment (4)
• This type of annuity is often called a deferred annuity, since
payments commence only after a deferred period.
• A symbol for an annuity-immediate deferred for m periods
with a term of n periods after the deferral period is
.
• And for annuity-due is
.
• Thus the above annuity could be labeled
or
.
46
Values after last payment (1)
• At t = 9, there exists 7 past end-of-year payments whose
accumulated value is represented by
. If this value is
accumulated forward to time t = 12, then the value of this series
of payments (3 periods after the last end-of-year payments) is:
• Alternatively, at t = 10, there exists 7 past beginning-of-year
payments whose accumulated value is represented by
. If
this value is accumulated forward to time t = 12, then the value
of this series of payments (2 periods after the last beginning-ofyear payments) is:
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Values after last payment (2)
• Another way to examine this situation is to add three more
payments to the existing 7 so that now there are 10 endof-year payments. Let the 3 additional payments be made
at t = 10, 11 and 12 (denoted as
).
• At t = 12, there now exists 10 end-of-year payments
whose accumulated value is represented by
. This
value would then be reduced by the value of the three
added-on payments, i.e.
.
• Therefore the accumulated value at t = 12 is
.
• This results in
• The general formula is:
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Values after last payment (3)
• To use the annuity due approach, we may add just 2 more
payments to the existing 7 so that now there are 9
beginning-of-year payments. Let the 2 additional
payments be made at t = 10 and 11 (denoted as
).
• At t = 12, there now exists 9 beginning-of-year
payments whose accumulated value is represented by
.
This value would then be reduced by the value of the two
added-on payments, i.e. .
• Therefore the accumulated value at t = 12 is
.
• This results in
.
• The general formula is
.
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Values between first and last dates (1)
• The 7 payments can be represented by an annuityimmediate or by an annuity-due depending on the time that
they are evaluated
• For example, at t = 2, the present value of the 7 end-ofyear payments is
. At t = 9, the future value of these
payments is
.
• At a point between t = 2 and t = 9, the present value and
future value can be accumulated to or discounted back
respectively.
• Take t = 6, the present value would be accumulated
forward 4 years, the accumulated value would be
discounted back 3 years.
• So we have
• The general formula is
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Values between first and last dates (2)
• Alternatively, at t = 3, we can view the payments as
beginning-of-year payments and its present value is
At t = 10, the future value of these payments would
be
.
• At t = 6, for example, the present value would be
accumulated forward 3 years and the accumulated
would be discounted back 4 years.
• So we have
• The general formula is
.
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Values between first and last dates (3)
•
•
•
•
•
•
•
At any time during the payments, there will exist a series of past
payments and a series of future payments.
We can view all payments as end-of-year payments. At t = 6, for
example, we have 4 past payments whose accumulated value is
and 3 future payments whose present value is
.
So the value at t = 6 of all 7 payments is
Alternatively, all the payments can be viewed as beginning-of-year
payments. Then at t = 6, we have 3 past payments and 4 future
payments with accumulated values and present values of
and
respectively.
So the value at t = 6 of all 7 payments is
This results in
The general formula is
,
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Values of annuities at other dates
• In general, the present value of an n-period
annuity where the first payment is m years from
now is
• The accumulated value of an n-period annuity
m periods after the last payment date is
• The current value of an n-period annuity
immediately after the first m ( ≤ n) payments
have been made is
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Perpetuity-immediate（永续年金） (1)
• A perpetuity-immediate makes payment of 1 at
the end of every year forever, i.e. n = .
• The present value of a perpetuity-immediate,
where the annual effective rate of interest is i,
shall be denoted by
, and is calculated as
follows:
54
Perpetuity-immediate (2)
• We can also derive the above formula by taking
the limit as n tends to infinity in the original
present value formula:
• Note that suppose an initial amount of
is
invested at t = 0. Then the interest payments,
payable at the end of each year, produced by this
investment is exactly 1.
•
is not defined because it would equal .
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Perpetuity-immediate (3)
• The present value formula for an annuityimmediate can be expressed as the difference of
two perpetuity-immediate:
• In this case, a perpetuity-immediate that is payable
forever is reduced by perpetuity-immediate
payments that start after n years. The present
value at t = 0, results end-of-year payments
remaining only for the first n years.
56
Perpetuity-due (1)
• A perpetuity-due makes payment of 1 at the
beginning of every year forever, i.e. n = .
• The present value of a perpetuity-due, where the
annual effective rate of interest is i, shall be
denoted by
, and is calculated as follows:
57
Perpetuity-due (2)
• We can also derive the above formula by taking the
limit as n tends to infinity in the original present
value formula:
• Suppose an initial amount of
is invested at t = 0.
The annual interest, payable at the beginning of the
year, produced by this investment would be 1.
58
Example 3.5
• A leaves an estate of $100,000. Interest on the
estate is paid to beneficiary B for the first 10
years, to beneficiary C for the second 10 years,
and to charity D thereafter. All payments are
made at the end of year. Find the relative shares
of B, C, and D in the estate, if it is assumed the
estate will earn a 7% annual effective rate of
interest.
59
Example 3.5….cont’d
• The present value of B’s share is
• The present value of C’s share is
• The present value of D’s share is
• From the above, we can find that the relative shares of B,
C, and D is approximately 49%, 25%,26%.
60
Example 3.5….cont’d
• Note that the sum of the shares of B,C, and D is equal
to $100,000 as expected.
• Also note that the present value of the estate at the end
of 20 years is
, which equals D’s
share. This confirms the fact that charity D continuing
to receive the interest into perpetuity or receiving the
estate value in a lump-sum at the end of 20 years are
equivalent in value.
61
Problem with unknown time
• When solving a problem in annuity with
unknown time, the solution we often get is
not an integer.
• An adjustment to the payments can be
made so that n does become an integer.
62
Example
• How long will it take to payoff a loan
of $1000 if $100 is paid at the end
of every year and the annual effective
rate of interest is 5%?
• We need to solve the equation:
• Or
.
63
Example ….cont’d
• By inspection of the interest tables, we find the
value of n lying between 14 and 15.
• The equation of value at t = 14 is:
• The equation of value at t = 15 (after only 14
payments are made) is:
• Solving, we have for R1 = 20.07 and R2=21.07
• We should note that R1 (1.05) = R2.
64
Example ….cont’d
• In our case, the equation of value is:
• Or
i.e.
• we can determine that k = 0.2067.
• So the final payment is 100(1.05k-1)/0.05 =
20.27, or estimated value of 100k=20.67 to be
made at the time 14.2067.
65
Unknown time
• From above argument, we see that it is very
inconvenient to determine the final time of
payment, which is t = 14.2067, with a
payment of $20.67.
• In practice, either a final payment at time t =
14 of $(100 + R1 = 120.27) is made, or
• a final payment at t = 15 of $ R2 = 21.07 is
made.
66
Unknown rate of interest
• Assuming that we do not have a calculator
• There are three approaches to solving for an
unknown rate of interest when
• They are:
– Algebraic Techniques
– Linear Interpolation
– Successive Approximation (Iteration)
67
Algebraic Techniques
• Note that
is an n
degree polynomial and can be easily solved if n is small.
• When n gets too big, we can use a series expansion
of
, or even better, of
. Then we can solve for a
low degree polynomial equation, considering only the
first few terms
68
Algebraic Techniques
• For a series expansion of
, if considering the first
two terms, we can derive an approximation for i.
2(n  k )
i
k (n  1)
69
Example 3.8
• At what rate of interest, convertible quarterly, is
$16,000 the present value of $1000 paid at the end of
every quarter for five years?
2(n  k )
• Use the approximate formula i 
to obtain the
k (n  1)
answer.
• We have n=20 and k=16, so i  2(20  16)  0.0238
16(20  1)
• That is, the interest rate is approximately 2.38% per
quarter.
70
Linear Interpolation
• In this method, we need to find the value of
at two different interest rates i1 and i2
such that
and
• Then we can obtain by linear interpolation:
.
71
Example on Linear Interpolation
• At what interest rate, convertible semi-annually, is
$32,000 the present value of $2,000 payable every
six month for 10 years?
• Let j = i(2)/2, so the equation of value is
• Define
• We need to find the value of j for which f (j) = 0.
72
Example on Linear Interpolation – cont’d
• By looking up the tables, we found that
f (0.0200) = 16.3514 – 16 = 0.3514.
• and f (0.0250) = 15.5892 – 16 = – 0.4108.
• Now performing a linear interpolation, we get:
• Which gives i(2) = 2j = 4.46%.
73