MTH6154: Financial Mathematics 1
Dr Dr Dudley Stark
Term 1, 2022-23
Contents
Preface 2
1 Interest rates and present value analysis 4
1.1 Interest rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Variable interest rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 The instantaneous interest rate . . . . . . . . . . . . . . . . . . . . . 7
1.2.2 The yield curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Present value analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4.1 Defining the present value . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4.2 Discount Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4.3 Balancing present values . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5 Rates of return . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.1 The annualised rate of return and the equivalent effective interest rate 14
1.5.2 Internal rate of return . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Immunisation of assets and liabilities 18
2.1 Measuring the effect of varying interest rates . . . . . . . . . . . . . . . . . . 18
2.2 Reddington immunisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3 Immunisation in practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4 Immunisation when interest is compounded yearly . . . . . . . . . . . . . . . 23
3 Bonds and the term structure of interest rates 25
3.1 Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.1.1 The present value of a bond . . . . . . . . . . . . . . . . . . . . . . . 26
3.1.2 The no-arbitrage principle and the fair price of a bond . . . . . . . . . 27
3.1.3 The bond yield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2 The term structure of interest rates . . . . . . . . . . . . . . . . . . . . . . . 30
3.2.1 Spot rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2.2 Forward rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4 Stochastic interest rates 36
4.1 A fixed interest rate model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.2 A varying interest rate model . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.2.1 The growth of a single deposit . . . . . . . . . . . . . . . . . . . . . 38
4.2.2 The growth of regular deposits . . . . . . . . . . . . . . . . . . . . . 39
4.3 Log-normally distributed interest rates . . . . . . . . . . . . . . . . . . . . . . 40
1
5 Equities and their derivatives 42
5.1 Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.2 Forwards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.3 Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
5.4 Pricing options: model-independent pricing principles . . . . . . . . . . . . . . 47
5.5 Pricing options via the risk-neutral distribution . . . . . . . . . . . . . . . . . 49
5.5.1 Betting strategies, the risk-neutral distribution, and the arbitrage theorem 49
5.5.2 Options pricing using the risk-neutral distribution . . . . . . . . . . . . 53
6 The binomial model 55
6.1 The single-period binomial model . . . . . . . . . . . . . . . . . . . . . . . . 55
6.1.1 The no-arbitrage condition and the risk-neutral distribution . . . . . . 55
6.1.2 Option pricing in the single-period binomial model . . . . . . . . . . . 57
6.1.3 Replicating portfolios and the delta-hedging formula . . . . . . . . . . 60
6.2 The two-period binomial model . . . . . . . . . . . . . . . . . . . . . . . . . 62
6.2.1 Options pricing in the two-period binomial model . . . . . . . . . . . . 62
6.2.2 Replicating portfolios in the two-period binomial model . . . . . . . . 68
6.2.3 Option pricing via the two-period risk-neutral distribution . . . . . . . 68
6.3 The multi-period binomial model . . . . . . . . . . . . . . . . . . . . . . . . 69
6.3.1 Options pricing in the multi-period binomial model . . . . . . . . . . . 70
7 From the multi-period binomial model to the Black-Scholes formula 73
7.1 General discrete-time models . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.2 The log-normal process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.3 The Black-Scholes formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
7.3.1 Properties of the Black-Scholes formula . . . . . . . . . . . . . . . . . 77
7.4 The log-normal process as an approximation of the bin. model . . . . . . . . . 78
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 Man-
agement1) 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 ‘fixed-income’), 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:
1Formally 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
trade-off 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 time-value of money and the no-arbitrage princi-
ple. These concepts aremodel-free, 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
Black-Scholes 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 fixed-income 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
1
n
-th of a year. This means that every 1
n
-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
P
(
1 +
r
n
)n
. (2)
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).
Example 1.1. Suppose you borrow an amount P , to be repaid after one year at interest
rate r, compounded semi-annually. 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
P
(
1 +
r
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
P
(
1 +
r
2
)(
1 +
r
2
)
= P
(
1 +
r
2
)2
.
2A 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
100
(
1 +
0.18
12
)12
= 100× 1.195 = £119.5.
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 1
n
-th of a year, we therefore have
reff =
P (1 + r/n)n − P
P
=
(
1 +
r
n
)n
− 1. (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
100
= 0.195 = 19.5%.
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
reff =
(
1 +
1
365
24.5%
)365
− 1 = 0.2775.
• For company B, the effective rate is:
reff =
(
1 +
1
12
24.7%
)12
− 1 = 0.2770.
3Effective 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 1
n
-th of a year for n very large. In the limiting case, i.e. as n→∞, we say that interest
is continuously-compounded, and in this case the amount owed after one year is
lim
n→∞
P
(
1 +
r
n
)n
= Per
(we prove this rigorously in Coursework 1). The effective rate for continuously-compounded
interest at nominal rate r is therefore
reff =
Per − P
P
= er − 1.
Example 1.5. Suppose that a bank offers continuously-compounded interest at nominal
rate 5%. The effective rate is
reff = e
0.05 − 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)︸ ︷︷ ︸
T times
= P (1 + reff)
T .
For example, if interest is compounded n times a year, this is
P
(
1 +
r
n
)nT
,
and if it is compounded continuously
P (er)T = PerT .
Example 1.6 (The doubling rule). If a bank offers continuously-compounded 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
PerT = 2P ⇒ erT = 2 ⇒ rT = ln 2,
and therefore
T =
ln 2
r
≈ 0.693
r
.
For example, if the nominal rate is r = 12%, it takes 5.78 years for the amount to double.
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 instanta-
neous interest rate. We always assume that the instantaneous interest is continuously-
compounded, 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
er(t)h = 1 + r(t)h+
1
2
r(t)2h2 + . . . ≈ 1 + r(t)h.
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 piece-wise continuous function. Then
P ′(t) = P (t)r(t).
Proof. Let h denote a small time-period. 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)
h
≈ P (t)r(t).
As h becomes smaller, the above approximation becomes more and more accurate, and so
lim
h→0
P (t+ h)− P (t)
h
= P (t)r(t).
Recognising the left-hand side as the definition of the derivative, we conclude that
P ′(t) = P (t)r(t),
The above equation is a separable first order differential equation, easily solved by writing
P ′(t)
P (t)
= r(t)
and integrating both sides: ∫ t
0
P ′(u)
P (u)
du =
∫ t
0
r(u) du.
Thus
[lnP (u)]u=tu=0 =
∫ t
0
r(u) du =⇒ lnP (t) = lnP (0) +
∫ t
0
r(u) du,
and so
P (t) = P (0) exp
(∫ t
0
r(u) du
)
, (4)
which is the desired relation between the amount P (t) in your bank at time t and the time-
varying 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 r¯(t) is defined to be
the average value of r(t) on the interval (0, t):
r¯(t) =
1
t
∫ t
0
r(u) du.
The yield curve allows us to write the formula for P (t) in (4) as
P (t) = P (0) exp
(∫ t
0
r(u) du
)
= P (0)er¯(t)t.
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 de-
posited/loaned, and is equal to
reff(T ) = e
r¯(T ) − 1.
Example 1.9. Suppose you deposit £100 in a bank offering the instantaneous interest rate
r(t) =
{
3% t < 2,
5% t ≥ 2.
Find the yield curve r¯(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,
r¯(t) =
1
t
∫ t
0
3% ds = 3%
whereas for t ≥ 2,
r¯(t) =
1
t
( ∫ 2
0
3% ds+
∫ t
2
5% ds
)
=
0.06
t
+
0.05(t− 2)
t
= 5%− 4%
t
.
The value of the deposit is
100er¯(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
r(t) =
3%
1 + t
+
4%× t
1 + t
.
Find the effective rate for a loan until time T = 5.
Proof. We can rewrite the interest rate as
r(t) =
3%
1 + t
+
4%× (1 + t)
1 + t
− 4%
1 + t
= 4%− 4%− 3%
1 + t
= 4%− 1%
1 + t
.
The yield curve is
r¯(t) =
1
t
∫ t
0
(
4%− 1%
1 + u
)
du =
1
t
(
4%× t− 1%× ln(1 + t)) = 4%− 1%
t
ln(1 + t).
The effective rate for the loan is
reff(5) = e
r¯(5) − 1 = e0.04− 0.015 ln(6) − 1 = 3.71%;
as expected, this lies somewhere between 3% and 4%.
8
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
1+rinf
P − P
P
=
1 + r
1 + rinf
− 1 = r − 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
r − rinf
1 + rinf
=
0.05− 0.03
1 + 0.03
= 0.0194 = 1.94% .
In the case of compounding interest, we must use the effective interest rate in place of the
nominal rate, i.e.,
ra =
reff − rinf
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
reff − rinf
1 + rinf
=
((1 + 0.05
12
)12 − 1)− 0.03
1 + 0.03
= 0.0205 = 2.05%.
Since rinf is usually much smaller than 100%, it is often sufficient to use the approximation
ra =
reff − rinf
1 + rinf
≈ reff − rinf.
1.4 Present value analysis
A cash-flow stream is a sequence of payments that are specified by their amount and
transaction time. Often, but not always, the cash-flow stream will consist of payments at
regular intervals (for instance, one per year), in which case we can write the cash-flow 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 cash-flow streams?
For this we introduce the concept of the present value of a cash-flow 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)−iV
pounds in the bank today. At the end of year i you will have (1+ r)−iV · (1+ r)i = V pounds
in the bank. Thus, £V received in i years’ time is worth (1 + r)−iV in today’s money. Given
this, we define the present value of £V at the end of year i to be
PV (V ) = (1 + r)−iV.
Exchanging V for its present value PV (V ) is sometimes referred to as discounting the
amount V , and the factor (1 + r)−i = PV (1) is referred to as the discount factor (more on
this later).
Similarly, if the interest is compounded every 1
n
-th of a year at nominal rate r, the present
value of V pounds in i years is
PV (V ) =
(
1 +
r
n
)−in
V.
We can generalise the notion of present value to a cash-flow stream a = (a1, a2, . . . , an)
that pays ai at the end of year i, for i = 1, . . . , n. The present value
4 of a is
PV (a) =
n∑
i=1
ai
(1 + r)i
.
To see this, observe that the cash-flow 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 PV (a1), PV (a2), . . . needed to replicate these payments. Since
PV (a) = PV (a1) + PV (a2) + . . .+ PV (an),
the total amount you need to deposit to replicate the cash-flow stream is PV (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 no-arbitrage assumption.
Example 1.14. You are offered three different jobs. The salary paid at the end of each year
(in thousands of pounds) is
Job\Year 1 2 3 4 5
A 32 34 36 38 40
B 36 36 35 35 35
C 40 36 34 32 30
Which job pays the best if the interest rate is (i) r = 10%, (ii) r = 20%, or (iii) r = 30%?
4The term net present value (NPV) is sometimes used when discussing cash-flow streams as opposed to
a single payment, but we will not use this term.
10
Solution. We shall compare the present values of the cash-flow streams. The present value
for job A is
PV (A) =
32
1 + r
+
34
(1 + r)2
+
36
(1 + r)3
+
38
(1 + r)4
+
40
(1 + r)5
,
while the present value for job B is
PV (B) =
36
1 + r
+
36
(1 + r)2
+
35
(1 + r)3
+
35
(1 + r)4
+
35
(1 + r)5
and the present value for job C is
PV (C) =
40
1 + r
+
36
(1 + r)2
+
34
(1 + r)3
+
32
(1 + r)4
+
30
(1 + r)5
.
Taking the different interest rates into account we get the following present values:
r PV A PV B PV C
0.1 135.03 134.41 132.14
0.2 105.51 106.20 105.50
0.3 85.20 86.61 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 cash-flow 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 ̸= 1,
b∑
i=a
ri =
ra − rb+1
1− r .
Hence, if |r| < 1,
∞∑
i=a
ri = lim
b→∞
b∑
i=a
ri = lim
b→∞
ra − rb+1
1− r =
ra
1− r .
These facts allow us to calculate the present value of perpetual cash-flow 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 cash-flow stream generated by the perpetuity is a = (a, a, a, . . .), and so its
present value is
PV (a) =
∞∑
i=1
a
(1 + r)i
= a× 1/(1 + r)
1− 1/(1 + r) =
a
(1 + r)− 1 =
a
r
.
11
Remark 1.17. Note that the present value a
r
of the perpetuity can also be justified by a
simple replication argument. Suppose that instead of purchasing a perpetuity you place a
r
in a bank account. After one year you have a
r
(1 + r) in the account. Suppose you withdraw
a to replicate the payment of the perpetuity. Then you are left with
a
r
(1 + r)− a = a
r
+ a− a = a
r
.
At the end of the second year, this will have grown to a
r
(1 + r), so if you again withdraw a
you are left with
a
r
(1 + r)− a = a
r
+ a− a = a
r
.
If you continue like this you can withdraw a from the bank account each year forever, which
replicates exactly the cash-flow stream generated by the perpetuity.
1.4.2 Discount Factors
It is often simpler to write the present value of a cash-flow 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
D(T ) =
1
(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
D(T ) =
1
(1 + reff)
T
.
For example, if interest is compounded monthly then
D(T ) =
1(
1 + r 1
12
)12T ,
if interest is continuously-compounded 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 cash-flow stream consists of payments a1, a2, . . . , an transacted at
times t1, t2, . . . , tn, then the present value of this cash-flow stream is
PV =
n∑
i=1
D(ti)ai.
12
1.4.3 Balancing present values
In many real-life 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 self-employed 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 cash-flow stream generated by these deposits. Similarly, let w =
£1,000, and let W denote the cash-flow 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.
PV (D) = PV (W ).
We first calculate PV (D). It is convenient to work with the monthly discount factor
β = D(1/12) =
1
1 + 0.06
12
=
1
1.005
.
Since there are 240 months in 20 years,
PV (D) = a+ aβ + aβ2 + · · ·+ aβ239 = a(1 + β + β2 + · · ·+ β239) = a1− β
240
1− β .
We next calculate PV (W ). Since her first withdrawal will take place in 240 months’ time,
and since there are 360 months in 30 years,
PV (W ) = wβ240 +wβ241 + · · ·+wβ240+359 = wβ240(1 + β + · · ·+ β359) = wβ2401− β
360
1− β .
Since PV (D) = PV (W ), we conclude that
a
1− β240
1− β = wβ
2401− β360
1− β ⇒ a = wβ
2401− β360
1− β240 = £361
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 cash-flow 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 + wβ2 + . . .+ wβ359 = w
1− β360
1− β .
Equating these values, we have
a β−1
β−240 − 1
β−1 − 1 = w
1− β360
1− β ⇒ a = w
1− β360
β−240 − 1 = £360.9 .
13
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
150,000
= 0.333 = 33.3%.
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 time-period.
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 =
(Pfinal
P
)1/T
− 1.
14
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
r =
(Pfinal
P
)1/T
− 1 =
(1750
1500
)1/2
− 1 = 0.0801 = 8.01%, .
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 cash-flow 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 cash-flow 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
P =
n∑
i=1
ai
(1 + r)ti
,
that is, the solution of the equation f(r) = 0, for the function
f(r) = −P +
n∑
i=1
ai(1 + r)
−ti .
The internal rate of return is the ‘hypothetical’ effective interest rate that makes the present
value of the cash-flow 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 cash-flow streams):
Theorem 1.25. Suppose P > 0 and ai, ti > 0 for i = 1, . . . , n. Then the equation
f(r) = 0 ⇔ P =
n∑
i=1
ai(1 + r)
−ti (6)
has a unique solution in (−1,∞).
15
Proof. Observe the following properties of the function f on (−1,∞):
• f is strictly decreasing, since
f ′(r) =
n∑
i=1
ai
−ti
(1 + r)ti+1
< 0.
• limr→−1 f(r) = +∞ since
lim
r→−1
f(r) = lim
r→−1
(
− P +
n∑
i=1
ai(1 + r)
−ti
)
= −P +
n∑
i=1
ai lim
r→−1
(1 + r)−ti = +∞.
• limr→+∞ f(r) = −P < 0 since
lim
r→+∞
f(r) = lim
r→+∞
(
− P +
n∑
i=1
ai(1 + r)
−ti
)
= −P +
n∑
i=1
ai lim
r→+∞
(1 + r)−ti = −P.
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 by-product of the proof of Theorem 1.25 is that the present value of a
cash-flow 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 =
10
(1 + r)
+
20
(1 + r)2
+
1,050
(1 + r)2
=
10
(1 + r)
+
1,070
(1 + r)2
.
Rewriting this, we have
1,000(1 + r)2 − 10(1 + r)− 1,070 = 0,
5Instead 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
(1 + r)
+
18 + 1,041
(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
cash-flow 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 cash-flow 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 cash-flow 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 cash-flow stream consists of a sequence of
payments Ci transacted at times ti. The present value V (r) of this cash-flow stream is then
V (r) =
m∑
i=1
CiD(ti; r).
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).
First-order properties. A simple way to quantify how much V (r) decreases with r is to
compute V ′(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 cash-flow stream is defined as
ν = −V
′(r)
V (r)
Remarks 2.2. Let us explain why ν is preferred to V ′(r):
1. The derivative V ′(r) is measured in the same monetary units (e.g. pounds, dollars, pence
etc.) as the present value V (r). By dividing V ′(r) by V (r), we standardise V ′(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 cash-flow stream, the minus sign is included
to ensure that ν is positive for such a stream (although it can be negative if the cash-flow
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 cash-flow stream is defined as
τ =
∑m
i=1 tiCiD(ti)
V (r)
=
∑m
i=1 tiCiD(ti)∑m
i=1CiD(ti)
.
18
In the case of positive payments Ci > 0, τ can be considered a weighted average of the
times ti
τ =
ω1t1 + · · ·+ ωmtm
ω1 + · · ·+ ωm
where the weights ωi = CiD(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 continuously-compounded interest, it turns out that they are exactly the same thing.
Proposition 2.4. If interest is continuously-compounded, 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∑
i=1
Cie
−rti and V ′(r) =
m∑
i=1
−tiCie−rti ,
we have
ν = −V
′(r)
V (r)
= −
∑m
i=1−tiCie−rti
V (r)
=
∑m
i=1 tiCiD(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
cash-flow stream, such that
ν = c0τ.
Remark 2.6. As we have seen, in the case of continuously-compounded 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)−1τ.
The equivalence between the effective duration and the duration leads to the following
useful general principle:
If cash-flow stream A has a larger duration than cash-flow 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 cash-flow stream with large duration depends
on the context, in particular whether the cash-flow 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 + 1,000β2 = 1,808,
whereas the present value of the liability is
2,215β3 = 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
τ(A) =
1× 1000β + 2× 1000β2
1808
= 1.483
and
τ(L) =
3× 2215β3
2215β3
= 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.
Second-order properties. We can refine our analysis of how V (r) varies with r by con-
sidering V ′′(r). As for the effective duration, it is more natural to considered the normalised
version V ′′(r)/V (r), which is called the convexity.
Definition 2.9. If V (r) > 0, the convexity of the cash-flow stream is defined as
c =
V ′′(r)
V (r)
.
Proposition 2.10. If interest is continuously-compounded, then the convexity is
c =
∑m
i=1 t
2
iCiD(ti)
V (r)
=
∑m
i=1 t
2
iCie
−rti∑m
i=1Cie
−rti . (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, . . . , t
2
n. 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
c =
∑m
i=1 ti(ti + 1/n)CiD(ti)
V (r)
× (1 + r/n)−2.
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 cash-flows (i.e. incoming payments), called assets, whose present value is
VA(r0), and a stream of negative future cash-flow (i.e. outgoing payments), called liabilities,
with present value −VL(r0). The present value of the combined cash-flow stream is therefore
V (r0) = VA(r0)− VL(r0).
The ‘immunisation’ of these cash-flow 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 cash-flows are said to be Reddington immune.
Zero-order immunisation: Matching assets and liabilities Zero-order immunisation
simply means that the present value of assets is matched by the present value of liabilities:
Definition 2.12. A cash-flow stream is said to be zero-order immune if the present value
of assets and liabilities is the same, i.e. VA(r0) = VL(r0).
Note that zero-order 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.
First-order immunisation: Matching the effective durations of assets and liabilities
Now assume that the cash-flow is zero-order 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) ̸= VL(r + ε), and we can use Taylor expansions to approximate the difference:
VA(r0 + ε) ≈ VA(r0) + V ′A(r0)ε = VA(r0)− VA(r0)νAε = VA(r0) (1− νAε) (8)
VL(r0 + ε) ≈ VL(r0) + V ′L(r0)ε = VL(r0)− VL(r0)νLε = VL(r0) (1− νLε) (9)
where νA = −V ′A(r0)/VA(r0) and νL = −V ′L(r0)/VL(r0) are the effective durations of the
cash-flow streams corresponding to the assets and liabilities. Hence
V (r0 + ε) = VA(r0 + ε)− VL(r + ε) ≈ VA(r0) (1− νAε)− VL(r0) (1− νLε) ,
which in the case of zero-order 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 cash-flow stream is said to be first-order immune if (i) it is immune to
zero-order, i.e. VA(r0) = VL(r0), and (ii) if the effective durations (equiv. durations) of the
cash-flow streams corresponding to the assets and liabilities are equal, i.e. if νA = νB (equiv.
τA = τB).
If a cash-flow stream is first-order immune, then the value of V (r0+ ε) is constant (up to
small errors) for small ϵ. 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):
VA(r0 + ε) ≈ VA(r0) + V ′A(r0)ε+
1
2
V ′′A(r0)ε
2 = VA(r0)− VA(r0)νAε+ 1
2
VA(r0)cAε
2
= VA(r0)
(
1− νAε+ 1
2
cAε
2
)
VL(r0 + ε) ≈ VL(r0) + V ′L(r0)ε+
1
2
V ′′L (r0)ε
2 = VL(r0)− VL(r0)νLε+ 1
2
VL(r0)cLε
2
= VL(r0)
(
1− νLε+ 1
2
cLε
2
)
where cA = V
′′
A(r0)/VA(r0) and cL = V
′′
L (r0)/VL(r0) are the convexity of the cash-flow
streams corresponding to the assets and liabilities respectively. Hence in the case of first-order
immunisation,
V (r0 + ε) ≈ VA(r0)
(
1− νAε+ 1
2
cAε
2
)
− VL(r0)
(
1− νLε+ 1
2
cLε
2
)
= VA(r0)
1
2
(cA − cL)ε2
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 cash-flow stream is Reddington immune if it is first-order 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 cash-flow 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 zero-
coupon 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 continuously-
compounded.
(a) Show that Reddington immunisation can be achieved by owning a combination of 2-year
and 7-year zero-coupon bonds.
(b) Explain why first-order immunisation cannot be achieved if the fund owns a combination
of 2-year and 3-year zero-coupon bonds.
Solution. (a) The yearly discount factor is β = e−0.08 = 0.923. Suppose we own 2-year
bonds paying £P and 7-year bonds paying £Q. Zero-order immunisation requires that
Pβ2 +Qβ7 = 20,000β4 + 18,000β6 = 25,661.
22
Moreover, first-order immunisation (i.e. τA = τL) requires that
2Pβ2 + 7Qβ7 = 4× 20,000β4 + 6× 18,000β6 = 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 continuously-compounded, we have
cA =
∑n
i=1 t
2
iCiβ
ti
VA(r)
=
22Pβ2 + 72Qβ7
25,661
= 29.8
and
cL =
∑n
i=1 t
2
iCiβ
ti
VL(r)
=
42 × 20,000β4 + 62 × 18,000β6
25,661
= 24.7.
Since cA > cL, Reddington immunisation is achieved.
(b) The duration of a combination of 2-year and 3-year 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 first-order 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∑
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
V ′(r) = −
n∑
i=1
tiCi
(1 + r)ti+1
and
V ′(r) =
n∑
i=1
ti(ti + 1)Ci
(1 + r)ti+2
.
With β defined as β = 1/(1 + r),
V (r) =
n∑
i=1
Ciβ
ti ,
V ′(r) = −
n∑
i=1
tiCiβ
ti+1,
and
V ′′(r) =
n∑
i=1
ti(ti + 1)Ciβ
ti+2,
23
The duration is
τ =
∑n
i=1 tiCiβ
ti∑n
i=1Ciβ
ti
while the effective duration is
ν = −V
′(r)
V (r)
=
∑n
i=1 tiCiβ
ti+1∑n
i=1Ciβ
ti
.
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 5-year zero-coupon bonds and 10-year zero-coupond bonds.
Solution. Suppose we purchase 5-year bonds paying P and 10-year bonds paying Q. Then,
with β = (1.07)−1,
VA(0.07) = Pβ
5 +Qβ10.
We also have
VL(0.07) = 50, 000(β
6 + β8) = 62, 418
and so zero order immunisation implies that
Pβ5 +Qβ10 = 62, 418. (10)
The effective duration of the assets is
νA =
5Pβ6 + 10Qβ11
62418
while the effective duration fo the liabilities is
νL =
50, 000(6β7 + 8β9)
62418
=
404398
62418
.
Setting νA = νL gives
5Pβ6 + 10Qβ11 = 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
cA =
V ′′A(0.07)
VA(0.07)
=
(5)(6)Pβ7 + (10)(11)Qβ12
62418
= 53.21
while the convexity of the liabilities is
cL =
V ′′L (0.07)
VL(0.07)
=
50000((6)(7)β8 + (8)(9)β10)
62418
= 48.90.
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 fixed-income security – these are tradable financial
instruments whose cash-flow 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
cash-flow is predetermined, the present value of a fixed-income security can fluctuate, since
their value also depends on the prevailing interest rates.
Fixed-income securities can be contrasted with equity, such as shares and equity indices.
These are securities whose cash-flow is not pre-determined. Other financial instruments of
this type include FX (foreign exchange) and commodities, and collectively they are sometimes
known as variable-income securities.
In this chapter we will use present value analysis and no-arbitrage 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 zero-coupon 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 pay-
ment, 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 zero-coupon 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 cash-flow stream generated by a typical bond might be as follows:6
Now
Issue price
1Y
Coupon
2Y
Coupon
3Y
Coupon
8Y
Coupon
9Y
Coupon
10Y
R × Face value + Coupon
Examples 3.1.
1. A 5-year 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 10-year bond with face value $150,000,000 and semi-annual
coupon paid at an annual rate of 2%? The bond will pay 20 coupon payments every
half-year for 10 years, each worth
$150,000,000× 2%× 1
2
= $1,500,000.
3.1.1 The present value of a bond
The present value of a bond is simply equal to the present value of the cash-flow stream that
the bond generates. By default we do not include the issue price in the cash-flow 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 zero-coupon 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
PV =
1
1 + r
C +
1
(1 + r)2
C + · · ·+ 1
(1 + r)T
C +
1
(1 + r)T
RP. (13)
Example 3.4. Suppose the interest rate is 1.5%. The present value of a 3-year bond with
face value £100,000, redeemable at par, with annual coupon payments of 1.7% is
PV =
0.017× 100,000
1 + 0.015
+
0.017× 100,000
(1 + 0.015)2
+
0.017× 100,000
(1 + 0.015)3
+
100,000
(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%.
6This picture reflects the cash-flow 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
PV = D(T1) · C1 +D(T2) · C2 + · · ·+D(Tn) · Cn +D(Tn) ·RP. (14)
3.1.2 The no-arbitrage 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 no-arbitrage principle, or the ’no-free-lunch 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 risk-free 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 non-negative 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 risk-free rate.
Definition 3.8. The no-arbitrage principle is the assumption, made about a financial mar-
ket, that no arbitrage opportunities exist.
The justification for the no-arbitrage 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 no-arbitrage principle may not be reflected in real-life financial markets,
especially those that are illiquid (meaning, in which trades are infrequent).
Along with the no-arbitrage 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 so-called ‘frictionless market’ hypothesis).
It turns out that the no-arbitrage principle (combined with the ‘frictionless market’ hy-
pothesis) is an extremely powerful tool with which to assign prices to financial instruments,
27
such as bonds; this is known as ‘arbitrage-free pricing’ or ‘rational pricing’. We will refer to
the resulting price as the ‘no-arbitrage’ price, or sometimes simply the ‘fair price’.
We begin with the the law of one price, a general expression of the no-arbitrage pricing
principle:
Theorem 3.9 (Law of one price). Under the no-arbitrage assumption, if two financial instru-
ments generate the same cash-flow streams then they have the same current price.
In particular, the current price of a financial instrument that generates a fixed cash-flow
stream is equal to the present value of this cash-flow stream.
Corollary 3.10. The no-arbitrage price of a bond is equal to its present value.
Proof. Suppose that financial instruments A and B generate the same cash-flow 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 cash-flow stream; since you sold
instrument A, you need to pay the buyer the associated cash-flow stream. As these cash-flow
streams are equal, the former can be used to pay off the latter, leaving you with a positive
profit at the expiry of the cash-flow streams. This is an arbitrage opportunity, so PA > PB
must be false. Reversing the roles of A and B, we conclude that
PA = PB.
Now suppose we have an instrument that generates a fixed cash-flow stream. We already
saw that if the present value of the cash-flow stream PV is placed in a bank account, it can be
used to exactly replicate the cash-flow 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 cash-flow 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 zero-coupon bonds (i.e. zero-coupon bonds that pay one unit
of cash at expiry). This is since, by the law of one price, the price VT of a unit zero-coupon
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 no-
arbitrage price, of any bond in terms of the discount factors.
Example 3.11. Calculate the no-arbitrage price of a 4-year bond with face value £200,
redeemable at par, with 4% annual coupons, if the price of unit zero-coupon 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 no-arbitrage price, of the bond is
200× 4%×
4∑
i=1
D(i) + 200×D(4),
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 no-arbitrage 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 no-arbitrage
principle requires that
1 + r1Y =
(
1 +
1
12
r1M
)12
.
First Solution. Assume for the sake of contradiction that
1 + r1Y <
(
1 +
1
12
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
in bank A. At the end of the year the account value would grow to
(
1 + 1
12
r1M
)12
, which after
paying back the loan gives us a profit of((
1 +
1
12
r1M
)12
−
(
1 + r1Y
))
.
Since by assumption this is strictly positive, we have made a risk-free profit which contradicts
the no-arbitrage principle. Therefore inequality (15) cannot hold.
Second Solution. Put 1(
1+ 1
12
r1M
)12 in bank A and 11+r1Y in bank B. Both invest investments
result in a unit pay off after 1 year. By The Law of One Price,
1(
1 + 1
12
r1M
)12 = 11 + r1Y ,
or
1 + r1Y =
(
1 +
1
12
r1M
)12
.
Exercise 3.13. Explain why, in the example above, the reverse inequality
1 + r1Y >
(
1 +
1
12
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 no-arbitrage 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
P0 =
n∑
i=1
Ci
(1 + r)ti
+
1
(1 + r)T
RP. (16)
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 2-year 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
1 + r
+
2,000
(1 + r)2
+
100,000
(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 20-year loans, then the the interest rate for 20-year
deposits will increase. Likewise, if there is a good supply of one month deposits the 1-month
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
7LIBOR 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 short-term interest rates; the total value of
instruments fixed by LIBOR is estimated to be in the hundreds of trillions of pounds.
8Source: 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 LIBOR
Overnight 0.67150%
1 Week 0.68825%
1 Month 0.71225%
2 Month 0.75775%
3 Month 0.78463%
6 Month 0.84050%
1 Year 0.96213%
The fact that interest rates are term-dependent is an additional complication when pricing
fixed-rate financial instruments. In this section we explore how these rates can be analysed.
In particular, we use the price of unit zero-coupon bonds to relate various interest rates,
via the no-arbitrage principle.
3.2.1 Spot rates
The first step in analysing term-dependent 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 term-dependent 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 + rTT ).
A fairer way to compare term-dependent rates is to compare their spot rates:
Definition 3.16. The spot rate, sT , for maturity T is defined as
sT = (1 + rTT )
1/T − 1 (17)
or, equivalently,
(1 + sT )
T = 1 + rTT.
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 1-year spot rate s1 is equal the nominal rate r1.
2. The 2-year spot rate s2 is equal to rate such that, compounding twice, has the same
effect as the nominal 2-year rate r2, i.e.
(1 + s2)
2 = 1 + r2 · 2.
3. The 6-month spot rate s0.5 has the same effect as the 6-month nominal rate compounded
twice, i.e.
1 + s0.5 =
(
1 +
1
2
r0.5
)2
⇔ (1 + s0.5)1/2 = 1 + r0.5 · 1
2
.
31
Example 3.19. Let r1/12 = 0.712% be the 1-month LIBOR rate quoted above. Then the
associated spot rate satisfies
1 + s1/12 =
(
1 + 0.712%
1
12
)12
= 1.00714
which yields a spot rate of sT = 0.714%.
In the case of term-dependent 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 + rTT )
−1 = (1 + sT )−T .
Example 3.20. Calculate the no-arbitrage price of a five-year 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
no-arbitrage price of
6,000
5∑
i=1
D(i) + 100,000D(5) = £92,247.
Another point of view is that, under the no-arbitrage assumption, spot rates are determined
by the price of unit zero-coupon bonds VT with expiry T . This is since
sT = D(T )
−1/T − 1 = V −1/TT − 1. (18)
Examples 3.21. The following table gives some prices for unit zero-coupon bonds with varying
maturities, alongside the implied spot rate calculated via the formula (18):
T =Year Price Spot rate sn
1 0.94 6.4%
5 0.70 7.4%
10 0.47 7.8%
15 0.30 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 s
′
T that is the equivalent continuously-
compounded rate. This satisfies
es
′
TT = 1 + rTT ⇔ s′T =
1
T
ln (1 + rTT ) .
3.2.2 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. no-arbitrage) 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 1-year loan for (1+s1)−1× £1m. Place this amount in a 2-year 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 no-arbitrage 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. no-arbitrage) interest rate for a FRA from time n1 years to
n2 years is
fn1,n2 =
((1 + sn2)n2
(1 + sn1)
n1
) 1
n2−n1 − 1,
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 no-arbitrage 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 =
(1 + sn2)
n2
(1 + sn1)
n1
,
33
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 risk-free 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 2-year LIBOR rate is r2 = 1.43% and the 5-year 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 + rTT )
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 + s5)
5
(1 + s2)2
=
1.0990
1.0286
= 1.0684,
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 zero-coupon bond prices with expiries n1 and n2 is given by
Vn1
Vn2
=
(1 + sn2)
n2
(1 + sn1)
n1
,
and therefore
Vn1
Vn2
= (1 + fn1,n2)
n2−n1 .
It follows that, just as for discount rates and bond prices, forward rates are also completely
determined by the prices of unit zero-coupon bonds.
Example 3.26. Suppose the price of 3-year and 5-year year unit zero-coupon bonds are 0.92
and 0.65 respectively. Calculate the 2-year forward rate f3,5 from year 2 to year 5.
Proof. The ratio of the zero-coupon bond prices satisfies
V3
V5
= (1 + f3,5)
5−3
and so
f3,5 =
(0.92
0.65
)1/2
− 1 = 18.97%.
34
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∏
i=1
(1 + fni,ni+1)
ni+1−ni =
j−1∏
i=1
(1 + sni+1)
ni+1
(1 + sni)
ni
=
(1 + snj)
nj
(1 + sn1)
n1
= (1 + fn1,nj)
nj−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 2-year deposit between T = 2 and T = 4, then in a 3-year deposit between
T = 4 and T = 7 (top of diagram below).
2 In a 5-year deposit between T = 2 and T = 7 (bottom of diagram below);
T=2 T=3 T=4 T=5 Y=6 T=7
(1 + f2,7)
5
(1 + f2,4)
2 × (1 + f4,7)3
Used in combination, these relationships can help calculate unknown forward or spot rates.
Example 3.27. Suppose the 3-year and 7-year year spot rates are 6% and 5% respectively,
and that the 3-year 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
1.063 × 1.0523 = 1.0148 =⇒ f3,4 = 1.48%.
T=0 T=1 T=2 T=3 T=4 T=5 Y=6 T=7
(1 + s7)
7
(1 + s3)
3 × (1 + f3,4) × (1 + f4,7)3
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 20-year 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
PE((1 +R)n) ≥ P (1 + E(R))n.
Example 4.1. Suppose the interest rate R has probability distribution
R =

0.06 with probability 0.2,
0.08 with probability 0.7,
0.10 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(S25)− 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 Exam-
ple 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(1 + E(R))
=
ln 2
ln(1 + 0.078)
= 9.23 years.
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 PSn. 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 PSn is
E (PSn) = PE
( n−1∏
i=0
(1 +Ri)
)
= P
n−1∏
i=0
E (1 +Ri) = P
n−1∏
i=0
(1 + E(Ri)),
with the second equality holding since we assume Ri are independent.
In order to calculate Var(PSn), we use the formula
Var(PSn) = P
2
(
E(S2n)− (E (Sn))2
)
.
We calculate E(S2n) as follows:
E
(
S2n
)
= E
( n−1∏
i=0
(1 +Ri)
2
)
=
n−1∏
i=0
E
(
(1 +Ri)
2
)
=
n−1∏
i=0
E
(
1 + 2Ri +R
2
i
)
=
n−1∏
i=0
(
1 + 2E(Ri) + E(R2i )
)
=
n−1∏
i=0
(
1 + 2E(Ri) + (E(Ri))2 +Var(Ri)
)
=
n−1∏
i=0
(
(1 + E(Ri))2 +Var(Ri)
)
where we again used the independence of the Ri, as well as the formula
E(R2i ) = (E(Ri))2 +Var(Ri).
Combining with our previous calculation of E(Sn),
Var(PSn) = P
2
(
E(S2n)− (E (Sn))2
)
= P 2
(
n−1∏
i=0
(
(1 + E(Ri))2 +Var(Ri)
)− n−1∏
i=0
(1 + E(Ri))2
)
.
If the interest rates Ri are also identically distributed with common mean µ and variance σ
2
then
E (PSn) = P
n−1∏
i=0
(1 + µ) = P (1 + µ)n
and
Var(PSn) = P
2
((
(1 + µ)2 + σ2
)n − (1 + µ)2n) . (19)
Remark 4.3. In the case that the interest rates are deterministic, we would expect that the
deposit value PSn is also deterministic. This is indeed the case, since setting σ
2 = 0 in the
formula (19) yields
Var(PSn) = P
2
((
(1 + µ)2 + σ2
)n − (1 + µ)2n) = P 2 (((1 + µ)2)n − (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 right-hand 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 σ2 = Var(Ri) via
σ2 = E(R2i )− 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) = Pn
In other words, if the expected interest rate is zero, the expected value of a fund where we
deposit £P every year is £Pn.
Solution. In the case µ = 0 the recursive formula reduces to
E(An) = E(An−1) + P,
which shows that E(An) = Pn by induction.
Remark 4.7. Note that in the case that the interest rate was zero (i.e. µ = 0 and σ2 = 0),
then An = Pn, so certainly E(An) = Pn 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 Log-normally distributed interest rates
In practice, the easiest stochastic interest rate model to work with is usually the log-normal
model, in which the interest rates are assumed to follow a log-normal process. This model
will appear again in the course when we discuss share prices.
A continuous random variable Y is said to be log-normally distributed with parameters
µ and σ2, written Y ∼ LogNormal(µ, σ2), if
lnY ∼ N (µ, σ2).
Equivalently, Y = eX , where X ∼ N (µ, σ2). The p.d.f. of Y is
fY (y) =
{
1√
2πσy
exp
(
− (ln y−µ)2
2σ2
)
, if y > 0,
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 log-normally 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 log-normal distributions behave well under multiplication. Sup-
pose that Y1 ∼ LogNormal(µ1, σ21) and Y2 ∼ LogNormal(µ2, σ22) are independent. Then
we can write Y1 = e
X1 and Y2 = e
X2 , where X1 ∼ N (µ1, σ21) and X2 ∼ N (µ2, σ22) are
independent. Therefore,
Y1Y2 = e
X1eX2 = eX1+X2 ,
and since
X1 +X2 ∼ N (µ1 + µ2, σ21 + σ22),
it follows that
Y1Y2 ∼ LogNormal(µ1 + µ2, σ21 + 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, σ2). Then
Sn =
n−1∏
i=0
(1 +Ri) ∼ LogNormal(nµ, nσ2).
Using the above theorem, the c.d.f. of Sn can be computed as
P(Sn ≤ x) = P(eX ≤ x) = P(X ≤ lnx) = P
(X − nµ√
nσ
≤ lnx− nµ√
nσ
)
where X ∼ N (nµ, nσ2). Therefore
P(Sn ≤ x) = Φ
( lnx− 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 log-normally
distributed with parameters µ = 0.075 and σ2 = 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, 5σ2).
The probability that An > 1500 is
P(An > 1500) = P(LogNormal(5µ, 5σ2) > 1.5) = 1− Φ
( ln 1.5− 5µ√
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
0.75 = Φ
(
ln(u/1000)− 5µ√
5σ
)
.
Looking up the normal c.d.f. we find that Φ(0.6745) = 0.75, so therefore
ln(u/1000)− 5µ√
5σ
= 0.6745,
that is,
u = 1000 exp
(
5µ+ 0.6745σ
√
5
)
= £1,510 .
Similarly, the lower quartile is
ℓ = 1000 exp
(
5µ− 0.6745σ
√
5
)
= £1,400 .
41
5 Equities and their derivatives
In previous chapters we considered fixed-income securities. Equities and derivatives differ
from fixed-income securities in that their cash-flow stream is not pre-determined, 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 over-the-counter (off-exchange). 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
4,520,000,000
= 0.0000000221%
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 non-dividend 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
short-selling, 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.
9These 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 no-arbitrage 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 no-
arbitrage 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 zero-coupon bonds with face-value 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 zero-coupon
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 continuously-compounded 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
1.13
= 885,
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
F =
1,000
1.13
( 1 + r£
1 + r€
)T
=
1,000
1.13
(1.005
1.01
)1/4
= £883.9.
44
Here we see that the relevant discount factor was
D(T ) =
( 1 + r£
1 + r€
)−T
,
which implies that value is ‘discounted’ with respect to the ‘currency adjusted’ rate
1 + r£
1 + r€
− 1 = r£ − r€
1 + r€
= −0.495%. (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 over-the-counter
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 Euro-
pean options (i.e., with the same expiry T and strike K), since they offer their holder more
optionality.
45
Exercise 5.9. Use a no-arbitrage 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 be bought for multiple shares simultaneously. For simplicity,
we will normally assume that the option refers to a single share.
Let S(t) denote the price of a share at time t. The owner of a European call option has only
one choice to make: at expiry T they must decide whether to exercise the option. A moment’s
reflection shows that:
• If S(T ) > K (known as the stock being ‘in-the-money’), then it is beneficial for the
owner to exercise the right, buy the stock, and make a profit by selling it immediately
on the market. This will yield a profit of S(T )−K.
• If S(T ) < K (known as the stock being ‘out-of-the-money’), then the owner will not
exercise the option because the stock can be bought cheaper on the market. This means
that the the investor earns zero from the option in this case.
Summarising this discussion, the payoff (another word for ‘return’) of a European call option
is equal to
Payoff(S(T )) =
{
0, if S(T ) < K,
(S(T )−K), if S(T ) ≥ K.
Using the function x+ defined by
x+ =
{
0, if x < 0,
x, if x ≥ 0,
the payoff can also be written as
Payoff(S(T )) = (S(T )−K)+.
Here is a plot of the payoff as a function of S(T ):
S(T )
Payoff
K
(S(T )−K)+
Exercise 5.11. Show that the payoff of a European put option is equal to (K − S(T ))+, and
plot its graph as a function of S(T ).
46
5.4 Pricing options: model-independent pricing principles
Despite the fact that options contracts have been traded for many centuries, assigning a fair
(i.e. no-arbitrage) price to an option is a very challenging task, and is the subject of much
ongoing research.
The simple reason for this complication is that, unlike for bonds or forwards, the no-
arbitrage price of an option depends on the model that is used to model the underlying asset.
A major milestone in the field of option pricing was the approach introduced by F. Black
and M.S. Scholes in 1973,10 for which M.S. Scholes was ultimately awarded the 1997 Nobel
Prize in Economics (along with R. Merton; F. Black had passed away by that point). We will
study the Black-Scholes approach later in this course, and in much more detail in Financial
Mathematics II. In the aftermath of Black and Scholes’ work, trading and academic interest
in options increased significantly, and nowadays trillions of pounds’ worth of options are traded
worldwide each year.11
Nevertheless, there are certain general model-independent pricing principles that are
simple to deduce from no-arbitrage arguments; these principles are the focus of this section.
Our first example is an important relationship between the prices of a European call and
put options with the same strike and expiry, known as put-call parity:
Theorem 5.12 (Put-call parity). Let C be the price of a European call option on a share
with strike price K and expiry T . Let P be the price of a European put option on the same
stock with the same strike and expiry. Suppose the underlying share has current price S(0).
Then
P + S(0) = C +K ·D(T )
where D(T ) is the discount factor at time T .
Proof. We argue using replicating portfolios. Consider the following two portfolios, both
formed at time 0, consisting of:
• Portfolio A. One share and one put option with strike K and expiry T .
• Portfolio B. One call option with strike K and expiry T , and the amount KD(T )
deposited in the bank.
Let S(T ) denote the price of the stock at time T . Then at time T , Portfolio A is worth
S(T ) + (K − S(T ))+,
whereas Portfolio B is worth
(S(T )−K)+ +KD(T )/D(T ) = (S(T )−K)+ +K.
Notice that
S(T ) + (K − S(T ))+ = max{S(T ), K} = (S(T )−K)+ +K
10F. Black and M.S. Scholes (1973). The Pricing of Options and Corporate Liabilities. Journal of Political
Economy, 81 (3), pp. 637–654.
11The overall derivative market is estimated to be worth in the quadrillions of pounds, more than the entire
worldwide GDP.
47
so the two portfolios have the same value at time T . Hence, by the law of one price, they also
have the same value at time 0. Since at time 0 the value of Portfolio A and Portfolio B are
respectively
S(0) + P and C +KD(T ),
we deduce that
S(0) + P = C +KD(T ).
We next show that it is never worth exercising an American call option early:
Theorem 5.13. An American call option should never be exercised before the expiry time.
Remarks 5.14. The above theorem has some important caveats:
• It only applies to American call options, and does not apply to American put options,
which in some circumstances should be exercised early.
• It only applies to non-dividend paying shares. If a share pays a dividend, then again in
some circumstances it is worth exercising an American call option early.
• It only applies when interest rates are positive.
Proof. Suppose you exercised the call option early, i.e. at some time t0 < T such that S(t0) >
K (if S(t0) ≤ K there would be no point exercising early). Then the payoff you receive at t0
is S(t0)−K. If you then deposit this money in the bank, your payoff at time T would be
(S(t0)−K)/D,
where D = D(T )/D(t0) is the discount factor that applies between t0 and T .
However, it turns out that there exists an alternative strategy that is more profitable than
exercising the option, namely, (short-)selling the share to receive S(t0), and depositing this
money in the bank. At time T the payoff of this alternative strategy is the sum of:
• The money in the bank, which has grown to S(t0)/D;
• The share you owe from the short-selling, which is worth S(T );
• The payoff of the option, which is (S(T )−K)+.
In other words, the total payoff of this strategy is
S(t0)/D − S(T ) + (S(T )−K)+.
It remains to argue that
S(t0)
D
− S(T ) + (S(T )−K)+ ≥ S(t0)−K
D
.
To show this, we first observe that
−S(T ) + (S(T )−K)+ =
{
−S(T ) + (S(T )−K) = −K if S(T ) ≥ K
−S(T ) + 0 ≥ −K if S(T ) < K
}
≥ −K,
so
S(t0)
D
− S(T ) + (S(T )−K)+ ≥ S(t0)
D
−K.
Then since D ≤ 1 (at least, if interest rates are positive),
S(t0)
D
−K ≥ S(t0)
D
− K
D
=
S(t0)−K
D
,
which proves the claim.
Exercise 5.15. What goes wrong if the argument is attempted for American put options?
48
5.5 Pricing options via the risk-neutral distribution
To develop model-independent pricing principles further we need to introduce a new kind of
no-arbitrage argument based on the risk-neutral distribution.
5.5.1 Betting strategies, the risk-neutral distribution, and the arbitrage theorem
Buying an option can be viewed as a kind of bet on the outcome of the financial markets, and
the general setting of betting strategies is a useful way to introduce the theory of risk-neutral
distributions.
Suppose there is an event (e.g. an experiment, sports event, financial outcome etc.) with
m different mutually exclusive outcomes {1, 2, . . . ,m}. A wager is a type of investment
such that, if you pay an initial amount P (called a bet), there is a function r(j), called the
return function, such that if the outcome of the event is j you are returned an amount
Pfinal(j) = P × r(j). The profit from such a transaction is
Pfinal(j)− P = P (r(j)− 1).
Example 5.16. Suppose that you buy 2 barrels of oil at the current price $50, and tomorrow
the oil price will either rise to $60 a barrel or fall to $20 a barrel. This transaction can be
viewed as a bet of 2× $50 = $100 on a wager with return function
r(price rises) =
60
50
= 1.2 , r(price falls) =
20
50
= 0.4.
Example 5.17. For example, consider the upcoming England vs Montenegro football match,
for which a bookmaker is quoting the following odds:
Outcome Odds
England win 1/16
Draw 14/1
Montenegro win 40/1
This this means that the bookmaker is offering three wagers rE, rD and rM with respective
return functions
rE(E wins) = £1.06 rD(E wins) = £0 rM(E wins) = £0
rE(Draw) = £0 rD(Draw) = £15 rM(Draw) = £0
rE(M wins) = £0 rD(M wins) = £0 rM(M wins) = £41
For instance, if you bet £2 on England winning and the match was a draw you would be
returned £0 so your profit would be −£2.
In general, if a bookmaker offers odds oi on outcome i this means they are offering the
wagers with return functions
ri(j) =
{
1 + oi if j = i,
0 if j ̸= i.
Let’s now suppose that there exists n different wagers available. A betting strategy is a
vector x = (x1, . . . , xn) in which xi is bet on wager i; these numbers can be negative, which
we interpret as offering (‘selling’) a bet to another person. Notice that the profit from the
betting strategy x when the outcome of the event is j is given by
Px(j) =
n∑
i=1
xi (ri(j)− 1).
49
From the point of view of the investor, it is clearly of interest whether there exists a
betting strategy that guarantees a profit no matter the outcome of the event (i.e. there is
an arbitrage opportunity). Let’s consider this question in some simple examples:
Example 5.18. Consider an NBA match12 between the Atlanta Hawks (A) and the Boston
Celtics (B), and a bookmaker quotes you odds of
Outcome Odds
Atlanta win 5/2
Boston Celtics win 2/3
Is there a betting strategy that guarantees a profit?
Solution. We seek x and y such that betting x on outcome A and y on outcome B leads to
a guaranteed profit no matter who wins. The profit from such a bet is
Px,y =
{
5
2
x− y if A wins,
2
3
y − x if B wins.
In order to make a guaranteed profit, it is enough that
5x/2− y > 0 and 2y/3− x > 0,
or equivalently
1.5x < y < 2.5x.
Hence for any x > 0 there is an interval (1.5x, 2.5x) from which we can choose y in order
to guarantee a profit. For example, we could choose x = £1 and y = £2, in which case the
profit is
P1,2 =
{
5
2
− 2 = 1
2
if A wins,
2
3
· 2− 1 = 1
3
if B wins.
Example 5.19. Consider the same set-up as in the previous example, except for the odds that
B wins have been reduced to 2/5. Is there still a betting strategy that guarantees a profit?
Solution. Now the profit is
Px,y =
{
5
2
x− y if A wins,
2
5
y − x if B wins,
and so a guaranteed profit is only possible if we can find x and y such that
2.5x− y ≥ 0 and 0.4y − x ≥ 0
and at least one of these inequalities is strict. Since this is equivalent to
2.5x ≥ y and y ≥ 2.5x
finding such an x and y is impossible.
We next state an important theorem in financial mathematics, often called simply ‘the
arbitrage theorem’, states that exactly one of the following is true:
12We consider basketball for the simple reason that there are no draws!
50
• Either there exists a betting strategy that generates a profit irrespective of the outcome
of the experiment (i.e. there is an arbitrage opportunity);
• Or, we can prescribe specific probabilities (called the ‘risk-neutral distribution’) to the
possible outcomes in such a way that the expected profit on any betting strategy will
be zero (in which case we say that all bets are fair).
Let us state this theorem more formally (we will not discuss the proof):
Theorem 5.20 (The arbitrage theorem). Suppose there is an event with m outcomes and
there are n possible wagers on these outcomes with respective return functions ri. Then
exactly one of the following two statements is true:
(i) There exists a betting strategy x = (x1, . . . , xn) such that the profit satisfies
Px(j) ≥ 0 for all j = 1, . . . , m
and
Px(j) > 0 for at least one j = 1, . . . , m;
(ii) There exists a probability vector p˜ = (p˜1, . . . , p˜m) over the comes, called the ‘risk-neutral
distribution’, such that p˜j > 0 for all j = 1, . . . , m, and
m∑
j=1
p˜j (ri(j)− 1) = 0 for all i = 1, . . . , n.
The up-shot of the arbitrage theorem is that, under the no-arbitrage principle, we are
guaranteed the existence of a ‘risk-neutral distribution’. This will be crucial in developing
the pricing principles in the following subsection.
Remark 5.21. One important distinction to make is that the ‘risk-neutral distribution’ does
not reflect anything about ‘real-world’ probabilities. Instead the ‘risk-neutral distribution’
should be thought of as artificial construct.
Remark 5.22. In general the risk-neutral distribution guaranteed by the arbitrage theorem
may not be unique, which poses problems for no-arbitrage pricing theory. Nevertheless, in
all the models that we consider in this course the risk-neutral distribution will be unique.
Example 5.23. Suppose there is an event with two outcomes, and there are two wagers with
return functions r1(1) = 2, r1(2) = −1, r2(1) = −1, and r2(2) = 5. Does there exist a
betting strategy that guarantees a profit?
Solution. We shall try to find a risk-neutral distribution, i.e. numbers p˜1, p˜2 > 0 with p˜1+ p˜2 =
1 such that the following system of linear equations
p˜1(r1(1)− 1) + p˜2(r1(2)− 1) = 0
p˜1(r2(1)− 1) + p˜2(r2(2)− 1) = 0
holds, that is,
p˜1 − 2p˜2 = 0
−2p˜1 + 4p˜2 = 0.
Note that, the second equation is a multiple of the first, thus p˜1 = 2p˜2. Using p˜1 + p˜2 = 1
now yields
2p˜2 + p˜2 = 1.
51
Thus
p˜1 =
2
3
and p˜2 =
1
3
is a risk-neutral distribution. Therefore, the arbitrage theorem implies that there is no betting
strategy that guarantees a profit.
Assume that, for a event with n outcomes, a bookmaker quotes odds of o1,. . . , on. What
can we say about when these odds permit a betting strategy that guarantees a profit?
Proposition 5.24. Define the numbers p˜i = 1/(1 + oi). Then the odds oi do not allow for
an arbitrage opportunity if and only if the numbers p˜i are a valid probability distribution on
the set of outcomes.
Proof. Recall that a bet on outcome i has return function
ri(j) =
{
1 + oi if j = i,
0 if j ̸= i.
Let (p1,. . . , pn) be a probability distribution over the outcomes. If this is a risk-neutral
distribution, then the expected profit on a bet on outcome i is equal to zero, i.e.,
0 =
∑
j
pj(ri(j)− 1) = pi × oi +
∑
j ̸=i
pj × (−1) = pioi − (1− pi) = pi(oi + 1)− 1,
which implies that
pi = 1/(1 + oi) = p˜i,
i.e. this distribution is ‘risk-neutral’ if and only if pi = p˜i for all i. Hence by Theorem 5.20
either (i) p˜i is a valid probability distribution and is also the (unique) risk-neutral distribution,
or else (ii) p˜i is not a valid probability distribution, in which case no ‘risk-neutral’ distribution
can exist.
Exercise 5.25. Let us recall the odds from the England vs Montenegro football match:
Outcome Odds
England win 1/16
Draw 14/1
Montenegro win 40/1
Does there exist an arbitrage opportunity?
Solution. We calculate p˜i = 1/(1 + oi):
p˜E =
1
1 + 1/16
= 0.941, p˜D =
1
1 + 14
= 0.0667, p˜M =
1
1 + 40
= 0.0244.
Since these numbers sum to 1.03 they do not define a valid probability distribution, there is
no risk-neutral distribution and so there does exist an arbitrage opportunity.
Remark 5.26. In the previous example we argued that there must exist a betting strategy
that guarantees a risk-free profit. Before you get excited, remember than in general such a
betting strategy will involve offering (‘selling’) bets as well as placing (‘buying’) them, and in
reality the only person likely to be able to exploit this arbitrage opportunity is the bookmaker.
Note that gambling does not satisfy the ‘frictionless market hypothesis’ !
52
5.5.2 Options pricing using the risk-neutral distribution
Recall that the no-arbitrage principle guarantees the existence of a probability distribution on
the set of outcomes, called the risk-neutral distribution, such that the expected profit of
any betting strategy is zero.
The purchase of a share can be interpreted as a betting strategy, since we are in effect
betting on the future value of the share. Therefore we deduce the following important principle:
Proposition 5.27. Under the no-arbitrage principle, there exists a probability distribution on
the future price of share, called the risk-neutral distribution, under which the expected profit
on the purchase of a share is zero.
The risk-neutral distribution will be denoted P˜, to distinguish it from the real-world prob-
abilities P. The corresponding expectation will be denoted E˜.
Remark 5.28. Our argument for the existence of the risk-neutral distribution required there
be only a finite number of outcomes of the experiment, whereas in reality a share price may
have an infinite number of future possible values. We will not dwell on this subtlety.
Let us analyse the implications of this principle by considering the expected profit on the
purchase of a share. To buy a share at price S(0) we need to borrow this amount to fund the
purchase. At a later time T we will posses the share, but the amount that we owe has grown
to S(0)/D(T ), where D(T ) is the discount factor to time T . Hence this transaction yields a
profit of
S(T )− S(0)/D(T ).
By the definition of the risk-neutral measure, the expected profit must be 0, so we conclude
that
0 = E˜
(
S(T )− S(0)/D(T )) = E˜(S(T ))− S(0)/D(T ).
This leads us the following important result:
Theorem 5.29. Under the no-arbitrage principle, the current price of a share is equal to its
(discounted) expected future price under the risk-neutral distribution, i.e.
S(0) = D(T )E˜(S(T )).
The above argument generalises to the purchase of any asset, and in the case of European
options leads to a general model-independent pricing formula for these options (the case of
American options is more complicated):
Theorem 5.30. Under the no-arbitrage principle, the current price of a European option is
equal to the (discounted) expected value of its payoff under the risk-neutral distribution.
Remark 5.31. We stress that the expected value of the payoff is being calculated under the
risk-neutral distribution, which does not necessarily equal the real-world probabilities. In fact,
real-world probabilities are irrelevant for the no-arbitrage pricing of financial instruments.
Example 5.32. Assume interest is continuously-compounded at nominal rate r. Then the
no-arbitrage price of a European call option with strike K and expiry T is
e−rT E˜
[
(S(T )−K)+] ,
where E˜ indicates that the expectation is calculated under the risk-neutral measure. Similarly,
the no-arbitrage price of a European put option is
e−rT E˜
[
(K − S(T ))+] .
53
We reiterate that the above formulas are model-independent. However, in the absence
of a model, these formulas cannot be used to compute an actual numerical ‘price’ of an
option. In order to compute a numerical price, we need to specify a model (or a risk-neutral
distribution) for the share price S(T ). Let’s see some simple examples:
Example 5.33. Assume interest is compounded annually at rate 2%. Suppose that a stock
has a risk-neutral distribution at T = 3 given by
S(3) =

20 with probability 0.3,
45 with probability 0.5,
60 with probability 0.2.
Calculate the no-arbitrage price of a European put option with strike 50 and expiry T = 3.
Solution. The price of the put is given by
P =
( 1
1.02
)3
E˜[(50− S(3))+]
where E˜ denotes expectation under the risk-neutral distribution. Hence
P =
( 1
1.02
)3(
(50− 20)+ × 0.3 + (50− 45)+ × 0.5 + (50− 60)+ × 0.2)
=
( 1
1.02
)3(
30× 0.3 + 5× 0.5) = 10.8.
Example 5.34. Assume interest is continuously-compounded at rate 3%. Suppose that a
stock has a uniform risk-neutral distribution at T = 5
S(5) ∼ U [20, 40],
Calculate the no-arbitrage price of a European call option on this stock with strike K = 35
and expiry T = 5.
Solution. The price of the call is C = e−0.03×5E˜[(S(5)− 35)+] where E˜ denotes expectation
under the risk-neutral distribution. Hence
Ce−0.03×5
∫ 40
20
1
20
(x− 35)+dx.
To calculate the integral it is simplest to split the domains into the regions (20, 35) and
(35, 40). On the domain (20, 35) the integrand is 0 (since the option is ‘out-of-the-money’),
so the integral contributes nothing. Hence we are left with
C = e−0.03×5
∫ 40
35
1
20
(x− 35)dx = e
−0.03×5
20
(
402/2− 352/2− 35× 5) = 0.538.
One useful feature of the general options pricing formula is that it can be used to infer
how the price depends on the option parameters, such as the strike price K.
Proposition 5.35. The price of a European call option is a decreasing function of the strike.
Proof. Since the payoff (S(T )−K)+ is a decreasing function of the strike K, so is the price
of the call option C = D(T ) E˜
[
(S(T )−K)+].
Exercise 5.36. Prove that the price of a European put option is a increasing function of K.
Exercise 5.37. Reprove the put-call parity relation using risk-neutral pricing formulae.
The pricing of American options is significantly more challenging than European options,
because we need to take into account all possible exercise strategies (i.e. whether to exercise
the option or not). We will briefly analyse American options in the next section.
54
6 The binomial model
In the previous chapter we showed that the no-arbitrage price of a European option is equal to
the discounted expected value of its payoff under the risk-neutral distribution. Unfortunately,
this does not tell us how to actually calculate the no-arbitrage price, since in order to determine
the risk-neutral distribution we will usually need to specify a model for the underlying share.
In this chapter we study the simplest model of a share in which the share price is assumed
to move exactly once per discrete time-period to one of two possible future states (called
the ‘up’ state and the ‘down’ state). This is known as the binomial model (‘bi’=‘two’).
Even though this is a crude model of a share, it is widely used by practitioners. The
reason is that the length of the discrete time-periods can be chosen to be very small (e.g. a
millisecond), which makes the model more powerful that it first appears. The downside is that
doing computations over many tiny time-periods is complex (necessitating computer aide).
6.1 The single-period binomial model
Let S(0) = S denote the current price of a share, and assume that price of the share at the
end of a single time-period is either S(1) = S u or S(1) = S d, where u and d are positive
real numbers satisfying 0 < d < u; we call the outcome S(1) = S u the ‘up’ state, and
the outcome S(1) = S d the ‘down’ state. This model is called the single-period binomial
model, and is illustrated in Figure 1.
r




>
Z
Z
Z
Z
Z
Z
Z
Z~
S
S u
S d
-
time
r r
0 1
Figure 1: A share price in the single-period binomial model
In order to fully specify the binomial model, we also introduce the nominal interest rate
r per time-period; this means that the discount factor to the end of the single time-period is
D(1) = 1/(1+ r) in the case of interest compounded once per time-period (‘simple interest’),
or D(1) = e−r in the case of continuously-compounded interest.
Remark 6.1. Sometimes the binomial model is specified by giving the values of S = S(0) and
the values of S(1) in the ‘up’ and ‘down’ state, rather than giving the multiplication factors
u and d. It is easy to convert between these specifications.
6.1.1 The no-arbitrage condition and the risk-neutral distribution
An important question that must be asked of any financial model is whether the no-arbitrage
principle applies (which could depend on the model parameters).
55
Proposition 6.2. The no-arbitrage principle is valid in the single-period binomial model if
and only if
d < 1/D(1) < u,
where D(1) = 1/(1 + r) in the case of simple interest and D(1) = e−r in the case of
continuously-compounded interest. If it exists, the risk-neutral distribution for the share price
at the end of the time-period is
S(1) =
{
uS with probability 1/D(1)−d
u−d ,
dS with probability u−1/D(1)
u−d .
Proof. We analyse the model in terms of betting strategies. There is only one possible wager
in the model: at time 0 you buy the share (with borrowed money), and then sell the share
at t = 1 (you could also short-sell the share, but really this is just the ‘negation’ of the first
wager). As we saw in the previous chapter, in order for this bet to be ‘fair’ the risk-neutral
distribution P˜ (if it exists) over S(1) must satisfy
S = D(1)E˜[S(1)] = D(1)(p˜u × uS + p˜d × dS),
where (p˜u, p˜d) are the risk-neutral probabilities of the ‘up’ and ‘down’ outcomes. Dividing
through by S, and since p˜u + p˜d = 1 (these define a probability distribution!), the above is
equivalent to
1 = D(1)
(
p˜uu+ (1− p˜u)d
)
=⇒ p˜u = 1/D(1)− d
u− d ,
and so
p˜d = 1− p˜u = 1− 1/D(1)− d
u− d =
u− 1/D(1)
u− d .
We conclude that the risk-neutral distribution, if it exists, must equal
p˜u =
1/D(1)− d
u− d and p˜d =
u− 1/D(1)
u− d .
To be a valid risk-neutral distribution, we also require that p˜u > 0 and p˜d > 0. The condition
p˜u > 0 implies that
1/D(1)− d > 0 =⇒ d < 1/D(1),
whereas the condition p˜d > 0 implies that
u− 1/D(1) > 0 =⇒ 1/D(1) < u.
Hence we conclude that the condition for the existence of a valid risk-neutral distribution, and
also the condition for no-arbitrage, is d < 1/D(1) < u.
Exercise 6.3. Prove Proposition 6.2 directly by showing that, if either 1/D(1) < d or
1/D(1) > u, then an arbitrage opportunity arises.
Exercise 6.4. Consider a single-period binomial model such that the current price of the
stock is S = S(0) = 40, and the price of the stock is either S(1) = 42 in the ‘up’ state or
S(1) = 30 in the ‘down’ state. The interest rate is r = 10% per time-period. Argue that
there is arbitrage in this model.
Solution. We have
u =
S(1) in ‘up’ state
S(0)
=
42
40
= 1.05 and d =
S(1) in ‘down’ state
S(0)
=
30
40
= 3/4.
Hence the condition for no-arbitrage is
d < 1 + r < u⇔ 0.75 < 1.1 < 1.05,
which is false, and so there is arbitrage in the model.
56
6.1.2 Option pricing in the single-period binomial model
Assume that a share price follows the single-period binomial model. What is the ‘fair’ price of
a European call option written on this share with expiry at the end of the single time-period?
To assign a fair price we must assume that the no-arbitrage principle holds. For simplicity
let us work with simple interest, in which case the condition for no-arbitrage is
d < 1 + r < u.
By the general no-arbitrage pricing principle in Theorem 5.30, the no-arbitrage price of the
call option C is
C =
1
1 + r
E˜((S(1)−K)+).
Evaluating the expectation with respect to the risk-neutral measure
p˜u =
1 + r − d
u− d and p˜u =
u− (1 + r)
u− d ,
yields the pricing formula
C =
1
1 + r
(
(Su−K)+1 + r − d
u− d + (Sd−K)
+u− (1 + r)
u− d
)
.
Exercise 6.5. Find the corresponding no-arbitrage price P for a European put option with
strike price K, and corresponding formulae in the case of continuously-compounded interest.
Remark 6.6. It will usually be the case that the strike price K satisfies
Sd < K < Su,
since otherwise either (i) K ≥ Su, in which case the option is always out-of-the money and
so is worthless, or (ii) K ≤ Sd, in which case the option is always in-the-money, and so is
equivalent to a portfolio consisting of a single share S minus K units of cash, and so is not
an interesting financial product. In this case the formula simplifies to
C =
(1 + r − d)(Su−K)
(1 + r)(u− d) .
Example 6.7. Consider a single-period binomial model with parameters S = S(0) = 40,
u = 3/2, d = 3/4 and continuously-compounded interest rate r = 5%. Verify the no-arbitrage
principle holds and compute the price C of a European call option with strike K = 45.
Solution. The no-arbitrage principle holds since
d < er < u⇔ 0.75 < 1.051 < 1.5.
and the risk-neutral distribution is
p˜u =
e0.05 − 0.75
1.5− 0.75 = 0.402 and p˜d = 1− p˜u = 0.598.
Since the payoff of the option is
Payoff =
{
(40× 3/2− 45)+ = 15 if S(1) = Su
(40× 3/4− 45)+ = 0 if S(1) = Sd ,
we calculate the no-arbitrage price of the option to be
C = e−0.05 (0.40169× 15 + 0.59831× 0) = 5.73.
57
Example 6.8. With the model as in the previous example, find the no-arbitrage price of a
European put option with strike price K = 45.
Solution. We could price the put option directly, but a simpler way is to use ‘put-call parity’
(since we have already computed the price of the corresponding call option).
By put-call parity, the no-arbitrage price P of the put option is equal to
P = C − S +KD(T ),
where S = 40, K = 45, D(T ) = e−0.05, and the price of the call option was previously
computed to be C = 5.73. Evaluating P , we see that
P = 5.73− 40 + 45× e−0.05 = 8.54
The same argument that we used to price European call options can be extended to price
any European derivative, i.e. any instrument that generates a pay-off at t = 1 depending on
the value of the stock. E.g. consider a European derivative with payoff
V (1) =
{
Vu(1) if S(1) = Su,
Vd(1) if S(1) = Sd.
Then by Theorem 5.30 the no-arbitrage price of this option is
V (0) = D(1) E˜(V (1)) = D(1)
(
p˜uVu(1) + p˜dVd(1)
)
.
Example 6.9. Consider a single-period binomial model with parameters S = S(0) = 60,
u = 2, d = 2/3 and nominal interest rate r = 4%. Verify that the no-arbitrage principle holds
and compute the no-arbitrage price of a European derivative with payoff
V (1) =
{
40 if S(1) = Su,
20 if S(1) = Sd.
Solution. The no-arbitrage principle holds since
d < 1 + r < u⇔ 2/3 < 1.04 < 2,
and the risk-neutral distribution is
p˜u =
1.04− 2/3
2− 2/3 = 0.28 and p˜d = 1− p˜u = 0.72.
Hence the no-arbitrage price of the option is
V (0) =
1
1.04
(0.28× 40 + 0.72× 20) = 24.6.
So far we have only considered European options. For American options the pricing for-
mulae have to be adapted slightly to take in account the ability to exercise the option early.
The basic idea is to interpret the European pricing formula in Theorem 5.30 as the value
of the American option were it forbidden to exercise the option early. Since this might be
less than the payoff of the option if it were exercised immediately, the value of the American
option at t = 0 is the maximum of the value computed via Theorem 5.30 and the value were
the option to be exercised immediately.
This analysis also allows us to determine the optimal exercise strategy – it is only optimal
to exercise early (at t = 0) if the value of exercising early exceeds the European value.
58
Example 6.10. Consider a single-period binomial model with parameters S = S(0) = 40,
u = 6/5, d = 4/5 and continuously-compounded interest rate r = 5%. Find the no-arbitrage
price P of an American put option with strike price K = 52. Is it optimal to exercise early?
Proof. The no-arbitrage condition holds since
4/5 < e0.05 < 6/5,
and the risk-neutral distribution is given by
p˜u =
er − d
u− d =
e0.05 − 4/5
6/5− 4/5 = 0.628 and p˜d = 1− p˜u = 0.372.
If the option were European (so must be exercised at t = 1) its payoff would be
Payoff =
{
(K − S(1))+ = (52− 40× 6/5)+ = 4 if S(1) = S(0)u
(K − S(1))+ = (52− 40× 4/5)+ = 20 if S(1) = S(0)d
Hence, according to the pricing formula in Theorem 5.30, the value of the option were it
forbidden to exercised early would be
e−r (p˜u × 4 + p˜d × 20) = e−0.05 (0.628× 4 + 0.372× 20) = 9.46 .
On the other hand, if the option were exercised at t = 0 it would generate a payoff of
(K − S(0))+ = (52− 40)+ = 12.
Since the option is worth more if exercised immediately (= 12) than its European value
(= 9.464), its no-arbitrage price equals the current payoff, i.e. P = 12, and it is optimal to
exercise at t = 0.
Example 6.11. Consider a single-period binomial model with parameters S = S(0) = 60,
u = 3/2, d = 2/3 and nominal interest rate r = 5%. Find the no-arbitrage price C of an
American call option with strike price K = 55. Is it optimal to exercise early?
Solution. The no-arbitrage condition holds since
2/3 < 1.05 < 3/2,
and the risk-neutral distribution is given by
p˜u =
1 + r − d
u− d =
1.05− 2/3
3/2− 2/3 = 0.46 and p˜d = 1− p˜u = 0.54.
If the option were European, its payoff at t = 1 would be
Payoff =
{
(S(1)−K)+ = (60× 3/2− 55)+ = 35 if S(1) = S(0)u
(S(1)−K)+ = (60× 2/3− 55)+ = 0 if S(1) = S(0)d
Hence, according to the pricing formula in Theorem 5.30, the value of the option were it
forbidden to be exercised at t = 0 would be
1
1.05
(p˜u × 35 + p˜d × 0) = 1
1.05
(0.46× 35) = 15.33 .
On the other hand, if the option were exercised immediately (at t = 0) it would generate a
payoff of
(S(0)−K)+ = (60− 55)+ = 5.
Since the option is worth less if exercised immediately (= 5) than its current value if it were
a European option (= 15.33), its no-arbitrage price equals its value were it to be exercised at
t = 1, i.e. C = 15.33, and it is not optimal to exercise early.
59
Remark 6.12. These examples are in agreement with the general phenomena that we proved
in the previous chapter: that American call options should never be exercised early (unlike
American put options, which should be exercised early in some situations).
6.1.3 Replicating portfolios and the delta-hedging formula
In the previous subsection we priced European options (and more general European derivatives)
in the single-period binomial model using the risk-neutral distribution. An alternate method
to price European derivatives in this model is via a ‘replicating portfolio’ argument; this is just
another version of the no-arbitrage principle.
Recall that the aim of ‘replicating portfolios’ arguments is to set up two portfolios that
have the same value at some future time. Suppose that at t = 0 we form a portfolio consisting
of Ψ units of share and Φ amount of money deposited in the bank; this portfolio is worth
P (0) = ΨS + Φ.
Now, at t = 1 this portfolio is worth
P (1) =
{
ΨS u+ (1 + r)Φ, if S(1) = Su,
ΨS d+ (1 + r)Φ, if S(1) = Sd.
Note that we are assuming the case of simple interest here.
Now suppose there is a certain European derivative with payoff at t = 1 equal to
V (1) =
{
Vu(1), if S(1) = Su,
Vd(1), if S(1) = Sd.
(22)
Is there a way to choose (Ψ,Φ) such that P (1) replicates the payoff V (1) at t = 1? For this
to occur we would need the following two equations in two unknowns to be satisfied:
ΨS u+ (1 + r)Φ = Vu(1) and ΨS d+ (1 + r)Φ = Vd(1).
Solving these equations for Ψ and Φ yields
Ψ =
Vu(1)− Vd(1)
S u− S d and Φ =
1
1 + r
(−Vu(1)S d+ Vd(1)S u)
S u− S d (23)
as the replicating portfolio for the derivative; this is known as the delta-hedging formula. If
Ψ is negative, the portfolio consists of a negative number of shares, and we interpret this as
shares being (short-)sold. If Φ is negative, the portfolio consists of a negative amount of
money deposited in the bank, and we interpret this as money borrowed from the bank.
There are two ways in which this formula might be applied in practice:
1. Since the initial value P (0) of the replicating portfolio is equal to the initial price V (0)
of the derivative (a consequence of ‘the law of one price’), we deduce that
V (0) = P (0) = ΨS + Φ,
which is an alternative method of pricing derivatives in the single-period binomial model.
60
2. The risk of selling options (i.e. for an investment bank) can be mitigated by constructing
the replicating portfolio as a counterbalance; this procedure is known as ‘hedging’. The
‘delta’ in the formula’s name refers to the fact that the formula for Ψ can be considered
the ‘discrete derivative’ of V (1) with respect to the share price S(1), and derivatives are
traditionally denoted by the symbol ∆.
Exercise 6.13. By plugging in the expressions for Ψ, Φ, p˜u and p˜d, verify that the single-
periodic binomial model pricing formula via replicating portfolios
V (0) = ΨS + Φ
coincides with the pricing formula via the risk-neutral distribution
V (0) =
1
1 + r
E˜[V (1)].
Exercise 6.14. Find the equivalent expression for Ψ and Φ in (23) in the case of continuously-
compounded interest.
Taking the point of view of an investment bank, for which the option price V (0) is known,
the delta-hedging formula takes the following simpler form:
Proposition 6.15 (Delta-hedging formula). A European derivative with current price V (0)
and payoff given by (22) can be replicated with a portfolio consisting of
Ψ =
Vu(1)− Vd(1)
S u− S d
shares (bought or short-sold, depending on if this is positive or negative), and the amount
Φ = V (0)− SΨ
deposited in the bank (or borrowed from the bank, if this is negative).
Example 6.16. Find the replicating portfolio for the European call option in Example 6.7.
Solution. Recall that S = S(0) = 40,K = 45, u = 3/2, d = 3/4, and interest is continuously-
compounded at nominal rate r = 5%. The pay-off of the option is (Su − K)+ = 15 if
S(1) = uS, and (Sd − K)+ = 0 if S(1) = dS. Applying the delta-hedging formula, the
replicating portfolio consists of
Ψ =
15− 0
60− 30 =
15
30
= 1/2
units of shares, and the remaining
Φ = C − ΦS = C − (1/2) · 40 = C − 20
deposited in, or loaned from, the bank. Since in Example 6.7 we calculated that C = 5.73,
Φ = 5.73− 20 = −14.27.
i.e., to construct the portfolio we need to loan 14.27 from the bank.
61
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S
Su0
Sd0
Su0u1(ωu)
Su0d1(ωu)
Sd0u1(ωd)
Su0d1(ωd)
- timer r r
0 1 2
Figure 2: A share price in the two-period binomial model
6.2 The two-period binomial model
We now extend the one-period binomial model to model the share over two successive time-
periods. This is known as the ‘two-period binomial model’, and is illustrated in Figure 2.
To help discuss the various nodes of the binomial tree, we introduce some notation. We
will denote by ωu and ωd the nodes at t = 1 corresponding to the event that the share moved
‘up’ and ‘down’ respectively, and by ωuu, ωud, ωdu and ωdd the nodes at t = 2 corresponding
to the events that the share moved ‘up’ and then ‘up’ again, ‘up’ and then ‘down’, etc.
6.2.1 Options pricing in the two-period binomial model
The no-arbitrage principle holds if and only if it holds for each of the single-period binomial
models that make up the two-period model, in other words, using the notation of Figure 2, if
and only if
d0 < 1/D(1) < d1 and max{d1(ωu), d1(ωd)} < 1/D(1; 2) < min{u1(ωu), u1(ωd)},
where D(1; 2) = D(2)/D(1). Observe that 1/D(1) = 1/D(1; 2) = 1 + r in the case of
interest compounded once per time-period, and 1/D(1) = 1/D(1; 2) = er in the case of
continuously-compounded interest.
Let us suppose that the no-arbitrage principle holds. Consider a European option in the
two-period binomial model whose payoff at time 2 given by
V (2) =

Vuu(2) if ω = ωuu
Vud(2) if ω = ωud
Vdu(2) if ω = ωdu
Vdd(2) if ω = ωdd.
The no-arbitrage price of this option at time 0 can be found by applying a backwards recur-
sion algorithm. To be precise, since the value of the option at time 2 is known (it is equal to
the payoff), we begin by calculating the no-arbitrage price of the option at time 1, at both the
62
ωu and ωd states, by solving the corresponding single-period binomial models between times 1
and 2. Then we calculate the no-arbitrage price at time 0 by solving the single-period binomial
tree between times 0 and 1. This algorithm is illustrated in the following example:
Example 6.17. Consider the two-period binomial model for a share price depicted in Figure 3.
Assume that interest is continuously-compounded at nominal rate (per time-period) of 5%.
Verify the no-arbitrage condition, and find the no-arbitrage price of a European call option
with strike price 45 and expiry T = 2.
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r
r
S(0) = 40
S(1) = 60
S(1) = 30
S(2) = 80
S(2) = 50
S(2) = 40
S(2) = 25
- timer r r
0 1 2
Figure 3: The share price tree in Example 6.17
Solution. In terms of the parameters in Figure 2, the model has parameters values S =
40, (u0, d0) = (3/2, 3/4) and (u1(ωu), d1(ωu)) = (u1(ωd), d1(ωd)) = (4/3, 5/6), since, for
example, considering states ωdu and ωdd we see that
u1(ωd) =
40
30
=
4
3
and d1(ωd) =
25
30
=
5
6
.
The no-arbitrage condition holds since
5/6 = max{3/4, 5/6} < e0.05 < min{3/2, 4/3} = 4/3.
Let V (n) denote the value of the option at time n. As described above, we work backwards
through the tree. The value of the option at time 2 is equal to its payoff, i.e.,
V (2) = (S(2)−K)+ =

(80− 45)+ = 35 if ω = ωuu,
(50− 45)+ = 5 if ω = ωud,
(40− 45)+ = 0 if ω = ωdu,
(25− 45)+ = 0 if ω = ωdd.
Figure 4 shows this information.
We next compute Vu(1) and Vd(1) by solving the respective single-period binomial model
between times 1 and 2. Here we view the option as a asset, existing between times 1 and
63
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r
V (0)
Vu(1)
Vd(1)
Vuu(2) = 35
Vud(2) = 5
Vdu(2) = 0
Vdd(2) = 0
- timer r r
0 1 2
Figure 4: The price tree after computing the terminal values
2, with payoff at time 2 given by the already computed values of V (2); this fits into our
framework for computing no-arbitrage prices of any European derivative in the single-period
binomial model.
Let us first consider Vu(1). We define (p˜u(ωu), p˜d(ωu)) to be the risk-neutral distribution
associated to the single-period binomial model from t = 1 to t = 2 in the state ω = ωu. We
can calculate this risk-neutral distribution to be
p˜u(ωu) =
er − d1(ωu)
u1(ωu)− d1(ωu) =
e0.05 − 5/6
4/3− 5/6 = 0.4359 and p˜d(ωu) = 1− p˜u(ωu) = 0.5641.
Hence,
Vu(1) = e
−r (p˜u(ωu)Vuu(2) + p˜d(ωu)Vud(2))
= e−0.05 (0.43588× 35 + 0.5641× 5)
= 17.195.
Similarly, for Vd(1) the risk-neutral distribution for the corresponding single-period model is
also
p˜u(ωd) =
er − d1(ωd)
u1(ωd)− d1(ωd) =
e0.05 − 5/6
4/3− 5/6 = 0.4359 and p˜d(ωd) = 1− p˜u(ωd) = 0.5641,
and hence
Vd(1) = e
−r (p˜u(ωd)Vdu(2) + p˜d(ωd)Vdd(2))
= e−0.05 (0.4359× 0 + 0.5641× 0)
= 0.
Figure 5 shows this information.
Finally we consider the single-period binomial model between times 0 and 1. The risk-
neutral measure corresponding to this model is
p˜u =
er − d0
u0 − d0 =
e0.05 − 3/4
3/2− 3/4 = 0.4017 and p˜d = 1− p˜u = 0.5983
64
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V (0)
Vu(1) = 17.195
Vd(1) = 0
Vuu(2) = 35
Vud(2) = 5
Vdu(2) = 0
Vdd(2) = 0
- timer r r
0 1 2
Figure 5: The price tree after computing the values at times 1 and 2
and so
V (0) = e−r (p˜uVu(1) + p˜dVd(1))
= e−0.05 (0.4017× 17.195 + 0.5983× 0)
= 6.570.
In conclusion, we have shown that the fair price of the option is 6.570.
Remark 6.18. The same backwards recursion algorithm can be used to find the value of any
European derivative in the two-period model.
To price American options we need to modify slightly the procedure explained above, since
we need to account for the possibility that it is optimal to exercise these options early. As
such, at each stage of the procedure (say at time t = i), we must consider whether the option
is in fact worth more if it is exercised immediately, i.e., whether its ‘exercise value’ exceeds its
‘European value’. If it does, then the ‘exercise value’ must be used when continuing
the backwards recursion algorithm to find the value of the option at time i− 1, since the
‘exercise value’, and not the ‘European value’, reflects the true value of the option.
As a by-product of this procedure, we also obtain the optimal exercise strategy for the
American option, i.e., we determine under what circumstances is it optimal to exercise the
American option early. These are simply the states at which the ‘exercise value’ was found to
exceed the ‘European value’.
Example 6.19. Consider the two-period binomial model for a share price depicted in Figure 6.
Assume that interest is continuously-compounded at nominal rate 5%. Verify the no-arbitrage
condition, find the no-arbitrage price of an American put option with strike 52 and expiry
T = 2, and determine the optimal exercise strategy.
Proof. In terms of the parameters in Figure 2, the model has parameters values S = 50 and
(u0, d0) = (u1(ωu), d1(ωu)) = (u1(ωd), d1(ωd)) = (6/5, 4/5). The no-arbitrage condition
holds since
4/5 < e0.05 < 6/5.
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S(0) = 50
S(1) = 60
S(1) = 40
S(2) = 72
S(2) = 48
S(2) = 48
S(2) = 32
- timer r r
0 1 2
Figure 6: The share price tree in Example 6.19
Hence the risk-neutral distribution is the same at each node of the tree, and is given by
p˜u =
er − d
u− d =
e0.05 − 4/5
6/5− 4/5 = 0.6282 and p˜d = 1− p˜u = 0.3718 .
The value of the option at time 2 is equal to its payoff, i.e.,
V (2) = (K − S(2))+ =

(52− 72)+ = 0 if ω = ωuu,
(52− 48)+ = 4 if ω = ωud,
(52− 48)+ = 4 if ω = ωdu,
(52− 32)+ = 20 if ω = ωdd.
.
Let us consider the value of the option at time 1, beginning with state ωu. Solving the
corresponding single-period binomial model, we see that the European value of the option
at this state is
e−r (p˜uVuu(2) + p˜dVud(2)) = e−0.05 (0.6282× 0 + 0.3718× 4) = 1.4147 .
On the other hand, the exercise value of the option at this state is
(K − Su(1))+ = (52− 60)+ = 0
which is less than the European value. Hence the value of the option Vd(1) at this state is
equal to its European value of 1.4147.
Turning to state ωd, the European value of the option at this state is
e−r (p˜uVdu(2) + p˜dVdd(2)) = e−0.05 (0.6282× 4 + 0.3718× 20) = 9.4639 .
On the other hand, the exercise value of the option at this state is
(K − Sd(1))+ = (52− 40)+ = 12,
66
and so in this case the exercise value exceeds the European value. Therefore it is optimal to
exercise at this state, and the value of the option Vd(1) is equal to its exercise value of 12.
Finally, we consider the value of the option at time 0. Solving the corresponding single-
period binomial model, we see that the European value of the option at this state is
e−r (p˜uVu(1) + p˜dVd(1)) = e−0.05 (0.6282× 1.4147 + 0.3718× 12) = 5.0894 .
(Note that we used the exercise value (= 12) in the second term!) On the other hand, the
exercise value of the option at this state is
(K − S(0))+ = (52− 50)+ = 2
which is less than the European value. Hence the value of the option V (0) at this state is
equal to its European value of 5.0894.
In conclusion, the no-arbitrage price of the option is 5.0894. We deduce the optimal
exercise strategy by considering at which nodes in the backwards recursion algorithm the
exercise value of the option exceeded the European value. In this case, this occurred only at
the state ωd, and so the optimal exercise strategy is to only exercise early at the state ωd, i.e.
if and only if the first movement of the share price is down.
Exercise 6.20. Assuming a two-period binomial model as in Exercise 6.19, find the no-
arbitrage price and optimal exercise strategy for an American call option with strike 40. Does
this agree with the general fact that American call options should never be exercised early?
Solution. We have already verified that no-arbitrage condition holds, and calculated that the
risk-neutral distribution is the same at each node of the tree and is given by
p˜u =
er − d
u− d =
e0.05 − 4/5
6/5− 4/5 = 0.6282 and p˜d = 1− p˜u = 0.3718 .
The value of the option at time 2 is equal to its payoff, i.e.,
V (2) = (S(2)− 40)+ =

(72− 40)+ = 32 if ω = ωuu,
(48− 40)+ = 8 if ω = ωud,
(48− 40)+ = 8 if ω = ωdu,
(32− 40)+ = 0 if ω = ωdd.
Let us consider the value of the option at time 1, beginning with state ωu. Solving the
corresponding single-period binomial model, we see that a European value of the option at
this state is
e−r (p˜uVuu(2) + p˜dVud(2)) = e−0.05 (0.6282× 32 + 0.3718× 8) = 21.95 .
On the other hand, the exercise value of the option at this state is
(Su(1)−K)+ = (60− 40)+ = 20
which is less than its European value. Hence the value of the option Vu(1) at this state is
equal to its European value of 21.95.
Turning to state ωd, the European value of the option at this state is
e−r (p˜uVdu(2) + p˜dVdd(2)) = e−0.05 (0.6282× 8 + 0.3718× 0) = 4.78 .
67
On the other hand, the exercise value of the option at this state is
(Sd(1)−K)+ = (40− 40)+ = 0,
and so again in this case the exercise value is less than its European value. Hence the value
of the option at this state is equal to its European value of 4.78.
Finally, we consider the value of the option at time 0. Solving the corresponding single-
period binomial model, we see that the European value of the option at this state is
e−r (p˜uVu(1) + p˜dVd(1)) = e−0.05 (0.6282× 21.95 + 0.3718× 4.78) = 14.81 .
On the other hand, the exercise value of the option at this state is
(S(0)− 40)+ = (50− 40)+ = 10
which is less than the European value. Hence the value of the option V (0) at this state is
equal to its European value of 14.81.
In conclusion, the no-arbitrage price of the option is 14.81, and the optimal exercise strategy
is to never exercise the option early.
6.2.2 Replicating portfolios in the two-period binomial model
Just like in the one-period model, we can also analyse European options in the two-period
binomial model by considering replicating portfolios. Once again we reduce the two-period
binomial model to two successive single-period binomial models, in each of which we apply
the ‘delta hedging’ formula.
Note that the replicating portfolio will be different at each time-step, and in the case
t = 1 it may also depend on the node ωu and ωd of the binomial tree. In practice this means
that the replicating portfolio needs to be adjusted dynamically in order to ensure that it
replicates the payoff of the option for all possible outcomes of the share price in the model.
6.2.3 Option pricing via the two-period risk-neutral distribution
Up until now we have been working with risk-neutral distributions that are defined for single-
period binomial models. In fact, it is possible to define a risk-neutral distribution directly
for the two-period binomial model. This leads to an alternative method to price European
options in the two-period binomial model that avoids the backward recursion algorithm.
Recall that (p˜u, p˜d) denotes the risk-neutral distribution that corresponds to the single-
period binomial model between times 0 and 1, and (p˜u(ωu), p˜d(ωu)) and (p˜u(ωd), p˜d(ωd))
denote the risk-neutral distributions that correspond to the single-period model between times
1 and 2 in the states ωu and ωd respectively. Then the risk-neutral distribution at t = 2 of
the two-period binomial model, written as P˜, is equal to
P˜(ω) =

p˜u × p˜u(ωu) if ω = ωuu,
p˜u × p˜d(ωu) if ω = ωud,
p˜d × p˜u(ωd) if ω = ωdu,
p˜d × p˜d(ωd) if ω = ωdd.
Recall that by the general no-arbitrage pricing principle we can express the no-arbitrage price
at t = 0 of any European option V (0) as
V0 = D(2) E˜(V (2)),
where V (2) is the payoff of the option at time 2.
68
Example 6.21. Consider the two-period binomial model as in Exercise 6.17. Find the risk-
neutral distribution for the share price S(2) at time 2, and use it to price a European put
option with strike 45 and expiry T = 2.
Solution. Recall that in Exercise 6.17 we calculated the risk-neutral distribution for the one-
period binomial model at time t = 0,
p˜u = 0.4017 and p˜d = 0.5983,
and the risk-neutral distributions for the one-period binomials models at time t = 1 to be
p˜u(ω) = 0.4359 and p˜d(ω) = 0.5641
for both ω = ωu or ω = ωd (i.e. the risk-neutral probabilities are the same at t = 1 for both
the ‘up’ and ‘down’ states). Hence we can calculate the overall risk-neutral distribution P˜ at
t = 2 to be
P˜(ω) =

p˜u × p˜u(ωu) = 0.175 if ω = ωuu,
p˜u × p˜d(ωu) = 0.227 if ω = ωud,
p˜d × p˜u(ωd) = 0.261 if ω = ωdu,
p˜d × p˜d(ωd) = 0.337 if ω = ωdd.
Notice that these probabilities sum to one, as they should. Via the general pricing formula,
the price of the put option is
P = e−2r E˜((K − S(2))+).
Considering the possible values of S(2) from Exercise 6.17, only the outcomes S(2) = 40
if ω = ωdu and S(2) = 25 if ω = ωdd are ‘in-the-money’. Evaluating the expectation with
respect to the risk-neutral measure, we have
P = e−0.05×2
(
0.261× (45− 40) + 0.337× (45− 25)) = 7.28.
Remark 6.22. We can verify that our answers to Exercises 6.17 and 6.21 satisfy put-call
parity. We found the prices of European call and put options with strike K = 45 and expiry
T = 2 to be C = 6.57 and P = 7.28. Since the initial stock price in the model was S = 40,
we indeed have put-call parity:
P + S = 7.28 + 40 = 47.3 and C +D(2)K = 6.57 + e−2×0.0545 = 47.3.
Remark 6.23. Although this method is perhaps simpler than the backwards recursion algo-
rithm, it is less general since it only works for European options. In particular, it cannot be
used to price American options.
6.3 The multi-period binomial model
The two-period binomial model can be extended to any number of time-periods; this is called
the multi-period binomial model. A major potential disadvantage of this model is that the
number of possible states grows exponentially; at t = n, the share price can be in 2n possible
states. For this reason we consider a simplified model in which the ‘up’ and ‘down’ factors u
and d are equal for every state. This can be interpreted as an independence assumption: the
potential range of future price movements does not depend on the current state.
69
Suppose a share has initial price S(0) = S. We model the share price S(i) for i = 1, . . . , n
by specifying that S(i) is either equal to uS(i − 1) or to d S(i − 1), where 0 < d < u are
parameters.
Let us consider the possible share prices S(n) at time n. Since the share price either moves
‘up’ or ‘down’ at each step, the possible values of S(n) are
S(n) = S ukdn−k
where k = {0, 1, . . . , n} represents the number of ‘up’ steps out of the n total steps (the
number of ‘down’ steps is therefore n − k). Notice that S(n) can take only n + 1 possible
values; for large n this is an enormous simplification over the general multi-period binomial
model, in which the share price has 2n possible states.
r




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Z
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Z
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Z
Z
Z
Z
Z
Z
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r
r
r
S
Su
Sd
Su2
Sud
Sd2
Sd3
Sud2
Su2d
Su3
- timer r r r
0 1 2 3
Figure 7: A share in the 3-period multi-period binomial model
6.3.1 Options pricing in the multi-period binomial model
Let D(1) be the discount factor per time-period. Just as in the single-period case, the no-
arbitrage principle holds if and only if
d < 1/D(1) < u,
in which case the risk-neutral distribution corresponding to each single-period model is
p˜u =
1/D(1)− d
u− d and p˜d =
u− 1/D(1)
u− d .
As we did in Section 6.2.3, we will analyse the multi-period model by considering the
risk-neutral distribution for S(n). Recall that
S(n) = S ukdn−k
70
where k = {0, 1, . . . , n} represents the number of ‘up’ steps. An equivalent way to write this
is to introduce the random variables
Xi =
{
1 , if S(i) = S(i− 1)u
0 , if S(i) = S(i− 1) d , K =
n∑
i=1
Xi,
and to write
S(n) = S uKdn−K .
Notice that the Xi are Bernoulli trials that record a ‘success’ if the share moved upwards, and a
‘failure’ if the share moved downwards. Since each Xi is an independent Bernoulli(p˜u) random
variable under the risk-neutral distribution, K =
∑n
i=1Xi is a Binomial random variable with
distribution
P˜(K = k) =
(
n
k
)
p˜ku(1− p˜u)n−k.
Hence we have:
Proposition 6.24. The risk-neutral distribution for S(n) is given by
S(n) = Sukdn−k with probability
(
n
k
)
p˜ku(1− p˜u)n−k, k = 0, 1, . . . , n .
We can now use the general no-arbitrage pricing theory to price European options in the model:
Theorem 6.25. Assume that a share follows a multi-period binomial model with parameters
u and d and interest rate r. Consider a European option with payoff at expiry time t = n
equal to
V (n) = g(S(n))
for g a given payoff function. Then the no-arbitrage price at time 0 of the option is
V (0) = D(n)
n∑
k=0
g
(
Sukdn−k
) (n
k
)
p˜ku(1− p˜u)n−k,
where p˜u = (1/D(1)− d)/(u− d).
Proof. By the general no-arbitrage pricing principle,
V (0) = D(n) E˜ [V (n)] = D(n) E˜[g(S(n)].
The result follows by evaluating the expectation using the risk-neutral distribution for S(n).
Example 6.26. Assume interest is compounded once per time-period. The no-arbitrage price
of a European call option with strike price K and expiry time n is
C =
1
(1 + r)n
n∑
k=0
(Sukdn−k −K)+
(
n
k
)
p˜ku(1− p˜u)n−k.
Example 6.27. Consider a share following a multi-period binomial model with parameters
S = 200, u = 1.1 and d = 0.9, and interest rate per time-period r = 1%. Find the
no-arbitrage price of a European call option with strike K = 425 and expiry time 10.
71
Solution. The no-arbitrage condition holds since
d < 1 + r < u⇔ 0.9 < 1.01 < 1.1
and so the the risk-neutral probabilities are
p˜u =
1 + r − d
u− d = 0.55 and p˜d = 1− p˜u = 0.45 .
The payoff of the call option is
(S(10)−K)+ = (Sukdn−k −K)+ = (200× 1.1k × 0.9 10−k − 425)+.
This is ‘in-the-money’ (i.e. has positive payoff) if and only if
200× 1.1k × 0.9 10−k > 425.
Examining the possible values of k, if k = 10 then the left-hand side is 519 > 425, whereas if
k ≤ 9 the left-hand side is at most 424 < 425. Hence the only state that is ‘in-the-money’ is
k = 10, which occurs with probability 0.5510. Therefore the price of the option is
C =
1
1.0110
E˜[(S(10)− 425)+]
=
1
1.0110
(519− 425)× 0.5510 = 0.215 .
Example 6.28. Suppose a share is modelled by assuming each month it either increases by
2% or decreases by 1%. The current price of the share is 130. Assume also that interest
is continuously-compounded at rate 12% per year. Find the no-arbitrage price of 1 million
European put options with strike 120 and expiry in one year.
Proof. The share follows a multi-period binomial model with parameters S = 130, u = 1.02
and d = 0.99, where each time-step is a month. The interest rate per time-step is
r = 12%/12 = 1%.
The no-arbitrage condition is satisfied since
d < er < u⇔ 0.99 < 1.01 < 1.02
and the risk-neutral probabilities are
p˜u =
er − d
u− d =
e0.01 − 0.99
1.02− 0.99 = 0.669 p˜d = 1− p˜d = 0.332 .
The payoff of each put option is
V (12) = (K − S(12))+ = (120− 130× 1.02k × 0.99 12−k)+,
which we calculate to be
V (12) =

4.77 if k = 0
1.28 if k = 1
0 if k ≥ 2.
Since the outcome k = 0 occurs with probability 0.331712 and the outcome k = 1 occurs with
probability
(
12
1
)× 0.6683× 0.331711, the no-arbitrage price of the 1 million put options is
P = 1,000,000× e−0.12E˜[V (12)]
= 1,000,000× e−0.12(4.77× 0.33212 + 1.28× 12× 0.668× 0.33211)
= 56.1 .
72
7 From the multi-period binomial model to the Black-
Scholes formula
In this final chapter of the course we derive the Black-Scholes formula, which is a celebrated
pricing formula for European options first published in 1973. This formula arises from the
Black-Scholes model, in which the stock price is assumed to follow a continuous-time stochastic
process known as geometric Brownian motion. While this is outside the scope of the course
(it will be studied in detail in FM II), we can nevertheless derive the Black-Scholes formula
via an approximation argument using the multi-period binomial model.
7.1 General discrete-time models
The most general discrete-time model for the stock price would be to define (S(i), i = 0, 1, . . .)
as a general sequence of random variables, which could be discrete or continuous; this is known
as a discrete time stochastic process. It is convention to take S(0) as deterministic (non-
random), since the stock price is assumed known at time 0. Note that the time-step in such
a model could be anything: a year, a week, a second etc.
One possible example of such a model is to take S(i) as an i.i.d. sequence. This will not
be a good model for the stock price for the same reasons that i.i.d. sequences are a poor model
for interest rates (see Section 4.2), namely since (i) the price of stock at one point in time
will generally be positively correlated with, and not independent to, the price at another point
in time, (ii) the variance in the stock price should grow over time rather than stay constant.
A more realistic model is to define
S(n) = S(0)×
n∏
i=1
Yi,
where Yi is a sequence of i.i.d. random variables, i.e. we assume that the multiplicative
increments S(n)/S(n−1) of the stock prices, rather than the prices themselves, are an i.i.d.
sequence. The multi-period binomial model is an example of such a model; in this case Yi are
discrete random variables taking values u and d with some (real-world) probabilities pu and pd.
7.2 The log-normal process
As we have seen, in order to price options we need to be able to analyse the (risk-neutral)
distribution of S(n). One choice of model that facilitates this is to let Yi be an i.i.d. sequence
of log-normal random variables, i.e., defining
Yi ∼ LogNormal(µ, σ2).
The resulting model for S(n) is called the log-normal process, and a remarkable property of
this model is that we can compute the distribution of S(n) explicitly:
Theorem 7.1. If S(n) is modelled by the log-normal process, then
S(n) ∼ S(0)× LogNormal(nµ, nσ2).
Proof. Since
S(n)
S(0)
= Y1Y2Y3 · · ·Yn,
and since Y1, Y2, . . . , Yn are independent and distributed as LogNormal(µ, σ
2) random vari-
ables, by Theorem 4.8 we have the result.
73
Example 7.2. Assume that the stock price follows a log-normal process with parameters
µ = 0.0165 and σ = 0.0730 and time-step equal to a week. Calculate the probability that
(a) The price increases over the first week,
(b) The price increases in each of the first two weeks;
(c) The price is higher at the end of the second week than at the starting time 0.
You may assume that Φ(0.23) = 0.591 and Φ(0.32) = 0.626.
Proof. Let S(n) denote the stock price at the end of the nth week.
(a) The desired probability is
P (S(1) > S(0)) = P
(S(1)
S(0)
> 1
)
= P
(
ln
S(1)
S(0)
> 0
)
.
Since
ln
S(1)
S(0)
∼ N (µ, σ2),
we deduce that
P (S(1) > S(0)) = P
(N (µ, σ2)− µ
σ
> −µ
σ
)
= 1− Φ
(
− µ
σ
)
= Φ
(µ
σ
)
= Φ(0.23) = 0.591 .
(b) Let p = 0.591 be the probability that the price increases over the first week. Since
the random variables S(1)/S(0) and S(2)/S(1) are i.i.d., the probability that the price
increases in each of the first two weeks is p2 = 0.349.
(c) Similarly to in (a), since
ln
S(2)
S(0)
∼ N (2µ, 2σ2),
the desired probability is
P (S(2) > S(0)) = P
(
ln
S(2)
S(0)
> 0
)
= P
(
ln S(2)
S(0)
− 2µ
√
2σ
> − 2µ√
2σ
)
= 1− Φ
(
−
√
2µ
σ
)
= Φ
(√2µ
σ
)
= Φ(0.32) = 0.626 .
7.3 The Black-Scholes formula
Suppose that a share is modelled using a log-normal process. The celebrated Black-Scholes
formula for the price of European options is the result of applying the general no-arbitrage
pricing theory that we developed in Chapter 5 to this model.
74
Theorem 7.3 (Black-Scholes formula). Suppose a share price is modelled by a log-normal
process with initial value S and log-normal parameters (µ, σ2). Assume interest is continuously-
compounded with nominal rate r. Then the no-arbitrage price of a European call option with
strike price K and maturity time T is
C = C(S, T,K, σ, r) = SΦ(η)−Ke−rTΦ(η − σ
√
T ), (24)
where
η =
ln S
K
+ rT
σ
√
T
+
1
2
σ
√
T .
Remark 7.4. By put-call parity, one can deduce a similar formula for European put options.
Remark 7.5. The parameters µ and σ are called the ‘drift’ and ‘volatility’ of the stock price
respectively. Remarkably, the price C does not depend on the drift µ. One explanation is
that µ affects only the ‘real-world’ probabilities, which are irrelevant in no-arbitrage pricing.
A more intuitive explanation is that the price of a call option reflects its usefulness as a way
for investors to manage risk, and since µ results in a deterministic average increase in the
stock price, it is irrelevant to the ability of the option to manage risk.
Example 7.6. Suppose a share is currently traded at a price of£30 and interest is continuously-
compounded at rate 8%. Assume that the volatility is σ = 0.2. Find the Black-Scholes price
C of a European call option with expiry in 3 months and strike £34, and the price P of the
European put option with the same strike and expiry (you may use Φ(1.00) = 0.841 and
Φ(1.10) = 0.864).
Solution. If time is measured in years, we have the parameter settings T = 3/12 = 0.25,
r = 0.08, σ = 0.2, K = 34, and S = 30. Hence
η = −1.00 and η − σ
√
T = −1.10,
and the Black-Scholes formula gives
C = 30Φ(−1.00)− 34e−0.02Φ(−1.10) = 30(1− 0.841)− 34e−0.02(1− 0.864) = 0.24.
By put-call parity, the price of put options is
P = C − S +Ke−rT = 0.24− 30 + 34e−0.02 = 3.57.
Remark 7.7. Note that the interest rate r is known to investors but the volatility σ is not,
which raises the question of how to implement the formula in practice. One view is that past
data can be used to approximate the volatility. Another view is that price data from options
that are traded on the open market can be plugged into the formula to derive an ‘implied
volatility’ (this requires a computer). In reality, practitioners use a mixture of these views.
To prove the Black-Scholes formula we will require an intermediate result about the risk-
neutral distribution in the model. We will prove this in the following section.
Proposition 7.8. Suppose a share price follows a log-normal process with initial value S and
parameters (µ, σ2), and assume that interest is continuously-compounded with nominal rate r.
Then there exists a risk-neutral measure P˜ under which the share price follows a log-normal
process with initial value S and parameters
µ˜ = r − 1
2
σ2 and σ˜2 = σ2.
75
Proof of Theorem 7.3. By general principles of no-arbitrage pricing,
C = e−rT E˜
[
(S(T )−K)+],
where E˜ denotes expectation under the risk-neutral distribution. Making use of the indicator
random variable 1S(T )>K (that equals 1 if S(T ) > K occurs and 0 otherwise), this can be
written as
C = e−rT E˜
[
(S(T )−K)1S(T )>K
]
= e−rT E˜
[
S(T )1S(T )>K
]− e−rT E˜[K1S(T )>K]
= e−rT E˜
[
S(T )1S(T )>K
]− e−rTK P˜(S(T ) > K), (25)
since E[1A] = P(A). By Proposition 7.8, under the risk-neutral distribution
S(T )
d∼ S × LogNormal(µ˜T, σ˜2T ) d∼ Se(r−σ2/2)T+σ
√
TZ
where Z
d∼ N (0, 1), and hence
S(T ) > K ⇔ Se (r−σ2/2)T+σ
√
T Z > K
⇔ (r − σ2/2)T + σ
√
T Z > lnK/S
⇔ Z > lnK/S − (r − σ
2/2)T
σ
√
T
⇔ Z > σ
√
T − lnS/K + r T
σ
√
T
− 1
2
σ
√
T
⇔ Z > σ
√
T − η.
Thus we simplify the two terms in (25) as
e−rT E˜
[
S(T )1S(T )>K
]
= e−rTE
[
Se (r−σ
2/2)T+σ
√
T Z1Z>σ
√
T−η
]
= S E
[
e(−σ
2/2)T+σ
√
T Z1Z>σ
√
T−η
]
(26)
and
e−rTK P˜
[
S(T ) > K
]
= e−rTK P
[
Z > σ
√
T − η]. (27)
To compute (26) we evaluate the expectation (using the p.d.f. of the standard normal):
S E
[
e(−σ
2/2)T+σ
√
T Z1Z>σ
√
T−η
]
= S
∫ ∞
σ
√
T−η
e(−σ
2/2)T+σ
√
T z 1√
2π
e−z
2/2 dz
= S
∫ ∞
σ
√
T−η
1√
2π
e−((σ
2/2)T−σ√T z+z2/2) dz
= S
∫ ∞
σ
√
T−η
1√
2π
e−(σ
√
T−z)2/2 dz.
Changing variables y 7→ σ√T − z, the above equals
−S
∫ −∞
η
1√
2π
e−y
2/2 dy = S
∫ η
−∞
1√
2π
e−y
2/2 dy
which we can write in terms of the standard normal c.d.f. as S Φ(η). The term (27) can also
be expressed in term of the standard normal c.d.f.:
e−rTKP
[
Z > σ
√
T − η] = e−rTK(1− Φ(σ√T − η)) = e−rTKΦ(η − σ√T ).
Plugging the expressions for (26) and (27) into (25), we get the desired formula:
C = SΦ(η)−Ke−rTΦ(η − σ
√
T ) .
76
7.3.1 Properties of the Black-Scholes formula
We briefly study the dependency of the Black-Scholes formula on the parameters S, T , K, σ,
and r.
Proposition 7.9. The partial derivatives of C = C(S, T,K, σ, r) are:
∂C
∂K
= −e−rTΦ(η − σ
√
T ) ,
∂C
∂S
= Φ(η) ,
∂C
∂r
= KTe−rTΦ(η − σ
√
T )
∂C
∂σ
= S
√
TΦ′(η) and
∂C
∂T
=
σ
2
√
T
SΦ′(η) +Kre−rTΦ(η − σ
√
T ).
Remark 7.10. In investment jargon these partial derivatives are known as the ‘greeks’ (since
they are traditionally denoted by Greek letters) and are an important investment tool with
which to analyse options. As an example, the continuous time analogue of the ‘delta-hedging’
formula states that the replicating portfolio should contain ∂C/∂S units of stock.
Remark 7.11. The above proposition shows that C is a decreasing function of K (when the
other parameters remain constant), and is an increasing function of S, T, σ, r. This can be
explained intuitively:
• We have already proven in Propositions 5.35 that the price of a European call option is
a decreasing function of the strike K. This is a model-free result, so applies equally to
the Black-Scholes formula.
• Similarly, the fact that C is an increasing function of S is a model-free result, as can
be seen from the fact that C = e−rT E˜[(S(T ) − K)+] and since (S(T ) − K)+ is an
increasing function of S.
• The fact that C increases with σ (and also T ) can be explained by considering that the
call option is a way to manage risk, and the more volatile the underlying asset (and the
longer time to which the option applies) the more it should cost to manage risk.
Proof. We only prove the formula for ∂C/∂K, since the others are similar. By the chain rule,
∂C
∂K
= SΦ′(η)
∂η
∂K
− e−rTΦ(η − σ
√
T )−Ke−rTΦ′(η − σ
√
T )
∂η
∂K
=
∂η
∂K
[SΦ′(η)−Ke−rTΦ′(η − σ
√
T )]− e−rTΦ(η − σ
√
T ).
It remains to show that
Ke−rTΦ′(η − σ
√
T ) = SΦ′(η).
First, rearranging the definition of η, we have that
η =
ln S
K
+ rT
σ
√
T
+
1
2
σ
√
T ⇔ σ
√
Tη − rT − 1
2
σ2T = ln
S
K
. (28)
Recalling also that Φ′(x) = 1√
2π
e−x
2/2, we have
K e−rT Φ′(η − σ
√
T ) = K e−rT
1√
2π
e−(η−σ
√
T )2/2
= K
1√
2π
e−η
2/2 eσ
√
Tη−rT− 1
2
σ2T
(28)
= K Φ′(η) eln(S/K)
= SΦ′(η) .
77
7.4 The log-normal process as an approximation of the bin. model
We finish the course by justifying, from first principles, our use of the log-normal process
in deriving the Black-Scholes formula. We also prove Proposition 7.8 on the risk-neutral
distribution for the log-normal process.
Recall that in the multi-period binomial model we assume that the stock price moves up or
down each time-period, and we can choose the time-period as small as we like. Even though
taking smaller time-periods leads to a more accurate model, a major problem arises in that
the pricing formulae become more and more complex.
The way out is to approximate a multi-period binomial model by a log-normal process;
as we will show, this approximation gets better as the time-period gets smaller.
Let us suppose that the time-period of the multi-period binomial model is 1/n, for n very
large. This means that the up/down multiplication factors should be very close to one, since
share prices should not move much in any single time-period. It turns out that, in order to
get convergence to a log-normal process, we need to choose
u = eσ/
√
n and d = e−σ/
√
n, (29)
where σ > 0 is the volatility parameter of the log-normal process.
Theorem 7.12. Fix parameters µ, r ∈ R and σ > 0. Consider a multi-period binomial model
for a share price S(t) with time-period 1/n , up/down multiplication factors (per time-period)
as in (29). Then the following hold for any t > 0:
• Suppose the ‘real world’ probability of up/down steps are
pu =
1
2
(
1 +
µ
σ
√
n
)
and pd =
1
2
(
1− µ
σ
√
n
)
.
Then, as n→∞,
S(t)
d⇒ S(0)LogNormal(µt, σ2t).
• Suppose interest is continuously-compounded at rate r. Then, as n→∞,
S(t)
d⇒ S(0)LogNormal(µ˜t, σ˜2t)
under the risk-neutral distribution P˜, where
µ˜ = r − 1
2
σ2 and σ˜2 = σ2.
Before proving this theorem let us remark that Proposition 7.8 is an immediate conse-
quence:
Proof of Proposition 7.8. Theorem 7.12 shows that a log-normal process with parameters
(µ, σ) can be approximated by a multi-period binomial model. Moreover, the risk-neutral
distribution of this multi-period binomial model also approximates a log-normal process with
parameter (µ˜, σ˜). Passing to the limit, the latter log-normal process must be a risk-neutral
distribution for the former log-normal process.
Proof of Theorem 7.12. Let us begin with the first claim of the theorem (the convergence
under the real-world probability distribution). There are two steps to the proof:
78
1) Express ln(S(t)/S(0)) as the sum of a sequence of i.i.d. random variables;
2) Apply a central limit theorem to deduce the result.
Step 1. The stock price at t in the multi-period binomial model with time-period 1/n can be
written
S(t) = S(0)
∏
i≤nt
Yn,i
where, for each n, Yn,i are an i.i.d. sequence of random variables with distribution
Yn,i ∼
{
u = exp(σ/
√
n) with probability pu =
1
2
(
1 + µ
σ
√
n
)
d = exp(−σ/√n) with probability pd = 12
(
1− µ
σ
√
n
)
.
Taking logarithms,
ln(S(t)/S(0)) =
∑
i≤nt
lnYn,i,
where
lnYn,i ∼
{
σ/
√
n with probability pu =
1
2
(
1 + µ
σ
√
n
)
,
−σ/√n with probability pd = 12
(
1− µ
σ
√
n
)
.
Step 2. Let us now apply a central limit theorem to this sum of i.i.d. random variables. Since
we are dealing with a triangular array of random variables, we are in the setting of Theorem 6.4
from the Review of Probability Theory (in fact, the setting is almost identical to Example 6.6
of that document), which states that if, as n→∞ ,
nE[lnYn,1]→ µ and nVar[lnYn,1]→ σ2,
(and a ‘technical condition’ nE[| lnYn,1|p]→ 0 holds for some p > 2), then∑
i≤n
Yn,i
d⇒ N (µ, σ2). (30)
Calculating
nE[lnYn,1] = n
(
(σ/
√
n)pu − (σ/
√
n)(1− pu)
)
= σ
√
n(2pu − 1) = σ
√
n
µ
σ
√
n
= µ
and
nVar[lnYn,1] = n
(
E[(lnYn,1)2]− E[lnYn,1]2
)
= n
(
(σ2/n)pu + (σ
2/n)(1− pu)− (µ/n)2
)
= σ2 − µ2/n→ σ2,
and applying the ‘triangular CLT’ (the ‘technical condition’
nE[| lnYn,1|4] ∼ nσ
4
n2
→ 0
clearly holds), we see that we have proven (30). Now, replacing n by nt in the above argument
shows that ln(S(t)/S(0)) =
∑
i≤nt lnYn,i converges, for large n, to the normal distribution
N (µt, σ2t), i.e. S(t)/S(0) converges to a log-normal random variable with the same parame-
ters. This completes the first claim of the theorem.
79
The proof of the second claim is similar, except we work under the risk-neutral probabilities.
Again we express
ln(S(t)/S(0)) =
∑
i≤n
lnYn,i,
but now, under the risk-neutral distribution,
lnYn,i ∼
{
σ/
√
n with probability p˜u,
−σ/√n with probability p˜d.
Let us calculate the risk-neutral probabilities. The discount factor for each period is e−r/n,
and so the risk-neutral probability for the up-state is
p˜u =
er/n − d
u− d =
er/n − e−σ/√n
eσ/
√
n − e−σ/√n .
Using the Taylor expansion of the exponential function
ex = 1 + x+
x2
2!
+
x3
3!
+ · · · ,
we can approximate p˜u, for large n, as
p˜u =
(1 + r/n+ · · · )− (1− σ/√n+ 1
2
σ2/n+ · · ·)(
1 + σ/
√
n+ 1
2
σ2/n+ · · ·)− (1− σ/√n+ 1
2
σ2/n+ · · ·)
≈ σ/
√
n+
(
r − 1
2
σ2
)
/n
2σ/
√
n
=
1
2
(
1 +
r − 1
2
σ2
σ
√
n
)
=
1
2
(
1 +
µ˜
σ
√
n
)
.
The CLT now applies in the same way, since all that has changed under the risk-neutral
distribution is that the risk-neutral parameters (µ˜, σ˜2) = (r − σ2/2, σ2) play the role of the
original parameters (µ, σ2). We conclude that for large n, ln(S(t)/S(0)) =
∑
i≤nt lnYn,i
converges to a normal distribution with parameters
µ˜t = (r − σ2/2)t and σ˜2t = σ2t,
i.e. S(t)/S(0) converges to a log-normal random variable with the same parameters.
Remark 7.13. There is a different proof of the Black-Scholes formula that uses a continuous
time stochastic process known as Brownian motion; this proof bypasses the multi-period
binomial model altogether.