Financial Mathematics and Option Pricing:
A Gentle Introduction
Klaus Volpert, PhD, Villanova University
CHAPTER 2: A THEORY OF INTEREST
In January 2014, billionaires Warren Buffett and Dan Gilbert of Quicken Loans teamed up to create a
contest that offered $1 billion to anyone that correctly predicted the winner of all 63 games of the 2014
NCAA men’s basketball tournament.
Of course, Buffett knew that the chances he would have to pay out the $1 billion were practically nil.
After all, there are 263 ways to fill out a prediction bracket. This number is unimaginably large. To put it
in perspective: If one could write down a different bracket every second, it would take 300 billion years
to get to the end. That’s 20 times the estimated age of the universe.1
But just in case, Buffett offered a choice to any winner: either accept the $1 billion in form of 40 annual
payments of $25 million, or a lump sum of $500 million instead.
Well, that begs the question: Imagine you were the winner, which option would you choose?
Your answer most likely depends on a number of personal preferences, so let us formulate the question
more mathematically:
Q: Suppose we could invest both the lump sum and the stream of payments at the same constant
interest rate r for the duration of the payments. For which value of r would the value of the two
investments at the beginning of the 40th year (when the last payment of $25 million is made) be the
same?
As we shall calculate later, this interest rate is 4.2%. One can consider this value an equilibrium rate. At
this value it does not matter which option you, the winner, chooses. After 40 years you’d come out the
same.
The same is true for Warren Buffett: if he can safely invest the $500 million at 4.2% for 40 years, it
would not matter to him which option you choose. Even if you chose the annual payments, he would set
aside the $500 million, invest it at 4.2% and from that fund pay out the annual payments. The fund
would be exactly at $0 after the last payment. So either way, it would cost him exactly $500 million now
to settle the claim.
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Of course not every bracket is equally likely. The most likely bracket (with every game won by the higher-seeded
team) I estimate (using historic data) to have a chance of occurring of one in 340 billion. . . Still, very long odds!
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So how was this equilibrium rate of 4.2% calculated?
Compounding Interest
When we invest an amount A in a certificate of deposit that pays interest at a rate r that is added to
the principal just once a year, then after one year we have
L  A  rA  A(1  r )
After two years we have
L  A(1  r ) (1  r )
After t years (with t being an integer), we have
L  A(1  r )t .
Many investments, though, calculate and add interest to the account more often than annually. If, for
example, interest is added monthly, it is understood that 1/12 of the interest rate is calculated and
added. So after one month our investment is
L  A(1 
r
)
12
After two months it is
L  A(1 
r 2
)
12
After one year it is
L  A(1 
r 12
)
12
And after t years it is
L  A(1 
r 12t
)
12
In general, if the interest rate is calculated and added n times per year, the future value of our
investment after t years is
r
L  A(1  ) nt
n
Example 1: if we invest A=$1000 at r=5%, compounded monthly, after 1 year, we will have
2
L  1000(1 
.05 12
)  1051.16
12
We see that we will actually earn more than the 5%, our nominal interest rate. That’s because the
interest added after one month itself earned interest over the remainder of the year, as did the next
interest payment, and so on. The effective interest rate for the year, in our example 5.116%, is called
the yield y, and needs to be distinguished from the nominal rate r of 5%. The relationship between the
yield and the nominal interest rate is
r
A(1  y )  A(1  ) n ,
n
and therefore
r
y  (1  ) n  1
n
Continuous Compounding
It may be counterintuitive, but the yield does not increase by much when we compound even more
often than monthly. In fact, we can, theoretically, compound infinitely often, and the yield will not go
beyond the following famous limit2:
r
lim y  lim(1  ) n  1  e r  1
n
n
n
In this case our investment after time t will be worth
L  Ae rt
Example 2: If the interest rate r=5.000% is compounded continuously, the yield is y=5.127%.
Investing a stream of payments
As in the case of the Buffett challenge, often we have to compare the time value of a lump sum with the
time value of a stream of payments. If these payments are constant at regular intervals, the formula for
a geometric series3
1  x  x 2  ...  x n 
x n1  1
x 1
helps us combine the future values of all the payments into one formula.
r n
)  e r is a classic and can be found in any calculus text.
n
2
n
n 1
3
Another classic: Let S  1  x  x  ...  x , then xS  S  x  1 . Now solve for S .
2
The proof that lim(1 
n
3
Example 3: In the Buffett challenge, there would be 40 payments of $25 million, commencing
immediately. The last payment would be made at the beginning of the 40th year. At that time the value
of the payments all invested at constant interest rate r, is
25(1  r )39  25(1  r )38  ...  25  25
(1  r )40  1
r
To find the value of r for which this future value is equal to the future value of the $500 million lump
sum we (numerically) solve the equation
500(1  r )39  25
(1  r )40  1
for r to find that r  4.2% .
r
Mortgages and other Loans
Suppose we borrow an amount A from a lender who demands a nominal interest rate r, compounded n
times a year, and suppose we make no payments, then the balance B(t) of what we owe at the end of
nt periods is
r
B (t )  A(1  ) nt , where nt is an integer.
n
If we make a payment P at the end of the first period (the same time that interest is compounded for
r
n
the first time), this would reduce the balance of what we owe at the end of nt periods by P(1  ) nt 1 ,
that is, by the amount of interest the portion P would have accrued over nt 1 periods. So the balance
on the loan would be
r
r
B(t )  A(1  ) nt  P(1  ) nt 1
n
n
Similarly, if we make a payment P at the end of every period, then after nt periods (and nt 1 payments),
the balance of the loan is
r
r
r
B(t )  A(1  ) nt  P[(1  ) nt 1  (1  ) nt 2  ...  1]
n
n
n
or
r
B(t )  A(1  ) nt  P[
n
r
(1  ) nt  1
n
].
r
n
4
1
Here we have again used the formula for the geometric series to simplify the sum on the right.4
The first term on the right-hand side is the future or time-value of the loan amount A , the second term
is the future or time- value of the stream of regular payments P, both after nt periods.
If the loan is to be paid off with a last payment P at time T , then the balance of the loan is zero, that is,
B(T )  0 , and
r
A(1  ) nT  P[
n
r
(1  ) nT  1
n
]
r
n
2
So at that instant the loan is paid off, the time-value of the loan amount and the time value of the
stream of payments are the same.
When we solve this equation for A we obtain a formula for what is called the present value of the
stream of payments.
n
r
A  P [1  (1  )  nT ]
r
n
3
We should think of A as the lump sum equivalent to the stream of future payments P . In other words,
A is the amount both lender and borrower are willing to exchange now for the promise of nT
payments P in the future.
Example 1: Suppose a car dealer is willing to loan us $10,000 for a car we wish to buy, and asks that we
pay it back over the course of 60 months, paying $200 a month. Is it a fair deal?
Answer: If we substitute A=10,000, n=12, T=5 and P=200 into equation 2 and numerically solve for the
interest rate r, we find that the underlying interest rate is 7.42%, a reasonable rate for a car loan.
Example 2: Suppose we are in the market for a house. We decide that we can afford to pay $1000 a
month. The best interest rate we can find for a 30-year mortgage is 6%. How much money can we
borrow?
Answer: If we substitute r=.06, n=12, P=1000, T=30 into equation 3, we find that we can borrow
A=$166,791.61.
4
Notice that the formula requires
nt to be an integer.
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Example 3: Suppose we borrow $100,000 at a 5% annual interest, compounded monthly. What would
our monthly payment be if
a) We pay the loan back over 15 years?
b) We pay the loan back over 30 years?
Answer:
(a) we substitute the values A=100,000, n=12, r=.05, T=15 into 2 and solve for P to find5 P=$790.79
(b) we substitute the values A=100,000, n=12, r=.05, T=30 into 2 and solve for P to find P=$536.82.
Notice that over the course of 30 years, the total payments amount to 360*$536.82=$193,255.20,
while for the 15-year loan the total comes to 180*$790.79=$142,342.20, - over $50,000 less!
Resets
Often the conditions of a loan change over the course of the repayments. The borrower may want to
refinance at a lower rate, for example, or pay the loan off sooner through higher payments. In such
cases one calculates the balance of the loan at the time of the change, redefines the loan amount to this
balance, resets the clock, and determines the remaining values with the same equations as before.
Example 4: Suppose we have a 15-year-mortgage of $100,000, paying $790.79 a month, as in Example
3(a). Suppose that after 5 years, our finances are looking up and we decide that we can afford paying
$1000 a month from here on out until the loan is paid off.
When will the loan be paid off, and how much is the last payment?
Answer: First we need to calculate how much we owe after 5 years (and 60 payments), so we substitute
A=100000, r=.05, n=12, P=790.79, t=5 into equation 1, and find that our balance on the loan is
B=$74,557.34.
Then we reset our loan to this loan amount: A=$74,557.34, reset P to P=1000, and solve equation 2 for T,
to find T= 7.46 years. That is a time between the 89th and 90th monthly payment after the reset.
Calculating the balance after the 90th payment to be B(7.5)= -$530.21 tells us that we only need to pay
$469.81 for the 90th and last payment after the reset.
Notice that paying the extra $209.21 every month allows us to pay the loan off early by 2.5 years. This
also saves us money in total payments made: Compare 180 payments of $790.79 with 60 payments of
$790.79, 89 payments of $1000 and one payment of $469.81, and we see that we save a total of $5425!
Example 5: Suppose we enter into a 30-year mortgage of $200,000 at a 5% annual interest, with a
monthly payment of $1073.64. Ten years later, the interest rate available in the mortgage market for a
20-year mortgage has dropped to 4%. Given that it costs $3000 to refinance, does it pay to refinance?
Answer: First we need to calculate how much we still owe. If we substitute A=$200,000, n=12, r=.05,
P=$1073.64, t=10 into equation 1, we see that our balance on the loan is still $162,684.63.
To evaluate whether a refinance is worth it we can take two approaches:
5
rounded to the nearest cent
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(a) we can decide to not touch the loan balance, but calculate the new monthly payment P:
So we use A=$162,684.63, and t=20 and r=0.04 to calculate P=$985.84. We’d be saving $87.80 a month!
It would take us about 3 years before we have saved more than the $3000 fee. This leaves us with a bit
of uncertainty as to whether it is worthwhile to refinance, since we are not quite sure, for example,
whether we might move before the three years are over.
(b) we can decide not to touch the monthly payment of P, and instead refinance a higher loan amount A:
So we use P=1073.64, r=.04, T=20 in 2 and calculate the new loan amount to be A=177,174.06. This
means that we can refinance, keeping the monthly payment constant, settle the old loan, and receive
$11,489.43 cash back even after having paid the $3000 fee. This gives us the clear answer of, yes, it is
worthwhile to refinance.
Example 6: (FM Practice Exam #1, problem #1)
Harvey is repaying a loan of $1000 with annual payments of $120 for 10 years, followed by annual
payments of $100 for as long as necessary. The amount of interest in the first payment is $100 and the
amount of interest in the 12th payment is X. Find X.
Answer: First we note that the interest rate must be 10%, since the interest portion on the first payment,
when the loan amount is still $1000, is given to be $100. So r=.10.
Then we find how much principal is left after 10 years (and 10 payments): substituting n=1, r=.10, t=10,
P=120 and A=1000 into equation 1, we find that the balance B=$681.25. Then we start a new loan with
new principal ($681.25) and new monthly payment P=$100. To find the interest in the 12th payment, we
calculate the difference in the balance after the 11th payment (B=$649.37) and 12th payment
(B=$614.31). This tells us that the 12th payment reduces the principal by $35.06, while X= $64.94 goes to
pay interest. (Double check: 10% of the balance of $649.38 is $64.94).
Exercises
Problem 1: Suppose you deposit $1000 in an account paying an annual rate of 6%. Find the value of
the account in 3 years if interest is compounded
(a) yearly:
(b) quarterly:
(c) monthly:
(d) daily:
(e) continuously:
Problem 2. What annual interest rate r would allow you to double your initial deposit in 6 years, if
interest is compounded
(a) quarterly
(b) continuously
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Problem 3: Find the effective interest rate if a nominal rate of 12% is compounded
(a) quarterly
(b) monthly
(c) continuously
Problem 4: If you receive 6% interest compounded monthly, about how many years will it take for
your investment to triple?
Problem 5: If you deposit $400 at the end of each month into an account earning 8% interest
compounded monthly, what is the value of the account at the end of
(a) 5 years?
(b) 10 years?
Problem 6: You deposit $700 at the end of each month into an account earning interest at an annual
rate of r compounded monthly. Find the value of r that produces an account value of $50,000 in 5 years
Problem 7: You deposit $400 at the end of each month into an account with an annual rate of 6%
compounded monthly. Determine the minimum number of payments required for the account to have a
value of at least $30000
Problem 8: A trust fund has an initial value of $300,000 and earns interest at an annual rate of 6%,
compounded monthly. If a withdrawal of $5000 is made at the end of each month, when will the account
fall below $150,000?
Problem 9: For a 30-year, $300,000 mortgage, determine the annual rate you will need to have
payments of $1800 per month.
Problem 10: How large a loan can you take out at an annual rate of 15% if you can afford to pay back
$1000 at the end of each month and you want to retire the loan after 5 years?
Problem 11: Suppose you take out a 20-year, $300,000 mortgage at an annual rate of 7%. Find your
monthly payment.
Decide after 15 years to pay off the mortgage. How much will you have to pay?
Problem 12: Suppose you take out a 30-year, $100,000 mortgage at 6%. After 10 years, interest rates
are down to 4%. You refinance the remainder of the loan by taking out a 20-year loan. Assume that the
cost of refinance is 3% of the new mortgage amount and it is rolled in with the new mortgage. How much
is your new monthly payment?
Below what interest rate does the refinancing make sense?
Problem 13: Suppose a loan of $20,000 is structured as follows: The interest rate is 10%,
compounded quarterly. Quarterly payments are $600 for the first 5 years. Then quarterly payments are X
for the next 10 years, paying the loan off after a total of 15 years. Find X.
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