1. MATH 3286 Mathematics of Finance
Alex Karassev

2. COURSE OUTLINE Theory of Interest Life Insurance
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

3. Chapter 1 INTEREST: THE BASIC THEORY
Accumulation Function
Simple Interest
Compound Interest
Present Value and Discount
Nominal Rate of Interest
Force of Interest

4. 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)

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 and Rate of Interest
Interest = Accumulated Value – Principal: Interest = A(t) – A(0)
Effective rate of interest i (per year):
Effective rate of interest in nth year in:

7. a(t)=t2+t+1 Example (p. 5) 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 Accumulation Functions)
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

9. a(t)=1+it, t ≥0 1.2 SIMPLE INTEREST
Amount function: A(t)=A(0) ∙a(t)=A(0)(1+it)
Effective rate is i
Effective rate in nth year:
(0,1) t
a(t) =1+it 1
1+i

10. a(t)=1+it i=0.15 A(t)=A(0)(1+it) =1000(1+0.15t) t=? A(0)=1000
Example (p. 5)
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
i=0.15
A(t)=A(0)(1+it) =1000(1+0.15t)
t=?

11. How to calculate t in practice?
Exact simple interest number of days
Ordinary simple interest (Banker’s Rule) number of days
t = t=
Number of days: count the last day but not the first

12. A(t)=1000(1+0.15t) Exact simple interest
Number of days (from Jan 1 to Jan 17) = 16
Exact simple interest
t=16/365
A(t)=1000( ∙ 16/365) =
Ordinary simple interest (Banker’s Rule)
t=16/360
A(t)=1000( ∙ 16/360) =

13. 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

14. a(t)=(1+i)t COMPOUND INTEREST Accumulation Function
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+it
(0,1) t
a(t)=(1+i)t 1
1+i

15. 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 1 t
a(t)=(1+i)t
1+i 2
(1+i)2

16. Example (p. 8) a(t)=(1+i)t A(t)=A(0)(1+i)t A(t)=1000(1+0.15)t
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 as in (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)=(1+i)t
A(t)=A(0)(1+i)t
A(0)=1000, i=0.15
A(t)=1000(1+0.15)t

17. 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

18. Calculation of present value
t=1, principal = 1
Let v denote the present value
v (1+i)=1
v=1/(1+i)

19. 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

20. (0,1) t
a(t)=(1+i)t
a(t)=(1+i)t
gives the value of one unit (at time 0) at any time t, past or future

21. If principal is not equal to 1…
present value = A(0) (1+i)t
t < 0
t = 0
t > 0
PRESENT VALUE A(0) (1+i)t
PRINCIPAL A (0)
ACCUMULATED VALUE A(0) (1+i)t

22. Example (p. 11) Solution a(t)=(1+i)t
The Kelly family buys a new house for 93,500 on May 1, 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 of A(0) = 93,500
= 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

23. 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

24. 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
a(t) =1+it
1 / (1 + it)
a(t) =1+it 1 t
1 / (1 - it)

25. Discount Look at 112 as a basic amount We invest 100
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

26. Rate of Discount d = i = 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) =

27. In nth year…

28. Identities relating d to i and v
Note: d < i

29. 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

30. 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, If the rate of compound interest is 7% per year, find the value of this deposit on August 1, 1994.

31. 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

32. 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 rates of interest (12% and 18%) were nominal rates of interest

33. …yet another example You have two credit card offers: Which is better?
17% convertible semi-annually
16% convertible monthly
Which is better?

34. 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

35. Note: If i is the effective rate of interest per year, it follows that
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
Equivalently:

36. Examples (p. 15)
Find the accumulated value of 1000 after three years at a rate of interest of 24 % per year convertible monthly
If i(6)=15% find the equivalent nominal rate of interest convertible semi-annually

37. Formula that relates nominal rates of interest

38. 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:

39. Formula relating nominal rates of interest and discount

40. Example
Find the nominal rate of discount convertible semiannualy which is equivalent to a nominal rate of interest of 12% convertible monthly

41. 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

42. Definition Nominal rate of interest equivalent to i:
Let m approach infinity:
We define the force of interest δ equal to this limit:

43. Formula
Force of interest δ = ln (1+i)
Therefore eδ = 1+i
and a (t) = (1+i)t =eδt
Practical use of δ: the previous formula gives good approximation to a(t) when m is very large

44. Example
A loan of 3000 is taken out on June 23, 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)

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

46. 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)= (1+i)t !

47. Example (p. 19)
Find in δt the case of simple interest
Solution

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

49. 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:
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

50. 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:

51. Chapter 2 INTEREST: BASIC APPLICATIONS
Equation of Value
Unknown Rate of Interest
Time-Weighted Rate of Return

52. 2.1 Equation of Value Four numbers: Time diagram
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

53. Examples
Find the accumulated value of 500 after 173 months at a rate of compound interest of 14% convertible quarterly (p. 30)
Alice borrows 5000 from FF Company at a rate of interest 18% per year convertible semi-annually. Two years later she pays the company Three years after that she pays the company How much does she owe seven years after the loan is taken out? (p. 31)
Eric deposits 8000 on Jan 1, and 6000 on Jan 1, 1997 and withdraws on Jan 1, Find the amount in Eric’s account on Jan 1, 2004 if the rate of compound interest is 15% per year (p. 31)

54. More Examples…
Unknown time
John borrows 3000 from FFC. Two years later he borrows another Two years after that he borrows an additional At what point in time would a single loan of 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)

55. 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)

56. Examples
Joan deposits 2000 in her bank account on January 1, 1995, and then deposits 3000 on January 1, 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.
Obtain a more exact answer to the previous question

57. 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:
The time-weighted rate of return is defined by i = (1+i1 ) (1+i2 ) … (1+in ) - 1

58. 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.

59. Unknown Time and Unknown Rate of Interest Continuous Annuities
Chapter 3 ANNUITIES
Basic Results
Perpetuities
Unknown Time and Unknown Rate of Interest
Continuous Annuities
Varying Annuities

60. 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

61. Series of n payments of 1 unit
Formulas
Series of n payments of 1 unit
present value of annuity
accumulated value of annuity
sn|
an| 1 1 1 1
….. n 1 2 3
The present (accumulated) value of the series of payments is equal to the sum of present (accumulated) values of each payment

62. Examples (p. 46)
(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
(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?
(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)
sn| 1 1 1 1
….. n 1 2 3
Annuity-due (payments made at the beginning)
än|
.. sn| 1 1 1 1
….. n
n+1 1 2 3

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 if i = 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

67. Formulas
present value of perpetuity
a∞| 1 1 1
….. 1 2 3

68. 3.4 Unknown Time and Unknown Rate of Interest
Examples – Unknown Time
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

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

70. Series of n∙m payments of 1/m
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 n∙m payments of 1/m
1/m
1/m
1/m
1/m
…..
1/m
2/m
3/m
n = nm(1/m)
Formula:
1/m
3/m
2/m
n = nm(1/m)
…..

71. Let m approach infinity…
Present value of continuous annuity
Accumulated value of continuous annuity
Formulas:

72. Arithmetic increasing perpetuities
3.6 Varying Annuities
Arithmetic annuities
increasing
decreasing
Arithmetic increasing perpetuities

73. 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

74. Series of n payments k-th payment = P+ (k-1)Q
Formulas
Series of n payments k-th payment = P+ (k-1)Q
k=1,2,…, n
accumulated value S
present value A P
P+Q
P+2Q
P+(n-1)Q
….. n 1 2 3

75. 1) Increasing annuity with P = 1, Q = 1
Series of n payments k-th payment = k
k=1,2,…, n
present value
(Ia)n|
accumulated value (Is)n| 1 2 3 n
Two Special Cases
….. n 1 2 3
We have:
P = 1
Q = 1

76. 2) Decreasing annuity with P = n, Q = -1
Series of n payments k-th payment = n – k + 1
k=1,2,…, n
present value
(Da)n|
accumulated value (Ds)n| n
n-1
n-2 1
….. n 1 2 3
We have:
P = n
Q = -1

77. Increasing perpetuity
k-th payment = k continues forever
present value
(Ia)∞| 1 2 3
….. 1 2 3

78. Examples (p )
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
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
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
Show both algebraically and verbally that (Da)n| = (n+1)an| - (Ia)n|

79. More examples: Geometric annuities
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%
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

80. Chapter 4 AMORTIZATION AND SINKING FUNDS
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 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

83. Examples (p )
(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. Outstanding Principal
Example
5,000 at 12 % per year repaid by 5 annual payments
Amortization schedule:
Duration
Payment
Interest
Principal Repaid
Outstanding Principal 1
600.00
787.05 2
505.55
881.50 3
399.77
987.28 4
281.30 5
148.61

87. Outstanding principal P
Interest earned during interval (t-1,t) is iP
Therefore interest portion of payment X is iP and principal portion is X - iP
Outstanding principal P
Payment X
t - 1 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

88. an-t| an| ….. ….. If each payment is X then
General method
outstanding principal at t
present value = outst. principal at 0
an-t|
an| 1 1 1 1 1
…..
….. n 1 2 t
t+1
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

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
Principal Repaid
Outstanding Principal 1
600.00
787.05 2
505.55
881.50 3
399.77
987.28 4
281.30 5
148.61
i = 12 %
Take the entry from “Outs. Principal” of the previous row, multiply it by i, and enter the result in “Interest”
“Payment” – “Interest” = “Principal Repaid”
“Outs. Principal” of prev. row - “Principal Repaid” = “Outs. Principal”
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 single lump-sum payment at some point in the future

93. ….. iL iL lump-sum payment L interest iL1 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

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 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. 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 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?

97. Amortization Schedule Other Topics
Chapter 5 BONDS
Price of a Bond
Book Value
Amortization Schedule
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) = 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. Equation to find yield rate i
redemption value C
Note: time is
measured in
half-years
coupon (interest) Fr Fr Fr
….. 1 2 n
purchase price P
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 < 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

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 < Bt < Bt+1< Bn = C or P = B0 > Bt > Bt+1 > Bn = C

105. Examples (p )
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.
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
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.
Assume simple interest at rate i per period between adjacent coupon payments

107. Alternative approach (1+i) Bt Bt Bt+1
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
flat price Fr Bt
Bt+1
market price t
t+1

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
Principal adjustment
Book Value 1 40
31.12
8.88 2
30.85
9.15 3
30.57
9.43 4
30.29
9.71
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 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 adjustment
Book Value 1 40
31.12
8.88 2
30.85
9.15 3
30.57
9.43 4
30.29
9.71

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

117. Chapter 6 PREPARATION FOR LIFE CONTINGENCIES
Introduction
Contingent Payments

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 , and the probability he will live for two years is 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
(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 7 LIFE TABLES AND POPULATION PROBLEMS
Introduction
Life Tables
The Stationary Population
Expectation of Life

123. 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

124. lx dx 7.2 Life Tables lx – number of lives survived to age x
1000 qx
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 – 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

125. Examples (p. 129 – p. 130) lx dx Find
Age lx dx
1000 qx
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 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

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 – lx+1) / lx and px = lx+1 / 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 – lx+n) / lx and n px = lx+n / lx
Formulas: lx – lx+n = dx + dx+1 + …+ dx+n n+mpx = mpx ∙ npx+m

129. 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 :

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
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 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

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 people aged 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 % 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: (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. Interpret verbally the expression Tx-Tx+n – nlx+n
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

140. Chapter 8 LIFE 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. t Ex = (t px ) (1 + t) – t = v t t px
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 if px = .95 for each x

145. Series of payments of 1 unit as long as individual is alive
Life annuity
Series of payments of 1 unit as long as individual is alive
present value (net single premium) of annuity ax 1 1 1
…..
…..
age x
x + 1
x + 2
x + n px
2px
npx
probability

146. Temporary life annuity
Series of n payments of 1 unit (contingent on survival)
present value ax:n|
last payment 1 1 1
…..
age x
x + 1
x + 2
x + n px
2px
npx
probability

147. 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 n|ax
first payment 1 1 … …
age x
x + 1
x + 2
x + n
x + n +1
x + n + 2
n+1px
n+2px
probability
Note:

148. äx ….. … äx:n| ….. n|äx … … Life annuities-due 1 1 1 1 px 2px npx 1 1

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 = Dx+n / Dx

151. Life annuity and commutation functions
Since nEx = Dx+n / Dx we 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 = 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

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 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

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, Margaret’s probability of survival is higher than that predicted by the life tables and, in fact, is equal to p , 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 = 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|

158. The following values are based on a unisex life table: N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865, N42 = 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. Annuity payable every 1/m part of the year
Usual life annuity ax 1 1 1
…..
…..
age x
x + 1
x + 2
x + n
Annuity payable every 1/m part of the year ax 1 1 1 1
…..
…..
age x
x + 1/m
x + 2/m
x + (m-1)/m
x + n