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MTH6154: Financial Mathematics 1
Dr Dr Dudley Stark
Term 1, 202122
Contents
Preface
2
1 Interest rates and present value analysis
1.1 Interest rates . . . . . . . . . . . . . . .
1.2 Variable interest rates . . . . . . . . . .
1.2.1 The instantaneous interest rate .
1.2.2 The yield curve . . . . . . . . . .
1.3 Inflation . . . . . . . . . . . . . . . . .
1.4 Present value analysis . . . . . . . . . .
1.4.1 Defining the present value . . . .
1.4.2 Discount Factors . . . . . . . . .
1.4.3 Balancing present values . . . . .
1.5 Rates of return . . . . . . . . . . . . . .
1.5.1 The annualised rate of return and
1.5.2 Internal rate of return . . . . . .
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equivalent effective interest rate
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2 Immunisation of assets and liabilities
2.1 Measuring the effect of varying interest rates . . .
2.2 Reddington immunisation . . . . . . . . . . . . .
2.3 Immunisation in practice . . . . . . . . . . . . . .
2.4 Immunisation when interest is compounded yearly
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3 Bonds and the term structure of interest rates
3.1 Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 The present value of a bond . . . . . . . . . . . . . .
3.1.2 The noarbitrage principle and the fair price of a bond
3.1.3 The bond yield . . . . . . . . . . . . . . . . . . . . .
3.2 The term structure of interest rates . . . . . . . . . . . . . .
3.2.1 Spot rates . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Forward rates . . . . . . . . . . . . . . . . . . . . .
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4 Stochastic interest rates
4.1 A fixed interest rate model . . . . . . .
4.2 A varying interest rate model . . . . .
4.2.1 The growth of a single deposit
4.2.2 The growth of regular deposits
4.3 Lognormally distributed interest rates .
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5 Equities and their derivatives
5.1 Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Forwards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Pricing options: modelindependent pricing principles . . . . . . . . . . . . . .
5.5 Pricing options via the riskneutral distribution . . . . . . . . . . . . . . . . .
5.5.1 Betting strategies, the riskneutral distribution, and the arbitrage theorem
5.5.2 Options pricing using the riskneutral distribution . . . . . . . . . . . .
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6 The binomial model
6.1 The singleperiod binomial model . . . . . . . . . . . . . . . . . .
6.1.1 The noarbitrage condition and the riskneutral distribution
6.1.2 Option pricing in the singleperiod binomial model . . . . .
6.1.3 Replicating portfolios and the deltahedging formula . . . .
6.2 The twoperiod binomial model . . . . . . . . . . . . . . . . . . .
6.2.1 Options pricing in the twoperiod binomial model . . . . . .
6.2.2 Replicating portfolios in the twoperiod binomial model . .
6.2.3 Option pricing via the twoperiod riskneutral distribution .
6.3 The multiperiod binomial model . . . . . . . . . . . . . . . . . .
6.3.1 Options pricing in the multiperiod binomial model . . . . .
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7 From the multiperiod binomial model to the BlackScholes formula
7.1 General discretetime models . . . . . . . . . . . . . . . . . . . . . .
7.2 The lognormal process . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 The BlackScholes formula . . . . . . . . . . . . . . . . . . . . . . . .
7.3.1 Properties of the BlackScholes formula . . . . . . . . . . . . .
7.4 The lognormal process as an approximation of the bin. model . . . . .
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Preface
This course is an introduction to financial mathematics, and is the first in a series of three
modules (along with Financial Mathematics II and Mathematical Tools for Asset Management1 ) that together give a thorough grounding in the theory and practice of modern
financial mathematics. While this course primarily deals with financial mathematics in discrete
time, FMII will cover financial mathematics in continuous time, and MTAM will focus on
risk management and portfolio theory.
So what exactly is financial mathematics? A simple definition would be that it is the use
of mathematical techniques to model and price financial instruments. These financial
instruments include:
• Debt (also known as ‘fixedincome’), such as cash, bank deposits, loans and bonds;
• Equity, such as shares/stock in a company; and
• Derivatives, including forwards, options and swaps.
As this definition suggests, there are two central questions in financial mathematics:
1
Formally known as Financial Mathematics III.
2
1. What is a financial instrument worth? To answer this, we will develop principled ways
in which to:
• Compare the value of money at different moments in time; and
• Take into account the uncertainty in the future value of an asset.
2. How can financial markets usefully be modelled? Note that modelling always involves a
tradeoff between accuracy and simplicity.
This course introduces you to the basics of financial mathematics, including:
• An overview of financial instruments, including bonds, forwards and options.
• An analysis of the fundamental ideas behind the rational pricing of financial instruments,
including the central ideas of the timevalue of money and the noarbitrage principle. These concepts are modelfree, meaning that they apply irrespective of the chosen
method of modelling the financial market;
• An introduction to financial modelling, including the use of the binomial model to
price European and American options.
The course culminates in a derivation, via an approximation procedure, of the celebrated
BlackScholes formula for pricing European options.
3
1
Interest rates and present value analysis
The value of money is not constant over time – a thousand pounds today is typically worth
more than a contract guaranteeing £1,000 this time next year. To explain this, consider that
• If £1,000 was deposited into a bank account today, it would accumulate interest by this
time next year;
• Inflation may reduce the purchasing power of £1,000 over the course of the year.
In this chapter we explore how to take into account interest rates and inflation when comparing
assets that generate cash at different moments of time, using the concept of present value;
this gives us a way to make a principled comparisons between investments. Later in the course
we will use these ideas to assign a fair price to fixedincome securities,2 such as bonds.
An important theme in this chapter is the notation of standardisation, i.e. adjusting
interest rates and rates of return so that different rates can be compared fairly.
1.1
Interest rates
Suppose we borrow an amount P , called the principal, at nominal interest rate r. This
means that if we repay the loan after 1 year, we will need to repay the principal plus an extra
sum rP called the interest, so in total
P + rP = P (1 + r).
(1)
Similarly, if you put an amount P in the bank at an annualised interest rate r, then in a year’s
time the account value will grow to P (1 + r). Note that here we have assumed that the
nominal interest rate is annualised, meaning that interest is calculated once per year; unless
explicitly specified otherwise interest rates will always be assumed to be annualised
in this course.
Sometimes the interest is not calculated once per year, but instead is compounded every
of a year. This means that every n1 th of a year you are charged (or, in the case of a bank
account, gain) interest at rate r/n on the principal as well as on the interest that has already
accumulated in previous periods. Continuing the example of the loan above, this would mean
that after one year we would owe
r n
.
(2)
P 1+
n
If the interest rate is ‘annualised’ then it is compounded annually (i.e. once per year), which
corresponds to n = 1 and gives the same result as in (1).
1
th
n
Example 1.1. Suppose you borrow an amount P , to be repaid after one year at interest
rate r, compounded semiannually. Then the following will happen sequentially throughout
the year. After half a year you will be charged interest at rate r/2, which is added on to the
principal. Thus, after 6 months you owe
r
.
P 1+
2
At the end of the year you are again charged interest at rate r/2, with the interest accumulating
on the entire sum owing, therefore at the end of the year you owe
r
r
r 2
P 1+
1+
=P 1+
.
2
2
2
2
A security is a tradable financial instrument.
4
This corresponds to choosing n = 2 in (2).
If n = 4 in (2) we say that interest is compounded quarterly, and if n = 12 we say that
interest is compounded monthly.
Example 1.2. Suppose that you borrow £100 at interest rate 18% compounded monthly.
After one year you owe
12
0.18
= 100 × 1.195 = £119.5.
100 1 +
12
In order to fairly compare interest rates we need to standardise to account for differences
in the compounding frequency. In this context the interest rate r is called the nominal rate.
To compare the effect of different compounding frequencies, we introduce the effective rate3
reff =
(amount after one year) − P
.
P
For an interest rate compounded every n1 th of a year, we therefore have
reff =
P (1 + r/n)n − P
r n
= 1+
− 1.
P
n
(3)
The effective rate quantifies the net effect of interest over the course of a year, and so is a
fairer measure of the true effect of interest. Under UK consumer law, all credit providers must
publish their effective rate.
Example 1.3. Suppose that you borrow £100 at nominal interest rate 18% compounded
monthly. What is the effective rate?
Solution. We know from Example 1.2 that after one year you owe £119.5. Hence
reff =
119.5 − 100
= 0.195 = 19.5%.
100
Example 1.4. Credit card company A charges a 24.5% nominal rate compounded daily
whereas credit card company B charges a 24.7% nominal rate compounded monthly. Which
company offers the better deal?
Solution. The nominal interest rates do not give a fair comparison, since they are compounded
at different frequencies. A fairer comparison is given by the effective rates for both deals:
• For company A, the effective rate is
365
1
reff = 1 +
24.5%
− 1 = 0.2775.
365
• For company B, the effective rate is:
12
1
reff = 1 + 24.7%
− 1 = 0.2770.
12
3
Effective rates are sometimes also called ‘annual percentage rates’ (APRs).
5
We conclude that company B offers the better deal even though the nominal rates advertised
might suggest the opposite.
Imagine now that interest is compounded at very small intervals, i.e. interest is compounded
every n1 th of a year for n very large. In the limiting case, i.e. as n → ∞, we say that interest
is continuouslycompounded, and in this case the amount owed after one year is
r n
lim P 1 +
= P er
n→∞
n
(we prove this rigorously in Coursework 1). The effective rate for continuouslycompounded
interest at nominal rate r is therefore
reff =
P er − P
= er − 1.
P
Example 1.5. Suppose that a bank offers continuouslycompounded interest at nominal
rate 5%. The effective rate is
reff = e0.05 − 1 = 0.05127 = 5.127%.
If a loan is taken out for T years, then the total amount owed after T years can be
calculated using the effective rate reff as
P (1 + reff ) × · · · × (1 + reff ) = P (1 + reff )T .

{z
}
T times
For example, if interest is compounded n times a year, this is
r nT
P 1+
,
n
and if it is compounded continuously
P (er )T = P erT .
Example 1.6 (The doubling rule). If a bank offers continuouslycompounded interest at
nominal rate r, how long does it take for the amount of money in the bank to double?
Solution. Let T denote the time in years by the amount P doubles. Then
P erT = 2P
⇒
erT = 2
⇒
rT = ln 2,
and therefore
ln 2
0.693
≈
.
r
r
For example, if the nominal rate is r = 12%, it takes 5.78 years for the amount to double.
T =
Remark 1.7. In investment folklore, the doubling rule is sometimes referred to as the ‘Rule
of 72’, since you can (roughly!) approximate ln 2 = 0.693 by 0.72, which often makes the
fraction easier to calculate, i.e. in the example above, 0.72/0.12 = 72/12 = 6 years. This
‘rule’ first appeared in print in 1494, long before logarithms were invented.
1.2
Variable interest rates
Interest rates may not be constant over time. To describe variable interest rates we introduce
the concepts of instantaneous interest rate and yield curve.
6
1.2.1
The instantaneous interest rate
Suppose that the interest rate at time t is equal to r = r(t); we call r(t) the instantaneous interest rate. We always assume that the instantaneous interest is continuouslycompounded, which means that for small h (and as long as r(t) is continuous at t), the
amount of interest that accrues between times t and t + h is equal to
1
er(t)h = 1 + r(t)h + r(t)2 h2 + . . . ≈ 1 + r(t)h.
2
We wish to calculate the amount P (t) that accumulates by time t, given that the amount
P (0) is deposited (or loaned) at time 0. It turns out that P (t) satisfies a simple differential
equation:
Theorem 1.8. Suppose that r(t) is a piecewise continuous function. Then
P 0 (t) = P (t)r(t).
Proof. Let h denote a small timeperiod. Since interest is continuously compounded, the
interest accruing between times t and t + h is approximately equal to r(t)h, i.e.,
P (t + h) ≈ P (t) + P (t)r(t)h,
or equivalently
P (t + h) − P (t)
≈ P (t)r(t).
h
As h becomes smaller, the above approximation becomes more and more accurate, and so
P (t + h) − P (t)
= P (t)r(t).
h→0
h
lim
Recognising the lefthand side as the definition of the derivative, we conclude that
P 0 (t) = P (t)r(t),
The above equation is a separable first order differential equation, easily solved by writing
P 0 (t)
= r(t)
P (t)
and integrating both sides:
Z
0
Thus
u=t
[ln P (u)]u=0
Z
=
t
P 0 (u)
du =
P (u)
t
Z
r(u) du.
0
t
Z
r(u) du =⇒ ln P (t) = ln P (0) +
0
t
r(u) du,
0
and so
P (t) = P (0) exp
Z
t
r(u) du ,
(4)
0
which is the desired relation between the amount P (t) in your bank at time t and the timevarying interest rate function r(t).
Note that if r(t) = r is constant, then P (t) = P (0)ert . Thus, the general formula (4)
reduces to the familiar formula in the special case of constant interest rate.
7
1.2.2
The yield curve
Suppose that the instantaneous interest rate is r(t). The yield curve r̄(t) is defined to be
the average value of r(t) on the interval (0, t):
Z
1 t
r̄(t) =
r(u) du.
t 0
The yield curve allows us to write the formula for P (t) in (4) as
Z t
P (t) = P (0) exp
r(u) du = P (0)er̄(t)t .
0
The effective rate can also be expressed in terms of the yield curve. Since the interest
rate is variable, this rate reff (T ) depends on the period of time T the money is deposited/loaned, and is equal to
reff (T ) = er̄(T ) − 1.
Example 1.9. Suppose you deposit £100 in a bank offering the instantaneous interest rate
(
3% t < 2,
r(t) =
5% t ≥ 2.
Find the yield curve r̄(t) and determine the value of the bank balance at time t = 3.
Proof. To calculate the yield curve we split the integral at t = 2. For t < 2,
Z
1 t
3% ds = 3%
r̄(t) =
t 0
whereas for t ≥ 2,
1
r̄(t) =
t
Z
2
Z
3% ds +
0
2
t
0.06 0.05(t − 2)
4%
5% ds =
+
= 5% −
.
t
t
t
The value of the deposit is
100er̄(3)3 = 100e(0.05−0.04/3)3 = 100e0.11 = £112.
Example 1.10. Suppose a bank offers a loan with instantaneous interest rate increasing slowly
from 3% to 4% via the function
3%
4% × t
r(t) =
+
.
1+t
1+t
Find the effective rate for a loan until time T = 5.
Proof. We can rewrite the interest rate as
3%
4% × (1 + t)
4%
4% − 3%
1%
r(t) =
+
−
= 4% −
= 4% −
.
1+t
1+t
1+t
1+t
1+t
The yield curve is
Z
1 t
1%
1
1%
r̄(t) =
4% −
du = 4% × t − 1% × ln(1 + t) = 4% −
ln(1 + t).
t 0
1+u
t
t
The effective rate for the loan is
reff (5) = er̄(5) − 1 = e0.04−
0.01
5
ln(6)
as expected, this lies somewhere between 3% and 4%.
8
− 1 = 3.71%;
1.3
Inflation
We finish the chapter by taking a quick look at inflation, the phenomenon of prices increasing
as a whole over time (or equivalently, the purchasing power of money eroding over time).
Inflation is most easily quantified by reference to an index, for example the Retail Price
Index (RPI), which tracks the price of a basket of goods. Let rinf be the (annualised) inflation
rate, which means that £1 today will have the purchasing power of £(1 + rinf ) in one year.
For example, if rinf = 3%, then a basket of goods costing £100 today, will cost £103 next
year. Equivalently, the purchasing power of £103 in one year is equivalent to £100 today.
How should we take inflation into account when assessing interest rates? Suppose you
deposit the amount P today. After one year the deposit will have grown to P (1 + r), where r
is the interest rate. On the other hand, the purchasing power of the amount P (1 + r) in one
year from now, taking into account inflation, is the same as the amount P (1 + r) (1 + rinf )−1
today. So the ‘real’ interest rate (real in the sense that it reflects the decrease of purchasing
power), also known as the inflation adjusted interest rate, is equal to
ra =
1+r
P
1+rinf
−P
P
=
1+r
r − rinf
−1=
.
1 + rinf
1 + rinf
Example 1.11. Suppose that the interest rate is 5% and the rate of inflation is 3%. Then
the inflation adjusted interest rate is
0.05 − 0.03
r − rinf
=
= 0.0194 = 1.94% .
1 + rinf
1 + 0.03
In the case of compounding interest, we must use the effective interest rate in place of the
nominal rate, i.e.,
reff − rinf
.
ra =
1 + rinf
Example 1.12. Suppose that the interest rate is 5% compounded monthly, and the rate of
inflation is 3%. Then the inflation adjusted interest rate is
((1 +
reff − rinf
=
1 + rinf
0.05 12
)
12
− 1) − 0.03
= 0.0205 = 2.05%.
1 + 0.03
Since rinf is usually much smaller than 100%, it is often sufficient to use the approximation
ra =
1.4
reff − rinf
≈ reff − rinf .
1 + rinf
Present value analysis
A cashflow stream is a sequence of payments that are specified by their amount and
transaction time. Often, but not always, the cashflow stream will consist of payments at
regular intervals (for instance, one per year), in which case we can write the cashflow stream
simply as a = (a1 , a2 , . . . , an ), where ai denotes the amount paid at the end of period i.
The question we consider in this section is:
How should we compare different cashflow streams?
For this we introduce the concept of the present value of a cashflow stream.
9
1.4.1
Defining the present value
Suppose that the interest rate is 10%. If somebody gives you £100 today, you could put it
in the bank and after a year you would have £110. Thus £100 received today has a value
of £110 in a year’s time, or conversely, £110 received in a year’s time has a present value of
£100.
More generally, suppose that the interest rate is r%, and suppose you put (1 + r)−i V
pounds in the bank today. At the end of year i you will have (1 + r)−i V · (1 + r)i = V pounds
in the bank. Thus, £V received in i years’ time is worth (1 + r)−i V in today’s money. Given
this, we define the present value of £V at the end of year i to be
P V (V ) = (1 + r)−i V.
Exchanging V for its present value P V (V ) is sometimes referred to as discounting the
amount V , and the factor (1 + r)−i = P V (1) is referred to as the discount factor (more on
this later).
Similarly, if the interest is compounded every n1 th of a year at nominal rate r, the present
value of V pounds in i years is
r −in
P V (V ) = 1 +
V.
n
We can generalise the notion of present value to a cashflow stream a = (a1 , a2 , . . . , an )
that pays ai at the end of year i, for i = 1, . . . , n. The present value4 of a is
P V (a) =
n
X
i=1
ai
.
(1 + r)i
To see this, observe that the cashflow stream a = (a1 , a2 , . . . , an ) can be replicated by
first splitting the stream into the individual payments a1 , a2 , . . ., and then depositing the
corresponding amounts P V (a1 ), P V (a2 ), . . . needed to replicate these payments. Since
P V (a) = P V (a1 ) + P V (a2 ) + . . . + P V (an ),
the total amount you need to deposit to replicate the cashflow stream is P V (a).
Remark 1.13. This is our first example of an argument that uses the idea of replication
to assign a value to a financial instrument. Later in the course we will formalise this type of
argument by using the noarbitrage assumption.
Example 1.14. You are offered three different jobs. The salary paid at the end of each year
(in thousands of pounds) is
JobYear
A
B
C
1
32
36
40
2
34
36
36
3
36
35
34
4
38
35
32
5
40
35
30
Which job pays the best if the interest rate is (i) r = 10%, (ii) r = 20%, or (iii) r = 30%?
4
The term net present value (NPV) is sometimes used when discussing cashflow streams as opposed to
a single payment, but we will not use this term.
10
Solution. We shall compare the present values of the cashflow streams. The present value
for job A is
P V (A) =
34
32
36
38
40
+
+
+
+
,
2
3
4
1 + r (1 + r)
(1 + r)
(1 + r)
(1 + r)5
while the present value for job B is
P V (B) =
36
35
35
35
36
+
+
+
+
2
3
4
1 + r (1 + r)
(1 + r)
(1 + r)
(1 + r)5
and the present value for job C is
P V (C) =
36
40
34
32
30
+
+
+
+
.
2
3
4
1 + r (1 + r)
(1 + r)
(1 + r)
(1 + r)5
Taking the different interest rates into account we get the following present values:
r
0.1
0.2
0.3
PV A
135.03
105.51
85.20
PV B
134.41
106.20
86.61
PV C
132.14
105.50
86.83
Thus,
• If r = 10%, then A pays best, then B, then C;
• If r = 20%, then B pays best, then A, then C;
• If r = 30%, then C pays best, then B, then A.
and so which job pays best depends on the current interest rate.
It is possible to consider cashflow streams which go on forever (‘in perpetuity‘). Before
doing so, recall some facts about geometric series:
Proposition 1.15. Let a ≤ b be positive integers. Then if r 6= 1,
b
X
i=a
ra − rb+1
r =
.
1−r
i
Hence, if r < 1,
∞
X
i
r = lim
i=a
b→∞
b
X
i=a
ra − rb+1
ra
=
.
b→∞
1−r
1−r
ri = lim
These facts allow us to calculate the present value of perpetual cashflow streams:
Example 1.16. A perpetuity entitles its owner to be paid an amount a > 0 at the end of
each year indefinitely. What is its present value if the interest rate is r?
Solution. The cashflow stream generated by the perpetuity is a = (a, a, a, . . .), and so its
present value is
P V (a) =
∞
X
i=1
a
1/(1 + r)
a
a
=a×
=
= .
i
(1 + r)
1 − 1/(1 + r)
(1 + r) − 1
r
11
Remark 1.17. Note that the present value ar of the perpetuity can also be justified by a
simple replication argument. Suppose that instead of purchasing a perpetuity you place ar
in a bank account. After one year you have ar (1 + r) in the account. Suppose you withdraw
a to replicate the payment of the perpetuity. Then you are left with
a
a
a
(1 + r) − a = + a − a = .
r
r
r
a
At the end of the second year, this will have grown to r (1 + r), so if you again withdraw a
you are left with
a
a
a
(1 + r) − a = + a − a = .
r
r
r
If you continue like this you can withdraw a from the bank account each year forever, which
replicates exactly the cashflow stream generated by the perpetuity.
1.4.2
Discount Factors
It is often simpler to write the present value of a cashflow stream using the concept of
discount factors. This is especially the case if interest rates vary over time.
Definition 1.18. The discount factor to T years, D(T ), is the present value of a payment
of 1 unit of cash in T years’ time. Equivalently, it is the reciprocal of what a bank account
holding 1 unit of cash now will be worth in T years’ time.
If interest rates are positive (which is most often the case) the discount factor is smaller
than 1 hence the phrase ‘discount’. This indicates that having money in the future is worth
less than having the same amount of money today.
Examples 1.19.
1. If the interest is compounded annually at nominal rate r then the discount factor to T
years is
1
.
D(T ) =
(1 + r)T
2. If the interest is compounded at a different frequency, we can write the discount factor
to T years in terms of the effective rate reff as
1
D(T ) =
.
(1 + reff )T
For example, if interest is compounded monthly then
1
D(T ) =
,
1 12T
1 + r 12
if interest is continuouslycompounded then
D(T ) = e−rT .
3. If r(t) is an instantaneous interest rate then
D(T ) = e−r(T ) T .
Proposition 1.20. If a cashflow stream consists of payments a1 , a2 , . . . , an transacted at
times t1 , t2 , . . . , tn , then the present value of this cashflow stream is
PV =
n
X
i=1
12
D(ti )ai .
1.4.3
Balancing present values
In many reallife situations it is essential that the present values of assets (i.e. incoming
payments) and liabilities (i.e. outgoing payments) are equal. The following is a practical
example of such a situation:
Example 1.21. Suppose that Alex is selfemployed and wants to save for her retirement. She
wants an income of £1,000 a month starting in 20 years’ time and lasting for 30 years. What
amount of money does she have to save every month (starting today) for the next 20 years to
fund her retirement? Suppose that the nominal rate is 6% compounded monthly.
There exist various possible solutions to this problem. Here are two of them:
Solution 1. Let a denote the as yet unknown amount of pounds she must deposit every month,
and let D denote the cashflow stream generated by these deposits. Similarly, let w =
£1,000, and let W denote the cashflow stream generated by her monthly withdrawals for her
retirement. In order to fund her retirement, the present value of her deposits must equal the
present value of her withdrawals, i.e.
P V (D) = P V (W ).
We first calculate P V (D). It is convenient to work with the monthly discount factor
β = D(1/12) =
1
1
.
0.06 =
1.005
1 + 12
Since there are 240 months in 20 years,
P V (D) = a + aβ + aβ 2 + · · · + aβ 239 = a(1 + β + β 2 + · · · + β 239 ) = a
1 − β 240
.
1−β
We next calculate P V (W ). Since her first withdrawal will take place in 240 months’ time,
and since there are 360 months in 30 years,
P V (W ) = wβ 240 + wβ 241 + · · · + wβ 240+359 = wβ 240 (1 + β + · · · + β 359 ) = wβ 240
1 − β 360
.
1−β
Since P V (D) = P V (W ), we conclude that
a
1 − β 240
1 − β 360
= wβ 240
1−β
1−β
⇒
a = wβ 240
1 − β 360
= £361
1 − β 240
that is, she has to deposit £361 every month to fund her retirement.
Solution 2. A alternative approach is to compare the value of the cashflow streams at the
end of year 20, instead of at the present time. Using the notation from the previous solution,
the value of her deposits at the end of year 20 is
aβ −240 + aβ −239 + … + aβ −2 + aβ −1 = aβ −1
β −240 − 1
β −1 − 1
since β −1 is the amount that £1 accumulates in a deposit of one month. Similarly, the value
at the end of year 20 of all her retirement withdrawals is
w + wβ + wβ 2 + . . . + wβ 359 = w
1 − β 360
.
1−β
Equating these values, we have
a β −1
β −240 − 1
1 − β 360
=
w
β −1 − 1
1−β
⇒
13
a=w
1 − β 360
= £360.9 .
β −240 − 1
1.5
Rates of return
Suppose you invest an amount P (the ‘principal’) which at a later times results in a value
of Pfinal . How can you assess the quality of this investment? Three possible quantifiers are:
• The return on the investment is the quantity Pfinal .
• The profit on the investment is the quantity Pfinal − P .
• The rate of return on the investment is the profit as a percentage of the principle
R=
Pfinal − P
.
P
(5)
Remark 1.22. Be careful when using the phrases ‘profit’ and ‘return’; practitioners are not
always consistent in their use.
Examples 1.23.
1. Suppose we invested P = £150,000 in a property which is now valued at Pfinal =
£200,000. What is the rate of return? This is given by
R=
200,000 − 150,000
= 0.333 = 33.3%.
150,000
2. What is the total return on an investment of $12,000 that guarantees a 20% rate of
return? What is the profit? Rearranging (5), the return is
Pfinal = P (1 + R) = $12,000 × (1 + 0.2) = $14,400
and the profit is
P × R = $12,000 × 0.2 = $2,400(= $14,400 − $12,000).
1.5.1
The annualised rate of return and the equivalent effective interest rate
The rate of return does not fairly evaluate the performance of an investment, since it does
not take into account how much time it took for the investment to accrue its value.
For example, a rate of a return of 33.3% as in Example 1.23 would be spectacular if gained
in two weeks, but would be less spectacular if the increase took 40 years. To fairly compare
rates of return, we need to standardise them to account for differences in timeperiod.
A naive standardisation would simply divide the rate of return R by the time T it took for
the investment to accrue its value, i.e.
R
r=
T
called annualised rate of return. A better standardisation is to consider the equivalent
effective interest rate r that a bank deposit would need to offer in order to match the return
on the investment. This is defined to satisfy
P (1 + r)T = Pfinal = P (1 + R).
Rearranging, this gives
r = (1 + R)
1/T
−1=
14
P
final
P
1/T
− 1.
Examples 1.24.
1. Suppose we invested £1,500 in shares two year’s ago, and these are now worth £1,750.
The equivalent effective interest rate is
P 1/T
1750 1/2
final
r=
−1=
− 1 = 0.0801 = 8.01%, .
P
1500
2. How much profit do we earn if we invest $2,200 for two years in an investment that
guarantees an equivalent effective interest rate of 3%? We have
Pfinal = P (1 + r)T = $2,200(1 + 0.03)2 = $2,330
and so the profit is
Pfinal − P = $2,330 − $2,200 = $130.
1.5.2
Internal rate of return
When considered the performance of an investment that generates a cashflow stream with
payments at multiple instances of time, we need a measure of the annualised rate of return
that takes into account all the payments. This is the internal rate of return (IRR).
Suppose that you invest amount P and after T years you get back amount Pfinal . Recall
that the equivalent effective interest rate of this investment satisfies
Pfinal = P (1 + r)T
⇔
P =
Pfinal
.
(1 + r)T
More generally, suppose the initial investment P generates the cashflow stream a = (a1 , a2 , . . . , an )
with payments at times t1 , t2 , . . . , tn . Then we define the internal rate of return to be the
number r ∈ (−1, ∞) such that
n
X
ai
,
P =
ti
(1
+
r)
i=1
that is, the solution of the equation f (r) = 0, for the function
f (r) = −P +
n
X
ai (1 + r)−ti .
i=1
The internal rate of return is the ‘hypothetical’ effective interest rate that makes the present
value of the cashflow stream equal to the principal of the investment. In other words, it is
the effective interest rate that a bank would need to offer in order to replicate the payment
stream of the investment. An investment can be considered to be ‘good’ if the IRR exceeds
the effective interest rate offered in the market, and ‘bad’ if the IRR is less than the effective
interest rate offered in the market.
The definition of the IRR is only meaningful if the equation f (r) = 0 has a unique solution
r ∈ (−1, ∞). We next proves this to be true (at least, for positive cashflow streams):
Theorem 1.25. Suppose P > 0 and ai , ti > 0 for i = 1, . . . , n. Then the equation
f (r) = 0
⇔
P =
n
X
i=1
has a unique solution in (−1, ∞).
15
ai (1 + r)−ti
(6)
Proof. Observe the following properties of the function f on (−1, ∞):
• f is strictly decreasing, since
f 0 (r) =
n
X
ai
i=1
−ti
< 0.
(1 + r)ti +1
• limr→−1 f (r) = +∞ since
lim f (r) = lim
r→−1
r→−1
−P +
n
X
−ti
ai (1 + r)
= −P +
n
X
i=1
ai lim (1 + r)−ti = +∞.
i=1
r→−1
• limr→+∞ f (r) = −P < 0 since
lim f (r) = lim
r→+∞
r→+∞
−P +
n
X
−ti
ai (1 + r)
= −P +
i=1
n
X
i=1
ai lim (1 + r)−ti = −P.
r→+∞
Since f is continuous on (−1, ∞) and strictly decreasing from +∞ to −P < 0, f attains the
value 0 at exactly one point in r ∈ (−1, ∞).
Remark 1.26. One byproduct of the proof of Theorem 1.25 is that the present value of a
cashflow stream is a decreasing function of the interest rate r. We will study this fact in
more detail in the next chapter.
In general calculating the IRR is difficult since it requires solving equation (6), which is
a polynomial equation of degree n for which no closed solutions are available in general.5
However the case n = 2 can be solved easily.
Example 1.27. Suppose that an investor bought shares in a company for £1,000 at the
beginning of 2015, and sold them at the beginning of 2017 for £1,050. At the end of 2015 and
2016 the investor also received dividends (i.e. cash payments) of £10 and £20, respectively.
(a) What is the IRR of this investment?
(b) What is the IRR of this investment if the investor had to pay 10% tax on the dividends
and 18% capital gains tax on the profit from the increase in share price?
(c) Suppose the nominal interest rate is 3.5%. By comparing the IRR to the nominal interest
rate, judge whether the purchase of these shares was a good investment.
Solution.
(a) The IRR is the solution in r ∈ (−1, ∞) of the equation
1,000 =
20
1,050
10
1,070
10
+
+
=
+
.
(1 + r) (1 + r)2 (1 + r)2
(1 + r) (1 + r)2
Rewriting this, we have
1,000(1 + r)2 − 10(1 + r) − 1,070 = 0,
5
Instead such equations must be solved numerically using trial and error, or sophisticated variations of this.
16
which is a quadratic equation in (1 + r) with solutions
1.0394
and
− 1.0294.
This implies that the solutions r are
0.0394
and
− 2.0294 (which we reject, since r > −1).
Thus the IRR of this investment is r = 3.94%.
(b) Since the two dividend payments were taxed at 10%, the investor received only 10 − 1 =
£9 at the end of 2015 and 20−2 = £18 at the end of 2016. As the investor made a profit
of £1,050 − £1,000 = £50 after selling the shares, they also had to pay 0.18 · 50 = £9
capital gains tax. This reduces the investor’s return to £1,050 − £9 = £1,041.
The IRR in this case is the solution in r ∈ (−1, ∞) of the equation
1,000 =
9
18 + 1,041
+
.
(1 + r)
(1 + r)2
Again, this is a quadratic equation in r with solutions
0.0335
and
− 2.0245 (which we reject, since r > −1).
Thus, in this case the IRR of this investment is r = 3.35%. Note that this is smaller
than the rate of return calculated in (a), which reflects the fact that the tax reduced
the profitability of the investment.
(c) Ignoring taxes (as in part (a)), the IRR exceeded the nominal interest rate, so the
purchase of the shares was a good investment. However, taking into account taxes, the
IRR was lower than the nominal interest rate (as in part (b)), so overall the purchase of
the shares was not a good investment.
17
2
Immunisation of assets and liabilities
A company that expects to receive payments in the future (or needs to make payments in
the future) is exposed to fluctuations in the interest rate. In this chapter we explore how the
cashflow of a company can be structured in a way that minimises this risk – this is referred
to as the immunisation of assets and liabilities.
Manipulating a cashflow stream so as to minimise exposure to fluctuations in financial
markets more generally is called risk management, and is the focus of Mathematical
Tools for Asset Management. The action of entering into transactions for the purpose of
managing risk is referred to as hedging.
2.1
Measuring the effect of varying interest rates
Let us first consider how the present value of a cashflow stream varies when the interest rate
r varies. Let r denote the current interest rate (which may or may not be compounded), and
let D(T ) be the discount factor to T years (sometimes we write D(T ; r) instead of D(T ) to
stress the dependence on r). Suppose that the cashflow stream consists of a sequence of
payments Ci transacted at times ti . The present value V (r) of this cashflow stream is then
V (r) =
m
X
Ci D(ti ; r).
i=1
We have already observed in Chapter 1.5.2 that V (r) is a decreasing function of the interest
rate r (at least, if all payments Ci are positive). We can analyse how rapidly V (r) decreases
with r by considering derivatives of V (r).
Firstorder properties. A simple way to quantify how much V (r) decreases with r is to
compute V 0 (r). For technical reasons explained below, practitioners prefer to use the following
slightly different quantity:
Definition 2.1. If V (r) > 0, the effective duration of the cashflow stream is defined as
ν=−
V 0 (r)
V (r)
Remarks 2.2. Let us explain why ν is preferred to V 0 (r):
1. The derivative V 0 (r) is measured in the same monetary units (e.g. pounds, dollars, pence
etc.) as the present value V (r). By dividing V 0 (r) by V (r), we standardise V 0 (r) to
remove the monetary units. In fact, the units in which ν is measured is years (why?).
2. Since V (r) is decreasing in r for a positive cashflow stream, the minus sign is included
to ensure that ν is positive for such a stream (although it can be negative if the cashflow
stream has negative payments).
Why is the effective duration called a ‘duration’ ? To answer this, let us first define a
second quantity, called simply the ‘duration’:
Definition 2.3. If V (r) > 0, the duration of the cashflow stream is defined as
Pm
Pm
ti Ci D(ti )
i=1 ti Ci D(ti )
τ=
= Pi=1
.
m
V (r)
i=1 Ci D(ti )
18
In the case of positive payments Ci > 0, τ can be considered a weighted average of the
times ti
ω1 t1 + · · · + ωm tm
τ=
ω1 + · · · + ωm
where the weights ωi = Ci D(ti ) > 0 are the present value of the payments.
What is the relationship between the effective duration ν and the duration τ ? In the case
of continuouslycompounded interest, it turns out that they are exactly the same thing.
Proposition 2.4. If interest is continuouslycompounded, then the effective duration is equal
to the duration, that is, ν = τ .
Proof. The discount factors are D(ti ) = e−rti . Calculating
V (r) =
m
X
Ci e
−rti
and
0
V (r) =
i=1
we have
V 0 (r)
=−
ν=−
V (r)
m
X
−ti Ci e−rti ,
i=1
Pm
i=1
−ti Ci e−rti
=
V (r)
Pm
i=1 ti Ci D(ti )
V (r)
= τ.
More generally, the duration is equal to the effective duration up to a constant.
Proposition 2.5. There is a constant c0 , depending on the interest rate but not on the
cashflow stream, such that
ν = c0 τ.
Remark 2.6. As we have seen, in the case of continuouslycompounded interest the constant
c0 = 1. The value of c0 in the general case is more complex (and not very important), but if
payments are made yearly and interest is compounded n times a year then (prove this!)
ν = (1 + r/n)−1 τ.
The equivalence between the effective duration and the duration leads to the following
useful general principle:
If cashflow stream A has a larger duration than cashflow stream B, then the
present value of A decreases in value faster than the present value of B when the
interest rate increases.
Remark 2.7. Whether it is ‘better’ to have a cashflow stream with large duration depends
on the context, in particular whether the cashflow stream consists of assets or liabilities and
whether the interest rate increases or decreases.
Example 2.8. You expect to be paid £1,000 at the end of this year and the end of the
following year, and need to make a payment of £2,215 in three years’ time. The interest rate
is currently 7% and is compounded yearly.
(a) Verify that the present value of your assets (incoming payments) and your liability
(outgoing payment) are equal.
(b) If the interest rate were to increase to 8%, would you be in a better or worse financial
position (in terms of the net present value of assets and liabilities)?
19
Solution.
(a) The discount factor to one year is β = 1/1.07. The present value of the assets is
1,000β + 1,000β 2 = 1,808,
whereas the present value of the liability is
2,215β 3 = 1,808.
(b) The present value of both the assets and liability will decrease if the interest rate increases
from 7% to 8%, so whether you are in a better financial position depends on which
decreases more steeply. This can be answered by comparing durations.
Calculating the durations explicitly gives
1 × 1000β + 2 × 1000β 2
= 1.483
τ (A) =
1808
and
3 × 2215β 3
τ (L) =
=3
2215β 3
Hence the duration of the liability is larger than the duration of the assets, meaning
that the present value of the liabilities will decrease more than the present value of the
assets, so the increase in interest rates would put you in a better financial position.
Note that we actually didn’t have to do the calculations explicitly: as a weighted average
of payment times, the duration of the assets must be between 1 and 2, whereas the
duration of the liability is equal to 3.
Secondorder properties. We can refine our analysis of how V (r) varies with r by considering V 00 (r). As for the effective duration, it is more natural to considered the normalised
version V 00 (r)/V (r), which is called the convexity.
Definition 2.9. If V (r) > 0, the convexity of the cashflow stream is defined as
c=
V 00 (r)
.
V (r)
Proposition 2.10. If interest is continuouslycompounded, then the convexity is
Pm 2
Pm 2 −rti
ti Ci e
i=1 ti Ci D(ti )
c=
= Pi=1
.
m
−rti
V (r)
i=1 Ci e
(7)
Again, in the case of positive payments Ci > 0, the convexity c can be interpreted as a
‘weighted average’ of the squared payment times t21 , . . . , t2n . However, this does not have a
direct financial meaning.
Remark 2.11. In general, the relationship between the convexity c and the weighted average
in (7) is more complex. For instance, if interest is compounded n times a year then
Pm
ti (ti + 1/n)Ci D(ti )
× (1 + r/n)−2 .
c = i=1
V (r)
20
2.2
Reddington immunisation
We now discuss how a company can minimise their exposure to fluctuations in the interest
rate. Let the current interest rate be r = r0 . We assume that a company has a stream
of positive future cashflows (i.e. incoming payments), called assets, whose present value is
VA (r0 ), and a stream of negative future cashflow (i.e. outgoing payments), called liabilities,
with present value −VL (r0 ). The present value of the combined cashflow stream is therefore
V (r0 ) = VA (r0 ) − VL (r0 ).
The ‘immunisation’ of these cashflow streams refers to a sequence of three conditions that
VA and VL need to satisfy in order to minimise expose to fluctuating interest rates; if all three
conditions are satisfied the cashflows are said to be Reddington immune.
Zeroorder immunisation: Matching assets and liabilities Zeroorder immunisation
simply means that the present value of assets is matched by the present value of liabilities:
Definition 2.12. A cashflow stream is said to be zeroorder immune if the present value
of assets and liabilities is the same, i.e. VA (r0 ) = VL (r0 ).
Note that zeroorder immunisation can in general only be achieved for the present value
of the interest rate r = r0 , and will fail as soon as r differs from r0 .
Firstorder immunisation: Matching the effective durations of assets and liabilities
Now assume that the cashflow is zeroorder immune, and consider the effect of a change in
the interest rate to r = r0 + ε, for ε a small number. As we previously observed, in general
VA (r0 + ε) 6= VL (r + ε), and we can use Taylor expansions to approximate the difference:
VA (r0 + ε) ≈ VA (r0 ) + VA0 (r0 )ε = VA (r0 ) − VA (r0 )νA ε = VA (r0 ) (1 − νA ε)
VL (r0 + ε) ≈ VL (r0 ) + VL0 (r0 )ε = VL (r0 ) − VL (r0 )νL ε = VL (r0 ) (1 − νL ε)
(8)
(9)
where νA = −VA0 (r0 )/VA (r0 ) and νL = −VL0 (r0 )/VL (r0 ) are the effective durations of the
cashflow streams corresponding to the assets and liabilities. Hence
V (r0 + ε) = VA (r0 + ε) − VL (r + ε) ≈ VA (r0 ) (1 − νA ε) − VL (r0 ) (1 − νL ε) ,
which in the case of zeroorder immunisation, simplifies to
V (r0 + ε) ≈ VA (r0 ) (νL − νA ) ε.
This quantity is zero if and only if the effective durations are equal, i.e. νA = νL . This is
equivalent to the durations being equal, i.e. τA = τB .
Definition 2.13. A cashflow stream is said to be firstorder immune if (i) it is immune to
zeroorder, i.e. VA (r0 ) = VL (r0 ), and (ii) if the effective durations (equiv. durations) of the
cashflow streams corresponding to the assets and liabilities are equal, i.e. if νA = νB (equiv.
τA = τB ).
If a cashflow stream is firstorder immune, then the value of V (r0 + ε) is constant (up to
small errors) for small . The relevant errors are very small since they correspond to terms of
order ε2 or higher, e.g., if ε = 0.1% = 0.001 then ε2 = 0.000001.
21
Reddington immunisation: Convexity We can use the notion of convexity to refine the
approximations in (8) and (9):
1
1
VA (r0 + ε) ≈ VA (r0 ) + VA0 (r0 )ε + VA00 (r0 )ε2 = VA (r0 ) − VA (r0 )νA ε + VA (r0 )cA ε2
2
2
1
= VA (r0 ) 1 − νA ε + cA ε2
2
1
1
VL (r0 + ε) ≈ VL (r0 ) + VL0 (r0 )ε + VL00 (r0 )ε2 = VL (r0 ) − VL (r0 )νL ε + VL (r0 )cL ε2
2
2
1
= VL (r0 ) 1 − νL ε + cL ε2
2
where cA = VA00 (r0 )/VA (r0 ) and cL = VL00 (r0 )/VL (r0 ) are the convexity of the cashflow
streams corresponding to the assets and liabilities respectively. Hence in the case of firstorder
immunisation,
1
1
V (r0 + ε) ≈ VA (r0 ) 1 − νA ε + cA ε2 − VL (r0 ) 1 − νL ε + cL ε2
2
2
1
= VA (r0 ) (cA − cL )ε2
2
Clearly it would be advantageous to demand that cA = cL so that V (r0 + ε) was constant for
small ε up to very small errors (of order ε3 ). In practice, it is often sufficient to demand that
cA ≥ cL , which ensures that, up to very small errors, V (r0 + ε) ≥ V (r0 ).
Definition 2.14. A cashflow stream is Reddington immune if it is firstorder immune, i.e.
VA (r0 ) = VL (r0 )
and νA (r0 ) = νL (r0 ),
and if, in addition, the convexities of the assets and liabilities satisfy cA ≥ cL .
If a cashflow stream is Reddington immune, up to very small errors any change in the
interest rates will increase V (r) (or, at least, V (r) will not decrease)
2.3
Immunisation in practice
Let us see how immunisation might be applied in practice. For this we will make use of zerocoupon bonds, which are simply assets that pay a fixed amount (say £1) at the end of a
fixed number of years (bonds will be discussed in more detail in the next chapter).
Example 2.15. An investor has a liability of £20,000 to be paid in 4 years and another of
£18,000 to be paid in 6 years. Suppose the interest rate is r = 8% and is continuouslycompounded.
(a) Show that Reddington immunisation can be achieved by owning a combination of 2year
and 7year zerocoupon bonds.
(b) Explain why firstorder immunisation cannot be achieved if the fund owns a combination
of 2year and 3year zerocoupon bonds.
Solution. (a) The yearly discount factor is β = e−0.08 = 0.923. Suppose we own 2year
bonds paying £P and 7year bonds paying £Q. Zeroorder immunisation requires that
P β 2 + Qβ 7 = 20,000β 4 + 18,000β 6 = 25,661.
22
Moreover, firstorder immunisation (i.e. τA = τL ) requires that
2P β 2 + 7Qβ 7 = 4 × 20,000β 4 + 6 × 18,000β 6 = 124,900.
Solving the two equations for P and Q gives the solution
P = 12,840,
Q = 25,770.
To determine whether Reddington immunisation is achieved for this solution, we calculate
the convexities. Since interest is continuouslycompounded, we have
Pn 2
t Ci β ti
22 P β 2 + 72 Qβ 7
cA = i=1 i
=
= 29.8
VA (r)
25,661
and
Pn
cL =
2
ti
i=1 ti Ci β
VL (r)
=
42 × 20,000β 4 + 62 × 18,000β 6
= 24.7.
25,661
Since cA > cL , Reddington immunisation is achieved.
(b) The duration of a combination of 2year and 3year bonds is at most 3 years, which
cannot be equal to the duration of the liabilities (which is at least 4 years), so in this
case firstorder immunisation is impossible.
2.4
Immunisation when interest is compounded yearly
If interest is compounded yearly at rate r, then the present value of a payment stream of
payments Ci made at times ti for i = 1, 2, . . . , n is
V (r) =
n
X
i=1
Ci
.
(1 + r)ti
Exactly the same definitions of ν and c can be made for this present value function as for
continuously compounded interest and the same immunisation conditions apply. However,
now
n
X
t i Ci
0
V (r) = −
(1 + r)ti +1
i=1
and
V 0 (r) =
n
X
ti (ti + 1)Ci
i=1
(1 + r)ti +2
.
With β defined as β = 1/(1 + r),
V (r) =
n
X
Ci β ti ,
i=1
0
V (r) = −
n
X
ti Ci β ti +1 ,
i=1
and
V 00 (r) =
n
X
ti (ti + 1)Ci β ti +2 ,
i=1
23
The duration is
Pn
ti Ci β ti
τ = Pi=1
n
ti
i=1 Ci β
while the effective duration is
V 0 (r)
=
ν=−
V (r)
Pn
ti +1
i=1 ti Ci β
P
.
n
ti
i=1 Ci β
As opposed to the case for continuously compounded interest, ν = βτ .
Example 2.16. A fund must make paments of £50, 000 at the end of the sixth and eigth
year from now. Show that immunisation can be acheived for 7% interest with a comination
of 5year zerocoupon bonds and 10year zerocoupond bonds.
Solution. Suppose we purchase 5year bonds paying P and 10year bonds paying Q. Then,
with β = (1.07)−1 ,
VA (0.07) = P β 5 + Qβ 10 .
We also have
VL (0.07) = 50, 000(β 6 + β 8 ) = 62, 418
and so zero order immunisation implies that
P β 5 + Qβ 10 = 62, 418.
(10)
The effective duration of the assets is
νA =
5P β 6 + 10Qβ 11
62418
while the effective duration fo the liabilities is
νL =
404398
50, 000(6β 7 + 8β 9 )
=
.
62418
62418
Setting νA = νL gives
5P β 6 + 10Qβ 11 = 404398.
(11)
Solving (10) and (11) for P and Q gives
P = £53, 710,
Q = £47, 454.
(12)
We must still show that the third immunisation condition is satisfied. The convexity of the
assets is
V 00 (0.07)
(5)(6)P β 7 + (10)(11)Qβ 12
cA = A
=
= 53.21
VA (0.07)
62418
while the convexity of the liabilities is
cL =
VL00 (0.07)
50000((6)(7)β 8 + (8)(9)β 10 )
=
= 48.90.
VL (0.07)
62418
Since cA ≥ cL , the third condition is satisfied and (12) gives immunisation.
24
3
Bonds and the term structure of interest rates
Bonds are an example of a debt or fixedincome security – these are tradable financial
instruments whose cashflow is fixed or predetermined. Other examples include bank deposits,
loans, interest rate futures, interest rate swaps, swaptions, caps, floors, etc. (often credit
derivatives are also included, e.g., CDSs (credit default swaps), CDOs (collateralised debt
obligations), convertible bonds, etc., since their structure is similar). Note that, although the
cashflow is predetermined, the present value of a fixedincome security can fluctuate, since
their value also depends on the prevailing interest rates.
Fixedincome securities can be contrasted with equity, such as shares and equity indices.
These are securities whose cashflow is not predetermined. Other financial instruments of
this type include FX (foreign exchange) and commodities, and collectively they are sometimes
known as variableincome securities.
In this chapter we will use present value analysis and noarbitrage arguments to assign a
fair price to bonds. We also consider the term structure of bonds, which allow us to define
a generalised notion of interest rates that depend, not just on the present time, but also on
the period of time in which money is invested. Using this idea, we will see that all bond prices
and all interest rates can be expressed in terms of the price of zerocoupon bonds.
3.1
Bonds
A bond is a contract issued by a government or a company in order to raise capital. Investors
that purchase a bond are promised a stream of fixed payments known as coupons, plus an
extra payment at the bond expiry, known as the redemption.
A bond is described by:
• An issue price, which is the amount the investor pays for the bond. It is sometimes
determined by an auction.
• Its face value, also called the principal. This will determine the final redemption payment, and for some bonds will also determine the coupon values.
• The expiry (or ‘maturity’) of the bond. This is the date of the last coupon payment
and/or when the investor receives the redemption payment.
• The coupon value and frequency, which expresses the value and frequency of the
coupon payments. The value is sometimes expressed as a fixed amount, and sometimes
as a (annualised) coupon rate, which is a percentage of the face value.
If there are m coupon payments in a year then each of them will be equal to the annual
coupon divided by m. A zerocoupon bond is a bond with no coupon payments.
Note that the coupon rate is in general different to the interest rate.
• The redemption value, which expresses the value of the redemption payment at expiry.
This is again sometimes expressed as a fixed amount, and sometimes as a fraction
R of the face value. If R = 1, then the bond is said to be ‘redeemable at par’, if
R > 1, ‘redeemable above par’, and if R < 1, ‘redeemable below par’. Most bonds are
redeemable at par.
25
The cashflow stream generated by a typical bond might be as follows:6
R × Face value + Coupon
Now
Coupon Coupon Coupon
1Y
2Y
Coupon Coupon
3Y
8Y
9Y
10Y
Issue price
Examples 3.1.
1. A 5year bond with a face value of £100,000,000 and an annual coupon rate of 3%
will pay 5 annual coupon payments of £3,000,000. Additionally, if the the bond is
redeemable at par (R = 1), at expiry the investor will be paid back the face value of
the bond: $100,000,000.
2. What coupons are paid by a 10year bond with face value $150,000,000 and semiannual
coupon paid at an annual rate of 2%? The bond will pay 20 coupon payments every
halfyear for 10 years, each worth
1
$150,000,000 × 2% × = $1,500,000.
2
3.1.1
The present value of a bond
The present value of a bond is simply equal to the present value of the cashflow stream that
the bond generates. By default we do not include the issue price in the cashflow stream, unless
specifically included (immediately after purchase, the initial payment is no longer relevant to
the bond value).
Example 3.2. Assuming an interest rate of r, the present value of a zerocoupon bond with
face value 1 that is redeemable at par in T years is equal to
1
.
(1 + r)T
Example 3.3. Assuming an interest rate of r, the present value of a bond maturing in T years
with yearly coupon payments of C, issue price P0 , face value P , and redemption price RP is
1
1
1
1
PV =
C+
C + ··· +
C+
RP.
(13)
2
T
1+r
(1 + r)
(1 + r)
(1 + r)T
Example 3.4. Suppose the interest rate is 1.5%. The present value of a 3year bond with
face value £100,000, redeemable at par, with annual coupon payments of 1.7% is
0.017 × 100,000 0.017 × 100,000 0.017 × 100,000
100,000
PV =
+
+
+
2
3
1 + 0.015
(1 + 0.015)
(1 + 0.015)
(1 + 0.015)3
= 1674.88 + 1650.12 + 1625.74 + 95631.70
= £100,582 .
Note that in this example the interest rate 1.5% is different to the coupon rate 1.7%.
6
This picture reflects the cashflow stream for the buyer: the buyer initially pays the issue price (negative,
downwards arrow), and then receives (positive, upwards arrow) a stream of payments. The picture for the
bond issuer would be the opposite: it starts with receiving the issue price, and continues with a stream of
outgoing payments.
26
Recall the definition of discount factors from Section 1.4.2; these allow us to express the
present value of a general bond:
Example 3.5. Let D(T ) be the discount factor for a payment after T years. Then the present
value of a bond that pays coupons C1 , . . . , Cn at times T1 , . . . , Tn and is redeemable at value
RP at the expiry time Tn , is equal to
P V = D(T1 ) · C1 + D(T2 ) · C2 + · · · + D(Tn ) · Cn + D(Tn ) · RP.
3.1.2
(14)
The noarbitrage principle and the fair price of a bond
How should we assign a fair price to a bond? To answer this, modern finance takes as its
starting point the assumption that there are no arbitrage opportunities in the market. This is
sometimes called the noarbitrage principle, or the ’nofreelunch principle’.
The definition of an arbitrage opportunity (or simply an ‘arbitrage’) is the following:
Definition 3.6. An arbitrage is an investment that guarantees a riskfree profit (relative
to discounted values). More precisely, it is the creation of a portfolio of assets and liabilities
worth P (0) = 0 at time 0 such that, at some time T in the future:
• The portfolio has a nonnegative value P (T ) under all possible outcomes, i.e.
P (T ) ≥ 0
in all possible states of the world at time T ;
• There is at least one outcome under which the portfolio has a strictly positive value,
i.e.
P (T ) > 0 in at least one possible state of the world at time T .
Remark 3.7. Equivalently, we could describe arbitrage in terms of a portfolio worth P (0) at
time 0, and then ask whether
P (T ) ≥ P (0)D(T )−1
in all states of the world. Since P (0)D(T )−1 is the amount that a bank account with P (0)
invested in it is worth at time T , arbitrage is essentially the condition that the investment is
guaranteed to outperform the riskfree rate.
Definition 3.8. The noarbitrage principle is the assumption, made about a financial market, that no arbitrage opportunities exist.
The justification for the noarbitrage principle is that, in an idealised version of the market,
the buying and selling activity of market participants will eliminate all arbitrage opportunities.
This is since, the moment an arbitrage opportunity arises, there will be such demand for the
relevant assets that it will increase their prices, eliminating the arbitrage opportunity.
Note that the noarbitrage principle may not be reflected in reallife financial markets,
especially those that are illiquid (meaning, in which trades are infrequent).
Along with the noarbitrage principal, we will also make a series of related assumptions,
namely that participants in the market may freely create, buy, and sell assets without
restriction or cost (the socalled ‘frictionless market’ hypothesis).
It turns out that the noarbitrage principle (combined with the ‘frictionless market’ hypothesis) is an extremely powerful tool with which to assign prices to financial instruments,
27
such as bonds; this is known as ‘arbitragefree pricing’ or ‘rational pricing’. We will refer to
the resulting price as the ‘noarbitrage’ price, or sometimes simply the ‘fair price’.
We begin with the the law of one price, a general expression of the noarbitrage pricing
principle:
Theorem 3.9 (Law of one price). Under the noarbitrage assumption, if two financial instruments generate the same cashflow streams then they have the same current price.
In particular, the current price of a financial instrument that generates a fixed cashflow
stream is equal to the present value of this cashflow stream.
Corollary 3.10. The noarbitrage price of a bond is equal to its present value.
Proof. Suppose that financial instruments A and B generate the same cashflow streams, and
are currently priced at PA and PB respectively. Assume for the sake of contradiction that
PA > PB ;
we will show that this creates an arbitrage opportunity. Consider selling instrument A and
simultaneously buying instrument B, making a profit of
PA − PB > 0.
Since you own instrument B you will receive its associated cashflow stream; since you sold
instrument A, you need to pay the buyer the associated cashflow stream. As these cashflow
streams are equal, the former can be used to pay off the latter, leaving you with a positive
profit at the expiry of the cashflow streams. This is an arbitrage opportunity, so PA > PB
must be false. Reversing the roles of A and B, we conclude that
PA = P B .
Now suppose we have an instrument that generates a fixed cashflow stream. We already
saw that if the present value of the cashflow stream P V is placed in a bank account, it can be
used to exactly replicate the cashflow stream. Hence, by the law of one price, the price of the
instrument equals the current value of the bank account, which is just the present value.
Two financial instruments that generate the same cashflow streams are sometimes called
replicating portfolios. Indeed, the law of one price is just a formal version of the ‘replication’
arguments we saw in previous chapters.
One consequence of the law of one price is that the price of any bond can be expressed
in terms of the price of unit zerocoupon bonds (i.e. zerocoupon bonds that pay one unit
of cash at expiry). This is since, by the law of one price, the price VT of a unit zerocoupon
bond with expiry T years is equal to the discount factor to T , that is,
VT = D(T ).
Then we can use formula (14), which expresses the present value, or equivalently the noarbitrage price, of any bond in terms of the discount factors.
Example 3.11. Calculate the noarbitrage price of a 4year bond with face value £200,
redeemable at par, with 4% annual coupons, if the price of unit zerocoupon bonds, with
respective expiries T = 1, 2, 3, 4, are
(0.9, 0.72, 0.65, 0.53) .
28
Solution. By (14) the present value, or equivalently the noarbitrage price, of the bond is
200 × 4% ×
4
X
D(i) + 200 × D(4),
i=1
where D(T ) is the discount factor to time T , or equivalently the price VT of a unit T year
ZCB. Inserting the values of VT into the formula yields £128.
More generally, the noarbitrage principle can also be used to fix relationships between
different financial quantities that are not strictly speaking financial instruments, such as interest
rates. We illustrate this with an example:
Example 3.12. Suppose that bank A offers us a monthly compounded interest rate r1M ,
and bank B offers a yearly compounded interest rate r1Y . Show that that the noarbitrage
principle requires that
12
1
1 + r1Y = 1 + r1M .
12
First Solution. Assume for the sake of contradiction that
12
1
1 + r1Y < 1 + r1M ,
12
(15)
(the reverse inequality is treated similarly). Then suppose we took out a loan of £1 in bank B,
which would require us to repay (1 + r1Y ) in one year’s time. We could then invest this money
12
1
in bank A. At the end of the year the account value would grow to 1 + 12
r1M , which after
paying back the loan gives us a profit of
12
1
− 1 + r1Y .
1 + r1M
12
Since by assumption this is strictly positive, we have made a riskfree profit which contradicts
the noarbitrage principle. Therefore inequality (15) cannot hold.
Second Solution. Put
1
1
1+ 12
r1M
12 in bank A and
1
1+r1Y
in bank B. Both invest investments
result in a unit pay off after 1 year. By The Law of One Price,
1
or
1+
1
r
12 1M
12 =
1
,
1 + r1Y
12
1
1 + r1Y = 1 + r1M .
12
Exercise 3.13. Explain why, in the example above, the reverse inequality
12
1
1 + r1Y > 1 + r1M
12
also leads to arbitrage.
29
3.1.3
The bond yield
If the interest rate is unknown, we cannot compute the present value. However, if we know
the issue price of a bond, we can use noarbitrage theory to deduce the ‘hypothetical’ interest
rate that would make the bond fairly price. This is called the bond yield, and is nothing more
than the ‘internal rate of return’ of an investment in the bond.
Definition 3.14. Suppose that a bond has issue price P0 , coupon payments of C1 , . . . , Cn at
times t1 , . . . , tn and redemption payment RP at expiry T . Then the bond yield is the unique
solution r ∈ (−1, ∞) to
n
X
1
Ci
+
RP.
(16)
P0 =
t
i
(1 + r)
(1 + r)T
i=1
Equation (16) has a unique solution for the same reason that the IRR is guaranteed to
exist (see Theorem 1.25). Just like the IRR, the bond yield can be used to determine whether
the bond is a good investment.
Example 3.15. A broker proposes to sell a 2year bond with face value £100,000, redeemable
at par, with annual coupon payments of 2%. The broker quotes an issue price of P0 =
£102,000. Find the yield of the bond, and decide whether the bond is a good investment if
the interest rate is 10%.
Solution. The yield is the (unique) solution r ∈ (−1, ∞) to the equation
102,000 =
2,000
100,000
2,000
+
+
.
2
1 + r (1 + r)
(1 + r)2
This is a quadratic in the variable x = 1 + r; solving this yields
x = −0.9902
or
1.0098.
Since x > 0 we can disregard the first solution, and we are left with
r = x − 1 = 0.98%.
Since this is far less than the interest rate, the bond is not a good investment.
As shown by the example, the yield does not equal the coupon rate in general.
3.2
The term structure of interest rates
So far it has been assumed that there is a single interest rate r = r(t) at each instant of
time, regardless of the type or duration of the deposit/loan. In practice, interest rates usually
depend on the term (i.e. the duration) of the deposit/loan. The main reason for this is supply
and demand: if many people are asking for 20year loans, then the the interest rate for 20year
deposits will increase. Likewise, if there is a good supply of one month deposits the 1month
interest rate will go down. This variation is referred to the term structure of interest rates.
For example, the sterling LIBOR rates7 on 17th September 2019 were:8
7
LIBOR stands for London Interbank Offered Rate, and is a set of benchmark interest rates fixed every day
at 11am London time. LIBOR is used by many institutions to set shortterm interest rates; the total value of
instruments fixed by LIBOR is estimated to be in the hundreds of trillions of pounds.
8
Source: The Intercontinental Exchange Benchmark Administration Ltd https://www.theice.com/
marketdata/reports/170. In this page you can consult current or past LIBOR rates for a number of
currencies, e.g. GBP, Euros, Yen and USD.
30
Term
Overnight
1 Week
1 Month
2 Month
3 Month
6 Month
1 Year
LIBOR
0.67150%
0.68825%
0.71225%
0.75775%
0.78463%
0.84050%
0.96213%
The fact that interest rates are termdependent is an additional complication when pricing
fixedrate financial instruments. In this section we explore how these rates can be analysed.
In particular, we use the price of unit zerocoupon bonds to relate various interest rates,
via the noarbitrage principle.
3.2.1
Spot rates
The first step in analysing termdependent interest rates is to convert them to effective
rates, analogous to the ‘effective rates’ we defined in Section 1.1 for different compounding
frequencies. The name for these effective rates, in the context of termdependent interest
rates, is spot rates.
By default, interest rates are quoted as (annualised) nominal rates; this is the case for
the LIBOR rates quoted above. This means that if you place P in a deposit for term T , at
interest rate rT , you will get back
P (1 + rT T ).
A fairer way to compare termdependent rates is to compare their spot rates:
Definition 3.16. The spot rate, sT , for maturity T is defined as
sT = (1 + rT T )1/T − 1
(17)
or, equivalently,
(1 + sT )T = 1 + rT T.
The function sT is called the spot rate curve.
Remarks 3.17. Equation (17) can be compared to (3), which was the formula that we used
to define ‘effective rates’ for various compounding frequencies. Just as for effective rates, spot
rates are also ‘artificial’, in the sense that they are not directly quoted on the market.
Examples 3.18.
1. The 1year spot rate s1 is equal the nominal rate r1 .
2. The 2year spot rate s2 is equal to rate such that, compounding twice, has the same
effect as the nominal 2year rate r2 , i.e.
(1 + s2 )2 = 1 + r2 · 2.
3. The 6month spot rate s0.5 has the same effect as the 6month nominal rate compounded
twice, i.e.
1
1 2
⇔ (1 + s0.5 )1/2 = 1 + r0.5 · .
1 + s0.5 = 1 + r0.5
2
2
31
Example 3.19. Let r1/12 = 0.712% be the 1month LIBOR rate quoted above. Then the
associated spot rate satisfies
1 12
1 + s1/12 = 1 + 0.712%
= 1.00714
12
which yields a spot rate of sT = 0.714%.
In the case of termdependent interest rates, the discount factors D(T ) that we use to do
present value analysis must be computed in terms of the nominal/spot rates
D(T ) = (1 + rT T )−1 = (1 + sT )−T .
Example 3.20. Calculate the noarbitrage price of a fiveyear bond with face value £100,000,
redeemable at par, with 6% annual coupons, if the spot rate curve is
(s1 , s2 , s3 , s4 , s5 ) = (7%, 7.25%, 7.5%, 7.75%, 8%) .
Solution. We use the discount factor D(i) = (1 + si )−i in the formula (14). This yields a
noarbitrage price of
6,000
5
X
D(i) + 100,000D(5) = £92,247.
i=1
Another point of view is that, under the noarbitrage assumption, spot rates are determined
by the price of unit zerocoupon bonds VT with expiry T . This is since
−1/T
sT = D(T )−1/T − 1 = VT
− 1.
(18)
Examples 3.21. The following table gives some prices for unit zerocoupon bonds with varying
maturities, alongside the implied spot rate calculated via the formula (18):
T =Year
1
5
10
15
Price
0.94
0.70
0.47
0.30
Spot rate sn
6.4%
7.4%
7.8%
8.4%
Remark 3.22. Observe that the spot rates sT we defined play the same role as the effective
rate reff , in the sense that they define an equivalent annualised interest rate. Some practitioners
instead convert rT to an instantaneous spot rate s0T that is the equivalent continuouslycompounded rate. This satisfies
0
esT T = 1 + rT T
3.2.2
s0T =
⇔
1
ln (1 + rT T ) .
T
Forward rates
How much interest should we expect to earn on money deposited or loaned between two
dates in the future? Surprisingly, the fair (i.e. noarbitrage) value of such a ‘forward rate’ is
completely determined by current spot rates.
Why is this? Suppose I expect to have £1m in one year’s time and that I wish to deposit it
for a further year. How much interest might I expect to receive? Well consider instead taking
the following sequence of actions:
32
t=0
• Take out a 1year loan for (1+s1 )−1 × £1m. Place this amount in a 2year deposit.
t=1
• Pay £1m that you expect to have at this time to pay off the loan. Think of this
money as an investment that will receive a return at time t = 2.
t=2
• The deposit will now be worth (1 + s2 )2 (1 + s1 )−1 ×£1m.
Under this procedure, the £1m we expect to have in one year’s time will grow to £1m
to (1 + s2 )2 /(1 + s1 )×£1m at the end of the second year. This implies that the money has
earned an ‘effective’ interest rate f1,2 satisfying
1 + f1,2 =
(1 + s2 )2
.
1 + s1
Let us now formalise this using noarbitrage principles.
Definition 3.23. A forward rate agreement (FRA) from time n1 to time n2 is a contract
whereby a party agrees to place a certain amount (the principal) in a deposit or a loan at time
n1 for an effective interest rate of f , and the deposit/loan expires at time n2 . In particular, a
deposit of P at time n1 will be worth, at the maturity time n2 ,
P (1 + f )n2 −n1 .
A signatory to a FRA is obliged to deposit (or loan) the money at the start time n1 (i.e. there
is no optionality).
Intuitively the fair value of f in a FRA might be thought of as both parties best guess
as to what the interest rate will be at time n1 . However this is incorrect, since as we argued
before f is completely determined by today’s spot rates.
Proposition 3.24. The fair (i.e. noarbitrage) interest rate for a FRA from time n1 years to
n2 years is
1
(1 + s )n2 n −n
n2
2
1
fn1 ,n2 =
− 1,
n
1
(1 + sn1 )
where sn1 and sn2 are the spot rates to n1 and n2 years. Equivalently,
(1 + fn1 ,n2 )n2 −n1 =
(1 + sn2 )n2
.
(1 + sn1 )n1
Proof. The proof consists of formalising the previous argument using the noarbitrage principle.
Suppose for the purposes of contradiction that a bank proposes a FRA with forward rate f
that is larger than the forward rate fn1 ,n2 . Then at time 0 we can (i) sign this FRA for a
deposit between times n1 and n2 , (ii) simultaneously take out a loan with term n2 of amount
(1 + sn1 )−n1 ,
and (iii) deposit this amount immediately into a bank account for term n1 . At time n1 , the
deposit will have grown to amount
(1 + sn1 )−n1 × (1 + sn1 )n1 = 1.
At this point we withdraw this money from the bank, and use it to fulfil the FRA. At time n2 ,
this will now be worth
(1 + f )n2 −n1 > (1 + fn1 ,n2 )n2 −n1 =
33
(1 + sn2 )n2
,
(1 + sn1 )n1
where the first inequality is by the assumption, and the second is the definition of the forward
rate fn1 ,n2 . At the same time, our initial loan has grown to value
(1 + sn2 )n2
(1 + sn1 )n1
so we can use the deposit to pay off the loan, generating a riskfree profit. A similar argument
(this time involving an FRA for a loan instead of a deposit) can be used to show that f < fn1 ,n2
also gives rise to arbitrage (details are left as an exercise).
Example 3.25. Suppose that the 2year LIBOR rate is r2 = 1.43% and the 5year LIBOR
rate is r5 = 1.98%. Find the fair forward rate f2,5 for a FRA to deposit cash in two years time
for a three year deposit.
Solution. The first step is to calculate spot rates using the formula
1 + sT = (1 + rT T )1/T .
This yields
1 + s2 = (1 + 1.43% · 2)1/2
and
1 + s5 = (1 + 1.98% · 5)1/5 ,
which gives s2 = 1.42% and s5 = 1.91%. Therefore we have
(1 + f2,5 )5−2 =
1.0990
(1 + s5 )5
=
= 1.0684,
(1 + s2 )2
1.0286
which gives f2,5 = 2.23%. For example, if a FRA is made to deposit £500 after 2 years at
the forward interest rates, then in 5 years the deposit will be worth
500(1 + f2,5 )3 = 500 × 1.0684 = £534.20
Observe that the ratio of unit zerocoupon bond prices with expiries n1 and n2 is given by
(1 + sn2 )n2
Vn1
=
,
Vn2
(1 + sn1 )n1
and therefore
Vn1
= (1 + fn1 ,n2 )n2 −n1 .
Vn2
It follows that, just as for discount rates and bond prices, forward rates are also completely
determined by the prices of unit zerocoupon bonds.
Example 3.26. Suppose the price of 3year and 5year year unit zerocoupon bonds are 0.92
and 0.65 respectively. Calculate the 2year forward rate f3,5 from year 2 to year 5.
Proof. The ratio of the zerocoupon bond prices satisfies
V3
= (1 + f3,5 )5−3
V5
and so
f3,5 =
0.92 1/2
0.65
34
− 1 = 18.97%.
Notice that f0,T = sT , i.e. the forward rate starting from time 0 is equal to the spot rate.
Moreovoer, for any sequence of years n1 < n2 < · · · nj , the following equality holds:
j−1
j−1
Y
Y
(1 + sni+1 )ni+1
ni+1 −ni
(1 + fni ,ni+1 )
=
(1 + sni )ni
i=1
i=1
=
(1 + snj )nj
= (1 + fn1 ,nj )nj −n1 .
(1 + sn1 )n1
As an example,
(1 + f2,4 )2 (1 + f4,7 )3 = (1 + f2,7 )5 ,
which can be understood as saying that the following two ways of depositing money between
times T = 2 and T = 7 must result in the same bank balance at T = 7:
2 First in a 2year deposit between T = 2 and T = 4, then in a 3year deposit between
T = 4 and T = 7 (top of diagram below).
2 In a 5year deposit between T = 2 and T = 7 (bottom of diagram below);
(1 + f2,4 )2
×
(1 + f4,7 )3
T=2 T=3 T=4 T=5 Y=6 T=7
(1 + f2,7 )5
Used in combination, these relationships can help calculate unknown forward or spot rates.
Example 3.27. Suppose the 3year and 7year year spot rates are 6% and 5% respectively,
and that the 3year forward rate f4,7 is 5.2%. Calculate the forward rate f3,4 .
Proof. We can calculate f3,4 via the relationship (see diagram below)
(1 + s3 )3 (1 + f3,4 )(1 + f4,7 )3 = (1 + s7 )7 .
This yields
1 + f3,4 =
1.057
= 1.0148 =⇒ f3,4 = 1.48%.
1.063 × 1.0523
(1 + s3 )3 × (1 + f3,4 ) × (1 + f4,7 )3
T=0 T=1 T=2 T=3 T=4 T=5 Y=6 T=7
(1 + s7 )7
35
4
Stochastic interest rates
Up until now interest rates have been considered to be deterministic (i.e. known in advance);
in this chapter we study a model of interest rates that allows for uncertainty.
Why is it useful to be able to model uncertain (or ‘stochastic’) interest rates? Often
financial contracts are of a long term nature (for example, a 20year mortgage), and usually
the initial interest rate is only valid for a short time; after that, the new prevailing interest rate
is applied. Since these future rates are uncertain, we need a model that allows for uncertainty.
4.1
A fixed interest rate model
The most elementary model of stochastic interest rates consists of a single interest rate R,
fixed throughout the period of the investment, that is a random variable (which could be
discrete or continuous). Suppose an investor places the amount P in the bank. Then the
value of the deposit after time n is
P (1 + R)n ,
which is itself a random variable. Without further information, we cannot say anything more
precise about this random variable. For instance, all we can say about its mean is that it is
E(P (1 + R)n ) = P E((1 + R)n ),
which is not equal to P (1 + E(R))n . In fact, using Jensen’s inequality (which is not a topic
of this module), one may show that
P E((1 + R)n ) ≥ P (1 + E(R))n .
Example 4.1. Suppose the interest rate
0.06
0.08
R=
0.10
R has probability distribution
with probability 0.2,
with probability 0.7,
with probability 0.1.
Find the mean and standard deviation of the accumulated value of a deposit of £5,000 for
five years.
Solution. Let S5 = (1 + R)5 , so that the unknown accumulated value is
P (5) = 5,000 × S5 .
Then
E(S5 ) = 0.2 × 1.065 + 0.7 × 1.085 + 0.1 × 1.105 = 1.457,
and we can use the formula Var(S5 ) = E(S52 ) − E(S5 )2 to compute
Var(S5 ) = E[(1 + R)10 ] − 1.4572
= 0.2 × 1.0610 + 0.7 × 1.0810 + 0.1 × 1.1010 − 1.4572
= 0.00528,
and so the standard deviation of S5 is
sd(S5 ) =
√
0.00528 = 0.0726.
36
Using linearity,
E[P (S5 )] = 5,000 × E(S5 ) = £7,285
and
sd(P (5)) = 5,000 × sd(S5 ) = £363.3.
On the other hand, notice that the mean rate of interest is
E(R) = 0.2 × 0.06 + 0.7 × 0.08 + 0.1 × 0.10 = 7.8%,
and so the accumulated value at the mean rate of interest is
5,000(1 + ER)5 = 5,000 × 1.0785 = £7,279,
which is indeed less than the mean accumulated value E[P (5)].
Example 4.2. Suppose you deposit P into a bank with random interest rate R as in Example 4.1. Calculate the expected time until the deposit value doubles to 2P .
Solution. Let T be the time until the deposit value doubles, which satisfies
P (1 + R)T = 2P
⇔
T =
ln 2
.
ln(1 + R)
We can then calculate E(T ) as
E(T ) = ln 2 × E[1/ ln(1 + R)]
= ln 2 × 0.2 × 1/ ln(1 + 0.06) + 0.7 × 1/ ln(1 + 0.08) + 0.1 × 1/ ln(1 + 0.10)
= 9.41 years.
Note that this is not equal to
ln 2
ln 2
=
= 9.23 years.
ln(1 + E(R))
ln(1 + 0.078)
4.2
A varying interest rate model
We next consider a model that allows for varying interest rates. Let Ri be the interest rate
applicable between year i and year i + 1, with each Ri a random variable. In order to make
calculations tractable, we will assume that Ri are mutually independent. Later on we will, in
addition, assume that the interest rates Ri are also identically distributed.
In reality the assumption that interest rates are i.i.d. may not be realistic:
1. Interest rates in different periods will not generally be independent. For instance, if
interest rates are low now then we expect interest rates in the immediate future to be
low as well, i.e. in general we expect that Ri and Ri+1 to be positively correlated.
2. The interest rates Ri for large i are more uncertain than the interest rate, say, next year,
and so we would expect Var(Ri ) to be larger than Var(R1 ).
Nevertheless, for simplicity in this course we will persist with these assumptions.
37
4.2.1
The growth of a single deposit
Suppose that we make a single deposit of £1 at time 0. In n years this will grow in value to
Sn = (1 + R0 )(1 + R1 ) · · · (1 + Rn−1 ).
More generally, if we deposit £P then this will grow in value to P Sn . In order to analyse
the performance of this investment, we calculate the mean and variance of the accumulated
value Sn .
The mean of the accumulated value P Sn is
n−1
n−1
n−1
Y
Y
Y
E (P Sn ) = P E
(1 + Ri ) = P
E (1 + Ri ) = P
(1 + E(Ri )),
i=0
i=0
i=0
with the second equality holding since we assume Ri are independent.
In order to calculate Var(P Sn ), we use the formula
Var(P Sn ) = P 2 E(Sn2 ) − (E (Sn ))2 .
We calculate E(Sn2 ) as follows:
E
Sn2
n−1
n−1
Y
Y
Y
n−1
2
2
=E
(1 + Ri ) =
E (1 + Ri ) =
E 1 + 2Ri + Ri2
i=0
=
=
n−1
Y
i=0
n−1
Y
i=0
i=0
1 + 2E(Ri ) + E(Ri2 ) =
n−1
Y
1 + 2E(Ri ) + (E(Ri ))2 + Var(Ri )
i=0
(1 + E(Ri ))2 + Var(Ri )
i=0
where we again used the independence of the Ri , as well as the formula
E(Ri2 ) = (E(Ri ))2 + Var(Ri ).
Combining with our previous calculation of E(Sn ),
Var(P Sn ) = P 2 E(Sn2 ) − (E (Sn ))2
= P2
n−1
Y
!
n−1
Y
(1 + E(Ri ))2 + Var(Ri ) −
(1 + E(Ri ))2 .
i=0
i=0
If the interest rates Ri are also identically distributed with common mean µ and variance σ 2
then
n−1
Y
E (P Sn ) = P
(1 + µ) = P (1 + µ)n
i=0
and
Var(P Sn ) = P 2
(1 + µ)2 + σ 2
n
− (1 + µ)2n .
(19)
Remark 4.3. In the case that the interest rates are deterministic, we would expect that the
deposit value P Sn is also deterministic. This is indeed the case, since setting σ 2 = 0 in the
formula (19) yields
n
n
Var(P Sn ) = P 2 (1 + µ)2 + σ 2 − (1 + µ)2n = P 2 (1 + µ)2 − (1 + µ)2n = 0.
38
Remark 4.4. If the variance of the interest rates goes up, then so does the variance of the
deposit value Sn . This can be seen by inspecting the formula (19) and checking that the
derivative of the righthand side with respect to σ is positive (exercise!).
Example 4.5. Suppose that Ri are all distributed as in Example 4.1, and suppose that we
again invest £5,000 for n = 5 years. Find the mean and variance of the accumulated value of
the investment.
Solution. In Example 4.1 we already calculated the mean accumulated value
P (1 + µ)5 = 5,000(1 + E[Ri ])5 = £7,279.
We can calculate σ 2 = Var(Ri ) via
σ 2 = E(Ri2 ) − E(Ri )2
= (0.2 × 0.062 + 0.7 × 0.082 + 0.1 × 0.102 ) − µ2 = 0.0062 − 0.0782
= 0.000116 .
Hence, applying formula (19), the variance of the accumulated value is
25,000,000 (1.0782 + 0.000116)5 − (1.078)10 = £26,500
4.2.2
The growth of regular deposits
We next consider an investment in which additional deposits are made every year. Assume
that at time i ∈ {0, 1, . . . , n − 1} we deposit £Pi .
Then the accumulated value of the deposit An at time n (by convention, just before the
new deposit is made) satisfies a recursion:
A0 = 0
and
An = (An−1 + Pn−1 )(1 + Rn−1 ) ,
n ≥ 1.
(20)
This recursion can be interpreted as follows: the value of the deposit at time n is equal to the
value at time n − 1, plus the £Pn−1 contribution from the previous year, all increased by the
interest rate applicable from time n − 1 to time n.
As before, in order to analyse the accumulated value An we can compute its expectation
and variance. The recursion formula (20) can be manipulated to give a recursion formula for
the expectation. Specifically, we compute the expectation of both sides of (20):
E (An ) = E ((An−1 + Pn−1 )(1 + Rn−1 ))
= E (An−1 + Pn−1 ) × E (1 + Rn−1 )
= (E(An−1 ) + Pn−1 ) (1 + E(Rn−1 ))
where we used the fact that An−1 + Pn−1 and (1 + Rn−1 ) are independent (since An−1 and
Rn−1 are independent – explain this).
In general, solving the resulting recursive formula for E(An ) is quite complicated. In the
special case that Pi = P (i.e. all deposit amounts are the same), and Ri are i.i.d. with common
mean µ we have, for example,
E (A1 ) = P (1 + µ)
E (A2 ) = (E (A1 + P )) (1 + µ) = P (2 + µ)(1 + µ)
E (A3 ) = (E (A2 ) + P ) (1 + µ) = P (1 + (2 + µ)(1 + µ)) (1 + µ).
39
Example 4.6. If Pi = P and the Ri are i.i.d. with common mean µ = 0, prove that
E(An ) = P n
In other words, if the expected interest rate is zero, the expected value of a fund where we
deposit £P every year is £P n.
Solution. In the case µ = 0 the recursive formula reduces to
E(An ) = E(An−1 ) + P,
which shows that E(An ) = P n by induction.
Remark 4.7. Note that in the case that the interest rate was zero (i.e. µ = 0 and σ 2 = 0),
then An = P n, so certainly E(An ) = P n as well.
A recursive formula for the variance can also be derived via a similar method. The formulae
are too complicated to manipulate by hand but could be easily programmed into Excel, for
example. We explore this in Coursework 3.
4.3
Lognormally distributed interest rates
In practice, the easiest stochastic interest rate model to work with is usually the lognormal
model, in which the interest rates are assumed to follow a lognormal process. This model
will appear again in the course when we discuss share prices.
A continuous random variable Y is said to be lognormally distributed with parameters
µ and σ 2 , written Y ∼ LogNormal(µ, σ 2 ), if
ln Y ∼ N (µ, σ 2 ).
Equivalently, Y = eX , where X ∼ N (µ, σ 2 ). The p.d.f. of Y is
(
(ln y−µ)2
√ 1
exp
−
, if y > 0,
2σ 2
2πσy
fY (y) =
0,
if y ≤ 0.
Let us consider the growth of a single deposit in the case that 1 + Ri are assumed to be
an i.i.d. sequence of lognormally distributed random variables. In that case the distribution
of
Sn = (1 + R0 )(1 + R1 ) · · · (1 + Rn−1 )
can be computed exactly (and not merely the mean and variance as we have previously done).
The underlying reason is that lognormal distributions behave well under multiplication. Suppose that Y1 ∼ LogNormal(µ1 , σ12 ) and Y2 ∼ LogNormal(µ2 , σ22 ) are independent. Then
we can write Y1 = eX1 and Y2 = eX2 , where X1 ∼ N (µ1 , σ12 ) and X2 ∼ N (µ2 , σ22 ) are
independent. Therefore,
Y1 Y2 = eX1 eX2 = eX1 +X2 ,
and since
X1 + X2 ∼ N (µ1 + µ2 , σ12 + σ22 ),
it follows that
Y1 Y2 ∼ LogNormal(µ1 + µ2 , σ12 + σ22 ).
By extending the previous results to n variables we can show the following:
40
Theorem 4.8. Suppose that 1 + Ri are a sequence of i.i.d. random variables with 1 + Ri ∼
LogNormal(µ, σ 2 ). Then
Sn =
n−1
Y
(1 + Ri ) ∼ LogNormal(nµ, nσ 2 ).
i=0
Using the above theorem, the c.d.f. of Sn can be computed as
X − nµ
ln x − nµ
X
P(Sn ≤ x) = P(e ≤ x) = P(X ≤ ln x) = P √
≤ √
nσ
nσ
where X ∼ N (nµ, nσ 2 ). Therefore
P(Sn ≤ x) = Φ
ln x − nµ
√
,
nσ
where Φ denotes the c.d.f. of the standard normal distribution.
Example 4.9. Suppose £1000 is deposited in a bank account for five years, and suppose
the interest rates Ri are a sequence of i.i.d. random variable such that 1 + Ri is lognormally
distributed with parameters µ = 0.075 and σ 2 = 0.0252 . What is the probability that the
accumulated value of the deposit exceeds £1,500? Find the upper and lower quartiles for the
accumulated value of the deposit.
Solution. The accumulated value An = 1000S5 has distribution
An ∼ 1000LogNormal(5µ, 5σ 2 ).
The probability that An > 1500 is
ln 1.5 − 5µ
√
P(An > 1500) = P(LogNormal(5µ, 5σ 2 ) > 1.5) = 1 − Φ
5σ
= 1 − Φ(0.545) = 1 − 0.7017 = 0.2929,
where the value of the normal c.d.f. has to be looked up (in a table for instance).
By definition of the quartile, the accumulated value An will exceed the upper quartile u
with probability 0.25, i.e.
0.75 = P(An ≤ u) = P(1000S5 ≤ u) = P(S5 ≤ u/1000).
Hence u satisfies
ln(u/1000) − 5µ
√
0.75 = Φ
.
5σ
Looking up the normal c.d.f. we find that Φ(0.6745) = 0.75, so therefore
ln(u/1000) − 5µ
√
= 0.6745,
5σ
that is,
√
u = 1000 exp 5µ + 0.6745σ 5 = £1,510 .
Similarly, the lower quartile is
√
` = 1000 exp 5µ − 0.6745σ 5 = £1,400 .
41
5
Equities and their derivatives
In previous chapters we considered fixedincome securities. Equities and derivatives differ
from fixedincome securities in that their cashflow stream is not predetermined, and instead
depends on market fluctuations. This complicates their analysis.
A derivative is a financial instrument (i.e. a contract between two or more parties) whose
value is ‘derived’ from one or more underlying assets (e.g. stocks, bonds, foreign exchange,
interest rates, commodities etc). There exist many types of derivatives; in this course we
study only the most common examples being forwards and options, but there are many
other ‘exotic’ derivatives. Derivatives can be bought and traded either through an exchange
such as the Chicago Mercantile Exchange (CME) or overthecounter (offexchange). The
most common use of derivatives is to reduce expose to risk (‘hedge’), or conversely to increase
exposure to risk (‘speculate’).
5.1
Shares
The basic instrument of equity is the share (or ‘stock’),9 which is a unit of ownership of a
company. Owning shares translates to partly owning the company which issues the shares.
Example 5.1. Apple currently has 4.52 billion shares. If you were to buy one share in Apple
(current price $240.24) you would own
1
= 0.0000000221%
4,520,000,000
of the company. The current market capitalisation of Apple (i.e. the total value its shares) is
4.52 billion × $240.24 = $1,086 trillion.
Sometimes shares entitle the holder to additional benefits, for instance to dividends (cash
payments made by the company to its shareholders). For simplicity, we will generally only
consider nondividend paying shares.
Shares are usually traded on a stock exchange, such as the London Stock Exchange or
the New York Stock Exchange. We will always assume that:
1. Shares can be bought and sold by an investor in any amount (even fractional amounts),
at any time, without transaction costs.
2. An investor can borrow a share and then subsequently sell it. This operation is called
shortselling, and can be viewed as owning a negative number of shares.
These assumptions are another instance of the ‘frictionless market hypothesis’.
5.2
Forwards
A forward (or a forward contract) is an agreement between two parties in which the seller
promises to sell a specified asset (the ‘underlying’) for a specified price F (the ‘forward price’)
at a specified time T (the ‘settlement date’) in the future, while the buyer promises to buy it.
Usually the underlying is either a commodity (e.g. oil, gas, gold etc.) or foreign exchange.
9
These terms are basically synonymous; shares is mainly used in the UK, while stock or common stock is
mainly used in the US.
42
Forward contract are not bought or sold, but are ‘entered into’. The investor can either
enter a long position in the forward, if they are buying the underlying asset at time T , or a
short position in the forward, if they are selling the underlying asset at time T .
As with all derivatives, forward contracts can be used to:
• Reduce exposure to risk, i.e. to hedge. Suppose that a company needs 1,000 barrels
of oil in 6 months’ time, and wants to eliminate the risk that the oil price will move
significantly higher by then. Or suppose they trade in the UK but need to buy oil in US
dollars. By entering into a forward contract they can fix the price of the future purchase,
and so eliminate their exposure to price/currency fluctuations.
• Increase exposure to risk, i.e. to speculate. Suppose that an investment bank expects
the price of oil to increase, but has no need for oil. By entering into a forward contract,
they can agree to buy oil at a fixed (low) price, and then at the settlement date, when
the oil is received, they can immediately sell for a profit (in practice, they would never
even take possession of the oil).
Example 5.2. Suppose a company enters into a forward contract with an oil supplier to
purchase 1,000 barrels of oil on 1st January 2020 for $42,000. Suppose that the price of oil
per barrel on 1 January 2020 is $50. Then the company, who agreed to buy 1,000 barrels on
1 January for $42,000, could immediately sell the oil for $50,000, yielding an instant profit of
1,000 · $50 − $42,000 = $8,000.
Conversely, if the price on 1 January 2020 is $35, then on 1 January the company takes an
instant loss of
1,000 · $35 − $42,000 = −$7,000.
Even though two companies are free to enter into a forward contract at any forward price F
they agree on, in reality the principles of noarbitrage pricing determine a single ‘fair’ forward
price for every asset:
Proposition 5.3. Suppose that an asset has current price S at time t = 0. By the noarbitrage principle, the forward price F agreed at t = 0 for the delivery of the underlying asset
at t = T must be equal to
F = S/D(T )
where D(T ) is the discount factor to time T .
Proof. We prove this result using ‘replicating portfolios’ (i.e. we apply the law of one price).
Consider the following two portfolios formed at t = 0:
• Portfolio A. A portfolio consisting of the asset.
• Portfolio B. A portfolio consisting of:
– a long position in the forward contract; and
– T year zerocoupon bonds with facevalue F (or, equivalently, D(T )F deposited
in a bank account).
43
We claim that both portfolios will have the same value at time T . To see this, observe that the
value of Portfolio A is equal to the value of the asset at time T . On the other hand, Portfolio
B also consists of the asset (guaranteed through the forward contract), less the forward price
F that needs to be paid, plus the F that the bond pays upon maturity. It total, we see that
Portfolio B has the same value as the asset at time T .
Since portfolios A and B are replicating, by the law of one price they have the same price
at t = 0. Let us now examine these prices. The price of Portfolio A at time 0 is equal to the
price of the asset, which is S. The price of Portfolio B is equal to the price of the zerocoupon
bond (remember that we do not pay anything to enter into the forward contract), which is
D(T )F . Equating these, we conclude that S = D(T ) F ⇔ F = S/D(T ).
The current price of an asset of often referred to as the spot price. It is equal to the
forward price of a contract with immediate delivery.
Example 5.4. Interest is continuouslycompounded at nominal rate r = 5%. What is the fair
forward price F for a contract to buy 1,000 barrels of oil in one year’s time if the spot price
is $37 per barrel.
Solution. Since D(1) = e−0.05 = 0.9512, the fair forward price is
F = 37, 000/0.9512 = $38, 897 .
Example 5.5 (Not covered in lectures). Suppose that the effective rate of interest in the UK
is r£ = 0.5%, and in the Eurozone is r€ = 1%. Suppose also that the current exchange rate
is €1.13 to £1. What is the fair forward price £F to buy €1,000 in three months’ time?
Solution. By the above proposition,
F = S/D(1/4),
where S is the current price (in pounds) of € 1,000
S=
1000
= 885,
1.13
and D(1/4) is the discount factor to 3 months. The complicating thing about this example is
determining the relevant discount factor D(T ), since there are two interest rates: r£ and r€ .
To resolve this, let us take a more direct approach using ‘replicating portfolios’. Consider
two portfolios, both of which will be worth €1,000 at time T = 1/4:
• Portfolio (A). A long position in a forward contract to buy € 1,000 at time T , and
£ F/(1 + r£ )T deposited in a UK bank until time T .
• Portfolio (B). € 1,000/(1 + r€ )T deposited in a bank in Germany until time T .
By the law of one price, since the two investments are worth the same amount at time T , they
are also worth the same amount at time 0. Thus, in pounds,
0 + F/(1 + r£ )T =
1,000/(1 + r€ )T
1.13
which implies
1,000 1 + r£ T
1,000 1.005 1/4
F =
=
= £883.9.
1.13 1 + r€
1.13 1.01
44
Here we see that the relevant discount factor was
1 + r −T
£
D(T ) =
,
1 + r€
which implies that value is ‘discounted’ with respect to the ‘currency adjusted’ rate
1 + r£
r£ − r€
−1=
= −0.495%.
1 + r€
1 + r€
(21)
Note that the expression (21) is mathematically the same as the ‘inflation adjusted’ interest
rate we saw in Section 1.3, with r€ playing the role of the inflation rate.
5.3
Options
An option is a derivative contract that gives the buyer the right, but not the obligation, to
buy or sell an underlying asset for a specific strike price K (also known as the ‘exercise price’)
on, and sometimes also prior to, the expiry time T (also known as the ‘maturity’). The seller
of the option is then obliged to fulfil the transaction. The buyer pays a certain fee to purchase
an option.
Options are traded at the Chicago Board Options Exchange (CBOE), or overthecounter
as customised contracts between a single buyer and seller.
Remark 5.6. Be sure you understand the difference between a forward contract and an option.
A forward contract gives you the obligation to buy or sell an underlying asset for a specified
price (the forward price F ), while an option gives you the right, but not the obligation, to
buy or sell an underlying asset for a specified price (the strike price K). On the other hand,
forward contracts are free to enter, whereas there is a cost to buy an option.
There are two main categories of options:
• Call options, which give the owner the right to buy the asset at the strike price;
• Put options, which give the owner the right to sell the asset at the strike price.
These are further divided into two main categories:
• European options which give the owner the right to buy or sell the asset at the strike
price at the expiry date.
• American options, which give the owner the right to buy or sell the underlying asset
at the strike price at any time up to and including the expiry.
Example 5.7. A European call option on a share, with strike K and expiry T , is a contract
that gives the right, but not the obligation, to purchase the share at time T for price K. A
European put option is similar, but the right to purchase is replaced by a right to sell.
Example 5.8. An American call option on a share, with strike K and expiry T , is a contract
that gives the right, but not the obligation, to purchase the share at any time up to and
including time T for price K. An American put option is similar, but the right to purchase is
replaced by a right to sell.
American options are more expensive (or at least as expensive) than the equivalent European options (i.e., with the same expiry T and strike K), since they offer their holder more
optionality.
45
Exercise 5.9. Use a noarbitrage argument to verify that an American option must cost at
least as much as the equivalent European option.
Remarks 5.10.
1. The options described above are actually only a small sample of the enormous variety
of options that are issued and traded. For this reason they are sometimes referred to as
‘vanilla’ options, with other variety of options known as ‘exotic’.
2. Although options can be written on any underlying asset, for simplicity we will work only
with options written on an underlying share.
3. In practice an option will …
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Stochastic Models
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