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Problems and Solutions
in Mathematical Finance
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For other titles in the Wiley Finance Series
please see www.wiley.com/finance
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Problems and Solutions
in Mathematical Finance
Volume 1: Stochastic Calculus
Eric Chin, Dian Nel and Sverrir Ólafsson
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This edition first published 2014
© 2014 John Wiley & Sons, Ltd
Registered office
John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, United Kingdom
For details of our global editorial offices, for customer services and for information about how to apply for
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expert assistance is required, the services of a competent professional should be sought.
Library of Congress Cataloging-in-Publication Data
Chin, Eric,
Problems and solutions in mathematical finance : stochastic calculus / Eric Chin, Dian Nel and Sverrir Ólafsson.
pages cm
Includes bibliographical references and index.
ISBN 978-1-119-96583-1 (cloth)
1. Finance – Mathematical models. 2. Stochastic analysis. I. Nel, Dian, II. Ólafsson, Sverrir, III. Title.
HG106.C495 2014
332.01′ 51922 – dc23
2013043864
A catalogue record for this book is available from the British Library.
ISBN 978-1-119-96583-1 (hardback) ISBN 978-1-119-96607-4 (ebk)
ISBN 978-1-119-96608-1 (ebk)
ISBN 978-1-118-84514-1 (ebk)
Cover design: Cylinder
Typeset in 10/12pt TimesLTStd by Laserwords Private Limited, Chennai, India
Printed in Great Britain by CPI Group (UK) Ltd, Croydon, CR0 4YY
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“the beginning of a task is the biggest step”
Plato, The Republic
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Contents
Preface
Prologue
About the Authors
ix
xi
xv
1
General Probability Theory
1.1 Introduction
1.2 Problems and Solutions
1.2.1 Probability Spaces
1.2.2 Discrete and Continuous Random Variables
1.2.3 Properties of Expectations
1
1
4
4
11
41
2
Wiener Process
2.1 Introduction
2.2 Problems and Solutions
2.2.1 Basic Properties
2.2.2 Markov Property
2.2.3 Martingale Property
2.2.4 First Passage Time
2.2.5 Reflection Principle
2.2.6 Quadratic Variation
51
51
55
55
68
71
76
84
89
3
Stochastic Differential Equations
3.1 Introduction
3.2 Problems and Solutions
3.2.1 Itō Calculus
3.2.2 One-Dimensional Diffusion Process
3.2.3 Multi-Dimensional Diffusion Process
95
95
102
102
123
155
4
Change of Measure
4.1 Introduction
4.2 Problems and Solutions
4.2.1 Martingale Representation Theorem
185
185
192
192
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Contents
4.2.2 Girsanov’s Theorem
4.2.3 Risk-Neutral Measure
5
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Poisson Process
5.1 Introduction
5.2 Problems and Solutions
5.2.1 Properties of Poisson Process
5.2.2 Jump Diffusion Process
5.2.3 Girsanov’s Theorem for Jump Processes
5.2.4 Risk-Neutral Measure for Jump Processes
194
221
243
243
251
251
281
298
322
Appendix A Mathematics Formulae
331
Appendix B Probability Theory Formulae
341
Appendix C Differential Equations Formulae
357
Bibliography
365
Notation
369
Index
373
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Preface
Mathematical finance is based on highly rigorous and, on occasions, abstract mathematical
structures that need to be mastered by anyone who wants to be successful in this field, be it
working as a quant in a trading environment or as an academic researcher in the field. It may
appear strange, but it is true, that mathematical finance has turned into one of the most advanced
and sophisticated field in applied mathematics. This development has had considerable impact
on financial engineering with its extensive applications to the pricing of contingent claims
and synthetic cash flows as analysed both within financial institutions (investment banks) and
corporations. Successful understanding and application of financial engineering techniques to
highly relevant and practical situations requires the mastering of basic financial mathematics.
It is precisely for this purpose that this book series has been written.
In Volume I, the first of a four volume work, we develop briefly all the major mathematical
concepts and theorems required for modern mathematical finance. The text starts with probability theory and works across stochastic processes, with main focus on Wiener and Poisson
processes. It then moves to stochastic differential equations including change of measure and
martingale representation theorems. However, the main focus of the book remains practical.
After being introduced to the fundamental concepts the reader is invited to test his/her knowledge on a whole range of different practical problems. Whereas most texts on mathematical
finance focus on an extensive development of the theoretical foundations with only occasional
concrete problems, our focus is a compact and self-contained presentation of the theoretical
foundations followed by extensive applications of the theory. We advocate a more balanced
approach enabling the reader to develop his/her understanding through a step-by-step collection of questions and answers. The necessary foundation to solve these problems is provided
in a compact form at the beginning of each chapter. In our view that is the most successful way
to master this very technical field.
No one can write a book on mathematical finance today, not to mention four volumes, without
being influenced, both in approach and presentation, by some excellent text books in the field.
The texts we have mostly drawn upon in our research and teaching are (in no particular order
of preference), Tomas Björk, Arbitrage Theory in Continuous Time; Steven Shreve, Stochastic
Calculus for Finance; Marek Musiela and Marek Rutkowski, Martingale Methods in Financial
Modelling and for the more practical aspects of derivatives John Hull, Options, Futures and
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Preface
Other Derivatives. For the more mathematical treatment of stochastic calculus a very influential text is that of Bernt Øksendal, Stochastic Differential Equations. Other important texts are
listed in the bibliography.
Note to the student/reader. Please try hard to solve the problems on your own before you
look at the solutions!
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Prologue
IN THE BEGINNING WAS THE MOTION. . .
The development of modern mathematical techniques for financial applications can be traced
back to Bachelier’s work, Theory of Speculation, first published as his PhD Thesis in 1900.
At that time Bachelier was studying the highly irregular movements in stock prices on the
French stock market. He was aware of the earlier work of the Scottish botanist Robert Brown,
in the year 1827, on the irregular movements of plant pollen when suspended in a fluid. Bachelier worked out the first mathematical model for the irregular pollen movements reported by
Brown, with the intention to apply it to the analysis of irregular asset prices. This was a highly
original and revolutionary approach to phenomena in finance. Since the publication of Bachelier’s PhD thesis, there has been a steady progress in the modelling of financial asset prices.
Few years later, in 1905, Albert Einstein formulated a more extensive theory of irregular molecular processes, already then called Brownian motion. That work was continued and extended
in the 1920s by the mathematical physicist Norbert Wiener who developed a fully rigorous
framework for Brownian motion processes, now generally called Wiener processes.
Other major steps that paved the way for further development of mathematical
finance included the works by Kolmogorov on stochastic differential equations, Fama
on efficient-market hypothesis and Samuelson on randomly fluctuating forward prices.
Further important developments in mathematical finance were fuelled by the realisation of the
importance of Itō’s lemma in stochastic calculus and the Feynman-Kac formula, originally
drawn from particle physics, in linking stochastic processes to partial differential equations
of parabolic type. The Feynman-Kac formula provides an immensely important tool for
the solution of partial differential equations “extracted” from stochastic processes via Itō’s
lemma. The real relevance of Itō’s lemma and Feynman-Kac formula in finance were only
realised after some further substantial developments had taken place.
The year 1973 saw the most important breakthrough in financial theory when Black and
Scholes and subsequently Merton derived a model that enabled the pricing of European call and
put options. Their work had immense practical implications and lead to an explosive increase
in the trading of derivative securities on some major stock and commodity exchanges. However, the philosophical foundation of that approach, which is based on the construction of
risk-neutral portfolios enables an elegant and practical way of pricing of derivative contracts,
has had a lasting and revolutionary impact on the whole of mathematical finance. The development initiated by Black, Scholes and Merton was continued by various researchers, notably
Harrison, Kreps and Pliska in 1980s. These authors established the hugely important role of
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Prologue
martingales and arbitrage theory for the pricing of a large class of derivative securities or,
as they are generally called, contingent claims. Already in the Black, Scholes and Merton
model the risk-neutral measure had been informally introduced as a consequence of the construction of risk-neutral portfolios. Harrison, Kreps and Pliska took this development further
and turned it into a powerful and the most general tool presently available for the pricing of
contingent claims.
Within the Harrison, Kreps and Pliska framework the change of numéraire technique plays
a fundamental role. Essentially the price of any asset, including contingent claims, can be
expressed in terms of units of any other asset. The unit asset plays the role of a numéraire. For
a given asset and a selected numéraire we can construct a probability measure that turns the
asset price, in units of the numéraire, into a martingale whose existence is equivalent to the
absence of an arbitrage opportunity. These results amount to the deepest and most fundamental
in modern financial theory and are therefore a core construct in mathematical finance.
In the wake of the recent financial crisis, which started in the second half of 2007, some
commentators and academics have voiced their opinion that financial mathematicians and their
techniques are to be blamed for what happened. The authors do not subscribe to this view. On
the contrary, they believe that to improve the robustness and the soundness of financial contracts, an even better mathematical training for quants is required. This encompasses a better
comprehension of all tools in the quant’s technical toolbox such as optimisation, probability,
statistics, stochastic calculus and partial differential equations, just to name a few.
Financial market innovation is here to stay and not going anywhere, instead tighter regulations and validations will be the only way forward with deeper understanding of models.
Therefore, new developments and market instruments requires more scrutiny, testing and validation. Any inadequacies and weaknesses of model assumptions identified during the validation process should be addressed with appropriate reserve methodologies to offset sudden
changes in the market direction. The reserve methodologies can be subdivided into model
(e.g., Black-Scholes or Dupire model), implementation (e.g., tree-based or Monte Carlo simulation technique to price the contingent claim), calibration (e.g., types of algorithms to solve
optimization problems, interpolation and extrapolation methods when constructing volatility
surface), market parameters (e.g., confidence interval of correlation marking between underlyings) and market risk (e.g., when market price of a stock is close to the option’s strike price
at expiry time). These are the empirical aspects of mathematical finance that need to be a core
part in the further development of financial engineering.
One should keep in mind that mathematical finance is not, and must never become, an esoteric subject to be left to ivory tower academics alone, but a powerful tool for the analysis of
real financial scenarios, as faced by corporations and financial institutions alike. Mathematical
finance needs to be practiced in the real world for it to have sustainable benefits. Practitioners
must realise that mathematical analysis needs to be built on a clear formulation of financial
realities, followed by solid quantitative modelling, and then stress testing the model. It is our
view that the recent turmoil in financial markets calls for more careful application of quantitative techniques but not their abolishment. Intuition alone or behavioural models have their
role to play but do not suffice when dealing with concrete financial realities such as, risk quantification and risk management, asset and liability management, pricing insurance contracts
or complex financial instruments. These tasks require better and more relevant education for
quants and risk managers.
Financial mathematics is still a young and fast developing discipline. On the other hand, markets present an extremely complex and distributed system where a huge number of interrelated
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financial instruments are priced and traded. Financial mathematics is very powerful in pricing
and managing a limited number of instruments bundled into a portfolio. However, modern
financial mathematics is still rather poor at capturing the extremely intricate contractual interrelationship that exists between large numbers of traded securities. In other words, it is only
to a very limited extent able to capture the complex dynamics of the whole markets, which is
driven by a large number of unpredictable processes which possess varying degrees of correlation. The emergent behaviour of the market is to an extent driven by these varying degrees of
correlations. It is perhaps one of the major present day challenges for financial mathematics to
join forces with modern theory of complexity with the aim of being able to capture the macroscopic properties of the market, that emerge from the microscopic interrelations between a
large number of individual securities. That this goal has not been reached yet is no criticism
of financial mathematics. It only bears witness to its juvenile nature and the huge complexity
of its subject.
Solid training of financial mathematicians in a whole range of quantitative disciplines,
including probability theory and stochastic calculus, is an important milestone in the further
development of the field. In the process, it is important to realise that financial engineering
needs more than just mathematics. It also needs a judgement where the quant should
constantly be reminded that no two market situations or two market instruments are exactly
the same. Applying the same mathematical tools to different situations reminds us of the
fact that we are always dealing with an approximation, which reflects the fact that we are
modelling stochastic processes i.e. uncertainties. Students and practitioners should always
bear this in mind.
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About the Authors
Eric Chin is a quantitative analyst at an investment bank in the City of London where he
is involved in providing guidance on price testing methodologies and their implementation,
formulating model calibration and model appropriateness on commodity and credit products.
Prior to joining the banking industry he worked as a senior researcher at British Telecom investigating radio spectrum trading and risk management within the telecommunications sector. He
holds an MSc in Applied Statistics and an MSc in Mathematical Finance both from University
of Oxford. He also holds a PhD in Mathematics from University of Dundee.
Dian Nel has more than 10 years of experience in the commodities sector. He currently works
in the City of London where he specialises in oil and gas markets. He holds a BEng in Electrical and Electronic Engineering from Stellenbosch University and an MSc in Mathematical
Finance from Christ Church, Oxford University. He is a Chartered Engineer registered with
the Engineering Council UK.
Sverrir Ólafsson is Professor of Financial Mathematics at Reykjavik University; a Visiting
Professor at Queen Mary University, London and a director of Riskcon Ltd, a UK based risk
management consultancy. Previously he was a Chief Researcher at BT Research and held
academic positions at The Mathematical Departments of Kings College, London; UMIST
Manchester and The University of Southampton. Dr Ólafsson is the author of over 95 refereed academic papers and has been a key note speaker at numerous international conferences
and seminars. He is on the editorial board of three international journals. He has provided an
extensive consultancy on financial risk management and given numerous specialist seminars to
finance specialists. In the last five years his main teaching has been MSc courses on Risk Management, Fixed Income, and Mathematical Finance. He has an MSc and PhD in mathematical
physics from the Universities of Tübingen and Karlsruhe respectively.
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1
General Probability Theory
Probability theory is a branch of mathematics that deals with mathematical models of trials
whose outcomes depend on chance. Within the context of mathematical finance, we will review
some basic concepts of probability theory that are needed to begin solving stochastic calculus
problems. The topics covered in this chapter are by no means exhaustive but are sufficient to
be utilised in the following chapters and in later volumes. However, in order to fully grasp the
concepts, an undergraduate level of mathematics and probability theory is generally required
from the reader (see Appendices A and B for a quick review of some basic mathematics and
probability theory). In addition, the reader is also advised to refer to the notation section (pages
369–372) on set theory, mathematical and probability symbols used in this book.
1.1 INTRODUCTION
We consider an experiment or a trial whose result (outcome) is not predictable with certainty.
The set of all possible outcomes of an experiment is called the sample space and we denote it
by Ω. Any subset A of the sample space is known as an event, where an event is a set consisting
of possible outcomes of the experiment.
The collection of events can be defined as a subcollection ℱ of the set of all subsets of Ω
and we define any collection ℱ of subsets of Ω as a field if it satisfies the following.
Definition 1.1 The sample space Ω is the set of all possible outcomes of an experiment
or random trial. A field is a collection (or family) ℱ of subsets of Ω with the following
conditions:
(a) ∅ ∈ ℱ where ∅ is the empty set;
(b) if A ∈ ℱ then Ac ∈ ℱ where Ac is the
⋃ complement of A in Ω;
(c) if A1 , A2 , . . . , An ∈ ℱ, n ≥ 2 then ni=1 Ai ∈ ℱ – that is to say, ℱ is closed under finite
unions.
It should be noted in the definition of a field that ℱ is closed under finite unions (as well
as under finite intersections). As for the case of a collection of events closed under countable
unions (as well as under countable intersections), any collection of subsets of Ω with such
properties is called a 𝜎-algebra.
Definition 1.2 If Ω is a given sample space, then a 𝜎-algebra (or 𝜎-field) ℱ on Ω is a family
(or collection) ℱ of subsets of Ω with the following properties:
(a) ∅ ∈ ℱ;
c
(b) if A ∈ ℱ then Ac ∈ ℱ where
⋃∞ A is the complement of A in Ω;
(c) if A1 , A2 , . . . ∈ ℱ then i=1 Ai ∈ ℱ – that is to say, ℱ is closed under countable unions.
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INTRODUCTION
We next outline an approach to probability which is a branch of measure theory. The reason
for taking a measure-theoretic path is that it leads to a unified treatment of both discrete and
continuous random variables, as well as a general definition of conditional expectation.
Definition 1.3 The pair (Ω, ℱ) is called a measurable space. A probability measure ℙ on a
measurable space (Ω, ℱ) is a function ℙ ∶ ℱ → [0, 1] such that:
(a) ℙ(∅) = 0;
(b) ℙ(Ω) = 1;
⋃∞
(c) if A1 , A2 , . . . ∈ ℱ and {Ai }∞
i=1 is disjoint such that Ai ∩ Aj = ∅, i ≠ j then ℙ( i=1 Ai ) =
∑∞
i=1 ℙ(Ai ).
The triple (Ω, ℱ, ℙ) is called a probability space. It is called a complete probability space
if ℱ also contains subsets B of Ω with ℙ-outer measure zero, that is ℙ∗ (B) = inf{ℙ(A) ∶ A ∈
ℱ, B ⊂ A} = 0.
By treating 𝜎-algebras as a record of information, we have the following definition of a
filtration.
Definition 1.4 Let Ω be a non-empty sample space and let T be a fixed positive number, and
assume for each t ∈ [0, T] there is a 𝜎-algebra ℱt . In addition, we assume that if s ≤ t, then
every set in ℱs is also in ℱt . We call the collection of 𝜎-algebras ℱt , 0 ≤ t ≤ T, a filtration.
Below we look into the definition of a real-valued random variable, which is a function that
maps a probability space (Ω, ℱ, ℙ) to a measurable space ℝ.
Definition 1.5 Let Ω be a non-empty sample space and let ℱ be a 𝜎-algebra of subsets of Ω.
A real-valued random variable X is a function X ∶ Ω → ℝ such that {𝜔 ∈ Ω ∶ X(𝜔) ≤ x}
∈ ℱ for each x ∈ ℝ and we say X is ℱ measurable.
In the study of stochastic processes, an adapted stochastic process is one that cannot “see
into the future” and in mathematical finance we assume that asset prices and portfolio positions
taken at time t are all adapted to a filtration ℱt , which we regard as the flow of information
up to time t. Therefore, these values must be ℱt measurable (i.e., depend only on information
available to investors at time t). The following is the precise definition of an adapted stochastic
process.
Definition 1.6 Let Ω be a non-empty sample space with a filtration ℱt , t ∈ [0, T] and let Xt
be a collection of random variables indexed by t ∈ [0, T]. We therefore say that this collection
of random variables is an adapted stochastic process if, for each t, the random variable Xt is
ℱt measurable.
Finally, we consider the concept of conditional expectation, which is extremely important
in probability theory and also for its wide application in mathematical finance such as pricing
options and other derivative products. Conceptually, we consider a random variable X defined
on the probability space (Ω, ℱ, ℙ) and a sub-𝜎-algebra 𝒢 of ℱ (i.e., sets in 𝒢 are also in ℱ).
Here X can represent a quantity we want to estimate, say the price of a stock in the future, while
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INTRODUCTION
3
𝒢 contains limited information about X such as the stock price up to and including the current
time. Thus, 𝔼(X|𝒢) constitutes the best estimation we can make about X given the limited
knowledge 𝒢. The following is a formal definition of a conditional expectation.
Definition 1.7 (Conditional Expectation) Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be
a sub-𝜎-algebra of ℱ (i.e., sets in 𝒢 are also in ℱ). Let X be an integrable (i.e., 𝔼(|X|) < ∞)
and non-negative random variable. Then the conditional expectation of X given 𝒢, denoted
𝔼(X|𝒢), is any random variable that satisfies:
(a) 𝔼(X|𝒢) is 𝒢 measurable;
(b) for every set A ∈ 𝒢, we have the partial averaging property
∫A
𝔼(X|𝒢) dℙ =
∫A
X dℙ.
From the above definition, we can list the following properties of conditional expectation.
Here (Ω, ℱ, ℙ) is a probability space, 𝒢 is a sub-𝜎-algebra of ℱand X is an integrable random
variable.
• Conditional probability. If 1IA is an indicator random variable for an event A then
𝔼(1IA |𝒢) = ℙ(A|𝒢).
• Linearity. If X1 , X2 , . . . , Xn are integrable random variables and c1 , c2 , . . . , cn are constants then
𝔼(c1 X1 + c2 X2 + . . . + cn Xn |𝒢) = c1 𝔼(X1 |𝒢) + c2 𝔼(X2 |𝒢) + . . . + cn 𝔼(Xn |𝒢).
• Positivity. If X ≥ 0 almost surely then 𝔼(X|𝒢) ≥ 0 almost surely.
• Monotonicity. If X and Y are integrable random variables and X ≤ Y almost surely then
𝔼(X|𝒢) ≤ 𝔼(Y|𝒢).
• Computing expectations by conditioning. 𝔼[𝔼(X|𝒢)] = 𝔼(X).
• Taking out what is known. If X and Y are integrable random variables and X is 𝒢 measurable then
𝔼(XY|𝒢) = X ⋅ 𝔼(Y|𝒢).
• Tower property. If ℋ is a sub-𝜎-algebra of 𝒢 then
𝔼[𝔼(X|𝒢)|ℋ] = 𝔼(X|ℋ).
• Measurability. If X is 𝒢 measurable then 𝔼(X|𝒢) = X.
• Independence. If X is independent of 𝒢 then 𝔼(X|𝒢) = 𝔼(X).
• Conditional Jensen’s inequality. If 𝜑 ∶ ℝ → ℝ is a convex function then
𝔼[𝜑(X)|𝒢] ≥ 𝜑[𝔼(X|𝒢)].
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Probability Spaces
PROBLEMS AND SOLUTIONS
Probability Spaces
1. De Morgan’s Law. Let Ai , i ∈ I where I is some, possibly uncountable, indexing set.
Show that
(⋃
)c ⋂
A =
Ac .
(a)
(⋂i∈I i )c ⋃i∈I ic
(b)
= i∈I Ai .
i∈I Ai
Solution:
(a) Let a ∈
(⋃
i∈I
Ai
)c
which implies a ∉
(
⋃
⋃
i∈I
Ai , so that a ∈ Aci for all i ∈ I. Therefore,
)c
⊆
Ai
⋂
i∈I
On the contrary, if we let a ∈
hence
i∈I
⋂
c
i∈I Ai
then a ∉ Ai for all i ∈ I or a ∈
)c
(
⋃
⋂
c
Ai ⊆
Ai .
i∈I
(⋃
)c
=
Therefore,
i∈I Ai
(b) From (a), we can write
Aci .
(⋃
i∈I
Ai
)c
and
i∈I
⋂
c
i∈I Ai .
(
⋃
)c
Aci
=
i∈I
⋂ ( )c ⋂
Aci =
Ai .
i∈I
i∈I
Taking complements on both sides gives
(
)c
⋃
⋂
Ai
=
Aci .
i∈I
i∈I
◽
2. ⋂
Let ℱ be a 𝜎-algebra of subsets of the sample space Ω. Show that if A1 , A2 , . . . ∈ ℱ then
∞
i=1 Ai ∈ ℱ.
⋃
c
Solution: Given that ℱ is a 𝜎-algebra then Ac1 , Ac2 , . . . ∈ ℱ and ∞
i=1 Ai ∈ ℱ. Further⋃∞ c (⋃∞ c )c
more, the complement of i=1 Ai is
∈ ℱ.
i=1 Ai
(⋃∞ c )c
Thus, from De Morgan’s law (see Problem 1.2.1.1, page 4) we have
=
i=1 Ai
⋂∞ ( c )c ⋂∞
= i=1 Ai ∈ ℱ.
i=1 Ai
◽
3. Show that if ℱ is a 𝜎-algebra of subsets of Ω then {∅, Ω} ∈ ℱ.
Solution: ℱ is a 𝜎-algebra of subsets of Ω, hence if A ∈ ℱ then Ac ∈ ℱ.
Since ∅ ∈ ℱ then ∅c = Ω ∈ ℱ. Thus, {∅, Ω} ∈ ℱ.
◽
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4. Show that if A ⊆ Ω then ℱ = {∅, Ω, A, Ac } is a 𝜎-algebra of subsets of Ω.
Solution: ℱ = {∅, Ω, A, Ac } is a 𝜎-algebra of subsets of Ω since
(i) ∅ ∈ ℱ.
(ii) For ∅ ∈ ℱ then ∅c = Ω ∈ ℱ. For Ω ∈ ℱ then Ωc = ∅ ∈ ℱ. In addition, for A ∈ ℱ
then Ac ∈ ℱ. Finally, for Ac ∈ ℱ then (Ac )c = A ∈ ℱ.
(iii) ∅ ∪ Ω = Ω ∈ ℱ, ∅ ∪ A = A ∈ ℱ, ∅ ∪ Ac = Ac ∈ ℱ, Ω ∪ A = Ω ∈ ℱ, Ω ∪
Ac = Ω ∈ ℱ, ∅ ∪ Ω ∪ A = Ω ∈ ℱ, ∅ ∪ Ω ∪ Ac = Ω ∈ ℱand Ω ∪ A ∪ Ac = Ω ∈ ℱ.
◽
5. Let {ℱ
⋂i }i∈I , I ≠ ∅ be a family of 𝜎-algebras of subsets of the sample space Ω. Show that
ℱ = i∈I ℱi is also a 𝜎-algebra of subsets of Ω.
Solution: ℱ =
⋂
i∈I
ℱi is a 𝜎-algebra by taking note that
(a) Since ∅ ∈ ℱi , i ∈ I therefore ∅ ∈ ℱ as well.
∈ I. Therefore, A ∈ ℱ and hence Ac ∈ ℱ.
(b) If A ∈ ℱi for all i ∈ I then Ac ∈ ℱi , i⋃
, A2 , . . . ∈ ℱi for all i ∈ I then ∞
(c) If A1⋃
k=1 Ak ∈ ℱi , i ∈ I and hence A1 , A2 , . . . ∈ ℱ
and ∞
A
∈
ℱ.
k=1 k
⋂
From the results of (a)–(c) we have shown ℱ = i∈I ℱi is a 𝜎-algebra of Ω.
◽
6. Let Ω = {𝛼, 𝛽, 𝛾} and let
ℱ1 = {∅, Ω, {𝛼}, {𝛽, 𝛾}}
and
ℱ2 = {∅, Ω, {𝛼, 𝛽}, {𝛾}}.
Show that ℱ1 and ℱ2 are 𝜎-algebras of subsets of Ω.
Is ℱ = ℱ1 ∪ ℱ2 also a 𝜎-algebra of subsets of Ω?
Solution: Following the steps given in Problem 1.2.1.4 (page 5) we can easily show ℱ1
and ℱ2 are 𝜎-algebras of subsets of Ω.
By setting ℱ = ℱ1 ∪ ℱ2 = {∅, Ω, {𝛼}, {𝛾}, {𝛼, 𝛽}, {𝛽, 𝛾}}, and since {𝛼} ∈ ℱand {𝛾} ∈
ℱ, but {𝛼} ∪ {𝛾} = {𝛼, 𝛾} ∉ ℱ, then ℱ = ℱ1 ∪ ℱ2 is not a 𝜎-algebra of subsets of Ω.
◽
7. Let ℱbe a 𝜎-algebra of subsets of Ω and suppose ℙ ∶ ℱ → [0, 1] so that ℙ(Ω) = 1. Show
that ℙ(∅) = 0.
Solution: Given that ∅ and Ω are mutually exclusive we therefore have
∅ ∩ Ω = ∅ and ∅ ∪ Ω = Ω.
Thus, we can express
ℙ(∅ ∪ Ω) = ℙ(∅) + ℙ(Ω) − ℙ(∅ ∩ Ω) = 1.
Since ℙ(Ω) = 1 and ℙ(∅ ∩ Ω) = 0 therefore ℙ(∅) = 0.
◽
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8. Let (Ω, ℱ, ℙ) be a probability space and let ℚ ∶ ℱ → [0, 1] be defined by ℚ(A) = ℙ(A|B)
where B ∈ ℱ such that ℙ(B) > 0. Show that (Ω, ℱ, ℚ) is also a probability space.
Solution: To show that (Ω, ℱ, ℚ) is a probability space we note that
ℙ(∅ ∩ B) ℙ(∅)
=
= 0.
ℙ(B)
ℙ(B)
ℙ(Ω ∩ B) ℙ(B)
=
= 1.
(b) ℚ(Ω) = ℙ(Ω|B) =
ℙ(B)
ℙ(B)
(c) Let A1 , A2 , . . . be disjoint members of ℱand hence we can imply A1 ∩ B, A2 ∩ B, . . .
are also disjoint members of ℱ. Therefore,
(a) ℚ(∅) = ℙ(∅|B) =
(
ℚ
∞
⋃
)
(
=ℙ
Ai
)
(⋃∞
)
∞
∞
|
∑
ℙ
ℙ(Ai ∩ B) ∑
i=1 (Ai ∩ B)
|
Ai | B =
ℚ(Ai ).
=
=
|
ℙ(B)
ℙ(B)
i=1
i=1
i=1
|
∞
⋃
i=1
Based on the results of (a)–(c), we have shown that (Ω, ℱ, ℚ) is also a probability space.
◽
9. Boole’s Inequality. Suppose {Ai }i∈I is a countable collection of events. Show that
(
⋃
ℙ
)
Ai
i∈I
≤
∑
( )
ℙ Ai .
i∈I
Solution: Without loss of generality we assume that I = {1, 2, . . .} and define B1 = A1 ,
Bi = Ai \ (A1 ∪ A2 ∪ . . . ∪ Ai−1 ), i ∈ {2, 3, . . .} such that {B1 , B2 , . . .} are pairwise disjoint and
⋃
⋃
Ai =
Bi .
i∈I
i∈I
Because Bi ∩ Bj = ∅, i ≠ j, i, j ∈ I we have
)
(
ℙ
⋃
Ai
(
=ℙ
i∈I
=
∑
⋃
)
Bi
i∈I
ℙ(Bi )
i∈I
=
∑
i∈I
=
ℙ(Ai \(A1 ∪ A2 ∪ . . . ∪ Ai−1 ))
∑{
(
)}
ℙ(Ai ) − ℙ Ai ∩ (A1 ∪ A2 ∪ . . . ∪ Ai−1 )
i∈I
≤
∑
i∈I
ℙ(Ai ).
◽
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10. Bonferroni’s Inequality. Suppose {Ai }i∈I is a countable collection of events. Show that
(
⋂
ℙ
)
≥1−
Ai
i∈I
∑
( )
ℙ Aci .
i∈I
Solution: From De Morgan’s law (see Problem 1.2.1.1, page 4) we can write
(
ℙ
⋂
)
Ai
((
=ℙ
i∈I
⋃
)c )
Aci
(
=1−ℙ
⋃
i∈I
)
.
Aci
i∈I
By applying Boole’s inequality (see Problem 1.2.1.9, page 6) we will have
(
)
⋂
∑ ( )
Ai ≥ 1 −
ℙ Aci
ℙ
i∈I
since ℙ
(⋃
)
c
i∈I Ai
≤
∑
i∈I
( c)
i∈I ℙ Ai .
◽
11. Bayes’ Formula. Let A1 , A2 , . . . , An be a partition of Ω, where
n
⋃
Ai = Ω, Ai ∩ Aj = ∅,
i=1
i ≠ j and each Ai , i, j = 1, 2, . . . , n has positive probability. Show that
ℙ(B|Ai )ℙ(Ai )
.
ℙ(Ai |B) = ∑n
j=1 ℙ(B|Aj )ℙ(Aj )
Solution: From the definition of conditional probability, for i = 1, 2, . . . , n
ℙ(Ai |B) =
ℙ(B|Ai )ℙ(Ai )
ℙ(B|Ai )ℙ(Ai )
ℙ(Ai ∩ B)
ℙ(B|A )ℙ(Ai )
= ∑n
.
= (⋃ i
) = ∑n
n
ℙ(B)
j=1 ℙ(B ∩ Aj )
j=1 ℙ(B|Aj )ℙ(Aj )
ℙ
(B
∩
A
)
j
j=1
◽
12. Principle of Inclusion and Exclusion for Probability. Let A1 , A2 , . . . , An , n ≥ 2 be a collection of events. Show that
ℙ(A1 ∪ A2 ) = ℙ(A1 ) + ℙ(A2 ) − ℙ(A1 ∩ A2 ).
From the above result show that
ℙ(A1 ∪ A2 ∪ A3 ) = ℙ(A1 ) + ℙ(A2 ) + ℙ(A3 ) − ℙ(A1 ∩ A2 ) − ℙ(A1 ∩ A3 )
− ℙ(A2 ∩ A3 ) + ℙ(A1 ∩ A2 ∩ A3 ).
Hence, using mathematical induction show that
(
ℙ
n
⋃
i=1
)
Ai
=
n
∑
i=1
ℙ(Ai ) −
n−1 ∑
n
∑
i=1 j=i+1
ℙ(Ai ∩ Aj ) +
n−2 ∑
n−1 ∑
n
∑
i=1 j=i+1 k=j+1
ℙ(Ai ∩ Aj ∩ Ak )
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− . . . + (−1)n+1 ℙ(A1 ∩ A2 ∩ . . . ∩ An ).
Finally, deduce that
(
ℙ
n
⋂
)
Ai
=
i=1
n
∑
ℙ(Ai ) −
i=1
n−1 ∑
n
∑
ℙ(Ai ∪ Aj ) +
i=1 j=i+1
n−2 ∑
n−1 ∑
n
∑
ℙ(Ai ∪ Aj ∪ Ak )
i=1 j=i+1 k=j+1
− . . . + (−1)n+1 ℙ(A1 ∪ A2 ∪ . . . ∪ An ).
Solution: For n = 2, A1 ∪ A2 can be written as a union of two disjoint sets
A1 ∪ A2 = A1 ∪ (A2 \ A1 ) = A1 ∪ (A2 ∩ Ac1 ).
Therefore,
ℙ(A1 ∪ A2 ) = ℙ(A1 ) + ℙ(A2 ∩ Ac1 )
and since ℙ(A2 ) = ℙ(A2 ∩ A1 ) + ℙ(A2 ∩ Ac1 ) we will have
ℙ(A1 ∪ A2 ) = ℙ(A1 ) + ℙ(A2 ) − ℙ(A1 ∩ A2 ).
For n = 3, and using the above results, we can write
ℙ(A1 ∪ A2 ∪ A3 ) = ℙ(A1 ∪ A2 ) + ℙ(A3 ) − ℙ[(A1 ∪ A2 ) ∩ A3 ]
= ℙ(A1 ) + ℙ(A2 ) + ℙ(A3 ) − ℙ(A1 ∩ A2 ) − ℙ[(A1 ∪ A2 ) ∩ A3 ].
Since (A1 ∪ A2 ) ∩ A3 = (A1 ∩ A3 ) ∪ (A2 ∩ A3 ) therefore
ℙ[(A1 ∪ A2 ) ∩ A3 ] = ℙ[(A1 ∩ A3 ) ∪ (A2 ∩ A3 )]
= ℙ(A1 ∩ A3 ) + ℙ(A2 ∩ A3 ) − ℙ[(A1 ∩ A3 ) ∩ (A2 ∩ A3 )]
= ℙ(A1 ∩ A3 ) + ℙ(A2 ∩ A3 ) − ℙ(A1 ∩ A2 ∩ A3 ).
Thus,
ℙ(A1 ∪ A2 ∪ A3 ) = ℙ(A1 ) + ℙ(A2 ) + ℙ(A3 ) − ℙ(A1 ∩ A2 ) − ℙ(A1 ∩ A3 )
−ℙ(A2 ∩ A3 ) + ℙ(A1 ∩ A2 ∩ A3 ).
Suppose the result is true for n = m, where m ≥ 2. For n = m + 1, we have
ℙ
(( m
⋃
)
Ai
)
∪ Am+1
=ℙ
i=1
(m
⋃
(
=ℙ
)
m
⋃
i=1
)
Ai
)
((
+ ℙ Am+1 − ℙ
Ai
i=1
(
(
)
+ ℙ Am+1 − ℙ
(
m
⋃
)
Ai
)
∩ Am+1
i=1
m
⋃
(
Ai ∩ Am+1
i=1
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.
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By expanding the terms we have
(m+1 )
m
m
m
m−1
m−2
⋃
∑
∑ ∑
∑ m−1
∑ ∑
ℙ
Ai =
ℙ(Ai ) −
ℙ(Ai ∩ Aj ) +
ℙ(Ai ∩ Aj ∩ Ak )
i=1
i=1
i=1 j=i+1
i=1 j=i+1 k=j+1
− . . . + (−1)m+1 ℙ(A1 ∩ A2 ∩ . . . ∩ Am )
)
(
)
(
+ℙ Am+1 − ℙ (A1 ∩ Am+1 ) ∪ (A2 ∩ Am+1 ) . . . ∪ (Am ∩ Am+1 )
∑
m+1
=
m
∑ ∑
m−1
ℙ(Ai ) −
i=1
m
∑ ∑ ∑
m−2 m−1
ℙ(Ai ∩ Aj ) +
i=1 j=i+1
ℙ(Ai ∩ Aj ∩ Ak )
i=1 j=i+1 k=j+1
− . . . + (−1)m+1 ℙ(A1 ∩ A2 ∩ . . . ∩ Am )
−
m
∑
ℙ(Ai ∩ Am+1 ) +
i=1
m−1
∑
m
∑
ℙ(Ai ∩ Aj ∩ Am+1 )
i=1 j=i+1
+ . . . − (−1)m+1 ℙ(A1 ∩ A2 ∩ . . . ∩ Am+1 )
=
m+1
∑
ℙ(Ai ) −
i=1
m m+1
∑
∑
ℙ(Ai ∩ Aj ) +
i=1 j=i+1
m−1
∑
m m+1
∑
∑
ℙ(Ai ∩ Aj ∩ Ak )
i=1 j=i+1 k=j+1
− . . . + (−1)m+2 ℙ(A1 ∩ A2 ∩ . . . ∩ Am+1 ).
Therefore, the result is also true for n = m + 1. Thus, from mathematical induction we
have shown for n ≥ 2,
( n )
n
n−1 n
n−2 n−1
n
⋃
∑
∑
∑
∑
∑ ∑
Ai =
ℙ(Ai ) −
ℙ(Ai ∩ Aj ) +
ℙ(Ai ∩ Aj ∩ Ak )
ℙ
i=1
i=1
i=1 j=i+1
i=1 j=i+1 k=j+1
− . . . + (−1)n+1 ℙ(A1 ∩ A2 ∩ . . . ∩ An ).
From Problem 1.2.1.1 (page 4) we can write
( n )
(( n
)c )
( n
)
⋂
⋃
⋃
ℙ
Ai = ℙ
Aci
Aci .
=1−ℙ
i=1
i=1
i=1
Thus, we can expand
( n )
n
n−1 ∑
n−2 ∑
n
n−1 ∑
n
⋂
∑
∑
∑
Ai = 1 −
ℙ(Aci ) +
ℙ(Aci ∩ Acj ) −
ℙ(Aci ∩ Acj ∩ Ack )
ℙ
i=1
i=1
i=1 j=i+1
i=1 j=i+1 k=j+1
+ . . . − (−1)n+1 ℙ(Ac1 ∩ Ac2 ∩ . . . ∩ Acn )
=1−
n
∑
(1 − ℙ(Ai )) +
i=1
−
n−2 ∑
n−1 ∑
n
∑
i=1 j=i+1 k=j+1
n
n−1 ∑
∑
(1 − ℙ(Ai ∪ Aj ))
i=1 j=i+1
(1 − ℙ(Ai ∪ Aj ∪ Ak ))
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+ . . . − (−1)n+1 (1 − ℙ(A1 ∪ A2 ∪ . . . ∪ An ))
=
n
∑
ℙ(Ai ) −
i=1
n−1 ∑
n
∑
ℙ(Ai ∪ Aj ) +
n−2 ∑
n−1 ∑
n
∑
i=1 j=i+1
ℙ(Ai ∪ Aj ∪ Ak )
i=1 j=i+1 k=j+1
− . . . + (−1)n+1 ℙ(A1 ∪ A2 ∪ . . . ∪ An ).
◽
13. Borel–Cantelli Lemma. Let (Ω, ℱ, ℙ) be a probability space and let A1 , A2 , . . . be sets in
ℱ. Show that
∞ ∞
∞
⋂
⋃
⋃
Ak ⊆
Ak
m=1 k=m
k=m
and hence prove that
(
ℙ
∞
∞ ⋃
⋂
)
Ak
m=1 k=m
Solution: Let Bm =
∞
⋃
k=m
∞
∑
⎧
ℙ(Ak ) = ∞
⎪1 if Ai ∩ Aj = ∅, i ≠ j and
k=1
⎪
=⎨
∞
⎪
∑
ℙ(Ak ) < ∞.
⎪0 if
⎩
k=1
Ak and since
∞
⋂
m=1
∞ ∞
⋂
⋃
Bm ⊆ Bm therefore we have
Ak ⊆
m=1 k=m
∞
⋃
Ak .
k=m
From Problem 1.2.1.9 (page 6) we can deduce that
)
(∞ )
(∞ ∞
∞
⋂⋃
⋃
∑
ℙ
Ak ≤ ℙ
Ak ≤
ℙ(Ak ).
m=1 k=m
For the case
and hence
∞
∑
k=1
k=m
k=m
ℙ(Ak ) < ∞ and given it is a convergent series, then lim
(
ℙ
∞ ∞
⋂
⋃
m→∞ k=m
)
Ak
∞
∑
ℙ(Ak ) = 0
=0
m=1 k=m
if
∞
∑
k=1
ℙ(Ak ) < ∞.
For the case Ai ∩ Aj = ∅, i ≠ j and
∞
∑
k=1
(( ∞
)c)
(∞ )
⋃
⋃
=
ℙ(Ak ) = ∞, since ℙ
Ak + ℙ
Ak
k=m
1 therefore from Problem 1.2.1.1 (page 4)
(∞
)
(( ∞ )c )
(∞
)
⋃
⋃
⋂
c
Ak = 1 − ℙ
Ak
Ak .
ℙ
=1−ℙ
k=m
k=m
k=m
k=m
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∞
∑
From independence and because
11
ℙ(Ak ) = ∞ we can express
k=1
(
ℙ
∞
⋂
)
Ack
=
k=m
∞
∏
∞
∞
∑∞
( ) ∏
(
( )) ∏
ℙ Ack =
e−ℙ(Ak ) = e− k=m ℙ(Ak ) = 0
1 − ℙ Ak ≤
k=m
k=m
(
for all m and hence
ℙ
∞
⋃
k=m
)
Ak
(
=1−ℙ
k=m
∞
⋂
)
Ack
=1
k=m
for all m. Taking the limit m → ∞,
(
ℙ
∞
∞ ⋃
⋂
)
Ak
(
= lim ℙ
m→∞
m=1 k=m
for the case Ai ∩ Aj = ∅, i ≠ j and
1.2.2
∞
∑
k=1
∞
⋃
)
Ak
=1
k=m
ℙ(Ak ) = ∞.
◽
Discrete and Continuous Random Variables
1. Bernoulli Distribution. Let X be a Bernoulli random variable, X ∼ Bernoulli(p), p ∈ [0, 1]
with probability mass function
ℙ(X = 1) = p,
ℙ(X = 0) = 1 − p.
Show that 𝔼(X) = p and Var(X) = p(1 − p).
Solution: If X ∼ Bernoulli(p) then we can write
ℙ(X = x) = px (1 − p)1−x ,
x ∈ {0, 1}
and by definition
𝔼(X) =
1
∑
xℙ(X = x) = 0 ⋅ (1 − p) + 1 ⋅ p = p
x=0
𝔼(X ) =
2
1
∑
x2 ℙ(X = x) = 0 ⋅ (1 − p) + 1 ⋅ p = p
x=0
and hence
𝔼(X) = p,
Var(X) = 𝔼(X 2 ) − 𝔼(X)2 = p(1 − p).
◽
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1.2.2 Discrete and Continuous Random Variables
2. Binomial Distribution. Let {Xi }ni=1 be a sequence of independent Bernoulli random variables each with probability mass function
ℙ(Xi = 1) = p,
ℙ(Xi = 0) = 1 − p,
and let
X=
n
∑
p ∈ [0, 1]
Xi .
i=1
Show that X follows a binomial distribution, X ∼ Binomial(n, p) with probability mass
function
( )
n k
ℙ(X = k) =
p (1 − p)n−k , k = 0, 1, 2, . . . , n
k
such that 𝔼(X) = np and Var(X) = np(1 − p).
Using the central limit theorem show that X is approximately normally distributed, X ∻
𝒩(np, np(1 − p)) as n → ∞.
Solution: The random variable X counts the number of Bernoulli variables X1 , . . . , Xn
that are equal to 1, i.e., the number of successes in the n independent trials. Clearly X takes
values in the set N = {0, 1, 2, . . . , n}. To calculate the probability that X = k, where k ∈ N
is the number of successes we let E be the event such that Xi1 = Xi2 = . . . = Xik = 1 and
Xj = 0 for all j ∈ N\ S where S = {i1 , i2 , . . . , ik }. Then, because the Bernoulli variables
are independent and identically distributed,
ℙ(E) =
∏
ℙ(Xj = 1)
j∈S
∏
ℙ(Xj = 0) = pk (1 − p)n−k .
j∈N \ S
( )
n
combinations to select sets of indices i1 , . . . , ik from N, which
k
are mutually exclusive events, so
However, as there are
ℙ(X = k) =
( )
n k
p (1 − p)n−k ,
k
k = 0, 1, 2, . . . , n.
From the definition of the moment generating function of discrete random variables (see
Appendix B),
( ) ∑ tx
MX (t) = 𝔼 etX =
e ℙ(X = x)
x
( )
n x
where t ∈ ℝ and substituting ℙ(X = x) =
p (1 − p)n−x we have
x
MX (t) =
n
∑
x=0
( )
n ( )
∑
n x
n
n−x
p (1 − p)
(pet )x (1 − p)n−x = (1 − p + pet )n .
e
=
x
x
tx
x=0
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13
By differentiating MX (t) with respect to t twice we have
M ′X (t) = npet (1 − p + pet )n−1
M ′′X (t) = npet (1 − p + pet )n−1 + n(n − 1)p2 e2t (1 − p + pet )n−2
and hence
𝔼(X) = M ′X (0) = np
Var(X) = 𝔼(X 2 ) − 𝔼(X)2 = M ′′X (0) − M ′X (0)2 = np(1 − p).
Given the sequence Xi ∼ Bernoulli(p), i = 1, 2, . . . , n are independent and identically distributed, each having expectation 𝜇 = p and variance 𝜎 2 = p(1 − p), then as n → ∞, from
the central limit theorem
∑n
i=1 Xi − n𝜇 D
−−→ 𝒩(0, 1)
√
𝜎 n
or
D
X − np
−−→ 𝒩(0, 1).
√
np(1 − p)
Thus, as n → ∞, X ∻ 𝒩(np, np(1 − p)).
◽
3. Poisson Distribution. A discrete Poisson distribution, Poisson(𝜆) with parameter 𝜆 > 0
has the following probability mass function:
ℙ(X = k) =
𝜆k −𝜆
e ,
k!
k = 0, 1, 2, . . .
Show that 𝔼(X) = 𝜆 and Var(X) = 𝜆.
For a random variable following a binomial distribution, Binomial(n, p), 0 ≤ p ≤ 1 show
that as n → ∞ and with p = 𝜆∕n, the binomial distribution tends to the Poisson distribution
with parameter 𝜆.
Solution: From the definition of the moment generating function
( ) ∑ tx
MX (t) = 𝔼 etX =
e ℙ(X = x)
x
where t ∈ ℝ and substituting ℙ(X = x) =
MX (t) =
∞
∑
etx
x=0
𝜆x −𝜆
e we have
x!
∞
∑
(𝜆et )x
t
𝜆x −𝜆
e = e−𝜆
= e𝜆(e −1) .
x!
x!
x=0
By differentiating MX (t) with respect to t twice
M ′X (t) = 𝜆et e𝜆(e −1) ,
t
M ′′X (t) = (𝜆et + 1)𝜆et e𝜆(e −1)
t
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we have
𝔼(X) = M ′X (0) = 𝜆
′′
′
Var(X) = 𝔼(X 2 ) − 𝔼(X)2 = MX (0) − MX (0)2 = 𝜆.
If X ∼ Binomial(n, p) then we can write
ℙ(X = k) =
n!
pk (1 − p)n−k
k!(n − k)!
n(n − 1) . . . (n − k + 1)(n − k)!
k!(n − k)!
(
)
(n − k)(n − k − 1) 2
k
× p 1 − (n − k)p +
p + ...
2!
(
)
n(n − 1) . . . (n − k + 1) k
(n − k)(n − k − 1) 2
=
p 1 − (n − k)p +
p + ... .
k!
2!
=
For the case when n → ∞ so that n ≫ k we have
(
)
(np)2
nk
ℙ(X = k) ≈ pk 1 − np +
+ ...
k!
2!
=
nk k −np
pe .
k!
By setting p = 𝜆∕n, we have ℙ(X = k) ≈
𝜆k −𝜆
e .
k!
◽
4. Exponential Distribution. Consider a continuous random variable X following an exponential distribution, X ∼ Exp(𝜆) with probability density function
fX (x) = 𝜆e−𝜆x ,
x≥0
1
1
where the parameter 𝜆 > 0. Show that 𝔼(X) = and Var(X) = 2 .
𝜆
𝜆
Prove that X ∼ Exp(𝜆) has a memory less property given as
ℙ(X > s + x|X > s) = ℙ(X > x) = e−𝜆x ,
x, s ≥ 0.
For a sequence of Bernoulli trials drawn from a Bernoulli distribution, Bernoulli(p), 0 ≤
p ≤ 1 performed at time Δt, 2Δt, . . . where Δt > 0 and if Y is the waiting time for the
first success, show that as Δt → 0 and p → 0 such that p∕Δt approaches a constant 𝜆 > 0,
then Y ∼ Exp(𝜆).
Solution: For t < 𝜆, the moment generating function for a random variable X ∼ Exp(𝜆) is
( )
MX (t) = 𝔼 etX =
∞
∫0
∞
etu 𝜆e−𝜆u du = 𝜆
∫0
e−(𝜆−t)u du =
𝜆
.
𝜆−t
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15
Differentiation of MX (t) with respect to t twice yields
′
MX (t) =
𝜆
,
(𝜆 − t)2
Therefore,
′
𝔼(X) = MX (0) =
2𝜆
.
(𝜆 − t)3
′′
MX (t) =
1
,
𝜆
′′
𝔼(X 2 ) = MX (0) =
and the variance of X is
Var(X) = 𝔼(X 2 ) − [𝔼(X)]2 =
By definition
ℙ(X > x) =
2
𝜆2
1
.
𝜆2
∞
∫x
𝜆e−𝜆u du = e−𝜆x
and
∞
−𝜆u
ℙ(X > s + x, X > s) ℙ(X > s + x) ∫s+x 𝜆e du
= e−𝜆x .
=
= ∞
ℙ(X > s + x|X > s) =
ℙ(X > s)
ℙ(X > s)
∫s 𝜆e−𝜆𝑣 d𝑣
Thus,
ℙ(X > s + x|X > s) = ℙ(X > x).
If Y is the waiting time for the first success then for k = 1, 2, . . .
ℙ(Y > kΔt) = (1 − p)k .
By setting y = kΔt, and in the limit Δt → 0 and assuming that p → 0 so that p∕Δt → 𝜆,
for some positive constant 𝜆,
(
( ) )
x
ℙ(Y > y) = ℙ Y >
Δt
Δt
≈ (1 − 𝜆Δt)y∕Δt
( )(
= 1 − 𝜆y +
≈ 1 − 𝜆y +
y
Δt
y
Δt
)
−1
2!
(𝜆Δt)2 + . . .
(𝜆y)2
+ ...
2!
= e−𝜆x .
In the limit Δt → 0 and p → 0,
ℙ(Y ≤ y) = 1 − ℙ(Y > y) ≈ 1 − e−𝜆y
and the probability density function is therefore
fY (y) =
d
ℙ(Y ≤ y) ≈ 𝜆e−𝜆y , y ≥ 0.
dy
◽
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5. Gamma Distribution. Let U and V be continuous independent random variables and let
W = U + V. Show that the probability density function of W can be written as
∞
fW (𝑤) =
∫−∞
∞
fV (𝑤 − u)fU (u) du =
∫−∞
fU (𝑤 − 𝑣)fV (𝑣) d𝑣
where fU (u) and fV (𝑣) are the density functions of U and V, respectively.
Let X1 , X2 , . . . , Xn ∼ Exp(𝜆) be a sequence of independent and identically distributed random variables, each following an exponential distribution with common parameter 𝜆 > 0.
Prove that if
n
∑
Y=
Xi
i=1
then Y follows a gamma distribution, Y ∼ Gamma (n, 𝜆) with the following probability
density function:
(𝜆y)n−1 −𝜆y
fY (y) =
𝜆e , y ≥ 0.
(n − 1)!
Show also that 𝔼(Y) =
n
n
and Var(Y) = 2 .
𝜆
𝜆
Solution: From the definition of the cumulative distribution function of W = U + V we
obtain
f (u, 𝑣) dud𝑣
FW (𝑤) = ℙ(W ≤ 𝑤) = ℙ(U + V ≤ 𝑤) =
∫ ∫ UV
u+𝑣≤𝑤
where fUV (u, 𝑣) is the joint probability density function of (U, V). Since U ⟂
⟂ V therefore
fUV (u, 𝑣) = fU (u)fV (𝑣) and hence
FW (𝑤) =
fUV (u, 𝑣) dud𝑣
∫ ∫
u+𝑣≤𝑤
=
fU (u)fV (𝑣) dud𝑣
∫ ∫
u+𝑣≤𝑤
∞
=
∫−∞
{
}
𝑤−u
∫−∞
fV (𝑣) d𝑣
fU (u) du
∞
=
∫−∞
FV (𝑤 − u)fU (u) du.
By differentiating FW (𝑤) with respect to 𝑤, we have the probability density function
fW (𝑤) given as
∞
fW (𝑤) =
∞
d
F (𝑤 − u)fU (u) du =
f (𝑤 − u)fU (u) du.
∫−∞ V
d𝑤 ∫−∞ V
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17
Using the same steps we can also obtain
∞
fW (𝑤) =
fU (𝑤 − 𝑣)fV (𝑣) d𝑣.
∫−∞
∑
To show that Y = ni=1 Xi ∼ Gamma (n, 𝜆) where X1 , X2 , . . . , Xn ∼ Exp(𝜆), we will prove
the result via mathematical induction.
For n = 1, we have Y = X1 ∼ Exp(𝜆) and the gamma density fY (y) becomes
fY (y) = 𝜆e−𝜆y ,
y ≥ 0.
Therefore, the result is true for n = 1.
Let us assume that the result holds for n = k and we now wish to compute the density for
the case n = k + 1. Since X1 , X2 , . . . , Xk+1 are all mutually independent and identically
∑
distributed, by setting U = ki=1 Xi and V = Xk+1 , and since U ≥ 0, V ≥ 0, the density of
∑k
Y = i=1 Xi + Xk+1 can be expressed as
y
fY (y) =
fV (y − u)fU (u) du
∫0
y
=
∫0
𝜆e−𝜆(y−u) ⋅
(𝜆u)k−1 −𝜆u
du
𝜆e
(k − 1)!
y
=
𝜆k+1 e−𝜆y
uk−1 du
(k − 1)! ∫0
=
(𝜆y)k −𝜆y
𝜆e
k!
∑
which shows the result is also true for n = k + 1. Thus, Y = ni=1 Xi ∼ Gamma (n, 𝜆).
Given that X1 , X2 , . . . , Xn ∼ Exp(𝜆) are independent and identically distributed with com1
1
mon mean and variance 2 , therefore
𝜆
𝜆
( n )
n
∑
∑
n
Xi =
𝔼(Xi ) =
𝔼(Y) = 𝔼
𝜆
i=1
i=1
and
Var(Y) = Var
( n
∑
)
Xi
i=1
=
n
∑
Var(Xi ) =
i=1
n
.
𝜆2
◽
6. Normal Distribution Property I. Show that for constants a, L and U such that t > 0 and
L < U,
U
(
1
a𝑤− 2
1
e
√
2𝜋t ∫L
𝑤
√
t
)
[ (
)2
d𝑤 = e
1 2
a t
2
Φ
U − at
√
t
(
−Φ
)]
L − at
√
t
where Φ(⋅) is the cumulative distribution function of a standard normal.
4:33 P.M.
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Solution: Simplifying the integrand we have
(
U
1
a𝑤− 2
1
e
√
∫
2𝜋t L
𝑤
√
t
)2
(
U
1
−
1
d𝑤 = √
e 2
∫
2𝜋t L
U −1
2
𝑤2 −2a𝑤t
t
[(
1
e
=√
2𝜋t ∫L
1 2
(
U
1
−
e2a t
e 2
=√
2𝜋t ∫L
By setting x =
U
∫L
𝑤−at
√
t
𝑤−at
√
t
)
d𝑤
]
)2
−a2 t
d𝑤
)2
d𝑤.
𝑤 − at
√ we can write
t
(
1
−
1
e 2
√
2𝜋t
U
𝑤−at
√
t
)2
d𝑤 =
1 2
1
e− 2 x dx = Φ
√
2𝜋
∫ L−at
√
t
(
1
a𝑤− 2
1
Thus, √
e
2𝜋t ∫L
𝑤
√
t
1 2
a t
2
Φ
U − at
√
t
)
[ (
)2
d𝑤 = e
)
(
U−at
√
t
U − at
√
t
(
−Φ
(
.
)]
L − at
√
t
−Φ
)
L − at
√
t
.
◽
7. Normal Distribution Property II. Show that if X ∼ 𝒩(𝜇, 𝜎 2 ) then for 𝛿 ∈ { − 1, 1},
𝔼[max{𝛿(e − K), 0}] = 𝛿e
X
𝜇+ 21 𝜎 2
(
Φ
𝛿(𝜇 + 𝜎 2 − log K)
𝜎
)
(
− 𝛿KΦ
𝛿(𝜇 − log K)
𝜎
)
where K > 0 and Φ(⋅) denotes the cumulative standard normal distribution function.
Solution: We first let 𝛿 = 1,
∞
𝔼[max{eX − K, 0}] =
∫log K
(ex − K) fX (x) dx
∞
=
−1
1
e 2
(ex − K) √
∫log K
𝜎 2𝜋
∞
=
By setting 𝑤 =
∫log K
−1
1
e 2
√
𝜎 2𝜋
(
)
x−𝜇 2
+x
𝜎
(
)
x−𝜇 2
𝜎
dx
∞
dx − K
∫log K
−1
1
e 2
√
𝜎 2𝜋
(
)
x−𝜇 2
𝜎
x−𝜇
and z = 𝑤 − 𝜎 we have
𝜎
∞
𝔼[max{eX − K, 0}] =
∫ log K−𝜇
𝜎
∞
1 2
1 2
1
1
e− 2 𝑤 +𝜎𝑤+𝜇 d𝑤 − K
e− 2 𝑤 d𝑤
√
√
log K−𝜇
∫
2𝜋
2𝜋
𝜎
dx.
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∞
1 2
= e𝜇+ 2 𝜎
∫ log K−𝜇
𝜎
19
∞
1
1 2
2
1
1
e− 2 (𝑤−𝜎) d𝑤 − K
e− 2 𝑤 d𝑤
√
√
log K−𝜇
∫
2𝜋
2𝜋
𝜎
∞
∞
1 2
1 2
1
1
e− 2 z dz − K
e− 2 𝑤 d𝑤
√
√
∫ log K−𝜇 2𝜋
∫
2𝜋
𝜎
(
)
(
)
2
𝜇 + 𝜎 − log K
𝜇 − log K
𝜇+ 12 𝜎 2
=e
Φ
− KΦ
.
𝜎
𝜎
=e
𝜇+ 12 𝜎 2
log K−𝜇−𝜎 2
𝜎
Using similar steps for the case 𝛿 = −1 we can also show that
)
)
(
(
1 2
log K − 𝜇
log K − (𝜇 + 𝜎 2 )
− e𝜇+ 2 𝜎 Φ
.
𝔼[max{K − eX , 0}] = KΦ
𝜎
𝜎
◽
8. For x > 0 show that
∞
(
)
1 2
1 2
1 2
1
1
1
1
1
− 3 √
e− 2 x ≤
e− 2 z dz ≤ √
e− 2 x .
√
∫
x x
x
2𝜋
2𝜋
x 2𝜋
∞
1 2
1
1 d𝑣
=
e− 2 z dz using integration by parts, we let u = ,
√
∫x
z dz
2𝜋
1 2
1
du
1
= − 2 and 𝑣 = − √
e− 2 z . Therefore,
so that
dz
z
2𝜋
Solution: Solving
1 2
z
e− 2 z
√
2𝜋
∞
∫x
|∞
∞
1 2
1
1
1
− 12 z2
− 12 z2 ||
e
e
e− 2 z dz
dz = − √
√
√
| −∫
|
x
2𝜋
z 2𝜋
z2 2𝜋
|x
1 2
1
e− 2 x −
= √
∫x
x 2𝜋
∞
1 2
1
e− 2 z dz
√
z2 2𝜋
1 2
1
e− 2 x
≤ √
x 2𝜋
∞
since
∫x
1 2
1
e− 2 z dz > 0.
√
z2 2𝜋
∞
1 2
1
e− 2 z dz by parts where we let u =
√
∫x z2 2𝜋
1 2
du
1
3
so that
= − 4 and 𝑣 = − √
e− 2 z and hence
dz
z
2𝜋
To obtain the lower bound, we integrate
1 2
1 d𝑣
z
,
= √
e− 2 z
3
z dz
2𝜋
∞
∫x
|∞
∞
1 2
1
3
1
− 12 z2
− 12 z2 ||
dz = − √
−
e
e
e− 2 z dz
√
√
|
∫x z4 2𝜋
|
z2 2𝜋
z3 2𝜋
|x
=
1 2
1
e− 2 x −
√
∫x
x3 2𝜋
∞
3
√
z4 2𝜋
1 2
e− 2 z dz.
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Therefore,
∞
∫x
∞
since
∫x
we have
1 2
1 2
1 2
1
1
1
e− 2 z dz = √
e− 2 z − √
e− 2 x +
√
∫x
2𝜋
x 2𝜋
x3 2𝜋
(
)
1 2
1
1
1
− 3 √
e− 2 x
≥
x x
2𝜋
∞
3
√
1 2
e− 2 z dz
z4 2𝜋
1 2
3
e− 2 z dz > 0. By taking into account both the lower and upper bounds
√
z4 2𝜋
(
1
1
−
x x3
)
1 2
1
e− 2 x ≤
√
∫x
2𝜋
∞
1 2
1 2
1
1
e− 2 z dz ≤ √
e− 2 x .
√
2𝜋
x 2𝜋
◽
9. Lognormal Distribution I. Let Z ∼ 𝒩(0, 1), show that the moment generating function of
a standard normal distribution is
1 2
( )
𝔼 e𝜃Z = e 2 𝜃
for a constant 𝜃.
Show that if X ∼ 𝒩(𝜇, 𝜎 2 ) then Y = eX follows a lognormal distribution, Y = eX ∼
log-𝒩(𝜇, 𝜎 2 ) with probability density function
fY (y) =
−1
1
e 2
√
y𝜎 2𝜋
(
)
log y−𝜇 2
𝜎
(
)
1 2
2
2
with mean 𝔼(Y) = e𝜇+ 2 𝜎 and variance Var(Y) = e𝜎 − 1 e2𝜇+𝜎 .
Solution: By definition
( )
𝔼 e𝜃Z =
∞
1 2
1
e𝜃z ⋅ √
e− 2 z dz
2𝜋
∫−∞
∞
=
1
2 1 2
1
e− 2 (z−𝜃) + 2 𝜃 dz
√
2𝜋
∫−∞
∞
1 2
= e2𝜃
∫−∞
1
2
1
e− 2 (z−𝜃) dz
√
2𝜋
1 2
= e2𝜃 .
For y > 0, by definition
(
)
ℙ e < y = ℙ(X < log y) =
X
log y
∫−∞
−1
1
e 2
√
𝜎 2𝜋
(
)
x−𝜇 2
𝜎
dx
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21
and hence
fY (y) =
)
−1
d ( X
1
ℙ e <y = √
e 2
dy
y𝜎 2𝜋
(
)
log y−𝜇 2
𝜎
.
Given that log Y ∼ 𝒩(𝜇, 𝜎 2 ) we can write log Y = 𝜇 + 𝜎Z, Z ∼ 𝒩(0, 1) and hence
1 2
(
)
( )
𝔼(Y) = 𝔼 e𝜇+𝜎Z = e𝜇 𝔼 e𝜎Z = e𝜇+ 2 𝜎
1 2
( )
since 𝔼 e𝜃Z = e 2 𝜃 is the moment generating function of a standard normal distribution.
Taking second moments,
(
)
)
( )
(
2
𝔼 Y 2 = 𝔼 e2𝜇+2𝜎Z = e2𝜇 𝔼 e2𝜎Z = e2𝜇+2𝜎 .
Therefore,
(
( 2
)
)
1 2 2
2
2
= e𝜎 − 1 e2𝜇+𝜎 .
Var(Y) = 𝔼(Y 2 ) − [𝔼(Y)]2 = e2𝜇+2𝜎 − e𝜇+ 2 𝜎
10. Lognormal Distribution II. Let X ∼ log-𝒩(𝜇, 𝜎 2 ), show that for n ∈ ℕ
1 2 2
𝜎
𝔼 (X n ) = en𝜇+ 2 n
.
Solution: Given X ∼ log-𝒩(𝜇, 𝜎 2 ) the density function is
fX (x) =
− 12
1
√
(
e
)
log x−𝜇 2
𝜎
,
x>0
x𝜎 2𝜋
and for n ∈ ℕ,
∞
𝔼(X n ) =
xn fX (x) dx
∫0
∞
=
By substituting z =
∫0
1
1
n −
√ x e 2
x𝜎 2𝜋
(
)
log x−𝜇 2
𝜎
dx.
log x − 𝜇
dz
1
so that x = e𝜎z+𝜇 and
=
,
𝜎
dx x𝜎
∞
𝔼(X n ) =
∫−∞
∞
=
∫−∞
1 2
1
en(𝜎z+𝜇) e− 2 z x𝜎 dz
√
x𝜎 2𝜋
1 2
1
e− 2 z +n(𝜎z+𝜇) dz
√
2𝜋
1 2 2
𝜎
= en𝜇+ 2 n
∞
∫−∞
1
2
1
e− 2 (z−n𝜎) dz
√
2𝜋
◽
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1.2.2 Discrete and Continuous Random Variables
1 2 2
𝜎
= en𝜇+ 2 n
∞
since
∫−∞
1
2
1
e− 2 (z−n𝜎) dz = 1.
√
2𝜋
◽
11. Folded Normal Distribution. Show that if X ∼ 𝒩(𝜇, 𝜎 2 ) then Y = |X| follows a folded
normal distribution, Y = |X| ∼ 𝒩f (𝜇, 𝜎 2 ) with probability density function
√
fY (y) =
√
with mean
𝔼 (Y) = 𝜎
(
1
2 −2
e
𝜋𝜎 2
2
𝜋
e
y2 +𝜇 2
𝜎2
( )2
− 12 𝜇𝜎
)
cosh
( 𝜇y )
𝜎2
[
( 𝜇 )]
+ 𝜇 1 − 2Φ −
𝜎
and variance
{ √
Var (Y) = 𝜇 + 𝜎 −
2
2
𝜎
2
𝜋
e
( )2
− 12 𝜇𝜎
}2
[
( 𝜇 )]
+ 𝜇 1 − 2Φ −
𝜎
where Φ(⋅) is the cumulative distribution function of a standard normal.
Solution: For y > 0, by definition
y
ℙ (|X| < y) = ℙ(−y < X < y) =
∫−y
−1
1
e 2
√
𝜎 2𝜋
(
)
x−𝜇 2
𝜎
dx
[ ( )2
(
) ]
y+𝜇
y−𝜇 2
− 12 𝜎
− 12 𝜎
d
1
+e
e
fY (y) = ℙ (|X| < y) = √
dy
𝜎 2𝜋
(
)
√
1 y2 +𝜇 2
( 𝜇y )
2 − 2 𝜎2
e
cosh 2 .
=
2
𝜋𝜎
𝜎
and hence
By definition
∞
𝔼 (Y) =
∫−∞
∞
=
∫0
y fY (y) dy
y
−1
e 2
√
𝜎 2𝜋
(
)
y+𝜇 2
𝜎
∞
dy +
∫0
y
−1
e 2
√
𝜎 2𝜋
(
)
y−𝜇 2
𝜎
dy.
By setting z = (y + 𝜇)∕𝜎 and 𝑤 = (y − 𝜇)∕𝜎 we have
∞
𝔼 (Y) =
∫𝜇∕𝜎
∞
1
1
− 1 z2
− 1 𝑤2
√ (𝜎z − 𝜇)e 2 dz +
√ (𝜎𝑤 + 𝜇)e 2 d𝑤
∫
−𝜇∕𝜎
2𝜋
2𝜋
Page 22
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Discrete and Continuous Random Variables
−1
𝜎
e 2
=√
2𝜋
( )2
𝜇
𝜎
( )2
23
( )2
[
( 𝜇 )]
[
( 𝜇 )]
𝜇
−1
𝜎
+√
e 2 𝜎 −𝜇 1−Φ −
−𝜇 1−Φ
𝜎
𝜎
2𝜋
( 𝜇 )]
[ (𝜇)
−Φ −
+𝜇 Φ
𝜎
𝜎
2𝜎 − 12 𝜇𝜎
=√
e
2𝜋
√
( )2
[
( 𝜇 )]
−1 𝜇
2
=𝜎
.
e 2 𝜎 + 𝜇 1 − 2Φ −
𝜋
𝜎
( )
To evaluate 𝔼 Y 2 by definition,
( )
𝔼 Y2 =
∞
∫−∞
∞
=
∫0
y2 fY (y) dy
y2
−1
e 2
√
𝜎 2𝜋
(
)
y+𝜇 2
𝜎
∞
dy +
∫0
y2
−1
e 2
√
𝜎 2𝜋
(
)
y−𝜇 2
𝜎
dy.
By setting z = (y + 𝜇)∕𝜎 and 𝑤 = (y − 𝜇)∕𝜎 we have
( )
𝔼 Y2 =
∞
∫𝜇∕𝜎
∞
1
1
1
1
2 − z2
2 − 𝑤2
√ (𝜎z − 𝜇) e 2 dz +
√ (𝜎𝑤 + 𝜇) e 2 d𝑤
∫−𝜇∕𝜎 2𝜋
2𝜋
∞
∞
∞
1 2
1 2
1 2
2𝜇𝜎
1
𝜎2
=√
z2 e− 2 z dz − √
ze− 2 z dz + 𝜇2
e− 2 z dz
√
∫
∫
∫
𝜇∕𝜎
2𝜋 𝜇∕𝜎
2𝜋 𝜇∕𝜎
2𝜋
∞
∞
1 2
1 2
2𝜇𝜎
𝜎2
+√
𝑤2 e− 2 𝑤 d𝑤 + √
𝑤e− 2 𝑤 d𝑤
∫
∫
2𝜋 −𝜇∕𝜎
2𝜋 −𝜇∕𝜎
∞
+ 𝜇2
1 2
1
e− 2 𝑤 d𝑤
√
2𝜋
∫−𝜇∕𝜎
]
[
( )2
( 𝜇 ))
( 𝜇 ) 1 ( 𝜇 )2 √ (
2𝜇𝜎 − 12 𝜇𝜎
−2 𝜎
𝜎2
+ 2𝜋 1 − Φ
e
=√
−√
e
𝜎
𝜎
2𝜋
2𝜋
[
( 𝜇 )]
+ 𝜇2 1 − Φ
𝜎
]
[
( )2
( )2
(
( 𝜇 ))
√ (
2𝜇𝜎 − 12 𝜇𝜎
𝜇 ) − 12 𝜇𝜎
𝜎2
e
+√
−
e
+√
+ 2𝜋 1 − Φ −
𝜎
𝜎
2𝜋
2𝜋
[
( 𝜇 )]
+ 𝜇2 1 − Φ −
𝜎
( 𝜇 )]
(𝜇)
( 2
)[
2
= 𝜇 +𝜎
−Φ −
2−Φ
𝜎
𝜎
= 𝜇2 + 𝜎 2.
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1.2.2 Discrete and Continuous Random Variables
Therefore,
{ √
Var(Y) = 𝔼(Y 2 ) − [𝔼(Y)]2 = 𝜇 2 + 𝜎 2 −
𝜎
2
𝜋
e
( )2
− 12 𝜇𝜎
}2
( 𝜇 )]
+ 𝜇 1 − 2Φ −
𝜎
[
.
◽
X−𝜇
∼ 𝒩(0, 1).
12. Chi-Square Distribution. Show that if X ∼ 𝒩(𝜇, 𝜎 2 ) then Y =
𝜎
Let Z1 , Z2 , . . . , Zn ∼ 𝒩(0, 1) be a sequence of independent and identically distributed random variables each following a standard normal distribution. Using mathematical induction show that
Z = Z12 + Z22 + . . . + Zn2 ∼ 𝜒 2 (n)
where Z has a probability density function
fZ (z) =
such that
Γ
n
1
−1 − z
( )z2 e 2 ,
2 Γ n2
z≥0
n
2
∞
( )
n
n
=
e−x x 2 −1 dx.
∫0
2
Finally, show that 𝔼(Z) = n and Var(Z) = 2n.
X−𝜇
then
𝜎
(
)
X−𝜇
ℙ(Y ≤ y) = ℙ
≤ y = ℙ(X ≤ 𝜇 + 𝜎y).
𝜎
Solution: By setting Y =
Differentiating with respect to y,
fY (y) =
d
ℙ(Y ≤ y)
dy
𝜇+𝜎y
=
d
dy ∫−∞
−1
1
e 2
√
𝜎 2𝜋
−1
1
e 2
= √
𝜎 2𝜋
(
)
𝜇+𝜎y−𝜇 2
𝜎
(
)
x−𝜇 2
𝜎
dx
𝜎
1 2
1
e− 2 y
= √
2𝜋
which is a probability density function of 𝒩(0, 1).
For Z = Z12 and given Z ≥ 0, by definition
ℙ(Z ≤ z) = ℙ
(
Z12
( √
√)
≤ z = ℙ − z < Z1 < z = 2
√
z
)
∫0
1
− 1 z2
√ e 2 1 dz1
2𝜋
Page 24
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Discrete and Continuous Random Variables
and hence
25
[
]
√
z
1 2
1
z
1
1
d
e− 2 z1 dz1 = √ z− 2 e− 2 ,
2
fZ (z) =
∫0 √2𝜋
dz
2𝜋
z≥0
which is the probability density function of 𝜒 2 (1).
For Z = Z12 + Z22 such that Z ≥ 0 we have
ℙ(Z ≤ z) = ℙ(Z12 + Z22 ≤ z) =
1 − 12 (z21 +z22 )
dz1 dz2 .
e
∫ ∫ 2𝜋
z21 +z22 ≤z
Changing to polar coordinates (r, 𝜃) such that z1 = r cos 𝜃 and z2 = r sin 𝜃 with the Jacobian determinant
| 𝜕z1
|
| 𝜕(z1 , z2 ) | || 𝜕r
|=|
|J| = ||
|
| 𝜕(r, 𝜃) | || 𝜕z2
|
| 𝜕r
𝜕z1 |
|
𝜕𝜃 || |cos 𝜃 −r sin 𝜃 |
|=r
| = ||
|
𝜕z2 || | sin 𝜃 r cos 𝜃 |
|
𝜕𝜃 |
then
2𝜋
√
z
1
1 − 12 (z21 +z22 )
1 − 12 r2
e
e
dz1 dz2 =
r drd𝜃 = 1 − e− 2 z .
∫ ∫ 2𝜋
∫𝜃=0 ∫r=0 2𝜋
z21 +z22 ≤z
Thus,
⎡
⎤
⎥
1 − 12 (z21 +z22 )
d ⎢
e
dz1 dz2 ⎥
fZ (z) =
⎢
dz ⎢ ∫ ∫ 2𝜋
⎥
⎣ z21 +z22 ≤z
⎦
=
1 − 12 z
e ,
2
z≥0
which is the probability density function of 𝜒 2 (2).
Assume the result is true for n = k such that
U = Z12 + Z22 + . . . + Zk2 ∼ 𝜒 2 (k)
and knowing that
2
V = Zk+1
∼ 𝜒 2 (1)
then, because U ≥ 0 and V ≥ 0 are independent, using the convolution formula the density
∑
2
of Z = U + V = k+1
i=1 Zi can be written as
z
fZ (z) =
fV (z − u)fU (u) du
∫0
z
=
∫0
{
1
− 1 − 1 (z−u)
√ (z − u) 2 e 2
2𝜋
} ⎧
⎫
1
1 ⎪
⎪
1
k−1
−
u
⋅ ⎨ k ( ) u 2 e 2 ⎬ du
⎪22 Γ k
⎪
2
⎩
⎭
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4:33 P.M.
1.2.2 Discrete and Continuous Random Variables
1
z
e− 2 z
1
k
=√
(z − u)− 2 u 2 −1 du.
( )
k
k ∫0
2𝜋2 2 Γ 2
By setting 𝑣 =
u
we have
z
z
∫0
1
1
k
(z − u)− 2 u 2 −1 du =
∫0
=z
1
and because
∫0
1
k
(1 − 𝑣)− 2 𝑣 2 −1 d𝑣 = B
fore
fZ (z) =
2
k+1
2
(
1
k
(z − 𝑣z)− 2 (𝑣z) 2 −1 z d𝑣
k+1
−1
2
1 k
,
2 2
)
1
∫0
1
k
(1 − 𝑣)− 2 𝑣 2 −1 d𝑣
√ (k)
𝜋Γ 2
=
( ) (see Appendix A) thereΓ k+1
2
k+1
1
−1 − 1 z
( )z 2 e 2 ,
Γ k+1
2
z≥0
which is the probability density function of 𝜒 2 (k + 1) and hence the result is also true for
n = k + 1. By mathematical induction we have shown
Z = Z12 + Z22 + . . . + Zn2 ∼ 𝜒 2 (n).
By computing the moment generation of Z,
( )
MZ (t) = 𝔼 etZ =
∞
∫0
etz fZ (z) dz =
1
( )
∫0
2 Γ n2
n
2
∞ n
1
z 2 −1 e− 2 (1−2t)
dz
and by setting 𝑤 = 12 (1 − 2t)z we have
(
) n2 ∞ n
n
2
1
MZ (t) = n ( )
𝑤 2 −1 e−𝑤 d𝑤 = (1 − 2t)− 2 ,
∫
n
1
−
2t
0
22 Γ 2
Thus,
(
)
− n2 +1
′
MZ (t) = n(1 − 2t)
such that
′
𝔼 (Z) = MZ (0) = n,
,
′′
MZ (t) = 2n
(
(
)
1 1
t∈ − ,
.
2 2
(
)
)
− n +2
n
+ 1 (1 − 2t) 2
2
(
)
( )
′′
n
𝔼 Z 2 = MZ (0) = 2n
+1
2
and
( )
Var (Z) = 𝔼 Z 2 − [𝔼 (Z)]2 = 2n.
◽
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27
13. Marginal Distributions of Bivariate Normal Distribution. Let X and Y be jointly normally
distributed with means 𝜇x , 𝜇y , variances 𝜎x2 , 𝜎y2 and correlation coefficient 𝜌xy ∈ (−1, 1)
such that the joint density function is
fXY (x, y) =
2𝜋𝜎x 𝜎y
−
1
√
e
1−
1
2(1−𝜌2xy )
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
𝜌2xy
.
Show that X ∼ 𝒩(𝜇x , 𝜎x2 ) and Y ∼ 𝒩(𝜇y , 𝜎y2 ).
Solution: By definition,
∞
fX (x) =
∫−∞
∞
=
∫−∞
∞
=
∫−∞
fXY (x, y) dy
−
1
e
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
×e
=
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
dy
1
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
[
−
1
2(1−𝜌2xy )
1
2(1−𝜌2
xy )
−1
1
√ e 2
𝜎x 2𝜋
(
(1−𝜌2xy )
(
x−𝜇x
𝜎x
x−𝜇x
𝜎x
)2
)2
+𝜌2xy
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
dy
∞
∫−∞
g(x, y) dy
where
−
1
g(x, y) = √
√ e
1 − 𝜌2xy 𝜎y 2𝜋
1
2(1−𝜌2xy )𝜎y2
[ (
(
))]2
x−𝜇
y− 𝜇y +𝜌xy 𝜎y 𝜎 x
x
(
)
)
(
is the probability density function for 𝒩 𝜇y + 𝜌xy 𝜎y x − 𝜇x ∕𝜎x , (1 − 𝜌2xy )𝜎y2 .
Therefore,
∞
∫−∞
g(x, y) dy = 1
and hence
fX (x) =
−1
1
√ e 2
𝜎x 2𝜋
(
x−𝜇x
𝜎x
)2
.
Thus, X ∼ 𝒩(𝜇x , 𝜎x2 ). Using the same steps we can also show that Y ∼ 𝒩(𝜇y , 𝜎y2 ).
◽
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1.2.2 Discrete and Continuous Random Variables
14. Covariance of Bivariate Normal Distribution. Let X and Y be jointly normally distributed
with means 𝜇x , 𝜇y , variances 𝜎x2 , 𝜎y2 and correlation coefficient 𝜌xy ∈ (−1, 1) such that the
joint density function is
[
fXY (x, y) =
2𝜋𝜎x 𝜎y
−
1
√
e
1−
1
2(1−𝜌2xy )
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
𝜌2xy
.
Show that the covariance of X and Y, Cov(X, Y) = 𝜌xy 𝜎x 𝜎y and hence show that X and Y
are independent if and only if 𝜌xy = 0.
Solution: By definition, the covariance of X and Y is
Cov(X, Y) = 𝔼[(X − 𝜇x )(Y − 𝜇y )]
∞
=
∞
∫−∞ ∫−∞
∞
=
(x − 𝜇x )(y − 𝜇y ) fXY (x, y) dx dy
∞
∫−∞ ∫−∞
∞
xyfXY (x, y) dx dy − 𝜇y
∞
∞
∫−∞ ∫−∞
∞
x fXY (x, y) dx dy
∞
∞
yfXY (x, y) dx dy + 𝜇x 𝜇y
f (x, y) dx dy
∫−∞ ∫−∞ XY
]
∞
∞
∞ [
∞
xyfXY (x, y) dx dy − 𝜇y
x
f (x, y) dy dx
=
∫−∞ ∫−∞ XY
∫−∞ ∫−∞
]
∞ [
∞
∞
∞
−𝜇x
y
fXY (x, y) dx dy +
𝜇 𝜇 f (x, y) dx dy
∫−∞ ∫−∞
∫−∞ ∫−∞ x y XY
−𝜇x
∫−∞ ∫−∞
∞
=
∞
∫−∞ ∫−∞
∞
+
∫−∞
yfY (y) dy
𝜇x 𝜇y fXY (x, y) dx dy
∞
∫−∞ ∫−∞
∞
=
∫−∞
∞
x fX (x) dx − 𝜇x
∞
∫−∞ ∫−∞
∞
=
∞
xyfXY (x, y) dx dy − 𝜇y
xyfXY (x, y) dx dy − 𝜇x 𝜇y − 𝜇x 𝜇y + 𝜇x 𝜇y
∞
∫−∞ ∫−∞
xyfXY (x, y) dx dy − 𝜇x 𝜇y
∞
∞
where fX (x) = ∫−∞ fXY (x, y) dy and fY (y) = ∫−∞ fXY (x, y) dx. Using the result of
Problem 1.2.2.13 (page 27) we can deduce that
(
) (
)
∞
∞
∞
∞
x−𝜇 2
− 12 𝜎 x
x
x
xyfXY (x, y) dx dy =
yg(x, y) dy dx
√ e
∫−∞
∫−∞ 𝜎 2𝜋
∫−∞ ∫−∞
x
where
1
−
g(x, y) = √
√ e
2
1 − 𝜌xy 𝜎y 2𝜋
1
2(1−𝜌2xy )𝜎y2
[
(
(
))]2
x−𝜇
y− 𝜇y +𝜌xy 𝜎y 𝜎 x
x
Page 28
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29
(
)
)
(
is the probability density function for 𝒩 𝜇y + 𝜌xy 𝜎y x − 𝜇x ∕𝜎x , (1 − 𝜌2xy )𝜎y2 .
Therefore,
(
)
∞
x − 𝜇x
.
yg(x, y) dy = 𝜇y + 𝜌xy 𝜎y
∫−∞
𝜎x
Thus,
∞
Cov(X, Y) =
− 12
x
√
∫−∞ 𝜎 2𝜋
x
= 𝜇x 𝜇y +
= 𝜇x 𝜇y +
(
e
𝜌xy 𝜎y
∞
𝜎x ∫−∞
𝜌xy 𝜎y (
𝜎x2
𝜎x
= 𝜌xy 𝜎x 𝜎y
where
∞
𝔼(X 2 ) =
∫−∞
x−𝜇x
𝜎x
)2
[
(
𝜇y + 𝜌xy 𝜎y
−1
x2
√ e 2
𝜎x 2𝜋
+
𝜇x2
)
x−𝜇x
𝜎x
𝜌xy 𝜎y 𝜇x2
−
−1
x2
√ e 2
𝜎x 2𝜋
(
𝜎x
(
x−𝜇x
𝜎x
x − 𝜇x
𝜎x
)2
dx −
)]
dx − 𝜇x 𝜇y
𝜌xy 𝜎y 𝜇x2
𝜎x
− 𝜇x 𝜇y
− 𝜇x 𝜇y
)2
dx = 𝜎x2 + 𝜇x2 .
To show that X and Y are independent if and only if 𝜌xy = 0 we note that if X ⟂
⟂ Y then
Cov(X, Y) = 0, which implies 𝜌xy = 0. On the contrary, if 𝜌xy = 0 then from the joint density of (X, Y) we can express it as
fXY (x, y) = fX (x) fY (y)
∞
where fX (x) =
∫−∞
−1
1
√ e 2
𝜎x 2𝜋
(
x−𝜇x
𝜎x
)2
∞
dx and fY (y) =
(
1
√
∫−∞ 𝜎 2𝜋
y
− 12
e
y−𝜇y
𝜎y
)2
dy and so
X⟂
⟂ Y.
Thus, if the pair X and Y has a bivariate normal distribution with means 𝜇x , 𝜇y , variances
𝜎x2 , 𝜎y2 and correlation 𝜌xy then X ⟂
⟂ Y if and only if 𝜌xy = 0.
◽
15. Minimum and Maximum of Two Correlated Normal Distributions. Let X and Y be jointly
normally distributed with means 𝜇x , 𝜇y , variances 𝜎x2 , 𝜎y2 and correlation coefficient 𝜌xy ∈
(−1, 1) such that the joint density function is
fXY (x, y) =
2𝜋𝜎x 𝜎y
1
√
−
e
1
2(1−𝜌2xy )
[
(
1 − 𝜌2xy
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
.
Show that the distribution of U = min{X, Y} is
𝜌xy 𝜎y
⎞
⎛
⎛ −u + 𝜇 + 𝜌xy 𝜎x (u − 𝜇 ) ⎞
x
y ⎟
𝜎y
⎜ −u + 𝜇y + 𝜎x (u − 𝜇x ) ⎟
⎜
fU (u) = Φ ⎜
√
√
⎟ fX (u) + Φ ⎜
⎟ fY (u).
⎟
⎟
⎜
⎜
𝜎y 1 − 𝜌2xy
𝜎x 1 − 𝜌2xy
⎠
⎠
⎝
⎝
4:33 P.M.
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1.2.2 Discrete and Continuous Random Variables
and deduce that the distribution of V = max{X, Y} is
𝜌xy 𝜎y
⎛
⎛ 𝑣 − 𝜇 − 𝜌xy 𝜎x (u − 𝜇 ) ⎞
⎞
x
y ⎟
𝜎y
⎜ 𝑣 − 𝜇y − 𝜎x (𝑣 − 𝜇x ) ⎟
⎜
fV (𝑣) = Φ ⎜
√
√
⎟ fX (𝑣) + Φ ⎜
⎟ fY (𝑣)
⎜
⎜
⎟
⎟
𝜎y 1 − 𝜌2xy
𝜎x 1 − 𝜌2xy
⎝
⎝
⎠
⎠
− 12
1
√
where fX (z) =
(
z−𝜇x
𝜎x
(
)2
− 12
1
√
, fY (z) =
e
e
𝜎x 2𝜋
𝜎y 2𝜋
cumulative standard normal distribution function (cdf).
z−𝜇y
𝜎y
)2
and Φ(⋅) denotes the
Solution: From Problems 1.2.2.13 (page 27) and 1.2.2.14 (page 28) we can show that
X ∼ 𝒩(𝜇x , 𝜎x2 ), Y ∼ 𝒩(𝜇y , 𝜎y2 ), Cov(X, Y) = 𝜌xy 𝜎x 𝜎y such that 𝜌xy ∈ (−1, 1).
For U = min{X, Y} then by definition the cumulative distribution function (cdf) of U is
ℙ(U ≤ u) = ℙ (min{X, Y} ≤ u)
= 1 − ℙ (min{X, Y} > u)
= 1 − ℙ (X > u, Y > u) .
To derive the probability density function (pdf) of U we have
d
ℙ(U ≤ u)
du
d
= − ℙ (X > u, Y > u)
du
fU (u) =
∞
=−
∞
d
du ∫x=u ∫y=u
−
×e
1
2(1−𝜌2xy )
[
(
1
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
dydx
= g(u) + h(u)
where
∞
g(u) =
∫y=u
−
1
e
√
2
2𝜋𝜎x 𝜎y 1 − 𝜌xy
1
2(1−𝜌2xy )
[
(
u−𝜇x
𝜎x
)2
(
−2𝜌xy
u−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
x−𝜇x
𝜎x
]
)( u−𝜇 ) ( u−𝜇 )2
y
y
+
𝜎
𝜎
y
y
dy
and
∞
h(u) =
∫x=u
2𝜋𝜎x 𝜎y
−
1
√
e
1−
𝜌2xy
1
2(1−𝜌2xy )
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
y
y
dx.
Page 30
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Discrete and Continuous Random Variables
31
By focusing on
∞
g(u) =
1
e
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
∫y=u
−
1
2(1−𝜌2xy )
[
(
u−𝜇x
𝜎x
{[(
∞
=
1
e
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
∫y=u
− 12
1
√
=
𝜎x 2𝜋
e
y − 𝜇y
and letting 𝑤 =
𝜎y
g(u) =
(
u−𝜇x
𝜎x
∞
∫y=u
(
√
(
−2𝜌xy
)
y−𝜇y
𝜎y
u−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
u−𝜇
−𝜌xy 𝜎 x
x
𝜎y
u − 𝜇x
𝜎x
√
−
1
e
y
dy
}
(
)]2 (
)
u−𝜇 2
+ 𝜎 x (1−𝜌2xy )
x
[(
)2
− 𝜌xy
1
−
2(1−𝜌2xy )
)2
1
2(1−𝜌2xy )
y−𝜇y
𝜎y
)
(
−𝜌xy
u−𝜇x
𝜎x
dy
)]2
dy
2𝜋(1 − 𝜌2xy )
)
we have
1 − 𝜌2xy
−1
1
e 2
√
𝜎x 2𝜋
(
u−𝜇x
𝜎x
)2
∞
∫𝑤=
( u−𝜇 )
u−𝜇y
x
𝜎y −𝜌xy 𝜎x
√
1−𝜌2xy
1 2
1
e− 2 𝑤 d𝑤
√
2𝜋
𝜌xy 𝜎y
⎛
⎞
⎜ −u + 𝜇y + 𝜎x (u − 𝜇x ) ⎟
=Φ⎜
√
⎟ fX (u).
⎜
⎟
𝜎y 1 − 𝜌2xy
⎝
⎠
In a similar vein we can also show
∞
h(u) =
∫x=u
−
1
e
√
2
2𝜋𝜎x 𝜎y 1 − 𝜌xy
(
=
1
e
√
𝜎y 2𝜋
− 12
u−𝜇y
𝜎y
)2
1
2(1−𝜌2xy )
[
(
∞
∫𝑤=
u−𝜇x
𝜎x −𝜌xy
√
(
x−𝜇x
𝜎x
u−𝜇y
𝜎y
1−𝜌2xy
)2
)
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( u−𝜇 ) ( u−𝜇 )2
y
y
+
𝜎
𝜎
y
y
dx
1 2
1
e− 2 𝑤 d𝑤
√
2𝜋
⎛ −u + 𝜇 + 𝜌xy 𝜎x (u − 𝜇 ) ⎞
x
y ⎟
𝜎y
⎜
=Φ⎜
√
⎟ fY (u).
⎜
⎟
𝜎x 1 − 𝜌2xy
⎝
⎠
Therefore,
𝜌xy 𝜎y
⎞
⎛
⎛ −u + 𝜇 + 𝜌xy 𝜎x (u − 𝜇 ) ⎞
x
y ⎟
𝜎y
⎜ −u + 𝜇y + 𝜎x (u − 𝜇x ) ⎟
⎜
fU (u) = Φ ⎜
√
√
⎟ fX (u) + Φ ⎜
⎟ fY (u).
⎟
⎟
⎜
⎜
𝜎y 1 − 𝜌2xy
𝜎x 1 − 𝜌2xy
⎠
⎠
⎝
⎝
4:33 P.M.
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1.2.2 Discrete and Continuous Random Variables
As for the case V = max{X, Y}, then by definition the cdf of V is
ℙ(V ≤ 𝑣) = ℙ (max{X, Y} ≤ 𝑣)
= ℙ (X ≤ 𝑣, Y ≤ 𝑣) .
The pdf of V is
d
ℙ(V ≤ 𝑣)
d𝑣
d
= ℙ (X ≤ 𝑣, Y ≤ 𝑣)
d𝑣
fV (𝑣) =
x=𝑣
=
y=𝑣
d
d𝑣 ∫−∞ ∫−∞
[
−
×e
1
2(1−𝜌2
xy )
y=𝑣
=
∫−∞
x=𝑣
+
∫−∞
(
2𝜋𝜎x 𝜎y
x−𝜇x
𝜎x
)2
1
√
(
−2𝜌xy
1 − 𝜌2xy
x−𝜇x
𝜎x
1
e
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
−
−
1
e
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
1
2(1−𝜌2xy )
1
2(1−𝜌2xy )
[
(
[
(
y
𝑣−𝜇x
𝜎x
x−𝜇x
𝜎x
)2
)2
(
−2𝜌xy
(
−2𝜌xy
dy dx
𝑣−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
x−𝜇x
𝜎x
]
)( 𝑣−𝜇 ) ( 𝑣−𝜇 )2
y
y
+
𝜎
𝜎
y
y
y
y
dy
dx.
Following the same steps as described above we can write
𝜌xy 𝜎y
⎞
⎛
⎛ 𝑣 − 𝜇 − 𝜌xy 𝜎x (𝑣 − 𝜇 ) ⎞
x
y ⎟
𝜎y
⎜ 𝑣 − 𝜇y − 𝜎x (𝑣 − 𝜇x ) ⎟
⎜
f
fV (𝑣) = Φ ⎜
(𝑣)
+
Φ
√
√
⎟ X
⎟ fY (𝑣).
⎜
⎟
⎟
⎜
⎜
𝜎y 1 − 𝜌2xy
𝜎x 1 − 𝜌2xy
⎠
⎠
⎝
⎝
◽
16. Bivariate Standard Normal Distribution. Let X ∼ 𝒩(0, 1) and Y ∼ 𝒩(0, 1) be jointly normally distributed with correlation coefficient 𝜌xy ∈ (−1, 1) where the joint cumulative
distribution function is
(
𝚽(𝛼, 𝛽, 𝜌xy ) =
𝛽
𝛼
∫−∞ ∫−∞
√
2𝜋
1
− 12
e
x2 −2𝜌xy xy+y2
1−𝜌2xy
dx dy.
1 − 𝜌2xy
By using the change of variables Y = 𝜌xy X +
√
1 − 𝜌2xy Z, Z ∼ 𝒩(0, 1), X ⟂
⟂ Z show that
⎞
⎛
⎜ 𝛽 − 𝜌xy x ⎟
𝚽(𝛼, 𝛽, 𝜌xy ) =
f (x)Φ ⎜ √
⎟ dx
∫−∞ X
⎜ 1 − 𝜌2xy ⎟
⎠
⎝
𝛼
)
Page 32
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Discrete and Continuous Random Variables
33
1 2
1
where fX (x) = √
e− 2 x and Φ(⋅) is the cumulative distribution function of a standard
2𝜋
normal.
Finally, deduce that 𝚽(𝛼, 𝛽, 𝜌xy ) + 𝚽(𝛼, −𝛽, −𝜌xy ) = Φ(𝛼).
Solution: Let y = 𝜌xy x +
√
1 − 𝜌2xy , and hence
√
dy
1 − 𝜌2xy z. Differentiating y with respect to z we have
=
dz
(
𝚽(𝛼, 𝛽, 𝜌xy ) =
𝛽
𝛼
∫−∞ ∫−∞
𝛽−𝜌xy x
√
=
∫−∞
1−𝜌2xy
√
2𝜋
− 12
1
e
√
2𝜋
⎛ x2 −2𝜌xy x(𝜌xy x+
× e
=
∫−∞
𝛼
=
∫−∞
1−𝜌2xy
𝛼
∫−∞
1 − 𝜌2xy
1−𝜌2xy z)+(𝜌xy x+
1−𝜌2xy
⎝
𝛽−𝜌xy x
dx dy
1
√
− 12 ⎜
⎜
√
)
1 − 𝜌2xy
𝛼
∫−∞
x2 −2𝜌xy xy+y2
1−𝜌2xy
√
1−𝜌2xy z)2 ⎞
⎟
⎟
⎠
√
1 − 𝜌2xy dx dz
1 − 12 (x2 +z2 )
dx dz
e
2𝜋
𝛽−𝜌 x
⎡ √ xy 2
⎤
1 2
1
1
1−𝜌xy
− 12 x2 ⎢
e
e− 2 z dz⎥ dx
√
√
⎢∫−∞
⎥
2𝜋
2𝜋
⎣
⎦
⎞
⎛
⎜ 𝛽 − 𝜌xy x ⎟
=
f (x)Φ ⎜ √
⎟ dx.
∫−∞ X
⎜ 1 − 𝜌2xy ⎟
⎠
⎝
𝛼
Finally,
⎛
⎞
⎜ 𝛽 − 𝜌xy x ⎟
f (x)Φ ⎜ √
𝚽(𝛼, 𝛽, 𝜌xy ) + 𝚽(𝛼, −𝛽, −𝜌xy ) =
⎟ dx
∫−∞ X
⎜ 1 − 𝜌2xy ⎟
⎝
⎠
𝛼
⎞
⎛
⎜ −𝛽 + 𝜌xy x ⎟
+
f (x)Φ ⎜ √
⎟ dx
∫−∞ X
⎜ 1 − 𝜌2xy ⎟
⎠
⎝
𝛼
⎞
⎛
⎞⎤
⎡ ⎛
⎜ −𝛽 + 𝜌xy x ⎟⎥
⎢ ⎜ 𝛽 − 𝜌xy x ⎟
=
f (x) Φ √
⎟ + Φ⎜√
⎟⎥ dx
∫−∞ X ⎢⎢ ⎜⎜
⎜ 1 − 𝜌2xy ⎟⎥
1 − 𝜌2xy ⎟
⎣ ⎝
⎠
⎝
⎠⎦
𝛼
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1.2.2 Discrete and Continuous Random Variables
𝛼
=
∫−∞
fX (x) dx
= Φ(𝛼)
⎛
⎞
⎞
⎛
⎜ −𝛽 + 𝜌xy x ⎟
⎜ 𝛽 − 𝜌xy x ⎟
since Φ ⎜ √
⎟ + Φ⎜√
⎟ = 1.
⎜ 1 − 𝜌2xy ⎟
⎜ 1 − 𝜌2xy ⎟
⎝
⎠
⎠
⎝
N.B. Similarly we can also show that
⎞
⎛
⎜ 𝛼 − 𝜌xy y ⎟
𝚽(𝛼, 𝛽, 𝜌xy ) =
f (y)Φ ⎜ √
⎟ dy
∫−∞ Y
⎜ 1 − 𝜌2xy ⎟
⎠
⎝
𝛽
1 2
1
where fY (y) = √
e− 2 y and 𝚽(𝛼, 𝛽, 𝜌xy ) + 𝚽(−𝛼, 𝛽, −𝜌xy ) = Φ(𝛽).
2𝜋
◽
17. Bivariate Normal Distribution Property. Let X and Y be jointly normally distributed with
means 𝜇x , 𝜇y , variances 𝜎x2 , 𝜎y2 and correlation coefficient 𝜌xy ∈ (−1, 1) such that the joint
density function is
[
fXY (x, y) =
2𝜋𝜎x 𝜎y
1
√
−
e
1
2(1−𝜌2xy )
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
1 − 𝜌2xy
y
.
Show that
⎞
⎛
𝜇 − 𝜇y + 𝜎x (𝜎x − 𝜌xy 𝜎y ) ⎟
]
[
𝜇x + 12 𝜎x2 ⎜ x
X
Y
Φ⎜ √
𝔼 max{e − e , 0} = e
⎟
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎠
⎝
−e
𝜇y + 12 𝜎y2
⎛
⎞
⎜ 𝜇x − 𝜇y − 𝜎y (𝜎y − 𝜌xy 𝜎y ) ⎟
Φ⎜ √
⎟.
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎝
⎠
Solution: By definition
]
[
𝔼 max{eX − eY , 0} =
y=∞
x=∞
∫y=−∞ ∫x=y
(ex − ey ) fXY (x, y) dx dy
= I1 − I2
y=∞
where I1 =
x=∞
∫y=−∞ ∫x=y
y=∞
ex fXY (x, y) dx dy and I2 =
x=∞
∫y=−∞ ∫x=y
ey fXY (x, y) dx dy.
Page 34
Chin
1.2.2
c01.tex
V3 - 08/23/2014
Discrete and Continuous Random Variables
y=∞
For the case I1 =
y=∞
I1 =
ex fXY (x, y) dx dy we have
x=∞
−
ex
2𝜋𝜎x 𝜎y
(
y=∞
=
x=∞
∫y=−∞ ∫x=y
∫y=−∞ ∫x=y
1
e
∫y=−∞ 𝜎 √2𝜋
y
35
− 12
√
e
1
2(1−𝜌2xy )
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
1 − 𝜌2xy
)2
y−𝜇y
𝜎y
[ (
(
))]2 ⎤
⎡
y−𝜇y
1
−
x− 𝜇x +𝜌xy 𝜎x 𝜎
⎢ x=∞
⎥
2
ex
y
e 2(1−𝜌xy )
×⎢
dx⎥ dy
√
∫
⎢ x=y 𝜎x 2𝜋(1 − 𝜌2xy )
⎥
⎣
⎦
y=∞
=
∫y=−∞
(
1
−
1
e 2
√
𝜎y 2𝜋
where g(x, y) =
)2
y−𝜇y
𝜎y
−
]
x=∞
x
e g(x, y) dx dy
∫x=y
1
√
[
1
2(1−𝜌2xy )
[ (
(
))]2
y−𝜇y
x− 𝜇x +𝜌xy 𝜎x 𝜎
y
which is the probability
2𝜋(1 −
(
)
]
[
y − 𝜇y
2
2
,
(1
−
𝜌
density function of 𝒩 𝜇x + 𝜌xy 𝜎x
xy ) 𝜎x . Thus, from Problem 1.2.2.7
𝜎y
(page 18) we can deduce
𝜎x
e
𝜌2xy )
(
x=∞
∫x=y
x
e g(x, y) dx = e
𝜇x +𝜌xy 𝜎x
y−𝜇y
𝜎y
)
+ 12 (1−𝜌2xy )𝜎x2
⎛
⎜ 𝜇x + 𝜌xy 𝜎x
× Φ⎜
⎜
⎜
⎝
(
)
⎞
+ (1 − 𝜌xy )2 𝜎x2 − y ⎟
⎟.
√
⎟
𝜎x 1 − 𝜌2xy
⎟
⎠
y−𝜇y
𝜎y
Thus, we can write
1
I1 = e𝜇x + 2 (1−𝜌xy )𝜎x
y=∞
2
2
[(
− 12
y−𝜇y
𝜎y
)2
(
−2𝜌xy 𝜎x
y−𝜇y
𝜎y
)]
1
e
∫y=−∞ 𝜎 √2𝜋
y
(
)
⎛
y−𝜇y
⎞
𝜇
+ (1 − 𝜌xy )2 𝜎x2 − y ⎟
+
𝜌
𝜎
xy x
⎜ x
𝜎y
⎟ dy
× Φ⎜
√
⎜
⎟
2
𝜎x 1 − 𝜌xy
⎜
⎟
⎝
⎠
×
4:33 P.M.
Page 35
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36
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4:33 P.M.
1.2.2 Discrete and Continuous Random Variables
(
y=∞
1
y−𝜇y −𝜌xy 𝜎x 𝜎y
)2
−
1
𝜎y
e 2
=e
√
∫y=−∞ 𝜎 2𝜋
y
(
)
y−𝜇y
⎞
⎛
+ (1 − 𝜌xy )2 𝜎x2 − y ⎟
⎜ 𝜇x + 𝜌xy 𝜎x
𝜎y
⎟ dy.
× Φ⎜
√
⎟
⎜
2
𝜎x 1 − 𝜌xy
⎟
⎜
⎠
⎝
𝜇x + 12 𝜎x2
Let u =
y − 𝜇y − 𝜌xy 𝜎x 𝜎y
𝜎y
𝜇x + 12 𝜎x2
I1 = e
𝜇x + 12 𝜎x2
=e
u=∞
∫u=−∞
then from the change of variables
⎛
⎞
𝜇 − 𝜇y + 𝜎x (𝜎x − 𝜌xy 𝜎y ) − u(𝜎y − 𝜌xy 𝜎x ) ⎟
1
− 12 u2 ⎜ x
Φ⎜
e
√
√
⎟ du
2
2𝜋
⎜
⎟
𝜎x 1 − 𝜌xy
⎝
⎠
𝑣=
u=∞
𝜇x −𝜇y +𝜎x (𝜎x −𝜌xy 𝜎y )
𝜎x
∫u=−∞ ∫𝑣=−∞
By setting 𝑤 = 𝑣 +
√
1−𝜌2xy
−
u(𝜎y −𝜌xy 𝜎x )
𝜎x
√
1−𝜌2xy
1 − 12 (u2 +𝑣2 )
d𝑣du.
e
2𝜋
u(𝜎y − 𝜌xy 𝜎x )
and 𝜎 2 = 𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ,
√
2
𝜎x 1 − 𝜌xy
1 2
I1 = e𝜇x + 2 𝜎x
u=∞
×
×
=e
𝑤=
𝜇x −𝜇y +𝜎x (𝜎x −𝜌xy 𝜎y )
∫u=−∞ ∫𝑤=−∞
𝜎x
√
1−𝜌2xy
1
2𝜋
(
)
⎛ 𝜎 −𝜌 𝜎 ⎞
⎡
⎤
(𝜎y −𝜌xy 𝜎x )2
y xy x ⎟
− 12 ⎢𝑤2 −2u𝑤⎜ √
+ 1+ 2
u2 ⎥
2 )
⎜𝜎
⎢
⎟
⎥
𝜎
(1−𝜌
2
x
xy
⎦
⎝ x 1−𝜌xy ⎠
e ⎣
d𝑤du
𝜇x + 12 𝜎x2
u=∞
×
𝑤=
𝜇x −𝜇y +𝜎x (𝜎x −𝜌xy 𝜎y )
∫u=−∞ ∫𝑤=−∞
(
− 12
×e
𝜎2
𝜎x2 (1−𝜌2xy )
𝜎x
√
1−𝜌2xy
1
2𝜋
⎤
√
)⎡
⎥
⎢
𝜎x (𝜎y −𝜌xy 𝜎x )
1−𝜌2xy
2
2⎥
⎢( 𝑤
) −2u𝑤
+u
2
𝜎
⎥
⎢
𝜎2
⎥
⎢ 𝜎 2 (1−𝜌2 )
⎦
⎣ x
xy
d𝑤du.
Page 36
Chin
1.2.2
c01.tex
V3 - 08/23/2014
Discrete and Continuous Random Variables
Finally, by setting 𝑤 =
I1 = e
𝜇x + 12 𝜎x2
where 𝜌xy =
Therefore,
37
𝑤
,
⎛
⎞
⎜
⎟
𝜎
⎜ √
⎟
⎜ 𝜎x 1 − 𝜌2xy ⎟
⎝
⎠
𝑤=
u=∞
𝜇x −𝜇y +𝜎x (𝜎x −𝜌xy 𝜎y )
𝜎
√
∫u=−∞ ∫𝑤=−∞
𝜎y − 𝜌xy 𝜎x
𝜎
I1 = e
𝜇x + 12 𝜎x2
=e
𝛽
⎞
⎛
⎜ 𝜇x − 𝜇y + 𝜎x (𝜎x − 𝜌xy 𝜎y ) ⎟
Φ⎜ √
⎟
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎠
⎝
(
𝛼
− 12
1
y=∞
Using similar steps for the case I2 =
I2 =
y=∞
=
x=∞
∫y=−∞ ∫x=y
∫y=−∞
−
2𝜋𝜎x 𝜎y
(
1
−
ey
e 2
√
𝜎y 2𝜋
y−𝜇y
𝜎y
e
)
dx dy is the cumulative distri-
x=∞
∫y=−∞ ∫x=y
ey
√
x2 −2𝜌xy+y2
1−𝜌2
1
2(1−𝜌2xy )
ey fXY (x, y) dx dy we have
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌xy
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
1 − 𝜌2xy
)2
[ (
(
))]2 ⎤
⎡
y−𝜇y
1
−
x− 𝜇x +𝜌xy 𝜎x 𝜎
⎢ x=∞
⎥
1
y
2(1−𝜌2xy )
e
× ⎢
dx⎥ dy
√
∫
⎢ x=y 𝜎x 2𝜋(1 − 𝜌2xy )
⎥
⎣
⎦
y=∞
=
∫y=−∞
d𝑤du
2
1 − 𝜌xy
⎞
⎛
⎟
⎜ 𝜇x − 𝜇y + 𝜎x (𝜎x − 𝜌xy 𝜎y )
𝚽⎜ √
, ∞, 𝜌xy ⎟
⎟
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2
⎠
⎝
e
√
∫−∞ ∫−∞ 2𝜋 1 − 𝜌2
bution function of a standard bivariate normal.
y=∞
e
1
(𝑤2 −2𝜌xy u𝑤+u2 )
2(1−𝜌2xy )
.
𝜇x + 12 𝜎x2
where 𝚽(𝛼, 𝛽, 𝜌) =
2𝜋
−
1
(
1
−
ey
e 2
√
𝜎y 2𝜋
y−𝜇y
𝜎y
)2
[
x=∞
∫x=y
]
g(x, y) dx dy
y
4:33 P.M.
Page 37
Chin
38
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4:33 P.M.
1.2.2 Discrete and Continuous Random Variables
[
where g(x, y) =
𝜎x
−
1
√
e
(
(
))]2
y−𝜇y
x− 𝜇x +𝜌xy 𝜎x 𝜎
1
2(1−𝜌2xy )
y
which is the probability
𝜌2xy )
2𝜋(1 −
[
(
)
]
y − 𝜇y
density function of 𝒩 𝜇x + 𝜌xy 𝜎x
, (1 − 𝜌xy )2 𝜎x2 .
𝜎y
Thus,
(
y=∞
I2 =
∫y=−∞
y−𝜇y
⎛
⎡
x−𝜇x −𝜌xy 𝜎x
𝜎y
1⎜
(
)2 ⎢
−
√
2⎜
1 y−𝜇y
x=∞
2
y
𝜎
1−𝜌
−
⎜
⎢
x
xy
e
1
e ⎝
e 2 𝜎y
√
√
⎢∫
𝜎y 2𝜋
⎢ x=y 𝜎x 2𝜋(1 − 𝜌2xy )
⎢
⎣
(
x − 𝜇x − 𝜌xy 𝜎x
and by setting z =
𝜎x
y=∞
I2 =
∫y=−∞
y=∞
=
∫y=−∞
By setting z =
𝜎y
⎞
⎟
⎟
⎟
⎠
⎤
⎥
⎥
dx⎥ dy
⎥
⎥
⎦
)
,
1 − 𝜌2xy
⎡
⎤
)
(
⎢ z=∞
⎥
1
− 12 z2
y−𝜇y
dz⎥ dy
e
⎢∫ y−𝜇x −𝜌xy 𝜎x 𝜎y √
√
2𝜋
⎢ z=
⎥
𝜎x
1−𝜌2xy
⎣
⎦
(
)
] ⎛
[(
y−𝜇y ⎞
)2
y−𝜇y
−y + 𝜇x + 𝜌xy 𝜎x
⎜
⎟
− 12
−2y
𝜎y
𝜎y
1
⎟ dy.
Φ⎜
e
√
√
⎜
⎟
2
𝜎y 2𝜋
𝜎x 1 − 𝜌xy
⎜
⎟
⎝
⎠
(
1
−
ey
e 2
√
𝜎y 2𝜋
y − 𝜇y
𝜎y
I2 = e
√
y−𝜇y
) 2
y−𝜇y
𝜎y
)2
therefore
𝜇y + 12 𝜎y2
z=∞
∫z=−∞
⎞
⎛
𝜇 − 𝜇y − z(𝜎y − 𝜌xy 𝜎x ) ⎟
1
− 12 (z−𝜎y )2 ⎜ x
Φ⎜
e
√
√
⎟ dz
2
2𝜋
⎟
⎜
𝜎x 1 − 𝜌xy
⎠
⎝
and substituting u = z − 𝜎y ,
I2 = e
=e
𝜇y + 12 𝜎y2
𝜇y + 12 𝜎y2
u=∞
∫u=−∞
u=∞
∫u=−∞
⎞
⎛
𝜇 − 𝜇y − (u + 𝜎y )(𝜎y − 𝜌xy 𝜎x ) ⎟
1
− 12 u2 ⎜ x
e
Φ⎜
√
√
⎟ du
2𝜋
⎟
⎜
𝜎x 1 − 𝜌2xy
⎠
⎝
⎞
⎛
𝜇 − 𝜇y − 𝜎y (𝜎y − 𝜌xy 𝜎x ) u(𝜎y − 𝜌xy 𝜎x ) ⎟
1
− 12 u2 ⎜ x
e
Φ⎜
−
√
√
√
⎟ du
2𝜋
⎜
𝜎x 1 − 𝜌2xy
𝜎x 1 − 𝜌2xy ⎟
⎠
⎝
Page 38
Chin
1.2.2
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V3 - 08/23/2014
Discrete and Continuous Random Variables
=e
𝜇y + 12 𝜎y2
𝑣=
u=∞
39
𝜇x −𝜇y −𝜎y (𝜎y −𝜌xy 𝜎x )
𝜎x
∫u=−∞ ∫𝑣=−∞
√
1−𝜌2xy
−
u(𝜎y −𝜌xy 𝜎x )
𝜎x
√
1−𝜌2xy
1 − 12 (u2 +𝑣2 )
d𝑣du.
e
2𝜋
Using the same steps as described before, we let 𝑤 = 𝑣 +
u(𝜎y − 𝜌xy 𝜎x )
and 𝜎 2 = 𝜎x2 −
√
2
𝜎x 1 − 𝜌xy
2𝜌xy 𝜎x 𝜎y + 𝜎y2 so that
1 2
I 2 = e𝜇y + 2 𝜎 y
×
𝜇x −𝜇y −𝜎y (𝜎y −𝜌xy 𝜎x )
𝑤=
u=∞
∫u=−∞ ∫𝑤=−∞
⎡
⎛
𝜇y + 12 𝜎y2
u=∞
𝜎x
√
1−𝜌2xy
⎞
(
𝜎y −𝜌xy 𝜎x
⎟+
− 12 ⎢𝑤2 −2u𝑤⎜
⎜𝜎
⎢
2 ⎟
⎣
⎝ x 1−𝜌xy ⎠
√
×e
=e
𝜎x
√
1−𝜌2xy
(
𝜎2
𝜎x2 (1−𝜌2
xy )
×e
By setting 𝑤 =
𝜇y + 12 𝜎y2
I2 = e
where 𝜌xy =
𝑤
⎛
⎞
⎜
⎟
𝜎
⎜ √
⎟
⎜ 𝜎x 1 − 𝜌2xy ⎟
⎝
⎠
𝑤=
u=∞
𝜎
⎤
u2 ⎥
⎥
⎦
d𝑤du
1
2𝜋
⎤
√
)⎡
⎢
⎥
𝜎x (𝜎y −𝜌xy 𝜎x )
1−𝜌2xy
2
2
⎢( 𝑤
) −2u𝑤
+u ⎥
2
𝜎
⎢
⎥
2
𝜎
⎢ 2
⎥
⎣ 𝜎x (1−𝜌2xy )
⎦
√
2𝜋
−
1
e
1
(𝑤2 −2𝜌xy u𝑤+u2 )
2(1−𝜌2xy )
2
1 − 𝜌xy
, thus
I2 = e
𝜇y + 12 𝜎y2
𝜇y + 12 𝜎y2
=e
d𝑤du.
,
𝜇x −𝜇y −𝜎y (𝜎y −𝜌xy 𝜎x )
𝜎
∫u=−∞ ∫𝑤=−∞
𝜎y − 𝜌xy 𝜎x
)
𝜇x −𝜇y −𝜎y (𝜎y −𝜌xy 𝜎x )
∫𝑤=−∞
− 12
(𝜎y −𝜌xy 𝜎x )2
1+ 2
𝜎x (1−𝜌2xy )
∫u=−∞
𝑤=
×
1
2𝜋
⎞
⎛
⎟
⎜ 𝜇x − 𝜇y − 𝜎y (𝜎y − 𝜌xy 𝜎x )
𝚽⎜ √
, ∞, 𝜌xy ⎟
⎟
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2
⎠
⎝
⎛
⎞
⎜ 𝜇x − 𝜇y − 𝜎y (𝜎y − 𝜌xy 𝜎x ) ⎟
Φ⎜ √
⎟.
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎝
⎠
d𝑤du
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By substituting I1 and I2 back into 𝔼[max{eX − eY , 0}] we have
[
{
}]
𝔼 max e − e , 0
X
Y
=e
−e
𝜇x + 12 𝜎x2
𝜇y + 12 𝜎y2
⎞
⎛
⎜ 𝜇x − 𝜇y + 𝜎x (𝜎x − 𝜌xy 𝜎y ) ⎟
Φ⎜ √
⎟
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎠
⎝
⎛
⎞
⎜ 𝜇x − 𝜇y − 𝜎y (𝜎y − 𝜌xy 𝜎y ) ⎟
Φ⎜ √
⎟.
⎜
𝜎x2 − 2𝜌xy 𝜎x 𝜎y + 𝜎y2 ⎟
⎝
⎠
◽
18. Markov’s Inequality. Let X be a non-negative random variable with mean 𝜇. For 𝛼 > 0,
show that
𝜇
ℙ (X ≥ 𝛼) ≤ .
𝛼
Solution: Since 𝛼 > 0, we can write
1I{X≥𝛼}
{
1 X≥𝛼
=
0 otherwise
and since X ≥ 0, we can deduce that
X
.
𝛼
1I{X≥𝛼} ≤
Taking expectations
and since
we have
(
) 𝔼(X)
𝔼 1I{X≥𝛼} ≤
𝛼
)
(
𝔼 1I{X≥𝛼} = 1 ⋅ ℙ (X ≥ 𝛼) + 0 ⋅ ℙ (X ≤ 𝛼) = ℙ (X ≥ 𝛼)
ℙ (X ≥ 𝛼) ≤
and
𝔼(X) = 𝜇,
𝜇
.
𝛼
N.B. Alternatively, we can also show the result as
𝔼(X) =
)
1 − Fx (u) du ≥
𝛼
∞(
∫0
∫0
)
(
)
1 − Fx (u) du ≥ 𝛼 1 − Fx (𝛼)
(
and hence it follows that ℙ(X ≥ 𝛼) = 1 − Fx (𝛼) ≤
𝔼(X)
.
𝛼
◽
19. Chebyshev’s Inequality. Let X be a random variable with mean 𝜇 and variance 𝜎 2 . Then
for k > 0, show that
𝜎2
ℙ (|X − 𝜇| ≥ k) ≤ 2 .
k
Solution: Take note that |X − 𝜇| ≥ k if and only if (X − 𝜇)2 ≥ k2 . Because (X − 𝜇)2 ≥ 0,
and by applying Markov’s inequality (see Problem 1.2.2.18, page 40) we have
]
[
(
) 𝔼 (X − 𝜇)2
𝜎2
2
2
=
ℙ (X − 𝜇) ≥ k ≤
k2
k2
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41
and hence the above inequality is equivalent to
ℙ (|X − 𝜇| ≥ k) ≤
1.2.3
𝜎2
.
k2
◽
Properties of Expectations
1. Show that if X is a random variable taking non-negative values then
∞
⎧∑
⎪ ℙ(X > x)
⎪
𝔼 (X) = ⎨ x=0
⎪ ∞
⎪∫ ℙ(X ≥ x) dx
⎩ 0
if X is a discrete random variable
if X is a continuous random variable.
Solution: We first show the result when X takes non-negative integer values only. By
definition
𝔼(X) =
∞
∑
yℙ(X = y)
y=0
=
y
∞ ∑
∑
ℙ(X = y)
y=0 x=0
=
∞ ∑
∞
∑
ℙ(X = y)
x=0 y=x+1
=
∞
∑
ℙ(X > x).
x=0
For the case when X is a continuous random variable taking non-negative values we have
∞
𝔼(X) =
=
=
∫0
∫0
yfX (y) dy
∞{
∫0
∞{
∫0
}
y
fX (y) dx
∞
∫x
dy
}
fX (y) dy dx
∞
=
∫0
ℙ(X ≥ x) dx.
◽
2. Hölder’s Inequality. Let 𝛼, 𝛽 ≥ 0 and for p, q > 1 such that
lowing inequality:
𝛼𝛽 ≤
𝛼p 𝛽 q
+
p
q
1 1
+ = 1 show that the folp q
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holds. Finally, if X and Y are a pair of jointly continuous variables, show that
𝔼 (|XY|) ≤ {𝔼 (|X p |)}1∕p {𝔼 (|Y q |)}1∕q .
Solution: The inequality certainly holds for 𝛼 = 0 and 𝛽 = 0. Let 𝛼 = ex∕p and 𝛽 = ey∕q
where x, y ∈ ℝ. By substituting 𝜆 = 1∕p and 1 − 𝜆 = 1∕q,
e𝜆x+(1−𝜆)y ≤ 𝜆ex + (1 − 𝜆)ey
holds true since the exponential function is a convex function and hence
𝛼𝛽 ≤
By setting
𝛼=
hence
𝛼p 𝛽 q
+ .
p
q
|X|
,
{𝔼 (|X p |)}1∕p
𝛽=
|Y|
{𝔼 (|Y q |)}1∕q
|X|p
|Y|q
|XY|
≤
+
.
p 𝔼 (|X p |) q 𝔼 (|Y q |)
{𝔼 (|X p |)}1∕p {𝔼 (|Y q |)}1∕q
Taking expectations we obtain
𝔼 (|XY|) ≤ {𝔼 (|X p |)}1∕p {𝔼 (|Y q |)}1∕q .
◽
3. Minkowski’s Inequality. Let X and Y be a pair of jointly continuous variables, show that
if p ≥ 1 then
)}1∕p
{ (
≤ {𝔼 (|X p |)}1∕p + {𝔼 (|Y p |)}1∕p .
𝔼 |X + Y|p
Solution: Since 𝔼(|X + Y|) ≤ 𝔼(|X|) + 𝔼(|Y|) and using Hölder’s inequality we can
write
(
)
𝔼 (|X + Y|p ) = 𝔼 |X + Y||X + Y|p−1
)
(
)
(
≤ 𝔼 |X||X + Y|p−1 + 𝔼 |Y||X + Y|p−1
)
)
(
(
≤ {𝔼(|X p |)}1∕p {𝔼 |X + Y|(p−1)q }1∕q + {𝔼(|Y p |)}1∕p {𝔼 |X + Y|(p−1)q }1∕q
= {𝔼 (|X p |)}1∕p {𝔼 (|X + Y|p )}1∕q + {𝔼 (|Y p |)}1∕p {𝔼 (|X + Y|p )}1∕q
1 1
+ = 1.
p q
Dividing the inequality by {𝔼 (|X + Y|p )}1∕q we get
since
(
)
{𝔼 |X + Y|p }1∕p ≤ {𝔼 (|X p |)}1∕p + {𝔼 (|Y p |)}1∕p .
◽
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43
4. Change of Measure. Let Ω be a probability space and let ℙ and ℚ be two probability
measures on Ω. Let Z(𝜔) be the Radon–Nikodým derivative defined as
ℚ(𝜔)
ℙ(𝜔)
Z(𝜔) =
such that ℙ(Z > 0) = 1. By denoting 𝔼ℙ and 𝔼ℚ as expectations under the measure ℙ and
ℚ, respectively, show that for any random variable X,
𝔼ℚ (X) = 𝔼ℙ (XZ),
𝔼ℙ (X) = 𝔼ℚ
( )
X
.
Z
Solution: By definition
𝔼ℚ (X) =
∑
𝜔∈Ω
Similarly
𝔼ℙ (X) =
∑
X(𝜔)ℚ(𝜔) =
∑
X(𝜔)Z(𝜔)ℙ(𝜔) = 𝔼ℙ (XZ).
𝜔∈Ω
X(𝜔)ℙ(𝜔) =
𝜔∈Ω
( )
∑ X(𝜔)
X
.
ℚ(𝜔) = 𝔼ℚ
Z(𝜔)
Z
𝜔∈Ω
◽
5. Conditional Probability. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra
of ℱ (i.e., sets in 𝒢 are also in ℱ). If 1IA is an indicator random variable for an event A
defined as
{
1 if 𝜔 ∈ A
1IA (𝜔) =
0 otherwise
show that
𝔼(1IA |𝒢) = ℙ(A|𝒢).
Solution: Since 𝔼(1IA |𝒢) is 𝒢 measurable we need to show that the following partial
averaging property:
∫B
𝔼(1IA |𝒢) dℙ =
∫B
1IA dℙ =
∫B
ℙ(A|𝒢) dℙ
is satisfied for B ∈ 𝒢. Setting
{
1IB (𝜔) =
and expanding
∫B
∫B
{
1 if 𝜔 ∈ B
0 otherwise
and
1IA∩B (𝜔) =
1
0
if 𝜔 ∈ A ∩ B
otherwise
ℙ(A|𝒢) dℙ we have
ℙ(A|𝒢) dℙ = ℙ(A ∩ B) =
∫Ω
1IA∩B dℙ =
∫Ω
1IA ⋅ 1IB dℙ =
∫B
1IA dℙ.
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Since 𝔼(1IA |𝒢) is 𝒢 measurable we have
∫B
𝔼(1IA |𝒢) dℙ =
∫B
1IA dℙ
and hence 𝔼(1IA |𝒢) = ℙ(A|𝒢).
◽
6. Linearity. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ(i.e., sets
in 𝒢 are also in ℱ). If X1 , X2 , . . . , Xn are integrable random variables and c1 , c2 , . . . , cn
are constants, show that
)
(
𝔼 c1 X1 + c2 X2 + . . . + cn Xn |𝒢 = c1 𝔼(X1 |𝒢) + c2 𝔼(X2 |𝒢) + . . . + cn 𝔼(Xn |𝒢).
(
)
Solution: Given 𝔼 c1 X1 + c2 X2 + . . . + cn Xn |𝒢 is 𝒢 measurable, and for any A ∈ 𝒢,
∫A
(
)
𝔼 c1 X1 + c2 X2 + . . . + cn Xn |𝒢 dℙ =
∫A
= c1
(c1 X1 + c2 X2 + . . . + cn Xn ) dℙ
∫A
X1 dℙ + c2
+ . . . + cn
Since
Xi dℙ =
∫A
∫A
X2 dℙ
Xn dℙ.
(
𝔼(Xi |𝒢) dℙ for i = 1, 2, . . . , n therefore 𝔼 c1 X1 + c2 X2 + . . . +
∫)A
∫A
cn Xn |𝒢 = c1 𝔼(X1 |𝒢) + c2 𝔼(X2 |𝒢) + . . . + cn 𝔼(Xn |𝒢).
◽
7. Positivity. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ(i.e., sets
in 𝒢 are also in ℱ). If X is an integrable random variable such that X ≥ 0 almost surely
then show that
𝔼(X|𝒢) ≥ 0
almost surely.
Solution: Let A = {𝑤 ∈ Ω ∶ 𝔼(X|𝒢) < 0} and since 𝔼(X|𝒢) is 𝒢 measurable therefore
A ∈ 𝒢. Thus, from the partial averaging property we have
∫A
𝔼(X|𝒢) dℙ =
∫A
X dℙ.
X dℙ ≥ 0 but 𝔼(X|𝒢) dℙ < 0, which is a con∫A
∫A
tradiction. Thus, ℙ(A) = 0, which implies 𝔼(X|𝒢) ≥ 0 almost surely.
◽
Since X ≥ 0 almost surely therefore
8. Monotonicity. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ(i.e.,
sets in 𝒢 are also in ℱ). If X and Y are integrable random variables such that X ≤ Y almost
surely then show that
𝔼(X|𝒢) ≤ 𝔼(Y|𝒢).
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Solution: Since 𝔼(X − Y|𝒢) is 𝒢 measurable, for A ∈ 𝒢 we can write
∫A
𝔼 (X − Y|𝒢) dℙ =
∫A
(X − Y) dℙ
and since X ≤ Y, from Problem 1.2.3.7 (page 44) we can deduce that
∫A
𝔼 (X − Y|𝒢) dℙ ≤ 0
and hence
𝔼 (X − Y|𝒢) ≤ 0.
Using the linearity of conditional expectation (see Problem 1.2.3.6, page 44)
𝔼 (X − Y|𝒢) = 𝔼(X|𝒢) − 𝔼(Y|𝒢) ≤ 0
and therefore 𝔼(X|𝒢) ≤ 𝔼(Y|𝒢).
◽
9. Computing Expectations by Conditioning. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢
be a sub-𝜎-algebra of ℱ (i.e., sets in 𝒢 are also in ℱ). Show that
𝔼[𝔼(X|𝒢)] = 𝔼(X).
Solution: From the partial averaging property we have, for A ∈ 𝒢,
∫A
𝔼(X|𝒢) dℙ =
∫A
X dℙ
or
𝔼[1IA ⋅ 𝔼(X|𝒢)] = 𝔼(1IA ⋅ X)
{
where
1IA (𝜔) =
1 if 𝜔 ∈ A
0 otherwise
is a 𝒢 measurable random variable. By setting A = Ω we obtain 𝔼[𝔼(X|𝒢)] = 𝔼(X).
◽
10. Taking Out What is Known. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a
sub-𝜎-algebra of ℱ (i.e., sets in 𝒢 are also in ℱ). If X and Y are integrable random
variables and if X is 𝒢 measurable show that
𝔼(XY|𝒢) = X ⋅ 𝔼(Y|𝒢).
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Solution: Since X and 𝔼(Y|𝒢) are 𝒢 measurable therefore X ⋅ 𝔼(Y|𝒢) is also 𝒢 measurable and it satisfies the first property of conditional expectation. By calculating the partial
averaging of X ⋅ 𝔼(Y|𝒢) over a set A ∈ 𝒢 and by defining
{
1 if 𝜔 ∈ A
0 otherwise
1IA (𝜔) =
such that 1IA is a 𝒢 measurable random variable we have
∫A
X ⋅ 𝔼(Y|𝒢) dℙ = 𝔼[1IA ⋅ X𝔼(Y|𝒢)]
= 𝔼[1IA ⋅ XY]
=
∫A
XY dℙ.
Thus, X ⋅ 𝔼(Y|𝒢) satisfies the partial averaging property by setting
∫A
∫A
XY dℙ =
𝔼(XY|𝒢) dℙ. Therefore, 𝔼(XY|𝒢) = X ⋅ 𝔼(Y|𝒢).
◽
11. Tower Property. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ
(i.e., sets in 𝒢 are also in ℱ). If ℋ is a sub-𝜎-algebra of 𝒢 (i.e., sets in ℋ are also in 𝒢)
and X is an integrable random variable, show that
𝔼[𝔼 (X|𝒢)|ℋ] = 𝔼(X|ℋ).
Solution: For an integrable random variable Y, by definition we know that 𝔼(Y|ℋ) is
ℋ measurable, and hence by setting Y = 𝔼 (X|𝒢), and for A ∈ ℋ, the partial averaging
property of 𝔼[𝔼 (X|𝒢)|ℋ] is
∫A
𝔼[𝔼 (X|𝒢)|ℋ] dℙ =
∫A
𝔼 (X|𝒢) dℙ.
Since A ∈ ℋ and ℋ is a sub-𝜎-algebra of 𝒢, A ∈ 𝒢. Therefore,
∫A
𝔼(X|ℋ) dℙ =
∫A
X dℙ =
∫A
𝔼(X|𝒢) dℙ.
This shows that 𝔼(X|ℋ) satisfies the partial averaging property of 𝔼[𝔼 (X|𝒢)|ℋ], and
hence 𝔼[𝔼 (X|𝒢)|ℋ] = 𝔼(X|ℋ).
◽
12. Measurability. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ(i.e.,
sets in 𝒢 are also in ℱ). If the random variable X is 𝒢 measurable then show that
𝔼(X|𝒢) = X.
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Solution: From the partial averaging property, for A ∈ Ω,
∫A
𝔼(X|𝒢) dℙ =
∫A
X dℙ
and if X is 𝒢 measurable then it satisfies
𝔼(X|𝒢) = X.
◽
13. Independence. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ(i.e.,
sets in 𝒢 are also in ℱ). If X = 1IB such that
{
1 if 𝜔 ∈ B
0 otherwise
1IB (𝜔) =
and 1IB is independent of 𝒢 show that
𝔼(X|𝒢) = 𝔼(X).
Solution: Since 𝔼(X) is non-random then 𝔼(X) is 𝒢 measurable. Therefore, we now need
to check that the following partial averaging property:
∫A
𝔼(X) dℙ =
∫A
is satisfied for A ∈ 𝒢.
Let X = 1IB such that
X dℙ =
∫A
𝔼(X|𝒢) dℙ
{
1 if 𝜔 ∈ B
0 otherwise
1IB (𝜔) =
and the random variable 1IB is independent of 𝒢. In addition, we also define
{
1IA (𝜔) =
1 if 𝜔 ∈ A
0 otherwise
where 1IA is 𝒢 measurable. For all A ∈ 𝒢 we have
∫A
X dℙ =
∫A
1IB dℙ =
∫A
ℙ(B) dℙ = ℙ(A)ℙ(B).
Furthermore, since the sets A and B are independent we can also write
∫A
X dℙ =
∫A
1IB dℙ =
∫Ω
1IA 1IB dℙ =
∫Ω
1IA∩B dℙ = ℙ(A ∩ B)
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where
{
1IA∩B (𝜔) =
1 if 𝜔 ∈ A ∩ B
0 otherwise
and hence
∫A
X dℙ = ℙ(A ∩ B) = ℙ(A)ℙ(B) = ℙ(A)𝔼(X) =
∫A
𝔼(X) dℙ.
Thus, we have 𝔼(X|𝒢) = 𝔼(X).
◽
14. Conditional Jensen’s Inequality. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a
sub-𝜎-algebra of ℱ (i.e., sets in 𝒢 are also in ℱ). If 𝜑 ∶ ℝ → ℝ is a convex function and
X is an integrable random variable show that
𝔼[𝜑(X)|𝒢] ≥ 𝜑[𝔼(X|𝒢)].
Deduce that if X is independent of 𝒢 then the above inequality becomes
𝔼[𝜑(X)] ≥ 𝜑[𝔼(X)].
Solution: Given that 𝜑 is a convex function,
𝜑(x) ≥ 𝜑(y) + 𝜑′ (y)(y − x).
By setting x = X and y = 𝔼(X|𝒢) we have
𝜑(X) ≥ 𝜑[𝔼(X|𝒢)] + 𝜑′ [𝔼(X|𝒢)][𝔼(X|𝒢) − X]
and taking conditional expectations,
𝔼[𝜑(X)|𝒢] ≥ 𝜑[𝔼(X|𝒢)].
If X is independent of 𝒢 then from Problem 1.2.3.13 (page 47) we can set y = 𝔼(X|𝒢) =
𝔼(X). Using the same steps as described above we have
𝜑(X) ≥ 𝜑[𝔼(X)] + 𝜑′ [𝔼(X)][𝔼(X) − X]
and taking expectations we finally have
𝔼[𝜑(X)] ≥ 𝜑[𝔼(X)].
◽
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15. Let (Ω, ℱ, ℙ) be a probability space and let 𝒢 be a sub-𝜎-algebra of ℱ (i.e., sets in 𝒢 are
also in ℱ). If X is an integrable random variable and 𝔼(X 2 ) < ∞ show that
]
( )
[
𝔼 𝔼(X|𝒢)2 ≤ 𝔼 X 2 .
Solution: From the conditional Jensen’s inequality (see Problem 1.2.3.14, page 48) we
set 𝜑(x) = x2 which is a convex function. By substituting x = 𝔼 (X|𝒢) we have
(
)
𝔼(X|𝒢)2 ≤ 𝔼 X 2 |𝒢 .
Taking expectations
]
[ (
)]
[
𝔼 𝔼(X|𝒢)2 ≤ 𝔼 𝔼 X 2 |𝒢
and from the tower property (see Problem 1.2.3.11, page 46)
)]
( )
[ (
𝔼 𝔼 X 2 |𝒢 = 𝔼 X 2 .
]
( )
[
Thus, 𝔼 𝔼(X|𝒢)2 ≤ 𝔼 X 2 .
◽
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2
Wiener Process
In mathematics, a Wiener process is a stochastic process sharing the same behaviour as Brownian motion, which is a physical phenomenon of random movement of particles suspended in a
fluid. Generally, the terms “Brownian motion” and “Wiener process” are the same, although the
former emphasises the physical aspects whilst the latter emphasises the mathematical aspects.
In quantitative analysis, by drawing on the mathematical properties of Wiener processes to
explain economic phenomena, financial information such as stock prices, commodity prices,
interest rates, foreign exchange rates, etc. are treated as random quantities and then mathematical models are constructed to capture the randomness. Given these financial models are
stochastic and continuous in nature, the Wiener process is usually employed to express the
random component of the model. Before we discuss the models in depth, in this chapter we
first look at the definition and basic properties of a Wiener process.
2.1 INTRODUCTION
By definition, a random walk is a mathematical formalisation of a trajectory that consists of
taking successive random steps at every point in time. To construct a Wiener process in continuous time, we begin by setting up a symmetric random walk – such as tossing a fair coin
infinitely many times where the probability of getting a head (H) or a tail (T) in each toss is
1
. By defining the i-th toss as
2
{
1
if toss is H
Zi =
−1 if toss is T
and setting M0 = 0, the process
Mk =
k
∑
Zi ,
k = 1, 2, . . .
i=1
is a symmetric random walk. In a continuous time setting, to approximate a Wiener process
for n ∈ ℤ+ we define the scaled symmetric random walk as
⌊nt⌋
1
1 ∑
Wt(n) = √ M⌊nt⌋ = √
Zi
n
n i=1
such that in the limit of n → ∞ we can obtain the Wiener process where
⌊nt⌋
D
1 ∑
lim W (n) = lim √
Zi −−→ 𝒩(0, t).
n→∞ t
n→∞ n
i=1
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Given that a Wiener process is a limiting distribution of a scaled symmetric random walk,
the following is the definition of a standard Wiener process.
Definition 2.1 (Standard Wiener Process) Let (Ω, ℱ, ℙ) be a probability space. A stochastic process {Wt ∶ t ≥ 0} is defined to be a standard Wiener process (or ℙ-standard Wiener
process) if:
(a) W0 = 0 and has continuous sample paths;
(b) for each t > 0 and s > 0, Wt+s − Wt ∼ 𝒩(0, s) (stationary increment);
⟂ Wt (independent increment).
(c) for each t > 0 and s > 0, Wt+s − Wt ⟂
A standard Wiener process is a standardised version of a Wiener process, which need not
begin at W0 = 0, and may have a non-zero drift term 𝜇 ≠ 0 and a variance term not necessarily
equal to t.
̂ t ∶ t ≥ 0} is called a Wiener process if it can
Definition 2.2 (Wiener Process) A process {W
be written as
̂ t = 𝜈 + 𝜇t + 𝜎Wt
W
where 𝜈, 𝜇 ∈ ℝ, 𝜎 > 0 and Wt is a standard Wiener process.
Almost all financial models have the Markov property and without exception the Wiener
process also has this important property, where it is used to relate stochastic calculus to partial
differential equations and ultimately to the pricing of options. The following is an important
result concerning the Markov property of a standard Wiener process.
Theorem 2.3 (Markov Property) Let (Ω, ℱ, ℙ) be a probability space. The standard Wiener
process {Wt ∶ t ≥ 0} is a Markov process such that the conditional distribution of Wt given
the filtration ℱs , s < t depends only on Ws .
Another generalisation from the Markov property is the strong Markov property, which is an
important result in establishing many other properties of Wiener processes such as martingales.
Clearly, the strong Markov property implies the Markov property but not vice versa.
Theorem 2.4 (Strong Markov Property) Let (Ω, ℱ, ℙ) be a probability space. If {Wt ∶ t ≥
0} is a standard Wiener process and given ℱt is the filtration up to time t, then for s > 0,
Wt+s − Wt ⟂
⟂ ℱt .
Once we have established the Markov properties, we can use them to show that a Wiener
process is a martingale. Basically, a stochastic process is a martingale when its conditional
expected value of an observation at some future time t, given all the observations up to some
earlier time s, is equal to the observation at that earlier time s. In formal terms the definition
of this property is given as follows.
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Definition 2.5 (Martingale Property for Continuous Process) Let (Ω, ℱ, ℙ) be a probability space. A stochastic process {Xt ∶ t ≥ 0} is a continuous-time martingale if:
(a) 𝔼(Xt |ℱs ) = Xs , for all 0 ≤ s ≤ t;
(b) 𝔼(|Xt |) < ∞;
(c) Xt is ℱt -adapted.
In addition to properties (b) and (c), the process is a submartingale if 𝔼(Xt |ℱs ) ≥ Xs and a
supermartingale if 𝔼(Xt |ℱs ) ≤ Xs for all 0 ≤ s ≤ t.
In contrast, we can also define the martingale property for a discrete process.
Definition 2.6 (Martingale Property for Discrete Process) A discrete process X = {Xn ∶
n = 0, 1, 2, . . .} is a martingale relative to (Ω, ℱ, ℙ) if for all n:
(a) 𝔼(Xn+1 |ℱn ) = Xn ;
(b) 𝔼(|Xn |) < ∞;
(c) Xn is ℱn -adapted.
Together with properties (b) and (c), the process is a submartingale if 𝔼(Xn+1 |ℱn ) ≥ Xn and a
supermartingale if 𝔼(Xn+1 |ℱn ) ≤ Xn for all n.
A most important application of the martingale property of a stochastic process is in the area
of derivatives pricing where under the risk-neutral measure, to avoid any arbitrage opportunities, all asset prices have the same expected rate of return that is the risk-free rate. As we
shall see in later chapters, by modelling an asset price with a Wiener process to represent the
random component, under the risk-neutral measure, the expected future price of the asset discounted at a risk-free rate given its past information is a martingale. In addition, martingales
do have certain features even when they are stopped at random times, as given in the following
theorem.
Theorem 2.7 (Optional Stopping (Sampling)) Let (Ω, ℱ, ℙ) be a probability space. A random time T ∈ [0, ∞) is called a stopping time if {T ≤ t} ∈ ℱt for all t ≥ 0. If T < ∞ is a
stopping time and {Xt } is a martingale, then 𝔼(XT ) = 𝔼(X0 ) if any of the following are true:
(a) 𝔼(T) < ∞;
(b) there exists a constant K such that 𝔼(|Xt+Δt − Xt |) ≤ K, for Δt > 0.
Take note that an important application of the optional stopping theorem is the first passage
time of a standard Wiener process hitting a level. Here one can utilise it to analyse American
options, where in this case the exercise time is a stopping time. In tandem with the stopping time, the following reflection principle result allows us to find the joint distribution of
(max0≤s≤t Ws , Wt ) and the distribution of max0≤s≤t Ws , which in turn can be used to price exotic
options such as barrier and lookback options.
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Theorem 2.8 (Reflection Principle) Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥
0} be a standard Wiener process. By setting T as a stopping time and defining
{
if t ≤ T
̃ t = Wt
W
2WT − Wt if t > T
̃ t ∶ t ≥ 0} is also a standard Wiener process.
then {W
Finally, another useful property of the Wiener process is the quadratic variation,
where if {Wt ∶ t ≥ 0} is a standard Wiener process then by expressing dWt as the
infinitesimal increment of Wt and setting ti = it∕n, i = 0, 1, 2, . . . , n − 1, n > 0 such that
0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, the quadratic variation of Wt is defined as
lim
n→∞
n−1
∑
t
(Wti+1 − Wti )2 =
i=0
∫0
dWu2 = t
which has a finite value. In addition, the cross-variation of Wt and t, the quadratic variation of
t and the p-order variation, p ≥ 3 of Wt can be expressed as
lim
n→∞
n−1
∑
(Wti+1 − Wti )(ti+1 − ti ) =
i=0
lim
n→∞
and
lim
n−1
∑
n→∞
n−1
∑
dWu du = 0,
∫0
t
(ti+1 − ti )2 =
i=0
t
(Wti+1 − Wti )p =
i=0
t
∫0
∫0
p
du2 = 0
dWu = 0,
p ≥ 3.
Informally, we can therefore write
(dWt )2 = dt,
dWt dt = 0,
(dt)2 = 0,
(dWt )p = 0,
p≥3
where dWt and dt are the infinitesimal increment of Wt and t, respectively. The significance
of the above results constitutes the key ingredients in Itō’s formula to find the differential of a
stochastic function and also in deriving the Black–Scholes equation to price options.
In essence, the many properties of the Wiener process made it a very suitable choice to
express the random component when modelling stock prices, interest rates, foreign currency
exchange rates, etc. One notable example is when pricing European-style options, where,
due to its inherent properties we can obtain closed-form solutions which would not be possible had it been modelled using other processes. In addition, we could easily extend the
one-dimensional Wiener process to a multi-dimensional Wiener process to model stock prices
that are correlated with each other as well as stock prices under stochastic volatility. However, owing to the normal distribution of the standard Wiener process it tends to provide stock
price returns that are symmetric and short-tailed. In practical situations, stock price returns are
skewed and have heavy tails and hence they do not follow a normal distribution. Nevertheless,
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even if there exist complex models to capture such attributes (for example, using a Lévy
process), a Wiener process is still a vital component to model sources of uncertainty in finance.
2.2
2.2.1
PROBLEMS AND SOLUTIONS
Basic Properties
1. Let (Ω, ℱ, ℙ) be a probability space. We consider a symmetric random walk such that the
j-th step is defined as
⎧
⎪1
Zj = ⎨
⎪−1
⎩
with probability
with probability
1
2
1
2
where Zi ⟂
⟂ Zj , i ≠ j. By setting 0 = k0 < k1 < k2 < . . . < kt , we let
Mki =
ki
∑
Zj ,
i = 1, 2, . . . , t
j=1
where M0 = 0.
Show that the symmetric random walk has independent increments such that the random
variables
Mk1 − Mk0 , Mk2 − Mk1 , . . . , Mkt − Mkt−1
are independent.
Finally, show that 𝔼(Mki+1 − Mki ) = 0 and Var(Mki+1 − Mki ) = ki+1 − ki .
Solution: By definition
Mki − Mki−1 =
ki
∑
Zj = si
j=ki−1 +1
and for m < n, m, n = 1, 2, . . . , t,
ℙ(Mkn − Mkn−1 = sn |Mkm − Mkm−1 = sm )
=
=
ℙ(Mkn − Mkn−1 = sn , Mkm − Mkm−1 = sm )
ℙ(Mkm − Mkm−1 = sm )
ℙ(sum of walks in [kn−1 + 1, kn ] ∩ sum of walks in [km−1 + 1, km ])
.
ℙ(sum of walks in [km−1 + 1, km ])
Because m < n and since Zi is independent of Zj , i ≠ j, there are no overlapping
events between the intervals [kn−1 + 1, kn ] and [km−1 + 1, km ] and hence for m < n,
m, n = 1, 2, . . . , t,
ℙ(Mkn − Mkn−1 = sn |Mkm − Mkm−1 = sm ) = ℙ(sum of walks in [kn−1 + 1, kn ])
= ℙ(Mkn − Mkn−1 = sn ).
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Thus, we can deduce that Mk1 − Mk0 , Mk2 − Mk1 , . . . , Mkt − Mkt−1 are independent.
Since 𝔼(Zj ) = 1 ⋅ 12 − 1 ⋅ 12 = 0 and Var(Zj ) = 𝔼(Zj2 ) = 1 ⋅ 12 + 1 ⋅ 12 = 1, and because Zj ,
j = 1, 2, . . . are independent, we have
𝔼(Mki − Mki−1 ) =
ki
∑
𝔼(Zj ) = 0
j=ki−1 +1
and
Var(Mki − Mki−1 ) =
ki
∑
Var(Zj ) =
j=ki−1 +1
ki
∑
1 = ki − ki−1 .
j=ki−1 +1
◽
2. Let (Ω, ℱ, ℙ) be a probability space. For a symmetric random walk
Mk =
k
∑
Zi
i=1
with starting point at M0 = 0, ℙ(Zi = 1) = ℙ(Zi = −1) = 12 , show that Mk is a martingale.
Solution: To show that Mk is a martingale we note that
(a) For j < k, j, k ∈ ℤ+ , using the independent increment property and 𝔼(Mk − Mj ) = 0
as shown in Problem 2.2.1.1 (page 55),
𝔼(Mk |ℱj ) = 𝔼(Mk − Mj + Mj |ℱj )
= 𝔼(Mk − Mj |ℱj ) + 𝔼(Mj |ℱj )
= 𝔼(Mk − Mj ) + Mj
= Mj .
(b) Given Mk =
∑k
i=1 Zi
it follows that
|
|∑
k
k
∑
| k | ∑
|Zi | =
1 = k < ∞.
|Mk | = || Zi || ≤
| i=1 | i=1
i=1
|
|
(c) Mk is clearly ℱk -adapted.
From the results of (a)–(c), we have shown that Mk is a martingale.
3. Donsker Theorem. Let (Ω, ℱ, ℙ) be a probability space. For a symmetric random walk
Mk =
k
∑
i=1
Zi
◽
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where M0 = 0, ℙ(Zi = 1) = ℙ(Zi = −1) =
1
2
and by defining
⌊nt⌋
1
1 ∑
Wt(n) = √ M⌊nt⌋ = √
Zi
n
n i=1
for a fixed time t, show that
⌊nt⌋
D
1 ∑
lim Wt(n) = lim √
Zi −−→ 𝒩(0, t).
n→∞
n→∞ n
i=1
Solution: Since for all i = 1, 2, . . . ,
𝔼(Zi ) = 0 and
Var(Zi ) = 1
and given Zi ⟂
⟂ Zj , i ≠ j,
⌊nt⌋
)
(
1 ∑
𝔼(Zi ) = 0
𝔼 Wt(n) = √
n i=1
and
⌊nt⌋
⌊nt⌋
)
(
⌊nt⌋
1∑
1∑
.
Var Wt(n) =
Var(Zi ) =
1=
n i=1
n i=1
n
For n → ∞ we can deduce that
)
(
lim 𝔼 Wt(n) = 0
n→∞
and
)
(
⌊nt⌋
lim Var Wt(n) = lim
= t.
n→∞
n→∞ n
Therefore, from the central limit theorem,
⌊nt⌋
lim W (n)
n→∞ t
D
1 ∑
= lim √
Zi −−→ 𝒩(0, t).
n→∞ n
i=1
◽
4. Covariance of Two Standard Wiener Processes. Let (Ω, ℱ, ℙ) be a probability space and
let {Wt ∶ t ≥ 0} be a standard Wiener process. Show that
Cov(Ws , Wt ) = min{s, t}
and deduce that the correlation coefficient of Ws and Wt is
√
min{s, t}
.
𝜌=
max{s, t}
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Solution: Since Wt ∼ 𝒩(0, t), Ws ∼ 𝒩(0, s) and by definition
Cov(Ws , Wt ) = 𝔼(Ws Wt ) − 𝔼(Ws )𝔼(Wt ) = 𝔼(Ws Wt ).
Let s ≤ t and because Ws ⟂
⟂ Wt − Ws ,
(
)
𝔼(Ws Wt ) = 𝔼 Ws (Wt − Ws ) + Ws2
( )
= 𝔼(Ws (Wt − Ws )) + 𝔼 Ws2
( )
= 𝔼(Ws )𝔼(Wt − Ws ) + 𝔼 Ws2
= s.
⟂ Ws − W t ,
For s > t and because Wt ⟂
)
(
𝔼(Ws Wt ) = 𝔼 Wt (Ws − Wt ) + Wt2
( )
= 𝔼(Wt )𝔼(Ws − Wt ) + 𝔼 Wt2
= t.
Therefore, Cov(Ws , Wt ) = s ∧ t = min{s, t}.
By definition, the correlation coefficient of Ws and Wt is defined as
𝜌= √
=
Cov(Ws , Wt )
Var(Ws )Var(Wt )
min{s, t}
.
√
st
For s ≤ t,
s
𝜌= √ =
st
whilst for s > t,
t
𝜌= √ =
st
√
Therefore, 𝜌 =
min{s, t}
.
max{s, t}
√
√
s
t
t
.
s
◽
5. Joint Distribution of Standard Wiener Processes. Let (Ω, ℱ, ℙ) be a probability space and
let {Wtk ∶ tk ≥ 0}, k = 0, 1, 2, . . . , n be a standard Wiener process where 0 = t0 < t1 <
t2 < . . . < tn .
Find the moment generating function of the joint distribution (Wt1 , Wt2 , . . . , Wtn ) and its
corresponding probability density function.
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Show that for t < T, the conditional distributions
(
)
yt t(T − t)
Wt |WT = y ∼ 𝒩
,
T
T
and
WT |Wt = x ∼ 𝒩(x, T − t).
Solution: By definition, the moment generating function for the joint distribution (Wt1 ,
Wt2 , . . . , Wtn ) is given as
(
)
(
)
MWt ,Wt , ... ,Wt 𝜃1 , 𝜃2 , . . . , 𝜃n = 𝔼 e𝜃1 Wt1 +𝜃2 Wt2 + ... +𝜃n Wtn , 𝜃1 , 𝜃2 , . . . , 𝜃n ∈ ℝ.
1
2
n
Given that Wt1 , Wt2 − Wt1 , . . . , Wtn − Wtn−1 are independent and normally distributed, we
can write
𝜃1 Wt1 + 𝜃2 Wt2 + . . . + 𝜃n Wtn = (𝜃1 + 𝜃2 + . . . + 𝜃n )Wt1 + (𝜃2 + . . . + 𝜃n )(Wt2 − Wt1 )
+ . . . + 𝜃n (Wtn − Wtn−1 )
and since for any 𝜃, s < t, e𝜃(Wt −Ws ) ∼ log-𝒩(0, 𝜃 2 (t − s)) therefore
[
(
)
(
)]
)
(
(𝜃1 +𝜃2 + ... +𝜃n )Wt1 +(𝜃2 + ... +𝜃n ) Wt2 −Wt1 + ... +𝜃n Wtn −Wtn−1
𝜃1 Wt +𝜃2 Wt + ... +𝜃n Wtn
1
2
=𝔼 e
𝔼 e
[
] [
]
= 𝔼 e(𝜃1 +𝜃2 + ... +𝜃n )Wt1 𝔼 e(𝜃2 + ... +𝜃n )(Wt2 −Wt1 )
[ (
)]
𝜃n Wtn −Wt
n−1
× ... × 𝔼 e
2
1
= e 2 (𝜃1 +𝜃2 + ... +𝜃n )
2
t1 + 12 (𝜃2 + ... +𝜃n ) (t2 −t1 )+ ... + 12 𝜃n2 (tn −tn−1 )
1 T
𝚺𝜽
= e2𝜽
where 𝜽T = (𝜃1 , 𝜃2 , . . . , 𝜃n ) and 𝚺 is the covariance matrix for the Wiener process. From
Problem 2.2.1.4 (page 57) we can express
⎡ 𝔼(Wt2 )
⎢ 𝔼(W1 W )
t2 t1
𝚺=⎢
⎢⋮
⎢ 𝔼(W W )
tn t1
⎣
𝔼(Wt1 Wt2 ) . . .
𝔼(Wt22 )
...
⋮
⋱
𝔼(Wtn Wt2 ) . . .
𝔼(Wt1 Wtn ) ⎤ ⎡t1
𝔼(Wt2 Wtn ) ⎥⎥ ⎢t1
=
⋮
⎥ ⎢⎢ ⋮
⎥ ⎣t1
𝔼(Wt2 )
⎦
n
t1
t2
⋮
t2
...
...
⋱
...
t1 ⎤
t2 ⎥
.
⋮⎥
⎥
tn ⎦
Using ∏elementary column operations we can easily show that the determinant
|𝚺| = ni=1 (ti − ti−1 ) ≠ 0. Therefore, 𝚺−1 exists and the probability density function for
the joint distribution (Wt1 , Wt2 , . . . , Wtn ) is given as
fWt
1
,Wt , ... ,Wtn (x)
where xT = (x1 , x2 , . . . , xn ).
2
=
1
n
2
(2𝜋) |𝚺|
1 T −1
𝚺 x
1
2
e− 2 x
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For the case of joint distribution of (Wt , WT ), t < T we can deduce that
[
1
exp −
2
1
fWt ,WT (x, y) =
√
(2𝜋) t(T − t)
(
Tx2 − 2txy + ty2
t(T − t)
)]
with conditional density function of Wt given WT = y
fWt |WT (x|y) =
fWt ,WT (x, y)
fWT (y)
= √
(
2𝜋
1
t(T − t)
T
)
(
yt )2 ⎤
⎡
x
−
⎢ 1
T ⎥
exp ⎢−
t(T
−
t) ⎥⎥
2
⎢
⎣
⎦
T
and conditional density function of WT given Wt = x
fWT |Wt (y|x) =
fWt ,WT (x, y)
fWt (x)
=√
1
2𝜋(T − t)
[
]
1 (y − x)2
exp −
.
2 T −t
Thus,
(
Wt |WT = y ∼ 𝒩
yt t(T − t)
,
T
T
)
and
WT |Wt = x ∼ 𝒩(x, T − t).
◽
6. Reflection. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener
process. Show that under reflection, Bt = −Wt is also a standard Wiener process.
Solution:
(a) B0 = −W0 = 0, and given that Wt has continuous sample paths we can deduce that Bt
has continuous sample paths as well.
(b) For t > 0, s > 0
Bt+s − Bt = −Wt+s + Wt = −(Wt+s − Wt ) ∼ 𝒩(0, s).
(c) Since Wt+s − Wt ⟂
⟂ Wt therefore
𝔼[(Bt+s − Bt )Bt ] = Cov(−Wt+s + Wt , −Wt ) = −Cov(Wt+s − Wt , Wt ) = 0.
Given Bt ∼ 𝒩(0, t) and Bt+s − Bt ∼ 𝒩(0, s) and the joint distribution of Bt and Bt+s −
Bt is a bivariate normal (see Problem 2.2.1.5, page 58), then if Cov(Bt+s − Bt , Bt ) = 0
so Bt+s − Bt ⟂
⟂ Bt .
From the results of (a)–(c) we have shown that Bt = −Wt is also a standard Wiener process.
◽
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61
7. Time Shifting. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. Show that under time shifting, Bt = Wt+u − Wu , u > 0 is also a standard
Wiener process.
Solution: By setting t̃ = t + u.
(a) B0 = Wu − Wu = 0, and given Wt has continuous sample paths therefore we can
deduce Bt has continuous sample paths as well.
(b) For t > 0, u > 0, s > 0
Bt+s − Bt = W̃t+s − W̃t+u ∼ 𝒩(0, s).
⟂ Wt therefore
(c) Since Wt+s − Wt ⟂
𝔼[(Bt+s − Bt )Bt ] = Cov(Wt+u+s − Wt+u , Wt+u ) = Cov(Wt̃ +s − Wt̃ , Wt̃ ) = 0.
Given Bt ∼ 𝒩(0, t) and Bt+s − Bt ∼ 𝒩(0, s) and the joint distribution of Bt and Bt+s −
Bt is a bivariate normal (see Problem 2.2.1.5, page 58), then if Cov(Bt+s − Bt , Bt ) = 0
so Bt+s − Bt ⟂
⟂ Bt .
From the results of (a)–(c) we have shown that Bt = Wt+u − Wu is also a standard Wiener
process.
◽
8. Normal Scaling. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. Show that under normal scaling, Bt = cW t , c ≠ 0 is also a standard
c2
Wiener process.
Solution:
(a) B0 = cW0 = 0 and it is clear that Bt has continuous sample paths for t ≥ 0 and c ≠ 0.
(b) For t > 0, s > 0
(
)
Bt+s − Bt = c W t+s − W t
c2
c2
with mean
(
)
𝔼(Bt+s − Bt ) = c𝔼 W t+s
c2
)
(
− c𝔼 W t = 0
c2
and variance
(
)
)
(
)
(
Var(Bt+s − Bt ) = Var cW t+s + Var cW t − 2Cov cW t+s , cW t
c2
Since both W t+s
c2
c2
c2
c2
= t + s + t − 2min{t + s, t}
= s.
(
(
)
)
t+s
t
∼ 𝒩 0, 2
and W t ∼ 𝒩 0, 2 , then Bt+s − Bt ∼ 𝒩(0, s).
c
c
c2
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(c) Finally, to show that Bt+s − Bt ⟂
⟂ Bt we note that since Wt has the independent increment property, so
𝔼
( )
) ]
Bt+s − Bt Bt = 𝔼(Bt+s Bt ) − 𝔼 B2t
( )
= Cov(Bt+s , Bt ) − 𝔼 B2t
[(
= min{t + s, t} − t
= 0.
Since Bt ∼ 𝒩(0, t) and Bt+s − Bt ∼ 𝒩(0, s) and the joint distribution of Bt and Bt+s −
Bt is a bivariate normal (see Problem 2.2.1.5, page 58), then if Cov(Bt+s − Bt , Bt ) = 0
so Bt+s − Bt ⟂
⟂ Bt .
From the results of (a)–(c) we have shown that Bt = cW t , c ≠ 0 is also a standard Wiener
c2
process.
◽
9. Time Inversion. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. Show that under time inversion,
{
0
Bt =
tW 1
t
if t = 0
if t ≠ 0
is also a standard Wiener process.
Solution:
(a) B0 = 0 and it is clear that Bt has continuous sample paths for t > 0. From continuity
at t = 0 we can deduce that tW 1 → 0 as t → 0.
t
(b) Since Wt ∼ 𝒩(0, t) we can deduce tW 1 ∼ 𝒩(0, t), t > 0. Therefore,
t
Bt+s − Bt = (t + s)W
1
t+s
− tW 1
t
with mean
(
𝔼(Bt+s − Bt ) = 𝔼 (t + s)W
)
1
t+s
(
)
− 𝔼 tW 1 = 0
t
and variance
(
Var(Bt+s − Bt ) = Var (t + s)W
)
1
t+s
(
)
(
+ Var tW 1 − 2Cov (t + s)W
t
1
t+s
, tW 1
)
t
= t + s + t − 2min{t + s, s}
= s.
Since the sum of two normal distributions is also a normal distribution, so Bt+s − Bt ∼
𝒩(0, s).
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63
(c) To show that Bt+s − Bt ⟂
⟂ Bt we note that since Wt has the independent increment
property, so
( )
𝔼[(Bt+s − Bt )Bt ] = 𝔼(Bt+s Bt ) − 𝔼 B2t
( )
= Cov(Bt+s , Bt ) − 𝔼 B2t
= min{t + s, t} − t
= 0.
Since Bt ∼ 𝒩(0, t) and Bt+s − Bt ∼ 𝒩(0, s) and the joint distribution of Bt and Bt+s −
Bt is a bivariate normal (see Problem 2.2.1.5, page 58), then if Cov(Bt+s − Bt , Bt ) = 0
⟂ Bt .
so Bt+s − Bt ⟂
{
0
if t = 0
From the results of (a)–(c) we have shown that Bt = tW 1 if t ≠ 0 is also a standard
t
Wiener process.
◽
10. Time Reversal. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. Show that under time reversal, Bt = W1 − W1−t is also a standard Wiener
process.
Solution:
(a) B0 = W1 − W1 = 0 and it is clear that Bt has continuous sample paths for t ≥ 0.
(b) Since Wt is a standard Wiener process, we have W1 ∼ 𝒩(0, 1) and W1−t ∼ 𝒩(0, 1 − t).
Therefore,
Bt+s − Bt = W1 − W1−(t+s) − W1 + W1−t = W1−t − W1−(t+s)
with mean
𝔼(Bt+s − Bt ) = 𝔼(W1−t ) − 𝔼(W1−(t+s) ) = 0
and variance
Var(Bt+s − Bt ) = Var(W1−t ) + Var(W1−(t+s) ) − 2Cov(W1−t , W1−(t+s) )
= 1 − t + 1 − (t + s) − 2min{1 − t, 1 − (t + s)}
= s.
Since the sum of two normal distributions is also a normal distribution, so Bt+s − Bt ∼
𝒩(0, s).
(c) To show that Bt+s − Bt ⟂
⟂ Bt we note that since Wt has the independent increment
property, so
( )
𝔼[(Bt+s − Bt )Bt ] = 𝔼(Bt+s Bt ) − 𝔼 B2t
( )
= Cov(Bt+s , Bt ) − 𝔼 B2t
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= min{t + s, t} − t
= 0.
Given Bt ∼ 𝒩(0, t) and Bt+s − Bt ∼ 𝒩(0, s) and the joint distribution of Bt and Bt+s −
Bt is a bivariate normal (see Problem 2.2.1.5, page 58), then if Cov(Bt+s − Bt , Bt ) = 0
⟂ Bt .
so Bt+s − Bt ⟂
From the results of (a)–(c) we have shown that Bt = W1 − W1−t is also a standard Wiener
process.
◽
11. Multi-Dimensional Standard Wiener Process. Let (Ω, ℱ, ℙ) be a probability space and let
{Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n be a sequence of independent standard Wiener processes.
)T
(
Show that the vector Wt = Wt(1) , Wt(2) , . . . , Wt(n) is an n-dimensional standard Wiener
process with the following properties:
(a) W0 = 𝟎 and Wt is a vector of continuous sample paths;
(b) for each t > 0 and s > 0, Wt+s −Wt ∼ 𝒩n (𝟎, sI) where I is an n × n identity matrix;
(c) for each t > 0 and s > 0, Wt+s − Wt ⟂
⟂ Wt .
Solution:
)T
(
(a) Since W0(i) = 0 for all i = 1, 2, . . . , n therefore W0 = W0(1) , W0(2) , . . . , W0(n) = 𝟎
where 𝟎 = (0, 0, . . . , 0)T is an n-vector of zeroes. In addition, Wt is a vector of continuous sample paths due to the fact that Wt(i) , t ≥ 0 has continuous sample paths.
(b) For s, t > 0 and for i, j = 1, 2, . . . , n we have
)
( (i)
− Wt(i) = 0, i = 1, 2, . . . , n
𝔼 Wt+s
and
[(
)(
)]
(j)
(j)
(i)
𝔼 Wt+s
− Wt(i) Wt+s − Wt
=
{
s
0
i=j
i ≠ j.
Therefore, Wt+s − Wt ∼ 𝒩n (𝟎, sI) where I is an n × n identity matrix.
(j)
(i)
(c) For s, t > 0 and since Wt+s
− Wt(i) ⟂
⟂ Wt(i) , Wt(i) ⟂
⟂ Wt , i ≠ j, i, j = 1, 2, . . . , n we can
(j)
(i)
(i)
⟂ Wt . Therefore, Wt+s − Wt ⟂
⟂ Wt .
deduce that Wt+s − Wt ⟂
(
)T
is an
From the results of (a)–(c) we have shown that Wt = Wt(1) , Wt(2) , . . . , Wt(n)
n-dimensional standard Wiener process.
◽
12. Let (Ω, ℱ, ℙ) be a probability space and let
( {Wt ∶ t ≥ 0}
) be a standard Wiener process.
t
Show that the pair of random variables Wt , ∫0 Ws ds has the following covariance
matrix
1 2
t
⎡ t
2 ⎤
⎥
𝚺=⎢
⎢ 1 2 1 3⎥
t
t
⎦
⎣2
3
√
3
with correlation coefficient
.
2
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65
(
)
t
Solution: By definition, the covariance matrix for the pair Wt , ∫0 Ws ds is
(
)
t
⎡
Var(Wt )
Cov Wt , ∫0 Ws ds ⎤
)
(
) ⎥.
(
𝚺=⎢
t
⎥
⎢Cov W , ∫ t W ds
∫
Var
W
ds
t
s
s
0
0
⎦
⎣
Given that Wt ∼ 𝒩(0, t) we know
Var(Wt ) = t.
)
(
t
For the case Cov Wt , ∫0 Ws ds we can write
(
Cov Wt ,
t
∫0
)
(
Ws ds = 𝔼 Wt
)
( t
)
Ws ds − 𝔼(Wt )𝔼
Ws ds
∫0
∫0
(
)
t
= 𝔼 Wt
W ds
∫0 s
( t
)
=𝔼
W W ds
∫0 t s
t
t
=
𝔼(Wt Ws ) ds
∫0
]
[
𝔼 Ws (Wt − Ws ) + Ws2 ds.
t
=
∫0
From the independent increment property of a Wiener process we have
(
)
t
Cov Wt ,
∫0
t
Ws ds =
∫0
t
=
t
𝔼(Ws )𝔼(Wt − Ws ) ds +
∫0
∫0
( )
𝔼 Ws2 ds
( )
𝔼 Ws2 ds
t
=
s ds
∫0
t2
.
2
=
Finally,
(
Var
t
∫0
[(
)
Ws ds = 𝔼
Ws ds
∫0
[(
=𝔼
)2 ]
t
)(
t
∫0
[ (
− 𝔼
Ws ds
t
∫0
)]2
t
∫0
Ws ds
)] [
Wu du −
]2
t
∫0
𝔼(Ws )
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=𝔼
s=t
=
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Basic Properties
]
u=t
Ws Wu duds
∫s=0 ∫u=0
s=t
V3 - 08/23/2014
u=t
𝔼(Ws Wu ) duds.
∫s=0 ∫u=0
From Problem 2.2.1.4 (page 57) we have
(
)
𝔼 Ws Wu = min{s, u}
and hence
(
Var
t
∫0
)
Ws ds =
s=t
u=t
s=t
=
u=s
∫s=0 ∫u=0
t
=
min{s, u} duds
∫s=0 ∫u=0
∫0
s=t
u duds +
u=t
∫s=0 ∫u=s
s duds
t
1 2
s ds +
s(t − s) ds
∫0
2
1
= t3 .
3
Therefore,
1 2
t
2 ⎤
⎡ t
𝚺=⎢
⎢1 2
⎣2t
⎥
1 3⎥
t ⎦
3
with correlation coefficient
(
Cov Wt ,
)
t
Ws ds
∫0
𝜌= √
(
Var(Wt )Var
t
∫0
Ws ds
√
t2 ∕2
3
.
=
√ =
)
2
t2 ∕ 3
◽
13. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
s
t
Show that the covariance of ∫0 Wu du and ∫0 W𝑣 d𝑣, s, t > 0 is
(
Cov
s
∫0
)
t
Wu du,
∫0
√
with correlation coefficient
W𝑣 d𝑣
=
1
1
min{s3 , t3 } + |t − s|min{s2 , t2 }
3
2
min{s3 , t3 } 3
+ |t − s|
max{s3 , t3 } 2
√
min{s, t}
.
max{s3 , t3 }
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67
Solution: By definition
(
s
Cov
∫0
[(
=𝔼
s
∫0
s
=
W𝑣 d𝑣
∫0
)(
Wu du
)]
t
∫0
[
𝔼(Wu W𝑣 ) dud𝑣 −
∫0 ∫0
s
∫0
[
s
−𝔼
W𝑣 d𝑣
t
s
=
)
t
Wu du,
] [
Wu du 𝔼
]
t
W𝑣 d𝑣
∫0
∫0
][ t
]
𝔼(Wu ) du
𝔼(W𝑣 ) d𝑣
∫0
t
∫0 ∫0
min{u, 𝑣} dud𝑣
since 𝔼(Wu W𝑣 ) = min{u, 𝑣} and 𝔼(Wu ) = 𝔼(W𝑣 ) = 0.
For the case s ≤ t, using the results of Problem 2.2.1.12 (page 64),
(
Cov
s
∫0
t
Wu du,
∫0
)
W𝑣 d𝑣 =
s
t
∫0 ∫0
s
min{u, 𝑣} dud𝑣
s
s
min{u, 𝑣} dud𝑣 +
=
∫0 ∫0
s
t
1
= s3 +
𝑣 dud𝑣
∫0 ∫ s
3
s
1
𝑣(t − s) d𝑣
= s3 +
∫0
3
1
1
= s3 + (t − s)s2 .
3
2
t
∫0 ∫s
min{u, 𝑣} dud𝑣
Following the same steps as described above, for the case s > t we can also show
(
Cov
s
t
Wu du,
∫0
∫0
)
1
1
W𝑣 d𝑣 = t3 + (s − t)t2 .
3
2
Thus,
(
Cov
s
∫0
)
1
1
W𝑣 d𝑣 = min{s3 , t3 } + |t − s|min{s2 , t2 }.
∫0
3
2
t
Wu du,
s
t
From the definition of the correlation coefficient between ∫0 Wu du and ∫0 W𝑣 d𝑣,
(
Cov
𝜌= √
Var
(
s
∫0
s
∫0
)
t
Wu du,
∫0
W𝑣 d𝑣
)
(
Wu du Var
)
t
∫0
W𝑣 d𝑣
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=
1
min{s3 , t3 }
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+ 12 |t − s|min{s2 , t2 }
√
1
s3 t 3
3
min{s3 , t3 } 3 |t − s|min{s2 , t2 }
+
.
√
√
2
s3 t 3
s3 t 3
For s ≤ t we have
√
𝜌=
s3 3
+ (t − s)
t3 2
√
s
t3
and for s > t
√
𝜌=
Therefore, we can deduce
√
𝜌=
2.2.2
√
t3 3
+ (s − t)
s3 2
t
.
s3
√
min{s3 , t3 } 3
+ |t − s|
max{s3 , t3 } 2
min{s, t}
.
max{s3 , t3 }
◽
Markov Property
1. The Markov Property of a Standard Wiener Process. Let (Ω, ℱ, ℙ) be a probability space
and let {Wt ∶ t ≥ 0} be a standard Wiener process with respect to the filtration ℱt , t ≥ 0.
Show that if f is a continuous function then there exists another continuous function g
such that
𝔼[f (Wt )|ℱu ] = g(Wu )
for 0 ≤ u ≤ t.
Solution: For 0 ≤ u ≤ t we can write
]
[
]
[
𝔼 f (Wt )|ℱu = 𝔼 f (Wt − Wu + Wu )|ℱu .
Since Wt − Wu ⟂
⟂ ℱu and Wu is ℱu measurable, by setting Wu = x where x is a constant
value
𝔼[f (Wt − Wu + Wu )|ℱu ] = 𝔼[f (Wt − Wu + x)].
Because Wt − Wu ∼ 𝒩(0, t − u) we can write 𝔼[f (Wt − Wu + x)] as
𝔼[f (Wt − Wu + x)] = √
1
∞
2𝜋(t − u) ∫−∞
2
𝑤
− 2(t−u)
f (𝑤 + x)e
d𝑤.
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69
By setting 𝜏 = t − u and y = 𝑤 + x, we can rewrite 𝔼[f (Wt − Wu + x)] = 𝔼[f (Wt − Wu +
Wu )] as
∞
(y−x)2
1
𝔼[f (Wt − Wu + Wu )] = √
f (y)e− 2𝜏 dy
2𝜋𝜏 ∫−∞
∞
=
∫−∞
f (y)p(𝜏, Wu , y) dy
where the transition density
2
(y−Wu )
1
e− 2𝜏
p(𝜏, Wu , y) = √
2𝜋𝜏
is the density of Y ∼ 𝒩(Wu , 𝜏). Since the only information from the filtration ℱu is Wu ,
therefore
𝔼[f (Wt )|ℱu ] = g(Wu )
where
∞
g(Wu ) =
∫−∞
f (y)p(𝜏, Wu , y) dy.
◽
2. The Markov Property of a Wiener Process. Let (Ω, ℱ, ℙ) be a probability space and let
{Wt ∶ t ≥ 0} be a standard Wiener process with respect to the filtration ℱt , t ≥ 0. By
considering the Wiener process
̂ t = a + bt + cWt ,
W
a, b ∈ ℝ,
c>0
show that if f is a continuous function then there exists another continuous function g such
that
̂ u)
̂ t )|ℱu ] = g(W
𝔼[f (W
for 0 ≤ u ≤ t.
Solution: For 0 ≤ u ≤ t we can write
[
]
[
]
̂ t )||ℱu = 𝔼 f (W
̂t − W
̂u + W
̂ u )||ℱu .
𝔼 f (W
|
|
̂t − W
̂u ⟂
Since Wt − Wu ⟂
⟂ ℱu we can deduce that W
⟂ ℱu . In addition, because Wu is ℱu
̂
̂ u = x where x is a constant
measurable, hence Wu is also ℱu measurable. By setting W
value,
]
[
[
]
̂t − W
̂t − W
̂u + W
̂ u )||ℱu = 𝔼 f (W
̂ u + x) .
𝔼 f (W
|
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[
]
̂t − W
̂ u ∼ 𝒩(b(t − u), c2 (t − u)), we can write 𝔼 f (W
̂t − W
̂ u + x) as
Taking note that W
[
]
[
]
2
− 12 (𝑤−b(t−u))
2
∞
1
c (t−u)
̂t − W
̂ u + x) = √
𝔼 f (W
f (𝑤 + x)e
d𝑤.
∫
c 2𝜋(t − u) −∞
[
]
[
̂t − W
̂ u + x) = 𝔼 f (W
̂t − W
̂u +
By setting 𝜏 = t − u and y = 𝑤 + x, we can rewrite 𝔼 f (W
]
̂ u ) as
W
[
]
[
∞
1
−
̂t − W
̂u + W
̂ u ) = √1
𝔼 f (W
f (y)e 2
∫
c 2𝜋𝜏 −∞
∞
=
∫−∞
̂ u )2
(y−b𝜏−W
c2 𝜏
]
dy
̂ u , y) dy
f (y)p(𝜏, W
where the transition density
[
1
p(𝜏, Wu , y) = √
2𝜋𝜏
− 12
e
̂ u )2
(y−b𝜏−W
c2 𝜏
]
̂ u , c2 𝜏). Since the only information from the filtration ℱu
is the density of Y ∼ 𝒩(b𝜏 + W
is Wu , therefore
]
[
̂ u)
̂ t )||ℱu = g(W
𝔼 f (W
|
where
̂ u) =
g(W
∞
∫−∞
̂ u , y) dy.
f (y)p(𝜏, W
◽
3. The Markov Property of a Geometric Brownian Motion. Let (Ω, ℱ, ℙ) be a probability
space and let {Wt ∶ t ≥ 0} be a standard Wiener process with respect to the filtration ℱt ,
t ≥ 0. By considering the geometric Wiener process
(
St = S 0 e
)
𝜇− 21 𝜎 2 t+𝜎Wt
,
𝜇 ∈ ℝ,
𝜎>0
where S0 > 0 show that if f is a continuous function then there exists another continuous
function g such that
𝔼[f (St )|ℱu ] = g(Su )
for 0 ≤ u ≤ t.
Solution: For 0 ≤ u ≤ t we can write
[ (
)| ]
St
|
𝔼[f (St )|ℱu ] = 𝔼 f
⋅ S |ℱ
Su u || u
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71
(
where
log
St
Su
)
∼𝒩
((
)
)
1
𝜇 − 𝜎 2 (t − u), 𝜎 2 (t − u) .
2
Since St ∕Su ⟂
⟂ ℱu and Su is ℱu measurable, by setting Su = x where x is a constant value
[ (
)| ]
)]
[ (
St
St
|
⋅ S |ℱ =𝔼 f
⋅x .
𝔼 f
Su u || u
Su
By setting 𝜈 = 𝜇 − 12 𝜎 2 so that St ∕Su ∼ log-𝒩(𝜈(t − u), 𝜎 2 (t − u)), we can write
[ (
)]
St
𝔼 f
⋅x
as
Su
[
]
)]
[ (
(log 𝑤−𝜈(t−u))2
∞
− 12
St
1
2
𝜎 (t−u)
⋅x =
f (𝑤 ⋅ x) e
d𝑤.
𝔼 f
√
Su
𝜎𝑤 2𝜋(t − u) ∫−∞
[ (
)]
)]
[ (
St
St
By setting 𝜏 = t − u and y = 𝑤 ⋅ x, we can rewrite 𝔼 f
as
⋅x
=𝔼 f
⋅ Su
Su
Su
[
]
(log ( yx )−𝜈𝜏 )2
[ (
)]
∞
− 12
St
𝜎2 𝜏
1
⋅S
f (y)e
dy
𝔼 f
= √
∫
Su u
𝜎y 2𝜋𝜏 −∞
∞
=
∫−∞
f (y)p(𝜏, Wu , y) dy
where the transition density
[
1
−
1
p(𝜏, Wu , y) = √
e 2
𝜎y 2𝜋𝜏
(log y−log Su −𝜈𝜏 )2
]
𝜎2 𝜏
is the density of Y ∼ log-𝒩(log Su + 𝜈𝜏, 𝜎 2 𝜏). Since the only information from the filtration ℱu is Su , therefore
𝔼[f (St )|ℱu ] = g(Su )
such that
∞
g(Su ) =
∫−∞
f (y)p(𝜏, Wu , y) dy.
◽
2.2.3
Martingale Property
1. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that Wt is a martingale.
Solution: Given Wt ∼ 𝒩(0, t).
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(a) For s ≤ t and since Wt − Ws ⟂
⟂ ℱs , we have
𝔼(Wt |ℱs ) = 𝔼(Wt − Ws + Ws |ℱs ) = 𝔼(Wt − Ws |ℱs ) + 𝔼(Ws |ℱs ) = Ws .
(b) Since Wt ∼ 𝒩(0, t), |Wt | follows a folded normal distribution such√that |Wt | ∼
𝒩f (0, t). From Problem 1.2.2.11 (page 22), we can deduce 𝔼(|Wt |) = 2t∕𝜋 < ∞.
In contrast, we can also√utilise Hölder’s inequality (see Problem 1.2.3.2, page 41) to
√
deduce that 𝔼(|Wt |) ≤ 𝔼(Wt2 ) = t < ∞.
(c) Wt is clearly ℱt -adapted.
From the results of (a)–(c) we have shown that Wt is a martingale.
◽
2. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that Xt = Wt2 − t is a martingale.
Solution: Given Wt ∼ 𝒩(0, t)
⟂ ℱs , we have
(a) For s ≤ t and since Wt − Ws ⟂
[(
( 2
)
)2 | ]
𝔼 Wt − t|ℱs = 𝔼 Wt − Ws + Ws | ℱs − t
|
[ (
(
)
[(
)2 | ]
)| ]
|
= 𝔼 Wt − Ws | ℱs + 2𝔼 Ws Wt − Ws | ℱs + 𝔼 Ws2 | ℱs − t
|
|
|
= t − s + 0 + Ws2 − t
= Ws2 − s.
(b) Since |Xt | = |Wt2 − t| ≤ Wt2 + t we can therefore write
)
(
(
)
( )
|
|
𝔼 |Wt2 − t| ≤ 𝔼 Wt2 + t = 𝔼 Wt2 + t = 2t < ∞.
|
|
(c) Since Xt = Wt2 − t is a function of Wt , hence it is ℱt -adapted.
From the results of (a)–(c) we have shown that Xt = Wt2 − t is a martingale.
◽
3. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
1 2
For 𝜆 ∈ ℝ show that Xt = e𝜆Wt − 2 𝜆 t is a martingale.
(
)
Solution: Given Wt ∼ 𝒩(0, t) we can write log Xt = 𝜆Wt − 12 𝜆2 t ∼ 𝒩 − 12 𝜆2 t, 𝜆2 t and
(
)
hence eXt ∼ log-𝒩 − 12 𝜆2 t, 𝜆2 t .
(a) For s ≤ t and since Wt − Ws ⟂
⟂ ℱs , we have
)
(
(
)
𝜆Wt − 12 𝜆2 t ||
− 1 𝜆2 t
𝜆Wt |
ℱ
𝔼 e
| s = e 2 𝔼 e || ℱs
|
]
[
1 2
|
= e− 2 𝜆 t 𝔼 e𝜆(Wt −Ws )+𝜆Ws | ℱs
|
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73
] [
]
[
1 2
|
|
= e− 2 𝜆 t 𝔼 e𝜆(Wt −Ws ) | ℱs 𝔼 e𝜆Ws | ℱs
|
|
1 2
1 2
(t−s)
= e− 2 𝜆 t ⋅ e 2 𝜆
=e
𝜆Ws − 12 𝜆2 s
⋅ e𝜆Ws
.
1 2 |
1 2
|
(b) By setting |Xt | = ||e𝜆Wt − 2 𝜆 t || = e𝜆Wt − 2 𝜆 t ,
|
|
)
(
1 2
1 2
1 2
1 2
𝔼(|Xt |) = 𝔼 e𝜆Wt − 2 𝜆 t = e− 2 𝜆 t 𝔼(e𝜆Wt ) = e− 2 𝜆 t ⋅ e 2 𝜆 t = 1 < ∞.
(c) Since Xt is a function of Wt , hence it is ℱt -adapted.
1 2
From the results of (a)–(c) we have shown that Xt = e𝜆Wt − 2 𝜆 t is a martingale.
◽
4. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that Xt = Wt3 − 3tWt is a martingale.
Solution:
⟂ ℱs , we have
(a) For s ≤ t and since Wt − Ws ⟂
)
(
)
(
|
𝔼 Xt || ℱs = 𝔼 Wt3 − 3tWt | ℱs
|
[ (
)| ]
(
)2
= 𝔼 Wt Wt − Ws + Ws − 3t || ℱs
|
[ (
[ (
)2 | ]
)| ]
= 𝔼 Wt Wt − Ws | ℱs + 2Ws 𝔼 Wt Wt − Ws | ℱs
|
|
)
(
)
(
+Ws2 𝔼 Wt || ℱs − 3t𝔼 Wt || ℱs
[(
)(
)2 | ]
= 𝔼 Wt − Ws + Ws Wt − Ws | ℱs
|
[(
)(
)| ]
+2Ws 𝔼 Wt − Ws + Ws Wt − Ws | ℱs + Ws3 − 3tWs
|
[(
[ (
)3 | ]
)2 | ]
= 𝔼 Wt − Ws | ℱs + 𝔼 Ws Wt − Ws | ℱs
|
|
[(
[ (
)2 | ]
)| ]
+2Ws 𝔼 Wt − Ws | ℱs + 2Ws 𝔼 Ws Wt − Ws | ℱs + Ws3 − 3tWs
|
|
= Ws (t − s) + 2Ws (t − s) + Ws3 − 3tWs
= Ws3 − 3sWs
where, from Problem 2.2.1.5 (page 58), we can deduce that the moment generating
1 2 2
|
d3
MWt (𝜃)||
= 0.
function of Wt is MWt (𝜃) = e 2 𝜃 t and 𝔼(Wt3 ) =
3
d𝜃
|𝜃=0
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(b) Since |Xt | = |Wt3 − 3tWt | ≤ |Wt3 | + 3t|Wt | and from Hölder’s inequality (see Problem 1.2.3.2, page 41),
√ (
) (
)
(
)
2
2
𝔼(|Xt |) ≤ 𝔼 ||Wt2 || 𝔼 ||Wt || + 3t𝔼 |Wt |
=
√ ( ) ( )
(
)
𝔼 Wt4 𝔼 Wt2 + 3t𝔼 |Wt | .
Since Wt2 ∕t ∼ 𝜒(1)
have 𝔼(Wt4 ) = 3t and utilising Hölder’s inequality again, we
√ we
√
√
( 2) √
have 𝔼(|Wt |) ≤ 𝔼 Wt = t < ∞. Therefore, 𝔼(|Xt |) ≤ t 2 + 3t t < ∞.
(c) Since Xt is a function of Wt , hence it is ℱt -adapted.
From the results of (a)–(c) we have shown that Xt = Wt3 − 3tWt is a martingale.
◽
5. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
For 𝜆 ∈ ℝ show that the following hyperbolic processes:
1 2
Xt = e− 2 𝜆 t cosh(𝜆Wt )
1 2
Yt = e− 2 𝜆 t sinh(𝜆Wt )
are martingales.
Solution: By definition we can write
1 2
Xt = e− 2 𝜆 t cosh(𝜆Wt ) =
1 2
(
)
1 2
1 𝜆Wt − 12 𝜆2 t
e
+ e−𝜆Wt − 2 𝜆 t .
2
1 2
Since for 𝜆 ∈ ℝ, Xt(1) = 12 e𝜆Wt − 2 𝜆 t and Xt(2) = 12 e−𝜆Wt − 2 𝜆 t are martingales we have the
following properties:
(
)
|
(a) For s ≤ t, 𝔼 Xt(1) + Xt(2) | ℱs = Xs(1) + Xs(2) .
|
)
(
)
(
)
(
|
|
| (1) |
|
|
(b) Because 𝔼 |Xt | < ∞ and 𝔼 |Xt(2) | < ∞, so 𝔼 |Xt(1) + Xt(2) | < ∞.
|
|
|
|
|
|
(c) Since Xt is a function of Wt , hence it is ℱt -adapted.
1 2
From the results of (a)–(c) we have shown that Xt = e− 2 𝜆 t cosh(𝜆Wt ) is a martingale.
1 2
For Yt = e− 2 𝜆 t sinh(𝜆Wt ) we note that
1 2
Yt = e− 2 𝜆 t sinh(𝜆Wt ) =
(
)
1 2
1 𝜆Wt − 12 𝜆2 t
e
− e−𝜆Wt − 2 𝜆 t
2
and similar steps can be applied to show that Yt is also a martingale.
◽
6. Let (Ω, ℱ, ℙ) be a probability space. Show that the sample paths of a standard Wiener
process {Wt ∶ t ≥ 0} are continuous but not differentiable.
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75
Solution: The path Wt ∼ 𝒩(0, t) is continuous in probability if and only if, for every
𝛿 > 0 and t ≥ 0,
lim ℙ(|Wt+Δt − Wt | ≥ 𝛿) = 0.
Δt→0
Given that Wt is a martingale and from Chebyshev’s inequality (see Problem 1.2.2.19,
page 40),
ℙ(|Wt+Δt − Wt | ≥ 𝛿) = ℙ(|Wt+Δt − 𝔼(Wt+Δt |ℱt )| ≥ 𝛿)
≤
=
=
Var(Wt+Δt |ℱt )
𝛿2
Var(Wt+Δt − Wt + Wt |ℱt )
𝛿2
Var(Wt+Δt − Wt )
𝛿2
+
Var(Wt |ℱt )
𝛿2
Δt
= 2
𝛿
since Wt+Δt − Wt ⟂
⟂ ℱt and Wt is ℱt measurable and hence Var(Wt |ℱt ) = 0.
Taking the limit Δt → 0, we have
ℙ(|Wt+Δt − Wt | ≥ 𝛿) → 0
and therefore we conclude that the sample path is continuous. By setting
√
ΔWt = Wt+Δt − Wt = 𝜙 Δt
where 𝜙 ∼ 𝒩(0, 1) and taking limit Δt → 0, we have
lim
Δt→0
ΔWt
𝜙
= lim √ = ±∞
Δt→0
Δt
Δt
depending on the sign of 𝜙. Therefore, Wt is not differentiable.
◽
7. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process
with respect to the filtration ℱt , t ≥ 0.
Show that if 𝜑 is a convex function and 𝔼[|𝜑(Wt )|] < ∞ for all t ≥ 0, then 𝜑(Wt ) is a
submartingale.
Finally, deduce that |Wt |, Wt2 , e𝛼Wt , 𝛼 ∈ ℝ and Wt+ = max{0, Wt } are non-negative submartingales.
Solution: Let s < t and because Wt is a martingale we therefore have 𝔼(Wt |ℱs ) = Ws
and hence 𝜑[𝔼(Wt |ℱs )] = 𝜑(Ws ). From the conditional Jensen’s inequality (see Problem 1.2.3.14, page 48),
𝔼[𝜑(Wt )|ℱs ] ≥ 𝜑[𝔼(Wt |ℱs )] = 𝜑(Ws ).
In addition, since 𝔼[|𝜑(Wt )|] < ∞ and 𝜑(Wt ) is clearly ℱt -adapted we can conclude that
𝜑(Wt ) is a submartingale.
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By setting 𝜑(x) = |x|, x ∈ ℝ and for 𝜃 ∈ (0, 1) and x1 , x2 ∈ ℝ such that x1 ≠ x2 ,
we have
|𝜃x1 + (1 − 𝜃)x2 | ≤ 𝜃|x1 | + (1 − 𝜃)|x2 |.
Therefore, |x| is a non-negative convex function. Because 𝔼(|Wt |) < ∞ for all t ≥ 0, so
|Wt | is a non-negative submartingale.
On the contrary, by setting 𝜑(x) = x2 , x ∈ ℝ and since 𝜑′′ (x) = 2 ≥ 0 for all x , so x2 is a
non-negative convex function. Since 𝔼(|Wt2 |) < ∞ for all t ≥ 0, so Wt2 is a non-negative
submartingale.
′′
2 𝛼x
Using the same steps we define 𝜑(x) = e𝛼x where 𝛼, x ∈ ℝ(and since
) 𝜑 (x) = 𝛼 e ≥ 0
𝛼x
𝛼W
|
|
t
<
∞
for
all
t
≥ 0 we
for all x, so e is a non-negative convex function. Since 𝔼 |e |
𝛼W
t
is a non-negative submartingale.
can conclude that e
Finally, by setting 𝜑(x) = max{0, x}, x ∈ ℝ and for 𝜃 ∈ (0, 1) and x1 , x2 ∈ ℝ such that
x1 ≠ x2 , we have
max{0, 𝜃x1 + (1 − 𝜃)x2 } ≤ max{0, 𝜃x1 } + max{0, (1 − 𝜃)x2 }
= 𝜃1 max{0, x1 } + (1 − 𝜃1 )max{0, x2 }.
Therefore, x+ = max{0, x} is a non-negative convex function and since 𝔼(|Wt+ |) < ∞ for
all t ≥ 0, so Wt+ = max{0, Wt } is a non-negative submartingale.
◽
2.2.4
First Passage Time
{
}
1. Doob’s Maximal Inequality. Let (Ω, ℱ, ℙ) be a probability space and let Xt ∶ 0 ≤ t ≤ T
be a continuous non-negative submartingale with respect to the filtration ℱt , 0 ≤ t ≤ T.
Given 𝜆 > 0 and 𝜏 = min{t ∶ Xt ≥ 𝜆}, show that
(
)
( )
( )
𝔼 X0 ≤ 𝔼 Xmin{𝜏,T } ≤ 𝔼 XT .
By writing
Xmin{𝜏,T } = X𝜏 1I{𝜏≤T } + XT 1I{𝜏>T }
show that
(
ℙ
)
sup Xt ≥ 𝜆
0≤t≤T
≤
( )
𝔼 XT
𝜆
.
Deduce that if {Yt ∶ 0 ≤ t ≤ T} is a continuous non-negative supermartingale with respect
to the filtration ℱt , 0 ≤ t ≤ T then
( )
)
(
𝔼 Y0
.
ℙ sup Yt ≥ 𝜆 ≤
𝜆
0≤t≤T
Solution: For 𝜆 > 0 we let 𝜏 = min{t ∶ Xt ≥ 𝜆} so that 0 ≤ min{𝜏, T} ≤ T. Because Xt
is a non-negative submartingale we have
)
(
𝔼 XT || ℱmin{𝜏,T } ≥ Xmin{𝜏,T }
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or
77
)
[ (
)]
(
( )
𝔼 Xmin{𝜏,T } ≤ 𝔼 𝔼 XT || ℱmin{𝜏,T } = 𝔼 XT .
Using the same steps we can deduce
(
)
( )
( )
𝔼 X0 ≤ 𝔼 Xmin{𝜏,T } ≤ 𝔼 XT .
By definition
Xmin{𝜏,T } = X𝜏 1I{𝜏≤T } + XT 1I{𝜏>T }
{
where
{
1 𝜏≤T
,
0 𝜏>T
1I{𝜏≤T } =
1 𝜏>T
0 𝜏 ≤ T.
1I{𝜏>T } =
Therefore,
(
(
)
(
)
(
)
)
(
)
𝔼 Xmin{𝜏,T } = 𝔼 X𝜏 1I{𝜏≤T } + 𝔼 XT 1I{𝜏>T } ≥ 𝜆ℙ 𝜏 ≤ T + 𝔼 XT 1I{𝜏>T } .
)
(
( )
Taking note that 𝔼 Xmin{𝜏,T } ≤ 𝔼 XT , we can write
(
(
)
(
)
)
(
)
( )
( )
𝜆ℙ 𝜏 ≤ T ≤ 𝔼 Xmin{𝜏,T } − 𝔼 XT 1I{𝜏>T } ≤ 𝔼 XT − 𝔼 XT 1I{𝜏>T } ≤ 𝔼 XT .
{
Since
}
{𝜏 ≤ T} ⇐⇒
sup Xt ≥ 𝜆
0≤t≤T
)
(
therefore
sup Xt ≥ 𝜆
ℙ
≤
0≤t≤T
𝔼(XT )
.
𝜆
If {Yt }0≤t≤T is a supermartingale then
(
)
( )
( )
𝔼 YT ≤ 𝔼 Ymin{𝜏,T } ≤ 𝔼 Y0
where in this case 𝜏 = min{t ∶ Yt ≥ 𝜆} and by analogy with the above steps, we have
(
ℙ
)
sup Yt ≥ 𝜆
0≤t≤T
≤
( )
𝔼 Y0
𝜆
.
◽
2. Doob’s Lp Maximal Inequality. Let (Ω, ℱ, ℙ) be a probability space and let Z be a continuous non-negative random variable where for m > 0, 𝔼(Z m ) < ∞. Show that
𝔼(Z m ) = m
∞
∫0
𝛼 m−1 ℙ(Z > 𝛼) d𝛼.
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First Passage Time
Let {Xt ∶ 0 ≤ t ≤ T} be a continuous non-negative submartingale
to the fil( with respect
p)
tration ℱt , 0 ≤ t ≤ T. Using the above result, for p > 1 and if 𝔼 sup0≤t≤T Xt < ∞ show
that
(
)p
) (
( p)
p
p
𝔼 XT .
𝔼 sup Xt ≤
p−1
0≤t≤T
Deduce that if {Yt }0≤t≤T is a continuous non-negative supermartingale with respect to the
filtration ℱt , 0 ≤ t ≤ T then
)
(
𝔼
p
sup Yt
(
≤
0≤t≤T
p
p−1
)p
( p)
𝔼 Y0 .
Solution: By defining the indicator function
{
1 Z>𝛼
0 Z≤𝛼
1I{Z>𝛼} =
we can prove
∞
∫0
∞
m𝛼 m−1 ℙ(Z > 𝛼) d𝛼 =
∫0
[
=𝔼
[
=𝔼
m𝛼 m−1 𝔼(1I{Z>𝛼} ) d𝛼
]
∞
m𝛼 m−1 1I{Z>𝛼} d𝛼
∫0
]
Z
∫0
m𝛼 m−1 d𝛼
= 𝔼(Z m ).
Let 𝜏 = min{t ∶ Xt ≥ 𝜆}, 𝜆 > 0 so that 0 ≤ min{𝜏, T} ≤ T. Therefore, we can deduce
{
}
{𝜏 ≤ T} ⇐⇒
sup Xt ≥ 𝜆
0≤t≤T
and from Problem 2.2.4.1 (page 76) we can show that
)
(
(
)
∞
p
𝔼 sup Xt =
p𝜆p−1 ℙ sup Xt > 𝜆 d𝜆
∫0
0≤t≤T
0≤t≤T
∞
)
(
≤
d𝜆
p𝜆p−2 𝔼 XT 1I{sup
0≤t≤T Xt ≥𝜆}
∫0
(
)
∞
p−2
𝜆 1I{sup
= p 𝔼 XT
d𝜆
0≤t≤T Xt ≥𝜆}
∫0
)
(
sup0≤t≤T Xt
p−2
𝜆
d𝜆
= p 𝔼 XT
∫0
})
{
(
p
p−1
=
.
𝔼 XT ⋅ sup Xt
p−1
0≤t≤T
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First Passage Time
79
(
p)
Using Hölder’s inequality and taking note that 𝔼 sup0≤t≤T Xt < ∞, we can write
(
𝔼
)
sup
0≤t≤T
or
p
Xt
)1
(
p p
≤
𝔼 sup Xt
0≤t≤T
)
(
and hence
𝔼
) p−1
(
( p)1
p
≤
𝔼 XT p 𝔼
p−1
sup
0≤t≤T
p
Xt
(
≤
sup
0≤t≤T
p−1
Xt
p
( p)1
p
𝔼 XT p
p−1
p
p−1
)p
( p)
𝔼 XT .
From Problem 2.2.4.1 (page 76), if {Yt }0≤t≤T is a supermartingale then
(
)
( )
( )
𝔼 YT ≤ 𝔼 Ymin{𝜏,T } ≤ 𝔼 Y0
where in this case 𝜏 = min{t ∶ Yt ≥ 𝜆}. Following the same steps as discussed before, we
have
) (
(
)p
( p)
p
p
𝔼 Y0 .
𝔼 sup Yt ≤
p−1
0≤t≤T
◽
3. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Using Doob’s maximal inequality show that for every T > 0 and 𝜆, 𝜃 > 0, we can express
)
(
1 2
𝜃Wt
𝜃𝜆
≤ e 2 𝜃 T−𝜃𝜆
≥e
ℙ sup e
0≤t≤T
and hence prove that
(
ℙ
)
𝜆2
≤ e− 2T .
sup Wt ≥ 𝜆
0≤t≤T
Finally, deduce that
(
ℙ
)
sup ||Wt || ≥ 𝜆
𝜆2
≤ 2e− 2T .
0≤t≤T
Solution: Since 𝜆 > 0 and for 𝜃 > 0 we can write
(
)
)
(
)
(
ℙ sup Wt ≥ 𝜆 = ℙ sup 𝜃Wt ≥ 𝜃𝜆 = ℙ sup e𝜃Wt ≥ e𝜃𝜆 .
0≤t≤T
0≤t≤T
0≤t≤T
From Problem 2.2.3.7 (page 75) we have shown that e𝜃Wt is a submartingale and it is also
non-negative. Thus, from Doob’s maximal inequality theorem,
( 𝜃W )
)
(
𝔼
e T
1 2
𝜃 T−𝜃𝜆
2
ℙ sup e𝜃Wt ≥ e𝜃𝜆 ≤
=
e
.
𝜃𝜆
e
0≤t≤T
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T−𝜃𝜆
By minimising e 2 𝜃
we have
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First Passage Time
we obtain 𝜃 = 𝜆∕T and substituting it into the inequality
)
(
𝜆2
ℙ sup Wt ≥ 𝜆 ≤ e− 2T .
0≤t≤T
Given that Wt ∼ 𝒩(0, t) we
) ℙ(|Wt | ≥ 𝜆) = ℙ(Wt ≥ 𝜆) + ℙ(Wt ≤ −𝜆) =
( can deduce that
2ℙ(Wt ≥ 𝜆) and hence ℙ
sup ||Wt || ≥ 𝜆
𝜆2
≤ 2e− 2T .
0≤t≤T
◽
4. Let (Ω, ℱ, ℙ) be a probability space and let Wt be the standard Wiener process with respect
to the filtration ℱt , t ≥ 0.
By writing Xt = x0 + Wt where x0 ∈ ℝ show that {Xt ∶ t ≥ 0} is a continuous martingale.
Let Ta = inf{t ≥ 0 ∶ Xt = a} and Tb = inf{t ≥ 0 ∶ Xt = b} where a < b. Show, using the
optional stopping theorem, that
ℙ(Ta < Tb ) =
b − x0
.
b−a
Solution: Let Xt = x0 + Wt where Wt is a standard Wiener process. Because Wt is a martingale (see Problem 2.2.3.1, page 71) and because x0 is a constant value, following the
same steps we can easily prove that Xt is also a martingale.
By writing T = inf{t ≥ 0 ∶ Xt ∉ (a, b)} as the first exit time from the interval (a, b), then
from the optional stopping theorem
)
(
𝔼(XT |X0 = x0 ) = 𝔼 x0 + WT |X0 = x0 = x0
and by definition
)
(
𝔼 XT |X0 = x0 = aℙ(XT = a|X0 = x0 ) + bℙ(XT = b|X0 = x0 )
(
)
(
)
x0 = aℙ Ta < Tb + bℙ Ta ≥ Tb
)
[
(
)]
(
x0 = aℙ Ta < Tb + b 1 − ℙ Ta < Tb .
Therefore,
ℙ(Ta < Tb ) =
b − x0
.
b−a
◽
5. Let (Ω, ℱ, ℙ) be a probability space and let Wt be the standard Wiener process with respect
to the filtration ℱt , t ≥ 0.
By writing Xt = x + Wt where x ∈ ℝ show that Xt and (Xt − x)2 − t are continuous martingales.
Let T = inf{t ≥ 0 ∶ Xt ∉ (a, b)} be the first exit time from the interval (a, b), a < x < b
and assuming T < ∞ almost surely show, using the optional stopping theorem, that
ℙ(XT = a|X0 = x) =
b−x
b−a
and
ℙ(XT = b|X0 = x) =
with
𝔼(T) = (b − x)(x − a).
x−a
b−a
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First Passage Time
81
Solution: Let Xt = x + Wt where Wt is a standard Wiener process. Because Wt and
Wt2 − t are martingales (see Problems 2.2.3.1, page 71 and 2.2.3.2, page 72) and
because x0 is a constant value, following the same steps we can easily prove that Xt and
(Xt − x)2 − t are also martingales.
For x ∈ (a, b) and because Xt is a martingale, from the optional stopping theorem
)
(
)
(
𝔼 XT |X0 = x = 𝔼 x + WT |X0 = x = x
and by definition
(
)
(
)
(
)
𝔼 XT |X0 = x = aℙ XT = a|X0 = x + bℙ XT = b|X0 = x
)
[
(
)]
(
x = aℙ XT = a|X0 = x + b 1 − ℙ XT = a|X0 = x .
Therefore,
ℙ(XT = a|X0 = x) =
b−x
b−a
and
ℙ(XT = b|X0 = x) = 1 − ℙ(XT = a|X0 = x) =
x−a
.
b−a
Since Yt = (Xt − x)2 − t is a martingale, from the optional stopping theorem we have
𝔼(YT |X0 = x) = 𝔼(Y0 |X0 = x) = 0
or
𝔼
]
[(
]
[(
)2
)2 |
|
XT − x − T | X0 = x = 𝔼 XT − x | X0 = x − 𝔼(T|X0 = x) = 0.
|
|
Therefore,
𝔼(T|X0 = x) = 𝔼
[(
]
)2 |
XT − x | X0 = x
|
= (a − x)2 ℙ(XT = a|X0 = x) + (b − x)2 ℙ(XT = b|X0 = x)
)
)
(
(
b−x
x−a
= (a − x)2
+ (b − x)2
b−a
b−a
= (b − x)(x − a).
◽
6. Let (Ω, ℱ, ℙ) be a probability space and let {Xt ∶ t ≥ 0} be a continuous martingale on ℝ.
Show that if T = inf{t ≥ 0 ∶ Xt ∉ (a, b)} is the first exit time from the interval (a, b) and
assuming T < ∞ almost surely then for 𝜃 > 0,
1 2
1 2
t
Zt = (e𝜃b − e−𝜃a )e𝜃Xt − 2 𝜃 t + (e−𝜃a − e𝜃b )e−𝜃Xt − 2 𝜃
is a martingale. Using the optional stopping theorem and if a < x < b, show that
[ (
)]
(
) cosh 𝜃 x − 1 (a + b)
1 2 |
2
𝔼 e− 2 𝜃 T || X0 = x =
, 𝜃 > 0.
[
]
|
cosh 12 𝜃(b − a)
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First Passage Time
Solution: Given that Xt is a martingale and both e𝜃b − e−𝜃a and e−𝜃a − e𝜃b are independent of Xt , then using the results of Problem 2.2.3.3 (page 72) we can easily show that
{Zt ∶ t ≥ 0} is a martingale.
From the optional stopping theorem
𝔼(ZT |X0 = x) = 𝔼(Z0 |X0 = x)
and using the identity sinh(x) + sinh(y) = 2 sinh
(x + y)
2
cosh
(x − y)
2
we have
𝔼(Z0 |X0 = x) = e𝜃(b−x) − e−𝜃(b−x) + e−𝜃(a+x) − e𝜃(a−x)
= 2 {sinh [𝜃(b − x)] + sinh [−𝜃(a − x)]}
[
= 4 sinh
]
[ (
)]
1
1
𝜃(b − a) cosh 𝜃 x − (a + b)
2
2
and
𝔼(ZT |X0 = x) = 𝔼(ZT |XT = a, X0 = x)ℙ(XT = a|X0 = x)
+𝔼(ZT |XT = b, X0 = x)ℙ(XT = b|X0 = x)
= 𝔼(ZT |XT = a, X0 = x)ℙ(XT = a|X0 = x)
+𝔼(ZT |XT = b, X0 = x)(1 − ℙ(XT = a|X0 = x))
= [𝔼(ZT |XT = a, X0 = x) − 𝔼(ZT |XT = b, X0 = x)]ℙ(XT = a|X0 = x)
+𝔼(ZT |XT = b, X0 = x)
)
(
[ 𝜃(b−a)
]
− 12 𝜃 2 T ||
−𝜃(b−a)
−e
= e
𝔼 e
| X0 = x
| )
(
− 12 𝜃 2 T ||
= 2 sinh[𝜃(b − a)]𝔼 e
| X0 = x
|
since 𝔼(ZT |XT = a, X0 = x) = 𝔼(ZT |XT = b, X0 = x). Finally,
(
𝔼
e
− 12 𝜃 2 T
)
[
|
|X = x =
| 0
|
4 sinh
1
𝜃(b
2
]
[ (
)]
− a) cosh 𝜃 x − 12 (a + b)
2 sinh[𝜃(b − a)]
[ (
)]
1
cosh 𝜃 x − 2 (a + b)
=
.
[
]
cosh 12 𝜃(b − a)
◽
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First Passage Time
83
7. Laplace Transform of First Passage Time. Let (Ω, ℱ, ℙ) be a probability space and let
{Wt ∶ t ≥ 0} be a standard Wiener process.
1 2
Show that for 𝜆 ∈ ℝ, Xt = e𝜆Wt − 2 𝜆 t is a martingale.
By setting Ta,b = inf{t ≥ 0 ∶ Wt = a + bt} as the first passage time of hitting the sloping
line a + bt, where a and b are constants, show that the Laplace transform of its distribution
is given by
√
)
(
2
𝔼 e−𝜃Ta,b = e−a(b+ b +2𝜃) , 𝜃 > 0
and hence show that
𝔼(Ta,b ) =
( )
a −2ab
e
b
and
ℙ(Ta,b < ∞) = e−2ab .
1 2
Solution: To show that Xt = e𝜆Wt − 2 𝜆 t is a martingale, see Problem 2.2.3.3 (page 72).
From the optional stopping theorem
( )
𝔼(XTa,b ) = 𝔼 X0 = 1
and because at t = Ta,b , WTa,b = a + bTa,b we have
]
[
1 2
𝔼 e𝜆(a+bTa,b )− 2 𝜆 Ta,b = 1
or
[
)
]
𝜆b− 12 𝜆2 Ta,b
(
𝔼 e
= e−𝜆a .
(
)
√
By setting 𝜃 = − 𝜆b − 12 𝜆2 we have 𝜆 = b ± b2 + 2𝜃. Since 𝜃 > 0 we must have
√
𝔼(e−𝜃Ta,b ) ≤ 1 and therefore 𝜆 = b + b2 + 2𝜃. Thus,
√
)
(
2
𝔼 e−𝜃Ta,b = e−a(b+ b +2𝜃) .
To find 𝔼(Ta,b ) we differentiate 𝔼(e−𝜃Ta,b ) with respect to 𝜃,
√
)
(
2
a
e−a(b+ b +2𝜃)
𝔼 Ta,b e−𝜃Ta,b = √
2
b + 2𝜃
and setting 𝜃 = 0,
) ( a ) −2ab
(
.
𝔼 Ta,b =
e
b
)
(
Finally, to find ℙ Ta,b < ∞ by definition
(
)
𝔼 e−𝜃Ta,b =
∞
∫0
e−𝜃t fTa,b (t) dt
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where fTa,b (t) is the probability density function of Ta,b . By setting 𝜃 = 0,
)|
)
(
(
= e−2ab .
ℙ Ta,b < ∞ = 𝔼 e−𝜃Ta,b |
|𝜃=0
)
(
)
(
N.B. Take note that if b = 0 we will have 𝔼 Ta,b = ∞ and ℙ Ta,b < ∞ = 1 (we say
Ta,b is finite almost surely).
◽
2.2.5
Reflection Principle
1. Reflection Principle. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process. By setting T as a stopping time and defining
{
̃ t = Wt
W
2WT − Wt
if t ≤ T
if t > T
̃ t ∶ t ≥ 0} is also a standard Wiener process.
show that {W
Solution: If T is finite then from the strong Markov property both the paths Yt = {Wt+T −
WT ∶ t ≥ 0} and −Yt = {−(Wt+T − WT ) ∶ t ≥ 0} are standard Wiener processes and independent of Xt = {Wt ∶ 0 ≤ t ≤ T}, and hence both (Xt , Yt ) and (Xt , −Yt ) have the same
distribution. Given the two processes defined on [0, T] and [0, ∞), respectively, we can
paste them together as follows:
𝜙 ∶ (X, Y) → {Xt 1I{t≤T} + (Yt−T + WT )1I{t≥T} ∶ t ≥ 0}.
Thus, the process arising from pasting Xt = {Wt ∶ 0 ≤ t ≤ T} to Yt = {Wt+T − WT ∶ t ≥
0} has the same distribution, which is {Wt ∶ t ≥ 0}. In contrast, the process arising from
̃ t ∶ t ≥ 0}. Thus,
pasting Xt = {Wt ∶ 0 ≤ t ≤ T} to −Yt = {−(Wt+T − WT ) ∶ t ≥ 0} is {W
̃ t ∶ t ≥ 0} is also a standard Wiener process.
{W
◽
2. Reflection Equality. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. By defining Tm = inf{t ≥ 0 ∶ Wt = m}, m > 0 as the first passage time,
then for 𝑤 ≤ m, m > 0, show that
ℙ(Tm ≤ t, Wt ≤ 𝑤) = ℙ(Wt ≥ 2m − 𝑤).
Solution: From the reflection principle in Problem 2.2.5.1 (page 84), since WTm = m,
ℙ(Tm ≤ t, Wt ≤ 𝑤) = ℙ(Tm ≤ t, 2WTm − Wt ≤ 𝑤)
= ℙ(Wt ≥ 2m − 𝑤).
◽
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85
3. First Passage Time Density Function. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶
t ≥ 0} be a standard Wiener process. By setting T𝑤 as a stopping time such that T𝑤 =
inf{t ≥ 0 ∶ Wt = 𝑤}, 𝑤 ≠ 0 show, using the reflection principle, that
(
|𝑤|
ℙ(T𝑤 ≤ t) = 1 + Φ − √
t
)
(
−Φ
|𝑤|
√
t
)
and the probability density function of T𝑤 is given as
|𝑤| − 𝑤2
fT𝑤 (t) = √
e 2t .
t 2𝜋t
Solution: We first consider the case when 𝑤 > 0. By definition
ℙ(T𝑤 ≤ t) = ℙ(T𝑤 ≤ t, Wt ≤ 𝑤) + ℙ(T𝑤 ≤ t, Wt ≥ 𝑤).
For the case when Wt ≤ 𝑤, by the reflection equality
ℙ(T𝑤 ≤ t, Wt ≤ 𝑤) = ℙ(Wt ≥ 𝑤).
On the contrary, if Wt ≥ 𝑤 then we are guaranteed T𝑤 ≤ t. Therefore,
ℙ(T𝑤 ≤ t, Wt ≥ 𝑤) = ℙ(Wt ≥ 𝑤)
and hence
∞
ℙ(T𝑤 ≤ t) = 2ℙ(Wt ≥ 𝑤) =
∫𝑤
1
√
2𝜋t
y2
e− 2t dy.
√
By setting x = y∕ 2t we have
∞
ℙ(T𝑤 ≤ t) = 2
∫ √𝑤
t
(
)
( )
1 − x22
𝑤
𝑤
dx = 1 + Φ − √ − Φ √ .
√ e
t
t
2𝜋
For the case when 𝑤 ≤ 0, using the reflection principle
ℙ(T𝑤 ≤ t) = ℙ(T𝑤 ≤ t, Wt ≤ 𝑤) + ℙ(T𝑤 ≤ t, Wt ≥ 𝑤)
= ℙ(T𝑤 ≤ t, −Wt ≥ −𝑤) + ℙ(T𝑤 ≤ t, −Wt ≤ −𝑤)
= ℙ(T𝑤 ≤ t, −Wt ≥ −𝑤) + ℙ(T𝑤 ≤ t, −Wt ≤ −𝑤)
̃ t ≥ −𝑤) + ℙ(T𝑤 ≤ t, W
̃ t ≤ −𝑤)
= ℙ(T𝑤 ≤ t, W
̃ t = −Wt is also a standard Wiener process. Therefore,
where W
(
̃ t ≥ −𝑤) = 1 + Φ
ℙ(T𝑤 ≤ t) = 2ℙ(W
𝑤
√
t
)
(
𝑤
− Φ −√
t
)
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86
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2.2.5
(
and hence
|𝑤|
ℙ(T𝑤 ≤ t) = 1 + Φ − √
t
)
(
−Φ
|𝑤|
√
t
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Reflection Principle
)
.
To find the density of T𝑤 we note that
[
(
)
(
)]
|𝑤|
|𝑤|
|𝑤| − 𝑤2
d
d
fT𝑤 (t) = ℙ(T𝑤 ≤ t) =
1+Φ −√
e 2t .
−Φ √
= √
dt
dt
t
t
t 2𝜋t
◽
4. Let (Ω, ℱ, ℙ) be a probability space and let Mt = max Ws be the maximum level reached
0≤s≤t
by a standard Wiener process {Wt ∶ t ≥ 0} in the time interval [0, t]. Then for a ≥ 0, x ≤ a
and for all t ≥ 0, show using the reflection principle that
(
)
( )
⎧
x − 2a
x
a ≥ 0, x ≤ a
√
⎪Φ √ − Φ
t
t
⎪
⎪
(
)
⎪ ( )
ℙ(Mt ≤ a, Wt ≤ x) = ⎨
a
a
a ≥ 0, x ≥ a
⎪Φ √ − Φ − √
t
t
⎪
⎪
⎪
⎩0
a≤0
and the joint probability density function of (Mt , Wt ) is
⎧ 2(2a − x) − 1
⎪
e 2
fMt ,Wt (a, x) = ⎨ √
t 2𝜋t
⎪
0
⎩
(
2a−x
√
t
)2
a ≥ 0, x ≤ a
otherwise.
Solution: For a ≥ 0, x ≤ a, let Ta = inf{t ≥ 0 ∶ Wt = a} then
{Mt ≥ a} ⇐⇒ {Ta ≤ t}.
Taking T = Ta in the reflection principle,
ℙ(Mt ≥ a, Wt ≤ x) = ℙ(Ta ≤ t, Wt ≤ x)
= ℙ(Ta ≤ t, Wt ≥ 2a − x)
= ℙ(Wt ≥ 2a − x)
(
)
2a − x
=1−Φ
√
t
)
(
x − 2a
.
=Φ
√
t
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87
Because
ℙ(Wt ≤ x) = ℙ(Mt ≤ a, Wt ≤ x) + ℙ(Mt ≥ a, Wt ≤ x)
then for a ≥ 0, x ≤ a,
ℙ(Mt ≤ a, Wt ≤ x) = ℙ(Wt ≤ x) − ℙ(Mt ≥ a, Wt ≤ x)
(
)
( )
x − 2a
x
.
=Φ √ −Φ
√
t
t
For the case when a ≥ 0, x ≥ a and because Mt ≥ Wt , we have
ℙ(Mt ≤ a, Wt ≤ x) = ℙ(Mt ≤ a, Wt ≤ a) = ℙ(Mt ≤ a)
)
(
and by substituting x = a into Φ
x
√
t
(
)
x − 2a
√
t
−Φ
(
)
a
√
t
ℙ(Mt ≤ a, Wt ≤ x) = Φ
we have
(
)
a
− Φ −√
t
.
Finally, for the case when a ≤ 0, and since Mt ≥ W0 = 0, then ℙ(Mt ≤ a, Wt ≤ x) = 0.
To obtain the joint probability density function of (Mt , Wt ), by definition
fMt ,Wt (a, x) =
𝜕2
ℙ(Mt ≤ a, Wt ≤ x)
𝜕a𝜕x
and since ℙ(Mt ≤ a, Wt ≤ x) is a function in both a and x only for the case when a ≥ 0,
x ≤ a, then
[ ( )
(
)]
x
x − 2a
𝜕2
Φ √ −Φ
fMt ,Wt (a, x) =
√
𝜕a𝜕x
t
t
⎡
−1
𝜕 ⎢ 1
=
e 2
√
𝜕a ⎢ 2𝜋t
⎣
(
x
√
)2
1
−√
e
2𝜋t
t
(
1
−2(x − 2a) − 2
e
=
√
t 2𝜋t
x−2a
√
t
(
1
2(2a − x) − 2
e
= √
t 2𝜋t
2a−x
√
t
(
− 12
x−2a
√
t
)2
⎤
⎥
⎥
⎦
)2
)2
.
◽
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Reflection Principle
5. Let (Ω, ℱ, ℙ) be a probability space and let mt = min Ws be the minimum level reached
0≤s≤t
by a standard Wiener process {Wt ∶ t ≥ 0} in the time interval
show that
(
)
(
)
⎧
x
2b − x
√
⎪Φ − √ − Φ
t
t
⎪
)
( )
⎪ (
ℙ(mt ≥ b, Wt ≥ x) = ⎨
b
b
⎪Φ − √ − Φ √
t
t
⎪
⎪
⎩0
[0, t]. Then for all t ≥ 0
b ≤ 0, b ≤ x
b ≤ 0, b ≥ x
b≥0
and the joint probability density function of (mt , Wt ) is
⎧ −2(2b − x) − 1
⎪
e 2
fmt ,Wt (b, x) = ⎨ √
t 2𝜋t
⎪
⎩0
(
2b−x
√
t
)2
b ≤ 0, b ≤ x
otherwise.
Solution: Given mt = − max {−Wt } we have for b ≤ 0, b ≤ x,
0≤s≤t
(
{
)
}
ℙ(mt ≥ b, Wt ≥ x) = ℙ − max −Wt ≥ b, Wt ≥ x
0≤s≤t
(
)
{
}
= ℙ − max −Wt ≥ b, −Wt ≤ −x
0≤s≤t
(
)
̃ t ≤ −b, W
̃ t ≤ −x ,
= ℙ max W
0≤s≤t
where the last equality comes from the symmetric property of the standard Wiener process
̃ t = −Wt ∼ 𝒩(0, t) is also a standard Wiener process.
such that W
Since
(
)
(
)
̃ t ≤ −b, W
̃ t ≥ −b, W
̃ t ≤ −x) = ℙ max W
̃ t ≤ −x + ℙ max W
̃ t ≤ −x
ℙ(W
0≤s≤t
0≤s≤t
and from Problem 2.2.5.4 (page 86), we can write
(
̃ t ≥ −b, W
̃ t ≤ −x
ℙ max W
(
)
0≤s≤t
)
2b − x
√
t
=Φ
so
)
(
)
(
̃ t ≤ −b, W
̃ t ≥ −b, W
̃ t ≤ −x = ℙ(W
̃ t ≤ −x) − ℙ max W
̃ t ≤ −x
ℙ max W
0≤s≤t
(
x
= Φ −√
t
0≤s≤t
)
(
−Φ
2b − x
√
t
)
.
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89
Consequently, if b ≤ 0, b ≤ x we have
)
(
x
ℙ(mt ≥ b, Wt ≥ x) = Φ − √
t
(
−Φ
)
2b − x
√
t
.
For the case when b ≤ 0, b ≥ x and since mt ≤ Wt ,
ℙ(mt ≥ b, Wt ≥ x) = ℙ(mt ≥ b, Wt ≥ b) = ℙ(mt ≥ b).
(
)
(
)
x
2b − x
Setting x = b in Φ − √ − Φ
would therefore yield
√
t
t
)
(
b
ℙ(mt ≥ b, Wt ≥ x) = Φ − √
(
−Φ
t
)
b
√
t
.
Finally, for b ≥ 0, since mt ≤ W0 = 0 so ℙ(mt ≥ b, Wt ≥ x) = 0.
To obtain the joint probability density function by definition,
𝜕2
ℙ(mt ≤ b, Wt ≤ x)
𝜕b𝜕x
𝜕2
=
ℙ(mt ≥ b, Wt ≥ x)
𝜕b𝜕x
fmt ,Wt (b, x) =
and since ℙ(mt ≥ b, Wt ≥ x) is a function in both b and x only for the case when b ≤ 0,
b ≤ x, so
[ (
)
(
)]
𝜕2
x
2b − x
Φ −√ − Φ
fmt ,Wt (b, x) =
√
𝜕b𝜕x
t
t
⎡
1
1 −2
𝜕 ⎢
−√
e
=
𝜕b ⎢
2𝜋t
⎣
(
x
√
)2
2.2.6
1
e
+√
2𝜋t
t
(
1
−2(2b − x) − 2
=
e
√
t 2𝜋t
2b−x
√
t
(
− 12
2b−x
√
t
)2
⎤
⎥
⎥
⎦
)2
.
◽
Quadratic Variation
1. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that it has finite quadratic variation such that
⟨W, W⟩t = lim
n→∞
n−1
∑
(Wti+1 − Wti )2 = t
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Finally, deduce that dWt ⋅ dWt = dt.
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2.2.6 Quadratic Variation
Solution: Since the quadratic variation is a sum of random variables, we need to show
that its expected value is t and its variance converges
) to zero as n → ∞.
(
Let ΔWti = Wti+1 − Wti ∼ 𝒩(0, t∕n) where 𝔼 ΔWt2i = t∕n then, by taking expectations
we have
)
(
n−1
n−1 (
)
∑
∑
2
𝔼 lim
ΔWti = lim
𝔼 ΔWt2i = t.
n→∞
n→∞
i=0
i=0
Because ΔWt2i ∕(t∕n) ∼ 𝜒 2 (1) we have 𝔼(ΔWt4i ) = 3(t∕n)2 . Therefore, by independence of
increments
)2
)
(
(
n−1
n−1
⎤
⎡
∑
∑
ΔWt2i = 𝔼 ⎢ lim
ΔWt2i − t ⎥
Var lim
n→∞
⎥
⎢ n→∞ i=0
i=0
⎦
⎣
[
n−1
(
) ]
∑
t 2
2
𝔼 ΔWti −
= lim
n→∞
n
i=0
n−1
∑
= lim
n→∞
(
i=0
3t2 2t2
t2
− 2 + 2
2
n
n
n
)
2t2
n→∞ n
= 0.
= lim
∑
n−1
Since lim
n→∞ i=0
t
t
(Wti+1 − Wti )2 = ∫0 dWs ⋅ dWs = t and ∫0 ds = t, then by differentiating
both sides with respect to t we have dWt ⋅ dWt = dt.
◽
2. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that the following cross-variation between Wt and t, and the quadratic variation of
t, are
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
ti+1 − ti = 0
i=0
lim
n→∞
n−1
∑
(
ti+1 − ti
)2
=0
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Finally, deduce that dWt ⋅ dt = 0 and dt ⋅ dt = 0.
Solution: Since Wti+1 − Wti ∼ 𝒩(0, t∕n) then taking expectation and variance we have
[
𝔼
lim
n→∞
n−1 (
∑
Wti+1 − Wti
i=0
]
)(
ti+1 − ti
)
= lim
n→∞
)
) (
ti+1 − ti 𝔼 Wti+1 − Wti = 0
n−1
∑
(
i=0
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Quadratic Variation
91
and
[
Var
lim
n→∞
n−1 (
∑
Wti+1 − Wti
]
)(
ti+1 − ti
)
= lim
n→∞
i=0
(
)
)2
ti+1 − ti Var Wti+1 − Wti
n−1
∑
(
i=0
∑ t )2 t
n→∞
n n
i=0
n−1 (
= lim
t3
n→∞ n
= 0.
= lim
n−1 (
Thus, lim
∑
n→∞ i=0
Wti+1 − Wti
)(
)
ti+1 − ti = 0.
In addition,
lim
n→∞
n−1
∑
(ti+1 − ti )2 = lim
n→∞
i=0
n−1 ( )
∑
t 2
n
i=0
= lim
n→∞
t2
= 0.
n
Finally, because
n−1 (
∑
lim
n→∞
Wti+1 − Wti
)(
)
ti+1 − ti =
i=0
lim
n→∞
n−1
∑
(
)2
ti+1 − ti
i=0
t
dWs ⋅ ds = 0
∫0
t
=
∫0
ds ⋅ ds = 0
then, by differentiating both sides with respect to t, we can deduce that dWt ⋅ dt = 0 and
dt ⋅ dt = 0.
◽
3. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that it has unbounded first variation such that
n−1
∑
|
|
|Wti+1 − Wti | = ∞
|
|
n→∞
lim
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Solution: Since
n−1
∑(
i=0
)2
Wti+1 − Wti
n−1
∑
n−1
|
| ∑|
|
≤ max |Wtk+1 − Wtk | ⋅
|W − Wti |, then
| i=0 | ti+1
|
0≤k≤n−1 |
n−1
∑(
)2
Wti+1 − Wti
i=0
|
|
.
|Wti+1 − Wti | ≥
|
|
|
|
max |Wtk+1 − Wtk |
i=0
|
0≤k≤n−1 |
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2.2.6 Quadratic Variation
n−1 (
)2
∑
|
|
Wti+1 − Wti = t < ∞,
As Wt is continuous, lim max |Wtk+1 − Wtk | = 0 and lim
|
n→∞ 0≤k≤n−1 |
n→∞ i=0
n−1
∑|
|
we can conclude that lim
|W − Wti | = ∞.
|
n→∞ i=0 | ti+1
◽
4. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Show that for p ≥ 3,
n−1 (
)p
∑
lim
Wti+1 − Wti = 0
n→∞
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Finally, deduce that (dWt )p = 0, p ≥ 3.
Solution: We prove this result via mathematical induction.
For p = 3, we can express
|∑
| n−1 (
)3 ||
)2 ||
| n−1 (
| ||∑
| ≤ max ||W
|
|
W
W
−
W
−
W
−
W
⋅
|
ti+1
ti
tk | |
ti+1
ti
| 0≤k≤n−1 | tk+1
|
|
|
|
| i=0
| i=0
|
|
|
|
n−1 (
)2
|
| ∑
= max |Wtk+1 − Wtk | ⋅
Wti+1 − Wti .
|
0≤k≤n−1 |
i=0
Because Wt is continuous, we have
|
|
lim max |Wtk+1 − Wtk | = 0.
|
n→∞ 0≤k≤n−1 |
n−1 (
In addition, because lim
∑
n→∞ i=0
)2
Wti+1 − Wti
= t < ∞ (see Problem 2.2.6.1, page 89), so
|
n−1 (
)3 ||
∑
|
| lim
| ≤ 0.
−
W
W
ti+1
ti
|n→∞
|
|
|
i=0
|
|
or
lim
n→∞
n−1 (
∑
)3
Wti+1 − Wti
Thus, the result is true for p = 3.
Assume the result is true for p = m, m > 3 so that lim
Then for p = m + 1,
= 0.
i=0
n−1 (
∑
n→∞ i=0
Wti+1 − Wti
)m
= 0.
| n−1 (
|∑
)m+1 ||
)m ||
| n−1 (
|
| ||∑
|
|
max |Wtk+1 − Wtk | ⋅ |
Wti+1 − Wti
Wti+1 − Wti ||
|
| ≤ 0≤k≤n−1
|
| |
| i=0
|
|
i=0
|
|
|
|
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Quadratic Variation
and because Wt
lim
n∑
−1 (
n→∞ i=0
Wti+1 − Wti
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93
is continuous such that
)m
lim
max
n→∞ 0≤k≤n − 1
(
)
Wtk+1 − Wtk = 0 and
= 0, so
|
n−1 (
)m+1 ||
∑
|
|≤0
| lim
−
W
W
ti+1
ti
|
|n→∞
|
|
i=0
|
|
or
lim
n→∞
n−1 (
∑
)m+1
Wti+1 − Wti
= 0.
i=0
Thus, the result is also true for p = m + 1. From mathematical induction we have shown
n−1 (
)p
∑
Wti+1 − Wti = 0, p ≥ 3.
lim
n→∞ i=0
t(
)p
)p
Wti+1 − Wti =
dWs = 0, by differentiating both sides
n→∞ i=0
∫0
)p
(
with respect to t we have dWt = 0, p ≥ 3.
◽
Since for p ≥ 3, lim
n−1
∑(
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3
Stochastic Differential Equations
A stochastic differential equation (SDE) is a differential equation in which one or more of
the terms has a random component. Within the context of mathematical finance, SDEs are
frequently used to model diverse phenomena such as stock prices, interest rates or volatilities
to name but a few. Typically, SDEs have continuous paths with both random and non-random
components and to drive the random component of the model they usually incorporate a Wiener
process. To enrich the model further, other types of random fluctuations are also employed in
conjunction with the Wiener process, such as the Poisson process when modelling discontinuous jumps. In this chapter we will concentrate solely on SDEs having only a Wiener process,
whilst in Chapter 5 we will discuss SDEs incorporating both Poisson and Wiener processes.
3.1 INTRODUCTION
To begin with, a one-dimensional stochastic differential equation can be described as
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
where Wt is a standard Wiener process, 𝜇(Xt , t) is defined as the drift and 𝜎(Xt , t) the volatility.
Many financial models can be written in this form, such as the lognormal asset random walk,
common spot interest rate and stochastic volatility models.
From an initial state X0 = x0 , we can write in integrated form
t
Xt = X0 +
∫0
t
𝜇(Xs , s) ds +
∫0
𝜎(Xs , s) dWs
[
]
|𝜇(Xs , s)| + 𝜎(Xs , s)2 ds < ∞ and the solution of the integral equation is called an
∫0
Itō process or an Itō diffusion.
In Chapter 2 we described the properties of a Wiener process and, given that it is
non-differentiable, there is a crucial difference between stochastic calculus and integral
calculus. We consider an integral with respect to a Wiener process and write
t
with
t
It =
∫0
f (Ws , s) dWs
where the integrand f (Wt , t) is ℱt measurable (i.e., f (Wt , t) is a stochastic process adapted to
the filtration of a Wiener process) and is square-integrable
( t
)
2
𝔼
f (Ws , s) ds < ∞
∫0
for all t ≥ 0.
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INTRODUCTION
Let t > 0 be a positive constant and assume f (Wti , ti ) is constant over the interval [ti , ti+1 ),
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t for n ∈ ℕ. Here we call such a process
f an elementary process or a simple process, and for such a process the Itō integral It can be
defined as
n−1
t
∑
f (Ws , s) dWs = lim
f (Wti , ti )(Wti+1 − Wti ).
It =
n→∞
∫0
i=0
We now give a formal result of the Itō integral as follows.
Theorem 3.1 Let {Wt ∶ t ≥ 0} be a standard Wiener process on the probability space
(Ω, ℱ, ℙ), and let ℱt , t ≥ 0 be the associated filtration. The Itō integral defined by
t
It =
∫0
f (Ws , s) dWs = lim
n→∞
n−1
∑
f (Wti , ti )(Wti+1 − Wti )
i=0
where f is a simple process and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ has
the following properties:
• The paths of It are continuous.
• For each t, It is ℱt measurable.
t
t
• If It = ∫0 f (Ws , s) dWs and Jt = ∫0 g(Ws , s) dWs where g is a simple process, then
t
It ± Jt =
∫0
[
]
f (Ws , s) ± g(Ws , s) dWs
and for a constant c
t
cIt =
∫0
cf (Ws , s) dWs .
• It is a martingale.
(
t
• 𝔼[It2 ] = 𝔼
∫0
)
f (Ws , s)2 ds .
t
• The quadratic variation ⟨I, I⟩t =
∫0
f (Ws , s)2 ds.
In mathematics, Itō’s formula (or lemma) is used in stochastic calculus to find the differential
of a function of a particular type of stochastic process. In essence, it is the stochastic calculus
counterpart of the chain rule in ordinary calculus via a Taylor series expansion. The formula is
widely employed in mathematical finance and its best-known application is in the derivation
of the Black–Scholes equation used to value options. The following is a formal result of Itō’s
formula.
Theorem 3.2 (One-Dimensional Itō’s Formula) Let {Wt ∶ t ≥ 0} be a standard Wiener
process on the probability space (Ω, ℱ, ℙ) and let ℱt , t ≥ 0 be the associated filtration. Consider a stochastic process Xt satisfying the following SDE
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
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or in integrated form,
t
Xt = X0 +
∫0
t
𝜇(Xs , s) ds +
∫0
𝜎(Xs , s) dWs
[
]
|𝜇(Xs , s)| + 𝜎(Xs , s)2 ds < ∞. Then for any twice differentiable function g(Xt , t),
∫0
the stochastic process Yt = g(Xt , t) satisfies
t
with
𝜕g
𝜕g
1 𝜕2 g
(Xt , t) dXt +
(X , t)(dXt )2
(Xt , t) dt +
𝜕t
𝜕Xt
2 𝜕Xt2 t
[
]
2g
𝜕
𝜕g
𝜕g
𝜕g
1
(X , t) + 𝜇(Xt , t)
=
(X , t) + 𝜎(Xt , t)2 2 (Xt , t) dt + 𝜎(Xt , t)
(Xt , t) dWt
𝜕t t
𝜕Xt t
2
𝜕X
𝜕Xt
t
dYt =
where (dXt )2 is computed according to the rule
(dWt )2 = dt,
(dt)2 = dWt dt = dt dWt = 0.
In integrated form,
t
t 2
𝜕g
𝜕g
𝜕 g
1
(Xs , s) dXs +
(X , s) d⟨X, X⟩s
(Xs , s) ds +
∫0 𝜕t
∫0 𝜕Xt
2 ∫0 𝜕Xt2 s
[
]
t
2
𝜕g
𝜕g
1
2𝜕 g
(X , s) + 𝜇(Xs , s)
= Y0 +
(X , s) + 𝜎(Xs , s)
(Xs , s) ds
∫0
𝜕t s
𝜕Xt s
2
𝜕Xt2
t
Yt = Y0 +
t
+
𝜎(Xs , s)
∫0
𝜕g
(X , s) dWs
𝜕Xt s
t
where ⟨X, X⟩t = ∫0 𝜎(Xs , s)2 ds.
The theory of SDEs is a framework for expressing the dynamical models that include both
the random and non-random components. Many solutions to SDEs are Markov processes,
where the future depends on the past only through the present. For this reason, the solutions
can be studied using backward and forward Kolmogorov equations, which turn out to be linear parabolic partial differential equations of diffusion type. But before we discuss them, the
following Feynman–Kac theorem describes an important link between stochastic differential
equations and partial differential equations.
Theorem 3.3 (Feynman–Kac Formula for One-Dimensional Diffusion Process) Let
{Wt ∶ t ≥ 0} be a standard Wiener process on the probability space (Ω, ℱ, ℙ) and let ℱt ,
t ≥ 0 be the associated filtration. Let Xt be the solution of the following SDE:
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
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and define r as a function of t. For t ∈ [0, T] where T > 0 and if V(Xt , t) satisfies the partial
differential equation (PDE)
𝜕2V
1
𝜕V
𝜕V
(X , t) − r(t)V(Xt , t) = 0
(Xt , t) + 𝜎(Xt , t)2 2 (Xt , t) + 𝜇(Xt , t)
𝜕t
2
𝜕Xt t
𝜕Xt
subject to the boundary condition V(XT , T) = Ψ(XT ), then under the filtration ℱt the solution
of the PDE is given by
]
[
|
T
V(Xt , t) = 𝔼 e− ∫t r(u)du Ψ(XT )|| ℱt .
|
The Feynman–Kac formula can be used to study the transition probability density function
of the one-dimensional stochastic differential equation
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt .
For t ∈ [0, T], T > 0 and conditioning Xt = x we can write
]
[
T
|
V(x, t) = 𝔼 e− ∫t r(u)du Ψ(XT )|ℱt
|
= e− ∫t
T
∞
r(u)du
∫−∞
Ψ(y)p(x, t; y, T) dy
where the random variable XT has transition probability density p(x, t; y, T) in the y variable.
The transition probability density function p(x, t; y, T) satisfies two parabolic partial differential equations, the forward Kolmogorov equation (or Fokker–Planck equation) and the
backward Kolmogorov equation. In the forward Kolmogorov function, it involves derivatives
with respect to the future state y and time T, whilst in the backward Kolmogorov function, it
involves derivatives with respect to the current state x and time t.
Theorem 3.4 (Forward Kolmogorov Equation for One-Dimensional Diffusion Process) Let {Wt ∶ t ≥ 0} be the standard Wiener process on the probability space (Ω, ℱ, ℙ).
Consider the stochastic differential equation
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
and for t ∈ [0, T], T > 0 and by conditioning Xt = x, the random variable XT having a transition probability density p(x, t; y, T) in the y variable satisfies
𝜕
1 𝜕2
𝜕
(𝜎(y, T)2 p(x, t; y, T)) − (𝜇(y, T)p(x, t; y, T)).
p(x, t; y, T) =
𝜕T
2 𝜕y2
𝜕y
Theorem 3.5 (Backward Kolmogorov Equation for One-Dimensional Diffusion Process) Let {Wt ∶ t ≥ 0} be the standard Wiener process on the probability space (Ω, ℱ, ℙ).
Consider the stochastic differential equation
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
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and for t ∈ [0, T], T > 0 and by conditioning Xt = x, the random variable XT having a transition probability density p(x, t; y, T) in the x variable satisfies
1
𝜕
𝜕2
𝜕
p(x, t; y, T) + 𝜎(x, t)2 2 p(x, t; y, T) + 𝜇(x, t) p(x, t; y, T) = 0.
𝜕t
2
𝜕x
𝜕x
In contrast, a multi-dimensional diffusion process can be described by a set of stochastic
differential equations
dXt(i) = 𝜇(i) (Xt(i) , t) dt +
n
∑
(j)
𝜎 (i,j) (X (i) , t) dWt ,
i = 1, 2, . . . , m
j=1
]T
[
where Wt = Wt(1) , Wt(2) , . . . , Wt(n) is the n-dimensional standard Wiener process such that
[
(j)
dWt(i) dWt = 𝜌ij dt, 𝜌ij ∈ (−1, 1) for i ≠ j, i, j = 1, 2, . . . , n, 𝝁(Xt , t) = 𝜇(1) (Xt(1) , t), 𝜇 (2) (Xt(2) ,
]T
t), . . . , 𝜇(m) (Xt(m) , t) is the m-dimensional drift vector and
. . . 𝜎 (1,n) (Xt(1) , t) ⎤
⎥
. . . 𝜎 (2,n) (Xt(2) , t) ⎥
⎥
⋱
⋮
. . . 𝜎 (m,n) (Xt(m) , t)⎥⎦
⎡ 𝜎 (1,1) (X (1) , t) 𝜎 (1,2) (X (1) , t)
t
⎢ (2,1) t(2)
(2,2) (X (2) , t)
𝜎
(X
,
t)
𝜎
t
t
𝝈(Xt , t) = ⎢
⎢
⋮
⋮
⎢𝜎 (m,1) (X (m) , t) 𝜎 (m,2) (X (m) , t)
⎣
t
t
is the m × n volatility matrix. Given the initial state X0(i) , we can write in integrated form
Xt(i) = X0(i) +
t
with
t
∫0
𝜇 (i) (Xs(i) , s) ds +
n
∑
j=1
t
∫0
(j)
𝜎 (i,j) (Xs(i) , s) dWs
[
]
|𝜇(i) (Xs(i) , s)| + |𝜎 (i,p) (Xs(i) , s)𝜎 (i,q) (Xs(i) , s)| ds < ∞ for i = 1, 2, . . . , m and p, q =
∫0
1, 2, . . . , n.
Definition 3.6 (Multi-Dimensional Itō’s Formula) Let {Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n be a
sequence of standard Wiener processes on the probability space (Ω, ℱ, ℙ), and let ℱt , t ≥ 0
be the associated filtration. Consider a stochastic process Xt(i) satisfying the following SDE
dXt(i) = 𝜇(i) (Xt(i) , t) dt +
n
∑
(j)
𝜎 (i,j) (X (i) , t) dWt ,
i = 1, 2, . . . , m
j=1
or in integrated form
Xt(i) = X0(i) +
t
t
∫0
𝜇 (i) (Xs(i) , s) ds +
n
∑
j=1
t
∫0
(j)
𝜎 (i,j) (Xs(i) , s) dWs
with ∫0 |𝜇(i) (Xs(i) , s)| + |𝜎 (i,p) (Xs(i) , s)𝜎 (i,q) (Xs(i) , s)| ds < ∞ for i = 1, 2, . . . , [m and p, q =
1, 2, . . . , n. Then, for any twice differentiable function h(Xt , t), where Xt = Xt(1) , Xt(2) , . . . ,
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, the stochastic process Zt = h(Xt(1) , Xt(2) , . . . , Xt(m) , t) satisfies
m
m m
∑
𝜕h
1 ∑ ∑ 𝜕2h
𝜕h
(j)
(i)
(X
,
t)
dX
+
(Xt , t) dXt(i) dXt
(Xt , t) dt +
t
t
(i)
(j)
(i)
𝜕t
2
𝜕X
𝜕X
𝜕X
i=1
i=1 j=1
t
t
t
[
m
∑
𝜕h
𝜕h
(Xt , t) +
=
𝜇(i) (Xt(i) , t) (i) (Xt , t)
𝜕t
𝜕Xt
i=1
(
)
]
m m
n
n
2h
∑
∑
𝜕
1 ∑∑
(j)
𝜌
𝜎 (i,k) (Xt(i) , t)𝜎 (j,l) (Xt , t)
(Xt , t) dt
+
(j)
2 i=1 j=1 ij k=1 l=1
𝜕Xt(i) 𝜕Xt
( n
)
m
∑
∑
𝜕h
(i,j) (i)
𝜎 (Xt , t)
(Xt , t) dWt(i)
+
(i)
𝜕Xt
i=1
j=1
dZt =
(j)
where dXt(i) dXt is computed according to the rule
(j)
dWt(i) dWt = 𝜌ij dt,
(dt)2 = dWt(i) dt = dt dWt(i) = 0
such that 𝜌ij ∈ (−1, 1) and 𝜌ii = 1. In integrated form
t
Zt = Z0 +
∫0
m
t∑
𝜕h
𝜕h
(Xs , s) dXs(i)
(Xs , s) ds +
(i)
∫0
𝜕t
i=1 𝜕Xt
1 ∑ ∑ 𝜕2h
+
(Xs , s) d⟨X (i) , X (j) ⟩s
(j)
(i)
∫0 2
i=1 j=1 𝜕Xt 𝜕Xt
[
m
t
∑
𝜕h
𝜕h
= Z0 +
𝜇 (i) (Xs(i) , s) (i) (Xs , s)
(Xs , s) +
∫0
𝜕t
𝜕X
i=1
t
( n n
)
]
m
m
∑∑
𝜕2 h
1 ∑∑
(i,k) (i)
(j,l) (j)
𝜌
𝜎 (Xs , s)𝜎 (Xs , s)
(Xs , s) ds
+
(j)
2 i=1 j=1 ij k=1 l=1
𝜕Xt(i) 𝜕Xt
)
( n
m
t∑
∑
𝜕h
𝜎 (i,j) (Xs(i) , s)
(Xs , s) dWs(i) ,
+
∫0
𝜕X (i)
m
t
i=1
m
j=1
t
where
⟨X (i) , X (j) ⟩
t
=
∫0
𝜌ij
n
n
∑
∑
t
(j)
𝜎 (i,k) (Xs(i) , s)𝜎 (j,l) (Xs , s) ds.
k=1 l=1
The Feynman–Kac theorem for a one-dimensional diffusion process also extends to a
multi-dimensional version.
Theorem 3.7 (Feynman–Kac Formula for Multi-Dimensional Diffusion Process) Let
{Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n be a sequence of standard Wiener processes on the probability
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space (Ω, ℱ, ℙ), and let ℱt , t ≥ 0 be the associated filtration. Let Xt(i) be the solution of the
following SDE
dXt(i) = 𝜇(i) (Xt(i) , t) dt +
n
∑
(j)
𝜎 (i,j) (X (i) , t) dWt ,
i = 1, 2, . . . , m
j=1
(j)
where dWt(i) dWt = 𝜌ij dt, 𝜌ij ∈ (−1, 1) for i ≠ j, 𝜌ii = 1, i, j = 1, 2, . . . , n and define r to be
]T
[
a function of t. By denoting Xt = Xt(1) , Xt(2) , . . . , Xt(m) , for t ∈ [0, T] where T > 0 and if
V(Xt , t) satisfies the PDE
( n n
)
m m
1 ∑ ∑ ∑ ∑ (i,k) (i)
𝜕2V
𝜕V
(j,l) (j)
𝜎 (Xt , t)𝜎 (Xt , t)
(Xt , t)
(Xt , t) +
(j)
𝜕t
2 i=1 j=1 k=1 l=1
𝜕X (i) 𝜕X
t
∑
m
+
𝜇 (i) (Xt(i) , t)
i=1
𝜕V
𝜕Xt(i)
t
(Xt , t) − r(t)V(Xt , t) = 0
subject to the boundary condition V(XT , T) = Ψ(XT ), then under the filtration ℱt , the solution
of the PDE is given by
]
[
|
T
− ∫t r(u)du
|
V(Xt , t) = 𝔼 e
Ψ(XT )| ℱt .
|
Similarly, we have the Kolmogorov forward and backward equations for multi-dimensional
diffusion processes as well.
Theorem 3.8 (Forward Kolmogorov Equation for Multi-Dimensional Diffusion Process) Let {Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n be a sequence of standard Wiener processes on the
probability space (Ω, ℱ, ℙ). Consider the stochastic differential equations
dXt(i) = 𝜇(i) (Xt(i) , t) dt +
n
∑
(j)
𝜎 (i,j) (X (i) , t) dWt ,
i = 1, 2, . . . , m
j=1
(j)
where dWt(i) dWt = 𝜌ij dt, 𝜌ij ∈ (−1, 1) for i ≠ j, 𝜌ii = 1, i, j = 1, 2, . . . , n. By denoting
[
]T
Xt = Xt(1) , Xt(2) , . . . , Xt(m) , and for t ∈ [0, T], T > 0 and by conditioning Xt = x where
]T
[
x = x(1) , x(2) , . . . , x(m) , the m-dimensional random variable XT having a transition
]T
[
probability density p(x, t; y, T) in the m-dimensional variable y = y(1) , y(2) , . . . , y(m)
satisfies
[ n n
]
m m
∑∑(
)
𝜕2
𝜕
1 ∑∑
𝜌 𝜎 (i,k) (y(i) , T)𝜎 (j,l) (y(j) , T) p(x, t; y, T)
p(x, t; y, T) =
𝜕T
2 i=1 j=1 𝜕y(i) 𝜕y(j) k=1 l=1 kl
−
m
∑
𝜕
(𝜇 (i) (y(i) , T)p(x, t; y, T)).
(i)
𝜕y
i=1
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Theorem 3.9 (Backward Kolmogorov Equation for Multi-Dimensional Diffusion Process) Let {Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n be a sequence of independent standard Wiener processes on the probability space (Ω, ℱ, ℙ). Consider the stochastic differential equations
dXt(i) = 𝜇(i) (Xt(i) , t) dt +
n
∑
(j)
𝜎 (i,j) (X (i) , t) dWt ,
i = 1, 2, . . . , m
j=1
(j)
where dWt(i) dWt = 𝜌ij dt, 𝜌ij ∈ (−1, 1) for i ≠ j, 𝜌ii = 1, i, j = 1, 2, . . . , n. By denoting
[
]T
Xt = Xt(1) , Xt(2) , . . . , Xt(m) , and for t ∈ [0, T], T > 0 and by conditioning Xt = x where
]T
[
x = x(1) , x(2) , . . . , x(m) , the m-dimensional random variable XT having a transition
]T
[
probability density p(x, t; y, T) in the m-dimensional variable y = y(1) , y(2) , . . . , y(m)
satisfies
)
( n n
m m
𝜕2
𝜕
1 ∑∑ ∑∑
(i,k) (i)
(j,l) (j)
𝜌kl 𝜎 (x , t)𝜎 (x , t)
p(x, t; y, T)
p(x, t; y, T) +
𝜕t
2 i=1 j=1 k=1 l=1
𝜕x(i) 𝜕x(j)
+
m
∑
𝜇 (i) (x(i) , t)
i=1
𝜕
p(x, t; y, T) = 0.
𝜕x(i)
3.2
3.2.1
PROBLEMS AND SOLUTIONS
Itō Calculus
1. Itō Integral. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener
process. Let the Itō integral of Wt dWt be defined as the following limit
n−1
∑
t
I(t) =
∫0
Ws dWs = lim
n→∞
(
)
Wti Wti+1 − Wti
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t for n ∈ ℕ.
Show that the quadratic variation of Wt is
⟨W, W⟩t = lim
n→∞
and hence
I(t) =
n−1 (
∑
)2
Wti+1 − Wti
=t
i=0
1( 2 )
Wt − t .
2
Finally, show that the Itō integral is a martingale.
Solution: For the first part of the solution, see Problem 2.2.6.1 (page 89).
)2
n−1 (
∑
Wti+1 − Wti = t and by expanding,
Given lim
n→∞ i=0
I(t) = lim
n→∞
n−1
∑
i=0
(
)
Wti Wti+1 − Wti
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{ (
)
)2 }
(
1
1
2
2
= lim
Wti+1 − Wti −
Wti+1 − Wti
n→∞
2
2
i=0
[
(
) ]
1
lim Wt2n − W02 − t
=
2 n→∞
1( 2 )
=
Wt − t .
2
n−1
∑
To show that I(t) is a martingale, see Problem 2.2.3.2 (page 72).
t
N.B. Without going through first principles, we can also show that ∫0 Ws dWs =
)
1( 2
1 2
Wt − t by using Itō’s formula on Xt = 2 Wt , where
2
dXt =
2
𝜕Xt
𝜕Xt
1 𝜕 Xt
dt +
dWt +
dWt2 + . . .
𝜕t
𝜕Wt
2 𝜕Wt2
1
= Wt dWt + dt.
2
Taking integrals,
t
∫0
t
dXs =
∫0
t
Xt − X0 =
t
and since W0 = 0, so
∫0
Ws dWs =
∫0
t
Ws dWs +
1
ds
2 ∫0
1
Ws dWs + t
2
1( 2 )
Wt − t .
2
◽
2. Stratonovich Integral. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process. Let the Stratonovich integral of Wt ∘ dWt be defined by the following
limit
n−1
t
(
)(
)
∑
1
Ws ∘ dWs = lim
Wti+1 + Wti Wti+1 − Wti
S(t) =
n→∞
∫0
2
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < . . . < tn−1 < tn = t, n ∈ ℕ.
Show that S(t) = 12 Wt2 and show also that the Stratonovich integral is not a martingale.
Solution: Expanding,
S(t) = lim
n→∞
= lim
n→∞
n−1
(
∑
1
i=0
2
Wti+1 + Wti
n−1
(
∑
1
i=0
(
2
Wt2i+1 − Wt2i
1
Wt2n − Wt20
n→∞ 2
1
= Wt2 .
2
= lim
)
)(
)
Wti+1 − Wti
)
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u
Let S(u) =
we have
∫0
Ws ∘ dWs , u < t and under the filtration ℱu and because Wt − Wu ⟂
⟂ ℱu ,
(
)
1
|
𝔼(S(t)|ℱu ) = 𝔼 Wt2 | ℱu
|
2
[(
)2 | ]
1
= 𝔼 Wt − Wu + Wu | ℱu
|
2
[
[
)2 | ]
(
)| ] 1 ( | )
1 (
= 𝔼 Wt − Wu | ℱu + 𝔼 Wu Wt − Wu | ℱu + 𝔼 Wu2 | ℱu
|
|
|
2
2
1
1
= (t − u) + Wu2
2
2
1 2
≠ Wu .
2
Therefore, S(t) is not a martingale.
◽
3. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Let the integral of Wt ∗ dWt be defined by the following limit
n−1
∑
t
J(t) =
Ws ∗ dWs = lim
∫0
n→∞
(
)
Wti+1 Wti+1 − Wti
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Show that the quadratic variation of Wt is
⟨W, W⟩t = lim
n→∞
and hence
J(t) =
n−1 (
∑
)2
Wti+1 − Wti
=t
i=0
1( 2 )
Wt + t .
2
Finally, show also that the integral is not a martingale.
n−1 (
Solution: To show that lim
By expanding,
∑
n→∞ i=0
J(t) = lim
n→∞
n−1
∑
)2
Wti+1 − Wti
= t, see Problem 2.2.6.1 (page 89).
(
)
Wti+1 Wti+1 − Wti
i=0
n−1 {
(
(
)
)2 }
1
1
Wt2i+1 − Wt2i +
Wti+1 − Wti
n→∞
2
2
i=0
[
(
) ]
1
lim Wt2n − W02 + t
=
2 n→∞
1( 2 )
=
Wt + t .
2
= lim
∑
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105
To show that J(t) is not a martingale, we note from Problem 3.2.1.2 (page 103) that under
the filtration ℱu , u < t,
( (
)
)
1
|
𝔼(J(t)|ℱu ) = 𝔼
Wt2 + t |ℱu
|
2
(
)
1
1
|
= t + 𝔼 Wt2 |ℱu
|
2
2
1
1
1
= t + (t − u) + Wu2
2
2
2
)
(
1
≠
Wu2 + u .
2
Therefore, J(t) is not a martingale.
◽
4. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
We define the Stratonovich integral of f (Wt , t) ∘ dWt as the following limit
t
f (Ws , s) ∘ dWs = lim
∫0
n→∞
n−1
∑
f
(W + W
ti
ti+1
2
i=0
, ti
)(
)
Wti+1 − Wti
and the Itō integral of f (Wt , t) dWt as
t
∫0
f (Ws , s) dWs = lim
n→∞
n−1
∑
f (Wti , ti )(Wti+1 − Wti )
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Show that
t
∫0
f (Ws , s) ∘ dWs =
t
t
𝜕f (Ws , s)
1
ds +
f (Ws , s) dWs .
∫0
2 ∫0
𝜕Ws
Solution: To prove this result we consider the difference between the two stochastic integrals and using the mean value theorem,
t
∫0
t
f (Ws , s) ∘ dWs −
= lim
n→∞
n−1
∑
i=0
f
∫0
(W + W
ti
ti+1
2
f (Ws , s) dWs
, ti
)(
)
Wti+1 − Wti − lim
n→∞
n−1
∑
)
(
f (Wti , ti ) Wti+1 − Wti
i=0
)
](
n−1 [ ( W + W
)
∑
ti
ti+1
= lim
f
, ti − f (Wti , ti ) Wti+1 − Wti
n→∞
2
i=0
[ (
](
)
Wti+1 − Wti+1 )
f Wti +
, ti − f (Wti , ti ) Wti+1 − Wti
n→∞
2
i=0
[
]
n−1
2
(
)2
(
)3
∑
1 𝜕f (Wti , ti )
1 𝜕 f (Wti , ti )
Wti+1 − Wti +
Wti+1 − Wti + . . .
= lim
n→∞
2 𝜕Wti
4 𝜕Wt2
i=0
i
= lim
n−1
∑
4:19 P.M.
Page 105
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106
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3.2.1
∑
n−1
since lim
n→∞ i=0
4:19 P.M.
Itō Calculus
(Wti+1 − Wti )p = 0 for p ≥ 3 and hence, for a simple process g(Wt , t),
|
n−1
(
)p ||
∑
|
| lim
g(Wti , ti ) Wti+1 − Wti ||
|n→∞
|
|
i=0
|
|
|∑
n−1 (
)p ||
|
|
Wti+1 − Wti ||
≤ lim max |g(Wtk , tk )| |
n→∞ 0≤k≤n−1
|
| i=0
|
|
= 0, p ≥ 3
or
lim
n−1
∑
n→∞
)p
(
g(Wti , ti ) Wti+1 − Wti = 0,
p ≥ 3.
i=0
Therefore,
t
t
f (Ws , s) ∘ dWs −
∫0
= lim
n→∞
= lim
n→∞
= lim
n→∞
=
1
2 ∫0
f (Ws , s) dWs
∫0
n−1 𝜕f (W , t ) (
∑
ti i
1
i=0
)2
Wti+1 − Wti
𝜕Wti
2
n−1 𝜕f (W
(
∑
ti , ti )
1
i=0
)
ti+1−ti + lim
𝜕Wti
2
n→∞
n−1 𝜕f (W , t ) (
∑
ti i
1
2 𝜕Wti
(
)
𝜕f Ws , s
[(
n−1 𝜕f (W
∑
ti , ti )
1
i=0
2
𝜕Wti
)2 (
)]
Wti+1 −Wti − ti+1 − ti
)
ti+1 − ti
i=0
t
𝜕Ws
ds
since, from Problem 2.2.6.1 (page 89),
lim
n→∞
[(
n−1 𝜕f (W
∑
ti , ti )
1
i=0
2
𝜕Wti
)2
Wti+1 − Wti
(
)]
− ti+1 − ti
[ n−1
]
n−1 (
(
)2 ∑
)
| 1 𝜕f (Wt , tk ) | ∑
|
|
k
Wti+1 − Wti −
ti+1 − ti
≤ lim max |
|
n→∞ 0≤k≤n−1 | 2
𝜕Wtk || i=0
i=0
|
= 0.
Thus,
t
∫0
f (Ws , s) ∘ dWs =
t
t
𝜕f (Ws , s)
1
ds +
f (Ws , s) dWs .
∫0
2 ∫0
𝜕Ws
◽
Page 106
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Itō Calculus
107
5. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process
such that ℱt is the associated filtration. The Itō integral with respect to the standard Wiener
process can be defined as
t
It =
f (Ws , s) dWs = lim
∫0
n→∞
n−1
∑
)
(
f (Wti , ti ) Wti+1 − Wti
i=0
where f is a simple process and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Show that the path of It is continuous and that it is also ℱt measurable for all t.
Solution: Given that Wti , ti = it∕n, n ∈ ℕ is both continuous and ℱti measurable, then
for a simple process, f ,
t
It =
f (Ws , s) dWs = lim
∫0
n→∞
n−1
∑
)
(
f (Wti , ti ) Wti+1 − Wti .
i=0
The path of the Itō integral is also continuous and ℱt measurable for all t.
◽
6. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process
such that ℱt is the associated filtration. The Itō integrals with respect to the standard
Wiener process can be defined as
t
It =
f (Ws , s) dWs = lim
∫0
n→∞
and
t
Jt =
g(Ws , s) dWs = lim
∫0
n→∞
n−1
∑
(
)
f (Wti , ti ) Wti+1 − Wti
i=0
n−1
∑
)
(
g(Wti , ti ) Wti+1 − Wti
i=0
where f and g are simple processes and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t,
n ∈ ℕ. Show that
t[
]
I t ± Jt =
f (Ws , s) ± g(Ws , s) dWs
∫0
and for constant c,
t
cIt =
∫0
t
cf (Ws , s) dWs ,
cJt =
∫0
cg(Ws , s) dWs .
Solution: Using the sum rule in integration,
t
I t ± Jt =
∫0
= lim
n→∞
t
f (Ws , s) dWs ±
n−1
∑
i=0
∫0
g(Ws , s) dWs
n−1
(
(
)
)
∑
f (Wti , ti ) Wti+1 − Wti ± lim
g(Wti , ti ) Wti+1 − Wti
n→∞
i=0
4:19 P.M.
Page 107
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108
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3.2.1
n−1 [
∑
= lim
n→∞
t
=
Itō Calculus
)
](
f (Wti , ti ) ± g(Wti , ti )
4:19 P.M.
Wti+1 − Wti
i=0
[
∫0
]
f (Ws , s) ± g(Ws , s) dWs .
For constant c,
t
cIt = c
f (Ws , s) dWs
∫0
= c lim
n−1
∑
n→∞
= lim
i=0
n−1
∑
n→∞
(
)
f (Wti , ti ) Wti+1 − Wti
(
)
c f (Wti , ti ) Wti+1 − Wti
i=0
t
=
∫0
c f (Ws , s) dWs .
t
The same steps can be applied to show that cJt =
∫0
cg(Ws , s) dWs .
◽
7. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process
such that ℱt is the associated filtration. The stochastic Itō integral with respect to the
standard Wiener process can be defined as
t
It =
∫0
f (Ws , s) dWs = lim
n−1
∑
n→∞
(
)
f (Wti , ti ) Wti+1 − Wti
i=0
where f is a simple function and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Using the properties of the standard Wiener process, show that It is a martingale.
Solution: Given that Wt is a martingale, we note the following:
(a) Under the filtration ℱu , u < t, by definition
t
u
f (Ws , s) dWs =
∫0
t
f (Ws , s) dWs +
∫0
= lim
m−1
∑
n→∞
f (Ws , s) dWs = lim
∑
n→∞
i=0
]
(
)
f (Wti , ti ) Wti+1 − Wti
i=m
m−1
u
∫0
𝔼(Iu |ℱu ) = Iu .
(
)
f (Wti , ti ) Wti+1 − Wti
[ n−1
∑
n→∞
Iu =
f (Ws , s) dWs
i=0
+ lim
where
∫u
)
(
f (Wti , ti ) Wti+1 − Wti ,
m<n−1
and
Page 108
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Itō Calculus
109
Finally, because {Wt ∶ t ≥ 0} is a martingale we have
]
[
m−1
)||
(
∑
|
f (Wti , ti ) Wti+1 − Wti | ℱu
𝔼(It |ℱu ) = 𝔼 lim
n→∞
|
i=0
|
[
[ n−1
]
]
(
) ||
∑
f (Wti , ti ) Wti+1 − Wti || ℱu
+𝔼 lim
n→∞
|
i=m
|
[ n−1
]
|
(
)|
∑
f (Wti , ti ) Wti+1 − Wti || ℱu
= Iu + lim 𝔼
n→∞
|
i=m
|
n−1 [
)
]
(
∑
|
𝔼 f (Wti , ti ) Wti+1 − Wti |ℱu
= Iu + lim
|
n→∞
i=m
n−1 [ [
(
)
]
]
∑
|
|
𝔼 𝔼 f (Wti , ti ) Wti+1 − Wti |ℱti |ℱu
|
|
n→∞
= Iu + lim
i=m
n−1 [
)
]
(
∑
|
= Iu + lim
𝔼 f (Wti , ti ) Wti − Wti |ℱu
|
n→∞
i=m
= Iu .
(b) Assuming | f (Wt , t)| < ∞, we have
|∑
)||
(
| n−1
|It | = lim || f (Wti , ti ) Wti+1 − Wti ||
n→∞ |
|
|
| i=0
n−1
(
)
∑|
|
|f (Wt , ti ) Wt − Wt |
≤ lim
|
|
i
i+1
i
n→∞
|
i=0 |
[
]
n−1
|
|
|∑|
max |W
≤ lim
− Wtk |
| f (Wti , ti )| .
|
|
|
n→∞ 0≤k≤n−1 | tk+1
i=0
|
|
Because Wt is continuous we have lim max |Wtk+1 − Wtk | = 0, therefore we can
|
n→∞ 0≤k≤n−1 |
deduce that 𝔼(|It |) < ∞.
(c) Since It is a function of Wt , hence it is ℱt -adapted.
From the results of (a)–(c) we have shown that It is a martingale.
N.B. From the above result we can easily deduce that if {Mt }t≥0 is a martingale with
continuous sample paths and by defining the following stochastic Itō integral:
t
It =
∫0
f (Ms , s) dMs = lim
n→∞
n−1
∑
)
(
f (Mti , ti ) Mti+1 − Mti
i=0
where f is a simple process and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ,
then It is a martingale.
◽
4:19 P.M.
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3.2.1
4:19 P.M.
Itō Calculus
8. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
The stochastic Itō integral with respect to a standard Wiener process can be defined as
t
s dWs = lim
∫0
n→∞
n−1 (
)
∑
ti Wti+1 − Wti
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ. Prove that
t
t
∫0
s dWs = tWt −
∫0
Ws ds.
Solution: By definition, W0 = 0 and we can expand
t
∫0
n−1 (
)
∑
s dWs = lim
ti Wti+1 − Wti
n→∞
= lim
n→∞
= lim
n→∞
i=0
n−1 ( (
)
)
∑
ti Wti+1 − Wti + ti+1 Wti+1 − ti+1 Wti+1
i=0
n−1 (
∑
)
)
(
ti+1 Wti+1 − ti Wti − ti+1 − ti Wti+1
i=0
n−1 (
= lim
n→∞
∑
)
ti+1 Wti+1 − ti Wti − lim
n→∞
i=0
= tWt − lim
n−1
∑
n→∞
n−1
∑
(
)
Wti+1 ti+1 − ti
i=0
(
)
Wti+1 ti+1 − ti .
i=0
By definition,
t
Ws ds = lim
∫0
and to show
lim
n→∞
n−1
∑
n−1
∑
n→∞
(
)
Wti ti+1 − ti
i=0
n−1
∑
(
)
(
)
Wti+1 ti+1 − ti − lim
Wti ti+1 − ti = 0
n→∞
i=0
i=0
we note from Problem 2.2.6.2 (page 90) that
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
ti+1 − ti = 0.
i=0
Therefore, we can deduce that
lim
n→∞
n−1
∑
i=0
(
)
Wti+1 ti+1 − ti =
t
∫0
Ws ds
Page 110
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Itō Calculus
111
and hence
t
∫0
t
s dWs = tWt −
∫0
Ws ds.
◽
9. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
The stochastic Itō integral with respect to a standard Wiener process can be defined as
n−1
∑
t
∫0
Ws2
dWs = lim
n→∞
(
)
Wt2i Wti+1 − Wti
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ. Prove that
t
∫0
t
1 3
W ds.
W −
3 t ∫0 s
Ws2 dWs =
Solution: By definition,
n−1
n−1
(
)
)
(
)2
(
∑
∑
1
Wt2i Wti+1 − Wti = lim
Wti Wti+1 − Wti
Wt3i+1 − Wt3i − lim
n→∞
n→∞
n→∞
3
i=0
i=0
i=0
lim
n−1
∑
− lim
n→∞
= lim
n→∞
n−1 (
∑
1
i=0
3
)3
Wti+1 − Wti
n−1
(
∑
1
i=0
− lim
n→∞
− lim
n→∞
n−1
)
∑
(
)
Wt3i+1 − Wt3i − lim
Wti ti+1 − ti
3
n−1
∑
n→∞
[(
Wti
)2
Wti+1 − Wti
i=0
(
− ti+1 − ti
i=0
n−1 (
∑
1
i=0
3
)3
Wti+1 − Wti
t
1
= Wt3 −
W ds
∫0 s
3
since from Problems 2.2.6.1 (page 89) and 2.2.6.4 (page 92),
[(
]|
|
n−1
)2
∑
|
|
| lim
Wti Wti+1 − Wti − (ti+1 − ti ) ||
|n→∞
|
|
i=0
|
|
|
|∑
n−1 (
n−1
)2 ∑
(
)|
|
|
ti+1 − ti ||
≤ lim max |Wtk | |
Wti+1 − Wti −
n→∞ 0≤k≤n−1
|
| i=0
i=0
|
|
= lim max |Wtk ||t − t|
n→∞ 0≤k≤n−1
=0
]
)
4:19 P.M.
Page 111
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112
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3.2.1
4:19 P.M.
Itō Calculus
and
n−1 (
∑
n→∞
t
Therefore,
∫0
)3
Wti+1 − Wti
lim
= 0.
i=0
t
Ws2 dWs =
1 3
W −
W ds.
3 t ∫0 s
t
N.B. Instead of going through first principles, we can also show
t
∫0
Ws ds by applying Itō’s formula on Xt =
dXt =
∫0
Ws2 dWs =
1 3
W −
3 t
1 3
W , where
3 t
2
𝜕Xt
𝜕Xt
1 𝜕 Xt
dWt +
dWt2 + . . .
dt +
𝜕t
𝜕Wt
2 𝜕Wt2
= Wt2 dWt + Wt dt.
Taking integrals,
t
∫0
t
dXs =
t
Ws2 dWs +
∫0
t
Xt − X 0 =
∫0
t
Ws2 dWs +
t
and since W0 = 0, therefore
∫0
Ws ds
∫0
∫0
Ws ds
t
Ws2 dWs =
1 3
W ds.
W −
3 t ∫0 s
◽
10. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
The stochastic Itō integral with respect to a standard Wiener process can be defined as
t
It =
∫0
f (Ws , s) dWs = lim
n→∞
n−1
∑
)
(
f (Wti , ti ) Wti+1 − Wti
i=0
where f is a simple process and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Using the properties of a standard Wiener process, show that
[
𝔼
t
]
f (Ws , s) dWs = 0
∫0
and deduce that if {Mt ∶ t ≥ 0} is a martingale then
[
𝔼
where g is a simple process.
t
∫0
]
g(Ms , s) dMs = 0
Page 112
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Itō Calculus
4:19 P.M.
113
Solution: By definition,
[
𝔼
]
n−1 [
(
)]
∑
f (Ws , s) dWs = lim
𝔼 f (Wti , ti ) Wti+1 − Wti
t
∫0
n→∞
i=0
)
]]
(
∑ [ [
|
𝔼 𝔼 f (Wti , ti ) Wti+1 − Wti |ℱti
|
n→∞
n−1
= lim
i=0
= lim
n→∞
n−1 [
(
)]
∑
𝔼 f (Wti , ti ) Wti − Wti
i=0
=0
)
(
|
since {Wt ∶ t ≥ 0} is a martingale and therefore 𝔼 Wti+1 |ℱti = Wti .
|
Using the same steps as described above, if {Mt ∶ t ≥ 0} is a martingale then
[
𝔼
]
t
∫0
g(Ms , s) dMs = 0
where g is a simple process.
◽
11. Itō Isometry. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process. The stochastic Itō integral with respect to a standard Wiener process can
be defined as
n−1
∑
t
It =
∫0
f (Ws , s) dWs = lim
n→∞
)
(
f (Wti , ti ) Wti+1 − Wti
i=0
where f is a simple process and ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ.
Using the properties of a standard Wiener process, show that
[(
]
[ t
)2 ]
t
2
f (Ws , s) dWs
f (Ws , s) ds .
=𝔼
𝔼
∫0
∫0
Solution: Using the property of independent increments of a standard Wiener process as
well as the martingale properties of Wt and Wt2 − t, we note that
[(
[
) ]
]
2
t
𝔼
∫0
{
⎡
= 𝔼⎢
⎢
⎣
= lim
n→∞
lim
n→∞
n−1
∑
i=0
n−1
∑
t
−𝔼
f (Ws , s) dWs
∫0
f (Ws , s)2 ds
}2
)
(
f (Wti , ti ) Wti+1 − Wti
i=0
[
(
𝔼 f (Wti , ti ) Wti+1 − Wti
2
)2 ]
⎤
⎥−𝔼
⎥
⎦
− lim
n→∞
[
lim
n→∞
n−1
∑
f (Wti , ti )
2
]
(
ti+1 − ti
i=0
n−1 [
∑
(
)]
𝔼 f (Wti , ti )2 ti+1 − ti
i=0
)
Page 113
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114
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3.2.1
= lim
n→∞
n−1
∑
[
𝔼 f (Wti , ti )
i=0
2
{(
)2
Wti+1 − Wti
[ [
4:19 P.M.
Itō Calculus
}]
)
(
− ti+1 − ti
]]
}
(
) ||
𝔼 𝔼 f (Wti , ti )
= lim
Wti+1 − Wti − ti+1 − ti | ℱti
|
n→∞
i=0
|
n−1 [ [
{
}| ]]
∑
2
2
2
= lim
𝔼 𝔼 f (Wti , ti ) Wti+1 − 2Wti+1 Wti + Wti − ti+1 + ti || ℱti
n→∞
|
i=0
]
n−1 { [ [
[
(
)|
]
∑
|
𝔼 𝔼 f (Wti , ti )2 Wt2i+1 − ti+1 || ℱti − 2𝔼 f (Wti , ti )2 Wti+1 Wti |ℱti
= lim
|
n→∞
|
i=0
]]}
[
(
)|
+ 𝔼 f (Wti , ti )2 Wt2i + ti || ℱti
|
n−1 [
(
)]
∑
𝔼 f (Wti , ti )2 Wt2i − ti − 2Wt2i + Wt2i + ti
= lim
n−1
∑
n→∞
2
{(
)2
i=0
= 0.
[(
Therefore, 𝔼
)2 ]
t
∫0
f (Ws , s) dWs
[
=𝔼
]
f (Ws , s)2 ds .
t
∫0
◽
12. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
The stochastic Itō integral with respect to a standard Wiener process can be defined as
t
It =
∫0
f (Ws , s) dWs
where f is a simple process. Show that the Itō integral has quadratic variation process
⟨I, I⟩t given by
t
⟨I, I⟩t =
∫0
Solution: By definition,
⟨I, I⟩t = lim
m→∞
f (Ws , s)2 ds.
m−1
∑(
)2
Itk+1 − Itk
k=0
where tk = kt∕m, 0 = t0 < t1 < t2 < . . . < tm−1 < tm = t, m ∈ ℕ. We first concentrate on
one of the subintervals [tk , tk+1 ) on which f (Ws , s) = f (Wtk , tk ) is a constant value for all
s ∈ [tk , t+1 ). Partitioning
tk = s0 < s1 < . . . < sn = tk+1 ,
we can write
si+1
Isi+1 − Isi =
∫si
(
)
f (Wtk , tk ) dWu = f (Wtk , tk ) Wsi+1 − Wsi .
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Therefore,
(Itk+1 − Itk )2 = lim
n→∞
= lim
n→∞
n−1 (
∑
)2
Isi+1 − Isi
i=0
(
)]2
f (Wtk , tk ) Wsi+1 − Wsi
n−1 [
∑
i=0
= f (Wtk , tk ) lim
2
n→∞
= f (Wtk , tk )
2
(
n−1 (
∑
)2
Wsi+1 − Wsi
i=0
tk+1 − tk
)
)2
∑ (
W
since limn→∞ n−1
−
W
converges to the quadratic variation of a standard
s
s
i=0
i+1
i
Wiener process over [tk , tk+1 ) which is tk+1 − tk . Therefore,
(
)
(Itk+1 − Itk )2 = f (Wtk , tk )2 tk+1 − tk =
tk+1
∫tk
f (Ws , s)2 ds
where f (Ws , s) is a constant value for all s ∈ [tk , tk+1 ).
Finally, to obtain the quadratic variation of the Itō integral It ,
⟨I, I⟩t = lim
m−1 (
∑
m→∞
)2
Itk+1 − Itk
∑
m−1
= lim
m→∞
k=0
(
)
f (Wtk , tk )2 tk+1 − tk =
k=0
t
∫0
f (Ws , s)2 ds.
N.B. In differential form we can write d⟨I, I⟩t = dIt dIt = f (Wt , t)2 dt. By comparing the
results of the quadratic variation ⟨I, I⟩t and 𝔼(It2 ), we can see that the former is computed
path-by-path and hence the result is random. In contrast, the variance of the Itō integral,
Var(It ) = 𝔼(It2 ) = 𝔼(⟨I, I⟩t ), is the mean value of all possible paths of the quadratic variation and hence is non-random.
◽
13. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Using integration by parts, show that
t
∫0
t
Ws ds =
∫0
(t − s) dWs
( 3)
t
Ws ds ∼ 𝒩 0,
.
∫0
3
t
and prove that
Is
a standard Wiener process?
⎧0
t=0
⎪√
Bt = ⎨ 3 t
Ws ds t > 0
⎪
⎩ t ∫0
4:19 P.M.
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Itō Calculus
Solution: Using integration by parts,
( )
( )
du
d𝑣
ds = u𝑣 −
𝑣
ds.
u
∫
∫
ds
ds
We set u = Ws and d𝑣∕ds = 1. Therefore,
t
t
|t
Ws ds = sWs | −
s dWs
|0 ∫ 0
∫0
t
= tWt −
s dWs
∫0
t
=
∫0
(t − s) dWs .
Taking expectations,
(
𝔼
[(
𝔼
)
t
∫0
Ws ds
=𝔼
)2 ]
t
∫0
(
(
=𝔼
Ws ds
)
t
∫0
)
t
(t − s)2 ds
∫0
=0
(t − s) dWs
=
t3
.
3
t
To show that
∫0
Ws ds follows a normal distribution, let
t
Mt =
t
Ws ds =
∫0
∫0
(t − s) dWs
and using the properties of the Itō integral (see Problems 3.2.1.7, page 108 and 3.2.1.12,
page 114), we can deduce that
t
Mt is a martingale
so that dMt ⋅ dMt = t2 dt.
By defining
and ⟨M, M⟩t =
( )
3
𝜃Mt − 21 𝜃 2 t3
Zt = e
,
∫0
(t − s)2 ds =
t3
3
𝜃∈ℝ
then expanding dZt using Taylor’s theorem and applying Itō’s lemma, we have
dZt =
𝜕Zt
𝜕Zt
1 𝜕2f
dt +
dMt +
(dMt )2 + . . .
𝜕t
𝜕Mt
2 𝜕Mt2
1
= − 𝜃 2 t2 ZT dt + 𝜃Zt dMt +
2
1
= − 𝜃 2 t2 ZT dt + 𝜃Zt dMt +
2
= 𝜃Zt dMt .
1 2
𝜃 Zt (dMt )2 + . . .
2
1 22
𝜃 t Zt dt
2
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117
Taking integrals, we can express
t
∫0
t
dZu =
∫0
𝜃Zu dMu
t
Z t − Z0 = 𝜃
∫0
Zu dMu
t
Zt = 1 + 𝜃
∫0
Zu dMu .
Finally, by taking expectations and knowing that Mt is a martingale, we have
(
𝔼(Zt ) = 1 + 𝜃𝔼
)
t
∫0
Zu dMu
and hence
(
𝔼(e
𝜃Mt
)=e
1 2
𝜃
2
t3
3
=1
)
which is the moment generating function
( 3 ) of a normal distribution with mean zero and
t
t
t3
Ws ds ∼ 𝒩 0,
.
variance . Thus,
∫
3
3
0
Even though Bt ∼ 𝒩(0, t) for t > 0, Bt is not a standard Wiener process since for t, u > 0,
𝔼(Bt+u − Bt ) = 𝔼(Bt+u ) − 𝔼(Bt )
(√
)
(√
)
t+u
3
3 t
=𝔼
Ws ds − 𝔼
W ds
t + u ∫0
t ∫0 s
=0
and using the result of Problem 2.2.1.13 (page 66),
Var(Bt+u − Bt ) = Var(Bt+u ) + Var(Bt ) − 2Cov(Bt+u , Bt )
(√
)
(√
)
t+u
3
3 t
= Var
Ws ds + Var
W ds
t + u ∫0
t ∫0 s
)
(√
√
t+u
3
3 t
Ws ds,
W ds
−2Cov
t + u ∫0
t ∫0 s
]
[
1 3 1 2
6
t + ut
=t+u+t−
t(t + u) 3
2
= 2t + u −
u2
t+u
≠u
=
t(2t + 3u)
t+u
4:19 P.M.
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which shows that Bt does not have the stationary increment property. Therefore, Bt is not
a standard Wiener process.
◽
14. Generalised Itō Integral. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a
standard Wiener process. Given that f is a simple process, show
]
[
t
t
𝜕f
𝜕f
𝜕2f
1
f (Ws , s) dWs = Wt f (Wt , t) −
(W , s) + Ws
(W , s) ds
Ws (Ws , s) +
∫0
∫0
𝜕t
𝜕Wt s
2 𝜕Wt2 s
t
−
and
t
∫0
∫0
Ws
𝜕f
(W , s) dWs
𝜕Wt s
]
𝜕f
1 𝜕2f
f (Ws , s) ds = tf (Wt , t) −
s
(W , s) ds
(W , s) +
∫0
𝜕t s
2 𝜕Wt2 s
t
t
−
∫0
s
[
𝜕f
(W , s) dWs .
𝜕Wt s
Solution: For the first result, using Taylor’s theorem on d(Wt f (Wt , t)) and subsequently
applying Itō’s formula we have
[
]
𝜕f
𝜕f
(Wt , t) dWt
d(Wt f (Wt , t)) = Wt (Wt , t) dt + f (Wt , t) + Wt
𝜕t
𝜕Wt
[
]
𝜕2f
𝜕f
1 𝜕f
+
(W , t) +
(W , t) + Wt
(Wt , t) dWt2 + . . .
2 𝜕Wt t
𝜕Wt t
𝜕Wt2
[
]
𝜕f
𝜕f
(Wt , t) dWt
= Wt (Wt , t) dt + f (Wt , t) + Wt
𝜕t
𝜕Wt
[
]
𝜕f
𝜕2f
1
2
+
(W , t) + Wt
(Wt , t) dt
2
𝜕Wt t
𝜕Wt2
]
[
𝜕f
𝜕2f
𝜕f
1
(W , t) + Wt
(W , t) dt
= Wt (Wt , t) +
𝜕t
𝜕Wt t
2 𝜕Wt2 t
]
[
𝜕f
(Wt , t) dWt .
+ f (Wt , t) + Wt
𝜕Wt
Taking integrals from 0 to t,
]
𝜕f
𝜕f
𝜕2 f
1
d(Ws f (Ws , s)) =
(W , s) + Ws
(W , s) ds
Ws (Ws , s) +
∫0
∫0
𝜕t
𝜕Wt s
2 𝜕Wt2 s
]
t[
𝜕f
+
f (Ws , s) + Ws
(W , s) dWs
∫0
𝜕Wt s
t
t
[
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119
and rearranging the terms, finally
t
∫0
t
f (Ws , s) dWs = Wt f (Wt , t) −
t
−
∫0
Ws
]
𝜕f
𝜕f
𝜕2f
1
(W , s) + Ws
(W , s) ds
Ws (Ws , s) +
𝜕t
𝜕Wt s
2 𝜕Wt2 s
[
∫0
𝜕f
(W , s) dWs
𝜕Wt s
since W0 = 0.
As for the second result, from Taylor’s theorem and Itō’s formula
d(tf (Wt , t)) = f (Wt , t) dt + t
𝜕f
𝜕f
1 𝜕2f
(Wt , t) dWt + t
(W , t) dWt2 + . . .
(Wt , t) dt + t
𝜕t
𝜕Wt
2 𝜕Wt2 t
𝜕f
𝜕f
1 𝜕2f
(Wt , t) dWt + t
(W , t) dt
(Wt , t) dt + t
𝜕t
𝜕Wt
2 𝜕Wt2 t
[
]
𝜕f
𝜕f
1 𝜕2f
= f (Wt , t) + t (Wt , t) + t
(Wt , t) dt + t
(W , t) dWt .
2
𝜕t
2 𝜕Wt
𝜕Wt t
= f (Wt , t) dt + t
Taking integrals from 0 to t we have
t
∫0
t
d(sf (Ws , s)) =
[
∫0
]
𝜕f
1 𝜕2f
(W , s) ds
f (Ws , s) + s (Ws , s) + s
𝜕t
2 𝜕Wt2 s
t
+
s
∫0
𝜕f
(W , s) dWs
𝜕Wt s
and hence
t
∫0
]
𝜕f
1 𝜕2f
(W , s) +
f (Ws , s) ds = tf (Wt , t) −
s
(W , s) ds
∫0
𝜕t s
2 𝜕Wt2 s
t
t
−
∫0
s
[
𝜕f
(W , s) dWs .
𝜕Wt s
◽
15. One-Dimensional Lévy Characterisation Theorem. Let (Ω, ℱ, ℙ) be a probability space
and let {Mt ∶ t ≥ 0} be a martingale with respect to the filtration ℱt , t ≥ 0. By assuming
M0 = 0, Mt has continuous sample paths whose quadratic variation
lim
n→∞
n−1 (
∑
)2
Mti+1 − Mti
=t
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ show that Mt is a standard
Wiener process.
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Itō Calculus
Solution: We first need to show that Mt ∼ 𝒩(0, t) or, using the moment generating func1 2
(
)
tion approach, we need to show that 𝔼 e𝜃Mt = e 2 𝜃t for a constant 𝜃.
1 2
Let f (Mt , t) = e𝜃Mt − 2 𝜃 t and since dMt ⋅ dMt = dt and (dt)𝜈 = 0, 𝜈 ≥ 2 from Itō’s formula,
𝜕f
𝜕f
1 𝜕2f
dMt +
(dMt )2 + . . .
dt +
𝜕t
𝜕Mt
2 𝜕Mt2
(
)
𝜕f
𝜕f
1 𝜕2f
dMt
+
=
dt +
𝜕t 2 𝜕Mt2
𝜕Mt
(
)
1
1
= − 𝜃 2 t + 𝜃 2 t f (Mt , t) dt + 𝜃f (Mt , t) dMt
2
2
= 𝜃f (Mt , t) dMt .
df (Mt , t) =
Taking integrals from 0 to t, and then taking expectations, we have
t
∫0
t
df (Ms , s) = 𝜃
f (Ms , s) dMs
∫0
t
f (Mt , t) − f (M0 , 0) = 𝜃
∫0
f (Ms , s) dMs
𝔼(f (Mt , t)) = 1 + 𝜃𝔼
[
t
∫0
]
f (Ms , s) dMs .
By definition of the stochastic Itō integral we can write
[
𝔼
t
∫0
]
n−1 [
(
)]
∑
f (Ms , s) dMs = lim
𝔼 f (Mti , ti ) Mti+1 − Mti
n→∞
i=0
)
]]
(
∑ [ [
|
𝔼 𝔼 f (Mti , ti ) Mti+1 − Mti |ℱti
|
n→∞
n−1
= lim
i=0
= lim
n→∞
n−1 [
(
)]
∑
𝔼 f (Mti , ti ) Mti − Mti
i=0
=0
since {Mt }t≥0 is a martingale and hence
𝔼(f (Mt , t)) = 1
or
1 2
(
)
𝔼 e𝜃Mt = e 2 𝜃 t
which is the moment generating function for the normal distribution with mean zero and
variance t. Therefore, Mt ∼ 𝒩(0, t).
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Since Mt is a martingale, for s < t
(
)
(
)
𝔼 Mt || ℱs = 𝔼 Mt − Ms + Ms || ℱs
)
(
)
(
= 𝔼 Mt − Ms || ℱs + 𝔼 Ms || ℱs
(
)
= 𝔼 Mt − Ms || ℱs + Ms
= Ms .
)
(
)
(
⟂ ℱs . So, we have
Therefore, 𝔼 Mt − Ms || ℱs = 𝔼 Mt − Ms = 0 and hence Mt − Ms ⟂
shown that Mt has the independent increment property.
Finally, to show that Mt has the stationary increment, for t > 0 and s > 0 we have
)
(
)
( )
(
𝔼 Mt+s − Mt = 𝔼 Mt+s − 𝔼 Mt = 0
and using the independent increment property of Mt
)
(
)
( )
(
)
(
Var Mt+s − Mt = Var Mt+s + Var Mt − 2Cov Mt+s , Mt
)
(
) ( )]
[ (
= t + s + t − 2 𝔼 Mt+s Mt − 𝔼 Mt+s 𝔼 Mt
(
)
= 2t + s − 2𝔼 Mt+s Mt
( (
)
)
= 2t + s − 2𝔼 Mt Mt+s − Mt + Mt2
)
( )
( ) (
= 2t + s − 2𝔼 Mt 𝔼 Mt+s − Mt − 2𝔼 Mt2
= 2t + s − 2t
= s.
Therefore, Mt+s − Mt ∼ 𝒩(0, s).
Because M0 = 0 and also Mt has continuous sample paths with independent and stationary
increments, so Mt is a standard Wiener process.
◽
16. Multi-Dimensional Lévy Characterisation Theorem. Let (Ω, ℱ, ℙ) be a probability space
and let {Mt(1) ∶ t ≥ 0}, {Mt(2) ∶ t ≥ 0}, . . . , {Mt(n) ∶ t ≥ 0} be martingales with respect to
the filtration ℱt , t ≥ 0. By assuming M0(i) = 0, Mt(i) has continuous sample paths whose
quadratic variation
m (
)2
∑
(i)
Mt(i)
lim
−
M
=t
tk
k+1
n→∞
k=0
(j)
and cross-variation between Mt(i) and Mt , i ≠ j, i, j = 1, 2, . . . , n
lim
m→∞
m (
∑
− Mt(i)
Mt(i)
k+1
k
)(
)
(j)
(j)
Mtk+1 − Mtk = 0
k=0
where tk = kt∕m, 0 = t0 < t1 < t2 < . . . < tm−1 < tm = t, m ∈ ℕ, show that Mt(1) ,
Mt(2) , . . . , Mt(n) are independent standard Wiener processes.
4:19 P.M.
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Solution: Following Problem 3.2.1.15 (page 119), we can easily prove that
Mt(1) , Mt(2) , . . . , Mt(n) are standard Wiener processes. In order to show Mt(1) , Mt(2) , . . . , Mt(n)
are mutually independent, we need to show the joint moment generating function of
Mt(1) , Mt(2) , . . . ,
Mt(n) is
(
)
𝔼 e𝜃
1
(1) M (1) +𝜃 (2) M (2) + ... 𝜃 (n) M (n)
t
t
t
= e 2 (𝜃
(1) )2 t
1
⋅ e 2 (𝜃
(2) )2 t
1
· · · e 2 (𝜃
(n) )2 t
for constants 𝜃 (1) , 𝜃 (2) , . . . , 𝜃 (n) .
(i) (i) 1 (i) 2
∏
Let f (Mt(1) , Mt(2) , . . . , Mt(n) , t) = ni=1 e𝜃 Mt − 2 (𝜃 ) t and since dMt(i) ⋅ dMt(i) = dt,
(j)
dMt(i) ⋅ dMt = 0, i, j = 1, 2, . . . , n, i ≠ j and (dt)𝜈 = 0, 𝜈 ≥ 2, from Itō’s formula
∑ 𝜕f
𝜕f
dt +
dMt(i)
(i)
𝜕t
𝜕M
i=1
n
df (Mt(1) , Mt(2) , . . . , Mt(n) , t) =
t
𝜕2f
1 ∑∑
(j)
dMt(i) dMt
2 i=1 j=1 𝜕M (i) 𝜕M (j)
n
+
n
t
t
n
∑
n
𝜕f
𝜕f
1 ∑ 𝜕2f
(i)
dM
+
(dMt(i )2
dt +
t
(i)
(i) 2
𝜕t
2
𝜕M
𝜕(M
)
i=1
i=1
t
t
(
)
n
n
∑ 𝜕f
𝜕f
1 ∑ 𝜕2f
=
dMt(i)
+
dt +
(i)
(i)
𝜕t 2 i=1 𝜕(M )2
i=1 𝜕M
=
t
t
∑ 𝜕f
n
=
=
dMt(i)
(i)
i=1 𝜕Mt
n
∑
𝜃 (i) f (Mt(1) , Mt(2) ,
i=1
. . . , Mt(n) , t) dMt(i)
since
]
[
n
n
n
𝜕f
1 ∑ 𝜕2f
1 ∑ (i) 2 1 ∑ (i) 2
+
= −
(𝜃 ) +
(𝜃 ) f (Mt(1) , Mt(2) , . . . , Mt(n) , t) = 0.
𝜕t 2 i=1 𝜕(M (i) )2
2 i=1
2 i=1
t
Integrating both sides from 0 to t and taking expectations, we have
t
∫0
df (Ms(1) , Ms(2) , . . . , Ms(n) , s) =
n
∑
i=1
f (Mt(1) , Mt(2) ,
...
t
∫0
, Mt(n) , t) = f (M0(1) , M0(2) ,
+
n
∑
i=1
𝔼[ f (Mt(1) , Mt(2) ,
...
𝜃 (i) f (Ms(1) , Ms(2) , . . . , Ms(n) , t) dMs(i)
, Mt(n) , t)] = 1
+
t
∫0
n
∑
i=1
. . . , M0(n) , 0)
𝜃 (i) f (Ms(1) , Ms(2) , . . . , Ms(n) , t) dMs(i)
[
𝜃 𝔼
(i)
t
∫0
]
f (Ms(1) , Ms(2) , . . . , Ms(n) , t) dMs(i) .
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123
Because Mt(i) is a martingale, we can easily show (see Problem 3.2.1.15, page 119) that
[
𝔼
t
∫0
]
(
)
f Ms(1) , Ms(2) , . . . , Ms(n) , t dMs(i) = 0
for i = 1, 2, . . . , n and hence
( (1) (1) (2) (2)
)
1 (1) 2
1 (2) 2
1 (n) 2
(n) (n)
𝔼 e𝜃 Mt +𝜃 Mt + ... 𝜃 Mt
= e 2 (𝜃 ) t ⋅ e 2 (𝜃 ) t · · · e 2 (𝜃 ) t
where the joint moment generating function of Mt(1) , Mt(2) , . . . , Mt(n) is a product of
moment generating functions of Mt(1) , Mt(2) , . . . , Mt(n) . Therefore, Mt(1) , Mt(2) , . . . , Mt(n)
are independent standard Wiener processes.
◽
3.2.2
One-Dimensional Diffusion Process
1. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Find the SDE for the random process Xt = Wtn , n ∈ ℤ+ .
Show that
t (
)
( ) 1
𝔼 Ws(n−2) ds
𝔼 Wtn = n(n − 1)
∫0
2
and using mathematical induction prove that
n
⎧ n!t 2
( n) ⎪ n ( n )
𝔼 Wt = ⎨ 2 2
!
2
⎪
⎩0
n = 2, 4, 6, . . .
n = 1, 3, 5, . . .
Solution: By expanding dXt using Taylor’s formula and applying Itō’s formula,
dXt =
2
𝜕Xt
𝜕Xt
1 𝜕 Xt
dWt +
dWt2 + . . .
dt +
𝜕t
𝜕Wt
2 𝜕Wt2
1
dWtn = nWt(n−1) dWt + n(n − 1)Wt(n−2) dt.
2
Taking integrals,
t
∫0
dWsn =
Wtn =
t
t
t
t
1
nWs(n−1) dWs + n(n − 1)
W (n−2) ds
∫0
∫0 s
2
1
nWs(n−1) dWs + n(n − 1)
W (n−2) ds.
∫0
∫0 s
2
Finally, by taking expectations,
t (
)
( ) 1
𝔼 Wtn = n(n − 1)
𝔼 Ws(n−2) ds.
∫0
2
4:19 P.M.
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To prove the final result, we will divide it into two sections, one for even numbers n = 2k,
k ∈ ℤ+ and another for odd numbers n = 2k + 1, k ∈ ℤ+ . We note that for n = 2 we have
( ) 2!t
=t
𝔼 Wt2 =
2
and because Wt ∼ 𝒩(0, t), the result is true for n = 2.
We assume that the result is true for n = 2k, k ∈ ℤ+ . That is
) (2k)!tk
(
𝔼 Wt2k = k .
2 k!
For n = 2(k + 1), k ∈ ℤ+ we have
t (
)
(
)
1
𝔼 Ws2k ds
𝔼 Wt2(k+1) = (2k + 2)(2k + 1)
∫0
2
t
(2k)!sk
1
= (2k + 2)(2k + 1)
ds
∫0 2k k!
2
=
(2k + 2)! t k
s ds
2k+1 k! ∫0
=
(2k + 2)!tk+1
2k+1 (k + 1)!
(2(k + 1))!t
=
2
2(k+1)
2
2(k+1)
2
.
(2(k + 1)∕2)!
Thus, the result is also true for n = 2(k + 1), k ∈ ℤ+ .
For n = 1, we have
( )
𝔼 Wt = 0
and because Wt ∼ 𝒩(0, t), the result is true for n = 1.
We assume the result is true for n = 2k + 1, k ∈ ℤ+ such that
)
(
𝔼 Wt2k+1 = 0.
For n = 2(k + 1) + 1, k ∈ ℤ+
)
t (
(
)
1
𝔼 Wt2(k+1)+1 = (2k + 3)(2k + 2)
𝔼 Ws2k+1 ds = 0
∫
2
0
and hence the result is also true for n = 2(k + 1) + 1, k ∈ ℤ+ . Therefore, by mathematical
induction
n
⎧ n!t 2
n = 2, 4, 6, . . .
(
)
( ) ⎪ n
𝔼 Wtn = ⎨ 2 2 n !
2
⎪
n = 1, 3, 5, . . .
⎩0
◽
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125
2. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
1 2
𝜃Wt − 2 𝜃 t
.
For a constant 𝜃 find the SDE for the random process
( 𝜃W X
)t = e
t
By writing the SDE in integral form calculate 𝔼 e
, the moment generating function
of a standard Wiener process.
Solution: Expanding dXt using Taylor’s theorem and applying Itō’s formula,
2
𝜕Xt
𝜕Xt
1 𝜕 Xt
dWt +
dWt2 + . . .
dt +
𝜕t
𝜕Wt
2 𝜕Wt2
(
)
𝜕Xt 1 𝜕 2 Xt
𝜕Xt
+
=
dt +
dWt
2
𝜕t
2 𝜕Wt
𝜕Wt
(
)
1
1
= − 𝜃 2 Xt + 𝜃 2 Xt dt + 𝜃Xt dWt
2
2
= 𝜃Xt dWt .
dXt =
Taking integrals, we have
t
t
dXs =
∫0
𝜃Xs dWs
∫0
t
Xt − X0 =
Since X0 = 1 and taking expectations,
( )
𝔼 Xt − 1 = 𝔼
so
( )
𝔼 Xt = 1
or
𝜃Xs dWs .
∫0
(
)
t
∫0
𝜃Xs dWs
=0
)
(
1 2
𝔼 e𝜃Wt − 2 𝜃 t = 1.
1 2
(
)
Therefore, 𝔼 e𝜃Wt = e 2 𝜃 t .
◽
3. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Consider the process
Zt = e𝜃Wt
where 𝜃 is a constant parameter.
Using Itō’s formula,
( find) an SDE for Zt .
By setting mt = 𝔼 e𝜃Wt show that the integrated SDE can be expressed as
dmt 1 2
− 𝜃 mt = 0.
dt
2
Given W0 = 0, solve the first-order ordinary differential equation to find mt .
4:19 P.M.
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Solution: Using Itō’s formula we expand Zt as
dZt =
2
𝜕Zt
1 𝜕 Zt
dWt +
(dWt )2 + . . .
𝜕Wt
2 𝜕Wt2
1
= 𝜃e𝜃Wt dWt + 𝜃 2 e𝜃Wt dt
2
1
= 𝜃Zt dWt + 𝜃 2 Zt dt.
2
Taking integrals and then expectations,
t
∫0
t
dZs =
t
𝜃Zs dWs +
∫0
t
Zt − Z0 =
1 2
𝜃 Zs ds
∫0 2
t
1 2
𝜃 Zs ds
∫0 2
)
( t
)
t
1 2
𝜃Zs dWs + 𝔼
𝜃 Zs ds
∫0 2
∫0
∫0
(
𝔼(Zt ) − 𝔼(Z0 ) = 𝔼
𝜃Zs dWs +
t
1 2
𝜃 𝔼(Zs ) ds
∫0 2
)
t
𝜃Zs dWs = 0. By differentiating the integral equation we have
𝔼(Zt ) − 1 =
(
where Z0 = 1 and 𝔼
∫0
t
d𝔼(Zt ) d
1 2
=
𝜃 𝔼(Zs ) ds
dt
dt ∫0 2
d𝔼(Zt ) 1 2
= 𝜃 𝔼(Zt )
dt
2
or
dmt 1 2
− 𝜃 mt = 0.
dt
2
1 2
1 2
Setting the integrating factor as I = e− ∫ 2 𝜃 dt = e− 2 𝜃 t and multiplying the differential
equation with I, we have
(
)
1 2
1 2
)
(
d
mt e− 2 𝜃 t = 0 or e− 2 𝜃 t 𝔼 e𝜃Wt = C
dt
(
)
where C is a constant. Since 𝔼 e𝜃W0 = 1, so C = 1 and hence we finally obtain
1 2
)
(
◽
𝔼 e𝜃Wt = e 2 𝜃 t .
t
4. Let Mt =
∫0
f (s) dWs and show that the SDE satisfied by
{
Xt = exp
t
1
𝜃Mt − 𝜃 2
f (s)2 ds
2 ∫0
is
dXt = 𝜃f (t)Xt dWt
}
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(
and show also that Mt ∼ 𝒩 0,
127
t
∫0
)
f (s)2 ds .
Solution: From Itō’s formula,
2
𝜕Xt
𝜕Xt
1 𝜕 Xt
dWt +
dWt2 + . . .
dt +
𝜕t
𝜕Wt
2 𝜕Wt2
)
(
(
𝜕X
𝜕Xt
𝜕Xt 𝜕Mt
1 𝜕
= t dt +
dWt +
⋅
𝜕t
𝜕Mt 𝜕Wt
2 𝜕Wt 𝜕Mt
[
(
)]
(
𝜕Xt 1 𝜕
𝜕Xt 𝜕Mt
𝜕Xt
=
dt +
⋅
⋅
+
𝜕t
2 𝜕Wt 𝜕Mt 𝜕Wt
𝜕Mt
dXt =
Since
)
𝜕Mt
dt
𝜕Wt
)
𝜕Mt
dWt .
𝜕Wt
⋅
{
}
t
𝜕Mt
𝜕Xt
1
2
= 𝜃 exp 𝜃Mt −
f (s) ds = 𝜃Xt ,
= f (t)
𝜕Mt
2 ∫0
𝜕Wt
(
)
𝜕Xt 𝜕Mt
𝜕Xt
𝜕
1
= 𝜃 2 f (t)2 Xt ,
⋅
= − 𝜃 2 f (t)2 Xt
𝜕Wt 𝜕Mt 𝜕Wt
𝜕t
2
so
dXt = 𝜃f (t)Xt dWt .
Writing in integral form and then taking expectations,
t
∫0
t
dXs =
𝜃f (s)Xs dWs
∫0
t
Xt − X0 =
∫0
(
( )
( )
𝔼 Xt − 𝔼 X0 = 𝔼
𝜃f (s)Xs dWs
)
t
𝜃f (s)Xs dWs
∫0
( )
( )
𝔼 Xt = 𝔼 X0 = 1.
=0
Therefore,
)
(
𝔼 e𝜃Mt = exp
(
t
1 2
f (s)2 ds
𝜃
2 ∫0
)
which is the moment generating function
( fort a normal
) distribution with mean zero and
t
2
2
f (s) ds. Hence Mt ∼ 𝒩 0,
f (s) ds .
variance
∫0
∫0
◽
5. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Suppose Xt follows the generalised SDE
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
where 𝜇 and 𝜎 are functions of Xt and t. Show that Xt is a martingale if 𝜇(Xt , t) = 0.
4:19 P.M.
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Solution: It suffices to show that under the filtration ℱs where s < t, for Xt to be a martingale
)
(
𝔼 Xt |ℱs = Xs .
By taking integrals of the SDE,
t
∫s
t
dX𝑣 =
t
𝜇(X𝑣 , t) d𝑣 +
∫s
t
Xt − Xs =
∫s
𝜎(X𝑣 , 𝑣) dW𝑣
∫s
t
𝜇(X𝑣 , t) d𝑣 +
∫s
𝜎(X𝑣 , 𝑣) dW𝑣 .
Taking expectations,
)
(
𝔼 Xt − X s = 𝔼
[
]
t
∫s
𝜇(X𝑣 , t) d𝑣
or
( )
( )
𝔼 Xt = 𝔼 Xs + 𝔼
[
]
t
∫s
𝜇(X𝑣 , t) d𝑣 .
Under the filtration ℱs , s < t
)
[
|
|
𝜇(X𝑣 , t) d𝑣| ℱs
|
∫s
|
[
]
t
|
|
𝜇(X𝑣 , t) d𝑣| ℱs .
= Xs + 𝔼
|
∫s
|
(
)
𝔼 Xt |ℱs = 𝔼 Xs |ℱs + 𝔼
(
t
Therefore, if 𝜇(Xt , t) = 0 then Xt is a martingale.
]
◽
6. Bachelier Model (Arithmetic Brownian Motion). Let (Ω, ℱ, ℙ) be a probability space and
let {Wt ∶ t ≥ 0} be a standard Wiener process. Suppose Xt follows the arithmetic Brownian motion with SDE
dXt = 𝜇dt + 𝜎dWt
where 𝜇 and 𝜎 are constants. Taking integrals show that for t < T,
XT = Xt + 𝜇(T − t) + 𝜎WT−t
where WT−t = WT − Wt ∼ 𝒩(0, T − t). Deduce that XT , given Xt = x, follows a normal
distribution with mean
)
(
𝔼 XT |Xt = x = x + 𝜇(T − t)
and variance
(
)
Var XT |Xt = x = 𝜎 2 (T − t).
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129
Solution: Taking integrals of dXt = 𝜇dt + 𝜎dWt we have
T
∫t
T
dXs =
T
𝜇 ds +
∫t
∫t
𝜎 dWs
XT − Xt = 𝜇(T − t) + 𝜎(WT − Wt )
or
XT = Xt + 𝜇(T − t) + 𝜎WT−t
where WT−t ∼ 𝒩(0, T − t). Since Xt , 𝜇 and 𝜎 are deterministic components therefore XT ,
given Xt = x follows a normal distribution with mean x + 𝜇(T − t) and variance 𝜎 2 (T − t).
◽
7. Black–Scholes Model (Geometric Brownian Motion). Let (Ω, ℱ, ℙ) be a probability space
and let {Wt ∶ t ≥ 0} be a standard Wiener process. Suppose Xt follows the geometric
Brownian motion with SDE
dXt = 𝜇Xt dt + 𝜎Xt dWt
where 𝜇 and 𝜎 are constants. By applying Itō’s formula to Yt = log Xt and taking integrals
show that for t < T,
(
)
𝜇− 12 𝜎 2 (T−t)+𝜎WT−t
XT = Xt e
where WT−t ∼ 𝒩(0, T − t). Deduce that XT , given Xt = x follows a lognormal distribution
with mean
)
(
𝔼 XT |Xt = x = xe𝜇(T−t)
and variance
( 2
)
(
)
Var XT |Xt = x = x2 e𝜎 (T−t) − 1 e2𝜇(T−t) .
Solution: From Taylor’s expansion and subsequently using Itō’s formula,
1
1
dX −
(dXt )2 + . . .
Xt t 2Xt2
1 ( 2 2 )
𝜎 Xt dt
= 𝜇dt + 𝜎dWt −
2Xt2
)
(
1
= 𝜇 − 𝜎 2 dt + 𝜎dWt .
2
d(log Xt ) =
Taking integrals,
)
T
1
𝜇 − 𝜎 2 du +
𝜎 dWu
∫t
∫t
∫t
2
(
)
1
log XT − log Xt = 𝜇 − 𝜎 2 (T − t) + 𝜎(WT − Wt )
2
T
(
T
d(log Xu ) =
(
XT = Xt e
)
𝜇− 12 𝜎 2 (T−t)+𝜎WT−t
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where WT − Wt = WT−t ∼ 𝒩(0, T − t). Therefore,
)
[
(
]
1
XT ∼ log-𝒩 log Xt + 𝜇 − 𝜎 2 (T − t), 𝜎 2 (T − t)
2
and from Problem 1.2.2.9 (page 20) the mean and variance of XT , given Xt = x are
(
𝔼 XT |Xt = x
)
(
)
log x+ 𝜇− 21 𝜎 2 (T−t)+ 12 𝜎 2 (T−t)
=e
= xe𝜇(T−t)
and
(
)
(
) ( 2
2 log x+2 𝜇− 21 𝜎 2
Var XT |Xt = x = e𝜎 (T−t) − 1 e
( 2
)
= x2 e𝜎 (T−t) − 1 e2𝜇(T−t)
)
(T−t)+𝜎 2 (T−t)
respectively.
◽
8. Generalised Geometric Brownian Motion. Let (Ω, ℱ, ℙ) be a probability space and let
{Wt ∶ t ≥ 0} be a standard Wiener process. Suppose Xt follows the generalised geometric
Brownian motion with SDE
dXt = 𝜇t Xt dt + 𝜎t Xt dWt
where 𝜇t and 𝜎t are time dependent. By applying Itō’s formula to Yt = log Xt and taking
integrals show that for t < T,
{
XT = Xt exp
T
∫t
(
)
1
𝜇s − 𝜎s2 ds +
∫t
2
}
T
𝜎s dWs
.
Deduce that XT , given Xt = x follows a lognormal distribution with mean
(
)
T
𝔼 XT |Xt = x = xe∫t 𝜇s ds
and variance
)
( T 2
)
(
T
Var XT |Xt = x = x2 e∫t 𝜎s ds − 1 e2 ∫t 𝜇s ds .
Solution: From Taylor’s expansion and using Itō’s formula,
1
1
dX −
(dXt )2 + . . .
Xt t 2Xt2
)
1 (
= 𝜇t dt + 𝜎dWt − 2 𝜎t2 Xt2 dt
2Xt
)
(
1
= 𝜇t − 𝜎t2 dt + 𝜎t dWt .
2
d(log Xt ) =
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131
Taking integrals,
T
)
1
𝜎s dWs
𝜇s − 𝜎s2 ds +
∫t
∫t
∫t
2
T(
)
T
1
𝜇s − 𝜎s 2 ds +
log XT − log Xt =
𝜎s dWs
∫t
∫t
2
}
{ T(
T
)
1
XT = Xt exp
𝜎 dWs .
𝜇s − 𝜎s2 ds +
∫t
∫t
2
T
T
(
d(log Xs ) =
Thus,
[
XT ∼ log-𝒩 log Xt +
T
∫t
(
)
1
𝜇s − 𝜎s2 ds,
∫t
2
]
T
𝜎s2 ds
and from Problem 1.2.2.9 (page 20) we have mean
T
)
(
log x+∫t
𝔼 XT |Xt = x = e
= xe∫t
T
(
)
𝜇s − 12 𝜎s2 ds+ 12 ∫t T 𝜎s2 ds
𝜇s ds
and variance
(
(
)
T
) ( T 2
(
2 log x+∫t 𝜇s − 12 𝜎s2
Var XT |Xt = x = e∫t 𝜎s ds − 1 e
)
( T 2
T
= x2 e∫t 𝜎s ds − 1 e2 ∫t 𝜇s ds .
) )
T
ds +∫t 𝜎s2 ds
◽
9. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Suppose Xt follows the geometric Brownian motion with SDE
dXt = 𝜇Xt dt + 𝜎Xt dWt
where 𝜇 and 𝜎 are constants. Show that if Yt = Xtn , for some constant n, then Yt follows
the geometric Brownian process of the form
)
(
dYt
1
= n 𝜇 + (n − 1)𝜎 2 dt + n𝜎dWt .
Yt
2
Deduce that given Yt , t < T, YT follows a lognormal distribution with the form
YT = Y t
(
)
n 𝜇− 12 𝜎 2 (T−t)+n𝜎WT−t
e
where WT−t ∼ 𝒩(0, T − t) with mean
(
)
𝔼 YT |Yt = y =
(
)
n 𝜇+ 12 (n−1)𝜎 2 (T−t)
ye
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and variance
(
)
(
2 2
2n
Var YT |Yt = y = y2 (en 𝜎 (T−t) − 1)e
)
𝜇+ 12 (n−1)𝜎 2 (T−t)
.
Solution: From Itō’s formula,
1
dYt = nXtn−1 dXt + n(n − 1)Xtn−2 dXt2
2
1
= nXtn−1 (𝜇Xt dt + 𝜎Xt dWt ) + n(n − 1)𝜎 2 Xtn dt
2
)
(
1
= n 𝜇 + (n − 1)𝜎 2 Xtn dt + n𝜎Xtn dWt .
2
By substituting Xtn = Yt we have
)
(
dYt
1
= n 𝜇 + (n − 1)𝜎 2 dt + n𝜎dWt .
Yt
2
by analogy with Problem 3.2.2.7 (page
Since Yt follows a geometric
)
( Brownian motion,
129) and by setting 𝜇 ← n 𝜇 + 12 (n − 1)𝜎 2 and 𝜎 ← n𝜎, we can easily show that
(
)
n 𝜇− 12 𝜎 2 (T−t)+n𝜎WT−t
YT = Yt e
follows a lognormal distribution where WT−t ∼ 𝒩(0, T − t). In addition, we can also
deduce
(
)
)
(
n 𝜇+ 12 (n−1)𝜎 2 (T−t)
𝔼 YT |Yt = y = ye
and
(
)
( 2 2
(
)
2n
Var YT |Yt = y = y2 en 𝜎 (T−t) − 1 e
)
𝜇+ 12 (n−1)𝜎 2 (T−t)
.
◽
10. Ornstein–Uhlenbeck Process. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0}
be a standard Wiener process. Suppose Xt follows the Ornstein–Uhlenbeck process with
SDE
dXt = 𝜅(𝜃 − Xt ) dt + 𝜎dWt
where 𝜅, 𝜃 and 𝜎 are constants. By applying Itō’s formula to Yt = e𝜅t Xt and taking integrals
show that for t < T,
]
[
XT = Xt e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
T
∫t
𝜎e−𝜅(T−s) dWs .
Using the properties of stochastic integrals on the above expression, find the mean and
variance of XT , given Xt = x.
Deduce that XT follows a normal distribution.
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133
Solution: Expanding Yt = e𝜅t Xt using Taylor’s formula and applying Itō’s formula,
we have
1
d(e𝜅t Xt ) = 𝜅e𝜅t Xt dt + e𝜅t dXt + 𝜅 2 e𝜅t Xt (dt)2 + . . .
2
= 𝜅e𝜅t Xt dt + e𝜅t (𝜅(𝜃 − Xt ) dt + 𝜎dWt )
= 𝜅𝜃e𝜅t dt + 𝜎e𝜅t dWt .
Integrating the above expression,
T
∫t
d(e𝜅t Xs ) =
T
∫t
𝜅𝜃e𝜅s ds +
T
∫t
𝜎e𝜅s dWs
[
]
XT = Xt e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
(
Given the fact that
𝔼
and
[(
𝔼
∫t
𝜎e
)2 ]
T
𝜎e
−𝜅(T−s)
dWs
∫t
𝜎e−𝜅(T−s) dWs .
)
T
∫t
T
−𝜅(T−s)
(
=𝔼
dWs
=0
)
T
∫t
𝜎 e
2 −2𝜅(T−s)
ds
=
]
𝜎2 [
1 − e−2𝜅(T−t)
2𝜅
the mean and variance of XT , given Xt = x are
)
[
(
]
𝔼 XT |Xt = x = xe−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t)
and
]
) 𝜎2 [
(
1 − e−2𝜅(T−t)
Var XT |Xt = x =
2𝜅
respectively.
T
Since
∫t
𝜎e−𝜅(T−s) dWs can be written in the form
T
∫t
𝜎e−𝜅(T−s) dWs = lim
n→∞
n−1
∑
(
)
𝜎e−𝜅(T−ti ) Wti+1 − Wti
i=0
where ti = t + i(T − t)∕n, t = t0 < t1 < t2 < . . . < tn−1 < tn = T, n ∈ ℕ then due to the
stationary increment of a standard Wiener process, we can see that each term of Wti+1 −
(
)
T −t
Wti ∼ 𝒩 0,
is normal multiplied by a deterministic exponential term. Thus, the
n
product is normal and given that the sum of normal variables is normal we can deduce
)
(
] 𝜎2 [
]
[
−𝜅(T−t)
−𝜅(T−t)
−2𝜅(T−t)
,
.
+𝜃 1−e
XT ∼ 𝒩 xe
1−e
2𝜅
◽
4:19 P.M.
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11. Geometric Mean-Reverting Process. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶
t ≥ 0} be a standard Wiener process. Suppose Xt follows the geometric mean-reverting
process with SDE
dXt = 𝜅(𝜃 − log Xt )Xt dt + 𝜎Xt dWt ,
X0 > 0
where 𝜅, 𝜃 and 𝜎 are constants. By applying Itō’s formula to Yt = log Xt show that the
diffusion process can be reduced to an Ornstein–Uhlenbeck process of the form
[
]
1
dYt = 𝜅(𝜃 − Yt ) − 𝜎 2 dt + 𝜎dWt .
2
Show also that for t < T,
(
)
)
𝜎2 (
log XT = (log Xt )e−𝜅(T−t) + 𝜃 −
1 − e−𝜅(T−t) +
∫t
2𝜅
T
𝜎e−𝜅(T−s) dWs .
Using the properties of stochastic integrals on the above expression, find the mean and
variance of XT , given Xt = x and deduce that XT follows a lognormal distribution.
Solution: By expanding Yt = log Xt using Taylor’s formula and subsequently applying
Itō’s formula, we have
d(log Xt ) =
1
1
dX −
(dXt )2 + . . .
Xt t 2Xt2
1
= 𝜅(𝜃 − log Xt ) dt + 𝜎dWt − 𝜎 2 dt
2
( (
) 1 2)
= 𝜅 𝜃 − log Xt − 𝜎 dt + 𝜎dWt
2
(
)
1
dYt = 𝜅(𝜃 − Yt ) − 𝜎 2 dt + 𝜎dWt .
2
and hence
Using the same steps in solving the Ornstein–Uhlenbeck process, we apply Itō’s formula
on Zt = e𝜅Yt such that
1
d(e𝜅t Yt ) = 𝜅e𝜅t Yt dt + e𝜅t dYt + 𝜅 2 e𝜅t Yt (dt)2 + . . .
2
[(
)
]
1
= 𝜅e𝜅t Yt dt + e𝜅t 𝜅(𝜃 − Yt ) − 𝜎 2 dt + 𝜎dWt
2
(
)
1 2 𝜅t
𝜅t
𝜅t
= 𝜅𝜃e − 𝜎 e dt + 𝜎e dWt .
2
Taking integrals from t to T, we have
T
∫t
d(e𝜅s Ys ) =
T
∫t
(
)
1
𝜅𝜃e𝜅s − 𝜎 2 e𝜅s ds +
∫t
2
T
𝜎e𝜅s dWs
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135
(
or
−𝜅(T−t)
YT = Yt e
𝜎2
+ 𝜃−
2𝜅
)
(
)
1 − e−𝜅(T−t) +
T
∫t
𝜎e−𝜅(T−s) dWs .
By analogy with Problem 3.2.2.10 (page 132), we can deduce that YT = log XT follows a
normal distribution and hence
)
(
)
(
( −𝜅(T−t) )
) 𝜎2 (
)
𝜎2 (
−𝜅(T−t)
−2𝜅(T−t)
1−e
1−e
+ 𝜃−
,
log XT ∼ 𝒩 log Xt e
2𝜅
2𝜅
or
)
(
)
(
(
)
) 𝜎2 (
)
𝜎2 (
1 − e−𝜅(T−t) ,
1 − e−2𝜅(T−t)
XT ∼ log −𝒩 log Xt e−𝜅(T−t) + 𝜃 −
2𝜅
2𝜅
with mean
(
)
𝔼 XT |Xt = x = exp
)
(
}
𝜎2
𝜎2
(1 − e−𝜅(T−t) ) +
(1 − e−2𝜅(T−t) )
e−𝜅(T−t) log x + 𝜃 −
2𝜅
4𝜅
{
and variance
}
)
𝜎2 (
1 − e−2𝜅(T−t) − 1
2𝜅
)
{
(
)
𝜎2 (
−𝜅(T−t)
1 − e−𝜅(T−t) +
× exp 2e
log x + 2 𝜃 −
2𝜅
}
2
)
𝜎 (
1 − e−2𝜅(T−t)
.
2𝜅
)
(
Var XT |Xt = x = exp
{
◽
12. Cox–Ingersoll–Ross (CIR) Model. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥
0} be a standard Wiener process. Suppose Xt follows the CIR model with SDE
√
)
(
dXt = 𝜅 𝜃 − Xt dt + 𝜎 Xt dWt ,
X0 > 0
where 𝜅, 𝜃 and 𝜎 are constants. By applying Itō’s formula to Zt = e𝜅t Xt and Zt2 = e2𝜅t Xt2
and taking integrals show that for t < T,
]
[
XT = Xt e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
T
∫t
√
𝜎e−𝜅(T−s) Xs dWs
and
)
(
XT2 = Xt2 e−2𝜅(T−t) + 2𝜅𝜃 + 𝜎 2
T
∫t
T
e−2𝜅(T−s) Xs ds + 2𝜎
∫t
3
e−2𝜅(T−s) Xs2 dWs .
Using the properties of stochastic integrals on the above two expressions, find the mean
and variance of XT , given Xt = x.
4:19 P.M.
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One-Dimensional Diffusion Process
Solution: Using Taylor’s formula and applying Itō’s formula on Zt = e𝜅t Xt , we have
(
)
1
d e𝜅t Xt = 𝜅e𝜅t Xt dt + e𝜅t dXt + 𝜅 2 e𝜅t Xt (dt)2 + . . .
2
( (
)
√
)
= 𝜅e𝜅t Xt dt + e𝜅t 𝜅 𝜃 − Xt dt + 𝜎 Xt dWt
√
= 𝜅𝜃e𝜅t dt + 𝜎e𝜅t Xt dWt .
Integrating the above expression,
T
∫t
)
(
d e𝜅t Xs =
T
∫t
𝜅𝜃e𝜅s ds +
T
∫t
√
𝜎e𝜅s Xs dWs
and therefore
[
]
XT = Xt e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
T
∫t
√
𝜎e−𝜅(T−s) Xs dWs .
For the case of Zt2 = e2𝜅t Xt2 , by the application of Taylor’s expansion and Itō’s formula
we have
)
(
1 ( 2 ) 2𝜅t 2
1 ( 2𝜅t ) ( )2
4𝜅 e Xt (dt)2 +
2e
dXt + . . .
d e2𝜅t Xt2 = 2𝜅e2𝜅t Xt2 dt + 2e2𝜅t Xt dXt +
2
2
( (
)
√
)
)
(
= 2𝜅e2𝜅t Xt2 dt + 2e2𝜅t Xt 𝜅 𝜃 − Xt dt + 𝜎 Xt dWt + e2𝜅t 𝜎 2 Xt dt
3
(
)
= e2𝜅t 2𝜅𝜃 + 𝜎 2 Xt dt + 2𝜎e2𝜅t Xt2 dWt .
By taking integrals,
T
∫t
(
) (
)
d e2𝜅s Xs2 = 2𝜅𝜃 + 𝜎 2
T
∫t
T
e2𝜅s Xs ds + 2𝜎
3
e2𝜅s Xs2 dWs
∫t
and we eventually obtain the following expression:
(
)
XT2 = e−2𝜅(T−t) Xt2 + 2𝜅𝜃 + 𝜎 2
T
∫t
T
e−2𝜅(T−s) Xs ds + 2𝜎
∫t
3
e−2𝜅(T−s) Xs2 dWs .
Given Xt = x, and by taking the expectation of the expression XT , we have
)
)
(
(
𝔼 XT |Xt = x = xe−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) .
(
)
To find the variance, Var XT |Xt = x we first take the expectation of XT2 ,
(
)
)
(
𝔼 XT2 |Xt = x = x2 e−2𝜅(T−t) + 2𝜅𝜃 + 𝜎 2
= x2 e−2𝜅(T−t)
T
∫t
(
)
e−2𝜅(T−s) 𝔼 Xs |Xt = x ds
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)
(
+ 2𝜅𝜃 + 𝜎 2
137
T
[
)]
(
e−2𝜅(T−s) xe−𝜅(s−t) + 𝜃 1 − e−𝜅(s−t) ds
∫t
)
(
2𝜅𝜃 + 𝜎 2
2 −2𝜅(T−t)
=x e
+
(x − 𝜃) (e−𝜅(T−t) − e−2𝜅(T−t) )
𝜅
+
)
𝜃(2𝜅𝜃 + 𝜎 2 ) (
1 − e−2𝜅(T−t) .
2𝜅
Therefore,
)
(
) [ (
)]2
(
Var XT |Xt = x = 𝔼 XT2 |Xt = x − 𝔼 XT |Xt = x
=
) 𝜃𝜎 2 (
)
x𝜎 2 ( −𝜅(T−t)
e
1 − 2e−𝜅(T−t) + e−2𝜅(T−t) .
− e−2𝜅(T−t) +
𝜅
2𝜅
◽
13. Brownian Bridge Process. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a
standard Wiener process. Suppose Xt follows the Brownian bridge process with SDE
dXt =
y − Xt
dt + dWt ,
1−t
X1 = y
where the diffusion is conditioned to be at y at time t = 1. By applying Itō’s formula to
Yt = (y − Xt )∕(1 − t) and taking integrals show that under an initial condition X0 = x and
for 0 ≤ t < 1,
)
(
t
1
dWs .
Xt = yt + (1 − t) x +
∫0 1 − s
Using the properties of stochastic integrals on the above expression, find the mean and
variance of Xt , given X0 = x and show that Xt follows a normal distribution.
Solution: By expanding Yt = (y − Xt )∕(1 − t) using Taylor’s formula and subsequently
applying Itō’s formula, we have
2
2
𝜕Yt
𝜕Y
1 𝜕 Yt
1 𝜕 Yt
2
(dt)
+
(dXt )2 + . . .
dt + t dXt +
𝜕t
𝜕Xt
2 𝜕t2
2 𝜕Xt2
)]
[(
) (y − X
y − Xt
1
t
dt + dWt
=
dt −
1−t
1−t
(1 − t)2
)
(
1
dWt .
=−
1−t
dYt =
Taking integrals,
t
∫0
t
dYs = −
1
dWs
∫0 1 − s
t
Yt − Y0 = −
1
dWs
∫0 1 − s
4:19 P.M.
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t
y − Xt
1
=y−x−
dWs .
∫0 1 − s
1−t
Therefore,
(
Xt = yt + (1 − t) x +
t
∫0
)
1
dWs .
1−s
Using the properties of stochastic integrals,
)
( t
1
dWs = 0
𝔼
∫0 1 − s
[(
( t
)2 ]
)
t
1
1
1
dWs
𝔼
=𝔼
ds
=
∫0 1 − s
∫0 (1 − s)2
1−t
therefore the mean and variance of Xt are
(
)
𝔼 Xt |X0 = x = yt + x(1 − t)
and
)
(
Var Xt |X0 = x =
1
1−t
respectively.
t
t
1
dWs is in the form
Since
f (s) dWs , from Problem 3.2.2.4 (page 126) we can
∫0 1 − s
∫0
t
1
dWs follows a normal distribution,
easily prove that
∫0 1 − s
t
∫0
(
)
1
1
.
dWs ∼ 𝒩 yt + x(1 − t),
1−s
1−t
◽
14. Forward Curve from an Asset Price Following a Geometric Brownian Motion. Let
(Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process.
Suppose the asset price St at time t follows a geometric Brownian motion
dSt
= 𝜇dt + 𝜎dWt
St
where 𝜇 is the drift parameter and 𝜎 is the volatility. We define the forward price F(t, T)
as an agreed-upon price set at time t to be paid or received at time T, t ≤ T and is given
by the relationship
)
(
F(t, T) = 𝔼 ST || ℱt .
Show that the forward curve follows
dF(t, T)
= 𝜎dWt .
F(t, T)
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139
Solution: Given that St follows a geometric Brownian motion then, following
Problem 3.2.2.7 (page 129), we can easily show
)
(
𝔼 ST || ℱt = St e𝜇(T−t) .
(
)
Because F(t, T) = 𝔼 ST || ℱt = St e𝜇(T−t) , and by expanding F(t, T) using Taylor’s theorem and applying Itō’s lemma,
dF(t, T) =
𝜕F
1 𝜕 2 F ( )2
𝜕F
dSt + . . .
dSt +
dt +
𝜕t
𝜕St
2 𝜕St2
= −𝜇St e𝜇(T−t) dt + e𝜇(T−t) dSt .
Since dSt = 𝜇St dt + 𝜎dWt we have
(
)
dF(t, T) = −𝜇St e𝜇(T−t) dt + e𝜇(T−t) 𝜇St dt + 𝜎St dWt
= 𝜎St e𝜇(T−t) dWt
= 𝜎F(t, T) dWt .
◽
15. Forward Curve from an Asset Price Following a Geometric Mean-Reverting Process. Let
(Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard Wiener process. Suppose the asset price St at time t follows a geometric mean-reverting process
)
(
dSt
= 𝜅 𝜃 − log St dt + 𝜎dWt
St
where 𝜅 is the mean-reversion rate, 𝜃 is the long-term mean and 𝜎 is the volatility parameter. We define the forward price F(t, T) as an agreed-upon price set at time t to be paid
or received at time T, t ≤ T and is given by the relationship
)
(
F(t, T) = 𝔼 ST || ℱt .
Show that the forward curve follows
dF(t, T)
= 𝜎e−𝜅(T−t) dWt .
F(t, T)
Solution: Using the steps described in Problem 3.2.2.11 (page 134), the mean of ST given
St is
(
)
{
}
(
)
) 𝜎2 (
)
𝜎2 (
𝔼 ST || ℱt = exp e−𝜅(T−t) log St + 𝜃 −
1 − e−𝜅(T−t) +
1 − e−2𝜅(T−t)
.
2𝜅
4𝜅
)
(
Because F(t, T) = 𝔼 ST || ℱt and expanding F(t, T) using Taylor’s theorem, we have
dF(t, T) =
𝜕F
1 𝜕2F
𝜕F
dSt +
(dSt )2 + . . .
dt +
𝜕t
𝜕St
2 𝜕St2
4:19 P.M.
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]
[
(
)
𝜎2
𝜎2
= 𝜅e−𝜅(T−t) log St − 𝜃 −
𝜅e−𝜅(T−t) − e−2𝜅(T−t) F(t, T) dt
2𝜅
2
[
]
e−𝜅(T−t) e−2𝜅(T−t)
e−𝜅(T−t)
1
−
+
F(t, T)(dSt )2 + . . .
F(t, T) dSt +
+
St
2
St2
St2
)
(
By substituting dSt = 𝜅 𝜃 − log St St dt + 𝜎St dWt and applying Itō’s lemma,
]
(
[
)
dF(t, T)
𝜎2
𝜎2
= 𝜅e−𝜅(T−t) log St − 𝜃 −
𝜅e−𝜅(T−t) − e−2𝜅(T−t) dt
F(t, T)
2𝜅
2
[ (
)
] 𝜎2
𝜎2
+e−𝜅(T−t) 𝜅 𝜃 − log St dt + 𝜎dWt − e−𝜅(T−t) dt + e−2𝜅(T−t) dt
2
2
= 𝜎e−𝜅(T−t) dWt .
◽
16. Forward–Spot Price Relationship I. Let {Wt ∶ t ≥ 0} be the standard Wiener process on
the probability space (Ω, ℱ, ℙ). We define the forward curve F(t, T) following the SDE
dF(t, T)
= 𝜎(t, T) dWt
F(t, T)
as an agreed-upon price of an
( asset )with current spot price St to be paid or received at time
T, t ≤ T where F(t, T) = 𝔼 ST || ℱt such that ST is the spot price at time T and 𝜎(t, T) > 0
is a time-dependent volatility.
Show that the spot price has the following SDE
]
[
t
t
dSt
𝜕 log F(0, t)
𝜕𝜎(u, t)
𝜕𝜎(u, t)
−
du +
dWu dt + 𝜎(t, t) dWt .
=
𝜎(u, t)
∫0
∫0
St
𝜕t
𝜕t
𝜕t
Solution: By expanding log F(t, T) using Taylor’s theorem and then applying Itō’s
lemma,
d log F(t, T) =
1
1
dF(t, T)2 + . . .
dF(t, T) −
F(t, T)
2F(t, T)2
1
= 𝜎(t, T) dWt − 𝜎(t, T)2 dt
2
and taking integrals,
t
∫0
so we have
t
d log F(u, T) =
∫0
t
𝜎(u, T) dWu −
1
t
1
𝜎(u, T)2 du
2 ∫0
2 du+∫ t
0
F(t, T) = F(0, T)e− 2 ∫0 𝜎(u,T)
𝜎(u,T)dWu
By setting T = t, the spot price St = F(t, t) can be expressed as
1
t
St = F(0, t)e− 2 ∫0 𝜎(u,t)
2 du+∫ t 𝜎(u,t)dW
u
0
.
.
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141
Expanding using Taylor’s theorem,
dSt =
2
2
𝜕St
𝜕 2 St
𝜕St
1 𝜕 St 2 1 𝜕 St
2
dWt +
dt
+
dW
+
dt dWt + . . .
dt +
t
𝜕t
𝜕Wt
2 𝜕t2
2 𝜕Wt2
𝜕t𝜕Wt
and applying Itō’s lemma again,
(
dSt =
𝜕St 1 𝜕 2 St
+
𝜕t
2 𝜕Wt2
)
dt +
𝜕St
dWt .
𝜕Wt
From the spot price equation we have
]
[
t
t
𝜕St
𝜕𝜎(u, t)
𝜕𝜎(u, t)
𝜕F(0, t)
1
=
F(0, t)−1 St −
𝜎(t, t)2 + 2𝜎(u, t)
du − 2
dWu St
∫0
∫0
𝜕t
𝜕t
2
𝜕t
𝜕t
]
[
t
t
𝜕 log F(0, t) 1
𝜕𝜎(u, t)
𝜕𝜎(u, t)
=
𝜎(u, t)
− 𝜎(t, t)2 −
du +
dWu St ,
∫0
∫0
𝜕t
2
𝜕t
𝜕t
[ t
]
𝜕St
𝜕
=
𝜎(u, t) dWu St = 𝜎(t, t)St
𝜕Wt
𝜕Wt ∫0
and
𝜕 2 St
𝜕Wt2
= 𝜎(t, t)2 St .
𝜕St 𝜕St
𝜕 2 St
,
By substituting the values of
and
into dSt =
𝜕t 𝜕Wt
𝜕Wt2
𝜕St
dWt , we finally have
𝜕Wt
(
𝜕St 1 𝜕 2 St
+
𝜕t
2 𝜕Wt2
)
dt +
]
[
t
t
dSt
𝜕 log F(0, t)
𝜕𝜎(u, t)
𝜕𝜎(u, t)
−
du +
dWu dt + 𝜎(t, t) dWt .
=
𝜎(u, t)
∫0
∫0
St
𝜕t
𝜕t
𝜕t
◽
17. Clewlow–Strickland 1-Factor Model. Let {Wt ∶ t ≥ 0} be the standard Wiener process on
the probability space (Ω, ℱ, ℙ). Suppose the forward curve F(t, T) follows the process
dF(t, T)
= 𝜎e−𝛼(T−t) dWt
F(t, T)
where t ≤ T, 𝛼 is the mean-reversion parameter
)and 𝜎 is the volatility. From the
(
forward–spot price relationship F(t, T) = 𝔼 ST || ℱt where ST is the spot price at time
T, show that
]
[
) 𝜎2 (
)
(
dSt
𝜕 log F(0, t)
−2𝛼t
dt + 𝜎dWt
=
1−e
+ 𝛼 log F(0, t) − log St +
St
𝜕t
4
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and the forward curve at time t is given by
[
F(t, T) = F(0, T)
St
F(0, t)
]e−𝛼(T−t)
𝜎 2 −𝛼T 2𝛼t
(e −1)(e−𝛼t −e−𝛼T )
e− 4𝛼 e
.
Finally, show that conditional on F(0, T), F(t, T) follows a lognormal distribution with
mean
𝔼 [F(t, T)| F(0, T)] = F(0, T)
and variance
{
Var [F(t, T)| F(0, T)] = F(0, T)2 exp
}
]
𝜎 2 [ −2𝛼(T−t)
− e−2𝛼T − 1
e
2𝛼
.
Solution: For the case when
dF(t, T)
= 𝜎(t, T) dWt
F(t, T)
with 𝜎(t, T) a time-dependent volatility, from Problem 3.2.2.16 (page 140) the corresponding spot price SDE is given by
]
[
t
t
dSt
𝜕 log F(0, t)
𝜕𝜎(u, t)
𝜕𝜎(u, t)
=
𝜎(u, t)
−
du +
dWu dt + 𝜎(t, t) dWt .
∫0
∫0
St
𝜕t
𝜕t
𝜕t
Let 𝜎(t, T) = 𝜎e−𝛼(T−t) and taking partial differentiation with respect to T, we have
𝜕𝜎(t, T)
= −𝛼𝜎e−𝛼(T−t) .
𝜕T
Thus,
t
∫0
𝜎(u, t)
t
)
𝜕𝜎(u, t)
𝜎2 (
1 − e−2𝛼t
du = −
𝛼𝜎 2 e−2𝛼(t−u) du = −
∫0
𝜕t
2
and
t
∫0
t
𝜕𝜎(u, t)
𝛼𝜎e−𝛼(t−u) dWu .
dWu = −
∫0
𝜕t
Using Itō’s lemma,
d log F(t, T) =
1
1
dF(t, T) −
dF(t, T)2 + . . .
F(t, T)
2F(t, T)2
1
= 𝜎e−𝛼(T−t) dWt − 𝜎 2 e−2𝛼(T−t) dt
2
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143
and taking integrals,
t
log F(t, T) = log F(0, T) −
t
1
𝜎 2 e−2𝛼(T−u) du +
𝜎e−𝛼(T−u) dWu .
∫0
2 ∫0
By setting T = t such that F(t, t) = St , we can write
t
t
1
𝜎 2 e−2𝛼(t−u) du +
𝜎e−𝛼(t−u) dWu
∫0
2 ∫0
log St = log F(0, t) −
therefore
t
∫0
) 1 t 2 −2𝛼(t−u)
(
𝜕𝜎(u, t)
dWu = 𝛼 log F(0, t) − log St −
𝛼𝜎 e
du
𝜕t
2 ∫0
) 𝜎2 (
)
(
1 − e−2𝛼t .
= 𝛼 log F(0, t) − log St −
4
t
t
𝜕𝜎(u, t)
𝜕𝜎(u, t)
du and
dWu into the spot
∫0
∫0
𝜕t
𝜕t
price SDE and taking note that 𝜎(t, t) = 𝜎, we eventually have
]
[
) 𝜎2 (
)
(
dSt
𝜕 log F(0, t)
=
1 − e−2𝛼t dt + 𝜎dWt .
+ 𝛼 log F(0, t) − log St +
St
𝜕t
4
𝜎(u, t)
By substituting the values of
Since
t
1
F(t, T) = F(0, T)e− 2 ∫0 𝜎
2 e−2𝛼(T−u) du+∫ t
0
𝜎e−𝛼(T−u) dWu
with
t
∫0
𝜎 2 e−2𝛼(T−u) du =
)
𝜎 2 −2𝛼T ( 2𝛼t
e
e −1
2𝛼
and using the spot price equation
t
∫0
t
𝜎e−𝛼(T−u) dWu = e−𝛼(T−t)
𝜎e−𝛼(t−u) dWu
∫0
{ [
}
]
t
St
1
−𝛼(T−t)
2 −2𝛼(t−u)
log
=e
𝜎 e
du
+
F(0, t)
2 ∫0
]
}
{ [
)
St
𝜎2 (
−𝛼(T−t)
−2𝛼t
1−e
+
=e
log
F(0, t)
4𝛼
therefore
[
F(t, T) = F(0, T)
St
F(0, t)
]e−𝛼(T−t)
𝜎 2 −𝛼T 2𝛼t
(e −1)(e−𝛼t −e−𝛼T )
e− 4𝛼 e
Finally, using
1
d log F(t, T) = 𝜎e−𝛼(T−t) dWt − 𝜎 2 e−2𝛼(T−t) dt
2
.
4:19 P.M.
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and taking integrals we have
t
log F(t, T) = log F(0, T) +
∫0
𝜎e−𝛼(T−u) dWu −
]
𝜎 2 [ −2𝛼(T−t)
− e−2𝛼T .
e
4𝛼
From the property of the Itō integral,
[
𝔼
]
t
∫0
𝜎e
−𝛼(T−u)
dWu = 0
and
[(
𝔼
)2 ]
t
∫0
𝜎e
−𝛼(T−u)
[
=𝔼
dWu
=
t
Since
∫0
t
∫0
]
𝜎 2 e−2𝛼(T−u) du
]
𝜎 2 [ −2𝛼(T−t)
− e−2𝛼T .
e
2𝛼
t
𝜎e−𝛼(T−u) dWu is in the form
t
∫0
f (u) dWu , from Problem 3.2.2.4 (page 126)
𝜎e−𝛼(T−u) dWu follows a normal distribution.
∫0
In addition, since we can also write the Itō integral as
we can easily show that
t
∫0
𝜎e−𝛼(T−u) dWu = lim
n→∞
n−1
∑
𝜎e−𝛼(T−ti ) (Wti+1 − Wti )
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ and due
( to the
) stationary
t
increment of a standard Wiener process, each term of Wti+1 − Wti ∼ 𝒩 0,
is normally
n
distributed multiplied by a deterministic term and we can deduce that
t
∫0
𝜎e
−𝛼(T−u)
)
(
]
𝜎 2 [ −2𝛼(T−t)
−2𝛼T
.
dWu ∼ 𝒩 0,
−e
e
2𝛼
Hence,
(
)
] 𝜎 2 [ −2𝛼(T−t)
]
𝜎 2 [ −2𝛼(T−t)
e
e
log F(t, T) ∼ 𝒩 log F(0, T) −
− e−2𝛼T ,
− e−2𝛼T
4𝛼
2𝛼
which implies
𝔼 [F(t, T)| F(0, T)] = F(0, T)
{
and
2
Var [F(t, T)| F(0, T)] = F(0, T) exp
}
]
𝜎 2 [ −2𝛼(T−t)
e
− e−2𝛼T − 1
2𝛼
.
◽
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145
18. Constant Elasticity of Variance Model. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶
t ≥ 0} be a standard Wiener process with respect to the filtration ℱt , t ≥ 0. Suppose St > 0
follows a constant elasticity of variance (CEV) model of the form
dSt = rSt dt + 𝜎(St , t)St dWt
where r is a constant and 𝜎(St , t) is a local volatility function. By setting 𝜎(St , t) = 𝛼St𝛽−1
with 𝛼 > 0 and 0 < 𝛽 < 1, show using Itō’s formula that 𝜎(St , t) satisfies
{[
]
}
d𝜎(St , t)
1
= (𝛽 − 1) r + (𝛽 − 2)𝜎(St , t)2 dt + 𝜎(St , t) dWt .
𝜎(St , t)
2
Finally, conditional on St show that for t < T,
[
ST = e
rT
1−𝛽
e−r(1−𝛽)t St
1
] 1−𝛽
T
+𝛼
∫t
−r(1−𝛽)u
e
dWu
.
Solution: From Itō’s formula,
d𝜎(St , t) =
2
𝜕𝜎(St , t)
1 𝜕 𝜎(St , t)
dSt +
(dSt )2 + . . .
𝜕St
2 𝜕St2
1
= 𝛼(𝛽 − 1)St𝛽−2 dSt + 𝛼(𝛽 − 1)(𝛽 − 2)St𝛽−3 (dSt )2 + . . .
2
1
= 𝛼(𝛽 − 1)St𝛽−2 (rSt dt + 𝛼St𝛽 dWt ) + 𝛼(𝛽 − 1)(𝛽 − 2)(𝛼 2 St2𝛽 dt)
2
[(
)
]
1
= (𝛽 − 1) r𝜎(St , t) + (𝛽 − 2)𝜎(St , t)3 dt + 𝜎(St , t)2 dWt .
2
Therefore,
{[
]
}
d𝜎(St , t)
1
= (𝛽 − 1) r + (𝛽 − 2)𝜎(St , t)2 dt + 𝜎(St , t) dWt .
𝜎(St , t)
2
To find the solution of the CEV model, let Xt = e−rt St and by Itō’s formula
2
𝜕Xt
𝜕X
1 𝜕 Xt
(dSt )2 + . . .
dt + t dSt +
𝜕t
𝜕St
2 𝜕St2
(
)
= −re−rt St dt + e−rt rSt dt + 𝛼St𝛽 dWt
dXt =
= 𝛼e−rt St𝛽 dWt
= 𝛼e−r(1−𝛽)t Xt𝛽 dWt .
Taking integrals
T
∫t
dXu
Xu𝛽
T
=𝛼
∫t
e−r(1−𝛽)u dWu
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we have
XT1−𝛽 = Xt1−𝛽 + 𝛼
T
∫t
[
XT =
or
Xt1−𝛽
e−r(1−𝛽)u dWu
+𝛼
−r(1−𝛽)u
e
∫t
[
rT
ST = e
1
] 1−𝛽
T
e−r(1−𝛽)t St1−𝛽
dWu
1
] 1−𝛽
T
+𝛼
∫t
−r(1−𝛽)u
e
dWu
.
◽
19. Geometric Average. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0} be a standard
Wiener process with respect to the filtration ℱt , t ≥ 0. Suppose St satisfies the following
geometric Brownian motion model:
dSt
= 𝜇dt + 𝜎dWt ,
St
where 𝜇 and 𝜎 are constant parameters. By considering a geometric average of St
1
Gt = e t
∫0t Su du
,
G0 = S0
show that Gt satisfies the following SDE:
(
)
dGt
1
1
𝜎 ̃
=
𝜇 − 𝜎 2 dt + √ dW
t
Gt
2
6
3
√
t
̃t = 3
where W
W du.
t ∫0 u
Under what condition is this SDE valid?
Solution: Using the steps described in Problem 3.2.2.7 (page 129) for u < t,
1 2
)u+𝜎Wu
Su = S0 e(𝜇− 2 𝜎
(
)
1
log Su = log S0 + 𝜇 − 𝜎 2 u + 𝜎Wu .
2
or
Taking the natural logarithm of Gt ,
t
1
log Su du
t ∫0
t[
)
]
(
1
1
log S0 + 𝜇 − 𝜎 2 u + 𝜎Wu du
=
t ∫0
2
log Gt =
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147
t
t(
t
)
1
1
1
𝜎
log S0 du +
W du
𝜇 − 𝜎 2 u du +
t ∫0
t ∫0
2
t ∫0 u
t
(
)
1
1
𝜎
W du.
= log S0 +
𝜇 − 𝜎2 t +
2
2
t ∫0 u
=
From Problem 3.2.1.13 (page 115), using integration by parts we can write
t
∫0
t
Wu du =
∫0
(t − u) dWu
)
(
1
Wu du ∼ 𝒩 0, t3 and hence
∫0
3
t
and we can deduce that
t
(
)
𝜎
1
Wu du ∼ 𝒩 0, 𝜎 2 t
t ∫0
3
̃t =
where W
Thus,
(
)
𝜎 ̃
1 2
√ W
t ∼ 𝒩 0, 𝜎 t
3
3
√
3 t
W du ∼ 𝒩(0, t).
t ∫0 u
log Gt = log S0 +
or
or
(
)
1
1
𝜎 ̃
𝜇 − 𝜎2 t + √ W
t
2
2
3
(
)
dGt
1
1
𝜎 ̃
=
𝜇 − 𝜎 2 dt + √ dW
t
Gt
2
6
3
with G0 = S0 .
̃ t does not have the stationary increment property (see Problem
However, given that W
3.2.1.13, page 115), this SDE is only valid if the geometric average starts at time t = 0.
◽
20. Feynman–Kac Formula for One-Dimensional Diffusion Process. We consider the following PDE problem:
𝜕V
𝜕V 1
𝜕2V
− r(t)V(St , t) = 0
+ 𝜎(St , t)2 2 + 𝜇(St , t)
𝜕t
2
𝜕St
𝜕St
with boundary condition V(ST , T) = Ψ(ST ) where 𝜇, 𝜎 are known functions of St and t,
r and Ψ are functions of t and ST , respectively with t < T. Using Itō’s formula on the
process
u
Zu = e− ∫t r(𝑣)d𝑣 V(Su , u)
where St satisfies the generalised SDE
dSt = 𝜇(St , t) dt + 𝜎(St , t) dWt
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such that {Wt ∶ t ≥ 0} is a standard Wiener process on the probability space (Ω, ℱ, ℙ),
show that under the filtration ℱt the solution of the PDE is given by
]
[
|
T
V(St , t) = 𝔼 e− ∫t r(𝑣)d𝑣 Ψ(ST )|| ℱt .
|
Solution: Let g(u) = e− ∫t
u
r(𝑣)d𝑣
and hence we can write
Zu = g(u)V(Su , u).
By applying Taylor’s expansion and Itō’s formula on dZu we have
2
𝜕Zu
𝜕Z
1 𝜕 Zu
(dSu )2 + . . .
du + u dSu +
𝜕u
𝜕Su
2 𝜕Su2
)
)
)
(
(
(
𝜕g
1
𝜕V
𝜕2V
𝜕V
dSu +
+ V(Su , u)
du + g(u)
g(u) 2 (dSu )2
= g(u)
𝜕u
𝜕u
𝜕Su
2
𝜕Su
)
(
(
)
)
)
𝜕V
𝜕V ( (
− r(u)g(u)V(Su , u) du + g(u)
𝜇 Su , u du + 𝜎(Su , u)dWu
= g(u)
𝜕u
𝜕Su
(
)
𝜕2V
1
g(u) 2 (𝜎(Su , u)2 du)
+
2
𝜕Su
(
)
2
𝜕V 1
𝜕V
2𝜕 V
+ 𝜎(Su , u)
= g(u)
+ 𝜇(Su , u)
− r(u)V(Su , u) du
𝜕u 2
𝜕Su
𝜕Su2
dZu =
+g(u)𝜎(Su , u)
= g(u)𝜎(Su , u)
since
𝜕V
dWu
𝜕Su
𝜕V
dWu
𝜕Su
𝜕V
𝜕V 1
𝜕2V
− r(u)V(Su , u) = 0.
+ 𝜎(Su , u)2 2 + 𝜇(Su , u)
𝜕t
2
𝜕Su
𝜕Su
Taking integrals we have
T
∫t
T
dZu =
g(u)𝜎(Su , u)
∫t
T
ZT − Zt =
∫t
e− ∫t
u
r(𝑣)d𝑣
𝜕V
dWu
𝜕Su
𝜎(Su , u)
𝜕V
dWu .
𝜕Su
By taking expectations and using the properties of the Itō integral,
)
( )
( )
(
𝔼 ZT − Zt = 0 or 𝔼 Zt = 𝔼 ZT
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149
and hence under the filtration ℱt ,
(
)
𝔼 Zt |ℱt = 𝔼(ZT |ℱt )
]
[
]
[
|
t
T
|
𝔼 e− ∫t r(𝑣)d𝑣 V(St , t)| ℱt = 𝔼 e− ∫t r(𝑣)d𝑣 V(ST , T)|| ℱt
|
|
]
[
|
T
V(St , t) = 𝔼 e− ∫t r(𝑣)d𝑣 Ψ(ST )|| ℱt .
|
◽
21. Backward Kolmogorov Equation for One-Dimensional Diffusion Process. Let {Wt ∶ t ≥
0} be a standard Wiener process on the probability space (Ω, ℱ, ℙ). For t ∈ [0, T], T > 0
consider the generalised stochastic differential equation
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
where 𝜇(Xt , t) and 𝜎(Xt , t) are functions dependent on Xt and t. By conditioning Xt = x
and XT = y, let p(x, t; y, T) be the transition probability density of XT at time T starting at
time t at point Xt . For any function Ψ(XT ), from the Feynman–Kac formula the function
[
T
− ∫t r(u)du
f (x, t) = 𝔼 e
= e − ∫t
T
r(u)du
∫
|
Ψ(XT )|| ℱt
|
]
Ψ(y)p(x, t; y, T) dy
satisfies the partial differential equation
𝜕2f
𝜕f
𝜕f
1
+ 𝜎(x, t)2 2 + 𝜇(x, t) − r(t)f (x, t) = 0
𝜕t 2
𝜕x
𝜕x
where r is a time-dependent function.
Show that the transition probability density p(t, x; T, y) in the y variable satisfies
1
𝜕2
𝜕
𝜕
p(x, t; y, T) + 𝜎(x, t)2 2 p(x, t; y, T) + 𝜇(x, t) p(x, t; y, T) = 0.
𝜕t
2
𝜕x
𝜕x
Solution: From the Feynman–Kac formula, for any function Ψ(XT ), the function
]
[
|
T
f (x, t) = 𝔼 e− ∫t r(u)du Ψ(XT )|| ℱt
|
= e − ∫t
T
r(u)du
∫
Ψ(y)p(x, t; y, T) dy
satisfies the following PDE:
𝜕f
𝜕f
𝜕2 f
1
+ 𝜎(x, t)2 2 + 𝜇(x, t) − r(t)f (x, t) = 0.
𝜕t 2
𝜕x
𝜕x
4:19 P.M.
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By differentiation, we have
T
T
𝜕f
𝜕
Ψ(y)p(x, t; y, T) dy + e− ∫t r(u)du
Ψ(y) p(x, t; y, T) dy
= r(t)e− ∫t r(u)du
∫
∫
𝜕t
𝜕t
= r(t)f (x, t) + e− ∫t
T
r(u)du
∫
Ψ(y)
𝜕
p(x, t; y, T) dy
𝜕t
T
𝜕f
𝜕
Ψ(y) p(x, t; y, T) dy and
= e− ∫t r(u)du
∫
𝜕x
𝜕x
T
𝜕2f
𝜕2
Ψ(y) 2 p(x, t; y, T) dy.
= e− ∫t r(u)du
2
∫
𝜕x
𝜕x
Substituting the above equations into the PDE, we obtain
e− ∫t
T
r(u)du
∫
×
Ψ(y)
[
]
𝜕
1
𝜕2
𝜕
p(x, t; y, T) + 𝜎(x, t)2 2 p(x, t; y, T) + 𝜇(x, t) p(x, t; y, T) dy = 0.
𝜕t
2
𝜕x
𝜕x
Finally, irrespective of the choice of Ψ(y) and r(t), the transition probability density function p(x, t; y, T) satisfies
𝜕2
𝜕
𝜕
1
p(x, t; y, T) + 𝜎(x, t)2 2 p(x, t; y, T) + 𝜇(x, t) p(x, t; y, T) = 0.
𝜕t
2
𝜕x
𝜕x
◽
22. Forward Kolmogorov Equation for One-Dimensional Diffusion Process. Let {Wt ∶ t ≥ 0}
be a standard Wiener process on the probability space (Ω, ℱ, ℙ). For t ∈ [0, T], T > 0
consider the generalised stochastic differential equation
dXt = 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt
where 𝜇(Xt , t) and 𝜎(Xt , t) are functions dependent on Xt and t. Using Itō’s formula on the
function f (Xt , t), show that
T
f (XT , T) = f (Xt , t) +
(
∫t
T
+
∫t
𝜎(Xs , s)
)
2
𝜕f
𝜕f
1
2𝜕 f
(X , s) + 𝜎(Xs , s)
(Xs , s) ds
(X , s) + 𝜇(Xs , s)
𝜕t s
𝜕Xt s
2
𝜕Xt2
𝜕f
(X , s) dWs
𝜕Xt s
and taking the expectation conditional on Xt = x, show that in the limit T → t
[
T
∫t
𝔼
]
|
2
𝜕f
𝜕f
1
|
2 𝜕 f
(X , s) + 𝜎(Xs , s)
(Xs , s)| Xt = x ds = 0.
(X , s) + 𝜇(Xs , s)
|
𝜕t s
𝜕Xt s
2
𝜕Xt 2
|
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151
Let XT = y and define the transition probability density p(x, t; y, T) of XT at time T starting
at time t at point Xt . By writing the conditional expectation in terms of p(x, t; y, T) and
integrating by parts twice, show that
𝜕
𝜕
1 𝜕2
(𝜎(y, T)2 p(x, t; y, T)) − (𝜇(y, T)p(x, t; y, T)).
p(x, t; y, T) =
2
𝜕T
2 𝜕y
𝜕y
Solution: For a suitable function f (Xt , t) and using Itō’s formula,
df (Xt , t) =
𝜕f
𝜕f
1 𝜕2f
(Xt , t) dt +
(Xt , t) dXt +
(X , t)(dXt )2 + . . .
𝜕t
𝜕Xt
2 𝜕Xt2 t
(
) 1
𝜕f
𝜕2f
𝜕f
(Xt , t) 𝜇(Xt , t) dt + 𝜎(Xt , t) dWt + 𝜎(Xt , t)2 2 (Xt , t) dt
(Xt , t) dt +
𝜕t
𝜕Xt
2
𝜕Xt
(
)
𝜕f
𝜕2f
𝜕f
1
(Xt , t) + 𝜇(Xt , t)
=
(Xt , t) + 𝜎(Xt , t)2 2 (Xt , t) dt
𝜕t
𝜕Xt
2
𝜕Xt
=
+ 𝜎(Xt , t)
𝜕f
(X , t) dWt .
𝜕Xt t
Taking integrals,
T
∫t
T
df (Xs , s) =
(
∫t
)
2
𝜕f
𝜕f
1
2𝜕 f
(X , s) + 𝜎(Xs , s)
(Xs , s) ds
(X , s) + 𝜇(Xs , s)
𝜕t s
𝜕Xt s
2
𝜕Xt2
T
+
𝜎(Xs , s)
∫t
f (XT , T) = f (Xt , t)
(
T
+
∫t
)
2
𝜕f
𝜕f
1
2𝜕 f
(X , s) + 𝜎(Xs , s)
(Xs , s) ds
(X , s) + 𝜇(Xs , s)
𝜕t s
𝜕Xt s
2
𝜕Xt2
T
+
∫t
𝜕f
(X , s) dWs
𝜕Xt s
𝜎(Xs , s)
𝜕f
(X , s) dWs .
𝜕Xt s
By taking conditional expectations given Xt = x we have
(
)
(
)
𝔼 f (XT , T)|| Xt = x = 𝔼 f (Xt , t)|| Xt = x
(
)
T
|
2
𝜕f
𝜕f
1
|
2𝜕 f
𝔼
(X , s) + 𝜎(Xs , s)
(Xs , s)| Xt = x ds
(X , s) + 𝜇(Xs , s)
+
|
∫t
𝜕t s
𝜕Xt s
2
𝜕Xt2
|
and since in the limit
(
)
(
)
lim 𝔼 f (XT , T)|| Xt = x = 𝔼 f (Xt , t)|| Xt = x
T→t
4:19 P.M.
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so
[
T
𝔼
∫t
]
|
2
𝜕f
𝜕f
1
|
2𝜕 f
(X , s) + 𝜎(Xs , s)
(Xs , s)| Xt = x ds = 0
(X , s) + 𝜇(Xs , s)
|
𝜕t s
𝜕Xt s
2
𝜕Xt2
|
or
T
∫t
∞
[
∫−∞
]
𝜕2f
𝜕f
𝜕f
1
(y, s) + 𝜇(y, s) (y, s) + 𝜎(y, s)2 2 (y, s) p(x, t; y, s) dyds = 0
𝜕s
𝜕y
2
𝜕y
where p(x, t; y, s) is the transition probability density function of Xt in the y-variable.
Integrating by parts,
T
∞
∫t ∫−∞
∞
𝜕f
(y, s)p(x, t; y, s) dyds =
∫−∞ ∫t
𝜕s
𝜕f
(y, s)p(x, t; y, s) ds dy
𝜕s
|T
f (y, s)p(x, t; y, s)|| dy
|t
∞
=
T
∫−∞
∞
−
T
∫−∞ ∫t
∞
= −
T
∫t
∞
∫−∞
𝜇(y, s)
∫t
𝜕f
(y, s)p(x, t; y, s) dyds =
∫t
𝜕y
T
−
∫t
T
∫t
∞
∫−∞
1
2
𝜎(y, s)2
T
=
∫t
f (y, s)
∞
f (y, s)
𝜕
(𝜇(y, s)p(x, t; y, s)) dyds
𝜕y
f (y, s)
𝜕
(𝜇(y, s)p(x, t; y, s)) dyds
𝜕y
∞
∫−∞
𝜕2f
(y, s)p(x, t; y, s) dyds
𝜕y2
|∞
1 𝜕
|
2
f (y, s)𝜎(y, s) p(x, t; y, s)| ds
|
2 𝜕y
|−∞
T
−
∫t
𝜕
p(x, t; y, s) ds dy
𝜕s
𝜕
p(x, t; y, s) dyds
𝜕s
|∞
|
f (y, s)𝜇(y, s)p(x, t; y, s)|
ds
|
|−∞
∫−∞
T
= −
f (y, s)
∞
∫−∞
T
𝜕
p(x, t; y, s) ds dy
𝜕s
T
∫−∞ ∫t
T
= −
f (y, s)
∫t
T
∞
𝜕
1 𝜕
f (y, s) (𝜎(y, s)2 p(x, t; y, s)) dyds
∫−∞ 2 𝜕y
𝜕y
∞
1 𝜕
𝜕
f (y, s) (𝜎(y, s)2 p(x, t; y, s)) dyds
∫t ∫−∞ 2 𝜕y
𝜕y
∞
T
)||
1
𝜕 (
=−
𝜎(y, s)2 p(x, t; y, s) | ds
f (y, s)
|
∫t 2
𝜕y
|−∞
=−
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T
+
∫t
∞
∫−∞
∫t
T
=
153
∞
∫−∞
1
𝜕2
f (y, s) 2 (𝜎(y, s)2 p(x, t; y, s)) dyds
2
𝜕y
1
𝜕2
f (y, s) 2 (𝜎(y, s)2 p(x, t; y, s)) dyds
2
𝜕y
and by substituting the above relationships into the integro partial differential equation,
we have
T
∫t
[
∞
∫−∞
f (y, s) ×
]
)
𝜕
1 𝜕2 (
𝜕
2
p(x, t; y, s) +
𝜎(y, s) p(x, t; y, s) dyds = 0.
(𝜇 (y, s) p (x, t; y, s)) −
𝜕s
𝜕y
2 𝜕y2
Finally, by differentiating the above equation with respect to T, we obtain
∞
f (y, T) ×
∫−∞
[
]
)
𝜕
1 𝜕2 (
𝜕
2
p(x, t; y, T) +
𝜎(y,
T)
(𝜇(y, T)p(x, t; y, T)) −
p(x,
t;
y,
T)
dy = 0
𝜕T
𝜕y
2 𝜕y2
and irrespective of the choice of f , the transition probability density function p(x, t; y, T)
satisfies
) 𝜕
1 𝜕2 (
𝜕
2
p(x, t; y, T) =
𝜎(y,
T)
(𝜇(y, T)p(x, t; y, T)) .
p(x,
t;
y,
T)
−
𝜕T
2 𝜕y2
𝜕y
◽
23. Backward Kolmogorov Equation for a One-Dimensional Random Walk. We consider a
one-dimensional symmetric random walk where at initial time t0 , a particle starts at x0 and
is at position x at time t. At time t + 𝛿t, the particle can either move to x + 𝛿x or x − 𝛿x each
with probability 12 . Let p(x, t; x0 , t0 ) denote the probability density of the particle position
x at time t starting at x0 at time t0 .
By writing the backward equation
in a discrete fashion and expanding it using Taylor’s
√
series, show that for 𝛿x = 𝛿t and in the limit 𝛿t → 0
2
𝜕p(x, t; x0 , t0 )
1 𝜕 p(x, t; x0 , t0 )
.
=−
𝜕t
2
𝜕x2
Solution: By denoting p(x, t; x0 , t0 ) as the probability density function of the particle position x at time t, hence the discrete model of the backward equation is
p(x, t; x0 , t0 ) =
1
1
p(x − 𝛿x, t + 𝛿t; x0 , t0 ) + p(x + 𝛿x, t + 𝛿t; x0 , t0 ).
2
2
Using Taylor’s series, we have
p(x − 𝛿x, t + 𝛿t; x0 , t0 ) = p(x, t + 𝛿t; x0 , t0 ) −
+
𝜕p(x, t + 𝛿t; x0 , t0 )
𝛿x
𝜕x
2
1 𝜕 p(x, t + 𝛿t; x0 , t0 )
(−𝛿x)2 + O((𝛿x)3 )
2
𝜕x2
4:19 P.M.
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and
p(x + 𝛿x, t + 𝛿t; x0 , t0 ) = p(x, t + 𝛿t; x0 , t0 ) +
+
𝜕p(x, t + 𝛿t; x0 , t0 )
𝛿x
𝜕x
2
1 𝜕 p(x, t + 𝛿t; x0 , t0 )
(𝛿x)2 + O((𝛿x)3 ).
2
𝜕x2
Substituting these two equations into the backward equation,
p(x, t; x0 , t0 ) = p(x, t + 𝛿t; x0 , t0 ) +
By setting 𝛿x =
[
− lim
𝛿t→0
2
1 𝜕 p(x, t + 𝛿t; x0 , t0 )
(𝛿x)2 + O((𝛿x)3 ).
2
𝜕x2
√
𝛿t, dividing the equation by 𝛿t and taking limits 𝛿t → 0
[ 2
]
√ ]
p(x, t + 𝛿t; x0 , t0 ) − p(x, t; x0 , t0 )
1 𝜕 p(x, t + 𝛿t; x0 , t0 )
+
O(
𝛿t)
= lim
𝛿t→0 2
𝛿t
𝜕x2
we finally have
2
𝜕p(x, t; x0 , t0 )
1 𝜕 p(x, t; x0 , t0 )
=−
.
𝜕t
2
𝜕x2
◽
24. Forward Kolmogorov Equation for a One-Dimensional Random Walk. We consider a
one-dimensional symmetric random walk where at initial time t0 , a particle starts at y0
and is at position y at terminal time T > 0. At time T − 𝛿T, the particle can either move to
y + 𝛿y or y − 𝛿y each with probability 12 . Let p(y, T; y0 , t0 ) denote the probability density
of the position y at time T starting at y0 at time t0 .
By writing the forward equation
in a discrete fashion and expanding it using Taylor’s
√
series, show that for 𝛿y = 𝛿T and in the limit 𝛿T → 0
𝜕p(y, T; y0 , t0 ) 1 𝜕 2 p(y, T; y0 , t0 )
=
.
𝜕T
2
𝜕y2
Solution: By denoting p(y, T; y0 , t0 ) as the probability density function of the particle
position y at time T, hence the discrete model of the forward equation is
p(y, T; y0 , t0 ) =
1
1
p(y − 𝛿y, T − 𝛿T; y0 , t0 ) + p(y + 𝛿y, T − 𝛿T; y0 , t0 ).
2
2
By expanding p(y − 𝛿y, T − 𝛿T; y0 , t0 ) and p(y + 𝛿y, T − 𝛿T; y0 , t0 ) using Taylor’s series,
we have
p(y − 𝛿y, T − 𝛿T; y0 , t0 ) = p(y, T − 𝛿T; y0 , t0 ) −
+
𝜕p(y, T − 𝛿T; y0 , t0 )
𝛿y
𝜕y
2
1 𝜕 p(y, T − 𝛿T; y0 , t0 )
(−𝛿y)2 + O((𝛿y)3 )
2
𝜕y2
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155
and
p(y + 𝛿y, T − 𝛿T; y0 , t0 ) = p(y, T − 𝛿T; y0 , t0 ) +
𝜕p(y, T − 𝛿T; y0 , t0 )
𝛿y
𝜕y
2
1 𝜕 p(y, T − 𝛿T; y0 , t0 )
(𝛿y)2 + O((𝛿y)3 ).
2
𝜕y2
+
Substituting the above two equations into the discrete forward equation,
p(y, T; y0 , t0 ) = p(y, T − 𝛿T; y0 , t0 ) +
By setting 𝛿y =
[
lim
𝛿T→0
√
𝛿T, dividing the equation by 𝛿T and taking limits 𝛿T → 0
[ 2
]
√ ]
p(y, T; y0 , t0 ) − p(y, T − 𝛿T; y0 , t0 )
1 𝜕 p(y, T − 𝛿T; y0 , t0 )
+
O(
𝛿T)
= lim
𝛿T→0 2
𝛿T
𝜕y2
we finally have
3.2.3
2
1 𝜕 p(y, T − 𝛿T; y0 , t0 )
(𝛿y)2 + O((𝛿y)3 ).
2
𝜕y2
𝜕p(y, T; y0 , t0 ) 1 𝜕 2 p(y, T; y0 , t0 )
.
=
𝜕T
2
𝜕y2
◽
Multi-Dimensional Diffusion Process
1. Let (Ω, ℱ, ℙ) be a probability space and consider n assets with prices St(i) , i = 1, 2, . . . , n
satisfying the SDEs
dSt(i) = 𝜇 (i) St(i) dt + 𝜎 (i) St(i) dWt(i)
(
)(
)
(j)
dWt(i) dWt = 𝜌(ij) dt
where {Wt(i) ∶ t ≥ 0}, i = 1, 2, . . . , n are standard Wiener processes, 𝜌(ij) ∈ (−1, 1), i ≠ j
and 𝜌(ii) = 1.
By considering the function f (St(1) , St(2) , . . . , St(n) ), show using Itō’s formula that
df (St(1) , St(2) ,
...
, St(n) ) =
n
∑
𝜇(i) St(i)
i=1
1 ∑ ∑ (ij) (i) (j) (i) (j) 𝜕 2 f
dt
+
𝜌 𝜎 𝜎 St St
dt
(j)
2 i=1 j=1
𝜕S(i)
𝜕S(i) 𝜕S
n
t
∑
n
+
n
𝜕f
𝜎 (i) St(i)
i=1
t
𝜕f
𝜕St(i)
t
dWt(i) .
Solution: By expanding df (St(1) , St(2) , . . . , St(n) ) using Taylor’s formula,
df (St(1) , St(2) ,
...
, St(n) )
=
n
∑
𝜕f
dSt(i)
(i)
i=1 𝜕St
1 ∑ ∑ 𝜕2f
(j)
+
dS(i) dSt + . . .
2 i=1 j=1 𝜕S(i) S(j) t
n
n
t
t
4:19 P.M.
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156
3.2.3
=
n
∑
𝜕f (
i=1
𝜕St(i)
𝜇(i) St(i) dt + 𝜎 (i) St(i) dWt(i)
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Multi-Dimensional Diffusion Process
)
n
n
(
)(
)
1 ∑ ∑ 𝜕2f
(j)
(i)
(i) (i)
(i) (i)
(j) (j)
(j) (j)
𝜇
𝜇
S
dt
+
𝜎
S
dW
S
dt
+
𝜎
S
dW
+
t
t
t
t
t
t
2 i=1 j=1 𝜕S(i) S(j)
t
t
+ ...
(j)
By setting (dt)2 = 0, (dWt(i) )2 = dt, (dWt(i) )(dWt ) = 𝜌(ij) dt, dWt(i) dt = 0, i, j = 1, 2, . . . , n
we have
df (St(1) , St(2) , . . . , St(n) ) =
n
∑
𝜇 (i) St(i)
i=1
+
n
∑
dt +
(i)
𝜕St
𝜎 (i) St(i)
i=1
1 ∑ ∑ (ij) (i) (j) (i) (j) 𝜕 2 f
𝜌 𝜎 𝜎 St St
dt
(j)
2 i=1 j=1
𝜕S(i) 𝜕S
n
𝜕f
n
t
𝜕f
𝜕St(i)
t
dWt(i) .
◽
2. Let (Ω, ℱ, ℙ) be a probability space. We consider two assets with prices St(1) , St(2) satisfying
the SDEs
dSt(1) = 𝜇(1) St(1) dt + 𝜎 (1) St(1) dWt(1)
dSt(2) = 𝜇(2) St(2) dt + 𝜎 (2) St(2) dWt(2)
dWt(1) dWt(2) = 𝜌dt
where 𝜇 (1) , 𝜇(2) , 𝜎 (1) , 𝜎 (2) are constants and {Wt(1) ∶ t ≥ 0}, {Wt(2) ∶ t ≥ 0} are standard
Wiener processes with correlation 𝜌.
By letting Ut = St(1) ∕St(2) show that the SDE satisfied by Ut is
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) − 𝜇(2) + (𝜎 (2) )2 − 𝜌𝜎 (1) 𝜎 (2) , 𝜎 =
√
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2) and
𝜎 (1) Wt(1) − 𝜎 (2) Wt(2)
.
Vt = √
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
Finally, show that {Vt ∶ t ≥ 0} is a standard Wiener process.
Solution: By using Taylor’s expansion and applying Itō’s formula,
)
(
dUt = d St(1) ∕St(2)
=
1
dSt(1) −
(2)
St
St(1)
(St(2) )2
dSt(2) +
St(1)
(St(2) )3
(dSt(2) )2 −
1
(St(2) )2
dSt(1) dSt(2) + . . .
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157
)
(
= 𝜇(1) Ut dt + 𝜎 (1) Ut dWt(1) − 𝜇 (2) Ut dt + 𝜎 (2) Ut dWt(2)
+(𝜎 (2) )2 Ut dt − 𝜌𝜎 (1) 𝜎 (2) Ut dt
(
)
)
(
= 𝜇(1) − 𝜇(2) + (𝜎 (2) )2 − 𝜌𝜎 (1) 𝜎 (2) Ut dt + Ut 𝜎 (1) dWt(1) − 𝜎 (2) dWt(2)
√
)
(
= 𝜇 (1) − 𝜇(2) + (𝜎 (2) )2 − 𝜌𝜎 (1) 𝜎 (2) Ut dt + (𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2) Ut dVt .
Therefore,
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) − 𝜇(2) + (𝜎 (2) )2 − 𝜌𝜎 (1) 𝜎 (2) , 𝜎 =
√
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2) and
𝜎 (1) Wt(1) − 𝜎 (2) Wt(2)
.
Vt = √
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
To show that {Vt ∶ t ≥ 0} is a standard Wiener process, we first show that Vt ∼ 𝒩(0, t).
Taking expectations,
𝜎 (1) 𝔼(Wt(1) ) − 𝜎 (2) 𝔼(Wt(2) )
=0
𝔼(Vt ) = √
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
]
[
(1) )2 (W (1) )2 + (𝜎 (2) )2 (W (2) )2 − 2𝜎 (1) 𝜎 (2) W (1) W (2)
(𝜎
t
t
t
t
𝔼(Vt2 ) = 𝔼
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
=
(𝜎 (1) )2 t + (𝜎 (2) )2 t − 2𝜌𝜎 (1) 𝜎 (2) t
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
= t.
Given that Wt(1) ∼ 𝒩(0, t), Wt(2) ∼ 𝒩(0, t) and a linear combination of normal variates is
also normal, therefore Vt ∼ 𝒩(0, t).
To show that {Vt ∶ t ≥ 0} is a standard Wiener process, we note the following:
(a) V0 = 0 and it is clear that Vt has continuous sample paths for t ≥ 0.
(b) For t > 0, s > 0 we have
(1)
(2)
− Wt(1) ∼ 𝒩(0, s),
Wt+s
− Wt(2) ∼ 𝒩(0, s)
Wt+s
)
(
(1)
(2)
Cov Wt+s
− Wt(1) , Wt+s
− Wt(2) = 𝜌s
and hence
(
)
(
)
(1)
(2)
𝜎 (1) Wt+s
− Wt(1) − 𝜎 (2) Wt+s
− Wt(2)
Vt+s − Vt =
√
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
4:19 P.M.
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3.2.3
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Multi-Dimensional Diffusion Process
with mean
(
(
)
)
(1)
(1)
(2)
(1)
− Wt(2)
− 𝜎 (2) 𝔼 Wt+s
(
) 𝜎 𝔼 Wt+s − Wt
=0
𝔼 Vt+s − Vt =
√
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
and variance
(
)
Var Vt+s − Vt =
)
)
(
(
⎡(𝜎 (1) )2 Var W (1) − W (1) + (𝜎 (2) )2 Var W (2) − W (2) ⎤
t+s
t+s
⎢
(t
) t ⎥
(1)
(1)
(2)
(2)
(1)
(2)
⎢
⎥
−2𝜎 𝜎 Cov Wt+s − Wt , Wt+s − Wt
⎣
⎦
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
= s.
Therefore, Vt+s − Vt ∼ 𝒩(0, s).
(c) For t > 0, s > 0, to show that Vt+s − Vt ⟂
⟂ Vt we note that
[(
) ]
(
)
( 2)
𝔼 Vt+s − Vt Vt = 𝔼 Vt+s Vt − 𝔼 Vt
(
)(
)
⎡ (1) (1)
(2) W (2)
(1) W (1) − 𝜎 (2) W (2) ⎤
W
−
𝜎
𝜎
𝜎
t
t
t+s
t+s
( 2)
⎢
⎥
= 𝔼⎢
⎥ − 𝔼 Vt
(1)
2
(2)
2
(1)
(2)
(𝜎 ) + (𝜎 ) − 2𝜌𝜎 𝜎
⎢
⎥
⎣
⎦
(
(
)
)
⎡(𝜎 (1) )2 𝔼 W (1) W (1) − 𝜎 (1) 𝜎 (2) 𝔼 W (1) W (2) ⎤
t+s
t+s t
⎥
⎢
(t
)
(1)
(2)
(2) )2 𝔼(W (2) W (2) )⎥
⎢−𝜎 (1) 𝜎 (2) 𝔼 W W
+
(𝜎
t
t
t+s
t+s ⎦
⎣
−t
=
(1)
2
(2)
2
(1)
(2)
(𝜎 ) + (𝜎 ) − 2𝜌𝜎 𝜎
]
[ (1) 2
(𝜎 ) min {t, t + s} − 𝜌𝜎 (1) 𝜎 (2) min{t, t + s}
−𝜌𝜎 (1) 𝜎 (2) min{t, t + s} + (𝜎 (2) )2 min{t, t + s}
−t
=
(𝜎 (1) )2 + (𝜎 (2) )2 − 2𝜌𝜎 (1) 𝜎 (2)
=t−t
= 0.
of Vt and) Vt+s − Vt
Since Vt ∼ 𝒩(0, t), Vt+s − Vt ∼ 𝒩(0, s) and the joint distribution
(
is a bivariate normal (see Problem 2.2.1.5, page 58), if Cov Vt+s − Vt , Vt = 0 then
Vt+s − Vt ⟂
⟂ Vt .
From the results of (a)–(c) we have shown that Vt is a standard Wiener process.
◽
3. Let (Ω, ℱ, ℙ) be a probability space. We consider two assets with prices St(1) , St(2) satisfying
the SDEs
dSt(1) = 𝜇(1) St(1) dt + 𝜎 (1) St(1) dWt(1)
dSt(2) = 𝜇(2) St(2) dt + 𝜎 (2) St(2) dWt(2)
dWt(1) dWt(2) = 𝜌dt
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Multi-Dimensional Diffusion Process
159
where 𝜇(1) , 𝜇(2) , 𝜎 (1) , 𝜎 (2) are constants and {Wt(1) ∶ t ≥ 0}, {Wt(2) ∶ t ≥ 0} are standard
Wiener processes with correlation 𝜌.
By letting Ut = St(1) St(2) , show that the SDE satisfied by Ut is
dUt = 𝜇Ut dt + 𝜎Ut dVt
√
where 𝜇 = 𝜇(1) + 𝜇(2) + 𝜌𝜎 (1) 𝜎 (2) , 𝜎 = (𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2) and
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2)
.
Vt = √
(𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2)
Show that {Vt ∶ t ≥ 0} is a standard Wiener process.
Solution: From Itō’s formula,
dUt = d(St(1) St(2) )
= St(1) dSt(2) + St(2) dSt(1) + (dSt(1) )(dSt(2) ) + . . .
(
)
(
)
= St(1) 𝜇(1) St(1) dt + 𝜎 (1) St(1) dWt(1) + St(2) 𝜇 (2) St(2) dt + 𝜎 (2) St(2) dWt(2)
+𝜌𝜎 (1) 𝜎 (2) St(1) St(2) dt
(
)
(
)
= 𝜇 (1) + 𝜇(2) + 𝜌𝜎 (1) 𝜎 (2) Ut dt + Ut 𝜎 (1) dWt(1) + 𝜎 (2) dWt(2)
√
(
)
= 𝜇 (1) + 𝜇(2) + 𝜌𝜎 (1) 𝜎 (2) Ut dt + (𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2) Ut dVt .
Therefore,
dUt = 𝜇Ut dt + 𝜎Ut dVt
√
where 𝜇 = 𝜇(1) + 𝜇(2) + 𝜌𝜎 (1) 𝜎 (2) , 𝜎 = (𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2) and
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2)
Vt = √
.
(𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2)
Following the same procedure as described in Problem 3.2.3.2 (page 156), we can show
that
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2)
∼ N(0, t)
Vt = √
(𝜎 (1) )2 + (𝜎 (2) )2 + 2𝜌𝜎 (1) 𝜎 (2)
and is also a standard Wiener process.
◽
4. Let (Ω, ℱ, ℙ) be a probability space. We consider three assets with prices St(1) , St(2) and St(3)
satisfying the SDEs
dSt(1) = 𝜇 (1) St(1) dt + 𝜎 (1) St(1) dWt(1)
dSt(2) = 𝜇 (2) St(2) dt + 𝜎 (2) St(2) dWt(2)
4:19 P.M.
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Multi-Dimensional Diffusion Process
dSt(3) = 𝜇 (3) St(3) dt + 𝜎 (3) St(3) dWt(3)
(j)
dWt(i) dWt = 𝜌ij dt,
i ≠ j and
i, j = 1, 2, 3
where 𝜇 (1) , 𝜇(2) , 𝜇 (3) , 𝜎 (1) , 𝜎 (2) , 𝜎 (3) are constants and {Wt(1) ∶ t ≥ 0}, {Wt(2) ∶ t ≥ 0},
(j)
{Wt(3) ∶ t ≥ 0} are standard Wiener processes with correlations dWt(i) dWt = 𝜌ij dt, i ≠ j
and (dWt(i) )2 = dt for i, j = 1, 2, 3.
By letting Ut = (St(1) St(2) )∕St(3) , show that the SDE satisfied by Ut is
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) + 𝜇(2) − 𝜇(3) + 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) − 𝜌23 𝜎 (2) 𝜎 (3) , 𝜎 2 = (𝜎 (1) )2 +
(𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3) and
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
Vt = √
.
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3)
Show that {Vt ∶ t ≥ 0} is a standard Wiener process.
Solution: From Itō’s lemma,
dUt =
𝜕Ut
𝜕St(1)
+
+
=
dSt(1) +
𝜕Ut
𝜕St(2)
dSt(2) +
𝜕Ut
𝜕St(3)
dSt(3)
2
2
2
1 𝜕 Ut
1 𝜕 Ut
1 𝜕 Ut
(1) 2
(2) 2
(dS
)
+
(dS
)
+
(dS(3) )2
t
t
2 𝜕(S(1) )2
2 𝜕(S(2) )2
2 𝜕(S(3) )2 t
t
t
t
𝜕 2 Ut
dSt(1) dSt(2) +
(2)
𝜕St(1) 𝜕St
𝜕 2 Ut
dSt(1) dSt(3) +
(3)
𝜕St(1) 𝜕St
𝜕 2 Ut
𝜕St(2) 𝜕St(3)
dSt(2) dSt(3) + . . .
St(2) (
St(3)
−
+
−
) S(1) (
)
𝜇 (1) St(1) dt + 𝜎 (1) St(1) dWt(1) + t(3) 𝜇(2) St(2) dt + 𝜎 (2) St(2) dWt(2)
St
St(1) St(2) (
(St(3) )2
) S(1) S(2) (
)
𝜇 (3) St(3) dt + 𝜎 (3) St(3) dWt(3) + t (3) t
(𝜎 (3) St(3) )2 dt
(St )3
1
St(3)
(
𝜌12 𝜎 (1) 𝜎 (2) St(1) St(2) dt
St(1) (
(St(3) )2
)
−
𝜌23 𝜎 (2) 𝜎 (3) St(2) St(3) dt
St(2) (
(St(3) )2
𝜌13 𝜎 (1) 𝜎 (3) St(1) St(3) dt
)
)
(
)
= 𝜇 (1) + 𝜇(2) − 𝜇(3) + 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) − 𝜌23 𝜎 (2) 𝜎 (3) Ut dt
(
)
+Ut 𝜎 (1) dWt(1) + 𝜎 (2) dWt(2) − 𝜎 (3) dWt(3)
(
)
= 𝜇(1) + 𝜇(2) − 𝜇(3) + 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) − 𝜌23 𝜎 (2) 𝜎 (3) Ut dt
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√
+
161
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3) Ut dVt
which therefore becomes
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) + 𝜇(2) − 𝜇(3) + 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) − 𝜌23 𝜎 (2) 𝜎 (3) , 𝜎 2 = (𝜎 (1) )2 +
(𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3) and
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
Vt = √
.
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3)
Following the steps described in Problem 3.2.3.2 (page 156), we can easily show that
𝜎 (1) Wt(1) + 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
Vt = √
∼ 𝒩(0, 1)
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 + 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) − 2𝜌23 𝜎 (2) 𝜎 (3)
and is also a standard Wiener process.
◽
5. Let (Ω, ℱ, ℙ) be a probability space. We consider three assets with prices St(1) , St(2) and St(3)
satisfying the SDEs
dSt(1) = 𝜇 (1) St(1) dt + 𝜎 (1) St(1) dWt(1)
dSt(2) = 𝜇 (2) St(2) dt + 𝜎 (2) St(2) dWt(2)
dSt(3) = 𝜇 (3) St(3) dt + 𝜎 (3) St(3) dWt(3)
(j)
dWt(i) dWt = 𝜌ij dt,
i ≠ j and
i, j = 1, 2, 3
where 𝜇 (1) , 𝜇(2) , 𝜇 (3) , 𝜎 (1) , 𝜎 (2) , 𝜎 (3) are constants and {Wt(1) ∶ t ≥ 0}, {Wt(2) ∶ t ≥ 0},
(j)
{Wt(3) ∶ t ≥ 0} are standard Wiener processes with correlations dWt(i) dWt = 𝜌ij dt, i ≠ j
and (dWt(i) )2 = dt for i, j = 1, 2, 3.
By letting Ut = St(1) ∕(St(2) St(3) ), show that the SDE satisfied by Ut is
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) − 𝜇(2) − 𝜇(3) + (𝜎 (2) )2 + (𝜎 (3) )2 − 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) + 𝜌23 𝜎 (2) 𝜎 (3) ,
𝜎 2 = (𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3) and
𝜎 (1) Wt(1) − 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
Vt = √
.
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3)
Show that {Vt ∶ t ≥ 0} is a standard Wiener process.
4:19 P.M.
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162
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Multi-Dimensional Diffusion Process
Solution: Applying Itō’s lemma,
dUt =
𝜕Ut
𝜕St(1)
+
𝜕Ut
𝜕St(2)
dSt(2) +
𝜕Ut
𝜕St(3)
dSt(3)
2
2
2
1 𝜕 Ut
1 𝜕 Ut
1 𝜕 Ut
(1) 2
(2) 2
(dS
)
+
(dS
)
+
(dSt(3) )2
t
2 𝜕(S(1) )2
2 𝜕(S(2) )2 t
2 𝜕(S(3) )2
t
t
t
+
=
dSt(1) +
𝜕 2 Ut
𝜕St(1) 𝜕St
St(2) St(3)
+
−
𝜕 2 Ut
dSt(1) dSt(3) +
(3)
𝜕St(1) 𝜕St
(
1
−
dSt(1) dSt(2) +
(2)
)
𝜇 (1) St(1) dt + 𝜎 (1) St(1) dWt(1) −
St(1)
St(1)
)
(𝜎 (3) St(3) )2 dt −
St(2) (St(3) )3
1
(
1
(St(2) )2 St(3)
)
𝜌13 𝜎 (1) 𝜎 (3) St(1) St(3) dt +
dSt(2) dSt(3) + . . .
(
𝜇(2) St(2) dt + 𝜎 (2) St(2) dWt(2)
(St(2) )2 St(3)
)
𝜇(3) St(3) dt + 𝜎 (3) St(3) dWt(3) +
(
𝜕St(2) 𝜕St(3)
St(1)
(
St(2) (St(3) )2
𝜕 2 Ut
St(1)
(St(2) )3 St(3)
(
(𝜎 (2) St(2) )2 dt
(
𝜌12 𝜎 (1) 𝜎 (2) St(1) St(2) dt
St(1)
(
)
)
)
𝜌23 𝜎 (2) 𝜎 (3) St(2) St(3) dt
)
St(2) (St(3) )2
(St(2) St(3) )2
)
( (1)
= 𝜇 − 𝜇(2) − 𝜇(3) + (𝜎 (2) )2 + (𝜎 (3) )2 − 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) + 𝜌23 𝜎 (2) 𝜎 (3) Ut dt
(
)
+Ut 𝜎 (1) dWt(1) − 𝜎 (2) dWt(2) − 𝜎 (3) dWt(3)
)
(
= 𝜇 (1) − 𝜇(2) − 𝜇(3) + (𝜎 (2) )2 + (𝜎 (3) )2 − 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) + 𝜌23 𝜎 (2) 𝜎 (3) Ut dt
√
+ (𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3) Ut dVt .
We can therefore write
dUt = 𝜇Ut dt + 𝜎Ut dVt
where 𝜇 = 𝜇(1) − 𝜇(2) − 𝜇(3) + (𝜎 (2) )2 + (𝜎 (3) )2 − 𝜌12 𝜎 (1) 𝜎 (2) − 𝜌13 𝜎 (1) 𝜎 (3) + 𝜌23 𝜎 (2) 𝜎 (3) ,
𝜎 2 = (𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3) and
𝜎 (1) Wt(1) − 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
Vt = √
.
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3)
As described in Problem 3.2.3.2 (page 156) we can easily show that
Vt = √
𝜎 (1) Wt(1) − 𝜎 (2) Wt(2) − 𝜎 (3) Wt(3)
(𝜎 (1) )2 + (𝜎 (2) )2 + (𝜎 (3) )2 − 2𝜌12 𝜎 (1) 𝜎 (2) − 2𝜌13 𝜎 (1) 𝜎 (3) + 2𝜌23 𝜎 (2) 𝜎 (3)
∼ 𝒩(0, t)
and is a standard Wiener process.
◽
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163
)
(
6. Bessel Process. Let (𝛀, ℱ, ℙ) be a probability space and let Wt = Wt(1) , Wt(2) , . . . , Wt(n)
be an n-dimensional standard Wiener process, with Wt(i) , i = 1, 2, . . . , n being independent
one-dimensional standard Wiener processes. By setting
√
Xt = (Wt(1) )2 + (Wt(2) )2 + . . . + (Wt(n) )2
show that
t
Wt =
∫0
t
t
Ws(1)
Ws(2)
Ws(n)
dWs(1) +
dWs(2) + . . . +
dWs(n)
∫0 Xs
∫0 Xs
Xs
is a standard Wiener process.
Show also that Xt satisfies the n-dimensional Bessel process with SDE
(
dXt =
n−1
2Xt
)
dt + dWt
where X0 = 0.
Solution: We first need to show that Wt ∼ 𝒩(0, t). Using the properties of the stochastic
Itō integral and the independence property of Wt(i) , i = 1, 2, . . . , n,
]
t
t
Ws(1)
Ws(2)
Ws(n)
(1)
(2)
(n)
dWs +
dWs + . . . +
dWs
𝔼(Wt ) = 𝔼
∫0 Xs
∫0 Xs
∫0 Xs
)
(
)
(
)
(
t
t
t
Ws(1)
Ws(2)
Ws(n)
(1)
(2)
(n)
+𝔼
+ ... + 𝔼
=𝔼
dWs
dWs
dWs
∫0 Xs
∫0 Xs
∫ 0 Xs
[
=0
t
(
⎡
𝔼(Wt2 ) = 𝔼 ⎢
⎢ ∫0
⎣
t
Ws(1)
dWs(1)
Xs
(
+
t
∫0
t
Ws(i)
dWs(i)
Xs
)(
Ws(2)
dWs(2)
Xs
)2
(
+ ... +
)⎤
(j)
Ws
(j) ⎥
+2
dWs ⎥
∫0
∫ 0 Xs
⎥
i=1 j=1
⎦
i≠j
(
(
)2
)2
⎡
(1)
⎤
⎡
(2)
⎤
t
t
W
W
(1)
(2)
s
s
⎥ + 𝔼⎢
⎥
=𝔼⎢
dWs
dWs
⎢ ∫0 Xs
⎥
⎢ ∫0 Xs
⎥
⎣
⎦
⎣
⎦
)2
(
⎡
⎤
t
Ws(n)
dWs(n) ⎥
+ ... + 𝔼⎢
⎢ ∫0 Xs
⎥
⎣
⎦
n
n
∑
∑
(
)2
t
t
∫0
Ws(n)
dWs(n)
Xs
)2
4:19 P.M.
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Multi-Dimensional Diffusion Process
(
)
(
)
(
)
⎡ t W (1) 2 ⎤
⎡ t W (2) 2 ⎤
⎡ t W (n) 2 ⎤
s
s
s
=𝔼⎢
ds⎥ + 𝔼 ⎢
ds⎥ + . . . + 𝔼 ⎢
ds⎥
⎢∫0
⎥
⎢∫ 0
⎥
⎢∫0
⎥
Xs
Xs
Xs
⎣
⎦
⎣
⎦
⎣
⎦
( t )
=𝔼
ds
∫0
= t.
Because Wt is a function of n independent normal variates, so Wt ∼ 𝒩(0, t). To show that
Wt is also a standard Wiener process, we note the following:
[
]
(2)
(n)
t
t
t
Ws(1)
W
W
s
s
dWs(1) +
dWs(2) + . . . +
dWs(n) = 0 and it is
(a) W0 = lim
∫0 Xs
∫0 Xs
t→0 ∫0
Xs
clear that Wt has continuous sample paths for t ≥ 0.
(b) For t > 0, u > 0
t+u
Wt+u − Wt =
∫0
t
t
Ws(1)
Ws(2)
Ws(n)
dWs(1) −
dWs(2) − . . . −
dWs(n)
∫ 0 Xs
∫0 Xs
Xs
t
−
∫0
t+u
=
t+u
t+u
Ws(1)
Ws(2)
Ws(n)
dWs(1) +
dWs(2) + . . . +
dWs(n)
∫0
∫0
Xs
Xs
Xs
∫t
Ws(1)
dWs(1) +
∫t
Xs
Ws(2)
dWs(2) + . . . +
∫t
Xs
t+u
t+u
Ws(n)
dWs(n) .
Xs
Taking expectations,
)
(
Ws(1)
(1)
dWs
+𝔼
∫t
∫t
Xs
)
(
t+u
Ws(n)
(n)
+ ... + 𝔼
dWs
∫t
Xs
(
)
𝔼 Wt+u − Wt = 𝔼
(
t+u
t+u
Ws(2)
dWs(2)
Xs
)
=0
and due to the independence of Wt(i) , i = 1, 2, . . . , n,
𝔼
[(
Wt+u − Wt
(
⎡
= 𝔼⎢
⎢ ∫t
⎣
t+u
+2
)2 ]
Ws(1)
dWs(1)
Xs
n ∑
n
∑
i=1 j=1
i≠j
(
t+u
∫t
(
)2
+
t+u
∫t
Ws(i)
dWs(i)
Xs
Ws(2)
dWs(2)
Xs
)(
∫t
t+u
(
)2
+ ... +
)⎤
(j)
Ws
(j) ⎥
dWs ⎥
Xs
⎥
⎦
t
∫0
Ws(n)
dWs(n)
Xs
)2
Page 164
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(
⎡
= 𝔼⎢
⎢ ∫t
⎣
(
)2
⎤
⎡
(2)
⎤
t+u
W
(2)
s
⎥ + 𝔼⎢
⎥
dWs
⎥
⎢ ∫t
⎥
Xs
⎦
⎣
⎦
)2
(
⎤
⎡
t+u
Ws(n)
(n)
⎥
⎢
dWs
+ ... + 𝔼
⎥
⎢ ∫t
Xs
⎦
⎣
(
)2
(
)
⎡ t+u W (1)
⎡ t+u W (2) 2 ⎤
⎡
⎤
s
s
= 𝔼⎢
ds⎥ + 𝔼 ⎢
ds⎥ + . . . + 𝔼 ⎢
⎢∫t
⎢∫ t
⎢∫t
⎥
⎥
Xs
Xs
⎣
⎣
⎣
⎦
⎦
[ t+u ]
ds
=𝔼
∫t
t+u
Ws(1)
dWs(1)
Xs
4:19 P.M.
165
)2
t+u
(
Ws(n)
Xs
)2
⎤
ds⎥
⎥
⎦
= u.
Thus, Wt+u − Wt ∼ 𝒩(0, u).
(c) To show that Wt+u − Wt ⟂
⟂ Wt we note that from the independent increment property
of Wt(i) , i = 1, 2, . . . , n,
𝔼
[(
) ]
Wt+u − Wt Wt
)
( )
(
= 𝔼 Wt+u Wt − 𝔼 Wt2
)
[(
t+u
t+u
t+u
Ws(1)
Ws(2)
Ws(n)
(1)
(2)
(n)
=𝔼
dWs +
dWs + . . . +
dWs
∫0
∫0
∫0
Xs
Xs
Xs
)]
(
t
t
t
( )
Ws(1)
Ws(2)
Ws(n)
(1)
(2)
(n)
− 𝔼 Wt2
×
dWs +
dWs + . . . +
dWs
∫ 0 Xs
∫0 Xs
∫0 Xs
(
[(
)2
)(
)]
n
n
⎡
(i)
⎤ ∑
(i)
(i)
t
t+u
t
∑
W
W
W
s
s
s
(i)
(i)
(i)
⎥+
=
𝔼⎢
dWs
𝔼
dWs
dWs
∫t
∫0 Xs
⎢ ∫0 Xs
⎥ i=1
Xs
i=1 ⎣
⎦
)(
)]
[(
n
n
(j)
t+u
t
( )
∑∑
Ws(i)
Ws
(j)
(i)
− 𝔼 Wt2
+
𝔼
dWs
dWs
∫ 0 Xs
∫t
Xs
i=1 j=1
i≠j
n
⎡ t
∑
=
𝔼⎢
⎢∫
i=1 ⎣ 0
=t−t
(
Ws(i)
Xs
)2
⎤
( )
ds⎥ − 𝔼 Wt2
⎥
⎦
= 0.
Since Wt ∼ 𝒩(0, t), Wt+s − Wt ∼ 𝒩(0, s) and the joint distribution of Wt and Wt+s −
Wt is a bivariate normal (see Problem 2.2.1.5, page 58), if Cov(Wt+s − Wt , Wt ) = 0
then Wt+s − Wt ⟂
⟂ Wt .
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Thus, from the results of (a)–(c) we have shown that Wt is a standard Wiener process.
(j)
From Itō’s formula and given dWt(i) dWt = 0, i ≠ j,
n
∑
𝜕Xt
n
2
1 ∑ 𝜕 Xt
(i)
dW
+
(dWt(i) )2
dXt =
t
(i)
(i) 2
2
i=1 𝜕Wt
i=1 𝜕(Wt )
n
n
2
∑ 𝜕Xt
1 ∑ 𝜕 Xt
(i)
=
dW
+
dt
t
(i)
2 i=1 𝜕(W (i) )2
i=1 𝜕Wt
t
(
)2
⎡ n
⎤
(i)
−1
X
−
X
W
t
t
⎥
1 ⎢∑ t
(i)
dWt + ⎢
=
⎥ dt
2
Xt
2 ⎢ i=1
Xt
⎥
i=1
⎣
⎦
(
)
2
∑n
⎡
⎤
Wt(i) ⎥
i=1
⎢
1 n
= dWt + ⎢ −
⎥
2 ⎢ Xt
Xt3
⎥
⎣
⎦
(
)
X2
1 n
− t3
= dWt +
2 Xt Xt
)
(
n−1
dt.
= dWt +
2Xt
n
∑
Therefore,
Xt =
Wt(i)
√
(Wt(1) )2 + (Wt(2) )2 + . . . + (Wt(n) )2
(
satisfies the SDE
dXt =
n−1
2Xt
)
dt + dWt .
◽
7. Forward–Spot Price Relationship II. Let (Ω, ℱ, ℙ) be a probability space. We consider
the forward price F(t, T) of an asset St satisfying the SDE
dF(t, T)
= 𝜎1 (t, T) dWt(1) + 𝜎2 (t, T) dWt(2)
F(t, T)
where 𝜎1 (t, T) > 0, 𝜎2 (t, T) > 0 are time-dependent volatilities and {Wt(1) ∶ t ≥ 0},
{Wt(2) ∶ t ≥ 0} are standard Wiener processes with correlation 𝜌 ∈ (−1, 1). Given the
relationship F(t, T) = St er(T−t) where r is the risk-free interest rate, show that
(
)
t[
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
𝜕𝜎1 (u, t)
𝜕 log F(0, t)
𝜎1 (u, t)
−
+ 𝜌 𝜎1 (u, t)
+ 𝜎2 (u, t)
∫0
𝜕t
𝜕t
𝜕t
𝜕t
}
]
t
t
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
𝜕𝜎 (u, t)
+ 𝜎2 (u, t) 2
du +
dWu(1) +
dWu(2) dt + 𝜎(t, t) dWt
∫0
∫0
𝜕t
𝜕t
𝜕t
dSt
=
St
{
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Multi-Dimensional Diffusion Process
where
𝜎(t, t) =
167
√
𝜎1 (t, t)2 + 2𝜌𝜎1 (t, t)𝜎2 (t, t) + 𝜎2 (t, t)2
and
Wt = √
𝜎1 (t, t)Wt(1) + 𝜎2 (t, t)Wt(2)
𝜎1 (t, t)2 + 2𝜌𝜎1 (t, t)𝜎2 (t, t) + 𝜎2 (t, t)2
is a standard Wiener process.
Solution: Expanding log F(t, T) using Taylor’s theorem and then applying Itō’s lemma,
we have
d log F(t, T) =
1
1
dF(t, T) −
dF(t, T)2 + . . .
F(t, T)
2F(t, T)2
= 𝜎1 (t, T) dWt(1) + 𝜎2 (t, T) dWt(2)
]
1[
−
𝜎1 (t, T)2 + 𝜎2 (t, T)2 + 2𝜌𝜎1 (t, T)𝜎2 (t, T) dt
2
and taking integrals,
(
log
F(t, T)
F(0, T)
)
t
=
∫0
−
𝜎1 (u, T) dWu(1) +
t
∫0
𝜎2 (u, T) dWu(2)
t[
]
1
𝜎1 (u, T)2 + 𝜎2 (u, T)2 + 2𝜌𝜎1 (u, T)𝜎2 (u, T) du
2 ∫0
We finally have
t
1
F(t, T) = F(0, T)e− 2 ∫0 [𝜎1 (u,T)
2 +𝜎 (u,T)2 +2𝜌𝜎 (u,T)𝜎 (u,T)]du+∫ t 𝜎 (u,T)dW (1) +∫ t 𝜎 (u,T)dW (2)
2
1
2
u
u
0 1
0 2
.
By setting T = t and taking note that St = F(t, t), the spot price St has the expression
t
1
St = F(0, t)e− 2 ∫0 [𝜎1 (u,t)
2 +𝜎
t
(1)
t
(2)
2
2 (u,t) +2𝜌𝜎1 (u,t)𝜎2 (u,t)]du+∫0 𝜎1 (u,t) dWu +∫0 𝜎2 (u,t) dWu
.
To find the SDE for St , we apply Itō’s lemma
dSt =
𝜕St
𝜕St
𝜕St
dt +
dWt(1) +
dWt(2)
(1)
(2)
𝜕t
𝜕Wt
𝜕Wt
2
2
𝜕 2 St
1 𝜕 St
1 𝜕 St
(1) 2
(2) 2
(dW
)
+
(dW
)
+
dWt(1) dWt(2) + . . .
t
t
(1)
(2)
2 𝜕(W (1) )2
2 𝜕(W (2) )2
𝜕Wt 𝜕Wt
t
t
(
)
2
𝜕St 1 𝜕 2 St
𝜕St
𝜕 2 St
1 𝜕 St
=
dt +
+
+𝜌
dWt(1)
+
(1)
(2)
(1)
𝜕t
2 𝜕(W (1) )2 2 𝜕(W (2) )2
𝜕W 𝜕W
𝜕W
+
t
+
𝜕St
𝜕Wt(2)
dWt(2) .
t
t
t
t
4:19 P.M.
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Multi-Dimensional Diffusion Process
Taking partial differentiations of St , we have
]
𝜕St 𝜕 log F(0, t) 1 [
𝜎1 (t, t)2 + 𝜎2 (t, t)2 + 2𝜌𝜎1 (t, t)𝜎2 (t, t)
=
−
𝜕t
𝜕t
2
(
)
t[
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
𝜕𝜎1 (u, t)
𝜎1 (u, t)
−
+ 𝜌 𝜎1 (u, t)
+ 𝜎2 (u, t)
∫0
𝜕t
𝜕t
𝜕t
]
𝜕𝜎 (u, t)
+ 𝜎2 (u, t) 2
du
𝜕t
t
+
𝜕St
𝜕Wt(1)
𝜕 2 St
𝜕(Wt(1) )2
∫0
t
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
dWu(1) +
dWu(2) ,
∫0
𝜕t
𝜕t
𝜕St
= 𝜎1 (t, t)St ,
𝜕Wt(2)
= 𝜎2 (t, t)St ,
𝜕 2 St
= 𝜎1 (t, t)2 St ,
𝜕(Wt(2) )2
= 𝜎2 (t, t)2 St ,
𝜕 2 St
𝜕Wt(1) 𝜕Wt(2)
= 𝜎1 (t, t)𝜎2 (t, t)St
and substituting them into the SDE we eventually have
dSt
=
St
(
)
t[
𝜕𝜎 (u, t)
𝜕𝜎 (u, t)
𝜕𝜎 (u, t)
𝜕 log F(0, t)
𝜎1 (u, t) 1
−
+ 𝜌 𝜎1 (u, t) 2
+ 𝜎2 (u, t) 1
∫0
𝜕t
𝜕t
𝜕t
𝜕t
}
]
t
t
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
𝜕𝜎 (u, t)
+ 𝜎2 (u, t) 2
du +
dWu(1) +
dWu(2) dt
∫0
∫0
𝜕t
𝜕t
𝜕t
{
+ 𝜎1 (t, t) dWt(1) + 𝜎2 (t, t) dWt(2)
(
)
{
t[
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
𝜕𝜎1 (u, t)
𝜕 log F(0, t)
−
+ 𝜌 𝜎1 (u, t)
+ 𝜎2 (u, t)
𝜎1 (u, t)
=
∫0
𝜕t
𝜕t
𝜕t
𝜕t
]
}
t
t
𝜕𝜎 (u, t)
𝜕𝜎1 (u, t)
𝜕𝜎2 (u, t)
+ 𝜎2 (u, t) 2
du +
dWu(1) +
dWu(2) dt
∫0
∫0
𝜕t
𝜕t
𝜕t
+ 𝜎(t, t) dWt
where
𝜎(t, t) =
√
𝜎1 (t, t)2 + 2𝜌𝜎1 (t, t)𝜎2 (t, t) + 𝜎2 (t, t)2
and from the steps discussed in Problem 3.2.3.2 (page 156) we can prove that
Wt = √
𝜎1 (t, t)Wt(1) + 𝜎2 (t, t)Wt(2)
𝜎1 (t, t)2 + 2𝜌𝜎1 (t, t)𝜎2 (t, t) + 𝜎2 (t, t)2
∼ 𝒩(0, t)
and is also a standard Wiener process.
◽
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169
8. Gabillon 2-Factor Model. Let {Wt(s) ∶ t ≥ 0} and {Wt(l) ∶ t ≥ 0} be standard Wiener processes on the probability space (Ω, ℱ, ℙ) with correlation 𝜌 ∈ (−1, 1). Suppose the forward curve F(t, T) follows the process
dF(t, T)
= 𝜎s e−𝛼(T−t) dWt(s) + 𝜎l (1 − e−𝛼(T−t) ) dWt(l)
F(t, T)
where t < T, 𝛼 is the mean-reversion parameter and 𝜎s and 𝜎l are the short-term and
long-term volatilities, respectively.
By setting
√
dWt(s) = dWt(1) and dWt(l) = 𝜌dWt(1) + 1 − 𝜌2 dWt(2)
where Wt(1) and Wt(2) are standard Wiener processes, Wt(1) ⟂
⟂ Wt(2) , show that
dF(t, T)
= 𝜎(t, T) dWt
F(t, T)
where
𝜎(t, T) =
√
)
)
(
(
𝜎l2 + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2 e−2𝛼(T−t) + 2 𝜌𝜎s 𝜎l − 𝜎l2 e−𝛼(T−t)
and
√
(
))
(
)
( −𝛼(T−t)
𝜎s e
+ 𝜌𝜎l 1 − e−𝛼(T−t) Wt(1) + 1 − 𝜌2 1 − e−𝛼(T−t) Wt(2)
Wt =
√
)
)
(
(
𝜎l2 + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2 e−2𝛼(T−t) + 2 𝜌𝜎s 𝜎l − 𝜎l2 e−𝛼(T−t)
is a standard Wiener process.
Finally, show that conditional on F(0, T), F(t, T) follows a lognormal distribution with
mean
𝔼 [F(t, T)| F(0, T)] = F(0, T)
and variance
{
)
(
𝜎l2 t + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2
(
e−2𝛼(T−t) − e−2𝛼T
2𝛼
( −𝛼(T−t)
)
}
) e
(
− e−𝛼T
+2 𝜌𝜎s 𝜎l − 𝜎l2
−1 .
𝛼
Var [F(t, T)| F(0, T)] = F(0, T)2 exp
Solution: By defining dWt(s) = dWt(1) and dWt(l) = 𝜌dWt(1) +
⟂ Wt(2) , the SDE of F(t, T) can be expressed as
Wt(1) ⟂
)
√
1 − 𝜌2 dWt(2) such that
)
√
(
)(
dF(t, T)
= 𝜎s e−𝛼(T−t) dWt(1) + 𝛼l 1 − e−𝛼(T−t) 𝜌dWt(1) + 1 − 𝜌2 dWt(2)
F(t, T)
√
(
)
(
)
= 𝜎s e−𝛼(T−t) + 𝜌𝜎l (1 − e−𝛼(T−t) ) dWt(1) + 1 − 𝜌2 𝜎l 1 − e−𝛼(T−t) dWt(2)
= 𝜎(t, T) dWt
4:19 P.M.
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Multi-Dimensional Diffusion Process
where, owing to the fact that dWt(1) ⋅ dWt(2) = 0, we have
(
))2 (
) (
)2
(
+ 1 − 𝜌2 𝜎l2 1 − e−𝛼(T−t)
𝜎(t, T)2 = 𝜎s e−𝛼(T−t) + 𝜌𝜎l 1 − e−𝛼(T−t)
(
)
(
)2
= 𝜎s2 e−2𝛼(T−t) + 2𝜌𝜎s 𝜎l e−𝛼(T−t) 1 − e−𝛼(T−t) + 𝜎l2 1 − e−𝛼(T−t)
)
)
(
(
= 𝜎l2 + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2 e−2𝛼(T−t) + 2 𝜌𝜎s 𝜎l − 𝜎l2 e−𝛼(T−t)
and using the steps described in Problem 3.2.3.2 (page 156), we can easily show that
√
(
))
(
)
( −𝛼(T−t)
𝜎s e
+ 𝜌𝜎l 1 − e−𝛼(T−t) Wt(1) + 1 − 𝜌2 1 − e−𝛼(T−t) Wt(2)
∼ 𝒩(0, t)
Wt =
√
)
)
(
(
𝜎l2 + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2 e−2𝛼(T−t) + 2 𝜌𝜎s 𝜎l − 𝜎l2 e−𝛼(T−t)
is also a standard Wiener process.
By expanding d log F(t, T) using Taylor’s theorem and applying Itō’s formula, we have
(
)2
dF(t, T) 1 dF(t, T)
+ ...
−
d log F(t, T) =
F(t, T)
2 F(t, T)
1
= 𝜎(t, T) dWt − 𝜎(t, T)2 dt.
2
Taking integrals from 0 to t,
t
∫0
t
d log F(u, T) =
∫0
t
𝜎(u, T) dWu −
1
𝜎(u, T)2 du
2 ∫0
t
log F(t, T) = log F(0, T) +
∫0
t
𝜎(u, T) dWu −
1
𝜎(u, T)2 du.
2 ∫0
From the property of Itō’s integral, we have
[ t
]
𝔼
𝜎(u, T) dWu = 0
∫0
[(
and
𝔼
)2 ]
t
∫0
𝜎(u, T) dWu
[
=𝔼
t
∫0
]
𝜎(u, T) du =
2
t
Since the Itō integral
∫0
t
t
and hence
∫0
𝜎(u, T)2 du.
t
𝜎(u, T) dWu is in the form
(page 126) we can easily prove that
∫0
t
∫0
∫0
f (u) dWu , from Problem 3.2.2.4
𝜎(u, T) dWu follows a normal distribution,
(
𝜎(u, T) dWu ∼ 𝒩 0,
t
∫0
)
𝜎(u, T) du
2
(
)
t
t
1
2
2
log F(t, T) ∼ 𝒩 log F(0, T) −
𝜎(u, T) du,
𝜎(u, T) du .
∫0
2 ∫0
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171
Solving the integral,
t
∫0
t
𝜎(u, T)2 du =
∫0
[
)
)
]
(
(
𝜎l2 + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2 e−2𝛼(T−u) + 2 𝜌𝜎s 𝜎l − 𝜎l2 e−𝛼(T−u) du
) e−2𝛼(T−u)
) e−𝛼(T−u) |t
(
(
|
= 𝜎l2 u + 𝜎s2 − 2𝜌𝜎s 𝜎l + 𝜎l2
+ 2 𝜌𝜎s 𝜎l − 𝜎l2
2𝛼
𝛼 ||0
)
(
( 2
) e−2𝛼(T−t) − e−2𝛼T
2
2
= 𝜎l t + 𝜎s − 2𝜌𝜎s 𝜎l + 𝜎l
2𝛼
( −𝛼(T−t)
)
−𝛼T
) e
(
−e
+ 2 𝜌𝜎s 𝜎l − 𝜎l2
.
𝛼
Since F(t, T) conditional on F(0, T) follows a lognormal distribution, we have
𝔼[F(t, T)|F(0, T)] = F(0, T)
and
{
(
)
(
e−2𝛼(T−t) − e−2𝛼T
Var [F(t, T)| F(0, T)] = F(0, T) exp
+
− 2𝜌𝜎s 𝜎l +
2𝛼
)
}
( −𝛼(T−t)
−𝛼T
(
) e
−e
+ 2 𝜌𝜎s 𝜎l − 𝜎l2
−1 .
𝛼
𝜎l2 t
2
𝜎s2
𝜎l2
)
◽
9. Integrated Square-Root Process. Let (Ω, ℱ, ℙ) be a probability space and let {Wt ∶ t ≥ 0}
be a standard Wiener process. Suppose Xt follows the CIR model with SDE
√
dXt = 𝜅(𝜃 − Xt ) dt + 𝜎 Xt dWt , X0 > 0
where 𝜅, 𝜃 and 𝜎 are constants. We consider the integral
t
Yt =
∫t0
Xu du,
Xt0 > 0
as the integrated square-root process of Xt up to time t from initial time t0 , t0 < t.
Show that for n ≥ 1, m ≥ 1, n, m ∈ ℕ, the process Xtn Ytm satisfies the SDE
)
]
) 1
d(Xtn Ytm ) [( (
−1
n𝜅 𝜃 − Xt + n(n − 1)𝜎 2 Xt−1 + mXt Yt−1 dt + n𝜎Xt 2 dWt
n m =
Xt Yt
2
and hence show that 𝔼(Xtn Ytm |Xt0 ) satisfies the following first-order ordinary differential
equation
(
)
(
)
(
)
d
|
𝔼 Xtn Ytm || Xt0 = n𝜅𝜃𝔼 Xtn−1 Ytm | Xt0 − n𝜅𝔼 Xtn Ytm || Xt0
|
dt
)
(
)
(
1
|
|
+ n(n − 1)𝜎 2 𝔼 Xtn Ytm |Xt0 + m𝔼 Xtn+1 Ytm−1 |Xt0 .
|
|
2
Finally, find the first two moments of Yt , given Xt0 .
4:19 P.M.
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Multi-Dimensional Diffusion Process
Solution: From Taylor’s theorem and subsequently applying Itō’s formula, we can write
1
d(Xtn ) = nXtn−1 dXt + n(n − 1)Xtn−2 (dXt2 ) + . . .
2
√
1
= nXtn−1 [𝜅(𝜃 − Xt ) dt + 𝜎 Xt dWt ] + n(n − 1)Xtn−2 (𝜎 2 Xt dt)
2
]
[ (
)
n− 1
1
= n𝜅 𝜃 − Xt Xtn−1 + n(n − 1)𝜎 2 Xtn−1 dt + n𝜎Xt 2 dWt
2
and
1
d(Ytm ) = mYtm−1 dYt + m(m − 1)Ytm−2 (dYt )2 + . . .
2
= mXt Ytm−1 dt.
Thus,
d(Xtn Ytm ) = Ytm d(Xtn ) + Xtn d(Ytm ) + d(Xtn )d(Ytm )
)
]
[( (
) 1
n− 1
= n𝜅 𝜃 − Xt + n(n − 1)𝜎 2 Xtn−1 Ytm dt + n𝜎Xt 2 Ytm dWt
2
+ mXtn+1 Ytm−1 dt
or
)
]
) 1
d(Xtn Ytm ) [( (
− 12
2
−1
−1
=
n𝜅
𝜃
−
X
+
mX
Y
n(n
−
1)𝜎
X
dt
+
n𝜎X
+
t
t
t
t
t dWt .
Xtn Ytm
2
Taking integrals,
t
∫t0
t
d(Xun Yum ) =
[
]
(
)
1
n𝜅 𝜃 − Xu Xun−1 Yum + n(n − 1)𝜎 2 Xun−1 Yum + mXun+1 Yum−1 du
2
∫t0
t
+
∫t0
n− 12
n𝜎Xu
Yum dWu
t[
]
(
)
1
n𝜅 𝜃 − Xu Xun−1 Yum + n(n − 1)𝜎 2 Xun−1 Yum + mXun+1 Yum−1 du
Xtn Ytm = Xtn0 Ytm0 +
∫t0
2
t
n− 12
+ n𝜎Xu
∫t0
Yum dWu
t0
and taking expectations given Xt0 and because Yt0 =
∫t0
t
𝔼[Xtn Ytm |Xt0 ] =
∫t0
t
𝔼[n𝜅(𝜃 − Xu )Xun−1 Yum |Xt0 ]du +
∫t0
𝔼
Xu du = 0, we have
[
]
|
1
n(n − 1)𝜎 2 Xun−1 Yum || Xt0 du
2
|
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]
[
t
]
|
[
n− 1
|
𝔼 mXun+1 Yum−1 | Xt0 du +
𝔼 n𝜎Xu 2 Yum || Xt0 dWu
|
∫t0
∫t0
|
4:19 P.M.
173
t
+
t [
[ (
]
]
)
1
|
|
𝔼 n𝜅 𝜃 − Xu Xun−1 Yum |Xt0 du +
𝔼 n(n − 1)𝜎 2 Xun−1 Yum |Xt0 du
|
|
∫t0
∫t0
2
t
=
[
]
|
𝔼 mXun+1 Yum−1 |Xt0 du.
|
∫t0
t
+
Differentiating with respect to t yields
[
]
[(
]
]
[
)
1
d
|
|
|
𝔼 Xtn Ytm |Xt0 = n𝜅𝔼 𝜃 − Xt Xtn−1 Ytm |Xt0 + n(n − 1)𝜎 2 𝔼 Xtn−1 Ytm |Xt0
|
|
|
dt
2
]
[
|
+ m𝔼 Xtn+1 Ytm−1 |Xt0
|
]
[
]
[
|
|
= n𝜅𝜃𝔼 Xtn−1 Ytm |Xt0 − n𝜅𝔼 Xtn Ytm |Xt0
|
|
[
]
[
]
1
|
|
+ n(n − 1)𝜎 2 𝔼 Xtn−1 Ytm |Xt0 + m𝔼 Xtn+1 Ytm−1 |Xt0 .
|
|
2
To find the mean of Yt given Xt0 , we let n = 0 and m = 1 so that
]
[
]
[
d
|
|
𝔼 Yt |Xt0 = 𝔼 Xt |Xt0 .
|
|
dt
From Problem 3.2.2.12 (page 135), we have
[
]
)
(
|
𝔼 Xt |Xt0 = Xt0 e−𝜅(t−t0 ) + 𝜃 1 − e−𝜅(t−t0 )
|
and therefore,
t[
[
]
(
)]
|
𝔼 Yt |Xt0 =
Xt0 e−𝜅(u−t0 ) + 𝜃 1 − e−𝜅(u−t0 ) du
|
∫t0
)(
(
)
1
Xt0 − 𝜃 1 − e−𝜅(t−t0 ) .
= 𝜃(t − t0 ) +
𝜅
For the second moment of Yt given Xt0 , we set n = 0 and m = 2 so that
[
]
[
]
d
|
|
𝔼 Yt2 |Xt0 = 2𝔼 Xt Yt |Xt0
|
|
dt
and to find 𝔼[Xt Yt |Xt0 ], we set n = 1 and m = 1 so that
[
]
[
]
[
]
[
]
d
|
|
|
|
𝔼 Xt Yt |Xt0 = 𝜅𝜃𝔼 Yt |Xt0 − 𝜅𝔼 Xt Yt |Xt0 + 𝔼 Xt2 |Xt0
|
|
|
|
dt
or
[
]
[
]
[
]
[
]
d
|
|
|
|
𝔼 Xt Yt |Xt0 + 𝜅𝔼 Xt Yt |Xt0 = 𝜅𝜃𝔼 Yt |Xt0 + 𝔼 Xt2 |Xt0 .
|
|
|
|
dt
Page 173
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Multi-Dimensional Diffusion Process
By setting the integrating factor I = e𝜅t , the solution of the first-order differential
equation is
]]
[
]
[
]
[
[
d 𝜅t
|
|
|
e 𝔼 Xt Yt |Xt0 = 𝜅𝜃𝔼 Yt |Xt0 + 𝔼 Xt2 |Xt0
|
|
|
dt
or
t{
[
]
[
]
[
]}
|
|
|
𝔼 Xt Yt |Xt0 = e−𝜅t
𝜅𝜃𝔼 Yu |Xt0 + 𝔼 Xu2 |Xt0 du.
|
|
|
∫t
0
Since
)(
(
[
]
)
1
|
Xt0 − 𝜃 1 − e−𝜅(t−t0 )
𝔼 Yt |Xt0 = 𝜃(t − t0 ) +
|
𝜅
and from Problem 3.2.2.12 (page 135),
)
(
]
[
)
(
2𝜅𝜃 + 𝜎 2
|
(Xt0 − 𝜃) e−𝜅(t−t0 ) − e−2𝜅(t−t0 )
𝔼 Xt2 |Xt0 = Xt20 e−2𝜅(t−t0 ) +
|
𝜅
(
)
𝜃 2𝜅𝜃 + 𝜎 2 (
)
1 − e−2𝜅(t−t0 )
+
2𝜅
we can easily show that
(
)
[
]
𝜎2
1
|
e−𝜅t
𝔼 Xt Yt |Xt0 = 𝜅𝜃 2 (t − t0 )2 e−𝜅t + 𝜃(t − t0 ) Xt0 +
|
2
2𝜅
)
(
)
(
𝜅𝜃 + 𝜎 2
(Xt0 − 𝜃) 1 − e−𝜅(t−t0 ) e−𝜅t
+
2
𝜅
(
[
)(
)] (
)
1
2𝜅𝜃 + 𝜎 2
1
2
+
Xt0 − 𝜃
Xt0 −
1 − e−2𝜅(t−t0 ) e−𝜅t .
2𝜅
𝜅
2
Finally, by substituting the result of 𝔼[Xt Yt |Xt0 ] into
[
]
t [
]
|
|
𝔼 Yt2 |Xt0 = 2
𝔼 Xu Yu |Xt0
|
|
∫t
0
and using integration by parts, we eventually arrive at
[
(
(
)
)]
[
]
)
(
𝜎2
𝜎2
1
2|
2
𝔼 Yt |Xt0 = 2 Xt0 + 4𝜃 +
1 − e−𝜅(t−t0 ) e−𝜅t0
Xt0 + 𝜃 𝜃 −
|
𝜅
2𝜅
𝜅
]
[
2
𝜃
𝜎
−
2(Xt0 + 1) +
+ t − t0 (t − t0 )e−𝜅t
𝜅
𝜅
)(
(
)(
)
𝜅𝜃 + 𝜎 2
X
−
−
𝜃
1 − e−2𝜅(t−t0 ) e−𝜅t0
t0
3
𝜅
)(
[
(
)] (
)
1
2𝜅𝜃 + 𝜎 2
1
2
Xt0 − 𝜃
1 − e−3𝜅(t−t0 ) e−𝜅t0 .
− 2 Xt0 −
𝜅
2
3𝜅
◽
Page 174
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175
10. Heston Model. Let (Ω, ℱ, ℙ) be a probability space and let {WtS ∶ t ≥ 0}, {Wt𝜎 ∶ t ≥ 0}
be two standard Wiener processes with correlation 𝜌 ∈ (−1, 1). Suppose the asset price St
takes the form
dSt = 𝜇St dt + 𝜎t St dWtS ,
= 𝜅(𝜃 −
d𝜎t2
𝜎t2 )
S0 > 0
dt + 𝛼𝜎t dWt𝜎 ,
𝜎0 > 0
dWtS dWt𝜎 = 𝜌dt
where 𝜇, 𝜅, 𝜃 and 𝛼 are constants, and 𝜎t is the stochastic volatility of St . By defining
⟂ Wt𝜎 , show that we can write
{Bt ∶ t ≥ 0} as a standard Wiener process where Bt ⟂
WtS = 𝜌Wt𝜎 +
√
1 − 𝜌2 Bt .
Using the above relation show also that for 0 ≤ t ≤ T we can write
(
log
ST ∕𝜉T
St ∕𝜉t
s
)
𝜎
1
= 𝜇(T − t) − (1 − 𝜌2 )
∫t
2
1 2
T
𝜎u2 du +
√
T
1 − 𝜌2
∫t
𝜎u dBu
s 2
where 𝜉s = e𝜌 ∫0 𝜎u dWu − 2 𝜌 ∫0 𝜎u du .
Prove that the relation of 𝜉T and 𝜉t can be expressed as
T
𝜉T = 𝜉t + 𝜌
∫t
𝜎u 𝜉u dWu𝜎 .
Conditional on ℱt and {𝜎u ∶ t ≤ u ≤ T}, show that
(
log
)|
[(
]
)
1 2
|
2
(T − t)
(T − t), 𝜎RMS
| ℱt , {𝜎u ∶ t ≤ u ≤ T} ∼ 𝒩 𝜇 − 𝜎RMS
|
2
|
√(
) T
1 − 𝜌2
=
𝜎u2 du.
T − t ∫t
ST ∕𝜉T
St ∕𝜉t
where 𝜎RMS
Solution: From WtS = 𝜌Wt𝜎 +
√
𝔼(WtS ) = 𝔼(𝜌Wt𝜎 +
1 − 𝜌2 Bt we have
√
1 − 𝜌2 Bt ) = 𝜌𝔼(Wt𝜎 ) +
√
1 − 𝜌2 𝔼(Bt ) = 0
and
Var (WtS ) = Var (𝜌Wt𝜎 +
√
1 − 𝜌2 Bt ) = 𝜌2 Var (Wt𝜎 ) + (1 − 𝜌2 )Var (Bt ) = t.
Given both Wt𝜎 ∼ 𝒩(0, t) and Bt ∼ 𝒩(0, t), therefore
𝜌Wt𝜎 +
√
1 − 𝜌2 Bt ∼ 𝒩(0, t).
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In addition, using Itō’s formula and taking note that Wt𝜎 ⟂
⟂ Bt ,
dWtS ⋅ dWt𝜎 = d(𝜌Wt𝜎 +
= (𝜌dWt𝜎 +
√
√
1 − 𝜌2 Bt ) ⋅ dWt𝜎
1 − 𝜌2 dBt ) ⋅ dWt𝜎
√
= 𝜌(dWt𝜎 )2 + 1 − 𝜌2 dBt ⋅ dWt𝜎
= 𝜌dt.
√
Thus, we can write WtS = 𝜌Wt𝜎 + 1 − 𝜌2 Bt .
Writing the SDEs of St and 𝜎t2 in terms of Wt𝜎 and Bt ,
dSt = 𝜇St dt + 𝜎t St (𝜌dWt𝜎 +
√
1 − 𝜌2 dBt )
d𝜎t2 = 𝜅(𝜃 − 𝜎t2 ) dt + 𝛼𝜎t dWt𝜎
and following Itō’s lemma,
)
(
dSt 1 dSt 2
−
+ ...
St
2 St
√
1
= 𝜇dt + 𝜎t St (𝜌dWt𝜎 + 1 − 𝜌2 dBt ) − 𝜎t2 (𝜌2 dt + (1 − 𝜌2 ) dt)
2
)
(
√
1
= 𝜇 − 𝜎t2 dt + 𝜌𝜎t dWt𝜎 + 1 − 𝜌2 𝜎t dBt
2
d log St =
and taking integrals,
T
∫t
T
d log Su =
∫t
(
)
1
𝜇 − 𝜎u2 du + 𝜌
∫t
2
log ST = log St + 𝜇(T − t) −
1
2 ∫t
T
𝜎u dWu𝜎 +
√
T
T
𝜎u2 du + 𝜌
∫t
T
1 − 𝜌2
𝜎u dBu
∫t
𝜎u dWu𝜎 +
√
T
1 − 𝜌2
∫t
𝜎u dBu .
By setting
t
𝜎
1 2 t 2
∫0 𝜎u du
𝜉t = e𝜌 ∫0 𝜎u dWu − 2 𝜌
T
and
𝜉T = e𝜌 ∫0
𝜎u dWu𝜎 − 12 𝜌2 ∫0 𝜎u2 du
T
we have
1
log ST = log St + 𝜇(T − t) − (1 − 𝜌2 )
∫t
2
or
(
log
ST ∕𝜉T
St ∕𝜉t
)
T
𝜎u2 du +
1
= 𝜇(T − t) − (1 − 𝜌2 )
∫t
2
√
T
1 − 𝜌2
T
𝜎u2 du +
∫t
√
𝜎u dBu + log 𝜉T − log 𝜉t
T
1 − 𝜌2
∫t
𝜎u dBu .
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Multi-Dimensional Diffusion Process
177
To show the relation between 𝜉T and 𝜉t , from Taylor’s theorem and then using Itō’s lemma,
d𝜉t =
2
2
𝜕𝜉t
𝜕𝜉t
1 𝜕 𝜉t
1 𝜕 𝜉t
𝜎
2
dt +
dW
+
(dt)
+
(dWt𝜎 )2 + . . .
t
𝜕t
𝜕Wt𝜎
2 𝜕t2
2 𝜕(Wt𝜎 )2
1
1
= − 𝜌2 𝜎t2 𝜉t dt + 𝜌𝜎t 𝜉t dWt𝜎 + 𝜌2 𝜎t2 𝜉t dt
2
2
= 𝜌𝜎t 𝜉t dWt𝜎 .
Taking integrals,
T
∫t
T
d𝜉u =
∫t
or
𝜌𝜎u 𝜉u dWu𝜎
T
𝜉T = 𝜉t + 𝜌
∫t
𝜎u 𝜉u dWu𝜎 .
Finally, conditional on ℱt and {𝜎u ∶ t ≤ u ≤ T},
[
𝔼 log
(
ST ∕𝜉T
St ∕𝜉t
]
)|
1
|
| ℱt , {𝜎u ∶ t ≤ u ≤ T} = 𝜇(T − t) − (1 − 𝜌2 )
|
∫t
2
|
(
)
1 2
= 𝜇 − 𝜎RMS
(T − t)
2
and using the properties of Itō’s integral,
[
]
)
ST ∕𝜉T ||
Var log
| ℱ , {𝜎 ∶ t ≤ u ≤ T}
St ∕𝜉t || t u
[
]
T
|
|
2
𝜎u dBu | ℱt , {𝜎u ∶ t ≤ u ≤ T}
= (1 − 𝜌 )Var
|
∫t
|
[(
]
)
2|
T
|
2
𝜎u dBu || ℱt , {𝜎u ∶ t ≤ u ≤ T}
= (1 − 𝜌 )𝔼
∫t
|
|
[
]2
T
|
|
𝜎u dBu | ℱt , {𝜎u ∶ t ≤ u ≤ T}
− (1 − 𝜌2 )𝔼
|
∫t
|
[
]
T
|
|
2
2
= (1 − 𝜌 )𝔼
𝜎u du| ℱt , {𝜎u ∶ t ≤ u ≤ T}
|
∫t
|
(
T
= (1 − 𝜌2 )
∫t
2
(T − t)
= 𝜎RMS
𝜎u2 du
T
𝜎u2 du
4:19 P.M.
Page 177
Chin
178
3.2.3
(
2
where 𝜎RMS
=
1 − 𝜌2
T −t
T
Since
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Multi-Dimensional Diffusion Process
T
∫t
𝜎u2 du.
𝜎u dBu is normally distributed, therefore
∫t
(
log
)
c03.tex
ST ∕𝜉T
St ∕𝜉t
)|
)
[(
]
1 2
|
2
(T − t), 𝜎RMS
(T − t) .
| ℱt , {𝜎u ∶ t ≤ u ≤ T} ∼ 𝒩 𝜇 − 𝜎RMS
|
2
|
N.B. Given that the stochastic volatility 𝜎t follows a CIR process, from the results of Prob2 .
lem 3.2.3.9 (page 171) we can easily find the mean and variance of 𝜎RMS
◽
11. Feynman–Kac Formula for Multi-Dimensional Diffusion Process. Let (Ω, ℱ, ℙ) be a probability space and let St = (St(1) , St(2) , . . . , St(n) ). We consider the following PDE problem:
1 ∑∑
𝜕2V
𝜕V
(j)
𝜌ij 𝜎(St(i) , t)𝜎(St , t)
(S , t)
(St , t) +
(j) t
𝜕t
2 i=1 j=1
𝜕S(i) 𝜕S
n
n
t
∑
n
+
𝜇(St(i) , t)
i=1
𝜕V
𝜕St(i)
t
(St , t) − r(t)V(St , t) = 0
with boundary condition V(ST , T) = Ψ(ST ) where 𝜇, 𝜎 are known functions of St(i) and t,
r and Ψ are functions of t and ST , respectively where t < T. Using Itō’s formula on the
process,
u
Zu = e− ∫t r(𝑣)d𝑣 V(Su , u)
where St(i) satisfies the generalised SDE
dSt(i) = 𝜇(St(i) , t) dt + 𝜎(St(i) , t) dWt(i)
(j)
such that {Wt(i) ∶ t ≥ 0} is a standard Wiener process and dWt(i) ⋅ dWt = 𝜌ij dt, 𝜌ij ∈
(−1, 1) for i ≠ j and 𝜌ij = 1 for i = j where i, j = 1, 2, . . . , n, show that under the filtration
ℱt , the solution of the PDE is given by
[
T
− ∫t r(𝑣) d𝑣
V(St , t) = 𝔼 e
]
|
|
Ψ(ST )| ℱt .
|
Solution: In analogy with Problem 3.2.2.20 (page 147), we let g(u) = e− ∫t
u
Zu = g(u)V(Su , u).
By applying Taylor’s expansion and Itō’s formula on dZu , we have
∑ 𝜕Zu (i) 1 ∑ ∑ 𝜕 2 Zu
𝜕Z
(j)
dSu +
dSu(i) dSu + . . .
dZu = u du +
(i)
(i) (j)
𝜕u
2
i=1 𝜕Su
i=1 j=1 𝜕S 𝜕S
n
n
n
u
u
r(𝑣) d𝑣
and set
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179
(
)
)
n
∑
𝜕g
𝜕V
𝜕V
+ V(Su , u)
du +
g(u) (i) dSu(i)
= g(u)
𝜕u
𝜕u
𝜕Su
i=1
(
)
n
n
𝜕2V
1 ∑∑
(j)
g(u)
dSu(i) dSu
+
(i) (j)
2 i=1 j=1
𝜕Su 𝜕Su
(
)
𝜕V
= g(u)
− r(u)g(u)V(Su , u) du
𝜕u
(
)
n
∑
𝜕V
g(u) (i) (𝜇(Su(i) , u) du + 𝜎(Su(i) , u) dWu(i) )
+
𝜕Su
i=1
(
)
n
n
)
(
1 ∑∑
𝜕2V
(j)
+
𝜌ij 𝜎(Su(i) , u)𝜎(Su , u) dt
g(u) (i) (j)
2 i=1 j=1
𝜕Su 𝜕Su
(
n
n
1 ∑∑
𝜕2V
𝜕V
(j)
𝜌ij 𝜎(Su(i) , t)𝜎(Su , u)
(S , u)
= g(u)
(Su , u) +
(j) u
𝜕u
2 i=1 j=1
𝜕Su(i) 𝜕Su
)
n
∑
𝜕V
(i)
𝜇(Su , u) (i) (Su , u) − r(u)V(Su , u) du
+
𝜕Su
i=1
(
+g(u)
n
∑
𝜕V
𝜎(Su(i) , u)
i=1
𝜕Su(i)
= g(u)
n
∑
𝜎(Su(i) , u)
i=1
𝜕V
𝜕Su(i)
dWu(i)
dWu(i)
since
1 ∑∑
𝜕2V
𝜕V
(j)
(Su , u) +
𝜌ij 𝜎(Su(i) , t)𝜎(Su , u)
(S , u)
(j) u
𝜕u
2 i=1 j=1
𝜕S(i) 𝜕S
n
n
u
u
+
n
∑
𝜇(Su(i) , u)
i=1
𝜕V
𝜕Su(i)
(Su , u) − r(u)V(Su , u) = 0.
By integrating both sides of dZu we have
T
∫t
dZu =
n
∑
i=1
ZT − Zt =
n
∑
i=1
{
T
∫t
{
T
∫t
g(u)𝜎(Su(i) , u)
u
− ∫t r(𝑣) d𝑣
e
}
𝜕V
dWu(i)
(i)
𝜕Su
𝜎(Su(i) , u)
𝜕V
dWu(i)
𝜕Su(i)
Taking expectations and using the property of Itō calculus,
(
)
( )
( )
𝔼 ZT − Zt = 0 or 𝔼 Zt = 𝔼 ZT .
}
.
4:19 P.M.
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180
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Multi-Dimensional Diffusion Process
Therefore, under the filtration ℱt ,
)
(
)
(
𝔼 Zt |ℱt = 𝔼 ZT |ℱt
[
]
[
]
|
t
T
|
− ∫t r(𝑣) d𝑣
− ∫t r(𝑣) d𝑣
|
𝔼 e
V(St , t)| ℱt = 𝔼 e
V(ST , T)| ℱt
|
|
]
[
|
T
− ∫t r(𝑣) d𝑣
|
V(St , t) = 𝔼 e
Ψ(ST )| ℱt .
|
◽
12. Backward Kolmogorov Equation for a Two-Dimensional Random Walk. We consider a
two-dimensional symmetric random walk where at initial time t0 , a particle starts at (x0 , y0 )
and is at position (x, y) at time t. At time t + 𝛿t, the particle can either move to (x + 𝛿x, y),
(x − 𝛿x, y), (x, y + 𝛿y) or (x, y − 𝛿y) each with probability 14 . Let p(x, y, t; x0 , y0 , t0 ) denote
the probability density of the particle position (x, y) at time t starting at (x0 , y0 ) at time t0 .
By writing the backward equation
√ in a discrete fashion and expanding it using Taylor’s
series, show that for 𝛿x = 𝛿y = 𝛿t and in the limit 𝛿t → 0,
𝜕p(x, y, t; x0 , y0 , t0 )
1
=−
𝜕t
4
(
𝜕 2 p(x, y, t; x0 , y0 , t0 ) 𝜕 2 p(x, y, t; x0 , y0 , t0 )
+
𝜕x2
𝜕y2
)
.
Solution: By denoting p(x, y, t; x0 , y0 , t0 ) as the probability density function of the particle
position (x, y) at time t, the discrete model of the backward equation is
1
1
p(x, y, t; x0 , y0 , t0 ) = p(x − 𝛿x, y, t + 𝛿t; x0 , y0 , t0 ) + p(x + 𝛿x, y, t + 𝛿t; x0 , y0 , t0 )
4
4
1
1
+ p(x, y − 𝛿y, t + 𝛿t; x0 , y0 , t0 ) + p(x, y + 𝛿y, t + 𝛿t; x0 , y0 , t0 ).
4
4
Expanding p(x − 𝛿x, y, t + 𝛿t; x0 , y0 , t0 ), p(x + 𝛿x, y, t + 𝛿t; x0 , y0 , t0 ), p(x, y − 𝛿y, t +
𝛿t; x0 , y0 , t0 ) and p(x, y + 𝛿y, t + 𝛿t; x0 , y0 , t0 ) using Taylor’s series, we have
p(x − 𝛿x, y, t + 𝛿t; x0 , y0 , t0 ) = p(x, y, t + 𝛿t; x0 , y0 , t0 ) −
𝜕p(x, y, t + 𝛿t; x0 , y0 , t0 )
𝛿x
𝜕x
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(−𝛿x)2 + O((𝛿x)3 )
2
𝜕x2
𝜕p(x, y, t + 𝛿t; x0 , y0 , t0 )
p(x + 𝛿x, y, t + 𝛿t; x0 , y0 , t0 ) = p(x, y, t + 𝛿t; x0 , y0 , t0 ) +
𝛿x
𝜕x
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(𝛿x)2 + O((𝛿x)3 )
+
2
𝜕x2
𝜕p(x, y, t + 𝛿t; x0 , y0 , t0 )
p(x, y − 𝛿y, t + 𝛿t; x0 , y0 , t0 ) = p(x, y, t + 𝛿t; x0 , y0 , t0 ) −
𝛿y
𝜕y
+
+
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(−𝛿y)2 + O((𝛿y)3 )
2
𝜕y2
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181
and
p(x, y + 𝛿y, t + 𝛿t; x0 , y0 , t0 ) = p(x, y, t + 𝛿t; x0 , y0 , t0 ) +
+
𝜕p(x, y, t + 𝛿t; x0 , y0 , t0 )
𝛿y
𝜕y
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(𝛿y)2 + O((𝛿y)3 ).
2
𝜕y2
By substituting the above equations into the discrete backward equation,
p(x, y, t; x0 , y0 , t0 ) = p(x, y, t + 𝛿t; x0 , y0 , t0 ) +
+
Setting 𝛿x = 𝛿y =
√
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(𝛿x)2
4
𝜕x2
2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
(𝛿y)2 + O((𝛿x)3 ) + O((𝛿y)3 ).
4
𝜕y2
𝛿t and dividing the equation by 𝛿t and in the limit 𝛿t → 0
[
p(x, y, t + 𝛿t; x0 , y0 , t0 ) − p(x, y, t; x0 , y0 , t0 )
lim
𝛿t→0
𝛿t
)
( 2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
= − lim
𝛿t→0 4
𝜕x2
)
( 2
1 𝜕 p(x, y, t + 𝛿t; x0 , y0 , t0 )
− lim
𝛿t→0 4
𝜕y2
√
+ lim O( 𝛿t)
]
𝛿t→0
we eventually arrive at
𝜕p(x, y, t; x0 , y0 , t0 )
1
=−
𝜕t
4
(
𝜕 2 p(x, y, t; x0 , y0 , t0 ) 𝜕 2 p(x, y, t; x0 , y0 , t0 )
+
𝜕x2
𝜕y2
)
.
◽
13. Forward Kolmogorov Equation for a Two-Dimensional Random Walk. We consider a
two-dimensional symmetric random walk where at initial time t0 , a particle starts at
(X0 , Y0 ) and is at position (X, Y) at terminal time T > 0. At time T − 𝛿T, the particle can
either move to (X + 𝛿X, Y), (X − 𝛿X, Y), (X, Y + 𝛿Y) or (X, Y − 𝛿Y) each with probability
1
. Let p(X, Y, T; X0 , Y0 , t0 ) denote the probability density of the position (X, Y) at time T
4
starting at (X0 , Y0 ) at time t0 .
By writing the forward equation√in a discrete fashion and expanding it using Taylor’s
series, show that for 𝛿X = 𝛿Y = 𝛿T and in the limit 𝛿T → 0
𝜕p(X, Y, T; X0 , Y0 , t0 ) 1
=
𝜕T
4
(
𝜕 2 p(X, Y, T; X0 , Y0 , t0 ) 𝜕 2 p(X, Y, T; X0 , Y0 , t0 )
+
𝜕X 2
𝜕Y 2
)
.
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Solution: By denoting p(X, Y, T; X0 , Y0 , t0 ) as the probability density function of the particle position (X, Y) at time T starting at (X0 , Y0 ) at initial time t0 , the discrete model of
the forward equation is
1
p(X, Y, T; X0 , Y0 , t0 ) = p(X − 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 )
4
1
+ p(X + 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 )
4
1
+ p(X, Y − 𝛿Y, T − 𝛿T; X0 , Y0 , t0 )
4
1
+ p(X, Y + 𝛿Y, T − 𝛿T; X0 , Y0 , t0 ).
4
Expanding p(X − 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 ), p(X + 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 ), p(X, Y −
𝛿Y, T − 𝛿T; X0 , Y0 , t0 ) and p(X, Y + 𝛿Y, T − 𝛿T; X0 , Y0 , t0 ) using Taylor’s series, we have
p(X − 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 ) = p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝜕p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿X
𝜕X
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
(−𝛿X)2 + O((𝛿X)3 )
+
2
𝜕X 2
p(X + 𝛿X, Y, T − 𝛿T; X0 , Y0 , t0 ) = p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
−
𝜕p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿X
𝜕X
+
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
(𝛿X)2 + O((𝛿X)3 )
2
𝜕X 2
p(X, Y − 𝛿Y, T − 𝛿T; X0 , Y0 , t0 ) = p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
+
−
𝜕p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿Y
𝜕Y
+
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
(−𝛿Y)2 + O((𝛿Y)3 )
2
𝜕Y 2
and
p(X, Y + 𝛿Y, T − 𝛿T; X0 , Y0 , t0 ) = p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
+
𝜕p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿Y
𝜕Y
+
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
(𝛿Y)2 + O((𝛿Y)3 ).
2
𝜕Y 2
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183
By substituting the above equations into the discrete forward equation,
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
(𝛿X)2
4
𝜕X 2
𝜕 2 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
+
(𝛿Y)2 + O((𝛿X)3 ) + O((𝛿Y)3 ).
𝜕Y 2
p(X, Y, T; X0 , Y0 , t0 ) = p(X, Y, T − 𝛿T; X0 , Y0 , t0 ) +
Setting 𝛿X = 𝛿Y =
√
𝛿T, dividing the equation by 𝛿T and in the limit 𝛿T → 0
[
lim
𝛿T→0
p(X, Y, T; X0 , Y0 , t0 ) − p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿T
]
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
𝛿T→0 4
𝜕X 2
2
1 𝜕 p(X, Y, T − 𝛿T; X0 , Y0 , t0 )
+ lim
𝛿T→0 4
𝜕Y 2
√
+ lim O( 𝛿T)
= lim
𝛿T→0
we eventually arrive at
𝜕p(X, Y, T; X0 , Y0 , t0 ) 1
=
𝜕T
4
(
𝜕 2 p(X, Y, T; X0 , Y0 , t0 ) 𝜕 2 p(X, Y, T; X0 , Y0 , t0 )
+
𝜕X 2
𝜕Y 2
)
.
◽
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4
Change of Measure
In finance, derivative instruments such as options, swaps or futures can be used for both
hedging and speculation purposes. In a hedging scenario, traders can reduce their risk exposure by buying and selling derivatives against fluctuations in the movement of underlying risky
asset prices such as stocks and commodities. Conversely, in a speculation scenario, traders can
also use derivatives to profit in the future direction of underlying prices. For example, if a trader
expects an asset price to rise in the future, then he/she can sell put options (i.e., the purchaser
of the put options pays an initial premium to the seller and has the right but not the obligation
to sell the shares back to the seller at an agreed price should the share price drop below it at the
option expiry date). Given the purchaser of the put option is unlikely to exercise the option,
the seller would be most likely to profit from the premium paid by the purchaser. From the
point of view of trading such contracts, we would like to price contingent claims (or payoffs
of derivative securities such as options) in such a way that there is no arbitrage opportunity
(or no risk-free profits). By doing so we will ensure that even though two traders may differ in
their estimate of the stock price direction, yet they will still agree on the price of the derivative
security. In order to accomplish this we can rely on Girsanov’s theorem, which tells us how
a stochastic process can have a drift change (but not volatility) under a change of measure.
With the application of this important result to finance we can convert the underlying stock
prices under the physical measure (or real-world measure) into the risk-neutral measure (or
equivalent martingale measure) where all the current stock prices are equal to their expected
future prices discounted at the risk-free rate. This is in contrast to using the physical measure,
where the derivative security prices will vary greatly since the underlying assets will differ in
degrees of risk from each other.
4.1 INTRODUCTION
From the seminal work of Black, Scholes and Merton in using diffusion processes and martingales to price contingent claims such as options on risky asset prices, the theory of mathematical finance is one of the most successful applications of probability theory. Fundamentally,
the Black–Scholes model is concerned with an economy consisting of two assets, a risky asset
(stock) whose price St , t ≥ 0 is a stochastic process and a risk-free asset (bond or money market account) whose value Bt grows at a continuously compounded interest rate. Here we can
assume that St and Bt satisfy the following equations
dBt
dSt
= 𝜇t dt + 𝜎t dWt ,
= rt dt
St
Bt
where 𝜇t is the stock price growth rate, 𝜎t is the stock price volatility, rt is the risk-free rate and
Wt is a standard Wiener process on the probability space (Ω, ℱ, ℙ). In the financial market we
would like to price a contingent claim Ψ(ST ) at time t ≤ T in such a way that no risk-free profit
or arbitrage opportunity exists. But before we define the notion of arbitrage we need some
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financial terminologies. The following first two definitions refer to the concepts of trading
strategy and self-financing trading strategy, which form the basis of creating a martingale
framework to price a contingent claim.
Definition 4.1(a) (Trading Strategy) In a continuous time setting, at time t ∈ [0, T] we consider an economy which consists of a non-dividend-paying risky asset St and a risk-free asset
Bt . A trading strategy (or portfolio) is a pair (𝜙t , 𝜓t ) of stochastic processes which are adapted
to the filtration ℱt , 0 ≤ t ≤ T holding 𝜙t shares of St and 𝜓t units invested in Bt . Therefore, at
time t the value of the portfolio, Πt is
Πt = 𝜙t St + 𝜓t Bt .
Definition 4.1(b) (Self-Financing Trading Strategy) At time t ∈ [0, T], the trading strategy
(𝜙t , 𝜓t ) of holding 𝜙t shares of non-dividend-paying risky asset St and 𝜓t units in risk-free asset
Bt having a portfolio value
Πt = 𝜙t St + 𝜓t Bt
is called self-financing (or self-financing portfolio) if
dΠt = 𝜙t dSt + 𝜓t dBt
which implies the change in the portfolio value is due to changes in the market conditions and
not to either infusion or extraction of funds.
Note that the change in the portfolio value is only attributed to St and Bt rather than 𝜙t and
𝜓t . However, given that we cannot deposit or withdraw cash in the portfolio there will be
restrictions imposed on the pair (𝜙t , 𝜓t ). In order to keep the trading strategy self-financing we
can see that if 𝜙t is increased then 𝜓t will be decreased, and vice versa. Thus, when we make
a choice of either 𝜙t or 𝜓t , the other can easily be found.
Definition 4.1(c) (Admissible Trading Strategy) At time t ∈ [0, T], the trading strategy
(𝜙t , 𝜓t ) of holding 𝜙t shares of non-dividend-paying risky asset St and 𝜓t units in risk-free
asset Bt having a portfolio value
Πt = 𝜙t St + 𝜓t Bt
is admissible if it is self-financing and if Πt ≥ −𝛼 almost surely for some 𝛼 > 0 and t ∈ [0, T].
From the above definition, the existence of a negative portfolio value which is bounded from
below almost surely shows that the investor cannot go too far into debt and thus have a finite
credit line. Once we have restricted ourselves to a class of admissible strategies, the absence
of arbitrage opportunities (“no free lunch”) can be ensured when pricing contingent claims.
But before we discuss its precise terminology we need to define first the concept of attainable
contingent claim.
Definition 4.1(d) (Attainable Contingent Claim) Consider a trading strategy (𝜙t , 𝜓t ) at
time t ∈ [0, T] of holding 𝜙t shares of non-dividend-paying risky asset St and 𝜓t units in
risk-free asset Bt having a portfolio value
Πt = 𝜙t St + 𝜓t Bt .
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The contingent claim Ψ(ST ) is attainable if there exists an admissible strategy worth ΠT =
Ψ(ST ) at exercise time T.
Definition 4.1(e) (Arbitrage) An arbitrage opportunity is an admissible strategy if the following criteria are satisfied:
(no net investment initially)
(i) Π0 = 0
(ii) ℙ(ΠT ≥ 0) = 1 (always win at time T)
(iii) ℙ(ΠT > 0) > 0 (making a positive return on investment at time T).
For risky assets which are traded in financial markets, the pricing of contingent claims based
on the underlying assets is determined based on the presence/absence of arbitrageable opportunities as well as whether the market is complete/incomplete. The following definition discusses
the concept of a complete market.
Definition 4.1(f) (Complete Market) A market is said to be complete if every contingent
claim is attainable (i.e., can be replicated by a self-financing trading strategy). Otherwise, it
is incomplete.
In general, when we have NA number of traded assets (excluding risk-free assets) and NR
sources of risk, the financial “rule of thumb” is as follows:
• If NA < NR then the market has no arbitrage and is incomplete.
• If NA = NR then the market has no arbitrage and is complete.
• If NA > NR then the market has arbitrage.
Within the framework of a Black–Scholes model, the following theorem states the existence
of a trading strategy pair (𝜙t , 𝜓t ) in a portfolio.
Theorem 4.2 (One-Dimensional Martingale Representation Theorem) Let {Wt ∶ 0 ≤ t ≤
T} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ) and let ℱt , 0 ≤ t ≤ T
be the filtration generated by Wt . If Mt , 0 ≤ t ≤ T is a martingale with respect to this filtration
then there exists an adapted process 𝛾t , 0 ≤ t ≤ T such that
t
M t = M0 +
∫0
𝛾u dWu ,
0 ≤ t ≤ T.
From the martingale representation theorem it follows that martingales can be represented
as Itō integrals. However, the theorem only states that an adapted process 𝛾t exists but does
not provide a method to find it explicitly. Owing to the complexity of the proof which involves
functional analysis, the details are omitted in this book.
To show the relationship between the martingale representation theorem and the trading
t
strategy pair (𝜓t , 𝜙t ), if we set B0 = 1 so that Bt = e∫0 ru du then the discounted portfolio value
can be written as
̃ t = B−1
Π
t Πt .
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Therefore, the trading strategy pair (𝜙t , 𝜓t ) is self-financing if and only if the discounted portfolio value can be written as a stochastic integral
̃0 +
̃t = Π
Π
t
∫0
𝜙𝑣 d(B−1
𝑣 S𝑣 ).
By assuming that (𝜙t , 𝜓t ) is a self-financing trading strategy which replicates the contingent
claim Ψ(ST ), it is hoped that the discounted risky asset value B−1
t St will be a martingale (or
ℙ-martingale), so that by taking expectations under the ℙ measure
𝔼ℙ
[
[
]
]
[
]
T
t
|
|
|
|
̃0 +
̃ T | ℱt = 𝔼ℙ Π
̃ 0 | ℱt + 𝔼ℙ
̃
Π
𝜙𝑣 d(B−1
S𝑣 )| ℱt = Π
𝜙 d(B−1
𝑣
𝑣 S𝑣 ) = Πt
|
|
|
∫0
∫0 𝑣
|
or
Πt = 𝔼
ℙ
[
e
T
− ∫t ru du
]
[
]
|
|
T
ℙ
− ∫t ru du
|
|
ΠT | ℱt = 𝔼 e
Ψ(ST )| ℱt
|
|
since ΠT = Ψ(ST ). Although B−1
t St is not a ℙ-martingale on the probability space (Ω, ℱ, ℙ),
there exists another equivalent martingale measure ℚ (or risk-neutral measure) such that B−1
t St
is a ℚ-martingale on the probability space (Ω, ℱ, ℚ). Before we state the theorem we first
present a few intermediate results which will lead to the main result.
Definition 4.3 Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be
another probability measure on (Ω, ℱ, ℚ). Assume that for every A ∈ ℱ satisfying ℙ(A) = 0,
we also have ℚ(A) = 0, then we say ℚ is absolutely continuous with respect to ℙ on ℱand we
write it as ℚ ≪ ℙ.
Theorem 4.4 (Radon–Nikodým Theorem) Let (Ω, ℱ, ℙ) be the probability space satisfying
the usual conditions. Let ℚ be another probability measure on (Ω, ℱ, ℚ). Under the assumption
that ℚ ≪ ℙ, there exists a non-negative random variable Z such that
dℚ
=Z
dℙ
and we call Z the Radon–Nikodým derivative of ℚ with respect to ℙ.
Take note that in finance we need a stronger statement – that is, ℚ to be equivalent to ℙ,
ℚ ∼ ℙ which is ℚ ≪ ℙ and ℙ ≪ ℚ. By imposing the condition ℚ to be equivalent to ℙ, if an
event cannot occur under the ℙ measure then it also cannot occur under the ℚ measure and
vice versa. By doing so, we can now state the following definition of the equivalent martingale
measure for asset prices which pay no dividends.
Definition 4.5 (Equivalent Martingale Measure) Let (Ω, ℱ, ℙ) be the probability space
satisfying the usual conditions and let ℚ be another probability measure on (Ω, ℱ, ℚ). The
probability measure ℚ is said to be an equivalent martingale measure (or risk-neutral measure) if it satisfies:
• ℚ is equivalent to ℙ, ℚ ∼ ℙ;
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(i)
• the discounted price processes {B−1
t St }, i = 1, 2, . . . , m are martingales under ℚ, that
is
[
]
(i) |
−1 (i)
𝔼ℚ B−1
u Su || ℱt = Bt St
for all 0 ≤ t ≤ u ≤ T.
Note that for the case of dividend-paying assets, the discounted values of the asset prices with
the dividends reinvested are ℚ-martingales.
Once the equivalent martingale measure is defined, the next question is can we transform a
diffusion process into a martingale by changing the probability measure? The answer is yes
where the transformation can be established using Girsanov’s theorem, which is instrumental
in risk-neutral pricing for derivatives.
Theorem 4.6 (One-Dimensional Girsanov Theorem) Let {Wt ∶ 0 ≤ t ≤ T} be a
ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ) and let ℱt , 0 ≤ t ≤ T be
the associated Wiener process filtration. Suppose 𝜃t is an adapted process, 0 ≤ t ≤ T and
consider
t
1 t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds .
If
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
then Zt is a positive ℙ-martingale for 0 ≤ t ≤ T. By changing the measure ℙ to a measure ℚ
such that
(
)
dℚ ||
dℚ ||
= Zt ,
𝔼ℙ
ℱ
t =
|
dℙ |
dℙ ||ℱt
then
̃ t = Wt +
W
t
∫0
𝜃u du
is a ℚ-standard Wiener process.
In probability theory, the importance of Girsanov’s theorem cannot be understated as it provides a formal concept of how stochastic processes change under changes in measure. The
theorem is especially important in the theory of financial mathematics as it tells us how to
convert from the physical measure ℙ to the risk-neutral measure ℚ. In short, from the application of this theorem we can change from a Wiener process with drift to a standard Wiener
process.
In contrast, the converse of Girsanov’s theorem says that every equivalent measure is given
by a change in drift. Thus, by changing the measure it is equivalent to changing the drift and
hence in the Black–Scholes model there is only one equivalent risk-neutral measure. Otherwise we would have multiple arbitrage-free derivative prices.
Corollary 4.7 If {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on the probability space
(Ω, ℱ, ℙ) and ℚ is equivalent to ℙ then there exists an adapted process 𝜃t , 0 ≤ t ≤ T such
that:
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̃ t = Wt + ∫ t 𝜃u du is a ℚ-standard Wiener process,
(i) W
0
(ii) the Radon–Nikodým derivative of ℚ with respect to ℙ is
(
)
t
1 t 2
dℚ ||
dℚ ||
ℙ
ℱt =
𝔼
= Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
|
|
dℙ |
dℙ |ℱt
for 0 ≤ t ≤ T.
By comparing the martingale representation theorem with Girsanov’s theorem we can see
that in the former the filtration generated by the standard Wiener process is more restrictive
than the assumption given in Girsanov’s theorem. By including this extra restriction on the
filtration in Girsanov’s theorem we have the following corollary.
Corollary 4.8 Let {Wt ∶ 0 ≤ t ≤ T} be a ℙ-standard Wiener process on the probability
space (Ω, ℱ, ℙ) and let ℱt , 0 ≤ t ≤ T be the filtration generated by Wt . By assuming the
̃ t , 0 ≤ t ≤ T is a ℚ-martingale, there exists
one-dimensional Girsanov theorem holds and if M
an adapted process {̃
𝛾t ∶ 0 ≤ t ≤ T} such that
t
̃t = M
̃0 +
M
∫0
̃ u,
̃
𝛾u d W
0≤t≤T
̃ t is a ℚ-standard Wiener process.
where W
By extending the one-dimensional Girsanov theorem to multiple risky assets where each of
the assets is a random component driven by an independent standard Wiener process, we state
the following multi-dimensional Girsanov theorem.
Theorem 4.9 (Multi-Dimensional Girsanov Theorem) Let Wt = (Wt(1) , Wt(2) , . . . , Wt(n) )T
be an n-dimensional ℙ-standard Wiener process, with {Wt(i) }0≤t≤T , i = 1, 2, . . . , n being an
independent one-dimensional ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ)
and let ℱt , 0 ≤ t ≤ T be the associated Wiener process filtration. Suppose we have an
n-dimensional adapted process 𝜽t = (𝜃t(1) , 𝜃t(2) , . . . , 𝜃t(n) )T , 0 ≤ t ≤ T and we consider
{
Zt = exp
If
𝔼
where
ℙ
t
−
∫0
(
𝜽u ⋅ dWu −
{
exp
T
}
t
1
‖𝜽 ‖2 du .
2 ∫0 ‖ u ‖ 2
1
‖𝜽 ‖2 dt
∫
2 0 ‖ t ‖2
})
<∞
√
||𝜽t ||2 = (𝜃t(1) )2 + (𝜃t(2) )2 + . . . + (𝜃t(n) )2
then Zt is a positive ℙ-martingale for 0 ≤ t ≤ T. By changing the measure ℙ to a measure ℚ
such that
(
)
dℚ ||
dℚ ||
ℱ
𝔼ℙ
= Zt
t =
|
dℙ |
dℙ ||ℱt
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INTRODUCTION
191
for all 0 ≤ t ≤ T then
̃ t = Wt +
W
t
∫0
𝜽u du
̃ t = (W
̃ (1) , W
̃ (2) , . . . , W
̃ (n) )T and
is an n-dimensional ℚ-standard Wiener process where W
t
t
t
t
̃t are
𝜃 (i) du, i = 1, 2, . . . , n such that the component processes of W
∫0 t
independent under ℚ.
̃ (i) = W (i) +
W
t
t
By amalgamating the results of the multi-dimensional Girsanov theorem we also state the
following multi-dimensional martingale representation theorem.
Theorem 4.10 (Multi-Dimensional Martingale Representation Theorem) Let Wt =
(Wt(1) , Wt(2) , . . . , Wt(n) )T be an n-dimensional ℙ-standard Wiener process, with {Wt(i) }0≤t≤T ,
i = 1, 2, . . . , n being an independent one-dimensional ℙ-standard Wiener process on the
probability space (Ω, ℱ, ℙ) and let ℱt , 0 ≤ t ≤ T be the filtration generated by Wt . If Mt ,
0 ≤ t ≤ T is a martingale with respect to this filtration, then there exists an n-dimensional
adapted process 𝜸 t = (𝛾t(1) , 𝛾t(2) , . . . , 𝛾t(n) )T , 0 ≤ t ≤ T such that
t
Mt = M0 +
∫0
𝜸 u ⋅ dWu ,
0 ≤ t ≤ T.
̃ t , 0 ≤ t ≤ T is a
By assuming the multi-dimensional Girsanov theorem holds and if M
𝛾t(1) , ̃
𝛾t(2) , . . . , ̃
𝛾t(n) )T ,
ℚ-martingale there exists an n-dimensional adapted process ̃
𝜸 t = (̃
0 ≤ t ≤ T} such that
t
̃u , 0 ≤ t ≤ T
̃0 +
̃t = M
̃
M
𝜸 ⋅ dW
∫0 u
)T
(
̃t = W
̃ (2) , . . . , W
̃ (n) is an n-dimensional ℚ-standard Wiener process.
̃ (1) , W
where W
t
t
t
In our earlier discussion we used the risk-free asset Bt to construct our trading strategy where,
under the risk-neutral measure ℚ the discounted risky asset price B−1
t St is a ℚ-martingale,
that is
[
]
|
−1
𝔼ℚ B−1
T ST || ℱt = Bt St .
In finance terminology the risk-free asset Bt which does the discounting is called the numéraire.
Definition 4.11 A numéraire is an asset with positive price process which pays no dividends.
Suppose there is another non-dividend-paying asset Nt with strictly positive price process and under the risk-neutral measure ℚ, the discounted price B−1
t Nt is a ℚ-martingale.
From Girsanov’s theorem we can define a new probability measure ℚN given by the
Radon–Nikodým derivative
/
Bt
Nt
dℚN ||
=
.
dℚ ||ℱt N0 B0
By changing the ℚ measure to an equivalent ℚN measure, the discounted price Nt−1 St is a
ℚN -martingale such that
[
]
N
|
𝔼ℚ NT−1 ST | ℱt = Nt−1 St .
|
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Martingale Representation Theorem
Thus, we can deduce that
Bt 𝔼
ℚ
[
]
[
]
ST ||
ST ||
ℚN
ℱ = Nt 𝔼
ℱ
BT || t
NT || t
and by returning to the discussion of the self-financing trading strategy, for a contingent claim
Ψ(ST ) we will also eventually obtain
[
]
[
]
|
|
ℚ Ψ(ST ) |
ℚN Ψ(ST ) |
ℱ = Nt 𝔼
ℱ .
Bt 𝔼
BT || t
NT || t
The idea discussed above is known as the change of numéraire technique and is often used
to price complicated derivatives such as convertible bonds, foreign currency and interest rate
derivatives.
4.2
4.2.1
PROBLEMS AND SOLUTIONS
Martingale Representation Theorem
1. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process and( let X) be a real-valued random variable on the probability space (Ω, ℱ, ℙ) such that 𝔼ℙ |X|2 < ∞. For the process
(
)
Mt = 𝔼ℙ X|ℱt
( )
show that 𝔼ℙ Mt2 < ∞ and that Mt is a ℙ-martingale with respect to the filtration ℱt ,
0 ≤ t ≤ T generated by Wt .
Solution: By definition,
[ (
( )
)2 ]
𝔼ℙ Mt2 = 𝔼ℙ 𝔼ℙ X|ℱt
.
2
From Jensen’s inequality (see Problem 1.2.3.14,
(
) page 48) we set 𝜑(x) = x , which is a
ℙ
convex function. By substituting x = 𝔼 X|ℱt we have
(
)2
(
)
𝔼ℙ X|ℱt ≤ 𝔼ℙ X 2 |ℱt .
Taking the expectation,
[ (
)2 ]
[ (
)]
𝔼ℙ 𝔼ℙ X|ℱt
≤ 𝔼ℙ 𝔼ℙ X 2 |ℱt
and from the tower property (see Problem 1.2.3.11, page 46)
[ (
)]
( )
𝔼ℙ 𝔼ℙ X 2 |ℱt = 𝔼ℙ X 2 .
Thus,
[ (
)2 ]
( )
≤ 𝔼ℙ X 2
𝔼ℙ 𝔼ℙ X|ℱt
[ (
(
)
( )
)2 ]
and because 𝔼ℙ |X|2 < ∞, so 𝔼ℙ Mt2 = 𝔼ℙ 𝔼ℙ X|ℱt
< ∞.
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193
To show that Mt is a ℙ-martingale we note that:
(a) Under the filtration ℱs , 0 ≤ s ≤ t and using the tower property (see Problem 1.2.3.11,
page 46),
[ (
(
)
)| ]
(
)
𝔼ℙ Mt || ℱs = 𝔼ℙ 𝔼ℙ X|ℱt | ℱs = 𝔼ℙ X|ℱs = Ms .
|
(b) Since 𝔼ℙ (|X|2 ) < ∞ we can deduce, using Hölder’s inequality, that
√ (
[
(
)|]
[ (
)]
)
)
| (
𝔼ℙ |Mt | = 𝔼ℙ |𝔼ℙ X|ℱt | ≤ 𝔼ℙ 𝔼ℙ |X| |ℱt = 𝔼ℙ (|X|) ≤ 𝔼ℙ |X|2 < ∞.
|
|
(c) Mt is clearly ℱt -adapted for 0 ≤ t ≤ T.
From the results of (a)–(c) we have shown that Mt is a ℙ-martingale with respect to ℱt ,
0 ≤ t ≤ T.
◽
2. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process and( let X) be a real-valued random variable on the probability space (Ω, ℱ, ℙ) such that 𝔼ℙ |X|2 < ∞. Given that the process
(
)
Mt = 𝔼ℙ X|ℱt
is a ℙ-martingale with respect to the filtration ℱt , 0 ≤ t ≤ T generated by Wt , then from the
martingale representation theorem there exists an adapted process 𝛾u , 0 ≤ u ≤ T such that
t
M t = M0 +
∫0
𝛾u dWu ,
0 ≤ t ≤ T.
Show that
(a) if X = WT then 𝛾t = 1,
(b) if X = WT2 then 𝛾t = 2Wt ,
(c) if X = WT3 then 𝛾t = 3(Wt2 + T − t),
1 2
(T−t)
(d) if X = e𝜎WT , 𝜎 ∈ ℝ then 𝛾t = 𝜎e𝜎Wt + 2 𝜎
,
for 0 ≤ t ≤ T.
Solution: To find the adapted process 𝛾t we note that from Itō’s formula,
dMt =
2
𝜕Mt
𝜕Mt
1 𝜕 Mt
dWt +
dWt2 + . . . = 𝛾t dWt
dt +
𝜕t
𝜕Wt
2 𝜕Wt2
for 0 ≤ t ≤ T.
(
)
(a) For Mt = 𝔼ℙ WT || ℱt and since Wt is a ℙ-martingale (see Problem 2.2.3.1, page 71),
(
)
Mt = 𝔼ℙ WT || ℱt = Wt
and hence, for 0 ≤ t ≤ T,
𝛾t =
𝜕Mt
= 1.
𝜕Wt
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Girsanov’s Theorem
(
)
|
(b) For Mt = 𝔼ℙ WT2 | ℱt and since Wt2 − t is a ℙ-martingale (see Problem 2.2.3.2,
|
page 72),
(
)
|
𝔼ℙ WT2 − T | ℱt = Wt2 − t
|
or
(
)
|
Mt = 𝔼ℙ WT2 | ℱt = Wt2 + T − t.
|
Thus,
𝛾t =
𝜕Mt
= 2Wt
𝜕Wt
for 0 ≤ t ≤ (
T.
)
|
(c) If Mt = 𝔼ℙ WT3 | ℱt and since Wt3 − 3tWt is a ℙ-martingale (see Problem 2.2.3.4,
|
page 73),
(
)
|
𝔼ℙ WT3 − 3TWT | ℱt = Wt3 − 3tWt
|
or
(
)
|
Mt = 𝔼ℙ WT3 | ℱt = Wt3 + 3(T − t)Wt .
|
Therefore,
𝛾t =
(
)
𝜕Mt
= 3 Wt2 + T − t
𝜕Wt
for 0 ≤ t ≤ T.
1 2
(
)
(d) If Mt = 𝔼ℙ e𝜎WT || ℱt and since e𝜎Wt − 2 𝜎 t is a ℙ-martingale (see Problem 2.2.3.3,
page 72),
(
)
𝜎WT − 12 𝜎 2 T ||
𝜎Wt − 12 𝜎 2 t
ℙ
𝔼
e
| ℱt = e
|
or
(
)
1 2
|
Mt = 𝔼ℙ e𝜎WT | ℱt = e𝜎Wt + 2 𝜎 (T−t) .
|
Thus,
𝛾t =
1 2
𝜕Mt
= 𝜎e𝜎Wt + 2 𝜎 (T−t)
𝜕Wt
for 0 ≤ t ≤ T.
◽
4.2.2
Girsanov’s Theorem
1. Novikov’s Condition I. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ) and let 𝜃t be an adapted process, 0 ≤ t ≤ T. By considering
t
1
t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
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Girsanov’s Theorem
195
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
and if
then, without using Itō’s formula, show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
Solution: Using the properties of stochastic integrals (see Problems 3.2.1.10, page 112;
3.2.1.11, page 113; and 3.2.2.4, page 126), we can deduce
(
)
t
t
2
𝜃 dWs ∼ 𝒩 0,
𝜃 ds .
∫0 s
∫0 s
In the same vein, for any u < t, the random variable
(
)
t
t
2
𝜃 dWs ∼ 𝒩 0,
𝜃 ds
∫u s
∫u s
u
t
⟂ ∫u 𝜃s dWs . To show that Zt is a ℙ-martingale, we note the following.
and ∫0 𝜃s dWs ⟂
(a) Under the filtration ℱu ,
ℙ
𝔼
(
)
Zt ||ℱu = 𝔼ℙ
(
e
1
t 2
1
t 2
1
t 2
1
t 2
= e− 2 ∫0 𝜃s ds
= e− 2 ∫0 𝜃s ds
= e− 2 ∫0 𝜃s ds
| ℱu
|
(
)
t
|
𝔼ℙ e− ∫0 𝜃s dWs | ℱu
|
(
)
u
t
|
𝔼ℙ e− ∫0 𝜃s dWs −∫u 𝜃s dWs | ℱu
|
(
)
(
)
u
t
|
|
𝔼ℙ e− ∫0 𝜃s dWs | ℱu 𝔼ℙ e− ∫u 𝜃s dWs | ℱu
|
|
u
= e− 2 ∫0 𝜃s ds ⋅ e− ∫0
u
= e − ∫0
u
)
− ∫0t 𝜃s dWs − 12 ∫0t 𝜃s2 ds ||
𝜃s dWs
1
𝜃s dWs
t 2
1
⋅ e 2 ∫u 𝜃s ds
t 2
1
⋅ e− 2 ∫0 𝜃s ds ⋅ e− 2 ∫t
1
u 2
𝜃s ds
= e− ∫0 𝜃s dWs − 2 ∫0
(
)
Therefore, 𝔼ℙ Zt ||ℱu = Zu .
(b) Taking the expectation of |Zt |,
u 2
𝜃s ds
.
)
(
t
1 t 2 |
( )
|
𝔼ℙ |Zt | = 𝔼ℙ ||e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds ||
|
|
(
)
t
1 t 2
= 𝔼ℙ e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
(
)
1 t 2
t
= e− 2 ∫0 𝜃s ds 𝔼ℙ e− ∫0 𝜃s dWs
1
t 2
1
t 2
= e− 2 ∫0 𝜃s ds ⋅ e 2 ∫0 𝜃s ds
=1<∞
1 2
since, for X ∼ log-𝒩(𝜇, 𝜎 2 ), 𝔼ℙ (X) = e𝜇+ 2 𝜎 .
(c) Because Zt is a function of Wt , it is ℱt -adapted.
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Girsanov’s Theorem
From the results of (a)–(c) we have shown that {Zt ∶ 0 ≤ t ≤ T} is a ℙ-martingale and
because Zt > 0, {Zt }0≤t≤T is a positive ℙ-martingale.
◽
2. Novikov’s Condition II. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ) and let 𝜃t be an adapted process, 0 ≤ t ≤ T. By considering
t
1
t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
and if
then, using Itō’s formula, show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
t
t
Solution: Let Zt = f (Xt ) = eXt where Xt = − ∫0 𝜃s dWs − 12 ∫0 𝜃s2 ds and by applying
Taylor’s expansion and subsequently Itō’s formula,
𝜕f
1 𝜕2f
dXt +
(dXt )2 +
𝜕Xt
2 𝜕Xt2
)
(
1
= eXt −𝜃t dWt − 𝜃t2 dt +
2
= −𝜃t Zt dWt .
dZt =
...
1 Xt 2
e 𝜃t dt
2
Integrating both sides of the equation from u to t, where u < t,
t
∫u
t
dZs = −
∫u
𝜃s Zs dWs
t
Zt = Zu −
∫u
𝜃s Zs dWs .
t
Under the filtration ℱu and because ∫u 𝜃s Zs dWs is independent of ℱu , we have
(
)
𝔼ℙ Zt |ℱu = Zu
(
)
( t
)
|
|
ℙ
where
𝜃 Z dWs | ℱu = 𝔼
𝜃 Z dWs = 0. Using the same steps as
|
∫u s s
∫u s s
|
( )
described in Problem 4.2.2.1 (page 194), we can also show that 𝔼ℙ |Zt | < ∞. Since Zt
is ℱt -adapted and Zt > 0, we have shown that {Zt ∶ 0 ≤ t ≤ T} is a positive ℙ-martingale.
◽
𝔼ℙ
t
3. Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be another
probability measure on (Ω, ℱ, ℚ) such that ℚ is absolutely continuous with respect to ℙ
dℚ ||
= Zt for all 0 ≤ t ≤ T,
on ℱ. Using the Radon–Nikodým theorem show that if
dℙ ||ℱt
then Zt is a positive ℙ-martingale.
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Girsanov’s Theorem
197
Solution: By definition, the probability measures ℙ and ℚ are functions
ℙ ∶ Ω → [0, 1] and ℚ ∶ Ω → [0, 1].
Because ℚ is absolutely continuous with respect to ℙ on ℱ then, from the
Radon–Nikodým theorem,
(
)
dℚ
dℚ
∶ Ω → (0, ∞) and ℙ
> 0 = 1.
dℙ
dℙ
Furthermore,
ℙ
𝔼
[
]
dℚ
dℚ
⋅ dℙ = dℚ = 1.
=
∫Ω
∫
dℙ
Ω dℙ
dℚ ||
= Zt we can therefore deduce that Zt > 0 for all 0 ≤ t ≤ T. To show that Zt
dℙ ||ℱt
is a positive martingale we have the following:
Given
(a) Under the filtration ℱt we define the Radon–Nikodým derivative as
Zt = 𝔼
ℙ
(
)
dℚ ||
ℱ .
dℙ || t
For 0 ≤ s ≤ t ≤ T, and using the tower property of conditional expectation (see Problem 1.2.3.11, page 46),
𝔼
(
ℙ
)
[
Zt || ℱs = 𝔼ℙ 𝔼ℙ
(
dℚ ||
ℱ
dℙ || t
)| ]
(
)
dℚ ||
|
ℙ
ℱ = Zs .
| ℱs = 𝔼
|
dℙ || s
|
|( dℚ | )| dℚ |
dℚ
|
|
|
| . Therefore, for 0 ≤ t ≤ T and using
> 0, so |Zt | = |
|=
| dℙ ||ℱt |
dℙ
dℙ ||ℱt
|
|
the tower property of conditional expectation (see Problem 1.2.3.11, page 46),
(b) Since
]
dℚ ||
ℱ
dℙ || t
[ (
)]
[ ( )]
dℚ ||
ℱ
𝔼ℙ 𝔼ℙ ||Zt || = 𝔼ℙ 𝔼ℙ
dℙ || t
𝔼ℙ (|Zt |) = 𝔼ℙ
we have
[
)
(
( )
dℚ
= 1 < ∞.
𝔼ℙ ||Zt || = 𝔼ℙ
dℙ
(c) Zt is clearly ℱt -adapted for 0 ≤ t ≤ T.
From the results of (a)–(c) we have shown that Zt is a positive ℙ-martingale.
◽
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Girsanov’s Theorem
4. Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be another
probability measure on (Ω, ℱ, ℚ) such that ℚ is absolutely continuous with respect to ℙ
dℚ ||
on ℱ. Let
= Zt for all 0 ≤ t ≤ T, so that Zt is a positive ℙ-martingale. By letting
dℙ ||ℱt
{Xt ∶ 0 ≤ t ≤ T} be an ℱt measurable random variable, show using the Radon–Nikodým
theorem that
[ (
)]
( )
dℚ ||
ℚ
ℙ
𝔼 Xt = 𝔼 Xt
dℙ ||ℱt
and hence deduce that for A ∈ ℱt ,
(
∫A
Xt dℚ =
∫A
dℚ ||
dℙ ||ℱt
Xt
)
dℙ.
dℚ
= Z where Z is a
Solution: From the definition of the Radon–Nikodým derivative
dℙ
positive random variable, we can write
∫Ω
Xt dℚ =
∫Ω
Xt Z dℙ
( )
( )
𝔼ℚ Xt = 𝔼ℙ Xt Z .
Using the tower property of conditional expectation (see Problem 1.2.3.11, page 46),
( )
( )
𝔼ℚ Xt = 𝔼ℙ Xt Z
[ (
)]
= 𝔼ℙ 𝔼ℙ Xt Z|ℱt
[
(
)]
= 𝔼ℙ Xt 𝔼ℙ Z|ℱt .
Under the filtration ℱt , we define the Radon–Nikodým derivative as
Zt = 𝔼
ℙ
(
dℚ ||
ℱ
dℙ || t
)
=
dℚ ||
dℙ ||ℱt
and therefore
( )
(
)
𝔼ℚ Xt = 𝔼ℙ Xt Zt
By defining
or
[ (
)]
( )
dℚ ||
𝔼ℚ Xt = 𝔼ℙ Xt
.
dℙ ||ℱt
{
1 if A ∈ ℱt
1IA =
0 otherwise
and if Xt is ℱt measurable, then for any A ∈ ℱt ,
(
)
(
)
𝔼ℚ 1IA Xt = 𝔼ℙ 1IA Xt Zt
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Girsanov’s Theorem
4:27 P.M.
199
which is equivalent to
∫A
Xt dℚ =
∫A
(
or
∫A
Xt dℚ =
∫A
Xt
Xt Zt dℙ
dℚ ||
dℙ ||ℱt
)
dℙ.
◽
5. Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be another
probability measure on (Ω, ℱ, ℚ) such that ℚ is absolutely continuous with respect to ℙ
dℚ ||
= Zt for all 0 ≤ t ≤ T, so that Zt is a positive ℙ-martingale. By letting
on ℱ. Let
dℙ ||ℱt
{Xt ∶ 0 ≤ t ≤ T} be an ℱt measurable random variable, show using the Radon–Nikodým
theorem that
[
)−1 ]
(
( )
dℚ ||
ℙ
ℚ
𝔼 Xt = 𝔼 Xt
dℙ ||ℱt
and hence deduce that for A ∈ ℱt ,
(
∫A
Xt dℙ =
∫A
Xt
dℚ ||
dℙ ||ℱt
)−1
dℚ.
dℚ
Solution: From the definition of the Radon–Nikodým derivative
= Z where Z is a
dℙ
positive random variable, we can write
∫Ω
Xt dℙ =
∫Ω
Xt Z −1 dℚ
( )
(
)
𝔼ℙ Xt = 𝔼ℚ Xt Z −1 .
Using the tower property of conditional expectation (see Problem 1.2.3.11, page 46),
( )
(
)
𝔼ℙ Xt = 𝔼ℚ Xt Z −1
[ (
)]
= 𝔼ℚ 𝔼ℚ Xt Z −1 |ℱt
[
(
)]
= 𝔼ℚ Xt 𝔼ℚ Z −1 |ℱt .
Under the filtration ℱt , we define the Radon–Nikodým derivative as
(
)
dℚ ||
dℚ ||
Z t = 𝔼ℙ
ℱ
t =
|
dℙ |
dℙ ||ℱt
and therefore
( )
(
)
𝔼 Xt = 𝔼ℚ Xt Zt−1
ℙ
or
[
)−1 ]
(
( )
dℚ ||
ℚ
.
𝔼 Xt = 𝔼 Xt
dℙ ||ℱt
ℙ
Page 199
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4.2.2
By defining
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Girsanov’s Theorem
{
1 if A ∈ ℱt
1IA =
0 otherwise
and if Xt is ℱt measurable, then for any A ∈ ℱt ,
(
)
(
)
𝔼ℙ 1IA Xt = 𝔼ℚ 1IA Xt Zt−1
which is equivalent to
∫A
Xt dℙ =
or
∫A
(
∫A
Xt dℙ =
∫A
Xt
Xt Zt−1 dℚ
dℚ ||
dℙ ||ℱt
)−1
dℚ.
◽
6. Let {Wt ∶ 0 ≤ t ≤ T} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ)
and let ℱt , 0 ≤ t ≤ T be the associated Wiener process filtration. By defining
t
1 t 2
dℚ ||
= Zt = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du
|
dℙ |ℱt
( 1 T 2 )
where for all 0 ≤ t ≤ T the adapted process 𝜃t satisfies 𝔼ℙ e 2 ∫0 𝜃t dt < ∞, show that ℚ
is a probability measure.
Solution: From Problem 4.2.2.4 (page 198) we can deduce that for A ∈ ℱt ,
∫A
or
dℚ =
∫A
Zt dℙ
(
)
ℚ(A) = 𝔼ℙ Zt 1IA
{
1 if A ∈ ℱt
1IA =
0 otherwise.
where
In addition, from Problem
( 4.2.2.1 (page
) 194) we can deduce that the adapted process 𝜃t
t
follows
∫0
t
𝜃s dWs ∼ 𝒩 0,
∫0
𝜃s2 ds
so that
(
)
1 t 2
1 t 2
1 t 2
t
𝔼ℙ (Zt ) = e− 2 ∫0 𝜃u du 𝔼ℙ e− ∫0 𝜃u dWu = e− 2 ∫0 𝜃u du ⋅ e 2 ∫0 𝜃u du = 1.
Since ℙ is a probability measure and based on the above results, ℚ is also a probability
measure because
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201
(
)
(a) ℚ(∅) = 𝔼ℙ Zt 1I∅ = 𝔼ℙ (Zt )ℙ(∅) = 0 since ℙ(∅) = 0.
(b) ℚ(Ω) = 𝔼ℙ (Zt 1IΩ ) = 𝔼ℙ (Zt )ℙ(Ω) = 1 since ℙ(Ω) = 1 and 𝔼ℙ (Zt ) = 1.
(c) For A1 , A2 , . . . ∈ ℱt , Ai ∩ Aj = ∅, i ≠ j, i, j ∈ {1, 2, . . .}
∞
⋃
∞
∞
(
) ⋃
( ) ( )
( ) ⋃
ℚ Ai =
𝔼ℙ Zt 1IAi =
𝔼ℙ Zt ℙ Ai
i=1
and since
⋃∞
i=1
∞
⋃
i=1
ℙ(Ai ) =
∑∞
i=1
i=1
ℙ(Ai ),
∞
∞
∞
(
) ∑
( ) ∑
( ) ( ) ∑
ℚ Ai =
𝔼ℙ Zt ℙ Ai =
𝔼ℙ Zt 1IAi =
ℚ(Ai ).
i=1
i=1
i=1
i=1
◽
7. Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be another
probability measure on (Ω, ℱ, ℚ) such that ℚ is absolutely continuous with respect to ℙ
dℚ ||
on ℱ. Let
= Zt for all 0 ≤ t ≤ T, so that Zt is a positive ℙ-martingale. By letting
dℙ ||ℱt
{Xt ∶ 0 ≤ t ≤ T} be an ℱt measurable random variable and using the partial averaging
property given as
∫A
𝔼ℙ (Y|𝒢) dℙ =
∫A
Ydℙ,
for all
A∈𝒢
where Y is a random variable on the probability space (Ω, ℱ, ℙ) and 𝒢 is a sub-𝜎-algebra
of ℱ show that for 0 ≤ s ≤ t ≤ T,
)−1 [ (
)| ]
(
|
|
(
)
dℚ
dℚ
|
|
|
𝔼ℙ X t
𝔼ℚ Xt |ℱs =
|ℱ .
dℙ ||ℱs
dℙ ||ℱt || s
)−1 [ (
)| ]
|
(
)
dℚ
dℚ ||
|
|
𝔼ℙ X t
Solution: First, note that
| ℱ = Zs−1 𝔼ℙ Xt Zt |ℱs is
dℙ ||ℱs
dℙ ||ℱt || s
ℱs -measurable. For any A ∈ ℱs , and using the results of Problem 4.2.2.4 (page 198) as
well as partial averaging, we have
(
∫A
(
)
Zs−1 𝔼ℙ Xt Zt |ℱs dℚ =
∫A
(
)
Zs−1 𝔼ℙ Xt Zt |ℱs ⋅ Zs dℙ
=
∫A
Xt Zt dℙ
=
∫A
Xt dℚ.
Using the partial averaging once again, we have
∫A
Xt dℚ =
∫A
(
)
𝔼ℚ Xt |ℱs dℚ.
4:27 P.M.
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Therefore,
∫A
(
)
Zs−1 𝔼ℙ Xt Zt |ℱs dℚ =
or
∫A
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Girsanov’s Theorem
(
)
𝔼ℚ Xt |ℱs dℚ
(
)
(
)
Zs−1 𝔼ℙ Xt Zt |ℱs = 𝔼ℚ Xt |ℱs .
By substituting Zt =
dℚ ||
we have
dℙ ||ℱt
(
ℚ
𝔼 (Xt |ℱs ) =
dℚ ||
dℙ ||ℱs
[
)−1
𝔼
ℙ
(
Xt
dℚ ||
dℙ ||ℱt
)| ]
|
| ℱs .
|
|
◽
8. Let (Ω, ℱ, ℙ) be the probability space satisfying the usual conditions. Let ℚ be another
probability measure on (Ω, ℱ, ℚ) such that ℚ is absolutely continuous with respect to ℙ
dℚ ||
= Zt for all 0 ≤ t ≤ T, so that Zt is a positive ℙ-martingale. By letting
on ℱ. Let
dℙ ||ℱt
{Xt ∶ 0 ≤ t ≤ T} be an ℱt measurable random variable and using the partial averaging
property given as
∫A
𝔼ℙ (Y|𝒢) dℙ =
∫A
Y dℙ,
for all
A∈𝒢
where Y is a random variable on the probability space (Ω, ℱ, ℙ) and 𝒢 is a sub-𝜎-algebra
of ℱ show that for 0 ≤ s ≤ t ≤ T,
[ (
)
)−1 | ]
(
|
|
(
)
dℚ
dℚ
|
|
|
𝔼ℙ Xt |ℱs =
𝔼ℚ Xt
| ℱs .
|
|
|
dℙ |ℱs
dℙ |ℱt
|
[
]
)
(
)−1 |
(
(
)
dℚ ||
dℚ ||
|
ℚ
Xt
𝔼
Solution: First, note that
| ℱs = Zs 𝔼ℚ Xt Zt−1 |ℱs is
|
dℙ ||ℱs
dℙ ||ℱt
|
ℱs measurable. For any A ∈ ℱs , and using the results of Problem 4.2.2.5 (page 199) as
well as partial averaging, we have
∫A
(
)
Zs 𝔼ℚ Xt Zt−1 |ℱs dℙ =
∫A
=
∫A
=
∫A
(
)
Zs 𝔼ℚ Xt Zt−1 |ℱs ⋅ Zs−1 dℚ
Xt Zt−1 dℚ
Xt dℙ.
Using the partial averaging once again, we have
∫A
Xt dℙ =
∫A
(
)
𝔼ℙ Xt |ℱs dℙ.
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Girsanov’s Theorem
203
Therefore,
∫A
or
(
)
Zs 𝔼ℚ Xt Zt−1 |ℱs dℙ =
∫A
(
)
𝔼ℙ Xt |ℱs dℙ
(
)
(
)
Zs 𝔼ℚ Xt Zt−1 |ℱs = 𝔼ℙ Xt |ℱs .
By substituting Zt =
ℙ
𝔼
dℚ ||
we have
dℙ ||ℱt
(
)
Xt |ℱs =
(
dℚ ||
dℙ ||ℱs
[
)
𝔼
(
ℚ
Xt
dℚ ||
dℙ ||ℱt
)−1 | ]
|
| ℱs .
|
|
◽
9. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ) and
suppose 𝜃t is an adapted process, 0 ≤ t ≤ T. We consider
t
1
t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
and if
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
By changing the measure ℙ to a measure ℚ such that
̃ t = Wt +
W
dℚ ||
= Zt , show that
dℙ ||ℱt
t
∫0
𝜃u du
is a ℚ-martingale.
Solution: The first part of the result is given in Problem 4.2.2.1 (page 194) or Problem
̃ t is a ℚ-martingale we note the following:
4.2.2.2 (page 196). To show that W
(a) Let 0 ≤ s ≤ t ≤ T, under the filtration ℱs and using the result of Problem 4.2.2.7 (page
201), we have
(
)
(
)
̃ t || ℱs = 1 𝔼ℙ W
̃ t Zt || ℱs .
𝔼ℚ W
|
|
Zs
From Itō’s formula and using the results of Problem 4.2.2.2 (page 196),
)
(
̃ t dZt + Zt dW
̃ t + dW
̃ t dZt
̃ t Zt = W
d W
(
)
(
)
) (
)(
̃ t −𝜃t Zt dWt + Zt dWt + 𝜃t dt + dWt + 𝜃t dt −𝜃t Zt dWt
=W
(
)
̃ t Zt dWt .
= 1 − 𝜃t W
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Girsanov’s Theorem
By integrating both sides of the equation from s to t, where s < t,
t
∫s
(
)
̃ u Zu =
d W
t
(
)
̃ u Zu dWu
1 − 𝜃u W
∫s
̃ s Zs +
̃ t Zt = W
W
t
Under the filtration ℱs and because
∫s
t
∫s
(
)
̃ u Zu dWu .
1 − 𝜃u W
(
)
̃ u Zu dWu ⟂
1 − 𝜃u W
⟂ ℱs ,
(
)
̃ s Zs
̃ t Zt || ℱs = W
𝔼ℙ W
|
(
)
( t(
)
t(
)
)
|
|
ℙ
̃ u Zu dWu | ℱs = 𝔼ℙ
̃ u Zu dWu = 0.
where 𝔼
1 − 𝜃u W
1 − 𝜃u W
|
∫s
∫s
|
Thus,
(
)
(
)
̃ Z =W
̃ s.
̃ t || ℱs = 1 𝔼ℙ W
̃ t Zt || ℱs = 1 W
𝔼ℚ W
|
|
Zs
Zs s s
(b) For 0 ≤ t ≤ T,
t
|
| t
|
|
|̃ | |
|
|
|
𝜃u du| ≤ ||Wt || + |
𝜃u du| .
|Wt | = |Wt +
| | |
|∫0
|
|
∫0
|
|
|
|
Taking expectations under the ℚ measure, using Hölder’s inequality (see
Problem 1.2.3.2,
page 41) and taking
note that the positive random variable
(
)
t
t
1
2
2
𝜃 ds,
𝜃 ds under ℙ, we have
Zt ∼ log-𝒩 − 2
∫0 s
∫0 s
)
(
|
(
) || t
|̃ |
|
𝔼ℚ |W
𝜃u du|
| ≤ 𝔼ℚ ||Wt || + |
t
| |
|∫ 0
|
|
|
t
|
|
(
) |
|
𝜃u du|
= 𝔼ℙ ||Wt || Zt + |
|
|∫0
|
|
t
|
|
(
) |
|
= 𝔼ℙ ||Wt Zt || + |
𝜃 du|
|∫0 u |
|
|
√ ( ) ( ) | t
|
|
|
≤ 𝔼ℙ Wt2 𝔼ℙ Zt2 + |
𝜃u du|
|∫0
|
|
|
√
t
|
|
t 2
|
|
= te∫0 𝜃u du + |
𝜃 du|
|∫0 u |
|
|
<∞
( 1 T 2 )
since 𝔼ℙ e 2 ∫0 𝜃u du < ∞.
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Girsanov’s Theorem
205
̃ t is a function of Wt , it is ℱt -adapted.
(c) Because W
̃ t , 0 ≤ t ≤ T is a ℚ-martingale.
From the results of (a)–(c), we have shown that W
◽
10. One-Dimensional Girsanov Theorem. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process
on the probability space (Ω, ℱ, ℙ) and suppose 𝜃t is an adapted process, 0 ≤ t ≤ T. We
consider
t
1 t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
and if
show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
By changing the measure ℙ to a measure ℚ such that
̃ t = Wt +
W
dℚ ||
= Zt , show that
dℙ ||ℱt
t
∫0
𝜃u du
is a ℚ-standard Wiener process.
Solution: The first part of the result is given in Problem 4.2.2.1 (page 194) or Problem
4.2.2.2 (page 196). As for the second part of the result, we can prove it using two methods.
̃ t ∼ 𝒩(0, t) and to do this we
Method 1. We first need to show that under the ℚ measure, W
rely on the moment generating function. By definition, for a constant 𝜓,
(
)
̃
MW̃ (𝜓) = 𝔼ℚ e𝜓 Wt
t
(
)
̃
= 𝔼ℙ e𝜓 Wt Zt
(
)
t
1 t 2
t
= 𝔼ℙ e𝜓Wt +𝜓 ∫0 𝜃s ds ⋅ e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
=e
(
)
∫0t 𝜓𝜃s − 12 𝜃s2 ds
(
)
t
𝔼ℙ e𝜓Wt −∫0 𝜃s dWs
=e
(
)
t
∫0 𝜓𝜃s − 12 𝜃s2 ds
( t
)
t
𝔼ℙ e∫0 𝜓dWs −∫0 𝜃s dWs
(
)
t
∫0 𝜓𝜃s − 12 𝜃s2 ds
( t
)
𝔼ℙ e∫0 (𝜓−𝜃s )dWs .
=e
Using the properties of the stochastic Itō integral under the ℙ-measure
(
)
t
t(
)2
𝜓 − 𝜃s ds
(𝜓 − 𝜃s ) dWs ∼ 𝒩 0,
∫0
∫0
therefore
( t
)
1 t
2
𝔼ℙ e∫0 (𝜓−𝜃s )dWs = e 2 ∫0 (𝜓−𝜃s )
ds
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4.2.2
and hence
1
MW̃ (𝜓) = e 2 𝜓
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Girsanov’s Theorem
2t
t
which is a moment generating function of a normal distribution with mean zero and variance t.
̃ t ∼ 𝒩(0, t).
Thus, under the ℚ measure, W
̃
Finally, to show that Wt , 0 ≤ t ≤ T is a ℚ-standard Wiener process we have the following:
̃ 0 = 0 and W
̃ t certainly has continuous sample paths for t > 0.
(a) W
̃ t ∼ 𝒩(0, s) since
̃ t+s − W
(b) For t > 0 and s > 0, W
(
)
(
)
( )
̃ t = 𝔼ℚ W
̃ t+s − W
̃ t+s − 𝔼ℚ W
̃t = 0
𝔼ℚ W
and from the results of Problem 2.2.1.4 (page 57),
(
)
(
)
( )
(
)
̃ t = Varℚ W
̃t
̃ t+s − W
̃ t+s + Varℚ W
̃ t − 2Covℚ W
̃ t+s , W
Varℚ W
= t + s + t − 2min{t + s, t}
= s.
̃ t is a ℚ-martingale (see Problem 4.2.2.9, page 203), then for t > 0, s > 0
(c) Because W
and under the filtration ℱt ,
(
)
̃ t.
̃ t+s || ℱt = W
𝔼ℚ W
|
Since we can write
(
)
(
)
̃t + W
̃ t || ℱt
̃ t+s ||ℱt = 𝔼ℚ W
̃ t+s − W
𝔼ℚ W
|
|
(
)
(
)
̃ t || ℱt + 𝔼ℚ W
̃ t+s − W
̃ t || ℱt
= 𝔼ℚ W
|
|
(
)
|
̃ t | ℱt + W
̃t
̃ t+s − W
= 𝔼ℚ W
|
(
)
(
)
̃ t+s − W
̃ t+s − W
̃ t || ℱt = 𝔼ℚ W
̃t = 0
𝔼ℚ W
|
then
̃ t+s − W
̃t ⟂
̃ t+s − W
̃t ⟂
̃ t.
which implies W
⟂ ℱt . Therefore, W
⟂W
̃ t , 0 ≤ t ≤ T is a ℚ-standard Wiener process.
From the results of (a)–(c) we have shown W
̃
Method 2. The process Wt , 0 ≤ t ≤ T is a ℚ-martingale with the following properties:
̃ 0 = 0 and has continuous paths for t > 0.
(i) W
(ii) By setting ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn−1 < tn = t, n ∈ ℕ the quadratic varĩ t is
ation of W
lim
n→∞
n−1 (
∑
̃t − W
̃t
W
i+1
i
i=1
)2
= lim
n→∞
= lim
n→∞
n−1
∑
(
ti+1
Wti+1 − Wti +
i=1
n−1 (
∑
)2
Wti+1 − Wti
i=1
∫0
𝜃u du −
ti
∫0
)2
𝜃u du
Page 206
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V3 - 08/23/2014
207
n−1 (
)(
∑
+ lim
2 Wti+1 − Wti
n→∞
+ lim
n→∞
i=1
n−1
∑
i=1
(
ti+1
∫ti
ti+1
∫ti
)
𝜃u du
)2
𝜃u du
=t
ti+1
since the quadratic variation of Wt is t and lim
n→∞ ∫t
𝜃u du = 0. Informally, we can
i
̃ t = dt.
̃ t dW
write dW
Based on properties (i) and (ii), and from the Lévy characterisation theorem (see Problem
̃ t , 0 ≤ t ≤ T is a ℚ-standard Wiener process.
3.2.1.15, page 119), we can deduce that W
◽
11. Multi-Dimensional Novikov Condition. Let Wt = (Wt(1) , Wt(2) , . . . , Wt(n) )T be an
n-dimensional ℙ-standard Wiener process, with {Wt(i) ∶ 0 ≤ t ≤ T}, i = 1, 2, . . . , n
an independent one-dimensional ℙ-standard Wiener process on the probability space
(Ω, ℱ, ℙ). Suppose we have an n-dimensional adapted process 𝜽t = (𝜃t(1) , 𝜃t(2) , . . . , 𝜃t(n) )T ,
0 ≤ t ≤ T. By considering
{
t
}
t
1
2
‖
‖
𝜽
−
𝜽 ⋅ dWu −
du
∫0 u
2 ∫0 ‖ u ‖2
(
{
Zt = exp
and if
𝔼
ℙ
exp
where
||𝜽t ||2 =
T
1
‖𝜽 ‖2 dt
2 ∫0 ‖ t ‖2
})
<∞
√
(𝜃t(1) )2 + (𝜃t(2) )2 + . . . + (𝜃t(n) )2
show that Zt is a positive ℙ-martingale.
Solution: To show that Zt is a positive ℙ-martingale, we have the following.
(a) We can write Zt as
{
Zt = exp
t
−
∫0
𝜽u ⋅ dWu −
}
t
1
‖𝜽 ‖2 du
2 ∫0 ‖ u ‖2
}
n
t∑
1
(i) 2
−
(𝜃 ) du
= exp −
∫0
2 ∫0 i=1 u
i=1
{ n (
)}
t
t
∑
1
(i)
(i)
(i) 2
−
𝜃 dWu −
(𝜃 ) du
= exp
∫0 u
2 ∫0 u
i=1
{
n
t∑
𝜃u(i) dWu(i)
4:27 P.M.
Page 207
Chin
208
c04.tex
4.2.2
=
{
n
∏
exp
t
−
i=1
=
n
∏
∫0
𝜃u(i) dWu(i)
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4:27 P.M.
Girsanov’s Theorem
}
t
1
(i) 2
−
(𝜃 ) du
2 ∫0 u
Zt(i)
i=1
t (i)
(i)
t
1
(i) 2
where Zt(i) = e− ∫0 𝜃u dWu − 2 ∫0 (𝜃u ) du . From Problem 4.2.2.1 (page 194), we have
shown that Zt(i) is a positive ℙ-martingale and since {Wt(i) ∶ 0 ≤ t ≤ T}, i = 1, 2, . . . , n
is an independent one-dimensional ℙ-standard Wiener process on the probability
space (Ω, ℱ, ℙ), so for 0 ≤ s ≤ t,
( n
)
n
n
(
) ∏
∏ (i) ||
∏
(
)
|
ℙ
ℙ
Zt | ℱs =
𝔼ℙ Zt(i) | ℱs =
Zs(i) = Zs .
𝔼 Zt |ℱs = 𝔼
|
|
i=1
i=1
i=1
|
(b) Furthermore, because 𝔼ℙ (|Zt(i) |) < ∞, hence
( n
( n
)
)
n
(
)
|∏
|
∏ | (i) |
∏
( )
| |
(i) |
ℙ | |
ℙ |
ℙ
𝔼 |Zt | = 𝔼
Zt | ≤ 𝔼
𝔼ℙ |Zt(i) | < ∞.
|
|Zt | ≤
| |
|
|
| |
| i=1
i=1
i=1
|
(c) Zt is clearly ℱt -adapted for 0 ≤ t ≤ T.
From the results of (a)–(c) and since Zt > 0, we have shown that Zt is a positive
ℙ-martingale.
◽
12. Multi-Dimensional Girsanov Theorem. Let Wt = (Wt(1) , Wt(2) , . . . , Wt(n) )T be an
n-dimensional ℙ-standard Wiener process, with {Wt(i) ∶ 0 ≤ t ≤ T}, i = 1, 2, . . . , n
an independent one-dimensional ℙ-standard Wiener process on the probability space
(Ω, ℱ, ℙ). Suppose we have an n-dimensional adapted process 𝜽t = (𝜃t(1) , 𝜃t(2) , . . . , 𝜃t(n) )T ,
0 ≤ t ≤ T. By considering
{
}
t
t
1
2
‖
‖
𝜽
𝜽 ⋅ dWu −
du
Zt = exp −
∫0 u
2 ∫0 ‖ u ‖2
and if
𝔼
where
ℙ
(
{
exp
T
1
‖𝜽 ‖2 dt
2 ∫0 ‖ t ‖2
})
<∞
√
||𝜽t ||2 = (𝜃t(1) )2 + (𝜃t(2) )2 + . . . + (𝜃t(n) )2
show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
By changing the measure ℙ to a measure ℚ such that
that
̃t = Wt +
W
dℚ ||
= Zt for all 0 ≤ t ≤ T, show
dℙ ||ℱt
t
∫0
𝜽u du
Page 208
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209
̃ t = (W
̃ (1) , W
̃ (2) , . . . , W
̃ (n) )T and
is an n-dimensional ℚ-standard Wiener process where W
t
t
t
t
(i)
(i)
(i)
̃t are indẽ = W + ∫ 𝜃 du, i = 1, 2, . . . , n such that the component processes of W
W
t
t
0 t
pendent under ℚ.
Solution: The first part of the result is given in Problem 4.2.2.11 (page 207).
̃t = (W
̃ (1) , W
̃ (2) , . . . , W
̃ (n) )T , and
As for the second part of the result, we note that given W
t
t
t
following the steps given in Problem 4.2.2.9 (page 203), we can easily show that for each
i = 1, 2, . . . , n,
̃ (i) = W (i) +
W
t
t
t
∫0
𝜃u(i) du
̃ (i) = 0, and has continuous sample paths with quadratic
is a ℚ-martingale. Because W
0
variation equal to t (see Problem 4.2.2.10, page 205), then from the multi-dimensional
̃ (i) ,
Lévy characterisation theorem (see Problem 3.2.1.16, page 121) we can show that W
t
(j)
(i)
⟂ Wt , i ≠ j, i, j =
i = 1, 2, . . . , n is a ℚ-standard Wiener process. Finally, since Wt ⟂
1, 2, . . . , n we have
(
)(
)
̃ (j) = dW (i) + 𝜃 (i) dt dW (j) + 𝜃 (j) dt
̃ (i) ⋅ dW
dW
t
t
t
t
t
t
(j)
(j)
(j)
(j)
= dWt(i) dWt + 𝜃t dWt(i) dt + 𝜃t(i) dWt dt + 𝜃t(i) 𝜃t dt2
= 0.
Thus, from the multi-dimensional Lévy characterisation theorem (see Problem 3.2.1.16,
̃t , 0 ≤ t ≤ T is an n-dimensional ℚ-standard Wiener propage 121) we can deduce that W
̃t is independent under ℚ.
cess such that each component process of W
◽
13. Convergence Issue of Equivalent Measures. Let Wt = (Wt(1) , Wt(2) , . . . , Wt(n) )T be an
n-dimensional ℙ-standard Wiener process, with {Wt(i) ∶ 0 ≤ t ≤ T}, i = 1, 2, . . . , n
an independent one-dimensional ℙ-standard Wiener process on the probability space
(Ω, ℱ, ℙ). Suppose we have an n-dimensional adapted process 𝜽t = (𝜃t(1) , 𝜃t(2) , . . . , 𝜃t(n) )T ,
0 ≤ t ≤ T. We consider
{
t
}
t
1
2
‖
‖
𝜽
−
𝜽 ⋅ dWu −
du
∫0 u
2 ∫0 ‖ u ‖2
(
{
Zt = exp
and if
𝔼
ℙ
where
||𝜽t ||2 =
exp
T
1
‖𝜽 ‖2 dt
2 ∫0 ‖ t ‖2
})
<∞
√
(𝜃t(1) )2 + (𝜃t(2) )2 + . . . + (𝜃t(n) )2
show that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
4:27 P.M.
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210
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4.2.2
By changing the measure ℙ to a measure ℚ such that
that
̃t = Wt +
W
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Girsanov’s Theorem
dℚ ||
= Zt for all 0 ≤ t ≤ T, show
dℙ ||ℱt
t
𝜽u du
∫0
̃ t = (W
̃ (1) , W
̃ (2) , . . . , W
̃ (n) )T and
is an n-dimensional ℚ-standard Wiener process where W
t
t
t
t
(i)
(i)
(i)
̃t are indẽ = W + ∫ 𝜃 du, i = 1, 2, . . . , n such that the component processes of W
W
t
t
0
t
t
𝜃 (i) dWu(i) , i = 1, 2, . . . , n be a sequence of random vari∫0 u
ables on the probability space (Ω, ℱ, ℙ) and assume 𝜃t(1) = 𝜃t(2) = . . . = 𝜃t(n) = 𝜃t where
𝜃t is an adapted process, 0 ≤ t ≤ T. For n → ∞ show that under the ℙ measure
pendent under ℚ. Let X (i) =
(
)
1 ∑ (i)
ℙ lim
X =0
n→∞ n
i=1
n
=1
and under the equivalent ℚ measure
(
)
1 ∑ (i)
X =0
ℚ lim
n→∞ n
i=1
n
= 0.
Discuss the implications on the equivalent measures ℙ and ℚ.
Solution: The first two results are given in Problem 4.2.2.12 (page 208).
As for the final result, because 𝜃t(1) = 𝜃t(2) = . . . = 𝜃t(n) = 𝜃t then under the ℙ measure we
can deduce that
(
)
t
X (i) =
∫0
𝜃u(i) dWt(i) ∼ 𝒩 0,
t
∫0
𝜃u2 du
for i = 1, 2, . . . , n. Because {Wt(i) ∶ 0 ≤ t ≤ T}, i = 1, 2, . . . , n are independent and identically, distributed,
1 ∑ (i) 1 ∑
X =
n i=1
n i=1
n
n
or
1 ∑ (i) 1 ∑
X =
n i=1
n i=1
n
n
{
t
∫0
{
t
∫0
}
𝜃u(i)
dWt(i)
)
t
1
2
∼ 𝒩 0,
𝜃 du
n ∫0 u
√
}
𝜃u(i) dWt(i)
(
t
1
𝜃 2 du,
n ∫0 u
=𝜙
Taking limits n → ∞, we have
(
)
1 ∑ (i)
X =0
ℙ lim
n→∞ n
i=1
n
= 1.
𝜙 ∼ 𝒩(0, 1).
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Girsanov’s Theorem
211
On the contrary, under the measure ℚ for i = 1, 2, . . . , n,
̃ (i) = W (i) +
W
t
t
t
∫0
𝜃u(i) du
̃ (2) , . . . , W
̃ (n) are independent and identically
̃ (1) , W
is a ℚ-standard Wiener process and W
t
t
t
distributed. Thus,
̃ (i) = dW (i) + 𝜃 (i) dt
dW
t
t
t
and in turn we can write
t
t
𝜃u(i) dWu(i) =
∫0
∫0
̃ u(i) −
𝜃u(i) dW
t
∫0
(𝜃u(i) )2 du.
Because 𝜃t(1) = 𝜃t(2) = . . . = 𝜃t(n) = 𝜃t and under the ℚ measure,
t
X (i) =
∫0
(
𝜃u(i) dWu(i) ∼ 𝒩 −
t
∫0
t
𝜃u2 du,
∫0
)
𝜃u2 du .
̃ (i) }0≤t≤T , i = 1, 2, . . . , n are independent and identically distributed,
Since {W
t
1 ∑ (i) 1 ∑
X =
n i=1
n i=1
n
n
{
t
∫0
}
𝜃u(i)
(
∼𝒩 −
dWt(i)
t
∫0
𝜃u2
)
t
1
2
du,
𝜃 du
n ∫0 u
or
n
1∑
n
i=1
X (i) =
n
1∑
n
i=1
By setting n → ∞,
{
t
∫0
}
𝜃u(i) dWt(i)
√
t
=−
(
∫0
𝜃u2 du + 𝜙
t
1
𝜃 2 du,
n ∫0 u
𝜙 ∼ 𝒩(0, 1).
)
1 ∑ (i)
X =0
ℚ lim
n→∞ n
i=1
n
= 0.
Therefore, in the limit n → ∞, the measures ℙ and ℚ fail to be equivalent.
◽
14. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ). By
considering a Wiener process with drift
̂ t = 𝛼t + Wt
W
̂ t is a standard Wiener
and by using Girsanov’s theorem to find a measure ℚ under which W
process, show that the joint density of
̂t
W
and
̂s
̂ t = max W
M
0≤s≤t
4:27 P.M.
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212
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4.2.2
under ℙ is
⎧ 2(2x−𝑤) 𝛼𝑤− 1 𝛼 2 t− 1 (2x−𝑤)2
2
2t
⎪ √ e
f ̂ℙ ̂ (x, 𝑤) = ⎨ t 2𝜋t
Mt ,Wt
⎪0
⎩
V3 - 08/23/2014
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Girsanov’s Theorem
x ≥ 0, 𝑤 ≤ x
otherwise.
Finally, deduce that the joint density of
̂t
W
under ℙ is
f ℙ ̂ (y, 𝑤)
m
̂ t ,Wt
̂s
m
̂ t = min W
and
0≤s≤t
⎧ −2(2y−𝑤) 𝛼𝑤− 1 𝛼 2 t− 1 (2y−𝑤)2
2
2t
e
⎪ √
= ⎨ t 2𝜋t
⎪0
⎩
y ≤ 0, y ≤ 𝑤
otherwise.
Solution: By defining
̂ t = 𝛼t + Wt
W
using Itō’s formula we can write
and
̂s
̂ t = max W
M
0≤s≤t
̃t
̂ t = dW
dW
̃ t = Wt + ∫ 𝛼 du. From Girsanov’s theorem there exists an equivalent probability
where W
0
measure ℚ on the filtration ℱs , 0 ≤ s ≤ t defined by the Radon–Nikodým derivative
t
s
Z s = e− ∫0
s
𝛼dWu − 12 ∫0 𝛼 2 du
1 2
s
= e−𝛼Ws − 2 𝛼
̂
1 2
s
= e−𝛼(Ws −𝛼s)− 2 𝛼
̂
1 2
s
= e−𝛼Ws + 2 𝛼
̂ t is a ℚ-standard Wiener process.
so that W
̂ t, W
̂ t)
From Problem 2.2.5.4 (page 86), under ℚ, the joint probability distribution of (M
can be written as
f ̂ℚ ̂ (x, 𝑤)
Mt ,Wt
⎧ 2(2x−𝑤) − 1 (2x−𝑤)2
⎪ √ e 2t
= ⎨ t 2𝜋t
⎪0
⎩
x ≥ 0, 𝑤 ≤ x
otherwise.
̂ t ) under ℙ, we use the
̂ t, W
In order to find the joint cumulative distribution function of (M
result given in Problem 4.2.2.5 (page 199)
)
(
(
)
̂ t ≤ 𝑤 = 𝔼ℙ 1I ̂ ̂
̂ t ≤ x, W
ℙ M
{Mt ≤x,Wt ≤𝑤}
(
)
= 𝔼ℚ Zt−1 1I{M̂ ≤x,Ŵ ≤𝑤}
t
t
Page 212
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213
)
( ̂ 1 2
= 𝔼ℚ e𝛼Wt − 2 𝛼 t 1I{M̂ ≤x,Ŵ ≤𝑤}
𝑤
=
x
∫−∞ ∫−∞
𝛼𝑣− 12 𝛼 2 t
e
t
t
f ̂ℚ
(u, 𝑣) dud𝑣.
̂t
Mt ,W
̂ 0 = 0, M
̂ t ≥ 0 therefore W
̂t ≤ M
̂ t . By definition, the joint density of (M
̂ t, W
̂ t ) under
Since W
ℙ is
)
(
𝜕2
̂t ≤ 𝑤
̂ t ≤ x, W
ℙ M
f ̂ℙ ̂ (x, 𝑤) =
Mt ,Wt
𝜕x𝜕𝑤
which implies
{
f ̂ℙ ̂ (x, 𝑤)
Mt ,Wt
=
1
2(2x−𝑤) 𝛼𝑤− 12 𝛼2 t− 2t
(2x−𝑤)2
√
e
t 2𝜋t
y ≥ 0, 𝑤 ≤ x
0
otherwise.
For the case of the joint density of
̂ t = 𝛼t + Wt
W
̂s
m
̂ t = min W
and
0≤s≤t
̂ t ) under ℚ is
from Problem 2.2.5.5 (page 88), the joint probability distribution of (̂
mt , W
f ℚ ̂ (y, 𝑤)
m
̂ t ,Wt
⎧ −2(2y−𝑤) − 1 (2y−𝑤)2
e 2t
⎪ √
= ⎨ t 2𝜋t
⎪0
⎩
y ≤ 0, y ≤ 𝑤
otherwise.
̂ t)
Using the same techniques as discussed earlier, the joint cumulative distribution of (̂
mt , W
under ℙ is
)
)
(
(
̂ t ≤ 𝑤 = 𝔼ℙ 1I
ℙ m
̂ t ≤ y, W
̂
{m̂ t ≤y,Wt ≤𝑤}
(
)
= 𝔼ℚ Zt−1 1I{m̂ ≤y,Ŵ ≤𝑤}
t
t
( ̂ 1 2
)
= 𝔼ℚ e𝛼Wt − 2 𝛼 t 1I{m̂ ≤x,Ŵ ≤𝑤}
t
t
𝑤
=
y
∫−∞ ∫−∞
1 2
e𝛼𝑣− 2 𝛼 t f ℚ
̂t
m
̂ t ,W
(u, 𝑣) dud𝑣.
Therefore,
fℙ
̂t
m
̂ t ,W
(y, 𝑤) =
)
(
𝜕2
̂t ≤ 𝑤
ℙ m
̂ t ≤ y, W
𝜕y𝜕𝑤
⎧ −2(2y−𝑤) 𝛼𝑤− 1 𝛼 2 t− 1 (2y−𝑤)2
2
2t
e
⎪ √
= ⎨ t 2𝜋t
⎪0
⎩
y ≤ 0, y ≤ 𝑤
otherwise.
◽
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Girsanov’s Theorem
15. Running Maximum and Minimum of a Wiener Process. Let {Wt ∶ t ≥ 0} be a ℙ-standard
Wiener process on the probability space (Ω, ℱ, ℙ) and let
Xt = 𝜈 + 𝜇t + 𝜎Wt
be a Wiener process starting from 𝜈 ∈ ℝ with drift 𝜇 ∈ ℝ and volatility 𝜎 > 0. Let the
running maximum of the process Xt up to time t be defined as
MtX = max Xs .
0≤s≤t
Using Girsanov’s theorem to find a measure ℚ under which Xt is a standard Wiener process, show that the cumulative distribution function of the running maximum is
)
)
(
(
2𝜇(x−𝜈)
( X
)
x − 𝜈 − 𝜇t
−x + 𝜈 − 𝜇t
2
−e 𝜎 Φ
, x≥𝜈
ℙ Mt ≤ x = Φ
√
√
𝜎 t
𝜎 t
where Φ(⋅) is the standard normal cumulative distribution function.
Finally, deduce that the cumulative distribution function of the running minimum
mXt = min Xs
0≤s≤t
is
ℙ
(
mXt
)
≤x =Φ
(
x − 𝜈 − 𝜇t
√
𝜎 t
)
(
+e
2𝜇(x−𝜈)
𝜎2
Φ
x − 𝜈 + 𝜇t
√
𝜎 t
)
,
x ≤ 𝜈.
Find the probability density functions for MtX and mXt .
Solution: Let Yt =
Xt − 𝜈
such that
𝜎
Yt = 𝛼t + Wt
where 𝛼 = 𝜇∕𝜎. Following this, we can define
MtY = max Ys
0≤s≤t
Xt − 𝜈
. From Problem 4.2.2.14 (page 211) we can
𝜎
Y and Y as
write the joint density of Mt
t
to be the running maximum of Yt =
f ℙY (y, 𝑤y )
Mt ,Yt
⎧ 2(2y−𝑤y ) 𝛼𝑤 − 1 𝛼2 t− 1 (2y−𝑤 )2
y
2t
e y 2
⎪ √
= ⎨ t 2𝜋t
⎪0
⎩
y ≥ 0, 𝑤y ≤ y
otherwise.
Given that the pair of random variables (MtY , Yt ) only take values from the set
{(y, 𝑤y ) ∶ y ≥ 0, y ≥ 𝑤y }
Page 214
4.2.2
Girsanov’s Theorem
215
Figure 4.1 Region of MtY ≤ y in shaded area.
then in order to get the cumulative distribution function of MtY we compute over the shaded
region given in Figure 4.1
ℙ(MtY ≤ y) =
0
∫−∞ ∫0
y
f ℙY (u, 𝑣) dud𝑣 +
Mt ,Yt
y
∫0 ∫𝑣
y
f ℙY (u, 𝑣) dud𝑣
Mt ,Yt
such that
y
0
∫−∞ ∫0
Mt ,Yt
0
y
2(2u − 𝑣) 𝛼𝑣− 1 𝛼 2 t− 1 (2u−𝑣)2
2t
e 2
dud𝑣
√
∫−∞ ∫0 t 2𝜋t
|u=y
0
1 𝛼𝑣− 12 𝛼2 t− 2t1 (2u−𝑣)2 ||
=−
e
√
| d𝑣
∫−∞ 2𝜋t
|
|u=0
f ℙY (u, 𝑣) dud𝑣 =
0
1 2
1
2
1
= −√
e𝛼𝑣− 2 𝛼 t− 2t (2y−𝑣) d𝑣
∫
2𝜋t −∞
0
1 2
1 2
1
+√
e𝛼𝑣− 2 𝛼 t− 2t 𝑣 d𝑣
∫
2𝜋t −∞
and using similar steps we have
y
∫0 ∫𝑣
y
y
1 2
1
2
1
f ℙY (u, 𝑣) dud𝑣 = − √
e𝛼𝑣− 2 𝛼 t− 2t (2y−𝑣) d𝑣
Mt ,Yt
2𝜋t ∫0
y
1 2
1 2
1
+√
e𝛼𝑣− 2 𝛼 t− 2t 𝑣 d𝑣.
2𝜋t ∫0
Chin
216
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4.2.2
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4:27 P.M.
Girsanov’s Theorem
Therefore,
y
y
1 2
1
1 2
1 2
2
1
1
ℙ(MtY ≤ y) = − √
e𝛼𝑣− 2 𝛼 t− 2t (2y−𝑣) d𝑣 + √
e𝛼𝑣− 2 𝛼 t− 2t 𝑣 d𝑣
2𝜋t ∫−∞
2𝜋t ∫−∞
and by completing the squares we have
1
1
1
𝛼𝑣 − 𝛼 2 t − (2y − 𝑣)2 = − (𝑣 − 2y − 𝛼t)2 + 2𝛼y
2
2t
2t
1
1
1
𝛼𝑣 − 𝛼 2 t − 𝑣2 = − (𝑣 − 𝛼t)2 .
2
2t
2t
Thus, for y ≥ 0,
y
y
1
2
1 − 2t1 (𝑣−𝛼t)2
e2𝛼y
e
e− 2t (𝑣−2y−𝛼t) d𝑣 +
d𝑣
ℙ(MtY ≤ y) = − √
√
∫−∞ 2𝜋t
2𝜋t ∫−∞
−y−𝛼t
√
t
y−𝛼t
√
t
1 − 1 z2
1 − 1 z2
= −e
√ e 2 dz + ∫
√ e 2 dz
∫−∞
−∞
2𝜋
2𝜋
(
)
(
)
−y − 𝛼t
y − 𝛼t
= −e2𝛼y Φ
+Φ
√
√
t
t
2𝛼y
(
or
ℙ(MtY
≤ y) = Φ
y − 𝛼t
√
t
)
(
2𝛼y
−e
Φ
−y − 𝛼t
√
t
)
.
By substituting
MtY =
MtX − 𝜈
,
𝜎
y=
(
)
x−𝜈
,
𝜎
and
𝛼=
𝜇
𝜎
we eventually have
ℙ(MtX ≤ x) = Φ
x − 𝜈 − 𝜇t
√
𝜎 t
(
−e
2𝜇(x−𝜈)
𝜎2
Φ
−x + 𝜈 − 𝜇t
√
𝜎 t
)
,
x ≥ 𝜈.
Finally, we focus on the minimum value of Xt . Take note that Y0 = 0 and hence by defining
mYt = min Ys
0≤s≤t
we have mYt ≤ 0. Therefore, for y ≤ 0 and following the symmetry of the standard Wiener
process,
(
)
( Y
)
ℙ mt ≤ y = ℙ min Ys ≤ y
(
0≤s≤t
{
}
= ℙ − max −Ys ≤ y
0≤s≤t
)
Page 216
Chin
4.2.2
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Girsanov’s Theorem
217
(
{
}
{
}
)
= ℙ − max −𝛼s − Ws ≤ y
(
0≤s≤t
)
= ℙ − max −𝛼s + Ws ≤ y
(
0≤s≤t
)
̃
= ℙ max Ys ≥ −y
0≤s≤t
̃t = −𝛼t + Wt . Thus,
where Y
ℙ
(
mYt
(
)
̃
≤ y = 1 − ℙ max Ys ≤ −y
)
0≤s≤t
(
)
)
y + 𝛼t
−y + 𝛼t
2𝛼y
=1−Φ
+e Φ
√
√
t
t
(
)
)
(
y + 𝛼t
y − 𝛼t
+ e2𝛼y Φ
.
=Φ
√
√
t
t
(
Substituting
mYt =
mXt − 𝜈
,
𝜎
y=
(
)
we have
ℙ(mXt
≤ x) = Φ
x − 𝜈 − 𝜇t
√
𝜎 t
x−𝜈
𝜎
and
(
+e
2𝜇(x−a)
𝜎2
Φ
𝛼=
x − 𝜈 + 𝜇t
√
𝜎 t
𝜇
𝜎
)
,
x ≤ 𝜈.
By denoting fMX (x) and fmX (x) as the probability density functions for MtX and mXt , respect
t
tively, therefore
)
d ( X
ℙ Mt ≤ x
dx
[ (
)
)]
(
2𝜇(x−𝜈)
x − 𝜈 − 𝜇t
−x
+
𝜈
−
𝜇t
d
=
Φ
− e 𝜎2 Φ
√
√
dx
𝜎 t
𝜎 t
fMX (x) =
t
(
1
−
1
e 2
= √
𝜎 2𝜋t
1
e
+ √
𝜎 2𝜋t
and
x−𝜈−𝜇t
√
𝜎 t
(
)2
2𝜇 2𝜇(x−𝜈)
− 2 e 𝜎2 Φ
𝜎
2𝜇(x−𝜈) 1
−2
𝜎2
(
1
−
2
e 2
= √
𝜎 2𝜋t
x−𝜈−𝜇t
√
𝜎 t
(
−x+𝜈−𝜇t
√
𝜎 t
−x + 𝜈 − 𝜇t
√
𝜎 t
)
)2
(
)2
2𝜇 2𝜇(x−𝜈)
− 2 e 𝜎2 Φ
𝜎
−x + 𝜈 − 𝜇t
√
𝜎 t
)
,
x≥𝜈
4:27 P.M.
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218
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4.2.2
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Girsanov’s Theorem
d
ℙ(mXt ≤ x)
dx
(
[ (
)
)]
2𝜇(x−a)
x
−
𝜈
+
𝜇t
x − 𝜈 − 𝜇t
d
=
Φ
+ e 𝜎2 Φ
√
√
dx
𝜎 t
𝜎 t
fmX (x) =
t
(
1
−
1
e 2
= √
𝜎 2𝜋t
(
1
−
2
e 2
= √
𝜎 2𝜋t
x−𝜈−𝜇t
√
𝜎 t
(
)2
x−𝜈−𝜇t
√
𝜎 t
2𝜇 2𝜇(x−𝜈)
+ 2 e 𝜎2 Φ
𝜎
x − 𝜈 + 𝜇t
√
𝜎 t
(
)2
2𝜇 2𝜇(x−𝜈)
+ 2 e 𝜎2 Φ
𝜎
)
x − 𝜈 + 𝜇t
√
𝜎 t
1
e
+ √
𝜎 2𝜋t
2𝜇(x−𝜈) 1
−2
𝜎2
(
x−𝜈+𝜇t
√
𝜎 t
)2
)
,
x ≤ 𝜈.
◽
16. First Passage Time Density of a Standard Wiener Process Hitting a Sloping Line. Let
{Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on {the probability space }(Ω, ℱ, ℙ). By setting T𝛼+𝛽t as a stopping time such that T𝛼+𝛽t = inf t ≥ 0 ∶ Wt = 𝛼 + 𝛽t , 𝛼, 𝛽 ∈ ℝ, 𝛼 ≠ 0
show that the probability density function of T𝛼+𝛽t is given as
|𝛼|
fT𝛼+𝛽t (t) = √
t 2𝜋t
1
2
e− 2t (𝛼+𝛽t) .
̃ t = Wt − ∫ t 𝛽 du such that
Solution: By defining W
0
{
}
̃t = 𝛼
T𝛼+𝛽t = inf t ≥ 0 ∶ W
then, from Girsanov’s theorem, there exists an equivalent probability measure ℚ on the
filtration ℱs , 0 ≤ s ≤ t defined by the Radon–Nikodým derivative
s
Zs = e∫0
s
𝛽 dWu − 12 ∫0 𝛽 2 du
1 2
s
= e𝛽Ws − 2 𝛽
̃
1 2
s
= e𝛽(Ws +𝛽s)− 2 𝛽
̃
1 2
s
= e𝛽 W s + 2 𝛽
̃ t is a ℚ-standard Wiener process. From Problem 2.2.5.3 (page 85), the probability
so that W
̃ t = 𝛼} under ℚ is therefore
density function of T𝛼+𝛽t = inf {t ≥ 0 ∶ W
|𝛼|
̃fT (t) = √
𝛼+𝛽t
t 2𝜋t
1
e− 2t 𝛼 .
2
To find the probability density of T𝛼+𝛽t under ℙ we note that
(
)
(
)
ℙ T𝛼+𝛽t ≤ t = 𝔼ℙ 1IT𝛼+𝛽t
Page 218
Chin
4.2.2
Girsanov’s Theorem
c04.tex
V3 - 08/23/2014
219
(
)
= 𝔼ℚ Zt−1 1IT𝛼+𝛽t
( ̃ 1 2
)
= 𝔼ℚ e−𝛽 Wt − 2 𝛽 t 1IT𝛼+𝛽t
(
)
1 2
= 𝔼ℚ e−𝛼𝛽− 2 𝛽 t 1IT𝛼+𝛽t
t
=
∫0
t
=
∫0
t
=
∫0
1 2
e−𝛼𝛽− 2 𝛽 u ̃fT𝛼+𝛽t (u) du
1 2
u
e−𝛼𝛽− 2 𝛽
1
e− 2u 𝛼 du
2
|𝛼| − 1 (𝛼+𝛽u)2
du.
e 2u
√
u 2𝜋u
Since
fT𝛼+𝛽t (t) =
we therefore have
|𝛼|
√
u 2𝜋u
)
d (
ℙ T𝛼+𝛽t ≤ t
dt
1
2
|𝛼|
e− 2t (𝛼+𝛽t) .
fT𝛼+𝛽t (t) = √
t 2𝜋t
◽
17. Let {Wt ∶ t ≥ 0} be a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ), and
let ℱt be the filtration generated by Wt . Suppose 𝜃t is an adapted process, 0 ≤ t ≤ T and
by considering
t
1 t 2
Zt = e− ∫0 𝜃s dWs − 2 ∫0 𝜃s ds
and if
( 1 T 2 )
𝔼ℙ e 2 ∫0 𝜃t dt < ∞
then Zt is a positive ℙ-martingale for 0 ≤ t ≤ T. By changing the measure ℙ to a measure
dℚ ||
ℚ such that
= Zt , from Girsanov’s theorem
dℙ ||ℱt
̃ t = Wt +
W
t
∫0
𝜃u du,
0≤t≤T
̃ t , 0 ≤ t ≤ T is a ℚ-martingale, then show that Mt =
is a ℚ-standard Wiener process. If M
̃
Zt Mt , 0 ≤ t ≤ T is a ℙ-martingale.
Using the martingale representation theorem, show that there exists an adapted process
{̃
𝛾t ∶ 0 ≤ t ≤ T} such that
̃t = M
̃0 +
M
t
∫0
̃ u,
̃
𝛾u dW
0 ≤ t ≤ T.
4:27 P.M.
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Chin
220
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4.2.2
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Girsanov’s Theorem
̃ t is a ℙ-martingale, we note the following:
Solution: To show that Mt = Zt M
(a) From Problem 4.2.2.8 (page 202),
(
)
(
)
|
̃ s = Ms .
𝔼ℙ Mt |ℱs = Zs 𝔼ℚ Mt Zt−1 | ℱs = Zs M
|
(b) From Problem 4.2.2.5 (page 199),
(
)
)
)
(
(
)
(
1
|
|
|̃ |
ℙ
ℚ
|Mt |
= 𝔼ℚ |Mt Zt−1 | = 𝔼ℚ |M
𝔼 |Mt | = 𝔼
<∞
|
|
|
| t|
Zt
̃ t is a ℚ-martingale.
since M
̃ t is clearly ℱt -adapted.
(c) Mt = Zt M
̃ t is a ℙ-martingale.
From the results of (a)–(c) we have shown that Mt = Zt M
−1
̃
Using Itō’s formula on dMt = d(Mt Zt ) and since dZt = −𝜃t Zt dWt (see Problem 4.2.2.2,
̃ t = dWt + 𝜃t dt, we have
page 196) and dW
(
) 1
M
M
1
d Mt Zt−1 = dMt − 2t dZt + 3t (dZt )2 − 2 dMt dZt
Zt
Zt
Zt
Zt
=
) M (
(
)
)
M (
1
1
dMt − 2t −𝜃t Zt dWt + 3t 𝜃t2 Zt2 dt − 2 dMt −𝜃t Zt dWt
Zt
Zt
Zt
Zt
=
)
𝜃
M𝜃 (
1
dMt + t dMt dWt + t t dWt + 𝜃t dt
Zt
Zt
Zt
=
𝜃
M𝜃
1
̃ t.
dMt + t dMt dWt + t t dW
Zt
Zt
Zt
Since Mt is a ℙ-martingale then, from the martingale representation theorem, there exists
an adapted process 𝛾t , 0 ≤ t ≤ T such that
t
Mt = M0 +
∫0
𝛾u dWu ,
0≤t≤T
or
dMt = 𝛾t dWt
and hence
̃t =
dM
𝛾t
𝛾𝜃
M𝜃
̃t
dWt + t t dt + t t dW
Zt
Zt
Zt
𝛾t
𝛾𝜃
M𝜃
̃ t − 𝜃t dt) + t t dt + t t dW
̃t
(dW
Zt
Zt
Zt
)
(
𝛾t + Mt 𝜃t
̃t
dW
=
Zt
=
̃t
=̃
𝛾t d W
Page 220
Chin
4.2.3
Risk-Neutral Measure
where ̃
𝛾t =
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221
𝛾t + Mt 𝜃t
. Integrating on both sides, we therefore have
Zt
̃t = M
̃0 +
M
4.2.3
c04.tex
t
∫0
̃ u,
̃
𝛾u dW
0 ≤ t ≤ T.
◽
Risk-Neutral Measure
1. Geometric Brownian Motion. Consider an economy consisting of a risk-free asset and a
stock price (risky asset). At time t, the risk-free asset Bt and the stock price St have the
following diffusion processes
dBt = rt Bt dt, dSt = 𝜇t St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price drift rate, 𝜎t is the stock price volatility
(which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on
the probability space (Ω, ℱ, ℙ).
From the following discounted stock price process
t
Xt = e− ∫0 ru du St
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent risk-neutral measure ℚ, Xt is a ℚ-martingale and by applying the martingale
representation theorem show also that
t
Xt = X0 +
̃ t = Wt +
where W
∫0
̃ u,
𝜎 u Xu d W
0≤t≤T
t
𝜆u du is a ℚ-standard Wiener process with 𝜆t =
∫0
the market price of risk.
Finally, show that under the ℚ measure the stock price follows
𝜇t − rt
defined as
𝜎t
̃ t.
dSt = rt St dt + 𝜎t St dW
Solution: By expanding dXt using Taylor’s theorem and then applying Itō’s formula,
dXt =
2
2
𝜕Xt
𝜕X
1 𝜕 Xt 2 1 𝜕 Xt 2
dt
+
dS + . . .
dt + t dSt +
𝜕t
𝜕St
2 𝜕t2
2 𝜕St2 t
t
= −rt Xt dt + e− ∫0 ru du dSt
= (𝜇t − rt )Xt dt + 𝜎t Xt dWt
[(
)
]
𝜇t − rt
dt + dWt
= 𝜎t Xt
𝜎t
̃t
= 𝜎t Xt dW
4:27 P.M.
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222
4.2.3
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Risk-Neutral Measure
𝜇t − rt
. From Girsanov’s theorem there exists
∫0
𝜎t
an equivalent martingale measure or risk-neutral measure on the filtration ℱs , 0 ≤ s ≤ t
defined by the Radon–Nikodým derivative
̃ t = Wt +
where W
t
𝜆u du such that 𝜆t =
s
Zs = e− ∫0
s
𝜆u du− 12 ∫0 𝜆2u dWu
̃ t is a ℚ-standard Wiener process. Given that under the risk-neutral measure ℚ,
so that W
the discounted stock price diffusion process
̃t
dXt = 𝜎t Xt dW
has no dt term, then Xt is a ℚ-martingale (see Problem 3.2.2.5, page 127). Thus, from the
martingale representation theorem, there exists an adapted process 𝛾u , 0 ≤ u ≤ T such that
t
Xt = X0 +
∫0
or
t
Xt = X0 +
∫0
̃ u,
𝛾u dW
̃ u,
𝜎 u Xu d W
0≤t≤T
0≤t≤T
t
̃ u is a ℚ-martingale.
𝜎 X dW
∫0 u u
̃ t + 𝜆t dt into dSt = 𝜇t St dt + 𝜎t St dWt , the stock price difFinally, by substituting dWt = dW
fusion process under the risk-neutral measure becomes
such that under ℚ, the process
̃ t.
dSt = rt St dt + 𝜎t St dW
◽
2. Arithmetic Brownian Motion. Consider an economy consisting of a risk-free asset and a
stock price (risky asset). At time t, the risk-free asset Bt and the stock price St have the
following diffusion processes
dBt = rt Bt dt, dSt = 𝜇t dt + 𝜎t dWt
where rt is the risk-free rate, 𝜇t is the stock price drift rate, 𝜎t is the stock price volatility
(which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on
the probability space (Ω, ℱ, ℙ).
From the following discounted stock price process
t
Xt = e− ∫0 ru du St
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent risk-neutral measure ℚ, Xt is a ℚ-martingale and by applying the martingale
representation theorem show also that
t
Xt = X0 +
∫0
̃
𝜎u B−1
u d Wu ,
0≤t≤T
Page 222
Chin
4.2.3
Risk-Neutral Measure
̃ t = Wt +
where W
c04.tex
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223
t
∫0
𝜆u du is a ℚ-standard Wiener process such that
𝜆t =
𝜇t − rt St
𝜎t
is defined as the market price of risk.
Finally, show that under the ℚ measure the stock price follows
̃ t.
dSt = rt St dt + 𝜎t dW
Solution: By expanding dXt using Taylor’s theorem and then applying Itō’s formula,
dXt =
2
2
𝜕Xt
𝜕X
1 𝜕 Xt 2 1 𝜕 Xt 2
dt
+
dS + . . .
dt + t dSt +
𝜕t
𝜕St
2 𝜕t2
2 𝜕St2 t
t
= −rt Xt dt + e− ∫0 ru du dSt
t
t
= e− ∫0 ru du (𝜇t − rt St )dt + 𝜎t e− ∫0 ru du dWt
[(
)
]
t
𝜇t − rt St
dt + dWt
= 𝜎t e− ∫0 ru du
𝜎t
̃
= 𝜎t B−1
t d Wt
𝜇t − rt St
. From Girsanov’s theo∫0
𝜎t
rem there exists an equivalent martingale measure or risk-neutral measure on the filtration
ℱs , 0 ≤ s ≤ t defined by the Radon–Nikodým derivative
t
t
̃ t = Wt +
where Bt = e∫0 ru du , W
𝜆u du such that 𝜆t =
s
Zs = e− ∫0
s
𝜆u du− 12 ∫0 𝜆2u dWu
̃ t is a ℚ-standard Wiener process. Given that under the risk-neutral measure ℚ,
so that W
the discounted stock price diffusion process
̃
dXt = 𝜎t B−1
t d Wt
has no dt term, then Xt is a ℚ-martingale (see Problem 3.2.2.5, page 127). Thus, from the
martingale representation theorem, there exists an adapted process 𝛾u , 0 ≤ u ≤ T such that
t
Xt = X0 +
∫0
or
t
X t = X0 +
t
such that under ℚ, the process
∫0
∫0
̃ u,
𝛾u dW
̃
𝜎u B−1
u d Wu ,
0≤t≤T
0≤t≤T
̃
𝜎u B−1
u d Wu is a ℚ-martingale.
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̃ t + 𝜆t dt into dSt = 𝜇t dt + 𝜎t dWt , the stock price diffuFinally, by substituting dWt = dW
sion process under the risk-neutral measure becomes
̃ t.
dSt = rt St dt + 𝜎t dW
◽
3. Discounted Portfolio. Consider an economy consisting of a risk-free asset and a stock
price (risky asset). At time t, the risk-free asset Bt and the stock price St have the following
diffusion processes
dBt = rt Bt dt, dSt = 𝜇t St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price drift rate, 𝜎t is the stock price volatility
(which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on
the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of stock
and 𝜓t units being invested in a risk-free asset. From the following discounted portfolio
value
t
Yt = e− ∫0 ru du Πt
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent
risk-neutral measure ℚ, then the discounted portfolio Yt is a ℚ-martingale.
Solution: At time t, the portfolio Πt is valued as
Πt = 𝜙t St + 𝜓t Bt
and the differential of the portfolio is
dΠt = 𝜙t dSt + 𝜓t rt Bt dt
(
)
= 𝜙t 𝜇t St dt + 𝜎t St dWt + rt 𝜓t Bt dt
(
)
= rt Πt dt + 𝜙t 𝜇t − rt St dt + 𝜙t 𝜎t St dWt
(
)
= rt Πt dt + 𝜙t 𝜎t St 𝜆t dt + dWt
where 𝜆t =
formula,
𝜇t − rt
. By expanding dYt using Taylor’s theorem and then applying Itō’s
𝜎t
dYt =
2
2
𝜕Yt
𝜕Y
1 𝜕 Yt 2 1 𝜕 Yt 2
dt
+
dΠt + . . .
dt + t dSt +
𝜕t
𝜕Πt
2 𝜕t2
2 𝜕Π2t
t
= −rt Yt dt + e− ∫0 ru du dΠt
(
)
t
t
= −rt e− ∫0 ru du Πt dt + e− ∫0 ru du rt Πt dt + 𝜙t 𝜎t St (𝜆t dt + dWt )
(
)
t
= 𝜙t d e− ∫0 ru du St
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Risk-Neutral Measure
where 𝜆t =
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225
𝜇t − rt
and from Problem 4.2.3.1 (page 221),
𝜎t
(
)
t
t
d e− ∫0 ru du St = 𝜎t e− ∫0 ru du St (𝜆t dt + dWt ))
t
̃t
= 𝜎t e− ∫0 ru du St dW
̃ t = Wt +
where W
t
𝜆 du.
∫0 u
From Girsanov’s theorem, there exists an equivalent martingale measure or risk-neutral
measure on the filtration ℱs , 0 ≤ s ≤ t defined by the Radon–Nikodým derivative
s
Zs = e− ∫0
s
𝜆u du− 12 ∫0 𝜆2u dWu
̃ t is a ℚ-standard Wiener process. Given that under the risk-neutral measure ℚ,
so that W
the discounted stock price diffusion process
t
̃t
dYt = 𝜙t 𝜎t e− ∫0 ru du St dW
has no dt term, then the discounted portfolio Yt is a ℚ-martingale (see Problem 3.2.2.5,
page 127).
◽
4. Self-Financing Trading Strategy. Consider an economy consisting of a risk-free asset and
a stock price (risky asset). At time t, the risk-free asset Bt and the stock price St have the
following diffusion processes
dBt = rt Bt dt, dSt = 𝜇t St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price drift rate, 𝜎t is the stock price volatility
(which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on
the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of stock
and 𝜓t units being invested in a risk-free asset. Under the risk-neutral measure ℚ, the
following discounted portfolio value
t
Yt = e− ∫0 ru du Πt
is a ℚ-martingale. Using the martingale representation theorem, show that the portfolio
(𝜙t , 𝜓t ) trading strategy has the values
𝜙t =
𝛾 t Bt
,
𝜎 t St
𝜓t =
𝜎t Πt − 𝛾t Bt
,
𝜎 t Bt
0≤t≤T
where 𝛾t , 0 ≤ t ≤ T is an adapted process.
Solution: Since Yt is a ℚ-martingale with respect to the filtration ℱt , 0 ≤ t ≤ T (see Problem 4.2.3.3, page 224) and
t
̃t
dYt = 𝜙t 𝜎t e− ∫0 ru du St dW
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𝜇t − rt
then, from the martingale representation theo∫0
𝜎t
rem, there exists an adapted process 𝛾t , 0 ≤ t ≤ T such that
̃ t = Wt +
where W
t
𝜆u du, 𝜆t =
t
Yt = Y0 +
∫0
̃ 𝑣,
𝛾𝑣 d W
0 ≤ t ≤ T.
t
̃ t,
Taking integrals of dYt = 𝜙t 𝜎t e− ∫0 ru du St dW
t
∫0
t
dY𝑣 =
∫0
𝑣
𝜙𝑣 𝜎𝑣 e− ∫0
t
Y t = Y0 +
ru du
̃𝑣
S𝑣 d W
𝑣
𝜙𝑣 𝜎𝑣 e− ∫0
∫0
ru du
̃ 𝑣.
S𝑣 d W
Therefore, for 0 ≤ t ≤ T, we can set
t
𝛾t = 𝜙t 𝜎t e− ∫0 ru du St
or
t
𝛾 e∫0 ru du
𝛾B
𝜙t = t
= t t
𝜎t St
𝜎 t St
t
where Bt = e∫0 ru du . Since 𝜓t Bt = Πt − 𝜙t St , therefore
𝜓t =
𝜎t Πt − 𝛾t Bt
.
𝜎 t Bt
◽
5. Self-Financing Portfolio. Consider an economy consisting of a risk-free asset and a stock
price (risky asset). At time t, the risk-free asset Bt and the stock price St have the following
diffusion processes
dBt = rt Bt dt, dSt = 𝜇t St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price drift rate, 𝜎t is the stock price volatility
(which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on
the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of stock
and 𝜓t units being invested in a risk-free asset. For each of the following choices of 𝜙t :
(a) 𝜙t = 𝛼, where 𝛼 is a constant
(b) 𝜙t = Stn , n > 0
t
(c) 𝜙t = ∫0 Sun du, n > 0
find the corresponding 𝜓t so that the trading strategy at time t, (𝜙t , 𝜓t ) is self-financing.
Solution: By definition, the portfolio at time t, Πt = 𝜙t St + 𝜓t Bt is self-financing if dΠt =
t
𝜙t dSt + 𝜓t dBt . Taking note that Bt = e∫0 ru du satisfies dBt = rt Bt dt and from Itō’s lemma,
we have (dSt )2 = 𝜎t2 St2 dt and (dSt )𝜈 = 0 for 𝜈 ≥ 3. Then, for each of the following cases:
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227
(a) If 𝜙t = 𝛼, we have Πt = 𝛼St + 𝜓t Bt and in differential form
dΠt = 𝛼dSt + 𝜓t dBt + Bt d𝜓t
and in order for the portfolio to be self-financing (i.e., dΠt = 𝛼dSt + 𝜓t dBt ) we have
Bt d𝜓t = 0 or 𝜓t = 𝛽 for some constant 𝛽.
(b) If 𝜙t = Stn , we have Πt = Stn+1 + 𝜓t Bt and in differential form
1
dΠt = (n + 1)Stn dSt + n(n + 1)Stn−1 (dSt )2 + 𝜓t dBt + Bt d𝜓t
2
and for the portfolio to be self-financing (i.e., dΠt = Stn dSt + 𝜓t dBt ) we therefore set
1
nStn dSt + n(n + 1)Stn−1 (dSt )2 + Bt d𝜓t = 0
2
or
t
𝜓t = −
∫0
t
t
Sun
n(n + 1)𝜎u2 Sun+1
dSu −
du
∫0
Bu
Bu
u
∫0
e− ∫0
t
u
e− ∫0 r𝑣 d𝑣 n(n + 1)𝜎u2 Sun+1 du.
∫)0
(
t
t
(c) If 𝜙t = ∫0 Sun du, we have Πt = ∫0 Sun du St + 𝜓t Bt and in differential form
=−
r𝑣 d𝑣 n
Su
dSu −
(
dΠt =
Stn+1 dt
+
t
∫0
Sun
)
du dSt + 𝜓t dBt + Bt d𝜓t
(
)
t
and for the portfolio to be self-financing (i.e., dΠt = ∫0 Su du dSt + 𝜓t dBt ) we
require
Stn+1 dt + Bt d𝜓t = 0
or
t
𝜓t = −
∫0
t
u
Sun+1
du = −
e− ∫0 r𝑣 d𝑣 Sun+1 du.
∫0
Bu
◽
6. Stock Price with Continuous Dividend Yield (Geometric Brownian Motion). Consider an
economy consisting of a risk-free asset and a dividend-paying stock price (risky asset). At
time t, the risk-free asset Bt and the stock price St have the following diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )St dt + 𝜎t St dWt
such that rt is the risk-free rate, 𝜇t is the stock price drift rate, Dt is the continuous dividend
yield, 𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T}
is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
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At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of stock
and 𝜓t units being invested in a risk-free asset. From the following discounted portfolio
value
t
Yt = e− ∫0 ru du Πt
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent
risk-neutral measure ℚ, then the discounted portfolio Yt is a ℚ-martingale.
Show also that under the ℚ-measure the stock price follows
̃t
dSt = (rt − Dt )St dt + 𝜎t St dW
̃ t = Wt +
where W
t
∫0
𝜆u du is a ℚ-standard Wiener process such that
𝜆t =
𝜇t − rt
𝜎t
is defined as the market price of risk.
Solution: At time t, the portfolio Πt is valued as
Πt = 𝜙t St + 𝜓t Bt
and since the trader will receive Dt St dt for every stock held and because the trader holds
𝜙t of the stock, then in differential form
dΠt = 𝜙t dSt + 𝜙t Dt St dt + 𝜓t dBt
[
]
= 𝜙t (𝜇t − Dt )St dt + 𝜎t St dWt + 𝜙t Dt St dt + 𝜓t rt Bt dt
= rt Πt dt + 𝜙t (𝜇t − rt )St dt + 𝜙t 𝜎t St dWt
= rt Πt dt + 𝜙t 𝜎t St (𝜆t dt + dWt )
where 𝜆t =
𝜇t − rt
. Following Problem 4.2.3.3 (page 224),
𝜎t
)
(
t
t
̃t
dYt = d e− ∫0 ru du Πt = 𝜙𝜎t e− ∫0 ru du St dW
such that
̃ t = Wt +
W
t
∫0
𝜆u du.
By applying Girsanov’s theorem to change the measure ℙ to an equivalent risk-neutral
̃ t is a ℚ-standard Wiener process, the discounted portfolio Yt
measure ℚ, under which W
is a ℚ-martingale.
̃ t + 𝜆t dt into dSt = (𝜇t − Dt )St dt + 𝜎t St dWt , the stock price difBy substituting dWt = dW
fusion process under the risk-neutral measure becomes
̃ t.
dSt = (rt − Dt )St dt + 𝜎t St dW
◽
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229
7. Stock Price with Continuous Dividend Yield (Arithmetic Brownian Motion). Consider an
economy consisting of a risk-free asset and a dividend-paying stock price (risky asset). At
time t, the risk-free asset Bt and the stock price St have the following diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )dt + 𝜎t dWt
such that rt is the risk-free rate, 𝜇t is the stock price drift rate, Dt is the continuous dividend
yield, 𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T}
is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of stock
and 𝜓t units being invested in a risk-free asset. From the following discounted portfolio
value
t
Yt = e− ∫0 ru du Πt
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent
risk-neutral measure ℚ, then the discounted portfolio Yt is a ℚ-martingale.
Show also that under the ℚ-measure the stock price follows
̃t
dSt = (rt − Dt )St dt + 𝜎t dW
̃ t = Wt +
where W
t
∫0
𝜆u du is a ℚ-standard Wiener process such that
𝜆t =
𝜇t − rt St
𝜎t
is defined as the market price of risk.
Solution: At time t, the portfolio Πt is valued as
Πt = 𝜙t St + 𝜓t Bt
and since the trader will receive Dt dt for every stock held, then in differential form
(
)
dΠt = 𝜙t dSt + Dt dt + 𝜓t dBt
[
]
= 𝜙t 𝜇t dt + 𝜎t dWt + 𝜓t rt Bt dt
(
)
= rt Πt dt + 𝜙t 𝜇t − rt St dt + 𝜙t 𝜎t dWt
(
)
= rt Πt dt + 𝜙t 𝜎t 𝜆t dt + dWt
𝜇t − rt St
.
𝜎t
Following Problem 4.2.3.3 (page 224), the discounted portfolio becomes
(
)
t
t
̃t
dYt = d e− ∫0 ru du Πt = 𝜙t 𝜎t e− ∫0 ru du dW
where 𝜆t =
such that
̃ t = Wt +
W
t
∫0
𝜆u du.
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By applying Girsanov’s theorem to change the measure ℙ to an equivalent risk-neutral
̃ t is a ℚ-standard Wiener process, the discounted portfolio Yt
measure ℚ, under which W
is a ℚ-martingale.
̃ t + 𝜆t dt into dSt = (𝜇t − Dt )dt + 𝜎t dWt , the diffusion process
By substituting dWt = dW
under the risk-neutral measure becomes
̃ t.
dSt = (rt − Dt )St dt + 𝜎t dW
◽
8. Commodity Price with Cost of Carry. Consider an economy consisting of a risk-free asset
and a commodity price (risky asset). At time t, the risk-free asset Bt and the commodity
price St have the following diffusion processes
dBt = rt Bt dt, dSt = 𝜇t St dt + Ct dt + 𝜎t St dWt
such that rt is the risk-free rate, 𝜇t is the commodity price drift rate, Ct > 0 is the cost
of carry for storage per unit time, 𝜎t is the commodity price volatility (which are all time
dependent) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on the probability space
(Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t units of commodity and 𝜓t units being invested in a risk-free asset. From the following discounted
portfolio value
t
Yt = e− ∫0 ru du Πt
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent
risk-neutral measure ℚ, then the discounted portfolio Yt is a ℚ-martingale.
Show also that under the ℚ-measure the commodity price follows
̃t
dSt = rt St dt + Ct dt + 𝜎t St dW
̃ t = Wt +
where W
t
∫0
𝜆u du is a ℚ-standard Wiener process such that
𝜆t =
𝜇t − rt
𝜎t
is defined as the market price of risk.
Solution: At time t, the portfolio Πt is valued as
Πt = 𝜙t St + 𝜓t Bt
and since the trader will pay Ct dt for storage, and because the trader holds 𝜙t of the commodity, then in differential form
dΠt = 𝜙t dSt − 𝜙t Ct dt + 𝜓t rt Bt dt
[
]
= 𝜙t 𝜇t St dt + Ct dt + 𝜎t St dWt − 𝜙t Ct dt + 𝜓t rt Bt dt
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231
(
)
= rt Πt dt + 𝜙t 𝜇t − rt St dt + 𝜙t 𝜎t St dWt
(
)
= rt Πt dt + 𝜙t 𝜎t St 𝜆t dt + dWt
where 𝜆t =
𝜇t − rt
. Following Problem 4.2.3.3 (page 224),
𝜎t
(
)
t
t
̃t
dYt = d e− ∫0 ru du Πt = 𝜙t 𝜎t e− ∫0 ru du St dW
such that
̃ t = Wt +
W
t
∫0
𝜆u du.
By applying Girsanov’s theorem to change the measure ℙ to an equivalent risk-neutral
̃ t is a ℚ-standard Wiener process, the discounted portfolio Yt
measure ℚ, under which W
is a ℚ-martingale.
̃ t + 𝜆t dt into dSt = 𝜇t St dt + Ct dt + 𝜎t St dWt , the commodity
By substituting dWt = dW
price diffusion process under the risk-neutral measure becomes
̃ t.
dSt = rt St dt + Ct dt + 𝜎t St dW
◽
9. Pricing a Security Derivative. Consider an economy consisting of a risk-free asset and a
stock price (risky asset). At time t, the risk-free asset Bt and the stock price St have the
following diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )St dt + 𝜎t St dWt
where rt is the risk-free rate, Dt is the continuous dividend yield, 𝜇t is the stock price
growth rate, 𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤
t ≤ T} is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt holding 𝜙t shares of
stock and 𝜓t units being invested in a risk-free asset. Let Ψ(ST ) represents the payoff of a
derivative security at time T, 0 ≤ t ≤ T such that Ψ(ST ) is ℱT measurable. Assuming the
trader begins with an initial capital Π0 and a trading strategy (𝜙t , 𝜓t ), 0 ≤ t ≤ T such that
the portfolio value at time T is
ΠT = Ψ(ST ) almost surely
show that under the risk-neutral measure ℚ,
[
]
|
T
ℚ
− ∫t ru du
|
Ψ(ST )| ℱt ,
Ψ(St ) = 𝔼 e
|
0 ≤ t ≤ T.
Solution: At time t, in order to calculate the price of a derivative security Ψ(St ), 0 ≤ t ≤ T
t
with payoff at time T given as Ψ(ST ), we note that because e− ∫0 ru du Πt is a ℚ-martingale
and ΠT = Ψ(ST ) almost surely,
[
]
[
]
|
|
t
T
T
− ∫0 ru du
ℚ
− ∫0 ru du
ℚ
− ∫0 ru du
|
|
e
Πt = 𝔼 e
ΠT | ℱt = 𝔼 e
Ψ(ST )| ℱt .
|
|
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Since the derivative security Ψ and the portfolio Π have identical values at T, and by
assuming the trader begins with an initial capital Π0 , the value of the portfolio Πt = Ψ(St )
for 0 ≤ t ≤ T. Thus, we can write
[
]
|
T
ℚ
− ∫t ru du
|
Ψ(ST )| ℱt , 0 ≤ t ≤ T.
Ψ(St ) = 𝔼 e
|
◽
10. First Fundamental Theorem of Asset Pricing. Consider an economy consisting of a
risk-free asset and a stock price (risky asset). At time t, the risk-free asset Bt and the stock
price St have the following diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price growth rate, Dt is the continuous dividend
yield, 𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T}
is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
At time t, 0 ≤ t ≤ T we consider a trader who has a self-financing portfolio valued at Πt
and we define an arbitrage strategy such that the following criteria are satisfied:
(i) Π0 = 0
(ii) ℙ(ΠT ≥ 0) = 1
(iii) ℙ(ΠT > 0) > 0.
Prove that if the trading strategy has a risk-neutral probability measure ℚ, then it does not
admit any arbitrage opportunities.
Solution: We prove this result via contradiction.
Suppose the trading strategy has a risk-neutral probability measure ℚ and it admits arbit
trage opportunities. Since the discounted portfolio e− ∫0 ru du Πt is a ℚ-martingale, by letting Π0 = 0, and from the optional stopping theorem,
(
)
T
𝔼ℚ e− ∫0 ru du ΠT = Π0 = 0.
Since ℙ(ΠT ≥ 0) = 1 and because ℙ is equivalent to ℚ, therefore ℚ(ΠT ≥ 0) = 1. In addiT
tion, since ΠT ≥ 0 therefore e− ∫0 ru du ΠT ≥ 0. By definition,
(
)
T
𝔼ℚ e− ∫0 ru du ΠT =
∞
∫0
(
)
T
ℚ e− ∫0 ru du ΠT ≥ x dx =
∞
∫0
)
(
T
ℚ ΠT ≥ xe∫0 ru du dx = 0
which implies
ℚ(ΠT ≥ 0) = 0.
Because ℚ is equivalent to ℙ, therefore
ℙ(ΠT ≥ 0) = 0
which is a contradiction to the fact that there is an arbitrage opportunity.
◽
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11. Second Fundamental Theorem of Asset Pricing. Consider an economy consisting of a
risk-free asset and a stock price (risky asset). At time t, the risk-free asset Bt and the stock
price St have the following diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock price growth rate, Dt is the continuous dividend
yield, 𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T}
is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
We consider a market model under the risk-neutral measure ℚ where at each time t, 0 ≤
t ≤ T, the portfolio Πt is constructed with 𝜓t number of stocks and 𝜙t units of non-risky
assets. Prove that if the market is complete (i.e., every derivative security can be hedged),
then the risk-neutral measure is unique.
Solution: Suppose the market model has two risk-neutral measures ℚ1 and ℚ2 . Let A ∈
ℱT , and we consider the payoff of a derivative security
{
Ψ(ST ) =
T
e∫0
0
ru du
A ∈ ℱT
A ∉ ℱT .
As the market model is complete, there is a portfolio value process Πt , 0 ≤ t ≤ T with the
same initial condition Π0 that satisfies ΠT = Ψ(ST ). Since both ℚ1 and ℚ2 are risk-neutral
t
measures, the discounted portfolio e− ∫0 ru du Πt is a martingale under ℚ1 and ℚ2 .
From the optional stopping theorem we can write
]
[
(
)
T
T
ℚ1 (A) = 𝔼ℚ1 e− ∫0 ru du Ψ(ST ) = 𝔼ℚ1 e− ∫0 ru du ΠT = Π0
and
]
[
(
)
T
T
ℚ2 (A) = 𝔼ℚ2 e− ∫0 ru du Ψ(ST ) = 𝔼ℚ2 e− ∫0 ru du ΠT = Π0 .
Therefore,
ℚ1 = ℚ2
since A is an arbitrary set.
◽
12. Change of Numéraire. Consider an economy consisting of a risk-free asset and a stock
price (risky asset). At time t, the risk-free asset Bt and the stock price St have the following
diffusion processes
dBt = rt Bt dt, dSt = (𝜇t − Dt )St dt + 𝜎t St dWt
where rt is the risk-free rate, 𝜇t is the stock drift rate, Dt is the continuous dividend yield,
𝜎t is the stock price volatility (which are all time dependent) and {Wt ∶ 0 ≤ t ≤ T} is a
ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
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Let Ψ(ST ) represent the payoff of a derivative security at time T, 0 ≤ t ≤ T such that Ψ(ST )
is ℱT measurable and under the risk-neutral measure ℚ, let the numéraire Nt(i) , i = 1, 2 be a
strictly positive price process for a non-dividend-paying asset with the following diffusion
process:
̃ (i)
dN (i) = rt N (i) dt + 𝜈 (i) N (i) dW
t
t
t
t
t
̃ (i) , 0 ≤ t ≤ T is a ℚ-standard Wiener process. Show that
where 𝜈t(i) is the volatility and W
t
for 0 ≤ t ≤ T,
(
(
| )
| )
Ψ(ST ) ||
Ψ(ST ) ||
(1) ℚ(1)
(2) ℚ(2)
Nt 𝔼
| ℱt = Nt 𝔼
| ℱt
NT(1) ||
NT(2) ||
and
Nt(1)
dℚ(1) ||
=
dℚ(2) ||ℱt N (1)
0
/
Nt(2)
N0(2)
where N (i) is a numéraire and ℚ(i) is the measure under which the stock price discounted
by N (i) is a martingale.
t
Solution: By solving dBt = rt Bt dt we have Bt = e∫0 ru du . For i = 1, 2 and given
t
Bt = e∫0 ru du , from Problem 4.2.3.1 (page 221) we can deduce that
(i)
−1 (i) (i) ̃ (i)
d(B−1
t Nt ) = Bt Nt 𝜈t d Wt
is a ℚ-martingale and from Itō’s formula we have
d(log (B−1
t Nt )) =
(i)
d(B−1
t Nt )
(i)
B−1
t Nt
) 2
(
⎛
−1 N (i) ⎞
d
B
t
t
⎟
1⎜
− ⎜
⎟ + ...
(i)
−1
2⎜ B N
⎟
t
t
⎠
⎝
̃ (i) − 1 (𝜈 (i) )2 dt.
= 𝜈t(i) dW
t
2 t
Taking integrals,
(
log
B−1
t Nt
B−1
N0
0
or
)
t
=
∫0
̃ u(i) −
𝜈u(i) dW
t (i)
(i)
(i) ∫0 𝜈u
B−1
t Nt = N0 e
t
1 (i) 2
(𝜈 ) du
∫0 2 u
̃ u(i) − 1 ∫ t (𝜈u(i) )2 du
dW
2 0
since B0 = 1. Using Girsanov’s theorem to change from the ℚ measure to an equivalent
ℚ(i) measure, the Radon–Nikodým derivative is
t
Nt(i)
̃ u(i) − 1 ∫ t (𝜈u(i) )2 du
dℚ(i) ||
− ∫0 (−𝜈u )dW
2 0
=
e
=
dℚ ||ℱt
N0(i) Bt
t
(i)
̃ (i) −
so that W t = W
t
∫0
𝜈u(i) du, 0 ≤ t ≤ T is a ℚ(i) -standard Wiener process.
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235
Under the risk-neutral measure ℚ, the stock price follows
̃t
dSt = (rt − Dt )St dt + 𝜎t St dW
)
t(
𝜇u − ru
̃ t = Wt +
du is a ℚ-standard Wiener process. By changing the ℚ
where W
∫0
𝜎u
measure to an equivalent ℚ(i) measure, the discounted stock price
{Nt(i) }−1 St
is a ℚ(i) -martingale. Thus, for a derivative payoff Ψ(ST ),
𝔼
(
)
Thus,
(
dℚ(i) ||
dℚ ||ℱt
ℚ(i)
)−1 |
| ⎤
| ℱt ⎥
| ⎥
| ⎦
|
(
| )
| )
Ψ(ST ) ||
Ψ(ST ) ||
(2) ℚ(2)
| ℱt = Nt 𝔼
| ℱt .
NT(1) ||
NT(2) ||
(1)
Nt(1) 𝔼ℚ
To find
)
( (i)
⎡
dℚ ||
⎢
𝔼
Ψ(ST )
Ψ(ST )|ℱt =
⎢
dℚ ||ℱ
T
⎣
(
)
N (i) BT ||
N (i)
(i)
ΨT 0 (i) || ℱt
= (i)t 𝔼ℚ
N 0 Bt
NT ||
(
| )
BT Nt(i) ℚ(i) Ψ(ST ) |
| ℱt .
=
𝔼
|
Bt
NT(i) ||
(
ℚ
dℚ(1)
on ℱt we note that
dℚ(2)
( (1)
)
( (2)
)
N0 Ψ(St ) ||
N0 Ψ(St ) ||
(2)
ℚ(1)
ℚ
|ℱ = 𝔼
|ℱ .
𝔼
| 0
| 0
Nt(1) ||
Nt(2) ||
By letting Xt , 0 ≤ t ≤ T be an ℱt measurable random variable, from Problem 4.2.2.4
(page 198)
[ (
)]
(1)
(2)
dℚ(1) ||
𝔼ℚ (Xt ) = 𝔼ℚ Xt
.
dℚ(2) ||ℱt
We can therefore deduce that
N0(1) Ψ(St )
Nt(1)
or
(
dℚ(1) ||
dℚ(2) ||ℱt
)
Nt(1)
dℚ(1) ||
=
dℚ(2) ||ℱt N (1)
0
N.B. An alternative method is to let
dℚ(1)
dℚ(2)
=
/
N0(2) Ψ(St )
Nt(2)
Nt(2)
.
N0(2)
/
dℚ(1) dℚ(2)
and the result follows.
=
dℚ
dℚ
◽
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13. Black–Scholes Equation. Consider an economy consisting of a risk-free asset and a stock
price (risky asset). At time t, the risk-free asset Bt and the stock price St have the following
diffusion processes
dBt = rBt dt, dSt = 𝜇St dt + 𝜎St dWt
where r is the risk-free rate, 𝜇 is the stock price drift rate, 𝜎 is the stock price volatility
(which are all constants) and {Wt ∶ 0 ≤ t ≤ T} is a ℙ-standard Wiener process on the
probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt , given as
Πt = 𝜙t St + 𝜓t Bt
where he holds 𝜙t shares of stock and 𝜓t units being invested in a risk-free asset. Show
that the portfolio (𝜙t , 𝜓t ) is self-financing if and only if Πt satisfies the Black–Scholes
equation
𝜕Πt 1 2 2 𝜕 2 Πt
𝜕Π
+ rSt t − rΠt = 0.
+ 𝜎 St
2
𝜕t
2
𝜕St
𝜕St
Solution: By applying Taylor’s theorem on dΠt ,
dΠt =
2
𝜕Πt
𝜕Πt
1 𝜕 Πt 2
dt +
dSt +
dS + . . .
𝜕t
𝜕St
2 𝜕St2 t
and since dSt2 = (𝜇St dt + 𝜎St dWt )2 = 𝜎 2 St2 dt such that (dt)𝜈 = 0, 𝜈 ≥ 2, we have
𝜕Πt
𝜕Πt
dS +
dt +
𝜕t
𝜕St t
(
𝜕Πt
𝜕Πt
dSt +
+
=
𝜕St
𝜕t
dΠt =
2
1 2 𝜕 Πt
dt
𝜎
2 𝜕St2
2
1 2 2 𝜕 Πt
𝜎 St
2
𝜕St2
)
dt.
In contrast, the portfolio (𝜙t , 𝜓t ) is self-financing if and only if
dΠt = 𝜙t dSt + 𝜓t dBt = 𝜙t dSt + rBt 𝜓t dt.
By equating both of the equations we have
rBt 𝜓t =
𝜕Πt 1 2 2 𝜕 2 Πt
+ 𝜎 St
𝜕t
2
𝜕St2
and
𝜙t =
𝜕Πt
𝜕St
and substituting the above two equations into
Πt = 𝜙t St + 𝜓t Bt
we have
𝜕Π
𝜕Πt 1 2 2 𝜕 2 Πt
+ rSt t − rΠt = 0.
+ 𝜎 St
2
𝜕t
2
𝜕St
𝜕St
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237
N.B. Take note that the Black–Scholes equation does not contain the growth parameter
𝜇, which means that the value of a self-financing portfolio is independent of how rapidly
or slowly a stock grows. In essence, the only parameter from the stochastic differential
equation dSt = 𝜇St dt + 𝜎St dWt that affects the value of the portfolio is the volatility.
◽
14. Black–Scholes Equation with Stock Paying Continuous Dividend Yield. Consider an economy consisting of a risk-free asset and a stock price (risky asset). At time t, the risk-free
asset Bt and the stock price St have the following diffusion processes
dBt = rBt dt, dSt = (𝜇 − D)St dt + 𝜎St dWt
where r is the risk-free rate, 𝜇 is the stock price drift rate, D is the continuous dividend
yield, 𝜎 is the stock price volatility (which are all constants) and {Wt ∶ 0 ≤ t ≤ T} is a
ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
At time t, we consider a trader who has a portfolio valued at Πt , given as
Πt = 𝜙t St + 𝜓t Bt
where he holds 𝜙t shares of stock and 𝜓t units being invested in a risk-free asset. Show
that the portfolio (𝜙t , 𝜓t ) is self-financing if and only if Πt satisfies the Black–Scholes
equation with continuous dividend yield
𝜕Π
𝜕Πt 1 2 𝜕 2 Πt
+ (r − D)St t − rΠt = 0.
+ 𝜎
2
𝜕t
2 𝜕St
𝜕St
Solution: By applying Taylor’s theorem on dΠt ,
dΠt =
2
𝜕Πt
𝜕Πt
1 𝜕 Πt 2
dSt +
dS + . . .
dt +
𝜕t
𝜕St
2 𝜕St2 t
and since dSt2 = ((𝜇 − D)St dt + 𝜎St dWt )2 = 𝜎 2 St2 dt such that (dt)𝜈 = 0, 𝜈 ≥ 2, we have
𝜕Πt
𝜕Πt
dt +
dS +
𝜕t
𝜕St t
(
𝜕Πt
𝜕Πt
dSt +
+
=
𝜕St
𝜕t
dΠt =
2
1 2 𝜕 Πt
𝜎
dt
2 𝜕St2
2
1 2 2 𝜕 Πt
𝜎 St
2
𝜕St2
)
dt.
Since the trader will receive DSt dt for every stock held, the portfolio (𝜙t , 𝜓t ) is
self-financing if and only if
dΠt = 𝜙t dSt + 𝜓t dBt + 𝜙t DSt dt = 𝜙t dSt + (rBt 𝜓t + 𝜙t DSt )dt.
By equating both of the equations we have
rBt 𝜓t + 𝜙t DSt =
𝜕Πt 1 2 2 𝜕 2 Πt
+ 𝜎 St
𝜕t
2
𝜕St2
and
𝜙t =
𝜕Πt
𝜕St
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and substituting the above two equations into
Πt = 𝜙t St + 𝜓t Bt
we have
𝜕Πt 1 2 2 𝜕 2 Πt
𝜕Π
+ (r − D)St t − rΠt = 0.
+ 𝜎 St
𝜕t
2
𝜕St
𝜕St2
◽
15. Foreign Exchange Rate under Domestic Risk-Neutral Measure. We consider a foreign
exchange (FX) market which at time t consists of a foreign-to-domestic FX spot rate Xt , a
f
risk-free asset in domestic currency Bdt and a risk-free asset in foreign currency Bt . Here,
f
Xt Bt denotes the foreign risk-free asset quoted in domestic currency. Assume that the
evolution of these values has the following diffusion processes
f
f
f
dXt = 𝜇t Xt dt + 𝜎t Xt dWtx , dBdt = rtd Bdt dt and dBt = rt Bt
where 𝜇t is the drift parameter, 𝜎t is the volatility parameter, rtd is the domestic risk-free
f
rate and rt is the foreign risk-free rate (which are all time dependent) and {Wtx ∶ 0 ≤ t ≤
T} is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
From the discounted foreign risk-free asset in domestic currency
̃t =
X
f
X t Bt
Bdt
show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent domestic
̃t is a ℚd -martingale.
risk-neutral measure ℚd then X
Finally, show that under the ℚd measure the FX rate follows
f
̃ td
dXt = (rtd − rt )Xt + 𝜎t Xt dW
̃ d = Wx +
where W
t
t
f
t
∫0
𝜆u du is a ℚd -standard Wiener process with 𝜆t =
𝜇t + rt − rtd
.
𝜎t
̃t ,
Solution: By applying Taylor’s theorem on dX
̃t =
dX
2̃
̃t
̃
̃
̃t
̃t
𝜕X
𝜕X
𝜕X
1𝜕 X
1 𝜕X
1 𝜕X
f
f
t
dXt + ft dBt + dt dBdt +
(dXt )2 +
(dBt )2 +
(dBdt )2
2
f
𝜕Xt
2 𝜕Xt
2 𝜕B
2 𝜕Bdt
𝜕Bt
𝜕Bt
t
+
̃t
𝜕2X
̃t
̃t
𝜕2 X
𝜕2 X
f
f
d
(dX
)(dB
)
+
(dX
)(dB
)
+
(dBt )(dBdt ) + . . .
t
t
t
t
f
d
f
𝜕Xt 𝜕Bt
𝜕Xt 𝜕Bt
𝜕Bt 𝜕Bdt
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239
and from Itō’s formula,
̃
̃t
̃
̃
𝜕2 X
𝜕X
𝜕X
𝜕X
1
f
dXt + ft dBt + dt dBdt + 𝜎t2 Xt2 2t dt
𝜕Xt
2
𝜕Bt
𝜕Xt
𝜕Bt
( )
(
)
( f)
f
Xt Bt
Bt
Xt
f f
x
(𝜇t Xt dt + 𝜎t Xt dWt ) +
Bt rt dt −
rtd Bdt dt
=
Bdt
Bdt
(Bdt )2
(
)
f
̃t dt + 𝜎t X
̃t dWtx
= 𝜇t + rt − rtd X
[(
)
]
f
𝜇t + rt − rtd
̃t
dt + dWtx
= 𝜎t X
𝜎t
̃t =
dX
̃t d W
̃ td
= 𝜎t X
f
𝜇t + rt − rtd
. From Girsanov’s theorem there
∫0
𝜎t
exists an equivalent domestic risk-neutral measure ℚd on the filtration ℱs , 0 ≤ s ≤ t
defined by the Radon–Nikodým derivative
̃ d = Wx +
where W
t
t
t
𝜆u du such that 𝜆t =
s
Zs = e− ∫0
̃ d is a ℚd -standard Wiener process.
so that W
t
Therefore, by substituting
(
dWtx
=
̃ td
dW
s
𝜆u du− 12 ∫0 𝜆2u dWux
−
f
𝜇t + rt − rtd
𝜎t
)
dt
into dXt = 𝜇t Xt dt + 𝜎t Xt dWtx , under the domestic risk-neutral measure ℚd , the FX spot
rate SDE is
f
̃ td .
dXt = (rtd − rt )Xt + 𝜎t Xt dW
◽
16. Foreign Exchange Rate under Foreign Risk-Neutral Measure. We consider a foreign
exchange (FX) market which at time t consists of a foreign-to-domestic FX spot rate Xt , a
f
risk-free asset in domestic currency Bdt and a risk-free asset in foreign currency Bt . Here,
d
Bt ∕Xt denotes the domestic risk-free asset quoted in foreign currency. Assume that the
evolution of these values has the following diffusion processes
f
f
f
dXt = 𝜇t Xt dt + 𝜎t Xt dWtx , dBdt = rtd Bdt dt and dBt = rt Bt
where 𝜇t is the drift parameter, 𝜎t is the volatility parameter, rtd is the domestic risk-free
f
rate and rt is the foreign risk-free rate (which are all time dependent) and {Wtx ∶ 0 ≤ t ≤
T} is a ℙ-standard Wiener process on the probability space (Ω, ℱ, ℙ).
From the discounted domestic risk-free asset in foreign currency
̃t =
X
Bdt
f
Xt Bt
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show, using Girsanov’s theorem, that by changing the measure ℙ to an equivalent foreign
̃t is a ℚf -martingale.
risk-neutral measure ℚf then X
Finally, show that under the ℚf measure the FX rate follows
f
̃f
dXt = (rtd − rt + 𝜎t2 )Xt + 𝜎t Xt dW
t
̃ f = Wx +
where W
t
t
f
t
𝜆u du is a ℚf -standard Wiener process with 𝜆t =
∫0
𝜇t + rt − rtd − 𝜎t2
.
𝜎t
̃t ,
Solution: By applying Taylor’s theorem on dX
̃t =
dX
2̃
̃t
̃
̃
̃t
̃t
𝜕X
𝜕X
𝜕X
1𝜕 X
1 𝜕X
1 𝜕X
f
f 2
t
2
dXt + ft dBt + dt dBdt +
(dX
)
+
(dB
)
+
(dBdt )2
t
t
f
d
2
𝜕Xt
2
2
2
𝜕B
𝜕X
𝜕B
𝜕Bt
𝜕Bt
t
t
t
+
̃t
𝜕2X
̃t
̃t
𝜕2 X
𝜕2 X
f
f
d
(dX
)(dB
)
+
(dX
)(dB
)
+
(dBt )(dBdt ) + . . .
t
t
t
t
f
d
f
d
𝜕X
𝜕B
𝜕Xt 𝜕Bt
𝜕Bt 𝜕Bt
t
t
and from Itō’s formula,
̃t
̃
̃
̃
𝜕X
𝜕X
𝜕X
𝜕2X
1
f
dXt + ft dBt + dt dBdt + 𝜎t2 Xt2 2t dt
𝜕Xt
2
𝜕Xt
𝜕Bt
𝜕Bt
)
)
)
(
(
(
d
Bdt
B
1
f
f
t
(𝜇t Xt dt + 𝜎t Xt dWtx ) −
Bt rt dt +
rtd Bdt dt
=−
f
f
f
Xt2 Bt
Xt (Bt )2
Xt Bt
)
(
Bdt
𝜎t2 Xt2 dt
+
f
Xt3 Bt
(
)
f
̃t dWtx
̃t dt − 𝜎t X
= rtd − rt + 𝜎t2 − 𝜇t X
[(
)
]
f
𝜇t + rt − rtd − 𝜎t2
̃t
= −𝜎t X
dt + dWtx
𝜎t
̃t =
dX
̃t d W
̃f
= −𝜎t X
t
f
𝜇t + rt − rtd − 𝜎t2
. From Girsanov’s theorem
∫0
𝜎t
there exists an equivalent foreign risk-neutral measure ℚf on the filtration ℱs , 0 ≤ s ≤ t
defined by the Radon–Nikodým derivative
̃ f = Wx +
where W
t
t
t
𝜆u du such that 𝜆t =
s
Zs = e− ∫0
𝜆u du− 12 ∫0s 𝜆2u dWux
̃ f is a ℚf -standard Wiener process.
so that W
t
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Risk-Neutral Measure
241
Therefore, by substituting
(
dWtx
=
̃f
dW
t
−
f
𝜇t + rt − rtd − 𝜎t2
𝜎t
)
dt
into dXt = 𝜇t Xt dt + 𝜎t Xt dWtx , under the foreign risk-neutral measure ℚf , the FX spot rate
SDE is
f
̃f.
dXt = (rtd − rt + 𝜎t2 )Xt + 𝜎t Xt dW
t
◽
17. Foreign-Denominated Stock Price under Domestic Risk-Neutral Measure. Let (Ω, ℱ, ℙ)
be a probability space and let {Wts ∶ 0 ≤ t ≤ T} and {Wtx ∶ 0 ≤ t ≤ T} be ℙ-standard
Wiener processes on the filtration ℱt , 0 ≤ t ≤ T. Let St and Xt denote the stock price
quoted in foreign currency and the foreign-to-domestic exchange rate, respectively, each
having the following SDEs
dSt = (𝜇s − Ds )St dt + 𝜎s St dWts
dXt = 𝜇x Xt dt + 𝜎x Xt dWtx
dWts ⋅ dWtx = 𝜌dt,
𝜌 ∈ (−1, 1)
where 𝜇s , Ds and 𝜎s are the stock price drift, continuous dividend yield and volatility,
respectively, whilst 𝜇x and 𝜎x are the exchange rate drift and volatility, respectively. Here,
we assume Wts and Wtx are correlated with correlation coefficient 𝜌 ∈ (−1, 1). In addition,
f
let Bt and Bdt be the risk-free assets in foreign and domestic currencies, respectively, having
the following differential equations
f
f
dBt = rf Bt dt and dBdt = rd Bdt dt
where rf and rd are the foreign and domestic risk-free rates.
Show that for the stock price denominated in domestic currency, Xt St follows the diffusion
process
√
d(Xt St )
= (𝜇s + 𝜇x + 𝜌𝜎s 𝜎s − Ds )dt + 𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 dWtxs
Xt S t
𝜎s Wts + 𝜎x Wtx
where Wtxs = √
is a ℙ-standard Wiener process.
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2
Using Girsanov’s theorem show that under the domestic risk-neutral measure ℚd , the stock
price denominated in domestic currency has the diffusion process
√
d(Xt St )
̃ txs
= (rd − Ds )dt + 𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 dW
Xt S t
̃ xs
where W
t
⎛
⎞
𝜇
+
𝜇
+
𝜌𝜎
𝜎
−
r
⎜
⎟
s
x
s
s
d
d
= Wtxs + ⎜ √
⎟ t is a ℚ -standard Wiener process.
⎜ 𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 ⎟
⎝
⎠
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Solution: From Problem 3.2.3.3 (page 158) we can easily show that for Xt St , its diffusion
process is
d(Xt St )
= (𝜇s + 𝜇x + 𝜌𝜎s 𝜎x − Ds )dt +
Xt S t
√
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 dWtxs
𝜎s Wts + 𝜎x Wtx
where Wtxs = √
is a ℙ-standard Wiener process.
2
2
𝜎s + 2𝜌𝜎s 𝜎x + 𝜎x
At time t, we let the portfolio Πt be valued as
Πt = 𝜙t Ut + 𝜓t Bdt
where 𝜙t and 𝜓t are the units invested in Ut = Xt St and the risk-free asset Bdt , respectively.
Taking note that the holder will receive Ds Ut dt for every stock held, then
dΠt = 𝜙t (dUt + Ds Ut dt) + 𝜓t rd Bdt dt
[
]
√
xs
2
2
= 𝜙t (𝜇s + 𝜇x + 𝜌𝜎s 𝜎x )Ut dt + 𝜎s + 2𝜌𝜎s 𝜎x + 𝜎x Ut dWt + 𝜓t rd Bdt dt
[
]
√
xs
2
2
= rd Πt dt + 𝜙t (𝜇s + 𝜇x + 𝜌𝜎s 𝜎x − rd )Ut dt + 𝜎s + 2𝜌𝜎s 𝜎x + 𝜎x Ut dWt
= rd Πt dt + 𝜙t
√
̃ txs
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 Ut dW
𝜇s + 𝜇x + 𝜌𝜎s𝜎x − rd
̃ xs = 𝜆t + W xs such that 𝜆 = √
where W
.
t
t
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2
By applying Girsanov’s theorem to change the measure ℙ to an equivalent risk-neutral
̃ xs is a ℚd -standard Wiener process, then the discounted portmeasure ℚd , under which W
t
−r
t
d
folio e d Πt is a ℚ -martingale.
̃ xs − 𝜆dt into
Finally, by substituting dWtxs = dW
t
d(Xt St )
= (𝜇s + 𝜇x + 𝜌𝜎s 𝜎s − Ds )dt +
Xt St
√
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 dWtxs
the stock price diffusion process under the risk-neutral measure ℚd becomes
d(Xt St )
= (rd − Ds )dt +
Xt S t
√
̃ txs .
𝜎s2 + 2𝜌𝜎s 𝜎x + 𝜎x2 dW
◽
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5
Poisson Process
In mathematical finance the most important stochastic process is the Wiener process, which
is used to model continuous asset price paths. The next important stochastic process is the
Poisson process, used to model discontinuous random variables. Although time is continuous,
the variable is discontinuous where it can represent a “jump” in an asset price (e.g., electricity
prices or a credit risk event, such as describing default and rating migration scenarios). In
this chapter we will discuss the Poisson process and some generalisations of it, such as the
compound Poisson process and the Cox process (or doubly stochastic Poisson process) that
are widely used in credit risk theory as well as in modelling energy prices.
5.1 INTRODUCTION
In this section, before we provide the definition of a Poisson process, we first define what a
counting process is.
Definition 5.1 (Counting Process) Let (Ω, ℱ, ℙ) be a probability space. A stochastic process
{Nt ∶ t ≥ 0} where Nt denotes the number of events that have occurred in the interval (0, t] is
said to be a counting process if it has the following properties:
(a) Nt ≥ 0;
(b) Nt is an integer value;
(c) for s < t, Ns ≤ Nt and Nt − Ns is the number of events occurring in (s, t].
Once we have defined a counting process we can now give a proper definition of a Poisson
process. Basically, there are three definitions of a Poisson process (or homogeneous Poisson
process) and they are all equivalent.
Definition 5.2(a) Let (Ω, ℱ, ℙ) be a probability space. A Poisson process (or homogeneous
Poisson process) {Nt ∶ t ≥ 0} with intensity 𝜆 > 0 is a counting process with the following
properties:
(a) N0 = 0;
(b) Nt has independent and stationary increments;
(c) the sample paths Nt have jump discontinuities of unit magnitude such that for h > 0
ℙ(Nt+h
⎧1 − 𝜆h + o(h)
⎪
= i + j|Nt = i) = ⎨𝜆h + o(h)
⎪o(h)
⎩
j=0
j=1
j > 1.
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Definition 5.2(b) Let (Ω, ℱ, ℙ) be a probability space. A Poisson process (or homogeneous
Poisson process) {Nt ∶ t ≥ 0} with intensity 𝜆 > 0 is a counting process with the following
properties:
(a) N0 = 0;
(b) Nt has independent and stationary increments;
(𝜆t)k e−𝜆t
(c) ℙ(Nt = k) =
, k = 0, 1, 2, . . .
k!
Definition 5.2(c) Let (Ω, ℱ, ℙ) be a probability space. A Poisson process (or homogeneous
Poisson process) {Nt ∶ t ≥ 0} with intensity 𝜆 > 0 is a counting process with the following
properties:
(a) N0 = 0;
(b) the inter-arrival times of events 𝜏1 , 𝜏2 , . . . form a sequence of independent and identically
distributed random variables where 𝜏i ∼ Exp(𝜆), i = 1, 2, . . . ;
(c) the sample paths Nt have jump discontinuities of unit magnitude such that for h > 0
ℙ(Nt+h
⎧1 − 𝜆h + o(h)
⎪
= i + j|Nt = i) = ⎨𝜆h + o(h)
⎪o(h)
⎩
j=0
j=1
j > 1.
Like a standard Wiener process, the Poisson process has independent and stationary increments and so it is also a Markov process.
Theorem 5.3 (Markov Property) Let (Ω, ℱ, ℙ) be a probability space. The Poisson process
{Nt ∶ t ≥ 0} is a Markov process such that the conditional distribution of Nt given the filtration
ℱs , s < t depends only on Ns .
In addition, the Poisson process also has a strong Markov property.
Theorem 5.4 (Strong Markov Property) Let (Ω, ℱ, ℙ) be a probability space. If {Nt ∶ t ≥
⟂ ℱt .
0} is a Poisson process and given ℱt is the filtration up to time t then for s > 0, Nt+s − Nt ⟂
By modifying the intensity parameter we can construct further definitions of a Poisson process and the following definitions (which are all equivalent) describe a non-homogeneous
Poisson process with intensity 𝜆t being a deterministic function of time.
Definition 5.5(a) Let (Ω, ℱ, ℙ) be a probability space. A Poisson process (or nonhomogeneous Poisson process) {Nt ∶ t ≥ 0} with intensity function 𝜆t ∶ ℝ+ → ℝ+ is a
counting process with the following properties:
(a) N0 = 0;
(b) Nt has independent and stationary increments;
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(c) the sample paths Nt have jump discontinuities of unit magnitude such that for h > 0
ℙ(Nt+h
⎧1 − 𝜆 h + o(h)
t
⎪
= i + j|Nt = i) = ⎨𝜆t h + o(h)
⎪o(h)
⎩
j=0
j=1
j > 1.
Definition 5.5(b) Let (Ω, ℱ, ℙ) be a probability space. A Poisson process (or nonhomogeneous Poisson process) {Nt ∶ t ≥ 0} with intensity function 𝜆t ∶ ℝ+ → ℝ+ is a
counting process with the following properties:
(a) N0 = 0;
(b) Nt has independent and stationary increments;
t
(Λ )k e−Λt
𝜆 du is known as the intensity
(c) ℙ(Nt = k) = t
, k = 0, 1, 2, . . . where Λt =
∫0 u
k!
measure (hazard function or cumulative intensity).
It should be noted that if 𝜆t is itself a random process, then {Nt ∶ t ≥ 0} is called a doubly
stochastic Poisson process, conditional Poisson process or Cox process. Given that 𝜆t is a
random process, the resulting stochastic process {Λt ∶ t ≥ 0} defined as
t
Λt =
∫0
𝜆u du
is known as the hazard process. In credit risk modelling, due to the stochastic process of the
intensity, the Cox process can be used to model the random occurrence of a default event, or
even the number of contingent claims in actuarial models (claims that can be made only if one
or more specified events occurs).
Given that the Poisson process Nt with intensity 𝜆 is an increasing counting process, therefore
it is not a martingale. However, in its compensated form, Nt − 𝜆t is a martingale.
Theorem 5.6 Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt . By defining the compensated Poisson
̂t as
process N
̂t = Nt − 𝜆t
N
̂t is a martingale.
then N
In the following we define an important generalisation of the Poisson process known as a
compound Poisson process, where the jump size (or amplitude) can be modelled as a random
variable.
Definition 5.7 (Compound Poisson Process) Let {Nt ∶ t ≥ 0} be a Poisson process with
intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt ,
and let X1 , X2 , . . . be a sequence of independent and identically distributed random variables
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with common mean 𝔼(Xi ) = 𝔼(X) = 𝜇 and variance Var (Xi ) = Var (X) = 𝜎 2 . Assume also that
X1 , X2 , . . . are independent of Nt . The compound Poisson process Mt is defined as
Mt =
Nt
∑
Xi ,
t ≥ 0.
i=1
From the definition we can deduce that although the jumps in Nt are always of one unit, the
jumps in Mt are of random size since Xt is a random variable. The only similarity between
the two processes is that the jumps in Nt and Mt occur at the same time. Furthermore, like the
Poisson process, the compound Poisson process Mt also has the independent and stationary
increments property.
Another important generalisation of the Poisson process is to add a drift and a standard
Wiener process term to generate a jump diffusion process of the form
dXt = 𝜇(Xt− , t) dt + 𝜎(Xt− , t)dWt + J(Xt− , t)dNt
where J(Xt− , t) is a random variable denoting the size of the jump and
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆 dt.
In the presence of jumps, Xt is a càdlàg process, where it is continuous from the right
lim Xu = lim+ Xu = Xt
u↑t
u→t
which is also inclusive of any jumps at time t. To specify the value just before a jump we write
Xt− , which is the limit from the left:
lim Xu = lim− Xu = Xt− .
u→t
u↓t
Let Xt be a càdlàg process with jump times 𝜏1 < 𝜏2 < . . . The integral of a function 𝜓t with
respect to Xt is defined by the pathwise Lebesgue–Stieltjes integral
t
∫0
𝜓s dXs =
∑
𝜓s ΔXs =
0<s≤t
∑
𝜓s (Xs − Xs− ) =
0<s≤t
Xt
∑
𝜓s
s=1
where the size of the jump, ΔXt is denoted by
ΔXt = Xt − Xt−
and given there is no jump at time zero we therefore set ΔX0 = 0.
Theorem 5.8 (One-Dimensional Itō Formula for Jump Diffusion Process) Consider a
stochastic process Xt satisfying the following SDE:
dXt = 𝜇(Xt− , t) dt + 𝜎(Xt− , t)dWt + J(Xt− , t)dNt
or in integrated form
t
X t = X0 +
∫0
t
𝜇(Xs− , s) ds +
∫0
t
𝜎(Xs− , s) dWs +
∫0
J(Xs− , s) dNs
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t
with
247
{
}
|𝜇(Xs , s)| + 𝜎(Xs , s)2 ds < ∞. Then, for a function f (Xt , t), the stochastic process
∫0
Yt = f (Xt , t) satisfies
𝜕f
𝜕f
1 𝜕2 f
1 𝜕3 f
2
(Xt− , t) dt +
(Xt− , t)dXt +
(X
(X − , t)(dXt )3 + . . .
− , t)(dXt ) +
t
𝜕t
𝜕Xt
2! 𝜕Xt2
3! 𝜕Xt3 t
]
[
2
𝜕f
𝜕f
1
2𝜕 f
(X − , t) + 𝜎(Xt− , t)
(Xt− , t) dt
(X − , t) + 𝜇(Xt− , t)
=
𝜕t t
𝜕Xt t
2
𝜕Xt2
dYt =
+ 𝜎(Xt− , t)
𝜕f
(X − , t)dWt + [ f (Xt− + Jt− , t) − f (Xt− , t)]dNt
𝜕Xt t
where Jt− = J(Xt− , t), (dXt )m , m = 1, 2, . . . are expressed according to the rule
( dt)2 = ( dt)3 = . . . = 0
(dWt )2 = dt,
(dWt )3 = (dWt )4 = . . . = 0
(dNt )2 = (dNt )3 = . . . = dNt
dWt dt = dNt dt = dNt dWt = 0.
In integrated form
t
Yt = Y0 +
∫0
t
𝜕f
𝜕f
(X − , s) dXs
(Xs− , s) ds +
∫0 𝜕Xt s
𝜕t
t 2
t 3
𝜕 f
𝜕 f
1
1
2
(X
(X − , s)(dXs )3 + . . .
− , s)(dXs ) +
s
2
∫
∫
2! 0 𝜕Xt
3! 0 𝜕Xt3 s
[
]
t
2f
𝜕
𝜕f
𝜕f
1
(X − , s) + 𝜇(Xs− , s)
= Y0 +
(X − , s) + 𝜎(Xs− , s)2 2 (Xs− , s) ds
∫0 𝜕t s
𝜕Xt s
2
𝜕Xt
+
t
+
∫0
𝜎(Xs− , s)
t
𝜕f
(Xs− , s) dWs +
[ f (Xs− + Js− , s) − f (Xs− , s)] dNs
∫0
𝜕Xt
where
t
∫0
[
∑ [
]
]
f (Xs− + Js− , s) − f (Xs− , s) dNs =
f (Xs , s) − f (Xs− , s) ΔNs
0<s≤t
=
Nt
∑
[ f (Xs , s) − f (Xs− , s)].
s=1
The following is the multi-dimensional version of Itō’s formula for a jump diffusion process.
Theorem 5.9 (Multi-Dimensional Itō Formula for Jump Diffusion Process) Consider the
stochastic processes Xt(i) , i = 1, 2, . . . , n each satisfying the following SDE:
(i)
(i)
dXt(i) = 𝜇(Xt(i)
− , t) dt + 𝜎(Xt− , t)dWt + J(Xt− , t)dNt
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or in integrated form
Xt(i) = X0(i) +
t
t
∫0
𝜇(Xs(i)− , s) ds +
{
|𝜇(Xs(i) , s)| + 𝜎(Xs(i) , s)2
t
∫0
𝜎(Xs(i)− , s) dWs +
t
∫0
J(Xs(i)− , s) dNs
}
ds < ∞. Then for a function f (Xt(1) , Xt(2) , . . . , Xt(n) , t), the
∫0
stochastic process Yt = f (Xt(1) , Xt(2) , . . . , Xt(n) , t) satisfies
with
∑ 𝜕f
𝜕f (1) (2)
(2)
(n)
(i)
(Xt(1)
(Xt− , Xt− , . . . , Xt(n)
− , t) dt +
− , Xt− , . . . , Xt− , t)dXt
(i)
𝜕t
𝜕X
i=1
n
dYt =
t
+
[
=
n
n ∑
∑
i=1 j=1
𝜕2f
1
(j)
(2)
(n)
(i)
(X (1)
− , Xt− , . . . , Xt− , t)dXt dXt + . . .
2! 𝜕X (i) 𝜕X (j) t
t
t
∑
𝜕f (1) (2)
𝜕f (1) (2)
𝜇(Xt(i)
(Xt− , Xt− , . . . , Xt(n)
(Xt− , Xt− , . . . , Xt(n)
− , t) +
− , t)
− , t)
𝜕t
𝜕X
t
i=1
]
n ∑
n
∑
𝜕2 f
1
(j)
(i)
(1)
(2)
(n)
+
𝜎(Xt− , t)𝜎(Xt− , t) (i) (j) (Xt− , Xt− , . . . , Xt− , t) dt
2
𝜕X 𝜕X
i=1 j=1
n
t
∑
n
+
𝜎(Xt(i)
− , t)
i=1
[
𝜕f
𝜕Xt(i)
t
(2)
(n)
(Xt(1)
− , Xt− , . . . , Xt− , t)dWt
]
(1)
(2)
(2)
(n)
(n)
(1)
(2)
(n)
+ f (Xt(1)
− + Jt− , Xt− + Jt− , . . . , Xt− + Jt− , t) − f (Xt− , Xt− , . . . , Xt− , t) dNt
(i)
m
where Jt(i)
− = J(Xt− , t), (dXt ) , m = 1, 2, . . . are expressed according to the rule
( dt)2 = ( dt)3 = . . . = 0
(dWt )2 = dt,
(dWt )3 = (dWt )4 = . . . = 0
(dNt )2 = (dNt )3 = . . . = dNt
dWt dt = dNt dt = dNt dWt = 0.
In integrated form
[ n
]
t ∑
𝜕f (1) (2)
𝜕f
(2)
(n)
(i)
(Xs(1)
Yt = Y0 +
(X − , Xs− , . . . , Xs(n)
− , s) ds +
− , Xs− , . . . , Xs− , s) dXs
(i)
∫0 𝜕t s
∫0
𝜕X
i=1
t
[ n n
]
t ∑
2
∑ 1
𝜕 f
(j)
(1)
(2)
(n)
(i)
(X − , Xs− , . . . , Xs− , s) dXs dXs + . . .
+
∫0
2! 𝜕X (i) 𝜕X (j) s
i=1 j=1
t
t
[
n
t
∑
𝜕f (1) (2)
𝜕f (1) (2)
(Xs− , Xs− , . . . , Xs(n)
= Y0 +
𝜇(Xs(i)− , s)
(Xs− , Xs− , . . . , Xs(n)
− , s) +
− , s)
∫0 𝜕t
𝜕X
t
i=1
t
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INTRODUCTION
+
n
n ∑
∑
1
i=1 j=1
t
+
n
∑
∫0
i=1
t
+
[
2
[
∫0
249
(j)
𝜎(Xs(i)− , s)𝜎(Xs− , s)
𝜎(Xs(i)− , s)
]
𝜕2 f
(j)
𝜕Xt(i) 𝜕Xt
𝜕f
(2)
(Xs(1)
− , Xs− ,
(i)
𝜕Xt
(2)
(n)
(Xs(1)
− , Xs− , . . . , Xs− , s)
...
ds
]
, Xs(n)
− , s) dWs
]
(1)
(2)
(2)
(n)
(n)
(1)
(2)
(n)
f (Xs(1)
− + Js− , Xs− + Js− , . . . , Xs− + Js− , s) − f (Xs− , Xs− , . . . , Xs− , s) dNs
where
t
∫0
=
[
]
(1)
(2)
(2)
(n)
(n)
(1)
(2)
(n)
f (Xs(1)
− + Js− , Xs− + Js− , . . . , Xs− + Js− , s) − f (Xs− , Xs− , . . . , Xs− , s) dNs
∑ [
]
(2)
(n)
f (Xs(1) , Xs(2) , . . . , Xs(n) s) − f (Xs(1)
− , Xs− , . . . , Xs− , s) ΔNs
0<s≤t
=
Nt [
∑
]
(2)
(n)
f (Xs(1) , Xs(2) , . . . , Xs(n) s) − f (Xs(1)
− , Xs− , . . . , Xs− , s) .
s=1
By recalling that we can use Girsanov’s theorem to change the measure so that a Wiener
process (standard Wiener process with drift) becomes a standard Wiener process, we can also
change the measure for both Poisson and compound Poisson processes. For a Poisson process,
the outcome of the change of measure affects the intensity whilst for a compound Poisson
process, the change of measure affects both the intensity and the distribution of the jump amplitudes. In the following we provide general results for the change of measure for Poisson and
compound Poisson processes.
Theorem 5.10 (Girsanov’s Theorem for Poisson Process) Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to the
filtration ℱt , 0 ≤ t ≤ T. Let 𝜂 > 0 and consider the Radon–Nikodým derivative process
( 𝜂 )Nt
Zt = e(𝜆−𝜂)t
.
𝜆
By changing the measure ℙ to a measure ℚ such that
(
)
dℚ ||
dℚ ||
ℙ
ℱt =
= Zt
𝔼
|
dℙ |
dℙ ||ℱt
then, under the ℚ measure, Nt ∼ Poisson(𝜂t) with intensity 𝜂 > 0.
Theorem 5.11 (Girsanov’s Theorem for Compound Poisson Process) Let {Nt ∶ 0 ≤ t ≤
T} be a Poisson process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with
respect to the filtration ℱt , 0 ≤ t ≤ T and let X1 , X2 , . . . be a sequence of independent and
identically distributed random variables where each Xi , i = 1, 2, . . . has a probability density
function f ℙ (Xi ) under the ℙ measure. Let X1 , X2 , . . . also be independent of Nt and Wt . From
the definition of the compound Poisson process
Mt =
Nt
∑
i=1
Xi ,
t≥0
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we let 𝜂 > 0 and consider the Radon–Nikodým derivative process
t
⎧ (𝜆−𝜂)t ∏
𝜂ℚ(Xi )
⎪e
𝜆ℙ(Xi )
⎪
i=1
Zt = ⎨
Nt
⎪ (𝜆−𝜂)t ∏
𝜂f ℚ (Xi )
⎪e
𝜆f ℙ (Xi )
⎩
i=1
N
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where ℚ(Xi ) (f ℚ (Xi )) is the probability mass (density) function of Xi , i = 1, 2, . . . under the ℚ
measure. For the case of continuous random variables Xi , i = 1, 2, . . . we also let ℚ be absolutely continuous with respect to ℙ on ℱ. By changing the measure ℙ to measure ℚ such that
(
)
dℚ ||
dℚ ||
ℱ
𝔼ℙ
= Zt
t =
|
dℙ |
dℙ ||ℱt
then, under the ℚ measure, Mt is a compound Poisson process with intensity 𝜂 > 0 and Xi ,
i = 1, 2, . . . is a sequence of independent and identically distributed random variables with
probability mass (density) functions ℚ(Xi ) (f ℚ (Xi )), i = 1, 2, . . .
By augmenting the Wiener process to the Poisson and compound Poisson processes we have
the following results.
Theorem 5.12 (Girsanov’s Theorem for Poisson Process and Standard Wiener Process) Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ 0 ≤ t ≤ T}
be a standard Wiener process defined on the probability space (Ω, ℱ, ℙ) with respect to the
filtration ℱt , t ≥ 0. Suppose 𝜃t , 0 ≤ t ≤ T is an adapted process and 𝜂 > 0. We consider the
Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1) = e(𝜆−𝜂)t
and
t
( 𝜂 )Nt
𝜆
1
t 2
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du .
By changing the measure ℙ to measure ℚ such that
(
)
dℚ ||
dℚ ||
ℙ
ℱt =
𝔼
= Zt
|
dℙ |
dℙ ||ℱt
̃t =
then, under the ℚ measure, Nt is a Poisson process with intensity 𝜂 > 0, the process W
t
Wt +
∫0
̃ t.
𝜃u du is a ℚ-standard Wiener process and Nt ⟂
⟂W
Theorem 5.13 (Girsanov’s Theorem for Compound Poisson Process and Standard
Wiener Process) Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 and
{Wt ∶ 0 ≤ t ≤ T} be a standard Wiener process defined on the probability space (Ω, ℱ, ℙ)
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251
with respect to the filtration ℱt , 0 ≤ t ≤ T. Suppose 𝜃t , 0 ≤ t ≤ T is an adapted process,
𝜂 > 0, X1 , X2 , . . . is a sequence of independent and identically distributed random variables
where each Xi , i = 1, 2, . . . has a probability mass (density) function ℙ(Xi ) > 0 (f ℙ (Xi ) > 0)
under the ℙ measure and assume also that the sequence X1 , X2 , . . . is independent of Nt . We
consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
=⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
Zt(1)
and
t
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
1
t 2
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du
where ℚ(Xi ) (f ℚ (Xi ))
probability
mass (density) function of Xi , i = 1, 2, . . . under the
( is1 the
)
T 2
ℚ measure and 𝔼ℙ e 2 ∫0 𝜃u du < ∞. For the case of continuous random variables Xi , i =
1, 2, . . . we also let ℚ be absolutely continuous with respect to ℙ on ℱ. By changing the
measure ℙ to measure ℚ such that
(
)
dℚ ||
dℚ ||
ℱ
𝔼ℙ
= Zt
t =
|
dℙ |
dℙ ||ℱt
∑Nt
then, under the ℚ measure, the process Mt = i=1
Xi is a compound Poisson process with
intensity 𝜂 > 0 where Xi , i = 1, 2, . . . is a sequence of independent and identically distributed
random variables with probability mass (density) functions ℚ(Xi ) (f ℚ (Xi )), i = 1, 2, . . . , the
̃ t = Wt +
process W
t
∫0
̃ t.
𝜃u du is a ℚ standard Wiener process and Mt ⟂
⟂W
5.2
5.2.1
PROBLEMS AND SOLUTIONS
Properties of Poisson Process
1. Let (Ω, ℱ, ℙ) be a probability
( space
) and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0. Show that Cov Ns , Nt = 𝜆min{s, t} and deduce that the correlation coefficient
of Ns and Nt is
√
min{s, t}
.
max{s, t}
( )
( )
Solution: Since Nt ∼ Poisson(𝜆t) with 𝔼 Nt = 𝜆t, 𝔼 Nt2 = 𝜆t + 𝜆2 t2 and by definition
)
(
)
( ) ( )
(
Cov Ns , Nt = 𝔼 Ns Nt − 𝔼 Ns 𝔼 Nt .
𝜌=
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For s ≤ t, using the stationary and independent properties of a Poisson process,
(
)
( (
)
)
𝔼 Ns Nt = 𝔼 Ns Nt − Ns + Ns2
))
( )
( (
= 𝔼 Ns Nt − Ns + 𝔼 Ns2
)
( )
( ) (
⟂ N t − Ns
= 𝔼 Ns 𝔼 Nt − Ns + 𝔼 Ns2 , Ns ⟂
= 𝜆s ⋅ 𝜆(t − s) + 𝜆s + 𝜆2 s2
= 𝜆2 st + 𝜆s
which implies
)
(
Cov Ns , Nt = 𝜆2 st + 𝜆s − 𝜆2 st = 𝜆s.
Similarly, for t ≤ s,
)
( (
)
)
(
𝔼 Ns Nt = 𝔼 Nt Ns − Nt + Nt2
))
( )
( (
= 𝔼 Nt Ns − Nt + 𝔼 Nt2
)
( )
( ) (
= 𝔼 Nt 𝔼 Ns − Nt + 𝔼 Nt2 ,
Nt ⟂
⟂ Ns − Nt
= 𝜆t ⋅ 𝜆(s − t) + 𝜆t + 𝜆2 t2
= 𝜆2 st + 𝜆t
and hence
Cov(Ns , Nt ) = 𝜆2 st + 𝜆t − 𝜆2 st = 𝜆t.
Thus, Cov(Ns , Nt ) = 𝜆min{s, t}.
By definition, the correlation coefficient of Ns and Nt is defined as
)
(
Cov Ns , Nt
𝜌= √
( )
( )
Var Ns Var Nt
=
𝜆min{s, t}
√
𝜆 st
=
min{s, t}
.
√
st
For s ≤ t,
s
𝜌= √ =
st
whilst for s > t,
t
𝜌= √ =
st
√
Therefore, 𝜌 =
min{s, t}
.
max{s, t}
√
√
s
t
t
.
s
◽
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253
2. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 satisfying the following conditions for small h > 0 and k ∈ ℕ:
(a) N0 = 0 and Nt is a non-decreasing counting process;
⎧1 − 𝜆h + o(h) j = 0
⎪
(b) ℙ(Nt+h = i + j|Nt = i) = ⎨𝜆h + o(h)
j=1
⎪o(h)
j > 1;
⎩
(c) Nt has independent stationary increments.
By setting pk (t) = ℙ(Nt = k), show that
{
−𝜆p0 (t)
′
pk (t) =
𝜆pk−1 (t) − 𝜆pk (t)
{
with boundary condition
pk (0) =
1
0
k=0
k≠0
k=0
k ≠ 0.
By solving the differential-difference equation using mathematical induction or the
moment generating function, show that
ℙ(Nt = k) =
(𝜆t)k −𝜆t
e ,
k!
k = 0, 1, 2, . . .
Solution: By definition, for k ≠ 0,
∑
ℙ(Nt+h = k|Nt = j)ℙ(Nt = j)
ℙ(Nt+h = k) =
j
= ℙ(Nt+h = k|Nt = k − 1)ℙ(Nt = k − 1)
+ ℙ(Nt+h = k|Nt = k)ℙ(Nt = k) + o(h)
= (𝜆h + o(h))ℙ(Nt = k − 1) + (1 − 𝜆h + o(h))ℙ(Nt = k).
By setting pk (t + h) = ℙ(Nt+h = k), pk−1 (t) = ℙ(Nt = k − 1) and pk (t) = ℙ(Nt = k), therefore for k ≠ 0,
pk (t + h) = 𝜆hpk−1 (t) + (1 − 𝜆h)pk (t) + o(h)
lim
h→0
pk (t + h) − pk (t)
o(h)
= 𝜆pk−1 (t) − 𝜆pk (t) + lim
h→0 h
h
and we will have
p′k (t) = 𝜆pk−1 (t) − 𝜆pk (t),
k ≠ 0.
For the case when k = 0,
ℙ(Nt+h = 0) = ℙ(Nt+h = 0|Nt = 0)ℙ(Nt = 0) + o(h)
= (1 − 𝜆h + o(h))ℙ(Nt = 0) + o(h)
or
p0 (t + h) = (1 − 𝜆h)p0 (t) + o(h)
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where p0 (t + h) = ℙ(Nt+h = 0) and p0 (t) = ℙ(Nt = 0). By subtracting p0 (t) from both
sides of the equation, dividing by h and letting h → 0, we will have
p′0 (t) = −𝜆p0 (t).
Knowing that ℙ(N0 = 0) = 1, hence
{
p′k (t) =
k=0
−𝜆p0 (t)
𝜆pk−1 (t) − 𝜆pk (t) k ≠ 0
with boundary condition
{
pk (0) =
1
0
k=0
k ≠ 0.
The following are two methods to solve the differential-difference equation.
Method I: Mathematical Induction. Let k = 0. To solve
p′0 (t) + 𝜆p0 (t) = 0
we let the integrating factor I = e𝜆t . By multiplying the integrating factor on both sides of
the equation and taking note that p0 (0) = ℙ(N0 = 0) = 1, we have
ℙ(Nt = 0) = e−𝜆t .
Therefore, the result is true for k = 0.
Assume the result is true for k = j, j = 0, 1, 2, . . . such that
ℙ(Nt = j) =
(𝜆t)j e−𝜆t
.
j!
For k = j + 1 the differential-difference equation is
p′j+1 (t) + 𝜆pj+1 (t) =
𝜆(𝜆t)j e−𝜆t
.
j!
Setting the integrating factor I = e𝜆t and multiplying it by both sides of the differential
equation,
(𝜆t)j+1
d 𝜆t
𝜆j+1 tj
(e pj+1 (t)) =
or e𝜆t pj+1 (t) =
+C
dt
j!
(j + 1)!
where C is a constant. Knowing that pj+1 (0) = ℙ(N0 = j + 1) = 0 we have C = 0 and
hence
(𝜆t)j+1 e−𝜆t
ℙ(Nt = j + 1) =
.
(j + 1)!
Therefore, the result is also true for k = j + 1. Thus, using mathematical induction we
have shown
(𝜆t)k −𝜆t
ℙ(Nt = k) =
e , k = 0, 1, 2, . . .
k!
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255
Method II: Moment Generating Function. Let
∞
∑
MNt (t, u) = 𝔼(euNt ) =
euk ℙ(Nt = k) =
k=0
∞
∑
euk pk (t)
k=0
be defined as the moment generating function of the Poisson process. Taking the first
partial derivative with respect to t,
𝜕MNt (t, u)
𝜕t
=
∞
∑
euk p′k (t)
k=0
=𝜆
∞
∑
euk pk−1 (t) − 𝜆
k=1
∞
∑
euk pk (t)
k=0
∑
∞
= 𝜆eu
eu(k−1) pk−1 (t) − 𝜆
∞
∑
k=1
or
𝜕MNt (t, u)
𝜕t
euk pk (t)
k=0
= 𝜆(eu − 1)MNt (t, u)
with boundary condition MN0 (0, u) = 1.
u
By setting the integrating factor I = e−𝜆(e −1)t and multiplying it by the first-order differential equation, we have
)
(
u
u
𝜕
MNt (t, u)e−𝜆(e −1)t = 0 or MNt (t, u)e−𝜆(e −1)t = C
𝜕t
where C is a constant. Since MN0 (0, u) = 1 we have C = 1 and hence
MNt (t, u) = e𝜆(e
u −1)t
By definition, MNt (t, u) =
∑∞
k=0
euk ℙ(Nt = k) and using Taylor’s series expansion we have
𝜆(eu −1)t
MNt (t, u) = e
.
−𝜆t
=e
∞
∑
(𝜆teu )k
k=0
and hence
ℙ(Nt = k) =
(𝜆t)k −𝜆t
e ,
k!
k!
=
∞
∑
(
uk
e
k=0
(𝜆t)k −𝜆t
e
k!
k = 0, 1, 2, . . .
)
◽
3. Pure Birth Process. Let (Ω, ℱ, ℙ) be a probability space. If {Nt ∶ t ≥ 0} is a Poisson
process with intensity 𝜆 > 0 then for small h > 0 and k ∈ ℕ, show that it satisfies the
following property:
ℙ(Nt+h
⎧1 − 𝜆h + o(h)
⎪
= k + j|Nt = k) = ⎨𝜆h + o(h)
⎪o(h)
⎩
j=0
j=1
j > 1.
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Solution: We first consider j = 0 where
ℙ(Nt+h = k|Nt = k) =
=
ℙ(Nt+h = k, Nt = k)
ℙ(Nt = k)
ℙ(zero arrival between (t, t + h], Nt = k)
.
ℙ(Nt = k)
Since the event {Nt = k} relates to arrival during the time interval [0, t] and the event
{zero arrival between (t, t + h]} relates to arrivals after time t, both of the events are independent and because Nt has stationary increments,
ℙ(Nt+h = k|Nt = k) =
ℙ(zero arrival between (t, t + h])ℙ(Nt = k)
ℙ(Nt = k)
= ℙ(zero arrival between (t, t + h])
= ℙ(Nt+h − Nt = 0)
= e−𝜆h
(𝜆h)2 (𝜆h)3
−
+ ...
2!
3!
= 1 − 𝜆h + o(h).
= 1 − 𝜆h +
Similarly,
ℙ(Nt+h = k + 1|Nt = k) = ℙ(1 arrival between (t, t + h])
= ℙ(Nt+h − Nt = 1)
= 𝜆he−𝜆h
)
(
(𝜆h)2 (𝜆h)3
−
+ ...
= 𝜆h 1 − 𝜆h +
2!
3!
= 𝜆h + o(h)
and finally
ℙ(Nt+h > k + 1|Nt = k) = 1 − ℙ(Nt+h = k|Nt+h = k) − ℙ(Nt+h = k + 1|Nt = k)
= 1 − (1 − 𝜆h + o(h)) − 𝜆h + o(h)
= o(h).
◽
4. Arrival Time Distribution. Let the inter-arrival times of “jump” events 𝜏1 , 𝜏2 , . . . be a
sequence of independent and identically distributed random variables where each 𝜏i ∼
Exp(𝜆), 𝜆 > 0, i = 1, 2, . . . has a probability density function f𝜏 (t) = 𝜆e−𝜆t , t ≥ 0. By
defining the arrival time of the n-th jump event as
Tn =
n
∑
i=1
𝜏i
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257
where 𝜏i = Ti − Ti−1 , show that the arrival time Tn , n ≥ 1 follows a gamma distribution,
Tn ∼ Gamma(n, 𝜆) with probability density function given as
fTn (t) =
(𝜆t)n−1 −𝜆t
𝜆e ,
(n − 1)!
t ≥ 0.
Let Nt be the number of jumps that occur at or before time t. Explain why for k ≥ 1,
ℙ(Nt ≥ k) = ℙ(Tk ≤ t)
and for k = 0,
ℙ(Nt = 0) = ℙ(T1 > t).
Using the above results show that Nt is a Poisson process with intensity 𝜆 having the
probability mass function
ℙ(Nt = k) =
(𝜆t)k −𝜆t
e ,
k!
k = 0, 1, 2, . . .
Solution: The first part of the proof is analogous to Problem 1.2.2.5 (page 16) and we
shall omit it. As for the remaining part of the proof we note that for k ≥ 1, Nt ≥ k if and
only if there are at least k jump events by time t, which implies that Tk (the time of the
k-th jump) is less than or equal to t. Thus,
ℙ(Nt ≥ k) = ℙ(Tk ≤ t),
k ≥ 1.
On the contrary, for k = 0, Nt = 0 if and only if there are zero jumps by time t, which
implies T1 (the time of the first jump) occurs after time t. Therefore,
ℙ(Nt = 0) = ℙ(T1 > t).
For k = 0,
∞
ℙ(Nt = 0) = ℙ(T1 > t) = ℙ(𝜏1 > t) =
For k ≥ 1,
∫t
ℙ(Nt = k) = ℙ(Nt ≥ k) − ℙ(Nt ≥ k + 1).
By solving
t
ℙ(Nt ≥ k + 1) = ℙ(Tk+1 ≤ t) =
b
and using integration by parts
set
𝜆e−𝜆s ds = e−𝜆t .
∫a
u=
u(x)
∫0
(𝜆s)k −𝜆s
𝜆e ds
k!
b
|b
d
d
𝑣(x) u(x) dx, we
𝑣(x) dx = u(x)𝑣(x)|| −
dx
dx
|a ∫a
(𝜆s)k
du
𝜆k sk−1
⇒
=
k!
ds
(k − 1)!
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d𝑣
= 𝜆e−𝜆s ⇒ 𝑣 = −e−𝜆s .
ds
Thus,
ℙ(Nt ≥ k + 1) = −
s=t
t
(𝜆s)k−1 −𝜆s
(𝜆s)k −𝜆s ||
(𝜆t)k −𝜆t
e | +
𝜆e ds = −
e + ℙ(Nt ≥ k).
|
∫0 (k − 1)!
k!
k!
|s=0
Therefore, for k ≥ 1
ℙ(Nt = k) =
(𝜆t)k −𝜆t
e .
k!
Since 0! = 1, in general we can express the probability mass function of a Poisson process
with intensity 𝜆 > 0 as
ℙ(Nt = k) =
(𝜆t)k −𝜆t
e ,
k!
k = 0, 1, 2, . . .
◽
5. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0. If 𝜏1 , 𝜏2 , . . . is a sequence of inter-arrival times, show that the random variables
are mutually independent and each follows an exponential distribution with parameter 𝜆.
Solution: We prove this result via mathematical induction.
Since
ℙ(𝜏1 > t) = ℙ(Nt = 0) = e−𝜆t
the density of 𝜏1 is
f𝜏1 (t) =
d
d
ℙ(𝜏1 < t) =
(1 − e−𝜆t ) = 𝜆e−𝜆t
dt
dt
and hence 𝜏1 ∼ Exp(𝜆).
By conditioning on 𝜏1 and since 𝜏1 < 𝜏2 ,
ℙ(𝜏2 > t|𝜏1 = t1 ) = ℙ(zero arrival in (t1 , t1 + t]|𝜏1 = t1 ).
Since the event {𝜏1 = t1 } is only related to arrivals during the time interval [0, t1 ] and
the event {zero arrival in (t1 , t1 + t]} relates to arrivals during the time interval (t1 , t1 + t],
these two events are independent. Thus, we have
ℙ(𝜏2 > t|𝜏1 = t1 ) = ℙ(zero arrival in (t1 , t1 + t]) = ℙ(𝜏2 > t) = e−𝜆t
with 𝜏2 ⟂
⟂ 𝜏1 and 𝜏2 ∼ Exp(𝜆).
We assume that for n > 1, 𝜏n ⟂
⟂ 𝜏i and each 𝜏i ∼ Exp(𝜆), i = 1, 2, . . . , n − 1 such that
ℙ(𝜏n > t|𝜏1 = t1 , 𝜏2 = t2 , . . . , 𝜏n−1 = tn−1 ) = ℙ(zero arrival in (Tn−1 , Tn−1 + t])
= e−𝜆t
where Tn−1 =
n−1
∑
i=1
ti .
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Conditional on 𝜏1 = t1 , 𝜏2 = t2 , . . . , 𝜏n = tn , such that 𝜏1 < 𝜏2 < . . . < 𝜏n < 𝜏n+1 we have
ℙ(𝜏n+1 > t|𝜏1 = t1 , 𝜏2 = t2 , . . . , 𝜏n = tn )
= ℙ(zero arrival in (Tn , Tn + t]|𝜏1 = t1 , 𝜏2 = t2 , . . . , 𝜏n = tn )
where Tn =
n
∑
i=1
ti . Since the events {zero arrival in (Tn , Tn + t]} and {𝜏i = ti }, i = 1,
2, . . . , n are mutually independent, then
ℙ(𝜏n+1 > t|𝜏1 = t1 , 𝜏2 = t2 , . . . , 𝜏n = tn ) = ℙ(zero arrival in (Tn , Tn + t])
= ℙ(𝜏n+1 > t)
= e−𝜆t
which implies 𝜏n+1 ⟂
⟂ 𝜏i , i = 1, 2, . . . , n and 𝜏n+1 ∼ Exp(𝜆).
Thus, from mathematical induction, the claim is true for all n ≥ 1.
◽
6. Stationary and Independent Increments. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶
t ≥ 0} be a Poisson process with intensity 𝜆 > 0. Suppose 0 = t0 < t1 < . . . < tn , using
the property of the Poisson process
ℙ(Nti +h
⎧1 − 𝜆h + o(h) 𝑣 = 𝑤
⎪
= 𝑣|Nti = 𝑤) = ⎨𝜆h + o(h)
𝑣=𝑤+1
⎪o(h)
𝑣>𝑤+1
⎩
show that the increments
Nt1 − Nt0 , Nt2 − Nt1 , . . . , Ntn − Ntn−1
are stationary and independent with the distribution of Nti − Nti−1 the same as the distribution of Nti −ti−1 .
Solution: To show that {Nt ∶ t ≥ 0} is stationary we note that for 0 ≤ ti−1 < ti , m ≥ 0
∑
ℙ(Nti = u + m, Nti−1 = u)
ℙ(Nti − Nti−1 = m) =
u
=
∑
ℙ(Nti = u + m|Nti−1 = u)ℙ(Nti−1 = u)
u
=
∑
pu,u+m (ti−1 , ti )ℙ(Nti−1 = u)
u
where pu,u+m (ti−1 , ti ) = ℙ(Nti = u + m|Nti−1 = u).
From the property of the Poisson process
ℙ(Nti +h
⎧1 − 𝜆h + o(h) 𝑣 = 𝑤
⎪
= 𝑣|Nti = 𝑤) = ⎨𝜆h + o(h)
𝑣=𝑤+1
⎪o(h)
𝑣>𝑤+1
⎩
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we can write
ℙ(Nti +h = 𝑣|Nti−1 = u) = ℙ(Nti +h = 𝑣|Nti = 𝑣 − 1)ℙ(Nti = 𝑣 − 1|Nti−1 = u)
+ ℙ(Nti +h = 𝑣|Nti = 𝑣)ℙ(Nti = 𝑣|Nti−1 = u)
∑
+
ℙ(Nti +h = 𝑣|Nti = 𝑤)ℙ(Nti = 𝑤|Nti−1 = u)
𝑤<𝑣−1
= (𝜆h + o(h))ℙ(Nti = 𝑣 − 1|Nti−1 = u)
+ (1 − 𝜆h + o(h))ℙ(Nti = 𝑣|Nti−1 = u) + o(h)
or
p′0,0 (ti−1 , ti ) = − 𝜆p0,0 (ti−1 , ti )
p′u,𝑣 (ti−1 , ti ) = −𝜆pu,𝑣 (ti−1 , ti ) + 𝜆pu,𝑣−1 (ti−1 , ti ),
where
p′u,𝑣 (ti−1 , ti ) = lim
𝑣 = u + 1, u + 2, . . .
ℙ(Nti +h = 𝑣|Nti−1 = u) − ℙ(Nti = 𝑣|Nti−1 = u)
h
h→0
.
Using the same steps as discussed in Problem 5.2.1.2 (page 253), we can deduce that
pu,𝑣 (ti−1 , ti ) =
𝜆𝑣−u (ti − ti−1 )𝑣−u e−𝜆(ti −ti−1 )
,
(𝑣 − u)!
𝑣 = u + 1, u + 2, . . .
By substituting the above result into ℙ(Nti − Nti−1 = m),
ℙ(Nti − Nti−1 = m) =
∑
pu,u+m (ti−1 , ti )ℙ(Nti−1 = u)
u
=
𝜆m (ti − ti−1 )m e−𝜆(ti −ti−1 ) ∑
ℙ(Nti−1 = u)
m!
u
=
𝜆m (ti − ti−1 )m e−𝜆(ti −ti−1 )
m!
which is a probability mass function for the Poisson (𝜆(ti − ti−1 )) distribution. Thus, for
all i = 1, 2, . . . , n, Nti − Nti−1 is a stationary process and has the same distribution as
Poisson(𝜆(ti − ti−1 )).
Finally, to show {Nt ∶ t ≥ 0} has independent increments we note for mi ≥ 0, mj ≥ 0,
i < j, i, j = 1, 2, . . . , n and from the stationary property of the Poisson process,
ℙ(Ntj − Ntj−1 = mj |Nti − Nti−1 = mi ) =
=
=
ℙ(Ntj − Ntj−1 = mj , Nti − Nti−1 = mi )
ℙ(Nti − Nti−1 = mi )
ℙ(Ntj −tj−1 = mj , Nti −ti−1 = mi )
ℙ(Nti −ti−1 = mi )
ℙ(arrivals in (tj−1 , tj ] ∩ arrivals in (ti−1 , ti ])
ℙ(arrivals in (ti−1 , ti ])
.
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Because i < j and since there are no overlapping events between the time intervals (ti−1 , ti ]
and (tj−1 , tj ], thus for i < j, i, j = 1, 2, . . . , n
ℙ(Ntj − Ntj−1 = mj |Nti − Nti−1 = mi ) = ℙ(arrivals in (tj−1 , tj ]) = ℙ(Ntj −tj−1 = mj ).
N.B. In general, by denoting ℱtk−1 as the 𝜎-algebra of information and observing the Poisson process Nt for 0 ≤ t ≤ tk−1 , we can deduce that Ntk − Ntk−1 is independent of ℱtk−1 .
◽
7. Superposition. Let Nt(1) , Nt(2) , . . . , Nt(n) be n independent Poisson processes with intensities
𝜆(1) , 𝜆(2) , . . . , 𝜆(n) , respectively, where the sequence of processes is defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt . Show that Nt(1) + Nt(2) + . . . + Nt(n)
is a Poisson process with intensity 𝜆(1) + 𝜆(2) + . . . + 𝜆(n) .
(
)
Solution: For Nt(i) ∼ Poisson 𝜆(i) t , i = 1, 2, . . . , n the moment generating function for
a Poisson process is
∞ ux (i) x
∞
( (i) )
(i) u x
(i) ∑ e (𝜆 t)
(i) ∑ (𝜆 te )
(i) u
MN (i) (t, u) = 𝔼 euNt = e−𝜆 t
= e−𝜆 t
= e𝜆 t(e −1)
t
x!
x!
x=0
x=0
where u ∈ ℝ.
(j)
⟂ Nt , i ≠ j, i, j = 1, 2, . . . , n the
By setting Nt = Nt(1) + Nt(2) + . . . + Nt(n) such that Nt(i) ⟂
moment generating function for Nt is
)
(
MNt (t, u) = 𝔼 euNt
( (1) (2)
)
(n)
= 𝔼 euNt +uNt + ... +uNt
( (1) )
( (2) )
( (n) )
= 𝔼 euNt ⋅ 𝔼 euNt · · · 𝔼 euNt
= e𝜆
(1) t(eu −1)
⋅ e𝜆
(2) t(eu −1)
· · · e𝜆
(1) +𝜆(2) + ... +𝜆(n) )t(eu −1)
= e(𝜆
(n) t(eu −1)
.
Therefore,
(
)
Nt(1) + Nt(2) + . . . + Nt(n) ∼ Poisson (𝜆(1) + 𝜆(2) + . . . + 𝜆(n) )t .
◽
8. Markov Property. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on
the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt . Show that if f is a continuous function then there exists another continuous function g such that
[
]
|
𝔼 f (Nt )|ℱu = g(Nu )
|
for 0 ≤ u ≤ t.
Solution: For 0 ≤ u ≤ t we can write
[
]
[
]
|
|
𝔼 f (Nt )|ℱu = 𝔼 f (Nt − Nu + Nu )|ℱu .
|
|
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Since Nt − Nu ⟂
⟂ ℱu and Nu is ℱu measurable, by setting Nu = x where x is a constant
value,
[
]
[
]
|
𝔼 f (Nt − Nu + Nu )|ℱu = 𝔼 f (Nt − Nu + x) .
|
Because Nt − Nu ∼ Poisson(𝜆(t − u)), we can write 𝔼[ f (Nt − Nu + x)] as
[
]
𝔼 f (Nt − Nu + x) =
∞
∑
f (k + x)
k=0
e−𝜆(t−u) (𝜆(t − u))k
.
k!
]
[
]
[
By setting 𝜏 = t − u and y = k + x, we can rewrite 𝔼 f (Nt −Nu +x) = 𝔼 f (Nt − Nu +Nu )
as
∞
] ∑
[
e−𝜆𝜏 (𝜆𝜏)y−Nu
f (y)
𝔼 f (Nt − Nu + Nu ) =
(y − Nu )!
y=N
u
∑
∞
=
f (y + Nu )
y=0
e−𝜆𝜏 (𝜆𝜏)y
.
y!
Since the only information from the filtration ℱu is Nu , then
[
]
|
𝔼 f (Nt )|ℱu = g(Nu )
|
where
g(Nu ) =
∞
∑
f (y + Nu )
y=0
e−𝜆𝜏 (𝜆𝜏)y
.
y!
◽
9. Compensated Poisson Process. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0
defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt . By defining the
̂t as
compensated Poisson process, N
̂t = Nt − 𝜆t
N
̂t is a martingale.
show that N
̂t is a martingale, for 0 ≤ s < t we note the following.
Solution: To show that N
(a) Since the increment Nt − Ns is independent of ℱs and has expected value 𝜆(t − s), we
can write
)
(
)
(
̂t − N
̂t || ℱs = 𝔼 N
̂s + N
̂s || ℱs
𝔼 N
|
|
)
(
)
(
̂s ||ℱs
̂t − N
̂s ||ℱs + 𝔼 N
=𝔼 N
|
|
)
(
|
̂s
= 𝔼 Nt − Ns − 𝜆(t − s)|ℱs + N
|
(
)
̂s
= 𝔼 Nt − Ns − 𝜆(t − s) + N
̂s .
=N
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)
(
(
)
)
( )
(
|̂ |
= 𝔼 ||Nt − 𝜆t|| ≤ 𝔼 Nt + 𝜆t < ∞ since Nt ≥ 0 and hence 𝔼 ||Nt || =
(b) 𝔼 |N
|
t
| |
2𝜆t < ∞.
̂t = Nt − 𝜆t is clearly ℱt -adapted.
(c) The process N
̂ t = Nt − 𝜆t is a martingale.
From (a)–(c) we have shown that N
◽
10. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt . By defining the compensated Poisson process,
̂ t as
N
̂t = Nt − 𝜆t
N
̂ 2 − 𝜆t is a martingale.
show that N
t
̂ 2 − 𝜆t is a martingale, for 0 ≤ s < t we note the following.
Solution: To show that N
t
⟂ ℱs and has expected value 𝜆(t − s), and because Nt −
(a) Since the increment Nt − Ns ⟂
Ns ⟂
⟂ Ns , we can write
)
[
]
(
̂t − N
̂t2 − 𝜆t||ℱs = 𝔼 (N
̂s + N
̂s )2 || ℱs − 𝜆t
𝔼 N
|
|
]
[
]
(
)
[
|
2
̂s (N
̂s2 || ℱs − 𝜆t
̂t − N
̂s ) | ℱs + 2𝔼 N
̂t − N
̂s )|| ℱs + 𝔼 N
= 𝔼 (N
|
|
|
[(
)2 | ]
= 𝔼 (Nt − Ns ) − 𝜆(t − s) | ℱs
|
) [
]
(
̂s2 − 𝜆t
+ 2𝔼 Ns − 𝜆s 𝔼 (Nt − Ns ) − 𝜆(t − s) + N
̂s2 − 𝜆t
= 𝜆(t − s) + 0 + N
̂s2 − 𝜆s.
=N
|̂2
|
(b) Since |N
− 𝜆t| ≤ (Nt − 𝜆t)2 + 𝜆t therefore
| t
|
)
[(
(
)2 ]
|
|̂2
− 𝜆t| ≤ 𝔼 Nt − 𝜆t
𝔼 |N
+ 𝜆t = 2𝜆t < ∞
t
|
|
[
]
( )
as 𝔼 (Nt − 𝜆t)2 = Var Nt = 𝜆t.
(
)
̂ 2 − 𝜆 = Nt − 𝜆t 2 − 𝜆t is clearly ℱt -adapted.
(c) The process N
t
̂ 2 − 𝜆t is a martingale.
From (a)–(c) we have shown that N
t
◽
11. Exponential Martingale Process. Let Nt be a Poisson process with intensity 𝜆 > 0 defined
on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt . Show that for u ∈ ℝ,
u −1)
Xt = euNt −𝜆t(e
is a martingale.
Solution: Given that Nt ∼ Poisson(𝜆t) then for u ∈ ℝ,
𝔼(euNt ) = e−𝜆t
∞ ux
∑
e (𝜆t)x
x=0
x!
= e−𝜆t
∞
∑
(𝜆teu )x
x=0
x!
= e𝜆t(e
u −1)
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and hence for 0 ≤ s < t and because Nt − Ns ∼ Poisson(𝜆(t − s)),
(
)
u
𝔼 eu(Nt −Ns ) = e𝜆(t−s)(e −1) .
u −1)
To show that Xt = euNt −𝜆t(e
is a martingale, for 0 ≤ s < t we have the following:
⟂ ℱs and has stationary increment, therefore eNt −Ns ⟂
⟂ ℱs and eNt −Ns ⟂
⟂
(a) Since Nt − Ns ⟂
N
s
e . Thus, we can write
)
(
)
(
u
|
𝔼 Xt |ℱs = 𝔼 euNt −𝜆t(e −1) | ℱs
|
)
(
u
|
= e−𝜆t(e −1) 𝔼 euNt | ℱs
|
)
(
u
|
= e−𝜆t(e −1) 𝔼 euNt −uNs +uNs | ℱs
|
) (
)
(
|
−𝜆t(eu −1)
uNs |
=e
𝔼 e | ℱs 𝔼 eu(Nt −Ns ) | ℱs
|
|
u −1)
= e−𝜆t(e
= euNs −𝜆s(e
⋅ euNs ⋅ e𝜆(t−s)(e
u −1)
u −1)
= Xs .
u
u
|
|
(b) Since |Xt | = |euNt −𝜆t(e −1) | = euNt −𝜆t(e −1) , we can write
|
|
)
)
(
)
(
(
u
u
u
u
𝔼 |Xt | = 𝔼 euNt −𝜆t(e −1) = e−𝜆t(e −1) 𝔼 euNt = e−𝜆t(e −1) ⋅ e𝜆t(e −1) = 1 < ∞.
u −1)
(c) The process Xt = euNt −𝜆t(e
is clearly ℱt -adapted.
u −1)
From (a)–(c) we have shown that Xt = euNt −𝜆t(e
is a martingale.
◽
12. Compound Poisson Process. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 >
0 defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt , and let
X1 , X2 , . . . be a sequence of independent and identically distributed random variables with
common mean 𝔼(Xi ) = 𝔼(X) = 𝜇 and variance Var (Xi ) = Var (X) = 𝜎 2 . Let X1 , X2 , . . . be
independent of Nt . By defining the compound Poisson process Mt as
Mt =
Nt
∑
Xi ,
t≥0
i=1
show that the moment generating function for Mt is
𝜑Mt (t, u) = e𝜆t(𝜑X (u)−1) ,
u∈ℝ
( )
where 𝜑X (u) = 𝔼 euX .
Further, show that 𝔼(Mt ) = 𝜇𝜆t and Var (Mt ) = (𝜇2 + 𝜎 2 )𝜆t.
Solution: By definition, for u ∈ ℝ
(
)
𝜑Mt (t, u) = 𝔼 euMt
( ∑Nt )
= 𝔼 eu i=1 Xi
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265
)]
[ ( ∑
Nt
|
= 𝔼 𝔼 eu i=1 Xi || Nt
|
∞
)
∑ ( ∑n |
=
𝔼 eu i=1 Xi | Nt = n ℙ(Nt = n).
|
n=0
Since X1 , X2 , . . . , Xn are independent and identically distributed, we can therefore write
𝜑Mt (t, u) =
∞
∑
( )n
𝔼 euX ℙ(Nt = n)
n=0
=
∞
∑
(𝜑X (u))n
n=0
= e−𝜆t
e−𝜆t (𝜆t)n
n!
∞
∑
(𝜆t𝜑X (t))n
n=0
n!
= e𝜆t(𝜑X (u)−1) .
By taking first and second partial derivatives of 𝜑Mt (t, u) with respect to u, we have
𝜕𝜑Mt (t, u)
𝜕u
and
𝜕 2 𝜑Mt (t, u)
𝜕u2
= 𝜆t𝜑′X (u)e𝜆t(𝜑X (u)−1)
{
}
= 𝜆t𝜑′′X (u) + (𝜆t)2 (𝜑′X (u))2 e𝜆t(𝜑X (u)−1) .
Since 𝜑X (0) = 1, therefore
𝔼(Mt ) =
𝜕𝜑Mt (t, 0)
𝜕u
= 𝜆t𝜑′X (0) = 𝜆t𝔼(X) = 𝜇𝜆t
and
)2
(
( ) 𝜕 2 𝜑Mt (t, 0)
= 𝜆t𝜑′′X (0) + (𝜆t)2 𝜑′X (0) = 𝜆t𝔼(X 2 ) + (𝜆t)2 𝔼(X)2 .
𝔼 Mt2 =
2
𝜕u
Thus, the variance of Y is
( )2
[
]
( )
( )
Var (Mt ) = 𝔼 Mt2 − 𝔼 Mt = 𝜆t𝔼 X 2 = 𝜆t Var (X) + 𝔼(X)2 = (𝜇 2 + 𝜎 2 )𝜆t.
◽
13. Martingale Properties of Compound Poisson Process. Let {Nt ∶ t ≥ 0} be a Poisson
process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect
to the filtration ℱt , and let X1 , X2 , . . . be a sequence of independent and identically
distributed random variables with common mean 𝔼(Xi ) = 𝔼(X) = 𝜇 and variance
Var (Xi ) = Var (X) = 𝜎 2 . Let X1 , X2 , . . . be independent of Nt . By defining the compound
Poisson process Mt as
Nt
∑
Mt =
Xi , t ≥ 0
i=1
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and assuming 𝔼(|X|) < ∞, show that for 0 ≤ s < t
⎧= M
s
⎪
𝔼(Mt |ℱs ) ⎨≥ Ms
⎪≤ M
s
⎩
if 𝜇 = 0
if 𝜇 > 0
if 𝜇 < 0.
Solution: To show that Mt can be a martingale, submartingale or supermartingale, for
0 ≤ s < t we have the following.
(a) 𝔼(Mt |ℱs ) = 𝔼(Mt − Ms + Ms |ℱs ) = 𝔼(Mt − Ms |ℱs ) + 𝔼(Ms |ℱs ) = 𝜇𝜆(t − s) + Ms
since the increment
independent
(∑
) of ℱs and has mean 𝜇𝜆(t − s).
( ∑ Mt)− Ms is
Nt | |
| Nt |
X = 𝜆t𝔼(|X|) < ∞ since 𝔼(|X|) < ∞.
(b) 𝔼|Mt |) = 𝔼 | i=1 Xi | ≤ 𝔼
i=1 | i |
|
∑N|t
(c) The process Mt = i=1 Xi is clearly ℱt -adapted.
We can therefore deduce that Mt is a martingale, submartingale or supermartingale by
setting 𝜇 = 0, 𝜇 > 0 or 𝜇 < 0, respectively.
◽
14. Compensated Compound Poisson Process. Let {Nt ∶ t ≥ 0} be a Poisson process with
intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration
ℱt . Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables with common mean 𝔼(Xi ) = 𝔼(X) = 𝜇 and variance Var (Xi ) = Var (X) = 𝜎 2 . Let
X1 , X2 , . . . be independent of Nt . By defining the compound Poisson process Mt as
Mt =
Nt
∑
Xi ,
t≥0
i=1
and assuming 𝔼(|X|) < ∞, show that the compensated compound Poisson process
̂ t = Mt − 𝜇𝜆t
M
is a martingale.
̂ t is a martingale, for 0 ≤ s < t we have the following.
Solution: To show that M
(a) Given the increment Mt − Ms is independent of ℱs and has mean 𝜇𝜆(t − s),
(
)
(
)
̂s + M
̂ s || ℱs
̂ t || ℱs = 𝔼 M
̂t − M
𝔼 M
|
|
)
(
|
= 𝔼 Mt − Ms − 𝜇𝜆(t − s) + Ms − 𝜇𝜆s|ℱs
|
)
(
|
= 𝔼 Mt − Ms |ℱs − 𝜇𝜆(t − s) + Ms − 𝜇𝜆s
|
= 𝜇𝜆(t − s) − 𝜇𝜆(t − s) + Ms − 𝜇𝜆s
= Ms − 𝜇𝜆s.
)
(∑
)
)
(∑
(
Nt | |
|
| Nt
|̂ |
X
+ 𝜇𝜆t = 𝜆t𝔼(|X|) + 𝜇𝜆t < ∞
Xi − 𝜇𝜆t| ≤ 𝔼
(b) 𝔼 |M
| = 𝔼 | i=1
t
i
i=1 | |
|
|
| |
since 𝔼(|X|) < ∞.
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267
̂ t = ∑Nt Xi − 𝜇𝜆t is clearly ℱt -adapted.
(c) The process M
i=1
̂ t = ∑Nt Xi − 𝜇𝜆t is a martingale.
From the results of (a)–(c) we have shown that M
i=1
◽
15. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt . Let X1 , X2 , . . . be a sequence
( ) of independent
and identically
distributed
random
variables
with
common
mean
𝔼
Xi = 𝔼(X) = 𝜇 and
( )
variance Var Xi = Var (X) = 𝜎 2 . Let X1 , X2 , . . . be independent of Nt . By defining the
compound Poisson process Mt as
Mt =
Nt
∑
Xi ,
t≥0
i=1
show that for 0 = t0 < t1 < t2 < . . . < tn the increments
Mt1 − Mt0 , Mt2 − Mt1 , . . . , Mtn − Mtn−1
are independent, stationary and the distribution of Mti − Mti−1 is the same as the distribution of Mti −ti−1 .
Solution: We first show that the increment Mtk − Mtk−1 , k = 1, 2, . . . , n is stationary and
has the same distribution as Mtk −tk−1 .
Using the technique of moment generating function, for u ∈ ℝ
)
(
𝜑Mt −Mt (u) = 𝔼 eu(Mtk −Mtk−1 )
k
k−1
))
( (∑Nt
∑Nt
=𝔼 e
(
=𝔼 e
= 𝜑Mt
k
i=1
u
u
Xi − i=1k−1 Xi
)
∑Ntk
Nt
k −tk−1
k−1
+1
Xi
(u).
Therefore, Mtk − Mtk−1 , k = 1, 2, . . . , n is stationary and has the same distribution as
Mtk −tk−1 .
Finally, to show that Mti − Mti−1 is independent of Mtj − Mtj−1 , i, j = 1, 2, . . . , n and i ≠ j,
using the joint moment generating function, for u, 𝑣 ∈ ℝ
𝜑(Mt −Mt
i
i−1
)(Mt −Mt
j
j−1
) (u, 𝑣) = 𝜑(Mt −t
i
i−1
)(Mt −t
j
j−1
) (u, 𝑣)
)
( u(M
)+𝑣(Mt −t )
j j−1
= 𝔼 e ti −ti−1
(
⎛ u
= 𝔼 ⎜e
⎜
⎝
(
=𝔼 e
∑Nti
k=Nt
(
u
i−1
) ( N
∑ tj
X
+𝑣
k
k=N
+1
k=Nt
tj−1
))
∑Nti
i−1
)
X
+1 k
X
+1 k
⎞
⎟
⎟
⎠
( N
∑ tj
⎛ 𝑣
𝔼 ⎜e
⎜
⎝
k=Nt
)
j−1
+1
Xk
⎞
⎟
⎟
⎠
4:29 P.M.
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i
i−1
(u)𝜑Mt −t
j
j−1
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(𝑣)
since X1 , X2 , . . . is a sequence of independent and identically distributed random variables
and also the intervals [Nti−1 + 1, Nti ] and [Ntj−1 + 1, Ntj ] have no overlapping events.
Thus, Mti − Mti−1 ⟂
⟂ Mtj − Mtj−1 , for all i, j = 1, 2, . . . , n, i ≠ j.
N.B. Since the compound Poisson process Mt has increments which are independent and
stationary, therefore from Problem 5.2.1.8 (page 261) we can deduce that Mt is Markov.
That is, if f is a continuous function then there exists another continuous function g
such that
]
[
𝔼 f (Mt )|ℱu = g(Mu )
for 0 ≤ u ≤ t.
◽
16. Decomposition of a Compound Poisson Process (Finite Jump Size). Let Nt be a Poisson
process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to
the filtration ℱt . Let X1 , X2 , . . . be a sequence of independent and identically distributed
random variables, and let X1 , X2 , . . . be independent of Nt such that
ℙ(X = xk ) = p(xk )
and
K
∑
ℙ(X = xk ) = 1
k=1
d
where X = Xi , i = 1, 2, . . . and x1 , x2 , . . . , xK is a finite set of non-zero numbers. By defining the compound Poisson process as
Mt =
Nt
∑
Xi
i=1
show that Mt and Nt can be written as
Mt =
K
∑
xk Nt(k)
and
Nt =
k=1
K
∑
Nt(k)
k=1
(
)
where Nt(k) ∼ Poisson 𝜆tp(xk ) , k = 1, 2, . . . , K is a sequence of independent and identical Poisson processes.
Solution: From Problem 5.2.1.12 (page 264), the moment generating function for Mt is
𝜑Mt (t, u) = e𝜆t(𝜑X (u)−1) ,
u∈ℝ
( )
where 𝜑X (u) = 𝔼 euX . Given ℙ(X = xk ) = p(xk ), k = 1, 2, . . . , K we have
( )
𝜑X (u) = 𝔼 euX
=
K
∑
euxk ℙ(X = xk )
k=1
=
K
∑
k=1
euxk p(xk ).
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Substituting 𝜑X (u) into the moment generating function for Mt ,
𝜆t
𝜑Mt (t, u) = e
Because
∑K
k=1 p(xk )
= 1,
𝜆t
𝜑Mt (t, u) = e
(∑
K
uxk p(x )−1
k
k=1 e
)
.
(∑
)
K
uxk p(x )−∑K p(x )
k
k
k=1 e
k=1
∑K
= e𝜆t k=1 (e k −1)p(xk )
K
∏
ux
=
e𝜆tp(xk )(e k −1)
ux
k=1
K
(
)
∏
(k)
=
𝔼 euxk Nt
k=1
which is a product of K independent
moment
generating functions of the random variable
(
)
xk Nt(k) where Nt(k) ∼ Poisson 𝜆tp(xk ) . Thus,
K
)
( ∑K
)
(
∏
(k)
(k)
= 𝔼 eu k=1 xk Nt
𝜑Mt (t, u) =
𝔼 euxk Nt
k=1
K
∑
xk Nt(k) .
k=1
(
)
of independent
and
Given that Nt(k) ∼ Poisson 𝜆tp(xk ) , k = 1, 2, . . . , K is a sequence
(
)
K
K
K
∑ (k)
∑
∑ (k)
identically distributed Poisson processes,
Nt ∼ Poisson
𝜆tp(xk ) or
Nt ∼
which implies Mt =
k=1
Poisson(𝜆t) and hence Nt =
K
∑
k=1
k=1
k=1
Nt(k) .
◽
17. Let {X1 , X2 , . . . ,} be a series of independent and identically distributed random variables
with moment generating function
( )
MX (𝜉) = 𝔼 e𝜉X , 𝜉 ∈ ℝ
d
where X = Xi , i = 1, 2, . . . and let {t1 , t2 , . . .} be its corresponding sequence of independent and identically distributed random jump times of a Poisson process {Nt ∶ t ≥ 0} with
intensity parameter 𝜆 > 0 where ti ∼ Exp(𝜆), i = 1, 2, . . . Let X1 , X2 , . . . be independent
of Nt .
Show that for n ≥ 1, 𝜅 > 0 the process {Yt ∶ t ≥ 0} with initial condition Y0 = 0 given as
Yt =
n
∑
e−𝜅(t−ti ) Xi
i=1
( )
has moment generating function MY (𝜉, t) = 𝔼 e𝜉Yt given by
MY (𝜉, t) =
n
∏
i=1
{
}
t
∫0
−𝜅s
MXi (𝜉e
) 𝜆e
−𝜆(t−s)
ds
4:29 P.M.
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(
)(
)(
)n−1
𝜆
e−𝜅t − e−𝜆t 1 − e−𝜆t
𝜆−𝜅
(
)(
)(
)n−1
𝜆
Var (Yt ) = n(𝜇2 + 𝜎 2 )
e−2𝜅t − e−𝜆t 1 − e−𝜆t
𝜆 − 2𝜅
)2 (
(
)2 (
)n−2
𝜆
e−2𝜅t − e−𝜆t 1 − e−𝜆t
+ n(n − 1)𝜇 2
𝜆 − 2𝜅
)2 (
(
)2 (
)2(n−1)
𝜆
− n2 𝜇 2
e−𝜅t − e−𝜆t 1 − e−𝜆t
𝜆−𝜅
𝔼(Yt ) = n𝜇
where 𝔼(Xi ) = 𝜇 and Var (Xi ) = 𝜎 2 for i = 1, 2, . . .
Solution: By definition,
n
n
( ∑n −𝜅(t−t ) ) ∏
( −𝜅(t−t ) ) ∏
( )
i Xi
i Xi
=
=
MY (𝜉, t) = 𝔼 e𝜉Yt = 𝔼 e𝜉 i=1 e
𝔼 e𝜉e
MXi (𝜉e−𝜅(t−ti ) ).
i=1
i=1
Using the tower property,
( −𝜅(t−t ) )
i Xi
MXi (𝜉e−𝜅(t−ti ) ) = 𝔼 e𝜉e
)]
[ ( −𝜅(t−t )
i Xi |
= 𝔼 𝔼 e𝜉e
| ti = s
|
t (
)
−𝜅(t−s) X
i 𝜆e−𝜆s ds
=
𝔼 e𝜉e
∫0
t
=
MXi (𝜉e−𝜅(t−s) )𝜆e−𝜆s ds
∫0
t
=
Therefore,
MY (𝜉, t) =
n
∏
∫0
{
i=1
MXi (𝜉e−𝜅s )𝜆e−𝜆(t−s) ds.
}
t
∫0
MXi (𝜉e−𝜅s ) 𝜆e−𝜆(t−s) ds
.
t
By setting 𝜉 = 𝜉e−𝜅s and M Xi (𝜉) =
with respect to 𝜉, we have
∫0
MXi (𝜉e−𝜅s )𝜆e−𝜆(t−s) ds, and differentiating MY (𝜉, t)
⎧
⎫
n
n
𝜕MY (𝜉, t) ∑ ⎪ 𝜕M Xi (𝜉) ∏
⎪
=
M Xj (𝜉)⎬
⎨
𝜕𝜉
𝜕𝜉
⎪
i=1 ⎪
j=1
⎩
⎭
j≠i
] n
⎫
⎧[
n
t 𝜕M (𝜉)
∑
∏
⎪
⎪
Xi
𝜕𝜉
−𝜆(t−s)
ds
M Xj (𝜉)⎬
⋅ 𝜆e
⋅
=
⎨ ∫
𝜕𝜉
0
𝜕𝜉
⎪
i=1 ⎪
j=1
⎭
⎩
j≠i
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271
] n [
⎧[
]⎫
n
t 𝜕M (𝜉)
t
∑
∏
⎪
⎪
Xi
−𝜆t (𝜆−𝜅)s
−𝜆(t−s)
=
ds
MXi (𝜉)𝜆e
ds ⎬
𝜆e e
⎨ ∫
∫
0
0
𝜕𝜉
⎪
i=1 ⎪
j=1
⎭
⎩
j≠i
and
⎫
⎫
⎧ 2
⎧
n
n
n
n
n
𝜕 2 MY (𝜉, t) ∑ ⎪ 𝜕 M Xi (𝜉) ∏
⎪ ∑ ∑ ⎪ 𝜕M Xi (𝜉) 𝜕M Xj (𝜉) ∏
⎪
=
M Xj (𝜉)⎬ +
M Xk (𝜉)⎬
⎨
⎨ 𝜕𝜉
2
2
𝜕𝜉
𝜕𝜉
𝜕𝜉
⎪ i=1 j=1 ⎪
⎪
i=1 ⎪
j=1
k=1
⎭
⎭
k≠i,j
⎩
⎩
j≠i
i≠j
] n [
⎧[
]⎫
n
t 𝜕 2 M (𝜉)
t
∑
∏
⎪
⎪
Xi
=
𝜆e−𝜆t e(𝜆−2𝜅)s ds
MXj (𝜉)𝜆e−𝜆(t−s) ds ⎬
⎨ ∫
2
∫
0
0
⎪
i=1 ⎪
j=1
𝜕𝜉
⎩
⎭
j≠i
]
⎧[
n ∑
n
⎡ 𝜕MX (𝜉)
⎤
∑
⎪ 𝜕MXi (𝜉) −𝜆t (𝜆−𝜅)s
j
ds ⎢
𝜆e e
𝜆e−𝜆t e(𝜆−𝜅)s ds⎥
+
⎨
⎢ 𝜕𝜉
⎥
𝜕𝜉
i=1 j=1 ⎪
⎣
⎦
⎩
i≠j
×
n
∏
[
t
∫0
k=1
k≠i,j
MXi (𝜉)𝜆e
−𝜆(t−s)
]⎫
⎪
ds ⎬ .
⎪
⎭
By substituting 𝜉 = 0 and taking note that MX (0) = 1,
𝜕 2 MX (0)
𝜕𝜉
2
𝜕MX (0)
𝜕𝜉
= 𝔼(X) = 𝜇 and
= 𝔼(X 2 ) = 𝜇2 + 𝜎 2 , we therefore have
𝔼(Yt ) =
𝜕MY (0, t)
𝜕𝜉
⎧[
] n [ t
]⎫
n
t
∑
∏
⎪
⎪
−𝜆t (𝜆−𝜅)s
−𝜆(t−s)
ds
𝜆e
ds ⎬
=
⎨ ∫ 𝜇𝜆e e
∫0
0
⎪
i=1 ⎪
j=1
⎭
⎩
j≠i
)(
(
)
(
)
n−1
𝜆
e−𝜅t − e−𝜆t 1 − e−𝜆t
= n𝜇
𝜆−𝜅
and
𝔼(Yt2 ) =
𝜕 2 MY (0, t)
𝜕𝜉 2
⎧[
] n [ t
]⎫
n
t(
∑
∏
) −𝜆t (𝜆−2𝜅)s
⎪
⎪
2
2
−𝜆(t−s)
=
ds
𝜆e
ds ⎬
⎨ ∫ 𝜇 + 𝜎 𝜆e e
∫
0
0
⎪
i=1 ⎪
j=1
⎭
⎩
j≠i
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⎧
]⎫
n
n [
n
t
∑
∑
][
]∏
⎪
⎪[
−𝜆t (𝜆−𝜅)s
−𝜆t (𝜆−𝜅)s
−𝜆(t−s)
+
ds 𝜇𝜆e e
ds
𝜆e
ds ⎬
⎨ 𝜇𝜆e e
∫0
⎪
i=1 j=1 ⎪
k=1
⎭
k≠i,j
⎩
i≠j
(
)(
)
𝜆
= n(𝜇2 + 𝜎 2 )
e−2𝜅t − e−𝜆t (1 − e−𝜆t )n−1
𝜆 − 2𝜅
)2 (
(
)2 (
)n−2
𝜆
e−2𝜅t − e−𝜆t 1 − e−𝜆t
+ n(n − 1)𝜇 2
.
𝜆 − 2𝜅
Therefore,
( )
( )
( )2
Var Yt = 𝔼 Yt2 − 𝔼 Yt
(
)(
)(
)n−1
𝜆
= n(𝜇2 + 𝜎 2 )
e−2𝜅t − e−𝜆t 1 − e−𝜆t
𝜆 − 2𝜅
(
)2 (
)2 (
)n−2
𝜆
e−2𝜅t − e−𝜆t 1 − e−𝜆t
+ n(n − 1)𝜇 2
𝜆 − 2𝜅
)2 (
(
)2 (
)2(n−1)
𝜆
e−𝜅t − e−𝜆t 1 − e−𝜆t
− n2 𝜇 2
.
𝜆−𝜅
◽
18. Let {X1 , X2 , . . . , } be a series of independent and identically distributed random variables
with moment generating function
( )
MX (𝜉) = 𝔼 e𝜉X ,
𝜉∈ℝ
d
where X = Xi , i = 1, 2, . . . and let {t1 , t2 , . . .} be its corresponding sequence of random
jump times of a Poisson process {Nt ∶ t ≥ 0} with intensity parameter 𝜆 > 0 where ti ∼
Exp(𝜆), i = 1, 2, . . . In addition, let the sequence X1 , X2 , . . . be independent of Nt . Show
that for 𝜅 > 0 the process {Yt ∶ t ≥ 0} with initial condition Y0 = 0 given as
Yt =
Nt
∑
e−𝜅(t−ti ) Xi
i=1
has moment generating function MY (𝜉, t) = 𝔼(e𝜉Yt ) given by
{
MY (𝜉, t) = exp
t
𝜆
∫0
(
)
MX (𝜉e−𝜅s ) − 1 ds
and
)
( ) 𝜆𝜇 (
𝔼 Yt =
1 − e−𝜅t
𝜅
( )
)(
)
𝜆 ( 2
𝜇 + 𝜎 2 1 − e−2𝜅t
Var Yt =
2𝜅
( )
( )
where 𝔼 Xi = 𝜇 and Var Xi = 𝜎 2 for i = 1, 2, . . .
}
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273
Solution: Given the mutual independence of the random variables Xi and Xj , i ≠ j the
conditional expectation of the process Yt given the first jump time, t1 = s is
(
{ N
)
}|
t
(
)
∑
|
𝜉Yt |
−𝜅(t−ti )
|
𝔼 e | t1 = s = 𝔼 exp
𝜉
e
Xi | t 1 = s
|
|
i=1
|
}|
(
{ N
)
t
(
)
∑
|
{ −𝜅(t−s) }|
= 𝔼 exp 𝜉e
X1 | t1 = s 𝔼 exp 𝜉
e−𝜅(t−ti ) Xi || t1 = s .
|
|
i=2
|
From the properties of the Poisson process, the sum conditioned on t1 = s has the same
unconditional distribution as the original sum, starting with the first jump i = 1 until time
t − s, and we therefore have
)
(
})
(
{
|
𝔼 e𝜉Yt | t1 = s = 𝔼 exp 𝜉e−𝜅(t−s) Xt MY (𝜉, t − s)
|
= MX (𝜉e−𝜅(t−s) )MY (𝜉, t − s).
Since the first jump time is exponentially distributed, t1 ∼ Exp(𝜆) and from the tower
property
[ (
)]
|
MY (𝜉, t) = 𝔼 𝔼 e𝜉Yt | t1 = s
|
t (
)
𝔼 e𝜉Yt |t1 = s 𝜆e−𝜆s ds
=
∫0
t
=
MX (𝜉e−𝜅(t−s) )MY (𝜉, t − s)𝜆e−𝜆s ds
∫0
t
=
∫0
MX (𝜉e−𝜅s )MY (𝜉, s)𝜆e−𝜆(t−s) ds.
Differentiating the integral with respect to t, we have
t
𝜕MY (𝜉, t)
M (𝜉e−𝜅s )MY (𝜉, s)𝜆e−𝜆(t−s) ds
= MX (𝜉e−𝜅t )MY (𝜉, t)𝜆 − 𝜆
∫0 X
𝜕t
= 𝜆(MX (𝜉e−𝜅t ) − 1)MY (𝜉, t)
and solving the first-order differential equation,
)
(
t{
}
MX (𝜉e−𝜅s ) − 1 ds .
MY (𝜉, t) = exp 𝜆
∫0
Setting 𝜉 = 𝜉e−𝜅s and differentiating MY (𝜉, t) with respect to 𝜉, we have
(
t{
( )
} )
𝜕MY (𝜉, t)
𝜕
MX 𝜉 − 1 ds MY (𝜉, t)
=
𝜆
∫0
𝜕𝜉
𝜕𝜉
{
}
t
−𝜅s 𝜕MX (𝜉)
e
=𝜆
ds MY (𝜉, t)
∫0
𝜕𝜉
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and differentiating the above equation with respect to 𝜉,
}
}
{
{
t
t
𝜕MY (𝜉, t)
𝜕2 MY (𝜉, t)
𝜕
−𝜅s 𝜕MX (𝜉)
−𝜅s 𝜕MX (𝜉)
ds MY (𝜉, t) + 𝜆
ds
=𝜆
e
e
2
∫
∫
𝜕𝜉
𝜕𝜉
𝜕𝜉
0
0
𝜕𝜉
𝜕𝜉
}
}
{
{
t
t
𝜕MY (𝜉, t)
𝜕 2 MX (𝜉)
𝜕MX (𝜉)
ds
e−2𝜅s
ds
M
(𝜉,
t)
+
𝜆
e−𝜅s
.
=𝜆
Y
2
∫0
∫0
𝜕𝜉
𝜕𝜉
𝜕𝜉
By substituting 𝜉 = 0, and since MY (0, t) = 1, we therefore have
( ) 𝜕MY (0, t) 𝜆𝜇 (
)
𝔼 Yt =
1 − e−𝜅t
=
𝜕𝜉
𝜅
( ) 𝜕 2 MY (0, t)
)(
)
𝜆 ( 2
=
𝔼 Yt2 =
𝜇 + 𝜎 2 1 − e−2𝜅t + 𝔼(Yt )2
2
2𝜅
𝜕𝜉
and hence
( )
𝜆 2
Var Yt =
(𝜇 + 𝜎 2 )(1 − e−2𝜅t )
2𝜅
( )
( )
where 𝔼 Xi = 𝜇 and Var Xi = 𝜎 2 for i = 1, 2, . . .
◽
19. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt , and let X1 , X2 , . . . be a sequence of independent
and identically distributed random variables which are also independent of Nt . By defining
the compound Poisson process Mt as
Mt =
Nt
∑
Xi ,
t≥0
i=1
show that its quadratic variation is
⟨M, M⟩t = lim
n→∞
n−1 (
∑
)2
Mti+1 − Mti
=
i=0
Nt
∑
Xi2
i=1
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and deduce that dMt ⋅ dMt =
Xt2 dNt where Xt is the jump size if N jumps at t.
Solution: From the definition of quadratic variation
⟨M, M⟩t = lim
n→∞
n−1 (
∑
)2
Mtk+1 − Mtk
k=0
where tk = tk∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ, we can set
2
Nt
N tn
Nt
n−1 ⎛Ntk+1
⎞
k
∑
∑
∑
∑
∑
⎜
⟨M, M⟩t = lim
Xi −
Xi ⎟ = lim
Xi2 =
Xi2
n→∞
n→∞
⎜
⎟
i=Nt
k=0 ⎝ i=1
i=1
i=1
⎠
0
since Nt0 = N0 = 0.
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275
From the definition
lim
n→∞
n−1 (
∑
)2
Mti+1 − Mti
t
=
i=0
and since
Nt
∑
(
dMs
∫0
)2
=
Nt
∑
Xi2
i=1
t
Xi2 =
∫0
i=1
Xu2− dNu
where Xt− is the jump size before a jump event at time t, then by differentiating both sides
with respect to t we finally have dMt ⋅ dMt = Xt2 dNt where Xt is the jump size if N jumps
at t.
◽
20. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt and let X1 , X2 , . . . be a sequence of independent
and identically distributed random variables which are also independent of Nt . By defining
the compound Poisson process Mt as
Mt =
Nt
∑
Xi ,
t≥0
i=1
show that the cross-variation between Mt and Nt is
⟨M, N⟩t = lim
n−1 (
∑
Mti+1
n→∞
Nt
)(
) ∑
− Mti Nti+1 − Nti =
Xi
i=0
i=1
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and deduce that dMt ⋅ dNt =
Xt dNt .
Solution: From the definition of cross-variation, for tk = kt∕n, 0 = t0 < t1 < t2 < . . . <
∑Nt
tn = t, n ∈ ℕ and since we can also write Nt = i=1
1,
⟨M, N⟩t = lim
n→∞
n−1 (
∑
Mtk+1 − Mtk
)(
)
Ntk+1 − Ntk
k=0
Nt
Nt
tk+1
tk+1
n−1 ⎛N∑
⎞
⎞ ⎛N∑
k
k
∑
∑
∑
⎜
⎟
⎜
Xi −
Xi
1−
1⎟
= lim
n→∞
⎟
⎜
⎟ ⎜ i=1
k=0 ⎝ i=1
i=1
i=1 ⎠
⎠⎝
N
= lim
n→∞
=
Nt
∑
tn
∑
i=Nt
Xi ⋅ 1
0
Xi
i=1
since Nt0 = N0 = 0.
Given
n−1 (
)(
)
∑
lim
Mti+1 − Mti Nti+1 − Nti =
n→∞
i=0
t
∫0
dMs ⋅ dNs =
Nt
∑
i=1
Xi
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Nt
∑
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t
Xi =
Xs− dNs
∫0
i=1
where Xt− is the jump size before a jump event at time t, then by differentiating the integrals
with respect to t we have dMt ⋅ dNt = Xt dNt where Xt is the jump size if N jumps at t.
◽
21. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt . Show that its quadratic variation is
⟨N, N⟩t = lim
n→∞
n−1 (
∑
)2
Nti+1 − Nti
= Nt
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and deduce that dNt ⋅ dNt = dNt .
Solution: Since Nt is a counting process, we can define the compound Poisson process
Mt as
Nt
∑
Xi
Mt =
i=1
where Xi is a sequence of independent and identically distributed random variables which
are also independent of Nt . Using the results in Problem 5.2.1.19 (page 274),
⟨M, M⟩t =
Nt
∑
Xi2 .
i=1
By setting Xi = 1 for all i = 1, 2, . . . , Nt , then Mt = Nt which implies
⟨N, N⟩t =
Nt
∑
1 = Nt .
i=1
By definition,
⟨N, N⟩t = lim
n→∞
n−1 (
∑
)2
Nti+1 − Nti
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and since
n−1 (
)2
∑
Nti+1 − Nti =
lim
n→∞
i=0
t
∫0
(
dNs
)2
= Nt
then, by differentiating both sides with respect to t, we have dNt ⋅ dNt = dNt .
◽
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277
22. Let {Wt ∶ t ≥ 0} be a standard Wiener process and {Nt ∶ t ≥ 0} be a Poisson process with
intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ). Show that the cross-variation
between Wt and Nt is
n−1 (
∑
⟨W, N⟩t = lim
n→∞
Wti+1 − Wti
)(
)
Nti+1 − Nti = 0
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and deduce that dWt ⋅ dNt = 0.
Solution: Since
| n−1
|∑
|
|
| n−1
|
|
| ||∑
| (W − W )(N − N )| ≤ max ||W
|
−
W
(N
−
N
)
|
ti+1
ti
ti+1
ti |
t
t
t
t
|
|
|
k+1
k
i+1
i
||
| i=0
| 0≤k≤n−1 |
|
i=0
|
|
|
|
and because Wt is continuous, we have
|
|
lim max |Wtk+1 − Wtk | = 0.
|
n→∞ 0≤k≤n−1 |
Therefore, we conclude that
|
n−1 (
)(
)||
∑
|
| lim
|
−
W
−
N
W
N
ti+1
ti
ti+1
ti | ≤ 0
|n→∞
|
|
i=0
|
|
and hence
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
Nti+1 − Nti = 0.
i=0
Finally, because
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
Nti+1 − Nti =
i=0
t
∫0
dWs ⋅ dNs = 0
then, by differentiating both sides with respect to t, we can deduce that dWt ⋅ dNt = 0.
◽
23. Let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt and let X1 , X2 , . . . be a sequence of independent
and identically distributed random variables which are also independent of Nt . By defining
the compound Poisson process Mt as
Mt =
Nt
∑
i=1
Xi ,
t≥0
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show that the cross-variation between Wt and Mt is
n−1 (
∑
⟨W, M⟩t = lim
n→∞
Wti+1 − Wti
)(
)
Mti+1 − Mti = 0
i=0
where ti = it∕n, 0 = t0 < t1 < t2 < . . . < tn = t, n ∈ ℕ and deduce that dWt ⋅ dMt = 0.
Solution: Since we can write
| n−1
|∑
|
)(
)||
| n−1 (
|
| ||∑
|
| ≤ max ||W
|
W
M
−
W
−
M
−
W
(M
−
M
)
|
ti+1
ti
ti+1
ti |
t
t
t
t
|
|
|
k+1
k
i+1
i
||
| i=0
| 0≤k≤n−1 |
|
i=0
|
|
|
|
and because Wt is continuous, we have
|
|
lim max |Wtk+1 − Wtk | = 0.
|
n→∞ 0≤k≤n−1 |
Thus, we conclude that
|
n−1 (
)(
)||
∑
|
| lim
|
−
W
−
M
W
M
ti+1
ti
ti+1
ti | ≤ 0
|n→∞
|
|
i=0
|
|
or
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
Mti+1 − Mti = 0.
i=0
Finally, because
lim
n→∞
n−1 (
∑
Wti+1 − Wti
)(
)
Mti+1 − Mti =
i=0
t
∫0
dWs ⋅ dMs = 0
then, by differentiating both sides with respect to t, we can deduce that dWt ⋅ dMt = 0.
◽
24. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a standard Wiener process relative to the same filtration
ℱt , t ≥ 0. By considering the process
Zt = eu1 Nt +u2 Wt
where u1 , u2 ∈ ℝ, (use)Itō’s formula to find an SDE for Zt .
By setting mt = 𝔼 Zt show that solution to the SDE can be expressed as
]
)
dmt [( u
1
− e 1 − 1 𝜆 + u22 mt = 0.
dt
2
( )
⟂ Wt .
Solve the differential equation to find 𝔼 Zt and deduce that Nt ⟂
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279
Solution: By expanding Zt using Taylor’s theorem and taking note that (dWt )2 = dt,
(dNt )2 = (dNt )3 = · · · = dNt , dWt dNt = 0,
dZt =
2
2
𝜕Zt
𝜕Zt
1 𝜕 Zt
1 𝜕 Zt
2
dNt +
dWt +
(dN
)
+
(dWt )2
t
𝜕Nt
dWt
2 𝜕Nt2
2 𝜕Wt2
3
𝜕 2 Zt
1 𝜕 Zt
(dNt dWt ) +
(dNt )3 + . . .
𝜕Nt 𝜕Wt
3! 𝜕Nt3
(
)
1
1
1
= u1 + u21 + u31 + . . . Zt dNt + u2 Zt dWt + u22 Zt dt
2
3!
2
1
= (eu1 − 1)Zt dNt + u2 Zt dWt + u22 Zt dt.
2
+
Taking integrals, we have
t
t
t
t
1
Z dWs + u22
Z ds
∫0 s
∫0
∫0
2 ∫0 s
t
t
] t
[
1
̂s + u2
Zs d N
Zs dWs + 𝜆 (eu1 − 1) + u22
Z ds
Zt − 1 = (eu1 − 1)
∫0 s
∫0
∫0
2
dZs = (eu1 − 1)
Zs dNs + u2
̂t = Nt − 𝜆t. Taking expectations,
where Z0 = 1 and N
] t
[
1
𝔼(Zt ) = 1 + 𝜆 (eu1 − 1) + u22
𝔼(Zs ) ds
∫0
2
̂t are martingales we therefore have
where since both Wt and N
)
( t
)
( t
̂
Z dWs = 0 and 𝔼
Z dN = 0.
𝔼
∫0 s s
∫0 s
By differentiating the integral equation,
]
[
d
1
𝔼(Zt ) = 𝜆 (eu1 − 1) + u22 𝔼(Zt )
dt
2
or
]
dmt [ u
1
− 𝜆 (e 1 − 1) + u22 mt = 0
dt
2
where mt = 𝔼(Zt ).
1 2
1 2
By setting the integrating factor to be I = e− ∫ (𝜆(e −1)+ 2 u2 ) dt = e−(𝜆(e −1)+ 2 u2 )t and multiplying the differential equation with I, we have
(
)
1 2
1 2
(
)
u1
u1
d
mt e−𝜆t(e −1)− 2 u2 t = 0 or e−𝜆t(e −1)− 2 u2 t 𝔼 eu1 Nt +u2 Wt = C
dt
)
(
where C is a constant. Since 𝔼 eu1 N0 +u2 W0 = 1 therefore C = 1, and hence we finally
obtain
1 2
)
(
u1
𝔼 eu1 Nt +u2 Wt = e𝜆t(e −1)+ 2 u2 t .
u1
u1
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Since the joint moment generating function of
1 2
(
)
u
𝔼 eu1 Nt +u2 Wt = e𝜆t(e 1 −1) ⋅ e 2 u2 t
can be expressed as a product of the moment generating functions for Nt and Wt , respectively, we can deduce that Nt and Wt are independent.
◽
25. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a standard Wiener process relative to the same filtration
ℱt , t ≥ 0. Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables which are also independent of Nt and Wt . By defining a compound Poisson
process Mt as
Nt
∑
Mt =
Xi , t ≥ 0
i=1
and by considering the process
Zt = eu1 Mt +u2 Wt
where u1 , u2 ∈ ℝ, (use)Itō’s formula to find an SDE for Zt .
By setting mt = 𝔼 Zt show that solution to the SDE can be expressed as
]
)
dmt [( 𝜑 (u )
1
− e X 1 − 1 𝜆 + u22 mt = 0
dt
2
( uX)
where 𝜑X (u1 ) = 𝔼 e 1 t is the moment generating function of Xt , which is the jump size
if N jumps at time t.
( )
⟂ Wt .
Finally, solve the differential equation to find 𝔼 Zt and deduce that Mt ⟂
Solution: By expanding Zt using Taylor’s theorem and taking note that (dWt )2 = dt,
dMt dWt = 0, (dNt )k = dNt and (dMt )k = Xtk dNt for k = 1, 2, . . . ,
2
2
𝜕Zt
𝜕Zt
1 𝜕 Zt
1 𝜕 Zt
2
dMt +
dWt +
(dM
)
+
(dWt )2
t
𝜕Mt
dWt
2 𝜕Mt2
2 𝜕Wt2
dZt =
3
𝜕 2 Zt
1 𝜕 Zt
(dMt dWt ) +
(dMt )3 + . . .
𝜕Mt 𝜕Wt
3! 𝜕Mt3
)
(
1
1
1
= u1 Xt + u21 Xt2 + u31 Xt3 + . . . Zt dNt + u2 Zt dWt + u22 Zt dt
2
3!
2
(
)
1
= eu1 Xt − 1 Zt dNt + u2 Zt dWt + u22 Zt dt.
2
+
Integrating, we have
t
∫0
dZs =
Zt − 1 =
t
(
t
(
)
eu1 Xt − 1 Zs dNs + u2
∫0
∫0
eu1 Xt
t
t
1
Z dWs + u22
Z ds
∫0 s
2 ∫0 s
t
t[ (
)
) 1 ]
̂s + u2
𝜆 eu1 Xt − 1 + u22 Zs ds
− 1 Zs d N
Zs dWs +
∫0
∫0
2
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281
̂t = Nt − 𝜆t. Taking expectations and because the jump size variable
where Z0 = 1 and N
Xt is independent of Nt and Wt , we have
[ ( (
)
) 1 ] t ( )
𝔼(Zt ) = 1 + 𝜆 𝔼 eu1 Xt − 1 + u22
𝔼 Zs ds
∫0
2
( t
)
( t
)
̂s = 0.
since 𝔼
Zs dWs = 0 and 𝔼
Zs dN
∫0
∫0
By differentiating the integral equation,
[ ( (
)
) 1 ] ( )
d ( )
𝔼 Zt = 𝜆 𝔼 eu1 Xt − 1 + u22 𝔼 Zt
dt
2
or
) 1 ]
dmt [ (
− 𝜆 𝜑X (u1 ) − 1 + u22 mt = 0
dt
2
( )
( uX)
where mt = 𝔼 Zt and 𝜑X (u1 ) = 𝔼 e 1 t .
1 2
1 2
By setting the integrating factor to be I = e− ∫ (𝜆(𝜑X (u1 )−1)+ 2 u2 ) dt = e−(𝜆(𝜑X (u1 )−1)+ 2 u2 )t and
multiplying the differential equation with I, we have
(
)
1 2
1 2
d
mt e−𝜆t(𝜑X (u1 )−1)− 2 u2 t = 0 or e−𝜆t(𝜑X (u1 )−1)− 2 u2 t 𝔼(eu1 Mt +u2 Wt ) = C
dt
(
)
where C is a constant. Since 𝔼 eu1 M0 +u2 W0 = 1 therefore C = 1, and hence we finally
obtain
1 2
(
)
𝔼 eu1 Mt +u2 Wt = e𝜆t(𝜑X (u1 )−1)+ 2 u2 t .
Since the joint moment generating function of
1 2
)
(
𝔼 eu1 Mt +u2 Wt = e𝜆t(𝜑X (u1 )−1) ⋅ e 2 u2 t
can be expressed as a product of the moment generating functions for Mt and Wt , respectively, we can deduce that Mt and Wt , are independent.
◽
5.2.2
Jump Diffusion Process
1. Pure Jump Process. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson
process with intensity 𝜆 > 0 relative to the filtration ℱt , t ≥ 0. Suppose St follows a pure
jump process
dSt
= (Jt − 1)dNt
St −
where Jt is the jump size variable if N jumps at time t, Jt ⟂
⟂ Nt and
{
1 with probability 𝜆dt
dNt =
0 with probability 1 − 𝜆dt.
Explain why the term in dNt is Jt − 1 and not Jt .
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Show that the above differential equation can also be written as
dSt
= dMt
St −
∑Nt
where Mt = i=1
(Ji − 1) is a compound Poisson process such that Ji , i = 1, 2, . . . is a
sequence of independent and identically distributed random variables which are also independent of Nt .
Solution: Let St− be the value of St just before a jump and assume there occurs an instantaneous jump (i.e., dNt = 1) in which St changes from St− to Jt St− where Jt is the jump
size. Thus,
dSt = Jt St− − St− = (Jt − 1)St−
or
dSt
= (Jt − 1).
St−
Therefore, we can write
where
dSt
= (Jt − 1)dNt
St −
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆 dt.
Given the compound Poisson process
Mt =
Nt
∑
(Ji − 1)
i=1
we let Mt− be the value of Mt just before a jump event. If N jumps at time t then
dMt = Mt − Mt− = Jt − 1.
Thus, in general, we can write
dMt = (Jt − 1)dNt
which implies the pure jump process can also be expressed as
dSt
= dMt .
St −
◽
2. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 relative to the filtration ℱt , t ≥ 0. Suppose St follows a pure jump process
dSt
= (Jt − 1)dNt
St −
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283
where Jt is the jump size variable if N jumps at time t and
{
1 with probability 𝜆 dt
0 with probability 1 − 𝜆 dt.
dNt =
Assume Jt follows a lognormal distribution such that log Jt ∼ 𝒩(𝜇J , 𝜎J2 ) and Jt is also
independent of Nt . By applying Itō’s formula on log St and taking integrals show that for
t < T,
NT−t
∏
ST = St
Ji
i=1
provided Ji ∈ (0, 2], i = 1, 2, . . . is a sequence of independent and identically distributed
jump size random variables which are independent of Nt .
Given St and NT−t = n, show that ST follows a lognormal distribution with mean
1 2
(
)
𝔼 ST || St , NT−t = n = St en(𝜇J + 2 𝜎J )
and variance
(
)
2
2
Var ST || St , NT−t = n = St2 (en𝜎J − 1)en(2𝜇J +𝜎J ) .
Finally, given only St , show that
{ (
)
}
1 2
(
)
𝔼 ST || St = St exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t)
and
[
{
(
)}
{
(
)}]
1 2
(
)
2
Var ST || St = St2 exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
.
Solution: By letting St− denote the value of St before a jump event, and expanding
d(log St ) using Taylor’s theorem and taking note that dNt ⋅ dNt = dNt ,
(
(
(
)
)
)
dSt 1 dSt 2 1 dSt 3 1 dSt 4
−
+
−
+ ...
d(log St ) =
St −
2 St −
3 St −
4 St −
{
}
1
1
1
= (Jt − 1) − (Jt − 1)2 + (Jt − 1)3 − (Jt − 1)4 + . . . dNt
2
3
4
= log Jt dNt
provided −1 < Jt − 1 ≤ 1 or 0 < Jt ≤ 2.
By taking integrals, we have
T
∫t
T
d(log Su ) =
(
log
ST
St
)
∫t
∑
log Ju dNu
NT−t
=
i=1
log Ji
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5.2.2
or
∏
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NT−t
ST = St
Ji
i=1
where Ji ∈ (0, 2] is the jump size occurring at time instant ti and NT−t = NT − Nt is the
total number of jumps in the time interval (t, T].
Since log Ji ∼ 𝒩(𝜇J , 𝜎J2 ), i = 1, 2, . . . , NT−t are independent and identically distributed,
and conditional on St and NT−t = n,
∑
NT−t
log ST = log St +
log Ji
i=1
follows a normal distribution. Therefore, the mean and variance of ST are
1 2
(
)
𝔼 ST ||St , NT−t = n = St en(𝜇J + 2 𝜎J )
and
( 2
)
(
)
2
Var ST ||St , NT−t = n = St2 en𝜎J − 1 en(2𝜇J +𝜎J )
respectively.
Finally, conditional only on St , by definition
)
∏ ||
Ji || St
St
i=1 ||
(N
| )
T−t
∏
|
Ji || St
= St 𝔼
i=1 ||
(N
)
T−t
∏
= St 𝔼
Ji
(
)
𝔼 ST || St = 𝔼
(
(
NT−t
i=1
∑NT−t
= St 𝔼 e
i=1
)
log Ji
.
By applying the tower property and from Problem 5.2.1.12 (page 264), we have
)]
[ ( ∑
NT−t
(
)
|
log
J
i
|
|
𝔼 ST | St = St 𝔼 𝔼 e i=1
| NT−t
|
]}
{
[ ( log J )
= St exp 𝜆(T − t) 𝔼 e t − 1
]}
{
[ ( )
= St exp 𝜆(T − t) 𝔼 Jt − 1 .
1 2
Since 𝔼(Jt ) = e𝜇J + 2 𝜎J therefore
{ (
)
}
1 2
𝔼(ST |St ) = St exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t) .
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285
For the case of variance of ST conditional on St , by definition
(
)
∏ ||
Ji || St
Var(ST |St ) = Var St
i=1 ||
)
(N
T−t
∏
2
= St Var
Ji
NT−t
i=1
(
= St2 Var
= St2
e
)
∑NT−t
log Ji
i=1
{ (
)
( ∑
)2 }
∑NT−t
NT−t
2 i=1
log Ji
log
J
i
𝔼 e
− 𝔼 e i=1
{ (
)
( ∑
)2 }
∑NT−t
NT−t
2
.
= St2 𝔼 e i=1 log Ji − 𝔼 e i=1 log Ji
By applying the tower property and from Problem 5.2.1.12 (page 264), we have
)]
{ [ ( ∑
)]}2
[ ( ∑
NT−t
NT−t
log Ji2 ||
log Ji ||
2
i=1
i=1
− St 𝔼 𝔼 e
𝔼 e
| NT−t
| NT−t
|
|
]}
[ ( 2)
[
{
{
]}
2
2
= St exp 𝜆 (T − t) 𝔼 Jt − 1 − St exp 2𝜆 (T − t) 𝔼(Jt ) − 1 .
Var(ST |St ) = St2 𝔼
1 2
Since 𝔼(Jt ) = e𝜇J + 2 𝜎J and 𝔼(Jt2 ) = e2(𝜇J +𝜎J ) therefore
2
)}
{
(
)}]
[
{
(
1 2
2
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
.
Var(ST |St ) = St2 exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
◽
3. Merton’s Model. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson
process with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a standard Wiener process relative to the
same filtration ℱt , t ≥ 0. Suppose St follows a jump diffusion process with the following
SDE
dSt
= (𝜇 − D) dt + 𝜎dWt + (Jt − 1)dNt
St −
{
where
dNt =
1 with probability 𝜆 dt
0 with probability 1 − 𝜆 dt
with constants 𝜇, D and 𝜎 being the drift, continuous dividend yield and volatility, respectively, and Jt the jump variable such that log Jt ∼ 𝒩(𝜇J , 𝜎J2 ). Assume that Jt , Wt and Nt are
mutually independent. In addition, we also let Ji , i = 1, 2, . . . be a sequence of independent and identically distributed jump size random variables which are also independent of
Nt and Wt .
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By applying Itō’s formula on log St and taking integrals show that for t < T,
(
ST = St e
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
Ji
i=1
provided Jt ∈ (0, 2] and WT−t ∼ 𝒩(0, T − t).
Given St and NT−t = n, show that ST follows a lognormal distribution with mean
1 2
(
)
𝔼 ST || St , NT−t = n = St e(𝜇−D)(T−t)+n(𝜇J + 2 𝜎J )
and variance
)
( 2
)
(
2
2
Var ST || St , NT−t = n = St2 e𝜎 (T−t)+n𝜎J − 1 e2(𝜇−D)(T−t)+n(2𝜇J +𝜎J ) .
Finally, conditional only on St , show that
{ (
)
}
1 2
)
(
𝔼 ST || St = St e(𝜇−D)(T−t) exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t)
and
)[
{
(
)}
( 2
)
(
2
Var ST || St = St2 e2(𝜇−D)(T−t) e𝜎 (T−t) − 1 exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
)}]
{
(
1 2
.
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
Solution: By letting St− denote the value of St before a jump event, and expanding
d(log St ) using both Taylor’s theorem and Itō’s lemma,
)
)
)
(
(
(
dSt 1 dSt 2 1 dSt 3 1 dSt 4
d(log St ) =
−
+
−
+ ...
St− 2 St−
3 St −
4 St −
1
= (𝜇 − D) dt + 𝜎dWt + (Jt − 1)dNt − (𝜎 2 dt + (Jt − 1)2 dNt )
2
1
1
3
4
+ (Jt − 1) dNt − (Jt − 1) dNt + . . .
3
4
)
(
1 2
= 𝜇 − D − 𝜎 dt + 𝜎dWt
2
{
}
1
1
1
+ (Jt − 1) − (Jt − 1)2 + (Jt − 1)3 − (Jt − 1)4 + . . . dNt
2
3
4
(
)
1 2
= 𝜇 − D − 𝜎 dt + 𝜎dWt + log Jt dNt
2
provided −1 < Jt − 1 ≤ 1 or 0 < Jt ≤ 2.
By taking integrals, we have
T
∫t
T
d(log Su ) =
∫t
(
1
𝜇 − D − 𝜎2
2
)
T
du +
∫t
T
𝜎 dWu +
∫t
log Ju dNu
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Jump Diffusion Process
(
log
ST
St
)
287
NT−t
)
(
∑
1
= 𝜇 − D − 𝜎 2 (T − t) + 𝜎WT−t +
log Ji
2
i=1
or
(
ST = St e
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
Ji
i=1
where Ji ∈ (0, 2] is the random jump size occurring at time ti and NT−t = NT − Nt is the
total number of jumps in the time interval (t, T].
Since WT−t ∼ 𝒩(0, T − t), log Ji ∼ 𝒩(𝜇J , 𝜎J2 ), i = 1, 2, . . . and by independence,
)
]
[
(
1
log ST |{St , NT−t = n} ∼ 𝒩 log St + 𝜇 − D − 𝜎 2 (T − t) + n𝜇J , 𝜎 2 (T − t) + n𝜎J2 .
2
Therefore, conditional on St and NT−t = n, ST follows a lognormal distribution with mean
1 2
(
)
𝔼 ST ||St , NT−t = n = St e(𝜇−D)(T−t)+n(𝜇J + 2 𝜎J )
and variance
( 2
)
(
)
2
2
Var ST ||St , NT−t = n = St2 e𝜎 (T−t)+n𝜎J − 1 e2(𝜇−D)(T−t)+n(2𝜇J +𝜎J ) .
Finally, given only St and from the mutual independence of a Wiener process and a compound Poisson process, we have
| ]
|
𝔼(ST |St ) = 𝔼 St e
Ji || St
i=1 ||
(N
)
(
)
T−t
∏
𝜇−D− 12 𝜎 2 (T−t)
= St e
𝔼(e𝜎WT−t )𝔼
Ji
[
(
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
(N
)
T−t
∏
(𝜇−D)(T−t)
𝔼
Ji .
= St e
i=1
i=1
1 2
(T−t)
Since 𝔼(e𝜎WT−t ) = e 2 𝜎
, from Problem 5.2.2.2 (page 282) we therefore have
{ (
)
}
1 2
𝔼(ST |St ) = St e(𝜇−D)(T−t) exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t) .
As for the variance of St conditional on St , we can write
| ]
|
Ji || St
Var(ST |St ) = Var St e
i=1 ||
)
(N
(
)
T−t
1 2
∏
( 𝜎W )
2 2 𝜇−D− 2 𝜎 (T−t)
T−t
= St e
Var e
Ji .
Var
[
(
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
i=1
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288
5.2.2
Since Var(e𝜎WT−t ) = (e𝜎 (T−t) − 1)e𝜎
(page 285), we finally have
2
2 (T−t)
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and using the results from Problem 5.2.2.3
[
{
(
)}
2
2
Var(ST |St ) = St2 e2(𝜇−D)(T−t) (e𝜎 (T−t) − 1) exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
)}]
{
(
1 2
.
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
◽
4. Ornstein–Uhlenbeck Process with Jumps. Let (Ω, ℱ, ℙ) be a probability space and
let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a
standard Wiener process relative to the same filtration ℱt , t ≥ 0. Suppose St follows an
Ornstein–Uhlenbeck process with jumps of the form
dSt = 𝜅(𝜃 − St− ) dt + 𝜎dWt + log Jt dNt
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆 dt
where
with 𝜅, 𝜃 and 𝜎 being constant parameters such that 𝜅 is the mean-reversion rate, 𝜃 is
the long-term mean and 𝜎 is the volatility. The random variable Jt is the jump amplitude
such that log Jt ∼ 𝒩(𝜇J , 𝜎J2 ), and Wt , Nt and Jt are mutually independent. Suppose the
sequence of jump amplitudes Ji , i = 1, 2, . . . is independent and identically distributed,
and also independent of Nt and Wt .
Using Itō’s lemma on e𝜅t St show that for t < T,
−𝜅(T−t)
ST = St e
[
+𝜃 1−e
−𝜅(T−t)
]
+
∫t
∑
NT−t
T
𝜎e
−𝜅(T−u)
dWu +
e−𝜅(T−t−ti ) log Ji
i=1
and conditional on St and NT−t = n,
)
[
(
]
𝔼 ST |St , NT−t = n = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t)
)[
(
][
]n−1
𝜆
e−𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
+n𝜇J
𝜆−𝜅
and
Var(ST |St , NT−t = n) =
]
𝜎2 [
1 − e−2𝜅(T−t)
2𝜅
(
) ( 𝜆 ) [ −2𝜅(T−t)
][
]n−1
− e−𝜆(T−t) 1 − e−𝜆(T−t)
e
+n 𝜇J2 + 𝜎J2
𝜆 − 2𝜅
)2 [
(
]2 [
]n−2
𝜆
e−2𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
+n(n − 1)𝜇J2
𝜆 − 2𝜅
(
)2 [
]2 [
]2(n−1)
𝜆
e−𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
−n2 𝜇J2
.
𝜆−𝜅
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289
Finally, conditional on St show also that
(
)
] 𝜆𝜇 [
]
[
𝔼 ST |St = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) + J 1 − e−𝜅(T−t)
𝜅
and
) 𝜎2 [
]
)[
]
(
𝜆 ( 2
1 − e−2𝜅(T−t) +
𝜇J + 𝜎J2 1 − e−𝜅(T−t) .
Var ST |St =
2𝜅
2𝜅
What is the distribution of ST given St ?
Solution: In order to solve the jump diffusion process we note that
d(e𝜅t St ) = 𝜅e𝜅t St dt + e𝜅t dSt + 𝜅 2 e𝜅t St ( dt)2 + . . .
and using Itō’s lemma,
d(e𝜅t St ) = 𝜅𝜃e𝜅t dt + 𝜎e𝜅t dWt + e𝜅t log Jt dNt .
For t < T the solution is represented by
]
[
ST = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
Since
T
e−𝜅(T−u) log Ju dNu =
∫t
T
∫t
NT
∑
T
𝜎e−𝜅(T−u) dWu +
∑
∫t
e−𝜅(T−u) log Ju dNu .
NT−t
e−𝜅(T−ti ) log Ji =
i=Nt
e−𝜅(T−t−ti ) log Ji
i=1
where Ji is the random jump size occurring at time ti and NT−t = NT − Nt is the total
number of jumps in the time interval (t, T], therefore
[
]
ST = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) +
∫t
∑
NT−t
T
𝜎e−𝜅(T−u) dWu +
e−𝜅(T−t−ti ) log Ji .
i=1
Given St , from the properties of Itō’s integral we have
[
𝔼
T
∫t
]
𝜎e−𝜅(T−u) dWu = 0
and
[(
𝔼
)2 ]
T
∫t
𝜎e
−𝜅(T−u)
dWu
[
=𝔼
T
∫t
𝜎 e
2 −2𝜅(T−u)
]
]
𝜎2 [
du =
1 − e−2𝜅(T−t) .
2𝜅
Thus, given St and NT−t = n, from Problem 5.2.1.17 (page 269) we can easily obtain
]
[N
|
T−t
)
(
∑
][
]n−1
|
𝜆 [ −𝜅(T−t)
e−𝜅(T−t−ti ) log Ji ||St , NT−t = n = n𝜇J
− e−𝜆(T−t) 1 − e−𝜆(T−t)
e
𝔼
𝜆
−
𝜅
|
i=1
|
Page 289
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290
5.2.2
and
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Jump Diffusion Process
]
|
|
e−𝜅(T−t−ti ) log Ji || St , NT−t = n
Var
|
i=1
|
(
)[
][
]n−1
𝜆
= n(𝜇J2 + 𝜎J2 )
e−2𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
𝜆 − 2𝜅
)2
(
𝜆
+ n(n − 1)𝜇J2
𝜆 − 2𝜅
)2
(
[ −2𝜅(T−t)
]2 [
]n−2
𝜆
× e
− e−𝜆(T−t) 1 − e−𝜆(T−t)
− n2 𝜇J2
𝜆−𝜅
] [
]
[ −𝜅(T−t)
−𝜆(T−t) 2
−𝜆(T−t) 2(n−1)
1−e
× e
−e
.
[N
T−t
∑
Therefore,
)
[
(
]
𝔼 ST |St , NT−t = n = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t)
(
)[
][
]n−1
𝜆
e−𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
+n𝜇J
𝜆−𝜅
and
(
]
) 𝜎2 [
1 − e−2𝜅(T−t)
Var ST |St , NT−t = n =
2𝜅
(
)[
][
]n−1
𝜆
e−2𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
+n(𝜇J2 + 𝜎J2 )
𝜆 − 2𝜅
)2 [
(
]2 [
]n−2
𝜆
+n(n − 1)𝜇J2
e−2𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
𝜆 − 2𝜅
)2 [
(
]2 [
]2(n−1)
𝜆
e−𝜅(T−t) − e−𝜆(T−t) 1 − e−𝜆(T−t)
−n2 𝜇J2
.
𝜆−𝜅
Finally, conditional on St , from Problem 5.2.1.18 (page 272) we have
[N
]
T−t
∑
]
𝜆𝜇J [
−𝜅(T−t−ti )
𝔼
e
log Ji =
1 − e−𝜅(T−t)
𝜅
i=1
and
Var
[N
T−t
∑
i=1
]
e−𝜅(T−t−ti ) log Ji =
)[
]
𝜆 ( 2
𝜇 + 𝜎J2 1 − e−2𝜅(T−t) .
2𝜅 J
Therefore,
(
[
)
] 𝜆𝜇 [
]
𝔼 ST |St = St e−𝜅(T−t) + 𝜃 1 − e−𝜅(T−t) + J 1 − e−𝜅(T−t)
𝜅
and
) 𝜎2 [
]
)[
]
(
𝜆 ( 2
1 − e−2𝜅(T−t) +
𝜇 + 𝜎J2 1 − e−𝜅(T−t) .
Var ST |St =
2𝜅
2𝜅 J
Without the jump component, ST given St is normally distributed (see Problem 3.2.2.10,
∑NT−t −𝜅(T−t−t )
i log J
e
page 132). With the inclusion of a jump component, the term i=1
i
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291
consists of summing over the product of random jump times in the exponent and jump
amplitudes where we do not have an explicit expression for the distribution. Thus, the
distribution of ST conditional on St is not known.
◽
5. Geometric Mean-Reverting Jump Diffusion Model. Let (Ω, ℱ, ℙ) be a probability space
and let {Nt ∶ t ≥ 0} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a
standard Wiener process relative to the same filtration ℱt , t ≥ 0. Suppose St follows a
geometric mean-reverting jump diffusion process of the form
dSt
= 𝜅(𝜃 − log St− ) dt + 𝜎dWt + (Jt − 1)dNt
St −
{
where
1 with probability 𝜆 dt
0 with probability 1 − 𝜆 dt
dNt =
with 𝜅, 𝜃 and 𝜎 being constant parameters. Here, 𝜅 is the mean-reversion rate, 𝜃 is the
long-term mean and 𝜎 is the volatility. The random variable Jt is the jump amplitude
such that log Jt ∼ 𝒩(𝜇J , 𝜎J2 ), and Wt , Nt and Jt are mutually independent. Suppose the
sequence of jump amplitudes Ji , i = 1, 2, . . . is independent and identically distributed,
and also independent of Nt and Wt .
By applying Itō’s formula on Xt = log St , show that
)
(
1
dXt = 𝜅(𝜃 − Xt− ) − 𝜎 2 dt + 𝜎dWt + log Jt dNt .
2
Using Itō’s lemma on e𝜅t Xt and taking integrals show that for t < T,
(
)
]
𝜎2 [
−𝜅(T−t)
log ST = (log St )e
+ 𝜃−
1 − e−𝜅(T−t)
2𝜅
+
∫t
∑
NT−t
T
𝜎e
−𝜅(T−s)
dWs +
e−𝜅(T−t−ti ) log Ji .
i=1
Conditional on St and NT−t = n, show that
(
}
{(
)
] 𝜎2 [
]
𝜎2 [
−𝜅(T−t)
−2𝜅(T−t)
+
𝜃−
1−e
1−e
2𝜅
4𝜅
]n
T
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
ds
)
−𝜅(T−t)
𝔼 ST || St , NT−t = n = Ste
exp
[
× 𝜆
∫t
and
)
}
{ (
]
)
𝜎2 [
2e−𝜅(T−t)
−𝜅(T−t)
|
1−e
Var ST | St , NT−t = n = St
exp 2 𝜃 −
2𝜅
[
{ 2
}
]
{ 2
}
]
]
𝜎 [
𝜎 [
−2𝜅(T−t)
−2𝜅(T−t)
1−e
1−e
× exp
− 1 exp
2𝜅
2𝜅
(
4:29 P.M.
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Chin
292
5.2.2
{[
T
𝜆
×
− 𝜆
e
∫t
[
T
∫t
2𝜇J e−𝜅s +𝜎J2 e−2𝜅s −𝜆(T−t−s)
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
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Jump Diffusion Process
]n
ds
]2n }
ds
.
Finally, given only St , show that
{
𝔼(ST |St ) = St exp
+𝜆
∫t
(
)
] 𝜎2 [
]
1 − e−𝜅(T−t) +
1 − e−2𝜅(T−t)
4𝜅
}
T
−𝜅s 1 2 −2𝜅s
e𝜇J e + 2 𝜎J e
ds − 𝜆(T − t)
−𝜅(T−t)
e
𝜎2
+ 𝜃−
2𝜅
[
and
{
Var(ST |St ) = St2
exp
(
)
}
]
𝜎2 [
−𝜅(T−t)
+2 𝜃−
1−e
2𝜅
}
]
{ 2
}
]
]
[
𝜎 [
− 1 exp
e−𝜆(T−t)
1 − e−2𝜅(T−t)
1 − e−2𝜅(T−t)
2𝜅
]
T
(
) }
−𝜅s 1 2 −2𝜅s
−𝜅s 1 2 −𝜅s
e𝜇J e + 2 𝜎J e − 2 ds − e−𝜆(T−t) .
e𝜇J e + 2 𝜎J e
−𝜅(T−t)
2e
[
{ 2
𝜎
× exp
2𝜅
[
{
× exp 𝜆
∫t
Solution: We first let St− denote the value of St before a jump event and by expanding
d(log St ) using Taylor’s theorem and subsequently applying Itō’s lemma, we have
)
)
)
(
(
(
dSt 1 dSt 2 1 dSt 3 1 dSt 4
−
+
−
+ ...
d(log St ) =
St −
2 St −
3 St −
4 St−
1
= 𝜅(𝜃 − log St− ) dt + 𝜎dWt + (Jt − 1)dNt − [𝜎 2 dt + (Jt − 1)2 dNt ]
2
1
1
+ (Jt − 1)3 dNt − (Jt − 1)4 dNt + . . .
3
4
]
[
1
= 𝜅(𝜃 − log St− ) − 𝜎 2 dt + 𝜎dWt + log Jt dNt
2
provided the random jump amplitude −1 < Jt − 1 ≤ 1 or 0 < Jt ≤ 2.
Setting Xt = log St− we can redefine the above SDE into an Ornstein–Uhlenbeck process
with jumps
dXt = 𝜅(𝜃 − Xt− ) dt + 𝜎dWt + log Jt dNt
and applying Taylor’s theorem and then Ito’s lemma on d(e𝜅t Xt ), we have
d(e𝜅t Xt ) = 𝜅e𝜅t Xt− dt + e𝜅t dXt + 𝜅 2 e𝜅t Xt− ( dt)2 + . . .
= 𝜅𝜃e𝜅t dt + 𝜎e𝜅t dWt + e𝜅t log Jt dNt .
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293
Taking integrals for t < T,
T
∫t
d(e𝜅s Xs ) =
T
(
1
𝜅𝜃e𝜅s − 𝜎 2 e𝜅s
2
∫t
)
T
ds +
𝜎e𝜅s dWs +
∫t
T
∫t
e𝜅s log Js dNs
or
)
(
]
𝜎2 [
1 − e−𝜅(T−t) +
XT = Xt e−𝜅(T−t) + 𝜃 −
∫t
2𝜅
∑
NT−t
T
𝜎e−𝜅(T−s) dWs +
e−𝜅(T−t−ti ) log Ji .
i=1
Substituting XT = log ST and Xt = log St , we finally have
(
)
]
𝜎2 [
−𝜅(T−t)
log ST = (log St )e
+ 𝜃−
1 − e−𝜅(T−t)
2𝜅
𝜎e−𝜅(T−s) dWs +
∫t
−𝜅(T−t)
exp
{(
)
]
𝜎2 [
𝜃−
1 − e−𝜅(T−t)
2𝜅
∫t
∑
NT−t
T
+
e−𝜅(T−t−ti ) log Ji
i=1
or
ST = Ste
∑
NT−t
T
+
𝜎e
−𝜅(T−s)
dWs +
−𝜅(T−t−ti )
e
log Ji
}
.
i=1
To find the expectation and variance of ST given St , we note that from Itō’s integral
)
(
T
]
𝜎2 [
−𝜅(T−u)
−2𝜅(T−t)
1−e
𝜎e
dWu ∼ 𝒩 0,
∫t
2𝜅
and hence
[
{
𝔼 exp
∫t
and
}]
T
𝜎e−𝜅(T−u) dWu
{
= exp
}
]
𝜎2 [
1 − e−2𝜅(T−t)
4𝜅
}]
[
{ T
−𝜅(T−u)
𝜎e
dWu
Var exp
∫t
}
]
{ 2
}
[
{ 2
]
]
𝜎 [
𝜎 [
− 1 exp
.
= exp
1 − e−2𝜅(T−t)
1 − e−2𝜅(T−t)
2𝜅
2𝜅
Conditional on NT−t = n, from Problem 5.2.1.17 (page 269) and by setting 𝜉 = 1, X =
log J and due to the independence of log Ji , i = 1, 2, . . . , n, we can write
[
{ n
}]
}
n {
T
∑
∏
−𝜅(T−t−ti )
−𝜅s
−𝜆(T−t−s)
𝔼 exp
=
e
log Ji
Mlog Ji (e ) 𝜆e
ds
∫t
i=1
i=1
[
= 𝜆
T
∫t
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
]n
ds
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Chin
294
5.2.2
and
[
{
𝔼 exp
2
n
∑
}]
−𝜅(T−t−ti )
e
=
log Ji
i=1
n
∏
{
[
T
= 𝜆
V3 - 08/23/2014
}
−𝜅s
Mlog Ji (2e
) 𝜆e
−𝜆(T−t−s)
2𝜇J e−𝜅s +𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
∫t
4:29 P.M.
Jump Diffusion Process
T
∫t
i=1
c05.tex
ds
]n
ds .
Therefore,
[
{ n
}]
∑
−𝜅(T−t−ti )
e
log Ji
Var exp
i=1
[
{
= 𝔼 exp
2
n
∑
}]
[
{
− 𝔼 exp
e−𝜅(T−t−ti ) log Ji
n
∑
i=1
[
= 𝜆
T
∫t
}]2
e−𝜅(T−t−ti ) log Ji
i=1
2𝜇J e−𝜅s +𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
]n
ds
[
T
− 𝜆
∫t
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
]2n
.
ds
Conditional on St and NT−t = n and due to the independence of Wt and Nt , we can easily
show that
{(
)
}
)
(
] 𝜎2 [
]
𝜎2 [
e−𝜅(T−t)
−𝜅(T−t)
−2𝜅(T−t)
|
exp
𝜃−
𝔼 ST | St , NT−t = n = St
1−e
1−e
+
2𝜅
4𝜅
]n
[
T
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s −𝜆(T−t−s)
e
ds
× 𝜆
∫t
and
{ (
}
)
(
)
−𝜅(T−t)
𝜎2
Var ST || St , NT−t = n = St2e
exp 2 𝜃 −
[1 − e−𝜅(T−t) ]
2𝜅
[
{ 2
}
]
{ 2
}
]
]
𝜎 [
𝜎 [
× exp
1 − e−2𝜅(T−t)
1 − e−2𝜅(T−t)
− 1 exp
2𝜅
2𝜅
{[
]
n
T
−𝜅s
2 −2𝜅s
e2𝜇J e +𝜎J e −𝜆(T−t−s) ds
×
𝜆
∫t
[
]2n }
T
− 𝜆
∫t
−𝜅s + 1 𝜎 2 e−2𝜅s −𝜆(T−t−s)
2 J
e𝜇J e
ds
.
Finally, by treating NT−t as a random variable, from Problem 5.2.1.18 (page 272) and by
setting 𝜉 = 1 and X = log J, we can express
}]
[
{N
{
}
T−t
T[
∑
]
−𝜅(T−t−ti )
−𝜅s
= exp 𝜆
Mlog J (e ) − 1 ds
e
log Ji
𝔼 exp
∫t
i=1
{
= exp
T
𝜆
∫t
−𝜅s + 1 𝜎 2 e−2𝜅s
2 J
e𝜇J e
}
ds − 𝜆(T − t)
Page 294
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295
and
[
{
𝔼 exp
∑
}]
NT−t
2
e
−𝜅(T−t−ti )
{
i=1
T
𝜆
= exp
log Ji
∫t
{
]
Mlog J (2e−𝜅s ) − 1 ds
T
𝜆
= exp
[
2𝜇J e−𝜅s +𝜎J2 e−2𝜅s
e
∫t
}
}
ds − 𝜆(T − t)
and subsequently
[
Var exp
{N
T−t
∑
}]
−𝜅(T−t−ti )
e
log Ji
i=1
[
{
= 𝔼 exp
}]
2
−𝜅(T−t−ti )
e
log Ji
[
− 𝔼 exp
i=1
{
= exp
∑
NT−t
T
𝜆
∫t
{
− exp 2𝜆
2𝜇J e−𝜅s +𝜎J2 e−2𝜅s
e
T
T
∫t
e
log Ji
ds − 𝜆(T − t)
}
ds − 2𝜆(T − t)
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s
e
}]2
−𝜅(T−t−ti )
i=1
}
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s
e
∫t
[
{
−𝜆(T−t)
exp 𝜆
=e
{N
T−t
∑
(
e
𝜇J e−𝜅s + 12 𝜎J2 e−𝜅s
)
− 2 ds
]
}
−𝜆(T−t)
−e
.
Conditional only on St and from the independence of Wt and Nt , we have
)
{(
] 𝜎2 [
]
𝜎2 [
1 − e−𝜅(T−t) +
1 − e−2𝜅(T−t)
𝜃−
2𝜅
4𝜅
}
T
−𝜅s 1 2 −2𝜅s
e𝜇J e + 2 𝜎J e
ds − 𝜆(T − t)
−𝜅(T−t)
𝔼(ST |St ) = Ste
+𝜆
∫t
exp
and
}
{ (
)
]
𝜎2 [
−𝜅(T−t)
exp 2 𝜃 −
1−e
2𝜅
[
{ 2
}
]
{ 2
}
]
]
𝜎 [
𝜎 [
−2𝜅(T−t)
−2𝜅(T−t)
× exp
−
1
exp
e−𝜆(T−t)
1−e
1−e
2𝜅
2𝜅
]
[
{
T
(
) }
𝜇J e−𝜅s + 12 𝜎J2 e−2𝜅s
𝜇J e−𝜅s + 12 𝜎J2 e−𝜅s
−𝜆(T−t)
e
.
× exp 𝜆
e
− 2 ds − e
∫t
−𝜅(T−t)
Var(ST |St ) = St2e
◽
6. Kou’s Model. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ t ≥ 0} be a Poisson process
with intensity 𝜆 > 0 and {Wt ∶ t ≥ 0} be a standard Wiener process relative to the same
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filtration ℱt , t ≥ 0. Suppose St follows a jump diffusion process with the following SDE
dSt
= (𝜇 − D) dt + 𝜎dWt + (Jt − 1)dNt
St −
where
{
1
dNt =
0
with probability 𝜆dt
with probability 1 − 𝜆dt
with constants 𝜇, D and 𝜎 > 0 being the drift, continuous dividend yield and volatility,
respectively. The jump variable is denoted by Jt , where Xt = log Jt follows an asymmetric
double exponential distribution with density function
{
x≥0
p𝛼e−𝛼x
fXt (x) =
(1 − p)𝛽e𝛽x x > 0
where 0 ≤ p ≤ 1, 𝛼 > 1 and 𝛽 > 0. Assume that Jt , Wt and Nt are mutually independent
and let Ji , i = 1, 2, . . . be a sequence of independent and identically distributed random
variables which are independent of Nt and Wt .
By applying Itō’s formula on log St and taking integrals show that for t < T,
(
ST = St e
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
Ji
i=1
provided the random variable Ji ∈ (0, 2], i = 1, 2, . . . , NT−t and WT−t ∼ 𝒩(0, T − t).
Given St and NT−t = n, show that the mean and variance of ST are
[
]n
)
(
𝛽
𝛼
(𝜇−D)(T−t)
p
𝔼 ST |St , NT−t = n = St e
+ (1 − p)
𝛼−1
𝛽+1
and
)
( 2
(
)
Var ST |St , NT−t = n = St2 e2(𝜇−D)(T−t) e𝜎 (T−t) − 1
{
[
]2 }n
𝛽
𝛽
𝛼
𝛼
.
× p
+ (1 − p)
− p
+ (1 − p)
𝛼−2
𝛽+2
𝛼−1
𝛽+1
Finally, conditional only on St , show that
[
]
)
}
{ (
1 2
𝛽
𝛼
(𝜇−D)(T−t)
p
𝔼(ST |St ) = St e
+ (1 − p)
exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t)
𝛼−1
𝛽+1
and
]
( 2
)[
𝛽
𝛼
Var(ST |St ) = St2 e2(𝜇−D)(T−t) e𝜎 (T−t) − 1 p
+ (1 − p)
𝛼−2
𝛽+2
)}
{
(
)}]
[
{
(
1 2
2
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
.
× exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
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297
Solution: Using the same steps as described in Problem 5.2.2.3 (page 285), for t < T we
can show the solution of the jump diffusion process is
(
ST = St e
)
NT−t
𝜇−D− 12 𝜎 2 (T−t)+𝜎WT−t ∏
Ji
i=1
provided Ji ∈ (0, 2], i = 1, 2, . . . , NT−t .
To find the mean and variance of Jt we note that
( )
( )
𝔼 Jt = 𝔼 eXt
∞
=
ex fXt (x) dx
∫−∞
0
=
∫−∞
=p
∞
(1 − p)𝛽e(1+𝛽)x dx +
p𝛼e(1−𝛼)x dx
∫0
𝛽
𝛼
+ (1 − p)
𝛼−1
𝛽+1
and
(
)
( )
𝔼 Jt2 = 𝔼 e2Xt
∞
=
e2x fXt (x) dx
∫−∞
0
=
∫−∞
=p
∞
(1 − p)𝛽e(2+𝛽)x dx +
∫0
p𝛼e(2−𝛼)x dx
𝛽
𝛼
+ (1 − p)
𝛼−2
𝛽+2
so that
Var (Jt ) = 𝔼(Jt2 ) − 𝔼(Jt )2
[
]2
𝛽
𝛽
𝛼
𝛼
+ (1 − p)
− p
+ (1 − p)
=p
.
𝛼−2
𝛽+2
𝛼−1
𝛽+1
Since WT−t ∼ 𝒩(0, T − t) so that
1 2
(
)
𝔼 e𝜎WT−t = e 2 𝜎 (T−t)
and
) 2
(
) ( 2
Var e𝜎WT−t = e𝜎 (T−t) − 1 e𝜎 (T−t)
and conditional on St , NT−t = n and from independence, we have
n
1 2
(
)∏
)
(
𝔼 ST |St , NT−t = n = St e(𝜇−D)(T−t)− 2 𝜎 (T−t) 𝔼 e𝜎WT−t
𝔼(Ji )
[
(𝜇−D)(T−t)
= St e
i=1
𝛽
𝛼
p
+ (1 − p)
𝛼−1
𝛽+1
]n
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and
n
(
)
(
)∏
2
Var ST |St , NT−t = n = St2 e2(𝜇−D)(T−t)−𝜎 (T−t) Var e𝜎WT−t
Var (Ji )
= St2 e2(𝜇−D)(T−t)
{
×
(
e
𝜎 2 (T−t)
i=1
)
−1
[
]2 }n
𝛽
𝛽
𝛼
𝛼
.
p
+ (1 − p)
− p
+ (1 − p)
𝛼−2
𝛽+2
𝛼−1
𝛽+1
Finally, conditional on St and from Problem 5.2.2.2 (page 282), we can show
)
(N
T−t
∏
(
)
(𝜇−D)(T−t)− 12 𝜎 2 (T−t) ( 𝜎WT−t )
𝔼 e
Ji
𝔼 ST |St = St e
𝔼
i=1
[
= St e(𝜇−D)(T−t) p
]
)
}
{ (
1 2
𝛽
𝛼
+ (1 − p)
exp 𝜆 e𝜇J + 2 𝜎J − 1 (T − t)
𝛼−1
𝛽+1
and
(
Var ST |St
)
2
= St2 e2(𝜇−D)(T−t)−𝜎 (T−t) Var
(
e
𝜎WT−t
)
)
(N
T−t
∏
Var
Ji
i=1
)[
]
𝛽
𝛼
+ (1 − p)
𝛼−2
𝛽+2
[
{
(
)}
{
(
)}]
1 2
2
× exp 𝜆(T − t) e2(𝜇J +𝜎J ) − 1
− exp 2𝜆(T − t) e𝜇J + 2 𝜎J − 1
.
(
= St2 e2(𝜇−D)(T−t) e𝜎
2 (T−t)
−1
p
◽
5.2.3
Girsanov’s Theorem for Jump Processes
1. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Let 𝜂 > 0 and consider the
Radon–Nikodým derivative process
( 𝜂 )Nt
.
Zt = e(𝜆−𝜂)t
𝜆
Show that
dZt
= (𝜆 − 𝜂) dt −
Zt−
(
𝜆−𝜂
𝜆
)
dNt
for 0 ≤ t ≤ T.
Solution: From Taylor’s theorem,
dZt =
2
3
𝜕Zt
𝜕 2 Zt
𝜕Z
1 𝜕 Zt
1 𝜕 Zt
2
(dNt dt) +
(dN
)
+
(dNt )3 + . . .
dt + t dNt +
t
𝜕t
𝜕Nt
𝜕t𝜕Nt
2! 𝜕Nt2
3! 𝜕Nt3
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299
Since dNt dt = 0, (dNt )2 = (dNt )3 = . . . = dNt , we have
𝜕Z
dZt = t dt +
𝜕t
From Zt = e(𝜆−𝜂)t
( 𝜂 )Nt
𝜆
(
2
3
𝜕Zt
1 𝜕 Zt
1 𝜕 Zt
+
+
+ ...
𝜕Nt 2! 𝜕Nt2 3! 𝜕Nt3
)
dNt .
we can express
( 𝜂 )Nt
𝜕Zt
= (𝜆 − 𝜂)Zt
= (𝜆 − 𝜂)e(𝜆−𝜂)t
𝜕t
𝜆
( 𝜂 ) ( 𝜂 )Nt
(𝜂)
𝜕Zt
= e(𝜆−𝜂)t log
= log
Z
𝜕Nt
𝜆
𝜆
𝜆 t
( 𝜂 ) 𝜕Z
[ ( 𝜂 )]2
𝜕 2 Zt
t
=
log
=
log
Zt .
𝜆 𝜕Nt
𝜆
𝜕Nt2
In general, we can deduce that
𝜕 m Zt [ ( 𝜂 )]m
= log
Zt ,
𝜕Ntm
𝜆
m = 1, 2, . . .
Therefore,
}
[ ( 𝜂 )]2
[ ( 𝜂 )]3
1
1
+
+ . . . Zt dNt
dZt = (𝜆 − 𝜂)Zt dt + log
+
log
log
𝜆
2!
𝜆
3!
𝜆
{ (𝜂)
}
log
= (𝜆 − 𝜂)Zt dt + e 𝜆 − 1 Zt dNt
= (𝜆 − 𝜂)Zt dt +
{
(𝜂 )
(
)
𝜂−𝜆
𝜆
Zt dNt .
By letting Zt− denote the value of Zt before a jump event, we have
dZt
= (𝜆 − 𝜂) dt −
Zt−
(
𝜆−𝜂
𝜆
)
dNt .
◽
2. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Let 𝜂 > 0 and consider the
Radon–Nikodým derivative process
Zt = e(𝜆−𝜂)t
( 𝜂 )Nt
𝜆
.
( )
Show that 𝔼ℙ Zt = 1 and Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
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Girsanov’s Theorem for Jump Processes
Solution: From Problem 5.2.3.1 (page 298), Zt satisfies
(
)
dZt
𝜆−𝜂
= (𝜆 − 𝜂) dt −
dNt
Zt −
𝜆
(
)
𝜂−𝜆
=
(dNt − 𝜆 dt)
𝜆
)
(
𝜂−𝜆
d(Nt − 𝜆t)
=
𝜆
)
(
𝜂−𝜆
̂t
dN
=
𝜆
̂t = Nt − 𝜆t. Since N
̂t = Nt − 𝜆t is a ℙ-martingale, therefore
where N
(
a ℙ-martingale.
Taking integrals,
t
∫0
𝜂−𝜆
𝜆
)
̂t is also
N
(
)
𝜂−𝜆
̂u
dN
∫0
𝜆
(
)
t
𝜂−𝜆
̂u
Zt = Z0 +
Z−
dN
∫0 u
𝜆
t
dZu =
Zu−
and taking expectations,
( )
( )
𝔼ℙ Zt = 𝔼ℙ Z0 + 𝔼ℙ
[
(
t
Zu−
∫0
𝜂−𝜆
𝜆
)
]
̂u = 1
dN
)
[ t
(
]
𝜂−𝜆
̂ u = 0. Thus,
̂t being a ℙ-martingale, 𝔼ℙ
dN
since Z0 = 1 and due to N
Zu−
∫0
𝜆
( )
𝔼ℙ Zt = 1 for all 0 ≤ t ≤ T.
To show that Zt is a positive ℙ-martingale, by taking integrals for s < t,
t
∫s
(
t
dZu =
Zu−
∫s
(
t
Zt − Zs =
∫s
Zu−
𝜂−𝜆
𝜆
𝜂−𝜆
𝜆
)
)
̂u
dN
̂u .
dN
Taking expectations under the filtration ℱs , s < t,
ℙ
𝔼
(
)
Zt − Zs |ℱs = 𝔼
ℙ
[
(
t
∫s
Zu−
𝜂−𝜆
𝜆
)
|
̂u || ℱs
dN
|
|
̂t is a ℙ-martingale, therefore
and because N
(
)
𝔼ℙ Zt − Zs |ℱs = 0 or
(
)
𝔼ℙ Zt |ℱs = Zs .
]
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301
( )
In addition,
since Zt > 0 for all 0 ≤ t ≤ T, therefore |Zt | = Zt and hence 𝔼ℙ |Zt | =
(
)
𝔼ℙ Zt = 1 < ∞ for all 0 ≤ t ≤ T. Finally, because Zt is ℱt -adapted, we can conclude
that Zt is a positive ℙ-martingale for all 0 ≤ t ≤ T.
◽
3. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Let 𝜂 > 0 and consider the
Radon–Nikodým derivative process
Zt = e(𝜆−𝜂)t
( 𝜂 )Nt
𝜆
.
By changing the measure ℙ to a measure ℚ such that
(
)
dℚ ||
dℚ ||
ℙ
= Zt
𝔼
ℱt =
|
dℙ |
dℙ ||ℱt
show that under the ℚ measure, Nt ∼ Poisson(𝜂t) with intensity 𝜂 > 0 for 0 ≤ t ≤ T.
Solution: To solve this problem we need to show
(
)
u
𝔼ℚ euNt = e𝜂t(e −1)
which is the moment generating function of a Poisson process, Nt ∼ Poisson(𝜂t),
0 ≤ t ≤ T.
For u ∈ ℝ, using the moment generating function approach,
(
)
|
( uN )
ℚ
ℙ
uNt dℚ |
t
=𝔼
𝔼 e
e
dℙ ||ℱt
[
( )Nt ]
ℙ
uNt (𝜆−𝜂)t 𝜂
=𝔼 e e
𝜆
[
]
𝜂
= e(𝜆−𝜂)t 𝔼ℙ e(u+log( 𝜆 ))Nt
{ (
)}
( )
𝜂
u+log 𝜆
(𝜆−𝜂)t
exp 𝜆t e
−1
=e
since the moment generating function of Nt ∼ Poisson(𝜆t) is
(
)
m
𝔼ℙ emNt = e𝜆t(e −1)
for all m ∈ ℝ and hence
(
)
u
𝔼ℚ euNt = e𝜂t(e −1)
which is the moment generating function of a Poisson process with intensity 𝜂. Therefore,
under the ℚ measure, Nt ∼ Poisson(𝜂t) for 0 ≤ t ≤ T.
◽
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4. Let {Nt ∶ 0 ≤ t ≤ 0} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , t ≥ 0. Suppose 𝜃t is an adapted process,
0 ≤ t ≤ T and 𝜂 > 0. We consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1) = e(𝜆−𝜂)t
and
( 𝜂 )Nt
𝜆
t
1
t 2
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du .
Show that 𝔼ℙ (Zt ) = 1 and Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
Solution: From( Problem
4.2.2.1
)
( (page
) 194) and Problem 5.2.3.2 (page 299), we have
shown that 𝔼ℙ Zt(1) = 1, 𝔼ℙ Zt(2) = 1 and both Zt(1) and Zt(2) are ℙ-martingales for
0 ≤ t ≤ T.
From Itō’s lemma,
(2)
d(Zt(1) Zt(2) ) = Zt(1)
+ Zt(2) dZt(1) + dZt(1) dZt(2)
− dZt
(1)
where Zt(1)
before a jump event. Since dZt(1) dZt(2) = 0 thus, by taking
− is the value of Zt
integrals,
t
∫0
d(Zu(1) Zu(2) ) =
t
∫0
(2)
Zu(1)
− dZu +
Zt(1) Zt(2) = Z0(1) Z0(2) +
t
∫0
t
∫0
Zu(2) dZu(1)
(2)
Zu(1)
− dZu +
t
∫0
Zu(2) dZu(1) .
Taking expectations under the ℙ measure,
(
)
(
)
𝔼ℙ Zt(1) Zt(2) = 𝔼ℙ Z0(1) Z0(2) = 1
since Z0(1) = 1, Z0(2) = 1 and both Zt(1) and Zt(2) are ℙ-martingales.
To show that Zt = Zt(1) ⋅ Zt(2) is a positive ℙ-martingale, by taking integrals for s < t of
(2)
+ Zt(2) dZt(1)
d(Zt(1) Zt(2) ) = Zt(1)
− dZt
we have
Zt(1) Zt(2) = Zs(1) Zs(2) +
t
∫s
(2)
Zu(1)
− dZu +
t
∫s
Zu(2) dZu(1)
and taking expectations with respect to the filtration ℱs , s < t
(
)
|
𝔼ℙ Zt(1) Zt(2) |ℱs = Zs(1) Zs(2)
|
)
)
( t
( t
|
|
(1)
(2) |
(2)
(1) |
ℙ
ℙ
since 𝔼
Z − dZu | ℱs = 𝔼
Z dZu | ℱs = 0.
|
|
∫s u
∫s u
|
|
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303
|
|
In addition, since Zt(1) > 0 and Zt(2) > 0 for all 0 ≤ t ≤ T, therefore |Zt(1) Zt(2) | = Zt(1) Zt(2) and
|
|
)
(
hence 𝔼ℙ ||Zt || = 𝔼ℙ (Zt ) = 1 < ∞ for 0 ≤ t ≤ T. Finally, because Zt is also ℱt -adapted
we can conclude that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
◽
5. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Suppose 𝜃t is an adapted
process, 0 ≤ t ≤ T and 𝜂 > 0. We consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1) = e(𝜆−𝜂)t
and
t
( 𝜂 )Nt
𝜆
1
t 2
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du .
By changing the measure ℙ to measure ℚ such that
(
)
dℚ ||
dℚ ||
ℙ
= Zt
𝔼
ℱt =
|
dℙ |
dℙ ||ℱt
show that, under the ℚ measure, Nt is a Poisson process with intensity 𝜂 > 0, the process
̃ t = Wt +
W
t
∫0
̃ t for 0 ≤ t ≤ T.
𝜃u du is a ℚ-standard Wiener process and Nt ⟂
⟂W
Solution: The first two results are given in Problem 5.2.3.3 (page 301) and Problem
4.2.2.10 (page 205).
To prove the final result we need to show that, under the ℚ measure,
̃
1 2
𝔼ℚ (eu1 Nt +u2 Wt ) = e𝜂t(e 1 −1) ⋅ e 2 u2 t
u
which is a joint product of independent moment generating functions of Nt ∼ Poisson(𝜂t)
̃ t ∼ 𝒩(0, t).
and W
From the definition
(
)
(
)
(
)
̃
̃
̃
𝔼ℚ eu1 Nt +u2 Wt = 𝔼ℙ eu1 Nt +u2 Wt Zt = 𝔼ℙ eu1 Nt Zt(1) ⋅ eu2 Wt Zt(2)
we let
(1)
(2)
Zt = Zt ⋅ Zt
where
(1)
Z t = eu1 Nt Zt(1)
and
(2)
̃
Z t = eu2 Wt Zt(2) .
Using Itō’s lemma, we have
( (1) (2) )
(1)
(2)
(2)
(1)
(1)
(2)
dZ t = d Z t Z t
= Z t dZ t + Z t dZ t + dZ t dZ t
(1)
(1)
where Z t− is the value of Z t before a jump event.
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(1)
For dZ t we can expand using Taylor’s theorem,
(1)
(1)
dZ t
(1)
𝜕Z
𝜕Z
= t dNt + t(1) dZt(1)
𝜕Nt
𝜕Zt
(1)
(1)
⎡ 2 (1)
⎤
𝜕2 Z t
𝜕2Zt
1 ⎢ 𝜕 Zt
(1)
(1) 2 ⎥
2
+
(dN
)
+
2
(dN
dZ
)
+
(dZ
)
t
t
t
t
⎥
2! ⎢ 𝜕Nt2
𝜕Nt 𝜕Zt(1)
𝜕(Zt(1) )2
⎣
⎦
(1)
⎡ 3 (1)
𝜕3 Z t
1 ⎢ 𝜕 Zt
3
(dNt ) + 3
(dNt )2 (dZt(1) )
+
2 𝜕Z (1)
3! ⎢ 𝜕Nt3
𝜕N
t
t
⎣
(1)
+3
(
Because dZt(1) =
(1)
𝜕3Zt
𝜕Nt 𝜕(Zt(1) )2
𝜂−𝜆
𝜆
)
⎤
(1) 3 ⎥
+ ...
(dZ
)
t
⎥
𝜕(Zt(1) )3
⎦
(1)
(dNt )(dZt(1) )2
+
𝜕3Zt
(1)
̂
̂t = dNt − 𝜆 dt, we can write
Zt(1)
= eu1 Nt Zt(1) and dN
− d Nt , Z t
(1)
dZ t = u1 Z t− dNt + eu1 Nt dZt(1)
(
[
)
]
(1)
𝜂−𝜆
1 2 (1)
̂
+
u Z − dNt + 2
u1 Z t− dNt dNt
2! 1 t
𝜆
]
(
[
)
(1)
𝜂−𝜆
1 3 (1)
̂t + . . .
u1 Z t− dNt + 3
u21 Z t− dNt dN
+
3!
𝜆
(
)
[
] (1)
𝜂 − 𝜆 u1 Nt (1) ̂
1
1
= u1 + u21 + u31 + . . . Z t− dNt +
e Zt dNt
2!
3!
𝜆
(
)
] (1)
𝜂−𝜆 [
1
1
̂t
+
u1 + u21 + u31 + . . . Z t− dNt dN
𝜆
2!
3!
)
)
(
(
(1)
(1)
(1)
𝜂−𝜆
𝜂−𝜆
u1
̂
̂t
(eu1 − 1) Z t− dNt dN
= (e − 1) Z t− dNt +
Z t − d Nt +
𝜆
𝜆
)
(
(𝜂)
(1)
(1)
𝜂−𝜆
̂t
(eu1 − 1)Z t− dNt +
=
Z t− d N
𝜆
𝜆
since (dNt )2 = dNt and dNt dt = 0.
(2)
For the case of dZ t , by applying Taylor’s theorem
(2)
(2)
dZ t
(2)
𝜕Z
𝜕Z
̃ t + t dZ (2)
= t dW
t
̃t
𝜕W
𝜕Zt(2)
(2)
(2)
⎡ 2 (2)
⎤
𝜕2Zt
𝜕2Z t
1 ⎢ 𝜕 Zt
(2)
(2) 2 ⎥
2
̃ t) + 2
̃ t )(dZ ) +
+
(dW
(d
W
(dZ
)
+ ...
t
t
⎥
2! ⎢ 𝜕(W
̃ t )2
̃ t 𝜕Z (2)
𝜕W
𝜕(Zt(2) )2
t
⎣
⎦
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305
̃ t = dWt + 𝜃t dt and dZ (2) = −𝜃t Z (2) dWt (see Problem 4.2.2.2, page 196), we
and since dW
t
t
can write
(2)
(2)
(2)
1
dZ t = u22 Z t dt + (u2 − 𝜃t )Z t dWt .
2
̂t = dNt − 𝜆 dt, dNt dWt = 0, dWt dt = 0, dNt dt = 0, (dWt )2 = dt and ( dt)2 = 0,
Since dN
we have
(1)
(2)
1
Z t− dZ t = u22 Z t dt + (u2 − 𝜃t )Z t dWt ,
2
(
)
(𝜂)
(2)
(1)
𝜂−𝜆
̂t
Z t dZ t =
Z t dN
(eu1 − 1)Z t dNt +
𝜆
𝜆
and
(1)
(2)
dZ t dZ t = 0.
̂t + 𝜆 dt, we have
Thus, by letting dNt = dN
(
)
(𝜂)
𝜂−𝜆
1
̂t
Z t dN
(eu1 − 1)Z t dNt +
dZ t = u22 Z t dt + (u2 − 𝜃t )Z t dWt +
2
𝜆
𝜆
( u
)
[
]
𝜂e 1 − 𝜆
1
̂t .
= 𝜂(eu1 − 1) + u22 Z t dt + (u2 − 𝜃t )Z t dWt +
Z t dN
2
𝜆
Taking integrals, we have
]
t
1
𝜂 (eu1 − 1) + u22 Z s ds +
(u − 𝜃s )Z s dWs
∫0
∫0 2
2
)
t ( u1
𝜂e − 𝜆
̂s
+
Z s dN
∫0
𝜆
t
[
Zt = 1 +
̂t and Wt are ℙ-martingales, thus by taking expectations under
where Z 0 = 1, and because N
the ℙ measure we have
( )
𝔼ℙ Z t = 1 +
t
∫0
[
] ( )
1
𝜂 (eu1 − 1) + u22 𝔼ℙ Z s ds.
2
By differentiating the above equation with respect to t,
( ) [
] ( )
d ℙ
1
𝔼 Z t = 𝜂 (eu1 − 1) + u22 𝔼ℙ Z t
dt
2
or
( )
where mt = 𝔼ℙ Z t .
]
dmt [ u
1
− 𝜂 (e 1 − 1) + u22 mt = 0
dt
2
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1 2
By setting the integrating factor to be I = e− ∫ (𝜂(e −1)+ 2 u2 ) dt = e−(𝜂(e
tiplying it by the first-order ordinary differential equation, we have
u1
u1 −1)+ 1 u2 )t
2 2
and mul-
)
(
1 2
1 2
u1
u1
d
mt e−𝜂t(e −1)− 2 u2 t = 0 or e−𝜂t(e −1)− 2 u2 t 𝔼ℙ (Z t ) = C
dt
( )
(
)
̃
where C is a constant. Since 𝔼ℙ Z 0 = 𝔼ℙ eu1 N0 Z0(1) ⋅ eu2 W0 Z0(2) = 1, therefore C = 1.
Thus, we will have
(
)
( )
1 2
u1
̃
𝔼ℚ eu1 Nt +u2 Wt = 𝔼ℙ Z t = e𝜂t(e −1)+ 2 u2 t .
Since the joint moment generating function of
(
)
1 2
u
̃
𝔼ℚ eu1 Nt +u2 Wt = e𝜂t(e 1 −1) ⋅ e 2 u2 t
can be expressed as a product of the moment generating functions for Nt ∼ Poisson(𝜂t)
̃ t are independent.
̃ t ∼ 𝒩(0, t), respectively, we can deduce that Nt and W
and W
◽
6. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , t ≥ 0. Let X1 , X2 , . . . be a sequence of
independent and identically distributed random variables with common probability mass
∑
function ℙ(X = xk ) = p(xk ) > 0, k = 1, 2, . . . , K and Kk=1 ℙ(X = xk ) = 1. Let X1 , X2 , . . .
also be independent of Nt . From the definition of a compound Poisson process
Mt =
Nt
∑
Xi ,
t≥0
i=1
show that Mt and Nt can be expressed as
Mt =
K
∑
xk Nt(k) ,
Nt =
k=1
K
∑
Nt(k)
k=1
(j)
where Nt(k) ∼ Poisson(𝜆tp(xk )) such that Nt(i) ⟂
⟂ Nt , i ≠ j.
Let 𝜂1 , 𝜂2 , . . . , 𝜂K > 0 and consider the Radon–Nikodým derivative process
Zt =
K
∏
Zt(k)
k=1
(
) (k)
𝜂k Nt
where
=
such that 𝜆k = 𝜆p(xk ), k = 1, 2, . . . , K.
𝜆k
Show that 𝔼ℙ (Zt ) = 1 and Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
Zt(k)
e(𝜆k −𝜂k )t
Solution: The first part of the result follows from Problem 5.2.1.16 (page 268).
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307
From Problem 5.2.3.1 (page 298), for each k = 1, 2, . . . , K, Zt(k) satisfies
(
dZt(k)
=
Zt(k)
−
𝜂k − 𝜆k
𝜆k
)
̂ (k)
dN
t
̂ (k) = N (k) − 𝜆k t, 𝜆k = 𝜆tp(xk ), therefore we can easily show that N
̂ (k) = N (k) −
where N
t
t
t
t
)
(
𝜂k − 𝜆k
(k)
̂
Nt is also a ℙ-martingale and subsequently,
𝜆k t is a ℙ-martingale. Thus,
𝜆k
( )
𝔼ℙ Zt(k) = 1 and Zt(k) is a ℙ-martingale for 0 ≤ t ≤ T.
For the case
K
∏
Zt(k)
Zt =
i=1
(j)
and because the Poisson processes Nt(i) and Nt , i ≠ j have no simultaneous jumps, we can
(j)
⟂ Zt , i ≠ j. Therefore,
deduce that Zt(i) ⟂
( )
𝔼ℙ Zt = 𝔼ℙ
( K
∏
)
Zt(k)
=
k=1
K
∏
K
( ) ∏
𝔼ℙ Zt(k) =
1=1
k=1
k=1
for 0 ≤ t ≤ T.
Finally, to show that Zt is a positive ℙ-martingale, 0 ≤ t ≤ T we prove this result via mathematical induction. Let K = 1, then obviously Zt is a positive ℙ-martingale. Assume that
∏
(k)
Zt is a positive ℙ-martingale for K = m, m ≥ 1. For K = m + 1 and because m
k=1 Zt and
Zt(m+1) have no simultaneous jumps, by Itō’s lemma
(
(
)
) (
)
(2)
(m)
d Zt(1) Zt(2) . . . Zt(m) + Zt(1)
d Zt(1) Zt(2) . . . Zt(m) Zt(m+1) = Zt(m+1)
dZt(m+1)
−
− Zt − . . . Zt −
(
)
+ d Zt(1) Zt(2) . . . Zt(m) dZt(m+1)
(k)
where Zt(k)
before
− is the value of Zt
( a jump event. )
(j)
(i)
Since Nt ⟂
⟂ Nt , i ≠ j therefore d Zt(1) Zt(2) . . . Zt(m) dZt(m+1) = 0, and by taking integrals
for s < t,
Zt(1) Zt(2) . . . Zt(m) Zt(m+1) = Zs(1) Zs(2) . . . Zs(m) Zs(m+1) +
t
+
Because
m
∏
k=1
∫s
(
(2)
Zu(1)
− Zu−
t
)
(
Zu(m+1)
d Zu(1) Zu(2) . . . Zu(m)
−
∫s
)
dZu(m+1) .
. . . Zu(m)
−
Zt(k) and Zt(m+1) are positive ℙ-martingales and the integrands are
left-continuous, taking expectations with respect to the filtration ℱs , s < t
(
)
|
𝔼ℙ Zt(1) Zt(2) . . . Zt(m) Zt(m+1) | ℱs = Zs(1) Zs(2) . . . Zs(m) Zs(m+1) .
|
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In addition, since Zt(i) > 0, i = 1, 2, . . . , m + 1 for all t ≥ 0, therefore
| ∏
|m+1
|∏ (k) | m+1
|
|
Z
Zt(k)
t |=
|
| k=1
| k=1
|
|
and hence for t ≥ 0,
(|m+1
(m+1
)
|)
∏ (k)
|∏ (k) |
ℙ
|
Zt || = 𝔼
Zt
𝔼
= 1 < ∞.
|
|
| k=1
k=1
|
|
ℙ
∏
m+1
Finally, because
k=1
Zt(k) is ℱt -adapted, the process
Thus, from mathematical induction, Zt =
0 ≤ t ≤ T.
K
∏
k=1
∏
m+1
k=1
Zt(k)
Zt(k) is a positive ℙ-martingale.
is a positive ℙ-martingale for
◽
7. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Let X1 , X2 , . . . be a sequence of
independent and identically distributed random variables with common probability mass
∑
function ℙ(X = xk ) = p(xk ) > 0, k = 1, 2, . . . , K and Kk=1 ℙ(X = xk ) = 1. Let X1 , X2 , . . .
also be independent of Nt . From the definition of a compound Poisson process
Mt =
Nt
∑
Xi ,
t≥0
i=1
we can set Mt and Nt as
Mt =
K
∑
xk Nt(k) ,
Nt =
k=1
K
∑
Nt(k)
k=1
( ( ))
where Nt(k) ∼ Poisson 𝜆tp xk is the number of jumps in Mt of size xk up to and includ(j)
⟂ Nt , i ≠ j.
ing time t such that Nt(i) ⟂
Let 𝜂1 , 𝜂2 , . . . , 𝜂K > 0 and consider the Radon–Nikodým derivative process
Zt =
K
∏
Zt(k)
k=1
(
)Nt(k)
𝜂
k
where
= e(𝜆k −𝜂k )t
such that 𝜆k = 𝜆p(xk ), k = 1, 2, . . . , K. By changing the
𝜆k
measure ℙ to a measure ℚ such that
(
)
dℚ ||
dℚ ||
ℱ
𝔼ℙ
= Zt
t =
|
dℙ |
dℙ ||ℱt
Zt(k)
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309
show that under the ℚ measure, Mt is a compound Poisson process with intensity 𝜂 =
∑K
k=1 𝜂k > 0 and X1 , X2 , . . . is a sequence of independent and identically distributed random variables with common probability mass function
𝜂k
𝜂1 + 𝜂2 + . . . + 𝜂K
ℚ(X = xk ) = q(xk ) =
such that
K
∑
k=1
ℚ(X = xk ) = 1.
Solution: From the independence of Nt(1) , Nt(2) , . . . , Nt(K) under ℙ, for u ∈ ℝ and using
the moment generating function approach
(
)
(
)
dℚ ||
𝔼ℚ euMt = 𝔼ℙ euMt
dℙ ||ℱt
=𝔼
=
⎡
ℙ⎢ u
e
∑K
⎢
⎣
K
∏
(k)
⋅
k=1 xk Nt
K
∏
(
(𝜆k −𝜂k )
e
k=1
(𝜆k −𝜂k )t ℙ
[
𝔼
e
(
e
𝜂k
𝜆k
)
]
𝜂
(k)
uxk +log( 𝜆k ) Nt
k
)Nt(k) ⎤
⎥
⎥
⎦
.
k=1
( (k) )
m
= e𝜆tp(xk )(e −1)
From Problem 5.2.1.14 (page 266), for a constant m > 0, 𝔼ℙ emNt
therefore
𝜂
K
uxk +log( k )
( uM ) ∏
𝜆k −1)
ℚ
(𝜆k −𝜂k )t
𝜆tp(xk )(e
t
e
⋅e
𝔼 e
=
k=1
=
K
∏
(𝜆k −𝜂k )t
e
⋅e
( t
)
𝜂
𝜆k t 𝜆k euxk −1
k
k=1
=
K
∏
e𝜂k t(e
uxk −1)
k=1
(∑
)
K
uxk −1
𝜂t
k=1 q(xk )e
=e
which is the moment generating function for a compound Poisson process with intensity
𝜂
𝜂 > 0 and jump size distribution ℚ(Xi = xk ) = q(xk ) = k , i = 1, 2, . . .
𝜂
◽
8. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Let X1 , X2 , . . . be a sequence
of independent and identically distributed random variables which are also independent
of Nt , and define the compound Poisson process as
Mt =
Nt
∑
i=1
Xi ,
0 ≤ t ≤ T.
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By changing the measure ℙ to a measure ℚ such that
(
)
dℚ ||
dℚ ||
𝔼ℙ
= Zt
ℱ
t =
|
dℙ |
dℙ ||ℱt
show that for 𝜂 > 0 the Radon–Nikodým derivative process Zt can be written as
⎧
Nt
⎪ (𝜆−𝜂)t ∏ 𝜂ℚ(Xi )
e
⎪
𝜆ℙ(Xi )
i=1
⎪
Zt = ⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where ℙ(X) (f ℙ (X)) and ℚ(X) (f ℚ (X)) are the probability mass (density) functions of ℙ
and ℚ, respectively. Note that for the case of continuous variables Xi , i = 1, 2, . . . , we
assume ℚ is absolutely continuous with respect to ℙ on ℱ.
Solution: We first consider the discrete case where we let X1 , X2 , . . . be a sequence of
independent and identically distributed random variables with common probability mass
∑
function ℙ(X = xk ) = p(xk ) > 0, k = 1, 2, . . . , K and Kk=1 ℙ(X = xk ) = 1.
From Problem 5.2.3.7 (page 308), the compound Poisson process Mt and the Poisson
process Nt can be expressed as
Mt =
K
∑
xk Nt(k) ,
Nt =
k=1
K
∑
Nt(k)
k=1
( ( ))
(j)
where Nt(k) ∼ Poisson 𝜆tp xk such that Nt(i) ⟂
⟂ Nt , i ≠ j. By setting 𝜂1 , 𝜂2 , . . . , 𝜂K > 0,
we consider the Radon–Nikodým derivative process
Zt =
K
∏
Zt(k)
k=1
(
)Nt(k)
𝜂k
and 𝜆k = 𝜆p(xk ) for k = 1, 2, . . . , K. Thus, under the
𝜆k
∑
ℚ measure, Mt is a compound Poisson process with intensity 𝜂 = Kk=1 𝜂k > 0 and
X1 , X2 , . . . is a sequence of independent and identically distributed random variables
with common probability mass function
𝜂k
ℚ(X = xk ) = q(xk ) =
.
𝜂1 + 𝜂2 + . . . + 𝜂K
where Zt(k) = e(𝜆k −𝜂k )t
Therefore,
Zt =
K
∏
Zt(k)
k=1
=
K
∏
k=1
(
(𝜆k −𝜂k )
e
𝜂k
𝜆k
)Nt(k)
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5.2.3
Girsanov’s Theorem for Jump Processes
=e
k=1 (𝜆k −𝜂k )t
K
∏
(
k=1
=e
K
∏
(
k=1
= e(𝜆−𝜂)t
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311
∑K
(𝜆−𝜂)t
c05.tex
𝜂q(xk )
𝜆p(xk )
Nt
∏
𝜂ℚ(Xi )
i=1
𝜂k
𝜆k
𝜆ℙ(Xi )
)Nt(k)
)Nt(k)
.
By analogy with the discrete case, we can deduce that when Xi , i = 1, 2, . . . are continuous
random variables the Radon–Nikodým derivative process can also be written as
Zt = e(𝜆−𝜂)t
Nt
∏
𝜂f ℚ (Xi )
i=1
𝜆f ℙ (Xi )
,
0≤t≤T
where f ℙ (X) and f ℚ (X) are the probability density functions of ℙ and ℚ measures, respectively, and ℚ is absolutely continuous with respect to ℙ on ℱ.
◽
9. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T and let X1 , X2 , . . . be a sequence
of independent and identically distributed random variables which are also independent of
Nt . Under the ℙ measure, each Xi , i = 1, 2, . . . has a probability mass (density) function
ℙ(Xi ) (f ℙ (Xi )). From the definition of the compound Poisson process
Mt =
Nt
∑
Xi ,
0≤t≤T
i=1
we let 𝜂 > 0 and consider the Radon–Nikodým derivative process
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
Zt = ⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where ℚ(Xi ) (f ℚ (Xi )) is the probability mass (density) function of Xi , i = 1, 2, . . . under
the ℚ measure. For the case of continuous random variables Xi , i = 1, 2, . . . we also let ℚ
be absolutely continuous with respect to ℙ on ℱ. By changing the measure ℙ to measure
ℚ such that
(
)
dℚ ||
dℚ ||
= Zt
𝔼ℙ
ℱ
t =
|
dℙ |
dℙ ||ℱt
show that under the ℚ measure for 0 ≤ t ≤ T, Mt is a compound Poisson process
with intensity 𝜂 > 0 and Xi , i = 1, 2, . . . is a sequence of independent and identically
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distributed random variables with probability mass (density) functions ℚ(Xi ) (f ℚ (Xi )),
i = 1, 2, . . .
Solution: To solve this problem we need to show
(
)
ℚ
𝔼ℚ euMt = e𝜂t(𝜑X (u)−1) ,
u∈ℝ
which is the moment generating function of a compound Poisson process with intensity 𝜂
such that
Nt
⎧∑
if X is discrete
⎪ euXi ℚ(Xi )
( uX ) ⎪ i=1
ℚ
ℚ
=⎨
𝜑X (u) = 𝔼 e
⎪ ∞
⎪
eux f ℚ (x) dx if X is continuous.
⎩∫−∞
Without loss of generality we consider only the continuous case where, by definition,
(
)
( uM )
|
ℚ
ℙ
uMt dℚ |
t
=𝔼 e
𝔼 e
dℙ ||ℱt
(
)
= 𝔼ℙ euMt Zt
)
(
Nt
∑Nt
∏
𝜂f ℚ (Xi )
u i=1
Xi (𝜆−𝜂)t
ℙ
e
=𝔼
e
𝜆f ℙ (Xi )
i=1
[ (
| )]
Nt
∑Nt
∏
𝜂f ℚ (Xi ) ||
u i=1
Xi
(𝜆−𝜂)t ℙ
ℙ
=e
e
N
𝔼 𝔼
𝜆f ℙ (Xi ) || t
i=1
|
)
(
Nt
∞
∑Nt
∑
∏
𝜂f ℚ (Xi )
u i=1
Xi
(𝜆−𝜂)t
ℙ
=e
ℙ(Nt = n).
𝔼
e
𝜆f ℙ (Xi )
n=0
i=1
Since X1 , X2 , . . . are independent and identically distributed random variables and
because Nt ∼ Poisson(𝜆t), we can express
(N
)
∞
t ( )
∑
∏
( uM )
𝜂 uXi f ℚ (Xi )
ℚ
(𝜆−𝜂)t
ℙ
t
e
𝔼 e
=e
ℙ(Nt = n)
𝔼
𝜆
f ℙ (Xi )
n=0
i=1
=e
(𝜆−𝜂)t
∞
∑
[
𝔼
ℙ
(
n=0
Because
𝔼ℙ
(
𝜂 uX f ℚ (X)
e ℙ
𝜆
f (X)
)
𝜂 uX f ℚ (X)
e ℙ
𝜆
f (X)
∞
=
∫−∞
∞
)]n
e−𝜆t (𝜆t)n
.
n!
𝜂 ux f ℚ (x) ℙ
⋅ f (x) dx
e
𝜆 f ℙ (x)
𝜂 ux ℚ
e f (x) dx
∫−∞ 𝜆
𝜂
(u)
= 𝜑ℚ
𝜆 X
=
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5.2.3
Girsanov’s Theorem for Jump Processes
where 𝜑ℚ
(u) =
X
∞
∫−∞
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313
eux f ℚ (x) dx, then
ℚ
𝔼 (e
uMt
(𝜆−𝜂)t
)=e
∞ (
∑
𝜂
𝜆
n=0
−𝜂t
=e
)n e−𝜆t (𝜆t)n
(u)
𝜑ℚ
X
n!
∞
∑
(𝜂t𝜑ℚ
(u))n
X
n!
n=0
ℚ
= e𝜂t(𝜑X (u)−1) .
Based on the moment generating function we can deduce that under the ℚ measure, Mt
is a compound Poisson process with intensity 𝜂 > 0 and X1 , X2 , . . . are also independent and identically distributed random variables with probability density function f ℚ (Xi ),
i = 1, 2, . . .
◽
10. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 defined on the probability
space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T and let X1 , X2 , . . . be a sequence
of independent and identically distributed random variables which are also independent
of Nt . Under the ℙ measure each Xi , i = 1, 2, . . . has a probability mass (density) function
ℙ(Xi ) (f ℙ (Xi )). From the definition of the compound Poisson process
Mt =
Nt
∑
Xi ,
0≤t≤T
i=1
we let 𝜂 > 0 and consider the Radon–Nikodým derivative process
⎧
Nt
⎪ (𝜆−𝜂)t ∏ 𝜂ℚ(Xi )
e
⎪
𝜆ℙ(Xi )
i=1
⎪
Zt = ⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where ℚ(Xi ) (f ℚ (Xi )) is the probability mass (density) function of Xi , i = 1, 2, . . . under
the ℚ measure. For the case of continuous random variables Xi , i = 1, 2, . . . we also let
ℚ be absolutely
with respect to ℙ on ℱ.
( continuous
)
Show that 𝔼ℙ Zt = 1 and Zt is a positive ℙ martingale for 0 ≤ t ≤ T.
Solution: Without loss of generality we consider only the case of continuous random
variables Xi , i = 1, 2, . . .
Nt 𝜂f ℚ (X )
∏
i
is defined as a pure jump process, and let Zt−
Let Zt = e(𝜆−𝜂)t Gt , where Gt =
ℙ
i=1 𝜆f (Xi )
denote the value of Zt before a jump event.
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From Itō’s lemma,
dZt =
2
𝜕Zt
𝜕Z
1 𝜕 Zt
(dGt )2 + . . .
dt + t dGt +
𝜕t
𝜕Gt
2 𝜕G2t
= (𝜆 − 𝜂)e(𝜆−𝜂)t Gt dt + e(𝜆−𝜂)t dGt
= (𝜆 − 𝜂)Zt− + e(𝜆−𝜂)t dGt .
Let Gt− be the value Gt just before a jump event and assume there occurs an instantaneous
jump at time t. Therefore,
dGt = Gt − Gt−
𝜂f ℚ (Xt )
G − − Gt −
𝜆f ℙ (Xt ) t
( ℚ
)
𝜂f (Xt )
=
−
1
Gt −
𝜆f ℙ (Xt )
=
𝜂f ℚ (Xt )
is the size of the jump variable if N jumps at time t.
𝜆f ℙ (Xt )
By defining a new compound Poisson process
where
Ht =
Nt
∑
𝜂f ℚ (Xi )
i=1
𝜆f ℙ (Xi )
we let Ht− be the value of Ht just before a jump. At jump time t (which is also the jump
time of Mt and Gt ), we have
dHt = Ht − Ht− =
𝜂f ℚ (Xt )
.
𝜆f ℙ (Xt )
Therefore, at jump times of N we can write
dGt
= dHt − 1
Gt −
and in general we can write
dGt
= dHt − dNt
Gt −
where
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆 dt
and
dHt =
𝜂f ℚ (Xt )
dN .
𝜆f ℙ (Xt ) t
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Thus,
dZt = (𝜆 − 𝜂)Zt− dt + e(𝜆−𝜂)t (dHt − dNt )Gt−
= (𝜆 − 𝜂)Zt− dt + Zt− dHt − Zt− dNt
or
dZt
= d(Ht − 𝜂t) − d(Nt − 𝜆t).
Zt−
̂ t = Ht − 𝜂t is also a
̂ t = Nt − 𝜆t is a ℙ-martingale we now need to show that H
Since N
ℙ-martingale. By definition,
𝔼ℙ
∞
since
∫−∞
[
]
𝜂f ℚ (X)
𝜂 ∞ f ℚ (x) ℙ
𝜂
=
⋅ f (x) dx =
𝜆 ∫−∞ f ℙ (x)
𝜆
𝜆f ℙ (X)
f ℚ (x) dx = 1. From Problem 5.2.1.14 (page 266), we can therefore deduce that
[ ℚ ]
ℙ 𝜂f (X)
̂
Ht = Ht − 𝔼
𝜆t = Ht − 𝜂t
𝜆f ℙ (X)
is a ℙ-martingale as well.
̂t = Nt − 𝜆t and H
̂ t = Ht − 𝜂t into the stochastic differential equation, we
Substituting N
have
dZt
̂ t − dN
̂t
= dH
Zt−
and taking integrals, for t ≥ 0
t
∫0
t
dZu =
∫0
̂u −
Zu− dH
t
Zt = Z0 +
∫0
t
∫0
̂u
Zu− dN
̂u −
Zu− dH
t
∫0
̂u .
Zu− dN
Taking expectations,
( )
( )
𝔼 Zt = 𝔼ℙ Z0 + 𝔼ℙ
ℙ
[
t
∫0
]
[
ℙ
̂
Zu− dHu − 𝔼
t
∫0
]
̂
Zu− dNu = 1
̂ t and N
̂t are both ℙ-martingales.
since Z0 = 1, H
To show that Zt is a positive ℙ-martingale, by taking integrals for s < t
t
∫s
t
dZu =
∫s
t
Z t − Zs =
∫s
̂u −
Zu− dH
̂u −
Zu− dH
t
∫s
t
∫s
̂u
Zu− dN
̂u
Zu− dN
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and taking expectations under the filtration ℱs ,
ℙ
𝔼
[
]
ℙ
Zt − Zs |ℱs = 𝔼
[
]
]
[ t
|
|
|
|
ℙ
̂ u | ℱs − 𝔼
̂u | ℱs .
Z − dH
Z − dN
|
|
∫s u
∫s u
|
|
t
̂ t and N
̂t are ℙ-martingales, therefore
Because H
[
]
𝔼ℙ Zt |ℱs = Zs .
( )
( )
In addition, since Zt > 0 for 0 ≤ t ≤ T, then |Zt | = Zt and hence 𝔼ℙ |Zt | = 𝔼ℙ Zt =
1 < ∞ for 0 ≤ t ≤ T. Because Zt is also ℱt -adapted, we can conclude that Zt is a positive
ℙ-martingale for 0 ≤ t ≤ T.
◽
11. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ 0 ≤ t ≤ T}
be a standard Wiener process defined on the probability space (Ω, ℱ, ℙ) with respect
to the filtration ℱt , 0 ≤ t ≤ T. Suppose 𝜃t is an adapted process, 0 ≤ t ≤ T and 𝜂 > 0.
Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables
where each Xi , i = 1, 2, . . . has a probability mass (density) function ℙ(Xi ) (f ℙ (Xi )) under
the ℙ measure. Let X1 , X2 , . . . also be independent of Nt and Wt .
From the definition of the compound Poisson process
Mt =
Nt
∑
Xi ,
0≤t≤T
i=1
we consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1)
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
=⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
and
t
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
1
t 2
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du
where ℚ(Xi ) (f ℚ (Xi )) is
mass (density) function of Xi , i = 1, 2, . . . under
( the probability
)
1
T 2
the ℚ measure and 𝔼ℙ e 2 ∫0 𝜃u du < ∞. For the case of continuous random variables Xi ,
i = 1, 2, . . . we
( also
) let ℚ be absolutely continuous with respect to ℙ on ℱ.
Show that 𝔼ℙ Zt = 1 and Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
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317
Solution: From
4.2.2.1
( Problem
)
( (page
) 194) and Problem 5.2.3.10 (page 313), we have
(1)
(2)
ℙ
ℙ
= 1, 𝔼 Zt
= 1, Zt(1) and Zt(2) are both ℙ-martingales. From Itō’s
shown that 𝔼 Zt
lemma,
(2)
d(Zt(1) Zt(2) ) = Zt(1)
+ Zt(2) d + dZt(1) dZt(2)
− dZt
where Zt− is the value of Zt(1) before a jump event. Since dZt(1) dZt(2) = 0, by taking
integrals,
t
∫0
d(Zu(1) Zu(2) ) =
t
∫0
(2)
Zu(1)
− dZu +
Zt(1) Zt(2) = Z0(1) Z0(2) +
t
∫0
t
∫0
Zu(2) dZu(1)
(2)
Zu(1)
− dZu +
t
∫0
Zu(2) dZu(1) .
Taking expectations under the ℙ measure,
(
)
(
)
𝔼ℙ Zt(1) Zt(2) = 𝔼ℙ Z0(1) Z0(2) = 1
since Z0(1) = 1, Z0(2) = 1 and both Zt(1) and Zt(2) are ℙ-martingales.
To show that Zt = Zt(1) ⋅ Zt(2) is a positive ℙ-martingale, by taking integrals for s < t of
(2)
+ Zt(2) dZt(1)
d(Zt(1) Zt(2) ) = Zt(1)
− dZt
we have
Zt(1) Zt(2) = Zs(1) Zs(2) +
t
∫s
(2)
Zu(1)
− dZu +
t
∫s
Zu(2) dZu(1)
and taking expectations with respect to the filtration ℱs , s < t
(
)
|
𝔼ℙ Zt(1) Zt(2) | ℱs = Zs(1) Zs(2)
|
)
)
( t
(
t
|
|
(1)
(2) |
(2)
(1) |
ℙ
ℙ
since 𝔼
Z − dZu | ℱs = 𝔼
Z dZu | ℱs = 0.
|
|
∫s u
∫s u
|
|
|
|
In addition, since Zt(1) > 0 and Zt(2) > 0 for all 0 ≤ t ≤ T, therefore |Zt(1) Zt(2) | = Zt(1) Zt(2) and
|
|
(
(
)
)
hence 𝔼ℙ |Zt | = 𝔼ℙ Zt = 1 < ∞ for 0 ≤ t ≤ T. Finally, because Zt is also ℱt -adapted,
we can conclude that Zt is a positive ℙ-martingale for 0 ≤ t ≤ T.
◽
12. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ 0 ≤ t ≤ T}
be a standard Wiener process defined on the probability space (Ω, ℱ, ℙ) with respect
to the filtration ℱt , 0 ≤ t ≤ T. Suppose 𝜃t is an adapted process, 0 ≤ t ≤ T and 𝜂 > 0.
Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables
where each Xi , i = 1, 2, . . . has a probability mass (density) function ℙ(Xi ) (f ℙ (Xi )) under
the ℙ measure. Let X1 , X2 , . . . also be independent of Nt and Wt .
From the definition of the compound Poisson process
Mt =
Nt
∑
i=1
Xi ,
0≤t≤T
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Girsanov’s Theorem for Jump Processes
we consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1)
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
=⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
and
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
t
t 2
1
Zt(2) = e− ∫0 𝜃u dWu − 2 ∫0 𝜃u du
where ℚ(Xi ) (f ℚ (Xi )) is
mass (density) function of Xi , i = 1, 2, . . . under
( the probability
)
1
T 2
the ℚ measure and 𝔼ℙ e 2 ∫0 𝜃u du < ∞. For the case of continuous random variables Xi ,
i = 1, 2, . . . we also let ℚ be absolutely continuous with respect to ℙ on ℱ.
By changing the measure ℙ to measure ℚ such that
(
)
dℚ ||
dℚ ||
ℙ
= Zt
ℱ =
𝔼
dℙ || t
dℙ ||ℱt
show that under the ℚ measure, the process Mt =
Nt
∑
i=1
Xi is a compound Poisson process
with intensity 𝜂 > 0 where Xi , i = 1, 2, . . . is a sequence of independent and identically
distributed random variables with probability mass (density) functions ℚ(Xi ) (f ℚ (Xi )), i =
̃ t = Wt +
1, 2, . . . , the process W
t
∫0
̃ t.
𝜃u du is a ℚ-standard Wiener process and Mt ⟂
⟂W
Solution: The first two results are given in Problem 5.2.3.9 (page 311) and Problem
4.2.2.10 (page 205).
To prove the final result we need to show that, under the ℚ measure,
(
)
1 2
ℚ
̃
𝔼ℚ eu1 Mt +u2 Wt = e𝜂t(𝜑X (u1 )−1) ⋅ e 2 u2 t
̃t ∼
which is a joint product of independent moment generating functions of Mt and W
𝒩(0, t).
By definition,
(
)
(
)
(
)
̃
̃
̃
𝔼ℚ eu1 Mt +u2 Wt = 𝔼ℙ eu1 Mt +u2 Wt Zt = 𝔼ℙ eu1 Mt Zt(1) ⋅ eu2 Wt Zt(2)
and we let
(1)
(2)
Zt = Zt ⋅ Zt
where
(1)
Z t = eu1 Mt Zt(1)
and
(2)
̃
Z t = eu2 Wt Zt(2) .
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(1)
4:29 P.M.
319
(1)
By setting Z t− to denote the value of Z t before a jump event and using Itō’s lemma, we
have
(1) (2)
(1)
(2)
(2)
(1)
(1)
(2)
dZ t = d(Z t Z t ) = Z t− dZ t + Z t dZ t + dZ t dZ t .
(1)
For dZ t we can expand, using Taylor’s theorem,
(1)
(1)
dZ t
(1)
𝜕Z
𝜕Z
= t dMt + t(1) dZt(1)
𝜕Mt
𝜕Zt
(1)
(1)
⎤
⎡ 2 (1)
𝜕2 Z t
𝜕2 Z t
1 ⎢ 𝜕 Zt
(1)
(1) 2 ⎥
2
+
(dM
)
+
2
(dM
dZ
)
+
(dZ
)
t
t
t
t
⎥
2! ⎢ 𝜕Mt2
𝜕Mt 𝜕Zt(1)
𝜕(Zt(1) )2
⎦
⎣
+
(1)
⎡ 3 (1)
𝜕3 Z t
1 ⎢ 𝜕 Zt
3
(dMt ) + 3
(dMt )2 (dZt(1) )
(1)
2
3! ⎢ 𝜕Mt3
𝜕Mt 𝜕Zt
⎣
(1)
+3
⎤
(1) 3 ⎥
(dZ
)
+ ...
t
⎥
𝜕(Zt(1) )3
⎦
(1)
𝜕3Zt
𝜕Mt 𝜕(Zt(1) )2
(dMt )(dZt(1) )2 +
𝜕3 Z t
̂
̂
Because dMt = Xt dNt , dZt(1) = Zt(1)
− (d Ht − d Nt ) where
⎧ 𝜂ℚ(Xt )
⎪ 𝜆ℙ(X ) dNt − 𝜂 dt
t
̂t = ⎪
dH
⎨
⎪ 𝜂f ℚ (Xt )
dNt − 𝜂 dt
⎪ ℙ
⎩ 𝜆f (Xt )
if Xt is discrete, ℙ(Xt ) > 0
if Xt is continuous
̂t = Nt − 𝜆 dt (see Problem 5.2.3.10, page 313), we can write
and dN
]
[
(1)
(1)
(1)
(1)
(1)
̂ t − dN
̂ t ) + 1 (u1 Xt )2 Z t− dNt + 2u1 Xt Z t− dNt (dH
̂ t − dN
̂t )
dZ t = (u1 Xt )Z t− dNt + Z t− (dH
2!
]
[
(1)
(1)
1
̂ t − dN
̂t ) + . . .
(u1 Xt )3 Z t− dNt + 3(u1 Xt )2 Z t− dNt (dH
+
3!
[
] (1)
(1)
1
1
̂ t − dN
̂t )
= (u1 Xt ) + (u1 Xt )2 + (u1 Xt )3 + . . . Z t− dNt + Z t− (dH
2!
3!
] (1)
[
1
1
̂ t − dN
̂t )
+ (u1 Xt ) + (u1 Xt )2 + (u1 Xt )3 + . . . Z t− dNt (dH
2!
3!
(1)
(1)
(1)
̂ t − dN
̂t ) + (eu1 Xt − 1)Z t− dNt (dH
̂ t − dN
̂t ).
= (eu1 Xt − 1)Z t− dNt + Z t− (dH
Since
[
]
⎧ 𝜂ℚ(Xt )
−
1
dNt − (𝜂 − 𝜆) dt
⎪ 𝜆ℙ(X )
t
⎪
̂t = ⎨
̂ t − dN
dH
]
[
⎪ 𝜂f ℚ (Xt )
−
1
dNt − (𝜂 − 𝜆) dt
⎪
⎩ 𝜆f ℙ (Xt )
if Xt is discrete, ℙ(Xt ) > 0
if Xt is continuous
Page 319
Chin
320
5.2.3
and
[
]
⎧ 𝜂ℚ(Xt )
−
1
dNt
⎪ 𝜆ℙ(X )
t
⎪
̂ t − dN
̂t ) = ⎨
dNt (dH
]
[
⎪ 𝜂f ℚ (Xt )
−
1
dNt
⎪
⎩ 𝜆f ℙ (Xt )
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Girsanov’s Theorem for Jump Processes
if Xt is discrete, ℙ(Xt ) > 0
if Xt is continuous
therefore
(1)
dZ t
(
)
) (1)
(1)
⎧ 𝜂ℚ(Xt ) ( u1 Xt
̂t
̂ t − dN
− 1 Z t− dNt + Z t− dH
if Xt is discrete, ℙ(Xt ) > 0
⎪ 𝜆ℙ(X ) e
t
⎪
=⎨
(
)
) (1)
⎪ 𝜂f ℚ (Xt ) ( u X
(1)
̂t
̂ t − dN
e 1 t − 1 Z t− dNt + Z t− dH
if Xt is continuous.
⎪ ℙ
⎩ 𝜆f (Xt )
(2)
For the case of dZ t , by applying Taylor’s theorem
(2)
(2)
dZ t
(2)
𝜕Z
𝜕Z
̃ t + t dZ (2)
= t dW
t
̃t
𝜕W
𝜕Zt(2)
(2)
(2)
⎡ 2 (2)
⎤
𝜕2Zt
𝜕2Z t
1 ⎢ 𝜕 Zt
(2)
(2) 2 ⎥
2
̃
̃
+
(dWt ) + 2
(d
W
)(dZ
)
+
(dZ
)
+ ...
t
t
t
⎥
2! ⎢ 𝜕(W
̃ t )2
̃ t 𝜕Z (2)
𝜕W
𝜕(Zt(2) )2
t
⎣
⎦
̃ t = dWt + 𝜃t dt and dZ (2) = −𝜃t Z (2) dWt (see Problem 4.2.2.2, page 196), we
and since dW
t
t
can write
(2)
(2)
(2)
1
dZ t = u22 Z t dt + (u2 − 𝜃t )Z t dWt .
2
Since dNt dWt = 0 we have
(1)
(2)
Z t dZ t =
(2)
(1)
Z t dZ t
and
1 2
u Z dt + (u2 − 𝜃t )Z t dWt ,
2 2 t
(
)
)
⎧ 𝜂ℚ(Xt ) ( u1 Xt
̂ t − dN
̂t
if Xt is discrete, ℙ(Xt ) > 0
− 1 Z t dNt + Z t dH
⎪ 𝜆ℙ(X ) e
t
⎪
=⎨
(
)
)
⎪ 𝜂f ℚ (Xt ) ( u X
̂t
̂ t − dN
if Xt is continuous
e 1 t − 1 Z t dNt + Z t dH
⎪ ℙ
⎩ 𝜆f (Xt )
(1)
(2)
dZ t dZ t = 0.
Without loss of generality, we let Xt be a continuous random variable and by letting dNt =
̂ t + 𝜆 dt we have
dN
(
)
)
𝜂f ℚ (X ) (
1
̂ t − dN
̂t
dZ t = u22 Z t dt + (u2 − 𝜃t )Z t dWt + ℙ t eu1 Xt − 1 Z t dNt + Z t dH
2
𝜆f (Xt )
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Girsanov’s Theorem for Jump Processes
[
=
321
]
) 1 2
𝜂f ℚ (Xt ) ( u X
1 t −1 +
u
e
Z dt + (u2 − 𝜃t )Z t dWt
2 2 t
f ℙ (Xt )
+
)
𝜂f ℚ (Xt ) ( u X
̂t + Z t (dH
̂ t − dN
̂t ).
e 1 t − 1 Z t dN
ℙ
𝜆f (Xt )
Taking integrals,
]
t
) 1 2
𝜂f ℚ (Xs ) ( u X
1 s −1 +
e
Z
ds
+
(u − 𝜃s )Z s dWs
u
∫0
∫0 2
2 2 s
f ℙ (Xs )
t
t
(
)
)
𝜂f ℚ (Xs ) ( u X
1 s − 1 Z dN
̂
̂
̂
e
+
+
+Z
−
d
N
d
H
s
s
s
s
s
∫0 𝜆f ℙ (Xs )
∫0
t
[
Zt = 1 +
̂t ,
where Z 0 = 1 and because the jump size variable Xt is independent of Nt and Wt , and N
̂
Ht and Wt are ℙ-martingales, by taking expectations under the ℙ measure we have
( )
𝔼ℙ Z t = 1 +
t
∫0
{
𝔼ℙ
[
} ( )
]
)
𝜂f ℚ (Xs ) ( u X
1 2
1 s −1
u
e
𝔼ℙ Z s ds.
+
2 2
f ℙ (Xs )
By differentiating the integrals with respect to t,
} ( )
]
( ) { [ 𝜂f ℚ (X ) (
)
d ℙ
1 2
t
u1 Xt
𝔼 Z t = 𝔼ℙ
u
e
𝔼ℙ Z t
−
1
+
dt
2 2
f ℙ (Xt )
{ [ (
( )
)
] 1 }
= 𝜂 𝔼ℚ eu1 Xt − 1 + u22 𝔼ℙ Z t
2
or
) 1 ]
dmt [ ( ℚ
− 𝜂 𝜑X (u1 ) − 1 + u22 mt = 0
dt
2
( )
(
)
(u ) = 𝔼ℚ eu1 Xt .
where mt = 𝔼ℙ Z t and 𝜑ℚ
X 1
1 2
1 2
By setting the integrating factor to be I = e− ∫ (𝜂(𝜑X (u1 )−1)+ 2 u2 ) dt = e−(𝜂(𝜑X (u1 )−1)+ 2 u2 )t and
multiplying the differential equation with I, we have
(
)
( )
1 2
1 2
d
mt e−𝜂t(𝜑X (u1 )−1)− 2 u2 t = 0 or e−𝜂t(𝜑X (u1 )−1)− 2 u2 t 𝔼ℙ Z t = C
dt
( )
(
)
̃
where C is a constant. Since 𝔼ℙ Z 0 = 𝔼ℙ eu1 M0 Z0(1) ⋅ eu2 W0 Z0(2) = 1, therefore C = 1.
Thus, we finally obtain
(
)
( )
1 2
̃
𝔼ℚ eu1 Mt +u2 Wt = 𝔼ℙ Z t = e𝜂t(𝜑X (u1 )−1)+ 2 u2 t .
Since the joint moment generating function of
(
)
1 2
̃
𝔼ℚ eu1 Mt +u2 Wt = e𝜂t(𝜑X (u1 )−1) ⋅ e 2 u2 t
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Risk-Neutral Measure for Jump Processes
can be expressed as a product of the moment generating functions for Mt (which is a com̃ t ∼ 𝒩(0, t), respectively, we can deduce
pound Poisson process with intensity 𝜂) and W
̃ t are independent.
that Mt and W
N.B. The same conclusion can also be obtained for the case of treating Xt as a discrete
random variable.
◽
5.2.4
Risk-Neutral Measure for Jump Processes
1. Simple Jump Process. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ 0 ≤ t ≤ T} be a
Poisson process with intensity 𝜆 > 0 relative to the filtration ℱt , 0 ≤ t ≤ T. Suppose the
asset price St follows a simple jump process
dSt
= (J − 1)dNt
St −
where J is a constant jump amplitude if N jumps at time t and
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆dt.
Let r be the risk-free interest rate.
By considering the Radon–Nikodým derivative process, for 𝜂 > 0
Zt = e(𝜆−𝜂)t
( 𝜂 )Nt
𝜆
,
0≤t≤T
show that by changing the measure ℙ to the risk-neutral measure ℚ, the above stochastic
differential equation is
dSt
= (J − 1)dNt
St −
where Nt ∼ Poisson(𝜂t), 𝜂 = r(J − 1)−1 > 0 provided J > 1.
Is the market arbitrage free and complete under the ℚ measure?
Solution: Let 𝜂 > 0 and from the Radon–Nikodým derivative process
Zt = e(𝜆−𝜂)t
( 𝜂 )Nt
𝜆
.
By changing the measure ℙ to the risk-neutral measure ℚ on the filtration ℱs , 0 ≤ s ≤ t
then under ℚ, Nt ∼ Poisson(𝜂t) and the discounted asset price e−rt St is a ℚ-martingale.
By setting St− to denote the value of St before a jump event and by expanding d(e−rt St ), we
have
d(e−rt St ) = −re−rt St dt + e−rt dSt
= −re−rt St dt + e−rt St (J − 1)dNt
= e−rtSt (−r + 𝜂(J − 1)) dt + e−rt St (J − 1)(dNt − 𝜂 dt).
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323
Since the compensated Poisson process Nt − 𝜂t is a ℚ-martingale (see Problem 5.2.1.9,
page 262), then, in order for e−rt St to be a ℚ-martingale, we can set
𝜂 = r(J − 1)−1 .
Since 𝜂 > 0 we require J > 1. Thus, under the risk-neutral measure ℚ,
dSt
= (J − 1)dNt
St −
where Nt ∼ Poisson(𝜂t).
The market is arbitrage free since we can construct a risk-neutral measure ℚ on the filtration
ℱs , 0 ≤ s ≤ t. Since ℚ is unique, the market is also complete.
◽
2. Pure Jump Process. Let (Ω, ℱ, ℙ) be a probability space and let {Nt ∶ 0 ≤ t ≤ T} be a
Poisson process with intensity 𝜆 > 0 relative to the filtration ℱt , 0 ≤ t ≤ T. Suppose St
follows a pure jump process
dSt
= (Jt − 1)dNt
St −
where Jt is the jump variable with mean 𝔼ℙ (Jt ) = J if N jumps at time t and
{
1 with probability 𝜆 dt
dNt =
0 with probability 1 − 𝜆dt.
Let r be the risk-free interest rate. Assume that Jt is independent of Nt and consider the
Radon–Nikodým derivative process
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
Zt = ⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where 𝜂 > 0, Xi = Ji − 1, ℙ(X) (f ℙ (X)) and ℚ(X) (f ℚ (X)) are the probability mass (density)
functions of ℙ and ℚ, respectively.
Show, by changing the measure ℙ to the risk-neutral measure ℚ, the above SDE can be
written as
dSt
= dMt
St −
∑Nt
where Mt = i=1
(Ji − 1) is a compound Poisson process with Ji , i = 1, 2, . . . a sequence of
independent and identically distributed jump amplitude random variables having intensity
𝜂=
provided 𝔼ℚ (Jt ) > 1.
r
>0
𝔼ℚ (Jt ) − 1
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324
5.2.4
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Risk-Neutral Measure for Jump Processes
Is the market arbitrage free and complete under the ℚ measure?
By assuming that the jump amplitude remains unchanged with the change of measure, find
the corresponding SDE under the equivalent martingale risk-neutral measure, ℚM .
Is the market arbitrage free and complete under the ℚM measure?
Solution: From Problem 5.2.2.1 (page 281) we can express the pure jump process as
dSt
= dMt
St −
∑Nt
where Mt = i=1
(Ji − 1) is a compound Poisson process with intensity 𝜆 > 0.
Under the risk-neutral measure ℚ on the filtration ℱs , 0 ≤ s ≤ t we let 𝜂 > 0 and by considering the Radon–Nikodým derivative process
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
Zt = ⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
where Xi = Ji − 1, ℙ(X) (f ℙ (X)) and ℚ(X) (f ℚ (X)) are the probability mass (density) functions of ℙ and, ℚ respectively, then by changing the measure ℙ to the risk-neutral measure
∑Nt
(Ji − 1) is a compound Poisson process with intensity
ℚ we have that, under ℚ, Mt = i=1
𝜂 > 0 and the discounted asset price e−rt St is a ℚ-martingale.
By letting St− = St to denote the value of St before a jump event and expanding d(e−rt St ),
we have
d(e−rt St ) = −re−rt St dt + e−rt dSt
= −re−rt St dt + e−rt St dMt
)
)
(
(
= e−rt St −r + 𝜂𝔼ℚ (Jt − 1) dt + e−rt St dMt − 𝜂𝔼ℚ (Jt − 1) dt .
Given that the compensated compound Poisson process Mt − 𝜂𝔼ℚ (Jt − 1)t is a
ℚ-martingale (see Problem 5.2.1.14, page 266), and in order for e−rt St to be a ℚ-martingale,
we can set
r
𝜂= ℚ
𝔼 (Jt ) − 1
provided 𝔼ℚ (Jt ) > 1.
Thus, under the risk-neutral measure ℚ we have
dSt
= (Jt − 1)dNt
St−
(
)−1
where Nt ∼ Poisson(𝜂t), 𝜂 = r 𝔼ℚ (Jt − 1)
> 0.
The market is arbitrage free since we can construct a risk-neutral measure ℚ on the filtration
ℱs , 0 ≤ s ≤ t. However, the market model is incomplete as ℚ is not unique given that we
can vary the jump distribution Jt .
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5.2.4
Risk-Neutral Measure for Jump Processes
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325
By assuming that the distribution of the jump amplitude Jt remains unchanged under the
change of measure, there exists an equivalent martingale risk-neutral measure ℚM ∼ ℚ
such that e−rt St is a ℚM -martingale. Thus, under the equivalent martingale risk-neutral
measure ℚM ,
)
)
(
(
𝔼ℚM Jt − 1 = 𝔼ℙ Jt − 1 = J − 1
and the corresponding SDE is
dSt
= (Jt − 1)dNt
St −
where Nt ∼ Poisson(𝜂t), 𝜂 = r(J − 1)−1 provided J > 1.
The market model is arbitrage free since we can construct an equivalent martingale
risk-neutral measure ℚM on the filtration ℱs , 0 ≤ s ≤ t. However, the market is incomplete
as ℚM is not unique given that we can still vary the jump size distribution Jt and hence J.
◽
3. Simple Jump Diffusion Process. Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity
𝜆 > 0 and {Wt ∶ 0 ≤ t ≤ T} be a standard Wiener process defined on the probability space
(Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Suppose the stock price St follows a
jump diffusion process
dSt
= (𝜇 − D) dt + 𝜎dWt + (J − 1)dNt
St −
where Wt ⟂
⟂ Nt , J is a constant jump amplitude if N jumps at time t and
{
1 with probability 𝜆dt
dNt =
0 with probability 1 − 𝜆dt.
The constant parameters 𝜇, D and 𝜎 are the drift, continuous dividend yield and volatility, respectively. In addition, let Bt be the risk-free asset having the following differential
equation:
dBt = rBt dt
where r is the risk-free interest rate.
By considering the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1) = e(𝜆−𝜂)t
and
t
( 𝜂 )Nt
𝜆
1
t 2
Zt(2) = e− ∫0 𝜃dWu − 2 ∫0 𝜃 du
( 1 T 2 )
where 𝜃 ∈ ℝ, 𝜂 > 0 and 𝔼ℙ e 2 ∫0 𝜃 du < ∞ show that, under the risk-neutral measure
ℚ, the SDE can be written as
dSt
̃ t + (J − 1)dNt
= (r − D − 𝜂(J − 1)) dt + 𝜎dW
St −
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Chin
326
5.2.4
̃ t = Wt +
where W
(
𝜇 − r + 𝜂(J − 1)
𝜎
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Risk-Neutral Measure for Jump Processes
)
t is the ℚ-standard Wiener process and
Nt ∼ Poisson(𝜂t).
Is the market arbitrage free and complete under the ℚ measure?
By assuming the jump component is uncorrelated with the market (i.e., the jump intensity
remains unchanged with the change of measure), find the corresponding SDE under the
equivalent martingale risk-neutral measure ℚM .
Is the market arbitrage free and complete under the ℚM measure?
Solution: Let 𝜃 ∈ ℝ and 𝜂 > 0 and consider the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1) = e(𝜆−𝜂)t
and
t
( 𝜂 )Nt
𝜆
1
t 2
Zt(2) = e− ∫0 𝜃dWu − 2 ∫0 𝜃
du
( 1 T 2 )
where 𝔼ℙ e 2 ∫0 𝜃 du < ∞. By changing the measure ℙ to the risk-neutral measure ℚ on
the filtration ℱs , 0 ≤ s ≤ t such that
(
)
dℚ ||
dℚ ||
ℙ
= Zt
𝔼
ℱt =
|
dℙ |
dℙ ||ℱt
then from Problem 5.2.3.5 (page 303), under the ℚ measure, the Poisson process Nt ∼
̃ t = Wt +
Poisson(𝜂t) has intensity 𝜂 > 0, the process W
t
∫0
𝜃 du is a ℚ-standard Wiener
̃ t.
⟂W
process and Nt ⟂
In the presence of dividends, the strategy of holding a single stock is no longer self-financing
as it pays out dividends at a rate DSt dt. By letting St− denote the value of St before a jump
event, at time t we let the portfolio Πt be valued as
Πt = 𝜙t St + 𝜓t Bt
where 𝜙t and 𝜓t are the units invested in St and the risk-free asset Bt , respectively. Given
that the holder will receive DSt dt for every stock held, then
dΠt = 𝜙t (dSt + DSt dt) + 𝜓t rBt dt
[
]
= 𝜙t 𝜇St dt + 𝜎St dWt + (J − 1)St dNt + 𝜓t rBt dt
[
]
= rΠt dt + 𝜙t St (𝜇 − r) dt + 𝜎dWt + (J − 1)dNt .
̃t −
By substituting Wt = W
t
𝜃 du and taking note that the compensated Poisson process
∫0
Nt − 𝜂t is a ℚ-martingale, we have
[
(
)]
̃ t + (J − 1) dNt − 𝜂 dt .
dΠt = rΠt dt + 𝜙t St (𝜇 − r − 𝜎𝜃 + 𝜂 (J − 1)) dt + 𝜎dW
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Risk-Neutral Measure for Jump Processes
327
̃ t and Nt − 𝜂t are ℚ-martingales and in order for the discounted portfolio e−rt Πt
Since both W
to be a ℚ martingale
d(e−rt Πt ) = −re−rt Πt dt + e−rt dΠt
[
(
)]
̃ t + (J − 1) dNt − 𝜂 dt
= e−rt 𝜙t St (𝜇 − r − 𝜎𝜃 + 𝜂 (J − 1)) dt + 𝜎dW
we set
𝜇 − r + 𝜂(J − 1)
.
𝜎
𝜃=
Therefore, the jump diffusion process under ℚ is
(
)
dSt
̃ t − 𝜃 dt + (J − 1)dNt
= (𝜇 − D) dt + 𝜎 dW
St −
̃ t + (J − 1)dNt
= (r − D − 𝜂 (J − 1)) dt + 𝜎dW
̃ t = Wt −
where W
(
𝜇 − r + 𝜂(J − 1)
𝜎
)
t is a ℚ-standard Wiener process and Nt ∼
Poisson(𝜂t).
The market is arbitrage free since we can construct a risk-neutral measure ℚ on the filtration
ℱs , 0 ≤ s ≤ t. The market model is incomplete as ℚ is not unique, since there are more
degrees of freedom in selecting J and 𝜂.
Under the assumption that the jump component is uncorrelated with the market, there exists
an equivalent martingale risk-neutral measure ℚM ∼ ℚ such that e−rt Πt is a ℚM -martingale.
Therefore, under ℚM ,
Nt ∼ Poisson(𝜆t)
and the SDE becomes
dSt
̃ t + (J − 1)dNt
= (r − D − 𝜆(J − 1)) dt + 𝜎dW
St −
̃ t = Wt −
where W
(
𝜇 − r + 𝜆(J − 1)
𝜎
)
t is a ℚM -standard Wiener process and
Nt ∼ Poisson(𝜆t).
The market is arbitrage free since we can construct an equivalent martingale risk-neutral
measure ℚM on the filtration ℱs , 0 ≤ s ≤ t. However, the market is not complete as ℚM is
not unique, since we can still vary the jump amplitude J.
◽
4. General Jump Diffusion Process (Merton’s Model). Let {Nt ∶ 0 ≤ t ≤ T} be a Poisson process with intensity 𝜆 > 0 and {Wt ∶ 0 ≤ t ≤ T} be a standard Wiener process defined on the
probability space (Ω, ℱ, ℙ) with respect to the filtration ℱt , 0 ≤ t ≤ T. Suppose the asset
price St follows a jump diffusion process
dSt
= (𝜇 − D) dt + 𝜎dWt + (Jt − 1)dNt
St −
4:29 P.M.
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Risk-Neutral Measure for Jump Processes
where Jt is the jump variable with mean 𝔼ℙ (Jt ) = J if N jumps at time t and
{
1 with probability 𝜆dt
dNt =
0 with probability 1 − 𝜆dt.
Assume that Wt , Nt and Jt are mutually independent. The constant parameters 𝜇, D and 𝜎
are the drift, continuous dividend yield and volatility, respectively. In addition, let Bt be the
risk-free asset having the following differential equation:
dBt = rBt dt
where r is the risk-free interest rate.
By considering the Radon–Nikodým derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1)
and
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
=⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
t
1
t 2
Zt(2) = e− ∫0 𝜃dWu − 2 ∫0 𝜃
du
where 𝜃 ∈ ℝ, 𝜂 > 0, ℚ(Xi ) (f ℚ (Xi )) is the probability
mass
(
) (density) function of Xi = Ji − 1,
1
T 2
𝜃
i = 1, 2, . . . under the ℚ measure and 𝔼ℙ e 2 ∫0
measure ℚ, the above SDE can be written as
du
< ∞ show that, under risk-neutral
dSt
̃ t + (Jt − 1)dNt
= (r − D − 𝜂𝔼ℚ (Jt − 1)) dt + 𝜎dW
St−
̃ t = Wt −
where W
(
𝜇 − r + 𝜂𝔼ℚ (Jt − 1)
𝜎
)
t is the ℚ-standard Wiener process and Nt ∼
Poisson(𝜂t), 𝜂 > 0.
Is the market model arbitrage free and complete under the ℚ measure?
By assuming the jump component is uncorrelated with the market (i.e., the jump intensity and jump size distribution remain unchanged with the change of measure), find the
corresponding SDE under the equivalent martingale risk-neutral measure ℚM .
Is the market model arbitrage free and complete under the ℚM measure?
Solution: From Problem 5.2.2.1 (page 281) we can express the jump diffusion process as
dSt
= (𝜇 − D) dt + 𝜎dWt + dMt
St −
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329
∑Nt
where Mt = i=1
(Ji − 1) is a compound Poisson process with intensity 𝜆 > 0 such that
the jump size (or amplitude) Ji , i = 1, 2, . . . is a sequence of independent and identically
distributed random variables.
Let 𝜃 ∈ ℝ, 𝜂 > 0 and Xi = Ji − 1, i = 1, 2, . . . be a sequence of independent and identically
distributed random variables where each Xi , i = 1, 2, . . . has a probability mass (density)
function ℙ(Xi ) > 0 (f ℙ (Xi ) > 0) under the ℙ measure. We consider the Radon–Nikodým
derivative process
Zt = Zt(1) ⋅ Zt(2)
such that
Zt(1)
and
Nt
⎧
∏
𝜂ℚ(Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆ℙ(Xi )
i=1
⎪
=⎨
Nt
⎪
∏
𝜂f ℚ (Xi )
⎪e(𝜆−𝜂)t
⎪
𝜆f ℙ (Xi )
i=1
⎩
if Xi is discrete, ℙ(Xi ) > 0
if Xi is continuous
t
1
t 2
Zt(2) = e− ∫0 𝜃dWu − 2 ∫0 𝜃
du
where ℚ(Xi ) (f ℚ (Xi )) is(the probability
mass (density) function of Xi , i = 1, 2, . . . under
)
1 T 2
∫0 𝜃 du
ℙ
2
< ∞. By changing the measure ℙ to the risk-neutral
the ℚ measure and 𝔼 e
measure ℚ on the filtration ℱs , 0 ≤ s ≤ t such that
𝔼ℙ
(
dℚ ||
ℱ
dℙ || t
)
=
dℚ ||
= Zt
dℙ ||ℱt
then from Problem 5.2.3.12 (page 317), under ℚ, the compound Poisson process Mt =
t
∑Nt
̃ t = Wt +
(J − 1) has intensity 𝜂 > 0, the process W
𝜃 du is a ℚ-standard Wiener
i=1 i
∫0
̃ t.
⟂W
process and Mt ⟂
In the presence of dividends, the simple strategy of holding a single risky asset is no longer
self-financing as the asset pays out dividends at a rate DSt dt. By letting St− denote the value
of St before a jump event at time t, we let the portfolio Πt be valued as
Πt = 𝜙t St + 𝜓t Bt
where 𝜙t and 𝜓t are the units invested in St and the risk-free asset Bdt , respectively. Given
that the holder will receive DSt dt for every risky asset held, then
dΠt = 𝜙t (dSt + DSt dt) + 𝜓t rBt dt
[
]
= 𝜙t 𝜇St dt + 𝜎St dWt + dMt + 𝜓t rBt dt
[
]
= rΠt dt + 𝜙t St (𝜇 − r) dt + 𝜎dWt + dMt .
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330
5.2.4
̃t −
By substituting Wt = W
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Risk-Neutral Measure for Jump Processes
t
𝜃 du and taking note that the compensated compound Pois∫0
son process Mt − 𝜂𝔼ℚ (Jt − 1)t is a ℚ-martingale, we have
[(
]
(
))
̃ t + dMt − 𝜂𝔼ℚ (Jt − 1) dt .
dΠt = rΠt dt + 𝜙t St 𝜇 − r − 𝜎𝜃 + 𝜂𝔼ℚ Jt − 1 dt + 𝜎dW
)
(
̃ t and Mt − 𝜂𝔼ℚ Jt − 1 t are ℚ-martingales and in order for the discounted
Since both W
portfolio e−rt Πt to be a ℚ-martingale
d(e−rt Πt ) = −re−rt Πt dt + e−rt dΠt
̃ t + dMt − 𝜂𝔼ℚ (Jt − 1) dt]
= e−rt 𝜙t St [(𝜇 − r − 𝜎𝜃 + 𝜂𝔼ℚ (Jt − 1)) dt + 𝜎dW
we set
𝜃=
𝜇 − r + 𝜂𝔼ℚ (Jt − 1)
.
𝜎
Therefore, the jump diffusion process under ℚ is
(
)
dSt
̃ t − 𝜃 dt + (Jt − 1)dNt
= (𝜇 − D) dt + 𝜎 dW
St −
(
(
))
̃ t + (Jt − 1)dNt
= r − D − 𝜂𝔼ℚ Jt − 1 dt + 𝜎dW
)
𝜇 − r + 𝜂𝔼ℚ (Jt − 1)
t and Nt ∼ Poisson(𝜂t).
𝜎
The market is arbitrage free since we can construct a risk-neutral measure ℚ on the filtration
ℱs , 0 ≤ s ≤ t. However, the market model is incomplete as ℚ is not unique, given that there
are more degrees of freedom in selecting 𝜂 and Jt .
Under the assumption that the jump component is uncorrelated with the market, there exists
an equivalent martingale risk-neutral measure ℚM ∼ ℚ such that e−rt Πt is a ℚM -martingale
and hence
̃ t = Wt −
where W
(
𝔼ℚM (Jt − 1) = 𝔼ℙ (Jt − 1) = J − 1
and
Nt ∼ Poisson(𝜆t).
Thus, under ℚM the jump diffusion process becomes
dSt
̃ t + (Jt − 1)dNt
= (r − D − 𝜆(J − 1)) dt + 𝜎dW
St −
(
)
𝜇
−
r
+
𝜆(J
−
1)
̃ t = Wt −
where W
t is a ℚM -standard Wiener process and Nt ∼
𝜎
Poisson(𝜆t).
The market is arbitrage free since we can construct an equivalent martingale risk-neutral
measure ℚM on the filtration ℱs , 0 ≤ s ≤ t. However, the market is incomplete as ℚM is
not unique, since we can still vary the jump size distribution Jt .
◽
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Appendix A
Mathematics Formulae
Indices
xa
= xa−b , (xa )b = (xb )a = xab
xb
( )a
1
x
xa
= a,
= a , x0 = 1.
x
y
y
xa xb = xa+b ,
x−a
Surds
√
x = a x,
√
a
1
a
√
a
√
x
a
x∕y = √
a y
(√ )b
b
a
x = xa .
√ √
xy = a x a y,
√√
√
a b
x = ab x,
(√ )a
a
x = x,
Exponential and Natural Logarithm
ex ey = ex+y ,
log (xy) = log x + log y,
(ex )y = (ey )x = exy , e0 = 1
( )
x
log
= log x − log y, log xy = y log x
y
log ex = x,
elog x = x,
ea log x = xa .
Quadratic Equation
For constants a, b and c, the roots of a quadratic equation ax2 + bx + c = 0 are
x=
−b ±
√
b2 − 4ac
.
2a
Binomial Formula
( )
( ) (
) (
)
n
n
n
n+1
n!
=
,
+
=
k
k
k+1
k+1
k!(n − k)!
(
)
n
n
∑
n n−k k ∑
n!
xn−k yk .
x y =
(x + y)n =
k
k!(n − k)!
k=0
k=0
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Appendix A
Series
Arithmetic: For initial term a and common difference d, the n-th term is
Tn = a + (n − 1)d
and the sum of n terms is
1
n[2a + (n − 1)d].
2
Sn =
Geometric: For initial term a and common ratio r, the n-th term is
Tn = arn−1
the sum of n terms is
Sn =
a(1 − rn )
,
1−r
and the sum of infinite terms is
lim S
n→∞ n
=
a
,
1−r
|r| < 1.
Summation
For n ∈ ℤ+ ,
n
∑
k=1
k=
1
n(n + 1),
2
n
∑
k2 =
k=1
n
∑
1
n(n + 1)(2n + 1),
6
k3 =
k=1
[
]2
1
n(n + 1) .
2
Let a1 , a2 , . . . be a sequence of numbers:
∑
an < ∞ ⇒ lim an = 0.
n→∞
∑
• If lim an ≠ 0 ⇒ an = ∞.
• If
4:36 P.M.
n→∞
Trigonometric Functions
sin(−x) = − sin x,
csc x =
1
,
sin x
cos2 x + sin2 x = 1,
cos(−x) = cos x,
sec x =
1
,
cos x
tan2 x + 1 = sec2 x,
tan x =
cot x =
1
tan x
cot2 x + 1 = csc2 x
sin(x ± y) = sin x cos y ± cos x sin y
cos(x ± y) = cos x cos y ∓ sin x sin y
tan(x ± y) =
sin x
cos x
tan x ± tan y
.
1 ∓ tan x tan y
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Appendix A
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333
Hyperbolic Functions
ex + e−x
sinh x
ex − e−x
, tanh x =
= x
2
cosh x e + e−x
1
1
1
, sech x =
, coth x =
csch x =
sinh x
cosh x
tanh x
sinh(−x) = − sinh x, cosh(−x) = cosh x, tanh(−x) = − tanh x
sinh x =
ex − e−x
,
2
cosh x =
cosh2 x − sinh2 x = 1,
coth2 x − 1 = csch2 x,
1 − tanh2 x = sech2 x
sinh(x ± y) = sinh x cosh y ± cosh x sinh y
cosh(x ± y) = cosh x cosh y ± sinh x sinh y
tanh(x ± y) =
tanh x ± tanh y
.
1 ± tanh x tanh y
Complex Numbers
Let 𝑤 = u + i𝑣 and z = x + iy where u, 𝑣, x, y ∈ ℝ, i =
√
−1 and i2 = −1 then
𝑤 ± z = (u ± x) + (𝑣 ± y)i, 𝑤z = (ux − 𝑣y) + (𝑣x + uy)i,
) (
)
(
𝑣x − uy
ux + 𝑣y
𝑤
+
i
=
z
x2 + y2
x 2 + y2
( )
𝑤
𝑤
= .
z = x − iy, z = z, 𝑤 + z = 𝑤 + z, 𝑤z = 𝑤 ⋅ z,
z
z
De Moivre’s Formula: Let z = x + iy where x, y ∈ ℝ and we can write
(y)
√
.
z = r(cos 𝜃 + i sin 𝜃), r = x2 + y2 , 𝜃 = tan−1
x
For n ∈ ℤ
[r(cos 𝜃 + i sin 𝜃)]n = rn [cos(n𝜃) + i sin(n𝜃)].
Euler’s Formula: For 𝜃 ∈ ℝ
ei𝜃 = cos 𝜃 + i sin 𝜃.
Derivatives
If f (x) and g(x) are differentiable functions of x and a and b are constants
Sum Rule:
d
(af (x) + bg(x)) = af ′ (x) + bg′ (x)
dx
Product/Chain Rule:
d
(f (x)g(x)) = f (x)g′ (x) + f ′ (x)g(x)
dx
4:36 P.M.
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4:36 P.M.
Appendix A
Quotient Rule:
d
dx
(
f (x)
g(x)
)
=
f ′ (x)g(x) − f (x)g′ (x)
,
g(x)2
g(x) ≠ 0
d
d
f (x) = f ′ (x) and g(x) = g′ (x).
dx
dx
If f (z) is a differentiable function of z and z = z(x) is a differentiable function of x, then
where
d
f (z(x)) = f ′ (z(x))z′ (x).
dx
If x = x(s), y = y(s) and F(s) = f (x(s), y(s)), then
𝜕f 𝜕x 𝜕f 𝜕y
d
F(s) =
⋅
+
⋅ .
ds
𝜕x 𝜕s 𝜕y 𝜕s
If x = x(u, 𝑣), y = y(u, 𝑣) and F(u, 𝑣) = f (x(u, 𝑣), y(u, 𝑣)), then
𝜕f 𝜕x 𝜕f 𝜕y
𝜕F
=
⋅
+
⋅ ,
𝜕u
𝜕x 𝜕u 𝜕y 𝜕u
𝜕f 𝜕x 𝜕f 𝜕y
𝜕F
=
⋅
+
⋅ .
𝜕𝑣
𝜕x 𝜕𝑣 𝜕y 𝜕𝑣
Standard Differentiations
If f (x) and g(x) are differentiable functions of x and a and b are constants
d
[f (x)]n = n[f (x)]n−1 f ′ (x)
dx
f ′ (x)
d
d f (x)
= f ′ (x)a f (x) log a
log f (x) =
,
a
dx
f (x)
dx
d
a = 0,
dx
d f (x)
= f ′ (x)e f (x) ,
e
dx
d
sin(ax) = a cos x,
dx
d
sinh(ax) = a cosh(ax),
dx
where f ′ (x) =
d
cos(ax) = −a sin(ax),
dx
d
cosh(ax) = a sinh(ax),
dx
d
tan(ax) = asec2 x
dx
d
tanh(ax) = asech2 (ax)
dx
d
f (x).
dx
Taylor Series
If f (x) is an analytic function of x, then for small h
f (x0 + h) = f (x0 ) + f ′ (x0 )h +
1 ′′
1
f (x0 )h2 + f ′′′ (x0 )h3 + . . .
2!
3!
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Appendix A
335
If f (x, y) is an analytic function of x and y, then for small Δx, Δy
f (x0 + Δx, y0 + Δy) = f (x0 , y0 ) +
𝜕f (x0 , y0 )
𝜕f (x0 , y0 )
Δx +
Δy
𝜕x
𝜕y
[ 2
]
𝜕 2 f (x0 , y0 )
𝜕 2 f (x0 , y0 )
1 𝜕 f (x0 , y0 )
2
2
ΔxΔy +
+
(Δx) + 2
(Δy)
2!
𝜕x𝜕y
𝜕x2
𝜕y2
[ 3
𝜕 3 f (x0 , y0 )
1 𝜕 f (x0 , y0 )
3
(Δx)2 Δy
+
(Δx)
+
3
3!
𝜕x3
𝜕x2 𝜕y
]
𝜕 3 f (x0 , y0 )
𝜕3 f (x0 , y0 )
2
3
Δx(Δy) +
(Δy) + . . .
+3
𝜕x𝜕y2
𝜕y3
Maclaurin Series
Taylor series expansion of a function about x0 = 0
1
= 1 − x + x2 − x3 + . . . , |x| < 1
1+x
1
= 1 + x + x2 + x3 + . . . , |x| < 1
1−x
1
1
ex = 1 + x + x2 + x3 + . . . , for all x
2!
3!
1
1
e−x = 1 − x + x2 − x3 + . . . , for all x
2!
3!
1
1
1
x ∈ (−1, 1]
log(1 + x) = x − x2 + x3 − x4 + . . . ,
2
3
4
1
1
1
log(1 − x) = −x − x2 − x3 − x4 + . . . , |x| < 1
2
3
4
1
1
1
sin x = x − x3 + x5 − x7 + . . . , for all x
3!
5!
7!
1
1
1
cos x = 1 − x2 + x4 − x6 + . . . , for all x
2!
4!
6!
2 5
17 7
𝜋
1 3
x + . . . , |x| <
tan x = x + x + x +
3
15
315
2
1 3 1 5 1 7
sinh x = x + x + x + x + . . . , for all x
3!
5!
7!
1 2 1 4 1 6
cosh x = 1 + x + x + x + . . . , for all x
2!
4!
6!
2 5
17 7
𝜋
1 3
tanh x = x − x + x −
x + . . . , |x| < .
3
15
315
2
Landau Symbols and Asymptotics
Let f (x) and g(x) be two functions defined on some subsets of real numbers, then as x → x0
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4:36 P.M.
Appendix A
• f (x) = O(g(x)) if there exists a constant K > 0 and 𝛿 > 0 such that |f (x)| ≤ K|g(x)| for
|x − x0 | < 𝛿.
f (x)
• f (x) = o(g(x)) if lim
= 0.
x→x0 g(x)
f (x)
• f (x) ∼ g(x) if lim
= 1.
x→x0 g(x)
L’Hospital Rule
Let f and g be differentiable on a ∈ ℝ such that g′ (x) ≠ 0 in an interval around a, except
possibly at a itself. Suppose that
lim f (x) = lim g(x) = 0
x→a
x→a
or
lim f (x) = lim g(x) = ±∞
x→a
then
x→a
f (x)
f ′ (x)
= lim ′ .
x→a g(x)
x→a g (x)
lim
Indefinite Integrals
If F(x) is a differentiable function and f (x) is its derivative, then
∫
f (x) dx = F(x) + c
d
F(x) = f (x) and c is an arbitrary constant.
dx
If f (x) is a continuous function then
′
where F (x) =
d
f (x) dx = f (x).
dx ∫
Standard Indefinite Integrals
If f (x) is a differentiable function of x and a and b are constants
∫
a dx = ax + c,
∫
(ax + b)n dx =
f ′ (x)
dx = log|f (x)| + c,
∫ f (x)
∫
∫
∫
e f (x) dx =
log(ax) dx = x log(ax) − ax + c,
1
sin(ax) dx = − cos(ax) + c,
a
∫
(ax + b)n+1
+ c,
a(n + 1)
∫
n ≠ −1
1 f (x)
+c
e
f ′ (x)
ax dx =
cos(ax) dx =
ax
+c
log a
1
sin(ax) + c
a
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Appendix A
337
∫
sinh(ax) dx =
1
cosh(ax) + c,
a
∫
cosh(ax) dx =
1
sinh(ax) + c
a
where c is an arbitrary constant.
Definite Integrals
If F(x) is a differentiable function and f (x) is its derivative and is continuous on a close interval
[a, b], then
b
∫a
f (x) dx = F(b) − F(a)
d
F(x) = f (x).
dx
If f (x) and g(x) are integrable functions then
′
where F (x) =
a
∫a
b
f (x) dx = 0,
b
∫a
∫a
a
f (x) dx = −
b
[𝛼f (x) + 𝛽g(x)] dx = 𝛼
b
∫a
c
f (x) dx =
∫a
f (x) dx
b
f (x) dx + 𝛽
∫a
∫b
∫a
g(x) dx,
𝛼, 𝛽 are constants
b
f (x) dx +
∫c
f (x) dx,
c ∈ [a, b].
Derivatives of Definite Integrals
If f (t) is a continuous function of t and a(x) and b(x) are continuous functions of x
b(x)
d
d
d
f (t) dt = f (b(x)) b(x) − f (a(x)) a(x)
dx ∫a(x)
dx
dx
b(x)
d
d
d
dt = b(x) − a(x).
dx ∫a(x)
dx
dx
If g(x, t) is a differentiable function of two variables then
b(x)
b(x)
𝜕g(x, t)
d
d
d
g(x, t) dt = g(x, b(x)) b(x) − g(x, a(x)) a(x) +
dt.
∫a(x)
dx ∫a(x)
dx
dx
𝜕x
Integration by Parts
For definite integrals
b
∫a
where u′ (x) =
b
|b
|
u(x)𝑣 (x) dx = u(x)𝑣(x)| −
𝑣(x)u′ (x) dx.
|
∫a
|a
′
d
d
u(x) and 𝑣′ (x) = 𝑣(x).
dx
dx
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Appendix A
Integration by Substitution
If f (x) is a continuous function of x and g′ is continuous on the closed interval [a, b] then
g(a)
b
f (x) dx =
∫g(b)
f (g(u))g′ (u) du.
∫a
Gamma Function
The gamma function is defined as
∞
Γ(z) =
such that
Γ(z + 1) = zΓ(z),
Γ
∫0
tz−1 e−t dt
( ) √
1
= 𝜋,
2
Γ(n) = (n − 1)! for n ∈ ℕ.
Beta Function
The beta function is defined as
1
B(x, y) =
∫0
tx−1 (1 − t)y−1 dt
for x > 0, y > 0
such that
B(x, y) = B(y, x),
In addition
B(x, y) =
Γ(x)Γ(y)
.
Γ(x + y)
u
∫0
tx−1 (u − t)y−1 dt = ux+y−1 B(x, y).
Convex Function
A set Ω in a vector space over ℝ is called a convex set if for x, y ∈ Ω, x ≠ y and for any
𝜆 ∈ (0, 1),
𝜆x + (1 − 𝜆)y ∈ Ω.
Let Ω be a convex set in a vector space over ℝ. A function f ∶ Ω → ℝ is called a convex
function if for x, y ∈ Ω, x ≠ y and for any 𝜆 ∈ (0, 1),
f (𝜆x + (1 − 𝜆)y) ≤ 𝜆f (x) + (1 − 𝜆)f (y).
If the inequality is strict then f is strictly convex.
If f is convex and differentiable on ℝ, then
f (x) ≥ f (y) + f ′ (y)(x − y).
′′
If f is a twice continuously differentiable function on ℝ, then f is convex if and only if f ≥ 0.
′′
If f > 0 then f is strictly convex.
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Appendix A
339
f is a (strictly) concave function if −f is a (strictly) convex function.
Dirac Delta Function
The Dirac delta function is defined as
{
𝛿(x) =
0 x≠0
∞ x=0
and for a continuous function f (x) and a constant a, we have
∞
∫−∞
∞
𝛿(x) dx = 1,
∫−∞
∞
f (x)𝛿(x) dx = f (0),
∫−∞
f (x)𝛿(x − a) dx = f (a).
Heaviside Step Function
The Heaviside step function, H(x) is defined as the integral of the Dirac delta function given as
{
x
H(x) =
∫−∞
𝛿(s) ds =
0 x<0
1 x > 0.
Fubini’s Theorem
Suppose f (x, y) is A × B measurable and if
(
∫A×B
f (x, y) d(x, y) =
∫A ∫B
∫A×B
|f (x, y)| d(x, y) < ∞ then
)
f (x, y) dy dx =
(
∫B ∫A
)
f (x, y) dx
dy.
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Appendix B
Probability Theory Formulae
Probability Concepts
Let A and B be events of the sample space Ω with probabilities ℙ(A) ∈ [0, 1] and ℙ(B) ∈
[0, 1], then
Complement:
ℙ(Ac ) = 1 − ℙ(A).
Conditional:
ℙ(A|B) =
ℙ(A ∩ B)
.
ℙ(B)
Independence: The events A and B are independent if and only if
ℙ(A ∩ B) = ℙ(A) ⋅ ℙ(B).
Mutually Exclusive: The events A and B are mutually exclusive if and only if
ℙ(A ∩ B) = 0.
Union:
ℙ(A ∪ B) = ℙ(A) + ℙ(B) − ℙ(A ∩ B).
Intersection:
ℙ(A ∩ B) = ℙ(A|B)ℙ(B) = ℙ(B|A)ℙ(A).
Partition:
ℙ(A) = ℙ(A ∩ B) + ℙ(A ∩ Bc ) = ℙ(A|B)ℙ(B) + ℙ(A|Bc )ℙ(Bc ).
Bayes’ Rule
Let A and B be events of the sample space Ω with probabilities ℙ(A) ∈ [0, 1] and ℙ(B) ∈
[0, 1], then
ℙ(B|A)ℙ(A)
ℙ(A|B) =
.
ℙ(B)
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Appendix B
Indicator Function
The indicator function 1IA of an event A of a sample space Ω is a function 1IA ∶ Ω → ℝ
defined as
{
1 if 𝜔 ∈ A
1IA (𝜔) =
0 if 𝜔 ∈ Ac .
Properties: For events A and B of the sample space Ω
1IAc = 1 − 1IA ,
𝔼(1IA ) = ℙ(A),
1IA∩B = 1IA 1IB ,
Var (1IA ) = ℙ(A)ℙ(A ),
c
1IA∪B = 1IA + 1IB − 1IA 1IB
Cov (1IA , 1IB ) = ℙ(A ∩ B) − ℙ(A)ℙ(B).
Discrete Random Variables
Univariate Case
Let X be a discrete random variable whose possible values are x = x1 , x2 , . . . , and let ℙ(X = x)
be the probability mass function.
Total Probability of All Possible Values:
∞
∑
ℙ(X = xk ) = 1.
k=1
Cumulative Distribution Function:
ℙ(X ≤ xn ) =
n
∑
ℙ(X = xk ).
k=1
Expectation:
𝔼(X) = 𝜇 =
∞
∑
xk ℙ(X = xk ).
k=1
Variance:
Var (X) = 𝜎 2
= 𝔼[(X − 𝜇)2 ]
=
∞
∑
(xk − 𝜇)2 ℙ(X = xk )
k=1
=
∞
∑
xk2 ℙ(X = xk ) − 𝜇 2
k=1
= 𝔼(X 2 ) − [𝔼(X)]2 .
Moment Generating Function:
MX (t) = 𝔼(e ) =
tX
∞
∑
k=1
etxk ℙ(X = xk ),
t ∈ ℝ.
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Appendix B
343
Characteristic Function:
𝜑X (t) = 𝔼(eitX ) =
∞
∑
eitxk ℙ(X = xk ),
i=
√
−1 and t ∈ ℝ.
k=1
Bivariate Case
Let X and Y be discrete random variables whose possible values are x = x1 , x2 , . . . and y =
y1 , y2 , . . . , respectively, and let ℙ(X = x, Y = y) be the joint probability mass function.
Total Probability of All Possible Values:
∞ ∞
∑
∑
ℙ(X = xj , Y = yk ) = 1.
j=1 k=1
Joint Cumulative Distribution Function:
ℙ(X ≤ xn , Y ≤ ym ) =
n ∑
m
∑
ℙ(X = xj , Y = yk ).
j=1 k=1
Marginal Probability Mass Function:
ℙ(X = x) =
∞
∑
ℙ(X = x, Y = yk ),
ℙ(Y = y) =
k=1
∞
∑
ℙ(X = xj , Y = y).
j=1
Conditional Probability Mass Function:
ℙ(X = x|Y = y) =
ℙ(X = x, Y = y)
,
ℙ(Y = y)
ℙ(Y = y|X = x) =
ℙ(X = x, Y = y)
.
ℙ(X = x)
Conditional Expectation:
𝔼(X|Y) = 𝜇x|y =
n
∑
xj ℙ(X = xj |Y = y),
𝔼(Y|X) = 𝜇y|x =
j=1
n
∑
k=1
Conditional Variance:
2
Var (X|Y) = 𝜎x|y
= 𝔼[(X − 𝜇x|y )2 |Y]
=
∞
∑
(xj − 𝜇x|y )2 ℙ(X = xj |Y = y)
j=1
=
∞
∑
j=1
2
xj2 ℙ(X = xj |Y = y) − 𝜇x|y
yk ℙ(Y = yk |X = x).
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Appendix B
2
Var (Y|X) = 𝜎y|x
= 𝔼[(Y − 𝜇y|x )2 |X]
=
∞
∑
(yk − 𝜇y|x )2 ℙ(Y = yk |X = x)
k=1
=
∞
∑
2
y2k ℙ(Y = yk |X = x) − 𝜇y|x
.
k=1
Covariance: For 𝔼(X) = 𝜇x and 𝔼(Y) = 𝜇y
Cov (X, Y) = 𝔼[(X − 𝜇x )(Y − 𝜇y )]
=
∞ ∞
∑
∑
(xj − 𝜇x )(yk − 𝜇y )ℙ(X = xj , Y = yk )
j=1 k=1
=
∞
∞ ∑
∑
xj yk ℙ(X = xj , Y = yk ) − 𝜇x 𝜇y
j=1 k=1
= 𝔼(XY) − 𝔼(X)𝔼(Y).
Joint Moment Generating Function: For s, t ∈ ℝ
MXY (s, t) = 𝔼(esX+tY ) =
∞ ∞
∑
∑
esxj +tyk ℙ(X = xj , Y = yk ).
j=1 k=1
Joint Characteristic Function: For i =
𝜑XY (s, t) = 𝔼(e
isX+itY
√
−1 and s, t ∈ ℝ
)=
∞ ∞
∑
∑
eisxj +ityk ℙ(X = xj , Y = yk ).
j=1 k=1
Independence: X and Y are independent if and only if
• ℙ(X = x, Y = y) = ℙ(X = x)ℙ(Y = y).
• MXY (s, t) = 𝔼(esX+tY ) = 𝔼(esX )𝔼(etY ) = MX (s)MY (t).
• 𝜑XY (s, t) = 𝔼(eisX+itY ) = 𝔼(eisX )𝔼(eitY ) = 𝜑X (s)𝜑Y (t).
Continuous Random Variables
Univariate Case
Let X be a continuous random variable whose values x ∈ ℝ and let fX (x) be the probability
density function.
Total Probability of All Possible Values:
∞
∫−∞
fX (x) dx = 1.
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Appendix B
345
Evaluating Probability:
b
ℙ(a ≤ X ≤ b) =
fX (x) dx.
∫a
Cumulative Distribution Function:
x
FX (x) = ℙ(X ≤ x) =
∫−∞
fX (x) dx.
Probability Density Function:
d
F (x).
dx X
fX (x) =
Expectation:
∞
𝔼(X) = 𝜇 =
∫−∞
xfX (x) dx.
Variance:
∞
Var (X) = 𝜎 2 =
∫−∞
∞
(x − 𝜇)2 fX (x) dx =
∫−∞
x2 fX (x) dx − 𝜇2 = 𝔼(X 2 ) − [𝔼(X)]2 .
Moment Generating Function:
∞
MX (t) = 𝔼(etX ) =
∫−∞
etx fX (x) dx,
t ∈ ℝ.
Characteristic Function:
𝜑X (t) = 𝔼(eitX ) =
∞
∫−∞
eitx fX (x) dx,
i=
√
−1 and t ∈ ℝ.
Probability Density Function of a Dependent Variable:
Let the random variable Y = g(X). If g is monotonic then the probability density function of
Y is
(
|−1
)| d
fY (y) = fX g−1 (y) || g−1 (y)||
| dy
|
where g−1 denotes the inverse function.
Bivariate Case
Let X and Y be two continuous random variables whose values x ∈ ℝ and y ∈ ℝ, and let
fXY (x, y) be the joint probability density function.
Total Probability of All Possible Values:
∞
∞
∫−∞ ∫−∞
∞
fXY (x, y) dx dy =
∞
∫−∞ ∫−∞
fXY (x, y) dydx = 1.
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Appendix B
Joint Cumulative Distribution Function:
x
FXY (x, y) = ℙ(X ≤ x, Y ≤ y) =
y
∫−∞ ∫−∞
y
fXY (x, y) dydx =
x
∫−∞ ∫−∞
fXY (x, y) dx dy.
Evaluating Joint Probability:
ℙ(xa ≤ X ≤ xb , ya ≤ Y ≤ yb ) =
xb
yb
yb
=
fXY (x, y) dydx
∫xa ∫ya
xb
fXY (x, y) dx dy
∫ya ∫xa
= FXY (xb , yb ) − FXY (xb , ya ) − FXY (xa , yb ) + FXY (xa , ya ).
Joint Probability Density Function:
fXY (x, y) =
𝜕2
𝜕2
FXY (x, y) =
F (x, y).
𝜕x𝜕y
𝜕y𝜕x XY
Marginal Probability Density Function:
∞
fX (x) =
∫−∞
∞
fXY (x, y) dy,
fY (y) =
∫−∞
fXY (x, y) dx.
Conditional Probability Density Function:
fX|Y (x|y) =
fXY (x, y)
,
fY (y)
fY|X (y|x) =
fXY (x, y)
.
fX (x)
Conditional Expectation:
∞
𝔼(X|Y) = 𝜇x|y =
∫−∞
∞
xfX|Y (x|y) dx,
𝔼(Y|X) = 𝜇y|x =
∫−∞
Conditional Variance:
2
Var (X|Y) = 𝜎x|y
= 𝔼[(X − 𝜇x|y )2 |Y]
∞
=
∫−∞
(x − 𝜇x|y )2 fX|Y (x|y) dx
∞
=
∫−∞
2
x2 fX|Y (x|y) dx − 𝜇x|y
2
Var (Y|X) = 𝜎y|x
= 𝔼[(Y − 𝜇y|x )2 |X]
∞
=
∫−∞
(y − 𝜇y|x )2 fY|X (y|x) dy
yfY|X (y|x) dy.
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Appendix B
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347
∞
=
∫−∞
2
y2 fY|X (y|x) dy − 𝜇y|x
.
Covariance: For 𝔼(X) = 𝜇x and 𝔼(Y) = 𝜇y
Cov (X, Y) = 𝔼[(X − 𝜇x )(Y − 𝜇y )]
∞
=
∫−∞ ∫−∞
∞
=
∞
(x − 𝜇x )(y − 𝜇y )fXY (x, y) dy dx
∞
∫−∞ ∫−∞
xyfXY (x, y) dy dx − 𝜇x 𝜇y
= 𝔼(XY) − 𝔼(X)𝔼(Y).
Joint Moment Generating Function: For t, s ∈ ℝ
∞
MXY (s, t) = 𝔼(esX+tY ) =
Joint Characteristic Function: For i =
∞
esx+ty fXY (x, y) dy dx.
∫−∞ ∫−∞
√
−1 and t, s ∈ ℝ
∞
𝜑XY (s, t) = 𝔼(eisX+itY ) =
∞
∫−∞ ∫−∞
eisx+ity fXY (x, y) dy dx.
Independence: X and Y are independent if and only if
• fXY (x, y) = fX (x)fY (y).
• MXY (s, t) = 𝔼(esX+tY ) = 𝔼(esX )𝔼(etY ) = MX (s)MY (t).
• 𝜑XY (s, t) = 𝔼(eisX+itY ) = 𝔼(eisX )𝔼(eitY ) = 𝜑X (s)𝜑Y (t).
Joint Probability Density Function of Dependent Variables: Let the random variables U =
g(X, Y), V = h(X, Y). If u = g(x, y) and 𝑣 = h(x, y can be uniquely solved for x and y in terms
of u and 𝑣 with solutions given by, say, x = p(u, 𝑣) and y = q(u, 𝑣) and the functions g and h
have continuous partial derivatives at all points (x, y) such that the determinant
| 𝜕g
|
| 𝜕x
J(x, y) = || 𝜕h
|
| 𝜕x
|
𝜕g ||
𝜕y || 𝜕g 𝜕h 𝜕g 𝜕h
𝜕h || = 𝜕x 𝜕y − 𝜕y 𝜕x ≠ 0
𝜕y ||
then the joint probability density function of U and V is
fUV (u, 𝑣) = fXY (x, y)|J(x, y)|−1
where x = p(u, 𝑣) and y = q(u, 𝑣).
Properties of Expectation and Variance
Let X and Y be two random variables and for constants a and b
𝔼(aX + b) = a𝔼(X) + b,
𝔼(aX + bY) = a𝔼(X) + b𝔼(Y),
Var (aX + b) = a2 Var (X)
Var (aX + bY) = a2 Var (X) + b2 Var (Y) + 2abCov (X, Y).
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Appendix B
Properties of Moment Generating and Characteristic Functions
If a random variable X has moments up to k-th order where k is a non-negative integer, then
𝔼(X k ) =
where i =
√
−1.
k
|
|
dk
| = i−k d 𝜑 (t)|
M
(t)
X
X |
|
k
k
dt
dt
|t=0
|t=0
If the bivariate random variables X and Y have moments up to m + n = k where m, n and k are
non-negative integers, then
|
|
dk
dk
MXY (s, t)||
= i−k m n 𝜑XY (s, t)||
m
n
ds dt
ds dt
|s=0,t=0
|s=0,t=0
𝔼(X m Y n ) =
where i =
√
−1.
Correlation Coefficient
Let X and Y be two random variables with means 𝜇x and 𝜇y and variances 𝜎x2 and 𝜎y2 . The
correlation coefficient 𝜌xy between X and Y is defined as
𝜌xy = √
Cov (X, Y)
=
𝔼[(X − 𝜇x )(Y − 𝜇y )]
𝜎x 𝜎y
Var (X)Var (Y)
.
Important information:
•
•
•
•
•
𝜌xy measures only the linear dependency between X and Y.
−1 ≤ 𝜌xy ≤ 1.
If 𝜌xy = 0 then X and Y are uncorrelated.
If X and Y are independent then 𝜌xy = 0. However, the converse is not true.
If X and Y are jointly normally distributed then X and Y are independent if and only if
𝜌XY = 0.
Convolution
If X and Y are independent discrete random variables with probability mass functions ℙ(X = x)
and ℙ(Y = y), respectively, then the probability mass function for Z = X + Y is
ℙ(Z = z) =
∑
ℙ(X = x)ℙ(Y = z − x) =
x
∑
ℙ(X = z − y)ℙ(Y = y).
y
If X and Y are independent continuous random variables with probability density functions
fX (x) and fY (y), respectively, then the probability density function for Z = X + Y is
∞
fZ (z) =
∫−∞
∞
fX (x)fY (z − x) dx =
∫−∞
fX (z − y)fY (y) dy.
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Discrete Distributions
Bernoulli: A random variable X is said to follow a Bernoulli distribution, X ∼ Bernoulli(p)
where p ∈ [0, 1] is the probability of success and the probability mass function is given as
P(X = x) = px (1 − p)1−x ,
x = 0, 1
where 𝔼(X) = p and Var (X) = p − p2 . The moment generating function is
MX (t) = 1 − p + pet ,
t∈ℝ
and the corresponding characteristic function is
𝜑X (t) = 1 − p + peit ,
i=
√
−1 and t ∈ ℝ.
Geometric: A random variable X is said to follow a geometric distribution, X ∼ Geometric(p),
where p ∈ [0, 1] is the probability of success and the probability mass function is given as
ℙ(X = x) = p(1 − p)x−1 ,
where 𝔼(X) =
x = 1, 2, . . .
1−p
1
. The moment generating function is
and Var (X) =
p
p2
MX (t) =
p
,
1 − (1 − p)et
t∈ℝ
and the corresponding characteristic function is
𝜑X (t) =
p
,
1 − (1 − p)eit
i=
√
−1 and t ∈ ℝ.
Binomial: A random variable X is said to follow a binomial distribution, X ∼ Binomial(n, p),
p ∈ [0, 1], where p ∈ [0, 1] is the probability of success and n ∈ ℕ0 is the number of trials and
the probability mass function is given as
( )
n x
ℙ(X = x) =
p (1 − p)n−x ,
x
x = 0, 1, 2, . . . , n
where 𝔼(X) = np and Var (X) = np(1 − p). The moment generating function is
MX (t) = (1 − p + pet )n ,
t∈ℝ
and the corresponding characteristic function is
𝜑X (t) = (1 − p + peit )n ,
i=
√
−1 and t ∈ ℝ.
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Appendix B
Negative Binomial: A random variable X is said to follow a negative binomial distribution,
X ∼ NB(r, p) where p ∈ [0, 1] is the probability of success and r is the number of successes
accumulated and the probability mass function is given as
(
)
x−1
ℙ(X = x) = r − 1 pr (1 − p)n−r , x = r, r + 1, r + 2, . . .
where 𝔼(X) =
r(1 − p)
r
. The moment generating function is
and Var (X) =
p
p2
)r
(
1−p
MX (t) =
, t < − log p
1 − pet
and the corresponding characteristic function is
)r
(
√
1−p
, i = −1 and t ∈ ℝ.
MX (t) =
it
1 − pe
Poisson: A random variable X is said to follow a Poisson distribution, X ∼ Poisson(𝜆), 𝜆 > 0
with probability mass function given as
ℙ(X = x) =
e−𝜆 𝜆x
,
x!
x = 0, 1, 2, . . .
where 𝔼(X) = 𝜆 and Var (X) = 𝜆. The moment generating function is
MX (t) = e𝜆(e −1) ,
t
t∈ℝ
and the corresponding characteristic function is
𝜑X (t) = e𝜆(e
it −1)
,
i=
√
−1 and t ∈ ℝ.
Continuous Distributions
Uniform: A random variable X is said to follow a uniform distribution, X ∼ 𝒰(a, b), a < b
with probability density function given as
fX (x) =
where 𝔼(X) =
1
,
b−a
a<x<b
(b − a)2
a+b
and Var (X) =
. The moment generating function is
2
12
MX (t) =
etb − eta
,
t(b − a)
t∈ℝ
and the corresponding characteristic function is
𝜑X (t) =
etb − eita
,
it(b − a)
i=
√
−1 and t ∈ ℝ.
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Appendix B
351
Normal: A random variable X is said to follow a normal distribution, X ∼ 𝒩(𝜇, 𝜎 2 ), 𝜇 ∈ ℝ,
𝜎 2 > 0 with probability density function given as
−1
1
fX (x) = √ e 2
𝜎 2𝜋
(
)
x−𝜇 2
𝜎
,
x∈ℝ
where 𝔼(X) = 𝜇 and Var (X) = 𝜎 2 . The moment generating function is
1 2 2
MX (t) = e𝜇t+ 2 𝜎 t ,
t∈ℝ
and the corresponding characteristic function is
1 2 2
t
𝜑X (t) = ei𝜇t− 2 𝜎
,
i=
√
−1 and t ∈ ℝ.
Lognormal: A random variable X is said to follow a lognormal distribution, X ∼ log-𝒩(𝜇, 𝜎 2 ),
𝜇 ∈ ℝ, 𝜎 2 > 0 with probability density function given as
fX (x) =
−1
1
√ e 2
x𝜎 2𝜋
1 2
(
)
log x−𝜇 2
𝜎
,
x>0
where 𝔼(X) = e𝜇+ 2 𝜎 and Var (X) = (e𝜎 − 1)e2𝜇+𝜎 . The moment generating function is
2
MX (t) =
2
∞ n
∑
t n𝜇+ 12 n2 𝜎 2
e
,
n!
n=0
t≤0
and the corresponding characteristic function is
𝜑X (t) =
∞
∑
(it)n
n=0
n!
1 2 2
𝜎
en𝜇+ 2 n
,
i=
√
−1 and t ∈ ℝ.
Exponential: A random variable X is said to follow an exponential distribution, X ∼ Exp(𝜆),
𝜆 > 0 with probability density function
fX (x) = 𝜆e−𝜆x ,
where 𝔼(X) =
x≥0
1
1
and Var (X) = 2 . The moment generating function is
𝜆
𝜆
MX (t) =
𝜆
,
𝜆−t
t<𝜆
and the corresponding characteristic function is
𝜑X (t) =
𝜆
,
𝜆 − it
i=
√
−1 and t ∈ ℝ.
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Appendix B
Gamma: A random variable X is said to follow a gamma distribution, X ∼ Gamma(𝛼, 𝜆), 𝛼, 𝜆 >
0 with probability density function given as
𝜆e−𝜆x (𝜆x)𝛼−1
,
Γ(𝛼)
fX (x) =
such that
∞
Γ(𝛼) =
where 𝔼(X) =
x≥0
e−x x𝛼−1 dx
∫0
𝛼
𝛼
and Var (X) = 2 . The moment generating function is
𝜆
𝜆
(
)
𝜆 𝛼
MX (t) =
, t<𝜆
𝜆−t
and the corresponding characteristic function is
𝜑X (t) =
(
𝜆
𝜆 − it
)𝛼
,
i=
√
−1 and t ∈ ℝ.
Chi-Square: A random variable X is said to follow a chi-square distribution, X ∼ 𝜒 2 (𝜈), 𝜈 ∈ ℕ
with probability density function given as
fX (x) =
such that
Γ
𝜈
1
−1 − x
( )x2 e 2 ,
2 Γ 𝜈2
𝜈
2
x≥0
∞
( )
𝜈
𝜈
e−x x 2 −1 dx
=
∫0
2
where 𝔼(X) = 𝜈 and Var (X) = 2𝜈. The moment generating function is
𝜈
MX (t) = (1 − 2t)− 2 ,
−
1
1
<t<
2
2
and the corresponding characteristic function is
𝜈
MX (t) = (1 − 2it)− 2 ,
i=
√
−1 and t ∈ ℝ.
Bivariate Normal: The random variables X and Y with means 𝜇x , 𝜇y , variances 𝜎x2 , 𝜎y2 ,
and correlation coefficient 𝜌xy ∈ (−1, 1) is said to follow a joint normal distribution,
[
] [
]
[ ]
𝜎x2 𝜌xy 𝜎x 𝜎y
𝜇x
Var(X) Cov (X, Y)
and 𝚺 =
=
(X, Y) ∼ 𝒩2 (𝝁, 𝚺) where 𝝁 =
𝜇y
Cov (X, Y) Var(Y)
𝜌xy 𝜎x 𝜎y 𝜎y2
with joint probability density function given as
fXY (x, y) =
1
√
2𝜋𝜎x 𝜎y 1 − 𝜌2xy
−
e
1
2(1−𝜌2xy )
[
(
x−𝜇x
𝜎x
)2
(
−2𝜌
x−𝜇x
𝜎x
]
)( y−𝜇 ) ( y−𝜇 )2
y
y
+
𝜎
𝜎
y
y
,
x, y ∈ ℝ.
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353
The moment generating function is
1
MXY (s, t) = e𝜇x s+𝜇y t+ 2 (𝜎x s
2 2 +2𝜌
2 2
xy 𝜎x 𝜎y st+𝜎y t )
,
s, t ∈ ℝ
and the corresponding characteristic function is
1
2 2 +2𝜌
𝜑XY (s, t) = ei𝜇x s+i𝜇y t− 2 (𝜎x s
2 2
xy 𝜎x 𝜎y st+𝜎y t )
,
i=
√
−1 and s, t ∈ ℝ.
Multivariate Normal: The random vector X = (X1 , X2 , . . . , Xn ) is said to follow a multivariate
normal distribution, X ∼ 𝒩n (𝝁, 𝚺) where
( )
Cov(X1 , X2 )
⎡ Var X1
⎡𝔼(X1 )⎤
⎢Cov(X1 , X2 ) Var(X2 )
⎢𝔼(X2 )⎥
and 𝚺 = ⎢
𝝁=⎢
⋮ ⎥
⋮
⋮
⎥
⎢
⎢
⎣Cov(X1 , Xn ) Cov(X2 , Xn )
⎣𝔼(Xn )⎦
. . . Cov(X1 , Xn )⎤
. . . Cov(X2 , Xn )⎥
⎥
⋱
⋮
⎥
. . . Var(Xn ) ⎦
with probability density function given as
fX (x1 , x2 , . . . , xn ) =
1
n
2
(2𝜋) |𝚺|
1
2
1 T −1
e− 2 X 𝚺 X .
The moment generating function is
MX (t1 , t2 , . . . , tn ) = e𝝁
𝚺t ,
t = (t1 , t2 , . . . , tn )T
𝚺t ,
t = (t1 , t2 , . . . , tn )T .
T t+ 1 tT
2
and the corresponding characteristic function is
𝜑X (t1 , t2 , . . . , tn ) = ei𝝁
T t− 1 tT
2
Integrable and Square Integrable Random Variables
Let X be a real-valued random variable
• If 𝔼(|X|) < ∞ then X is an integrable random variable.
• If 𝔼(X 2 ) < ∞ then X is a square integrable random variable.
Convergence of Random Variables
Let X, X1 , X2 , . . . , Xn be a sequence of random variables. Then
a.s.
(a) Xn −−−→ X converges almost surely if
)
(
ℙ lim Xn = X = 1.
n→∞
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Appendix B
r
(b) Xn −−→ X converges in the r-th mean, r ≥ 1, if 𝔼(|Xnr |) < ∞ and
lim 𝔼(|Xn − X|r ) = 0.
n→∞
P
(c) Xn −−→ X converges in probability, if for all 𝜀 > 0,
lim ℙ(|Xn − X| ≥ 𝜀) = 0.
n→∞
D
(d) Xn −−→ X converges in distribution, if for all x ∈ ℝ,
lim ℙ(Xn ≤ x) = ℙ(X ≤ x).
n→∞
Relationship Between Modes of Convergence
For any r ≥ 1
{
a.s
Xn −−−→ X
r
Xn −−→ X
If r > s ≥ 1 then
}
P
D
⇒ {Xn −−→ X} ⇒ {Xn −−→ X}.
r
s
{Xn −−→ X} ⇒ {Xn −−→ X}.
Dominated Convergence Theorem
a.s.
If Xn −−−→ X and for any n ∈ ℕ we have |Xn | < Y for some Y such that 𝔼(|Y|) < ∞, then
𝔼(|Xn |) < ∞ and
lim 𝔼(Xn ) = 𝔼(X).
n→∞
Monotone Convergence Theorem
a.s.
If 0 ≤ Xn ≤ Xn+1 and Xn −−−→ X for any n ∈ ℕ then
lim 𝔼(Xn ) = 𝔼(X).
n→∞
The Weak Law of Large Numbers
Let X1 , X2 , . . . , Xn be a sequence of independent and identically distributed random variables
with common mean 𝜇 ∈ ℝ. Then for any 𝜀 > 0,
(
ℙ
)
| X + X2 + . . . + Xn
|
lim || 1
− 𝜇 || ≥ 𝜀
n→∞ |
n
|
= 0.
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The Strong Law of Large Numbers
Let X1 , X2 , . . . , Xn be a sequence of independent and identically distributed random variables
with common mean 𝜇 ∈ ℝ. Then
)
(
X1 + X2 + . . . + Xn
= 𝜇 = 1.
ℙ lim
n→∞
n
The Central Limit Theorem
Let X1 , X2 , . . . , Xn be a sequence of independent and identically distributed random variables
with common mean 𝜇 ∈ ℝ and variance 𝜎 2 > 0 and denote the sample mean as
X=
X1 + X2 + . . . + Xn
n
( )
( )
𝜎2
where 𝔼 X = 𝜇 and Var X =
. By defining
n
( )
X−𝔼 X
X−𝜇
Zn = √
( ) = 𝜎∕√n
Var X
then for n → ∞,
X−𝜇 D
√ −−→ 𝒩(0, 1).
n→∞ 𝜎∕ n
lim Zn = lim
n→∞
That is, Zn follows a standard normal distribution asymptotically.
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Appendix C
Differential Equations Formulae
Separable Equations
The form
dy
= f (x)g(y)
dx
has a solution
1
dy =
f (x) dx.
∫
∫ g(y)
If g(y) is a linear equation and if y1 and y2 are two solutions, then y3 = ay1 + by2 is also a
solution for constant a and b.
First-Order Ordinary Differential Equations
General Linear Equation: The general form of a first-order ordinary differential equation
dy
+ f (x)y = g(x)
dx
has a solution
y = I(x)−1
where I(x) = e∫
f (x)dx
∫
I(u)g(u) du + C
is the integrating factor and C is a constant.
Bernoulli Differential Equation: For n ≠ 1, the Bernoulli differential equation has the form
dy
+ P(x)y = Q(x)yn
dx
which, by setting 𝑤 =
1
, can be transformed to a general linear ordinary differential
yn−1
equation of the form
d𝑤
+ (1 − n)P(x)𝑤 = (1 − n)Q(x)
dx
with a particular solution
𝑤 = (1 − n)I(x)−1
∫
I(u)Q(u) du
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358
where I(x) = e(1−n) ∫
equation becomes
4:37 P.M.
Appendix C
P(x)dx
is the integrating factor. The solution to the Bernoulli differential
{
y=
−1
(1 − n)I(x)
∫
}− 1
n−1
I(u)Q(u) du
+C
where C is a constant value.
Second-Order Ordinary Differential Equations
General Linear Equation: For a homogeneous equation,
a
d2 y
dy
+ b + cy = 0.
dx
dx2
By setting y = eux the differential equation has a general solution based on the characteristic
equation
au2 + bu + c = 0
such that m1 and m2 are the roots of the quadratic equation, and if
• m1 , m2 ∈ ℝ, m1 ≠ m2 then y = Aem1 x + Bem2 x
• m1 , m2 ∈ ℝ, m1 = m2 = m then y = emx (A + Bx)
• m1 , m2 ∈ ℂ, m1 = 𝛼 + i𝛽, m2 = 𝛼 − i𝛽 then y = e𝛼x [A cos(𝛽x) + B sin(𝛽x)]
where A, B are constants.
Cauchy–Euler Equation: For a homogeneous equation,
ax2
d2 y
dy
+ bx + cy = 0.
dx
dx2
By setting y = xu the Cauchy–Euler equation has a general solution based on the characteristic
equation
au2 + (b − a)u + c = 0
such that m1 and m2 are the roots of the quadratic equation, and if
• m1 , m2 ∈ ℝ, m1 ≠ m2 then y = Axm1 + Bxm2
• m1 , m2 ∈ ℝ, m1 = m2 = m then y = xm (A + B log x)
• m1 , m2 ∈ ℂ, m1 = 𝛼 + i𝛽, m2 = 𝛼 − i𝛽 then y = x𝛼 [A cos(𝛽 log x) + B sin(𝛽 log x)]
where A, B are constants.
Variation of Parameters: For a general non-homogeneous second-order differential equation,
a(x)
dy
d2 y
+ b(x) + c(x) = f (x)
dx
dx2
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Appendix C
359
has the solution
y = yc + y p
where yc , the complementary function, satisfies the homogeneous equation
a(x)
d2 yc
dy
+ b(x) c + c(x) = 0
2
dx
dx
and yp , the particular integral, satisfies
a(x)
d 2 yp
+ b(x)
dx2
dyp
+ c(x) = f (x).
dx
(2)
Let yc = C1 y(1)
c (x) + C2 yc (x) where C1 and C2 are constants, then the particular solution to
the non-homogeneous second-order differential equation is
yp = −y(1)
c (x)
y(2)
c (x)f (x)
∫ a(x)W(y(1) (x), y(2) (x))
c
c
dx + y(2)
c (x)
y(1)
c (x)f (x)
∫ a(x)W(y(1) (x), y(2) (x))
c
c
dx
(2)
where W(y(1)
c (x), yc (x)) is the Wronskian defined as
| y(1) (x)
|
y(2)
| c
c (x) |
|
|
|
|
d (2)
d (1)
(2)
(1)
(2)
|
W(y(1)
d (2) || = yc (x) yc (x) − yc (x) yc (x) ≠ 0.
c (x), yc (x)) = | d (1)
y (x)|
dx
dx
| yc (x)
dx c
| dx
|
|
|
|
|
Homogeneous Heat Equations
Initial Value Problem on an Infinite Interval: The diffusion equation of the form
𝜕u
𝜕2 u
= 𝛼 2,
𝜕t
𝜕x
𝛼 > 0,
−∞ < x < ∞,
t>0
with initial condition u(x, 0) = f (x) has a solution
∞
(x−z)2
1
f (z)e− 4𝛼t dz.
u(x, t) = √
2 𝜋𝛼t ∫−∞
Initial Value Problem on a Semi-Infinite Interval: The diffusion equation of the form
𝜕u
𝜕2u
= 𝛼 2,
𝜕t
𝜕x
𝛼 > 0,
0 ≤ x < ∞,
t>0
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Appendix C
with
• initial condition u(x, 0) = f (x) and boundary condition u(0, t) = 0 has a solution
[
]
∞
2
(x+z)2
1
− (x−z)
−
f (z) e 4𝛼t − e 4𝛼t dz
u(x, t) = √
2 𝜋𝛼t ∫0
• initial condition u(x, 0) = f (x) and boundary condition ux (0, t) = 0 has a solution
]
[
∞
2
(x+z)2
1
− (x−z)
−
u(x, t) = √
f (z) e 4𝛼t + e 4𝛼t dz
2 𝜋𝛼t ∫0
• initial condition u(x, 0) = 0 and boundary condition u(0, t) = g(t) has a solution
t
2
1
x
− x
u(x, t) = √
g(𝑤)e 4𝛼(t−𝑤)
√
∫
2 𝜋𝛼 0
t−𝑤
d𝑤.
Stochastic Differential Equations
Suppose Xt , Yt and Zt are Itō processes satisfying the following stochastic differential
equations:
dXt = 𝜇(Xt , t)dt + 𝜎(Xt , t)dWtx
y
dYt = 𝜇(Yt , t)dt + 𝜎(Yt , t))dWt
dZt = 𝜇(Zt , t)dt + 𝜎(Zt , t)dWtz
y
where Wtx , Wt and Wtz are standard Wiener processes.
Reciprocal:
)
1
d
)
(
Xt
dXt 2
dXt
+
.
( ) =−
Xt
Xt
1
Xt
(
Product:
d(Xt Yt ) dXt dYt dXt dYt
=
+
+
.
Xt Yt
Xt
Yt
Xt Yt
Quotient:
)
Xt
d
(
)
Yt
dYt 2
dXt dYt dXt dYt
−
−
+
.
( ) =
Xt
Yt
Xt Yt
Yt
Xt
Yt
(
Product and Quotient I:
(
)
Xt Yt
d
)
(
Zt
dXt dYt dZt dXt dYt dXt dZt dYt dZt
dZt 2
=
+
−
+
−
−
+
.
(
)
Xt
Yt
Zt
Xt Yt
Xt Zt
Yt Zt
Zt
Xt Y t
Zt
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4:37 P.M.
361
Product and Quotient II:
(
)
Xt
d
Yt Zt
dXt dYt dZt dXt dYt dXt dZt dYt dZt
−
−
−
−
+
.
(
) =
Xt
Yt
Zt
Xt Y t
Xt Zt
Yt Zt
Xt
Yt Zt
Black–Scholes Model
Black–Scholes Equation (Continuous Dividend Yield): At time t, let the asset price St follows
a geometric Brownian motion
dSt
= (𝜇 − D)dt + 𝜎dWt
St
where 𝜇 is the drift parameter, D is the continuous dividend yield, 𝜎 is the volatility parameter
and Wt is a standard Wiener process. For a European-style derivative V(St , t) written on the
asset St , it satisfies the Black–Scholes equation with continuous dividend yield
𝜕V 1 2 2 𝜕 2 V
𝜕V
+ 𝜎 St 2 + (r − D)St
− rV(St , t) = 0
𝜕t
2
𝜕St
𝜕St
where r is the risk-free interest rate. The parameters 𝜇, r, D and 𝜎 can be either constants,
deterministic functions or stochastic processes.
European Options: For a European option having the payoff
Ψ(ST ) = max{𝛿(ST − K), 0}
where 𝛿 ∈ { − 1, 1}, K is the strike price and T is the option expiry time, and if r, D and 𝜎 are
constants the European option price at time t < T is
V(St , t; K, T) = 𝛿St e−D(T−t) Φ(𝛿d+ ) − 𝛿Ke−r(T−t) Φ(𝛿d− )
log(St ∕K) + (r − D ± 12 𝜎 2 )(T − t)
where d± =
and Φ(⋅) is the cumulative distribution function
√
𝜎 T −t
of a standard normal.
Reflection Principle: If V(St , t) is a solution of the Black–Scholes equation then for a constant
B > 0, the function
(
U(St , t) =
St
B
)2𝛼 ( 2 )
B
V
,t ,
St
also satisfies the Black–Scholes equation.
(
1
𝛼=
2
)
1−
r−D
1 2
𝜎
2
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4:37 P.M.
Appendix C
Black Model
Black Equation: At time t, let the asset price St follows a geometric Brownian motion
dSt
= (𝜇 − D)dt + 𝜎dWt
St
where 𝜇 is the drift parameter, D is the continuous dividend yield, 𝜎 is the volatility parameter
and Wt is a standard Wiener process. Consider the price of a futures contract maturing at time
T > t on the asset St as
F(t, T) = St e(r−D)(T−t)
where r is the risk-free interest rate. For a European option on futures V(F(t, T), t) written on
a futures contract F(t, T), it satisfies the Black equation
𝜕2 V
𝜕V 1 2
+ 𝜎 F(t, T)2 2 − rV(F(t, T), t) = 0.
𝜕t
2
𝜕F
The parameters 𝜇, r, D and 𝜎 can be either constants, deterministic functions or stochastic
processes.
European Options on Futures: For a European option on futures having the payoff
Ψ(F(T, T)) = max{𝛿(F(T, T) − K), 0}
where 𝛿 ∈ { − 1, 1}, K is the strike price and T is the option expiry time, and if r and 𝜎 are
constants the price of a European option on futures at time t < T is
V(F(t, T), t; K, T) = 𝛿e−r(T−t) [F(t, T)Φ(𝛿d+ ) − KΦ(𝛿d− )]
log(F(t, T)∕K) ± 12 𝜎 2 (T − t)
where d± =
and Φ(⋅) is the cumulative distribution function of a
√
𝜎 T −t
standard normal.
Reflection Principle: If V(F(t, T), t) is a solution of the Black equation then for a constant
B > 0, the function
(
)
F(t, T)
B2
V
,t
U(F(t, T), t) =
B
F(t, T)
also satisfies the Black equation.
Garman–Kohlhagen Model
Garman–Kohlhagen Equation: At time t, let the foreign-to-domestic exchange rate Xt follows
a geometric Brownian motion
dXt
= 𝜇dt + 𝜎dWt
Xt
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4:37 P.M.
363
where 𝜇 is the drift parameter, 𝜎 is the volatility parameter and Wt is a standard Wiener
process. For a European-style derivative V(Xt , t) which depends on Xt , it satisfies the
Garman–Kohlhagen equation
𝜕V
𝜕V 1 2 2 𝜕 2 V
+ (rd − rf )Xt
− rd V(Xt , t) = 0
+ 𝜎 Xt
𝜕t
2
𝜕Xt
𝜕Xt2
where rd and rf are the domestic and foreign currency risk-free interest rates. The parameters
𝜇, rd , rf and 𝜎 can be either constants, deterministic functions or stochastic processes.
European Options: For a European option having the payoff
Ψ(XT ) = max{𝛿(XT − K), 0}
where 𝛿 ∈ { − 1, 1}, K is the strike price and T is the option expiry time, and if rd , rf and 𝜎
are constants the European option price (domestic currency in one unit of foreign currency) at
time t < T is
V(Xt , t; K, T) = 𝛿Xt e−rf (T−t) Φ(𝛿d+ ) − 𝛿Ke−rd (T−t) Φ(𝛿d− )
log(Xt ∕K) + (rd − rf ± 12 𝜎 2 )(T − t)
where d± =
and Φ(⋅) is the cumulative distribution func√
𝜎 T −t
tion of a standard normal.
Reflection Principle: If V(Xt , t) is a solution of the Garman–Kohlhagen equation then for a
constant B > 0 the function
(
)
( )2𝛼 ( 2 )
rd − rf
Xt
B
1
V
,t , 𝛼 =
1− 1
U(Xt , t) =
B
Xt
2
𝜎2
2
also satisfies the Garman–Kohlhagen equation.
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4:38 P.M.
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both01.tex
Notation
SET NOTATION
∈
∉
Ω
ℰ
∅
A
Ac
|A|
ℕ
ℕ0
ℤ
ℤ+
ℝ
ℝ+
ℂ
A×B
a∼b
⊆
⊂
∩
∪
\
Δ
sup
inf
[a, b]
[a, b)
(a, b]
(a, b)
ℱ, 𝒢, ℋ
is an element of
is not an element of
sample space
universal set
empty set
subset of Ω
complement of set A
cardinality of A
set of natural numbers, {1, 2, 3, . . .}
set of natural numbers including zero, {0, 1, 2, . . .}
set of integers, {0, ±1, ±2, ±3, . . .}
set of positive integers, {1, 2, 3, . . .}
set of real numbers
set of positive real numbers, {x ∈ ℝ ∶ x > 0}
set of complex numbers
cartesian product of sets A and B, A × B = {(a, b) ∶ a ∈ A, b ∈ B}
a is equivalent to b
subset
proper subset
intersection
union
difference
symmetric difference
supremum or least upper bound
infimum or greatest lower bound
the closed interval {x ∈ ℝ ∶ a ≤ x ≤ b}
the interval {x ∈ ℝ ∶ a ≤ x < b}
the interval {x ∈ ℝ ∶ a < x ≤ b}
the open interval {x ∈ ℝ ∶ a < x < b}
𝜎-algebra (or 𝜎-fields)
V3 - 08/23/2014
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Page 369
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V3 - 08/23/2014
370
Notation
MATHEMATICAL NOTATION
x+
x−
⌊x⌋
⌈x⌉
x∨y
x∧y
i
∞
∃
∃!
∀
≈
p =⇒ q
p ⇐= q
p ⇐⇒ q
f ∶ X → Y
f (x)
lim f (x)
max{x, 0}
min{x, 0}
largest integer not greater than and equal to x, max{m ∈ ℤ | m ≤ x}
smallest integer greater than and equal to x, min{n ∈ ℤ | n ≥ x}
max{x, y}
min{x,
y}
√
−1
infinity
there exists
there exists a unique
for all
approximately equal to
p implies q
p is implied by q
p implies and is implied by q
f is a function where every element of X has an image in Y
the value of the function f at x
limit of f (x) as x tends to a
𝛿x, Δx
f −1 (x)
′
′′
f (x), f (x)
2
dy d y
,
dx dx2
b
∫ y dx, ∫a y dx
2
𝜕f 𝜕 f
,
𝜕xi 𝜕x2
increment of x
the inverse function of the function f (x)
the first and second-order derivative of the function f (x)
x→a
i
𝜕2f
𝜕xi 𝜕xj
loga x
log x
n
∑
ai
i=1
n
∏
ai
i=1
4:38 P.M.
first and second-order derivative of y with respect to x
the indefinite and definite integral of y with respect to x
first and second-order partial derivative of f with respect to xi
where f is a function on (x1 , x2 , . . . , xn )
second-order partial derivative of f with respect to xi and xj
where f is a function on (x1 , x2 , . . . , xn )
logarithm of x to the base a
natural logarithm of x
a1 + a2 + . . . + an
a1 × a2 × . . . × an
|a| )
(
m
√
n
a
modulus of a
n!
( )
n
k
𝛿(x)
H(x)
n factorial
n!
for n, k ∈ ℤ+
k!(n − k)!
Dirac delta function
Heaviside step function
m
an
Page 370
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both01.tex
Notation
Γ(t)
B(x, y)
a
|a|
a⋅b
a×b
M
MT
M−1
|M|
V3 - 08/23/2014
371
gamma function
beta function
a vector a
magnitude of a vector a
scalar or dot product of vectors a and b
vector or cross-product of vectors a and b
a matrix M
transpose of a matrix M
inverse of a square matrix M
determinant of a square matrix M
PROBABILITY NOTATION
A, B, C
1IA
ℙ, ℚ
ℙ(A)
ℙ(A|B)
X, Y, Z
X, Y, Z
ℙ(X = x)
fX (x)
FX (x), ℙ(X ≤ x)
MX (t)
𝜑X (t)
P(X = x, Y = y)
fXY (x, y)
events
indicator of the event A
probability measures
probability of event A
probability of event A conditional on event B
random variables
random vectors
probability mass function of a discrete random variable X
probability density function of a continuous random variable X
cumulative distribution function of a random variable X
moment generating function of a random variable X
characteristic function of a random variable X
joint probability mass function of discrete variables X and Y
joint probability density function of continuous random
variables X and Y
FXY (x, y), ℙ(X ≤ x, Y ≤ y) joint cumulative distribution function of random variables X
and Y
joint moment generating function of random variables X and Y
MXY (s, t)
joint characteristic function of random variables X and Y
𝜑XY (s, t)
p(x, t; y, T)
transition probability density of y at time T starting at time t at
point x
∼
is distributed as
≁
is not distributed as
∻
is approximately distributed as
a.s
−−−→
converges almost surely
−−→
converges in the r-th mean
r
P
−−→
D
−−→
d
X=Y
converges in probability
converges in distribution
X and Y are identically distributed random variables
4:38 P.M.
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372
X⟂
⟂Y
∕ Y
X⟂
⟂
𝔼(X)
𝔼ℚ (X)
𝔼[g(X)]
𝔼(X|ℱ)
Var(X)
Var(X|ℱ)
Cov(X, Y)
𝜌xy
Bernoulli(p)
Geometric(p)
Binomial(n, p)
BN(n, r)
Poisson(𝜆)
Exp(𝜆)
Gamma(𝛼, 𝜆)
𝒰(a, b)
𝒩(𝜇, 𝜎 2 )
log-𝒩(𝜇, 𝜎 2 )
𝜒 2 (k)
𝒩n (𝝁, 𝚺)
Φ(⋅), Φ(x)
𝚽(x, y, 𝜌xy )
Wt
Nt
V3 - 08/23/2014
4:38 P.M.
Notation
X and Y are independent random variables
X and Y are not independent random variables
expectation of random variable X
expectation of random variable X under the probability
measure ℚ
expectation of g(X)
conditional expectation of X
variance of random variable X
conditional variance of X
covariance of random variables X and Y
correlation between random variables X and Y
Bernoulli distribution with mean p and variance p(1 − p)
geometric distribution with mean p−1 and variance (1 − p)p−2
binomial distribution with mean np and variance np(1 − p)
negative binomial distribution with mean rp−1 and variance
r(1 − p)p−2
Poisson distribution with mean 𝜆 and variance 𝜆
exponential distribution with mean 𝜆−1 and variance 𝜆−2
gamma distribution with mean 𝛼𝜆−1 and variance 𝛼𝜆−2
uniform distribution with mean 12 (a + b) and variance
1
(b
12
− a)2
normal distribution with mean 𝜇 and variance 𝜎 2
1 2
lognormal distribution with mean e𝜇+ 2 𝜎 and variance
2
2
(e𝜎 − 1)e2𝜇+𝜎
chi-square distribution with mean k and variance 2k
multivariate normal distribution with n-dimensional mean
vector 𝜇 and n × n covariance matrix 𝚺
cumulative distribution function of a standard normal
cumulative distribution function of a standard bivariate normal
with correlation coefficient 𝜌xy
standard Wiener process, Wt ∼ 𝒩(0, t)
Poisson process, Nt ∼ Poisson(𝜆t)
Page 372
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bindex.tex
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Index
adapted stochastic processes, 2, 303
admissible trading strategy, 186
American options, 53
appendices, 1, 331–63
arbitrage, 53, 185–7, 232, 322–30
arithmetic Brownian motion, 128–9, 222–3,
229–30
see also Bachelier model
stock price with continuous dividend
yield, 229–30
arithmetic series, formulae, 332
arrival time distribution, Poisson
process, 256–8
see also stock...
fundamental theories, 232–5
attainable contingent claim, 186–7
Bachelier model
see also arithmetic Brownian motion
definition and formulae, 128–9
backward Kolmogorov equation, 97, 98–102,
149–50, 153–4,
180–1
see also diffusion; parabolic...
definition, 97, 98–9, 149–50, 153–4, 180–1
multi-dimensional diffusion
process, 99–102, 155–83
one-dimensional diffusion process, 87–8,
123–55
one-dimensional random walk, 153–4
two-dimensional random walk, 180–1
Bayes’ Formula, 7
Bayes’ rule, 341
Bernoulli differential equation, 357–8
Bernoulli distribution, 11, 12, 13, 14–15, 349
Bessel process, 163–6
beta function, 338
binomial distribution, 12–14, 349–50
bivariate continuous random variables, 345–7
see also continuous...
bivariate discrete random variables, 343–4
see also discrete...
bivariate normal distribution, 27–9, 34–40,
60–2, 158, 165, 352–3
see also normal distribution
covariance, 28–9, 57–9, 158, 165
marginal distributions, 27
Black equation, 362
Black model, 362
Black–Scholes equation, 54, 96, 236–8, 361
see also partial differential equations
reflection principle, 54–5, 361
Black–Scholes model, 129–30, 185, 236–8,
361
see also geometric Brownian motion
Bonferroni’s inequality, 7
Boole’s inequality, 6, 7, 10
Borel–Cantelli lemma, 10–11
Brownian bridge process, 137–8
Brownian motion, 51–93, 95–183, 185–92,
221–4, 227–8, 361, 362–3
see also arithmetic...; diffusion...;
geometric...; random walks; Wiener
processes
definitions and formulae, 51, 128–32,
138–9, 221–4, 227–30
càdlàg process, 246
Cauchy–Euler equation, 358–9
cdf see cumulative distribution function
central limit theorem, 12, 13, 57, 355
CEV see constant elasticity of variance model
change of measure, 43, 185–242, 249–51,
301–30
see also Girsanov’s theorem
definitions, 185–92, 249–51
Chebyshev’s inequality, 40–1, 75
chi-square distribution, 24–6, 352
CIR see Cox–Ingersoll–Ross model
Clewlow–Strickland 1-factor model, 141–4
4:38 P.M.
Page 373
Chin
374
compensated Poisson process, 245, 262–3,
266–7, 323–6, 330
see also Poisson...
complement, probability concepts, 1, 4–5, 341
complete market, 187, 233, 322–30
compound Poisson process, 243, 245–6,
249–51, 264–88, 306–22, 323–30
see also Poisson...
decomposition, 268–9, 306–8
Girsanov’s theorem, 249–51
martingales, 265–7, 323–5
concave function, 339
conditional, probability concepts, 341
conditional expectation, 2–3, 43–9, 52–5,
75–6, 150–3, 192–3, 197–200, 273–4,
284–5, 341, 343, 346
conditional Jensen’s inequality, properties of
conditional expectation, 3, 48–9, 75–6,
192–4
conditional probability
density function formulae, 346
mass function formulae, 343
properties of conditional expectation, 3,
43–5
conditional variance, 343, 346
constant elasticity of variance model
(CEV), 145–6
contingent claims, 185–92, 245
see also derivatives
continuous distributions, 350–3
convergence of random variables, 10–11,
209–11, 353–4
convex function, 3, 42, 48–9, 75–6, 192–4,
338–9
convolution formulae, 25–6, 348
countable unions, 1–2
counting process, 243–330
see also Poisson...
definition and formulae, 243–4
covariance, 28–9, 57–60, 64–8, 117–18,
157–8, 165–6, 251–2, 344, 347
matrices, 59, 64–8
covariance of bivariate normal
distribution, 28–9, 57–60, 158, 165–6
covariance of two standard Wiener
processes, 57–60, 64–8
Cox process (doubly stochastic Poisson
process), 243, 245
see also Poisson process
Cox–Ingersoll–Ross model (CIR), 135–7,
171–4, 178
cumulative distribution function (cdf), 16–20,
22–4, 30–2, 37–40, 214–18, 342, 343,
345, 361–3
cumulative intensity, 245
see also intensity...
bindex.tex
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Index
De Moivre’s formula, 333
De Morgan’s law, 4, 7, 10
decomposition of a compound Poisson
process, 268–9, 306–8
differential equations, 52, 90–3, 95–183,
224–42, 253–5, 305–30, 357–63
see also ordinary...; partial...; stochastic...
differential-difference equations, 253–5
Dirac delta function, 339
see also Heaviside step function
discounted portfolio value, 187–92, 224–7,
228–32, 242, 324–30
discrete distributions, 349–50
discrete-time martingales, 53
dominated convergence theorem, 354
Donsker theorem, 56–7
Doob’s maximal inequality, 76–80
elementary process, 96
equivalent martingale measure, 185, 188–9,
209–11, 222–5, 230–1, 234–5, 238–42,
325–30
see also risk-neutral...
Euler’s formula, 333
events, definition, 1, 243–51
exclusion for probability, definition, 7–10
expectation, 2–3, 40–9, 52–5, 75–6, 90–1,
122–3, 125–6, 148–9, 157–8, 164–6,
192–4, 197–205, 270–2, 279–81,
305–30, 342, 343–7
exponential martingale process, 263–4
see also martingales
Feynman–Kac theorem, 97–102, 147–9,
178–80
see also diffusion
multi-dimensional diffusion process, 178–80
one-dimensional diffusion process, 147–9,
178
fields, definition, 1–2
filtration, 2–3, 52–5, 68–71, 75–82, 95–123,
128, 145–9, 178–80, 186–242, 244–51,
261–330
first fundamental theorem of asset pricing,
definition, 232
first passage time density function, 85–9,
218–19
first passage time of a standard Wiener
processes, 53, 76–89, 218–19
hitting a sloping line, 218–19
Laplace transform, 83–4
first-order ordinary differential
equations, 125–6, 171–74, 255, 265,
273–4, 306, 357–8
Fokker–Planck equation see forward
Kolmogorov equation
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folded normal distribution, 22–4, 72
see also normal distribution
foreign exchange (FX), 51, 54, 192, 238–42, 362
foreign-denominated stock price under
domestic risk-neutral measure, 241–2
risk-neutral measure, 238–42
forward curve from an asset price following a
geometric Brownian motion, 138–9
forward curve from an asset price following a
geometric mean-reverting
process, 139–40, 169–71
forward curves, 138–44
forward Kolmogorov equation, 97, 98–102,
150–3, 154–5, 181–3
see also diffusion; parabolic...
multi-dimensional diffusion process, 102
one-dimensional diffusion process, 150–3
one-dimensional random walk, 154–5
two-dimensional random walk, 180–3
Fubini’s theorem, 339
FX see foreign exchange
Gabillon 2-factor model, 169–71
gamma distribution, 16–17, 257, 352
Garman–Kohlhagen equation/model, 362–3
general probability theory, concepts, 1–49,
185–92, 341–55
generalised Brownian motion, 130–2
generalised Itō integral, 118–19
geometric average, 146–7
geometric Brownian motion, 70–1, 129–32,
138–9, 146–7, 221–2, 227–8, 361, 362–3
see also Black–Scholes options pricing
model; Brownian motion
definition and formulae, 129–32, 138–9,
146–7, 221–2, 227–8
forward curve from an asset price, 138–9
Markov property, 70–1
stock price with continuous dividend
yield, 227–8
geometric distribution, 349
geometric mean-reverting process, 134–5,
139–40, 169–71, 291–5
definition and formulae, 134–5, 139–40,
169–71, 291–5
forward curve from an asset price, 139–40,
169–71
jump-diffusion process, 291–5
geometric series, formulae, 332
Girsanov’s theorem, 185, 189–92, 194–242,
249–51, 298–322
see also real-world measure; risk-neutral
measure
corollaries, 189–90
definitions, 185, 189–92, 194–225, 249–51,
298–322
375
formulae, 189–92, 194–225, 249–51,
298–322
jump processes, 298–322
Poisson process, 249–51, 298–322
running maximum and minimum of a Wiener
process, 214–18
hazard function, 245
see also intensity...
hazard process, 245
heat equations, 359–60
Heaviside step function, 339
see also Dirac delta function
Heston stochastic volatility model, 175–8
Hölder’s inequality, 41–2, 72–4, 79, 193,
204–5
homogeneous heat equations, 359–60
homogeneous Poisson process see Poisson
process
hyperbolic functions, 74, 333
inclusion and exclusion for probability,
definition, 7–10
incomplete markets, 187, 324–30
independence, properties, 3, 11, 47–8, 163–6,
210, 259–61
independent events, probability
concepts, 259–61, 341, 344, 347
integrable random variable, 3, 44, 45–9, 353
see also random variables
integral calculus, definition, 95–6
integrated square-root process, 171–4
integration by parts, 19–20, 115–18, 147, 151,
174, 257–8, 337
Itō integral, 102–7, 108–116, 118–20, 144,
148–9, 163–6, 177–8, 187–92, 195–6,
205–7
Itō isometry, 113–14
Itō processes, 54, 95–123, 360–1
Itō’s formula see Itō’s lemma
Itō’s lemma, 96–7, 99–100, 102–27, 129–33,
134–7, 139–46, 147–53, 155–62, 166–8,
170–80, 196, 203–5, 212–13, 220–7,
234–41, 246–51, 278–80, 283–98,
307–8, 314–22
see also stochastic differential equations;
Taylor series
multi-dimensional Itō formulae for
jump-diffusion process, 247–9
one-dimensional Itō formulae for
jump-diffusion process, 246–7
Jensen’s inequality, properties of conditional
expectation, 3, 48–9, 75–6, 192–4
joint characteristic function, 344, 347
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376
joint cumulative distribution function, 16–17,
32, 212–13, 343, 345
joint distribution of standard Wiener
processes, 58–60, 64, 158, 165–6
joint moment generating function, 122–3, 267,
280–1, 306, 318–19, 321–2, 344, 347
joint probability density function, 16–17,
27–34, 86–9, 211–18, 345–6, 352
joint probability mass function, definition, 343
jump diffusion process, 246–51, 281–98,
298–30
see also diffusion; Poisson...; Wiener...
concepts, 246–51, 281–98, 325–30
geometric mean-reverting process, 291–5
Merton’s model, 285–8, 327–30
multi-dimensional Itō formulae, 247–9
one-dimensional Itō formulae, 246–7
pure jump process, 281–5, 323–5, 327–30
simple jump-diffusion process, 325–7
jumps, 95, 243–330
see also Poisson process
Girsanov’s theorem, 298–322
risk-neutral measure, 322–30
Kou’s model, 295–8
Laplace transform of first passage time, 83–4
laws of large numbers, formulae, 354–5
Lebesgue–Stieltjes integral, 246
Lévy processes, 55, 119–23, 207, 209
L’Hospital rule, formulae, 336
linearity, properties of conditional
expectation, 3, 44, 45
lognormal distribution, 20–2, 129–32, 134–5,
142–4, 169–71, 283–8, 351
Maclaurin series, 335
marginal distributions of bivariate normal
distribution, 27
marginal probability density function, 27–9,
346
marginal probability mass function, 343
Markov property of a geometric Brownian
motion, 70–1
Markov property of Poisson process, 244–5,
261–2, 268–9
Markov property of Wiener processes, 52,
68–71, 97
definition and formulae, 51–2, 68–71
Markov’s inequality, definition and
formulae, 40
martingale representation theorem, 187–8,
190, 192–4, 219–26
martingales, 52–3, 71–84, 96–123, 127–8,
185–242, 245, 262–7, 279–81, 299–330
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Index
see also equivalent...; stochastic processes;
Wiener processes
compound Poisson process, 265–8, 322–4
continuous processes property, 53
discrete processes property, 53
theorems, 53–5, 187–92
maximum of two correlated normal
distributions, definition, 29–32
mean value theorem, 105–6
mean-reversion, 134–5, 139–44, 169–71,
288–95
see also geometric mean-reverting process
measurability, properties of conditional
expectation, 3, 43–4, 45, 46, 199–200
measurable space, definition, 2
measure theory, definition, 2
measures, 2, 185–242, 249–51,
301–30
see also Girsanov’s theorem; real-world...;
risk-neutral...
change of measure, 185–242, 249–51,
301–30
Merton’s model, definition and
formulae, 285–8, 327–30
minimum and maximum of two correlated
normal distributions, 29–32
Minkowski’s inequality, 42
monotone convergence theorem, 354
monotonicity, properties of conditional
expectation, 3, 44–45
multi-dimensional diffusion process
see also diffusion
backward Kolmogorov equation, 101–2
definition and formulae, 99–102,
155–83
Feynman–Kac theorem, 178–80
forward Kolmogorov equation, 101
problems and solutions, 155–83
multi-dimensional Girsanov theorem, 190–1,
208–9, 249–51
multi-dimensional Itō formulae, 99–100
multi-dimensional Itō formulae jump-diffusion
process, 247–9
multi-dimensional Lévy characterisation
theorem, 121–3, 209
multi-dimensional martingale representation
theorem, 191–2
multi-dimensional Novikov condition,
207–8
multi-dimensional Wiener processes, 54, 64–8,
99–102, 163–6
multiplication, probability concepts, 341
multivariate normal distribution, 353
see also normal distribution
mutually exclusive events, probability
concepts, 12, 341
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negative binomial distribution, 350
normal distribution, 17–24, 29–32, 54, 62–3,
117–18, 128–9, 132–3, 135–8, 144,
170–1, 178, 206–7, 284–5, 348, 351,
361–3
see also bivariate...; folded...; log...;
multivariate...
minimum and maximum of two correlated
normal distributions, 29–32
Novikov’s conditions, 194–6, 207–8
numéraire, definition, 191–2
one-dimensional diffusion process, 97–9,
123–55, 178
see also diffusion
backward Kolmogorov equation, 149–50
definition and formulae, 97–9, 147–53
Feynman–Kac formulae, 147–9, 178
forward Kolmogorov equation, 150–53
one-dimensional Girsanov theorem, 189–90,
205–7
one-dimensional Itō formulae for
jump-diffusion process, 246–7
one-dimensional Lévy characterisation
theorem, 119–21, 207
one-dimensional martingale representation
theorem, 187–8, 190
one-dimensional random walk
see also random walks
backward Kolmogorov equation, 153–4
forward Kolmogorov equation, 154–5
optional stopping (sampling) theorem, 53,
80–4, 218–19, 232
ordinary differential equations, 125–6, 357–9
Ornstein–Uhlenbeck process,
definition, 132–3, 134–5, 288–91, 292–5
parabolic partial differential equations, 97
see also backward Kolmogorov...;
Black–Scholes...; diffusion...; forward
Kolmogorov...
partial averaging property, 3, 43, 45, 46–7,
201–3
partial differential equations (PDEs), 52,
97–102, 149–53, 178–80
see also backward Kolmogorov...;
Black–Scholes...; forward
Kolmogorov...; parabolic...
concepts, 97–102, 147–53, 168, 178–80
stochastic differential equations, 97–102,
168, 178–80
partition, probability concepts, 341
PDEs see partial differential equations
pdf see probability density function
physical measure see real-world measure
Poisson distribution, 13–14, 350
377
Poisson process, 95, 243–330
see also compensated...; compound...; Cox...;
jump...
Girsanov’s theorem, 249–51, 298–322
Markov property, 244–51, 261–3, 268
positivity, properties of conditional
expectation, 3, 44
principle of inclusion and exclusion for
probability, definition, 7–10
probability density function (pdf), 14–17,
20–2, 24–30, 35–40, 58–60, 84–9, 98,
150–3, 180–3, 214–18, 249–51, 256–9,
344, 345, 348–53
probability mass function, 11–13, 250, 257–8,
306–30, 342, 348, 349
probability spaces, 2–3, 4–11, 43–9, 52–93,
187–242
definition, 2
probability theory, 1–49, 185, 189–92, 341–55
formulae, 341–55
properties of characteristic function, 348
properties of conditional expectation, 3, 41–9,
192–4, 197–202, 269–72, 284–5
properties of expectation, 3, 40–9, 75–6,
192–4, 197–202, 270–2, 284–5, 347
definition and formulae, 40–9, 347
problems and solutions, 40–9
properties of moment generating, 348
properties of normal distribution, 17–20,
34–40
properties of the Poisson process, problems and
solutions, 243–51, 251–81
properties of variance, 347
pure birth process, 255–6
pure jump process, definition and
formulae, 281–85, 323–5
quadratic variation property of Wiener
processes, 54, 89–93, 96–123, 206–7,
209, 274–7
Radon–Nikodým derivative, 43, 188, 190,
191–2, 196–200, 212–13, 218–19,
222–5, 234–5, 239–42, 249–51, 298–330
random walks, 51–93, 95, 153–5, 180–3
see also Brownian motion; continuous-time
processes; symmetric...; Wiener
processes
definition, 51–5, 153–5, 180–3
real-world measure, 185, 189–242
see also Girsanov’s theorem
definition, 185
reflection principle, 53–5, 84–9, 361–3
Black equation, 362
Black–Scholes equation, 54, 361
definition, 53–4, 60, 84–9
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378
reflection principle (continued)
Garman–Kohlhagen equation, 362–3
Wiener processes, 53–5, 60, 84–9
risk-neutral measure, 53, 185, 188–242,
322–30
see also equivalent martingale...; Girsanov’s
theorem
definition, 185, 188–9, 221–42, 322–30
FX, 238–42
jump processes, 322–30
problems and solutions, 221–42,
322–30
running maximum and minimum of a Wiener
process, Girsanov’s theorem, 214–18
sample space, definition, 1, 4, 341–2
scaled symmetric random walk, 51–5
see also random walks
SDEs see stochastic differential equations
second fundamental theorem of asset pricing,
definition, 233
second-order ordinary differential equations
formulae, 358–9
variation of parameters, 358–9
self-financing trading strategy, 186–8, 192,
225–7, 236–8, 326–30
sets, 1, 2–11
definition, 1
𝜎-sigma-algebra, 1–5, 43–9, 201–5, 261
simple jump process, 322–3
simple jump-diffusion process, 325–7
simple process see elementary process
skew, 54
speculation uses of derivatives, 185
square integrable random variable, 95–6, 353
standard Wiener processes, 51–93, 95–183,
185–242, 244–51, 277–81, 285–98,
303–30
covariance of two standard Wiener
processes, 57–60, 64–8, 117–18
definition, 52, 54, 189–90, 250–1, 325–6
joint distribution of standard Wiener
processes, 58–63, 64, 158–9, 165–6
stationary and independent increments, Poisson
process, 259–81
stochastic differential equations
(SDEs), 95–183, 237–42, 246–51,
315–30, 360–1
definition, 95–102, 246–9
integral calculus contrasts, 95–6
partial differential equations, 97–102, 168,
178–80
stochastic processes, 2, 51, 52, 185–242,
243–330
see also martingales; Poisson...; Wiener...
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definitions, 2, 51, 52
stochastic volatility, 54, 95, 175–8
stopping times, Wiener processes, 53–5, 80–9,
218–19, 232
Stratonovich integral, 103–6
the strong law of large numbers, formulae, 355
strong Markov property, 52, 84–9
submartingales, 53, 75–80
supermartingales, 53, 76–9
symmetric random walk, 51–68, 88–9, 153–5,
180–3
see also random walks
time inversion, Wiener processes, 62–3
time reversal, Wiener processes, 63–4
time shifting, Wiener processes, 61
total probability of all possible values,
formulae, 342–5
tower property, conditional expectation, 3, 46,
49, 192–4, 197–9, 270–4, 284–5
trading strategy, 186–8, 191–2, 225–7, 236–8,
326–30
transition probability density function, 98–102,
149–53
see also backward Kolmogorov...; forward
Kolmogorov...
two-dimensional random walk
see also random walks
backward Kolmogorov equation, 180–1
forward Kolmogorov equation, 181–3
uniform distribution, 350
union, probability concepts, 341
univariate continuous random variables, 344–7
see also continuous...
univariate discrete random variables, 342–4
see also discrete...
variance, 13–49, 52–93, 115–83, 264–330,
342, 343, 345, 346, 347
see also covariance...
constant elasticity of variance model, 145–6
volatilities, 54, 95–183, 185–242, 285–330,
361–3
see also local...; stochastic...
the weak law of large numbers, formulae, 354
Wiener processes, 51–93, 95–183, 185–242,
242, 246–51, 277–81, 285–98, 303–30,
360–63
see also Brownian motion; diffusion...;
martingales; random walks
covariance of two standard Wiener
processes, 57–60, 64–8, 115–18
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Index
first passage time, 53, 76–89, 218–19
joint distribution of standard Wiener
processes, 58–60, 64, 158, 165–6
Markov property, 52, 68–71, 97–102
multi-dimensional Wiener processes, 54,
64–8, 99–102, 163–6
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quadratic variation property, 54, 89–93,
96–123, 206–7, 209, 274–5
reflection principle, 54–5, 60, 84–9
running maximum and minimum of a Wiener
process, 214–18
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