introductory mathematics and statistics for islamic finance...as an introductory text, no...

3GFFIRS 05/15/2014 13:39:21 Page ii

3GFFIRS 05/15/2014 13:39:21 Page i

IntroductoryMathematics and

Statistics forIslamic Finance

3GFFIRS 05/15/2014 13:39:21 Page ii

3GFFIRS 05/15/2014 13:39:21 Page iii

IntroductoryMathematics and

Statistics forIslamic Finance

ABBAS MIRAKHORNOUREDDINE KRICHENE

3GFFIRS 05/15/2014 13:39:21 Page iv

Cover image: iStockphoto.com/amir_npCover design: Wiley

Copyright 2014 by Abbas Mirakhor and Noureddine Krichene/John Wiley & Sons Singapore Pte. Ltd.

Published by John Wiley & Sons Singapore Pte. Ltd.

1 Fusionopolis Walk, #07-01, Solaris South Tower, Singapore 138628

All rights reserved.

No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form orby any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as expresslypermitted by law, without either the prior written permission of the Publisher, or authorization throughpayment of the appropriate photocopy fee to the Copyright Clearance Center. Requests for permissionshould be addressed to the Publisher, John Wiley & Sons Singapore Pte. Ltd., 1 Fusionopolis Walk, #07-01,Solaris South Tower, Singapore 138628, tel: 65-6643-8000, fax: 65-6643-8008, e-mail: [email protected].

Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts inpreparing this book, they make no representations or warranties with respect to the accuracy orcompleteness of the contents of this book and specifically disclaim any implied warranties ofmerchantability or fitness for a particular purpose. No warranty may be created or extended by salesrepresentatives or written sales materials. The advice and strategies contained herein may not be suitablefor your situation. You should consult with a professional where appropriate. Neither the publisher northe author shall be liable for any damages arising herefrom.

Other Wiley Editorial Offices

John Wiley & Sons, 111 River Street, Hoboken, NJ 07030, USA

John Wiley & Sons, The Atrium, Southern Gate, Chichester, West Sussex, P019 8SQ, United Kingdom

JohnWiley& Sons (Canada) Ltd., 5353 Dundas Street West, Suite 400, Toronto, Ontario, M9B 6HB, Canada

John Wiley& Sons Australia Ltd., 42 McDougall Street, Milton, Queensland 4064, Australia

Wiley-VCH, Boschstrasse 12, D-69469Weinheim, Germany

ISBN 978-1-118-77969-9 (Paperback)ISBN 978-1-118-77970-5 (ePDF)ISBN 978-1-118-77972-9 (ePub)

Typeset in 10/12pt, SabonLTStd-Roman by Thomson Digital, Noida, India.

Printed in Singapore by C.O.S. Printers Pte Ltd.

10 9 8 7 6 5 4 3 2 1

http://www.iStockphoto.com/amir_np

mailto:[email protected]

3GFFIRS 05/15/2014 13:39:21 Page v

To the memory of our respective parents

3GFFIRS 05/15/2014 13:39:21 Page vi

3GFTOC 05/15/2014 13:50:7 Page vii

Contents

Preface ix

Acknowledgments xi

About the Authors xiii

PART ONEMathematics

CHAPTER 1Elementary Mathematics 3

CHAPTER 2Functions and Models 25

CHAPTER 3Differentiation and Integration of Functions 41

CHAPTER 4Partial Derivatives 61

CHAPTER 5Logarithm, Exponential, and Trigonometric Functions 77

CHAPTER 6Linear Algebra 91

CHAPTER 7Differential Equations 117

CHAPTER 8Difference Equations 141

CHAPTER 9Optimization Theory 167

CHAPTER 10Linear Programming 195

vii

3GFTOC 05/15/2014 13:50:7 Page viii

PART TWOStatistics

CHAPTER 11Introduction to Probability Theory: Axioms and Distributions 227

CHAPTER 12Probability Distributions and Moment Generating Functions 251

CHAPTER 13Sampling and Hypothesis Testing Theory 271

CHAPTER 14Regression Analysis 301

CHAPTER 15Time Series Analysis 327

CHAPTER 16Nonstationary Time Series and Unit-Root Testing 355

CHAPTER 17Vector Autoregressive Analysis (VAR) 369

CHAPTER 18Co-Integration: Theory and Applications 381

CHAPTER 19Modeling Volatility: ARCH-GARCH Models 397

CHAPTER 20Asset Pricing under Uncertainty 413

CHAPTER 21The Consumption-Based Pricing Model 439

CHAPTER 22Brownian Motion, Risk-Neutral Processes, and the Black-Scholes Model 451

References 473

Index 475

viii CONTENTS

3GFLAST01 05/15/2014 13:42:9 Page ix

Preface

T he objective of this book is to provide an introductory and unified training inmathematics and statistics for students in Islamic finance. Students enrolled in

Islamic finance programs may have had different training in mathematical andstatistical methods. Some students may have advanced training in some mathematicalor statistical topics; however, they may not have been sufficiently exposed to sometopics that are highly relevant in Islamic finance or to the applications of quantitativemethods in this field. Other students may have had less advanced quantitativetraining. It will be therefore necessary to provide a homogenous quantitative trainingin mathematics and statistics for students, with a view to enhancing their command ofthe theory and practice of Islamic finance.

In view of the nature of Islamic finance, students or professionals should acquireadequate skills in computational mathematics and statistics in order to accomplishtheir duties in any financial or nonfinancial institutions where they might beemployed. Without computational skills, students or professionals may not be ableto manipulate economic and financial data; they may not meet the challenges of theirfinancial career. In fact, the finance industry has reached an extremely advanced stagein terms of the quantitative methods, computerization, product innovations, andarbitrage and trading programs that are used. Many institutions such as hedge funds,pension funds, investment corporations, insurance companies, and asset managementcompanies require advanced knowledge in actuaries, and models of investment andrisk management. Professionals have to satisfy the standards required by theseinstitutions and be able to use software, such as Microsoft Excel, EViews, Mathe-matica,MATLAB, andMaple, to process data and carry out computational tasks. TheInternet is rich in the use of computational tools. A student can plug in data and getinstantaneous answers; however, it is important for a student to understand the theoryunderlying the computational procedures.

While existing books on finance cover the topics of mathematics or statistics only,this book covers fundamental topics in both mathematics and statistics that areessential for Islamic finance. The book is also a diversified and up-to-date statisticaltext and prepares students for more advanced concepts in mathematics, statistics, andfinance.

Although most of the mathematical and statistical books concentrate ontraditional mathematics or statistics, this book uses examples and sample problemsdrawn from finance theory to illustrate applications in Islamic finance. For instance,a student will be exposed to financial products, asset pricing, portfolio selectiontheory, duration and convexity of assets, stock valuation, exchange rate pricing, andefficient market hypothesis. Examples are provided for illustrating these importanttopics.

ix

3GFLAST01 05/15/2014 13:42:9 Page x

A special feature of the book is that it starts from elementary notions in mathe-matics and statistics before advancing to more complex concepts. As an introductorytext, no prerequisite in mathematics or statistics is required. In mathematics, this bookstarts from elementary notions such as numbers, vectors, and matrices, before itadvances to topics in calculus and linear algebra. The same approach is applied instatistics; the book covers basic concepts in probability theory, such as events,probabilities, and distributions, and advances progressively to econometrics, timeseries analysis, and continuous time finance. Each chapter is aimed at an introductorylevel and does not go into detailed proofs or advanced concepts.

The questions at the end of each chapter repeat examples discussed in the chapterand students should be able to carry out computations using widely availablesoftware, such as Excel, Matlab, and Mathematica, online formulas, and othercalculators. Internet presentations that illustrate many procedures in the book arealso available. The successful resolution of these questions means that a student has agood understanding of the contents of the chapter. For self-checking, the answers havebeen made readily available online at www.wiley.com.

x PREFACE

http://www.wiley.com

3GFLAST02 05/15/2014 13:43:54 Page xi

Acknowledgments

T he authors acknowledge the valuable contribution of Jeremy Chia, editor at JohnWiley & Sons, who added considerable value to the manuscript. They express

deep gratitude to Kimberly Monroe-Hill for her hard work in the copyediting andproduction of the book. The authors also extend a special appreciation to NickWallwork and are thankful for the continuing support of John Wiley & SonsSingapore in promoting the development of Islamic finance.

Professor Abbas Mirakhor would like to thank Datuk Professor Syed Othman AlHabshi, the dean of faculty, and Professor Obiyathulla Ismath Bacha, the director ofgraduate studies, at INCEIF for their support. He also thanks Dr.Mohamed Eskandar,Ginanjar Dewandaru, Sayyid Aun Rizvi, and Fatemeh Kymia for their assistance.

xi

3GFLAST02 05/15/2014 13:43:54 Page xii

3GFLAST03 05/15/2014 13:48:24 Page xiii

About the Authors

Abbas Mirakhor is currently the First Holder of the Chair of Islamic Finance at theInternational Center for Education in Islamic Finance. He has served as the dean ofthe executive board of the International Monetary Fund from 1997 to 2008, and asthe executive director representing Afghanistan, Algeria, Ghana, Iran, Morocco,Pakistan, and Tunisia from 1990 to 2008. He has authored numerous publicationsand research papers on Islamic finance; among them are the Introduction to IslamicFinance (John Wiley & Sons, 2011), Risk Sharing in Islamic Finance (John Wiley &Sons, 2011), and The Stability of Islamic Finance (John Wiley & Sons, 2010).

Noureddine Krichene received his PhD in economics from the University of California,Los Angeles in 1980. He taught Islamic finance at the Global University, InternationalCenter for Education in Islamic Finance, Malaysia. He was an economist with theInternational Monetary Fund from 1986 to 2009. From 2005 to 2007, he was anadvisor at the Islamic Development Bank in Saudi Arabia. His areas of expertise are inthe international payments system, macroeconomic policies, finance, and energy andwater economics.

He is based in Laurel, Maryland.

xiii

3GFLAST03 05/15/2014 13:48:24 Page xiv

WEBC01 05/20/2014 10:3:50 Page 1

PART

OneMathematics

WEBC01 05/20/2014 10:3:50 Page 2

WEBC01 05/20/2014 10:3:50 Page 3

CHAPTER 1Elementary Mathematics

T his chapter covers the measurement and presentation of economic and financialdata. Data consists of numbers and graphics, which are essential for recording

and understanding financial data. All financial transactions are represented bynumbers. For instance, the price of a commodity A in terms of commodity B is anumber; it is the number of units of commodity B that is paid to get one unit ofcommodity A. Usually, unknown amounts are expressed as variables, designated bysymbols such as x, y, or any other symbol, and the equations that contain thesevariables may be expressed in the form of monomials, binomials, or polynomials.The applications of equations, sequences, and series are important concepts tounderstand in finance.

BASIC MATHEMATICAL OBJECTS

Numbers play a fundamental role in economics and finance. There are real numbersand complex or imaginary numbers. Real numbers are a subset of complex numbers.This section covers real numbers, complex numbers, the absolute value of a number,vectors and arrays, angles and directions, graphics, and the reporting of economic andfinancial data.

Real Numbers

The set of real numbers, denoted by R, is represented by a real line �∞;∞� � wherethe symbols �∞ and ∞ stand for minus infinity and plus infinity, respectively(Figure 1.1a). A real number is a value that represents a quantity along a continuousline. Real numbers include all the rational numbers, such as the integer �5 and thefraction 4/3, and all the irrational numbers such as

ffiffiffi2

p � 1:41421356 . . . (anirrational algebraic number) and π � 3:14159265 . . . (a transcendental number).Real numbers can be thought of as points on an infinitely long line called the numberline or real line, where the points corresponding to integers are equally spaced. Anoninteger real number has a decimal representation such as that of 8.632, whereeach consecutive digit is measured in units one tenth the size of the previous one. Thereal numbers are uncountable.

3

WEBC01 05/20/2014 10:3:50 Page 4

The set of real numbers,R, is a field, meaning that addition and multiplication aredefined and have the usual properties. The field R is ordered, meaning that there is atotal order � such that, for all real numbers x; y, and z:

■ If x � y then x � z � y � z:■ If x � 0 and y � 0 then xy � 0.

Complex Numbers

A complex number is written in the form

z � x � iy (1.1)

where x and y are real numbers and i is the imaginary unit, where

i2 � �1 (1.2)

In this expression, x is the real part of z denoted by Re z� �, and y is a real numbercalled the imaginary part of z and is denoted by Im z� �. The set of all complex numbersis denoted by C. Complex numbers extend the idea of the one-dimensional numberline to the two-dimensional complex plane by using the horizontal axis for the realpart and the vertical axis for the imaginary part. The complex number

z � x � iy

can be identified with the point x; y� � in the complex plane as shown in Figure 1.1b.A complex number whose real part is zero is said to be purely imaginary, whereas acomplex number whose imaginary part is zero is a real number. In this way thecomplex numbers contain the ordinary real numbers while extending them in order tosolve problems that cannot be solved with real numbers alone. The real line can bethought of as a part of the complex plane, and, correspondingly, complex numbersinclude real numbers as a special case.

a. Real line

1 2–1 0–2∞–∞

Real axis0

Imaginary axis

Origin

z = x + iyy

x

b. Complex number

FIGURE 1.1 Real and Complex Numbers

4 MATHEMATICS

WEBC01 05/20/2014 10:3:51 Page 5

The set C of complex numbers is a field. Briefly, this means that the followingfacts hold: first, any two complex numbers can be added and multiplied to yieldanother complex number. Second, for any complex number z, its additive inverse �zis also a complex number; and third, every nonzero complex number has a reciprocalcomplex number. Moreover, these operations satisfy a number of laws, for example,the law of commutativity of addition and multiplication for any two complexnumbers z1 and z2:

z1 � z2 � z2 � z1 and z1z2 � z2z1

These two laws and the other requirements on a field can be proven using the factthat the real numbers themselves form a field.

In Islamic finance, we make use of logarithm numbers, exponential numbers, andtrigonometric numbers. Trigonometric numbers are important in studying the slope ofa curve. These numbers will be introduced later in the book. Nonetheless, we mayprovide some examples.

Example: Consider number 1. Its natural logarithm is zero. Its exponential is2.718282. Consider the number π/2; its cosine, cosπ/2; is zero; its sine, sinπ/2 is 1.

Absolute Value of a Number

Numbers are also described by their absolute value or moduli. If a number isrepresented by a point on the real line, then the absolute value is a measure of thelength of the distance between the number and point zero. The numbers 5 and –5 havethe same absolute value: 5j j � �5j j � 5. In other words, when we see a number 2j j, thecorresponding number to this distance is either 2 or –2. The moduli of a complexnumber is the distance between the origin zero and the point represented by this

number (Figure 1.1b). If z � �4 � 3i, then zj j �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�4� �2 � 32

q� 5.

Vectors and Arrays

Economic data may be represented by a point on the real line. For instance, realgross domestic product is $14 billion in 2012. We show it by a point in the real line.However, economic information may have many dimensions. We need to go fromR, the real line, to higher dimension Euclidian space such as R2, R3, . . . , Rn. Forinstance, an Islamic stock has an expected return and a risk. Traditionally, if a sharehas an expected return of 7 percent and a risk of 9 percent, this share is representedwith a point whose coordinates are 7 percent and 9 percent (see Figure 1.2a).Further, we may be interested in the beta and alpha of the share. In this case, wehave to go to R4. If beta is 1.2 and alpha is 3.5 percent, then we describe the share bya vector in R4:

V �7%9%1:23:5%

26664

37775 (1.3)

Elementary Mathematics 5

WEBC01 05/20/2014 10:3:51 Page 6

Economic and financial information may need to be presented in the form of anarray. A portfolio may have four Islamic shares. We present information about theseshares in the form of a matrix:

A �

Share1 Share2 Share4 Share5Return 7% 12% 3% 7%Risk 9% 16% 6% 8%Beta 1:2 2 1:1 0:9Alpha 3:5 5:4% 2% 1:1

2666664

3777775

(1.4)

A matrix is also used to describe the structure of international trade such asexports and imports, the structure of an economy, or the production processes of afarm or an industry.

Angles and Directions

The notions of an angle and direction are important in economics and finance. We areinterested in the slope of a curve as well as the direction of economic motion. An angleshows the slope of a tangent line to a curve. A vector shows the direction of a motionalong the curve (see Figure 1.2b).

Graphics

Graphics are essential tools in reporting economic and financial information and inteaching economics and finance. They facilitate economic and financial analysis. Infact, graphics are essential in all fields. For instance, Google maps show us directionsin the form of a graph. A building or a house is designed in graphics before it is actuallyconstructed. A contractor cannot build any house before he has the mapping ofthe house.

Consider an Islamic bank that has a portfolio composed of Murabaha (26percent), Mudarabha (19 percent), Musharaka (25 percent), and Islamic funds (30percent). This information is shown in Figure 1.3.

A portfolio manager uses graphics to track the market value of his portfolio.Figure 1.4 shows the value of the portfolio over a period of 30 weeks.

Return0 0

Risk

A = (7%,9%)

Angle

Curve Direction

a. Vector b. Angle and direction

FIGURE 1.2 Vector, Angle, and Direction

6 MATHEMATICS

WEBC01 05/20/2014 10:3:51 Page 7

Reporting Economic and Financial Data

Besides graphics, economic and financial data is reported in special ways. If we say thereal gross domestic product (GDP) of Malaysia rose by 2 billion (RM) and that ofBurundi by 550 million (BF), this information is not easy to interpret. All we can say isthat real GDP did not fall in either country. However, if we say real GDP rose by 7percent in Malaysia and 2 percent in Burundi, this information is easier to interpretbecause it is placed in context of the existing GDP.

Economic and financial information is reported in the form of indicators; these arepercent changes, ratios, indices, elasticities, and other specific indicators. For instance,the balance sheet of a company is described in terms of ratios such as liquidity ratio,solvency ratio, and equity ratio.Macroeconomic indicators use ratios such as externaldeficit ratio, debt ratio, and fiscal deficit ratio. Indices are important. An indexnormalizes data to a base of 100, then compares the evolution of data in relationto this base. For instance, the price index measures the price of a basket ofcommodities in reference to a base of 100, called base year, and computes the period

26

1925

30

Murabaha Mudharab Musharaka Islamic funds

FIGURE 1.3 Portfolio of an Islamic Bank

80

85

90

95

100

105

110

115

120

125

130

302928272625242322212019181716151413121110987654321

FIGURE 1.4 Market Value of an Islamic Portfolio


WEBC01 05/20/2014 10:3:52 Page 8

change in relation to this base. Elasticities are away to describe economic and financialvariables. We say that demand for bread is inelastic, implying that consumers areunable to change their demand for bread whether prices of bread go skyward or dropsubstantially. In contrast, the demand for apples is elastic, implying that when theprice of apples increases, demand may decrease.

A percent change is defined as

Δx=x (1.5)

where Δx denotes change in a variable x. A percent change can be positive, zero, ornegative. A ratio involves two variables, one is the numerator and the other is thedenominator:

x=y (1.6)

For instance, per capita income is the ratio of real GDP in money terms divided bythe size of the population, that is, the number of citizens of a country. An index isreferred to by its abbreviation. For instance, CPI means consumer price index. S&P500 refers to Standard & Poor’s stock price index. Elasticity is computed as the ratioof two percent changes:

Δx=x� �= Δy=y� � (1.7)

It could be positive, zero, or negative. If it is close to zero, there is inelasticity of xin relation to y; if it is �∞, then there is high elasticity of x in relation to y.

VARIABLES, MONOMIALS, BINOMIALS, AND POLYNOMIALS

This section covers monomials, binomials, polynomials, polynomial lags, identities,and factorization of a polynomial.

A variable is designated by the symbol x. We perform algebraic operations on thevariable x. We may multiply x by any number a ∈ R, and we obtain ax. We maycompute powers of x such as

x�2; x�1; x0; x1; x2; . . . ; xn

We may perform operation such as

x � a; x � a� �2; . . . ; x � a� �n

For instance, we perform the following multiplications:

x � a� �2 � x2 � 2ax � a2 (1.8)

x � a� �3 � x3 � 3ax2 � 3a2x � a3 (1.9)

x � a� �4 � x4 � 4ax3 � 6a2x2 � 4a3x � a4 (1.10)

8 MATHEMATICS

WEBC01 05/20/2014 10:3:52 Page 9

Monomials, Binomials, and Polynomials

Amonomial is the product of nonnegative integer powers of variables. Consequently,a monomial has no variable in its denominator. It has one term (mono implies one):

6; 2x; �18; x2; 5y2; �2xy; or 97x2y3

We notice that there are no negative exponents and no fractional exponents. Thenumber 6 is a monomial since it can be written as 6x0 � 6.

A binomial is the sum of two monomials. It has two unlike terms (bi implies two):

3x � 1; x7 � 4x2; 2x � y; or y � y2

A trinomial is the sum of three monomials; it has three unlike terms (tri impliesthree):

x2 � 2x � 15x4 � 3x3 � 102x � y � 9

A polynomial is the sum of one or more terms (poly implies many):

x2 � 2x2x3 � x2 � 5x � 24x � 6y � 5

The degree of a polynomial is the highest exponent of its monomials. Polynomialsare in the simplest form when they contain no like terms. For instance, the polynomial

x2 � 2x � 1 � 3x2 � 4x

when simplified becomes

4x2 � 2x � 1

Polynomial Lags

In statistics we use a lag operator, denoted by L (lag). For instance, the price oftomatoes today is denoted as xt, the price of tomatoes last month is xt�1, the price twomonths past is xt�2, three months past is xt�3, . . . . , n months past is xt�n. Wepresent this information as xt�1 � Lxt, xt�2 � L2xt, xt�3 � L3xt, . . . ., xt�n � Lnxt.Our notation of tomato prices xt; xt�1; xt�2; xt�3; . . . ; xt�nf g can be written in apolynomial lag as

1; L; L2; L3; . . . :; Ln� �xt (1.11)


WEBC01 05/20/2014 10:3:53 Page 10

An example of a polynomial lag is

2 � 0:5L � 1:2L2 � 1:8L3 � 0:88L4 � 2:6L5

A polynomial lag is very useful in performing operations on a time series, suchas tomato prices, or any other time series such as the daily values of the DowJones Islamic stock index. We may perform operations on polynomial lag in the sameway as on any regular polynomial. For instance: 1 � 2L � L2 may be written as1 � L� � 1 � L� �.

Identities

Often in Islamic finance, we need to use identities; we provide some useful identities:

x � 1� �2 � x2 � 2x � 1 (1.12)

x � 1� � x � 1� � � x2 � 1 (1.13)

x2 � x � 1� �

x � 1� � � x3 � 1 (1.14)

xn�1 � xn�2 � ∙ ∙ ∙ � x � 1� �

x � 1� � � xn � 1 (1.15)

The binomial identity is an important one. It is stated as

x�y� �n�xn�nxn�1y�n n�1� �2!

xn�2y2� ∙ ∙∙�n n�1� �∙ ∙ ∙ n�k�1� �k!

xn�kyk� ∙ ∙ ∙�yn

(1.16)

Here k! is called factorial of k; it is written as k! � k � k � 1� � � k � 2� � �. . . � 2 � 1.

Factorization of a Polynomial

Let us consider the following product:

P x� � � x � 1� � x � 5� � x � 2� � x � 1� � � x4 � 3x3 � 11x2 � 3x � 10 (1.17)

We may reverse the path and start from the polynomial P x� � � x4 � 3x3 � 11x2 �3x � 10 and try to factorize it into x � 1� � x � 5� � x � 2� � x � 1� �. The values x � 1,x � �5, x � 2, and x � �1 are called the roots of the polynomials P x� �. If we replacex � 1 into the polynomial we find

P 1� � � 1 � 3 � 11 � 3 � 10 � 0

If we replace x � �5 into the polynomial we find

P �5� � � 625 � 375 � 275 � 15 � 10 � 0

10 MATHEMATICS

WEBC01 05/20/2014 10:3:53 Page 11

We observe that P x� � is different from 0 for any value of x different from1;�5; 2;�1. For instance, for x � 0 we have P 0� � � 10, and for x � 0:4 weP 0:4� � � 7:2576.

EQUATIONS

Equations are basic notions of finance. A large part of Islamic finance consists ofsolving equations such as computing internal rates of return, replicating portfolios,structuring products, pricing assets, and computing costs or break-even points. Asimple equation is of the form

x � 2 � 1 (1.18)

Obviously the solution is x � 3.Wemay have to deal with equations of the seconddegree, called quadratic equations, such as

ax2 � bx � c � 0 (1.19)

This equation has two roots. Completing the square can be used to derive ageneral formula for solving quadratic equations. Dividing the quadratic equation by a(which is allowed because a is nonzero), gives

x2 � bax � c

a� 0 (1.20)

Or x2 � bax � � c

a(1.21)

The quadratic equation is now in a form to which the method of completing thesquare can be applied. To “complete the square” is to add a constant to both sides ofthe equation such that the left hand side becomes a complete square

x2 � bax � 1b

2a

� �2

� � ca� 1b

2a

� �2

(1.22)

that produces

x � b2a

� �2

� � ca� 1b

2a

� �2

(1.23)

The right side can be written as a single fraction with common denominator 4a2.This gives

x � b2a

� �2

� b2 � 4ac4a2

(1.24)


WEBC01 05/20/2014 10:3:53 Page 12

Taking the square root of both sides yields

x � b2a

� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

(1.25)

Isolating x gives

x � � b2a

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

� �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

(1.26)

The plus-minus symbol “±” indicates that

x � �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

and x � �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

(1.27)

In the above formula, the expression underneath the square root sign is called thediscriminant of the quadratic equation, and is often represented using an uppercaseGreek delta, Δ:

Δ � b2 � 4ac (1.28)

A quadratic equation with real coefficients can have either one or two distinct realroots, or two distinct complex roots. In this case the discriminant determines thenumber and nature of the roots. There are three cases:

1. If the discriminant is positive, then there are two distinct roots

x1 � �b � ffiffiffiffiΔ

p2a

and x2 � �b � ffiffiffiffiΔ

p2a

(1.29)

both of which are real numbers.2. If the discriminant is zero, then there is exactly one real root, sometimes called a

double root:

x � � b2a

(1.30)

3. If the discriminant is negative, then there are no real roots. Rather, there are twodistinct (nonreal) complex roots, which are complex conjugates of each other. Inthese expressions i is the imaginary unit:

x1 � �b � iffiffiffiffiffiffiffi�Δp

2aand x2 � �b � i

ffiffiffiffiffiffiffi�Δp2a

(1.31)

12 MATHEMATICS

WEBC01 05/20/2014 10:3:54 Page 13

Thus the roots are distinct if and only if the discriminant is nonzero, and the rootsare real if and only if the discriminant is nonnegative.

Example: Complete the square for 3x2 � 6x � 11 � 0.

We apply the formula x � b2a

� �2 � b2� 4ac4a2 ; we obtain x � 1� �2 � 14

3 .Example: Solve the following quadratic function:

x2 � x � 6 � 0

We compute Δ � b2 � 4ac � 1 � 24 � 25,ffiffiffiffiΔ

p � ffiffiffiffiffiffi25

p � 5,

x1 � �b � ffiffiffiffiΔ

p2a

� �1 � 52

� 2; x1 � �b � ffiffiffiffiΔ

p2a

� �1 � 52

� �3

Example: Solve the following quadratic equation:*

x2 � x � 6 � 0

We compute Δ � b2 � 4ac � 1 � 24 � �23; ffiffiffiffiffiffiffi�Δp � ffiffiffiffiffiffi23

p,

We have two complex roots: x1 �� b2a� i

ffiffiffiffiffi�Δp2a ��1

2� iffiffiffiffi23

p2 and x2 �� b

2a� iffiffiffiffiffi�Δp2a �

�12� i

ffiffiffiffi23

p2 .

Example:Youwant to invest in two Islamic mutual fundsA and B, in proportionsx1 and x2, respectively, with x1 � x2 � 1, and x1 � 0; x2 � 0. The risk of mutual fundsA and B are σ1 � 9 percent� � and σ2 � 14 percent� �, respectively; the correlationcoefficient between expected returns is ρ � 0:6. Find the composition that will achievea portfolio risk of 11.2 percent.

The portfolio variance is

V x1; x2� � � σ21x21 � 2ρσ1σ2x1x2 � σ22x

22 � 81x21 � 151:2x1x2 � 196x22

We equate the portfolio variance to 11:22 � 125:44. We obtain

81x21 � 151:2x1x2 � 196x22 � 125:44

*Many online calculators are available for solving quadratic equations; for example, www.math.com/students/calculators/source/quadratic.htm.


http://www.math.com/students/calculators/source/quadratic.htm

http://www.math.com/students/calculators/source/quadratic.htm

WEBC01 05/20/2014 10:3:56 Page 14

We note that x2 � 1 � x1; hence, x22 � 1 � x1� �2 � 1 � 2x1 � x22. We replace x2into the equation, we obtain

81x21 � 151:2x1x2 � 196x22� 81x21 � 151:2x1 1 � x1� � � 196 1 � 2x1 � x21

� �� 125:8x21 � 240:8x1 � 196 � 125:44

Or

125:8x21 � 240:8x1 � 70:56 � 0

This is an equation of the form ax2 � bx � c � 0; its roots are

x � �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

and x � �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4ac

p2a

; we find x1 � 0:36 and x1 � 1:56:

The latter root means 156 percent of savings are invested in mutual fund A; sincewe cannot invest more than 100 percent in any fund, we discard this root. The rootx1 � 0:36 is acceptable. It implies x2 � 1 � 0:36 � 0:64. Hence 36 percent of thesavings are invested in fund A and 64 percent are invested in fund B.

EQUATIONS OF HIGHER ORDER

Entrepreneurs and planners undertake investment projects with a life span of manyyears.Theyare interested indetermining theprofitabilityof theproject inwhich theywillinvest considerable money. By deciding to invest in a specific project such as a textileplant, they have to renouncemany other alternative or competing projects. They tend tochoose the project that has the highest returns. The investment selection problemnecessitates solving an equation of the order n � 2, where n is an integer equal to the lifespan of the project or the investment horizon. Assuming a discount rate r, we maysummarize the cash flow (CF) of the project in table form, as shown here.

0 1 2 3 4 5 . . . n

Net cash flow CF0 CF1 CF2 CF3 CF4 CF5 . . . CFn

Discounted cashflow

CF0

1 � r� �0CF1

1 � r� �1CF2

1 � r� �2CF3

1 � r� �3CF4

1 � r� �4CF5

1 � r� �5 . . . CFn

1 � r� �n

The entrepreneur is interested in solving an equation of the form

CF0

1 � r� �0 �CF1

1 � r� �1 �CF2

1 � r� �2 �CF3

1 � r� �3 �CF4

1 � r� �4 �CF5

1 � r� �5 � ∙ ∙ ∙ � CFn

1 � r� �n � 0

(1.32)

14 MATHEMATICS

WEBC01 05/20/2014 10:3:57 Page 15

The rate of return r that solves this equation is called the internal rate of return(IRR). The equation can be rewritten in polynomial form as

a0 � a1x � a2x2 � a3x3 � a4x4 � a5x5 � ∙ ∙ ∙ � anxn � 0 (1.33)

where x � 11 � r

.

The equation admits n roots (real and complex). Unfortunately, there are noformulas that can be readily employed for finding the roots of the equation. We haveto proceed by iteration. We may draw a chart for the equation and see the pointswhere it intersects the horizontal line. We may also start from an initial value for xsuch as x0 and iterate until the polynomial becomes very close to zero. Since there are nroots, we may have to discard negative roots and consider only positive roots.Nonetheless, there is software such as Microsoft Excel that can provide the solutionsfor an equation of order n.

Example: Malay Palm Oil Corporation contemplates an investment project,which has the cash flow shown in the following table. We want to find the internalrate of return (IRR). We apply Excel formula IRR, we find r � 11 percent.

0 1 2 3 4 5 6 7 8 9 10Net cash flow �6,796 �500 �750 200 1,350 1,570 2,037 2,500 2,560 3,200 3,500

SEQUENCES

A sequence is an ordered list of objects (or events). Like a set, it contains members (alsocalled elements or terms). The number of ordered elements (possibly infinite) is calledthe length of the sequence. Unlike in a set, order matters, and exactly the sameelements can appear multiple times at different positions in the sequence. Mostprecisely, a sequence can be defined as a function whose domain is a countable,totally ordered set, such as the natural numbers.

An example of a sequence is (1, 2, 3, 5, 8). Sequences can be finite, as in thisexample, or infinite, such as the sequence of all even positive integers (2, 4, 6, . . . ).Finite sequences are sometimes known as strings or words and infinite sequences asstreams. The empty sequence (∅) is included in most notions of sequence, but may beexcluded depending on the context.

There are a number of ways to denote a sequence, some of which are moreuseful for specific types of sequences. One way to specify a sequence is to list theelements. For example, the first four odd numbers form the sequence (1, 3, 5, 7).This notation can be used for infinite sequences as well. For instance, the infinitesequence of positive odd integers can be written (1, 3, 5, 7, . . . ). Listing is mostuseful for infinite sequences with a pattern that can be easily discerned from the firstfew elements.

There are many important integer sequences. The prime numbers are numbersthat have no divisors but 1 and themselves. Taking these in their natural


WEBC01 05/20/2014 10:3:57 Page 16

order gives the sequence (2, 3, 5, 7, 11, 13, 17, 19, . . . ). The study of primenumbers has important applications for mathematics and specifically numbertheory.

The Fibonacci numbers are the integer sequence whose elements are the sum of theprevious two elements. The first two elements are either 0 and 1 or 1 and 1 so that thesequence is (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . ).

The terms of a sequence are commonly denoted by a single variable, say xn,where the index n indicates the nth element of the sequence. The sequence xn iswritten as

x1; x2; x3; . . . ; xn�1; xn; xn�1; . . . (1.34)

Indexing notation is used to refer to a sequence in the abstract. It is also a naturalnotation for sequences whose elements are related to the index n (the element’sposition) in a simple way. For instance, the sequence of the first 10 square numberscould be written as

xk � k2; k � 1; 2; . . . ; 10 (1.35)

This represents the sequence (1, 4, 9, . . . , 100). The length of a sequence isdefined as the number of terms in the sequence. A sequence of a finite length n is alsocalled an n-tuple. Finite sequences include the empty sequence (∅) that has noelements.

Normally, the term infinite sequence refers to a sequence that is infinite in onedirection, and finite in the other—the sequence has a first element, but no final element(a singly infinite sequence). A sequence that is infinite in both directions—it has neithera first nor a final element—is called a bi-infinite sequence, two-way infinite sequence,or doubly infinite sequence. For instance, a function from all integers into a set, such asthe sequence of all even integers ( . . . , �4, �2, 0, 2, 4, 6, 8, . . . ), is bi-infinite. Thissequence could be denoted 2n� �n�∞n��∞.

A sequence is said to be monotonically increasing if each term is greater than orequal to the one before it. For a sequence xn� � ∞n�1 this can be written as xn � xn�1 forall n ∈ N, where N is the set of all integers. If each consecutive term is strictlygreater (>) than the previous term then the sequence is called strictly monotonicallyincreasing. A sequence is monotonically decreasing if each consecutive term is lessthan or equal to the previous one, and strictly monotonically decreasing if each isstrictly less than the previous. If a sequence is either increasing or decreasing it iscalled a monotone sequence. This is a special case of the more general notion of amonotonic function.

The terms nondecreasing and nonincreasing are often used in place of increasingand decreasing in order to avoid any possible confusion with strictly increasing andstrictly decreasing, respectively.

Example: Generate the first 10 terms of the sequence: xn � xn�1 � xn�2 withx0 � 0 and x1 � 1. We obtain xn� �10n�0 � 0; 1; 1; 2; 3; 5; 8; 13; 21; 34f g.

16 MATHEMATICS

WEBC01 05/20/2014 10:4:0 Page 17

A Cauchy sequence is a sequence whose terms become arbitrarily close together asn gets very large. A sequence x1; x2; x3; . . . , of real numbers is called a Cauchysequence, if for every positive real number ε, there is a positive integerN such that forall natural numbers m; n > N

xm � xnj j < ε (1.36)

where the vertical bars denote the absolute value. In a similar way one can defineCauchy sequencesof rational or complexnumbers.Cauchy formulated sucha conditionby requiring xm � xnj j to be infinitesimal for every pair of infinitem; n > N. Figure 1.5depicts an example of a Cauchy sequence.

One of the most important properties of a sequence is convergence. Informally, asequence converges if it has a limit. Informally, a (singly-infinite) sequence has a limit ifit approaches some value ℓ, called the limit, as n becomes very large. That is, for anabstract sequence (xn) (with n running from 1 to infinity) the value of xn approaches ℓas n !∞, denoted limn!∞ xn � ℓ.

More precisely, the sequence converges if there exists a limit ℓ such that theremaining xn’s are arbitrarily close to ℓ for some n large enough. If a sequenceconverges to some limit, then it is convergent; a sequence is convergent if and only if itis a Cauchy sequence. If xn gets arbitrarily large as n !∞wewrite limn!∞ xn �∞ andsay that the sequence diverges to infinity.

Sequences arise in finance. For instance, if we consider any budget such as thebudget of a state, corporation, or household, we obtain sequences over time forreceipts, expenditure, and the balance, as shown in the following table:

Time 1 2 3 4 5 6 7 8 9

Receipts x1 x2 x3 x4 x5 x6 x7 x8 x9Expenditure y1 y2 y3 y4 y5 y6 y7 y8 y9Balance b1 b2 b3 b4 b5 b6 b7 b8 b9

0n

xn

FIGURE 1.5 Example of a Cauchy Sequence


WEBC01 05/20/2014 10:4:1 Page 18

Income from an asset such as a stock or a sukuk describes a sequence over periodsof time during which the income accrues. Likewise, the cash flow of a companymay bedescribed by the following sequence:

Time 1 2 3 . . . n . . .

Cash flow CF1 CF2 CF3 . . . CFn . . .

The present value of $1 to be received in future periods forms the followingsequence of discounted values:

Time 1 2 3 . . . n . . .

Present value of a future $1 $1

1 � r� �1$1

1 � r� �2$1

1 � r� �3$1

1 � r� �n . . .

We observe that the sequence $11 � r� �n is a convergent sequence for r > 0.

SERIES

The sum of the first n terms of a (one-sided) sequence forms the nth term in a sequencecalled a series. That is, the series of the sequence (xn) is the sequence (Sn) given by

S1 � x1S2 � x1 � x2

S3 � x1 � x2 � x3Sn � x1 � x2 � x3 � ∙ ∙ ∙ � xn

We can also write the nth term of the series as

SN � XNn�1

xn (1.37)

Then the concepts used to talk about sequences, such as convergence, carry overto series (the sequence of partial sums) and the properties can be characterized asproperties of the underlying sequences (such as xn). The limit, if it exists, of an infiniteseries (the series created from an infinite sequence) is written as

limn!∞

SN � X∞n�1

xn (1.38)

To sum up, given an infinite sequence of numbers {xn}, a series is informally theresult of adding all those terms together: x1 � x2 � x3 �∙ ∙ ∙. These can be written morecompactly using the summation symbol

P.

18 MATHEMATICS

WEBC01 05/20/2014 10:4:2 Page 19

Example:X ∞

n�112n

� 12� 14� 18�∙ ∙ ∙

The terms of the series are often produced according to a certain rule, such asby a formula. As there are an infinite number of terms, this series is often called aninfinite series. Unlike finite summations, infinite series need tools from mathemat-ical analysis, and specifically the notion of limits, to be fully understood andmanipulated. In addition to their ubiquity in mathematics, infinite series are alsowidely used in other quantitative disciplines such as physics, computer science,and finance.

Convergence of a Series

If a series Sn � X ∞n�0xn is convergent as n !∞, then it has a limit S � X ∞

n�0xn, andlimn !∞ xn � 0.

We consider the geometric series

Sn � a � ax � ax2 � ax3 � ∙ ∙ ∙ � axn�1 � ∙ ∙ ∙ (1.39)

This series converges and has the sum S � a1 � x if xj j < 1; it diverges if xj j � 1.

Example: Prove that the following series converges, and find its sum:

Sn � 0:6 � 0:06 � 0:006 � ∙ ∙ ∙ � 610n

� ∙ ∙ ∙

This is a geometric series; a � 0:6 and x � 0:1. Since xj j < 1, we conclude that theseries converges and has the sum S � a

1 � x � 0:61 � 0:1 � 2

3 � 0:666666.Example: Prove that the following series converges, and find its sum:

Sn � 2 � 23� 2

32� 2

33� ∙ ∙ ∙ � 2

3n�1� ∙ ∙ ∙

This is a geometric series; a � 2 and x � 1=3. Since xj j < 1, we conclude that theseries converges and has the sum S � a

1 � x � 21 � 1

3� 3:

Power Series A power series is a series of the form

X∞n�0

an x � c� �n (1.40)

The Taylor series at a point c of a function is a power series that, in many cases,converges to the function in a neighborhood of c. For example, the series

X ∞n�0

xnn! is

the Taylor series of ex at the origin 0 and converges to it for every x.

ex � X∞n�0

xn

n!(1.41)

where e denotes the exponential number.


WEBC01 05/20/2014 10:4:2 Page 20

Laurent Series Laurent series generalize power series by admitting terms into theseries with negative as well as positive exponents. A Laurent series is thus any series ofthe form

X∞n��∞

anxn (1.42)

APPLICATIONS OF SERIES TO PRESENT VALUE OF ASSETS

In Islamic finance sequences arise in relation to investment, production, budgeting,and many other economic and financial operations.

Applications of Series to Present Value Computation

The value V today (t = 0) of an asset maturing at date n is the present value (PV) of thefuture cash flows expressed as

V � CF1

1 � R� � �CF2

1 � R� �2 � ∙ ∙ ∙ � CFn

1 � R� �n (1.43)

Using the summation symbolP

we can rewrite V as

V � Xnt�1

CFt

1 � R� �t (1.44)

If the cash flows are constant and equal to CF, the value of the asset becomes

V � Xnt�1

CF1 � R� �t (1.45)

It can be rewritten as

V � CF �1 � 1

1 � R� �n� �

R(1.46)

If the number of future periods n !∞, the asset is called a perpetuity and its valuetoday becomes

V � CFR

(1.47)

20 MATHEMATICS

WEBC01 05/20/2014 10:4:2 Page 21

Example: Computing the price of an asset.Using the Microsoft Excel present value function, compute the price of an asset

today that provides a constant cash flow of $100 per month for 20 years, assuming adiscount rate of 5.77 percent. Compute the price of a perpetuity that pays a constantmonthly cash flow of $100, assuming a discount rate of 5.77 percent.

Here CF � $100, R � 5:77 percent × 1/12 = 0.4808 percent, andn � 12 � 20 � 240. Applying directly Equation (1.46) or using Excel’s present valuefunction PV 0:004808; 240; 100� �, we find V � $14; 220:18. The price of a perpetuityis V � 100=0:004808 � $20; 797:23.

Example: Constant payment.Compute the monthly paymentCF on a 30-year mortgage loan of $300,000 and a

mortgage rate equal to 7 percent per year. Here V � $300; 000; n � 12 � 30 � 360;and R � 7 percent � 1=12 � 0:00583. Inserting these values into the PV formula(Equation 1.46), we find CF � $1; 999:9, or CF � ∼$2; 000.

Application of series to stock valuation: Expected dividends serve as the basis forstock valuation. Like all financial assets, equilibrium stock prices are the present valueof a stream of cash flows. The value of stock,V0, is the present value of expected futuredividends; it is expressed as

V0 � D1

1 � rs� �1 �D2

1 � rs� �2 �D3

1 � rs� �3 � ∙ ∙ ∙ � D ∞1 � rs� � ∞ � X∞

t�1Dt

1 � rs� �t (1.48)

HereDt is the dividend the stockholder expects to receive at the end of the year t;rs is the minimum acceptable rate of return on the stock. A simplified approach tostock valuation is the constant growth stock model. In practice, it is difficult to get anaccurate forecast of the future dividends. However, in many cases, the stream ofdividends is expected to grow at a constant rate g. If this is the case, the above equationmay be written as

V0 � D0�1 � g�1�1 � rs�1 �D0�1 � g�2

�1 � rs�2 � ∙ ∙ ∙ �D0�1 � g� ∞�1 � rs� ∞

� D0

X ∞t�1

�1 � g�t�1 � rs�t �

D0�1 � g�rs � g

� D1

rs � g(1.49)

Here D0 is the most recent dividend, which has already been paid. As illustrationof the constant growth stock model, assume that Corporation XYZ just paid adividend of $1.15, so D0 = $1.15. Its stock has a required rate of return rs = 13.4percent and investors expect the dividend to grow at a constant 8 percent in the future.

The stock’s intrinsic value is $23, computed as V0 � $1:15 1:08� �0:134 � 0:08 � $23.

SUMMARY

This chapter deals with elementary mathematics applied in Islamic finance. It intro-duces basic mathematical objects such as real numbers, complex numbers, absolutevalue of a number, vectors, arrays, graphics, and the reporting of economic and


WEBC01 05/20/2014 10:4:4 Page 22

financial data using ratios and elasticities. It introduces the notions of variables,monomials, binomials, polynomials, polynomial lags, identities, and factorization of apolynomial. The chapter covers equations and their methods of solutions, and thenotions of sequences and series and their convergence to a finite limit. The chapterillustrates application of equations and series to Islamic finance.

By becoming familiar with the content of this chapter, the reader will be able toanalyze, in terms of ratios and graphically, Islamic financial data; compute the returnsto investment projects; and compute price of stocks as well as mortgage payments. Thechapter provides notions that are basic for Islamic finance.

QUESTIONS

1. An Islamic bank has its assets distributed as following: farming (28 percent),manufacturing (15 percent), mining (24 percent), commerce (25 percent), andconstruction (8 percent). Using Microsoft Excel, show a graph of the assets.

2. The price of bread increased by 25 percent, and the demand for bread fell by0.1 percent. Compute the elasticity of bread. The price of an exotic fruit rose by5 percent, and the demand for this fruit fell by 25 percent. Compute the priceelasticity. The price of gold rose by 400 percent, and the supply of gold rose by 0.5percent. Compute the supply price elasticity of gold.

3. The consumer price index in a country increased from 145.2 to 153.7. Computethe rate of inflation.

4. The population of the United States rose by 10 million, and that of Burundi by300,000. The population is growing faster in the United States compared toBurundi. Do you agree?

5. Complete the square for3x2 � 6x � 14 � 0, 7x2 � 2x � 14 � 0, and x23 � x

7 � 9 � 0.

6. Solve x2 � x � 4 � 0, �x2 � x � 4 � 0, and x2 � x � 210 � 0.

7. You want to invest your savings in two Islamic mutual funds, A and B, inproportions x1 and x2, respectively, with x1 � x2 � 1, and x1 � 0; x2 � 0. Therisk of mutual funds A and B are σ1 � 9 percent� � and σ2 � 14 percent� �, respec-tively; the correlation coefficient between expected returns is ρ � 0:35. Find thecomposition that will achieve a portfolio risk of 10.8 percent. Note that theportfolio variance is

V x1; x2� � � σ21x21 � 2ρσ1σ2x1x2 � σ22x

22

8. Malay Palm Oil Corporation contemplates an investment project that has thecash flow shown in the following table. Using Microsoft Excel, find the project’sinternal rate of return.

0 1 2 3 4 5 6 7 8 9 10 11 12

Net cashflow

�7,796 �1,500 �750 1,200 1,350 1,570 2,037 2,500 2,560 2,800 2,600 3,200 3,500

22 MATHEMATICS

WEBC01 05/20/2014 10:4:7 Page 23

9. A farm is expected to provide a net income of $330,000 per year; what is its worthtoday if the market rate of return is 7 percent? What is its worth if the rate ofreturn is 2 percent? What is its worth if the rate of return is 0 percent? Compareand explain the difference in the farm’s value.

10. An asset promises to pay a net cash flow as shown below; using Microsoft Excel,compute its value today if the market rate of return is 6.23 percent per year.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Net cashflow

796 1,500 750 1,200 1,350 1,570 2,037 2,500 2,560 2,800 2,600 3,200 1,211 1,768 2,030

11. Compute the monthly payment CF on a 30-year mortgage loan of $300,000 anda mortgage rate equal to 4.7 percent per year.

12. A stock promises to pay a net dividend as shown below; using Microsoft Excel,compute its value today if the minimum acceptable rate of return is 7.23 percentper year.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Net cashflow

796 856 904 1,020 1,350 1,570 1,590 1,500 1,560 1,800 1,700 1,720 1,211 1,768 1,600

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Net cashflow

13. A stock paid today has a dividend of $53. The minimum acceptable rate of returnis 8.5 percent. The dividend is expected to grow at 7 percent per year. How muchis the stock worth today?

14. Consider the sequence an � n3n � 2; write the terms of this sequence for

n � 1; 2; 3; 4; 5. Compute the limit of an when n !∞.

15. Consider the sequence an � 2ffiffiffiffiffiffiffiffiffiffin2 � 9

p ; write the terms of this sequence for

n � 1; 2; 3; 4; 5. Compute the limit of an when n !∞.

16. Consider the sequence an � 1 � 1n

� �n; write the terms of this sequence for

n � 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; do you expect an to converge to a limit asn !∞?

17. Find the values for which the series converges, and find the sum of series:

a. Sn � 1 � x � x2 � x3 � x4 � x5 � ∙ ∙ ∙ � �1� �n�1xnb. Sn � 1 � x2 � x4 � ∙ ∙ ∙ � x2n

c. Sn � 12 � x � 3� �

4 � x � 3� �28 � ∙ ∙ ∙ � x � 3� �n

2n � 1 � ∙ ∙ ∙


WEBC01 05/20/2014 10:4:7 Page 24

3GC02 05/15/2014 9:4:51 Page 25

CHAPTER 2Functions and Models

F unctions are essential concepts in economics and finance. They describe a relation-ship between two or more variables. For instance, the quantity of corn produced is

a function of acreage, machinery, labor, fertilizers, rainfall, and so on. Each of thesevariables has a significant effect on the output of corn.

An economic or financial function is called a model. The price of an asset is afunction of risk and return. The capital asset pricing model (CAPM) is an example offunction that includes risk and return.

The payoff to a futures contract is a function of the futures price and the price ofthe asset at maturity. The price of a call option is a function of the underlying assetprice, the strike price, the volatility of the asset price, the risk-free rate, and maturity.

Practitioners in finance should be able to define a function and present examplesof functions in economics and finance. Some of the aspects of functions they shouldunderstand include the parametric representation of a function and the notion ofvector-valued function. Level curves (i.e., contours) are also important analytical toolsin economics and finance.

This chapter shows the application of functions in Islamic finance and the role offunctions in describing economic and financial models. In particular, it coverscomputing the present value of an asset, expected rate of return in the CAPM,forward exchange rates, and payoffs of futures, options, and swaps contracts.

DEFINITION OF A FUNCTION

The notion of a function is basic in Islamic finance. We define a function as follows.Definition: A function f from a set D to a set E is a mapping that assigns to each

element x ∈ D one element y of E. We write a function as

y � f x� � (2.1)

where y is a dependent variable and x is a driving variable. The element y of E is thevalue of f at x and is denoted by f x� �. The set D is the domain of the function. Therange of f consists of all possible function values f x� � in E. A function is also arelationship between a set of inputs with a set of permissible outputs. It is like amachine; we put in one number and it transforms it into another number according toa preset rule. The notation is a follows: x ! f x� �; here x is the input and f x� � is theoutput.

25

3GC02 05/15/2014 9:4:52 Page 26

Definition: A domain is closed if the boundary of the domain is included in thedomain. A domain is open if the boundary of the domain is not included in thedomain.

Example: The interval �4; 10� � is closed; the boundary points are in the interval.The interval �2; 6� � is an open interval; the boundary points are excluded from theinterval. The intervals [1, 6.7) and (2.3, 7.2] are half-open (or half-closed).

Example: Find the domain of the function f x� � � ffiffiffiffiffiffiffiffiffiffiffix � 2

p. Its domain is 2; ∞� �.

Example: Find the domain of the function f x� � �ffiffiffiffiffiffi4�xp1�x . The function f x� � exists if

and only if 4 � x � 0 and 1 � x ≠ 0. Hence, the domain of definition is�4,1�∪ 1;∞� �� .

We may have a piecewise defined function. For instance, we may have

f x� � �2x � 3 if x < 0

x2 if 0 � x < 2

1 if x � 2

8>><>>:

If f and g are functions, we define the sum f � g, the difference f � g, the productf g, and the quotient f=g as

f � g� � x� � � f x� � � g x� �f � g� � x� � � f x� � � g x� �

f g� � x� � � f x� �g x� �f=g� � x� � � f x� �=g x� �

Example: Let f x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffi4 � x2

pand g x� � � 3x � 1; then

f � g� � x� � � f x� � � g x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffi4 � x2

p � 3x � 1

f � g� � x� � � f x� � � g x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffi4 � x2

p � 3x � 1

f g� � x� � � f x� �g x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffi4 � x2

p� �3x � 1� �

f=g� � x� � � f x� �g x� � �

ffiffiffiffiffiffiffiffiffiffiffiffiffi4 � x2

p� �= 3x � 1� �

A function is a polynomial function if f x� � is a polynomial:

f x� � � anxn � an�1xn�1 � ∙ ∙ ∙ � a1x � a0 (2.2)

where the coefficients a0; a1; . . . ; an are real numbers and the exponents are non-negative integers. If an ≠ 0, then f has degree n. The following are special cases, wherea ≠ 0:

Degree 0: constant function f x� � � aDegree 1: linear function f x� � � ax � bDegree 2: quadratic function f x� � � ax2 � bx � c

26 MATHEMATICS

3GC02 05/15/2014 9:4:53 Page 27

We may have composite function. The composite function f � g of f and g isdefined by

y � f � g� �

x� � � f g x� �� (2.3)

The domain of f � g is the set of all x in the domain of g such that g x� � is in thedomain of f (Figure 2.1).

Example: Let f x� � � x2 � 1 and g x� � � 3x � 5.

i. y � f g x� �� f 3x � 1� � � 3x � 1� �� 2 � 1 � 9x2 � 30x � 24ii. z � g f x� �� g x2 � 1

� � � 3 x2 � 1� � � 5 � 3x2 � 2

Example: Let y � ffiffiffiffiffiffiffiffiffiffiffiffiffix2 � 4

p; express y � f x� � as a composite function.

We let u � x2 � 4, then we have y � ffiffiffiu

p.

Example: Let y � 23x � 5; express y � f x� � as a composite function.

We let u � 3x � 5, then we have y � 2=u.We describe a function with a graph. In Figure 2.2, we present different forms of

functions. Figure 2.2a shows a linear function, to each x corresponds a single value y.Figure 2.2b shows a nonlinear function, a single ymay be obtained from two differentvalues of x. Figure 2.2c shows a discontinuous function; a jump occurs at x1; thisphenomena occurs often in financial markets where prices make sudden jumpsfollowing some news. Figure 2.2d shows a function where x may yield two differentvalues of y such as y � ffiffiffi

xp

.

Parametric Form of a Function

A function y � f x� � may be expressed in terms of a single parameter t, using theproperty of composition function: y � f g x� �� . For instance, if y � f x� � � ffiffiffi

xp

, andx � t, then using the composition property of function with x � g t� �, we obtain y t� � �f g t� �� ffiffi

tp

: A point A on the parametric curve y t� � � f x t� �� is now represented withthe coordinates x t� �; y t� �� . In Figure 2.3, point A x t� �; y t� �� is the end-point of vector~OA. We denote this vector by r t� � and we call it a location vector because it providesthe location of every point on the curve: y � f x� �.

Domain of gg

Domain of f

g(x)

f

( f g) f (g(x))

FIGURE 2.1 Composite Function

Functions and Models 27

3GC02 05/15/2014 9:4:53 Page 28

Hence, provided x may be expressed as a function from the real line R into D ofthe form x � x t� � and y � f x� � is defined, y may be expressed as

y t� � � f x t� �� (2.4)

Each point on the curve y t� � � f x t� �� is now described by a location vector. Wehave now a function of the form

t ∈ R ! r t� � � x t� �y t� �

� �∈ R2 (2.5)

This new function is called a vector-valued function.Example: y � 3x4 � 2x � 5, and x � t; we obtain r t� � � t

3t4 � 2t � 5

� �.

0

A

y(t)

y(t)r(t)

x(t)x(t)

FIGURE 2.3 Parametric Vector-Valued Function

a. Linear function

0x

x1

y1

y = f (x)

0x

x1

y1

y2

y = f (x)

0x

x2 x1

y1

y = f (x)

b. Parabolic function

c. Discontinuous function at x1 d. Multi-valued function

0x

x1

y1

y2

y = f (x)

FIGURE 2.2 Graphs of Functions

28 MATHEMATICS

3GC02 05/15/2014 9:4:53 Page 29

Reciprocally, assumewe have x � x t� � and y � y t� �; then, by eliminating t, we finda function of the form y � f x� �.

Example: Let x t� � � 1 � 2t and y t� � � �3t; then we can express y in terms of x,because both depend on the same variable t. We have t � � y

3 � � x � 1� �2 , or

y � 32 x � 1� �.Example: Let x � 2t and y � t2 � 1; we obtain: y � 1

4 x2 � 1.

FUNCTIONS AND MODELS IN ECONOMICS

Economics makes considerable use of functions and models. A model represents atheory and simplifies a real-world situation by describing it as an equation. It alsosaves on words by providing a clear formula instead of pages of description. Someimportant economic functions are reviewed: the demand and supply functions, thebudget constraint function, the production possibility frontier (PPF), the utilityfunction, the production function, and other functions in economics.

The Market Model: Demand and Supply Functions

Markets are based on supply and demand theory. This is the theory of classicaleconomics. Demand and supply functions provide a market model. Demand for aproduct is a function of its price and many other variables such as income andpreferences. We may express the demand function as

q � f p; y; z� � (2.6)

Here q is the quantity demanded (in physical units), p is the price in dollars perunit of product, y is the income (in constant dollars), z is any other variable that mayinfluence demand. We assume that demand varies in an opposite way with prices;when prices rise, demand falls and inversely. The market has also a supply function,which we state as

q � g p; w� � (2.7)

Here w is any other variable that may influence supply. Supply is an increasingfunction of prices. It is also influenced by other factors such as entrepreneurship,technology, weather, taxation, and so on.

From the demand and supply functions we obtain an excess demand function,defined as the difference between demand and supply,

e � g p; w� � � f p; y; z� � (2.8)

If supply exceeds demand, prices drop; if demand exceeds supply, pricesincrease. In Figure 2.4a, we portray the demand and supply curves and showthat equilibrium is obtained at the intersection of the two curves. In Figure 2.4b, weshow an increase in the demand; the new equilibrium has a higher price and a higherquantity. In Figure 2.4c we show an increase in supply; the new equilibrium has alower price and a higher quantity. The operations of the market assume nointerference of forces that will prevent market clearing.


3GC02 05/15/2014 9:4:53 Page 30

The Budget Constraint

The budget constraint is important in economics. Assuming no borrowing, income is alimit to what we may consume. A rich person has a different budget constraint than apoor person. Households, corporations, and governments face budget constraints.Weformalize a budget constraint in a two-dimensional diagram (Figure 2.5a). We assumeconsumers have a fixed income denoted by M; they may buy two commodities,oranges x� � and tomatoes y� �, at market prices that they cannot influence, px and py,respectively. Their budget constraint is

pxx � pyy � M (2.9)

The budget constraint depicts the baskets of oranges and tomatoes that consum-ers may buy given their fixed income M. Consumers may choose to spend all theirincome on oranges or on tomatoes; or they may choose a mix x; y� � that satisfies theirbudget constraint. They can afford only the combinations of oranges and tomatoeswithin their budget constraint; they cannot purchase the combinations outside thebudget constraint. If the price of oranges rises, the budget constraint moves inward

a. Unhampered market equilibrium

b. Increase in demand c. Increase in supply

Demand Supply

Excess

0

p0 E0

p0 q

p Demand

p1

p0

E1

q0 q1 q

p Supply

p2

p0

E2

q0 q2q

p

FIGURE 2.4 The Market Model: Demand and Supply Functions

Oranges Oranges Oranges

a. The budgetconstraint

b. The budget constraintafter a rise in px

c. The budget constraintafter a drop in px

0

Tomatoes

Attainableset

pxx + pyy = M

y

x 0

Tomatoes

Gain in realincome

y

x0

Tomatoes

Loss in realincome

y

x

FIGURE 2.5 The Budget Constraint

30 MATHEMATICS

3GC02 05/15/2014 9:4:54 Page 31

(Figure 2.5b), and the consumer is poorer; if the price of oranges drops, the budgetconstraint moves outward (Figure 2.5c), and the consumer is richer.

The Production Possibility Frontier (PPF)

The production possibility frontier is the equivalent of a budget constraint inproduction theory. Producers have limited resources in capital, land, and money.They have to allocate their resources to the production of some selected commodities.A nation also faces a similar constraint; it has fixed resources that has to allocatebetween the production of butter and guns. It has to trade off more butter for fewerguns, or less butter for more guns. We describe the production possibility frontier intwo-dimensions. We assume the producers have a fixed amount of resources, denotedby K; they may produce two commodities, corn x� � and wheat y� �. The PPF describesthe different combinations of corn and wheat the producers may choose to producegiven their constraint in machinery, land, and money capital. They may decide toallocate all their resources to produce only corn or only wheat, or they may alsochoose a mix of corn and wheat.

Assume ax units of resources are needed to produce one unit of x and ay units ofresources are needed to produce one unit of y; the PPF may be expressed as

axx � ayy � K (2.10)

We show this relationship in Figure 2.6a. If the producer becomes less efficient inthe production of x, we have a new technical coefficient ax > ax; the PPF movesinward as shown in Figure 2.6b. However, if the producer becomes more efficient inthe production of x, we have a new technical coefficient ax < ax; the PPF movesoutward as shown in Figure 2.6c.

Example:A producer has 10 hectares (ha) of land; he or she may produce corn x� �,wheat y� �, or both x; y� �. The technical coefficients are 4 tons of corn per hectare and 5tons of wheat per hectare. The technical coefficients are ax � 0:25ha=ton anday � 0:20ha=ton. The equation for the PPF is 0:25x � 0:2y � 10; this equation canbe written as y � �1:25x � 50.

a. Production possibility frontier (PPF)

b. PPF after loss of efficiency in producing

0

Attainableproduction

x

y

x

c. PPF after gain in efficiency in producing

x

Loss in output

x

y

0

Gain in output

x

0

FIGURE 2.6 Production Possibility Frontier (PPF)


3GC02 05/15/2014 9:4:54 Page 32

The Utility Function

Economic theory often uses the notion of consumer utility function. The lattermeasures the utility enjoyed by the consumer when consuming products or services.If you are hungry, you derive high utility by consuming the first quantities of yourmeal. If you are satiated, then additional food on your plate has very low utility. Youdecline any addition of food to your plate. The utility function may be formulated interms of one commodity or many commodities.

If the consumer consumes one commodity x, then the utility function isU � U x� �.If the consumer consumes two commodities, oranges x� � and tomatoes y� �, his or herutility is U � U x; y� �.

We are interested in the notion of indifference curve, defined as a level of constantutility U to be achieved by different combinations of commodities. If consuming threeoranges and one tomato provides the same utility as consuming one orange and fourtomatoes, we say that consumers are indifferent between these two combinations.They cannot rank them.On the contrary, if consuming two oranges and four tomatoesprocures higher utility than consuming one orange and four tomatoes, there is a rankorder; one combination is preferred to the other. Figure 2.7 shows an indifferencecurve; bundle A1 procures the same utility as bundle A2.

Production Function

The production function is an essential concept in growth theory. It is different fromthe production possibility frontier. We assume production Y is a function of capital(machinery), labor, technical progress, raw material, land, and entrepreneurship.Often, production Y is formulated in terms of capital K� � and L� � as

Y � F K; L� � (2.11)

We may decide to use output per labor, y � Y=L, which we relate to capital perworker as k � K=L; then we have

y � f k� � (2.12)

y1

y2

y

x1

A1

A2

x2x

0

FIGURE 2.7 The Indifference Curve

32 MATHEMATICS

3GC02 05/15/2014 9:4:54 Page 33

We show in Figure 2.8 an example of a production function. Output per worker isan increasing function of capital per worker. Point A0 describes the case of a poorcountry that has a little amount of capital per worker; consequently, it has a lowoutput per worker. PointA1 describes the case of a rich country that has a lot of capitalper worker; consequently, it has a high output per worker. The poor country has noother path except to invest and build more capital per worker in order to enhanceincome per worker. If it does not increase capital per worker it remains poor.

Example: If a country has the following production function y � k0:9, its capitalper worker is $500. Howmuch is output per worker? If the country wishes to attain anincome of $2,000 per worker, how much should its capital per worker be?

We use the production functionwith k � $500, andwe find y � 5000:9 � $268:60.If the country wishes to attain $2,000 per worker, it has to have a k such that2; 000 � k0:9; thus we find k � $4; 563:80. This shows the scope of capital accumula-tion for the country to better itself.

Other Functions in Economics

Many other functions are used in economics. We cite some here, without beingexhaustive.

Consumption function: C � aY � b; C is consumption, Y is income, a is marginalpropensity to consume, and b is minimum consumption. Here, consumption dependson income.

The investment function: I � I r� �; I is investment, and r is rate of return.Money demand function: M � M r; y; P� �; M is money, y real income, and P is

price level.The quantity theory of money: MV � Py; M is money, y real income, P is price

level, and V is money velocity.The sales function: R � p � q; R is revenue, p is price, and q is quantity sold.The cost function: C � C q� �; C is cost, and q is quantity sold.The profit function: π � R � C; π is profit, R is revenue, and C is cost.Besides functions, we use also identities. One important identity is national

income identity in money units:

Y � C � I �G �X �M (2.13)

y1

y0

y

0

A1

A0

k0 k1k

y = f (k)

FIGURE 2.8 Production Function


3GC02 05/15/2014 9:4:55 Page 34

Here Y is gross domestic product (in $), C is private consumption (in $), I is grossinvestment (in $),G is government spending (in $), X is exports of goods and services(in $), and M is imports of goods and services (in $). The identity is an accountingidentity and explains how national income has been used during the accounting year.This identity has a dual identity that shows the sources of Y, which are agriculture,mining, industry, construction, commerce, and services.

FUNCTIONS AND MODELS IN FINANCE

Many models arise in Islamic finance. Here are some examples of financial functionsthat include the present value function, the capital asset pricing model (CAPM), thepayoffs of futures, options, and swaps contracts, and the model of the forwardexchange rate.

The Present Value Function

The present value formula is an important function in finance; it relates the price of anasset to its cash flow and the rate of discount. The formula is

V � CF1

1 � R� � �CF2

1 � R� �2 � ∙ ∙ ∙ � CFn

1 � R� �n (2.14)

V is the value of an asset; CF1; CF2; . . . ; CFn is the cash flow until the maturity ofthe asset at time n, and R is the discount rate.

Value of a zero-sukuk: A zero-sukuk pays $A in three years; the rate of returnis r percent. The price V of the zero-sukuk is given by the following function:V � A

1 � r� �3Example: A Sharia-compliant asset has a maturity of 10 years and a cash flow

shown in the table below.What is its price today if the rate of return in the economy is6.5 percent per year?

Year 1 2 3 4 5 6 7 8 9 10

Cash flow in $ 500 575 605 776 819 856 905 920 960 995

We apply the present value formula with R � 6:5 percent, we findV � $5; 477:90.

The Capital Asset Pricing Model (CAPM)

The CAPM provides a price of an asset, for example, a stock, as a function of itsnondiversifiable risk:

Ri � Rf � β RM � Rf� �

(2.15)

Ri is the expected return of stock i, Rf is the riskless rate of return, RM is the expectedrate of return of the market portfolio, and β is the risk of stock i.

34 MATHEMATICS

3GC02 05/15/2014 9:4:56 Page 35

Example: We let Rf � 5:5 percent, RM � 7:8 percent, and β � 1:45. We want tofind Ri. We apply the CAPM

Ri � 5:5 � 1:45 � 7:8 � 5:5� � � 8:835%

Payoff of a Futures Contract

A futures contract derives its payoff from the value of the underlying asset at thematurity date. If today the futures price at the initiation of a futures contract is FT , andthe price of the asset at delivery time T is ST , then the payoff at the maturity date of thefutures contract at T would be

∏T � ST � FT� � (2.16)

The payoff function is shown in Figure 2.9a. If ST > FT , then the trader who takesa long position on the contract will make a profit equal to ∏T , and the trader whotakes a short position will make a loss equal to �∏T . If ST < FT , the long trader willlose �∏T and the short trader will make a profit ∏T .

Example: Today, you buy an oil futures contract for delivery in three months.Let the price agreed on today between you and the seller of the contract be $85/barrel.If at the delivery date, the price of oil is $100/barrel, then youwill make a profit of $15/barrel and the seller of the contract will have a loss of $15/barrel. If at the delivery datethe price of oil is $70/barrel, then you will lose $15/barrel and the seller of the contractwill make a profit of $15/barrel.

Payoff of an Option Contract

The payoff of an option contract is derived from a change in the value of theunderlying asset or liability before or at the expiration date of the option. Considera European call option (one that can be exercised on a specific day) with a strike priceK and maturity date T. Its payoff at maturity T is defined as

∏T � Max ST � K; 0� � (2.17)

a. Payoff of a futures contract at maturity T

b. Payoff of a call option at maturity T

c. Payoff of a put option at maturity T

ΠT

0 FT ST 0

ΠT

STK0

ΠT

STK

FIGURE 2.9 Payoff Functions


3GC02 05/15/2014 9:4:56 Page 36

This payoff function is shown inFigure2.9b. IfST > K, then the long trader realizes again equal to∏T � ST � K� � and the short trader incurs a loss equal to�∏T � K � ST� �.If ST < K, the option is worthless and the payoff is zero for both traders. Likewise, for aEuropean put option with strike K and maturity T, the payoff is defined as

∏T � Max K � ST ;0� � (2.18)

This payoff is shown in Figure 2.9c. If ST < K, the long trader realizes a gain equalto ∏T � K � ST� � and the short trader incurs a loss of �∏T � ST � K� �. If ST > K,the option is worthless and the payoff is zero for both traders.

Example: (i) You bought a call option on crude oil at a strike of $90/barrel. Atmaturity, the crude oil was $110/barrel. What is your payoff per barrel? The answer is$110 – $90 = $20/barrel.

(ii) You bought a put option on the euro at a strike of $1.30/euro. At maturity,the exchange rate was $1.25/euro. What is your payoff per one euro? The answer is$1.30 – $1.25 = $0.05/euro.

Payoff to a Swap

The swap is initiated today, t � 0. It has a notional principal in dollars equal to N. Itstipulates that fixed cash flows are computed at a fixed rate of returnRf ix, known today,and thefloating cashflowswill be computed at settlement dates that are the beginning ofeach reset period t = 1, t = 2, . . . t = T, using a reference floating rate Rf l, often, butnecessarily, the LIBOR. The payoffs will depend on Rf l that will prevail at each resetperiod t = 1, t = 2, . . . t = T. Hence the payoff at time t = i, denoted by∏i is equal to

∏i � N � Rf ix � Rf l; t�i�1� � � Tenor=360 (2.19)

where the tenor is the number of days between t � i and t � i � 1. If Rfix > Rf l; t�i�1,the fixed paying party has to make a net cash flow payment∏i to the floating party. IfRf ix < Rf l; t�i�1 the fixed paying party receives a net cash flow payment ∏i from thefloating party.

Example: You bought a swap for one year on a nominal of $100,000; you pay afixed rate of 6 percent and receive a floating rate at settlement. What is your payoff fora floating rate of 5 percent?

Π � $100; 000 � 5% � 6%� � � �$1; 000

Price of an Option

The price of an option, say a call option, is an example of function. It is stated as

C � f S; K; r; σ; T� � (2.20)

where C is the price of the option, S is the price of the underlying asset, K is the strikeprice, r is the riskless rate of return, σ is the volatility of the price of the asset, and T isthe maturity of the option.

36 MATHEMATICS

3GC02 05/15/2014 9:4:57 Page 37

The Forward Exchange Rate

The forward price of foreign exchange rate is an example of function in finance. Theforward exchange rate is stated as

FT � S01 � rd1 � rf

(2.21)

where FT is the forward exchange rate of local currency in terms of foreign currency,rd is the domestic rate of return, and rf is the foreign rate of return.

Example: Let S0 � $2=£, rd � 4 percent, and rf � 6 percent; then the one-yearforward rate is FT � $1:9622=£.

MULTIVARIATE FUNCTIONS IN ECONOMICS AND FINANCE

In economics and finance, often we have to deal with functions that involve manyvariables. A classic example is a consumer utility function where utility is a function ofmany goods. For instance, we may have:

U � U x1; x2� �; here utility is a function of two goods.U � U x1; x2; x3� �; here utility is a function of three goods.U � U x1; x2; x3; . . . ; xn� �; here utility is a function of n goods.

In production theory, output may be a function of many inputs. We may have:

Y � F x1; x2� �; here output is a function of two inputs.Y � F x1; x2; x3� �; here output is a function of three inputs.Y � F x1; x2; x3; . . . ; xn� �; here output is a function of n inputs.

Hence, a multivariate function is a function of a vector of variablesx1; x2; x3; . . . ; xn� �. Its graph is a surface inRn. If we have a one-dimensional functiony � f x� �, its graph is a curve in R2, as illustrated in Figure 2.10. A point on this curve is

given by a position vectorxy

� �. We have a two-dimensional function:

z � f x; y� � (2.22)

b. Level curvea. Location vector

0

y

x

f (x, y) = C

0

z (t)

r (t)

x (t)

y (t)

FIGURE 2.10 Multivariate Function


3GC02 05/15/2014 9:4:57 Page 38

This function describes a surface. A point on this surface is given by a position

vectorxyz

24

35 (Figure 2.10).

Parametric Representation

If x and y may be expressed as x � x t� � and y � y t� � in terms of a single parametert ∈ R, then we may express the function z � f x; y� � in parametric form:

z t� � � f x t� �; y t� �� (2.23)

We obtain a new function that is a mapping from the real line R into R3 of theform

t ∈ R ! r t� � �x t� �y t� �z t� �

24

35 (2.24)

A point on the surface of the curve is now the end-point of a location vector r t� �(Figure 2.10a).

Example: We consider a utility function: z � U x; y� � � x0:4y0:6. Assume x � t,y � t, then z � U x t� �; y t� �� t0:4t0:6 � t. A parametric representation of the utility

function would be a vector valued function of the form: t ! r t� � �ttt

24

35.

Conversely, if we have x � x t� �; y � y t� �; z � z t� �, we may eliminate t and expressz � f x; y� �.

Example: x � t, y � t2, and z � t3; then eliminating t, we find z � xy2.

Level Curves

Level curves, called also contours, are used in economics and finance. Instead ofrepresenting the surface of a multivariate function, we often deal with a level curve,that is, a contour, of the surface. It is much simpler to deal with one curve or few curvesinstead of dealing with the whole surface. A level curve is obtained as a cross-section ofa horizontal plane with the surface.

Definition: The level curves (contour curves) of z � f x; y� � are the curves in thexy-plane where the function is constant (Figure 2.10b). They have the equationsf x; y� � � C with constant C.

Example: Assume we have a Cobb-Douglas production function of the formz � xαyβ. This production describes a surface inR3. If we are interested in a productionlevel C, then we deal only with the equation xαyβ � C. This equation is represented bya contour shown in Figure 2.10b.

38 MATHEMATICS

3GC02 05/15/2014 9:4:58 Page 39

SUMMARY

This chapter describes the role of functions in Islamic finance and their applications asanalytical and computation tool of financial variables. It illustrates analytical notionssuch as the demand and supply functions, the budget constraint, the productionpossibility frontier (PPF), the utility function, the production function. It showsapplications in Islamic finance that include the present value function, the capitalasset pricing model (CAPM), payoffs of futures, options, and swaps contracts, and thedetermination of the forward exchange rate. The chapter introduces multivariatefunctions, their parametric representation, and the concept of contours or level curves.

QUESTIONS

1. Determine whether the interval is open or closed for 0;3� �; �4; 7� �; �8;�3� �,�100; 10; 000� �; �3; 7� �; 5; 12� �.

2. If f x� � � ffiffiffiffiffiffiffiffiffiffiffix � 4

p � 3x, find f 4� �; f 8� �; f 13� �.3. If f x� � � x

x � 3, find f �2� �; f 0� �; f 3:01� �.4. Find the domain of f x� � � x � 1

x3� 4x.

5. Find the domain of f x� � �ffiffiffiffiffiffiffiffiffiffi2x � 3

px2�5x � 4.

6. Let f x� � � x3 � 3x and g x� � � ffiffiffiffiffiffiffiffiffiffiffix � 2

p; find f � g

� �x� � and the domain of f � g

� �x� �;

find g � f� �

x� � and the domain of g � f� �

x� �.7. Let f x� � � ffiffiffiffiffiffiffiffiffiffiffi

3 � xp

andg x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 � 16

p;find f � g

� �x� � and thedomainof f � g

� �x� �;

find g � f� �


x� �.8. Let f x� � � x

x � 2 and g x� � � 3=x; find f � g� �

x� � and the domain of f � g� �

x� �; findg � f� �


x� �.9. Let y � 1= x � 3� �4, and find a composite function for y.

10. Let y � 4 � ffiffiffiffiffiffiffiffiffiffiffiffiffix2 � 1

p, and find a composite function for y.

11. Let x � 2t and y � 8 � 2t2, and find y � f x� �.12. Let x � 1 � 2t2, y � 4t, z � 3 � 2t � 2t2, and find z � f x; y� �.13. a. The demand for a product is q � �2p � 4, and the supply function is q � 1:2p.

Drawthese two functions in the plane q; p� � and compute themarket equilibrium.b. The demand function becomes q � �2p � 8; show graphically the new

demand function, compute the new equilibrium, and compare with (a).c. The demand function remains as in (a); however, the supply function becomes

q � 1:2p � 2. Show graphically the new supply curve, compute the newequilibrium, and compare with (a).

d. The market model is as in (a), then the government decides to fix the price atp � 0:7. Compute the excess demand and show graphically the distortioncreated by the government.

14. The consumer income is $50, the price of milk x is $5/gallon, and the price of meaty is $10/pound. Show graphically the consumer’s budget constraint; what is themaximum quantity of meat the consumer may consume? Assume the price of


3GC02 05/15/2014 9:4:59 Page 40

meat rises to $12.5/pound; show the new budget constraint. What happened tothe real income of the consumer?

15. Producers have 15 hectares of land; they may produce corn, sorghum, or both.The technical coefficients are 4 tons of corn per hectare and 5 tons of sorghum perhectare. Show graphically their production possibility frontier (PPF) and providean equation for the PPF. Assume producers have become more efficient inproducing corn with 6 tons of corn per hectare; show their new PPF and providethe new equation for the PPF.

16. If a country has the following production function y � k0:92, its capital per workeris $600. How much is the output per worker? If the country wishes to attain anincome of $2,000 per worker, how much should its capital per worker be?

17. A Sharia-compliant asset has a maturity of 10 years and a cash flow shown in thefollowing table. What is its price today if the rate of return in the economy is7.5 percent per year?

Year 1 2 3 4 5 6 7 8 9 10

Cash flow in $ 500 575 655 776 839 856 915 920 960 995

18. The riskless rate of return is Rf � 5:75 percent, the market rate of return isRM � 6:8 percent, and the stock risk is β � 1:54. Find the expected stockreturn Ri.

19. You sold a gold futures contract at $1,500/ounce. The price at maturity was$1,250/ounce. Compute your payoff if the size of the contract is 100 ounces. Areyou a winner or a loser?

20. You bought a call option on crude oil at a strike of $95/barrel. At maturity, thecrude oil was $115/barrel. What is your payoff per barrel?

21. You bought a put option on the euro at a strike of $1.35/euro. At maturity, theexchange rate was $1.22/euro. What is your payoff per one euro?

22. You bought a swap for one year on a nominal of $100,000; you pay a fixed rate of5.6 percent and receive a floating rate at settlement. What is your payoff for afloating rate of 6.5 percent?

23. Youwant to buy forward British sterling to be delivered eight months from today.The spot rate today is $1:49=£. The rate of return in the United States is 3 percentper year; in Britain, 5 percent per year. Compute the forward dollar/sterling rate.

40 MATHEMATICS

3GC03 05/15/2014 9:24:45 Page 41

CHAPTER 3Differentiation and Integration

of Functions

D ifferentiation and integration of a function are the main analytical tools in Islamicfinance. Differentiation and integration are related. The derivative of a function, if

integrated, returns the function itself; the integral of a function, if differentiated,returns the function itself. The process of finding a derivative is called differentiation.The reverse process is called integration. Basically, integration is a summationoperation; it is summing areas or volumes of the form f dx� �, where dx is a smallinterval.

This chapter covers the techniques of differentiation and integration, introducingthe notion of derivatives of functions, the notions of maxima and minima of afunction, Taylor expansion for a differentiable function, the mean-value theorem,integration of functions, and presents applications in Islamic finance related toduration and convexity of a sukuk.*

DIFFERENTIATION

Derivatives of functions are essential in economics and finance. We have to computederivatives of functions such as marginal utility, marginal productivity, or marginalconditions in optimization theory. The derivative is a measure of how a functionchanges as its input changes. Loosely speaking, a derivative can be thought of as howmuch one quantity is changing in response to changes in some other quantity. Let thefunction be y � f x� �; its derivative at point x is denoted by f ´ x� � and is defined as

f ´ x� � � limh!0

f x � h� � � f x� �h

(3.1)

Informally, the derivative is the ratio of an infinitesimal change of the output overan infinitesimal change of the input producing that change of output. Often we use thenotation dy

dx to express this ratio.For a real-value function of a single real variable, the derivative at a point equals

the slope of the tangent line to the graph of the function at that point. If x and y are realnumbers, and if the graph of y � f x� � is plotted against x, the derivative measures the

*Mathematica is a useful software for differentiation and integration of functions.

41

3GC03 05/15/2014 9:24:45 Page 42

slope of this graph at each point. Figure 3.1a shows a secant to the curve y � f x� �withintersection points x, f x� �� and x � h, f x � h� �� . The secant intersects the curve atpoint x � h´, f x � h´

� �� for h´ < h (Figure 3.1b). As h ! 0, the secant becomes a

tangent line to the curve at the point x, f x� �� . The derivative dy/dx is the slope of thetangent line at the point x, f x� �� (Figure 3.1c).

Example: Compute the derivative of x2. We apply the formula

f ´ x� � � limh!0

f x � h� � � f x� �h

We find

f ´ x� � � limh!0

x � h� �2 � x2

h� x2 � 2xh � h2 � x2

h� 2x � h � 2x

Example: Compute the derivative of: y � 3x � 8. We find dydx � 3. This gives an

exact value for the slope of a straight line.If the function f is not linear (i.e., its graph is not a straight line), then the change in

y divided by the change in x varies; differentiation is a method to find an exact valuefor this rate of change at any given value of x.

Example: Let f x� � � x2. Find an equation of the tangent line at point P �2; 4� �.The slope of the tangent line is f ´ x� � � 2x. Evaluated at x � �2, f ´ x� � � 2x � �4.

Using the definition of the slope, we have dydx � y � 4

x � 2 � �4, y � 4 � �4 x � 2� �, ory � �4x � 4.

Example: Let us consider a utility function of the consumer U � U x� � wherex is the quantity consumed. We assume U � x0.75. The marginal utility isdUdx � 0.75.x�0.25 � 0.75

x0.25.Example: Consider the production function y � k0.80 where y is output per

worker and k is capital per worker. The marginal product of capital per worker isdydk � 0.8k�0.2.

a. Secant to the curve b. Secant as h becomes smaller

c. Secant becomes tangent line as h 0

Tangentline

Slope

xx

f (x)

Secant

f (x)

xx

f (x)

f (x + h)

x + h x x

f (x)

f (x + h' )

x + h'

FIGURE 3.1 Derivative as the Slope of the Tangent Line

42 MATHEMATICS

3GC03 05/15/2014 9:24:46 Page 43

It is necessary to define an open interval of the x – axis. If a and b are numberssuch that a < b, all numbers x such that a < x < b form an open interval from a to b,or more briefly, the open interval a, b� �. The end points do not belong to the interval.By contrast, the set of all numbers x such as that a � x � b is called a closed interval,denoted by a, b� �; here the end points belong to the interval. By a neighborhood of x0we mean an open interval containing x0.

Definition: A function f is differentiable on an open interval a, b� � if f ´ existsfor every x in a, b� �. A function f is differentiable on a closed interval a, b� � if f isdifferentiable on an open interval a, b� � and if the following limits exist:

limh!0�

f a � h� � � f a� �h

and limh!0�

f b � h� � � f b� �h

(3.2)

The one-sided limits in the definition are referred to as the right-hand derivativeand left-hand derivative of f at a and b, respectively. Note that for the right-handderivative, h ! 0�, and a � h approaches a from the right. For the left-hand derivative,h ! 0�, and b � h approaches b from the right.

If f is defined on a closed interval a, b� � and is undefined elsewhere, then the right-hand and the left-hand derivatives define the slopes of the tangent lines at the pointsA a, f a� �� and B b, f b� �� , as illustrated in Figure 3.2a.

If f is defined on an open interval containing a, then f ´ a� � exists if and only if boththe right-hand and left-hand derivatives at a exist and are equal. The function whosegraph is sketched in Figure 3.2b has right-hand and left-hand derivatives at a that givethe slopes of the lines l1 and l2, respectively. Since the slopes of l1 and l2 are unequal,f ´ a� � does not exist. The graph of f has a corner at A a, f a� �� if f is continuous at a andif the right-hand and left-hand derivatives at a exist and are unequal; or if one of thederivatives exists at a and f ´ x� �j j !∞ as x ! a� or x ! a�.

Definition:A point C c, f c� �� on the graph of a function is a cusp if f is continuousat c and if the following two conditions hold:

(i) f ´ x� � !∞ as x approaches c from one side; and(ii) f ´ x� � ! �∞ as x approaches c from the other side.

Point C c, f c� �� in Figure 3.2b illustrates a cusp where the tangent lines arevertical.

a. Existence of derivatives b. Inexistence of derivatives at corner points

0 ca x

C(c, f (c))

y l1l2Slope

f(a+h) – f(a)

A(a, f (a))

a + hh > 0

0+ 0–

h < 0b + ha b

B(b, f (b))

f(b+h) – f(b)h h

limh

lim=h

FIGURE 3.2 Two-Sided Derivatives

Differentiation and Integration of Functions 43

3GC03 05/15/2014 9:24:47 Page 44

DIFFERENTIATION RULES

Some of the most basic differentiation rules are:

■ Constant rule: if f(x) is constant, then f ´ � 0■ Sum rule: (af � bg) � af ´ � bg´

for all functions f and g and all real numbers a and b.■ Product rule: f g� �´ � f ´g � f g´

for all functions f and g. By extension, this means that the derivative of a constant

times a function is the constant times the derivative of the function:d ax2� �dx � 2ax.

■ Quotient rule: fg

� �´ � f ´g�f g´g2

for all functions f and g at all inputs where g ≠ 0.■ Chain rule: If f x� � � h g x� �� then f ´ x� � � g´(x)h´ g x� ��

Example: f x� � � 2x4 � 5x3 � x2 � 4x � 1, find dy/dx.

dydx

� 8x3 � 15x2 � 2x � 4

Example: f x� � � x3 � 1� �

2x2 � 8x � 5� �

, find dy/dx.We use the product rule; we have

3x2 2x2 � 8x � 5� � � 4x � 8� � x3 � 1

� � � 10x4 � 32x3 � 15x2 � 8.

Example: Find dydx if y � 3x2�x�2

4x2�5 .By the quotient rule,

dydx

� 4x2 � 5� �

6x � 1� � � 8x� � 3x2 � x � 2� �

4x2 � 5� �2 � 4x2 � 14x � 5� �

4x2 � 5� �2

Example: f g� � � h g� � � g17, g x� � � x � x2.

We obtain f x� � � x � x2� �17, df

dx � 17 x � x2� �16 1 � 2x� �.

MAXIMUM AND MINIMUM OF A FUNCTION

In Islamic finance, we are interested in studying the existence of extremum values of afunction over a given domain of definition for the function.We need to study the shapeof the function over the domain. Derivatives help us to locate the relative maxima andminima of a function.

Definition: Let f be a function defined on an open interval a, b� �, and let x0 be apoint of the interval. We say that f has a relative (local) maximum at x0 if there is someneighborhood of x0, say a1,b1� � contained in a, b� � and containing x0, wherea1 < x0 < b1, such that

f x� � � f x0� � if a1 < x0 < b1 (3.3)

44 MATHEMATICS

3GC03 05/15/2014 9:24:47 Page 45

The same definition is made for a relative (local) minimum, the inequality beingreversed,

f x� � � f x0� � if a1 < x0 < b1 (3.4)

Both relative maximum and relative minimum are covered by the term relativeextremum.

Extremum value theorem: If a function f is continuous on a closed interval a, b� �,then f takes on a minimum value and a maximum value at least once in a, b� �.

The importance of this theorem is that it guarantees the existence of extrema if f iscontinuous on a closed interval. However, extremamay occur on intervals that are notclosed and for functions that are not continuous.

Several examples of local extrema are illustrated in Figure 3.3. At a pointcorresponding to a local extremum of the function graphed in Figure 3.3, the tangentline is horizontal or the graph has a corner. The x-coordinates of these points arenumbers at which the derivative is zero or does not exist.

Theorem: If a function f has a local extremum at a number x0 in an open interval,then either f ´ x0� � � 0 or f ´ x0� � does not exist. If f ´ x0� � exists and f ´ x0� � ≠ 0, then f x0� �is not a local extremum of the function f .

Theorem: If a function f is continuous on a closed interval a, b� � and has itsmaximum or minimum value at a number x0 in the open interval a, b� �, thenf ´ x0� � � 0, or f ´ x0� � does not exist.

The mere fact that f ´ x0� � � 0 does not guarantee that f has a relative extremum atx0. This is illustrated by f x� � � x3 at x � 0, where the graph has a horizontal tangentbut the function has neither a relative minimum nor a relative minimum (Figure 3.4c).If f is differentiable at x0, f

´ x0� � � 0 is a necessary, but not a sufficient, condition for fto have a relative extremum at x0.

Definition: A number x0 in the domain of a function f is a critical number of f ifeither f ´ x0� � � 0 or f ´ x0� � does not exist.

Example: If f x� � � x3 � 12x, find the maximum and minimum values of fon theclosed interval �3, 5� � and sketch the graph of f .

Local min

0

Local max

Local min

Local max Local max

Local min

y = f (x)

x1 x2xx3 x4 x5

FIGURE 3.3 Local Maxima and Minima


3GC03 05/15/2014 9:24:48 Page 46

We begin by finding the critical numbers of f . Differentiating yields

f ´ x� � � 3x2 � 12 � 3 x2 � 4� � � 3 x � 2� � x � 2� �

Since the derivatives exists for every x, the only critical numbers are those forwhich the derivative is zero, that is, –2 and 2. Since f is continuous on �3; 5� �, itfollows that the maximum and minimum values are among the numbers f �2� �, f 2� �,f �3� �, and f 5� �. Calculating these values we find �2� � � 16, f 2� � � �16, f �3� � � 9,and f 5� � � 65. The minimum value of f on �3; 5� � is the smallest value f 2� � � �16,and the maximum value is the endpoint extremum f 5� � � 65.

The graph of f is sketched in Figure 3.4a. The tangent line is horizontal at thepoint corresponding to each of the critical numbers, –2 and 2. It will follow that thatf �2� � � 16 is a local maximum for f as indicated by the graph.

Example: If f x� � � x � 1� �2/3 � 2, find the maximum and minimum values of fonthe closed interval 0, 9� � and sketch the graph of f .

We differentiate f x� �; we find f ´ x� � � 23 x � 1� ��1/3 � 2

3 x � 1� �1/3.

To find the critical numbers, we note that f ´ x� � ≠ 0 for every x and f ´ x� � does notexist at x � 1. Hence x � 1 is the only critical number in 0, 9� �. Thus, f has a minimumvalue f 1� � � 2 and a maximum value f 9� � � 6 on the interval 0, 9� �. The graph of f issketched in Figure 3.4b; note that

limx!1�

f ´ x� � � �∞ and limx!1�

f ´ x� � � ∞

Since f is continuous at x � 1, the graph has a cusp at 1; 2� �.Example: If f x� � � x3, prove that f has no local extremum.The graph of f is sketched in Figure 3.4c. The derivative is f ´ x� � � 3x2, which

exists for every x and is zero only if x � 0. Consequently, 0 is the only critical number.However, if x < 0, then f x� � is negative; and if x > 0, then f x� � is positive. Thus, f 0� � isneither a local maximum nor a local minimum. Since a local extremummust occur at acritical number, it follows that f has no local extrema. Note that the tangent line ishorizontal and crosses the graph at the point 0; 0� �.

a. f (x) = x3 – 12x b. f(x) = (x–1)2/3 + 2 c. f (x) = x3

0

(5, 65)

–3 5

(2, – 16)

y

x 0 9

(0, 3)

(1, 2)

(9, 6)

y

x0

y

x

FIGURE 3.4 Existence of Extrema

46 MATHEMATICS

3GC03 05/15/2014 9:24:50 Page 47

Example: Let f x� � � 4 � x2. Find the extrema of f on the following intervals: (i)�2, 1� �; (ii) �2, 1� �; (iii) 1, 2� �; and (iv) 1, 2� �.

The function is continuous; its derivative is dydx � �2x, which vanishes at x � 0. On

the interval �2, 1� �, we have f �2� � � 0, f 0� � � 4, and f 1� � � 3; the extrema are Maxf 0� � � 4 and Min f �2� � � 0. On the interval �2; 1� �, we have Max f 0� � � 4 and nominimum. On the interval 1, 2� �, we have Min f 2� � � 0 and no maximum; on theinterval 1, 2� �; we have no maximum and no minimum.

Example: Let f x� � � 1/x2; find the extrema of f on (i) �1, 2� �, and (ii) [–1, 2).The function f is not continuous at x � 0. Its derivative does not exist at this point.

On �1, 2� �, we have f �1� � � 1, and f 2� � � 1/4; we have no maximum and a minimumof f 2� � � 1/4. On [–1,2), we have no minimum and no maximum.

Example: Find the critical points of f x� � � x � 5� �2 � ffiffiffiffiffiffiffiffiffiffiffix � 43

pDifferentiating x � 5� �2 � x � 4� �1/3, we obtain

f ´ x� � � 13

x � 5� �2 x � 4� ��2/3 � 2 x � 5� � x � 4� �1/3 � x � 5� � 7x � 19� �3 x � 4� �2/3

Hence f ´ x� � � 0 if x � �5 or x � 19/7. The derivative f ´ x� � does not exist at x � 4.Thus, f has three critical numbers x � �5, x � 19/7, and x � 4.

Example: If f x� � � x1/3 8 � x� �, find the local extrema of f .

f ´ x� � � x1/3 � �1 � 8 � x� � � 13x�2/3 � 4 2 � x� �

3x2/3

Hence, the critical points of f are 0 and 2. This suggests we consider the sign off ´ x� � in each of the intervals �∞ ,0� �, 0, 2� �, 2, ∞� �. Since f ´ x� � is continuous and hasno zeros on each interval, we may determine the sign of f ´ x� � by using a suitabletest value f ´ a� �; all we need to know is its sign. Thus, if we choose a � 4 in 2, ∞� �,then

f ´ 4� � � 4 2 � 4� �3 4

23

� � < 0

Interval �∞ ,0� � 0; 2� � 2, ∞� �a �1 1 4Test value f ´ a� � f ´ �1� � > 0 f ´ 1� � > 0 f ´ 4� � < 0Sign of f ´ x� � + + �f x� � f increasing on �∞ ,0� � f increasing on 0, 2� � f decreasing on 2, ∞� �

By considering the sign of the first derivative, f has a local maximum at 2,since f ´ changes from positive to negative at 2. Thus we have a local maximum,

f 2� � � 21/3 8 � 2� � ∼ 7.6.The function does not have an extremum at 0, since f ´ does not change

sign at 0.


3GC03 05/15/2014 9:24:50 Page 48

MEAN VALUE THEOREM

In Islamic finance, we are often interested in computing means for a function. Themean-value theorem explains how to compute means of a function. If the function f iscontinuous on the closed interval a, b� � and f is differentiable on the open intervala, b� �, then a point c, a < c < b exists such that

f ´ c� � � f b� � � f a� �b � a

(3.5)

Figure 3.5 illustrates the mean-value theorem.Taylor’s formula can be viewed as being a generalization of themean-value theorem.

Namely, the mean-value formula can be written as the Taylor expansion of f x� �:f b� � � f a� � � f ´ c� � b � a� � (3.6)

Example: Consider the function f x� � � x � 4� �2 � 1 on the interval 3; 6� �. Let usfind f a� � and f b� � for a � 3 and b � 6. We replace in f x� � to findf a� � � 3 � 4� �2 � 1 � 2, f b� � � 6 � 4� �2 � 1 � 5. Now let us use the mean-value theo-rem to find our derivative at some point c.

f ´ c� � � f b� � � f a� �b � a

� 5 � 26 � 3

� 1

This tells us that the derivative at c is 1. This is also the average slope from a to b.Now that we know f ´ c� �) and the slope, we can find the coordinates for c. Let us plug cinto the derivative of the original equation and set it equal to the result of the mean-value theorem. We have

f c� � � c � 4� �2 � 1

f ´ c� � � 2c � 8

1 � 2c � 8

c � 9/2 and f c� � � 5/4

0

f(x)

f(b)

f(a)

a c bx

FIGURE 3.5 Mean-Value Theorem

48 MATHEMATICS

3GC03 05/15/2014 9:24:50 Page 49

POLYNOMIAL APPROXIMATIONS OF A FUNCTION:TAYLOR ’S EXPANSION

We observe in Figure 3.6 that f x � h� � is represented by the segment A0A3 and f x� � bythe segment A0A1. Furthermore, the slope of the tangent line at point A is thederivative of f x� � with respect to x; it is equal to f ´ x� � � A1A2

h .The segment A1A2 is equal to f ´ x� � � h. The segment A2A3 measures the curva-

ture of the function f x� �; it is a measure of convexity of the curve; for a linear functionf x� � the segment A2A3 is zero. We note that the segment A0A3 is the sum of threesegments:

A0A3 � A0A1 � A1A2 � A2A3 (3.7)

We may therefore write f x � h� � asf x � h� � � f x� � � f ´ x� �h � A2A3 (3.8)

As h becomes very small, the convexity segment A2A3 also becomes very small. Ifwe wish a linear approximation of f x� � in a small neighborhood h, we may simplywrite the function as

f x � h� � � f x� � � f ´ x� �h (3.9)

The derivatives of a function f at a point x provide polynomial approximation tothat function near x. For example, if f is twice differentiable, then

f x � h� � � f x� � � f ´ x� �h � 12f ´´ x� �h2 (3.10)

in the sense that

limh!0

f x � h� � � f x� � � f ´ x� �h � 12f ´´ x� �h2

h2� 0 (3.11)

f(x+h)

x+h

A3

A2

A1

A0h

Af(x )

x0

FIGURE 3.6 Principle of Linear Approximation


3GC03 05/15/2014 9:24:51 Page 50

If f is infinitely differentiable, then this is the beginning of the Taylor series for fevaluated at x.

A Taylor series is a representation of a function as an infinite sum of terms that arecalculated from the values of the function’s derivatives at a single point. The concept ofa Taylor series was formally introduced by the Englishmathematician Brook Taylor in1715. If the Taylor series is centered at zero, then that series is also called a Maclaurinseries, named after the Scottish mathematician Colin Maclaurin, who made extensiveuse of this special case of the Taylor series in the eighteenth century. It is commonpractice to approximate a function by using a finite number of terms of its Taylorseries. Taylor’s theorem gives quantitative estimates on the error in this approxima-tion. Any finite number of initial terms of the Taylor series of a function is calleda Taylor polynomial. The Taylor series of a function is the limit of that function’sTaylor polynomials, provided that the limit exists. A function may not be equal toits Taylor series, even if its Taylor series converges at every point.

For a function f that has n continuous derivatives in the neighborhood of a, theTaylor expansion of f is the power series

f x� � � f a� � � f ´ a� �1!

x � a� � � f ´´ a� �2!

x � a� �2 � f 3� �3!

x � a� �3 � . . . (3.12)

which can be written in the more compact sigma notation as

f x� � � X∞n�0

f n� � a� �n!

x � a� �n (3.13)

where n! denotes the factorial of n and f n� � a� � denotes the nth derivative of f evaluatedat the point a. In the case that a = 0, the series is also called a Maclaurin series.

f a � h� � � f a� � � f ´ a� �1!

h � f ´´ a� �2!

h2 � f 3� � a� �3!

h3 � . . .f n� � a� �n!

hn � . . . (3.14)

The Maclaurin series for any polynomial is the polynomial itself. The Maclaurinseries for 1 � x� ��1 at x � 0 is the geometric series

11 � x

� 1 � x � x2 � x3 � . . . (3.15)

so the Taylor series for x�1 at a � 1 is

1x� 1 � x � 1� � � x � 1� �2 � x � 1� �3 � . . . (3.16)

We apply Equation (3.12), f ´ � � 1x2, f

´´ � 2xx4 � 2

x3, f´´´ � � 6x2

x6 � � 6x4, and so on. At

point a � 1 we have f ´ � � 1x2 � �1, f ´´ � 2

x3 � 2, f ´´´ � � 6x4 � �6, . . . . Substituting

these derivatives in Equation (3.12), we obtain Equation (3.16).

50 MATHEMATICS

3GC03 05/15/2014 9:24:51 Page 51

INTEGRATION

The integration of a function has several uses. It can be used in finding a function’santi-derivative, in analyzing the fundamental theorems of calculus, in finding thechange in variables in indefinite integrals, and in finding the double integral of afunction.

Integration

Integration is often introduced as the reverse process to differentiation, and has wideapplications in finding areas under curves and volumes of solids. Integration with itsinverse, differentiation, is one of the two main operations in calculus. The simplestcase, the integral over x of a real-valued function f x� � is written as

∫f x� �dx (3.17)

The integral sign ∫ represents integration. The dx indicates that we are integratingover x; x is called the variable of integration. The function f x� �, the expression to beintegrated, is called the integrand. Because there is no domain specified for integration,the integral is called an indefinite integral.

Given a function f of a real variable x and an interval a, b� � of the real line, thedefinite integral

∫b

a

f x� �dx (3.18)

is defined informally to be the signed area of the region in the xy-plane bounded bythe graph of f, the x-axis, and the vertical lines x = a and x = b, such that the areaabove the x-axis adds to the total, and that below the x-axis subtracts from the total(see Figure 3.7a). When integrating over a specified domain a, b� � of x, we speak of adefinite integral.

a. Definite integral b. Anti derivative

Area= ab f (x) dx

0

f (x)

a b x

AreaF(x)=

ax f (t) dt

0 a x t

y = f (t)

FIGURE 3.7 Integral of a Function as Area under the Curve


3GC03 05/15/2014 9:24:52 Page 52

The term integral may also refer to the notion of the anti-derivative, a function Fwhose derivative is the given function f . In this case, it is called an indefinite integraland is written

F x� � � C � ∫f x� �dx (3.19)

Here C is an arbitrary constant.Example: (i) Let f x� � � x2; its anti-derivatives are 1

3 x3, 1

3 x3 � 2; 550, 1

3 x3 � C.

(ii) Let f x� � � 8x3; its anti-derivatives are 2 x4, 2 x4 � ffiffiffi77

p, 2 x4 � C.

Example: Evaluate ∫x2� 1� �2x2 dx.

We change the form of the integrand:

∫x2 � 1� �2

x2dx � ∫

x4 � 2x2 � 1x2

dx � ∫ x2 � 2 � x�2� �

dx

� x3

3� 2x � 1

x� C

Example: Evaluate ∫x3� 1� �x � 1 dx, x ≠ 1.

We change the form of the integrand:

∫x3 � 1� �x � 1

dx � ∫x � 1� � x2 � x � 1

� �x � 1

dx � ∫ x2 � x � 1� �

dx

� 13x3 � 1

2x2 � x � C

Example: Evaluate ∫3x � 5ffiffiffi

x3p

dx, x ≠ 0

∫3x � 5ffiffiffi

x3p dx � ∫ 3x � x�1/3 � 5 � x�1/3

� �dx

� ∫3x2/3dx � ∫5x�1/3dx � 95x5/3 � 15

2x2/3 � C

If f is a continuous real-valued function defined on a closed interval [a, b], then,once an antiderivative F of f is known, the definite integral of f over that interval isgiven by

∫b

a

f x� �dx � F b� � � F a� � (3.20)

Integrals and derivatives became the basic tools of calculus, with numerousapplications in science and engineering. The founders of calculus thought of theintegral as an infinite sum of rectangles of infinitesimal width.

52 MATHEMATICS

3GC03 05/15/2014 9:24:52 Page 53

Example: Let f x� � � 1x3; we compute its primitive (i.e., integral) and the area under

the curve between a � 1 and b � 4.

∫1x3

dx � ∫x�3dx � � 12x2

� C

The area under the curve between a � 1 and b � 4 is

∫4

1

x�3dx � � 12x2

��4

1� � 1

2 � 42� 1

2 � 12� 0.46875

The fundamental theorem of calculus is the statement that differentiation andintegration are inverse operations: if a continuous function is first integrated and thendifferentiated, the original function is retrieved. An important consequence, some-times called the second fundamental theorem of calculus, allows one to computeintegrals by using an anti-derivative of the function to be integrated.

The First Fundamental Theorem of Calculus

Let f be a continuous real-value function defined on a closed interval [a, b]. Let F bethe function defined, for all x in a, b� �, by

F x� � � ∫x

a

f t� �dt (3.21)

F is defined by the area shown in Figure 3.7b. Then, F is continuous on [a, b],differentiable on the open interval (a, b), and

F´ x� � � f x� � (3.22)

for all x in (a, b).Example: Consider f t� � � t3 on the interval 0, x� �. We have

F x� � � ∫x

a

f t� �dt � ∫x

0

t3dt � 14x4

F´ x� � � f x� � � x3

Second Fundamental Theorem of Calculus

Let f be a real-value function defined on a closed interval [a, b] that admits an anti-derivative g on [a, b]. That is, f and g are functions such that for all x in [a, b],

f x� � � g´ x� � (3.23)


3GC03 05/15/2014 9:24:53 Page 54

If f is integrable on [a, b] then

∫b

a

f x� �dx � g b� � � g a� � (3.24)

Change in Variables in Indefinite Integrals

If F is anti-derivative of f , then

∫f g x� �� g´ x� �dx � F g x� �� C (3.25)

If u � g x� � and du � g´ x� �dx, then

∫f u� �du � F u� � � C (3.26)

Example: Evaluate ∫ffiffiffiffiffiffiffiffiffiffiffiffiffiffi5x � 7

pdx.

We let u � 5x � 7 and du � 5dx; we write the integral as

∫ffiffiffiffiffiffiffiffiffiffiffiffiffiffi5x � 7

pdx � 1

5 ∫ffiffiffiffiffiffiffiffiffiffiffiffiffiffi5x � 7

p5dx� � � 1

5 ∫ffiffiffiu

pdu �

15u

32

32

� C

� 215

u3/2 � C � 212

5x � 7� �3/2 � C

Double Integral

Often in economics and finance, we deal with functions with many variables such asz � f x, y� �. Assuming f x, y� � is continuous in both x and y, we need to find an integralof the form

∫∫zdydx � ∫∫f x, y� �dydx (3.27)

This integral may be written as

∫∫f x, y� �dydx � ∫ ∫f x, y� �dy�

dx � ∫ ∫f x, y� �dx�

dy (3.28)

We may perform integration with respect to y considering x fixed, then integratethe result with respect to x. We also perform integration with respect to x consideringy fixed, then integrate the result with respect to y. The double integral will be the samewhether we integrate first with respect to y then x or integrate first with respect to xthen y. If we have a definite integral such as

∫b

a∫d

c

f x, y� �dy24

35dx (3.29)

54 MATHEMATICS

3GC03 05/15/2014 9:24:53 Page 55

we may write it as

∫b

a∫d

c

f x, y� �dy24

35dx � ∫

d

c∫b

a

f x, y� �dx24

35dy (3.30)

Example: Evaluate: (i) ∫4

1∫2

�12x � 6x2y� �

dydx; (ii) ∫2

�1∫4

12x � 6x2y� �

dxdy.

(i) We write the integral as

∫4

1∫2

�12x � 6x2y� �

dy

24

35dx � ∫

4

1

2xy � 6x2y2

2

�� 2�1dx

� ∫4

1

4x � 12x2� � � �2x � 3x2� ��

dx

� ∫4

1

6x � 9x2� �

dx � 3x2 � 3x3� 4

1 � 234

(ii) We write the integral as

∫2

�1∫4

1

2x � 6x2y� �

dx

24

35dy � ∫

2

�12

x2

2

�� 6

x3

3

�y

� 41dy

� ∫2

�116 � 128y� � � 1 � 2y� �

h idy

� ∫2

�1126y � 15� �dy � 63y2 � 15y

� 2�1 � 234

APPLICATIONS IN FINANCE: DURATION AND CONVEXITYOF A SUKUK

The duration and convexity of sukuks can be found using the applications discussednext.

Duration of a Sukuk

An application of differentiation in risk analysis is the concept of duration of a sukuk.Duration is used as an immunization technique of a portfolio of sukuks, meaning itneutralizes changes in the rate of return on the capital value of the portfolio.


3GC03 05/15/2014 9:24:53 Page 56

We let V be the value of a sukuk; we assume that the rate of return r changes by 1percent point, i.e.,Δr � 1 percent, and we would like to find the corresponding changeΔV in the value of the sukuk. We use the concept of modified duration of a sukuk to

compute this change. Modified duration of a sukuk is defined as: DM � 1VΔVΔr

. The

value of a sukuk is the present value of its cash flows:

V � CF1

1 � r� � �CF2

1 � r� �2 � ∙ ∙ ∙ � CFn

1 � r� �n � Xnt�1

CFt

1 � r� �t (3.31)

CFt is the cash flow in period t and n is the maturity of the sukuk. The derivativeΔV/Δr is

ΔVΔr

� � CF1

1 � r� �2 �2 � CF2

1 � r� �3 � ∙ ∙ ∙ � n � CFn

1 � r� �n�1

� � 11 � r� �

CF1

1 � r� �1 �2 � CF2

1 � r� �2 � ∙ ∙ ∙ � n � CFn

1 � r� �n�

� 11 � r� �

Xnt�1

t � CFt

1 � r� �t

(3.32)

DM � 1VΔVΔr

� 1V

� 11 � r� �

Xnt�1

t � CFt

1 � r� �t (3.33)

Duration requires the knowledge of the slope of a tangent line at a point E of theprice-yield curve as shown in Figure 3.8. The steeper the slope, the higher the durationis. We note that the Macaulay duration, D, is defined as

D � 1V

�Xnt�1

t � CFt

1 � r� �t (3.34)

Yield

Sukuk price

0

V0

V11

V12

r0

E0

E11

E12

rr1

FIGURE 3.8 Duration and Convexity of Sukuks

56 MATHEMATICS

3GC03 05/15/2014 9:24:54 Page 57

The relationship between modified duration and Macaulay duration is

DM � 1VΔVΔr

� 1V

� 11 � r� �

Xnt�1

t � CFt

1 � r� �t �1

1 � rD (3.35)

Example: Using Microsoft Excel, calculate the Macaulay duration and modifiedduration of a 10-year sukuk paying a coupon rate of 6 percent per year on a semi-annual basis; it has a yield rate of 8 percent, a settlement date on January 2, 2014, anda maturity date on December 31, 2023.

Macaulay duration � DURATION(''1/2/2014'', ''12/31/2023'', 0.06; 0:08; 2)

� 7.45 years.

This means that if the yield rate increases by 1 percent point the price of the sukukwill drop by 7.45 percent.

Modified duration � MDURATION(''1/2/2014'', ''12/31/2023'', 0.06; 0:08; 2)

� 7.16 years.

This means that if the yield rate increases by 1 percent point the price of the sukukwill drop by 7.16 percent.

Application of Taylor Expansion to the Convexity of Sukuk ’s Price

Analysis of the convexity of sukuk’s price uses Taylor expansion. As sukuk yield goeshigher, the price goes lower. The relationship between price and yield is not linear andhas a convex structure in nature. Figure 3.8 displays a convex curve relating yield toprice. A tangent line is drawn at an initial point E0 where yield r0 entails a sukuk priceV0. This tangent line is very similar to the concept of duration and represents the rate ofchange in price as yield changes. For small changes in yield, duration approximatesactual price change; however, when yields move farther away from the initial yield r0,duration becomes less reliable. In Figure 3.8, when yield changes from r0 to r1, durationpredicts a change of price from V0 toV12 whereas the actual change is fromV0 toV11;V11 > V12 duration overstates the change in price compared to actual change. Thisdiscrepancyowes to convexity of the yield-price curve. Themore this curve is convex theless accurate the price approximation that is derived from the tangent line, or duration.

Formally, the second-order differential of the sukuk price equation with respect toyield r is*

ΔV � ΔVΔr

Δr � 12Δ2VΔr2

Δr� �2 (3.36)

*We note that the Taylor expansion of V y� � in a neighborhood of ro is

V y� � � V r0� � � ΔVΔy

Δr � 12Δ2VΔr2

Δr� �2

where Δr � r � r0. We let ΔV � V r� � � V r0� �, and we obtain ΔV � ΔVΔr Δr � 1

2Δ2VΔr2 Δr� �2.


3GC03 05/15/2014 9:24:54 Page 58

Dividing through by V, we get

ΔVV

� 1VΔVΔr

Δr � 12V

Δ2VΔr2

Δr� �2

� � Modified duration� �Δy � Convexity2

Δr� �2 (3.37)

From the equation, duration can provide a good approximation of a price changein response to a yield change only if the price relationship is linear, displays very smallcurvature, that is, Δ2V

Δr2 ∼ 0, or the yield change Δr is very small with Δr� �2∼ 0.Otherwise, we have to account for convexity as stated earlier to get a betterapproximation of the price change when yield changes. Convexity is calculated usingthe following formula:

Convexity � 1VΔ2VΔr2

� 1

V � 1 � r� �2Xnt�1

t2 � t� �

CFt

1 � r� �t�

(3.38)

Convexity can also be estimated with a simpler formula, similar to the approxi-mation formula for duration,


� V� � V� � 2V0

V0 Δr� �2 (3.39)

where V0 = initial sukuk price, V– = sukuk price when the rate of return isincremented, V+ = sukuk price when the rate of return is decremented, and Δr =change in the rate of return in decimal form.

Example:We consider a 15-year maturity sukukwith a coupon rate of 6.5 percentper year, semi-annual payments, yield rate 8.2 percent per year, and a face value of$1,000. The settlement date is January 2, 2013, and the maturity date is December 31,2027. Using Excel’s financial price function, we find

V0 � PRICE(''1/2/2013'', ''12/31/2027'', 0.065; 0:082; 1000; 2) � $355.2015

V� � PRICE(''1/2/2013'', ''12/31/2027'', 0.065; 0:092; 1000; 2) � $311.8888

V� � PRICE(''1/2/2013'', ''12/31/2027'', 0.065; 0:072; 1000; 2) � $405.2599


� V� � V� � 2V0

V0 Δr� �2 � 405.2 � 311.9 � 2 � 355.2

355.2 � 0.01� �2 � 189.9

SUMMARY

The chapter covered differentiation, maximum and minimum of a function, mean-value theorem, polynomial approximations of a function, that is, Taylor expansion,integration, and applications in Islamic finance that included duration and convexityof a sukuk.

58 MATHEMATICS

3GC03 05/15/2014 9:24:55 Page 59

The techniques covered in this chapter are essential for practicing Islamic finance,managing portfolios, and hedging risk. They are needed for understanding manycomputational methods of Islamic finance.

QUESTIONS

1. Find the derivative of f and the equation of the tangent line to the graph of f at P.

a. f x� � � �5x2 � 8x � 2, P �1, � 11� �b. f x� � � x3 � 4x, P 2; 0� �

2. If y � 3x2 � 12x � 8, (i) find the slope of the tangent line at the point A 3, � 1� �,and (ii) find the point on the graph at which the tangent line is horizontal.

3. If f x� � � xj j, show that f is not differentiable at 0.

4. Compute the derivative offfiffiffiffiffix25

p.

5. Compute the derivative of (i) f x� � � x1/3 x2 � 3x � 2� �

and (ii) f x� � � x3� 3xffiffiffiffix23

p .

6. Compute derivative dydx if y � ffiffiffi

up

and u � x2 � 1.

7. Compute dydx if f x� � � 4x � 5

3x � 2.

8. Compute dydx if f x� � � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

5x2 � x � 43p

.

9. Compute dy/dx if y � 1u and u � ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3x � 2p

.

10. Find an equation of the tangent line to the graph of y � 6ffiffiffiffiffix23

p � for if 4ffiffix

p atA 1; 2� �.11. Find the extrema of f on each interval if x� � � 1

2 x2 � 2x; (a) 0; 5� �, (b) 0; 2� �,

(c) 0; 4� �, and (d) 2; 5� �.12. Find the critical numbers of the function f x� � � 4x3 � 5x2 � 42x � 7.

13. Let f x� � � x3 � 8x � 5; show that f satisfies the hypotheses of the mean-valuetheorem, and find the number c in the open interval 1;4� � that satisfies theconclusion of the theorem.

14. Let f x� � � 1/ x � 1� �2; determine whether f satisfies the hypotheses of the mean-value theorem; if so, find the numbers c in 0; 2� � that satisfy the conclusion of thetheorem.

15. Consider f x� � � 14x2 � 1, then show that f satisfies the hypotheses of the mean-

value theorem on the interval �1; 4� � and find a number c in �1; 4� � that satisfiesthe theorem. Illustrate the results graphically.

16. Let x� � � x � 3x � 2; determine whether f satisfies the hypotheses of the mean-value

theorem; if so, find the numbers c in �2; 3� � that satisfy the conclusion of thetheorem.

17. Find the local extrema of f x� � � x2/3 x2 � 8� �

.

18. Find the local extrema of f x� � � 10x3 x � 1� �2.19. A company has a monthly cost function C x� � � x3 � 3x2 � 80x � 500, where x is

the number of units produced each month. Each item produced is sold for


3GC03 05/15/2014 9:24:56 Page 60

$2; 800. Determine the production x that maximizes the profit.What is the largestpossible profit per week?

20. Compute ∫ffiffiffiffiffix23

pdx.

21. Evaluate ∫x4�1� �x�1 dx, x ≠ 1.

22. Evaluate ∫ 8x � 5ffiffix3

pdx.

23. Evaluate ∫ 2x3 � 1� �7

x2dx.

24. Evaluate ∫xffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 � 6x23

pdx.

25. Evaluate ∫ 5xffiffiffiffiffiffiffiffiffix2� 3

p dx.

26. Evaluate ∫ 31 � x� � dx.

27. Evaluate ∫2

1∫2

�112xy2 � 8x3� �

dydx.

28. Evaluate ∫3

0∫�1

�24xy3 � y� �

dxdy.

29. UsingMicrosoft Excel, calculate theMacaulay duration andmodified duration ofa 15-year sukuk paying a coupon rate of 6.5 percent per year on a semiannualbasis; it has a yield rate of 8.2 percent, a settlement date of January 2, 2014, and amaturity date of December 31, 2028.

30. Consider a 16-year maturity sukuk with a coupon rate of 6.75 percent per yearwith semiannual payments, yield rate of 8.2 percent per year, and a face value of$1,000. The settlement date is January 2, 2013, and the maturity date isDecember 31, 2028. Using the Microsoft Excel price function, compute theconvexity of the sukuk.

60 MATHEMATICS

3GC04 05/15/2014 9:37:10 Page 61

CHAPTER 4Partial Derivatives

E conomic and financial relations feature interdependence of many variables. Forinstance, the utility function of a consumer may include many products and is

written as

U � U x1, x2, x3� � (4.1)

where x1, x2, and x3 denote the quantities of oranges, tomatoes, and grape juice,respectively, consumed by the consumer. In economics, we are interested in how adependent variable responds when only one variable in the relationship changes; theassumption of maintaining all other fixed variables is called the ceteris paribuscondition. In this case we may be interested in computing the change in utility dueto a change in the consumption of oranges, keeping the consumption of tomatoes andgrape juice fixed. This rate of change is called the partial derivative of utility in respectto a change in oranges consumed. It is denoted as @U/@x1 to indicate it is a partialderivative. We call it marginal utility of oranges.

Likewise, any production process involves the interplay of many variables.Production depends on capital, labor, raw materials, entrepreneurship, roads, com-munications, taxes, and security. We may add other variables as we wish, such asresearch, banking system, and contracts reenforcement. Suppose that production Y isformulated as

Y � F K, L,M,Z� � (4.2)

where K,L, M, and Z stand for capital, labor, raw materials, and taxes, respectively.The producer may increase the quantity of labor holding everything else the same, thatis, ceteris paribus. He is interested in monitoring the partial derivative of Y in respectto labor L, @Y/@L, called the marginal productivity of labor.

In option pricing, the price of a call option C is a function of the price of theunderlying asset S, the strike price K, the volatility σ, the riskless rate of return r, andthe maturity of the option T. We have

C � C S,K, σ, r, T� � (4.3)

61

3GC04 05/15/2014 9:37:11 Page 62

We are often interested in the Greeks of the option. For instance, the partial derivativeof the option in respect to S is called the delta of the option and is written as

Δ � @C/@S (4.4)

Partial differentiation is an important technique in Islamic finance. Often, wedeal with a function of many variables. Geometrically, this function describes ahyper-surface in the Euclidian space Rn, n � 3. We study the definition andcomputation of its partial derivatives, its total differential, directional derivatives,gradients, tangent planes, and extrema. We discuss also its extrema when con-straints are imposed on its variables, particularly the method of Lagrange for findingconstrained extrema.

DEFINITION AND COMPUTATION OF PARTIAL DERIVATIVES

In general, the partial derivative of a function f x1, x2, . . . , xn� � in the direction xi at thepoint a1, a2, . . . , an� � is defined as

@f@xi

a1, a2, . . . , an� � � limh!0

f a1, a2, . . . a i � h, . . . , an� � � f a1, a2, . . . ai, . . . , an� �h

(4.5)

In the previous difference quotient, all the variables except xi are held fixed. Thechoice of fixed values determines a function of one variable. This expression alsoshows that the computation of partial derivatives reduces to the computation of one-variable derivatives. Partial derivative is denoted as f xi .

Example: Consider the function f x, y� � � x2 � xy � y2 � 4y3.We want to compute the partial derivative with respect to x holding y fixed.

We find

@f@x

� 2x � y

Example: The consumer utility function is U � U x1, x2� � � x0.71 x0.32 . The mar-ginal utility with respect to x1 is

@U@x1

� 0.7x�0.31 x0.32

Example: Consider the production Y � K0.6L0.4; the marginal product of labor is

@Y@L

� 0.4K0.6L�0.6

Example: The consumer utility function is U � U x1, x2� � � x0.71 x0.32 . The con-sumer is consuming three oranges x1 � 3� � and three tomatoes x2 � 3� �. His utility is

U � U x1, x2� � � x0.71 x0.32 � 30.730.3 � 3

62 MATHEMATICS

3GC04 05/15/2014 9:37:12 Page 63

He decides to increase his tomatoes by two tomatoes. His utility becomes

U � U x1, x2� � � x0.71 x0.32 � 30.750.3 � 3.4968

The marginal utility is

@U � 3.4968 � 3 � 0.4968

The marginal utility per one tomato is

@U@x2

� 0.49682

� 0.2484

Wemay compute this marginal utility by taking the partial derivative with respectto x2,

@U@x2

� 0.3x0.71 x�0.72

Evaluated at x1 � 3, x2 � 3� �,@U@x2

� 0.3 � 30.7 � 3�0.7 � 0.3.

We observe a small difference between the two methods for computing marginalutility:

0.3 � 0.2486 � 0.05158

The Chain Rule

Ifw � f u, v� �, with u � g x, y� � and v � h x, y� �, and if f , g, and h are differentiable, then

@w@x

� @w@u

@u@x

� @w@v

@v@x

(4.6)

@w@y

� @w@u

@u@y

� @w@v

@v@y

(4.7)

Example: Use the chain rule to find dw/dt if w � x2 � yz, with x � 3t2 � 1,y � 2t � 4, z � t3.

We apply the chain rule:

dwdt

� @w@x

dxdt

� @w@y

dydt

� @w@z

dzdt

� 2x� � 6t� � � z� � 2� � � y� � 3t2� �

� 2 3t2 � 1� ��

6t� � � t3� �

2� � � 2t � 4� � 3t2� � � 44t3 � 12t2 � 12t

Partial Derivatives 63

3GC04 05/15/2014 9:37:12 Page 64

The problem could also be solved without a chain rule by writing

w � 3t2 � 1� �2 � 2t � 4� �t3

and then finding dw/dt by single-variable method.

Derivatives of Implicit Functions

Partial derivatives can be used to find derivatives of functions that are determinedimplicitly. Suppose an equation F x, y� � � 0 determines a differentiable function f suchthat y � f x� �; that is, F x, f x� �� 0 for every x in the domain D of f . Then

dydx

� � @F x, y� �@x

/@F x, y� �

@y(4.8)

Example: Find dy/dx if f x� � is determined implicitly by

F x, y� � � y4 � 3y � 4x3 � 5x � 1 � 0.

We compute the derivative dy/dx as

dydx

� � Fx x, y� �Fy x, y� � �

12x2 � 54y3 � 3

TOTAL DIFFERENTIAL OF A FUNCTION WITH MANY VARIABLES

If all the partial derivatives of f x, y� � exist and are continuous at x, y, then we maywrite the total differential of f as

df � @f@x

dx � @f@y

dy (4.9)

Example: Let f x, y� � � 3x2 � xy, find df and use it to approximate the change in fif x, y� � changes from 1; 2� � to 1.01; 1:98� �. How does this compare with the exactchange in f ?

We have

df � @f@x

dx � @f@y

dy � 6x � y� �dx � xdy � 6 � 2� � 0.1� � � 1 �0.02� � � 0.06

We may compute df as

f x � dx, y � dy� � � f x, y� � � 3 � 1.012 � 1.01 � 1.98� � � 3 � 12 � 1 � 2

� � � 0.0605

The difference between the two methods is 0.0605 � 0.06 = 0.0005.

64 MATHEMATICS

3GC04 05/15/2014 9:37:13 Page 65

DIRECTIONAL DERIVATIVES

We let z � f x, y� �. The partial derivative @z@x � @f

@x measures the rate of change of z in the

direction of x, when x changes by one unit, holding y fixed. In Figure 4.1a, the initialpoint A x, y� � moves in the horizontal direction to B x � dx, y� �, since y is fixed.

Likewise, the partial derivative @z@y � @f

@y measures the rate of change of z in the directionof y, when y changes by one unit, holding x fixed. In Figure 4.1a the initial pointA x, y� � moves in the vertical direction to C x, y � dy� �, since x is fixed. We generalizethe rate of change of z � f x, y� � in any direction.

Let u � u1

u2

� �be the unit vector, meaning the length of u, denoted by u, is equal to

unity:

u �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu21 � u22

q� 1 (4.10)

Figure 4.1b shows a unit vector u. We consider a real scalar h; then the vector~AE � h � u is parallel to the vector u as illustrated in Figure 4.1c.Moreover, the lengthof~AE is jj~AEjj � jjhujj � hj ju � hj j.

We are interested in the change in z � f x, y� � when there is motion from pointA x, y� � to point E x � hu1, y � hu2� � as illustrated by Figure 4.1c. We have

Δz � f x � hu1, y � hu2� � � f x, y� � (4.11)

The average change is

Δzh

� f x � hu1, y � hu2� � � f x, y� �h

(4.12)

Definition: Let z � f x, y� � and let u � u1u2

� �be a unit vector with jjujj � 1. The

directional derivative of f at A x, y� � in the direction of u, denoted by Duf x, y� � is

Duf x, y� � � limh!0

f x � hu1, y � hu2� � � f x, y� �h

(4.13)

a. Partial directions c. Directional vector hub. Directional unit vector u

0

C(x, y + dy)

A(x, y) B(x + dx, y)

y

x

D(x + u1,y + u2)

A(x, y)

0

y

x

E(x + hu1,y + hu2)

A(x, y)

0

y

x

FIGURE 4.1 Directional Derivative


3GC04 05/15/2014 9:37:13 Page 66

If v is any vector that has the same direction as u, we shall refer toDuf x, y� � as thedirectional derivative of f in the direction of v.

Let z � f x, y� � be a differentiable function of two variables and let u � u1u2

� �be a

unit vector with u � 1, then

Duf x, y� � � f x x, y� �u1 � f y x, y� �u2 (4.14)

Example: Let f x, y� � � x3y2.Find the directional derivative of f at pointA �1; 2� � in the direction of v � 4

�3� �

.

The length of v is v �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi42 � �3� �2

q� 5. To obtain a unit vector u in the direction

of v, we scale back v by v � 5. We obtain u �45

�35

264

375.

The directional derivative of f at point A �1; 2� � in the direction of v is

Duf x, y� � � @f x, y� �@x

u1 � @f x, y� �@y

u2 � 3x2y2 � 45� 2x3y � �3/5

Duf �1; 2� � � 3 �1� �222 � 45� 2 �1� �32 � � 3

5� 12

GRADIENTS

Let f be a function of two variables. The gradient of f is the vector given by

rf x, y� � � f x x, y� �f y x, y� �

" #(4.15)

The directional derivative at A x, y� � may be expressed as

Duf x, y� � � f x x, y� �u1 � f y x, y� �u2 � rf x, y� �.u (4.16)

Example: Let f x, y� � � x2 � 4xy; compute the gradient at A 1; 2� �, and use thegradient to find the directional derivative at A 1; 2� � in the direction to B 2; 5� �.

By definition,

rf x, y� � � f x x, y� �f y x, y� �

" #� 2x � 4y

�4x" #

66 MATHEMATICS

3GC04 05/15/2014 9:37:14 Page 67

At A 1; 2� �,

rf 1; 2� � � 2x � 4y

�4x" #

� �6�4

" #

We let v �~AB � 2 � 15 � 2

� �� 1

3

� �; the length of v is v � ffiffiffiffiffiffiffiffiffiffiffiffiffi

1 � 32p � ffiffiffiffiffiffi

10p

. We

obtain a unit vector parallel to v by scaling back v by v.

u � vv�

1ffiffiffiffiffiffi10

p3ffiffiffiffiffiffi10

p

26664

37775

Applying the formula for the directional derivative, we have

Duf x, y� � � rf x, y� �.uDuf 1; 2� � � rf x, y� �.u � 1ffiffiffiffiffiffi

10p �6 � 4 � 3� � � � 18ffiffiffiffiffiffi

10p � �5.69.

Let A x, y� � be a fixed point, and consider the direction derivative Duf x, y� � asu � u1, u2 varies. For a given unit vector u, the directional derivative may be positive,that is, f x, y� � may increase; or negative, that is, f x, y� � may decrease; or 0. In manyapplications it is important to find the direction in which f x, y� � increases most rapidlyand the maximum rate.

Gradient theorem: Let f be a function of two variables that is differentiable atA x, y� �:i. The maximum value of Duf x, y� � at A x, y� � is rf x, y� �; andii. The maximum rate of increase of f x, y� � at A x, y� � occurs in the direction of

rf x, y� �.Let f be a function of two variables that is differentiable at A x, y� �:

i. The minimum value of Duf x, y� � at A x, y� � is �rf x, y� �; andii. The maximum rate of decrease of f x, y� � at A x, y� � occurs in the direction of

�rf x, y� �.Example: Let f x, y� � � 2 � x2 � 1

4 y2.

Find the direction in which f x, y� � increases most rapidly at pointA 1; 2� �, and findthe maximum rate of increase of f x, y� � at A 1; 2� �.

The gradient of f is rf x, y� � � 2xy2

" #; at point A 1; 2� �, rf 1; 2� � � 2

1

� �. The

maximum rate of increase of f at A 1; 2� � is rf x, y� � � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi22 � 12

p � ffiffiffi5

p � 2.2.


3GC04 05/15/2014 9:37:15 Page 68

TANGENT PLANES AND NORMAL LINES

Tangent planes and normal lines can be found using the techniques that follow.

Tangent Planes

Suppose a surface S is the graph of an equation F x, y, z� � � 0 and F has continuous firstpartial derivatives. LetA0 x0, y0, z0

� �be a point on S at which Fx, Fy, and Fy are not all

zero. A tangent line to S at A0 is, by definition, a tangent line l to any curve C that lieson S and contains A0. If C has a parametrization

x � f t� �, y � g t� �, x � h t� � (4.17)

for t in some interval I and if r t� � is the position vector of A x, y, z� �, then

r t� � �f t� �g t� �h t� �

264

375 (4.18)

Hence r´ t� � �f ´ t� �g´ t� �h´ t� �

264

375 is a tangent vector to C at A x, y, z� � as indicated in

Figure 4.2a.

For each t, the point f t� �, g t� �, h t� �� on C is also on S, and thereforeF f t� �, g t� �, h t� �� 0.

We let q � F x, y, z� �, with x � f t� �, y � g t� �, x � h t� �; then using a chain rule andthe fact that q � 0 for every t, we have

dqdt

� @q@x

dxdt

� @q@y

dydt

� @q@z

dzdt

� 0 (4.19)

a. Tangent vector r'(t) b. Tangent vector r'(t0)

r'(t)

r (t)

AC

A0

l

z

x

y

f(x0, y0)

A(x , y , z)

r'(t0)

C

A0

l

y

z

x

Δ

FIGURE 4.2 Tangent Planes and Orthogonal Gradients

68 MATHEMATICS

3GC04 05/15/2014 9:37:16 Page 69

Thus, for every point A x, y, z� � on C,

Fx x, y, z� �f ´ t� � � Fy x, y, z� �g´ t� � � Fz x, y, z� �h´ t� � � 0 (4.20)

Or equivalently,

rF x, y, z� �.r´ t� � � 0 (4.21)

In particular, if A0 x0, y0, z0� �

corresponds to t � t0, then

rF x0, y0, z0� �

.r´ t0� � � 0 (4.22)

Since r´ t0� � is a tangent vector to C at A0, this implies that the vector rFjA0 isorthogonal to every tangent line l to S at A0 (Figure 4.2b). The plane through A0, withnormal vectorrFjA0 , is the tangent plane to S atA0. We have shown that every tangentline l to S at A0 lies in the tangent plane at A0. We may state the following: SupposeF x, y, z� � has continuous partial derivatives and S the graph of F x, y, z� � � 0. If A0 is apoint on S and if Fx, Fy, and Fz are not all 0 at A0, then the vector rFjA0 is orthogonalto the tangent plane to S at A0.

We will refer to the gradient rFjA0 as a vector that is orthogonal to the surface Sat A0.

Assuming that Fx, Fy, and Fz are not all 0 at A0, an equation for the tangent planeof F x, y, z� � � 0 at the point A0 x0, y0, z0

� �is

Fx x0, y0, z0� �

x � x0� � � Fy x0, y0, z0� �

y � y0� � � Fz x0, y0, z0

� �z � z0� � � 0 (4.23)

Example: Find an equation for the tangent plane to the surface described by

F x, y, z� � � 34x2 � 3y2 � z2 � 12 � 0 atA0 2; 1; 2� �.

The partial derivatives are Fx x, y, z� � � 32 x, Fy x, y, z� � � 6y, and Fz x, y, z� � � 2z;

and hence, at A0 2; 1; 2� �: Fx 2; 1;2� � � 32 x � 3, Fy 2; 1; 2� � � 6y � 6, and

Fz 2; 1; 2� � � 2z � 4. The equation of the tangent plane is

Fx x0, y0, z0� �

x � x0� � � Fy x0, y0, z0� �

y � y0� � � Fz x0, y0, z0

� �z � z0� � � 0

3 x � 2� � � 6 y � 1� � � 4 z � 2� � � 3x � 6y � 4z � 20 � 0

If z � f x, y� � is an equation for S and we let F x, y, z� � � f x, y� � � z, the planeequation takes the form

f x x0, y0� �

x � x0� � � f y x0, y0� �

y � y0� � � �1� � z � z0� � � 0 (4.24)

An equation for the tangent plane to the graph of z � f x, y� � at pointA0 x0, y0, z0� �

is

z � z0 � f x x0, y0� �

x � x0� � � f y x0, y0� �

y � y0� �

(4.25)


3GC04 05/15/2014 9:37:17 Page 70

Normal Line

The line orthogonal to the tangent plane at a point A0 x0, y0, z0� �

on a surface S is anormal line to S atA0 x0, y0, z0

� �. If S is the graph of F x, y, z� � � 0, then the normal line

is parallel to the vectorrF x0, y0, z0� �

. In Figure 4.2b, the normal lineAA0 is parallel torF x0, y0, z0� �

. This means for a t ∈ R, we have

AA0 � t � rF x0, y0, z0� �

(4.26)

Or equivalently,

x � x0y � y0z � z0

24

35 �

tf x x0, y0, z0� �

tf y x0, y0, z0� �

tf z x0, y0, z0� �

264

375 (4.27)

We obtain the following parametric equations for the line through A0 x0, y0, z0� �

parallel to rF x0, y0, z0� �

:

x � x0 � tf x x0, y0, z0� �

; y � y0 � tf y x0, y0, z0� �

; z � z0 � tf z x0, y0, z0� �

(4.28)

Example: Find an equation for the normal line to the surface described by

F x, y, z� � � 34x2 � 3y2 � z2 � 12 � 0 atA0 2; 1; 2� �.

rF x0,y0,z0� ��rF 2;1;2� ��Fx 2;1;2� ��3,Fy 2;1;2� ��6,Fz 2;1;2� � �6; the parame-

tric equations for the normal line are: x�2�2t; y�1�6t; z�2�6t.

EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

A function f of two variables has a local maximum at a, b� � if there is an open disk Rcontaining a, b� � such that f x, y� � � f a, b� � for every x, y� � in R. Similarly, the functionhas a local minimum at c, d� � if there is an open disk R containing a, b� � such thatf x, y� � � f c, d� � for every x, y� � in R.

A region in the xy-plane is bounded if it is a subregion of a closed disk. If f iscontinuous on a closed and bounded region R, then f has a maximum f a, b� � and aminimum f c, d� � for some a, b� � and c, d� � in R; that is,

f c, d� � � f x, y� � � f a, b� � (4.29)

for every x, y� � in R.The local maxima and minima are the local extrema of f . If f has continuous first

partial derivatives at x0, y0� �

and if f x0, y0� �

is a local extremum of f , thenf x x0, y0� � � 0 and f y x0, y0

� � � 0.Local extrema can also occur at x0, y0

� �if either f x x0, y0

� �or f y x0, y0

� �does not

exist.

70 MATHEMATICS

3GC04 05/15/2014 9:37:18 Page 71

Definition: Let f be a function of two variables. A pair a, b� � is a critical point of fif either:

i. f x a, b� � � 0 and f y a, b� � � 0; orii. f x a, b� � or f y a, b� � does not exist.

When searching for local extrema of a function, we usually begin by finding thecritical points. We then test each point in some way to determine if it yields a localmaximum or minimum. A maximum or minimum of a function may occur at aboundary point of its domain R.

Example: Let f x, y� � � 1 � x2 � y2, with x2 � y2 � 4. Find the extrema of f .The restriction x2 � y2 � 4means that the points x, y� � belong to the closed diskC

centered at 0 and with radius 2 in the xy-plane. By definition, the critical points aresolutions for f x x, y� � � 2x � 0 and f y x, y� � � 2y � 0.

The only pair that satisfies these conditions is x � 0, y � 0� �; hence f 0; 0� � � 1 isthe only possible extremum. Moreover, f x, y� � � 1 � x2 � y2 > 1 if x, y� � ≠ 0. Itfollows that f has a local minimum 1 at 0; 0� �. To find possible boundary extrema,we evaluate points x, y� � that are located on the boundary of C. We see that any suchpoint satisfies x2 � y2 � 4 and leads to a maximum f x, y� � � 1 � 4 � 5.

To determine the extrema of a function f of two variables, it is convenient to usethe following function D, called the discriminant of f .

Definition: Let f be a function of two variables that has continuous second partialderivatives. The discriminant D of f is given by

D x, y� � � f xx x, y� �f yy x, y� � � f xy x, y� �h i2

(4.30)

The discriminant is in fact the determinant of the following matrix:

f xx f xyf yx f yy

� �(4.31)

Note that f xy � f yx.We propose tests for local extrema. Let f be a function of two variables that has

continuous second partial derivatives throughout an open disk R containing a, b� �. Iff x a, b� � � 0 and f y a, b� � � 0 and D a, b� � > 0, then f a, b� � isi. A local maximum of f if f xx a, b� � < 0; andii. A local minimum of f if f xx a, b� � > 0

A point A a, b, f a, b� �� on the graph of f is a saddle point if f x a, b� � � 0 andf y a, b� � � 0, and if there is an open disk R containing a, b� � such that f x, y� � > f a, b� �for some points in R and f x, y� � < f a, b� � for other points.

Let f have continuous second partial derivatives throughout an open disk Rcontaining a, b� �. If f x a, b� � � 0 and f y a, b� � � 0 and D a, b� � < 0, then the pointA a, b, f a, b� �� is a saddle point on the graph of f .

Example: If f x, y� � � x2 � 4xy � y3 � 4y, find the local extrema and saddle pointsof f .


3GC04 05/15/2014 9:37:19 Page 72

The first partial derivatives of f are f x x, y� � � 2x � 4y andf y x, y� � � �4x � 3y2 � 4. Since f x and f y exist for every x, y� �, the only criticalpoints are the solution of the system

f x x, y� � � 2x � 4y � 0

f y x, y� � � �4x � 3y2 � 4 � 0

Solving the system, we find that f has two critical points 4; 2� � and 43 ,

23

� �. The

second derivatives of f are f xx x, y� � � 2, f xy x, y� � � �4, and f yy x, y� � � 6y. Hence, thediscriminant is given by D x, y� � � 12y � 16. We present the findings as follows:

Critical Point Value of Discriminant Value of f xx Finding

4; 2� � D 4; 2� � � 8 > 0 f xx 4; 2� � � 2 > 0 f 4; 2� � � 0, local minimum43,23

� �D

43,23

� �� 8 < 0

Irrelevant 43,23

� �saddle point

EXTREMAL PROBLEMS WITH CONSTRAINTS

Many interesting maximum or minimum problems arise in such a form that we arerequired to find an extremal value of a function, say f x, y, z� �, where the variables x, y,and z are not independent of each other, but are restricted by some relationshipexisting between them; this relationship is expressed by an equation:

g x, y, z� � � 0 (4.32)

The equation g x, y, z� � � 0 is called the constraint on the variables x, y, z. It isimmaterial whether the equation constraint has the form g x, y, z� � � 0 or g x, y, z� � � Cwhere C is a specified constant, for the latter form of constraint can be writteng x, y, z� � � C � 0. An extremal problem with constraint may occur with any numberof variables and there may be more than one equation of constraint.

Example: Find the minimum value of f x, y, z� � � x � 3� �2 � y � 2� �2 � z � 1� �2subject to g x, y, z� � � 2x � 3y � 4z � 25 � 0.

In this problem, there is one constraint on the three variables.Example. Find the extrema of f x, y, z� � � x2 � y2 � z2 subject to the conditions

x � y � z � 0

x2

16� y2 � z2 � 1

In this problem, there are two constraints on the three variables, so that there isactually only one independent variable.

Example: Find the minimum value of f x, y, z, u, v� � � x � u� �2 � y � v� �2 � z2

subject to 3x2 � y2 � 6x � 4y � 12z � 43 � 0.In this problem there are five variables and one constraint, which happens to

involve only three of the five variables.

72 MATHEMATICS

3GC04 05/15/2014 9:37:20 Page 73

Elimination Method

There are various methods for dealing with extremal problems with constraints. Onemethod is to use the equation or equations of constraint to express some of thevariables in terms of remaining variables. These latter variables are chosen as theindependent variables, and the function whose extremal value is sought is thenexpressed in terms of the independent variables only. The solution is then carriedby standard methods. This method is the method of direct elimination.

Example: Find the point of the plane 2x � 3y � 4z � 25, which is nearest to thepoint (3,2,1).

If d is the distance from point x, y, z� � of the plane to 3; 2; 1� � we haveMinimize d2 � x � 3� �2 � y � 2� �2 � z � 1� �2 subject to z � 1

4 2x � 3y � 25� �.We eliminate z:

d2 � f x, y� � � x � 3� �2 � y � 2� �2 � 12x � 3

4y � 29

2

� �2

We seek the minimum of d2 as x, y range through all possible values. We look

therefore for points at which @f@x � @f

@y � 0. The equations to be considered are

2 x � 3� � � 212x � 3

4y � 29

2

� �.12� 0

2 y � 2� � � 212x � 3

4y � 29

2

� �. � 3

4

� �� 0

On simplifying, we obtain

10y � 3y � 53�6x � 25y � �55

The solution is x � 5, y � �1. Substituting in the equation of the plane, we findz � �3.

Lagrange Method

A simpler method to finding the extrema of a function subject to constraints on itsvariables is the method of the Lagrange multiplier. We develop the geometry of thismethod in Figure 4.3 for a problem of maximizing wheat production z � f x, y� � and aconstraint function on machinery and labor g x, y� � � 0. The production function isdescribedby level curves or contours. Themaximumoutput ofwheat is achievedwhenacontour of the production function is tangent to the constraint curve at A x*,y*� �.

The Lagrange method assumes that f and g have continuous first partialderivatives and g2x � g2y ≠ 0, that is, gx or gy are not simultaneously zero, meaningrg ≠ 0; if f has an extremum f x*,y*� � subject to the constraint g x, y� � � 0, then there isa real number λ such that

rf x*,y*� � � λrg x*,y*� � (4.33)


3GC04 05/15/2014 9:37:21 Page 74

Most important, we notice that the gradients rf x*,y*� � and rg x*,y*� � areorthogonal to the same tangent line at point A x*,y*� � and therefore are linearlyrelated. The multiplier λ measures the increase in f per one unit increase in gat A x*,y*� �.

The points at which a function f of two variables has relative extrema subject tothe constraint g x, y� � � 0 satisfy the system of equations:

f x x, y� � � λgx x, y� � (4.34)

f y x, y� � � λgy x, y� � (4.35)

g x, y� � � 0 (4.36)

These equations derive from the first-order conditions of the Lagrangian function,

L x, y, λ� � � f x, y� � � λg x, y� � (4.37)

Example: Maximize consumer utility U x, y� � � x0.6y0.4 subject to the budgetconstraint 3x � y � $100. Check the answer using the Microsoft Excel solver.

We form the Lagrangian: L x, y, λ� � � x0.6y0.4 � λ 100 � 3x � y� �@L@x

� 0.6x�0.4y0.4 � 3λ � 0

@L@y

� 0.4x0.6y�0.6 � λ � 0

@L@λ

� 100 � 3x � y � 0

From the first two equations we obtain

0.6x�0.4y0.40.4x0.6y�0.6 � 3λ

λ

y

x*x0

y*

g(x , y ) = 0

A(x*, y*)

f(x, y) = C

g(x*, y*)Δ f(x*, y*)Δ

FIGURE 4.3 Geometry of the Lagrange Method

74 MATHEMATICS

3GC04 05/15/2014 9:37:22 Page 75

Simplifying, we obtain0.6y0.4x

� 3, or y � 2x. Replacing in the budget constraint,we find x � 20, y � 40.We compute λ as λ � 0.4 � 200.640�0.6 � 0.2639. An increasein income by $1 increases utility by 0.2639. The marginal utility of $1 is 0.2369.

Lagrange’s method can be extended to functions of more than two variables. Wepresent extension to three variables, but we can state similar conditions for anynumber of variables.

Suppose that f and g are functions of three variables having continuous firstpartial derivatives and g2x � g2y � g2z ≠ 0, that is, gx, gy, and gz are not simultaneouslyzero, meaning rg ≠ 0. If f has an extremum f x*,y*,z*� � subject to the constraintg x, y, z� � � 0, then there is a real number λ such that

rf x*,y*,z*� � � λrg x*,y*,z*� � (4.38)

The points at which a function f of three variables has relative extrema subject tothe constraint g x, y, z� � � 0 satisfy the system of equations:

f x x, y, z� � � λgx x, y, z� � (4.39)

f y x, y, z� � � λgy x, y, z� � (4.40)

f z x, y, z� � � λgz x, y, z� � (4.41)

g x, y, z� � � 0 (4.42)

These equations derive from the first-order conditions of the Lagrangian function,

L x, y, z, λ� � � f x, y, z� � � λg x, y, z� � (4.43)

Example: Find the extrema of f x, y, z� � � x � y � z subject to x2 � y2 � z2 � 27.Check the answer using the Microsoft Excel solver.

We form the Lagrangian function,

L x, y, z, λ� � � x � y � z � λ 27 � x2 � y2 � z2� �

We compute first-order conditions:

@L@x

� 1 � 2λx � 0;@L@y

� 1 � 2λy � 0;@L@z

� 1 � 2λz � 0;@L@λ

� 27 � x2 � y2 � z2 � 0

We obtain x � y � z. Replacing in the constraint, we get 3x2 � 27, or x � �3.The solution is: x � y � z � 3 and λ � 1/6; f � 9; x � y � z � �3 and λ � �1/6;

f � �9.

SUMMARY

This chapter covered partial differentiation, an important technique used in Islamicfinance. It introduced the definition and computation of partial derivatives, the chainrule, the derivatives of implicit functions, total differential of a function with manyvariables, directional derivatives, gradients, tangent planes and normal lines, extrema


3GC04 05/15/2014 9:37:23 Page 76

of functions of several variables, and extremal problems with constraints; it illustratedthe elimination method and the Lagrange method in the computation of extremalvalues for a function under a set of constraints.

The chapter also provides essential tools underlying computational methods inIslamic finance, particularly in optimization of portfolios and in studying multi-dimensional functions.

QUESTIONS

1. Find the partial derivatives of f x, y� � � x3 � y2� �5

.

2. Find dw/dt if w � x3 � y3, with x � 1t � 1, y � t

t � 1.

3. Use partial derivatives to find dy/dx if y � f x� � is determined implicitly by theequation

6x � ffiffiffiffiffiffixy

p � 3y � 4 � 0

4. Consider the consumer utility function U � U x1, x2� � � x0.71 x0.32 . Compute themarginal utility with respect to x1 and x2 at point A x1 � 50, x2 � 70� � andcompute the total differential dU at point A x1 � 50, x2 � 70� �. What is themeaning of dU?

5. Consider the production Y � K0.6L0.4. Compute the marginal product of laborand capital at point A K � 50, L � 70� � and compute total differential dY at pointA K � 50, L � 70� �. What is the interpretation of dY?

6. Find the directional derivative of f x, y� � � x2 � 5xy � 3y2 at A 3, � 1� � in the

direction of u �ffiffiffi2

p/2ffiffiffi

2p

/2

" #.

7. Find the directional derivative of f x, y� � � x � y� �/ x � y� � at A 2, � 1� � in the

direction of v � 34

� �.

8. Let f x, y� � � 2 � 14 x

2 � y2; find the direction in which f x, y� � increases mostrapidly at the point A 2; 1� � and find the maximum rate of increase of f x, y� �at A 2; 1� �.

9. We consider 4x2 � y2 � 3z2 � 10 and A 2, � 3; 1� �. Find an equation for thetangent plane as well as an equation of a normal line to the graph of the equationat the point A.

10. If f x, y� � � 13 x

3 � 43 y

3 � x2 � 4y � 3, find the local extrema and saddle points of f .

11. Find the extrema and saddle points of f x, y� � � �x2 � 4x � y2 � 2y � 1.

12. Find the extrema and saddle points of f x, y� � � x2 � 3xy � y2 � 2y � 6x.

13. Find the extrema of f x, y, z� � � x2 � y2 � z2 subject to x � y � z � 1. Check youranswer using the Microsoft Excel solver.

14. Find the extrema of f x, y� � � 4x2 � 4xy � y2 subject to x2 � y2 � 1.

15. Find the extrema of f x, y, z� � � x2 � y2 � z2 subject to x � y � 1 and y2 � z2 � 1.

76 MATHEMATICS

3GC05 05/15/2014 9:45:48 Page 77

CHAPTER 5Logarithm, Exponential, and

Trigonometric Functions

T he logarithm, exponential, and trigonometric functions play a fundamental role inall fields of science and are particularly useful in Islamic finance. The logarithm and

exponential functions are related; one implies the other. The trigonometric functionsplay an important role inmeasuring slopes, direction, and rates of change of a functionand have wide applications in statistics such as in time series analysis.

LOGARITHM FUNCTIONS

Consider the quantity 24 � 16. Obviously, we have three numbers involved: themultiplicand 2, the exponent 4, and the product 16. From this operation we maymake the following statement: the exponent 4 is the logarithm of 16 with base 2.Likewise, we may consider the following quantity: 105 � 100; 000. We can makethe statement that the exponent 5 is the logarithm of 100,000 with base 10. Moregenerally, if

x � by (5.1)

then y is the logarithm of x with base b, and is written

y � logb x� � (5.2)

Accordingly, we write

4 � log2 16� �and 5 � log10 100; 000� �:

The logarithm of a number is the exponent to which another fixed value, the base,must be raised to produce that number. For example, the logarithm of 1,000 to base10 is 3, because 1,000 is 10 to the power 3: 1,000 = 10 × 10 × 10 = 103, solog101; 000 � 3.

77

3GC05 05/15/2014 9:45:49 Page 78

Logarithm Identities

If we consider the product 25 � 24 � 29 we obtain 32 � 16 � 512.We decompose it into: 5 � log2 32� �, 4 � log2 16� �, and 9 � log2 512� �; in other

words,

9 � 5 � 4 � log2 512� � � log2 32� � � log2 16� �More generally,

logb xz� � � logb x� � � logb z� � (5.3)

logb x=w� � � logb x� � � logb w� � (5.4)

logb xp� � � p � logb x� � (5.5)

logb x1=p� �

� 1plogb x� � (5.6)

Change of Base

The logarithm logax can be computed from the logarithms of x and a with respect toan arbitrary base b using the following formula

logax � logbxlogba

(5.7)

where logax � y implies x � ay. Now we want to solve this equation for y, using onlybase b logs, not base a logs. To do this, we take the log of each side: logbx � logb ay� �.Now we simplify the right side:

logbx � ylogba

To get y by itself, we just have to divide both sides by logba:

logbx=logba � y

Substituting logax back in for y we have

logax � logbxlogba

Example: 25 � 32, 24 � 16; 5 � log232, 4 � log216; log232=log216 � 54 � 1:25;

32 � 161:25; or 1:25 � log1632.Example: log23 � log103=log102≅0:47712=0:30103≅1:585; note that 21:585 ≅3.

The Natural Logarithmic Function

Let f be a function that is continuous on a closed interval a; b� �; we can define afunction:

F � ∫x

a

f t� �dt (5.8)

78 MATHEMATICS

3GC05 05/15/2014 9:45:50 Page 79

For x in a; b� �. If f t� � � 0 throughout a; b� �, then F x� � is the area under the graph off from a to x, as illustrated in Figure 5.1a. For the special case f t� � � tn, where n is arational number and n ≠ �1, we can find an explicit form for F. Thus, by the powerrule for integrals,

F � ∫x

a

f t� �dt � tn�1n � 1

� �xa� 1n � 1

xn�1 � an�1� �

if n ≠ � 1 (5.9)

As indicated, we cannot use t�1 � 1=t for the integrand, since 1n�1 is undefined for

n � �1. Hence, we are unable to determine an anti-derivative of 1=x, that is a functionF such that F´ x� � � 1=x. The introduction of the notion of logarithm provides asolution to this situation.

The natural logarithmic function, denoted by ln, is defined by

ln x� � � ∫x

1

1tdt for every x > 0: (5.10)

This expression is called the natural logarithm of x. The restriction x > 0 isnecessary because if x � 0, the integrand 1=t has an infinite discontinuity between xand 1 and hence ∫x1

1t dt does not exist.

If x > 1, the definite integral ∫x11t dt may be interpreted as the area of the region of

the graph of y � 1=t from t � 1 to t � x (Figure 5.1b).If 0 < x < 1, then since

∫x

1

1tdt � � ∫

1

x

1tdt (5.11)

a. The primitive function F b. Logarithm function

c. Negative of logarithm function

Area=

0

y

1

1

11

t

tx

xt

y =

dt = lnx

0

y

1

11tx

xt

1ty =

dt = – lnx

0

y

a x bt

y = f (t)

F(x) =

f (t)dta

x

FIGURE 5.1 Definition of the Natural Logarithm Function

Logarithm, Exponential, and Trigonometric Functions 79

3GC05 05/15/2014 9:45:50 Page 80

the integral is the negative of the area of the region under the graph 1=t from t � x tot � 1 (see Figure 5.1c). This shows that ln x� � is negative for 0 < x < 1 and positive forx > 1. Also note that by definition,

ln 1� � � ∫1

1

1tdt � 0 (5.12)

We note also that the derivative of the function F x� � � ∫x

1

1t dt with respect to x,

that is, dF x� �dx , is 1=x for every x > 0. Substituting ln x� � for ∫

x

1

1t dt yields

dlnxdx

� 1x

(5.13)

Hence, lnx is the primitive function, or the anti-derivative of 1=x. Since lnx isdifferentiable and its derivative 1=x is positive for everyx > 0, it follows that the naturallogarithm function is continuous and increasing throughout its domain. Also note that

d2ln x� �dx2

� ddx

1x

� � � 1

x2(5.14)

which is negative for every x > 0. Hence, the graph of the natural logarithm is concavedownward on 0; ∞� �.

Let us sketch the graph of y � ln x� �. If 0 < x < 1, then ln x� � < 0 and the graph isbelow the x-axis. If x > 1, the graph is above the x-axis. Since ln 1� � � 0, thex-intercept is 1 (see Figure 5.2a).

If u � g x� � and g is differentiable, then

dln u� �dx

� 1ududx

if g x� � > 0 (5.15)

dlnuj jdx

� 1ududx

if g x� � ≠ 0 (5.16)

We may use differentiation formulas for logarithm function ln to obtain formulasfor integration. In particular,

dlnuj jdx

� 1ududx

if g x� � ≠ 0 (5.17)

which gives, by integration,

∫1

g x� � g´ x� �dx � ln g x� �j j � C (5.18)

Hence, if u � g x� � and g is differentiable, then

∫1udu � lnuj j � C (5.19)

Example: ∫ 1x dx � lnxj j � C

80 MATHEMATICS

3GC05 05/15/2014 9:45:51 Page 81

Example: Evaluate ∫ x3x2� 5 dx:

We rewrite the integral as

∫x

3x2 � 5dx � ∫

13x2 � 5

xdx

Let u � 3x2 � 5, and du � 6xdx; we obtain

∫x

3x2 � 5dx � 1

6 ∫1

3x2 � 56xdx � 1

6 ∫1udu

� 16ln uj j � C � 1

6ln 3x2 � 5 � C

Example: Evaluate ∫421

9 � 2x dx:

Since 19 � 2x is continuous on 2; 4� �, the definite integral exists. One method of

evaluation consists of using an indefinite integral to find an anti-derivative of 19 � 2x. We

let u � 9 � 2x, du � �2dx:We proceed as follows:

∫1

9 � 2xdx � � 1

2 ∫1

9 � 2x�2� �dx

� � 12 ∫

1udu � � 1

2lnuj j � C � � 1

2ln 9 � 2xj j � C

Applying the fundamental theorem of calculus yields

∫4

2

19 � 2x

dx � � 12ln 9 � 2xj j�42 � �1

2ln1 � ln5� � � 1

2ln5:

b. Graph of the natural exponential function

a. Graph of the natural logarithm function

0

y = ln(x)

x10

y = ln(x)

y = exp(x)

y = x

x1

1

FIGURE 5.2 Graphs of the Natural Logarithm and Exponential Functions


3GC05 05/15/2014 9:45:51 Page 82

The natural logarithms satisfy the laws of logarithms. If p > 0 and q > 0, then

ln pq� � � ln p� � � ln q� � (5.20)

lnpq� ln p� � � ln q� � (5.21)

ln pr� � � r ln p� � for every rational number r (5.22)

THE EXPONENTIAL FUNCTION

The natural exponential function, denoted by exp, is the inverse of the naturallogarithm function.

Since exp is the inverse of ln, its domain is R, the real line, and its range is 0; ∞� �.y � exp x� � if and only if x � ln y� � (5.23)

where x is any real number and y > 0. We may write

ln expx� � � x and exp lny� � � y (5.24)

If two functions are the inverse of each other, then their graphs are reflectionsthrough the line y � x. Hence, the graph of y � exp x� � can be obtained by reflectingthe graph of y � ln x� � through the line y � x. The graph of the exponential function isshown in Figure 5.2b. Note that

limx! ∞

exp x� � �∞ and limx!� ∞

exp x� � � 0 (5.25)

There exists exactly one positive real number whose natural logarithm is 1. Thisnumber is denoted by e.

The letter e denotes the positive number such that

ln e� � � 1 (5.26)

The number e is equal to

e � limh!0

1 � h� �1=h � 2:71828 (5.27)

The number e is an irrational number.We shall define ex as the real number y suchthat

ln y� � � x (5.28)

If x is any real number, then

ex � y if and only if ln y� � � x (5.29)

82 MATHEMATICS

3GC05 05/15/2014 9:45:52 Page 83

The fact that ln expx� � � x and exp lny� � � y for every x > 0 may now be written as

ln ex� � � x for every x (5.30)

eln x� � � x for every x > 0 (5.31)

If p and q are real numbers and r is a rational number, then

epeq � ep�q; ep

eq� ep�q; and ep

� �r � epr (5.32)

The exponential function is ex; we may state that ex is its own derivative,

dex

dx� ex (5.33)


deu

dx� dudx

eu (5.34)


∫eudu � eu � C (5.35)

Example: Evaluate (i) ∫ e3x

x2 dx and (ii) ∫21e3x

x2 dx:

i. We rewrite the integral as ∫ e3x

x2 dx � ∫e3x 1x2 dx:

We use u � 3=x; du � � 3x2 dx.

∫e

3x

x2dx � � 1

3 ∫e3x � 3

x2

� dx

� � 13 ∫eudu � �1

3eu � C � � 1

3e3=x � C

ii. Using the anti-derivative found in (i) and applying the fundamental theorem ofcalculus yields

∫2

1

e3=x

x2dx � � 1

3e3x

i21� �1

3e32 � e3

� �� 5:2


3GC05 05/15/2014 9:45:52 Page 84

POWER SERIES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS

The Taylor expansion of f at point a is the power series

f x� � � f a� � � f ´ a� �1!

x � a� � � f ´´ a� �2!

x � a� �2 � f 3� �3!

x � a� �3 � ∙ ∙ ∙ (5.36)

Consider the Maclaurin series for 1 � x� ��1 at x � 0; it is the geometric series

11 � x

� 1 � x � x2 � x3 � ∙ ∙ ∙ (5.37)

By integrating this Maclaurin series, we find the Maclaurin series for ln 1 � x� �,where ln denotes the natural logarithm

ln 1 � x� � � �x � 12x2 � 1

3x3 � 1

4x4 � 1

5x5 � ∙ ∙ ∙

In fact, we compute ∫ 11�x dx; we let u � 1 � x, du � �dx, our integral becomes

∫1

1 � xdx � �∫

1udu � �ln u� � � �ln 1 � x� �

The integral of the right-hand side of Equation (5.37) is � 12 x

2 � 13 x

3�14 x

4 � 15 x

5 � ∙ ∙ ∙. Equating the two sides, we have

�ln 1 � x� � � x � 12x2 � 1

3x3 � 1

4x4 � 1

5x5 � ∙ ∙ ∙

or ln 1 � x� � � �x � 12x2 � 1

3x3 � 1

4x4 � 1

5x5 � ∙ ∙ ∙ (5.38)

The corresponding Taylor series for ln x� � at a = 1 is

ln x� � � x � 1� � � 12

x � 1� �2 � 13

x � 1� �3 � 14

x � 1� �4∙ ∙ ∙ (5.39)

By applying Taylor’s expansion formula (5.36), the Taylor series for the expo-nential function ex at a = 0 is

ex � 1 � x1!

� x2

2!� x3

3!� x4

4!� x5

5!� ∙ ∙ ∙ � X∞

n�0xn

n!(5.40)

84 MATHEMATICS

3GC05 05/15/2014 9:45:53 Page 85

GENERAL EXPONENTIAL AND LOGARITHMIC FUNCTIONS

We consider real positive number a. We define ax for every real number x. If theexponent is a rational number r, then

ar � eln�ar� � erln a� � (5.41)

This formula is the motivation for the following definition of ax. We have, forevery a > 0 and every real number x,

ax � exln a� � (5.42)

If f x� � � ax, then f is the exponential function with base a. Since ex is positive forevery x, so is ax.

We have the following differentiation rules for ax:

dax

dx� axln a� � (5.43)

For u � g x� � where g is differentiable we have

dau

dx� auln a� � du

dx(5.44)

If a ≠ 1 and f x� � � ax, then f is a one-to-one function. Its inverse function isdenoted by loga and is called the logarithm function with base a. Another way ofstating this relationship is

y � logax if and only if x � ay (5.45)

The logarithm logax is called the logarithm of x with base a. In this terminology,natural logarithms are logarithms with base e; that is,

ln x� � � logex (5.46)

To obtain the relationship between loga and ln, consider y � logax, or equiv-alently, x � ay. Taking the natural logarithm of both sides of the last equation yields

ln x� � � yln a� �or y � ln�x�=ln�a� (5.47)

This proves that

logax � ln x� �ln a� � (5.48)

Example: log5x � ln x� �=ln 5� �


3GC05 05/15/2014 9:45:53 Page 86

We show that the number e is a limit of the following expressions:

�i� limh!0

1 � h� �1=h � e; �ii� limn! ∞

1 � 1n

� n

� e (5.49)

Applying the definition of derivative to f x� � � ln x� � and using the laws oflogarithms yields

f ´ x� � � limh!0

ln x � h� � � ln x� �h

� limh!0

1hlnx � h

x� lim

h!0

1hln 1 � h

x

� � lim

h!0ln 1 � h

x

� 1=h

(5.50)

Since f ´ x� � � 1=x, we have for x � 1,

1 � limh!0

ln 1 � h� �1=h (5.51)

We next observe that

1 � h� �1=h � eln 1�h� �1=h (5.52)

Since the natural exponential function is continuous at 1, it follows that

limh!0

1 � h� �1=h � limh!0

eln 1�h� �1=h � elimh!0 eln 1�h� �1=h � e1 � e (5.53)

SOME APPLICATIONS OF LOGARITHM AND EXPONENTIALFUNCTIONS IN FINANCE

Logarithm and exponential function have several uses in finance; among them arefinding the simple compounding and continuous compounding of returns, the presentvalue formula, and the normal distribution.

Simple Compounding and Continuous Compounding of Returns

Rates of return on an asset are usually defined on an annual basis. Nonetheless, thepassage from annual rates of return to rates of return per an interval of time is simple.For instance, annual rates can be transformed into semiannual, quarterly, monthly,weekly, or daily rates. If the annual rate of return is R, the semiannual rate of returnwould be R=2. In general, if the number of days in a period of time is d, the rate ofreturn for that period would be d � R=365. Compounding returns on a $1 bankdeposit on a semiannual basis for n years would cumulate to $1 � 1 � R

2

� �2n. In

general, if the frequency at which returns are paid ism, then the compounded returnson $1 for n years would be

$1 � 1 � Rm

� m�n(5.54)

86 MATHEMATICS

3GC05 05/15/2014 9:45:54 Page 87

Returns may be assumed to accrue on a continuous basis. The compoundingfactor for returns on a $1 bank deposit for a one-year period on a continuous basis isexpressed as

exp R� � � limm! ∞

1 � Rm

� m

(5.55)

Let Rc denote the annual rate of return on a continuous basis; the relationshipbetween Rc and R is

eRc � 1 � Rm

� m

(5.56)

This means

Rc � m � ln 1 � Rm

� andR � m eRc=m � 1

� �(5.57)

Let R � 8 percent, m � 1, then Rc � 7:696 percent. If m � 4, then Rc �7:921 percent.

The growth formula: assume real gross domestic product (GDP � y) is increasingon a continuous basis at r percent per year, then GDP at time t is equal to yt � erty0where y0 is GDP at time 0.

The Present Value Formula

We may assume that an asset has a cash flow CFt that arises continuously over time;the value of the asset is the present value of this cash flow:

V0 � ∫T

0

e�rtCFtdt (5.58)

The Normal Distribution

In statistics the probability density function for the normal distribution isdefined by

f x� � � 1

σffiffiffiffiffiffi2π

p e�12

x � μσ

� �2

(5.59)

for real numbers μ and σ σ > 0� � where μ is the mean and σ2 is the variance of thedistribution.

We observe that f x� � has a maximum at x � π; f x� � ! 0 as x ! � ∞; moreover,

∫∞

� ∞

1


p e�12

x � μσ

� �2

dx � 1


3GC05 05/15/2014 9:45:54 Page 88

INTEGRATION BY PARTS

Integration by parts is a theorem that relates the integral of a product of functions tothe integral of their derivative and anti-derivative. It is frequently used to transformthe anti-derivative of a product of functions into an anti-derivative for which asolution can be more easily found. The rule can be derived in one line simply byintegrating the product rule of differentiation.

Let u � u x� �, v � v x� �, and the differentials du � u´ x� �dx and dv � v´ x� �dx, thenintegration by parts states that

∫u x� �v´ x� �dx � u x� �v x� � � ∫u´ x� �v x� �dx (5.60)

or more compactly,

∫udv � uv � ∫vdu (5.61)

Example: Evaluate ∫lnxdx:

Let dv � dx, u � ln x� �, v � x, du � 1x dx. We integrate by parts as follows:

∫ln x� �dx � ln x� �x � ∫ x� � 1xdx � ln x� �x � ∫dx � xln x� � � x � C

Example: Evaluate ∫xe2xdx.

Let dv � e2xdx, u � x, v � 12 e

2x, du � dx.We integrate by parts as

∫xe2xdx � x12e2x

� � ∫

12e2xdx

This gives

∫xe2xdx � x12e2x

� � 14e2x � C

TRIGONOMETRIC FUNCTIONS

Trigonometric functions play an essential role in many scientific fields. They are basedon the concept of an angle; the latter is determined by two rays, or segments, havingthe same vertex. Angles are measured in degrees or in radians. We have the followingcorrespondence between radians and degrees:

2π radians � 360°

or equivalently,

180° � π radians

88 MATHEMATICS

3GC05 05/15/2014 9:45:56 Page 89

We present examples of correspondence between degrees and radians.

Radians 0 π=6 π=6 π=2 2π=3 π 5π=3 11π=6 2π

Degrees 0° 30° 60° 90° 120° 180° 300° 330° 360°

We define some important trigonometric functions; these are the sine, cosine,tangent, and cotangent. We use a right triangle for these definitions as illustrated inFigure 5.3a. We consider the angle θ; it has an adjacent side with length a, an oppositeside with length b, and a hypotenuse with length c. We define the following functions:

sin θ � b=c (5.62)

cos θ � a=c (5.63)

tan θ � b=a (5.64)

cotang � 1tangent

� a=b (5.65)

We have the following properties of the sine and cosine functions:

sin θj j � 1

cos θj j � 1

sin0 � 0

sinπ2

� �� 1

cos0 � 1

cos π=2� � � 1

sin2θ � cos2θ � 1

a. Geometry of a angle

0

b. Graph of sine function

c (hyp)b (opp)

a (adj)

1

–1

y = sinθ

–π/2π/2

3π/2–2π –π π 2π

θ

θ

FIGURE 5.3 Trigonometric Functions


3GC05 05/15/2014 9:45:56 Page 90

In Figure 5.3b, we show the graph of y � sin θ. The graph displays periodicmovements, with a period of 2π.

Example. If a > 0, expressffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

pin terms of a trigonometric function of θ

without radicals by making the trigonometric substitution x � a sin θ for � π2 � θ � π

2.

We let x � asinθ;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � x2

p �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � asin θ� �2

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 � a2sin2θ

p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2 1 � sin2θ� �q

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia2cos2θ

p � a cos θ

SUMMARY

This chapter covers basic functions in Islamic functions: the logarithm, exponential,and trigonometric functions. These functions are essential for carrying out computa-tions in Islamic finance. The chapter describes logarithm functions, logarithm identi-ties, change of base, the natural logarithmic function, the exponential function, thepower series of logarithmic and exponential functions, the general exponential andlogarithmic functions, some applications of logarithm and exponential functions infinance, the integration by parts, and trigonometric functions.

The logarithm, exponential, and trigonometric functions are applied in manyareas of Islamic finance. They are highly valuable in computational methods related tocapital markets, asset pricing, and risk management.

QUESTIONS

1. Express log3x into base 5.

2. Find f ´ x� � if f x� � � ln 9x � 4� �.3. Find f ´ x� � if f x� � � ln 3 � 2xj j.4. Find f ´ x� � if f x� � � ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 � 2x3

p.

5. Compute ek�lnx.6. If y � e

ffiffiffiffiffiffiffiffiffiffix2 � 1

p, find dy=dx.

7. Let f x� � � e�x2=2 and find the local extrema of f .

8. Find f ´ x� � if f x� � � lnffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 � 2x3

p.

9. Find f ´ x� � if f x� � � x2e�2x.

10. Evaluate ∫421

9 � 2x dx.

11. Use integration by parts to evaluate ∫xe�xdx.

90 MATHEMATICS

3GC06 05/15/2014 9:56:28 Page 91

CHAPTER 6Linear Algebra

I slamic finance makes extensive use of linear algebra. In fact, linear algebra is one ofthe central disciplines in mathematics. It deals with vectors, matrices, and linear

equations, and plays a major role in econometrics, linear programming, and theallocation of resources.

The chapter provides essential notions of linear algebra in the form of definitionsand theorems. These notions are widely applied in Islamic finance and includeoperations using vectors and matrices, solutions of linear equations, computationof determinants and inverses of square matrices, and computation of characteristicequations, characteristic roots, and eigenvectors. It also addresses the notion of thestability of a linear system and Cholesky decomposition of a symmetric matrix.*

VECTORS

This section covers the addition of vectors, multiplication of vectors, vector space,linear combinations of vectors, linear dependence and linear independence of vectors,and bases of a vector space.

Scalars, vectors, and matrices are components of each other. A scalar is a one-dimensional vector, or a one-dimensional matrix; a vector is an ensemble of orderedscalars; and a matrix is a collection of vectors or scalars. In linear algebra, realnumbers are called scalars. A scalar, generally speaking, is another name for a realnumber. Scalars are quantities that are fully described by a magnitude (or numericalvalue) alone. Examples of scalars are the following numbers: –1.5, 0, π, e, and

ffiffiffiffiffiffi�93p

.A vector of dimension n is an ordered collection of n elements, which are called

components. Vectors are quantities that are fully described by both a magnitude and adirection. A vector is a specific mathematical structure. It has numerous physical andgeometric applications, which result mainly from its ability to represent magnitudeand direction simultaneously. The location of a point on a Cartesian coordinate planeis usually expressed as an ordered pair x, y� �, which is a specific example of a vector(Figure 6.1a). Being a vector, point x, y� � has a certain distance (magnitude) and angle(θ) relative to the origin 0, 0� �. Vectors apply to three-dimensional geometry as well asto higher dimension Euclidian space.

*Linear algebra calculators are widely available on the Internet, for example, www.bluebit.gr/matrix-calculator/. Software such asMATLAB,Microsoft Excel, andMaple are handy as linearalgebra calculators.

91

http://www.bluebit.gr/matrix-calculator/

http://www.bluebit.gr/matrix-calculator/

3GC06 05/15/2014 9:56:29 Page 92

00

y

z

y

x

e3

e1

e2

x

V

a. Vector V

b. Unit vectors

θ

FIGURE 6.1 Vectors

We often represent a vector by some letter V, or, equivalently, ~V . We write it in arow form such as

V � v1, v2, . . . , vn� � (6.1)

or column form such as

V �v1

v2

∙ ∙ ∙vn

26664

37775 (6.2)

A vector V � v1, v2� � is said to be an element of the Euclidian space R2; a vectorV � v1, v2, v3� � is an element of the Euclidian space R3; and V � v1, v2, . . . , vn� � is anelement of Rn. A zero vector is a vector whose components are all zeros.

Example: (2,�5), (�1, 0, 2), (4.5), and (π, a, b, 2/3) are all examples of vectors ofdimension 2, 3, 1, and 4 respectively. The first vector has components 2 and �5.

The magnitude of a vector V of dimension n, denoted V, is defined as

V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv21 � v22 � ∙ ∙ ∙ � v2n

q(6.3)

Geometrically speaking, magnitude is synonymous with “length,” “distance,” or“speed.” It is also calledmodulus or norm. In the two-dimensional case, according tothe Pythagorean theorem, the point represented by the vector V � v1, v2� � has adistance from the origin (0, 0) of

V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv21 � v22

q(6.4)

92 MATHEMATICS

3GC06 05/15/2014 9:56:29 Page 93

In the three-dimension case, the point represented by the vector ~V � v1, v2, v3� �has a distance from the origin 0, 0, 0� � of

V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv21 � v22 � v23

q(6.5)

With vectors of dimension n greater than three, the algebraic definition remains

V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiv21 � v22 � ∙ ∙ ∙ � v2n

q(6.6)

Example: Let ~V � �3, � 5, 6, � 2, 7� �, ~V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�3� �2 � �5� �2 � 62 � �2� �2 � 72

q�

11.09Unit vectors are defined as vectors whose magnitude is equal to 1. A unit vector u

in Rn is a vector that has unit length

u �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiu21 � u22 � ∙ ∙ ∙ � u2n

q

Example: Let ~V � �2, � 1, 4� �, V �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�2� �2 � �1� �2 � 42

q� ffiffiffiffiffiffi

21p

; we obtain aunit vector u:

~u � �2/ ffiffiffiffiffiffi21

p, � 1/

ffiffiffiffiffiffi21

p, 4/


p� �, with u � 1

Among the unit vectors there are vectors in Rn whose jth component is 1 and allremaining component are zeros. These vectors are usually denoted as e1, e2, . . . , enwhere the subscript refers to the unit component. The unit vectors in R3 aree1 � 1, 0, 0� �, e2 � 0, 1, 0� �, and e3 � 0, 0, 1� �. A geometric representation of unitvectors in R3 is shown in Figure 6.1b. We observe that they are orthogonal to eachother.

Often vectors are represented in terms of the unit vectors. We may write ~V inRn as

~V � v1e1 � v2e2 � ∙ ∙ ∙ � vnen (6.7)

Addition of Vectors

The sum of two vectors ~V � v1, v2, . . . , vn� � and ~W � w1,w2, . . . ,wn� � is defined as

V �W � v1 �w1, v2 �w2, . . . , vn �wn� � (6.8)

Figure 6.2a shows the graph of vector addition. Figure 6.2b shows the graph ofvector subtraction. The addition of vectors is only defined if both vectors have thesame dimension.

Linear Algebra 93

3GC06 05/15/2014 9:56:30 Page 94

Example: Let V � 2, � 3� � and W � 0, 1� �, then V �W � 2, � 2� �.Let V � 0.1, 2� � and W � �1,π� �, then V �W � 1.1, 2 � π� �.The scalar product of a scalar α by a vector ~V � v1, v2, . . . , vn� � is defined as

α~V � αv1,αv2, . . . ,αvn� � (6.9)

Example: Let V � 5, � 4� � and α � 2, then αV � 2 5, � 4� � � 2 � 5, 2 � �4� � �10, � 8� �

In general, 0 � V � 0, 0, . . . , 0� � and 1 � V � V, just as in the algebra ofscalars. The vector of any dimension n with all zero elements (0, 0, . . . , 0) is calledthe zero vector and is denoted 0.

Two vectors are equal if their corresponding components are equal.Example: If V � �2, 1� � and W � �2, 1� �, then V � W since �2 = �2 and 1 = 1.

However, V � 5, 3� � is not equal to W � 3, 5� � because even though they have thesame components, 3 and 5, the components do not occur in the same order.

Multiplication of Vectors

Multiplication of vectors plays an important role in economics and finance. Forinstance, let x be the quantity of oranges and y the quantity of tomatoes purchased; letpx be the price of oranges and py the price of tomatoes. We denote the quantity vectorby q � x, y� � and the price vector by p � (px, py). The expenditure is the product of twovectors p and q:

pq � pxx � pyy

The dot product of two vectors V and W (sometimes called the inner product,or, because its result is a scalar, the scalar product) is denoted by V.W and is definedas

V � W � VWcos θ (6.10)

where V means the magnitude (length) of vector V, W means the magnitude ofvector W, θ is the measure of the angle between V and W, and cos θ is the cosine of θ

a. Vector V + W b. Vector V – W

W

V

V + W

W

V

V – W

FIGURE 6.2 Vectors: Sum and Subtraction

94 MATHEMATICS

3GC06 05/15/2014 9:56:31 Page 95

0

0 0c. Orthogonal unit vectors d. Orthogonal vectors

V V

V

W

1

1

W

W

w2

w1

v2

v1

b. Vector productV W = v1w1 + v2w2

θ θ

a. Vector productV VW = W

cosθ

cosθ

W

FIGURE 6.3 Vector Multiplication

(Figure 6.3a). Geometrically, this means that V and W are drawn with a commonstarting point and then the length ofV is multiplied with the length of that componentofW that points in the same direction as V. We may also compute the dot product asthe sum of products of their respective coordinates (Figure 6.3b):

V � W � v1w1 � v2w2 � ∙ ∙ ∙ � vnwn (6.11)

Both methods yield the same answer.Example: Compute the dot product of V � �6, 8� � and W � 5, 12� �.

V � W � v1w1 � v2w2 � �6 � 5 � 8 � 12 � 66.

Two vectors are orthogonal to each other when they form a right angle. In thiscase, their dot product is zero because the cosine of a right angle is zero. In Figure 6.3c,we show the orthogonal unit vectors e1 � 1, 0� � and e2 � 0, 1� �. The producte1e2 � 1 � 0 � 0 � 1 � 0.

In Figure 6.3d, we show two orthogonal vectors V and W. Their dot productV � W � v1w1 � v2w2 � 0.

Example: Compute the dot product of V � �12; 16� � and W � 12; 9� �.

V � W � v1w1 � v2w2 � �12 � 12 � 16 � 9 � 0

Here V and W are orthogonal to each other.An inner product (or scalar product, dot product) of two vectors is a row vector

y 1, n� � times a column x n, 1� � vector, yielding a scalar:

yx � Xnj�1

yjxj (6.12)

Linear Algebra 95

3GC06 05/15/2014 9:56:32 Page 96

An outer product of two vectors is a column vector x n, 1� � times a row vectory 1, n� �, yielding a matrix:

xy �x1y1 . . . x1yn. . . . . . . . .

xny1 . . . xnyn

0@

1A (6.13)

Vector Space

A vector space over R (the set of real numbers) is a nonempty set V such that:

■ There is defined an operation of addition; that is, for all x, y ∈ V we havex � y ∈ V.

■ There is defined an operation of scalar multiplication; that is, for all α ∈ R , x ∈ Vwe have αx ∈ V.

■ There exists a vector 0 ∈ V such that for all x ∈ V we have x � 0 � x � 0 � x.

Example: The set of two-dimensional vectors R2 is a vector space.

Linear Combinations of Vectors

Definition: Let x1, x2, . . . , xn� �

be a family of vectors in a vector space V. We call x inV a linear combination of x1, x2, . . . , xn

� �if and only if there exist scalars

α1, α2, . . . , αn in R for which

x � α1x1 � α2x2 � ∙ ∙ ∙ � αnxn � Xni�1

αixi (6.14)

Example: Let x1 �20�1

0@

1A, x2 �

�427

0@

1A, α1 � 10, and α2 � �3.5.

Then, we have the following linear combination:

x � α1x1 � α2x2 � 1020�1

0@

1A � 3.5

�427

0@

1A �

34�7

�34.5

0@

1A

Example: Let x1 � 31

� �, x2 � �1

5

� �, x � �1

37

� �; then there exists α1 � 2, and

α2 � 7 such that

x � 2x1 � 7x2

96 MATHEMATICS

3GC06 05/15/2014 9:56:33 Page 97

Linear Dependence and Linear Independence of Vectors

Let V be a vector space and let x1, x2, . . . , xn� �

be a family of vectors in a vector spaceV. The family x1, x2, . . . , xn

� �is called linearly dependent if and only if there are

scalars αi, not all zero, such that

α1x1 � α2x2 � ∙ ∙ ∙ � αnxn � 0 (6.15)

If x1, x2, . . . , xn� �

is not linearly dependent, we call x1, x2, . . . , xn� �

linearlyindependent.

The family x1, x2, . . . , xn� �

is linearly independent if and only if

α1x1 � α2x2 � ∙ ∙ ∙ � αnxn � 0 (6.16)

implies that α1 � α2 � . . . � αn � 0

Example: x1 � 10

� �, x2 � 0

1

� �, and x3 � 1

1

� �

α1 � 1, α2 � 1, and α3 � �1, we have

α1x1 � α2x2 � α3x3 � 10

� �� 0

1

� �� 1

1

� �� 0

0

� �

Hence, x1, x2, and x3 are linearly dependent. In contrast, x1, x2� �

are linearlyindependent:

α1x1 � α2x2 � α110

� �� α2

01

� �� α1

α2

� �� 0

0

� �

We have α1 � 0 and α2 � 0.

Bases of a Vector Space

Definition: Let V be a vector space and let x1, x2, . . . , xn� �

be a family of vectors. Wecall x1, x2, . . . , xn

� �a basis for V if and only if:

i. x1, x2, . . . , xn� �

is linearly independent; and

ii. x1, x2, . . . , xn� �

spans V.

An easy but illuminating consequence of this definition is: Let V be a vector spaceand let x1, x2, . . . , xn

� �be a family of vectors in V. Then x1, x2, . . . , xn

� �is a basis if

and only if for each y ∈ V there are unique scalars α1, α2, . . . , αn such that

y � α1x1 � α2x2 � ∙ ∙ ∙ � αnxn � Xni�1

αixi (6.17)

Linear Algebra 97

3GC06 05/15/2014 9:56:33 Page 98

Example: Let e1 �100

0@

1A, e2 �

010

0@

1A, and e3 �

001

0@

1A; then it is easy to check

that e1, e2, e3� �

is basis for R3, three-dimensional vector space.

Example: Let x1 �121

0@

1A, x2 �

132

0@

1A, and x3 �

2�30

0@

1A

Then it is easy to check that α1x1 � α2x2 � α3x3 � 0 has a trivial solution α1 � α2 �α3 � 0:

α1121

0@

1A � α2

132

0@

1A � α3

2�30

0@

1A � 0

Therefore, x1, x2, and x3 form another basis of R3.

MATRICES

This section covers the definition of a matrix, the transpose of a matrix, and matrixmultiplication.

Matrices are important in finance and economics. They portray relationshipsbetween variables and technical structures in the economy. AmatrixA is a rectangulararray of numbers. It is extremely useful in describing the interaction of two variables,such as in joint probabilities, or the structure of the economy. The numbers in thematrix A are called the entries of A. The general form of a matrix is

A �a11 a12 . . . a1n

a21 a22 . . . a2n

. . . . . . . . . . . . . . . .am1 am2 . . . amn

26664

37775 (6.18)

Each aij is a real or complex number. The horizontal array

ai* � ai1 ai2 . . . ain� � (6.19)

is called the ith row of A. Similarly, the vertical array

a*j�

a1ja2j. . .

amj

26664

37775 (6.20)

is called the jth column of A.

98 MATHEMATICS

3GC06 05/15/2014 9:56:34 Page 99

If the matrix hasm rows and n columns it is called anm � nmatrix. In particular,ifm � n the matrix is called a squarematrix. The combination of scalars, vectors, andmatrices provides a linear system of the form

Ax � b (6.21)

where A is anm � nmatrix; x is n � 1 vector, and b ism � 1 vector, which may bezero-vector.

Example: Find the matrices A, b, x� � corresponding to the following system ofequations:

7x1 � 3x2 � x3 � 23

x1 � x2 � 5

19x2 � x3 � 80

7 3 �11 1 00 19 �1

24

35 x1

x2x3

24

35 �

23580

24

35

The solution to this system is a vector (x1 � 1, x2 � 4, x3 � �4).The sum of any two m � n matrices A and B is defined as

A � B � C (6.22)

The matrix C is again an m � n matrix. There exists an m � n matrix, each ofwhose entries is zero, with the property that A � 0 � A. Given any matrix A thereexists a unique X such that A �X � 0.

Transposes of Matrices

Definition: For anym � nmatrix A, the matrix B � A´ is the n � mmatrix given by:bij � aji i � 1, . . . , n and j � 1, . . . ,m.The matrix A´ is called the transpose of A.

Example: Find the transpose of A � 1 2 �10 3 7

.

We find A´ �1 02 3�1 7

24

35.

Definition of scalar multiplication: LetA � aij� �

be anm � n matrix with entries,and let α be a real or complex number. Then B � αA � Aα is defined by

bij � αaij, i � 1, . . . ,m, j � 1, . . . , n (6.23)

Linear Algebra 99

3GC06 05/15/2014 9:56:35 Page 100

Matrix Multiplication

Let B be an m � n matrix and let A be an n � p matrix, so that B has as manycolumns as A has rows. Then the product BA � C is defined as an m � p matrix.

B �b11 . . . b1n

. . . . . . . . . . .

. . . . . . . . . . . . .

bm1 . . . bmn

266664

377775A �

a11 . . . a1p

. . . . . . . . . . .

. . . . . . . . . . . . .an1 . . . anp

266664

377775

BA � C �c11 . . . c1p

. . . . . . . .

. . . . . . . . . .cm1 . . . cmp

266664

377775

(6.24)

We have cij � bi1a1j � bi2a2j � . . . binanj.

Example: Let B �1 1 21 2 31 4 9

24

35A �

0 11 �12 0

24

35, then C � BA �

5 08 �122 �3

24

35.

For all matrices A, B, and C, and any scalar α:

i. A BC� � � AB� �C;ii. A B � C� � � AB � AC;iii. A � B� �C � AC � BC;iv. α AB� � � αA� �B.

SQUARE MATRICES

This section covers symmetric matrices, positive definite matrices, quadratic forms,and orthogonal matrices.

Let A be an n � n matrix. The trace of A, tr A� �, is defined to be the sum of thediagonal entries, that is,

tr A� � � a11 � a22 � ∙ ∙ ∙ � ann (6.25)

Definition: A square matrix whose only nonzero entries occur in positions aii,i � 1, 2, . . . , n is called a diagonal matrix (that is, akl � 0 if k ≠ l). In general,a11, a22, . . . , ann is called the main diagonal.

Example: The matrix1 0 00 2 00 0 �6

24

35 is an example of a diagonal matrix.

100 MATHEMATICS

3GC06 05/15/2014 9:56:35 Page 101

Definition: Square matrices whose only nonzero entries occur on or abovethe main diagonal are called upper triangular matrices (those for whichaij � 0 if i > j).

Example: The matrix�1 �1 20 2 �10 0 3

24

35 is an upper triangular matrix.

A strictly upper triangular matrix is a triangular matrix whose diagonal entriesare zero (that is aij � 0 if i � j).

Example: The matrix0 4 �90 0 �10 0 0

24

35 is a strictly upper triangular matrix.

Definition: Let In be an n � n matrix; it is called the identity matrix if all itsdiagonal entries are equal to 1 and all off diagonal entries are zero.

Example: I3 �1 0 00 1 00 0 1

24

35 I4 �

1 0 0 0

0 1 0 0

0 0 1 00 0 0 1

26664

37775

In has the property that for any n � n matrix A,

InA � A � AIn (6.26)

Symmetric Matrix

A symmetric matrix is a square matrix that is equal to its transpose. Formally, matrixA is symmetric if

A � A´ (6.27)

Because the definition of matrix equality demands equality of their dimensions,only square matrices can be symmetric.

The entries of a symmetric matrix are symmetric with respect to the maindiagonal. So if the entries are written as A � aij

� �, then aij � aji for all indices i

and j.Symmetry plays an important role in finance. Consider a portfolio withm stocks;

the matrix of variance and covariance of returns, denoted byP

, is a symmetric matrixof the form

X �VarR1 Cov(R1,R2) . . . Cov(R1,Rm)

Cov(R2,R1) VarR2 . . . Cov(R2,Rm). . . . . . . . . . . .

Cov(Rm,R1) Cov(Rm,R2) . . . VarRm

26664

37775

Linear Algebra 101

3GC06 05/15/2014 9:56:36 Page 102

Example: The following 3 × 3 matrix is symmetric: A �1 7 37 4 �53 �5 6

24

35

Every diagonal matrix is symmetric, because all off-diagonal entries are zero.Similarly, each diagonal element of a skew-symmetric matrix must be zero, since eachis its own negative.

Positive Definite Matrix

An n � n real symmetric matrix is said to be positive semi-definite if for any realn � 1� � vector x,

xÁx � 0 (6.28)

We may make a stronger statement that a real symmetric matrix A is positivedefinite if for any real n � 1� � vector x,

xÁx > 0 (6.29)

Quadratic Forms

Given a square symmetric matrix A and a column vector x, the quadratic form of A is

QA x� � � xÁx (6.30)

Example: A � 1 33 4

then QA x� � � x21 � 4x22 � 6x1x2.

The quadratic form is positive definite if QA x� � > 0 for all x ≠ 0, is negativedefinite if QA x� � < 0 for all x ≠ 0, is positive semi-definite if QA x� � � 0 for all x andQA x� � � 0 for some x ≠ 0, and is negative semi-definite if QA x� � � 0 for all x andQA x� � � 0 for some x ≠ 0.

Example: A portfolio contains three Islamic shares A,B, andC in proportions ofx1, x2, and x3. The variance-covariance matrix is

81 75.6 19.875.6 196 61.619.8 61.6 121

24

35

The variance of the portfolio is a quadratic form:

V x1, x2, x3� � � x1 x2 x3� � 81 75.6 19.8

75.6 196 61.619.8 61.6 121

24

35 x1

x2x3

24

35

V x1, x2, x3� � � 81x21 � 196x22 � 121x23 � 151.2x1x2 � 39.6x1x3 � 123.2x2x3

102 MATHEMATICS

3GC06 05/15/2014 9:56:36 Page 103

Orthogonal Matrix

The square matrix A is orthogonal if and only if each column (row) vector of A isnormalized and orthogonal to any other column (row) vector so that

AA´ � A´A � In (6.31)

Example: A �3ffiffiffiffiffiffi10

p 2ffiffiffiffiffiffi40


p �0ffiffiffiffiffiffi40

p

0BBB@

1CCCA,A´ �





p

0BBB@

1CCCA,

AA´ �3ffiffiffiffiffiffi10




p

0BBB@

1CCCA





p

0BBB@

1CCCA � 1 0

0 1

� �

.A�1 � A´ if and only if A is orthogonal.A matrix A is idempotent if A2 � A.If A is idempotent, then Ah � A for all h � 1.

THE RANK OF A MATRIX

The column rank of a matrix A is the maximum number of linearly independentcolumn vectors of A. The row rank of A is the maximum number of linearlyindependent row vectors of A. Equivalently, the column rank of A is the dimensionof the column space ofA, while the row rank of A is the dimension of the row space ofA. A result of fundamental importance in linear algebra is that the column rank andthe row rank are always equal. This number (i.e., the number of linearly independentrows or columns) is simply called the rank of A.

Definition:The leading entry of a row ai* ≠ 0 is the first nonzero entry in that row.Definition: A matrix is in row echelon form (ref) when it satisfies the following

conditions.

■ The first nonzero element in each row, called the leading entry, is 1.■ Each leading entry is in a column to the right of the leading entry in the previous

row.■ Rows with all zero elements, if any, are below rows having a nonzero element.

A matrix is in reduced row echelon form (rref) when it satisfies the followingconditions.

■ Thematrix is in row echelon form (i.e., it satisfies the three conditions listed above).■ The leading entry in each row is the only nonzero entry in its column.

Linear Algebra 103

3GC06 05/15/2014 9:56:37 Page 104

A matrix in echelon form is called an echelon matrix. Matrix A and matrix B areexamples of echelon matrices.

Example: A �1 2 3 4

0 0 1 3

0 0 0 10 0 0 0

26664

37775 B �

1 2 0 0

0 0 1 0

0 0 0 10 0 0 0

26664

37775

Matrix A is in row echelon form, and matrix B is in reduced row echelon form.The row-echelon form of a matrix is a very handy tool; it can be used to

geometrically interpret different vectors and find out properties such as lineardependence and span.

The matrix A below describes a row-echelon form. The row-echelon form iswhere the leading (first nonzero) entry of each row has only zeroes below it.

A �1 1 20 1 10 0 �2

24

35

We describe the steps for obtaining the row-echelon form of a matrix. Startingwith a matrix of any size such as matrix B,

B �1 1 21 2 33 4 5

24

35

we do not tamper with the top row. We look at the first entry in each row and decidewhat factor of Row 1 needs to be added or subtracted from each row to get zero as thefirst term. For matrix B, we can see that Row 2 � Row 1 would give us a zero, andRow 3 � 3 � Row 1 would also

B1 �1 1 20 1 10 1 �1

24

35

So, after computing the row operations in Step 2, the matrix now looks like B1.We can then see what row operation would give us another zero in the bottom row; itis Row 3 � Row 2. Now our final matrix looks like B2, which is in row-echelon formas the first entry in each row, and has only zeroes below it.

B2 �1 1 20 1 10 0 �2

24

35

104 MATHEMATICS

3GC06 05/15/2014 9:56:37 Page 105

We may remark that a square matrix in row echelon form is obviously in uppertriangular form.*

Definition: Let A be an m � n matrix and let AR be the row echelon form of A.The rank of A is the number of nonzero rows of AR.

We note that

rank A� � � rank A´� � � rank A´A� � � rank AA´� � (6.32)

DETERMINANT OF A SQUARE MATRIX

A determinant is a scalar associated with a square matrix. The notion of determinantapplies only to square matrix. Consider two-dimensional vectors V � a, b� � andW � c, d� �. In matrix form we have

A � a bc d

We form a parallelogramwith sidesV andW. In two-dimensional space there is asimple formula for the area of a parallelogram bounded by vectors V and W withV � a, b� � and W � c, d� �, namely: ad � bcj j

We consider a 2 � 2 matrix A:

A � a11 a12a21 a22

Associated with the square matrix A is a scalar called the determinant of thematrix that is denoted by Aj j.

Aj j � a11 a12a21 a22

� a11a22 � a12a21 (6.33)

The determinant conveys the notion of an area of the parallelogram formed by thevectors V � a11, a12� � and W � a21, a22� �.

Example: Compute the determinant of A � 5 36 4

.

Applying the determinant formula we find Aj j � 5 � 4� � � 3 � 6� � �20 � 18 � 2.

*The reduction of a matrix A m � n� � into an echelon form may be performed by going to theLinear Algebra Toolkit website at www.math.odu.edu/∼bogacki/cgi-bin/lat.cgi and clicking onthe link called “Transforming a matrix to reduced row echelon form.”

Linear Algebra 105

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi

3GC06 05/15/2014 9:56:37 Page 106

We extend the notion of area in two-dimensional space to that of a volume inthree-dimensional space. Consider the matrix

A �a11 a12 a13a21 a22 a23a31 a32 a33

24

35

The vectors V � a11 a12 a13� �

, W � a21 a22 a23� �

, and Z� a31 a32 a33� �

form a parallelepiped. The volume associated with this parallelepiped is the determi-nant of the third order of the matrix A. It can be obtained by choosing a given row (orcolumn) and multiplying each element by the second order determinant remainingafter deletion of the row and column intersecting in that element, and finally summingthe three products.

a11a22 a23a32 a33

� a12

a21 a23a31 a33

� a13

a21 a22a31 a32

(6.34)

Note that the sign of the chosen elements changes alternately, beginning with thepositive sign.

Example: Consider the following matrix A:

A �1 6 50 3 14 8 9

24

35

Its determinant is computed as

1 6 50 3 14 8 9

� 1

3 18 9

� 6

0 14 9

� 5

0 34 8

� 27 � 8� � � 6 0 � 4� � � 5 0 � 12� � � �17

Example: Using the Microsoft Excel function MDETERM, compute the deter-minant of

A �

�1 0 1 �8 70 1 3 �4 1211 8 0 �6 55 2 0 17 15�5 �7 2 8 10

2666664

3777775

We find Aj j � 17; 909.The determinant of In � 1. If A and B are both n � nmatrices, then ABj j � Aj jBj j.

106 MATHEMATICS

3GC06 05/15/2014 9:56:38 Page 107

HOMOGENOUS SYSTEMS OF EQUATIONS

A system of linear equations is homogeneous if all of the constant terms are zero.

a11x1 � a12x2 � ∙ ∙ ∙ � a1nxn � 0

a21x1 � a22x2 � ∙ ∙ ∙ � a2nxn � 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

am1x1 � am2x2 � ∙ ∙ ∙ � amnxn � 0

A homogeneous system is equivalent to a matrix equation of the form

Ax � 0 (6.35)

where A is an m � n matrix, x is a column vector with n entries, and 0 is the zerovector with m entries.

Every homogeneous system has at least one solution, known as the zero solution(or trivial solution), which is obtained by assigning the value of zero to each of thevariables. If the system has a nonsingular matrix det A� � ≠ 0, then the trivial solution isalso the only solution. If the system has a singular matrix then there is a solution setwith an infinite number of solutions. This solution set has the following additionalproperties:

■ If u and v are two vectors representing solutions to a homogeneous system, thenthe vector sum u � v is also a solution to the system.

■ If u is a vector representing a solution to a homogeneous system, and α is anyscalar, then αu is also a solution to the system.

These are exactly the properties required for the solution set to be a linearsubspace of Rn. In particular, the solution set to a homogeneous system is the same asthe null space of the corresponding matrixAwhere the null space is the set of x vectorssuch that Ax � 0.

Example: Consider the following equation: 2x1 � 5x2 � 0.It is a homogeneous equation. It has an infinity of solutions of the form

x1 � 2.5x2. For instance, x1 � 0 and x2 � 0 is a solution; x1 � �5 and x2 � �1 isanother solution.

Example: Reduce the matrix of the following homogeneous system to rowechelon:

3x1 � 2x2 � 3x3 � 0

2x1 � x2 � x3 � 0

x1 � x2 � x3 � 0

We have A �3 2 �32 �1 11 1 1

24

35.

Linear Algebra 107

3GC06 05/15/2014 9:56:38 Page 108

Transforming a matrix to reduced row echelon form we find

Ar �1 0 00 1 00 0 1

24

35

The homogeneous admits a unique trivial solution: x1 � x2 � x3 � 0.

INVERSE AND GENERALIZED INVERSE MATRICES

Definition: LetA be an n � nmatrix. The n � nmatrixX is said to be an inverse ofAif and only if

XA � In andAX � In (6.36)

Definition: The square matrixA is called nonsingular if and only if A possesses aninverse denoted by A�1. If A has no inverse, A is called singular.

If the matrices A and B are nonsingular, then the product AB is nonsingular and

AB� ��1 � B�1A�1 (6.37)

Let A be an n � n matrix. Then the following are equivalent:

■ A is nonsingular.■ Rank A � n.■ The echelon row form AR � In.

Example: Using the Microsoft Excel MINVERSE function, find the inverse of

A �2 1 43 1 52 3 5

24

35

We find A�1 ��3.33 2.33 0.33�1.66 0.66 0.662.33 �1.33 �0.33

24

35.

Let V be an n-dimensional vector space, and letB1 � x1, . . . , xn� �

be a basis for V.LetB2 � y1, . . . , yn

� �be a family of vectors in V. ThenB2 is a basis for V if and only if

there is an n � n nonsingular matrix such that B2 � PB1.

Example: Consider B1 � x1 � 13

, x2 � �3

2

� �; let P � 1 �2

8 1

; then

B2 � y1 � 711

, y2 � 1

�22 � �

is a basis for R2.

108 MATHEMATICS

3GC06 05/15/2014 9:56:39 Page 109

Generalized Inverse of a Matrix

The generalized inverse of a matrix plays an important role in econometrics. Given thesquare or nonsquare matrix A of order m � n the generalized inverse matrix A� oforder n � m is the matrix satisfying the properties

AA�A � A

A�AA� � A�

Symmetry AA� � AA�� ´Symmetry A�A � A�A

� �´

If A is square and nonsingular, then A� is unique and is given by A�1, so A� is ageneralization of the concept of an inverse matrix. If A is m � n, where m > n andrank A � n, then

A� � A´A� ��1A´ (6.38)

Example: Let A �1 2�1 12 2

24

35. Applying Microsoft Excel matrix functions, we

find

A� � �0.03448 �0.48276 0.275860.241379 0.37931 0.068966

EIGENVALUES AND EIGENVECTORS

This section covers eigenvalues and eigenvectors of a square matrix, similarity ofsquare matrices, diagonable square matrices, and the Cholesky decomposition.

Let A be an n � n matrix. An eigenvector for a square matrix, A, is a nonzerovector x, which, when transformed by A, yields the same vector except for the scalefactor

Ax � λx (6.39)

Where the scale factor, λ, is an eigenvalue, or a characteristic root of A, theprevious equation can be written as

A � λI� �x � 0 (6.40)

which is a homogeneous system of equations; a necessary condition for the nontrivialsolution is that the coefficient matrix be of less than full-rank, so that

A � λIj j � 0 (6.41)

Linear Algebra 109

3GC06 05/15/2014 9:56:40 Page 110

The resulting equation for λ is the characteristic equation. IfA is an n � nmatrix,the characteristic equation is an nth order polynomial equation in λ.

A � λIj j � �λ� �n � α1 �λ� �n�1 � ∙ ∙ ∙ � αn�1 �λ� � � αn � 0 (6.42)

Example: Let A be an n � n matrix:

A � λIj j � a11 � λ a12a21 a22 � λ

� λ2 � a11 � a22� �λ � a11a22 � a12a21� � � 0

The solution to the characteristic equation consists of n roots, λ1, λ2, . . . , λn,which are not necessarily all distinct or real. To each of these characteristic roots therecorresponds a characteristic vector that is determined up to a constant.

Example: A � 6 10�2 �3

� �

The characteristic equation is λ2 � 3λ � 2 � 0 yielding λ1 � 1, λ2 � 2.To find the eigenvector belonging to λ1 � 1, we must solve the equation

A � I� �x � 5 10�2 �4

x1x2

� 0

0

We find that apart from scalar multiples1�2

is the unique eigenvector

belonging to 1.In case of λ2 � 2 we must solve the equation

A � 2I� �x � 4 10�2 �5

x1x2

� 0

0

We find that apart from scalar multiples1

�2.5

is the unique eigenvectorbelonging to 2.

Properties of the eigenvalues λ1, λ2, . . . , λn of any square matrix A include

i. λ1 � λ2 � ∙ ∙ ∙ � λn � trace A� � � α1ii. λ1 � λ2 � . . . � λn � Aj j � αn

where α1 and αn are coefficients of the characteristic equation.

Similarity of Square Matrices

Let A and B be n � n matrices. Then A and B are similar if and only if there is anonsingular matrix such that

B � P�1AP (6.43)

110 MATHEMATICS

3GC06 05/15/2014 9:56:40 Page 111

Example: Let A � �3 �64 7

.

If P � 1 12 3

, then by direct computation P�1AP � 1 0

0 3

� B.

Let A and B be similar matrices. Then

detA � detB (6.44)

Let A and B be similar matrices; then the family of eigenvalues of A and B are thesame. Further, ifB � P�1AP, then y is an eigenvector ofB belonging to λ if and only Pyis an eigenvector of A belonging to λ.

Example: Let A � 11 30�4 �11

.

An eigenvalue of A is λ � �1. An eigenvector of A belonging to �1 is

�5/21

� x

If P � 3 �5�1 2

,

then B � 1 00 �1

� 2 5

1 3

11 30�4 �11

3 �5�1 2

� P�1AP.

Clearly λ � �1 is an eigenvalue for B, and a computation shows that

P�1x � 01/2

is an eigenvector of B belonging to �1.

Diagonable Matrix

Let A be an n � n matrix. The matrix A is called diagonable if and only if there is adiagonal matrix Λ that is similar to A. We shall denote the diagonal matrix bydiag λ1, . . . , λn� �:

Λ �λ1 . . . . . . 0. . . . . . . . . . . .

. . . . . . . . . . . .

0 . . . . . . λn

26664

37775 (6.45)

Let A be an n � n matrix. If P�1AP � Λ, where Λ � diag λ1, . . . , λn� �, thenλ1, . . . , λn� � is the family of eigenvalues of A and p

*jis an eigenvector belonging to

λj for j � 1, . . . , n.

Linear Algebra 111

3GC06 05/15/2014 9:56:41 Page 112

Cholesky Decomposition

If the symmetric matrixP

is a positive-definite matrix, we can decompose it in termsof a lower triangular matrix, L, and its transpose, L´, which is an upper triangularmatrix:

X � LL´ (6.46)

Consider a lower triangular matrix L �3 0 0�2 �1 02 5 3

24

35; we obtain

P � LL´ �9 �6 6�6 5 �96 �9 38

24

35.

If we let L � a 0b c

, then

P � LL´ � a2 abab c2

.

Example: LetP � LL´ � a2 ab

ab c2

� 49 21

21 1

, then we have a2 � 49 and

a � 7; ab � 21 yielding b � 3; c2 � 1; and c � 1. We find L � 7 03 1

.

For L �a 0 0b d 0c e f

24

35, we have

P � LL´ �a2 ab acab b2 � d2 bc � deac bc � de c2 � e2 � f 2

264

375.

Example: Let

X �16 8 128 29 2612 26 61

24

35

Then by matching the terms, we find a � 4, b � 2, c � 3, d � 5, e � 4, and f � 6.This gives

L �4 0 02 5 03 4 6

24

35

STABILITY OF A LINEAR SYSTEM

Let A be an n � n matrix. Consider the sum

ST � In � A � A2 � ∙ ∙ ∙ � AT (6.47)

112 MATHEMATICS

3GC06 05/15/2014 9:56:42 Page 113

Premultiplying both sides by A we see that

AST � A � A2 � ∙ ∙ ∙ � AT�1 (6.48)

Subtracting the sum AST from ST , we find

In � A� �ST � In � AT�1 (6.49)

We recall that if A � λInj j � 0, then λ is an eigenvalue of A. We deduce that ifA � Inj j � 0, then λ � 1 is an eigenvalue ofA. Assuming that none of the eigenvalues ofA is equal to unity, the matrix In � A� � is nonsingular and we obtain

ST � In � A� ��1 In � AT�1� �(6.50)

If all the eigenvalues ofA are strictly less than one inmodulus, it can be shown thatAT ! 0 as T ! ∞ , implying that

ST � In � A � A2 � ∙ ∙ ∙ � AT � In � A� ��1 (6.51)

Definition: A linear system A n, n� � is stable if all the eigenvalues of A are strictlyless than one in modulus.

Example: Let A ��1 3 52 3 85 1 �3

24

35.

A is stable since ST �0.118 �0.125 �0.103�0.353 0.125 �0.1910.059 �0.125 0.074

24

35.

APPLICATIONS IN ECONOMETRICS

When there are more equations than unknowns, the system of linear equations isrewritten as

y � Xβ (6.52)

n � 1� � n � k� � k � 1� �

where X is an n � k coefficients matrix, β is a k � 1 vector of unknowns, and y is ann � 1 vector of constants. It is assumed that n, the number of equations, exceeds k,the number of unknowns. One approach to “solving” this system is that of the leastsquares fit. This approach minimizes the sum of squared deviations between theelements of y and the elements of Xβ, given as

S � y �Xβ2 � y �Xβ� �´ y �Xβ� � (6.53)

Linear Algebra 113

3GC06 05/15/2014 9:56:42 Page 114

The solution for β obtained from the problem of minimizing S is

β � X�y (6.54)

where X� is the generalized inverse. If X matrix has maximal rank, then

X� � X´X� ��1X´ when rankX � k (6.55)

In this case

β � X´X� ��1X´y (6.56)

This is called the least squares estimators of β.

SUMMARY

Linear algebra is used extensively in Islamic finance. This chapter covers the basicnotions of linear algebra and describes the notions of vectors, including operationsand linear combinations of vectors, the concept of vector space, and the bases of avector space. It introduces the notions of matrices, transposes of matrices, matrixmultiplication, square matrices, symmetric matrix, positive-definite matrix, quadraticforms, orthogonal matrix, the rank of a matrix, and determinant of a square matrix.The chapter also covers the topics of homogenous systems of equations, inverse andgeneralized inverse matrices, eigenvalues and eigenvectors, similarity of squarematrices, diagonable matrix, and Cholesky decomposition of a positive-definitematrix. It addresses the stability of a linear system and applications in econometrics.

Linear algebra can be applied in many areas, including asset pricing, deriving risk-neutral probability distributions, estimating linear models, optimizing portfoliosusing linear programming techniques, hedging risk, and structuring products.

QUESTIONS

1. Show graphically the vectors corresponding to the following: A x � �1, y � 3� �and B x � 3, y � �3� �.

2. Consider the following linear equations:

3x1 � x2 � x3 � 7

2x1 � 5x2 � 2x3 � �9�x1 � 3x2 � 7x3 � �25

Show the matrix form of these equations; use the Microsoft Excel solver tofind the solution vector x1, x2, x3� �.

3. Consider the vector V � x1 � 1, x2 � �1, x3 � 3, x4 � �7, x5 � 10� �. Find thelength Vj j.

114 MATHEMATICS

3GC06 05/15/2014 9:56:43 Page 115

4. Compute the dot product of V � 3, 6� � and W � �1, 0.5� �. Show their graph.What do you conclude?

5. Let x1 � 31

� �, x2 � �1

5

� �, and x � 5

7

� �; find α1 and α2 for which

x � α1x1 � α2x2.6. Which of the following families spans R3?

e1 �100

24

35, e2 �

010

24

35, e3 �

001

24

35

0@

1A

x1 ��110

24

35, x2 �

�101

24

35, x3 �

111

24

35

0@

1A

7. Let x1 �2�10

24

35, x2 �

121

24

35, x3 �

02�1

24

35

0@

1A

and e1 �100

24

35, e2 �

010

24

35, e3 �

001

24

35

0@

1A.

Show that x1, x2, x3� �

spansR3. Express each of the unit vectors e1, e2, and e3

as a linear combination of x1, x2, x3� �

.

8. Let B �3 �5 �9�1 14 172 1 0

24

35 and A �

2 00 4�1 �5

24

35; compute C � BA.

9. Reduce the following matrices to row echelon form

A �4 �8 161 �3 62 1 1

24

35B �

0 2 1 40 0 2 61 0 �3 2

24

35

10. Reduce the matrix of the following homogeneous system to row echelon:

4x � 8y � 12z � 0

2x � 2y � 2z � 0

5x � 5y � 5z � 0

Find the solutions for this system.

Linear Algebra 115

3GC06 05/15/2014 9:56:43 Page 116

11. Using the Microsoft Excel matrix inverse function, compute the inverse of thefollowing matrices:

A �1 1 1�1 2 10 0 3

24

35 B �

2 0 5 00 7 0 23 0 7 00 0 0 6

26664

37775

12. Let A �1 2�1 10 2

24

35. Applying the Microsoft Excel matrix function, compute the

generalized inverse A�.

13. Let � 3 21 �5

. Find the characteristic equation and the eigenvalues and

eigenvectors of A.

14. Find the Cholesky decomposition ofP �

16 20 1220 29 2712 27 61

24

35.

15. A portfolio contains three Islamic shares A, B, andC in proportions ofx1, x2, and x3. The variance-covariance matrix of returns is

121 61.6 19.861.6 196 75.619.8 75.6 81

24

35

Compute the variance of the portfolio’s return.

116 MATHEMATICS

3GC07 05/15/2014 10:18:43 Page 117

CHAPTER 7Differential Equations

D ifferential equations play an important role in Islamic finance. The rate of changeof one variable with respect to another is called a derivative. A differential equation

describes the relationship between a variable and its derivative. In economics andfinance, many problems give rise to differential equations. Economic and financialvariables are indexed on time t. For instance, the real gross domestic product GDP� � attime t is denoted GDPt and real GDP at time t � dt is denoted GDPt�dt; the rate ofchange of real GDP per unit of time is dGDPt

dt . A differential equation may be used to

describe the dynamics of real GDP by relating dGDPtdt toGDPt. A company may change

the price of its product based on its current price, that is, dpdt � f p� �. If the price p is high

and sales are slow, the company may lower its price, that is, dpdt < 0. Inversely, if the

price p is low and sales are high, the company may raise its price, that is, dpdt > 0.

In economics and finance, we may have a system of differential equations. In thiscase, companyA changes the price of its product not only in relation to its own currentprice, but in relation to its competitor B’s price. Likewise, Company B changes itsprice in relation to its own price and A’s price. We may imagine a situation wherecompany A decides to change its investment in research and development (R&D) inrelation to its current R&D investment and company B’s R&D. If company Bincreases its R&D investment, company A may be compelled to do the same inorder to remain in the market. Company B behaves likewise. If company A increasesits R&D investment, B responds by increasing its R&D investment.

In many economic models, we are interested in the dynamics of prices andquantities, in the dynamics of the economic growth process, or in strategies andsystems. We are also interested in stationary states of a differential equation or adifferential equation’s system, namely a state where the derivatives, that is, thechanges, become zero and the system is at rest. This state is defined by computingthe critical points of a differential equation or a system of equations, that is, points atwhich derivatives are zero. Differential equations are used to describe and solveproblems of motion, growth, competition, stationary states, and other types offinancial phenomena that involves variables and their rates of change.

EXAMPLES OF DIFFERENTIAL EQUATIONS

To convey an intuitive meaning of a differential equationwe describe an example calledthe banker’s equation. In this example, a differential equation is an equation that

117

3GC07 05/15/2014 10:18:44 Page 118

contains an unknown function y and its derivative dy/dt. More specifically, we have

dydt

� 0.03y (7.1)

Here we may think of t as time. The equation might represent a bank account,where the balance is y at a time t years after the account has been established, and theaccount is earning a 3 percent return per year. Regardless of the specific interpretation,

let’s see what the equation says. Since we see the term dydt we can tell that y is a function

of t, and that the rate of change is a multiple, namely 0.03, of the value of y itself. Wedefinitely should write y t� � instead of just y; however, there is no loss in meaning by

simply writing y. Likewise, dydt may be written as y´ or _y.

For example, if y happens to be $3; 000 at a particular time t, the rate of change of yis then 0.03 � 3;000 � $90 per year; the units of this rate in the bank account case aredollars/year. Thus y is increasing whenever y is positive. The balance will be $3; 090 ayear from now. Later when y is $7; 000, its rate of change will be 0.03 � 7; 000 � $210per year, which is much faster than when the balance was equal to y � $3; 000.

The banker’s equation shows the useful information we get from a differentialequation by just reading the equation carefully. One of the most important skills tolearn about differential equations is how to read them. For example, in the equation

_y � 0.03y � 15 (7.2)

there is a new negative influence on the rate of change, due to the �15. Thisamount of �15 could represent withdrawals from the account equal to $15 peryear. Whether the resulting value of dy

dt is actually negative depends on the currentvalue of y. This is an example of reading a differential equation. As a result of thisreading skill, we can recognize that the banker’s equation is idealized. It did notaccount for deposits or changes in the rate of return. It did not account forwithdrawals until we appended the �15.

A differential equation is an equation that involves derivatives of an unknownfunction f x� �. A function f x� � is a solution of a differential equation if it satisfies theequation; that is, if substitutionof f for theunknown functionproduces a true statement.To solve a differential equationmeans tofind all solutions. Sometimes, in addition to the

differential equation, we may know certain values of f or f ´ � dfdx, called initial condi-

tions. These initial conditions provide an explicit solution of the differential equation.As amatter of notation, often the derivative of y, dy/dx or dy/dt, is denoted y´ or _y.

If y is a function of x, the derivatives are y´ � dydx, y

´´ � d2ydx3, . . . ,y

n� � � dnydxn where n is a

positive integer. An equation that involves x, y, y´, y´´, . . . , y n� � is called an ordinaryequation of order n. Since time plays an important role and variables move in relationto time, we often replace x by t to indicate change in relation to time.

Let us compare an equation with a differential equation. A simple equation is x2 �1;we see the solution is x � 1 and x � �1.Hence, the solution consists of specific valuesof x. In contrast, the solution to a differential equation consists of a function y � f x� �.

Example: Solve the equations:

i. x2 � 9 � 0ii. y´ � x2 � 9 with initial condition y 0� � � 5

118 MATHEMATICS

3GC07 05/15/2014 10:18:45 Page 119

The solutions to the first equation are points x � �3. The solution to the secondequation is a function y � x3

3 � 9x � CwhereC is a constant. For the solution to satisfythe initial condition y 0� � � 5, C must be equal to 5. The explicit solution of thedifferential equation is y � x3

3 � 9x � 5. We note that the critical points of thedifferential equations are x � 3 and x � �3. At these points the slope of y is horizontaland y´ � 0. Later, we have to decide whether these critical points are stable or unstableequilibria; that is, if the system displaced by a shock away from these points will returnback to a critical point or will err away with no return.

Example: Prove that y � Ce�3x where C is a real number and a solution of thedifferential equation y´ � 3y � 0.

We compute the derivative of y � Ce�3x; we find y´ � �3Ce�3x. We substitute intothe differential equation; we find �3Ce�3x � 3Ce�3x � 0. Hence, y � Ce�3x solves thedifferential equation.

Many examples of differential equations may be provided:

y´ � dydt

� ky t� � (7.3)

This equation describes certain simple cases of population growth; it says that thechange in the population y is proportional to the population. The underlyingassumption is that each organism in the current population reproduces at a fixedrate, so the larger the population the more new organisms are produced. Although thisis too simple to model most real populations, it is useful in some cases over a limitedtime. When k > 0, the differential equation describes a quantity that increases inproportion to the current value; when k < 0, the differential equation describes aquantity that decreases in proportion to the current value.

i. The rate of change of real gross domestic product per unit of time, dydt, may be

related to the level of real gross domestic product y, according to the equation

y´ � αy (7.4)

The equation implies that the change of GDP is a constant fraction of y t� �.ii. A sum deposited at the bank has a rate of return r per unit of time; its rate of

change is related to the deposited amount y as follows:

_y � ry (7.5)

iii. The marginal cost of a manufacturer is a function of output x as follows:

C´ x� � � 30 � 0.05x (7.6)

iv. In economic growth, the rate of change of the capital per worker _k is related tocapital per worker k as follows:

_k � skα � n � δ� �k (7.7)

where s is the rate of saving, n is the rate of population growth, and δ is the rate ofcapital depreciation.

Differential Equations 119

3GC07 05/15/2014 10:18:46 Page 120

a. Linear differential equation b. Nonlinear differential equation

Critical point

0

y'

Ey

0

Critical points, E1 , E2 , E3

y'

yE1 E2 E3

FIGURE 7.1 Linear and Nonlinear Differential Equations

All theseexamplesareexamplesofordinarydifferential equations.Examples (i)–(iii)provide linear differential equations, that is, the relation between y´ and y is linear(Figure 7.1a). Example (iv) provides a nonlinear differential equationwhere the relationbetween y´ and y is not linear (Figure 7.1b).

SOLUTION METHODS FOR THE DIFFERENTIAL EQUATION

Differential equations can be solved in two ways. One method uses indefinite integralsand the other uses separable integrals.

Method of Indefinite Integrals

Indefinite integrals are useful for solving certain differential equations, because if weare given a derivative f ´ x� �, we can integrate and use the following relation involvingthe unknown f :

∫f ´ x� �dx � f x� � � C (7.8)

If we are also given an initial condition for f , it may be possible to find f x� �explicitly.

Example: The gross income of a Sharia-compliant bank, y, is related to itsoutstanding assets, x, that include Musharaka, Murabaha, and Mudarabah as

dydx

� f ´ x� � � x2 � 3x � 7

Express y in terms of x subject to the initial condition f 0� � � 2. We proceed asfollows:

∫f ´ x� �dx � ∫ x2 � 3x � 7� �

dx

We obtain

f x� � � x3

3� 3

x2

2� 7x � C

120 MATHEMATICS

3GC07 05/15/2014 10:18:47 Page 121

for some number C. Letting x � 0 and using the given initial condition f 0� � � 2 yieldsC � 2. Hence the solution f of the differential equation with the initial conditionf 0� � � 2 is

y � f x� � � x3

3� 3

x2

2� 7x � 2

Example: The output of palm oil of a Malay palm oil company, y, is related to its

capital stock, x, as follows; solve dydx � 4x1/2.

Solve y in terms of x with conditions y � 21 and x � 4. We proceed as follows:

∫dy � ∫4x1/2dx � 83x3/2 � C

We determine C using the conditions y � 21 and x � 4. We obtain

21 � 83 4

3/2 � C, which yields C � �1/3. The explicit solution is

y � 83x3/2 � 1/3

Method of Separable Variables

One of the simplest types of differential equations is

M x� � �N y� �y´ � 0 (7.9)

M x� � �N y� � dydx

� 0 (7.10)

Where M and N are continuous functions. If y � f x� � is a solution, thenM x� � �N f x� �� f ´ x� � � 0.

If f ´ x� � is continuous, then indefinite integration leads to

∫M x� �dx � ∫N f x� �� f ´ x� �dx � C (7.11)

∫M x� �dx � ∫N y� �dy � C (7.12)

The last equation is an implicit solution to the differential equation. Thedifferential equation

M x� � �N y� �y´ � 0 (7.13)

is separable, because the variables x and ymay be separated as indicated. An easy wayto remember the method of separating the variables is to change the equation M x� � �


3GC07 05/15/2014 10:18:48 Page 122

N y� � dydx � 0 to the differential form (7.14) and then integrate each term:

M x� �dx �N y� �dy � 0 (7.14)

Definition: A first-order differential function is separable if it can be written in theform

y´ � h x� �g y� � (7.15)

We can attempt to solve the equation by converting it to the form

∫1

g y� � dy � ∫h x� �dx (7.16)

This technique is called separation of variables. The simplest type of separableequation is one in which g y� � � 1, in which case we attempt to solve

∫1dy � ∫h x� �dx (7.17)

We can do this if we can find an anti-derivative of h x� �.Example: The likelihood of a portfolio loss for an Islamic bank, y, is related to the

size of the loss, x, as follows:

yx � dydx

� 0

Solve for y in terms of x.We first express the equation in differential form

yxdx � dy � 0

If y ≠ 0, we may separate the variables by dividing by y as follows:

xdx � dyy

� 0

Integrating each term, we obtain the implicit solution

12x2 � ln yj j � C

Solving for y we find

yj j � e� x22 �C � eCe� x2

2

Since exponential is always a positive function, we may write

y � Ce� x22

122 MATHEMATICS

3GC07 05/15/2014 10:18:49 Page 123

FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS

A simple but important and useful type of separable equation is the first-order homoge-neous linear equation. This linear equation occurs frequently in economics and finance.

Definition: A first-order homogeneous linear differential equation is of the form

y´ � P x� �y � 0 (7.18)

or equivalently,

y´ � �P x� �y (7.19)

In this definition, both y´ and y occur to the first power, that is, they are linear;homogeneous refers to the zero on the right-hand side of the first form of the equation.The equation

y´ � ky, or y´ � ky � 0 (7.20)

is linear and homogeneous, with a particularly simple P x� � � �k. Because first-orderhomogeneous linear equations are separable, we can solve them in the usual way:

y´ � �P x� �y

∫1ydy � �∫P x� �dx � C

ln yj j � �∫P x� �dx � C

y � �e�∫P x� �dx�C � Ae�∫P x� �dx

(7.21)

Example: The rate of change of wheat output, y, is related to fertilizers, x, asfollows: y´ � ky � 0.

Solve for y in terms of x.

The solution is y � Ae�∫P x� �dx � Aekx. A is a real constant.Definition: A first-order linear nonhomogeneous differential equation is an

equation of the form

y´ � P x� �y � Q x� � (7.22)

where P and Q are continuous functions. If in the definition Q x� � � 0 for every x, wemay separate the variables and then integrate as follows (provided y ≠ 0):

dydx

� P x� �y � 0

1ydydx

� �P x� �1ydy � �P x� �dx


3GC07 05/15/2014 10:18:50 Page 124

Integrating, we find

ln yj j � �∫P x� �dx � lnCj j

ln yj j � lnCj j � �∫P x� �dx

lnyC

�� e�∫P x� �dx

yC� e

�∫P x� �dxor ye∫

P x� �dx � C (7.23)

We note that

ddx

ye∫P x� �dx

" #� y´e∫

P x� �dx � yP x� �e∫P x� �dx � y´ � P x� �y� �e∫P x� �dx

If we multiply by e∫P x� �dx

both sides of y´ � P x� �y � Q x� �, we obtain

y´ � P x� �y� �e∫P x� �dx � Q x� �e∫P x� �dx

or equivalently,

ddx

ye∫P x� �dx

" #� Q x� �e∫P x� �dx

Integrating both sides gives us the following implicit solution of the first-orderdifferential equation:

ye∫P x� �dx � ∫Q x� �e∫P x� �dx

dx � C (7.24)

for a constant C. The expression

e∫P x� �dx

(7.25)

is an integrating factor of the differential equation. Hence, the first-order lineardifferential equation y´ � P x� �y � Q x� � may be transformed into a separable differen-

tial equation by multiplying both sides by the integrating factor e∫P x� �dx

.Example: The revenues of an Islamic bank, y, are related to its outstanding assets,

x, as follows: y´ � αy � b, where α and b are constants. Solve for y as a function of x.

124 MATHEMATICS

3GC07 05/15/2014 10:18:50 Page 125

This is a nonhomogeneous differential equation with P x� � � �α and Q x� � � b.The solution is

ye∫P x� �dx � ∫Q x� �e∫P x� �dx

dx � C

ye�αx � ∫be�αxdx � C � � bke�αx � C

y � � bα� Ceαx

Example: Solve the differential equation: dydx � 2xy � x.

The differential equation has the form y´ � P x� �y � Q x� � with P x� � � �2x andQ x� � � x.

An integrating factor is e∫�2xdx � e�x2 . Multiplying both sides of the given

differential equation by the integrating factor e�x2 we obtain

e�x2 dydx

� 2xe�x2y � xe�x2

or equivalently,

ddx

e�x2y

� �� xe�x2

Integrating both sides of the last equation gives us

e�x2y � ∫xe�x2dx � � 1

2e�x2 � C

Finally, multiplying by ex2gives

y � � 12� Cex

2

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

A second-order differential equation is one containing the second derivative. These arein general quite complicated, but one fairly simple type is useful: the second-orderlinear equation with constant coefficients.

Homogeneous Linear Differential Equation

The general second-order homogeneous linear differential equation with constantcoefficients has the form

y´´ � by´ � cy � 0 (7.26)


3GC07 05/15/2014 10:18:51 Page 126

where b and c are constants. If y � f x� � and y � g x� � are solutions of Equation (7.26),then

y � C1f x� � � C2g x� � (7.27)

is a solution for all real numbersC1 and C2. In our search for a solution of the second-order differential equation we use as a trial solution:

y � erx (7.28)

Since y´ � rerx and y´´ � r2erx, it follows that y � erx is a solution if and only if

r2erx � brerx � cerx � 0 (7.29)

or since erx ≠ 0, if and only if

r2 � br � c � 0 (7.30)

The last equation is very important in finding solutions. It is called the character-istic or auxiliary equation of the differential equation. If the roots r1 and r2 are real andunequal, then the general solution of y´´ � by´ � cy � 0 is

y � C1er1x � C2er2x (7.31)

Example: The cotton output of a farm, y, is related to the quantity of insecticides,x, as follows: y´´ � 3y´ � 10y � 0.

Solve for y as a function of x.The auxiliary equation is

r2 � 3r � 10 � 0 or r � 5� � r � 2� � � 0

Because the roots r1 � 5 and r2 � �2 are real and unequal, the general solution is

y � C1e5x � C2e�2x

If the auxiliary equation has a double root r, then the general solution of y´´ �by´ � cy � 0 is

y � C1erx � C2xerx

Example: The dynamics of the assets of an Islamic bank, y, are related to time x asfollows: y´´ � 6y´ � 9y � 0.

Solve y as a function of time x.The auxiliary equation r2 � 6r � 9 � 0 or r � 3� �2 � 0 has a double root equal

to 3. Hence, the general solution is

y � C1e3x � C2xe3x

126 MATHEMATICS

3GC07 05/15/2014 10:18:52 Page 127

Nonhomogeneous Linear Differential Equations

We consider second-order nonhomogeneous linear differential equations with con-stant coefficients, that is, equations of the form

y´´ � by´ � cy � k x� � (7.32)

where b and c are constants and k is a continuous function of x. Suppose that y1 and y2are solutions to y´´ � by´ � cy � k x� � and consider the function

h � y1 � y2 (7.33)

We substitute this function into the left-hand side of the differential equation andsimplify

y1 � y2� �´´ � b y1 � y2

� �´ � c y1 � y2� �

� y´1 � by1 � cy1� � � y´2 � by2 � cy2

� � � k x� � � k x� � � 0

So h is a solution to the homogeneous equation y´´ � by´ � cy � 0. Because weknow how to find all such h, then with just one particular solution y2 we can expressall possible solutions y1, namely,

y1 � h � y2 (7.34)

where now h is the general solution to the homogeneous equation. This is exactly howwe approached the first-order linear equation. To make use of this observation weneed a method to find a single solution y2. This turns out to be somewhat moredifficult than the first-order case, but if k x� � is of a certain simple form, we can find asolution using the method of undetermined coefficients.

Example: Solve the differential equation

y´´ � y´ � 6y � 18x2 � 5

The general solution of the homogeneous equation is yh � C1e3x � C2e�2x. Weguess that a solution to the nonhomogeneous equation might look like k x� � itself,namely, a quadratic:

y � ax2 � bx � c

Substituting this guess into the differential equation we get

y´´ � y´ � 6y � 2a � 2ax � b� � � 6 ax2 � bx � c� �

� �6ax2 � �2a � 6b� �x � 2a � b � 6c� �We want this to equal 18x2 � 5, so we need

�6a � 18;�2a � 6b � 0; and 2a � b � 6c � 5


3GC07 05/15/2014 10:18:53 Page 128

This is a system of three equations in three unknowns and is not hard tosolve: a � �3, b � 1, and c � �2. Thus the general solution to the differentialequation is

y � C1e3x � C2e�2x � 3x2 � x � 2

So the “judicious guess” is a function with the same form as k x� � but withundetermined (or better, yet to be determined) coefficients. This works whenever k x� �is a polynomial.

Example: Solve the differential equation

y´´ � 2y´ � 8y � e3x

The auxiliary equation r2 � 2r � 8 � 0 has roots 2 and –4. The general solution ofthe homogeneous equation is yh � C1e2x � C2e�4x.

Since k x� � � e3x, we seek a particular solution of the form yp � Ae3x. Since yp �3Ae3x and y´p � 9Ae3x, substitution in the given equation leads to

9Ae3x � 6Ae3x � 8Ae3x � e3x

Dividing both sides by e3x, we obtain 9A � 6A � 8A � 1 or A � 1/7.Thus, yp � 1

7 e3x; the general solution is

y � C1e2x � C2e�4x � 17e3x

If

y´´ � by´ � cy � k x� � (7.35)

k x� � is of the form xeδx, and δ is not a root of the auxiliary solution r2 � br � c � 0,then there is a particular solution of the form

yp � A � Bx� �eδx (7.36)

LINEAR DIFFERENTIAL EQUATION SYSTEMS

A linear differential equation system arises when there is interaction between vari-ables. For instance, country A increases its tariffs in retaliation to an increase of tariffsby a trading partner. Or a country depreciates its exchange rate in response toexchange rate depreciation by a competing country. A homogeneous system of lineardifferential equations has the form

dydt

� Ay t� � � 0 (7.37)

where y is a n, 1� � vector of variables y1 t� �, y2 t� �, . . . , yn t� �, and A is an n � nmatrix ofconstant terms. We consider simple linear systems of two equations and develop the

128 MATHEMATICS

3GC07 05/15/2014 10:18:53 Page 129

solutions and stability analysis for these systems. The results are easily generalized toan n-equation system. We consider the following system:

dx/dtdy/dt

� A

x t� �y t� �

(7.38)

where both x and y are functions of t. It is very important to understand the meaningof a differential equation system. While x and y depend on a common exogenousvariable t, they also influence each other. The variable t may be time or any othercommon driving variable. We let x´ � dx/dt and y´ � dy/dt; we write the system as

x´y´

� A

xy

(7.39)

We let A � a bc d

; then the system becomes

x´ � ax � byy´ � cx � dy

�(7.40)

The interpretation of this system is simple. The variable x changes in response toits own current level and to the level of y. For instance, company A increasesadvertising spending if its current spending on advertising is low and that of itscompetitor company B is high. In the same vein, a companyAmay lay off personnel ifits competitor is doing the same to improve productivity.

Transforming the System into a Second-Order HomogeneousDifferential Equation

The system can be easily reduced to one second-order homogeneous differentialequation. Taking the derivatives of the first equation we find

x´´ � ax´ � by´ or by´ � x´´ � ax´ (7.41)

We eliminate y from the system by multiplying the equation of x´ by d and theequation of y´ by b; subtracting one equation from the other yields

dx´ � by´ � ad � bc� �xSubstituting for by´ � x´´ � ax´, we find

dx´ � x´´ � ax´ � ad � bc� �xAfter rearranging the terms, we have

x´´ � a � d� �x´ � ad � bc� �x � 0 (7.42)

We note that the trace of A is Tr A� � � a � d� � and the determinant of A isdet A� � � ad � bc.


3GC07 05/15/2014 10:18:54 Page 130

The second-order homogeneous difference equation is therefore

x´´ � Tr A� �x´ � det A� �x � 0 (7.43)

We note that the same equation is obtained if we tried to eliminate x from thesystem; namely, we find

y´´ � Tr A� �y´ � det A� �y � 0 (7.44)

We proceed for the solution of the system by solving the characteristic equation.The characteristic roots λ1 and λ2 may be real or complex. In the real case, the rootsmay be distinct, that is, λ1 ≠ λ2 or equal, that is, λ1 � λ2. If the roots are distinct, thegeneral solution is

x � C1eλ1t � C2eλ2t (7.45)

If the roots are equal, the general solution is

x � C1eλt � C2teλt (7.46)

Example: The Musharaka portfolio, x, and Murabaha portfolio, y, of an Islamicbank influence each other according to the following system:

x´ � 3x � 2y

y´ � 4x � y

Find a solution for x.The second-order homogeneous equation corresponding to this system is

x´´ � 4x´ � 5x � 0

The characteristic equation is λ2 � 4λ � 5 � 0.

λ � �b �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib2 � 4c

p2

� 4 � 62

We find λ1 � �1 and λ2 � 5 and the general solution is

x t� � � C1e�t � C2e5t

Method of Eigenvalues and Eigenvectors

An alternative approach to solving the system is to compute eigenvalues andeigenvectors of the system. Let us assume that the solutions are of the form x �c1eλt and y � c2eλt, then x´ � λc1eλt and y´ � λc2eλt. Replacing the solutions in the

130 MATHEMATICS

3GC07 05/15/2014 10:18:55 Page 131

system we get

λc1eλt

λc2eλt

� A

c1eλt

c2eλt

(7.47)

The terms in eλt cancel out; we have

Ac1c2

� λ

c1c2

(7.48)

Ac � λc (7.49)

where c is an eigenvector with coordinates c1 and c2, that is: c � c1c2

.

In other words, if there exists a solution to the system of the form ceλt, thelinear system leads to the following equation after simplification of the exponentialterms:

Av � λv (7.50)

Here λ and v are respectively the eigenvalues and eigenvector of the matrix A. Asearch for the solution of the linear differential equation system amounts to computingthe eigenvalues and eigenvectors of A. The eigenvector satisfies the condition

A � λI� �v � 0 (7.51)

Hence A � λI� � is singular, and its determinant is null: A � λIj j � 0. The determi-nant is

a � λ� � d � λ� � � bc � 0 (7.52)

We obtain, after rearranging terms, the characteristic equation

λ2 � a � d� �λ � ad � bc � 0 (7.53)

The eigenvalues may be real or complex. We consider only real eigenvalues; wemay have two distinct eigenvalues λ1 ≠ λ2 or two eigenvalues of equal value λ1 � λ2.Let us consider the case λ1 ≠ λ2; let the eigenvectors corresponding to λ1 and λ2 be,respectively

v1 � v11v21

and v2 � v12

v22

(7.54)

Let us consider the eigenvector v1, the solution may be written as

xy

� v11eλ1t

v21eλ1t

(7.55)


3GC07 05/15/2014 10:18:55 Page 132

If we replace this solution in the system

dx/dtdy/dt

� A

x t� �y t� �

(7.56)

we find Av1 � λ1v1. In similar fashion, the solution corresponding to the eigenvectorv2 is

xy

� v12eλ2t

v22eλ2t

(7.57)

If we replace this solution in the system we findAv2 � λ2v2. The general solution is

x � C1v11eλ1t � C2v12eλ2t

y � C1v21eλ1t � C2v22eλ2t

�(7.58)

Example: The wheat x� � and soybean y� � crops of a farm are related by thefollowing system:

x´y´

� 3 2

4 1

xy

Solve for both x and y in terms of the common variable t.

The characteristic equation is A � λIj j � 3 � λ 24 1 � λ

�� 0.

3 � λ� � 1 � λ� � � 8 � λ2 � 4λ � 5 � 0

The eigenvalues are λ1 � �1 and λ2 � 5.

The eigenvector corresponding to λ1 � �1 is3 � λ1 2

4 1 � λ1

v11v21

� 0.

We have4 24 2

v11v21

� 0.

4v11 � 2v21 � 0 or 4v11 � �2v21A solution would be v11 � 1, v21 � �2, and v1 � 1

�2

.

The eigenvector corresponding to λ2 � 5 is3 � λ2 2

4 1 � λ2

v12v22

� 0.

We have�2 24 �4

v12v22

� 0.

�2v12 � 2v22 � 0 or v12 � v22

A solution would be v12 � 1, v22 � 1, and v2 � 11

.

The general solution is x � C1e�t � C2e5t and y � �2C1e�t � C2e5t.

132 MATHEMATICS

3GC07 05/15/2014 10:18:56 Page 133

PHASE DIAGRAMS AND STABILITY ANALYSIS

Stability analysis is important in economics andfinance.Weare interested indiscoveringwhether a market or an economy is stable or unstable. If a market is stable it will returnto equilibrium following a shock to demand or supply. If the market is unstable, it maygo throughbubbles followedbycrashes and rarelyfinds an equilibrium.There are forcesalien to themarket that create distortions and cause instability. Phase diagrams are usedto analyze stability of markets and are a feature of dynamic systems. They involve time-derivatives dx/dt, where x is a continuous function of time. They provide a way tovisualize solutions toanautonomousordinarydifferential equations systemandanalyzethe equilibriumandstabilityof thedynamic system.We studyphasediagrams in the caseof a simple ordinary differential equation and in the case of a linear system of twodifferential equations.Weanalyzephasediagramsof autonomousdifferential equationswhere time does not appear in the right-hand side of the equation.

Phase Line of an Ordinary Differential Equation

Example: Consider the following differential equation:

dxdt

� x 1 � x� �This is an autonomous equation because time does not appear in the right-hand side

of the equation. We note the critical values of dxdt are x � 0 and x � 1. At these values

dxdt � 0, and therefore there is nochange inx.Wedistinguish three cases:x > 1,0 < x < 1,and x < 0. In the first case, if the initial value of x exceeds 1, that is, x > 1, we have

dxdt

� x 1 � x� � < 0

As time evolves, x decreases because its time-derivative is negative until it reachesthe equilibrium point x � 1. This is illustrated in Figure 7.2a by a downward-movingtrajectory. In the second case, 0 < x < 1,

dxdt

� x 1 � x� � > 0

As time moves, x increases because its time-derivative is positive until it reachesthe equilibrium point x � 1. This is illustrated in Figure 7.2a by an upward movingtrajectory. In the third case, x < 0,

dxdt

� x 1 � x� � < 0

As time moves, x decreases and tends to �∞ as shown in Figure 7.2a. Themovements of x are summarized by a phase line as illustrated in Figure 7.2b. Theequilibrium point x � 1 is stable; displacement from this point will trigger dynamicsthat will push back x to equilibrium. The equilibrium point x � 0 is unstable;displacement from this point will push x from this equilibrium to a new equilibriumx � 1 or to �∞ if x < 0.


3GC07 05/15/2014 10:18:56 Page 134

Phase Diagram of a Linear Differential Equation System

Consider the linear differential equation system

x´ � ax � by

y´ � cx � dy

)(7.59)

The general solution is of the form

x � C1v11eλ1t � C2v12eλ2t

y � C1v21eλ1t � C2v22eλ2t

)(7.60)

For each value of t, we get a pair x t� �, y t� �� . If we plot these pairs in a diagramwith coordinates x t� � and y t� �, we get curves that relate the variables x t� � and y t� �together and describe their behavior as t moves. These curves are called trajectories ofthe solution x t� �, y t� �� , or phase diagrams, or phase planes. As in the case of onevariable, a phase diagram of a system of differential equations allows us to visualizethe solution trajectories and study the qualitative behavior of the system aroundequilibrium points. It provides insights of the solution without necessarily computingthe exact form of the solution. We observe that the behavior of x t� � and y t� � willdepend on the eigenvalues λ1 and λ2, which are the coefficients of the time variable t inthe solutions x and y as well as the coefficients C1 and C2. Two important referencephase lines are

x´ � ax � by � 0

y´ � cx � dy � 0

)(7.61)

We study the behavior of the trajectories x t� �, y t� �� in relation to these two lines.We consider examples to illustrate phase diagrams.

b. Phase line of the solutiona. Trajectories of the solution xt

t0

x

1

<0

0

dxdt

<0dxdt

>0dxdt

x

1

FIGURE 7.2 Phase Line

134 MATHEMATICS

3GC07 05/15/2014 10:18:57 Page 135

Example: The sukuks x� � and equity y� � portfolios of an Islamic bank are relatedby the following system:

x´ � �3x � y

y´ � x � 3y

Analyze the stability of bank’s overall portfolio.

The equilibrium point corresponds to dxdt � 0 and dy

dt � 0. By solving the equa-tions, the equilibrium point is E x � 0, y � 0� �. It would be useful to plot the phaseline (Figure 7.3a)

dxdt

� 0;�3x � y � 0 or y � 3x

We examine the signs of dxdt above and below the line. We note that for any point

x, y� � above the line y � 3x we have

�3x � y > 0

that is dxdt > 0, which implies that xt is increasing as time moves. We show the

movement of xt by a horizontal arrow pointing toward the phase line y � 3x. Wenote also that below the line y � 3x we have

�3x � y < 0

that is dxdt < 0, which implies that xt is decreasing as time moves. We show the

movement of xt by a horizontal arrow pointing toward the phase line y � 3x.In similar fashion, we draw the line dy

dt � 0 (Figure 7.3a)

x � 3y � 0 or y � x/3

We examine the signs of dydt above and below the line. We note that above the

line we have

x � 3y < 0

a. Phase lines b. Eigenvectors

0

y

x

v2v1

y' = 0

x' = 0

I

II

0

III

IV y

x

x' = 0

x' = 0

y' = 0

y' = 0

FIGURE 7.3 Phase Diagram and Eigenvectors: Stable Equilibrium


3GC07 05/15/2014 10:18:58 Page 136

that is, dydt < 0, which implies that yt is decreasing as time moves. We show the

movement of yt by a vertical arrow pointing toward the phase line y � x/3. We notealso that below the line we have

x � 3y > 0

that is, dydt > 0, which implies that yt is increasing as time moves. We show the

movement of yt by a vertical arrow pointing toward the phase line y � x/3. The phaselines form four regions: I, II, III, and IV. No matter where the initial point is, it will befound that each trajectory x t� �, y t� �� will converge to the fixed point E 0; 0� � over time.This equilibrium point E 0; 0� � is stable; any displacement from this equilibrium willtrigger forces that will bring back the system to this point.

Let us solve the eigenvalues and eigenvectors of the system. The characteristicequation is

λ2 � 6λ � 8 � 0

We find two distinct negative roots: λ1 � �2 and λ2 � �4. We solve the eigen-vectors according to the equation Av � λv or equivalently: A � λI� �v � 0. For λ1 � �2we have

�3 � 2 11 �3 � 2

v11v21

� 0

�v11 � v21 � 0

v11 � v21 � 0 or v11 � v21

An eigenvector would be

v1 � 11

For λ2 � �4 we have

�3 � 4 11 �3 � 4

v12v22

� 0

v12 � v22 � 0 v12 � �v22An eigenvector would be

v2 � 1�1

The general solution is

x � C1e�2t � C2e�4t

y � C1e�2t � C2e�4t

136 MATHEMATICS

3GC07 05/15/2014 10:18:59 Page 137

We show these two eigenvectors in Figure 7.3b. If we choose C2 � 0, then wehave x � C1e�2t and y � C1e�2t, or x � y. The initial point is therefore located on theline that passes through the eigenvector v1; the trajectory x t� �, y t� �� remains on theline that goes through the eigenvector v1 and converges directly to the fixedpoint x � 0, y � 0� � as t ! ∞. If we choose C1 � 0, then we have x � C2e�4t andy � �C2e�4t, or x � �y. The initial point is therefore located on the line that goesthrough the eigenvector v2; the trajectory x t� �, y t� �� remains on the line that goesthrough the eigenvector v2 and converges directly to the fixed point x � 0, y � 0� � ast ! ∞. If the initial point is not on a line that passes through an eigenvector, then thetrajectory x t� �, y t� �� will move according to the arrows in the phase plane and willeventually approach the fixed point x � 0, y � 0� � as t ! ∞. The critical pointx � 0, y � 0� � is a stable long-run equilibrium. If the system is shocked and movedaway from the equilibrium, it will return to equilibrium as time evolves.

Example:We consider the case of unstable equilibrium, which corresponds to twodistinct eigenvalues: one is positive and the other is negative.

x´ � x � y

y´ � 4x � y

In Figure 7.4a we show the phase lines x´ � 0 and y´ � 0. Above the line x´ � 0,that is, y � �x, we have x´ � dx

dt > 0; as time moves, xt increases and moves away fromthe line. This is shown by a rightward horizontal arrow. Below the line x´ � 0,x´ � dx

dt < 0; as time moves, xt decreases and moves away from the line. This is shown

by a leftward horizontal arrow. Above the line y´ � 0, that is, y � �4x, y´ � dydt > 0; as

time moves, yt increases and moves away from the line. This is shown by an upward

vertical arrow. Below the line y´ � 0, y´ � dydt < 0; as time moves, yt decreases. This is

shown by a downward vertical horizontal arrow.The two phase lines x´ � 0 and y´ � 0 intersect at the fixed point x � 0, y � 0� � and

form four regions: I, II, III, and IV. If the initial point x0, y0� �

starts in regions I or III, then

a. Phase lines b. Eigenvectors

0

v1 v2

y

x

I

II

III

IV

0

y

y' = 0

x' = 0

x

FIGURE 7.4 Phase Diagram and Eigenvectors: Saddle Path


3GC07 05/15/2014 10:18:59 Page 138

the trajectories x t� �, y t� �� will diverge away from the fixed point toward infinity. If theinitial point x0, y0

� �starts in regions II or IV, thenthe trajectories x t� �, y t� �� ,move toward

the fixed point but diverge awayas they get closer to it. They never reach the critical point.Wehave anunstable equilibrium.More specifically, if the system is shockedanddisplacedaway from x � 0, y � 0� �, it will never converge back to this critical point.

To study the dynamics of the system, we compute the eigenvalues and eigenvec-tors associated with the matrix A

x´y´

� 1 1

4 1

xy

The characteristic equation is

A � λIj j � 1 � λ 14 1 � λ

�� 0

1 � λ� � 1 � λ� � � 4 � λ2 � 2λ � 3 � 0

The eigenvalues are λ1 � �1 and λ2 � 3.The eigenvector corresponding to λ1 � �1is

1 � λ1 14 1 � λ1

v11v21

� 0

2 14 2

v11v21

� 0

2v11 � v21 � 0 or 2v11 � �v21

A solution would be v11 � 1 and v21 � �2; v1 � 1�2

.

The eigenvector corresponding to λ2 � 3 is

1 � λ2 14 1 � λ2

v12v22

� 0

�2 14 �2

v12v22

� 0

�2v12 � v22 � 0 or 2v12 � v22

A solution would be v12 � 1 and v22 � 2; v2 � 12

.

The general solution is

x � C1v11eλ1t � C2v12eλ2t and y � C1v21eλ1t � C2v22eλ2t

Replacing the values of the eigenvalues and eigenvectors we find the followinggeneral solution:

x � C1e�t � C2e3t; y � �2C1e�t � 2C2e3t

138 MATHEMATICS

3GC07 05/15/2014 10:19:0 Page 139

We show the eigenvectors in Figure 7.4b. Let us assume that C2 is zero; thesolution becomes x � C1e�t and y � �2C1e�t. The trajectory is located on theeigenvector v1 because

dydx

� 2C1e�t�C1e�t

� �2

As time t ! ∞, x ! 0 and y ! 0 since e�t ! 0. If we assume that C1 � 0, then thesolution becomes x � C2e3t; y � C2e3t. The trajectory xt, yt

� �is necessarily located on

the eigenvector v2 because

dydx

� 3 � 2C2e3t

3C2e3t� 2

As time t ! ∞, x ! ∞ and y ! ∞ because e3t ! ∞. For initial points off theeigenvectors, the direction of the trajectory is determined by the dominance of oneroot. In this example, the positive root will dominate the system. Hence for pointsabove v1 and v2, the solution path will veer toward the line through v2, whichcorresponds to the dominant root λ2 � 3. The same is true for any initial pointbelow v2 and above v1. An initial point below the line through v1 will be dominatedby the larger root and the system will veer toward minus infinity. In the same vein,an initial point above the line through v1 will be dominated by the larger root andthe system will veer toward infinity. In this case the critical point is called a saddlepoint. The line through v1 is called the stable arm, while the line through v2 is calledthe unstable arm. Saddle path equilibria are common in economics and one shouldlook out for them in terms of real distinct roots of opposite sign and the fact thatdet A� � is negative. It will also be important to establish the stable and unstable armsof the saddle point, which are derived from the eigenvectors associated with thecharacteristic roots.

SUMMARY

Islamic finance applies differential equations in wide range of applications. Forinstance, the price of assets evolves over time. To compute the price of an asset,we may require a differential equation as the case for the Black-Scholes option pricingmodel. This chapter introduces the definition of a differential equation throughexamples. It describes solution methods for differential equations that use the methodof indefinite integrals and the method of separable variables. It describes first-orderand second linear differential equations, homogeneous and nonhomogenous lineardifferential equations. The chapter analyzes linear differential equation systems andmethods of solutions based on eigenvalues and eigenvectors. It also addresses stabilityanalysis of a linear system of differential equations using the method of phasediagrams.

Differential equations help to understand dynamics of financial variables and areessential in dynamic optimization and in pricing of assets.


3GC07 05/15/2014 10:19:1 Page 140

QUESTIONS

1. Solve the differential equations dydt � 0.03y and dy

dt � 0.03y � 15 with initialcondition y0 � 1; 000.

2. Solve the equations x2 � 4 � 0 and y´ � x � 4. Compare the solutions.

3. Solve the differential equation

dydx

� f ´ x� � � x2 � x � 7

subject to the initial condition f 0� � � 4.

4. Show that the differential equation y´´ � 25y � 0 has the solution y � C1e5x �C2e�5x for all real numbers C1 and C2.

5. Solve the differential equations (a) xdy � ydx � 0 and (b) yx2 � dydx � 0.

6. Solve the differential equation dydx � 3x2y � x2.

7. Solve the homogeneous differential equations:(a) y´´ � 5y´ � 6y � 0, (b) y´´ � 5y´ � 0, and (c) y´´ � 4y´ � 4y � 0.

8. Solve the nonhomogeneous differential equations: d2ydx2 � 3 dy

dx � 4y � 2.

9. Find the general solutions of the following systems:

(a)xý´

� �3 1

4 �2

xy

(b)

xý´

� 2 3

1 1

xy

10. Determine the critical (equilibrium) points and classify as stable or unstable thefollowing differential equations:

(a) dxdt � x x � 1� � x � 2� � (b) dx

dt � x2 4 � x2� �

.

11. Find the general solutions of the following linear systems; analyze their stabilityand sketch few trajectories.

(a)xý´

� 3 �2

2 �2

xy

(b)

xý´

� 1 1

4 �2

xy

140 MATHEMATICS

3GC08 05/30/2014 1:14:13 Page 141

CHAPTER 8Difference Equations

I n Islamic finance, many models have a dynamic feature; variables may be related totheir past values and give rise to difference equations. In the cobweb model, a

supplier’s decision is based on lagged prices. High prices result in high output; thelatter induces a fall in prices, which in turn induces lower output, followed by higherprices. At the corporate level, low levels of investment may cause higher investmentspending in the future. A low performance may lead a firm to improve performance inthe future. In econometrics, difference equations characterize the autoregressivemodels. A variable is influenced by its past levels. Models of economic growth usedifference equations. In the same vein, macroeconomic models rely on differenceequations. More specifically, in inflation models, inflationary expectations are deter-mined by past price inflation rates.

In the forward-looking model, the expected future level of a variable may havean effect on the present level of the variable. For instance, if farmers anticipate adrought next year, they may decide to produce more this year. Likewise, if a carcompany anticipates high demand for cars in the future, it may decide to increase itspresent car production. If consumers anticipate a shortage of food, they may decideto store food now.*

This chapter covers the theory of difference equations. In difference equations,time is discrete; variables are dated at equal time intervals (e.g., year, month, week,day). The chapter discusses first-order and second-order difference equations, thesolutions methods, the notions of steady-state equilibrium, and the stability ofdifference equations or a system of difference equations. Stability is defined asautomatic convergence to equilibrium; once reached, the system remains in theequilibrium state. If displaced from equilibrium, the system will return back to thesame state. Instability is defined as an absence of equilibrium, where the system is inperpetual movement without attaining a steady state.

DEFINITION OF A DIFFERENCE EQUATION

An approach to define a difference equation is to compare a difference equation with asequence. A sequence has an explicit formula for its terms. More specifically, in asequence, the nth term xn, can be written as a function,

xn � f n� � (8.1)

*In a difference equation, we have xt � f xt�1; xt�2; . . .� �. In a forward-looking differenceequation, we have xt � g xt�1; xt�2; . . .� �.

141

3GC08 05/15/2014 10:34:10 Page 142

for some known function f . For example,

xn � 1 � 12

� �n

(8.2)

Here n is an integer. Then it is an easy matter to compute explicitly, say, x5 or x15.In such cases, we are able to compute any given term in the sequence without referenceto any other term in the sequence. However, it is often the case in applications that wedo not begin with an explicit formula for the terms of a sequence; rather, we mayknow only some relationship between the various terms. An equation that expresses avalue of a sequence as a function of the other terms in the sequence is called adifference equation.

In particular, an equation that expresses the value xn of a sequence xnf g as afunction of the term xn�1 is called a first-order difference equation. If we can find afunction f such that xn � f n� �, n � 1; 2; 3; . . ., then we will have solved the differenceequation.

Example: Suppose an investment account at an Islamic bank is growing at the rateof 2 percent per year. If we let x0 represent the size of the initial investment and xn thesize of deposit n years later, then

xn�1 � xn � 0:02xn � 1:02xn for n � 0; 1; 2; . . .

That is, the value in any given year is equal to the value in the previous year plus 2percent of the value in the previous year. We have an example of a first-orderdifference equation; it relates investment in a given year with the investment in theprevious year. Hence we know the value of a specific xn once we know the value ofxn�1. To get the sequence started we have to know the value of x0. For example, ifinitially we have a value of x0 � $100 and we want to know what the value will beafter four years, we may compute

x1 � 1:02x0 � 1:02 � 100 � 102

x2 � 1:02x1 � 1:022x0 � 104:4

x3 � 1:02x2 � 1:023x0 � 106:1208

x4 � 1:02x3 � 1:024x0 � 108:243216

We may work backward to find x4 explicitly in terms of x0. This is interestingbecause it indicates that we can compute x4 without reference to the values of x1, x2,and x3 provided that we know the value of x0. If we do this in general, then we havesolved the difference equation xn�1 � 1:02xn. Namely, we have, for any n �1; 2; 3; . . .;

xn � 1:02xn�1 � 1:022xn�2 � ∙ ∙ ∙ � 1:02nx0

142 MATHEMATICS

3GC08 05/15/2014 10:34:11 Page 143

If we consider the difference equation xn � αxn�1, where α is a constant, we arriveat the general result that the solution of the difference equation is

xn � αnx0 n � 1; 2; . . . (8.3)

Note that this difference equation, and its solution, are useful whenever we areinterested in a sequence of numbers where the n � 1� � term is a constant proportionof the nth term. Our first example, where a bank deposit was assumed to grow at aconstant rate, is a common example of this type of behavior. Another commonexample is when a quantity decreases at a constant rate over time. This behavior isdiscussed in the next example in the context of inflation of prices.

Example: Assume you have savings of $1,000 at the bank. Let the real rate ofreturn adjusted for inflation be r � �5 pecent. Then the real purchasing power of thesavings is expressed as xn � 0:95xn�1. The real purchasing power of the savings in fiveyears will be x5 � 0:955x0 � 0:955 � 1; 000 � $773:7809.

It is interesting to note that the first example is an example of exponential growth,whereas the second example is an example of exponential decay. The differenceequation will always lead to exponential growth when α > 1 and to exponential decaywhen 0 < α < 1.

Besides formulating difference equations, a main topic is to solve these equations.Namely, a solution is a function that, once incorporated in the difference equation,verifies this equation.

Example:

i. Consider the difference equation

yn�1 � yn � 1

Obviously yn � n is a solution. In fact, yn�1 � n � 1 and yn�1 � yn �n � 1� � � n � 1. Likewise, yn � n � C, where C is a real number, is a solution.

ii. Consider the difference equation

yn�1 � yn � n

We may easily verify that a solution is

yn � n n � 1� �2

� C

In fact,

yn�1 � yn � n n � 1� �2

� C � n n � 1� �2

� C � n

iii. Consider the difference equation

yn�2 � 3yn�1 � 2yn � 1

We may verify that a solution is

yn � C1 � C22n � n

Difference Equations 143

3GC08 05/15/2014 10:34:11 Page 144

In fact, yn�2 � C1 � C22n�2 � n � 2 and yn�1 � C1 � C22n�1 � n � 1. Wereplace in the difference equation; we find

C1 � C22n�2 � n � 2 � 3 C1 � C22

n�1 � n � 1� � � 2 C1 � C22

n � n� � � 1

FIRST-ORDER LINEAR DIFFERENCE EQUATIONS

This section describes solutions of the first-order difference equation, the impulseresponse function, and the cobweb model.

Solutions of the First-Order Difference Equation

Given constants real α and β, a difference equation of the form

xn�1 � αxn � β n � 0;1; 2; . . . (8.4)

is called a first-order linear difference equation. Note that the difference equation

xn�1 � αxn (8.5)

is of this form with β � 0. A procedure of recursive substitution, analogous to themethod we used to solve xn�1 � αxn, will enable us to solve this equation as well.Namely,

xn�1 � αxn � β; n � 0; 1; 2; . . .

xn�1 � α αxn�1 � β� � � β � α2xn�1 � β 1 � α� �xn�1 � α2 αxn�2 � β� � � β 1 � α� � � α3xn�2 � β 1 � α � α2

� �

xn�1 � αnx0 � β 1 � α � α2 � ∙ ∙ ∙ � αn�1� �

(8.6)

If α � 1 we have xn � x0 � nβ as the solution to the difference equationxn�1 � xn � β. If α ≠ 1, we have as the solution for the difference equation

xn�1 � αnx0 � β 1 � α � α2 � ∙ ∙ ∙ � αn�1� � � αnx0 � β

1 � αn

1 � α

� �(8.7)

The pattern of the solution depends on the coefficient α. Using Microsoft Excel,we illustrate the solution pattern for the difference equation

xn�1 � αxn � 3

with x0 � 1. Figure 8.1 illustrates four patterns corresponding to α � 0:8, α � �0:8,α � 1:2, and α � �1:2. Figure 8.1a illustrates α � 0:8. We observe that the solutionconverges to the upper limit equal to x* � β

1�α� � � 15; x* is called long-run equilibrium.

144 MATHEMATICS

3GC08 05/15/2014 10:34:12 Page 146

Figure 8.1b illustrates α � �0:8; we observe that the solution oscillates and convergesto a limit x* � β

1�α� � � 31:8 � 1:6666. Figure 8.1c illustrates α � 1:2. We observe that

there is not limit; the solution is explosive. Figure 8.1d illustrates α � �1:2; we observethat the solution oscillates in an explosive manner and has no limit.

The Impulse Response Function

The impulse function is an important concept in difference equation theory. Thedifference equation

xn � αxn�1 � β n � 0; 1; 2 . . . (8.8)

says that xn is equal to a fraction α of xn�1 augmented by a constant β.Wemay date theterm β and rewrite it as βn. Accordingly, the difference equation is rewritten as

xn � αxn�1 � βn n � 0; 1; 2 . . . (8.9)

We solve for xn in a recursive manner; we replace xn�1 by its difference equationxn�1 � αxn�2 � βn�1 to obtain

xn � αxn�1 � βn � α αxn�2 � βn�1� � � βn

xn � α2xn�2 � βn � αβn�1

We may replace xn�2 � αxn�3 � βn�2; we obtain

xn � α3xn�3 � βn � αβn�1 � α2βn�2

If we continue the process of substitution until the initial time, we obtain

xn � αnx0 � βn � αβn�1 � α2βn�2 � ∙ ∙ ∙ � αn�1β1

The coefficient αn�1 is called a dynamic multiplier; it describes the impact of theshock β1 on xn. If we rewrite the difference equation as

xn�1 � αn�1x0 � βn�1 � αβn � α2βn�1 � ∙ ∙ ∙ � αnβ1 (8.10)

then αn is a dynamic multiplier that describes the impact of the shock β1 on xn�1. Wemay state that α describes the impact of β1 on x2, α2, the impact of β1 on x3, and so on.We call the sequence

1; α; α2; . . . ; αj; . . . ; αn� �

; j � 1; 2; . . . ; n (8.11)

the impulse response of xj to a shock β1, j � 1; . . . ; n. The impulse response functiondescribes the impact of single shock on successive values of the variable xj.

146 MATHEMATICS

3GC08 05/15/2014 10:34:13 Page 147

Example:We let xn � 0:9xn�1 � βn; we position ourselves in time 1. We assume ashock β1. Its impulse response function is 1; 0:9; 0:92; 0:93; . . . ;

� �. The impact of the

shock β1dies off as n ! ∞ .

The Cobweb Model

The cobweb model in economics illustrates patterns of solutions of a simple first-orderdifference equation. Farmers decide on the basis of this year’s price for a certaincommodity the acreage they will plant with that crop. Anticipating that the price levelwill be maintained, if the price is high one year, farmers tend to plant heavily. Thefollowing year, when the crop is harvested and brought to market, the supply exceedsthe demand, prices fall, and farmers cut acreage devoted to this particular crop. Whenthe next year’s crop is harvested, supply may be below demand, prices increase,farmers plant more, the next year’s crop exceeds demand, prices fall, and so on.

We study three functions defined for time t � 0; 1; 2; . . .. We let St be the numberof units supplied in period t,Dt be the number of units demanded in period t, and pt bethe price per unit in period t. We make the following three assumptions:

i. A price-demand relationship is specified in which quantity demanded is deter-mined by the price at the time of purchase

Dt � �βdpt � ωd and βd > 0; ωd > 0 (8.12)

ii. A price-supply relationship relates the supply in any period with the price oneperiod before

St�1 � βspt � ωs and βs > 0; ωs > 0 (8.13)

iii. The market price is determined by the available supply; pt is determined as thesolution of the equation

St � Dt (8.14)

Now suppose p0 is known. If the functionsDt and St are known, wemay calculateS1 from (ii) and D1 from (iii) and so obtain p1 from the price-demand curve. Theprocess may be repeated starting from p1 to obtain p2. The sequence of prices pt

� � �p0; p1; p2; . . . may have an oscillatory behavior. By hypothesis (iii), price is deter-mined by the equality of supply and demand, or, writing these conditions for periodt � 1,

St�1 � Dt�1βspt � ωs � �βdpt�1 � ωd

orpt�1 � Apt � B (8.15)


3GC08 05/15/2014 10:34:13 Page 148

where

A � � βsβd

< 0 and B � ωd � ωs

βd

we obtain a first-order difference equation in pt. Since A is negative, it follows thatthe sequence pt

� �is always oscillatory. If �1 < A < 0, oscillation is damped and

converges to the limit

p* � B= 1 � A� �

There are finite oscillations if A � �1, and infinite oscillations if A < �1. Since Ais the ratio of the slopes of the supply and demand curves, this ratio is seen todetermine the behavior of the price sequence.

The reason for calling this cobweb behavior becomes clear with graphicalanalysis. In Figure 8.2, we illustrate the three cases: �1 < A < 0, A � 1, andA < �1. Starting with p0 we find S1 by moving vertically to the supply line, thenmoving horizontally (since D1 � S1) to find D1, which determines the price p1 on theprice axis. The supply S2 is found on the supply line directly above p1, and againS2 � D2 is found by moving horizontally to the demand line, and so on; theintersection of the demand and supply lines determines the equilibrium price p*. InFigure 8.2a, we find a cobweb moving toward this equilibrium point, prices alternat-ing above and below, but converging to p*. In Figure 8.2b, the cobweb consists ofone square endlessly repeated, prices oscillating finitely between just two values. InFigure 8.2c, the cobweb moves away from the intersection point, prices oscillatinginfinitely about the equilibrium value p*. The equilibrium p* is stable only in the first ofthese three cases.

Example: Derive the cobweb model Dt � �3pt � 10, St�1 � pt � 2, and St�1 �Dt�1. Compute p*.

The cobweb model is

pt�1 � Apt � B

Quantity Dt, St+1

0

Dt

p1pt

p0p*

St+1

Price 0pt

p1 p0p*

Dt St+1

Quantity Dt, St+1

Price 0pt

p1 p0p*

DtSt+1

Quantity Dt, St+1

Price

a. Damped oscillations b. Finite oscillations c. Infinite oscillations

FIGURE 8.2 Cobweb Model of a Difference Equation

148 MATHEMATICS

3GC08 05/15/2014 10:34:14 Page 149

where A � � βsβd< 0 and B � ωd � ωs

βd. We have A � �1=3 and B � 10 � 2

3 � 8=3; we find

pt�1 � � 13pt � 8

3

since �1 < A < 0, p* � B1�A� � � 8=3

1� 13� 2.

SECOND-ORDER LINEAR DIFFERENCE EQUATIONS

This section describes solutions of homogenous and nonhomogenous second-orderdifference equations; it illustrates the multiplier accelerator model as an application ofsecond-order difference equations.

Homogeneous Second-Order Difference Equations

Second-order homogeneous linear equations are of the form

xn�2 � bxn�1 � cxn � 0 (8.16)

where b and c are real constants. We proceed by trial and assume the solution is of aform similar to the first-order homogeneous difference equation, namely,

xn � Crn (8.17)

where C and r are constants to be found. We replace the guess solution into thedifference equation

Crn�2 � bCrn�1 � cCrn � 0 (8.18)

We assume r ≠ 0; we simplify by Crn, and the equation reduces to

r2 � br � c � 0 (8.19)

The quadratic equation is called the auxiliary equation (or the characteristicequation) of the difference equation. If r is a number satisfying the auxiliary equation,Crn is a solution to the difference equation; r is called a root of the equation. Theauxiliary equation is a quadratic algebraic equation and therefore has two nonzeroroots, say r1 and r2. Zero roots are excluded because c ≠ 0 is a requirement that isneeded for the difference equation to be of the second order. To these roots therecorrespond the solutions

x 1� �n � C1rn1 and x 2� �

n � C2rn2 (8.20)

A general solution of the difference equation is given by

xn � x 1� �n � x 2� �

n � C1rn1 � C2rn2 (8.21)


3GC08 05/15/2014 10:34:14 Page 150

Example: Consider the homogeneous difference equation

xn�2 � 3xn�1 � 2xn � 0

It has an auxiliary equation,

r2 � 3r � 2 � 0

It has two roots r1 � 1 and r2 � 2; the roots are unequal. A general solution is ofthe form

xn � C1rn1 � C2rn2 � C11n � C22

n

When the auxiliary has two roots that are equal, that is, r1 � r2 � r, the generalsolution becomes of the form

xn � C1 � C2n� �rn

Nonhomogeneous Second-Order Difference Equations

A nonhomogeneous second-order difference equation is of the form

xn�2 � bxn�1 � cxn � f n� � (8.22)

where f n� � is a function of n.Having already found the general solution of the corresponding homogeneous

equation, if we add to it any solution of xn�2 � bxn�1 � cxn � f n� � the sum will be thegeneral solution of the complete equation. Hence,

yc � yh � yp (8.23)

whereyc = complete solution;yh = solution of the homogeneous part; andyp = particular solution of nonhomogeneous part,

A number of special techniques exist for finding particular solutions of non-homogeneous difference equations. The most useful of these is the method of theundetermined coefficients. The following examples illustrate these techniques.

Example: Consider the equation

xn�2 � 3xn�1 � 2xn � 3n

150 MATHEMATICS

3GC08 05/15/2014 10:34:15 Page 151

We try a solution of the form

x*n � A3n

The constant coefficient A is as yet undetermined. We seek a value of A, if oneexists, for which x* will be a solution of the difference equation. Now,

x*n�2 � 3x*n�1 � 2x*n � A3n�2 � 3A3n�1 � 2A3n

A3n 9 � 9 � 2� � � 2A3n � 3n

In order that x* satisfies the difference equation, we must choose A � 12. Then

x*n � 12 3

n is the desired particular solution and because the auxiliary equation r2 �3r � 2 � 0 has two distinct roots r1 � 1 and r2 � 2, the general solution is given by

xn � C1 � C22n � 123n

where C1 and C2 are arbitrary constants.Example: Find a particular solution for the equation

xn�2 � 3xn�1 � 2xn � an

where a is some constant. The previous example is a special case with a � 3. As in theprevious example we attempt a trial solution of the form Aan. We obtain

x*n�2 � 3x*n�1 � 2x*n � Aan�2 � 3Aan�1 � 2Aan

� Aan a2 � 3a � 1� � � Aan a � 1� � a � 2� � � an

Hence, we require that

A a � 1� � a � 2� � � 1

We can solve forA provided that a is not equal to 1 or 2, that is, provided that a isnot a root of the auxiliary equation. We have A � 1= a � 1� � a � 2� �. Thus if a is notequal to 1 or 2, the solution is

x*n � 1a � 1� � a � 2� � a

n

Separate considerations are required for a � 1 and a � 2. We modify the trialsolution by n; then we determine the coefficient A so that the new function givenby

x*n � Anan


3GC08 05/15/2014 10:34:15 Page 152

becomes a solution of xn�2 � 3xn�1 � 2xn � an. We perform the calculation in thecase a � 1, that is, the difference equation becomes

xn�2 � 3xn�1 � 2xn � 1

And the trial function is x*n � An. Now,

x*n�2 � 3x*n�1 � 2x*n � A n � 2� � � 3A n � 1� � � 2An � �A � 1

So the choice A � �1 yields a particular solution given by

x*n � �nThus, if a � 1, the general solution of xn�2 � 3xn�1 � 2xn � an is

xn � C1 � C22n � n

For a � 2, the difference equation is

xn�2 � 3xn�1 � 2xn � 2n

The trial function becomes x*n � An2n; the coefficientAmust be equal to 1=2. Thegeneral solution is

xn � C1 � C22n � n2n�1

If the function f n� � is of the form nk, the trial guess is of the form

A0 � A1n � A2n2 � . . . � Aknk (8.24)

If the function f n� � is of the form nkan, the trial guess is of the form

an A0 � A1n � A2n2 � . . . � Aknk

� (8.25)

Example: Find the general solution of

yn�2 � 4yn�1 � 4yn � 3n � 2n

The auxiliary equation r2 � 4r � 4 � 0 has the equal roots r1 � r2 � 2. Thegeneral solution of the homogeneous equation is therefore given by

yn � C1 � C2n� �2n

To this we must add a particular solution of the complete equation. The right-hand term 3n leads to the trial solution y*n � A0 � A1n; the term 2n leads to the trialsolution A2n. But 2n is a solution of the reduced equation. Therefore, we multiply

152 MATHEMATICS

3GC08 05/15/2014 10:34:16 Page 153

by n and try An2n, but because this, too, is a solution of the reduced equation wemust again multiply by n and so arrive at the trial solution An22n. Combiningterms, we attempt a trial solution of the form

y*n � A0 � A1n � An22n

To determine the coefficients A0;A1; andAwe perform the necessary calculationsto find

y*n�2 � 4y*n�1 � 4y*n � A0 � 2A1� � � A1n � 8A2n � 3n � 2n

This if y* is to satisfy the difference equation, wemust haveA0 � 2A1 � 0,A1 � 3,8A � 1.

OrA0 � 6,A1 � 3, andA � 1=8. The general solution of the complete equation istherefore given by

yn � C1 � C2n� �2n � 6 � 3n � n22n=8

The Multiplier-Accelerator Model

The multiplier-accelerator macroeconomic model illustrates an application of second-order difference equation. The model was designed to demonstrate how the interac-tion of the multiplier and the accelerator can generate cycles. The model is one of aclosed economy and consists of a consumption function, an investment functionwhich incorporates the accelerator idea, and the income identity:

Ct � aYt�1; 0 < a < 1 �Consumption�It � b Yt�1 � Yt�2� �; b > 0 �Investment�Yt � Ct � It �Gt �Income identity�

where Ct , It Yt, and Gt denote private consumption expenditure, investment expen-diture, income, and government consumption, respectively. We let Gt � 1. Theparameter a is called the marginal propensity of consumption and is assumed tobe between zero and one. The parameter b bears no restriction beyond having to bepositive. Inserting the consumption and the investment equation into the incomeidentity leads to the following nonhomogeneous second order difference equation

Yt � aYt�1 � b Yt�1 � Yt�2� � � 1

Algebraic simplification produces this difference equation for national income:

Yt � a � b� �Yt�1 � bYt�2 � 1 (8.26)

The pattern of Yt is influenced by Yt�1 and Yt�2 and by the values of theparameters a and b. The multiplier-accelerator model is very sensitive to the value


3GC08 05/15/2014 10:34:16 Page 154

of the parameters a and b. The higher these parameters are, the more explosive themodel becomes. In Figure 8.3, we illustrate two simulations of the model, taking initialconditionsY1 � 1 andY2 � 2. The first simulation is for a � 0:5 and b � 1; the secondsimulation is for a � 0:8 and b � 1:5. We note the pattern of national income is highlycyclical in the first simulation (Figure 8.3a) and is cyclical and highly explosive in thesecond simulation (Figure 8.3b).

SYSTEM OF LINEAR DIFFERENCE EQUATIONS

In economics and finance, often systems arise from interaction of variables. A countrymay decide to invest more in education not only in response to its current educationinvestment, but also in response to its competitor country’s high education investment.A country may decide to reduce its income tax not only in response to its presenthigh tax rate but also in response to a lower tax rate in competing countries.

0

0.5

1

1.5

2

2.5

3

3.5

4

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

–200

–100

0

100

200

300

400

500

600

700

800

900

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

Time

Yt

Yt

Time

Rea

l Ou

tpu

tR

eal O

utp

ut

a. Simulation α = 0.5 and b = 1

b. Simulation α = 0.8 and b = 1.5

FIGURE 8.3 The Multiplier-Accelerator Model

154 MATHEMATICS

3GC08 05/15/2014 10:34:16 Page 155

In econometrics, we deal with autoregressive models where a variable is explained byits past levels as well as by past levels of some relevant variables.

In Islamic finance, examples of interactive systems are numerous. For instance,there may be interactions between one Islamic bank and another when attractinginvestors. Within an Islamic bank, there may be interactions between types ofinvestments such as sukuks, equities, Musharaka, Murabaha, and so on. A linearhomogeneous system may be formulated as

xt�1 � axt � bytyt�1 � cxt � dyt

In matrix form, the system is written as

xt�1yt�1

�� a b

c d

�xtyt

�

or

ut�1 � Aut (8.27)

where ut�1 � xt�1yt�1

�, A � a b

c d

�, and ut � xt

yt

�:

The system ut�1 � Aut is a homogeneous system. An autonomous system is of theform

ut�1 � Aut � b (8.28)

where b is a vector of constants; time does not appear explicitly in the system. Thesolution of the system ut�1 � Aut is

ut � Atu0; t � 0; 1; 2; . . . (8.29)

where u0 � x0y0

�. The problem of solving the system of difference equations has now

been reduced to the problem of finding the tth power of matrixA. For withAt known,we can perform the matrix multiplication Atu0 and thus find ut � Atu0.

Unfortunately, the elements of the tth power of an arbitrary matrix, and inparticular matrix A, are not easy to find directly. But the powers of a diagonal matrixare very easy to obtain. For if

D � λ1 00 λ2

�; thenDt � λt1 0

0 λt2

�; t � 1; 2; 3; . . . (8.30)

In view of the ease with whichDt may be found, it is of some practical importanceto investigate the possibility of reducing the problem of calculating At to that ofcalculating Dt for some appropriate choice of the diagonal matrix D.


3GC08 05/15/2014 10:34:17 Page 156

We use the notions of eigenvectors and eigenvalues. Matrix A is diagonalizable ifit can be written in the form

A � VDV�1 (8.31)

where V is a matrix of eigenvectors and D is a diagonal matrix of eigenvalues. Itfollows that

At � VDtV�1 (8.32)

We let the eigenvector corresponding to λ1 be v1 � v11v21

�and the eigenvector

corresponding to λ2 be v2 � v12v22

�. We note that the solution is

ut � Atu0 � VDtV�1u0 (8.33)

We perform the product

V�1u0 � C (8.34)

The vector C � C1

C2

�is a vector of constants determined by the initial vector u0.

The solution can be written as

ut � VDtC � v11 v12v21 v22

�λt1 00 λt2

�C1

C2

�(8.35)

We find the general solutions

xt � C1v11λt1 � C2v12λt2yt � C1v21λt1 � C2v22λt2

�(8.36)

Example: We assume that the portfolio value of an Islamic bank, xt, follows thesecond-order difference equation

xt�2 � 3xt�1 � 2xt � 0; t � 0; 1; 2; . . .

with initial values x0 � 0 and x1 � 1. Find a solution for xt.This is a homogeneous difference equation with constant coefficients. Since the

auxiliary equation r2 � 3r � 2 � 0 has roots 1 and 2, the general solution isxt � C11t � C22t � C1 � C22t. Taking into consideration the initial values, we findC1 � �1 and C2 � 1.

156 MATHEMATICS

3GC08 05/15/2014 10:34:17 Page 157

We want to show that this second-order equation is equivalent to a system of twofirst-order equations and may be solved using the matrix method. We introduce a newfunction y, defined only for t � 1; 2; 3; . . . , in order to write the given differenceequation as a system of two first-order equations,

xt�1 � 3xt � 2yt

yt�1 � xt

The system may be written in matrix form as

ut�1 � Aut; t � 1; 2; 3; . . .

where A � 3 �21 0

�and ut � xt

yt

�:

The solution of the system is

ut�1 � Atu1; t � 0; 1; 2; 3; . . .

The vector u1 is known and corresponds to the initial values

u1 � x1y1

�� x1

x0

�� 1

0

�

The characteristic equation is defined by

det A � λI� � � 3 � λ �21 �λ

� λ2 � 3λ � 2 � λ � 1� � λ � 2� � � 0

So the characteristic roots are λ1 � 1 and λ2 � 2. The eigenvector associated withλ1 � 1 is

Av1 � λ1v1; or A � λ1I� �v1 � 0

where v1 � v11v21

�.

The equations for v1 are

2v11 � 2v21 � 0

v11 � v21 � 0

We find v1 � 11

�.


3GC08 05/15/2014 10:34:18 Page 158

The eigenvector associated with λ2 � 2 is

Av2 � λ2v2; or A � λ2I� �v2 � 0

where v2 � v12v22

�.

The equation for v2 is

v12 � 2v22 � 0

We find v2 � 21

�.

The matrix of eigenvectors is

V � 1 21 1

�

Its inverse is

V�1 � �1 21 �1

�

The matrix of eigenvalues is

D � 1 00 2

�

We compute At as At � VDtV�1. We find

At � �1 � 2t�1 2 � 2t�1�1 � 2t 2 � 2t

�

The solution is

ut�1 � Atu1 � �1 � 2t�1 2 � 2t�1�1 � 2t 2 � 2t

�10

�� 1 � 2t�1

�1 � 2t

�; t � 0; 1; 2; . . .

Since xt � yt�1 is the second component of this column vector, we find the samesolution as for the homogeneous second-order difference equation given byxt � �1 � 2t.

EQUILIBRIUM AND STABILITY

Islamic finance is known for its stability. The stability of a market or an economy is ofparamount importance. We want to know whether the economy or the market will beable to return to long-term equilibrium following an exogenous shock. If the economy

158 MATHEMATICS

3GC08 05/15/2014 10:34:18 Page 159

deviates from its long-run equilibrium and cannot return to this equilibrium, then it isunstable. Stability conditions are necessary for a market to clear and establishequilibrium. If an economy cannot restore full employment of labor following arecession, then it is unstable. A financial system that undergoes bubbles and crashes isbasically unstable and oscillates between booms and crashes. It is always in cycles.

This section establishes conditions for stability of difference equations andstability of a system of difference equations; it describes phase planes as tools foranalyzing stability of a system of difference equations.

Conditions for Stability

Let us consider the first-order linear difference equation

xt � αxt�1 � b (8.37)

where α and b are both constants. The steady state equilibrium is defined as a statewhere xt remains invariant as time evolves; that is, xt � xt�1 � x*. From the differenceequation x* satisfies the condition

x* � αx* � b; or x* � b1 � α� �

The solution to the difference equation is

xt � αtx0 � b 1 � α � α2 � ∙ ∙ ∙ � αt�1� � � αtx0 � b

1 � αt

1 � α

� �(8.38)

The solution will converge to long-run equilibrium only if αj j < 1. If αj j > 1, thenthe solution will diverge to infinity. Figure 8.4a illustrates stable steady state equi-librium with α < 1, and Figure 8.4b illustrates unstable steady state equilibrium withα > 1.

Stability analysis is also conducted in the neighborhood of the steady-stateequilibrium x* by computing deviations from this equilibrium and examining whether

a. Stable steady state b. Unstable steady state

0

xt+1

xt+1 = αxt + b

E

45°xt

x*

x* 0

xt+1 = αxt + b

xt+1

E

xt

x*

x*

FIGURE 8.4 Stability of a First Difference Equation


3GC08 05/15/2014 10:34:19 Page 160

the variable xt once displaced from x* will converge or diverge away from x*. Thesteady state equilibrium is expressed as

x* � αx* � b (8.39)

The difference equation is xt � αxt�1 � b. The deviation from equilibrium x* iscomputed as the difference of these two equations

xt � x* � αxt�1 � b� � � αx* � b� � � α xt�1 � x*� �xt � x* � α xt�1 � x*� � (8.40)

If α < 1, the deviation xt � x* narrows at each step and xt converges back to x*following a displacement from this equilibrium. In contrast, if α > 1, the deviationxt � x* widens at each step and xt diverges away from x* following a displacementfrom this equilibrium.

From the preceding analysis, the nature of the solution of a linear differenceequation with constant coefficients, especially its limiting behavior, is dependent onboth the initial values prescribed for the solution and the roots of the auxiliaryequation. The same principle applies for the second-order homogeneous equation

xt�2 � bxt�1 � cxt � 0 (8.41)

The steady state equilibrium x* verifies

x* � bx* � cx* � 0 (8.42)

Hence, x* � 0 is the steady-state equilibrium. A necessary and sufficient conditionfor the solution to converge to 0 independently of the initial values x0 and x1 requiresthat both roots of the auxiliary equation r2 � br � c � 0 be less than 1 in absolutevalue. This is a key condition for the stability of a difference equation. Consider thecomplete difference equation

xt�2 � bxt�1 � cxt � A (8.43)

where A is a constant. If this equation has a constant function as a solution, then thevalue of this function is called an equilibrium (or stationary) value of x. Puttingxt � x*, a constant, in the equation, we find

x* � bx* � cx* � A (8.44)

Hence, if 1 � b � c ≠ 0, then x* � A1�b�c is an equilibrium value of x. The

equilibrium is said to be stable, or the difference equation is stable, if every solutionof the equation, independently of the initial values x0 and x1, converges to x*, that is, iflimt! ∞ xt � x*, for all x0 and x1.

Because a displacement from the equilibrium value is equivalent to considering anew solution with different initial conditions, we may alternatively define a stableequilibrium as one for which any displacement from equilibrium is followed by a

160 MATHEMATICS

3GC08 05/15/2014 10:34:19 Page 161

sequence of values of x, which again converges to equilibrium. It is convenient todefine a new function z, which measures the deviation of a solution x from itsequilibrium value x*; that is, we let

zt � xt � x* (8.45)

Since x is a solution, we have

zt�2 � bzt�1 � czt � xt�2 � bxt�1 � cxt� � � 1 � b � c� �x*� xt�2 � bxt�1 � cxt� � � A � 0 (8.46)

Hence, z is a solution of the homogeneous difference equation

zt�2 � bzt�1 � czt � 0 (8.47)

The definition of stability requires that a deviation of x from its equilibrium valueconverges to 0 for every pair of initial values z0 and z1. The auxiliary equation of thehomogeneous equation is

r2 � br � c � 0 (8.48)

A necessary and sufficient condition for the equilibrium value x* to be stable isρ < 1, where ρ � max r1j j; r2j j� � and r1 and r2 are the roots of the auxiliary equation.

Stability of the Linear Difference System

We analyze the stability of the linear system

xt � axt�1 � byt�1yt � cxt�1 � dyt�1

�(8.49)

We write the system in a matrix form as Xt � AXt�1, where Xt � xtyt

�and

A � a bc d

�.

If we have a system of nonhomogeneous differential equations

xt � axt�1 � byt�1 � B1

yt � cxt�1 � dyt�1 � B2

�(8.50)

whereB1 andB1 are real constant, then the system can be reduced into a homogeneoussystem by taking deviations from the steady-state equilibrium. More specifically, thesystem is written as

ut � Aut�1 � B (8.51)


3GC08 05/15/2014 10:34:20 Page 162

We let u* be the equilibrium vector; then we have

u* � Au* � B; or u* � I � A� ��1B (8.52)

We let ut be a solution to ut � Aut�1 � B; its deviation from u* is zt � ut � u*; thenwe may write

zt � Aut�1 � B � Au* � B � A ut�1 � u*� � � Azt�1 (8.53)

Hence, we have transformed the nonhomogeneous system in ut into a homoge-neous system in zt.

The characteristic equation of the system is

λ2 � trace A� �λ � det A� � � 0 (8.54)

The solution has two roots, λ1 and λ2. To understand the qualitative behavior ofut, we distinguish two cases:

1. λ1 and λ2 are real and distinct; and2. λ1 and λ2 are real and equal.

When the roots are real and distinct, the general solution is given by

ut � C1λt1 � C2λt2 � λt1 C1 � C2λ1λ2

� �t �(8.55)

Suppose without loss of generality that λ1j j > λ2j j so that λ1λ2

� t ! 0 as t ! ∞ . This

implies that the behavior of ut is asymptotically governed by the larger root λ1:

limt! ∞

ut � limt! ∞

C1λt1 (8.56)

Depending on the values of λ1, the following cases emerge:

λ1 > 1; C1λt1 diverges to ∞ as t ! ∞ ; the system is unstable.λ1 � 0; C1λt1 remains constant and equal to C1.λ1 < 1; C1λt1 converges to 0 as t ! ∞ ; the system is stable.�1 < λ1 < 0; C1λt1 oscillates around zero, alternating in sign, but converges

to zero.λ1 � �1; C1λt1 alternates between the values C1 and �C1.λ1 < �1; C1λt1 alternates in sign, but diverges in absolute value to ∞ .

In the case of λ � λ1 � λ2, the solution is given by

ut � C1 � C2t� �λt (8.57)

Clearly, if λ � 1, ut diverges monotonically; or, if λ � �1, ut diverges by alter-nating signs. For λ < 1, the solution converges to zero, because limt! ∞ tλt � 0.

162 MATHEMATICS

3GC08 05/15/2014 10:34:21 Page 163

Phase Plane

Phase diagrams offer a graphical device for studying the qualitative behavior of thesolution trajectories and their convergence or divergence away from equilibrium. Westudy the phase plane of a system of two first-order linear homogeneous differentialequations. By definition the phase line corresponds to a situation where Δxt � xt �xt�1 � 0 and Δyt � yt � yt�1 � 0. These conditions require that

xt � xt�1 � axt�1 � byt�1 � xt�1 � 0yt � yt�1 � cxt�1 � dyt�1 � yt�1 � 0

�(8.58)

The phase lines are therefore

a � 1� �xt�1 � byt�1 � 0cxt�1 � d � 1� �yt�1 � 0

�(8.59)

We study the sign of Δxt on each side of the phase line. If Δxt > 0, then xt isincreasing as time evolves. In contrast, ifΔxt < 0, then xt is decreasing as time evolves.Likewise, we analyze the sign ofΔyt on each side of the phase line. IfΔyt > 0, then yt isincreasing as time evolves. In contrast, ifΔyt < 0, then yt is decreasing as time evolves.

Example: TheMusharaka portfolio, xt, andMurabaha portfolio, yt, of an Islamicbank are related by the following system

xt � xt�1 � yt�1yt � 0:1xt�1 � 0:3yt�1

Analyze the stability of this system.The phase lines are

Δxt � �yt�1 � 0

Δyt � 0:1xt�1 � 0:7yt�1 � 0

We display the phase lines in Figure 8.5. The phase line Δxt � �yt�1 � 0 happensto be the x-axis. Above the x-axis, �yt�1 < 0 and therefore x is decreasing; below thex-axis, �yt�1 > 0 and therefore x is increasing. Above the phase line Δyt � 0,0:1xt�1 � 0:7yt�1 < 0; therefore y is decreasing. Below the phase line Δyt � 0,0:1xt�1 � 0:7yt�1 > 0; therefore y is increasing. The phase lines divide the planeinto four quadrants, I, II, III, and IV. In each quadrant there is an arrow pointing to theorigin; this means that any trajectory will arrive at the origin, however in a counter-clockwise direction.

We compute the eigenvalues and eigenvectors associated with the system. Thematrix of the system is

A � 1 �10:1 0:3

�

The characteristic equation is det A � λI� � � 1 � λ� � 0:3 � λ� � � 0:1 � λ2 � 1:3λ�0:4 � 0.


3GC08 05/15/2014 10:34:21 Page 164

The roots are λ1 � 0:8 and λ2 � 0:5. The eigenvector associated with λ1 � 0:8

satisfies the relation Av1 � 0:8v1; we find v1 � 15

�; the eigenvector associated with

λ2 � 0:5 satisfies the relationAv2 � 0:5v2; we find v2 � 12

�. A general solution of the

system isxt � C10:8

t � C20:5t

yt � 5C10:8t � 2C20:5

t

The solution is dominated by the larger root λ1 � 0:8 < 1. Both xt and yt willconverge to zero, irrespective of the initial values.

Example: We consider the following system:

xt � 2xt�1 � 2:1yt�1yt � 2xt�1 � 2:7yt�1

The phase lines (Figure 8.6) are

Δxt � xt�1 � 2:1yt�1 � 0Δyt � xt�1 � 3:7yt�1 � 0

We compute the eigenvalues and eigenvectors associated with the system. Thematrix of the system is

A � 2 �2:12 �2:7

�

The characteristic equation is det A � λI� � � 2 � λ� � �2:7 � λ� � � 4:4 � λ2 � 0:7λ�1:2 � 0:

I

II

III

IV

0

yt–1

xt–1

xt = xt–1 –yt–1yt = 0.1xt–1 + 0.3yt–1

Δyt < 0

Δxt < 0

Δyt > 0

Δxt > 0Δxt = 0

Δyt = 0

FIGURE 8.5 Phase Diagram

164 MATHEMATICS

3GC08 05/15/2014 10:34:22 Page 165

The roots are λ1 � �1:5 and λ2 � 0:8. An eigenvector associated with λ1 � �1:5 isv1 � 0:6

1

�; an eigenvector associated with λ2 � 0:8 is v2 � 7

4

�. A general solution

for the system is

xt � 0:6C1 �1:5� �t � 7C20:8t

yt � C1 �1:5� �t � 4C20:8t

The solution is dominated by the larger root λ1j j � 1:5 > 1. Both xt and yt willconverge to infinity in oscillatory manner.

SUMMARY

The chapter defines difference equations and describes solution methods of the first-order difference equations, the impulse response function, and the cobweb model. Italso covers second-order homogenous and nonhomogenous linear difference equa-tions, systems of linear difference equations, and stability conditions of a lineardifference equations system.

Islamic finance uses difference equations as a tool of analysis. Difference equa-tions arise in finance, econometrics, and economics and describe a variable in terms ofits lagged values. For instance, inflation dynamics depend on past price development.Islamic stock indices are propelled by recent market momentum. A financial variableinteracts within a system with other relevant variables. In this case, Islamic stockindices may interact with returns on sukuks or on commodities, or both.

QUESTIONS

1. A depositor has an amount of $1,000 at the bank; the rate of inflation is 5 percentper year. Provide a difference equation for the real value of the deposit. Using

I

II

III

IV

xt = 2xt–1 – 2.1yt–1

yt = 2xt–1 – 2.7yt–1

Δxt < 0

Δyt < 0

Δyt > 0

Δxt = 0

Δyt = 0

xt–1

0

Δxt > 0

yt–1

FIGURE 8.6 Phase Diagram


3GC08 05/15/2014 10:34:23 Page 166

Microsoft Excel, plot the graph of the difference equation. Compute the real valueof the deposit 20 years from the initial time.

2. Consider the difference equation yn�2 � yn � 0; show that yn � C1 � C2 �1� �n is asolution. C1 and C2 are arbitrary real numbers. Find C1 and C2 that satisfy theinitial conditions y0 � 1 and y1 � 2.

3. Using Microsoft Excel, display the pattern of the difference equation yn �αyn�1 � 5, n � 0; 1; . . . ; 50: Select α � 0:7, α � �0:7, α � 1:5, and α � �1:5.Compute the steady-state equilibrium x* for α � 0:7 and α � �0:7.

4. Consider the difference equation yn � �0:85yn�1 � βn; compute the impulseresponse function. Using Microsoft Excel, draw its diagram.

5. Derive the cobweb model for:

(a) Dt � �4pt � 8; St�1 � pt � 2; and St�1 � Dt�1.(b) Dt � �4pt � 25; St�1 � 4pt � 3; and St�1 � Dt�1.(c) Dt � �2:5pt � 25; St�1 � 7:5pt � 5; and St�1 � Dt�1.Using Microsoft Excel, draw the graph for each cobweb model. Compute p*.

6. Solve the second-order difference equation:

(a) yn�2 � 2yn�1 � 4yn � 1 with initial conditions y0 � 0 and y1 � 1.(b) yn�2 � 2yn�1 � 12yn � 2 with initial conditions y0 � 0 and y1 � 1.(c) 2yn�2 � 3yn�1 � 1yn � 3n � 1 with initial conditions y0 � 0 and y1 � 1.

7. Find the general solution of

yn�2 � 4yn�1 � 4yn � n � 2n

8. Solve the following second-order difference equations by replacing each with asystem of two first-order equations.

(a) yn�2 � 2yn�1 � 3yn � 0, t � 1; 2; 3; . . . ; y0 � 0 and y1 � 1.(b) yn�2 � yn�1 � 1yn � 0, t � 1; 2; 3; . . . ; y0 � 1 and y1 � 0.

9. Find the diagonal matrix corresponding to A � 2 �13 6

�.

10. Consider the systemxt � xt�1 � yt�1 � 1yt � 0:1xt�1 � 0:3yt�1 � 2

Compute the long-run equilibrium.

11. Consider the system

xt � xt�1 � yt�1yt � 0:03xt�1 � 0:6yt�1

with x0 � 5, and y0 � 2.

(a) Using Microsoft Excel, compute recursively the system for t � 1; 2; . . . ; 50.Plot the trajectory for xt and yt.

(b) Plot the phase diagram of the system.(c) Compute the solution of the system, taking into account the initial values.

166 MATHEMATICS

3GC09 05/15/2014 10:44:4 Page 167

CHAPTER 9Optimization Theory

I slamic finance is concerned with maximum and minimum problems known asoptimization problems. Resources are costly and scarce, with pressing priorities.

Optimization reconciles resource scarcity with pressing priorities. Modern “neo-classical marginalism” represents the culmination of optimization theory in econom-ics and finance. Value theory in economics derives demand functions from consumers’optimizing behavior and supply functions from producers’ optimizing behavior. Thetheory of international trade and comparative advantage is also based on theoptimization principle. Optimal allocation of resources requires that each countryspecializes in a product for which it has a comparative advantage; trade takes placewhen a country exports its product and imports the product of its trading counterpart.Optimization aims at achieving efficient use of resources—maximizing profits, max-imizing utility, exploiting price arbitrage opportunities, and enhancing economicgrowth. Firms or countries that fail to optimize the use of resources and achieveefficiency fall into decay and have problems. This is the case in countries with debtcrises, inflation, and stagnation.

Optimization is a key element of Islamic business management. Firms makeproduction, hiring, and investment plans based on the optimization principle.They try to save on expensive resources. For instance, if labor is expensive andhighly unionized, a company may resort to mechanization and capital-intensivetechniques to reduce the cost of labor. Transport companies have network modelsand optimize shipments between destinations according to importance and cost oftraffic.

In Islamic finance, optimization is a principle of portfolio diversification; aninvestor may select a portfolio that minimizes risk for a given return or maximizesreturns for a given risk.

Optimization underlines the selection of infrastructure projects by state and localgovernments. Capital resources are invested in projects with the highest social returns.

Optimization may have a dynamic aspect. For instance, workers can optimizetheir life-time consumption stream. They contribute to a pension fund by savingpart of their income during their youth to be able to consume during the retirementage when they are no longer able to work. A country optimizes over time. Forinstance, it may predict that population will grow rapidly and that it needs to startconstructing more water and power facilities, more transport infrastructure, andmore urban development to avoid future congestion. Optimal growth theory seeks

167

3GC09 05/15/2014 10:44:4 Page 168

the appropriate saving and capital accumulation path that will achieve higherincome and consumption in the future. Dynamic optimization relies on the princi-ples of static optimization.

Optimization may be unconstrained or constrained. Unconstrained optimizationmeans the absence of constraints. By definition, the optimum will be far better thanwhen there are constraints. The more constraints there are, the smaller the opportu-nity set becomes and the poorer the optimum becomes. Some constraints may benatural, such as limited land for agriculture or urban development. Other constraintsmay be imposed by legislation. For instance, a firm that faces many regulatory andinstitutional constraints, such as high taxes, rigid wages, or strong labor unions, willhave poorer performance andmay consider relocating to a country that imposes fewerconstraints. Similarly, a firm that faces poor labor qualifications and underdevelopedinfrastructure will have constraints that will hurt its performance. Many countriesface constraints that weaken economic growth. For example, governments that refuseto cut extra spending or the size of the civil service consume resources that could beredeployed to agriculture and industry.

In this chapter, we deal with static optimization. Optimization may be nonlinear(nonlinear programming) or linear (linear programming). In the former the objectivefunction or the constraint function, or both, is nonlinear. In the latter, all functions arelinear.

We formulate the principle of optimization by formulating the objective functionas well as the opportunities set. The latter is determined by prevailing resourceconstraints. We also discuss general principles of optimal solutions under no con-straint and under constraint.

THE MATHEMATICAL PROGRAMMING PROBLEM

In this section, we formulate the programming problem in terms of the objectivefunction, the constraints function, and the control instruments.Wewill distinguish thethree types of optimization problems: classical programming, nonlinear program-ming, and linear programming. And finally we’ll describe the geometry of theoptimization problem in terms of opportunity set and contours.

Formulation of the Programming Problem

Resources are scarce. There are many competing ends for using resources. Mathe-matically, optimization aims at selecting values of instrument variables, subject to adefined set of constraints, so as to maximize an objective function. A formal statementof the mathematical problem includes instruments, opportunity set, and an objectivefunction. The problem is choosing values for n variables x1; x2; . . . ; xn calledinstruments. The instruments are summarized by a column vector x, called theinstrument vector, x ∈ Rn:

x �x1. . .

xn

24

35 � x1; . . . ; xn� �´ (9.1)

168 MATHEMATICS

3GC09 05/15/2014 10:44:5 Page 169

The opportunity set X is a subset of Rn defined by m constraints,

g1 x� � � g1 x1; . . . ; xn� � � b1g2 x� � � g2 x1; . . . ; xn� � � b2

gm x� � � g1 x1; . . . ; xn� � � bm (9.2)

where the functions g1 x� �; . . . ; gm x� � arem given continuously differentiable functionsof the instruments, called constraint functions, and the parameters b1; . . . ; bm, calledconstraint constants, which are generally resource constraints. In vector form, theconstraints can be written as

g x� � � b (9.3)

where g x� � and b are m-dimensional column vectors. An instrument is feasible if itsatisfies all the constraints of the problem, and the set of all feasible vectors is theopportunity set X, a subset of Rn, that is,

X � x ∈ Rn : g x� � � bf g (9.4)

The objective function, assumed given and continuously differentiable, summa-rizes the objective of the problem. It is a scalar real-valued function of the instruments

F x� � � F x1; . . . ; xn� � (9.5)

The programming problem is that of choosing an instrument vector x from theopportunity set so as to maximize the value of the objective function

maxx F x� � subject to g x� � � b (9.6)

We distinguish three types of optimization problems:

1. Classical programming2. Nonlinear programming3. Linear programming

In classical programming the constraints are of the equality type, consisting ofm equalities. Thus classical programming is that of maximizing a given objectivefunction subject to given equality constraints


In nonlinear programming the constraints are of two types:

1. Nonnegativity constraints:

x1 � 0; . . . ; xn � 0 (9.8)

Optimization Theory 169

3GC09 05/15/2014 10:44:5 Page 170

2. Inequality constraints:

g x� � � b (9.9)

Thus the nonlinear programming problem is that of maximizing a given functionby choice of non-negative variables subject to inequality constraints

maxx F x� � subject to g x� � � b; x � 0 (9.10)

In linear programming the objective function is the linear form

F x� � � c1x1 � ∙ ∙ ∙ � cnxn � cx (9.11)

where c is the row vector of n given constants

c � c1; . . . ; cn� � (9.12)

The constraints are two types: linear inequality constraints and non-negativityconstraints.

1. Non-negativity constraints:

x1 � 0; . . . ; xn � 0 (9.13)

2. Inequality constraints:

a11x1 � ∙ ∙ ∙ : � a1nxn � b1a21x1 � ∙ ∙ ∙ : � a2nxn � b2

am1x1 � ∙ ∙ ∙ : � amnxn � bm (9.14)

In vector form the constraints can be written as

Ax � b; x � 0 (9.15)

where A is the given m � n m � n� � matrix:

A �a11 . . . . . . a1n. . . . . . . . . . . .

am1 . . . . . . amn

24

35 (9.16)

Thus the linear programming problem is that of maximizing a given linear formby choice of non-negative variables subject to linear inequality constraints

maxx F x� � � cx subject toAx � b; x � 0 (9.17)

170 MATHEMATICS

3GC09 05/15/2014 10:44:5 Page 171

The linear programming problem is thus a special case of the nonlinear pro-gramming for which the objective function and the constraints functions are all linear:

g x� � � b (9.18)

Thus the nonlinear programming problem is that of maximizing a given functionby choice of non-negative variables subject to inequality constraints

maxxF x� � subject to g x� � � b; x � 0 (9.19)

The Geometry of Optimization

In this section we illustrate the geometry of the optimization problem. We define thenotions of contours (i.e., level curves) and gradients.

A contour of the objective function is the set of points in Rn for which the value ofthe objective function is constant

x ∈ Rn : F x� � � C � constantf g (9.20)

In Figure 9.1a, we show contours of F x� � denoted by C1; C2; C3; andC4 withC1 < C2 < C3 < C4. These contours may represent utility indifference curves orproduction isoquants. In Figure 9.1b, we show contours of F x� � denoted byC1 andC2 with C1 < C2. These contours may describe the altitude of a mountain.

The preference direction is the direction in which the value of the objectivefunction, the constant C, is increasing fastest. This preference direction is given by thedirection of the gradient vector of first-order derivatives of the objective function

rF � @F@x

x� � � @F@x1

x� �; . . . ; @F@xn

x� ��

(9.21)

0

a. b.

C4

∂F/∂x

∂F/∂x

∂F/∂x

C3C2

C2

C1

C1

x2 x2

x1 x10

FIGURE 9.1 Contours and Gradients of the Objective Function


3GC09 05/15/2014 10:44:7 Page 172

In Figure 9.2, we represent examples of convex opportunity sets. In Figure 9.2a,pointE1 is not optimal; the objective function is notmaximized. PointE2 is not feasible;it lies outside the opportunity set. PointE* is an optimumpoint given by the tangency ofthe opportunity set contour with the objective function contour. It is feasible andsuperior to any other point within the opportunity set. In Figure 9.2b, the optimum isobtained at a boundary point of the opportunity set corresponding to the highestcontour that makes contact with the opportunity set. Figure 9.2c and Figure 9.2d showthe geometry of linear programming. The opportunity set and the objective function aredefined by linear relations. In Figure 9.2c, the optimum is given by a unique vertexpoint; however, in Figure 9.2d, the solution set is given by a line, implying manysolutions.

Figure 9.3 describes the types of solutions of an optimization problem. Theopportunity set X is shown by a dashed line. In Figure 9.3a, we have local interior

d. Bounding face solution

∂x∂F

c

C3

C2C10

x2

x1

X

a. Tangent solution E*

0

∂F/∂x

C3

E2

C2

C1

x2

x*2

x*1 x1

E1

X

E*

b. Boundary solution E*

∂F/∂x

C3

C2

C1

0

x2

x*2

x*1 x1

X

E*

c. Vertex solution E*

∂x∂F

c

C3C2

C1

0

x2

x*2

x*1 x1

X

E*

FIGURE 9.2 Opportunity Sets, Contours, and Optimum Solutions

b. Boundary solution c. Global maximum

Opportunity set = X

a. Interior solutions

F(x)

x*X

x**

F(x)

Xx*

F(x)

Xx*

FIGURE 9.3 Types of Solutions

172 MATHEMATICS

3GC09 05/15/2014 10:44:8 Page 173

solutions. In Figure 9.3b, we have a boundary solution. In Figure 9.3c, we have aglobal maximum solution.

UNCONSTRAINED OPTIMIZATION

Unconstrained optimization consists of finding optimal values of a scalar functionsuch as y � F x� �, or z � F x; y� �, or v � F x1; x2; . . . ; xn� �. We assume that F iscontinuous and differentiable. The function y � F x� � is represented by a curve inthe plane R2, and the function z � F x; y� � is represented by a surface in the three-dimensional space R3, whereas v � F x1; x2; . . . ; xn� � is represented by a hypersurfacein Rn�1.

One Variable Function y = F (x)

We consider the Taylor expansion of the function y � F x� � in the neighborhood of apoint x*,

F x� � � F x*� � � dF x*� �dx

x � x*� � � 12!d2F x*� �dx2

x � x*� �2 (9.22)

or equivalently,

F x* � h� � � F x*� � � dF x*� �dx

h � 12!d2F x*� �dx2

h2 (9.23)

If x* is a local maximum, then @F x*� �@x � 0 (Figure 9.4a). The Taylor approximation

becomes

F x� � � F x*� � � 12!d2F x*� �dx2

x � x*� �2 (9.24)

a. Maximum F(x*)

b. Minimum F(x*)

c. Saddle point F(x*)

Tangentline

0

F(x)

F(x*)

x* x

Tangentline

0

F(x*)

x* x

F(x)

Tangentline

0 x* x

F(x)

F(x*)

FIGURE 9.4 Optimums of a Smooth Function


3GC09 05/15/2014 10:44:8 Page 174

This implies that

F x� � � F x*� � andd2F x*� �dx2

� 0 (9.25)

If x* is a local minimum, then dF x*� �dx � 0 (Figure 9.4b). The Taylor approximation

becomes

F x� � � F x*� � � 12!d2F x*� �dx2

x � x*� �2 (9.26)

which implies that

F x� � � F x*� � andd2F x*� �dx2

� 0 (9.27)

If @F x*� �@x � 0 and d2F x*� �

dx2 ≷0, we have a saddle point (Figure 9.4c); the function isincreasing from one direction and decreasing from an opposite direction. This is incontrast to a local minimumwhere the function is decreasing from any direction to theminimum point or a local maximum where the function is increasing from anydirection to the maximum point.

Example:

i. The function y � x2 � 2x has a minimum at x � 1 since dydx � 2x � 2 � 0, and

d2Fdx2 � 2 > 0.

ii. The function y � �x2 � 2x has a maximum at x � �1 since dydx � �2x � 2 vanishes

at x � �1 and d2Fdx2 � �2 < 0.

iii. The function y � x3 � 27 has a saddle point at x � 0 since dydx � 3x2 vanishes at

x � 0 and d2Fdx2 � 6x≷0.

Example:You want to invest in two Islamic mutual fundsA and B, in proportionsx1 and x2, respectively, with x1 � x2 � 1 and x1 � 0; x2 � 0. The risks of mutualfunds A and B are σ1 � 9% and σ2 � 14%, respectively; the correlation coefficientbetween expected returns is ρ � 0:6. The portfolio variance is

V x1; x2� � � σ21x21 � 2ρσ1σ2x1x2 � σ22x

22 � 81x21 � 151:2x1x2 � 196x22

Assume you want a diversified portfolio with minimum risk. Using the MicrosoftExcel solver, find the composition of this portfolio.

Since x2 � 1 � x1, we may express the portfolio variance as

H x1� � � 81x21 � 151:2x1x2 � 196x22 � 81x21 � 151:2x1 1 � x1� � � 196 1 � x1� �2� 81x21 � 1 � x1� � 151:2x1 � 196 1 � x1� �� 81x21 � 1 � x1� � 196 � 44:8x1� �� 125:8x21 � 240:8x1 � 196:

174 MATHEMATICS

3GC09 05/15/2014 10:44:9 Page 175

We set dHdx1

� 0, 251:6x1 � 240:8 � 0, x1 � 240:8251:5 � 0:957, and x2 � 1 � 0:957 �

0:0429.

Function of Two Variables z = F (x,y)

We consider a function of two variables, z � F x; y� �. This function is represented by asurface in R3. A Taylor expansion at a point P x*; y*� � is

F�x; y� � F�x*; y*� � @F�x*; y*�@x

�x � x*� � @F�x*; y*�@y

�y � y*� � 12!

@2F�x*; y*�@x2

�x � x*�2�

�2@2F�x*; y*�

@x@y�x � x*��y � y*� � @2F�x*; y*�

@y2�y � y*�2

�(9.28)

An equivalent form for this expansion is

F x* � h; y* � k� � � F x*; y*� � � hFx x*; y*� � � kFy x*; y*� �� 12!

h2Fxx x*; y*� � � 2hkFxy x*; y*� � � k2Fyy x*; y*� �h i

(9.29)

where Fx � @F@x, Fy � @F

@y, Fxx � @2F@x2, Fyy � @2F

@y2, and Fxy � @2F@x@y :

We define the gradient of F, denoted asrF as the vector of first partial derivatives,

rF � Fx

Fy

� �(9.30)

The gradient at a point P x*; y*� � is expressed as

rF x*; y*� � � Fx x*; y*� �Fy x*; y*� ��

(9.31)

We define the Hessian H of F as the matrix of second partial derivatives,

H � Fxx FxyFyx Fyy

� �(9.32)

The discriminant of the Hessian is defined as

D � Fxx � Fyy � Fxy� �2

(9.33)

The Hessian at a point P x*; y*� � is expressed as

H � Fxx x*; y*� � Fxy x*; y*� �Fyx x*; y*� � Fyy x*; y*� ��

(9.34)


3GC09 05/15/2014 10:44:10 Page 176

Critical points, called also stationary points, on the surface z � F x; y� � are definedas points where

Fx � @F@x

� 0 and Fy � @F@y

� 0 (9.35)

The tangent plane at a critical point is horizontal. Critical points may be classifiedas local minima, maxima, or saddle points.

Definitions

i. A function z � F x; y� � has a local minimum at the point x*; y*� � if F x; y� � �F x*; y*� � for all points x; y� � in some region around x*; y*� �.

ii. A function z � F x; y� � has a local maximum at the point x*; y*� � if F x; y� � �F x*; y*� � for all points x; y� � in some region around x*; y*� �.

iii. A function z � F x; y� � has a local saddle at the point x*; y*� � if F x; y� �≷F x*; y*� �for all points x; y� � in some region around x*; y*� �. The function is increasing fromone direction to x*; y*� � and decreasing from the opposite.

To identify local maximum points, local minimum points, and saddle points ofz � F x; y� �, we use the second partial derivatives. Suppose that x*; y*� � is a criticalpoint of F x; y� � and that the second-order partial derivatives are continuous in someregion that contains x*; y*� �. We then have the following classifications of the criticalpoint:

If D > 0 and Fxx x*; y*� � > 0, then there is a local minimum at x*; y*� �.If D > 0 and Fxx x*; y*� � < 0, then there is a local maximum at x*; y*� �.If D < 0, then the point x*; y*� � is a saddle point.IfD � 0, then thepoint x*; y*� �maybea localminimum, localmaximumora saddle

point. Other techniques would need to be used to classify the critical point.

Example: Find and classify all the critical points of F x; y� � � 4 � x3 � y3 � 3xy.We need all the first-order conditions to find the critical points and the second-

order partial derivatives to classify the critical points. We compute first-order partialderivatives; critical points will be solutions to the system of equations,

Fx � 3x2 � 3y � 0Fy � 3y2 � 3x � 0

We can solve the first equation for y; we find y � x2. Plugging this into the secondequation gives

3x4 � 3x � 3x x3 � 1� � � 0

We find x � 0 and x � 1. The critical points are A � 0; 0� � and B � 1; 1� �. So, wehave two critical points. All we need to do now is classify them. To do this, we willneed the discriminant D,

D � Fxx � Fyy � Fxy� �2

176 MATHEMATICS

3GC09 05/15/2014 10:44:11 Page 177

Fxx � 6x, Fyy � 6y, and Fxy � �3. Plugging these values in the formula for D,we find

D � 36xy � 9

Evaluating D at A � 0; 0� �, we find D 0; 0� � � �9 < 0. Hence A is a saddle point.EvaluatingD atB � 1; 1� �, we findD 1; 1� � � 36 � 9 � 27 > 0. Further, Fxx > 0, pointB � 1; 1� � is a local minimum.

Example: Find and classify all the critical points for F x; y� � � 3x2y � y3�3x2 � 3y2 � 2.

To compute the critical points we need first-order partial derivatives

Fx � 6xy � 6x � 0

Fy � 3x2 � 3y2 � 6y � 0

From the first equation, we get x � 0 and y � 1. From the second equation, forx � 0 we get y � 0 and y � 2; and for y � 1, we get x � 1 and x � �1. We have fourcritical points: A x � 0; y � 0� �, B x � 0; y � 2� �, C x � 1; y � 1� �, and D x � �1;�y � 1�. To classify these critical points we need the second-order partial derivatives.These are: Fxx � 6y � 6, Fyy � 6y � 6, and Fxy � 6x.

We compute the discriminant

D � Fxx � Fyy � Fxy� �2 � 6y � 6� � 6y � 6� � � 36x2

At point A x � 0; y � 0� �, we haveD � 36 > 0 and Fxx � �6 < 0. Therefore, A isa local maximum.

At point B x � 0; y � 2� �, we haveD � 36 > 0 and Fxx � 6 > 0. Therefore, B is alocal minimum.

At point C x � 1; y � 1� �, we have D � �36 < 0. Therefore, C is a saddle point.At pointD x � �1; y � 1� �, we haveD � �36 < 0. Therefore,D is a saddle point.Example: You want to invest in three Islamic mutual funds A, B, and C in

proportions x1, x2, and x3, respectively, with x1 � x2 � x3 � 1, and x1 � 0;x2 � 0; x3 � 0. The risks of mutual funds A, B, and C are σ1 � 9 percent,σ2 � 14 percent, and σ3 � 11 percent, respectively; the correlation coefficient betweenexpected returns are ρ12 � 0:6, ρ13 � 0:2, and ρ23 � 0:4. Assume you want a diversi-fied portfolio with minimal risk. Find the composition of this portfolio.

We compute the variance-covariance matrix of returns; by definition, it is asymmetric matrix:

92 0:6 � 9 � 14 0:2 � 9 � 110:6 � 9 � 14 142 0:4 � 14 � 110:2 � 9 � 11 0:4 � 14 � 11 112

264

375 �

81 75:6 19:875:6 196 61:619:8 61:6 121

24

35


3GC09 05/15/2014 10:44:11 Page 178

The variance of the portfolio is in quadratic form:

V x1; x2; x3� � � x1 x2 x3� 81 75:6 19:8

75:6 196 61:619:8 61:6 121

24

35 x1

x2x3

24

35

V x1; x2; x3� � � 81x21 � 196x22 � 121x23 � 151:2x1x2 � 39:6x1x3 � 123:2x2x3

Since x3 � 1 � x1 � x2, we express the portfolio in terms of x1 and x2 as

F x1; x2� � � 162:4x21 � 193:8x22 � 230:4x1x2 � 202:4x1 � 118:8x2

@F@x1

� 324:8x1 � 230:4x2 � 202:4 � 0

@F@x2

� 230:4x1 � 387:6x2 � 118:8 � 0

The solution is x1 � 0:70, x2 � 0:11, and x3 � 0:19.

CONSTRAINED OPTIMIZATION

Constrained optimization arises when optimization is subject to a set of constraints.Constraints vary in nature; they may be resource constraints, regulatory and institu-tional constraints, time constraints, and so on. Formally, constrained optimization isstated as


We may state in details the constrained optimization problem as

maxx1;...; xnF x1; . . . ; xn� � subject to (9.37)

g1 x1; . . . ; xn� � � b1g2 x1; . . . ; xn� � � b2

gm x1; . . . ; xn� � � bm

gi x1; . . . ; xn� � � bi i � 1; . . . ; m (9.38)

The n variables x1; . . . ; xn are the instruments, summarized by the column vectorx. The function F x� � is the objective function, and the m functions g1 x� �;g2 x� �; . . . ; gm x� � are the constraint functions summarized by the column vectorg x� �. The constraints b1; b2; . . . ; bm are the constraint constants, summarized bythe vector b. The functions F x� �; g1 x� �; g2 x� �; . . . ; gm x� � are given and continuouslydifferentiable, b consists of given real numbers, and x can be any real vector, subjectonly to the m constraints. It is assumed that the number of instruments n and thenumber of constraints m are finite and that n > m, where the difference n �m is thenumber of degrees of freedom.

178 MATHEMATICS

3GC09 05/15/2014 10:44:12 Page 179

Geometrically, each of the m equality constraints defines a set of points inEuclidian n space, Rn, and the intersection of all m sets is the opportunity set(Figure 9.2),

X � x ∈ Rn jg x� � � b �

(9.39)

Contours of the objective function and the preference direction are shown inFigure 9.2. Geometrically, the problem is to find a point (or set of points) in theopportunity set at which the highest contour of the objective function is attained.

The Method of Lagrange Multipliers

A main property of constrained programming, be it linear or nonlinear, is that itintroduces a new set of variables called the Lagrange multipliers and establishes a dualform of the optimization program. More specifically, if we denote the Lagrangemultipliers by

y � y1; . . . ; ym� �

that is, a multiplier yi for each constraint gi, i � 1; . . . ;m, the search for an optimalvector of instruments x* is simultaneously a search for a set of dual variables y*. Themaximization of the objective function over x is equivalent to a minimization of theconstraint function for a given objective function. If we define the Lagrangian functionas L x; y� �, the optimum x*; y*� � is a saddle point satisfying

L x; y*� �� L x*; y*� � � L x*; y� � (9.40)

The method of Lagrange multipliers is used as a basic approach to almost alloptimization problems; it yields valuable information on the sensitivities of theoptimal value of the objective function to changes in the constraint constants,sensitivities that have important economic interpretations.

As an introduction to the method of the Lagrange multipliers, consider the onedegree of freedom problem in which n � 2, m � 1,

maxx1;x2F x1; x2� � subject to g x1; x2� � � b (9.41)

Assume a local solution exists that x* � x*1; x*2

� �, and at this point one of the

partial derivatives of the constraint function g x1;x2� � does not vanish. We assume,therefore, @g

@x2x*� � ≠ 0. Given this assumption, the total differential is

dg � @g@x1

dx1 � @g@x2

dx2 � 0 (9.42)

Hence, in the neighborhood of x* we obtain

dx2

dx1� �

dgdx1@g@x2

(9.43)


3GC09 05/15/2014 10:44:12 Page 180

From the constraint g x1; x2� � � b, we may solve for x2 as a function of x1; weobtain

x2 � h x1� �;wheredhdx1

� �dgdx1@g@x2

(9.44)

The optimization problem can be written as the unconstrained optimizationproblem in the single variable x1,

maxx1H x1� � � F x1; h x1� �� (9.45)

By the results of the unconstrained problem, a first-order condition for a localmaximum is

dHdx1

� @F@x1

� @F@x2

dx2dx1

� 0 or@F@x1

� � @F@x2

dx2dx1

(9.46)

We also have

@g@x1

� @g@x2

dx2dx1

� 0 or@g@x1

� � @g@x2

dx2

dx1(9.47)

Accordingly, combining (9.46 and (9.47)), the ratio @F@x1

� @g@x1

may be written as

@F@x1@g@x1

�@F@x2

dx2dx1

@g@x2

dx2dx1

�@F@x2@g@x2

(9.48)

We let y equal

y �@F@x2@g@x2

(9.49)

We obtain

@F@x1@g@x1

� y or@F@x1

� y@g@x1

� 0 (9.50)

180 MATHEMATICS

3GC09 05/15/2014 10:44:13 Page 181

From the definition of the ratio y � @F@x2@g@x2

we obtain

@F@x2

� y@g@x2

or@F@x2

� y@g@x2

� 0 (9.51)

A local maximum necessarily implies that

@F@xj

� y@g@xj

� 0; j � 1; 2 (9.52)

orrF � yrg (9.53)

Equation (9.53) means that the gradients of F and g are parallel. We eliminate thevariable y by taking the ratios of (9.52); we find

@F@x1@F@x2

�@g@x1@g@x2

(9.54)

The solution is shown geometrically in Figure 9.5. Each contour of F takes theform F x1; x2� � � constant. So from the total differential,

dF � @F@x1

dx1 � @F@x2

dx2 � 0 (9.55)

it follows that the slope of the contour is

dx2dx1

contour

� � @F@x1

�@F@x2

(9.56)

0

C2

∇gC1

g(x) = b

x2

x1

E1

E*

∇F

∇F

∇g

FIGURE 9.5 Geometry of the ConstrainedOptimization


3GC09 05/15/2014 10:44:13 Page 182

The slope of the constraint curve is

dx2dx1

constraint

� � @g@x1

�@g@x2

(9.57)

The first-order condition for a maximum therefore implies the tangency solutionat which the slope of the contour equals the slope of the constraint,

dx2dx1

contour

� dx2dx1

constraint

(9.58)

Now comes the critical observation. Note that the necessary conditions (9.52)plus the original constraint can be obtained as the conditions for a stationary point ofthe function,

L x1; x2; y� � � F x1;x2� � � y b � g x1; x2� �� (9.59)

Namely, the first partial derivatives of L x1; x2; y� � yield the conditions

@L@xj

� @F@xj

� y@g@xj

� 0 (9.60)

@L@y

� b � g x� � � 0 (9.61)

The variable y is known as the Lagrangianmultiplier and the function L x1; x2; y� �is known as the Lagrangian function.

Example: A consumer has the utility functionU � U x; y� � � x0:7y0:3, where x andy are two consumable goods, and faces a budget constraint of 4x � y � 100. Find thebasket x; y� � that maximizes the consumer utility.

We substitute y � 100 � 4x into U x; y� �. We obtain x� � � x0:7 100 � 4x� �0:3. Wetake the derivative

dHdx

� 0:7x�0:3 100 � 4x� �0:3 � 4 � 0:2x0:7 100 � 4x� ��0:7 � 0

We multiply by x0:3 100 � 4x� �0:7; we find 0:7 100 � 4x� � � 1:2x � 0. The solutionis x � 17:5, y � 82:5, U � 27:865.

Example: You want to invest in three Islamic mutual funds A, B, and C inproportions x1, x2, and x3, respectively, with x1 � x2 � x3 � 1, and x1 � 0;x2 � 0; x3 � 0. The expected return of mutual funds A, B, and C arer1 � 8 percent, r2 � 11 percent, and r3 � 9 percent, respectively. The risks areσ1 � 9 percent, σ2 � 14 percent, and σ3 � 11 percent, respectively; the correlationcoefficients between expected returns are ρ12 � 0:6, ρ13 � 0:2, and ρ23 � 0:4. Assumeyou want a diversified portfolio with a return of 10 percent but with minimum risk.Find the composition of this portfolio. Check the answer using the Microsoft Excelsolver.

182 MATHEMATICS

3GC09 05/15/2014 10:44:14 Page 183

We use the portfolio variance as computed above; we solve the problem:

MinimizeV x1; x2; x3� � � 81x21 � 196x22 � 121x23 � 151:2x1x2 � 39:6x1x3 � 123:2x2x3

subject to 8x1 � 11x2 � 9x3 � 10:

Since x3 � 1 � x1 � x2, we express the problem in terms of x1 and x2.

Minimize F x1; x2� � � 162:4x21 � 193:8x22 � 230:4x1x2 � 202:4x1 � 118:8x2

subject to � x1 � 2x2 � 1:

We form the Lagrangian:

L x1; x2; λ� � � 162:4x21 � 193:8x22 � 230:4x1x2 � 202:4x1 � 118:8x2 � λ 1 � x1 � 2x2� �@L@x1

� 324:8x1 � 230:4x2 � 202:4 � λ � 0

@L@x2

� 230:4x1 � 387:6x2 � 118:8 � 2λ � 0

@L@λ

� 1 � x1 � 2x2 � 0

The solution is x1 � 0:076, x2 � 0:538, x3 � 0:386, and λ � 53:7. The minimumrisk of the portfolio is V x1; x2; x3� � � 108:156, that is, σp � 10:4 percent.

THE GENERAL CLASSICAL PROGRAM

The general classical problem is


The first step is to introduce a row vector of m new variables called theLagrangian multiplier,

y � y1; . . . ; ym� �

(9.63)

The second step is to define the Lagrangian function,

L x; y� � � F x� � � y b � g x� �� (9.64)

or written in full,

L x1; . . . ; xn; y1; . . . ; ym� � � F x1; . . . ; xn� � �Xm

i�1 yi bi � g x1; . . . ; xn� �� (9.65)


3GC09 05/15/2014 10:44:14 Page 184

The final step is to find the point x*; y*� � at which all first order partial derivativesof the Lagrangian vanish:

@L@x

� @F@x

x*� � � y*@g@x

x*� � � 0 (9.66)

@L@y

x*; y*� � � b � g x*� � � 0 (9.67)

Simultaneously solving the m � n equations yields solutions for the m � nunknowns: the instruments x* � x*1; . . . ; x

*n

� �and the Lagrangian multipliers

y* � y*1; . . . ; y*m

� �. The value of the Lagrangian at point x*; y*� � is simply the value

of the objective function since the constraints are satisfied, so

L x*; y*� � � F x*� � (9.68)

The Geometry of Constrained Optimization

The geometry of constrained optimization enables us to understand the nature ofthis optimization and derive the analytical method, called the Lagrangian method,or the Lagrangian multiplier, for computing maxima, minima, or saddle points. InFigure 9.5, the constraint function g x� � � b provides a contour that defines theopportunity set X. The objective function is described by contours or level curves.At point E1, the gradients rF and rg point to different directions. This means thatwe may increase F without violating the constraint g x� � � b. At point E*, rF and rgpoint to the same direction. They are parallel and perpendicular to the same tangentline. We cannot further increase Fwithout violating the constraint g x� � � b. At pointE*, we have*

rF x*; y*� � � y � rg x*; y*� � (9.69)

Interpretation of the Lagrangian Multipliers

The values of the Lagrangian multipliers are not extraneous. They yield valuableinformation about the problem, which in part accounts for the usefulness of theLagrangian multiplier technique. The Lagrangian multipliers at the solution measurethe sensitivity of the optimal value of the objective function F* � F x*� � to variations inthe constraint constants b,

y* � @F=@b (9.70)

*A well-known property of the gradient of a curve is that the gradient at a point of the curve isperpendicular to the tangent line to the curve at this point. Hence, rF at a given point isperpendicular to the tangent to F at this point. Likewise, rg at a given point is perpendicular tothe tangent to g curve at this point. At the optimum point, ΔF and rg are perpendicular to thesame tangent line; they are therefore parallel and satisfy rF x*� � � yrg x*� �.

184 MATHEMATICS

3GC09 05/15/2014 10:44:15 Page 185

Therefore, the Lagrange multiplier equals the rate of the change in the optimalvalue of F resulting from the change of the constant b. If F is the profit function of theinputs, and b denotes the value of these inputs, then the derivative is the rate of changeof the profit from the change in the value of the inputs; that is, the Lagrange multiplieris the “marginal profit of money.” Likewise, if F is the utility function of consumptionand b denotes consumer’s income, the Lagrange multiplier is the “marginal utility perunit of money.” The Lagrange multipliers are also interpreted as the imputed value orshadow prices of constraint constants bi, i � 1; . . .m:

Example: Find the smallest value of

f x; y� � � x2 � y2 subject to the constraint g x; y� � � y � 3x � 3

The Lagrangian function is

L x; y; λ� � � f x; y� � � λ b � g x; y� �� x2 � y2 � λ 3 � 3x � y� �

The three first-order conditions are

i. @L@x � @f

@x � λ @g@x � 2x � 3λ � 0

ii. @L@y � @f

@y � λ @g@y � 2y � λ � 0

iii. @L@λ � b � g x; y� � � 3 � 3x � y � 0

We solve for x and y in (i) and (ii); respectively, we have x � 32 λ and y � λ

2.Replacing these into (iii), we have λ

2 � 3 � 3λ2 � 10λ

2 � 3. So λ � 35. It follows that x � 9

10

and y � 310. The point (0.9, 0.3) is a global minimum.

Example: Find the optimal values of the function

f x; y� � � xy subject to the constraint g x; y� � � x2

8� y2

2� 1

We construct the Lagrangian function and find its gradient,

L x; y; λ� � � f x; y� � � λ b � g x; y� �� xy � λ 1 � x2

8� y2

2

� �

i. @L@x � @f

@x � λ @g@x � y � λx

4 � 0

ii. @L@y � @f

@y � λ @g@y � x � λy � 0

iii. @L@λ � b � g x; y� � � x2

8 � y2

2 � 1 � 0

Combining (i) and (ii) yields λ2 � 4 and λ � �2. Thus x � �2y. Substituting thisequation into (iii) gives us y � �1 and x � �2. So there are four extremal points of f


3GC09 05/15/2014 10:44:16 Page 186

subject to the constraint g: A 2; 1� �, B �2;�1� �, C 2;�1� �, andD �2; 1� �. The maximumvalue 2 is achieved at the first two points, A and B, while the minimum value �2 isachieved at the last two points, C and D.

Nonlinear Programming

The nonlinear programming (NLP) is that of choosing nonnegative values of certainvariables so as to maximize or minimize a given function subject to a given set ofinequality constraints. The NLP problem is


or stated in detailed form,

maxx1;...; xnF x1; . . . ; xn� � subject to (9.72)

g1 x1; . . . ; xn� � � b1g2 x1; . . . ; xn� � � b2gm x1; . . . ; xn� � � bmx1 � 0; . . . ; xn � 0

The n variables x1; . . . ; xn are the instruments, summarized by the columnvector x. The function F x� � is the objective function, and the m functionsg1 x� �; g2 x� �; . . . ; gm x� � are the constraint functions summarized by the column vectorg x� �. The constraints b1; b2; . . . ; bm are the constraint constants, summarized by thevector b. The functions F x� �; g1 x� �; g2 x� �; . . . ; gm x� � are given and continuouslydifferentiable, b consists of given real numbers, and x can be any real vector, subjectonly to the m � n constraints.

We note that the direction of the inequalities �� is only a convention. Forexample, the inequality �3x1 � x2 � 6 can be converted to the � inequality bymultiplying by �1, yielding 3x1 � x2 � 6. We note that an equality constraint, forexample, �3x1 � 5x2 � 2 can be replaced by two inequality constraints:�3x1 � 5x2 � 2 and 3x1 � 5x2 � �2. We note that the non-negativity constraintson the instruments are not restrictive. If a particular variable, say xi, is unrestricted(i.e., could be positive, negative, or zero), then it could be replaced by the differencebetween two non-negative variables,

xi � xi � x´´i ; where xi � 0; x´´i � 0

The Case of No Inequality Constraints

In the case of no inequalities, m � 0, the basic problem becomes that of maximizing afunction by choosing non-negative values of the instruments:

maxx F x� � subject to x � 0 (9.73)

186 MATHEMATICS

3GC09 05/15/2014 10:44:17 Page 187

One approach to this problem is expansion by a Taylor’s series. Assuming a localmaximum of F exists at x*, then for all neighboring points x* � Δx,

F x*� � � F x* � hΔx� � (9.74)

Δx is a direction of movement in the opportunity set and h is an arbitrary, smallpositive number. Assuming F x� � is twice differentiable, we obtain a Taylor’s seriesexpansion about x* as

F x* � hΔx� � � F x*� � � h@F@x

x*� �Δx � 12!h2 Δx� �´ @2F

@x2x* � αhΔx� � Δx� � (9.75)

0 < α < 1

We obtain the fundamental inequality

h@F@x

x*� �Δx � 12!h2 Δx� �´ @2F

@x2x* � αhΔx� � Δx� � � 0 (9.76)

which is a necessary condition for a local maximum at x*. If x* is an interior solution,x* > 0, then the fundamental inequality must hold for all directions Δx, leading to thesame first-order conditions as in classical programming, namely the vanishing of allfirst-order partial derivatives. Suppose, however, one of the instruments is at theboundary x*j � 0. Assuming all other variations equal to zero, since at x*j � 0 the onlypermissible direction is that for which Δxj � 0, the fundamental inequality implies*

@F@xj

x*� �Δxj � 0 (9.77)

The fundamental inequality therefore requires as a first-order condition that

@F@xj

x*� � � 0 if x*j � 0 (9.78)

Thus, while the first derivative with respect to xj necessarily vanishes at an interior

solution x*j > 0� �

, at a boundary solution x*j � 0� �

the first derivative is less than or

equal to zero. But since the derivative takes the zero value (at an interior solution) orthe corresponding instrument takes the zero value (at a boundary solution), theproduct of the two always vanishes:

@F@xj

x*� �x*j � 0 (9.79)

*We have h @F@x x*� �Δx � 1

2! h2 Δx� �´ @2F

@x2 x* � αhΔx� � Δx� � � 0. Since h > 0, we divide by h to obtain

@F@x

x*� �Δx � 12!h Δx� �´ @2F

@x2x* � αhΔx� � Δx� � � 0

as h ! 0, 12!h Δx� �´ @2F

@x2 x* � αhΔx� � Δx� � � 0. We obtain, therefore, @F@x x*� �Δx � 0:


3GC09 05/15/2014 10:44:18 Page 188

a. Interior solution Zero slope at solution

b. Boundary solution Negative slope at solution

c. Boundary solution Zero slope at solution

0

F(x)

∂x∂F

(x*) = 0

x* > 0

F(x)

∂x∂F

(x*) < 0

0 x* = 0x x

F(x)

∂x∂F

(x*) = 0

0 x* = 0x

FIGURE 9.6 Alternative Possible Solutions of the Nonlinear Programming

Summing these conditions on the vanishing of the products yields

@F@x

x*� �x* �Xnj�1

@F@xj

x*� �x*j � 0 (9.80)

Thus a local maximum at x* is characterized by the following first-orderconditions:

@F@x

x*� � � 0 (9.81)

@F@x

x*� �x* � 0 (9.82)

x* � 0 (9.83)

These conditions state that the first partial derivative vanishes if the correspond-ing instrument is positive, and is nonpositive if the instrument is zero.

@F@xj

x*� � � 0 if x*j > 0; j � 1; . . . ; n (9.84)

@F@xj

x*� � � 0 if x*j � 0; j � 1; . . . ; n (9.85)

Figure 9.6 illustrates the alternative possible solutions of the NLP in onedimensional case. Figure 9.6a shows an interior solution at which the slope iszero, Figure 9.6b shows a boundary solution at which the slope is negative, andFigure 9.6c shows a boundary solution at which the slope is zero.

The Kuhn-Tucker (K-T) Conditions

The conditions for an optimum established for the case of no constraints are applied tothe case with constraints. The general NLP problem is


188 MATHEMATICS

3GC09 05/15/2014 10:44:18 Page 189

The inequality constraints can be converted to equality constraints by adding avector of m slack variables:

s � b � g x� � � s1; . . . ; sm� �´ (9.87)

So the NLP problem can be written as

maxxF x� � subject to g x� � � s � b; x � 0; s � 0 (9.88)

where the nonnegativity of the slack variables ensures the inequality constraints aremet. If theNLP did not contain m � n� � non-negativity constraints, x � 0, s � 0, then itwould be a classical programming problem for which the Lagrangian functionwould be

Ls � F x� � � y b � g x� � � s� � (9.89)

where y � y1; . . . ; ym� �

is a vector of Lagrange multipliers.The first-order necessary conditions would then be obtained as the conditions that

all first-order partial derivatives of Ls with respect to x; y, and s vanish. Because of thenon-negativity of x and s, however, the conditions on the first-order derivatives withrespect to these m � n� � variables are replaced by the conditions (9.81 and (9.82)).Thus, the first-order conditions for a local maximum are

@Ls

@x� @F@x

� y@g@x

� 0 (9.90)

@Ls

@xx � @F

@x� y

@g@x

� �x � 0 (9.91)

x � 0

@Ls

@y� b � g x� � � s � 0 (9.92)

@Ls

@s� �y � 0 (9.93)

@Ls

@ss � �ys � 0 (9.94)

s � 0

where all variables, functions, and derivatives are evaluated at x*; y*, and s*.Eliminating the vector of slack variables s by replacing it by b � g x� � yields theKuhn-Tucker conditions

@F@x

� y@g@x

� 0 (9.95)


3GC09 05/15/2014 10:44:19 Page 190

@F@x

� y@g@x

� �x � 0 (9.96)

x � 0

b � g x� � � 0 (9.97)

y b � g x� �� 0 (9.98)

y � 0

The same conditions result from defining the Lagrangian function for the initialproblem NLP as

L � L x; y� � � F x� � � y b � g x� �� (9.99)

The Kuhn-Tucker conditions are then

@L@x

x*; y*� � � @F@x

x*� � � y*@g@x

x*� � � 0 (9.100)

@L@x

x*; y*� �x* � @F@x

x*� � � y*@g@x

x*� ��

x* � 0 (9.101)

x* � 0

@L@y

x*; y*� � � b � g x*� � � 0 (9.102)

y*@L@y

x*; y*� � � y* b � g x*� �� 0 (9.103)

y* � 0

These conditions are necessary for a strict local maximum if the objective isstrictly concave and the constraint functions are convex, assuming a certain qualifi-cation condition holds. The constraint qualification conditions assume the existence ofx0 that satisfies all inequality constraints as strict inequalities, x0 � 0 and g x0� � < b.The Kuhn-Tucker conditions can be written out in full as

@L@xj

� @F@xj

�Xmi�1

yi@gi@xj

� 0; j � 1; . . . ; n (9.104)

Xnj�1

@L@xj

xj �Xnj�1

@F@xj

�Xmi�1

yi@gi@xj

!xj � 0 j � 1; . . . ; n (9.105)

xj � 0; j � 1; . . . ; n

190 MATHEMATICS

3GC09 05/15/2014 10:44:19 Page 191

@L@yi

� bi � gi x� � � 0; i � 1; . . . ; m (9.106)

Xmi�1

yi@L@yi

�Xmi�1

yi bi � gi x� �� 0; i � 1; . . . ; m (9.107)

yi � 0; i � 1; . . . ; m

where it is assumed that all variables, functions, and derivatives are evaluated atx*; y*� �.

Thus we state the conditions as

@F@xj

�Xmi�1

yi@gi@xj

� 0; but@F@xj

�Xmi�1

yi@gi@xj

� 0 if x*j > 0 j � 1; 2; . . . n (9.108)

x*j � 0; but x*j � 0 if@F@xj

�Xmi�1

yi@gi@xj

< 0; j � 1; 2; . . . n (9.109)

gi x*� � � bi; but gi x*� � � bi if y*i > 0; i � 1; 2; . . .m (9.110)

y*i � 0; but y*i � 0 if gi x*� � < bi; i � 1; 2; . . .m (9.111)

Conditions (9.108)–(9.111) are known as the complementary slackness condi-tions. Finally, since y* b � g x*� �� 0 the Lagrangian at the solution is simply theoptimal value of the objective function

L x*; y*� � � F x*� � � y* b � g x*� �� F x*� � (9.112)

Example: Maximize x3 � 3x subject to x � 2. Check answer using the MicrosoftExcel solver.

The Lagrangian is L x; y� � � x3 � 3x � y 2 � x� �:We need the Kuhn-Tucker conditions:

@L@x

� 3x2 � 3 � y � 0

x � 2

@L@y

� 2 � x � 0

y@L@y

� y 2 � x� � � 0

y � 0

Typically, at this point we must break the analysis into cases depending on thecomplementarity conditions.


3GC09 05/15/2014 10:44:20 Page 192

If y � 0, then 3x2 � 3 � 0 so x � 1 or x � �1. Both are feasible; F 1� � � �2 andF �1� � � 2:

If x � 2 then y � 9, which again is feasible. Since F 2� � � 2, we have two solutions:x � �1 and x � 2.

Example: Minimize x � 2� �2 � 2 y � 1� �2 subject to

x � 4y � 3

x � y

First we convert to standard form, to get

Maximize F x� � � � x � 2� �2 � 2 y � 1� �2 subject tox � 4y � 3

�x � y � 0

We form the Lagrangian:

L x; y; μ1; μ2� � � � x � 2� �2 � 2 y � 1� �2 � μ1 3 � x � 4y� � � μ2 0 � x � y� �

which gives the optimality conditions,

@L@x

� �2 x � 2� � � μ1 � μ2 � 0

@l@y

� �4 y � 1� � � 4μ1 � μ2 � 0

μ1@L@μ1

� μ1 3 � x � 4y� � � 0

μ2@L@μ2

� μ2 0 � x � y� � � 0

μ1 � 0; μ2 � 0

Since there are two complementarity conditions, there are four cases to check:

μ1 � 0; μ2 � 0 gives x � 2, y � 1; which is not feasible.μ1 � 0, x � y � 0 gives x � 4=3, y � 4=3 μ2 � �4=3, which is not feasible.μ2 � 0, 3 � x � 4y � 0 gives x � 5=3 y � 1=3, μ1 � 2=3, which is feasible.μ1 > 0, μ2 > 0, 3 � x � 4y � 0, x � y � 0 gives x � 3=5, y � 3=5, μ1 � 22=25,μ2 � �48=25, which is not feasible.

The solution is x � 53 ; y � 1

3, and F � �1:

192 MATHEMATICS

3GC09 05/15/2014 10:44:21 Page 193

SUMMARY

Optimization is a fundamental topic in Islamic finance. The chapter illustrates themathematical programming problem and the geometry of optimization. It deals withunconstrained optimization and introduces constrained optimization, as well as themethod of the Lagrange multipliers. It also introduces the general classical programand provides an interpretation of the Lagrangian multipliers. Also presented arenonlinear programming, in the case of no inequality constraints and in the case ofconstraints, and the formulation of Kuhn-Tucker (K-T) conditions of nonlinearprogramming problems.

Islamic finance uses optimization techniques in the selection of portfolios and inthe trade-off between risk and return. Islamic banks have resource constraints; theyhave to choose optimal investments. Optimization techniques allow them to makeefficient choices.

QUESTIONS

1. Find and classify all the critical points of F x; y� � � x2 � 2y2 � 4xy � 4y.

2. Find and classify all the critical points of F x; y� � � x4 � y4 � 4xy.

3. Find and classify all the critical points of F x; y� � � x2 � y2.

4. A consumer has the utility functionU � U x; y� � � x0:7y0:3, where x and y are twoconsumable goods, and faces a budget constraint of 4x � y � 100. Find the basketx; y� � that maximizes the consumer utility.

5. The total cost of production of items x and y is given by C � 5x2�3xy � 3y2 � 800. A total of 39 items must be manufactured. How many ofeach item should be manufactured to minimize the cost? Estimate the additionalproduction cost if 40 items are manufactured.

6. Using the Microsoft Excel solver, find the smallest value of f x; y� � � x2 � y2

subject to the constraint g x; y� � � y � 3x � 3.

7. Determine the minimizers/maximizers of the following five functions subject tothe given constraints:

a. f x; y� � � xy3 subject to 2x � 3y � 4.b. f x; y� � � 2x � 3y subject to x2 � y2 � 25.c. f x; y� � � y subject to x3 � y3 � 3xy � 0.d. f x; y� � � x3 � y3 subject to 2x � y � 1.e. f x; y� � � 5 � x � 2� �2 � 2 y � 1� �2 subject to 2x � y � 12.

8. Minimize x � 5� �2 � y � 5� �2 subject to

x2 � y2 � 5 � 0;12x � y � 2 � 0; x � 0; y � 0


3GC09 05/15/2014 10:44:22 Page 194

9. Minimize f � x21 � x22 � 60x1 subject to

g1 � x1 � 80 � 0

g2 � x1 � x2 � 120 � 0

x1 � 0; x2 � 0

10. Maximize f � x � y2 subject to x2 � y2 � 4, x � 0, and y � 0, using the Kuhn-Tucker method.

11. Maximize U � xy subject to 100 � x � y and x � 40, x � 0, and y � 0.

12. Maximize f � x subject to 1 � x� �3 � y � 0, x � 0; and y � 0.

13. You want to invest in two Islamic mutual funds, A and B, in proportions x1

and x2, respectively, with x1 � x2 � 1 and x1 � 0; x2 � 0. The expectedyield rates and risks of mutual funds A and B are r1 � 8%; σ1 � 9 percent andr2 � 10 percent; σ2 � 14 percent, respectively; the correlation coefficient betweenthe expected returns is ρ � 0:6. The expected portfolio return is

R x1; x2� � � r1x1 � r2x2 � 0:08x1 � 0:10x2

The portfolio variance is

V x1; x2� � � σ21x21 � 2ρσ1σ2x1x2 � σ22x

22 � 81x21 � 151:2x1x2 � 196x22

Using the Microsoft Excel solver, solve the problem

maxx1; x2R x1; x2� � � 8x1 � 10x2 subject to 81x21 � 151:2x1x2 � 196x22 � 121

14. You want to invest your saving in three Islamic mutual funds, A, B, and C, inproportions x1, x2, and x3, respectively, with x1 � x2 � x3 � 1 andx1 � 0; x2 � 0; x3 � 0. The expected returns of mutual funds A, B, and C arer1 � 8 percent, r2 � 11 percent, and r3 � 9 percent, respectively. The risks areσ1 � 9 percent, σ2 � 14 percent, and σ3 � 11 percent, respectively; the correla-tion coefficients between the expected returns are ρ12 � 0:6, ρ13 � 0:2, andρ23 � 0:4. Assume you want a diversified portfolio with a return of 10 percent.Using the Microsoft Excel solver, find the portfolio that minimizes the riskassociated with this return.

15. You want to invest your savings in three Islamic mutual funds, in proportions x1,x2, and x3, with x1 � x2 � x3 � 1, x1 � 0; x2 � 0; and x3 � 0, with expectedreturns of 10 percent, 10 percent, and 15 percent, respectively, so as to minimizerisk while achieving an overall expected return of 12 percent. The variance of thereturn has been calculated as 400x21 � 800x22 � 1; 600x23 � 200x1x2 � 400x2x3.Find the portfolio that minimizes risk for the targeted return.

194 MATHEMATICS

3GC10 05/15/2014 11:6:30 Page 195

CHAPTER 10Linear Programming

L inear programming (LP) is an important field of Islamic finance. LP is basic tounderstanding economic efficiency, the optimal allocation of resources, and the

valuation of these resources. Learning LP allows us to understand how scarceresources ought to be used in the most efficient way and how they ought to bevalued. If scarce resources are allocated to sectors where they are wasted, then there istremendous loss in growth. For instance, cheap money policy forces allocation ofresources to wasteful uses and causes inflation and significant loss of real income.

Allocation and valuation of resources are carried out simultaneously in LP. Manycorporations in Islamic finance, transportation, oil refining, agriculture, mining,industrial sectors, and utilities rely on LP in managing their investment, production,and distribution activities. LP has a dual form called allocation and valuation ofresources; it consists of finding a set of instruments:

■ x that maximize a linear function F x� � � cx subject to linear constraints Ax � b,x � 0.

■ Or in dual form, y that minimize a linear function G y� � � yb subject to linearconstraints yA � c, y � 0.

The nature of instruments, linear objective function, and linear constraints variesaccording to the programming problem. The instruments may be a set of quantitiesexpressed in physical units or prices expressed in dollar terms. The linear objectivefunction may be a revenue function expressed in dollars or a cost function expressed indollars. The constraints may be a resource constraint expressed in physical units orcost constraint expressed in dollars. The knowledge of the units (physical or moneyunits) in which each parameter of the LP is expressed is essential for understanding theLP program.

Often F x� � � cx is a revenue function; its dual, G y� � � yb, is a cost function.The solution of one form, for example, x*, simultaneously entails the solution of itsdual, y*, and reciprocally, if we start by finding a solution y*, we obtainsimultaneously a solution x*. Economically, the choice of outputs cannot be splitfrom the cost of inputs, and the choice of inputs cannot be split from theprofitability of outputs. Outputs and inputs are intimately related in terms ofpricing and quantities.

Duality is an essential feature of LP. If we solve a maximization LP called primal,then this LP has a dual called a minimization LP; inversely, a minimization LP calledprimal is a dual for the maximization LP. A primal and its dual are closely interrelated.

195

3GC10 05/15/2014 11:6:30 Page 196

A solution of the LP is a saddle point. The optimal value of the objective function is thesame in both programs, that is,

F x*� � � G y*� �, or cx* � by*

For a firm, this means that the value of the output is equal to the value of the costand the excess profit is zero. If there is excess profit, new producers will enter the fieldand the excess profit will be exploited. If there is loss, existing producers will closedown and loss will disappear. Themarginal value of an output is equal to the marginalvalue of its inputs. In either form, we solve simultaneously both the primal and thedual programs and we obtain a solution for both, as in a typical Lagrangian problemwhere we obtain solutions for the variables as well as the Lagrange multipliers.Duality is important in economic theory. It means that the search for an optimalallocation of resources cannot be dissociated from the search for equilibrium pricesthat balance demand and supply for these resources. Inversely, the search forequilibrium prices cannot be dissociated from the search for an optimal allocationof resources. Intuitively, the price of a resource, for example, wheat or fuel, cannot beindependent from its marginal utility or marginal productivity. The higher themarginal utility or productivity is, the more expensive the resource will be. Forinstance, high productivity labor is paid a higher salary than low productivity labor.

A large number of optimization problems is not linear in terms of objectivefunction as well as constraint functions. For instance, some production activitiesmay exhibit increasing returns to scale and therefore linearity assumption is notappropriate. Labor may exhibit marginal diminishing return. Utility of consump-tion may exhibit marginal diminishing utility. In these cases, we apply the tools ofnonlinear programming. However, many activities are characterized by linearrelationships. For instance, it takes a fixed amount of fuel to generate a fixednumber of electricity units. It takes a fixed quantity of flour to make a fixed quantityof bread, a given quantity of fertilizers for a given acre of land, a given quantity ofmilk for a given quantity of cheese, and a given quantity of chemicals for a drug. Inthese cases where linearity holds, we are entitled to apply LP to solve economic andfinancial problems.

An LP may be a maximization problem. Here a revenue function is to be amaximized subject to linear constraints on inputs. The solution is an optimal vector ofoutputs that maximizes the revenue function. Once we find this vector, there is nofurther increase in the output given the constraints. The LP may be a minimizationproblem. Here a cost function is to be minimized subject to constraints on output. Thesolution is an optimal vector of inputs that minimizes the cost function. Once we findthis vector, there is no further reduction of the cost given the constraints. Intuitively, ina maximization problem, we wish to maximize the value of output as much aspossible. However, we run into resource constraint, which prevents any furtherincrease in this value. For instance, farmers wish to produce as much wheat aspossible, but they will hit a constraint dictated by the size of their farm. Likewise, anairline company wishes to sell as many tickets as possible; however, it will hit aconstraint dictated by the number of seats in a jetliner. In a minimization problem, wewish to reduce inputs as much as possible; however, we cannot go further withoutimperiling the output constraint. For instance, a farmer cannot reduce the quantity offertilizers beyond a limit that violates his output constraint. Likewise, an airline

196 MATHEMATICS

3GC10 05/15/2014 11:6:30 Page 197

company cannot cut fuel use below a point that disturbs its flight schedule and reducesthe number of passengers to be flown.

LP problems can be solved instantaneously using the Microsoft Excel solver,MATLAB, and many online solvers.* We present in this chapter the simplex methodfor solving LP problems, which is a direct application of the elimination technique forsolving linear equations.

FORMULATION OF THE LP

The LP can be expressed in two forms, the standard form and the canonical form. Thegeometry of LP can be used to show minimizing and maximizing problems.

Standard Form and Canonical Form of the LP

The standard form of an LP involves inequality constraints. In this case the LP isstated as

maxx F � cx subject toAx � b, x � 0 (10.1)

Stated in detail, the LP is

maxx1,..., xn F x1, . . . , xn� � � c1x1 � . . . � cnxn (10.2)

subject toa11x1 � . . . � a1nx1n � b1ai1x1 � . . . � ainx1n � bi

am1x1 � . . . � amnx1n � bmx1 � 0, . . . , xn � 0

In vector form the constraints are written as

a1x1 � . . . � amxm � . . . � anxn � b

The canonical form of an LP is stated in the form of equalities constraints. The LPis formulated as


Each constraint is written as

Xn

j�1 aijxj � bi; xj � 0; i � 1, . . . ,m (10.4)

An LP in standard form is easily transformed into a canonical form by adding toeach constraint inequality a nonnegative slack variable in order to transform the

* http://www.zweigmedia.com/RealWorld/simplex.html.

Linear Programming 197

http://www.zweigmedia.com/RealWorld/simplex.html

3GC10 05/15/2014 11:6:31 Page 198

inequality constraint into an equality constraint. We have to transform the LP fromthe standard form into a canonical form in order to solve it.

Example: LP standard form: Maximize F x� � � x1 � 3x2 subject to

2x1 � x2 � 50x1 � 2x2 � 22x1, x2 � 0

To obtain the canonical form we add a slack variable x3 � 0 to the first inequalityconstraint and another slack variable x4 � 0 to the second inequality constraint.

LP canonical form: Maximize F x� � � x1 � 3x2 subject to

2x1 � x2 � x3 � 50x1 � 2x2 � x4 � 22x1, x2, x3, x4 � 0

The LP is a special case of the nonlinear programming problem for which both theobjective function and the constraint functions are linear. The variables x1, . . . , xn arethe instruments; they are unknown and have to be solved. They are summarized by acolumn vector x. The row vector c � c1, . . . , cn� �, the matrix A m � n� �, and thecolumn vector b � b1, . . . , bm� �´ are fixed real data, that is, c, A, and b consist ofgiven real numbers. The matrixA has columns a1, . . . , am, . . . , anf g and is a technicalmatrix; the coefficients aij may describe the quantity of input i per one unit of output j.Each column aj may describe a technical process. We assumem � n. Generally, cmaydescribe prices (e.g., price of wheat, price of milk, price of tomatoes), and b maydescribe resource constraints (e.g., fixed land area, fixed quantity of quantity of labor,fixed quantity of material input).

Example: The activity analysis problem: there are n activities, a1, . . . , an, that acompany may employ, using the available supply of m resources, R1, . . . ,Rm (land,labor, fuel, steel, etc.). Let bi be the available supply of resource Ri. Let aij be theamount of resource Ri used in operating activity aj at unit intensity. Let cj be the netvalue in dollars to the company of operating activity aj at unit intensity. Theproblem is to choose the intensities at which the various activities are to be operatedto maximize the value of the output to the company subject to the given resources.Let xj be the intensity at which aj is to be operated. The value of such an activityallocation is

F x� � � Xn

j�1 cjxj (10.5)

The amount of resource Ri used in this activity allocation must be no greater thanthe supply, bi, that is, Xn

j�1 aijxj � bi for i � 1, . . . ,m (10.6)

It is assumed that we cannot operate an activity at negative intensity, that is,

x1 � 0, . . . , xn � 0 (10.7)

Our problem is: maximize (10.5) subject to (10.6) and (10.7). This is exactly thestandard maximum problem.

198 MATHEMATICS

3GC10 05/15/2014 11:6:31 Page 199

Example: The diet problem: there are n different types of food, F1, . . . , Fn (meat,butter, eggs, oranges, etc.) that supply varying quantities of the m nutrientsN1, . . . ,Nm, that are essential to good health (proteins, vitamins, etc.). Let γi bethe minimum daily requirement of nutrient,Ni. Let πj be the price per unit of food, Fj.Let aij be the amount of nutrientNi contained in one unit of food Fj. The problem is tosupply the required nutrients at minimum cost. Let xj be the number of units of food Fj

to be purchased per day. The cost per day of such a diet is

F x� � � π1x1 � . . . � πnxn (10.8)

The amount of nutrient Ni contained in this diet is

a1ix1 � . . . � anixn for i � 1, . . . ,m (10.9)

We do not consider such a diet unless all the minimum daily requirements are met,that is, unless

a1ix1 � . . . � anixn � γi for i � 1, . . . ,m (10.10)

We cannot purchase a negative amount of food, so we automatically have theconstraints

x1 � 0, . . . , xn � 0 (10.11)

Our problem is: minimize (10.8) subject to (10.10) and (10.11). This is exactly thestandard minimum.

The Geometry of the LP

We describe the geometry of LP in R2. In Figure 10.1, we show the opportunity setsof a maximum LP problem. The opportunity set X is described by the constraintsAx � b. The objective function is described by contours of the form

F x� � � cx � constant (10.12)

a. Vertex solution b. Bounding face solution

0

x2

cx = constant

x1

E3E4

E2

∇F

X

E10

x2

cx = constant

x1

E3

E2x*

2

x*1

∇F

X

E1

FIGURE 10.1 Geometry of a Maximum LP


3GC10 05/15/2014 11:6:31 Page 200

Geometrically, the optimum is reached at a feasible point inX for which F x� � � cxachieves a maximum. In Figure 10.1a, we drew the opportunity set for two linearconstraints. Each of the vertices E1, E2, and E3 qualifies for being a solution, depend-ing on the slope of F x� �. At E1 or E3, only one constraint is active. At E2, bothconstraints are active. In Figure 10.1b, we drew the opportunity set for three linearconstraints. Each of the vertices E1, E2, E3, and E4 qualifies for being a solution,depending on the slope of F x� �. At vertices E1 and E4 only one constraint is active; theother two constraints are not. Assuming the slope of F x� � is such that we have abounding face, then the solution could be vertex E2 or E3; either vertex provides thesame maximum value of F x� �. However, at E2 or E3, only two constraints are active;the remaining constraint is not active.

Figure 10.2 shows the geometry of a minimization problem. The minimumsolution could be a vertex solution (Figure 10.2a) or a bounding face solution(Figure 10.2b). In the latter case, the solution is not unique. Vertex E2 and vertexE3 provide the same minimum value of the objective function.

Example:(i) A farmer produces tomatoes. The sale price is $1.1/kilogram of tomatoes. The

farmer has a cultivable plot of 800 squared meters. The tomatoes’ yield is 20 kilogramper squared meter. The LP problem is to maximize the sales proceeds subject to theland constraint. The answer is to produce 16,000 kilograms of tomatoes and reap asales revenue of $17,600. The farmer cannot do any better, because no land isavailable for higher tomato production.

(ii) We assume the farmer wishes to produce potatoes. Assume the potatoes’ yieldis 25 kilogram per squared meter and the price of potatoes is $0.9 per kilogram. Thefarmer now wants to maximize

F x� � � 1.1x1 � 0.9 � x2 subject to

120

x1 � 125

x2 � 800 and x1 � 0, x2 � 0

where x1 and x2 are the quantities of tomatoes and potatoes, respectively. We proceedby iteration. If farmers produce tomatoes only, their sales revenues are $17,600. Ifthey produce potatoes only, output will be 20,000 kilograms and sales revenues,

a. Vertex solution b. Bounding face solution

0

x2

cx = C

E2

E1 x1

X

∇F

0

x2

cx = C

E4

E3

E2

E1 x1

X

∇F

FIGURE 10.2 Geometry of the Minimum LP

200 MATHEMATICS

3GC10 05/15/2014 11:6:32 Page 201

$18,000. They will choose a corner solution consisting of producing onlypotatoes. This is illustrated in Figure 10.3a, which shows the constraint lineas well as the objective function line. The optimum is achieved at a corner pointA x1 � 0, x2 � 20; 000� �.

Now assume that the farmer faces a labor constraint in addition to a landconstraint. Assume a labor unit produces 2,000 kilograms of tomatoes and 1,600kilograms of potatoes. Assume the available labor input is equal to 10. We have nowtwo constraints. The LP can be stated as maximize,

F x� � � 1.1x1 � 0.9 � x2 subject to

120

x1 � 125

x2 � 800

12;000

x1 � 11;600

x2 � 10

The two constraints define an opportunity set drawn in Figure 10.3b. The interplayof the two constraintsmakes productionplanswithin one constraint unattainable underthe other constraint. We are therefore limited only to the frontier permissible underboth constraints. If farmers produce tomatoes only, their sales will be 1.1 � 16; 000 �$17; 600. If they produce potatoes only, their sales will be 0.9 � 16; 000 � $14; 400.The intersection point of the two constraints is E x1 � 8; 889; x2 � 8; 889� �. If theyproduce at point E, their sales will be 1.1 � 8; 889 � 0.9 � 8; 889 � $17; 778. Theoptimal point is therefore E x1 � 8; 889; x2 � 8; 889� �, at which sales are maximized.The solution is a corner solution provided by the vertex E x1 � 8;889; x2 � 8; 889� �.

THE ANALYTICAL APPROACH TO SOLVING AN LP:THE SIMPLEX METHOD

The solving of an LP is easily understood by explaining the meaning of economicconcepts involved in it.

a. Land constraint b. Land and labor constraints

x2

E3

Xcx = C

∇F

0 x1E1 0

x2

E4

E3

E3

X

cx = C

x*2

x*1

∇F

x1E1

FIGURE 10.3 Farmer’s LP Problem


3GC10 05/15/2014 11:6:33 Page 202

Notion of Technical Equivalence

We explain the notion of technical equivalence and opportunity cost. Assume thatfarmers have one acre of land. If they devote it to wheat only (activity a1), they producetwo metric tons of wheat; if they devote to tomatoes only (activity a2), they producefour metric tons of tomatoes; if they devote it to fodder only (activity a3), theyproduce five metric tons of fodder; if they devote it to cotton only (activity a4), theyproduce one metric of cotton. Our matrix of technical coefficients is:

Activity Wheat a1 Tomatoes a2 Fodder a3 Cotton a4

Land 0.5 0.25 0.2 1

One unit (i.e., metric ton) of wheat requires 0.5 unit of land, one unit of tomatoesrequires 0.25 unit of land, one unit of fodder requires 0.2 unit of land, and one unit ofcotton requires one unit of land. If we use wheat as a reference commodity, ourtechnical matrix becomes:


Wheat a1 1 0.5 0.4 2

We say technically one unit of tomatoes is equivalent to 0.5 unit of wheat;implying an additional unit of tomatoes requires renouncing to 0.5 unit of wheat. Oneunit of fodder is equivalent to 0.4 units of wheat, and one unit of cotton is equivalentto two units of wheat. If we change the reference to cotton, then our matrix becomes:


Cotton a4 0.5 0.25 0.2 1

We say that one unit of wheat is equivalent to 0.5 unit of cotton, one unit oftomatoes is equivalent to 0.25 unit of cotton, and one unit of fodder is equivalent to0.2 units of cotton. In terms of opportunity cost, if we want to produce one unit oftomatoes, we have to give up 0.25 units of cotton.* This will free enough land forproducing one unit of tomatoes.

Let us introduce prices, that is, cj. Let the price of wheat be $420/ton; the price oftomatoes, $200/ton; the price of fodder, $120/ton; and the price of cotton, $1,000/ton. The production of one unit of tomatoes earns $200 and renounces to $250 worthof cotton. In these conditions, the farmer evidently will not opt for tomatoes, sincecotton is more remunerative.

*The notion of technical equivalence is basic in international trade. For instance, in Ricardo’smodel, Portugal transforms one unit of wheat into one unit of clothing, and England transformsone unit of wheat into two units of clothing. England has comparative advantage in clothing,and Portugal has a comparative advantage in wheat.

202 MATHEMATICS

3GC10 05/15/2014 11:6:35 Page 203

The notion of technical equivalence is easy to illustrate in case of one productioninput. The reference commodity is called basic commodity, and the remainingcommodities are called nonbasic commodities. We normalize in respect to the basiccommodity; the nonbasic commodities are expressed in terms of the basic commodity.Let us add another input, labor. The matrix becomes:


Land 0.5 0.25 0.2 1Labor 2 5 1 6

With two inputs, we need two basic activities. Let us consider wheat and tomatoesas basic activities. We have a partition of the technical matrix into two matrices; abasic matrix Am,

Activity Wheat a1 Tomatoes a2

Land 0.5 0.25Labor 2 5

and a nonbasic matrix An�m,

Activity Fodder a3 Cotton a4

Land 0.2 1Labor 1 6

We compute the inverse of the basic matrix, that is, A�1m . We premultiply both Am

and An�m by A�1m ; we obtain:


Wheat a1 1 0 0.375 1.75Tomatoes a2 0 1 0.05 0.5

We say that one unit of fodder is technically equivalent to 0.375 � a1 � 0.05 � a2,and one unit of cotton is technically equivalent to 1.75 � a1 � 0.5 � a2. In other words,if we renounce 1.75 � a1 � 0.5 � a2, we will liberate enough land and labor forproducing one unit of cotton.

In terms of profit, if we decide to produce one unit of fodder, we earn $120.However, we have to renounce to 0.375 � 420 � 0.05 � 200 � $167.5. Hence, foddermakes a loss of $120 � $167.5 = �$47.5; fodder is therefore not profitable in relationto wheat and tomatoes. If we decide to produce one unit of cotton, we earn $1,000; werenounce to 1.75 � 420 � 0.5 � 200 � $835 of wheat and tomatoes. Hence, cottonmakes a unit profit of $165; cotton is therefore more profitable in relation to wheatand tomatoes. Hence, cotton has to become a basic activity and replace either wheat ortomatoes.


3GC10 05/15/2014 11:6:35 Page 204

The Simplex Method

The solving of an LP relies on the simplex method. In geometry, a simplex is ageneralization of the notion of a triangle or tetrahedron to arbitrary dimension. An n-simplex is an n-dimensional polytope, which is the convex hull of its n+ 1 vertices. Forexample, a 2-simplex is a triangle, and a 3-simplex is a tetrahedron. A single point maybe considered a 0-simplex, and a line segment may be considered a 1-simplex. Asimplex may be defined as the smallest convex set containing the given vertices. InFigure 10.4a the line AB illustrates a one-dimensional simplex; in Figure 10.4b thetriangle ABC illustrates a two-dimensional simplex.

To illustrate the simplex method, we have to transform the LP from the standardform to the canonical form by adding m nonnegative slack variables. Let us considerthe canonical form of the LP problem:

Maximize cx subject toAx � b; x � 0

The LP is written in a tableau form as

u1

um

266666664

377777775

a1 am ana11 a1m a1n

am1 amn amn

266666664

377777775

bb1

bm

266666664

377777775

u1 u2 um1 0 00 1

00 0 1

266666664

377777775� T0

where a1, . . . , am, . . . , an are column vectors describing the technical coefficients ofactivities j � 1, . . . ,m, . . . , n. For instance, the vector a1 describes the technicalcoefficients of inputs i � 1, . . . ,m or input requirements for producing one unit ofproduct 1. The unit vectors are described by column vectors u1, . . . , un. The canonicalform of the LP may be written in vector form as

a1x1 � ∙ ∙ ∙ � amxm � . . . � anxn � b (10.13)

a. One-dimensional simplex b. Two‐dimensional simplex

x2

A

B

x1A

B

C

x2

x3

x1

FIGURE 10.4 Geometry of the Simplex

204 MATHEMATICS

3GC10 05/15/2014 11:6:36 Page 205

The first step in the solution process is to reorganize the tableau T0. We assumethe rank of A is m. This implies that m column vectors are independent. We select abasis ofm independent vectors. We partition the matrix A into a square matrix Am ofm independent vectors and a matrix An�m of m � n dependent vectors. We partitionthe vector of unknown variables x into basic variables associated with matrix Am,denoted x1, . . . , xmf g, and nonbasic (redundant) variables associated with the matrixAn�m, denoted xm�1, . . . , xnf g.

The operation of constructing a basis of unit vectors is accomplished by themethod of successive elimination. The first row is normalized by a11, then it issubtracted from each other row in such a way to obtain a unit vector, u1, in thefirst column. The second row is normalized by the coefficient in cell 2; 2� �; then it issubtracted from each other row in such a way to obtain a unit vector u2 with 1 in cell2; 2� �, and zeros elsewhere. The third row is normalized by the coefficient in cell 3; 3� �;then it is subtracted from each other row in such a way to obtain a unit vector u3 with1 in cell 3; 3� �, and zeros elsewhere. The elimination operation is carried for all rows soas to obtain a unit vector in each elimination operation. This elimination procedureamounts to an inversion of the matrix Am. Once completed, we obtain a new tableauTs useful for initiating the simplex procedure.

a1a2∙ ∙ ∙

∙ ∙ ∙

∙ ∙ ∙

zjPj

266666666664

377777777775

a1 a2 ∙ ∙ ∙ am am�1 ∙ ∙ ∙ as ∙ ∙ ∙ an1 0 ∙ ∙ ∙ 0 ρ1m�1 ∙ ∙ ∙ ρ1s ∙ ∙ ∙ ρ1n0 1 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

0 0 ∙ ∙ ∙ 1 ρmm�1 ∙ ∙ ∙ ρms ∙ ∙ ∙ ρmn

z1 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ zs ∙ ∙ ∙ znP1 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ Ps ∙ ∙ ∙ Pn

26666666666664

37777777777775

bμ1

μm

26666666666664

37777777777775

v1 ∙ ∙ ∙ ∙ ∙ ∙ vmδ11 δ1m

δm1 δmm

26666666666664

37777777777775

�Tb

(10.14)

where: zs �Pmi�1 ρisci; Ps � cs� zs

The tableau Tb shows that the basic variables are now expressed in terms ofnonbasic variables and the columns v1, . . . , vm are the columns of the inverse matrixof Am, that is, A

�1m � v1, . . . , vmf g. It provides us with a initial solution,

x1 � μ1, x2 � μ2, . . . , xm � μm, xm�1 � 0, . . . , xn � 0 (10.15)

We replace this solution in the objective function and obtain

F x� � � c1x1 � . . . cmxm � . . . � cnxn � c1μ1 � . . . � cmμm (10.16)

Can we improve the basic solution? We consider a nonbasic activity s, as; thisactivity is technically equivalent to a combination of basic activities

as �Xm

i�1 ρisai (10.17)


3GC10 05/15/2014 11:6:36 Page 206

In other words, one unit of as absorbs as many resources asPm

i�1 ρisai units ofbasic activities. If we want to increase the production of commodity s by one unit, weneed resources by disallowing the production of

Pmi�1 ρisai units of basic commodities.

To decide about the profitability of producing one unit of commodity s, we comparethe marginal revenue cs we gain versus the marginal revenue we renounce

Pmi�1 ρisci. If

cs >Pm

i�1 ρisci, it is profitable to produce commodity s; if cs <Pm

i�1 ρisci, we dropcommodity s. We let

zs �Xm

i�1 ρisci (10.18)

and Ps � cs � zs (10.19)

where zs is the opportunity cost of commodity s and Ps is net profit from producingcommodity s. If Ps > 0, we produce commodity s; if Ps < 0, we drop commodity s.

Assume Ps is strictly positive for some activity s. This activity is now profitableand becomes a basic activity. Since we cannot have more thanm basic activities, oneexisting basic activity has to become nonbasic or redundant. Our query now iswhich basic activity to remove. Since by producing one unit of commodity s, wewant to free resources by renouncing a combination of

Pmi�1 ρisas; we want to

remove a basic variable that frees more of the most constraining resource on theproduction of s.

For instance, farmers are producing soybeans and wheat using as inputs givenquantities of labor and land. They realize that it is most profitable now to produce rice.The question is which product to remove: soybeans or wheat? The answer is simple.Assume the production of a metric ton of rice requires 2 acres and 2.5 labor units;assume the farmers have 100 acres and 50 labor units. Hence, the maximum quantityof rice allowed by land is 50 tons; that allowed by labor is 20 tons. Obviously, thefarmers remove the product that frees more labor for the production of rice.

From tableau Tb the basic variables are expressed in terms of nonbasic variablesas

x1 � μ1 � ρ1m�1xm�1 � ∙ ∙ ∙. � ρ1nxn � μ1 �Xni�m�1

ρ1jxj

x2 � μ2 � ρ2m�1xm�1 � ∙ ∙ ∙. � ρ2nxn � μ2 �Xnj�m�1

ρ2jxj

xi � μi � ρim�1xm�1 � ∙ ∙ ∙. � ρinxn � μi �Xni�m�1

ρijxj

(10.20)

Since xs is a new basic variable, all other nonbasic variables remain at zero. Oursystem of equations (20) becomes

x1 � μ1 � ρ1sxs

x2 � μ2 � ρ2sxs

xi � μi � ρisxs

(10.21)

206 MATHEMATICS

3GC10 05/15/2014 11:6:40 Page 207

We want to set the value of the exiting variable to zero without violating the non-negativity of all variables. For column as, we consider the minimum of the ratios,

Minμiρis

� �, i � 1, . . . ,m (10.22)

We remove the variable i that has the minimum of these ratios. We set xi � 0 andxs � μi

ρis. If we remove the variable i for which μi

ρis> μr

ρrs, then the new base variable xs

violates the non-negativity of a basic variable xr, r ≠ i; the latter variable becomesnegative,

xr � μr � ρrsxs � μr � ρrsμiρis

< 0 (10.23)

The simplex method is an iterative procedure. We start from an initial vertex andmove to another vertex based on two rules:

i. Choose a new basic activity as which improves the objective function, i.e., Ps > 0.

ii. Remove from the basis the activity with minimum ratio Min μiρis

� �, i � 1, . . . ,m.

The iteration continues until no activity has Ps > 0; an optimum value is thenreached for the objective function.

Example: Find the maximum of F x� � � 10x1 � 8x2 subject to

10x1 � 5x2 � 505x1 � 5x2 � 355x1 � 15x2 � 80

We rewrite the LP in canonical form by adding non-negative slack variables:x3 � 0, x4 � 0, and x5 � 0.

10x1 � 5x2 � x3 � 505x1 � 5x2 � x4 � 355x1 � 15x2 � x5 � 80

cj 10 8 0 0 0

Current vertex

ci xi b a1 a2 a3 a4 a5 Min

a3 0 x3 50 10 5 1 0 0 5a4 0 x4 35 5 5 0 1 0 7a5 0 x5 80 5 15 0 0 1 16F x� � 0 zj 0 0 0 0 0

Pj � cj � zj 10 8 0 0 0


3GC10 05/15/2014 11:6:45 Page 208

The LP is set in the form of a tableau. The initial vertex (basis) is provided bya3, a4, and a5.Thecorrespondingsolution isx3 � 50, x4 � 35, x5 � 80, x1 � 0, x2 � 0,and F x� � � 0. We observe that the indicator Pj � cj � zj shows that activity a1 or a2 isprofitable. We decide to enter activity a1. The minimum ratio μi

ρis� 5 indicates that we

have to remove activity a3 from the basis. The tableau becomes:

cj 10 8 0 0 0

Current vertex


a1 10 x1 5 1 0.5 0.1 0 0 10a4 0 x4 4 0 2.5 �0.5 1 0 4a5 0 x5 55 0 12.5 �0.5 0 1 4.4F x� � 50 zj 10 5 1 0 0

Pj � cj � zj 0 3 �1 0 0

The new solution is x1 � 5, x4 � 4, x5 � 55, x2 � 0, x3 � 0, and F x� � � 50. Theindicator Pj shows that activity a2 is profitable and ought to be brought into the basis.Theminimum ratio indicates that activity a4 is to be removed.We obtain a new simplextableau:

cj 10 8 0 0 0

Current vertex

ci xi b a1 a2 a3 a4 a5

a1 10 x1 3 1 0 0.2 �0.2 0a2 8 x2 4 0 1 �0.2 0.4 0a5 0 x5 5 0 0 2 �5 1F x� � 62 zj 10 8 0.4 1.2 0

Pj � cj � zj 0 0 �0.4 �1.2 0

Thesolutionisx1 � 3, x2 � 4, x5 � 5, x3 � 0, x4 � 0, and F x� � � 62.The indicatorline shows that all coefficients are negative; there is no longer any activity that can bebrought profitably into the basis. The solution is optimal. We observe that z3 �0.4, z4 � 1.2,andz5 � 0, tosolve thedualof thisLP.Thevalueof theminimumfunctionis

z3b1 � z4b2 � z5b3 � 0.4 � 50 � 1.2 � 35 � 62

It is the same as the maximum value in the primal problem.Example: Consider the following standard maximum problem:Find non-negative numbers x1, x2, x3, and x4 such that

MaxF x� � � 2x1 � 4x2 � x3 � x4 subject to

x1 � 3x2 � x4 � 4

2x1 � x2 � 3

x2 � 4x3 � x4 � 3

x1 � 0, x2 � 0, x3 � , x4 � 0

208 MATHEMATICS

3GC10 05/15/2014 11:6:55 Page 209

In order to convert this to a problem with equations instead of inequalities, weadjoin the unit vectors a5, a6, a7 and write

x1a1 � x2a2 � x3a3 � x4a4 � x5a5 � x6a6 � x7a7 � b

where all xi are to be non-negative. Now it is a trivial matter to find our initial feasiblesolution for we simply take the unit vectors as our starting basis. We set our initialtableau. An initial solution is

x1 � 0, x2 � 0, x3 � 0, x4 � 0, x5 � 4, x6 � 3, x7 � 3, and F � 0

cj 2 4 1 1 0 0 0

Current vertex

ci xi b a1 a2 a3 a4 a5 a6 a7 Min

a5 0 x5 4 1 3 0 1 1 0 0 4a6 0 x6 3 2 1 0 0 0 1 0 ∞a7 0 x7 3 0 1 4 1 0 0 1 3F x� � 0 zj 0 0 0 0 0 0 0

Pj 2 4 1 1 0 0 0

The indicator line Pj shows positive coefficients. We may bring any of the vectorsai into the basis. We consider a4. The minimum ratio indicates that we have to removea7. The tableau becomes:

cj 2 4 1 1 0 0 0

Current vertex


a5 0 x5 1 1 2 �4 0 1 0 �1 1a6 0 x6 3 2 1 0 0 0 1 0 1.5a4 1 x4 3 0 1 4 1 0 0 1 ∞F x� � 3 zj 0 1 4 1 0 0 1

Pj 2 3 �3 0 0 0 �1

We may next bring in either a1 or a2. To avoid fractions, we choose a1; theminimum ratio shows that we have to evict a5. The tableau becomes:

cj 2 4 1 1 0 0 0

Current vertex


a1 2 x1 1 1 2 �4 0 1 0 �1 �0.25a6 0 x6 1 0 �3 8 0 �2 1 2 0.125a4 1 x4 3 0 1 4 1 0 0 1 0.75F x� � 5 zj 2 5 �4 1 2 0 �1

Pj 0 �1 5 0 �2 0 1


3GC10 05/15/2014 11:7:2 Page 210

We may now bring in either a3 or a7. We choose a3 since it has greater impact onthe objective function. We replace a6. The tableau becomes:

cj 2 4 1 1 0 0 0

Current vertex


a1 2 x1 1.5 1 0.5 0 0 0 0.5 0 3a3 1 x3 0.125 0 �0.375 1 0 �0.25 0.125 0.25 �0.33a4 1 x4 2.5 0 2.5 0 1 1 �0.5 0 1F x� � 5.625 zj 2 3.125 1 1 0.75 0.625 0.25

Pj 0 0.875 0 0 �0.75 �0.625 �0.25

At this point we have no choice. We have to introduce a2 and remove a4. Thetableau becomes:

cj 2 4 1 1 0 0 0

Current vertex

ci xi b a1 a2 a3 a4 a5 a6 a7

a1 2 x1 1 1 0 0 �0.2 �0.2 0.6 0a3 1 x3 0.5 0 0 1 0.15 �0.1 0.05 0.25a2 4 x2 1 0 1 0 0.4 0.4 �0.2 0F x� � 6.5 zj 2 2 4 1 1.35 1.1 0.45 0.25

Pj 0 0 0 0 �0.35 �1.1 �0.45 �0.25

The line Pj shows that there is no further improvement in the objective function.The optimal solution is x1 � 1, x2 � 1, x3 � 0.5, x4 � 0, x5 � 0, x6 � 0, x7 � 0, andF x� � � 6.5. We observe that z5 � 1.1, z6 � 0.45, and z7 � 0.25 are solutions of thedual of the present LP. The value of the minimum function is

z5b1 � z6b2 � z7b3 � 1.1 � 4 � 0.45 � 3 � 0.25 � 3 � 6.5

It is the same as the maximum value in the primal problem.Example:Minimize: F x� � � �x1 � 2x2 subject to

�2x1 � x2 � x3 � 2

�x1 � 2x2 � x4 � 7

x1 � x5 � 3

x1, x2, x3, x4, x5 � 0

210 MATHEMATICS

3GC10 05/15/2014 11:7:10 Page 211

The initial tableau is:

cj − 1 −2 0 0 0

Current vertex


a3 0 x3 2 �2 1 1 0 0 2a4 0 x4 7 �1 2 0 1 0 3.5a5 0 x5 3 1 0 0 0 1 ∞F x� � zj 0 0 0 0 0

Pj �1 �2 0 0 0

An initial solution is x3 � 2, x4 � 7, x5 � 3, x4 � 0, x5, and F x� � � 0. The line Pj

shows that we can introduce a2; the minimum ratio leads to remove a3. The tableaubecomes:

cj −1 − 2 0 0 0

Current vertex


a2 �2 x2 2 �2 1 1 0 0 ∞a4 0 x4 3 3 0 �2 1 0 1a5 0 x5 3 1 0 0 0 1 3F x� � �4 zj 4 �2 �2 0 0

Pj �5 0 2 0 0

The line Pj shows that we can introduce a1; the minimum ratio evicts a4. Thetableau becomes:

cj − 1 − 2 0 0 0

Current vertex


a2 �2 x2 4 0 1 �0.33 0.66 0a1 �1 x1 1 1 0 �0.66 0.33 0a5 0 x5 2 0 0 0.66 �0.33 1F x� � �9 zj �1 �2 1.33 �1.66 0

Pj 0 0 �1.33 1.66 0


3GC10 05/15/2014 11:7:13 Page 212

The line Pj shows that a3 has to move in; activity a5 moves out. The tableaubecomes:

cj −1 − 2 0 0 0

Current vertex

ci xi B a1 a2 a3 a4 a5

a2 �2 x2 5 0 1 0 0.5 0.5a1 �1 x1 3 1 0 0 0 1a3 0 x3 3 0 0 1 �0.5 1.5F x� � �13 zj �1 �2 0 �1 �2

Pj 0 0 0 1 2

The line Pj shows that there is no further improvement in the objective function.The optimal solution is x1 � 3, x2 � 5, x3 � 3, x4 � 0, x5 � 0, and F x� � � �13.

THE DUAL PROBLEM OF THE LP

One of themost important facts about LP is that to every LP problem there correspondsa dual problem. Allocation of resources implies valuation of these resources. Dualityconsists of introducing pricing in the LP. A firm knows the breakdown of its sales byproduct, such as wheat, rice, and oats; it wants also to impute the value of these sales toits inputs, such as land, labor, machinery, and fertilizers. To an optimal allocation ofresources there corresponds an equilibrium price vector that equates marginal revenuesand marginal costs and eliminates excess profits. In a perfectly competitive market, aproducer should be able to make no profit by selling his resources instead of selling theoutput of these resources. For instance, if a bakery is using flour, oil, and yeast to make100 loaves of bread, it should make exactly the same revenue, if it chooses to sell tothe market its inputs, that is, flour, oil, and yeast, instead of selling bread. The solutionto the dual provides a price vector y* that values inputs in a fair way. If the originalproblem, called primal, is the LP maximum problem


the dual problem is the LP minimum problem

miny G � yb subject to yA � c, y � 0 (10.25)

where y is the row vector y � y1, y2, . . . , ym� �

Written in full, the dual problem is

miny1, ..., ymG y1, . . . , ym� � � b1y1 � ∙ ∙ ∙ � bmym (10.26)

212 MATHEMATICS

3GC10 05/15/2014 11:7:15 Page 213

subject to

a11y1 � a21y2 � . . . � am1ym � c1. . . . . . . . . . . . . . . . . . . . . . . . . . .

a1jy1 � a2jy2 � . . . � amjym � cj. . . . . . . . . . . . . . . . . . . . . . . . . . .

a1ny1 � a2ny2 � . . . � amnym � cny1 � 0, . . . , ym � 0.

(10.27)

The similarities and differences between the primal and dual should be evident.Both problems involve finding extremum of a linear function by choice of non-negative variables subject to linear inequality constraints; both use the same parame-ters, namely the matrix A, the column vector b, and the row vector c; and both use atotal of m � n inequality constraints.

However, the primal involves choosing n variables summarized by the columnvector x, whereas the dual involves choosing m variables summarized by the rowvector y; the original problem is one of maximization, whereas the dual is one ofminimization. The constraints constants of each problem become the objectiveconstants of the other. By applying the same transformation one more time it is clearthe original problem would reappear. The geometry of the dual problem is illustratedin Figure 10.5. Figure 10.5a shows a maximization LP; Figure 10.5b shows its dual,which is a minimization LP.

The dual problems can be described in the form of a tableau as shown here:

x1 x2 xn

y1 a11 a12 a1n � b1ym am1 am2 amn � b2� � �

c1 c2 cn

0a. Maximum LP b. Minimum LP

cx = C

yb = C

x2 y2

y1x10

X∇F

∇G

FIGURE 10.5 Geometry of the Dual LP


3GC10 05/15/2014 11:7:15 Page 214

THE LAGRANGIAN APPROACH: EXISTENCE, DUALITY, ANDCOMPLEMENTARY SLACKNESS THEOREMS

The nature of the dual problems can be understood using Lagrange multiplier analysissince the dual variables can be considered the Lagrange multipliers of the primalproblem. Assuming the primal is the maximum problem

maxx F x� � � cx subject toAx � b, x � 0 (10.28)

we define the Lagrangian as

L x, y� � � cx � y b � Ax� � � cx � yb � yAx (10.29)

According to the Kuhn-Tucker theorem, x* is a solution of the LP if there exists arow vector y* such that the following Kuhn-Tucker conditions hold at x*, y*

@L@x

� c � yA � 0 (10.30)

@L@x

x � c � yA� �x � 0 (10.31)

x � 0

@L@y

� b � Ax � 0 (10.32)

@L@y

y � y b � Ax� � � 0 (10.33)

y � 0

On the other hand if the primal has been the minimum problem

miny G � yb subject to yA � c, y � 0 (10.34)

we define the Lagrangian as

L y, x� � � yb � c � yA� �x � yb � cx � yAx (10.35)

The Kuhn-Tucker theorem implies that y* is a solution if there exists a columnvector x* such that the following Kuhn-Tucker conditions hold at x*, y*

@L@y

� b � Ax � 0 (10.36)

y@L@y

� y b � Ax� � � 0 (10.37)

y � 0

214 MATHEMATICS

3GC10 05/15/2014 11:7:16 Page 215

@L@x

� c � yA � 0 (10.38)

@L@x

x � c � yA� �x � 0 (10.39)

x � 0

The Lagrangian and the Kuhn-Tucker conditions are the same for both problems.The fundamental theorems of LP are based on these conditions.

The first fundamental theorem of LP is the existence theorem, which states that anecessary and sufficient condition for the existence of a solution to an LP is that theopportunity sets of both the problem and its dual are nonempty.

To show that if feasible vectors exist for both problems then there also existsolutions for both, consider the inequality constraints of the dual problems,

Ax � b (10.40)

yA � c (10.41)

Premultiplying the first set of inequalities by the non-negative vector y yields

yAx � yb � G y� � (10.42)

while postmultiplying the second set of inequalities by the non-negative vector xyields

yAx � cx (10.43)

Thus, if x and y are feasible

F x� � � G y� � (10.44)

That is, the value of the objective function in the maximizing problem cannotexceed the value of the objective function in the dual minimizing problem.

Suppose that feasible vectors x0, y0 exist for both problems. Then, since theopportunity set for the primal is nonempty, containing x0, and since the objectivefunction is bounded,

F x� � � G y0� �

for any feasible x (10.45)

It follows that a solution exists for the primal. Similarly for the dual, theopportunity set contains y0 and the objective function is bounded

F x0� � � G y� � for any feasible y (10.46)

so the dual has a solution.


3GC10 05/15/2014 11:7:16 Page 216

The second fundamental problem theorem of LP is the duality theorem, whichstates that a necessary and sufficient condition for a feasible vector to represent asolution to an LP is that there exists a feasible vector for the dual problem for whichthe values of the objective functions of both problems are equal.

To show that if x* is a solution for the maximum problem then there exists a y*,which is feasible for the dual problem and for which the values of the objectivefunctions are equal, consider the Kuhn-Tucker conditions. The vector y* is feasiblesince, as seen y*A � c, y* � 0 and the conditions

c � y*A� �x* � 0 (10.47)

y* c � y*A� � � 0 (10.48)

imply that

F x*� � � cx* � y*Ax* � y*b � G y*� � (10.49)

demonstrates the equality of the values of the objective functions.The third fundamental theorem of LP is the complementary slackness theorem

that states that a necessary and sufficient condition for feasible vectors x* and y* tosolve the dual problems is that they satisfy the complementary slackness conditions,

c � y*A� �x* � 0 (10.50)

y* c � y*A� � � 0 (10.51)

The sufficiency follows directly from the duality theorem, since assuming x*and y* are feasible, then, from the complementary slackness conditions:

F x*� � � cx* � y*Ax* � y*b � G y*� � (10.52)

So, since the values of the objective functions are equal, x* and y* are solutions.Written out in full, the complementary slackness conditions require that

cj �Xm

i�1 aijy*i

� �x*j � 0, j � 1, 2, . . . , n (10.53)

y*i bi �Xn

j�1 aijx*j

� �� 0, i � 1, 2, . . . ,m (10.54)

Combining with the feasibility restrictions,

x*j � 0; x*j � 0 ifXm

i�1 aijy*i > cj; j � 1, 2, . . . , n (10.55)

Xm

i�1 aijy*i � cj;

Xm

i�1 aijy*i � cj if x*j > 0 j � 1, 2, . . . , n (10.56)

y*i � 0; y*i � 0 ifXn

j�1 aijx*j < bi; i � 1, 2, . . . ,m (10.57)

Xn

j�1 aijx*j � bi;

Xn

j�1 aijx*j � bi if y*i > 0 i � 1, 2, . . . ,m (10.58)

216 MATHEMATICS

3GC10 05/15/2014 11:7:17 Page 217

Thus, if a certain constraint is satisfied at the solution as a strict inequality then thecorresponding dual variable is zero at the solution, and if a variable is positive at thesolution then the corresponding inequality constraint in the dual problem is satisfiedas an equality. These conditions are extremely useful in solving LP problems. Forexample, the solution to the dual problem would indicate which primal variables arezero at the solution and which primal inequality constraints are satisfied at thesolution as equalities.

Interpretation of the Dual Variables

Since the variables of the dual problem are Lagrangian multipliers for the primalproblem they can be interpreted as the sensitivity of the optimal value of the objectivefunction with respect to changes in the constraint constants. Thus,

y* � @F*

@b(10.59)

Similarly, the effect of changing the constraint constant in the dual problem is

x* � @G*

@c(10.60)

Thus, the sensitivity of the optimal value of the objective function to changes inthe constraint constant is measured by the optimal value of the corresponding dualvariable. This interpretation is identical to that of nonlinear programming. In certainproblems of economic allocation the dual variables have the natural interpretation ofimputed prices, being the change in economic values (e.g., profit, revenue, utility, orcost) as an economic quantity changes. These prices are called shadow prices.

Example: Solve the following linear programming problem by solving the dualproblem.

miny1, y2, y3, y4 G � 6y1 � 20y2 � 3y3 � 20y4 subject to3y1 � 6y2 � y3 � 2y4 � 4� 4y1 � 2y2 � y3 � 5y4 � 2y1, y2, y3, y4 � 0

We write the dual LP as

maxx1, x2 F � 4x1 � 2x23x1 � 4x2 � 66x1 � 2x2 � 20� x1 � x2 � 3

2 x1 � 5x2 � 20x1, x2 � 0

We write the LP in canonical form with slack variables

3x1 � 4x2 � x3 � 66x1 � 2x2 � x4 � 20�x1 � x2 � x5 � 32x1 � 5x2 � x6 � 20

x1, x2, x3, x4, x5, x6 � 0


3GC10 05/15/2014 11:7:27 Page 218

The initial tableau is:

cj 4 2 0 0 0 0

Current vertex

ci xi b a1 a2 a3 a4 a5 a6 Min

a3 0 x3 6 3 �4 1 0 0 0 �1.5a4 0 x4 20 6 2 0 1 0 0 10a5 0 x5 3 �1 1 0 0 1 0 3a6 0 x6 20 2 5 0 0 0 1 4F x� � 0 zj 0 0 0 0 0 0

Pj 2 4 1 1 0 0

Based on Pj and minimum ratio, we enter a2 and remove a5; the tableaubecomes:

cj 4 2 0 0 0 0

Current vertex


a3 0 x3 18 �1 0 1 0 4 0a4 0 x4 14 8 0 0 1 �2 0 1.75a2 2 x2 3 �1 1 0 0 1 0a6 0 x6 5 7 0 0 0 �5 1 0.714F x� � 6 zj �2 2 0 0 2 0

Pj 6 0 0 0 �2 0

Based on Pj and minimum ratio, we introduce a1 and remove a6; the tableaubecomes:

cj 4 2 0 0 0 0

Current vertex


a3 0 x3 18.71 0 0 1 0 3.29 0.14 5.70a4 0 x4 8.29 0 0 0 1 3.71 �1.14 2.23a2 2 x2 3.71 0 1 0 0 0.29 0.14 13.00a1 4 x1 0.71 1 0 0 0 �0.71 0.14F x� � 10.29 zj 4 2 0 0 �2.29 0.86

Pj 0 0 0 0 2.29 �0.86

218 MATHEMATICS

3GC10 05/15/2014 11:7:31 Page 219

The line Pj indicates that we have to introduce a5; the minimum ratio suggestsremoving a4. The tableau becomes:

cj 4 2 0 0 0 0

Current vertex


a3 0 x3 11.38 0 0 1 �0.88 0.00 1.15 5.70a5 0 x5 2.23 0 0 0 0.27 1.00 �0.31 2.23a2 2 x2 3.08 0 1 0 �0.08 0.00 0.23 13.00a1 4 x1 2.31 1 0 0 0.19 0.00 �0.08F x� � 15.38 zj 4 2 0 0.62 0.00 0.15

Pj 0 0 0 �0.62 0.00 �0.15

The solution of the dual is x1 � 2.31, x2 � 3.08, and F � 15.38.Thesolutionoftheprimal isx3 � 0, x4 � 0.62, x5 � 0, x6 � 0.15, andG � 15.38.

ECONOMIC THEORY AND DUALITY

To study the economics of duality let us consider a firm that produces three outputs,O1, O2, and O3, using three inputs, I1, I2, and I3. The unit profit contribution of theproducts O1, O2, and O3 are $40, $25, and $50, respectively. The number of units ofraw materials available are 36, 60, and 45 for I1, I2, and I3, respectively. Let thecompany produce x1, x2, x3 � 0 units of the products O1, O2, and O3, respectively.Then the problem can be expressed mathematically as

Maximize F x� � � 40x1 � 25x2 � 50x3 subject to :

x1 � 2x2 � x3 � 362x1 � x2 � 4x3 � 602x1 � 5x2 � 1x3 � 45x1, x2, x3 � 0

The optimal solution is x1 � 20, x2 � 0, x3 � 5, and F x� � � $1; 050.The dual is minimize G y� � � 36y1 � 60y2 � 45y3

y1 � 2y2 � 2y3 � 402y1 � y2 � 5y3 � 25y1 � 4y2 � y3 � 50

The optimal solution is y1 � 0, y2 � $10, y3 � $10, andG y� � � $1; 050.We inter-pret the dual variables as shadow prices imputed to the firm’s resources by allocatingthe profit (contribution) of the firm to its resources at the margin. We know already,the contribution of O1, O2, and O3 to the objective function. The dual tells us themarginal contribution of I1 to the objective function is zero, the marginal contribution


3GC10 05/15/2014 11:7:31 Page 220

of I2 is $10, and that of I3 is also $10. This information was not available until wesolved the dual problem.

Suppose the manager of the company wants to sell the three raw materials I1, I2,and I3 instead of using them for making products O1, O2, andO3 and, then, by sellingthe products earn a profit of $1,050. Suppose the selling prices were y1 � 0, y2 �$10, and y3 � $10 per unit of raw materials I1, I2, and I3, respectively. Then the costto the purchaser of all the three raw materials will be

36y1 � 60y2 � 45y3 � 36 � 0 � 60 � 10 � 45 � 10 � 1; 050

The purchaser will like to set the selling prices of I1, I2, and I3 so that the total costis minimum subject to the constraint that the producer will earn at least as much byselling the inputs used in activity j as by selling the product of that activity. So theobjective function of the purchaser will be to minimize

36y1 � 60y2 � 45y3

subject to the constraint that the marginal cost of purchase of inputs used in anyactivity j exceeding or equal the unit profit of that activity,

Pmi�1 aijyi � cj,

y1 � 2y2 � 2y3 � 402y1 � y2 � 5y3 � 25y1 � 4y2 � y3 � 50

9=; (10.61)

The marginal values of raw materials I1, I2, and I3 is y1 � 0, y2 � $10, and y3 �$10 per unit, respectively. Thus if managers sell the raw materials I1, I2, and I3 atprice y1 � 0, y2 � $10, and y3 � $10 per unit, respectively, they will get the sameincome $1,050 if they sell the products O1, O2, and O3 made from these resources.

In a perfectly competitive economy, it is assumed that, if a firm could make profitsin excess of the value of its resources, then some other firm would enter the marketwith a lower price, thus tending to eliminate these excess profits. The duality theory oflinear programming has had a significant impact on mathematical economics throughthe interpretation of the dual as the price-setting mechanism in a perfectly competitiveeconomy. Let us consider the Lagrangian of the LP maximization problem

L x, y� � � cx � y b � Ax� � � cx � yb � yAx

L x, y� � � Xn

j�1 cjxj �Xm

i�1 yi bi �Xn

j�1 aijxj� � (10.62)

The firm generates revenues and incurs costs by engaging in production activitiesand by buying and selling resources. Note that if

bi >Xn

j�1 aijxj (10.63)

the firm sells bi �Pnj�1 aijxj units of resources i to the marketplace at a price yi. The

market reacts to excess supply by reducing prices. If, however,

bi <Xn

j�1 aijxj (10.64)

220 MATHEMATICS

3GC10 05/15/2014 11:7:32 Page 221

the firm demandsPn

j�1 aijxj � bi units of resources i from the marketplace at a price yi.The market reacts to excess demand by hiking up prices. Equilibrium in the market forresource i requires

yi bi �Xn

j�1 aijxj� �

� 0 (10.65)

Now consider the Lagrangian of the LP minimization problem

L y, x� � � yb � c � yA� �x � yb � cx � yAx

L y, x� � � Xm

i�1 yibi �Xn

j�1 cj �Xm

i�1 aijyi� �

xj(10.66)

Note that the termPm

i�1 aijyi is the market opportunity cost for the firm using theresources a1, a2j, . . . , amj, in order to engage in the jth activity at unit level. Twoconsequences immediately follow. First, if the market sets the prices yi so that therevenue from engaging in an activity exceeds the market cost, that is,

cj >Xm

i�1 aijyi (10.67)

then the firm would be able to make arbitrarily large profits by engaging in the activityat an arbitrarily high level, a clearly unacceptable situation from the standpoint of themarket. The market instead will always choose to set its prices yi such that

Xm

i�1 aijyi � cj, j � 1, . . . , n (10.68)

Second, if the market sets the price of a resource so that the revenue from engagingin that activity does not exceed the potential revenue from the sale of the resourcesdirectly to the market, that is,

cj <Xm

i�1 aijyi (10.69)

then the firm will not engage in that activity at all. In this latter case, the opportunitycost associated with engaging in the activity is in excess of the revenue produced byengaging in the activity. Hence, equilibrium requires

cj �Xm

i�1 aijyi� �

xj � 0 (10.70)

Summarizing, equilibrium is established when each output or resource market isin balance:

bi �Xm

i�1 aijx*j

� �y*i � 0, i � 1, . . . ,m (10.71)

cj �Xn

j�1 aijy*i

� �x*j � 0, j � 1, . . . , n (10.72)

These equations are the complementary-slackness conditions of linear program-ming. The first condition implies that either the amount of resource i that is unused


3GC10 05/15/2014 11:7:32 Page 222

(slack in the ith constraint of the primal) is zero, or the price of resource i is zero. This isintuitively appealing because if a firm has excess of a particular resource, then themarket should not be willing to pay anything for the surplus of that resourcebecause the market wishes to minimize the firm’s profit. There may be a nonzeromarket price on a resource only if the firm is consuming all of that resource that isavailable. The second condition implies that either the amount of excess profit onthe jth activity (slack in the jth constraint of the dual) is zero or the level of activity jis zero. This is also appealing from the standpoint of the perfectly competitivemarket, which acts to eliminate any excess profits. If we had an equilibriumsatisfying the complementary-slackness conditions, then, by equating the Lagran-gians for this equilibrium, we can quickly conclude that the extreme values of theprimal and dual problems are equal, that is,

Xn

j�1 cjx*j �

Xm

i�1 biy*i (10.73)

Observe that this condition has the usual interpretation for a firm operating in aperfectly competitive market. It states that the maximum profit that the firm can makeequals the market evaluation of its initial endowment of resources. That is, the firmmakes no excess profits.

SUMMARY

Islamic finance uses LP for optimizing portfolios of assets and making efficient use ofscarce resources. The chapter covers the formulation of the LP, the standard form andcanonical form of the LP, the simplex method, the dual problem of the LP, theLagrangian approach to LP, the interpretation of the dual variables, and the economictheory and duality.

The application of the principles of LP enables Islamic financial corporations toattain considerably efficiency. At the level of investment projects, LP principles lead tothe best use of scarce capital resources by avoiding inefficient projects. At the level of anation, LP principles lead to full employment of labor resources; at the level ofinternational trade, LP principles lead to the most advantageous pattern of tradeamong nations.

QUESTIONS

1. Try to solve manually, then check your answer with the Microsoft Excel solver.

a. Find y � x1, . . . , x5� � � 0, which minimizes x1 � 6x2 � 7x3 � x4 � 5x5subject to

5x1 � 4x2 � 13x3 � 2x4 � x5 � 20

x1 � x2 � 5x3 � x4 � x5 � 8

b. Show the dual solution and the value of dual objective function.c. Analyze the sensitivity of the solution if b1 � 20 is increased by 5.

222 MATHEMATICS

3GC10 05/15/2014 11:7:32 Page 223

2. Use the Microsoft Excel solver for the following problem:

a. Consider the LP maximum problem

maxx1, x2 F � 3x1 � 2x2 subject to

2x1 � x2 � 6

x1 � 2x2 � 8

x1 � 0, x2 � 0

b. Show the dual solution and the value of dual objective function.c. Analyze the sensitivity of the solution if b2 � 8 is increased by 2.d. Analyze the sensitivity of the solution if c1 � 3 becomes c1 � 1.

3. Use the Microsoft Excel solver for the following:

a. Max F � �6x1 � 3x2 subject to

x1 � x2 � 12x1 � x2 � 13x2 � 2x1, x2 � 0

b. Show the dual solution and the value of dual objective function.

4. a. Minimize F � �5x1 � 7x2 � 12x3 � x4 subject to

2x1 � 3x2 � 2x3 � x4 � 383x1 � 2x2 � 4x3 � x4 � 55x1, x2, x3, x4 � 0

b. Show the dual solution and the value of dual objective function.c. Analyze the sensitivity of the solution if b2 � 55 is decreased by 5.

5. a. Maximize F � 5x1 � 3x2 � 2x3 subject to

4x1 � 5x2 � 2x3 � x4 � 203x1 � 4x2 � x3 � x4 � 30x1, x2, x3, x4 � 0

b. Show the dual solution and the value of dual objective function.c. Analyze the sensitivity of the solution if b1 � 20 is decreased by 5.d. Analyze the sensitivity of the solution if c3 � 2 becomes c3 � 3.

6. a. Solve the dual of:

Minimize F � 3x1 � 9x2 � 5x3 � 6x4 subject to4x1 � 3x2 � 5x3 � 8x4 � 242x1 � 7x2 � 4x3 � 6x4 � 17x1, x2, x3, x4 � 0


3GC10 05/15/2014 11:7:33 Page 224

b. Show the primal solution and the value of primal objective function.

7. a. Solve the dual of

Minimize F � �2x1 � 4x2 � 3x3 subject to9x1 � 2x2 � 8x3 � 53x1 � 3x2 � 3x3 � 77x1 � 5x2 � 2x3 � 9

x1, x2, x3 � 0

b. Show the primal solution and the value of primal objective function.

8. a. Solve the dual of

Minimize F � �101x1 � 87x2 � 23x3 subject to6x1 � 13x2 � 3x3 � 116x1 � 11x2 � 2x3 � 45x1 � 5x2 � x3 � 12x1, x2, x3 � 0

b. Show the primal solution and the value of primal objective function.c. What is the solution of the LP obtained by decreasing b2 � 45 to b2 � 30?d. By how much can b2 � 45 increase or decrease without changing the optimal

basis?e. What is the solution of the LP obtained by increasing the coefficient of x1 on

the objective function by 25?f. By how much can the coefficient of x3 in the objective function increase ordecrease without changing the optimal basis?

224 MATHEMATICS

3GC11 05/15/2014 11:31:16 Page 225

PART

TwoStatistics

3GC11 05/15/2014 11:31:16 Page 226

3GC11 05/15/2014 11:31:16 Page 227

CHAPTER 11Introduction to Probability Theory:

Axioms and Distributions

I slamic finance uses probability theory and statistics, including econometrics. Thesedisciplines are widely used inmany other sciences such as physics, medicine, biology,

engineering, communications, economics, management, and insurance. We encounterdeterministic as well as probabilistic events. You may predict with certainty that thesun will rise tomorrow, or the sun will set at 7.21 p.m. on March 20 of every year inWashington, D.C. However, you may not be as certain whether it will rain next yearon March 20 in Washington, D.C. The best you can do is to look at the record of therainfall in Washington, D.C., in the past, say during the past 100 years, and if yourealize it rained on 42 out of 100 days, then you may state that based on pastexperience it may rain on March 20 next year with a probability of 42 percent.

Probability theory is the science of randomness. It formulates models to studyrandom events and random variables, and emit probabilities as regards the occur-rences of random events. Statistics is the science of collection of data about specifiedvariables; it formulates models for analyzing sampled data.

Finance involves risk and uncertainty. Assets are risky and may be highly volatile.Each return has a risk. Investors form expectations regarding the payoffs of theirinvestment or hedging strategies. A firm may invest in new plants or products;however, actual payoffs are contingent on the state of the world at the time ofmaturity of the investment and may turn out totally different from expected payoffs.Investors try to hedge their positions in the stocks or commodity markets by shortingassets when prices start falling, or buying or selling derivatives. Moreover, traderswould like to buy stocks; however, they require a risk-premium in relation to risklessassets. Some traders want to evaluate the performance of their portfolios or run astress test, such as the value-at-risk (Var) test, on their portfolios and assess howmuchvalue of their portfolio is at risk in case some large drops in the price of their assets takeplace. Traders want to forecast the value of their portfolio and their profits or losses atsome time-horizon (next month, quarter, or year). Often, traders need to know howmuch they are prepared to pay for a derivative, such as a call or a put option, howmuch return they require on a particular stock, or how to manage the risk of theirportfolio. Companies want to make forecasts about their sales. Banks would like toassess lines and add tellers at peak hours. Probabilistic and statistical models provideadequate tools for making educated management or investment decisions.

227

3GC11 05/15/2014 11:31:16 Page 228

THE EMPIRICAL BACKGROUND: THE SAMPLE SPACE AND EVENTS

This section covers the notions of probabilistic experiment, sample space, and events.

Experiment

Before we speak of probabilities, we must agree on an idealized model of a particularconceptual experiment such as tossing a coin or counting the number of telephonecalls. At the outset we must agree on the possible outcomes of the experiment:the sample space and the probabilities associated with them. An experiment is theoperation of establishing certain conditions that may produce one of several outcomesor results. The mathematical theory of probability gains practical value and anintuitive meaning in connection with real or conceptual experiments such as tossinga coin once, tossing a coin 100 times, or throwing three dice. A theory necessarilyinvolves idealization and our first idealization concerns the possible outcomes of an“experiment” or “observation.”

Sample Space

The sample space Ω associated with an experiment is the collection of all possibleoutcomes of the experiment. If the experiment is throwing a die then the sample spaceis Ω � 1; 2; 3; 4; 5; 6f g. If the experiment is throwing a coin the sample space isH head� �, T tail� �f g. If the experiment is measuring a stock price, the sample spacemaybe written as x; 0 � x <∞f g. If the experiment is measuring the rate of return on astock, the sample space may be written as x;� ∞ < x <∞f g. If the experiment ismeasuring the payoff of a call option, the sample space may be written asx; 0 � x <∞f g.

Events

For uniform terminology, the results of experiments or observations will be calledevents. Thus, we speak of the event that of five coins tossed, more than three showheads. We distinguish between compound (or decomposable) and simple (indecom-posable) events. For example, saying that a throw with two dice resulted in “sum six”amounts to saying that it resulted in {(1, 5) or (2, 4), or (3, 3), or (4, 2), or (5, 1)} andthis enumeration decomposes the event “sum six” into five simple events. A particularoutcome, that is, an element inΩ, is called a sample point or sample. An eventA is a setof outcomes or, in other words, a subset of the sample space Ω. The event ωf gconsisting of a single sample ω ∈ Ω is called an elementary event. The individualoutcomes denoted by ω are called elementary events. The empty set ∅ and Ω itself areevents; ∅ is sometimes called the impossible event, and Ω the certain or sure event.

Example: Coin tossing: for the experiment of tossing a coin three times, thesample space consists of eight points,Ω � HHH, HHT, HTH, THH, HTT, THT,fTTH, TTTg. The event A, “two or more heads,” is the aggregate of four points,that is, A � HHH, HHT, HTH, THHf g. The event B, “just one tail,” means eitherHHT, or HTH, or THH; we say that B contains three points, that is, B �HHT, HTH, THHf g.

228 STATISTICS

3GC11 05/15/2014 11:31:16 Page 229

Example: Suppose that a sample of 100 stocks is taken in order to estimate howmany stocks are Sharia compliant. The only property of the sample of interest is thenumber x of Sharia-compliant stocks; this may be an integer between 0 and 100. Inthis case wemay agree that our sample space consists of the 101 “points” 0, 1, 2, . . . ,100. Every particular sample or observation is completely described by stating thecorresponding point x. An example of a compound event is the result that “themajority of the stocks sampled are Sharia compliant.” This means that the experimentresulted in one of the 50 simple events 51, 52, . . . , 100.

It should be clear that we shall never speak of probabilities except in relation to agiven sample space. We start with the notion of a sample space and its points. Thesample space provides a model of an ideal experiment in the sense that, by definition,every thinkable outcome of the experiment is completely described by one, and onlyone, sample point. It is meaningful to talk about an event A only when it is clear forevery outcome of the experiment whether the event A has occurred or has notoccurred. The collection of all those sample points representing outcomes where Ahas occurred completely describes the event. We therefore define the event to signifythe same as an aggregate of sample points. We say an event A consists of (or contains)certain points, namely those representing outcomes of the ideal experiment in which Aoccurs.

We can combine events to form new events using the various set operations(Figure 11.1):

i. A∪B is the event that occurs if A occurs or B occurs, or both.ii. A∩B is the event that occurs if A occurs and B occurs.iii. A�B is the event that occurs if A occurs and B does not occur.iv. Ac, the complement of A, is the event that occurs if A does not occur.v. Two events, A1 and A2, are mutually exclusive if they are disjoint, that is, if

A1 ∩A2 � ∅. In other words, A1 and A2 are mutually exclusive if they cannotoccur simultaneously.

A

A ∪ B

A/ B

B

Sample space Ω

i.

iii. iv. v.

ii.

Sample space Ω Sample space Ω Sample space Ω

A B

Sample space Ω

AA

B

A ∩ B

A1 ∩ A2 = ∅

A2

A1

Ac

FIGURE 11.1 Combination of Events

Introduction to Probability Theory 229

3GC11 05/15/2014 11:31:17 Page 230

Example: Toss a die and observe the number that appears on the top. The samplespace consists of six possible numbers: Ω � 1; 2; 3; 4;5; 6f g.

Let A be the event that an even number occurs, B that an odd number occurs, andC that a prime number occurs:

A � 2; 4; 6f g, B � 1; 3; 5f g,C � 2; 3; 5f gThen A∪C � 2; 3; 4; 5; 6f g is the event that an even or a prime number occurs,

B∩C � 3;5f g is the event that an odd prime number occurs, and Cc � 1; 2; 6f g is theevent that a prime number does not occur.

DEFINITION OF PROBABILITY

Probability theory is the study of random experiments. If a coin is tossed in the air, it iscertain that the coin will land; but it is not certain that a “Head” will appear.Probability has been studied empirically. For instance, we toss a coin n times, and wecall “Head” success s; we define the ratio

f � s/n (11.1)

This ratio is called relative frequency; it becomes stable and approaches a limit asn becomes large. This stability is the basis of probability theory. In probability theory,we define a mathematical model for the random experiments by assigning “probabil-ities” (or the limit values of the relative frequencies) to the events connected with theexperiment. The probability p of an event A is defined as follows: if A can occur in sways out of a total of n likely ways, then

p � P A� � � Number of favorable eventsNumber of possible events

� s/n (11.2)

The outcomes of a random trial are called random events. Let the frequency of anoutcome A in n repeated trials be the ratio nA/n of the number nA of occurrences of Ato the total n of trials. The number measured by the observed frequencies of a randomevent A is called the probability of A,

P A� � � nAn

(11.3)

Axioms of Probability

Let A be a subset of Ω and let A be a collection of such subsets of Ω. If we observe theoutcome ω, and ω is in A, we say that A has occurred. Intuitively, it is possible tospecify P A� �, the probability that A will occur based on repeating the experiment alarge number of times. It is reasonable to require that the function P A� � satisfy:

Axiom 1: For every A in A, 0 � P A� � � 1.Axiom 2: P Ω� � � 1.

230 STATISTICS

3GC11 05/15/2014 11:31:17 Page 231

Axiom 3: If A and B are mutually exclusive events, then P A∪B� � � P A� � � P B� �.Axiom 4: If A1,A2, . . . is a countable sequence from A and Ai∩Aj is the null set

for all i ≠ j, then PS ∞

i�1 Ai� � � P ∞

i�1 P Ai� �.Some basic results follow directly from our axioms (Figure 11.2).

i. If A � B, then P A� � � P B� �.ii. If A and B are two events, then P A B� � � P A� � � P A∩B� �.iii. If A and B are two events, then P A∪B� � � P A� � � P B� � � P A∩B� �.iv. If ∅ is the empty set, then P ∅� � � 0.v. If Ac is the complement of an event A, then P Ac� � � 1 � P A� �.

Example: Let three coins be tossed and the number of heads observed; then thesample space is Ω � HHH, HHT, HTH, THH, HTT, THT, TTH, TTTf g. Thenumber of heads can be A � 0; 1; 2; 3f g. We obtain the following probabilities:

P 0� � � 1/8;P 1� � � 3/8;P 2� � � 3/8;P 3� � � 1/8

Let A be the event that at least one head appears and let B be the event that allheads or all tails appear: A � 1; 2; 3f g and B � 0; 3f g.

P A� � � P 1� � � P 2� � � P 3� � � 38� 38� 18� 7/8

P B� � � P 0� � � P 3� � � 18� 18� 1/4

RANDOM VARIABLE

Although it is conceptually possible to enumerate all possible outcomes of anexperiment, it may be a practical impossibility to do so and, for most purposes, itis unnecessary to do so. It is usually enough to record the outcome by some functionthat assumes values on the real line. That is, we assign to each outcome ω a realnumber X ω� � and, if ω is observed, we record X ω� �. Such an assignment is called arandom variable. If we toss a die we could take X ω� � � 1 if the player wins and �1 ifthe gambling house wins.

AB BA A B

Sample space Ω Sample space Ω Sample space Ω

B/AA/B A/B

A ∩ B

FIGURE 11.2 Implications of the Probability Axioms


3GC11 05/15/2014 11:31:18 Page 232

Definition:A random variableX on a sample spaceΩ is a function fromΩ into theset R of real numbers; it maps outcomes of random experiments to numbers. Randomvariables can be classified as discrete, which are variables that have specific values, orcontinuous, which are variables that can have any values within a continuous range.Formally, a random variableX is a real valued function defined on Ω such that the setω : X ω� � � xf g is a member of A for every real number x. The function

FX x� � � P ω : X ω� � � xf g� �� (11.4)

is called the distribution function of the random variable X.Example: Payoff to a contingent contract is a random variable; hence, dividends

from a share, payoffs of a futures contract, call option, put option, and swap are allrandom variables.

TECHNIQUES OF COUNTING: COMBINATORIAL ANALYSIS

This section covers the notions factorial notation, permutations, ordered samples,combinations, and tree diagrams.

In a random experiment we often have to count the number of events. Techniquesof counting called combinatorial analysis have been used. If some experiment can beperformed in n1 different ways, and if, following this experiment, a second experimentcan be performed in n2 different ways, and if, following the second experiment, a thirdexperiment can be performed in n3 different ways, then the number of ways theexperiments can be performed in the order indicated is the product n1n2n3.

Example: Suppose a license plate contains two distinct letters followed by threedigits with the first digit not zero. How many different license plates can be printed?The answer is

26 � 25 � 9 � 10 � 10 � 585,000

Factorial Notation

The product of the positive integers from 1 to n occurs very often in probability theoryand is denoted by the special symbol n! (read “n factorial”):

n! � 1 � 2 � 3 . . . � n � 2� � � n � 1� �nIt is convenient to define 0! � 1.Example: 5! � 1 � 2 � 3 � 4 � 5 � 120,

8!6!

� 8 � 7 � 6!6!

� 8 � 7 � 56.

Permutations

An arrangement of a set of n objects in a given order is called permutation of theobjects (taken all at a time). An arrangement of any r � n of these objects in a givenorder is called a permutation of n objects taken r at a time. The number of

232 STATISTICS

3GC11 05/15/2014 11:31:18 Page 233

permutations of n objects taken r at a time is denoted by P n, r� �. It is given by thefollowing formula:

P n, r� � � n n � 1� � n � 2� � . . . n � r � 1� � � n!n � r� �! (11.5)

Example: Find the number of permutations of six objects, a, b, c, d, e, and f ,taken three at a time. P 6; 3� � � 120.

Ordered Samples

Many problems in combinatorial analysis and, in particular, probability are con-cerned with choosing a ball from an urn containing n balls (or a card from a deck, or aperson from a population). When we choose one ball after another from the urn, say rtimes, we call the choice an ordered sample of size r. Next we consider two cases.

Sampling with replacement: The ball is replaced in the urn before the next ball ischosen. Now since there are n different ways to choose each ball, the number ofordered samples with replacement of size r is

n � n ∙ ∙ ∙ � n � nr (11.6)

Sampling without replacement: Here the ball is not replaced in the urn before thenext ball is chosen. Thus there are no repetitions in the ordered sample. In otherwords, an ordered sample of size r without replacement is simply an r permutation ofthe objects in the urn. Thus there are

P n, r� � � n n � 1� � n � 2� � ∙ ∙ ∙ n � r � 1� � � n!n � r� �! (11.7)

P n, r� � is the number of different ordered samples without replacement from apopulation of n objects.

Combinations

Suppose we have a collection of n objects. A combination of these n objects taken r at atime is any subset of r elements. In other words, an r–combination is any selection ofr of the n objects where the order does not count. The number of combinations ofn objects taken r at a time is denoted by

C n, r� � � nr

� �(11.8)

Since each combination of n objects taken r at a time determines r! permutationsof the objects, we have the following relation:

P n, r� � � r!C n, r� � (11.9)


3GC11 05/15/2014 11:31:19 Page 234

Thus we obtain

C n, r� � � P n, r� �r!

� n!r! n � r� �! (11.10)

Example:

i. The combinations of the letters a, b, c, and d taken three at a time areabc, abd, acd, and bcd. Observe that the following combinations are equal:abc, acb, bac, bca, cab, and cba. Each is denoted by the set abc.

C 4; 3� � � P 4; 3� �3!

� 4.3:23.2

� 4

ii. Howmany committees of three can be formed from eight people? Each committeeis a combination of eight people taken three at a time.

C 8; 3� � � 83

� �� 8.7:6

3.2� 56

Tree Diagrams

Tree diagrams are used in Islamic finance. A tree diagram is a device to enumerate allthe possible outcomes of a sequence of experiments where each experiment can occurin a finite number of ways. The construction of tree diagrams is illustrated in thefollowing example.

Example: Find the product set A � B � C where A � 1; 2f g, B � a, b, cf g, andC � 3; 4f g. With the help of a tree diagram (Figure 11.3) we find

A � B � C � 1, a, 3� �, 1, a, 4� �, 1, b, 3� �, 1, b, 4� �, 1, c, 3� �, 1, c, 4� �, 2, a, 3� �,f2, a, 4� �, 2, b, 3� �, 2, b, 4� �, 2, c, 3� �, 2, c, 4� �g

CONDITIONAL PROBABILITY AND INDEPENDENCE

This section introduces the notions of conditional probability of an event, Bayes’Theorem, and independence of events.

Conditional Probability

Let E be an arbitrary event in a sample spaceΩwith P E� � > 0. The probability that anevent A occurs once E has occurred or, in other words, the conditional probability ofA given E, written P AjE� �

, is defined as

P AjE� � � P A∩E� �P E� � (11.11)

234 STATISTICS

3GC11 05/15/2014 11:31:19 Page 235

From the definition of conditional probability, we obtain the multiplicationtheorem

P A∩E� � � P E� � ?P AjE� �(11.12)

Example: Let a pair of fair dice be tossed. If the sum is 6, find the probability thatone of the dice is a 2.

E � sum is 6f g � 1; 5� �, 2; 4� �, 3; 3� �, 4; 2� �, 5; 1� �f g and

A � 2 appears on at least one dief g

Now E consists of five elements and two of them, 2; 4� �, 4; 2� �f g, belong to A.Hence, A∩E � 2; 4� �, 4; 2� �f g. Then P AjE� � � P A∩E� �

P E� � � 2/5.On the other hand, since A consists of 11 elements,

A � 2; 1� �, 2; 2� �, 2; 3� �, 2; 4� �, 2; 5� � 2; 6� �, 1; 2� �, 3; 2� �, 4; 2� �, 5; 2� �, 6; 2� �f g

and Ω consists of 36 elements, P A� � � 11/36. We note that the conditioning of A on Ehas increased the probability of A.

1

2

a

b

c

a

b

c

3

4

3

4

3

4

3

43

3

4

4

(1,a,3)

(1,a,4)

(1,b,3)

(1,b,4)

(1,c,3)

(1,c,,4)

(2,a,3)

(2,a,4)

(2,b,3)

(2,b,4)

(2,c,3)

(2,c,4)

FIGURE 11.3 Tree Diagram


3GC11 05/15/2014 11:31:20 Page 236

Bayes ’ Theorem

Suppose the events A1,A2, . . .An form a partition of the sample space Ω; that is,the events Ai are mutually exclusive and their union is Ω. Now let B be any otherevent. Then

B � Ω∩B � A1 ∪A2 ∪ . . . ∪A3� �∩B � A1 ∩B� �∪ A2 ∩B� �∪ . . . ∪ An ∩B� �(11.13)

where Ai ∩B� � are also mutually exclusive. Accordingly,

P B� � � P A1 ∩B� � � P A2 ∩B� �∙ ∙ ∙ � P An ∩B� � (11.14)

Thus by the multiplication theorem,

P B� � � P A1� �P BjA1� � � P A2� �P BjA2

� � � ∙ ∙ ∙ � P An� �P BjAn� �

(11.15)

On the other hand, for any i, the conditional probability ofAi givenB is defined by

P AijB� � � P Ai ∩B� �P B� � (11.16)

We know that P Ai∩B� � � P Ai� �P BjAi� �

Bayes’ Theorem (Figure 11.4): Suppose A1,A2,∙ ∙ ∙An form a partition of thesample space Ω and B is any event. Then for any i,

P AijB� � � P Ai ∩B� �P B� � � P Ai� �P BjAi

� �P A1� �P BjA1

� � � P A2� �P BjA2� � � ∙ ∙ ∙ � P An� �P BjAn

� �(11.17)

Bayes’ Theorem is a formula for determining conditional probability named after18th-century British mathematician Thomas Bayes. The theorem provides a way torevise existing predictions or theories given new or additional evidence. In finance,

B

Sample space Ω

A1 A2

An

A3

FIGURE 11.4 Bayes’ Theorem: Intersectionof B with a Partition of Ω

236 STATISTICS

3GC11 05/15/2014 11:31:20 Page 237

Bayes’ Theorem can be used to rate the risk of lending money to potential borrowers.The formula is

P AjB� � � P A∩B� �P B� � � P A� �.P BjA� �

P B� � (11.18)

Example: In a portfolio 51 percent of the sukuks are issued by Malaysiancompanies. One sukuk is selected.

i. Find the probability that the selected sukuk is issued by a Malaysian company.ii. It was later learned that the selected sukuk was rated AAA. Also 9.5 percent of

Malaysian sukuks are AAA, whereas 1.7 percent of non-Malaysian sukuks arerated AAA. Use this additional information to find the probability that theselected sukuk is Malaysian sukuk.

Use the following notation:A =Malaysia sukuk,Ac = non-Malaysia sukuk, B =rated AAA, and Bc = not rated AAA.

i. Before using the information in part (ii) we know only that 51 percent of thesukuks in the portfolio are Malaysian, so the probability of randomly selecting asukuk and getting a Malaysian sukuk is given by P A� � � 51 percent.

ii. Based on the additional information, we have P BjA� � � 0.095 and P BjAc� � �0.017.

Let us now apply Bayes’ Theorem:

P AjB� � � P A� �.P BjC� �P A� �.P BjC� � � P Ac� �.P BjAc� �

� 0.51 � 0.0950.51 � 0.095 � 0.49 � 0.017

� 0.85329341

Independence of Events

An eventB is said to be independent of an eventA if the probability thatB occurs is notinfluenced by whether A has or has not occurred. In other words, the probability of Bequals the conditional probability of B given A,

P B� � � P BjA� �(11.19)

Now substituting P B� � for P BjA� �in the multiplication theorem, P A∩B� � �

P A� �P BjA� �, we obtain

P A∩B� � � P A� �P B� � (11.20)


3GC11 05/15/2014 11:31:21 Page 238

Definition: Events A and B are independent if

P A∩B� � � P A� �P B� � (11.21)

Otherwise they are dependent.Example: A fair coin is tossed three times; we obtain the equiprobable space

Ω � HHH,HHT,HTH,HTT, THH, THT, TTH,TTTf g

Consider the events A � f irst toss is headf g and B � second toss is headf g.Clearly A and B are independent events. We have

P A� � � P HHH,HHT,HTH,HTTf g � 48� 1/2

P B� � � P HHH,HHT,THH, THTf g � 48� 1/2

P A∩B� � � P HHH,HHTf g� � � 28� 14� P A� �P B� �

PROBABILITY DISTRIBUTION OF A FINITE RANDOM VARIABLE

This section introduces the notions of probability distribution, histogram, cumulativedistribution function, and continuous random variables.

Probability Distribution and Histogram

Let X be a random variable on a sample space Ω assuming valuesX Ω� � � x1, x2, . . . , xnf g. The probability of xi is P X � xi� �; we write it as f xi� �.This function f onX Ω� �, that is, defined by f xi� � � P X � xi� �, is called the distributionor probability function ofX and is usually given in the form of a table (see Table 11.1).

The distribution f satisfies the conditions (i) f xi� � � 0 and (ii)Pn

i�1 f xi� � � 1.The graph of the distribution is a histogram that has on the horizontal axis the

values of the random variable and on the vertical axis the associated probabilities(Figure 11.5).

Example: The rate of return on a stock portfolio may assume any of the values,with respective probabilities, shown in Table 11.2.

The histogram for this distribution is shown in Figure 11.5.

TABLE 11.1 Probability Distribution

x1 x2 . . . xnf x1� � f x2� � . . . f xn� �

238 STATISTICS

3GC11 05/15/2014 11:31:22 Page 239

Cumulative Distribution Function

Let X be a random variable (discrete or continuous). The cumulative distributionfunction F of X is a function F : R ! R defined by

F a� � � P X � a� � (11.22)

IfX is a discrete random variable with distribution f , then F is the “step function”defined by

F x� � � Xxi�x

f xi� � (11.23)

F is monotonic increasing, that is, F a� � � F b� �whenever a � b and the limit of F tothe left is 0 and to the right is 1, limx!� ∞ F x� � � 0 and limx! ∞ F x� � � 1.

Continuous Random Variables

Suppose thatX is a random variable whose image setX Ω� � is a continuum of numberssuch as an interval. Recall from the definition of random variables that the set

0

2

4

6

8

10

12

14

16

18

–20 –15 –9.6 –5.2 –3 2015.712.3107.55.541.5

FIGURE 11.5 Histogram of a Probability Distribution

TABLE 11.2 Distribution of the Rate of Return on a Stock Portfolio

Rate of return(percentage)

�20 �15 �9.6 �5.2 �3 1.5 4 5.5 7.5 10 12.3 15.7 20

Probability(percentage)

2.0 4.2 2.5 12.0 5.0 11.0 14.0 12.0 17.1 5.0 4.5 3.7 7.0


3GC11 05/15/2014 11:31:23 Page 240

a � X � bf g is an event in Ω and therefore the probability P a � X � b� � is well-defined. We assume that there is a piecewise continuous function f : R ! R such thatP a � X � b� � is equal to the area under the graph of f between x � a and x � b (asshown in Figure 11.6a):

P a � X � b� � � ∫b

af x� �dx (11.24)

In this case X is said to be a continuous random variable. The function f is calledthe distribution function or the density function of X; it satisfies the conditions

i. f x� � � 0; and

ii. ∫ ∞� ∞ f x� �dx � 1.

That is, f is non-negative and the total area under its graph is 1.The cumulative distribution function F of a continuous random variable X is a

function F : R ! R defined by

F a� � � P X � a� � (11.25)

Since X is a continuous random variable with distribution f , then,

F x� � � ∫x

� ∞f t� �dt (11.26)

F is monotonic increasing, that is, F a� � � F b� �whenever a � b and the limit of F tothe left is 0 and to the right is 1, limx!�∞ F x� � � 0 and limx!∞ F x� � � 1 (Figure 11.6b).

1

b. Cumulative distribution function

a. Density function

0

0

f (x)

x

x

a b

F(x)

FIGURE 11.6 Continuous Density and Cumulative Distribu-tion Functions

240 STATISTICS

3GC11 05/15/2014 11:31:24 Page 241

MOMENTS OF A PROBABILITY DISTRIBUTION

The moments of a probability distribution are key parameters that describe orsummarize the distribution. Investors and financial analysts rely considerably onthese parameters in assessing risk and returns. This sections covers the first, second,third, and fourth moment of a probability distribution, called expected mean,variance, skewness, and kurtosis, respectively.

First Moment of the Random Variable

The first moment of the random variable is called the expected mean of the randomvariable. If x is a random variable with a probability density f x� �, then the mean orexpectation (or expected value) of x, denoted by E x� � � μ is defined as

μ � E x� � � x1f x1� � � x2f x2� � � ∙ ∙ ∙ � xnf xn� � � Xn

j�1 xif xi� � (11.27)

That is, E x� � is a weighted average of the possible values of x, each weighted by itsprobability. The expectation of a continuous variable is defined as

μ � E X� � � ∫∞

� ∞xf x� �dx (11.28)

The mean μ is also called the location parameter of the distribution.Example: A pair of fair dice is tossed. We obtain the finite equi-probable space Ω

consisting of the 36 ordered pairs of numbers between 1 and 6:

Ω � 1; 1� �, 1; 2� �, . . . , 6; 6� �f g

Let X assign to each point a, b� � in Ω the maximum of its numbers, that is,X a, b� � � max a, b� �. Then X is a random variable with image set X Ω� � �1; 2; 3; 4; 5; 6f g.

We compute the distribution f of X:

f 1� � � P X � 1� � � P 1; 1� �f g � 1/36f 2� � � P X � 2� � � P 1; 2� �, 2; 1� �, 2; 1� �f g � 3/36f 3� � � P X � 3� � � 5/36; f 4� � � P X � 4� � � 7/36f 5� � � P X � 5� � � 9/36; f 6� � � P X � 6� � � 11/36

We may describe the probability distribution in Table 11.3.

TABLE 11.3 Probability Distribution of X a, b� � � max a, b� �xi 1 2 3 4 5 6f xi� � 1/36 3/36 5/36 7/36 9/36 11/36


3GC11 05/15/2014 11:31:25 Page 242

We next compute the mean of x:

μ � E x� � � Pni�1 xif xi� �

� 1 � 136

� 2 � 336

� 3 � 536

� 4 � 736

� 5 � 936

� 6

� 1136

� 16136

� 4.47

Example: Compute the expected return on the portfolio of stocks given theprobability distribution of returns in Table 11.2.

E(x) � Pni�1 xif (xi) � �20 � 0.02 � 15 � 0.042 � 9.6 � 0.025

�3 � 0.05 � 1.5 � 0.11 � 4 � 0.14 � 5.5 � 0.12 � 7.5 � 0.171�10 � 0.05 � 12.4 � 0.045 � 15.7 � 0.037 � 20 � 0.07 � 3.6579

Let X be a random variable and c a real number. Then

i. (i) E cX� � � cE X� �; andii. (ii) E X � c� � � E X� � � c.

Let X and Y be random variables on the same sample space Ω. Then,E X � Y� � � E X� � � E Y� �.

Second Moment of the Random Variable: Variance and StandardDeviation

The mean of a random variable xmeasures, in a certain sense, the average value of x.The next concept, that of thevarianceofx,measures the“spread,”or“dispersion,”ofx.Then the variance of x, denoted by Var x� � is defined by

σ2x � Var x� � � Xn

i�1 xi � E x� �� 2f xi� � � E x � μ� �2� �(11.29)

where μ is the mean of X, that is, μ � E x� �. We note that the variance can beexpressed as

Var x� � � Xn

i�1 xi2f xi� � � μ2 � E x2

� � � μ2 (11.30)

The variance is expressed in squared units of the random variable. To be able touse the same unit of measurement as the random variable, we compute the standarddeviation of x; it is denoted by σx as

σx � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar x� �p

(11.31)

The mean and standard deviation are expressed in the same unit as the randomvariable; the standard deviation analyzes the spread of the statistical distributionaround the mean.

242 STATISTICS

3GC11 05/15/2014 11:31:25 Page 243

The variance of a continuous random variable is defined by

Var X� � � E X � μ� �2� � � ∫∞

� ∞x � μ� �2f x� �dx (11.32)

The variance Var X� � may be written as

Var X� � � E X2� � � μ2 � ∫∞

� ∞x2f x� �dx � μ2 (11.33)

The standard deviation σX is defined by σX � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar X� �p

when Var X� � exists.Example: The variance of returns described in Table 11.2 is computed as

Var x� � � E x2� � � μ2x � 88.11 � 13.38 � 74.733

The standard deviation of returns is σx � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi74.733

p � 8.645Example: We compute the variance and the standard deviation of x in the

example of die tossing in Table 11.3. Its mean is μx � 4.47. First, we compute E x2� �

.

E x2� �� Xn

i�1xi2f xi� �

� 12 ?136

� 22 ?336

� 32 ?536

� 42 ?736

� 52 ?936

� 62 ?1136

� 79136� 21.97

Hence Var x� � � E x2� � � μ2x � 21.97 � 19.98 � 1.99 and σx � ffiffiffiffiffiffiffiffiffiffi

1.99p � 1.4.

Let X be a random variable and c a real number. Then

i. Var X � c� � � Var X� �;ii. Var cX� � � c2Var X� �; andiii. Let X be a random variable with mean μ and standard deviation σ > 0. The

standardized random variable z corresponding to X is defined by

z � X � μσ

(11.34)

We may note that E z� � � 0 and Var z� � � 1.

Third Moment of a Random Variable: Skewness

The skewness of the probability distribution of a random variable X is the thirdstandardized moment, denoted by γ1 and defined as

γ1 � Ex � μσ

3� �

� μ3σ3

� E x � μ� �3� �E x � μ� �2� �� 3/2 (11.35)


3GC11 05/15/2014 11:31:26 Page 244

where μ3 is the thirdmoment about themean μ, σ is the standard deviation, andE is theexpectation operator. The skewness is also sometimes denoted as skew x� �. Skewnessdescribes the shape of the distribution. A symmetric distribution has γ1 � 0. Adistribution with γ1 < 0 is negatively skewed; it has long right tail (Figure 11.7a). Adistribution with γ1 > 0 is positively skewed; it has long left tail (Figure 11.7b).

Example: Compute the skewness of the returns’ distribution in Table 11.2.

γ1 � Ex � μσ

3� �

� �0.42221

Fourth Moment of a Random Variable: Kurtosis

The fourth standardized moment is defined as

γ2 � Ex � μ

σ

4� �

� μ4σ4

� E x � μ� �4� �E x � μ� �2� �� 2 (11.36)

where μ4 is the fourth moment about the mean and σ is the standard deviation. This issometimes used as the definition of kurtosis. Kurtosis measures the fatness of the tailsof the distribution. We define excess kurtosis as

γe2 � μ4σ4

� 3

The “minus 3” at the end of this formula is often explained as a correction tomake the kurtosis of the normal distribution equal to zero. Figure 11.7c shows a fat-tailed distribution.

a. Long right-tailed distribution

0

c. Fat‐tailed distribution

b. Long left-tailed distribution

f (x)

f (x)

f (x)

x

x

x

0

0

FIGURE 11.7 Skewed and Fat-Tailed Distributions

244 STATISTICS

3GC11 05/15/2014 11:31:28 Page 245

Example: The kurtosis of the returns’ distribution in Table 11.2 is

γ2 � Ex � μσ

4� �

� 3.42

In finance, besides mean and variance, we are often interested in skewness andfatness of the distribution.

JOINT DISTRIBUTION OF RANDOM VARIABLES

Let X and Y be random variables on a sample space Ω with respective image sets:

X Ω� � � x1, x2, . . . , xnf g and Y Ω� � � y1, y2, . . . , ymf gWe define the probability of the ordered pair xi, yj

to be P X � xi, Y � yj

,

which we write as h xi, yj

.

P X � xi, Y � yj

� h xi, yj

(11.37)

The function h on X Ω� � � Y Ω� � is called the joint probability of X and Y and isshown in Table 11.4.

The functions f and g are defined by

f xi� � � Xm

j�1 h xi, yj

(11.38)

and g yj

� Xn

i�1 h xi, yj

(11.39)

f xi� � is the sum of the entries in the ith row and g yj

is the sum of the entries in the jthcolumn. They are called the marginal distributions and are, in fact, the individualdistributions of X and Y, respectively. The joint distribution h satisfies the twoconditions:

i. h xi, yj

� 0; and

ii. andPn

i�1Pm

j�1 h xi, yj

� 1.

TABLE 11.4 Joint Distribution of Random Variables

y1 y2 ym Marginaldistribution of x

x1 h x1, y1� �

h x1, y2� �

. . . h x1, ym� �

f x1� �x2 h x2, y1

� �h x2, y2� �

. . . h x2, ym� �

f x2� �. . . . . . . . . . . . . . . . . .xn h xn, y1

� �h xn, y2� �

. . . h xn, ym� �

f xn� �Marginaldistribution of y

g y1� �

g y2� �

. . . g ym� �


3GC11 05/15/2014 11:31:30 Page 246

Now if X and Y are random variables with the joint distribution h xi, yj

andrespective means μX and μY , then the covariance of X and Y, denoted by Cov X, Y� � isdefined by

Cov X, Y� � � Xn

i�1Xm

j�1 xi � μX� �

yj � μY

h xi, yj

� E X � μX� �

Y � μY� ��

(11.40)

or equivalently by

Cov X, Y� � � Xi, jxiyjh xi, yj

� μXμY � E XY� � � E X� �E Y� � (11.41)

The correlation of X and Y, denoted by ρ X, Y� �, is defined as

ρ X, Y� � � Cov X, Y� �σXσY

(11.42)

The correlation ρ is dimensionless and has the following properties:

i. �1 � ρ � 1; andii. ρ aX � b, cY � d� � � ρ X, Y� � if a, c ≠ 0.

Example: The returns of stock X and Y at the end of the year are distributedjointly as shown in Table 11.5.

The mean of X is μX � Pni�1

xif xi� � � �0.5235The variance of X is Var X� � � Pn

i�1f xi� � xi � μX� �2 � 119.4262

The standard deviation of X is σX � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar X� �p � 10.92823

The mean of Y is μY � Pmj�1

yig yj

� �0.0729

The variance of Y is Var Y� � � Pmj�1

g yj

yj � μy

2 � 90.44616

The standard deviation of Y is σY � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar Y� �p � 9.510318

Cov X,Y� � � Xi,j

xiyjh xi, yj

� μXμY � 14.47284

The correlation of X and Y, denoted by ρ X,Y� �, is defined as

ρ X,Y� � � Cov X,Y� �σXσY

� 0.139622

TABLE 11.5 Joint Distribution of Stock X and Y Returns

y1 � −20 y2 � �11 y3 � 1 y4 � 4.5 y5 � 6.7 y6 � 9x1 � �17 0.05 0.06 0.021 0.012 0.05 0.023 0.216x2 � �8 0.011 0.012 0.033 0.04 0.06 0.025 0.181x3 � 4 0.024 0.041 0.07 0.045 0.09 0.076 0.346x4 � 12.5 0.023 0.052 0.061 0.047 0.023 0.051 0.257

0.108 0.165 0.185 0.144 0.223 0.175

246 STATISTICS

3GC11 05/15/2014 11:31:31 Page 247

Independent Random Variables

X and Y are independent variables if

P � PP

Now if X and Y have respective distributions f and g, and joint distribution h,then independence implies

h xi, yj

� f xi� �g yj

(11.44)

In other words, X and Y are independent if each entry h xi, yj

is the product ofits marginal entries.

We establish some important properties of independent random variables.Let X and Y be independent random variables. Then

i. E XY� � � E X� �E Y� �;ii. Var X � Y� � � Var X� � � Var Y� �; andiii. Cov X, Y� � � 0.

CHEBYSHEV ’S INEQUALITY AND THE LAW OF LARGE NUMBERS

This section covers Chebyshev’s inequality, the law of large numbers, and the CentralLimit Theorem.

The intuitive idea of probability is the so-called law of averages, that is, if an eventA occurs with probability p then the “average number of occurrences of A”approaches p as the number of independent trials increases. This concept is madeprecise by the law of large numbers stated later in this section.

Chebyshev ’s Inequality

Let X be a random variable with mean μ and standard deviation σ. Then for everyr > 0, we have the Chebyshev’s inequality:

P X � μj j � rσ� � � 1r2

(11.45)

Note that we can rewrite the Chebyshev’s inequality as

P X � μj j � rσ� � � 1 � 1r2

(11.46)

If we let ε � rσ so that r � ε/σ we have

P X � μj j � ε� � � σ2

ε2(11.47)


3GC11 05/15/2014 11:31:31 Page 248

and

P X � μj j � ε� � � 1 � σ2

ε2(11.48)

We see that

P μ � rσ � X � μ � rσ� � � 1 � 1r2

(11.49)

Example: X is a random variable with mean 11 and variance 9. Use Chebyshev’sinequality to find a lower bound for P 6 � X � 16� �. We have

P 6 � X � 16� � � P 11 � 5 � X � 11 � 5� � � P X � μj j � 5� � � 1 � 925

� 1625

� 0.64

Example: A gold mint company produces coins with an average diameter of 0.5inch and a standard deviation of 0.01 inch. Using Chebyshev’s inequality, find a lowerbound for the number of coins in a batch of 400 having diameter between 0.48 and0.52 inches.

We use

P X � xi,Y � yj

� P X � xi� �P Y � yj

P X � μj j � rσ� � � 1 � 1r2

with μ � 0.5, σ � 0.01, and r � 2. So,

P X � 0.5j j � 2 � 0.01� � � 1 � 1

22� 3/4

Therefore, at least 300 of the 400 coins will lie between 0.48 and 0.52 inches.

Law of Large Numbers

Let X1,X2, ∙ ∙ ∙ be a sequence of independent random variables with the samedistribution with mean μ and variance σ2. The sample mean is

Sn � X1 �X2 � ∙ ∙ ∙ �Xn� �/n (11.50)

Then for any ε > 0,

limn! ∞ P Sn � μ � ε� � � 0 (11.51)

The Central Limit Theorem

Let X1,X2, ∙ ∙ ∙ be a sequence of independent random variables with the samedistribution with mean μ and variance σ2. Let Sn � X1 � ∙ ∙ ∙ �Xn, then

Sn � nμσ

ffiffiffin

p ! N 0; 1� � (11.52)

248 STATISTICS

3GC11 05/15/2014 11:31:31 Page 249

where N 0; 1� � denotes the standard normal distribution with mean 0 and variance 1.Its expression is

P a � z � b� � � 1ffiffiffiffiffiffi2π

p ∫b

ae�z2/2dx (11.53)

Approximately the central limit theorem says that in a sequence of repeated trialsthe standardized sample mean approaches the standard normal distribution as thenumber of trials increases.

SUMMARY

This chapter introduced basic principles of probability theory and statistics. It coversthe notions of the sample space and events, the definition of probability, the notion ofrandom variable, the techniques of counting, the conditional probability and inde-pendence, the probability distribution of a finite random variable, the moments of aprobability distribution, the joint distribution of random variables, and Chebyshev’sinequality and the law of large numbers.

Islamic finance uses basic probability theory and statistics tools; these toolsanalyze returns, portfolios’ performance, and asset pricing, and they contribute tofinancial decision making.

QUESTIONS

1. You hold a Shariah-compliant stock in an efficient market, with equal probabilityof the stock’s price going up or down each day over a period of three days.What isthe possible sample space?

2. Referring to Question 1, what is the probability of the stock price going up in twoof the three days?

3. There are eight stocks available under the Shariah screening. You can have threestocks in your portfolio. Howmany possible combinations of stocks can you havein your portfolio?

4. There are six stocks available at the following prices:A � $10, $20, $30, $40, $50, $60f g.You draw randomly two stocks. What is the probability of drawing a

portfolio that is equal to $60 and that contains stock priced at $20?

5. There are six stocks available at the following prices:A � $10, $20, $30, $40, $50, $60f g.You draw randomly two stocks. What is the probability of drawing a

portfolio that is equal to $60 and that contains stock priced at $30?

6. In a portfolio, 60 percent of the sukuks are Musharakah sukuks; the rest areIjarah Sukuks. If one sukuk is selected at random, what is the probability it isan Ijarah sukuk?


3GC11 05/15/2014 11:31:34 Page 250

7. In a portfolio, 60 percent of the sukuks are Musharakah sukuks; the rest areIjarah sukuks. Further, 10 percent of the Ijarah sukuks are AA rated while2 percent ofMusharakah sukuks are AA rated. If one sukuk is selected at random,what is the probability it is an AA rated Ijarah sukuk?

8. You are an equity fund manager. Your portfolio comprises 51 percent growthstocks, and the rest are value stocks; 9.5 percent of growth stocks and 1.7 percentof value stocks are oil sector–related. If one stock is selected at random, usingBayes’ Theorem, what is the probability it is an oil growth stock?

9. While analyzing the past returns of ABC Islamic fund, we derive the followingprobabilities of returns:

Rate of return (percentage) − 20 − 15 −10 − 5 0 5 10 15 20

Probability (percentage) 12 8 5 11 5 14 15 12 18

Compute the mean, the variance and standard deviation, the skewness, and thekurtosis of the returns.

10. You hold two securities in your portfolio, X and Y, with joint probability ofreturns as displayed in the table below. Calculate the mean and variance ofX andY; calculate the correlation of X and Y.

Return of security Y

Return on security X �20 �15 �10 �5 0 5�20 0.04 0.05 0.011 0.002 0.04 0.013�15 0.001 0.002 0.015 0.03 0.05 0.015�10 0.014 0.031 0.054 0.02 0.08 0.066�5 0.013 0.042 0.04 0.037 0.013 0.0410 0.03 0.02 0.01 0.015 0.02 0.0255 0.04 0.01 0.02 0.025 0.03 0.035

11. In the Alpha Stock Exchange, the average price of the Islamic stocks is $5, with astandard deviation of 0.01. Using Chebyshev’s inequality, find a lower bound forthe number of stocks, ranging from $4.80 to $5.20.

250 STATISTICS

3GC12 05/15/2014 11:36:10 Page 251

CHAPTER 12Probability Distributions and Moment

Generating Functions

T his chapter describes some probability distributions that are widely used inIslamic finance. These are the uniform distribution, the Bernoulli distribution,

the binomial distribution, the Poisson distribution, the normal distribution, the chi-square distribution, the student’s t-distribution, and the F distribution. We displaythe principle underlying each distribution and show how probability computationsmay be carried out using Microsoft Excel. Many distributions are approximated bya normal distribution as the sample size or degrees of freedom become large.The probability distributions covered here are not exhaustive. There are manyimportant probability distributions used in finance that are not covered in this chapter.A probability distribution is characterized by its moments. We discuss the momentgenerating function (MGF), which provides an easier way to compute moments.We show how the MGF is used to derive moments of a random variable.

EXAMPLES OF PROBABILITY DISTRIBUTIONS

This section covers the uniform distribution, the Bernoulli distribution, the binomialdistribution, the Poisson distribution, the normal distribution, the chi-square distri-bution, the t distribution, and the F distribution.

The Uniform Distribution

The uniform distribution plays a crucial role in simulation analysis and in MonteCarlo methods. A continuous random variable with a uniform distribution isthe simplest random variable; its density is constant over some interval a, b� �and is zero elsewhere. The uniform distribution is called the rectangular distri-bution. The probability density function of the continuous uniform distribution is

f x� � �1

b � afor a � x � b

0 for x < a or x > b

8<: (12.1)

251

3GC12 05/15/2014 11:36:10 Page 252

The graph of a typical rectangular distribution is shown in Figure 12.1. For a randomvariable following this distribution, the expected value μ is

μ � E X� � � ∫b

a

xb � a

dx � 12

x2

b � a� �� b

a� a � b

2(12.2)

The second moment is computed as

E x2� � � ∫

b

a

x2

b � adx � 1

3x2

b � a� �� b

a� a2 � ab � b2

3(12.3)

The variance is

Var X� � � E x2� � � μ2 � b � a� �2

12(12.4)

Example: Let the random variable x be distributed as

f x� � �1

3 � 1for 1 � x � 3

0 for x < 1 or x > 3

8<: (12.5)

Given P x < u� � � 0.6, find the value of u. We have 12 u � a� � � 1

2 u � 1� � � 0.6, whichyields u � 2.2.

The Bernoulli Distribution

The Bernoulli distribution, named after Swiss scientist Jacob Bernoulli (1654–1705),is a discrete probability distribution, which takes value 1 with success probability pand value 0 with failure probability q � 1 � p. So if X is a random variable with thisdistribution, we have

P X � 1� � � 1 � P X � 0� � � 1 � q � p (12.6)

f (x)

a

1b – a

bx

FIGURE 12.1 A Rectangular Distribution

252 STATISTICS

3GC12 05/15/2014 11:36:11 Page 253

The mean ofX is p and the variance is p 1 � p� �. A classical example of a Bernoulliexperiment is a single toss of a coin. The coin might come up heads with probability pand tails with probability 1 � p. The experiment is called fair if p � 0.5, indicating theorigin of the terminology in betting (the bet is fair if both possible outcomes have thesame probability). The sample space of each individual trial is formed by two points,success (S) and failure (F). The sample space of n Bernoulli trials contains 2n points orsuccession of n symbols S and F, each representing one possible outcome of thecompound experiment. Since trials are independent, the probabilities multiply.

The Binomial Distribution

We consider repeated and independent trials of an experiment with two outcomes; wecall one of the outcomes success and the other outcome failure. Let p be the probabilityof success, so that q � 1 � p is the probability of failure. If we are interested in thenumber of successes and not in the order in which they occur, then the probability ofexactly x successes in n repeated trials is denoted b x; n, p� � and given by

b x; n, p� � � nx

� �px 1 � p� �n�x (12.7)

The three properties of the binomial distribution are:(i) Mean: μ � np, (ii) variance σ2 � npq, and (iii) standard deviation σ � ffiffiffiffiffiffiffiffi

npqp

Example: A fair die is tossed seven times; call a toss a success if a 2 or a 6appears. Then n � 7,

p � P 2; 6f g � 1/3 and q � 1 � p � 2/3

The probability that a 2 or a 6 occurs exactly three times i.e., x � 3� � is

b 3; 7; 1/3� � � 73

� �1/3� �3 2/3� �4 � 560/2187

Example: A fair die is tossed 120 times. The expected number of “5” is

μ � 1206 � 20; the standard deviation is σ �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1206 � 5/6

q� 4.02.

Example: (i) Using Microsoft Excel compute the probability of x � 23 for abinomial distribution with n � 70 and p � 0.20.

b 23; 70; 0.20� � � 7023

� �0.2230.847 � BINOM.DIST(23;70; 0.2,FALSE)

(ii) Using Microsoft Excel compute the cumulative probability P x � 23� �,that is, the number of success is at least x � 23 for a binomial distributionwith n � 70 and p � 0.20.

P x � 23� � � BINOM.DIST 23; 70; 0.2, TRUE� � � 0.996332

Probability Distributions and Moment Generating Functions 253

3GC12 05/15/2014 11:36:11 Page 254

The Poisson Distribution

If the number of trials n in a binomial distribution is large, the computations involvedin using the formula of the binomial distribution become quite lengthy; therefore, aconvenient approximation to the binomial distribution would be very useful. It turnsout that for large n there are two well-known density functions that give goodapproximations to the binomial density function: one when p is very small, that is, wehave a rare event; and the other when this is not the case. The approximation thatapplies when p is very small and n large is known as the Poisson density function withmean λ � np and variance λ � np and it defines the Poisson distribution.When p is notvery small, the approximation that applies is the normal distribution with mean μ �np and variance σ2 � np 1 � p� �.

The Poisson distribution is defined as

f x; λ� � � λxe�λx!

, x � 0; 1; 2, . . . (12.8)

where λ > 0 is some constant. The Poisson distribution appears in many naturalphenomena, such as the number of road accidents that occur with small probability inhigh car traffic or the number of droughts in weather patterns. The Poisson distribu-tion has mean μ � λ, variance σ2 � λ, and standard deviation σ � ffiffiffi

λp

. In Figure 12.2,we display shapes of the Poisson distribution for different values of the parameter λ.The Poisson distribution is approximated by a normal distribution. The Poissondistribution approaches a normal distribution with standardized variable z � x�λ� �ffiffi

λp as λ

increases indefinitely.Example: Experience shows that the mean number of Murabaha applications of

an Islamic bank executed during one day is 5. If the Murabaha department of thebank can handle a maximum of 8 Murabaha applications per day, what is theprobability that it will be unable to handle all the Muarabaha applications thatcome in during a period of one day? The desired probability can be obtained bycalculating the probability of receiving 8 or fewer Murabaha applications and thensubtracting this probability from 1. Using Microsoft Excel for λ � 5 in the Poissondensity, we have

P X � 8� � � POISSON.DIST 8; 5, TRUE� � �X8x�0

e�55xx!

� 0.932

f (x)

x

λ = 1

λ = 4

λ = 10

0

FIGURE 12.2 Poisson Distribution

254 STATISTICS

3GC12 05/15/2014 11:36:12 Page 255

Consequently the probability that the Murabaha department will be overtaxed isP X > 8� � � 0.068.

Example: An Islamic bank faces a default of 10 transactions among 500Murabaha transactions; what is the probability that the number of defaults willexceed 17 transactions?

P X � 17� � � 1 � P X � 17� � � 1 � POISSON.DIST 17; 10, TRUE� � � 0.0142

Example: Suppose that 2 percent of the Murabaha contracts made by an Islamicbank fall in default. Find the probability that there are 3 defaulted contracts in asample of 100 contracts.

The binomial distribution with n � 100 and p � 0.02 applies. However, sincep is small, we use the Poisson approximation with λ � np � 2. Thus, P � p 3; 2� � �e�2233! � 0.18.

The Normal Distribution

The normal (or Gaussian) distribution is defined as

f x� � � 1


p e� x�μ� �2/2σ2 (12.9)

where μ and σ are arbitrary constants. This function is certainly one of the mostimportant examples of a continuous probability distribution. The graph of thedistribution is portrayed in Figure 12.3. We observe that a normal distribution isrepresented by a bell-shaped curve that is symmetric about x � μ; the curve becomestaller and narrower as the standard deviation becomes smaller. It is important to notethat x enters in the function with second power, that is, x2. This implies that theprobability f x� � declines rapidly as x increases. In other words, the probability of an

0

Smaller variance

Larger variance

f (x)

x

x

f (x)

μ

FIGURE 12.3 Normal Distributions


3GC12 05/15/2014 11:36:12 Page 256

extreme event is almost zero in a normal distribution. In finance, major crashes orbooms are not rare; this limits the ability of the normal distribution to representfinancial random variables.

The properties of the normal distribution are (i) mean μ, (ii) variance σ2, and(iii) standard deviation σ.

We denote the normal distribution with mean μ and variance σ2 by

N μ, σ2� �

(12.10)

If we make the substitution z � x � μ� �/σ in the normal distribution N μ,σ2� �

weobtain the standard normal distribution,

f z� � � 1ffiffiffiffiffiffi2π

p e�z2/2 (12.11)

with mean μ � 0 and σ2 � 1.Let X be a continuous random variable with a normal distribution. We compute

the probability that X lies between a and b, denoted by P a � X � b� � as follows: wechange a and b into standard units a´ � a � μ� �/σ and b´ � b � μ� �/σ, respectively.We compute

P a � X � b� � � P a´ � z � b´� �

(12.12)

Here z is the standardized random variable corresponding to X; the standardizedz has the standard normal distribution, z∼N 0; 1� �.

The binomial distribution P k� � � b k; n, p� � is closely approximated by the normaldistribution when n is large, n � 40. The normal distribution has mean μ � np,variance σ2 � npq, and standard deviation σ � ffiffiffiffiffiffiffiffi

npqp

.Example: Let X be a random variable that has a binomial distribution with p �

1/5 and n � 40; then the distribution of X can be approximated by a normaldistribution with mean μ � np � 40 � 1

5 � 8, σ2 � npq � 40 � 15 � 4

5 � 325 � 6.4, and

σ � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi40 � 1/4 � 4/5

p � 2.53.Example: LetX∼N μ � 8; σ2 � 6.4

� �, computeprobabilityP X � 7� �.UsingMicro-

soft Excel, we have P X � 7� � � NORM.DIST 7; 8,6.40.5, TRUE� � � 0:346316.

Example: Let z follow a standard normal distribution: (i) givenP z < u� � � 0.95, findthe value of u; and (ii) given P z < u� � � 0.975, find the value of u. UsingMicrosoft Excel,wefind (i)u � NORM.S.INV 0.95� � � 1.644854, and (ii)u � NORM.S.INV 0.975� � �1:959964.

Example: Value-at-Risk (VaR) of a portfolio. We assume that a portfolio ofIslamic stocks has a value today equal to $100; we assume the monthly portfolio valueis distributed as a normal distribution with mean μ � $100 and a standard deviationσ � $10/month. We consider a time horizon=1 month. We compute the 95 percentVaR as VaR0.95 � 1.645σ � $10 � 1.645 � $16.45.

Example: Value-at-Risk (VaR) of a portfolio. We assume that a portfolio ofIslamic stocks has a value today equal to $1,200; we assume the daily portfolio value isdistributed as a normal distribution with mean μ � $1; 200 and a standard deviationσ � $87/day. We consider a time horizon=1 day. We compute the 95 percent VaR asVaR0.95 � 1.645σ � $87 � 1.645 � $143.12.

256 STATISTICS

3GC12 05/15/2014 11:36:13 Page 257

The Chi-Square Distribution

If z1, z2, . . . , zk are independent, standard normal random variables, then the sum oftheir squares,

X �Xk

i�1 z2i (12.13)

is distributed according to the chi-squared distributionwith k degrees of freedom. Thisis usually denoted as

X∼χ2 k� � or X∼χ2k (12.14)

The chi-squared distribution has one parameter, k, a positive integer that specifiesthe number of degrees of freedom (i.e., the number of zi’s). The chi-square distribution(also χ2 distribution) with k degrees of freedom is the distribution of a sum of thesquares of k independent standard normal random variables. It is one of the mostwidely used probability distributions in inferential statistics, for example, in hypothe-sis testing or in construction of confidence intervals.

The chi-square distribution is useful for testing hypotheses that deal with vari-ances of random variables. It is used in the common chi-squared tests for the goodnessof fit of an observed distribution to a theoretical one, the independence of two criteriaof classification of qualitative data, and in confidence interval estimation for apopulation standard deviation of a normal distribution from a sample standarddeviation.

The chi-square starts at the origin, is skewed to the right, and has a tail that extendsinfinitely far to the right. The exact shape of the distribution depends on the number ofdegrees of freedom, with the distribution becoming more and more symmetric as thenumber of degrees of freedomgets larger (Figure 12.4).When the degrees of freedomgetvery large, the chi-square distribution approximates the normal.

Example: The graph of the chi-square with 5 degrees of freedom is shown inFigure12.5.UsingMicrosoft Excel,find theprobability that (i) χ2 > 8, and (ii) χ2 < 1.5.

(i) P χ2 > 8� � � CHISQ.DIST.RT 8; 5� � � 0:156235628

(ii) P χ2 < 1.5� � � CHISQ.DIST 1.5; 5, TRUE� � � 0:086930185

Density

0

fk(x)

k = 2

k = 4k = 6

k = 10

x

FIGURE 12.4 Chi-Square Distribution forVarious Values of Degrees


3GC12 05/15/2014 11:36:13 Page 258

Example:

(i) We consider the same chi-square with 5 degrees of freedom; we are givenP u � χ2� � � 0.05 on the right tail, find u. We use Microsoft Excel.

u � CHISQ.INV.RT 0.05; 5� � � 11:0705

(ii) We are given P u � χ2� � � 0.10 on the left tail, find u. We useMicrosoft Excel.

u � CHISQ.INV 0.1; 5� � � 1:610308

The t Distribution

Student’s t-distribution with ν degrees of freedom is defined as the distribution of therandom variable t with

t � XffiffiffiffiffiffiffiZ/ν

p (12.15)

where X is normally distributed with expected value 0 and variance 1; Z has a chi-squared distribution with ν degrees of freedom; and X and Z are independent. The tdistribution has mean 0 for ν > 1.

The student’s t-distribution (or simply the t distribution) is a family of continuousprobability distributions that arise when estimating the mean of a normally distrib-uted population in situations where the sample size is small and population standarddeviation is unknown. It plays a role in a number of widely used statistical analyses,including the student’s t-test for assessing the statistical significance of the differencebetween two sample means, the construction of confidence intervals for the differencebetween two population means, and in linear regression analysis.

In statistics the variance of a random variable is sometimes assumed to be known.How do we test hypotheses when the variance is not known? The answer lies in the tdistribution. We assume X is normally distributed with mean 0 and variance 1 andthat Z is distributed as chi-square with N degrees of freedom. Then X and Z areindependent, X/

ffiffiffiffiffiffiffiffiffiZ/N

phas a t distribution with N degrees of freedom.

Density

0 χ2 χ2χ2

FIGURE 12.5 Chi-Square Distribution

258 STATISTICS

3GC12 05/15/2014 11:36:14 Page 259

Figure 12.6 illustrates the t distribution. Like the normal, the t distribution issymmetric and it approximates the normal for large sample sizes. But the t distributionhas fatter tails than the normal, an occurrence that is especially pronounced for samplesizes of roughly 30 or fewer, meaning that it is more prone to producing values that fallfar from its mean.

A student’s t-distribution shares some characteristics of the normal distributionand differs from it on others. Characteristics of the t distribution similar to the normaldistribution are:

■ It is bell-shaped.■ It is symmetric about the mean.■ The mean, median, and mode are equal to 0 and are located at the center of the

distribution.■ The curve never touches the x axis.

Characteristics of the t distribution that differ from the normal distribution:

■ The variance is greater than 1.■ The t distribution is a family of curves based on the concept of degrees of freedom,

which is related to sample size.■ As the sample size increases, the t distribution approaches the standard normal

distribution.

Example: We assume a random variable x has a t distribution with degrees offreedom ν � 20; compute P x < 2.2� �.

Using Microsoft Excel, we have P x < 2.2� � � T.DIST 2.2; 20, TRUE� � �0.980135705.

Example: We assume a random variable x has a t distribution with degrees offreedom ν � 23; givenP x � u� � � 0.95, we compute u � T.INV 0.95; 23� � � 1.713872.

The F Distribution

If X and Z are independent and distributed as chi-square with N1 and N2 degrees offreedom, respectively, then X/N1� �/ Z/N2� � is distributed according to an F distribution

Normal distribution

0

t distribution

f (x)

x

FIGURE 12.6 The t Distribution


3GC12 05/15/2014 11:36:14 Page 260

with N1 and N2 degrees of freedom. Hence, the F distribution is the ratio of two chi-square distributions:

F �XN1

� �

ZN2

� � � χ2N1

χ2N2

(12.16)

There are occasions when we wish to test joint hypotheses involving two or moreregression parameters, for example, the hypothesis that the intercept and slope areboth zero against the alternative that one or both are nonzero. The proper test statisticis based on the F distribution and is characterized by two parameters, the first beingassociated with the number of estimated parameters and the second being associatedwith the number of degrees of freedom. The F distribution can be used to test theequality of two variances.

TheF distribution, like the chi-square, has a skewed shape and ranges in value from0 to infinity. It is an asymmetric distribution that has a minimum value of 0, but nomaximum value. The curve reaches a peak not far to the right of 0, and then graduallyapproaches the horizontal axis the larger the F value is. The F distribution approaches,but never quite touches the horizontal axis. The F distribution has two degrees offreedom,N1 for the numerator, andN2 for the denominator. For each combination ofthese degrees of freedom there is a different F distribution. The F distribution is mostspread out when the degrees of freedom are small. As the degrees of freedom increase,the F distribution is less dispersed. Figure 12.7 shows the shape of the distribution fordifferent degrees of freedomN1 andN2. The F value is on the horizontal axis, with theprobability for each F value being represented by the vertical axis.

Example: Let x be a random variable that has an F distribution withN1 � 16 andN2 � 25, find P x < 2� �. Using Microsoft Excel, we have

P x < 2� � � F.DIST 2; 16; 25, TRUE� � � 0:941698

EMPIRICAL DISTRIBUTIONS

Financial time-series track data on financial variables such as stock prices, commodityprices, and yield rates. We analyze empirical data without necessarily postulating

0

f (x)N1 = 2,

N1 = 100,N2 = 100

N1 = 5,N2 = 2

N2 = 2

x

FIGURE 12.7 Shapes of the F Distribution

260 STATISTICS

3GC12 05/15/2014 11:36:14 Page 261

assumptions regarding the probability distribution of the generating process of thisdata. We may draw a sample of data of size n, x1, x2, . . . , xn of the random variable x,without specifying the theoretical distribution of x. The data set is represented by ahistogram. By analogy to theoretical distributions, we define empirical moments.

The kth moment about the origin of an empirical distribution is defined as

mk � 1n

Xni�1

xki (12.17)

Empirical moments are also called sample moments because they are based onsample values. By analogy with the definition for probability distributions, empiricalmoments about the mean are defined as follows: the kth moment about the mean of anempirical distribution is given by

mk � 1n

Xni�1

xi � x� �k (12.18)

The first moment,m1, is traditionally denoted by the symbol x. It gives the centerof gravity of an empirical distribution just as μ does for a theoretical distribution and itserves to measure where the empirical distribution is centered. It is called the samplemean and will be used to estimate the theoretical mean μ.

Since σ2 is the second moment about the mean for a theoretical distribution, itwould be natural to use the sample second moment around the sample mean toestimate σ2. In estimating the second moment around the mean, we divide by n � 1 inplace of n. The resulting quantity, denoted by s2, is called the sample variance. Hence,

s2 � 1n � 1

Xni�1

xi � x� �2 (12.19)

For convenience of calculation the sample variance is written as

s2 � 1n � 1

Xni�1

x2i � nx2� �

(12.20)

If the observational values x1, x2, . . . , xn have been classified into a frequencytable with yi representing the ith interval, f i representing the number of observation inthe ith interval, and H denotes the number of intervals, then the moments definitionswill assume the following forms:

mk � 1n

XHi�1

yki f i (12.21)

and

mk � 1n

XHi�1

yi � y� �kf i (12.22)

The value of y is computed as

y � m1 � 1n

XHi�1

yif i (12.23)


3GC12 05/15/2014 11:36:15 Page 262

The sample variance is

s2 � 1n � 1

XHi�1

yi � y� �2f i (12.24)

It is often useful to test whether a given data series approximates the normaldistribution. A formal test of normality is given by the Jarque-Bera statistic,

JB � n6

S2 � 1

4K � 3 2� �

(12.25)

where S2is skewness, S

2 � 1n

Pni�1 xi � x� �3/σ3; K2

is kurtosis, K2 � 1

n

Pni�1 xi � x� �4/σ4,

and σ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n

Pni�1 xi � x� �2

q.

Under the null hypothesis of independent normally distributed observations, theJarque-Bera statistic is distributed as a chi-squared random variable with 2 degrees offreedom. If the JB statistic is greater than the critical value of the chi-square, we rejectthe null-hypothesis of normality.

Example: We consider the weekly returns on the stock price index S&P 500during January 2008 to May 2013. The histogram for these returns is shown inFigure 12.8.* We compute the empirical moments: (i) the sample mean x � 0.053744;(ii) the sample standard deviation s � 3.167912; skewness=�0.857999; andkurtosis � 9.369166. The Jarque-Bera JB� � statistic is 507.6275; the critical valueof chi-square with 2 degrees of freedom and 5 percent significance level is 7.377. Sincethe JB exceeds the critical chi-square; we reject the normality hypothesis.

0

10

20

30

40

50

60

–20 –15 –10 –5 0 5 10

Series: RETURNSSample: 1/03/2008–5/15/2013Observations: 280

Mean 0.053744Median 0.186969Maximum 11.35590Minimum –20.08375Std. Dev. 3.167912Skewness –0.857999Kurtosis 9.369166

Jarque-Bera 507.6275Probability 0.000000

FIGURE 12.8 Weekly Returns on S&P 500 Stock Index, January 2008 to May 2013

*Histograms may be easily generated with Microsoft Excel.

262 STATISTICS

3GC12 05/15/2014 11:36:16 Page 263

MOMENT GENERATING FUNCTION (MGF)

Let X be a discrete random variable assuming the distinct values x1, . . . ., xn, withrespective probabilities p1, . . . ., pn. Its moment of order k is written as

E xk

�Xni�1

xki pi (12.26)

We define a moment generating function for this distribution, which provides aconvenient way for computing moments as*

MX t� � � E etX� � �Xn

i�1pie

txi (12.27)

Considering t as a constant here, at least from the perspective of taking expect-ations, we use the linearity of expectation to conclude that

MX t� � � E eXt� � �X∞

k�0tk

k!E Xkh i

(12.28)

For a continuous random variable X with a probability density f x� �, the kthmoment is defined analogously to the discrete case,

E xk

� ∫∞

� ∞xkf x� �dx (12.29)

The moment generating function of a continuous variable is defined as in the discretecase,

MX t� � � E eXt� � � E

X ∞k�0

Xktk

k!

!�X ∞

k�0tk

k!E Xkh i

(12.30)

*Themoment generating function is based on the expansion of the exponential function ex into aMaclaurin power series as

ex �X∞k�0

xk

k!� 1 � x � x2

2!� x3

3!� x4

4!� ∙ ∙ ∙

For etx, the power expansion becomes

etx �X∞k�0

tkxk

k!� 1 � tx � t2x2

2!� t3x3

3!� t4x4

4!� ∙ ∙ ∙

The parameter t plays a key role in computing moments of the random variable x.


3GC12 05/15/2014 11:36:16 Page 264

We observe clearly that the coefficient of tk gives us kth moment divided by k!. It isfairly easy to see that if we take the kth derivative of the moment generating function,and set t � 0, the result will be the kth moment.* In symbols, this is

ddt

� �k

MX t� �jt�0 � E Xk

� kth moment of X (12.31)

It may be that the moment generating function does not exist, because some of themoments may be infinite (or may not have a definite value, due to integrability issues).Also, even if the moments are all finite and have definite values, the generatingfunction may not converge for any value of t other than 0.

Examples of Moment Generating Functions

The moment generating functions (MGF) of uniform, Bernoulli, binomial, Poisson,and normal distributions are shown here.

The MGF of the Uniform Distribution The probability density function of the continu-ous uniform distribution is

f x� � �1

b � afor a � x � b

0 for x < a or x > b

8<: (12.32)

Applying the definition of MGF,

MX t� � � E ext� � � ∫

b

a

ext

b � adx � 1

tetb � eta

b � a(12.33)

We expand eta and etb as

eta � 1 � ta � t2a2

2!� t3a3

3!� ∙ ∙ ∙

etb � 1 � tb � t2b2

2!� t3b3

3!� ∙ ∙ ∙

*For example, let us consider the function

mx t� � � a0 � tx � t2x2/2! � t3x3/3!

The second derivative with respect to t is

d2mx t� �dt2

� x2 � tx3

To get rid of the remaining terms that are multiplied by powers of t and obtain only x2 we sett � 0. Hence,

d2mx t� �dt2

jt�0 � x2

264 STATISTICS

3GC12 05/15/2014 11:36:16 Page 265

We rewrite the MGF as

1tetb � eta

b � a� 1t b � a� � 1 � tb � t2b2

2!� t3b3

3!� ∙ ∙ ∙

!� 1 � ta � t2a2

2!� t3a3

3!∙ ∙ ∙

� �" #

� 1b � a

b � tb2

2!� t2b3

3!� ∙ ∙ ∙

!� a � ta2

2!� t2a3

3!� ∙ ∙ ∙

� �" #

The first moment of X is computed as

dMX t� �dt

� 1b � a

b2

2!� 2tb3

3!� ∙ ∙ ∙

!� a2

2!� 2ta3

3!� ∙ ∙ ∙

� �" #(12.34)

Setting t � 0 we obtain

dMX t� �dt

� 1b � a

b2

2

!� a2

2

� �" #� a � b

2(12.35)

The second moment is computed as

d2MX t� �dt2

� 1b � a

b3

3

!� a3

3

� �" #� a2 � ab � b2

3(12.36)

The variance of the uniform distribution is

E X2� � � μ2 � a2 � ab � b2

3� a � b

2

� �2

� b � a� �212

(12.37)

The MGF of the Bernoulli Distribution If X is a random variable with Bernoullidistribution, we have

P X � 1� � � 1 � P X � 0� � � 1 � q � p (12.38)

The MGF of the Bernoulli distribution is expressed as

MX t� � � E eXt� � � e0t 1 � p� � � e1tp � 1 � p� � � etp (12.39)


dMX t� �dt

jt�0 � etp � p (12.40)

The second moment of X is computed as

d2MX t� �dt2

jt�0 � etp � p (12.41)


3GC12 05/15/2014 11:36:17 Page 266

The variance of X is defined as

E X2� � � μ2 � p � p2 � p 1 � p� � (12.42)

The MGF of the Binomial Distribution The probability of exactly x successes in nrepeated trials is given by

b x; n, p� � � nx

� �px 1 � p� �n�x (12.43)

The MGF for a binomial distribution is defined as

MX t� � � E ext� ��Xn

x�0 ext n

x

� �pxqn�x�Xn

x�0nx

� �pet� �xqn�x� pet � q

� �n (12.44)

The last term is obtained from the binomial theorem. The first moment of X iscomputed as

dMX t� �dt

� netp pet � q� �n�1

(12.45)

Setting t � 0, we obtain

dMX t� �dt

jt�0 � np p � q� �n�1 � np p � q� �n�1 � np (12.46)


d2MX t� �dt2

� netp pet � q� �n�1 � n n � 1� �etp2 pet � q

� �n�2 (12.47)

Setting t � 0, we find

d2MX t� �dt2

t�0 � np � n n � 1� �p2 (12.48)


E X2� � � μ2 � np � n n � 1� �p2 � n2p2 � np � np2 � np 1 � p� � � npq (12.49)

The MGF of the Poisson Distribution The Poisson distribution is defined as

p x; λ� � � λxe�λx!

, x � 0; 1; 2, ∙ ∙ ∙ (12.50)

The MGF for the Poisson distribution is defined as

MX t� � � E ext� � �X ∞

x�0 ext λ

xe�λx!

�X ∞x�0

etλ� �xe�λx!

� e�λX ∞

x�0etλ� �xx!

(12.51)

266 STATISTICS

3GC12 05/15/2014 11:36:18 Page 267

Recall that ey �P ∞x�0

yx

x!; we let y � etλ, and we obtainP ∞

x�0etλ� �xx! � ee

tλ.The MGF is

MX t� � � e�λeetλ � exp λ et � 1� ��

(12.52)


dMX t� �dt

� λetexp λ et � 1� ��

(12.53)

Setting t � 0, we obtain dMX t� �dt � λ.


d2MX t� �dt2

� λ2e2texp λ et � 1� �� λetexp λ et � 1

� �� (12.54)


d2MX t� �dt2

� λ2 � λ (12.55)


E X2� � � μ2 � λ2 � λ � λ2 � λ (12.56)

The MGF of the Normal Distribution Let us compute the MGF for a normal randomvariable having variance σ2 and mean μ � 0. Note that the probability densityfunction for such a random variable is just

f x� � � 1


p e�12x2

σ2 (12.57)

The MGF is defined as

MX t� � � E ext� � � ∫

∞

� ∞extf x� �dx � ∫

∞

� ∞ext

1


p e�12x2

σ2dx (12.58)

We may rewrite the last integral as

∫∞

� ∞ext

1


p e�12x2

σ2dx � ∫∞

� ∞

1


p e�12

x2�2σ2 tx� �σ2 dx (12.59)

Now we complete x2 � 2σ2tx� �

to a square by adding to it (and then subtracting)σ4t2; we obtain

MX t� � � ∫∞

� ∞

1


p e�12

x�σ2 t� �2�σ4 t2� �σ2 dx � e

12σ

2t2∫∞

� ∞

1


p e�12

x�σ2 t� �2σ2 dx (12.60)


3GC12 05/15/2014 11:36:18 Page 268

Now we recognize that the last integral equals 1, because it is the integral of theprobability density function for the normal random variable N σ2t,σ2

� �over the full

interval �∞ , ∞� �. The moment generation function is therefore

MX t� � � e12σ

2t2 (12.61)


dMX t� �dt

� σ2te12σ

2t2 (12.62)

Setting t � 0, we obtain dMX t� �dt

t�0 � 0, that is, μ � 0.


d2MX t� �dt2

� σ2e12σ

2t2 � σ4t2e12σ

2t2 (12.63)


d2MX t� �dt2

jt�0 � σ2

The variance is defined as

E X2� � � μ2 � σ2 � 0 � σ2 (12.64)

We note that proceeding along the same reasoning we obtain the momentgeneration function of a normal distribution N μ, σ2

� �as

MX t� � � E ext� � � ∫

∞

� ∞extf x� �dx � ∫

∞

� ∞ext

1


p e�12x�μ� �2σ2 dx � e μt�1

2σ2t2� � (12.65)

SUMMARY

Islamic finance relies on statistical tools for assessing and managing risk. This chapterintroduced examples of probability distributions applied in Islamic finance, whichincluded the uniform distribution, the Bernoulli distribution, the Poisson distribution,the normal distribution, the chi-square distribution, the t distribution, and the Fdistribution. The chapter covered empirical distributions and the moment generatingfunction.

Probability distributions underlie finance models; Islamic banks, corporations,investment funds, and pension funds use probability distributions in their investmentdecisions.

268 STATISTICS

3GC12 05/15/2014 11:36:19 Page 269

QUESTIONS

1. Let the random variable x be distributed as

f x� � �1

4 � 1for 1 � x � 4

0 for x < 1 or x > 4

8<:

Given P x < u� � � 0.8, find the value of u.

2. a. Using Microsoft Excel, compute the probability of x � 25 for a binomialdistribution with n � 80 and p � 0.22.

b. Using Microsoft Excel compute the cumulative probability P x � 25� �, that is,the number of success is at least x � 25 for a binomial distribution with n � 80and p � 0.22.

3. Explain the relationship between a binomial and a Poisson distribution.

4. Show that the Poisson distribution f x; λ� � � λxe�λx! is a probability distribution, that

is,

X∞x�0

λxe�λx!

� 1

5. Ten percent of theMurabaha transactions for an Islamic bank fall in default. Findthe probability that in a sample of 10 transactions chosen at random exactly twowill be in default by using (i) the binomial distribution and (ii) the Poissonapproximation to the binomial distribution.

6. If the probability that an investor suffers loss from a bad sukuk is 0.001,determine the probability that out of 2,000 investors (i) exactly 3 and(ii) more than 2 investors will suffer loss from bad sukuks.

7. Suppose that 300 Malay Palm Oil Corporation shares are distributed randomlyamong 500 investors. Using the Poisson distribution, find the probability that agiven investor holds (i) exactly two shares and (ii) two or more shares of MalayPalm Oil Corporation.

8. Provide a definition of a chi-square probability distribution.

9. Consider a chi-square random variable with 6 degrees of freedom. Using Micro-soft Excel, find the probability that (i) χ2 > 7.8 and (ii) χ2 < 1.5.

10. Consider a chi-square random variable with 4 degrees of freedom; we are givenP u � χ2� � � 0.05 on the right tail, find u.

11. Provide a definition of a t probability distribution.

12. We assume a random variable x has a t distribution with degrees of freedomν � 24; compute P x < 2.0� �.

13. We assume a random variable x has a t distribution with degrees of freedomν � 20; given P x � u� � � 0.95, compute u.

14. Provide a definition of a F probability distribution.

15. Let x be random variable that has an F distribution with N1 � 17 and N2 � 23;find P x < 2� �.


3GC12 05/15/2014 11:36:19 Page 270

16. Download data on the S&P 500 stock index on a weekly basis during January2010 to June 2013 from Yahoo! Finance. Analyze the empirical distribution ofthe returns. Test for the normality hypothesis of the returns.

17. Define the moment generating function. Explain the role of the parameter t incomputing moments.

18. Apply the definition of the moment generation function to derive the mean andthe variance of the binomial, Poisson, and standard normal distributions.

270 STATISTICS

3GC13 05/15/2014 17:55:38 Page 271

CHAPTER 13Sampling and Hypothesis

Testing Theory

S ampling theory is a study of the relationship existing between a population andsamples drawn from the population. It is useful in estimating unknown population

parameters such as population mean μ and variance σ2 from knowledge of corre-sponding sample mean and variance, often called sample statistics. Sampling theory isalso useful in determining whether the observed differences between two samples aredue to chance variation or they are really significant. Such questions arise for testingdifferences in returns among assets. For instance, is the difference between returns onS&P 500 and the Nikkei 225 significant? Likewise, is the volatility of one indexdifferent from the volatility of another index? The analysis of differences in samplesinvolves the formulation of hypotheses and applications of tests of significance that areimportant in the theory of decisions. In order for the conclusions of sampling theoryand statistical inference to be valid, samples must be chosen so as to be representativeof the population. One way in which a representative sample may be obtained is by theprocess called random sampling, according to which eachmember of a population hasan equal chance of being included in the sample.

SAMPLING DISTRIBUTIONS

This section covers the sampling distribution of the mean, the sampling distribution ofproportions, and the sampling distribution of differences.

Consider all possible samples of size n that can be drawn from a given population.For each sample we can compute a statistic (such as the mean x and the standarddeviation s) that will vary from sample to sample. In this manner we obtain adistribution of the statistic that is called its sampling distribution. If, for example,the particular statistic used is the sample mean x, then the distribution is called thesampling distribution of the mean. Similarly, we could have sampling distributions ofvariance, standard deviation, median, and proportion. For each sampling distribu-tion, we can compute the mean and standard deviation. Thus, we speak of the meanand the standard deviation of the sampling distribution of the mean or the samplingdistribution of the variance.

271

3GC13 05/15/2014 17:55:40 Page 272

Sampling Distribution of the Mean

Suppose we draw random samples of size n from a population of finite size N. Eachrandom sample has random variables X1;X2; . . . ;Xn. The sample mean is

X � X1 �X2 � ∙ ∙ ∙ �Xn� �=n (13.1)

The random variable X may have a different value with each new sample. Wedenote its mean by μX and its standard deviation by σX. We denote the populationmean and standard deviation by μ and σ, respectively. Then we have

μX � μ (13.2)

and σX � σffiffiffin

pffiffiffiffiffiffiffiffiffiffiffiffiffiN � nN � 1

r(13.3)

If the population is infinite or if the sampling is with replacement, the previousresults reduce to

μX � μ (13.4)


p (13.5)

For large values of n n � 30� �, the sampling distribution of X is approximately anormal distribution with mean μX and standard deviation σX. This result derives fromthe central limit theorem. The accuracy of approximation improves as n gets larger. Incase the population is normally distributed, the sampling distribution of X is normaleven for small values of n.

Example: A population consists of five numbers 2; 3; 6; 8; 11f g. Consider allpossible samples of size 2 that can be drawn with replacement from this population.Find (i) the mean of the population, (ii) the standard deviation of the population,(iii) the mean of the sampling distribution of means, and (iv) the standard deviation ofthe sampling distribution of means.

i. μ � 2 � 3 � 6 � 8 � 115

� 6:0

ii. σ2 � 2 � 6� �2 � 3 � 6� �2 � 6 � 6� �2 � 8 � 6� �2 � 11 � 6� �25

� 10:8 and σ � 3:29

iii. There are 25 samples of size 2 that can be drawn with replacement, shown in thefollowing chart:

(2,2) (2,3) (2,6) (2,8) (2,11)(3,2) (3,3) (3,6) (3,8) (3,11)(6,2) (6,3) (6,6) (6,8) (6,11)(8,2) (8,3) (8,6) (8,8) (8,11)(11,2) (11,3) (11,6) (11,8) (11,11)

272 STATISTICS

3GC13 05/15/2014 17:55:42 Page 273

The corresponding sample means are

2.0 2.5 4 5 6.52.5 3 4.5 5.5 74 4.5 6 7 8.55 5.5 7 8 9.56.5 7 8.5 9.5 11

The mean of the sampling distribution of means is

μX � sum of all sample means25

� 15025

� 6:0

We note the fact that μX � μ.iv. The variance σ2

Xof the sampling distribution of means is obtained by subtract-

ing the mean 6 from each sample mean, squaring the result, adding all 25

numbers thus obtained, and dividing by 25. We find σ2X� 135

25� 5:40 and thus

σX � ffiffiffiffiffiffiffi5:4

p � 2:32. This illustrates the fact that for finite populations involving

sampling with replacement the sample variance is σ2X� σ2

n� 10:8

2� 5:40.

Sampling Distribution of Proportions

Suppose that a population is infinite and that the probability of occurrence ofan event (called success) is p, while the probability of nonoccurrence of the eventis q � 1 � p. Consider all possible samples of size n drawn from this population,and for each sample determine the proportion P of successes. We have thus asampling distribution of proportion Pwhose mean μP and standard deviation σP aregiven by

μP � p (13.6)

and σP �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip 1 � p� �

n

r(13.7)

For large values of n n � 30� � the sampling distribution is closely normallydistributed.

Example: Find the probability that in 120 tosses of a fair coin between 40percent and 60 percent will be heads. We consider the 120 tosses of the coin to be asample from the infinite population of all possible tosses of the coin. In thispopulation the probability of heads is p � 1=2 and the probability of tails isq � 1 � p � 1=2.

μP � p � 0:5 and σP �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:5 � 0:5

120

r� 0:0456

Sampling and Hypothesis Testing Theory 273

3GC13 05/15/2014 17:55:43 Page 274

40 percent in standard units � zp�0:4 � 0:4 � 0:50:0456

� �2:19


� 2:19

The required probability = area under normal curve between z � �2:19 andz � 2:19 � 2 � 0:4857 � 0:9714.

Example: It has been found that 2 percent of the Murabaha transactions of anIslamic bank made losses. What is the probability that in a sample of 400transactions 3 percent or more will prove loss-making?

μP � p � 0:02 and σP �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:02 � 0:98

400

r� 0:14

20� 0:007


� 1:43

Required probability = area under normal curve to the right of z � 1:43� 0:076359.

Sampling Distribution of Differences

Suppose we are given two populations with mean and standard deviation μ1; σ1� �

and μ2; σ2� �

, respectively. For each sample of size n1 drawn from the firstpopulation, we compute statistic T1, whose mean and standard deviation aredenoted by μT1

and σT1 , respectively. Similarly, for each sample of size n2 drawnfrom the second population, we compute statistic T2, whose mean and standarddeviation are denoted by μT2

and σT2 , respectively. From all possible combinationsof these samples from the two populations we can obtain a distribution of thedifferences, T1 � T2, which is called the sampling distribution of differences of thestatistics. Assuming that the samples are independent, the mean and standarddeviation of this sampling distribution, denoted respectively by μT1�T2

and σT1�T2 ,are given by

μT1�T2� μT1�μT2

(13.8)

and σT1�T2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2T1

� σ2T2

q(13.9)

If T1 and T2 are the sample means from the two populations, which we denote byX1 and X2, respectively, then the sampling distribution of the differences of meansX1 �X2� �

is given for infinite populations by

μX1�X2� μX1

� μX2� μ1 � μ2 (13.10)

274 STATISTICS

3GC13 05/15/2014 17:55:44 Page 275

and σX1�X2� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

σ2X1

� σ2X2

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ21n1

� σ22n2

s(13.11)

Corresponding results can be obtained for the sampling distribution of differencesin proportions from two binomially distributed populations with parameters p1; q1

� �and p2; q2

� �, respectively. In this case, T1 and T2 correspond to the proportion of

successes P1 and P2. We obtain

μP1�P2� μP1

� μP2� p1 � p2 (13.12)

and σP1�P2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2P1

� σ2P2

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip1q1n1

� p2q2n2

r(13.13)

If n1 and n2 are large n1; n2 � 0� � the sampling distributions of differences ofmeans or proportions are closely normally distributed.

Example: The stock index A has a mean annual return of 8 percent with astandard deviation of 11.5, while the stock index B has a mean return of 5.7 percentwith a standard deviation of 14.5. If random samples of 100 returns of each stockindex are tested, what is the probability that the stock index A returns will have amean return that is at least 1.3 percent more than the stock index B?

Let XA and XB denote the mean return of samples A and B, respectively. Then

μXA�XB� μXA

� μXB� 8 � 5:7 � 2:3

σXA�XB� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

σ2XA

� σ2XB

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2AnA

� σ2BnB

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi8� �2100

� 5:7� �2100

s� 0:982293

And the standardized variable for the difference in mean returns is

z � XA �XB� � � μXA�XB

� �σXA�XB

� XA � XB� � � 2:3

0:982293

The difference 1.3 percent in standard units is (1.3–2.3)/0.982293=�1.01803.Using Microsoft Excel, we have P z � �1:01803� � � 1 � P z � �1:01803� �� 1 �NORM:S:DIST 1:01803;TRUE� � � 0:845667.

Example: The Murabaha transactions of Bank A have a mean lifetime of140 days with a standard deviation of 20 days, while those of Bank B have amean lifetime of 120 days with a standard deviation of 10 days. If random samplesof 125 Murabaha transactions of each bank are tested, what is the probability thatthe Bank A transactions will have a mean lifetime that is at least 16 more than theBank B transactions?

Let XA and XB denote the mean lifetimes of samples A and B, respectively. Then

μXA�XB� μXA

� μXB� 140 � 120 � 20


3GC13 05/15/2014 17:55:45 Page 276

σXA�XB� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

σ2XA

� σ2XB

q �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2AnA

� σ2BnB

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10� �2125

� 20� �2125

s� 2

And the standardized variable for the difference in means is

z � XA �XB� � � μXA�XB

� �σXA�XB

� XA �XB� � � 20

2

The difference of 16 days in standard units is (16–20)/2=�2. Thus the requiredprobability is the area under normal curve to right of z � �2 is 0.97725.

Example: A and B play a game of “heads or tails,” each tossing 50 coins. A willwin the game if he tosses five or more heads than B; otherwise, B wins. Determine theodds against A winning any particular game.

Let PA and PB denote the proportion of heads obtained by A and B. If we assumethat the coins are all fair, the probability p of heads is 1/2. Then

μPA�PB� μPA

� μPB� 0

σPA�PB �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2PA

� σ2PB

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip1q1n1

� p2q2n2

r�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:5 � 0:5

50� 0:5 � 0:5

50

r� 0:1

The standardized variable for the difference in proportions is z � PA�PB�00:1 .

Five heads represent a proportion of 5/50=0.1. In a standardized unit we havez � 0:1�0

0:1 � 1. The required probability is the area under the normal curve to the rightof z � 1, which is 0.158655.

ESTIMATION OF PARAMETERS

We saw how sampling theory can be employed to obtain information aboutsamples drawn at random from a known population. We often want to inferinformation about a population from samples drawn from it. Statistical inferenceuses principles of sampling theory. One important problem of statistical inferenceis the estimation of population parameters (such as population mean μ andvariance σ2) from corresponding sample statistics (such as sample mean x andvariance s2).

Unbiased Estimates

If the mean of the sampling distribution of a statistic equals the correspondingpopulation parameter, the statistic is called an unbiased estimator of the parameter;otherwise, it is called a biased estimator. The corresponding values of such statisticsare called unbiased or biased estimates, respectively.

The mean of the sampling distribution of means X, μX, is μ, the population mean.Hence, the sample mean X is an unbiased estimate of μ.

276 STATISTICS

3GC13 05/15/2014 17:55:46 Page 277

The sample variance is defined as*

s2 �Pn

i�1 Xi �X� �2n

(13.14)

And unbiased estimator of σ is s given by

s2 �Pn

i�1 Xi �X� �2n � 1

� nn � 1

s2 (13.15)

The mean of the sampling distribution of variances s2 is

μs2 � n � 1n

σ2 (13.16)

*We show that s2 is a biased estimator of σ2. We write

s2 �Pn

i�1 xi � x� �2n

�Pn

i�1 x2i � 2xPn

i�1 xi � nx2

n

s2 �Pn

i�1 x2i � nx2

n

We compute E s2� � �

Pni�1 E�x2i �n

� E�x2�.It is well known that σ2 � E x2

i

� � � μ2. Hence,Pn

i�1 E�x2i � � nσ2 � nμ2.E�x2� can be written as

E�x2� � E

Pni�1 xin

� �2

� E

Pni�1 xi � μ� � � nμ� �

n

2

E�x2� � EPn

i�1 xi � μ� �2 � 2μPn

i�1 xi � μ� � � n2μ2

n2

" #

We observe that σ2 � E xi � μ� �2� �and E xi � μ� � � 0; we find

E�x2� � 1nσ2 � μ2

We collect terms, and we find

E s2� � � σ2 � μ2 � 1

nσ2 � μ2 � n � 1

n

� �σ2 ≠ σ2


3GC13 05/15/2014 17:55:47 Page 278

where σ2 is the population and n is the sample size. Thus the sample variance s2 is abiased estimate of the population σ2. By using the modified variance,

s2 � nn � 1

s2 (13.17)

We find μs2 � σ2, so that s2 is an unbiased estimator of σ2. In the language ofexpectation, we could say that a statistic is unbiased if its expectation equals thecorresponding parameter. Thus X and s2 are unbiased estimators because E X

� � � μand E s2

� � � σ2.

Efficient Estimates

If the sampling distributions of two statistics have the same mean (or expectation),then the statistic with the smaller variance is called an efficient estimator ofthe mean, while the other statistic is called an inefficient estimator. The correspond-ing values of the statistics are called efficient estimates, respectively. If weconsider all possible statistics whose sampling distributions have the same mean,the one with the smallest variance is called themost efficient or best estimator of thismean.

Point Estimates and Interval Estimates

An estimate of a population parameter given by a single number is called the pointestimate of the parameter. An estimate of a population parameter given by twonumbers between which the parameter may be considered to lie is called an intervalestimate of the parameter.

CONFIDENCE-INTERVAL ESTIMATES OF POPULATIONPARAMETERS

This section covers confidence intervals for means, confidence intervals for propor-tions, confidence intervals for differences, and confidence intervals for standarddeviations.

Let μT and σT be the mean and the standard deviation, respectively, of thesampling distribution of statistic T. Then if the sampling distribution of T isapproximately normal, which is true for many statistics if the sample size n � 40,we can construct a confidence interval within which the actual sample statistic Tis expected to lie. For instance, if the confidence level is 95 percent, the statistic T isexpected to lie 95 percent of the time within the interval

μT � 1:96σT ; μT � 1:96σTf g (13.18)

If we change the confidence level to 90 percent, the statistic T is expected to lie 90percent of the time within the interval

μT � 1:645σT ; μT � 1:645σTf g (13.19)

278 STATISTICS

3GC13 05/15/2014 17:55:47 Page 279

The numbers 1.645, 1.96, and so on are called confidence coefficients or criticalvalues and are denoted by zc. They are read from the standard normal distributiontable or using Microsoft Excel for the normal distribution. In Figure 13.1, theconfidence level is portrayed by the area under the normal curve.

Confidence Intervals for Means

If the statistic T is the sample X, then the 95 percent and 99 percent confidence limitsfor estimating the population mean μ are given by

X � 1:96σX (13.20)

X � 2:58σX (13.21)

respectively. More generally, the confidence limits are given by

X � zcσX (13.22)

where zc is a critical value that depends on the confidence level and can be read fromthe standard normal distribution table. Using the values of σX, we see that theconfidence limits for the population mean are given by

X � zcσffiffiffin

p (13.23)

Generally, the population standard deviation σ is unknown; thus, we obtain theabove confidence limits by using the sample estimate s or s. This will prove satisfactorywhen n � 30.

Confidence level = area under thecurve between critical values

Confidence interval Critical valueCritical value

0 Critical value

Standard normaldistribution N(0,1)

Confidence interval

Normal distributionN(μT ,σT)

μT

f (T)

f (z)

z–zc zc

T

FIGURE 13.1 Normal Distributions


3GC13 05/15/2014 17:55:48 Page 280

Example:Measurement of annual returns of a random sample of 200 Islamic stocksmade on January 15 showed a mean return of 6.26 percent and a standard deviationof 9.45 percent. Using Microsoft Excel find the (i) 90 percent, (ii) 95 percent, and(iii) 99 percent confidence limits for the mean return.

i. At the confidence level 90 percent, the critical z is

z0:05 � NORM:S:INV 0:95� � � 1:644854

The 90 percent confidence limits are

X � 1:64σffiffiffin

p � X � 1:64sffiffiffin

p � 6:26 � 1:64 � 9:45ffiffiffiffiffiffiffiffi200

p � 6:26 � 1:099117 � 5:16; 7:36� �

ii. At the confidence level 95 percent, the critical z is

z0:025 � NORM:S:INV 0:975� � � 1:96





p � 6:26 � 1:31 � 4:95; 7:56� �

iii. At the confidence level 99 percent, the critical z is

z0:005 � NORM:S:INV 0:995� � � 2:58





p � 6:26 � 1:72 � 4:54; 7:98� �

Confidence Intervals for Proportions

If the statistic T is the proportion of “successes” in a sample of size n drawn from abinomial population in which p is the proportion of successes (i.e., probability ofsuccess), then the confidence limits for p are given by

P � zcσP (13.24)

where P is the proportion of successes in the sample of size n. Using the values of σP,we see that the confidence limits for the population proportion are given by

P � zc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip 1 � p� �

n

r(13.25)

280 STATISTICS

3GC13 05/15/2014 17:55:48 Page 281

To compute these confidence limits, we can use the sample estimate P for p, whichwill generally prove satisfactory if n � 0.

Example: A sample poll of 100 investors chosen at random indicated that55 percent of them thought the Islamic stock index will outperform the conventionalstock index. Find the 95 percent confidence limits for the proportion of all investors infavor of the Islamic stock index.

The 95 percent confidence limits for the population p are

P � zcσP � P � 1:96

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip 1 � p� �

n

r� 0:55 � 1:96

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:55 � 0:45=100

p

� 0:55 � 0:10 � 0:45; 0:65� �

Confidence Intervals for Differences

If T1 and T2 are two sample statistics with approximately normal distributions,confidence limits for the difference of the population parameters corresponding to T1

and T2 are given by

T1 � T2� � � zcσT1�T2 � T1 � T2� � � zcffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2T1

� σ2T2

q(13.26)

For instance, the confidence limits for the difference of two population means, inthe case where the populations are infinite, are given by

X1 � X2 � zc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ21n1

� σ22n2

s(13.27)

Example: A sample of 150 Islamic stocks listed in Kuala Lampur stock exchangeshowed a mean return of 8.1 percent and a standard deviation of 7.5 percent. Asample of 200 Islamic stocks listed in Bahrain stock exchange showed amean return of7.2 percent and a standard deviation of 9.4 percent. Find the 95 percent confidencelimits for the difference of the mean returns of the populations of Islamic stocks listedin Kuala Lampur and Bahrain.

Confidence limits for the difference in mean returns are given by

X1 �X2 � zc


� σ22n2

s� 8:1 � 7:2 � 1:96

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7:52

150� 9:42

200

s� 0:9 � 1:77 � �0:87; 2:67� �

The confidence interval for difference in proportions is given as

P1 � P2 � zc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip1 1 � p1� �n1

� p2 1 � p2� �n2

s(13.28)


3GC13 05/15/2014 17:55:49 Page 282

Example: In a random sample of 400 Islamic stocks listed on stock exchange A,100 stocks outperformed themarket stock index A; in a random sample of 600 Islamicstocks listed on stock exchange B, 300 stocks outperformed the market index B.Construct a 95 percent confidence limits for the difference in proportions of all Islamicstocks that outperformed the market index A and B, respectively.

The confidence limits for the difference in proportions of the two markets aregiven by

P1 � P2 � zc

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip1 1 � p1� �n1

� p2 1 � p2� �n2

s(13.29)

Here P1 � 300600

� 0:5 and P2 � 100400

� 0:25 are the proportions of Islamic stocks

that outperformed the market index B and A, respectively. The 95 percent confi-dence limits are

0:5 � 0:25 � 1:96

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:5 � 0:5

600� 0:25 � 0:75

400

r� 0:25 � 0:06 � 0:31; 0:19� �

Confidence Intervals for Standard Deviations

The confidence limits for the standard deviation σ of a normally distributed popula-tion, as estimated from a sample with standard deviation s, are

s � zcσs � s � zcσffiffiffiffiffiffi2n

p (13.30)

To compute these confidence limits, we use s or s to estimate σ.Example: The standard deviation of the annual returns of a random sample of 200

stocks listed on amarket exchangewas computed to be 8.4 percent. Find the 95 percentconfidence limits for the standard deviation of all stocks listed on the market exchange.

The confidence limits for the population standard deviation σ are given by

s � zcσs � s � zcσffiffiffiffiffiffi2n

p (13.31)

We use the sample standard deviation to estimate σ. The confidence limits at95 percent are

8:4 � 1:96 � 8:4ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 � 200

p � 8:4 � 0:8232 � 7:5768; 9:2232� �

HYPOTHESIS TESTING

Very often in practice we are called to make decisions about populations on the basisof sample information. Such decisions are called statistical decisions. For example,

282 STATISTICS

3GC13 05/15/2014 17:55:50 Page 283

we may wish to decide on the basis of sample data whether a stock is performing orif the market is efficient.

This section covers the specification of statistical hypotheses, type I and type IIerrors, level of significance of a test, probability value: p-value, and special testsregarding the means and proportions.

Statistical Hypotheses

In attempting to reach decisions, it is useful tomake assumptions about the populationinvolved. Such assumptions, which may or may not be true, are called statisticalhypotheses. They are generally statements about the probability distributions of thepopulations.

Null Hypotheses

In many instances we formulate a statistical hypothesis for the sole purpose ofrejecting it. For example, if we want to decide whether a coin is loaded, we makethe hypothesis that the coin is fair (i.e., p � 0:5 where p is the probability of heads).Similarly, if we want to decide whether a product is better than another, we formulatethe hypothesis that there is no difference between the two products (i.e., any observeddifferences are duemerely to fluctuations in sampling from the same population). Suchhypotheses are often called null hypotheses and are denoted by H0.

Alternative Hypothesis

Any hypothesis that differs from a given hypothesis is called an alternativehypothesis. For example, if one hypothesis is p � 0:5, alternative hypotheses mightbe p � 0:6, p ≠ 0:5, or p > 0:5. A hypothesis alternative to the null hypothesis isdenoted by H1.

If we suppose that a particular hypothesis is true but we find the results observedin a random sample differ markedly from the results expected under the hypothesis(i.e., expected on the basis of pure chance, using sampling theory), then we would saythat the observed differences are significant and would be inclined to reject thehypothesis. Tests of hypotheses are procedures that enable us to determine whetherobserved samples differ significantly from the results expected and help us to decidewhether to accept or reject hypotheses.

Type I and Type II Errors

If we reject a hypothesis when it should be accepted, we commit a type I error. If weaccept a hypothesis when it should be rejected, we say that a type II error has beenmade. In either case, a wrong decision or error in judgment has occurred (see thefollowing table).

Truth

Decision H0 is true H1 is true

RejectH0 Type I error No errorDo not rejectH0 No error Type II error


3GC13 05/15/2014 17:55:50 Page 284

Probabilities for making test errors are depicted in Figure 13.2. A test’s probabil-ity of making a type I error is denoted by α. A test’s probability of making a type IIerror is denoted by β. All statistical hypothesis tests have a probability of making type Iand type II errors. For example, all blood tests for a disease will falsely detect thedisease in some proportion of people who don’t have it, and will fail to detect thedisease in some proportion of people who do have it.

Hence, a type I error is the incorrect rejection of a true null hypothesis. A type IIerror is the failure to reject a false null hypothesis. A type I error is a false positive.Usually a type I error leads one to conclude that a thing or relationship exists whenreally it doesn’t, for example, that a patient has a disease being tested for when reallythe patient does not have the disease, or that a medical treatment cures a disease whenreally it doesn’t. A type II error is a false negative. Examples of type II errors would bea blood test failing to detect the disease it was designed to detect in a patient who reallyhas the disease or a clinical trial of a medical treatment failing to show that thetreatment works when really it does. When comparing two means, concluding themeans were different when in reality they were not different would be a type I error;concluding the means were not different when in reality they were different would be atype II error.

Level of Significance

In testing a given hypothesis, the maximum probability with which we would bewilling to risk a type I error is called the level of significance of the test. Thisprobability is often denoted by α. In practice, a significance level of 0.05 (5 percent)or 0.01 (1 percent) is customary, although other values are used. If, for example,5 percent significance level is chosen in designing a decision rule, then there areabout 5 chances in 100 that we would reject the hypothesis when it should beaccepted; that is, we are about 95 percent confident that we have made the rightdecision. In such case we say the hypothesis has been rejected at the 5 percentsignificance level, which means that the hypothesis has a 5 percent probability ofbeing wrong.

To illustrate the ideas presented earlier, suppose that under a given hypothesis thesampling distribution of a statistic T is a normal distribution with mean μT andstandard deviation σT . Thus, the distribution of the standardized variable (or z-score),

Probability Type II error = β

β

Probability Type I error = α

αH0 H1

FIGURE 13.2 Significance Levels: Probabilities of Type I andType II Errors

284 STATISTICS

3GC13 05/15/2014 17:55:51 Page 285

given byz � T � μT

� �=σT (13.32)

is the standard normal distribution N 0; 1� � as shown in Figure 13.3. We can be95 percent confident that if the hypothesis is true, then the z-score of an actualsample statistic T will lie between –1.96 and 1.96 (because the area under thenormal curve between these values is 0.95). However, if on choosing a single sampleat random we find that the z-score lies outside the range –1.96 to 1.96, we wouldconclude that such event could happen with a probability of only 0.05 if the givenhypothesis were true. We would then say that this z-score differed significantly fromwhat would be expected under the hypothesis, and we would then be inclined toreject the hypothesis.

The total critical region area 0.05 is the significance level of the test. Itrepresents the probability of our being wrong in rejecting the hypothesis (i.e.,the probability of making a type I error). Thus we say that the hypothesis isrejected at 0.05 significance level or that the z-score of the given sample statistic issignificant at 0.05 level. The set of z-scores outside the range �1.96 to 1.96constitutes what is called the critical region of the hypothesis, the region of rejectionof the hypothesis, or the region of significance. The set of z-scores inside the range�1.96 to 1.96 is thus called the region of acceptance of the hypothesis, or the regionof insignificance.

We can formulate the following decision rule:

1. Reject the hypothesis at 0.05 significance level if the z-score of the statistic T liesoutside the range –1.96 to 1.96. This is equivalent to saying that the observedsample statistic is significant at the 0.05 level.

2. Accept the hypothesis otherwise.

0Confidence interval

Critical regionCritical region

0

Critical value = ± zc

zcz

–zc

Standard normaldistribution N(0,1)

Confidence interval

0.95

f (z)

f (z)

z = –1.96 z = 1.96z

FIGURE 13.3 Hypothesis Testing


3GC13 05/15/2014 17:55:51 Page 286

Because the z-score plays such an important role in tests of hypotheses, it is calleda test statistic.

In the previous test we were interested in extreme values of statistic T or itscorresponding z-score on both sides of the mean (i.e., in both tails of the distribution).Such tests are thus called two-sided tests or two-tailed tests. Often, however, we maybe interested only in extreme values to one side of the mean (i.e., one tail of thedistribution), such as when we are testing the hypothesis that one asset has higherreturn than another asset. Such tests are called one-sided tests or one-tailed tests. Insuch cases the critical region is a region to one side of the distribution, with area equalto the level of significance.

Probability Value: p-Value

In statistical significance testing the p-value is the probability of obtaining a teststatistic at least as extreme as the one that was actually observed, assuming that thenull hypothesis is true. This can be expressed as conditional probability (data|H0).The statistic should not be interpreted as the probability ofH0 being true. One often“rejects the null hypothesis” when the p-value is less than the predeterminedsignificance level, which is often 0.05 or 0.01, indicating that the observed resultwould be highly unlikely under the null hypothesis. Many common statistical tests,such as chi-squared tests or Student’s t-test, produce test statistics that can beinterpreted using p-values.

The classical approach to hypothesis testing is to compare a test statistic and acritical value. It is best used for distributions that give areas and requires the reading ofthe critical value (like the Student’s t-distribution) rather than distributions thatrequire finding a test statistic to find an area (like the normal distribution). Theclassical approach also has three different decision rules, depending on whether it is aleft tail, right tail, or two-tail test. One problem with the classical approach is that if adifferent level of significance is desired, a different critical value must be read from theprobability table.

The p-value approach analyzes hypothesis testing from a different manner.Instead of comparing z-scores or t-scores as in the classical approach, we compareprobabilities, or areas. The level of significance (alpha) is the area in the critical region.That is, the area in the tails to the right or left of the critical values. The p-value is thearea to the right or left of the test statistic. If it is a two-tail test, then we look up theprobability in one tail and double it. If the test statistic is in the critical region, thenthe p-value will be less than the level of significance. It does not matter whether it is aleft tail, right tail, or two-tail test. This rule always holds: Reject the null hypothesis ifthe p-value is less than the level of significance.

We will fail to reject the null hypothesis if the p-value is greater than or equal tothe level of significance. A benefit of the p-value is that the statistician immediatelyknows at what level the testing becomes significant. That is, a p-value of 0.06would berejected at an 0.10 level of significance, but it would fail to reject at an 0.05 level ofsignificance. The p-value measures the likelihood of a type I error, the probability ofincorrectly rejecting a correct null hypothesis. The higher the p-value, the more likelyit is that we will err in rejecting the null hypothesis; the lower the p-value, the morecomfortable we feel in rejecting it.

286 STATISTICS

3GC13 05/15/2014 17:55:52 Page 287

Example: Measurement of annual returns of a random sample of 200 Islamicstocks made on January 15 showed a mean return of 6.26 percent and a standarddeviation of 9.45 percent. We want to test the hypothesis H0: μ � 7:3 percent at5 percent significance level. Using Microsoft Excel find the p-value of the test.

We compute the z-score assuming H0 is true. We have

z � X � μσ=

ffiffiffin

p � 6:26 � 7:3

9:45=ffiffiffiffiffiffiffiffi200

p � �1:55

p-value = NORM.S.DIST(–1.55,TRUE)= 0.0598Based on the p-value we accept the null hypothesis, since 0.0598>0.05.

Special Tests

For large samples, the sampling distributions of many statistics are normal distribu-tions and we can use the z-scores of the standard normal distribution. The followingspecial cases are just a few of the statistics of practical interest.

Means

Here T � X, the sample mean; μT � μX � μ, the population mean; and σT � σX �σ=

ffiffiffin

p, where σ is the population standard deviation and n is the sample size. The

z-score is given by

z � X � μσ=

ffiffiffin

p (13.33)

When necessary, the sample deviation s or s is used to estimate σ.Example: The mean rate of return of a sample of 100 Islamic stocks is 7.0 percent

per year with a standard deviation of 8.2 percent. If μ is the mean return of all Islamicstocks listed on the exchange, test the hypothesis of μ � 8:7 percent against thealternative of μ ≠ 8:7 percent using significance level of 5 percent.

We must decide between two hypotheses:

H0 : μ � 8:7%

H1 : μ ≠ 8:7%

SinceH1: μ ≠ 8:7 percent includes values both larger and smaller than 8.7 percent,a two-tailed test should be used. For a two-tailed test at the 5 percent significance level,we have the following decision rule: Reject H0 if the z-score of the sample mean isoutside the range �1.96 to 1.96. Accept H0 otherwise. The statistic under considera-tion is the sample mean X. The sampling distribution of X has mean μX � μ andstandard deviation σX � σ=

ffiffiffin

p, where μ and σ are the mean and standard deviation of

the population of all Islamic stocks. Under the hypothesisH0, we have μ � 8:7 percent


p � 8:2ffiffiffiffiffiffiffiffi100

p � 0:82 using the sample standard deviation as an estimate of σ.


3GC13 05/15/2014 17:55:53 Page 288

We compute the observed z-score as

z � X � μσ=

ffiffiffin

p � 7:0 � 8:70:82

� �2:07

Since z � �2:07 lies outside the range �1.96 to 1.96, we reject H0 at the 0.05significance level.

Proportions

Here T � P, the proportion of successes in a sample; μT � μP � p, where p is thepopulation proportion of successes and n is the sample size; and σT � σP �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip 1 � p� �=np

.The z-score is given by

z � P � pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip 1 � p� �=np (13.34)

TESTS INVOLVING SAMPLE DIFFERENCES

Let X1 and X2 be the sample means obtained in large independent samples of sizes n1and n2 drawn from respective populations having means μ1 and μ2 and standarddeviations σ1 and σ2. Consider the null hypothesis that there is no difference betweenthe population means (i.e., μ1 � μ2), which is to say the samples are drawn from twopopulations having the same mean. Making the hypothesis μ1 � μ2 we see that thesampling distribution of differences in means is approximately normally distributed,with mean and standard deviation given by

μX1�X2(13.35)

σX1�X2�


� σ22n2

s(13.36)

where we can, if necessary, use the sample standard deviations s1 and s2 (or s1 and s2)as estimates of σ1 and σ2. By using the standardized variable, or z-score, given by

z � X1 �X2 � 0σX1�X2

� X1 � X2

σX1�X2

(13.37)

we can test the null hypothesis against alternative hypotheses (or the significance of anobserved difference) at an appropriate level of significance.

Example: A random sample of 156 Islamic stocks listed in Labuan stockexchange yielded a mean annual return of 9.1 percent and a standard deviationof 12.3 percent. A random sample of 242 conventional stocks listed in Labuan stockexchange yielded a mean annual return of 7.9 percent and a standard deviation of

288 STATISTICS

3GC13 05/15/2014 17:55:54 Page 289

16.2 percent. Is there a significant difference between the performance of the Islamicand conventional stocks at 0.05 significance level?

Suppose that the two classes of stocks come from two populations having therespective means μ1 and μ2. We thus need to decide between the hypotheses:

H0: μ1 � μ2, and the difference is due merely to chance.H1: μ1 ≠ μ2, and there is a significant difference between the classes of stocks.

Under the hypothesis H0, both classes come from the same population. We usethe sample standard deviations as estimates of σ1 and σ2. The mean and standarddeviation of the difference in means are given by

μX1�X2� 0

σX1�X2�


� σ22n2

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12:32

156� 16:22

242

s� 1:433

We compute the z-statistics as

z � X1 �X2 � 0σX1�X2

� 9:1 � 7:91:433

� 0:8372

We notice that z � 0:8372 lies within the interval [�1.96,196]. We thereforecannot reject H0.

SMALL SAMPLING THEORY

In the previous sections we often made use of the fact that for samples of size n � 30,called large samples, the sampling distributions of many statistics are approximatelynormal, the approximation becomes better with increasing n. For samples of sizen < 30, called small samples, this approximation is not good and becomes worse withdecreasing n, so that appropriate modifications must be made. A study of samplingdistributions of statistics for small samples is called small sampling theory. In thissection we study three important distributions: student’s t-distribution, the chi-squaredistribution, and the F distribution.

Tests Based on the Student ’s t-Distribution

Let us define the statistic

t � X � μs=

ffiffiffiffiffiffiffiffiffiffiffin � 1

p � X � μs=

ffiffiffin

p (13.38)

Suppose a small random sample of size n is taken from a population. If thepopulation from which the sample is drawn forms a normal distribution, the distribu-tion of t follows student’s t-distribution with n � 1 degrees of freedom. For large values


3GC13 05/15/2014 17:55:54 Page 290

of the degree of freedom, the student’s t-distribution is closely approximated by thestandardized normal distribution.

Confidence Intervals for t Distribution

As is done with normal distributions, we can define 95 percent, 99 percent, or otherconfidence intervals by using the t distribution table. In this manner we can estimatewithin specified limits of confidence the population mean μ. Figure 13.4 portrays theconfidence level and interval for the t distribution. For example, if�t0:975 and t0:975 arethe values of t for which 2.5 percent of the area lie in each tail of the t distribution, thenthe 95 percent confidence interval for t is

�t0:975 <X � μ

s=ffiffiffiffiffiffiffiffiffiffiffin � 1

p < t0:975 (13.39)

from which we see that μ is estimated to lie in the interval

X � t0:975sffiffiffiffiffiffiffiffiffiffiffin � 1

p < μ < X � t0:975sffiffiffiffiffiffiffiffiffiffiffin � 1

p (13.40)

with 95 percent confidence (i.e., probability 0.95). Note that t0:975 represents the97.5 percentile value, while t0:025 � �t0:975 represents the 2.5 percentile value. Ingeneral, we can represent confidence limits for population mean by

X � tcsffiffiffiffiffiffiffiffiffiffiffin � 1

p (13.41)

where �tc, called critical values or confidence coefficients, depend on the level ofconfidence desired and the sample size (degree of freedom). Tests of hypotheses andsignificance are easily extended to problems involving small samples, the onlydifference being that the z-score, or z-statistic, is replaced by a suitable t-score, ort-statistic.

Confidence level

Confidence interval

–tα /2 tα /2t

0

FIGURE 13.4 Distribution: Confidence Level andInterval

290 STATISTICS

3GC13 05/15/2014 17:55:55 Page 291

Example: A portfolio manager wants to track the market return of about6.8 percent. To determine whether his portfolio is in line with the market, a sampleof 10 stocks is chosen for which the mean return was 8.2 percent and standarddeviation was 11.2 percent. Test the hypothesis that the portfolio is in line with themarket at significance level of 5 percent.

We wish to decide between the hypotheses

H0: μ � 6:8 percent, the portfolio is in line with the market.H1: μ ≠ 6:8 percent, the portfolio is out of line with the market.

Thus a two-tailed test is required. Under hypothesis H0, we have

X � μs=

ffiffiffiffiffiffiffiffiffiffiffin � 1

p � 8:2 � 6:8

11:2=ffiffiffiffiffiffiffiffiffiffiffiffiffiffi10 � 1

p � 0:375

For a two-tailed test at the 0.05 significance level, we adopt the decision rule.Accept H0 if t lies inside the interval ��t0:975; t0:975�, which for 10 – 1 = 9

degrees of freedom is the interval [–2.26,2.26].* Reject otherwise. Since t � 0:375lies inside the interval, we conclude that the portfolio is in line with the marketportfolio.

We can apply student’s t-distribution to test differences of means. Suppose thattwo samples of sizes n1 and n2 are drawn from normal populations who standarddeviations are equal σ1 � σ2� �. Suppose further that these two samples have meansgiven by X1 and X2 and standard deviations given by s1 and s2, respectively. To testthe hypothesis H0 that the samples come from the same population (i.e., μ1 � μ2 aswell as σ1 � σ2) we use the t-score given by

t � X1 �X2� �

= σ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n1

� 1n2

s !(13.42)

where

σ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin1s21 � n2s22n1 � n2

s(13.43)

The distribution of t has degree of freedom equal to n1 � n2 � 2. We note that anestimate of σ2 is given by the weighted mean

σ2 � n1 � 1� �s21 � n2 � 1� �s22n1 � 1� � � n2 � 1� � � n1s21 � n2s22

n1 � n2 � 2(13.44)

where s21 and s22 are the unbiased estimates of σ21 and σ22.

*The limits are computed using Microsoft Excel: T.INV.2T(0.05,9)=2.26.


3GC13 05/15/2014 17:55:56 Page 292

Example: The risk-return profile of 16 Islamic stocks listed in Labuan showed amean return of 8.2 percent and a standard deviation of 12.3 percent. The risk-returnprofile of 14 Islamic stocks listed in Singapore showed a mean return of 6.5 percentand a standard deviation of 9.5 percent. Is there a significant difference between themean returns of the two groups of stocks at 0.05 significance level?

If μ1 and μ2 denote the population mean returns in Labuan and Singapore,respectively, we have to decide between the hypotheses

H0: μ1 � μ2, and there is essentially no difference between the groups of Islamicstocks.

H1: μ1 ≠ μ2, and there is a significant difference between the groups of Islamicstocks.

Under hypothesis H0, t � X1 �X2� �

= σ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n1

� 1n2

s !where σ �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin1s21 � n2s22n1 � n2 � 2

s.

Thus,

σ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffin1s21 � n2s22n1 � n2

s�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi16 � 12:32 � 14 � 9:52

16 � 14 � 2

s� 11:47

and; t � X1 � X2� �

σ �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n1

� 1n2

r � 8:2 � 6:5

11:47 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi116

� �� 1

14

� �s � 0:4049

Using a two-tailed test at 0.05 significance level, we would reject H0 if t wereoutside the range �t0:975 and t0:975, which for 28 degrees of freedom is the range�2:05; 2:05� �. Thus we cannot reject H0 at 0.05 significance level.

Tests Based on the Chi-Square Distribution

Let us define the statistic

χ2 � ns2

σ2� X1 �X� �2 � X2 �X

� �2 � ∙ ∙ ∙ � Xn �X� �2

σ2(13.45)

If we consider samples of size n drawn from a normal population with standarddeviation σ and, if for each sample we compute χ2, a sampling distribution for χ2 canbe obtained. This distribution is called the chi-square distribution with degree offreedom equal to n � 1.

Confidence Intervals for χ2

As is done with the normal and t distributions, we can define 95 percent, 99 percent,or other confidence limits and intervals for χ2 by using the table of the χ2distribution or Microsoft Excel. In this manner we can estimate within specifiedlimits of confidence the population standard deviation σ in terms of a samplestandard deviation s. Figure 13.5 describes the confidence level and interval for the

292 STATISTICS

3GC13 05/15/2014 17:55:57 Page 293

chi-square. For example, if χ20:025 and χ20:975 are the values of χ2, called critical values,

for which 2.5 percent of the area lies in each tail of the distribution, then the 95percent confidence interval is

χ20:025 <ns2

σ2< χ20:975 (13.46)

From which we see that σ is estimated with 95 percent confidence to lie in theinterval

sffiffiffin

pχ0:975

< σ <sffiffiffin

pχ0:025

(13.47)

The values of χ0:025 and χ0:975 represent the 2.5 and 97.5 percentile values,respectively.

For large values of ν ν � 30� � we can use the fact thatffiffiffiffiffiffiffiffi2χ2

p � ffiffiffiffiffiffiffiffiffiffiffiffiffi2ν � 1

p� �is

approximately normally distributed with mean 0 and standard deviation 1. That is,

z � ffiffiffiffiffiffiffiffi2χ2


p� �(13.48)

Thus, the normal distribution table can be used if ν � 30. Then if χ2α and zα arethe αth percentiles of the chi-square and normal distributions, we have

χ2α � 12

zα �ffiffiffiffiffiffiffiffiffiffiffiffiffi2ν � 1

p� �2(13.49)

Degrees of freedom are a key parameter of the chi-square. In order to computestatistics under chi-square distribution (or t distribution), it is necessary to useobservations obtained from a sample as well as certain population parameters. Ifthese parameters are unknown, they must be estimated from the sample. The numberof degrees of freedom of a statistic, denoted by ν, is defined as the number ofindependent observations in the sample (i.e., the sample size) minus the number k

Density

0

Confidencelevel

Confidenceinterval

χ0.0252 χ

0.975

χ2

2

FIGURE 13.5 Chi-Square Distribution:Confidence Level and Interval


3GC13 05/15/2014 17:55:58 Page 294

of population parameters, which must be estimated from sample observations, thatis, ν � k. In the case of statistic

χ2 � ns2

σ2� X1 �X� �2 � X2 �X

� �2 � ∙ ∙ ∙ � Xn �X� �2

σ2(13.50)

the number of independent observations in the sample is n, from which we mustcompute s. The degrees of freedom is obtained as follows: because we must estimate σ,we have k � 1 and so ν � n � 1.

Example: We consider a chi-square distribution with 5 degrees. Using MicrosoftExcel find the critical values of χ2 for which (i) the critical area on the right is 5 percent,and (ii) the total critical area is 0.05.

i. χ20:95 � CHISQ:INV:RT 0:05; 5� � � 11:0705.ii. Since the distribution is not symmetrical, there are many critical values for which

the total critical area is 0.05. It is customary to choose the two areas to be equal.In this case, each area has is 0.025. We have

χ20:975 � CHISQ:INV:RT 0:025; 5� � � 12:8325

χ20:025 � CHISQ:INV 0:025; 5� � � 0:831212

Example: The standard deviation of the returns of 16 Islamic stocks chosen atrandom in a portfolio of 1,000 stocks is 2.4. Find the 95 percent confidence limits ofthe standard deviation for all Islamic stocks in the portfolio.

The 95 percent confidence limits are given by

sffiffiffin

pχ0:975

< σ <sffiffiffin

pχ0:025

Using Microsoft Excel, for degrees of freedom ν � 16 � 1 � 15, we have

χ20:975 � CHISQ:INV:RT 0:025; 15� � � 27:48839 or χ0:975 � 5:24

We also have

χ20:025 � CHISQ:INV 0:025; 15� � � 6:262138 or χ0:025 � 2:50

Then, the confidence limits are 2:4ffiffiffiffiffiffi16

p=5:24 � 1:83 and 2:4


p=2:5 � 3:84.

Thus we have with 95 percent confidence level a population standard deviation σ thatlies between 1.83 and 3.84.

Example. Find χ20:95 for (i) ν � 50 and (ii) ν= 100 degrees of freedom.For ν>30, we can use the fact that

ffiffiffiffiffiffiffiffi2χ2


p� �is approximately normally

distributed with mean 0 and standard deviation 1. That is,

z � ffiffiffiffiffiffiffiffi2χ2


p� �

294 STATISTICS

3GC13 05/15/2014 17:55:59 Page 295

Thus, the normal distribution table can be used if ν � 30. Then if χ2α and zα arethe αth percentiles of the chi-square and normal distributions, we have

χ2α � 12

zα �ffiffiffiffiffiffiffiffiffiffiffiffiffi2ν � 1

p� �2

i. If ν � 50, χ20:95 � 12

z0:95 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 50� � � 1

p� �2 � 12

1:64 � ffiffiffiffiffiffi99

p� �2 � 67:2.

ii. If ν � 100, χ20:95 � 12

z0:95 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 100� � � 1

p� �2 � 12

1:64 � ffiffiffiffiffiffiffiffi199

p� �2 � 124:0.

Example: The standard deviation of the annual returns of a sample of 200Islamic stocks listed in Kuala Lampur exchange is 8.3 percent. Find the 95 percentconfidence limits for the standard deviation of all Islamic stocks.


sffiffiffin

pχ0:975

< σ <sffiffiffin

pχ0:025

For ν � 200 � 1 � 199 degrees of freedom, we find

χ20:975 � 12

z0:975 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 199� � � 1

p� �2 � 12

1:96 � ffiffiffiffiffiffiffiffi397

p� �2 � 239

χ20:025 � 12

z0:025 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 199� � � 1

p� �2 � 12

�1:96 � ffiffiffiffiffiffiffiffi397

p� �2 � 161

We obtain χ0:975 � 15:5 and χ0:025 � 12:7. Then the 95 percent confidence limits

are 8:3 �ffiffiffiffiffiffi200

p15:5 � 7:57 and 8:3 �


p12:7 � 9:24. Thus, we can be 95 percent confident

that the population standard deviation will lie between 7.57 and 9.24.The chi-square is applied in the analysis of observed and theoretical frequencies.

Often the results obtained in samples do not agree exactly with the theoreticalresults according to the rule of probability. For example, although theoreticalconsiderations lead us to expect 50 heads and 50 tails when we toss a fair coin100 times, it is rare that these results are obtained exactly. Suppose that in aparticular sample a set of possible events, E1;E2; . . . ;Ek are observed to occur withfrequencies x1; x2; . . . ; xk, called observed frequencies, and that according to theprobability rule they are expected to occur with frequencies f 1; f 2; . . . ; f k, calledexpected or theoretical frequencies (Table 13.1).

TABLE 13.1 Observed and Theoretical Frequencies

Events E1 E2 . . . Ek

Observed frequency x1 x2 . . . xkExpected frequency f 1 f 2 . . . f k


3GC13 05/15/2014 17:56:0 Page 296

Often we wish to know whether the observed frequencies differ significantly fromthe expected frequencies. A measure of the discrepancy between observed andexpected frequencies is supplied by the statistic χ2 given by

χ2 � x1 � f 1� �2

f 1� x2 � f 2� �2

f 2� ∙ ∙ ∙ � xk � f k

� �2f k

�Xkj�1

xj � f j� �2

f j(13.51)

The number of degrees of freedom is k � 1.Example: In 200 tosses of a coin 115 heads and 85 tails are observed. Test

the hypothesis that the coin is fair, using significance level of 5 percent. The statisticχ2 is

χ2 � x1 � f 1� �2

f 1� x2 � f 2� �2

f 2� 115 � 100� �2

100� 85 � 100� �2

100� 4:5

The number of degrees of freedom is ν � k � 1 � 2 � 1 � 1. The critical valueχ20:95 for one degree of freedom is 3.84. Thus, since 4.5>3.84 we reject the hypothesisthe coin is fair at 5 percent significance level.

Tests Based on the F Distribution

As we have seen, it is important in some applications to know the samplingdistribution of the difference X1 �X2

� �of two samples. Similarly, we may need

the sampling distribution of the difference in variances s21 � s22� �

. It turns out that thisdistribution is rather complicated. Because of this, we consider instead the statistics21=s

22 because a large ratio would indicate a large difference, while a ratio nearly equal

to 1 would indicate a small difference. The sampling distribution in such case can becomputed and is called the F distribution. More precisely, suppose that we have twosamples of sizes n1 and n2, respectively, drawn from two normal (or nearly normal)populations having variances σ21 and σ22. Let us define the statistic

F � s21=σ21s22=σ22

� n1s21= n1 � 1� �σ21n2s22= n2 � 1� �σ22 (13.52)

where s21 � n1s21n1�1� � and s22 � n2s22

n2�1� �.Then the sampling distribution of F is called the F distribution with ν1 � n1 � 1

and ν2 � n2 � 1 degrees of freedom. The confidence level and the critical region aredescribed in Figure 13.6.

Example: Two samples of sizes 9 and 12 are drawn from two normally distribu-tions having variances 16 and 25, respectively. If the sample variances are 20 and 8,determine whether the first sample has a significantly larger variance than the secondsample at significance levels of (i) 0.05 and (ii) 0.01.

i. For the two samples, 1 and 2, we have n1 � 9 and n2 � 12, σ21 � 16, σ22 � 25,s21 � 20, s22 � 8. Thus,

296 STATISTICS

3GC13 05/15/2014 17:56:1 Page 297

F � s21=σ21s22=σ22

� n1s21= n1 � 1� �σ21n2s22= n2 � 1� �σ22 � 9 � 20= 9 � 1� � 16� �

12 � 8= 12 � 1� � 25� � � 4:03

The degrees of freedom for the numerator and denominator of F are ν1 �n1 � 1 � 9 � 1 � 8 and ν2 � n2 � 1 � 12 � 1 � 11. Using Microsoft Excel, wehave F0:95 � F:INV:RT 0:05; 8; 11� � � 2:94799. Since the calculated F � 4:03 isgreater than F0:95 � F:INV:RT 0:05; 8; 11� � � 2:95, we conclude that the varianceof sample 1 is significantly larger than that for sample 2 at the 0.05 significancelevel.

ii. For ν1 � 8 and ν2 � 11, we haveF0:99 � F:INV:RT 0:01; 8; 11� � � 4:744468. In this case F � 4:03 is less than

F0:99 � 4:74, we cannot conclude that the variance of sample 1 is significantlylarger than that for sample 2 at the 0.01 significance level.

Example: The mean annual returns of Islamic stocks listed in Kuala Lampur andSingapore stock exchanges are nearly normally distributed, with standard deviationsof 6.8 percent and 8.4 percent, respectively. We select 16 Islamic stocks from KualaLampur and 20 Islamic stocks from Singapore with standard deviation of 7.4 percentand 9.1 percent, respectively. Test at 5 percent significance level whether the stockvariability in Kuala Lampur is less than that of Singapore.

For the two samples, 1 and 2, we have n1 � 16 and n2 � 20, σ1 � 6:8%,σ2 � 8:4%, s1 � 7:4%, s2 � 9:1%. Thus,

F � s21=σ21

s22=σ22� n1s21= n1 � 1� �σ21n2s22= n2 � 1� �σ22 � 16 � 7:42= 16 � 1� � 6:82

� �20 � 9:12= 20 � 1� � 8:42

� � � 1:02

The degrees of freedom for the numerator and denominator of F are ν1 � n1 � 1 �16 � 1 � 15 and ν2 � n2 � 1 � 20 � 1 � 19. Using Microsoft Excel, we haveF0:95 � F:INV:RT 0:05; 15; 19� � � 2:23. Since the calculated F � 1:03 is less thanF0:95 � F:INV:RT 0:05; 15; 19� � � 2:23, we conclude that the variance of sample 1is significantly less than that for sample 2 at the 0.05 significance level.

Density

0

Confidencelevel

Critical region

FF

FIGURE 13.6 F Distribution: ConfidenceLevel and Critical Region


3GC13 05/15/2014 17:56:2 Page 298

SUMMARY

Islamic finance uses sampling theory to draw inferences from a sample’s statisticssuch as sample mean, proportion, and standard deviation to population’s statisticsand to build confidence intervals about the latter statistics. The chapter coveredsampling distributions of the mean, proportion, and standard deviation, and confi-dence-interval estimates of population parameters that includedmeans, proportions,differences in means and proportions, and standard deviations. The chapterdescribed hypothesis testing, and tests involving sample differences. The chapterdiscussed small sampling theory based on the student’s t-distribution, chi-squaredistribution, and F distribution. Sampling and hypothesis testing theory is applied ineconometric estimations and in wide areas such as medicine, communications, andfinancial management.

QUESTIONS

1. The shares that are Sharia compliant represent 70 percent of total listed shares onthe stock exchange. Find the probability that in a random sample of 145 sharesbetween 45 percent and 60 percent will be Sharia compliant.

2. It has been found that 2 percent of the Murabaha transactions of an Islamicbank are defaulted. What is the probability that in a sample of 250 transactions4 percent or more will fall in default?

3. The stock index A has a mean annual return of 7.8 percent with a standarddeviation of 10.5, while the stock index B has a mean return of 5.7 percent with astandard deviation of 16.5. If random samples of 110 returns of each stock indexare tested, what is the probability that the stock index A returns will have a meanreturn that is at least 1.1 percent more than the stock index B?

4. The Murabaha transactions of Islamic bank A have a mean maturity of 400 dayswith a standard deviation of 25 days, while those of Islamic bank B have a meanmaturity of 350 days with a standard deviation of 15. If random samples of 120Murabaha contracts of each bank are tested, what is the probability that the bankA Murabaha contracts will have a mean maturity that is at least 14 days morethan those of bank B?

5. A and B play a game of “heads or tails,” each tossing 65 coins. A will win thegame if he tosses 8 or more heads than B; otherwise, B wins. Determine the oddsagainst A winning any particular game.

6. Measurement of annual returns of a random sample of 190 Islamic stocks madeon January 10 showed a mean return of 7.2 percent and a standard deviation of10.5 percent. Using Microsoft Excel find the (i) 90 percent, (ii) 95 percent, and(iii) 99 percent confidence limits for the mean return.

7. A sample poll of 110 investors chosen at random indicated that 57 percent ofthem thought the Islamic stock index will outperform the conventional stockindex. Find the 95 percent confidence limits for the proportion of all investors infavor of the Islamic stock index.

298 STATISTICS

3GC13 05/15/2014 17:56:2 Page 299

8. A sample of 140 Islamic stocks listed in Kuala Lampur showed a mean return of8.4 percent and a standard deviation of 7.9 percent. A sample of 210 Islamicstocks listed in Luxembourg showed a mean return of 6.8 percent and a standarddeviation of 9.4 percent. Find the 95 percent confidence limits for the difference ofthe mean returns of the populations of Islamic stocks listed in Kuala Lampur andLuxembourg.

9. In a random sample of 500 Islamic stocks listed on stock exchange A 140 stocksoutperformed themarket stock index A; in a random sample of 700 Islamic stockslisted on stock exchange B 300 stocks outperformed the market index B.Construct a 95 percent confidence limits for the difference in proportions ofall Islamic stocks that outperformed the market index A and B, respectively.

10. The standard deviation of the annual returns of a random sample of 180 stockslisted on amarket exchange was computed to be 10.4 percent. Find the 95 percentconfidence limits for the standard deviation of all stocks listed on the marketexchange.

11. Measurement of annual returns of a random sample of 250 Islamic stocks madeon January 15 showed a mean return of 7.26 percent and a standard deviation of9.3 percent. We want to test the hypothesis H0: μ � 7:0 percent at 5 percentsignificance level.

12. The mean rate of return of a sample of 100 Islamic stocks is 7.7 percent per yearwith a standard deviation of 9.2 percent. If μ is the mean return of all Islamicstocks listed on the exchange, test the hypothesis of μ = 8.2 percent against thealternative of μ ≠ 8.2 percent using significance level of 5 percent.

13. A random sample of 170 Islamic stocks listed in Labuan stock exchange yielded amean annual return of 9.5 percent and a standard deviation of 14.0 percent. Arandom sample of 260 conventional stocks listed in Labuan yielded a meanannual return of 7.6 percent and a standard deviation of 16.5 percent. Is there asignificant difference between the performance of the Islamic and conventionalstocks at significance level of 5 percent?

14. A portfolio manager wants to track the market return of about 4.8 percent. Todetermine whether his portfolio is in line with the market, a sample of 10 stocks ischosen for which the mean return was 6.2 percent and standard deviation was12.1 percent. Test the hypothesis that the portfolio is in line with the market at asignificance level of 5 percent.

15. The risk-return profile of 18 Islamic stocks listed in Labuan stock exchangeshowed a mean return of 9.2 percent and a standard deviation of 12.0 percent.The risk-return profile of 20 Islamic stocks listed in Singapore showed a meanreturn of 6.8 percent and a standard deviation of 9.5 percent. Is there a significantdifference between the mean returns of the two groups of stocks at a significancelevel of 5 percent?

16. We consider a chi-square distribution with 8 degrees. Using Microsoft Excel findthe critical values of χ2 for which (i) the critical area on the right is 5 percent, and(ii) the total critical area is 5 percent.

17. An Islamic bank has 1,000 Murabaha contracts. The standard deviation of thematurity of 18 Murabaha contracts chosen at random is 2.4 weeks. Find the


3GC13 05/15/2014 17:56:2 Page 300

95 percent confidence limits of the standard deviation for all Murabaha contractsat the Islamic Bank.

18. Find χ20:975 for (i) ν � 60 and (ii) ν = 110 degrees of freedom.

19. The standard deviation of the annual returns of a sample of 300 Islamic stockslisted in Kuala Lampur exchange is 10.2 percent. Find the 95 percent confidencelimits for the standard deviation of all Islamic stocks.

20. The mean annual returns of Islamic stocks listed in Kuala Lampur and Singaporeare nearly normally distributed, with standard deviations of 7.6 percent and8.2 percent, respectively. We select 17 Islamic stocks from Kuala Lampur and 22Islamic stocks from Singapore with standard deviation of 7.1 percent and9.0 percent, respectively. Test at 5 percent significance level whether the stockvariability in Kuala Lampur is less than that of Singapore.

300 STATISTICS

3GC14 05/15/2014 17:43:45 Page 301

CHAPTER 14Regression Analysis

I slamic finance uses regression analysis. Often a relationship is found to exist betweentwo (or more variables). For example, there is a close relationship between invest-

ment in human capital and economic growth, and between education and income.Similarly, there is a relationship between price and demand or price and supply of aproduct; for instance, the high price of oranges may reduce consumption of oranges.Similarly, the high price of oranges may increase the supply of oranges. In Islamic,finance, there is a relationship between the returns of a security and the marketportfolio return, called the capital asset pricing model (CAPM). Many other relation-ships exist between financial variables.

It is frequently desirable to postulate an equation that connects the variables andthen use data to estimate the parameters of the equation.* In this chapter we exploretechniques for estimating linear regression based on empirical data. The method ofestimation is called the ordinary least squares method (OLS). It is the most usedmethod in many fields of science. We present the principle of estimation based onchoosing parameters of the model that minimize the sum of squared errors (SSE)between actual data and fitted data. We present the techniques for deriving theestimated parameters, the probability distributions and significance tests of theseparameters, and we discuss the goodness of fit of the regression. We show how aregression is used to forecast values of the endogenous variable based on assumedvalues for the exogenous variables.

CURVE FITTING

This section covers the equations of approximating curves, the regression line, and thegeneral principle of estimation, which is the method of least squares.

To determine an equation that connects variables, a first step is to collect datathat show corresponding values of the variables under consideration. For example,suppose thatX and Y are the variables under study; they may denote the quantity offertilizers and the yield per hectare in a farming region. We collect a sample of nobervations: X1; Y1� �; X2; Y2� �; . . . ; Xn; Yn� �. A next step is to plot the points

*There are many statistical packages that run regression analysis. Microsoft Excel runsregression analysis. EViews, MATLAB, TSP, RATS, PC Give, STATA, and many otherpackages are tools for regression analysis. The R statistical package is available online andmay be downloaded and used for estimating regressions.

301

3GC14 05/15/2014 17:43:46 Page 302

X1; Y1� �; X2; Y2� �; . . . ; Xn; Yn� � in a diagram. The resulting set of points (graph) iscalled a scatter diagram. From the scatter diagram it is often possible to visualize asmooth curve that approximates the data. Such a curve is called an approximationcurve. In Figure 14.1a, the data appear to be approximated by a straight line, and sowe say that a linear relationship exists between the variables. In Figure 14.1b,however, although a relationship exists between the variables, it is not a linearrelationship, and so we call it a nonlinear relationship. The general problem offinding equations of approximating curves that fit given sets of data is called curvefitting. The distance between the observation and the fitted curve is called error. Thegeneral principle in fitting a curve to the data is to minimize the sum of the squarederrors, that is, to find a curve (linear or nonlinear) that has the least sum of squarederrors.

Equations of Approximating Curves

Several common types of approximating curves and their equations are listed belowfor reference purposes. All letters other than X and Y represent parameters. Ourobjective is to estimate these parameters. The variablesX andY are often referred to asindependent and dependent variables, respectively.

Straight line: Y � a0 � a1XParabola: Y � a0 � a1X � a2X2

Hyperbola: Y � 1a0�a1X

Exponential curve: Y � abX or log Y � log a �X:log b � a0 � a1XGeometric curve: Y � aXb or log Y � log a � b:logXLogistic curve: Y � 1

abX�d or 1Y � abX � d

To decide which curve should be used, it is helpful to obtain scatter diagrams oftransformed variables. For example, if a scatter diagram of log Y versus X shows alinear relationship, the equation has the exponential form Y � abX, while if log Yversus logX shows a linear relationship, the equation has the geometricform Y � aXb.

a. Linear relationship b. Nonlinear relationship

0

..

..

.

.

.

..

..

.

.

Error εi

Linear relationshipY

X 0

....

...

.

.. ....

.

..

..Error εi

Nonlinear relationshipY

X

FIGURE 14.1 Curve Fitting

302 STATISTICS

3GC14 05/15/2014 17:43:46 Page 303

The Linear Regression Line

The simplest type of approximating curve is a straight line, whose equation is

Y � a0 � a1X � ε (14.1)

The variable Y is the dependent variable,X is the independent variable, and ε is arandom error that has a mean and a variance.* The constant a0 is called the interceptof the line; it is the value of Y � a0 when X � 0. The coefficient a1 is the slope ofthe line; it is equal to the change in Y, ΔY, divided by the change in X, ΔX, that is,a1 � ΔY

ΔX.Nonetheless, this approximation may have to include more variables. For

instance, crops per hectare do not only increase with fertilizers but also with rainfalland machinery. Imagine we have crops related only to fertilizers. We may increasefertilizers but crops may decline sharply against the prediction of the model; we realizethere may be a severe drought that damages the crop, or cold weather that damages acoffee crop. In this case, the model has to include variables that act on the dependentvariable Y. The omission of these variables causes a mis-specification of the model;once relevant variables are included, the model becomes as follows

Y � a0 � a1X � a2Z � a3W � ε (14.2)

In this case each regression coefficient measures only the partial effect of therespective variable, assuming that the other variables are fixed. We have a1 � @Y

@X,a2 � @Y

@Z, and a3 � @Y@W.

We should stress that a model is by definition a simplification of the reality andshould not include many variables. A model with too many variables displaysoverfitting and is not adequate for use. For instance, supply of wheat may bedetermined by many factors including research, acreage, fertility of soil, farmingsystem, taxation, transport, storage, roads, and producers’ associations. Includingmany variables in a regression makes the regression useless. We limit our attentiononly to variables in the effects of which we are interested.

Example: The Fama-French multifactor model explains security return as

R � a � b1 Rm � Rf� � � b2SMB � b3HML (14.3)

Here R is the portfolio’s expected rate of return, Rf is the risk-free return rate,and Rm is the return of the whole stock market, SMB stands for the difference ofreturns between small minus big market capitalization, and HML for the differenceof returns between high minus low book-to-market ratio; SMB and HML measurethe historic excess returns of small caps over big caps and of value stocks overgrowth stocks.

*The probability distribution of ε is not known. However, there is no objection to assuming aparticular distribution for ε:

Regression Analysis 303

3GC14 05/15/2014 17:43:47 Page 304

The Principle of Estimation: Method of Least Squares

The principle of estimation is the method of least squares. Let us assume that therelationship to be fitted is of the form

y � f x� � (14.4)

We collect a sample of observation (data): x1; y1� �

; x2; y2� �

; . . . ; xn; yn� �

. Let usassume that the fitted relationship is

y � f x� � (14.5)

By substituting the actual data of the independent variable x1; x2; ::; xn� � into f x� �we obtain fitted data for the dependent variable y1; y2; . . . ; yn

� �. The differences

ε1; ε2; . . . εn defined as

y1 � y1 � ε1� �

; y2 � y2 � ε2� �

; . . . ; yn � yn � εn� ��

(14.6)

are called estimated error terms or residuals. In Figure 14.1, the error termsε1; ε2; . . . ; εnf g are measured by the vertical distance between each data point andthe fitted curve. The error terms can be positive, negative, or even zero. A measureof the goodness of fit of the relation y � f x� � to the given data is provided by the sum ofthe squared errors SSE,

SSE � ε21 � ε22 � ∙ ∙ ∙ � ε2n �Xni�1

ε2i ; i � 1; 2; . . . ; n (14.7)

Obviously, we want to minimize the sum SSE. We choose the parameters of y �f x� � for which SSE is at its minimum value. The so-obtained curve is called the best-fitting curve. A curve having this property is said to fit the data in the least-squaressense and is called a least-square curve.

LINEAR REGRESSION ANALYSIS

This section covers the formulation of the regression model, the estimation of theregression model, the method of moments, the method of the maximum likelihood,the estimate of the variance of the error term, and the coefficient of determination andthe coefficient of correlation of a regression.

Formulation of the Regression Model

Regression analysis proceeds with a single regression variable. The extension tomultivariable regression is straightforward once results of the simple model areestablished. We have a data set of size n formed by the observations

x1; y1� �

; x2; y2� �

; . . . ; xn; yn� �

304 STATISTICS

3GC14 05/15/2014 17:43:47 Page 305

Our objective is to fit this data by a least-square line of the form

y � a � bx (14.8)

Our linear regression model can be formulated as

y � a � bx � ε (14.9)

Or equivalently, for each data observation

yi � a � bxi � εi (14.10)

The term ε � ε1; ε2; . . . ; εnf g is called a random error vector. We assume thatε1; ε2; . . . ; εnf g are independent, and identically distributed iid� � error terms with thefollowing four properties:

i. E εi� � � 0*;ii. Var εi� � � σ2;iii. Cov εi; εj

� � � 0 for all i ≠ j; andiv. E εx� � �Xn

i�1εixi � 0.

The last property means that the random variable ε is uncorrelated with x. Theregression model is described in Figure 14.2. The fitted least-square equation is shownby the line

y � a � bx (14.11)

The value y is called the fitted value of y. The difference y � y � ε is called theestimated error or residual. The actual data y can be decomposed therefore into twocomponents as

y � y � ε � a � bx � ε (14.12)

Hence actual data y is equal to the sum of the fitted data y plus an estimated errorterm ε. We observe that y is a linear relation of x; it belongs to the same plane as x. Wesay that y is a projection of y into the axis (or plane) of x. The geometric representationof this projection is shown in Figure 14.2 by a rectangular triangle. The vector y is thesum of two perpendicular vectors y and ε that we immediately obtain by thePythagorean theorem,

jjyjj2 � jjyjj2 � jjεjj2 (14.13)

*Recall the definition of expected value. If the density function of εi is f εi� � thenE εi� � �Pni�1

εif εi� �.


3GC14 05/15/2014 17:43:48 Page 306

Estimation of the Regression Model

The estimation of the regression model consists of estimating the regression coef-ficients a and b and consequently other estimates that depend on these coefficientssuch as ε. There are three equivalent methods for estimating the regression coefficients.These are:

1. The ordinary least-squares method (OLS)2. The normal equations methods (the methods of moments)3. The maximum likelihood method

We explore the method of least-squares, then we show its equivalence to themethod of moments as well as the maximum likelihood method. In the OLS, we chosethe parameters a and b that minimize the sum of squared errors (SSE) denoted byL a; b� �,

mina; b

L a; b� � � SSE �Xni�1

ε2i �Xni�1

yi � a � bxi� �2 (14.14)

A necessary condition for L to attain a minimum in respect to a and b is

@L@a

� 0 and@L@b

� 0 (14.15)

The above derivatives can be written as

@L@a

� 2Xni�1

yi � a � bxi� � � 0 (14.16)

@L@b

� 2Xni�1

xi yi � a � bxi� � � 0 (14.17)

.. .

.

.. .. .

.. . . . .

..

..

0

Projection of y onto the axis x

y

y =

yi

yi

xix

y

a + bx

y =

a = intercept

a + bx

ε

εi

FIGURE 14.2 Linear Regression Model

306 STATISTICS

3GC14 05/15/2014 17:43:48 Page 307

These two equations can be rearranged to yield

Xni�1

yi � n:a � bXni�1

xi (14.18)

Xni�1

xiyi � aXni�1

xi � bXni�1

x2i (14.19)

The estimated coefficients of a and b are denoted by a and b; they are derived fromEquations (14.18) and (14.19) and are expressed as

a �Pn

i�1 yi� � Pn

i�1 x2i� � � Pn

i�1 xi� � Pn

i�1 xiyi� �

nPn

i�1 x2i � Pni�1 xi

� �2 (14.20)

b � nPn

i�1 xiyi � Pni�1 xi

� � Pni�1 yi

� �nPn

i�1 x2i � Pni�1 xi

� �2 (14.21)

Let us define the sample means as

y �Pn

i�1 yin

and x �Pn

i�1 xin

(14.22)

We define deviations from the mean as

~yi � yi � y and ~xi � xi � x (14.23)

We can express a as

a � y � bx (14.24)

Furthermore we have

~y � b~x � ε (14.25)

By minimizing the sum of squared residuals,

SSE � ~L b� � �Xni�1

~yi � b~xi� �2 (14.26)

we obtain

d~Ldb

�Xni�1

~xi ~yi � b~xi� � � 0 (14.27)

b �Pn

i�1 ~xi~yiPni�1 ~x2i

(14.28)


3GC14 05/15/2014 17:43:53 Page 308

Example: We illustrate the application of the OLS method with the data inTable 14.1. We obtain

b �Pn


� 8:2013:1465

� 2:6064

a � y � bx � 18:62 � 2:6064 � 5:05 � 5:4577

The regression relation may be written as

y � a � bx � 5:4577 � 2:6064x

We note that we can perform the above computations using Microsoft Excel,EViews, or any regression software. Using Excel, we obtain the coefficient a and b as

a � INTERCEPT B3 : K3; B2 : K2� � and b � SLOPE B3 : K3; B2 : K2� �where the first range of data refers to the dependent variable y and the second range tothe independent variable x.

The Method of Moments

The method of moments uses simply conditions on the moments of the random errorterms εi.

The first moment condition isPn

i�1 εi � 0, which can be written as

Xni�1

yi � a � bxi� � � 0 (14.29)

The secondmoment condition is the absence of correlation between the error termand the variable x, that is, E εX� � � 0, which can be written as

Xni�1

xi yi � a � bxi� � � 0 (14.30)

TABLE 14.1 Illustrating Computations of a Simple Regression

xi 4.5 3 3.2 5.7 7.1 5.3 8 6.2 5.4 2.1 x � 5:05yi 16 13.2 14.2 22 23 19 26 23 19 10.8 y � 18:62~xi � xi � x �0.55 �2.05 �1.85 0.65 2.05 0.25 2.95 1.15 0.35 �2.95~yi � yi � y �2.62 �5.42 �4.42 3.38 4.38 0.38 7.38 4.38 0.38 �7.82~xi~yi 1.441 11.111 8.177 2.197 8.979 0.095 21.771 5.037 0.133 23.069

Pni�1 ~xi~yi� 8:201

~x2i 0.3025 4.2025 3.4225 0.4225 4.2025 0.0625 8.7025 1.3225 0.1225 8.7025Pn

i�1 ~x2i� 3:1465

308 STATISTICS

3GC14 05/15/2014 17:43:54 Page 309

These two equations are called the normal equations and are identical to theequation under the least-square method.

The Method of the Maximum Likelihood

It rests on two assumptions: (i) the random error terms are independent, and (ii) theyhave an assumed probability distribution f εi� �. More specifically, if we make theassumption that the error terms εi are normally distributed with mean 0 and standarddeviation σ, the density function for εi can be written as

f εi� � � 1


p e�12

ε2iσ2 � 1


p e�12

yi�a�bxi� �2σ2 (14.31)

Each of these density functions refer to one observation. Since the εi areindependent, the product of these density functions over the entire sample givesthe likelihood of obtaining the joint occurrence of the residuals ε1; ε2; . . . ; εnf g. Thisproduct, called the likelihood function ℓ, is given as

ℓ � ∏n

i�1f εi� � � 1


p !n

e� 12σ2

Pni�1

yi�a�bxi� �2(14.32)

The likelihood function, as its name implies, measures the likelihood of obtainingthe sample of points actually obtained, namely the data on yi and xi given the values ofa, b, and σ. The maximum likelihood method calls for choosing values of parametersthat maximize the likelihood function, that is, parameters maximizing the likelihoodof observing the sample that was in fact observed. The maximum-likelihood estima-tors are thus obtained in this case as the solutions to the problem

maxa; b;σ

ℓ � ℓ a; b; σ� � � ∏n

i�1f εi� � � 1


p !n

e� 12σ2

Pni�1

yi�a�bxi� �2(14.33)

It is convenient to work with the logarithm of the likelihood function ℓ, sincemaximizing lnℓ is equivalent tomaximizing ℓ, the logarithm being a strictly increasingfunction. Here,

lnℓ � �n2ln2π � nlnσ � 1

2σ2Xni�1

yi � a � bxi� �2 (14.34)

Since � 12σ2 is always negative, maximizing lnℓ by choice of a and b requires

minimizing

Xni�1

yi � a � bxi� �2

(14.35)

which is exactly the same problem as that of minimizing the sum of squared residuals.Thus, under the assumptions of independent, identically, and normally distributed


3GC14 05/15/2014 17:43:54 Page 310

random error terms the maximum likelihood estimators of a and b are identical tothose of the least squares.

Estimate of the Variance of the Error Term

The regression analysis does not end with the computation of the parameters. Wewant to assess the goodness of fit and run hypotheses testing on the regressioncoefficients. We assume that the error terms are independent and identically distrib-uted random variables with mean 0 and standard deviation σ. The estimation of theregression coefficients enables us to obtain an estimate for σ. The estimated regressioncan be written now as

y � a � bx (14.36)

For a given observation xi the corresponding fitted value of the dependentvariable is

yi � a � bxi (14.37)

The difference between the actual observation yi and the fitted value yi is called theestimated error or residual and is denoted by εi. We therefore express yi as

yi � yi � εi � a � bxi � εi (14.38)

The estimated error

εi � yi � yi � yi � a � bxi (14.39)

based on the sample ε1; ε2; . . . ; εnf g gives an estimate of σ

σ2 �Xni�1

ε2in�Xn

i�1

yi � a � bxi� �2

n� mean of squared errors (14.40)

The estimate σ2 is called themean of squared errors (MSE). The estimate σ2 is notunbiased. An unbiased estimator is provided by

s2 �Xni�1

ε2in � k � 1

�Xni�1

yi � a � bxi� �2

n � k � 1(14.41)

where k is the number of independent variables in the regression. In the simpleregression with one dependent variable the coefficient k is k � 1.

We note also that s2 can be written in terms of ~y and ~x as

s2 �Xni�1

ε2in � k � 1

�Xni�1

~yi � b~xi� �2n � k � 1

(14.42)

310 STATISTICS

3GC14 05/15/2014 17:43:55 Page 311

Example: Based on the above estimated regression,

y � a � bx � 5:4577 � 2:6064x

we estimate

s2 �Xni�1

ε2in � k � 1

�Xni�1

~yi � b~xi� �2n � k � 1

� 7:72611610 � 2

� 0:9658

The Coefficient of Determination and the Coefficientof Correlation

An important statistic for assessing the goodness of a regression is the R2, called thecoefficient of determination. The simple regression equation can be rewritten as

y � y � ε � a � bx � ε (14.43)

We have decomposed the actual observation yi into two components called theexplained part of yi measured by yi � a � bxi and the unexplained part of yi measuredby εi. By subtracting y on both sides we obtain

y � y � y � y � ε � a � bx � y � ε (14.44)

The deviation from the mean y � y is decomposed into an explained component,y � y, and unexplained component, ε. By the Pythagorean theorem we have

jjy � yjj2 � jjy � yjj2 � jjεjj2 (14.45)

Or equivalently,

Xni�1

yi � y� �2 �Xn

i�1yi � y� �2 �Xn

i�1ε2i (14.46)

We define the coefficient of determination for a regression R2 as the proportion ofthe total variance of y that is explained by the regression

R2 � jjy � yjj2jjy � yjj2 �

Pni�1 �yi � y�2Pni�1 �yi � y�2 � 1 �

Pni�1 ε2iPn

i�1 �yi � y�2 � 1 �Pn

i�1 ε2iPni�1 ~y2i

(14.47)

We note that 0 � R2 � 1.The higher the R2 the lower the sum of squared errors relative to the total sum of

squaresPn

i�1 yi � y� �2. R2 is a measure of the explanatory power of the regression; a

measure of how well the model, as estimated, fits the available data. If for example,


3GC14 05/15/2014 17:43:55 Page 312

R2 � 0:9, then 90 percent of the variance of the dependent variable is explained by theregression, with 10 percent left unexplained.

The coefficient of correlation is defined as

R � �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPn

i�1 yi � y� �2

Pni�1 yi � y�2�

vuut (14.48)

We may verify that R can be written as

R � Cov y; y� �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar y� �:Var y� �p �

Pni�1 ~yi~xiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiPn

i�1 ~x2i� � Pn

i�1 ~y2i� �q (14.49)

Note that R is independent of the choice of units of x and y.Example: The coefficient of determination associated with the regression

y � 5:4577 � 2:6064x

is computed as

R2 � 1 �Pn

i�1 ε2iPni�1 ~y2i

� 0:965

We define the following moments:

sxy �Pn

i ~xi~yin

; sx �ffiffiffiffiffiffiffiffiffiffiffiffiffiPn

i ~x2i

n

s; sy �

ffiffiffiffiffiffiffiffiffiffiffiffiffiPni ~y

2i

n

s(14.50)

Given y � a � bx, the regression line of y on x, can be written as

y � y � Rsysx

x � x� � (14.51)

The estimate of b is b � R sysx� Cov y; x� �

Var x� � .Example: Based on data in Table 14.1, we have sx �

ffiffiffiffiffiffiffiffiffiffiffiPn

i~x2i

n

r� 1:7738, sy �ffiffiffiffiffiffiffiffiffiffiffiPn

i~y2i

n

r� 4:7061, R � ffiffiffiffiffiffiffiffiffiffiffiffi

0:965p � 0:982. We compute b � R

sysx

� 0:982 � 4:70611:7738

� 2:60639.

THE PROBABILITY DISTRIBUTION OF THE ESTIMATEDREGRESSION COEFFICIENTS â AND b

The estimation of regression coefficients yields

a � y � bx (14.52)

312 STATISTICS

3GC14 05/15/2014 17:43:56 Page 313

b �Pn


(14.53)

Each estimated coefficient is a random variable with a statistical distribution thathas a mean and a variance. Since each coefficient is a linear combination of yi andtherefore of εi, its distribution is the same as that of εi. In particular, if the random termεi has a normal distribution N 0; σ2

� �, the coefficients a and b will in turn each have

a normal distribution. We establish therefore the mean and variance of thesedistributions.

The Distribution of bThe expected value of E b

� �is given by

E b� �

�Pn

i�1 ~xiE�~yi�Pni�1 ~x2i

�Pn

i�1 ~xiE b~xi � εi� �Pni�1 ~x2i

(14.54)

We note that E ~yi� � � E b~xi � εi� � � b~xi � E εi� �.

Because E εi� � � 0, E ~yi� � � b~xi; we obtain

E b� �

� bPn

i�1 ~x2iPni�1 ~x2i

� b (14.55)

The variance of b can be computed as

Var b� �

� VarPn


!� 1Pn

i�1 ~x2i

!2

VarXni�1

~xi b~xi � εi� � !

(14.56)

Since the term b~x2i is fixed and Var εi� � � σ2, we obtain

VarXni�1

~xi b~xi � εi� � !

� VarXni�1

b~x2i � ~xiεi� � !

� VarXni�1

~xiεi� � !

� σ2Xni�1

~x2i

(14.57)

By replacing in the formula for Var b� �

, the latter turns out

Var b� �

� σ2Pni�1 ~x2i

(14.58)

Assuming εi are normally distributed, we may state that b is normally distribu-tion as

b ∼N b;σ2Pni�1 ~x2i

!(14.59)


3GC14 05/15/2014 17:43:56 Page 314

The Distribution of â

From the relation

a � y � bx (14.60)

we may compute the expected mean E a� � asE a� � � E y � bx

� �� y � xE b

� �� y � bx � a (14.61)

We may also compute the variance,

Var a� � � Var y � bx� �

� Var y� � � Var bx� �

� 2Cov y; bx� �

(14.62)

We note that

y �Pn

i�1 yin

�Pn

i�1 a � bxi � εi� �n

Since a and b are constant, xi is fixed, and E xiεi� � � 0 and εi are independent, itfollows that

Var y� � � VarPn

i�1 εi� �n2

�Pn

i�1 Var εi� �n2

� nσ2

n2� σ2

n(14.63)

Var bx� �

� x2Var b� �

� x2σ2Pni�1 ~x2i

(14.64)

Cov�y; bx� � xCov�y; b� � xCov y;

Pni�1 ~xi~yiPni�1 ~x2i

!� x

Pni�1 ~xiCov�y; ~yi; �Pn

i�1 ~x2i� x

Pni�1 ~xiPni�1 ~x2i

σ2

n

(14.65)

Note thatPn

i�1 ~xi � 0; hence, Cov y; bx� �

� 0.

Var�a� � σ2

n� x2σ2Pn

i�1 ~x2i� σ2

Pni�1 ~x2i � nx2σ2

nPn

i�1 ~x2i� σ2

Pni�1 x2i

nPn

i�1 ~x2i(14.66)

Assuming εi are normally distributed we may state that a follows a normaldistribution

a∼N a;σ2Pn

i�1 x2inPn

i�1 ~x2i

!(14.67)

We may note that the covariance of a and b is

Cov a; b� �

� � σ2xPni�1 ~x2i

(14.68)

314 STATISTICS

3GC14 05/15/2014 17:43:57 Page 315

We observe that the variance of each coefficient involves the parameter σ2. Thelatter is not known and has to be estimated as indicated earlier. An unbiasedestimator for σ2 is

s2 �Xni�1

ε2in � k � 1

(14.69)

HYPOTHESIS TESTING OF â AND b

We would like to ascertain the significance of each coefficient a and b based on theknowledge of the distribution of these two coefficients. One way to test forthe statistical significance of each coefficient is to formulate the null hypothesisthat the coefficient is zero and the alternative hypothesis that it is differentfrom zero.

Test of Significance of the Regression Intercept â

The distribution of a is a ∼N a;σ2Pn

i�1 x2i

nPn

i�1 ~x2i

; the z statistic associated with this

distribution is

z � a � a� �=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2Xni�1

x2i =nXni�1

~x2i

s(14.70)

Since σ2 is unknown and must be estimated by s2 �Pni�1

ε2in � k � 1, the distribu-

tion of a changes into a student’s t-distribution with n � k � 1� � degrees of freedom.Since in our simple regression k � 1, we have the following student’s t-statistic fortesting hypotheses regarding a

t � a � affiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2Pn

i�1 x2i =nPn

i�1 ~x2iq � a � a

sa(14.71)

where sa �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2Pn

i�1 x2i =nPn

i�1 ~x2iq

. We may test the significance of the intercept

coefficient at a confidence level α as:Null hypothesis is H0: a � 0.The alternative hypothesis is H1: a ≠ 0.We replace a by its value in the above statistic, we obtain

t � a � 0sa

� asa

(14.72)

The critical value for tc � t α2 ; n � 2� �

is read from the student’s t-distributiontable or using Microsoft Excel. The confidence interval CI� � is


3GC14 05/15/2014 17:43:57 Page 316

CI � a � tcsa � 0 � tcsa (14.73)

If a falls in the confidence interval, or equivalently t � tc, we accept the nullhypothesis with a confidence level α and decide that the intercept is statisticallyinsignificant at this confidence level. Otherwise, we do not reject the hypothesis thatH1: a ≠ 0 and decide that the intercept is statistically significant at α confidencelevel.

Test of Significant of the Regression Slope bThe distribution of the slope b is b∼N b; σ2Pn

i�1 ~x2i

, the z-statistic associated with this

distribution is

z � b � b� �

=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2=

Xn

i�1 ~x2i

q(14.74)

Since σ2 is unknown and must be estimated by s2 �Pni�1

ε2in�k�1, the distribution of

a changes into student’s t-distribution with n � k� � degrees of freedom. Since in oursimple regression k � 1, we have the following student’s t-statistic for testing hypoth-eses regarding a,

t � b � b� �

=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2=

Xn

i�1 ~x2i

q� b � b� �

sb(14.75)

where sb � σ2=Pn

i�1 ~x2i . We may test the significance of the slope coefficient at aconfidence level α as:

Null hypothesis is H0: b � 0.The alternative hypothesis is H1: b ≠ 0.We replace a by its value in the above statistic, we obtain

t � b � 0sb

� bsb

(14.76)

The critical value for tc � t α2 ; n � 2� �

is read from the student’s t-distributiontable or Microsoft Excel. The confidence interval CI� � is

CI � a � tcsb � 0 � tcsb (14.77)

If b falls in the confidence interval, or equivalently t � tc, we accept the nullhypothesis with a confidence level α and decide that the slope is statisticallyinsignificant at this confidence level. Otherwise, we do not reject the hypothesisthat H1: b ≠ 0 and decide that the slope is statistically significant at α confidencelevel.

316 STATISTICS

3GC14 05/15/2014 17:44:0 Page 317

Test of the Simultaneous Significance of the RegressionCoefficients

In general, an analysis of variance (ANOVA) (Table 14.2) describes the differentsources of variation of the dependent variable of the regression Y. The “totalvariation” of the variable is the sum of squared deviations about the mean

SST �Xni�1

yi � y� �2 (14.78)

TheANOVA table splits this total into twoparts: the part explained by the regressionequation named as the sum of squares due to the regression (SSR) and the partunexplained by the regression named as the sum of squares due to the errors (SSE)

Xni�1

yi � yi� �2 �Xn

i�1yi � y� �2 �Xn

i�1yi � yi� �2 � SSR � SSE (14.79)

We are interested in comparing SSR and SSE using the ratio SSR=SSE. If theexplained part is large and the unexplained part is small, this ratio tends to be high. Ifthe explained part is small and the unexplained part is large, this ratio tends to besmall. It may be noted that SSR=σ2∼χ2k and SSE=σ2∼χ2n�k�1 where k is the number ofexogenous variables. The ratio SSR=SSE is the ratio of two chi-square distribution. Itis therefore a random variable distributed according to an F distribution withk; n � k � 1� � degrees of freedom.

F k; n � k � 1� � � SSR=kSSE= n � k � 1� � �

R2=k

1 � R2� �= n � k � 1� � (14.80)

The F statistic can be used to test the significance of the R2. The F statistic withk; n � k � 1� � degrees of freedom allows us to test the hypothesis that none of theexplanatory variables helps explain the variation of Y about its mean. In other words,the F statistic tests the hypothesis that

H0 : b � 0H1 : b ≠ 0

If the null hypothesis is true, then we would expect SSR, R2, and therefore F to beclose to zero. Thus, a high value of F statistic is a rationale for rejecting the null

TABLE 14.2 Analysis of Variance (ANOVA) for a Simple Regression

Source of Variation Degrees of Freedom df Sum of Squares SS F � Ratio

Regression k SSR �Pni�1 yi � y� �2 F k; n � k � 1� � � SSR=k

SSE= n�k�1� �Error n � k � 1 SSE �Pn

i�1 yi � yi� �2

Total n � 1 Xni�1

yi � yi� �2


3GC14 05/15/2014 17:44:2 Page 318

hypothesis. An F not significantly different from 0 lets us conclude that the explanatoryvariables do little to explain the variation ofY about itsmean. In the one variablemodel,for example, the F statistic tests whether the regression line is horizontal. In such a case,R2 � 0 and the regression explains none of the variation in the dependent variable.Notethat we do not test whether the regression line passes through the origin a � 0� �; ourobjective is simply to see whether we can explain any variation around the mean Y.

Consider a multiple regression model

y � a � b1x1 � b2x2 � ∙ ∙ ∙ � bkxk � ε (14.81)

where x1; x2; . . . ; xk are exogenous variables uncorrelated with ε. The F statistic isapplied to test the joint hypothesis that

H0 : b1 � b2 � ∙ ∙ ∙ � bk � 0

H1: at least one coefficient is not 0.Instead of testing each coefficient significance, we decide whether all regression

coefficients are all significant. This is equivalent to testing the significance of theregression itself. The null hypothesis is that the equation has absolutely no powermeans that the same value of y will be predicted regardless of the values of x.

Example: Capital asset pricing model (CAPM).An Islamic stock has an annual mean return y; the market portfolio has a mean

return x (Table 14.3).Using Microsoft Excel, data analysis, regression, we obtain the following

output:

Regression Statistics

Multiple R 0.977502R Square 0.955511Adjusted R Square 0.952088Standard Error 0.744595Observations 15ANOVA

df SS MS F Significance F

Regression 1 154.7973 154.7973 279.205 3.62E–10Residual 13 7.207482 0.554422Total 14 162.0048

CoefficientsStandardError t-stat P-value

Lower95%

Upper95%

Lower95.0%

Upper95.0%

Intercept 1.075261 0.503132 2.1371 0.0521 �0.01169 2.1622 �0.01169 2.1622X Variable 1 1.478886 0.088506 16.709 3.62E-10 1.28768 1.6700 1.28768 1.6700

318 STATISTICS

3GC14 05/15/2014 17:44:4 Page 319

We can write the estimated CAPM as

y � 1:075 � 1:479x

�ta � 2:137��tb � 16:709�;R2 � 0:955

We test the significance of the coefficient a and b based on the t-distribution.Thecritical value of the statistic t at 5 percent significant level for a two-tailed test isprovided by Microsoft Excel:

tc � tα2; n � 2

� �� T:INV:2T 0:05; 13� � � 2:16

The t-stat for the coefficient a is 2.137; it is less than 2.16; we cannot reject thehypothesis H0 : a � 0 at 5 percent significance level. The t-stat for the coefficient b is16.709; it is greater than 2.16; we reject the hypothesis H0 : b � 0 at 5 percentsignificance level. The F statistic is computed as

MSRMSE

� 154:7970:554

� 279:20

Using Microsoft Excel, we compute the critical F at 5 percent significance level as

0:05; 1; 13� � � F:INV:RT 0:05; 1; 13� � � 4:667

The computed F � 279:2 exceeds the critical F. The overall regression is thereforehighly significant in explaining the returns of the Islamic stock.

DIAGNOSTIC TEST OF THE REGRESSION RESULTS

We have to perform a diagnostic check on each regression we run. Besides thesignificance tests of the regression coefficients, we have to assess the goodness of fit andtest for the presence of serial correlation in the error terms. This section covers thediagnosis of the estimated residuals that include the standard error of the regression(SER), the R-squared, the serial autocorrelation test (Durbin-Watson statistic), andthe normality test of the errors.

Standard Error of Regression (SER) and the R-Squared

If we know the parameters, a; b, of the model then our estimation errors would bethe εi values with variance σ2. We would like to estimate σ2 because it tells us

TABLE 14.3 Estimation of the Capital Asset Pricing Model (CAPM)

x 4 4.2 7 8 5.5 3.5 2.1 0.5 6.6 7.2 8.3 4.8 5.8 3.9 7.4y 5.70 7.05 11.94 13.56 8.98 6.06 4.86 1.69 9.45 11.91 12.88 9.24 9.24 7.49 12.60


3GC14 05/15/2014 17:44:5 Page 320

whether our estimation errors are likely to be large or small. The estimated residuals εivalues are effectively estimates of the unobserved population disturbances, the εi values.Thus the sample variance of the εi values, which we denote s2, is an estimator of σ2:

s2 �Xni�1

ε2i = n � 2� � (14.82)

s2 is an estimate of the dispersion of the regression disturbance and hence is usedto assess goodness of fit of the model as well as the magnitude of the estimation errorsthat are likely to be made. The standard error of the regression is

SER � s �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiXni�1

ε2i = n � 2� �s

(14.83)

R2 is the percent of the variance of y explained by the variables included in theregression. A high R2 indicates a good fit. A low R2 indicates a poor fit.

Serial Correlation: Durbin-Watson Statistic

It is important to examine whether there are patterns in the estimation errors, becauseerrors from a good estimation model should be uncorrelated, Cov εi; εj

� � � 0 for alli ≠ j. The Durbin-Watson statistic tests for correlation in errors, called serial correla-tion in regressing disturbances. If the errors made by an estimation model are seriallycorrelated, then the OLS estimates are inefficient. If the error serial correlation ispositive the estimated residual variance σ2 may be underestimated and the R2 may beoverestimated. In the general case, the variances of the OLS estimates for regressioncoefficients will be biased. The estimated standard errors of the coefficients will besmaller than the true standard errors, and hence the estimated standard errors will beunderestimated. Therefore, the t-statistics will be overestimates; a regressioncoefficient that appears to be significant may not really be so. In sum, the effect ofserial correlation in errors is the estimated variances of the parameters will be biasedand inconsistent. Thus the t-test and F-test are no longer valid.

The Durbin-Watson test applies to the regression model as

yi � a � bxi � εi (14.84)

εi � θεi�1 � vi (14.85)

vi ∼ iid 0; σ2ν� �

(14.86)

The regression disturbance is serially correlated when θ ≠ 0. The hypothesisunder consideration is θ � 0.When θ � 0, the model is adequate for economic analysisand forecast, but when θ ≠ 0 the disturbance is serially correlated; we say that theεi follows an autoregressive process of order one, or AR 1� �. The formula for theDurbin-Watson test DW� � is

DW �Xnt�2

εt � εt�1� �2=Xnt�1

ε2t (14.87)

320 STATISTICS

3GC14 05/15/2014 17:44:5 Page 321

The statistic DW takes values in the interval 0;4� �. If θ � 0, DW should bearound 2.

Example: The regression above yields SER � 0:744595, R2 � 0:955, andDW � 1:952. The DW statistic is close to 2 indicating absence of serial correlationin the errors. The model is therefore adequate for economic analysis.

Normality Test

The diagnostic checkmay be extended to assess the normality of the error variable andascertain the shape of its distribution based on skewness and kutosis. Skewnessmeasures the amount of asymmetry in the distribution of errors. If a distribution issymmetric, skewness equals zero; the larger the absolute size of the skewness statistic,the more asymmetric is the distribution. A large positive value indicates a long righttail, and a large negative value indicates a long left tail. We estimate skewness of theestimated error terms as

S � 1n � k

Xnt�1

εi� �3=s3 (14.88)

The kurtosis of a random variable is a measure of the thickness of the tails of itsdistribution relative to those of a normal variable. A normal distribution has a kurtosisof 3; a kurtosis above 3 indicates “fat tails” or leptokurtosis. That is, the distributionhas more probability mass in the tails than the normal distribution. We estimatekurtosis of the estimated error terms as

K � 1n � k

Xnt�1

εi� �4=s4 (14.89)

The Jarque-Bera test statistic JB� � effectively aggregates the information in thedata about both skewness and kurtosis to produce an overall test for normality. Thestatistic is

JB � n � k6

S2 � 1

4K � 3� �2

(14.90)

Where n is the number of observations and k is the number of parametersestimated. Under the null hypothesis of independent normally distributed observa-tions, the Jarque-Bera statistic is distributed as a chi-square random variable with2 degrees of freedom in large sample. If JB < χ2c , where χ2c is the critical value of thechi-square random variable, the errors are normally distributed; otherwise, we rejectthe normality assumption.

PREDICTION

The purpose of prediction is to estimate a value of the dependent variable for a certainvalue of the explanatory variable. The least-squares predictor y is obtained by setting


3GC14 05/15/2014 17:44:6 Page 322

all coefficients equal to their least squares estimators, the explanatory variable at isgiven level x0. We have

y � a � bx0 (14.91)

Such a predictor is the unique best linear unbiased predictor of the dependentvariable y, given that x � x0. Specifically, among the class of linear and unbiasedpredictors, the least-squares predictor has minimum variance. This variance isgiven by

Var y� � � Var a � bx0� �

� σ21n� x0 � x� �2Pn

i�1 ~x2i

!(14.92)

From this result the variance of the predicted value increaseswith the error varianceσ2, increaseswith thevalues takenby the explanatoryvariablex0, anddecreaseswith the

spread of the data on the explanatory variablesPni�1

~x2i . Assuming normally distributed

error terms εi implies that the least-squares y is distributed normally as

y∼N a � bx0; σ21n� x0 � x� �2Pn

i�1 ~x2i

! !(14.93)

In terms of the standardized normal,

y � a � bx0ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiσ2

1n� x0 � x� �2Pn

i�1 ~x2i

!vuut∼N 0; 1� � (14.94)

Replacing σ2 by its unbiased estimator s2 �Pn

i�1 yi�yi� �2n�1 , the ratio

y � a � bx0

s

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n

� x0 � x� �2=Xn

i�1 ~x2i

r ∼t n � 1� � (14.95)

is distributed as a t distribution with n � 1 degrees of freedom. From the probabilitystatement,

P �tα=2 < t < tα=2� � � 1 � α (14.96)

where tα=2 is the value of the t distribution for a level of significance α=2 and n � 1degrees of freedom, it follows that

P �tα=2 < y � a � bx0� �=sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1=n � x0 � x� �2=Xn

i�1~x2i

s< tα=2

!� 1 � α (14.97)

322 STATISTICS

3GC14 05/15/2014 17:44:6 Page 323

Thus, a 1 � α� � confidence interval for the prediction y is given:

a � bx0 � s � tα=2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n� x0 � x� �2Pn

i�1 ~x2i

!vuut (14.98)

Example: We use the CAPM model in Table 14.3 to forecast the Islamic stockreturn assuming the market return will be at 5.4 percent. We solve our model forx0 � 5:4 percent; we find y � 1:479 � 5:4 � 1:075 � 9:061 percent.

We construct a 95 percent confidence interval for our forecast. We have

a � bx0 � s � tα=2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1n� x0 � x� �2Pn

i�1 ~x2i

!vuut � 9:061 � 0:7446 � 2:1448

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

115

� 5:4 � 5:2533� �270:777

!vuut

The 95 percent confidence interval is therefore 8:648; 9:475� �.

MULTIPLE CORRELATION

A dependent variable may be influenced by more than one variable. For instance, thesales of coffee at a coffee shop in a small tourist site are an increasing function of thenumber of tourists arriving in the site; however, during a summer week, the sales ofcoffee have fallen despite a rising number of arriving tourists. The manager of thecoffee shop realizes it was very hot during that week and tourists wanted coolbeverages instead of coffee.

We formulate a multiple correlation model as

y � a0 � a1x � a2z � ε (14.99)

The correlation coefficients between y; x� �; y; z� � and x; z� � are Ryx, Ryz, and Rxz,respectively. The regression coefficient a1 is a partial regression coefficient; it measuresthe effect of x on y, holding z fixed. Similarly, the regression coefficient a2 is a partialregression coefficient; it measures the effect of z on y, holding x fixed. The estimates ofa1 and a2 are, respectively,

a1 � Ryx � RyzRxz

1 � R2xz

!sysx

(14.100)

a2 � Ryz � RyxRxz

1 � R2xz

!sysz

(14.101)

where sy; sx, and sz are the standard deviations of y; x and, z, respectively. Intuitively,the coefficient a1 is corrected for the correlation between y; z� � and x; z� �; similarly,the coefficient a2 is corrected for the correlation between y; x� � and x; z� �.


3GC14 05/15/2014 17:44:10 Page 324

Example:Malay Palm Oil Corporation tracked data on its palm oil sales, real percapita growth, and palm oil prices as shown in Table 14.4. It wanted to estimate amulticorrelation between its sales of palm oil, real per capita income, and its salesprices of the form

y � a0 � a1x � a2z � ε (14.102)

Using Microsoft Excel we compute the partial correlation coefficients; these areRyx � 0:0427, Ryz � �0:9677, and Rxz � 0:1790. We also compute sy � 4:1664,sx � 0:8817, and sz � 2:9396.

a1 � Ryx � RyzRxz

1 � R2xz

!sysx

� 0:0427 � 0:9677 � 0:179

1 � 0:1792

4:16640:8817

� 1:05

a2 � Ryz � RyxRxz

1 � R2xz

!sysz

� �0:9677 � 0:0427 � 0:179

1 � 0:1792

4:16642:9396

� �1:43

An estimate of a0 is a0 � y � a1x � a2z � 0:1401. The estimated model is there-fore

y � 0:1401 � 1:05x � 1:43z

The elasticity of sales with respect to per capita income is 1.05; a 1 percentincrease in per capita income induces a 1.05 percent increase in sales. The elasticity ofsales with respect to price is �1.43; a decrease in price by 1 percent increases sales by1.43 percent.

SUMMARY

Islamic finance applies regression analysis for the evaluation of financial models.This chapter describes basic linear regression methods. It covers the notions ofcurve fitting, the estimation and testing of the linear regression model, thediagnosis test of the estimated model’s errors, the prediction exercise, and multipleregression. Regressions are estimated instantly by software packages. The chapterprovides the methods for the analysts to ascertain the significance of the estimatedmodels.

TABLE 14.4 Multiple Correlation between Palm Oil Sales, Per Capita Income, and Palm OilPrice (changes in percentages)

Sales, y 3.5 4.8 �0.1 2.6 �0.2 2.3 5.1 8.7 4.5 0.6 2.0 10.4 15.5 9.0 0.1 1.8 3.8Income, x 2.4 2.9 1.2 3.5 2 1 0.5 3.2 1.6 1.8 2.1 3.2 0.8 2.3 2.7 1 2.0Price, z �0.8 �1.1 0.8 0.5 1.5 �0.3 �3 �4 �2 0.7 0.8 �4 �10 �5 2.5 �0.6 �1.2

324 STATISTICS

3GC14 05/15/2014 17:44:12 Page 325

QUESTIONS

1. Explain the principle of fitting data by a curve.

2. State the assumptions underlying the estimation of a linear regression model.

3. Derive estimates of the parameters a and b of the model

yi � a � bxi � εi

based on the method of moments.

4. Compare the method of moments with the maximum likelihood method.

5. You are provided with the following data:

xi 4.2 3 3.2 5.7 7.1 5.3 7.0 6.2 5.4 2.1 3.5yi 16 13.0 14.2 22 23 19 27 23 19 10.8 15.0

Using Microsoft Excel or EViews, estimate the model

yi � a � bxi � εi

Compute the variance of the residuals, the variance of a and b, and

Cov a; b� �

. Compute the sum of squared residuals SSE� � and the sum of the

squares due to regression SSR� �. Compute R2 and the F statistic.

6. An Islamic stock has an annual mean return y; the market portfolio has a meanreturn x as shown in the below chart.

x 4 4.2 7.0 8.0 5.5 3.5 2.1 0.5 6.6 7.2 8.3 4.8 5.8 3.9 7.4 5.6 4.1y 5.7 7.0 11.9 13.5 8.9 6.0 4.8 1.7 9.5 11.4 12.0 9.2 9.2 7.5 12.6 9.2 8.7

Using Microsoft Excel, estimate a CAPM for the Islamic stock. Test thesignificance of its alpha and beta.

Display the estimated residuals; estimate the Durbin-Watson statistics andtest for the serial correlation in the error terms.

7. Based on your estimate of the CAPM, make an interval forecast for the return ofthe Islamic stock under the projection of a return of 6.4 percent for the marketportfolio at 95 percent confidence level.

8. You are provided with the following model:

Y � a0 � a1X � a2Z � ε

Explain the notion of partial correlation between Y; X� �; Y; Z� �, and X; Z� �.Compute the partial correlation coefficients. Provide estimates of a1 and a2 basedon the partial correlation coefficients.


3GC14 05/15/2014 17:44:14 Page 326

9. The sales of a shoe factory are related to per-capita income and shoe prices asfollows:

Sales, y 3.5 4.9 �0.2 2.6 �0.3 2.9 7.1 10.7 5.5 0.6 2.0 10.9 15.5 9.0 0.1 1.9 3.8

Income, x 2.4 2.9 1.2 3.5 1.2 1.8 0.5 3.2 1.6 1.8 2.1 3.2 0.8 2.3 2.7 1 2.0

Price, z �0.8 �1.1 0.8 0.5 1.5 �0.3 �3.8 �4.9 �2.3 0.7 0.8 �3.8 �10 �5 2.5 �0.7 �1.2

Estimate a multiregression model that relates shoe sales to income per capita andshoe prices. If the shoe factory decides to reduce prices by 5 percent what would be theimpact on shoe sales?

326 STATISTICS

3GC15 05/15/2014 12:0:8 Page 327

CHAPTER 15Time Series Analysis

I slamic finance applies time series analysis to financial data. A time series is a set ofobservations taken at specific times, usually at equal intervals. Examples of time

series are the gross domestic product of a nation observed over a number of years; theprice of a stock observed at an interval of five minutes; and the exchange rate of acurrency observed each hour, or each day. The interval of time at which a variable isobserved is called the frequency of the variable. If the variable is observed veryfrequently such as every month, day, hour, or five-minutes, it is called high-frequencyvariable. In contrast, if it is observed at long intervals such as every year or every fiveyears it is called low-frequency variable.Mathematically, a time series is defined by thevalues y1; y2; . . . of a variable y (e.g., closing price of a share, volume of traded shares,gold price) at times t1; t2; . . . Thus, y is a variable indexed on time and is representedas: yt1 ; yt2 ; . . . ytn .

A time series involving a variable y is represented by constructing a graph of yversus time. In Figure 15.1, we illustrate an example of a time series, which is themonthly S&P 500 stock price index spanning January 2000 to May 2013.

This chapter covers the component movements of a time series, stationary timeseries, the autocorrelation function, linear time-series models and Wold decomposi-tion of a stationary process, moving average MA� � linear models, autoregressive AR� �linear models, mixed autoregressive-moving average ARMA� � linear models, thepartial autocorrelation function, and forecasting based on time series.

COMPONENT MOVEMENTS OF A TIME SERIES

This section covers the components of a time series and the modeling the trend oftime series.

Components of a Time Series

Experience with many examples of time series has revealed certain characteristicmovements, or variations, some or all of which are present to varying degrees.Analysis of such movements is of great value in many connections, one of which isthe problem of forecasting future movements. Time series analysis consists of adescription of the component movements of the series. We assume that a timeseries variable y is the product of the variables T;C; S, and R that produce the

327

3GC15 05/15/2014 12:0:8 Page 328

trend, cyclic, seasonal, and random movements, respectively. More specifically y isexpressed as

y � T � C � S � R (15.1)

Time series analysis amounts to investigating the factors T;C; S, and R and isoften referred as a decomposition of a time series.

Trend movements: Some time series display a trend that may be an upward,stationary, or downward trend. The trend may be approximated by a trend line or acurve. Figure 15.2a displays an upward trend.

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

1/3/

2000

5/3/

2000

9/3/

2000

1/3/

2001

5/3/

2001

9/3/

2001

1/3/

2002

5/3/

2002

9/3/

2002

1/3/

2003

5/3/

2003

9/3/

2003

1/3/

2004

5/3/

2004

9/3/

2004

1/3/

2005

5/3/

2005

9/3/

2005

1/3/

2006

5/3/

2006

9/3/

2006

1/3/

2007

5/3/

2007

9/3/

2007

1/3/

2008

5/3/

2008

9/3/

2008

1/3/

2009

5/3/

2009

9/3/

2009

1/3/

2010

5/3/

2010

9/3/

2010

1/3/

2011

5/3/

2011

9/3/

2011

1/3/

2012

5/3/

2012

9/3/

2012

1/3/

2013

FIGURE 15.1 S&P 500 Stock Index, January 2000 to March 2013Source: Yahoo.finance.

a. Long-term trend b. Long-term trend and cyclical movements

c. Long-term trend and seasonal movements

0

yt

t 0

yt

t

yt

t0

FIGURE 15.2 Components of Time Series

328 STATISTICS

3GC15 05/15/2014 12:0:8 Page 329

Cyclical movements: These refer to oscillations about a trend line or curve. Thesecycles may or may not be periodic; that is, they may or may not follow exactly similarpatterns after equal intervals of time. An important example of cyclic movements is theso-called business cycle representing intervals of prosperity, recession, depression, andrecovery. Figure 15.2b displays cyclical movements around a trend line.

Seasonal movements: These refer to identical or almost identical patterns that atime series appears to follow during corresponding months of successive years. Forinstance, in each country, the sales of school supplies (books, notebooks, pens, etc.)rise at the beginning of the school year; the consumption of food and clothing may riseduring religious holidays; energy consumption may rise during winter; electricityconsumption may rise during summer; and air travel may rise during holiday orvacations. Figure 15.2c displays seasonal movements around the trend.

Irregular or randommovements:These refer to the sporadic motions of time seriesdue to chance events.

Modeling the Trend of Time Series

The series that we want to forecast vary over time, and we often attribute variations tounobserved underlying components, such as trends, seasons, and cycles. Often, wedeal with trends. We consider deterministic trends in which the trend evolves in aperfectly predictable way. Later, we broaden to allow for a stochastic trend. Trendmodels can be linear, quadratic, or exponential as a function of time. A linear trendmodel can be expressed as

yt � β0 � β1t; t � 1;2; . . . ;T (15.2)

A quadratic trend can be formulated as

yt � β0 � β1t � β2t2; t � 1; 2; . . . ;T (15.3)

An exponential, or log linear trend, can formulated as

yt � β0eβ1t (15.4)

or in logarithms as

lnyt � lnβ0 � β1t; t � 1; 2; . . . ;T (15.5)

A trend can be estimated in several ways. The least square method can be used tofind the equation of an appropriate trend line or curve. From this equation we cancompute the trend value β1t. The moving average method is used to estimate a trend.By usingmoving averages of appropriate orders, we can eliminate cyclic, seasonal, andrandom patterns, thus leaving only trend movement.

Example:We consider data onweekly S&P 500 stock index spanningMarch 2009toMay 2013. Using the ordinary least squares method, we estimate a trend of the form

ln�yt� � ln β0� � � β1t � νt (15.6)

Time Series Analysis 329

3GC15 05/15/2014 12:0:9 Page 330

where νt is a random error. We find

ln yt� � � 6:867 � 0:0022 � t

The stock index has been increasing at 0.22 percent perweek or 52�0:22 percent �11:44 percent per year during March 2009 to May 2013.

STATIONARY TIME SERIES

Wemay assume that a time series has been generated by a stochastic (random) processwith a structure that can be described. In other words, we assume that each valuey1; y2; . . . ; yT in the series is drawn randomly from a probability distribution;equivalently, we assume the series has been generated by a stochastic process. Inmodeling such a process, we attempt to describe the characteristics of its randomness.To be completely general, we assume that the observed series y1; y2; . . . ; yT is drawnfrom a set of jointly distributed random variables.

We address the concept of stationarity of the stochastic time series. As we developmodels for time series, we want to know whether the underlying stochastic processthat generated the series can be assumed to be invariant with respect to time. If theparameters of the stochastic process change over time, that is, the process is nonsta-tionary, so it will often be difficult to represent the time series over past and futureintervals of time by a simple algebraic model. By contrast, if the stochastic process isfixed in time, that is, if it is stationary, then one can model the process via an equationwith fixed coefficients that can be estimated from past data.

Properties of Stationary Processes

Any stochastic time series y1; y2; . . . ; yT can be thought of as having been generated bya set of jointly distributed random variables; that is, the set of data points y1; y2; . . . ; yTrepresents a particular outcome of the joint probability distribution

p y1; y2; . . . ; yT� �

(15.7)

The outcome is called a realization. Thus, y1; y2; . . . ; yT represents one particularrealization of the stochastic process represented by the probability distributionp y1; y2; . . . ; yT� �

. Similarly, a future observation yT�1 can be thought of as beinggenerated by a conditional probability distribution

p yT�1 jy1; y2; . . . ; yT� �(15.8)

that is, a probability distribution for yT�1 given the past observations y1; y2; . . . ; yT .We define a stationary process, then, as one whose joint distribution and conditionaldistribution both are invariant with respect to displacement in time. In other words, ifthe series yt is stationary, then

p yt; . . . ; yt�h� � � p yt�m; . . . ; yt�h�m

� �(15.9)

330 STATISTICS

3GC15 05/15/2014 12:0:9 Page 331

and

p yt� � � p yt�m

� �for any t; h; andm (15.10)

Note that if the series yt is stationary, the mean of the series, which is defined as

μy � E yt� �

(15.11)

must also be stationary, so that

E yt� � � E yt�m

� � � μ � constant for any t andm (15.12)

Furthermore, the variance of the series must be stationary, so that

σ2y � E yt � μy� �2� �

� E yt�m � μy� �2� �

� σ2 � constant (15.13)

and finally, for any lag h, the autocovariance of the series must be stationary, so that

Cov yt; yt�h� � � Cov yt � μy

� �yt�h � μy� �h i

(15.14)

Cov yt; yt�h� � � Cov yt�m; yt�h�m

� � � γh � constant (15.15)

In sum, a stationary process has a constant mean, constant variance, and for adisplacement h periods, a constant covariance γh. A stationary series is denoted I 0� �. Anestimate of themean μy of the process can be obtained from the samplemean of the series

y � 1T

XTt�1

yt (15.16)

An estimate of the variance is provided by the sample variance

σ2y � 1T

XTt�1

yt � y� �2 (15.17)

and an estimate of the autocovariance is provided by the sample covariance

γh �PT�h

t�1 yt � y� �

yt�h � y� �

T(15.18)

Example: A simple I 0� � process isyt � εt

where εt is a white noise with constant mean E εt� � � 0, constant variance E ε2t� � � σ2ε ,

and E εtεt�j� � � 0 for any j ≠ t. We use the Microsoft Excel function NORM.S.INV

(RAND()) to generate 100 observations of εt and portray the series in Figure 15.3. Weobserve the series displays no trend and reverts frequently to its mean, which is zero in


3GC15 05/15/2014 12:0:10 Page 333

the case of a white noise. Every stationary process displays similar behavior andreverts often to its mean μ:

Example: We consider the weekly percent change of the S&P 500 stock indexduring March 2009 to May 2013. We compute the sample mean at 0.386 percent perweek, or by multiplying by 52 at 20.1 percent per year. The standard deviation iscomputed at 2.375 percent per week, or by multiplying by the square root of 52 at17.128 percent per year.

CHARACTERIZING TIME SERIES: THEAUTOCORRELATION FUNCTION

While it is usually impossible to obtain a complete description of a stochastic process(i.e., to actually specify the underlying probability distributions), the autocorrelationfunction will prove extremely useful because it provides a partial description of theprocess for modeling purposes. The autocorrelation function tells us how muchcorrelation there is (and by implication how much interdependency there is) betweenneighboring data points in the series yt. We define the autocorrelation with lag h as

ρh � E yt � μy� �

yt�h � μy� �h i

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE yt � μy� �2� �

E yt�h � μy� �2� �s � Cov yt; yt�h

� �σyσt�h

(15.19)

For an I 0� � process the variance at time t in the denominator of ρh is the same asthe variance at time t � h; thus, the denominator is just the variance of the stochasticprocess, and

ρh � E yt � μy� �

yt�h � μy� �h iσ2y

(15.20)

Note the numerator in ρh is the covariance between yt and yt�h, γh, so that

ρh � γhγ0

(15.21)

Suppose the stochastic process is simply

yt � εt (15.22)

where εt is an independently distributed random variable with zero mean. Then it iseasy to see that the autocorrelation function for the process is given by ρ0 � 1, ρh � 0for h > 0. The process yt � εt is called white noise, and there is no model that canprovide a forecast any better than yT�h � 0. Thus the autocorrelation is zero for allh > 0, and there is little or no value in using a model to forecast the series.

The autocorrelation function ρh is purely theoretical. In practice, wemust calculate an estimate of the autocorrelation function, called the sample


3GC15 05/15/2014 12:0:10 Page 334

autocorrelation function:

ρh �PT�h

t�1 yt � y� �

yt�h � y� �

PTt�1 yt � y� �2 (15.23)

Example: Using the EViews, we compute the sample autocorrelation of theweekly returns on the S&P 500 index during March 2009 to May 2013 at lags 1,2, 3, 4, and 5. We find ρ1 � �0:078, ρ2 � �0:024, ρ3 � 0:008, ρ4 � �0:073, andρ5 � �0:028.

Based on the process yt � εt , it is often of interest to assess whether a series isreasonably approximated by a white noise, which is to say all its autocorrelations arezero in population. A key result that we simply assert is that if a series is white noise,then the distribution of the sample autocorrelation in large samples is

ρ h� �∼N 0; 1=T� � (15.24)

Thus, if the series is white noise, approximately 95 percent of the sampleautocorrelations should fall in the interval

�z0:025 � 1=ffiffiffiffiT

p(15.25)

z0:025 is the critical standard normal variable at significance level 0.025.Example:Using EViews with data on the weekly returns of the S&P 500 index for

March 2009 toMay 2013, we test the significance of autocorrelation of returns at lags1, 2, 3, 4, and 5. More specifically we test the null hypothesis

H0 : ρh � 0

against the alternative H1: ρh ≠ 0.The test is a two-tailed test with 2.5 percent significance level at each tail. The

number of observation is T � 218; the z-score at 0.025 is provided by the EXCELfunction z0:025 � NORM:S:INV 0:975� � � 1:96. The confidence interval is�1:96 � 1ffiffiffiffiffiffi

218p � �0:1327; 0:1327� �. We consider the sample autocorrelation:

ρ1 � �0:078, ρ2 � �0:024, ρ3 � 0:008, ρ4 � �0:073, and ρ5 � �0:028. Each of thesecoefficients falls in the interval �0:1327;0:1327� �. Hence, we cannot reject thehypothesis H0: ρh � 0 at the lags 1, 2, 3, 4, and 5.

The two-standard error bands, although very useful, provide 95 percent boundsonly for the sample autocorrelations taken one at a time. Ultimately, we often areinterested in whether a series is white noise, that is, whether all its autocorrelations arejointly 0. A simple extension lets us test this hypothesis. Rewrite the expression

ρ h� �∼N 0; 1=T� � (15.26)

as

ffiffiffiffiT

pρ h� �∼N 0; 1� � (15.27)

334 STATISTICS

3GC15 05/15/2014 12:0:11 Page 335

We recall that the square of a standard normal variable is a χ2 random variablewith one degree of freedom. Squaring both sides yields

Tρ2 h� �∼ χ21 (15.28)

It can be shown that, in addition to being approximately normally distributed, thesample autocorrelations at various displacements are approximately independent ofone another. It is known that the sum of independent χ2 is also χ2 with degrees offreedom equal to the sum of the degrees of freedom of the variables summed.Accordingly, the Box-Pierce Q-statistic

QBP � TXqh�1

ρ2 h� � (15.29)

is approximately distributed as χ2q, where q is a maximum displacement selected by theuser. A slight modification of this, designed to follow more closely the χ2 in smallsamples, is

QLB � T T � 2� �Xq

h�11

T � h

ρ2 h� � (15.30)

The QLB is called the Ljung-Box Q-statistic. Under the null hypothesis that y iswhite noise, QLB is approximately distributed as a χ2q random variable.

Example: Market efficiency hypothesis: test the joint hypothesis that all theautocorrelation coefficients of the weekly returns of the S&P 500 stock index duringMarch 2009 to May 2013 are zero.

We use the Q-statistic introduced by Box and Pierce,

QBP � TXqh�1

ρ2h

Q is approximately distributed as a chi-square with q degrees of freedom. Thus, ifthe calculated value of Q is greater than, say, the critical 5 percent level, we can be95 percent sure that the true autocorrelation coefficients ρ1; . . . ; ρh are not all zero.Weconsider the sample autocorrelation: ρ1 � �0:078, ρ2 � �0:024, ρ3 � 0:008,ρ4 � �0:073, and ρ5 � �0:028. We compute

QBP � 218 � �0:078� �2 � �0:024� �2 � 0:008� �2 � �0:073� �2 � �0:028� �2� � � 2:798:

We compute the critical χ2q at 5 degrees of freedom using the Microsoft Excelfunction

χ2q � CHISQ:INV:RT 0:05; 5� � � 11:0705

By comparing the critical value of the chi-square, 11:0705, to its estimated valueat 2.798, we fail to reject the joint null hypothesis that ρ1 � ρ2 � ∙ ∙ ∙ � ρh � 0. Thereturns on the S&P 500 stock index can thus be approximated by a white noiseprocess confirming the market efficiency hypothesis.


3GC15 05/15/2014 12:0:11 Page 336

Note that EViews performs the same test of the significance of the autocorrelation;however, it uses the QLB statistic.

Stationarity and the Autocorrelation Function

How canwe determine that a series is I 0� �?We can look at a plot of an autocorrelationfunction, called correlogram. Figure 15.4 shows the autocorrelation functions forstationary and nonstationary series. The autocorrelation function for a stationaryseries shown in Figure 15.4a drops off as h, the number of lags, becomes large, but asshown in Figure 15.4b, this usually is not the case for nonstationary series.

LINEAR TIME SERIES MODELS

We turn to the construction of time series models. Our objective is to develop modelsthat explain themovements of a time series by relating it to its own past values and to aweighted sum of current and lagged random disturbances.We examine simple movingaverage MA� � and autoregressive AR� � models for stationary processes. In a movingaverage, the process is described completely by a weighted sum of current and laggedrandom disturbances. In the autoregressive model, the process depends on a weightedsum of its past values and a random disturbance term. We introduce mixed auto-regressive-moving average ARMA� �models. In these models, the process is a functionof both its past values and lagged random disturbances as well as current disturbances.We show how the autocorrelation function can be used to help determine how manylagged disturbance terms and autoregressive terms should be included in the model.

Wold Decomposition of a Stationary Process

Let yt �

be any zero-mean covariance-stationary process. Then we can write it as

yt � εt � θ1εt�1 � θ2εt�2 � θ3εt�3 � ∙ ∙ ∙ :: �X∞i�0

θiεt�i (15.31)

εt ∼white noise 0; σ2ε� �

(15.32)

a. Correlogram of a stationary series

b. Correlogram of a nonstationary series

h

ρh

0h0

ρh

FIGURE 15.4 Correlograms of (a) Stationary and(b) Nonstationary Time Series

336 STATISTICS

3GC15 05/15/2014 12:0:12 Page 337

where θ0 � 1, andP∞i�0

θ2i < ∞ . The model for any covariance stationary series is some

infinite distributed lag of white noise, called theWold representation. The εt are calledinnovations; they represent that part of the evolution of y that is linearly unpredictableon the basis of the past of y.

The Wold decomposition can also be presented as

yt � μ �X∞i�0

θiεt�i (15.33)

In our statement of Wold’s theorem we assumed a zero mean. That may seemrestrictive, but it is not. We think of yt as yt � μ, so that the process is expressed indeviations from its mean. The deviation from the mean has a zero mean, byconstruction.Working with zero-mean process therefore involves no loss of generalitywhile facilitating notational economy. We use this device frequently.

Wold’s theorem tells us that when formulating forecasting models for I 0� �covariance stationary time series, we need to consider only models of the form

yt �X∞i�0

θiεt�i (15.34)

We compute the unconditional and conditional mean and variance of thispresentation. We obtain the following results.

The unconditional mean:

E�yt� � EX∞i�0

θiεt�i !

�X∞i�0

θiE�εt�i� �X∞i�0

θi � 0 � 0 (15.35)

The unconditional variance:

Var yt� � � Var

X∞i�0

θiεt�i !

�X∞i�0

θ2i Var εt�i� � �X∞i�0

θ2i σ2ε � σ2ε

X∞i�0

θ2i (15.36)

The conditional moments are computed based on the information set Ωt�1; whichcontains past and known innovations; that is, Ωt�1 � εt�1; εt�2; . . . :f g.

E yt j Ωt�1� � � E εt jΩt�1

� � � θ1E εt�1 jΩt�1� � � θ2E εt�2 jΩt�1

� � � ∙ ∙ ∙

� 0 � θ1εt�1 � θ2εt�2 � ∙ ∙ ∙ �X∞i�1

θiεt�i(15.37)

The conditional variance is

Var yt jΩt�1� � � E yt � E yt jΩt�1

� �� 2 j Ωt�1h i

� E εt� �2jΩt�1� � � σ2ε (15.38)

The key insight is that the conditional mean moves over time in response to theevolving information set. The model captures the dynamics of the process, and the


3GC15 05/15/2014 12:0:13 Page 338

evolving conditional mean is one crucial way of summarizing them. An important goalof time series modeling is capturing such conditional mean dynamics; theunconditional mean is constant as a requirement of stationarity, but the conditionalmean varies in response to the evolving set of information.

MOVING AVERAGE (MA) LINEAR MODELS

In the moving average process of order q each observation yt of an I 0� � process isgenerated by a weighted average of random disturbances going back q periods. Wedenote this process as MA q� �:

yt � μ � εt � θ1εt�1 � θ2εt�2 � ∙ ∙ ∙ � θqεt�q (15.39)

εt ∼N 0; σ2ε� �

(15.40)

Each disturbance term εt is assumed to be normal random variable with meanE εt� � � 0, variance E ε2t

� � � σ2ε , E εtεt�h� � � 0 for h ≠ 0, and covariance γh � 0 forh ≠ 0. The mean of the process yt is E yt

� � � μ and is independent of time. The processMA q� � is described by exactly q � 2 parameters, the mean μ, the disturbance varianceσ2ε , and the parameters θ1; θ2; . . . ; θq that determine the weights in themoving average.Let us now look at the variance, γ0, of the MA q� �:

Var yt� � � γ0 � E yt � μ

� �2h i

� E ε2t � θ21ε2t�1 � ∙ ∙ ∙ � θ2qε2t�q � 2θ1εtεt�1 � ∙ ∙ ∙� �

Var yt� � � σ2ε � θ21σ

2ε � ∙ ∙ ∙ � θ2qσ

2ε � σ2ε 1 � θ21 � ∙ ∙ ∙ � θ2q

� �(15.41)

Thus, if yt is the realization of a stationary random process, we must have

θ21 � ∙ ∙ ∙ � θ2q� �

< ∞ .

An example of a simple moving average process is the moving average process oforder 1 MA 1� �:

yt � μ � εt � θ1εt�1 (15.42)

This process has mean μ and variance γ0 � σ2ε 1 � θ21� �

. The covariance for one-lagdisplacement, γ1, is

γ1 � E yt � μ� �

yt�1 � μ� �� E εt � θ1εt�1� � εt�1 � θ1εt�2� �� θ1σ2ε (15.43)

In general, we can determine the covariance for h-lag displacement to be

γh � E yt � μ� �

yt�h � μ� �� E εt � θ1εt�1� � εt�h � θ1εt�h�1� �� 0 for h > 1 (15.44)

Thus, the MA 1� � process has a covariance of 0 when the displacement is morethan one period. We say that the process has a memory of only one period; any value

338 STATISTICS

3GC15 05/15/2014 12:0:13 Page 339

yt is correlated with yt�1 and yt�1 and no other time-series values. We can nowdetermine the autocorrelation function for the process MA 1� �:

ρh � γhγ0

�θ1σ2ε

σ2ε 1 � θ21� �� θ1

1 � θ21� � ; h � 1

0 h > 1

8><>: (15.45)

Example: The returns of a Sharia-compliant stock are described by a first ordermoving average, or MA 1� �, process:

yt � εt � θ1εt�1 � εt � 0:42εt�1; εt ∼N 0; 1� �We compute the parameters of the process. We find μ � E yt

� � � 0,γ0 � σ2ε 1 � θ21

� � � 1:1764, and

ρ1 � θ11 � θ21� � � 0:357; ρh � 0; h > 1

Now let us examine the moving average process of order 2,MA 2� �; its equation is

yt � μ � εt � θ1εt�1 � θ2εt�2 (15.46)

This process has mean μ, variance γ0 � σ2ε 1 � θ21 � θ22� �

, and covariances given by

γ1 � E yt � μ� �

yt�1 � μ� �� E εt � θ1εt�1 � θ2εt�2� � εt�1 � θ1εt�2 � θ2εt�3� ��

� θ1σ2ε � θ1θ2σ2ε(15.47)

γ2 � E yt � μ� �

yt�2 � μ� �� E εt � θ1εt�1 � θ2εt�2� � εt�2 � θ1εt�3 � θ2εt�4� ��

� θ2σ2ε(15.48)

And γh � 0 for h > 2.The autocorrelation function is

ρ1 � θ1σ2ε � θ1θ2σ2εσ2ε 1 � θ21 � θ22� � � θ1 � θ1θ2

1 � θ21 � θ22� � (15.49)

ρ2 � θ2σ2εσ2ε 1 � θ21 � θ22� � � θ2

1 � θ21 � θ22� � (15.50)

And γh � 0 for h > 2.Example:We consider theweekly returns on the S&P500 stock index duringMarch

2009 to May 2013. Using EViews, we represent the I 0� � process as MA 2� �. We find

yt � 0:383 � 0:082εt�1 � 0:034εt�2; σε � 2:3828


3GC15 05/15/2014 12:0:14 Page 340

We compute μ � 0:383, γ0 � σ2ε 1 � θ21 � θ22� � � 5:7233, ρ1 � θ1�θ1θ2

1�θ21�θ22� � � 0:084,

and ρ2 � θ21�θ21�θ22� � � 0:034.

Invertible MA Process

We note that the requirements of covariance-stationary process are constantunconditional mean, constant and finite unconditional variance, and the auto-correlation depends only on displacement; these requirements are met for anyMA 1� � process, regardless of the values of its parameters. If, moreover, θj j < 1, wecan invert the MA(1) process and express the current value of the series in terms of acurrent shock and lagged values of the series. This is called an autoregressiverepresentation. An autoregressive representation has a current shock and laggedobservable values of the series on the right. A moving average representation has acurrent shock and lagged unobservable shocks on the right. Let us compute theautoregressive representation:

yt � εt � θεt�1 (15.51)

We solve for the innovation as

εt � yt � θεt�1 (15.52)

Lagging by successively more periods gives expressions for innovations at variousdates:

εt�1 � yt�1 � θεt�2εt�2 � yt�2 � θεt�3εt�3 � yt�3 � θεt�4

And so forth. Making use of these expressions for lagged innovations, we cansubstitute backward in the MA(1) process, yielding an AR ∞� � for yt:

yt � εt � θyt�1 � θ2yt�2 � θ3yt�3 � ∙ ∙ ∙ (15.53)

Example:We consider theMA 1� � process: yt � εt � 0:35εt�1. We may invert it toan AR process as

yt � εt � 0:35yt�1 � 0:1225yt�2 � 0:0429yt�3 � 0:015yt�4 � 0:005yt�5 . . .

AUTOREGRESSIVE (AR) LINEAR MODELS

In the autoregressive process of order p the current observation yt is generated by aweighted average of past observations going back p periods, together with a randomdisturbance in the current period. We denote this process as AR p� �:

yt � η � β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � εt (15.54)

340 STATISTICS

3GC15 05/15/2014 12:0:15 Page 341

Here η is a constant term that relates to the mean of the stochastic process.If the autoregressive process is stationary, then, its mean, μ, must be invariant

with respect to time; that is, E yt� � � E yt�1

� � � E yt�2� � � ∙ ∙ ∙ � μ. The mean is thus

given by

μ � β1μ � β2μ � ∙ ∙ ∙ � βpμ � η (15.55)

μ � η1 � β1 � β2 � ∙ ∙ ∙ � βp

(15.56)

This formula for the mean of the process gives a condition for stationarity. If theprocess is I 0� �, the mean μmust be finite. If this is not the case, the process would driftfarther and farther away from any given reference point and could not stationary.Consider for instance the random walk with drift:

yt � yt�1 � η � εt (15.57)

Here β1 � 1 and μ � ∞ and if η > 0 the process continually drifts upward. If μ isto be finite, it is necessary that

β1 � β2 � ∙ ∙ ∙ � βp < 1 (15.58)

We consider a simple autoregressive process AR 1� �:yt � β1yt�1 � η � εt (15.59)

This process has a mean of

μ � η1 � β1

(15.60)

and is stationary if β1j j < 1. Let us calculate γ0, the variance of the process about itsmean, assuming stationarity, so that the variance is constant for β1j j < 1. Setting η � 0to scale the process to one that has zero mean, we have

γ0 � E β1yt�1 � εt� �2h i

� E β21y2t�1 � ε2t � 2β1yt�1εt

� � � β21γ0 � σ2ε (15.61)

so that

γ0 � σ2ε1 � β21

(15.62)

We can calculate the covariance of yt about its mean:

γ1 � E ytyt�1� � � E yt�1 β1yt�1 � εt

� �� β1γ0 � β1σ2ε1 � β21

(15.63)

γ2 � E ytyt�2� � � E yt�2 β21yt�2 � β1εt�1 � εt

� �� β21γ0 � β21σ2ε1 � β21

(15.64)


3GC15 05/15/2014 12:0:15 Page 342

Similarly, the covariance for h-lag displacement is

γh � βh1γ0 � βh1σ2ε1 � β21

(15.65)

The autocorrelation function for AR 1� � is thus particularly simple; it begins atρ0 � 1 and declines geometrically:

ρh � γhγ0

� βh1 (15.66)

Note that this process has an infinite memory. The current value of the processdepends on all past values, although the magnitude of this dependence declines withtime.

Example: Consider an AR 1� � I 0� � process, yt � β1yt�1 � εt � 0:57yt�1 � εt withεt ∼N 0; 1� �.

We find γ0 � σ2ε1 � β21

� 0:844 and ρh � 0:57h. For instance, ρ1 � 0:57 and

ρ2 � 0:3249.Let us now look at the second order autoregressive process AR 2� �:

yt � β1yt�1 � β2yt�2 � η � εt (15.67)

The process has a mean of

μ � η1 � β1 � β2

(15.68)

A necessary condition for stationarity is that β1 � β2 < 1.Let us now calculate the variances and covariances of yt (when yt is measured in

deviations form):

γ0 � E yt β1yt�1 � β2yt�2 � εt� �� β1γ1 � β2γ2 � σ2ε (15.69)

γ1 � E yt�1 β1yt�1 � β2yt�2 � εt� �� β1γ0 � β2γ1 (15.70)

γ2 � E yt�2 β1yt�1 � β2yt�2 � εt� �� β1γ1 � β2γ0 (15.71)

In general, for h � 2,

γh � E yt�h β1yt�1 � β2yt�2 � εt� �� β1γh�1 � β2γh�2 (15.72)

Solving for γ0 and γ1, we obtain the following equations:

γ1 � β1γ01 � β2

(15.73)

γ0 � 1 � β2� �

σ2ε1 � β2� �

1 � β2� �2 � β21h i (15.74)

342 STATISTICS

3GC15 05/15/2014 12:0:16 Page 343

The equations enable us to derive the autocorrelation function ρh

ρ1 � β11 � β2

(15.75)

and

ρ2 � β2 � β211 � β2

(15.76)

and for h � 2

ρh � β1ρh�1 � β2ρh�2 (15.77)

Example: Using EViews, we represent the weekly returns on the S&P 500 stockindex during March 2009 to May 2013 as an AR(2):

yt � 0:357 � 0:086yt�1 � 0:031yt�2; σε � 2:3587

We compute the parameters of the process:

μ � η1 � β1 � β2

� 0:3571 � 0:086 � 0:031

� 0:319

γ0 � 1 � β2� �

σ2ε1 � β2� �

1 � β2� �2 � β21h i � 5:9086; ρ1 � β1

1 � β2� �0:0838

ρ2 � β2 � β211 � β2

� �0:02395

Invertible AR(1) Process

The condition for invertibility of anMA(1) process is the counterpart to the conditionof stationarity of an AR(1) process; if

AR 1� � : yt � β1yt�1 � εt (15.78)

then β1j j < 1 implies

yt � εt � β1εt�1 � β21εt�2 � β31εt�3 � β41εt�4 � ∙ ∙ ∙ : (15.79)

an MA ∞� � representation with coefficients β1; β21; β

31; β

41; . . . :

�. More generally,

invertibility of an MA q� � process is the flip side of stationarity of an AR p� � process.Example: We invert the following AR 1� �: yt � 0:84yt�1 � εt.We note that β1j j � 0:84 < 1. We write the MA ∞� � representation as

yt � εt � 0:84εt�1 � 0:7056εt�2 � 0:5927εt�3 � 0:4979εt�4 � ∙ ∙ ∙


3GC15 05/15/2014 12:0:17 Page 344

MIXED AUTOREGRESSIVE-MOVING AVERAGE (ARMA)LINEAR MODELS

Many stationary random processes cannot be modeled as purely moving averages oras purely autoregressive. We may explore models called autoregressive-movingaverage process of order p; q� �, denoted by ARMA p; q� �:yt � β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � η � εt � θ1εt�1 � θ2εt�2 � ∙ ∙ ∙ � θqεt�q (15.80)

We assume that the process is stationary, so that its mean is constant over timeand is given by

μ � β1μ � β2μ � ∙ ∙ ∙ � βpμ � η (15.81)

μ � η1 � β1 � β2 � ∙ ∙ ∙ � βp

(15.82)

This gives a necessary condition for the stationarity of the process, that is,

β1 � β2 � ∙ ∙ ∙ � βp < 1 (15.83)

Let us consider the simplest autoregressive-moving average process, ARMA 1; 1� �,yt � β1yt�1 � η � εt � θ1εt�1 (15.84)

We set η � 0. The variances and covariances of this process are determined jointlyas

γ0 � E yt β1yt�1 � εt � θ1εt�1� �� E β1yt�1 � εt � θ1εt�1

� �2h i

γ0 � β21γ0 � 2β1θ1E yt�1εt�1� � � σ2ε � β21σ2ε

Since E yt�1εt�1� � � σ2ε , we have

γ0 1 � β21� � � σ2ε 1 � β21 � 2β1θ1

� �

so that the variance is given by

γ0 � 1 � β21 � 2β1θ1� �

1 � β21� � σ2ε (15.85)

We can now determine the covariances y1; y2; . . . ; recursively,

γ1 � E yt�1 β1yt�1 � εt � θ1εt�1� �� β1γ0 � θ1σ2ε � 1 � β1θ1

� �β1 � θ1� �

1 � β21� � σ2ε (15.86)

γ2 � E yt�2 β1yt�1 � εt � θ1εt�1� �� β1γ1 (15.87)

344 STATISTICS

3GC15 05/15/2014 12:0:17 Page 345

and similarly,

γh � β1γh�1 for h � 2 (15.88)

The autocorrelation function is given by

ρ1 � γ1γ0

� 1 � β1θ1� �

β1 � θ1� �

1 � β21 � 2β1θ1(15.89)

For displacement h greater than 1,

ρh � β1ρh�1 for h � 2 (15.90)

Thus, the autocorrelation function begins at its starting value ρ1, which is afunction of both β1 and θ1, and then starts decaying geometrically from the startingvalue. This reflects the fact that the moving average part of the process has a memoryof only one period.

Example: Using EViews we represent the weekly returns on the S&P 500 stockindex during March 2009 to May 2013 as an ARMA 1; 1� �: yt � 0:234� 0:884yt�1 �εt � 0:9908εt�1 and σε � 2:304:

Invertibility of the ARMA(1,1)

Let us consider the autoregressive-moving average process, ARMA 1; 1� �,yt � β1yt�1 � εt � θ1εt�1 (15.91)

We assume β1j j < 1. Invertibility means that we can express the ARMA 1; 1� �process as a MA ∞� �. By substituting backward, we can write

yt � β21yt�2 � β1εt�1 � β1θ1εt�2 � εt � θ1εt�1� β31yt�3 � β21εt�2 � β21θ1εt�3 � β1εt�1 � β1θ1εt�2 � εt � θ1εt�1� β31yt�3 � εt � β1 � θ1

� �εt�1 � β1 β1 � θ1

� �εt�2 � β21θ1εt�3

� β41yt�4 � εt � β1 � θ1� �

εt�1 � β1 β1 � θ1� �

εt�2 � β21 β1 � θ1� �

εt�3 � β31θ1εt�4

since β1j j < 1, βj1 ! 0 as j ! ∞ ; by continuing the backward substitution, the terms

βj1yt�j and βj�11 θ1εt�j will vanish. The ARMA process can therefore be expressed asMA ∞� �,yt � εt � β1 � θ1

� �εt�1 � β1 β1 � θ1

� �εt�2 � β21 β1 � θ1

� �εt�3 � β31 β1 � θ1

� �εt�4 � ∙ ∙ ∙ ∙ ∙ ∙

�X∞i�0

ψiεt�i(15.92)

where ψ0 � 1 and ψi � βi�11 β1 � θ1� �

. We have a Wold representation of the process yt.


3GC15 05/15/2014 12:0:18 Page 346

Example: Weekly returns of a Sharia-compliant stock has an ARMA 1; 1� �:yt � 0:6yt�1 � εt � 0:22εt�1. We provide a Wold representation as

yt � εt � 0:38εt�1 � 0:228εt�2 � 0:1368εt�3 � 0:082εt�4 � 0:049εt�5 � ∙ ∙ ∙ ∙ ∙ ∙

THE PARTIAL AUTOCORRELATION FUNCTION

One problem in constructing autoregressive models is identifying the order of theunderlying process. For moving average models this is less of a problem, since if theprocess is of order q, the sample autocorrelations should all be close to zero for lagsgreater than q. For the order of an autoregressive process information from the partialautocorrelation function may be required. To understand what the partial auto-correlation function is and how it can be used, let us consider the covariances andautocorrelation function for the autoregressive process of order p. The covariancewith displacement h is determined from

γh � E yt�h β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � εt� �h i

(15.93)

Letting h � 0; 1; . . . ; p, we obtain the following p � 1 difference equations in γ´s:

γ0 � β1γ1 � β2γ2 � ∙ ∙ ∙ � βpγp � σ2ε (15.94)

γ1 � β1γ0 � β2γ1 � ∙ ∙ ∙ � βpγp�1 (15.95)

γp � β1γp�1 � β2γp�2 � ∙ ∙ ∙ � βpγ0 (15.96)

For displacement h > p the covariances are determined by

γh � β1γh�1 � β2γh�2 � ∙ ∙ ∙ � βpγh�p (15.97)

By dividing on both sides by γ0 we can derive

ρ1 � β1 � β2ρ1 � ∙ ∙ ∙ � βpρp�1 (15.98)

ρp � β1ρp�1 � β2ρp�2 � ∙ ∙ ∙ � βp (15.99)

For displacement h > p the covariances are determined by

ρh � β1ρh�1 � β2ρh�2 � ∙ ∙ ∙ � βpρh�p (15.100)

The equations ρ1; ρ2; . . . ; ρp are called the Yule-Walker equations; if ρ1; ρ2; . . . ; ρpare known, the equations can be solved for β1; β2; . . . ; βp. The solution of the Yule-Walker equations requires the knowledge of p, the order of the autoregressive process.Therefore, we solve the Yule-Walker equations for successive values of p.

Suppose p � 1, then we have ρ1 � β1, and β2 � β3 � ∙ ∙ ∙ � 0. Using the sampleautocorrelations, we have ρ1 � β1. Thus, if the calculated value β1 is significantly

346 STATISTICS

3GC15 05/15/2014 12:0:19 Page 347

different from zero, we know the autoregressive process is at least of order 1. Let usdenote this value β1 by b1. Now consider the hypothesis that p � 2. The Yule-Walkerequations become

ρ1 � β1 � β2ρ1 (15.101)

ρ2 � β1ρ1 � β2 (15.102)

Using the sample autocorrelations we obtain new estimates for β1 and β2. If β2 issignificantly different from zero we can assume that process is at least of order 2; ifvalue for β2 is approximately zero, we can conclude that p � 1. Let us denote β2 by b2.We now repeat this process for successive values of p. For p � 3 we obtain an estimateof β3; which we denote b3. For p � 4 we obtain β4; which we denote by b4. We call theseries b1; b2; b3; . . . : the partial autocorrelation function. We can infer the order of theautoregressive process from its behavior. In particular, if the true order is p, we shouldobserve that bj � 0 for j > p.

The partial autocorrelation function at lag h is denoted by p h� �. It is interpreted inthe same manner as the regression coefficients of multivariable regression. A partialautocorrelation is just the coefficient on yt�h of a linear regression of yt onyt�1; yt�2; . . . ; yt�h. The sample partial autocorrelation at displacement h is estimatedby regression coefficient of yt�h in the regression of yt on yt�1; yt�2; . . . ; yt�h. The fittedregression is

yt � α � β1yt�1 � ∙ ∙ ∙ � βhyt�h (15.103)

The sample partial autocorrelation at displacement h is p h� � � βh. We pick uponly βh and disregard the coefficients of order less than h. We call this regression anautoregression because the variable is regressed on lagged values of itself. It is easy tosee that the autocorrelations and partial autocorrelations, although related, differ inan important way. The autocorrelations are just the “simple” or “regular” correla-tions between yt and yt�h. The partial autocorrelations, on the other hand, measure theassociation between yt and yt�h after controlling for the effects of yt�1; yt�2; . . . ; yt�h�1;that is, they measure the partial correlation between yt and yt�h.

As with the autocorrelations, we often graph the partial autocorrelations as afunction of h. Like the autocorrelation function, the partial autocorrelation functionprovides a summary of a series’ dynamics. All of the covariance stationary processeshave autocorrelation and partial autocorrelation functions that approach zero, oneway or another, as the displacement gets large.

Distributional results identical to those for the sample autocorrelations hold aswell for the sample partial autocorrelations. To test whether a particular bj is zero, wecan use the fact that it is approximately normally distributed, with mean 0 andvariance 1=T. Hence, we can check whether it is statistically significant at, say, the 5

percent level by determining whether it exceeds 2=ffiffiffiffiT

pin magnitude. If the series is a

white noise, approximately 95 percent of the sample partial autocorrelations should

fall in the interval �2 ffiffiffiffiT

p.

Example: Using EViews, we compute the partial autocorrelation for the weeklyreturns on the S&P 500 stock index duringMarch 2009 toMay 2013 at lags 1, 2, 3, 4,


3GC15 05/15/2014 12:0:19 Page 348

and 5. We find p 1� � � �0:078, p 2� � � �0:031, p 3� � � 0:003, p 4� � � �0:074, andp 5� � � �0:040.

Example: Using EViews, we compute the partial autocorrelation for the weeklyreturns on the S&P 500 stock index duringMarch 2009 toMay 2013 at lags 1, 2, and3 by estimating the following AR 1� �, AR 2� �, and AR 3� �:

yt � 0:411 � 0:078yt�1; which yields b1 � �0:078yt � 0:399 � 0:0:086yt�1 � 0:031yt�2; which yields b2 � �0:031yt � 0:399 � 0:1035yt�1 � 0:0359yt�2 � 0:003yt�3; which yields b3 � 0:003

We notice that we obtain approximately the same partial autocorrelation coef-ficients as in the preceding example.

FORECASTING BASED ON TIME SERIES

This section covers minimum mean-squared-error forecasts, the forecast confidenceinterval (CI), the forecast of the AR 1� � process, the MA 1� � process, and theARMA 1;1� � process.Minimum Mean-Squared-Error Forecasts

We turn now to forecasting. Our objective is to predict future values of a time seriessubject to as little error as possible. For this reason, we consider the optimum forecastto be that forecast that has a minimum mean-square forecast error. Since the forecasterror is a random variable, we minimize the expected value. Thus, we wish to chooseour forecast yT�h so that

s2T�h � E e2T�h� � � E yT�h � yT�h

� �2 (15.104)

is minimized. We show that this forecast is given by the conditional expectation ofyT�h, that is, by

yT�h � E yT�h jyT ; yT�1; . . . ; y1� �(15.105)

We examine the properties of the forecasts derived from some ARMAmodels. Inall cases that followwe assume that the parameters of the particular ARMAmodel areestimated as explained earlier.

Forecast Confidence Interval (CI)

Before we can calculate a confidence interval for our forecast yT�h, we need anestimate for σ2ε for the variance of the disturbance term. This estimate would be basedon the sum of squared residuals obtained after final estimates of the parameters havebeen obtained:

σ2ε �XTt�1

ε2t = T � p � q� � (15.106)

348 STATISTICS

3GC15 05/15/2014 12:0:20 Page 349

Here T � p � q is the number of degrees of freedom in the linear regression. Wesee that a confidence interval around a forecast h-periods ahead would be given by

CI 95%� � � yT�h � z0:025sT�h (15.107)

This interval gets larger as the lead time h becomes larger, although the exactpattern depends on the parameters of the time series process.

Forecast of the AR(1) Process

Let us begin with the stationary first-order autoregressive process, AR 1� �,yt � β1yt�1 � η � εt (15.108)

For this process the one-period forecast is

yT�1 � E yT�1 jyT ; yT�1; . . . ; y1� � � β1yT � η (15.109)

Similarly,

yT�2 � β1yT�1 � η � β21yT � β1 � 1� �

η (15.110)

yT�3 � β1yT�2 � η � β1 β21yT � β1 � 1� �

η� � � η � β31yT � β21 � β1 � 1

� �η (15.111)

and the h-period forecast is

yT�h � βh1yT � βh�11 � βh�21 � ∙ ∙ ∙ � β1 � 1� �

η (15.112)

Note in the limit as h becomes large, βh1 ! 0 and

limh! ∞

yT�h � ηX∞j�0

βj1 � η1 � β1

� μy (15.113)

We see, then, that the forecast tends to the mean of the series as h ! ∞ . This isnot surprising, because the series is stationary. As the lead time h becomes very large,there is essentially no useful information in recent values of the time series yT ; yT�1; . . .that can be used to influence the forecast away from the mean value. Thus, for a verylarge lead time the best forecast is the stationary mean of the series.

Let us now calculate the forecast error for this process. The forecast error hperiods ahead is given by

eT�h � yT�h � yT�h � β1yT�h�1 � η � εT�h � yT�h

eT�h � β21yT�h�2 � β1 � 1� �

η � εT�h � β1εT�h�1 � yT�h

eT�h � βh1yT � βh�11 � βh�21 � ∙ ∙ ∙ � β1 � 1� �

η � εT�h� β1εT�h�1 � ∙ ∙ ∙ � βh�11 εT�1 � yT�h


3GC15 05/15/2014 12:0:21 Page 350

Now substituting the equation of yT�h, we get

eT�h � εT�h � β1εT�h�1 � ∙ ∙ ∙ � βh�11 εT�1 (15.114)

which has a variance

E e2T�h� � � 1 � β21 � β41 � ∙ ∙ ∙ � β2h�21

� �σ2ε (15.115)

Note that this forecast error increases as h becomes larger.Example: Weekly returns of a Sharia-compliant share follows an AR 1� �:

yt � 0:45 � 0:36yt�1 � εt; εt ∼N 0; 1� �We are given yT � 0:85. We make one-period, two-period, and three-period

forecasts.We compute the variance of the forecast error for each forecast as well as theconfidence interval at 95 percent confidence level.

One-period forecast : yT�1 � 0:45 � 0:36 � 0:85 � 0:7475;E e2T�1� � � σ2ε � 1:0

CI 95%� � � yT�1 � z0:025 sT�1 � 0:7475 � 1:96 � 1

Two-period forecast : yT�2 � 0:45 � 0:36 � 0:7475 � 0:7191;E e2T�2� �

� 1 � β21� �

σ2ε � 1:1296

CI 95%� � � yT�2 � z0:025sT�2 � 0:7191 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:1296

p

Three-period forecast : yT�1 � 0:45 � 0:36 � 0:7191 � 0:7089

E e2T�3� � � 1 � β21 � β41

� �σ2ε � 1:1464

CI 95%� � � yT�2 � z0:025sT�2 � 0:7089 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:1464

p

Forecast of the MA(1) Process

Let us examine the simple first-order moving average process MA 1� �:yt � α � εt � θ1εt�1 (15.116)

The one-period forecast is

yT�1 � E yT�1 jyT ; yT�1; . . . ; y1� � � α � θ1εT (15.117)

where εT is the actual residual from the current and most recent observation yT . Bycomparison, the h-period forecast, for h > 1, is just

yT�h � E yT�h jyT ; yT�1; . . . ; y1� � � E α � εT�h � θ1εT�h�1� � � α (15.118)

350 STATISTICS

3GC15 05/15/2014 12:0:21 Page 351

This is also as expected, since theMA 1� � has a memory of only one period. Thus,recent data are of no help inmaking a forecast two ormore periods ahead, and the bestforecast is the mean of the series, α.

The variance of the forecast error forMA 1� � is σ2ε for one-period forecast, and for

h-period forecast, h > 1; it is given by

E e2T�h� � � E yT�h � yT�h

� �2 � E α � εT�h � θ1εT�h�1 � α� �2 � 1 � θ21� �

σ2ε (15.119)

Thus, the forecast error variance is the same for a forecast two periods ahead,three periods ahead, and so on, the forecast confidence intervals would have the samespread.

Example: Weekly returns of a Sharia-compliant stock follows an MA 1� �:yt � 0:77 � εt � 0:55εt�1; εt ∼N 0; 1� �

We propose to forecast yT�20; compute the variance of the forecast error andconstruct a 95 percent confidence interval for the forecast. We find yT�20 � 0:77and E e2T�h

� � � 1 � θ21� �

σ2ε � 1:3025. The confidence interval is CI 95 percent� � �0:77 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1:3025p

.

Forecast of the ARMA(1,1) Process

Let us calculate and examine the forecasts generated by the simplest mixed autor-egressive-moving average process, ARMA 1; 1� �:

yt � β1yt�1 � η � εt � θ1εt�1 (15.120)

The one-period forecast for the ARMA 1; 1� � model is given by

yT�1 � E yT�1 jyT ; yT�1; . . . ; y1� � � E β1yT � η � εT�1 � θ1εT� � � β1yT � η � θ1εT

(15.121)

The two-period forecast is

yT�2 � E β1yT�1 � η � εT�2 � θ1εT�1� � � β1yT�1 � η � β21yT � β1 � 1

� �η � β1θ1εT

(15.122)

Finally, the h-period is

yT�h � β1yT�h�1 � η � βh1yT � βh�11 � ∙ ∙ ∙ � β1 � 1� �

η � βh�11 θ1εT (15.123)

Note that the limiting value of the forecast as h becomes large is again the mean ofthe series,

limh! ∞

yT�h � η1 � β1

� μy (15.124)


3GC15 05/15/2014 12:0:22 Page 352

Examining these forecasts for different lead times, we see that the currentdisturbance helps determine the one-period forecast and in turn serves as a startingpoint from which the remainder of the forecast profile, which is autoregressive incharacter, decays toward the mean η

1�β1.The forecast error one period ahead is given by

eT�1 � yT�1 � yT�1 � β1yT � η � εT�1 � θ1εT � β1yT � η � θ1εT� � � εT�1 (15.125)

E e2T�1� � � E ε2T�1

� � � σ2ε (15.126)

The forecast error two periods ahead is given by

eT�2 � yT�2 � yT�2 � β1yT�1 � η � εT�2 � θ1εT�1� � � β1yT�1 � η

� ��

eT�2 � β1 yT�1 � yT�1� � � εT�2 � θ1εT�1 � β1εT�1 � εT�2 � θ1εT�1

� εT�2 � β1 � θ1� �

εT�1(15.127)

E e2T�2� � � E εT�2 � β1 � θ1

� �εT�1

� �2 � σ2ε � β1 � θ1� �2σ2ε (15.128)

The forecast error three periods ahead is given by

eT�3 � yT�3 � yT�3 � β1yT�2 � η � εT�3 � θ1εT�2� � � β1yT�2 � η

� �� β1 yT�2 � yT�2

� � � εT�3 � θ1εT�2

Note that eT�2 � yT�2 � yT�2 was computed as equal to εT�2 � β1 � θ1� �

εT�1;we have

eT�3 � β1 εT�2 � β1 � θ1� �

εT�1� � � εT�3 � θ1εT�2

� εT�3 � β1 � θ1� �

εT�2 � β1 β1 � θ1� �

εT�1(15.129)

E e2T�3� � � σ2ε � β1 � θ1

� �2σ2ε � β21 β1 � θ1� �2σ2ε (15.130)

Example:We use theARMA 1; 1� � estimated for the returns on the S&P 500 stockindex:

yt � 0:234 � 0:884yt�1 � εt � 0:9908εt�1; σε � 2:304; yT � 1:026504; εT � 1:39772

We propose to make one-period, two-period, and three-period forecasts.Compute the variance of the forecast error, and construct a 95 percent confidenceinterval for the forecast.

One-period forecast: yT�1 � 0:234 � 0:884yT � 0:9908εT � �0:24343; E e2T�1� � �

σ2ε � 5:308416, CI 95%� � � �0:24343 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5:308416

p

Two-period forecast: yT�2 � 0:234 � 0:884yT�1 � 0:018807

E e2T�2� � � σ2ε � β1 � θ1

� �2σ2ε � 5:368965

CI 95%� � � 0:018807 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5:368965

p

352 STATISTICS

3GC15 05/15/2014 12:0:23 Page 353

Three-period forecast : yT�3 � 0:234 � 0:884yT�2 � 0:250625

E e2T�3� � � σ2ε � β1 � θ1

� �2σ2ε � β21 β1 � θ1� �2σ2ε � 5:416281

CI 95%� � � 0:250625 � 1:96 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5:416281

p:

SUMMARY

Islamic finance applies time series models in analyzing financial data and in forecastingexercises. In fact, most financial markets data is reported in stock exchanges and newsmedia websites every five minutes or hour or day in time series forms. Analysts inIslamic finance follow market trends, assess risk and returns, price assets, and makeinvestment decisions based on market information as conveyed by time series data.This chapter provides some basic time series models that analyze market data andguide financial decisions. It covers component movements of a time series, stationarytime series, the autocorrelation function, Wold decomposition of a stationary process,moving average MA� � linear models, autoregressive AR� � linear models, mixedautoregressive-moving average ARMA� � linear models, the partial autocorrelationfunction, and the forecasting based on time series models.

QUESTIONS

1. Download the weekly S&P 500 stock index spanning January 2009 to June 2013;using ordinary least squares method, estimate a trend of the form

ln yt� � � ln�β0� � β1t

2. Define a stationary stochastic process I 0� �. What is the simplest stationaryprocess?

3. Using Microsoft Excel or EViews simulate 200 observations for a white noise.Plot the graph of the simulation.

4. Using the weekly returns of S&P 500 stock index January 2009 to June 2013compute the sample mean, variance, and standard deviation.

5. Using the weekly returns of S&P 500 stock index January 2009 to June 2013compute the sample autocorrelation at lags 1, 2, 3, 4, and 5. Test for thesignificance of each autocorrelation coefficient at 95 percent confidence level.

6. Using EViews with data on the returns on the weekly S&P 500 stock indexJanuary 2009 to June 2013, test for the joint significance of the autocorrelationcoefficients at lags 1, 2, 3, 4, and 5, at 5 percent significance level.

7. The returns of a Sharia-compliant stock is represented by a first-order movingaverage, or MA 1� �, process:

yt � εt � θ1εt�1 � εt � 0:82εt�1; εt ∼N 0; 1� �


3GC15 05/15/2014 12:0:23 Page 354

Compute the parameters of the process: μ, γ0, γ1, ρ1, and ρ2. Using MicrosoftExcel or EViews simulate 200 observations of the process and plot a graph of thesimulation.

8. We consider the returns on the weekly S&P 500 index during January 2009 toJune 2013. Using EViews, estimate anMA 2� � process. Compute μ, γ0, γ1, ρ1, andρ2. Test the significance of ρ1 and ρ2 at 5 percent significance level.

9. We consider the MA 1� � process yt � εt � 0:45εt�1. Invert it into an AR 5� �.10. Returns of a Sharia-compliant share may be represented by an AR 1� � process,

yt � β1yt�1 � εt � 0:85yt�1 � εt with εt ∼N 0; 1� �. Compute γ0, ρ1, and ρ2. Simu-late 200 observations of the process; plot the graph of the simulation.

11. Estimate an AR(2) for the weekly returns on the S&P 500 stock index, January2009 to June 2013.

Compute the parameters of the process: σε, μ; γ0, ρ1, and ρ2. Simulate 200observations of the process.

12. Invert the following AR 1� �, yt � 0:74yt�1 � εt, into MA 4� �.13. Using EViews, estimate an ARMA 1; 1� � for the weekly returns of the S&P 500

stock index, January 2009 to June 2013.

14. Returns of a Sharia-compliant stocks are described by an ARMA 1; 1� �,yt � 0:76yt�1 � εt � 0:42εt�1. Provide a Wold representation with six lags.

15. Using EViews, compute the partial autocorrelation for the weekly returns on theS&P 500 stock index during January 2009 to June 2013 at lags 1, 2, 3, 4, and 5.

16. Using EViews, compute the partial autocorrelation for the weekly returns on theS&P 500 stock index during January 2009 to June 2013 at lags 1 and 2, and byestimating AR 1� �, AR 2� �, and AR 3� �. Compare with partial autocorrelations inQuestion 15.

17. Estimate an AR 1� � for the weekly returns on the S&P 500 stock index duringJanuary 2009 to June 2013. Use yT , the latest observation, and one-period, two-period, and three-period forecasts. Compute the variance of the forecast error foreach forecast as well as the confidence interval at 95 percent confidence level.

18. Returns of a Sharia-compliant share follow an AR 1� � :yt � 0:72 � εt � 0:75εt�1; εt ∼N 0; 1� �

Make a forecast yT�20; compute the variance of the forecast error andconstruct a 95 percent confidence interval for the forecast.

19. Use the ARMA 1; 1� � estimated for the returns on the S&P 500 stock index. Makeone-period, two-period, and three-period forecasts, compute the variance of theforecast error, and construct a 95 percent confidence interval for the forecast.

354 STATISTICS

3GC16 05/15/2014 12:35:26 Page 355

CHAPTER 16Nonstationary Time Series

and Unit-Root Testing

I slamic finance deals with nonstationary time series such as share prices, sales, andcommodity prices. Often, financial time series, such as stock indices, tend to display a

trend over time and do not have a constant mean or variance; they are nonstationary. Ifthefirst differenceof anonstationaryvariable yields a stationaryvariable I 0� �,wedenotethe nonstationary time series by I 1� � andwe call it integrated time series of order one. Atheoretical nonstationary time series I 1� � is the randomwalk.Many nonstationary timeseries resemble the behavior of a random walk and can be approximated by a randomwalk. We cover the definition of random walk with and without drift; we apply therandom walk model to obtain a decomposition of nonstationary time series into trendcomponent, called also permanent component, and a transitory, or cyclical component.Thepermanent componentmeans thatwhen a series is exposed to a shock, the impact ofthe shock will persist and does not vanish. An example of such shock is a technologicalshock such as the invention of automobiles, airplanes, computers, and the Internet.These shocks increase productivity and remain forever.

We cover the nonstationarity test, called also unit root-test, to ascertain if a timeseries is nonstationary or stationary. The test is called the Dickey-Fuller test. We testnonstationary in three settings: (i) a random walk against a simple stationary series,(ii) a random walk against a stationary series with an intercept, and (iii) a randomwalk against a stationary series with an intercept and a time trend. A series may betrend stationary, which means its nonstationarity arises from the presence of adeterministic trend; once the latter is removed, the series becomes stationary. Wecover the notion of augmented Dickey-Fuller test to remedy for the presence of serialcorrelation in the error term.

THE RANDOM WALK

Let us take a simple example of a nonstationary process, the random walk, which isnothing more than an AR(1) with unit coefficient:

yt � yt�1 � εt (16.1)

The random variable εt is called white noise, with E εt� � � 0, E ε2t� � � σ2ε , and

E εtεs� � � 0 for t ≠ s. A random walk is a nonstationary process denoted as I 1� �

355

3GC16 05/15/2014 12:35:27 Page 356

because the first difference Δyt � yt � yt�1 is I 0� �, that is, each successive change in yt,Δyt � εt, is drawn independently from a probability distribution with mean 0 andvariance σ2ε . A random walk process could be generated by successive flips of a coinwhere a head receives a value of +1 and a tail receives a value of �1. The randomwalkis not covariance stationary, because the AR(1) coefficient is not less than one. Inparticular, it does not display mean reversion; in contrast to a stationary AR(1), itwanders up and down randomly, as its name suggests, with no tendency to return toany particular point. Although the randomwalk is ill-behaved, its first difference is theultimate well-behaved series: zero-mean white noise.

Now let us consider a random walk with drift:

yt � δ � yt�1 � εt (16.2)

Note that the random walk with drift is effectively a model of trend, because onaverage it grows each period by a drift δ. Thus the drift parameter plays the same roleas the slope parameter of a linear deterministic trend. We call the random walk with adrift a model of stochastic trend because the trend is driven by stochastic shocks, incontrast to deterministic trend.

Let us study the properties of random walks in greater detail. The random walk is

yt � yt�1 � εt (16.3)

Assuming the process starts at time 0 with value y0, we can write it as

yt � y0 �Xt

i�1 εi (16.4)

E yt� � � y0 and Var yt

� � � tσ2. In particular, note that limt! ∞ Var yt� � � ∞ so

that the variance grows continuously rather than converging to some finiteunconditional variance. Now consider the random walk with drift:

yt � δ � yt�1 � εt (16.5)

Assuming the process starts at time 0 with value y0, we can write it as

yt � y0 � tδ �Xt

i�1 εi (16.6)

E yt� � � y0 � tδ and Var yt

� � � tσ2. The random walk is decomposed into adeterministic trend y0 � tδ and a stochastic trend

Pti�1 εi.

Just as white noise is the simplest I(0) process, the randomwalk is the simplest I(1)process. And just as I(0) processes with richer dynamics than white noise can beconstructed by transforming white noise, so, too, can I(1) processes with richerdynamics than the random walk be obtained by transforming the random walk.

Example: We simulate 100 observations of a random walk with no drift; wesimulate also 100 observations of a random walk with a drift δ � 0:2. We display thesimulations in Figure 16.1.

Example: In Figure 16.2,weportray theweekly S&P500 stock indexduringMarch2009 to May 2013. We observe that its behavior depicts a random walk with a drift.

356 STATISTICS

3GC16 05/15/2014 12:35:28 Page 359

DECOMPOSITION OF A NONSTATIONARY TIME SERIES

We consider a nonstationary time series yt integrated of first-order I 1� �, meaningthat its first difference is stationary. We apply theWold decomposition theorem toΔytto obtain the following representation,*

Δyt � μ � C L� �εt; εt ∼ iid 0; σ2ε� �

(16.7)

where C L� � is a polynomial of order q in the lag operator that describes the MAprocess of Δyt:

C L� � � 1 � γ1L � γ2L2 � ∙ ∙ ∙ � γqL

q (16.8)

We note that C L� � may be decomposed as

C L� � � C 1� � � C* L� � 1 � L� � (16.9)

where C 1� � � 1 � γ1 � γ2 � ∙ ∙ ∙ � γq and C* L� � is a polynomial of order q � 1.By applying the decomposition of C L� �, we write Δyt as

Δyt � μ � C 1� � � C* L� � 1 � L� �� εt � μ � C 1� �εt � C* L� �Δεt (16.10)

By solving the difference equation we find

yt � y0 � μt � C 1� �zt � C* L� �εt (16.11)

where zt � Pti�1 εt and Δzt � εt. The variable zt is called a stochastic trend. The

component μt is a deterministic trend; and C* L� �εt is a temporary shock that affects ytaccording to a time lag defined by the lag order ofC* L� �, which is q � 1. For instance ifq � 1 � 2, then yt is affected by εt; εt�1, and εt�2 only. The trend in yt is made of adeterministic trend and a stochastic trend.

Example: Consider the process Δyt � C L� �εt � εt � γεt�1; 0 < γ < 1.In this case C L� � � 1 � γL, C 1� � � 1 � γ, and C* L� � � C L� ��C 1� �

1�L � 1�γL�1�γ1�L � �γ.

The decomposition of yt yields the following process:

yt � C 1� �zt � C* L� �εt � 1 � γ� �zt � γεt

Example: Consider the weekly S&P 500 stock index during March 2009 to May2013. We want to decompose it into a trend component and a transitory component.

*The lag operator L is used to represent lagged variables. For instance, εt�1 is written as Lεt , εt�2is written as L2εt, and εt�q is written as Lqεt . The notation C L� � designates the lag polynomial1 � γ1L � γ2L

2 � ∙ ∙ ∙ γqLq, where q is the order the moving average MA� � lag and γ1; γ2; . . . :; γq

are the coefficients of the polynomial. Hence,

C 1� � � 1 � γ1 � γ2 � ∙ ∙ ∙ � γq

Nonstationary Time Series and Unit-Root Testing 359

3GC16 05/15/2014 12:35:28 Page 360

We estimate an MA 2� � of the weekly difference of the form Δyt �μ � εt � γ1εt�1 � γ2εt�2.

We observe that C L� � � 1 � γ1L � γ2L2, C 1� � � 1 � γ1 � γ2,

C* L� � � C L� � � C 1� �1 � L

� 1 � γ1L � γ2L2 � 1 � γ1 � γ2

1 � L

� γ1 L � 1� � � γ2 L � 1� � L � 1� �1 � L

� �γ1 � γ2 � γ2L

The estimation yields

Δyt � 4:0327 � εt � 0:0973εt�1 � 0:0052εt�2C 1� � � 1 � γ1 � γ2 � 1 � 0:0973 � 0:0052 � 0:1025

C* L� � � �γ1 � γ2 � γ2L � 0:1025 � 0:0052L

The solution is

yt � y0 � μt � C 1� �zt � C* L� �εt � y0 � 4:0327t � 0:8975zt � 0:1025εt � 0:0052εt�1

We show a graph of the decomposition of the weekly S&P 500 stock index(Figure 16.3). We note a trend component y0 � 4:0327t, a stochastic trend component0:8975zt, and small transitory component 0:1025εt � 0:0052εt�1.

FORECASTING A RANDOM WALK

Suppose we wanted to make a forecast for a random walk with no drift. The one-period forecast is given by

yT�1 � E yT�1 jyT ; . . . ; y1� � � yT � E εT�1� � � yT (16.12)

The forecast two periods ahead is

yT�2 � E yT�2 jyT ; . . . ; y1� � � E yT�1 � εT�2� � � E yT � εT�1 � εT�2

� � � yT (16.13)

Similarly, the forecast h periods ahead is yT . Although the forecast yT�1 will be thesame no matter how large h is, the variance of the forecast error will grow as hbecomes larger. For a one-period forecast the error is given by

e1 � yT�1 � yT�1 � yT � εT�1 � yT � εT�1 (16.14)

And its variance is just E ε2T�1� � � σ2ε . For the two-period forecast,

e2 � yT�2 � yT�2 � yT � εT�1 � εT�2 � yT � εT�1 � εT�2 (16.15)

And its variance is

E εT�1 � εT�2� �2� � � E ε2T�1� � � E ε2T�2

� � � 2E εT�1εT�2� � (16.16)

360 STATISTICS

3GC16 05/15/2014 12:35:29 Page 362

Since εT�1 and εT�2 are independent, the term E εT�1εT�2� � � 0 and the errorvariance is 2σ2ε . Similarly, for h-period forecast, the variance of forecast error is hσ2ε .Thus, the standard error of forecast increases with the square root of h. We can thusobtain confidence intervals for our forecasts, and these intervals will become wider asthe forecast horizon increases. This is illustrated in Figure 16.4. Note that the forecastsare all equal to the last observation yT , but the confidence intervals represented by onestandard error in the forecast error increase as the square root of h increases.

MEANING AND IMPLICATIONS OF NONSTATIONARY PROCESSES

Assume a trending variable Yt, upward or downward; we want to model its behaviorover time. We may assume two possible models:

Model 1: Yt � Y0eβt � ηt (16.17)

Model 2: Yt � Yt�1eβ � ηt (16.18)

We assume the shock ηt to be a white noise, ηt ∼ iid 0; σ2η� �

. If we set yt � log Yt,

Models 1 and 2 may be rewritten as

Model 1: yt � y0 � βt � εt (16.19)

Model 2: yt � yt�1 � β � εt (16.20)

where εt ∼ iid 0; σ2ε� �

; if we solve Model 2 backward, we obtain

Model 2: yt � y0 � βt �Xt

i�1 εi (16.21)

In Model 1, an error term εt affects what is happening in the current time periodbut has no effect on what happens in succeeding time periods. In contrast, Model 2 is atypical random walk; it is called unit-root process because the coefficient of yt�1 isunity. An error term εt affects what happens in the current time period and also inevery succeeding time period. In Model 2, a shock to yt persists, but in Model 1, it

Forecastconfidence interval

Time0

yt

yT

T T + 1 T + 2

FIGURE 16.4 Forecasting a RandomWalk

362 STATISTICS

3GC16 05/15/2014 12:35:29 Page 363

disappears after the current time period. This has profound implications for financetheory: are shocks permanent or transitory? InModel 1, yt is not stationary. It followsa deterministic trend βt. However if we make the transformation, xt � yt � y0 � βt, weobtain a stationary model of the form

xt � εt (16.22)

If we take the first difference of Model 2, we obtain a stationary process:

Δyt � yt � yt�1 � β � εt (16.23)

Model 1 is called trend stationary; by removing the trend we obtain a stationaryprocess. Model 2 is called difference stationary; by taking first difference we obtain astationary process. In sum, many economic time series are trending. It is important todistinguish between two cases:

i. A stationary process with a deterministic trend and where shocks have transitoryeffects.

ii. A process with a stochastic trend where shocks have permanent effects.

We may combine Models 1 and 2. Suppose we believe that variable yt, which hasbeen growing over time, can be described by the following equation:

yt � α � δt � βyt�1 � εt (16.24)

One possibility is that yt has been growing because it has a positive trend δ > 0� �but would be stationary after detrending, that is, β < 1: In this case yt could be used ina regression, and all the results and tests of linear regression would apply. Anotherpossibility is that yt has been growing because it follows a randomwalkwith a positivedrift (i.e., α > 0; δ � 0; and β � 1). In this case one would want to work with Δyt.Detrending would not make the series stationary, and the inclusion of yt in aregression (even if detrended) could lead to spurious results.

In order to decide which model represents a trending variable, we need to studynonstationarity based on unit-root test. More specifically, we want to know if shockshave permanent or transitory effects. The persistence of shocks will be infinite fornonstationary series. For instance, a technical change such as new computer or newmedication will improve productivity and have lasting effect. In forecasting, we wantto know if the process has an attractor. For unit-root processes many asymptoticdistributions change. If the variables in the regression model are trending over time,that is, they are not stationary, we obtain spurious regressions. A regression of one onthe other could have a high R2 even if the two are totally unrelated. The standardassumptions for asymptotic analysis will not be valid. In other words, the usual “t-ratios”will not follow a t distribution, so we cannot validly undertake hypothesis testsabout the regression parameters. Unit-root tests generally have nonstandard andnonnormal asymptotic distributions. These distributions are functions of standardBrownian motions, and do not have convenient closed form expressions. Conse-quently, critical values must be calculated using simulationmethods. The distributionsare affected by the inclusion of deterministic terms, for example, constant, time trend,


3GC16 05/15/2014 12:35:30 Page 364

dummy variables, and so different sets of critical values must be used for testregressions with different deterministic terms.

DICKEY-FULLER UNIT-ROOT TESTS

Let us formulate the unit-root test with the simple AR 1� � process:yt � βyt�1 � εt; εt ∼N 0; σ2

� �(16.25)

We want to test H0 : unitroot : β � 1: yt � yt�1 � εt against the hypothesisH1 : stationary process : β < 1: yt � βyt�1 � εt, βj j < 1.

We have a left-tail test. We can regress yt on yt�1 and then use the standard t-testfor testing β � 1.

τ � β � 1

SE β� � (16.26)

Where SE β� �

is the standard error of β. Note the statistic τ is not the t-statisticcomputed by the regression software. The standard t-statistic is for the null of zerocoefficient, whereas τ is the t-statistic for a unit coefficient. A simple trick, however,makes standard software print τ automatically. Simply rewrite the first-order autor-egression as

yt � yt�1 � β � 1� �yt�1 � εt (16.27)

or

Δyt � πyt�1 � εt (16.28)

where π � β � 1. The null hypothesis becomesH0 : π � 0 againstH1 : π < 0. Thus τ isthe usual t-statistic in a regression of the first difference of y on the first lag of y,

τ � π

SE π� � (16.29)

A key result is that, in the unit-root case, τ does not have the t distribution.Instead, it has a special distribution called theDickey-Fuller distribution. Fuller (1976)presented tables of the percentage points of the distribution of τ, which we call theDickey-Fuller (DF) statistic, under the null hypothesis of a unit root. Statisticalpackages such as EViews, Pc Give, and Rats run instantaneously unit-root tests.

Example:We test the hypothesis of a random walk for the weekly S&P 500 stockindex during March 2009 to May 2013 at a confidence level of 95 percent.

We run a regression of the form Δyt � πyt�1 � εt.

We findΔyt � 0:0003yt�1 and τ � π

SE π� � � 2:007. The Dickey-Fuller critical value

at 5 percent is �1.94. We fail to reject the hypothesis of a unit root, π � 0.

364 STATISTICS

3GC16 05/15/2014 12:35:31 Page 365

Thus far we have shown how to test the null hypothesis of a randomwalk with nodrift against the alternative of a zero-mean, covariance stationary AR 1� �. Now weallow for a nonzero mean, μ, under the alternative hypothesis, which is of potentialimportance because business and economic data can rarely be assumed to have zero-mean. Under the alternative hypothesis, the process becomes a covariance stationaryAR 1� � process in deviations from the mean,

yt � μ � β yt�1 � μ� � � εt (16.30)

which we can rewrite as

yt � α � βyt�1 � εt (16.31)

Δyt � α � πyt�1 � εt (16.32)

where α � μ 1 � β� �. We want to test H0 : unitroot : β � 1: yt � yt�1 � εt against thehypothesis: H1 : stationary process with drift : β < 1: yt � α � βyt�1 � εt, βj j < 1.

If we knew μ, we could simply center the data and proceed as before. In practice, μmust be estimated along with other parameters. Although α vanishes under the unitroot null hypothesis of β � 1, it is nevertheless present under the alternative hypothe-sis, and so we include an intercept in the regression. The distribution of the

corresponding Dickey-Fuller statistic, τμ � π

SE π� �, has been tabulated under the null

hypothesis of α � 0; β � 1� �.Example:We test the hypothesis of a random walk for the weekly S&P 500 stock

index during March 2009 to May 2013 against the hypothesis of a stationary processwith drift at a confidence level of 95 percent.

We run a regression of the form Δyt � α � πyt�1 � εt.

We find Δyt � �0:0019yt�1 � 2:826, τμ � π

SE π� � � �2:21297. The Dickey-Fuller

critical value at 5 percent is �2.8747. We fail to reject the hypothesis of a unit root:π � 0.

Finally, let us allow for deterministic linear trend under the alternative hypothesisby writing the AR 1� � in deviations from a linear trend

yt � a � bt � β yt�1 � a � b t � 1� �� εt (16.33)

or

yt � α � δt � βyt�1 � εt (16.34)

where α � a 1 � β� � � bβ and δ � b 1 � β� �. Under the unit-root hypothesis that β � 1,we have a random walk with drift:

yt � b � yt�1 � εt (16.35)

We want to testH0 : unitroot : β � 1: yt � yt�1 � εt against the hypothesis: H1 : stationary

process with drift and trend : β < 1

yt � α � δt � βyt�1 � εt; βj j < 1 (16.36)


3GC16 05/15/2014 12:35:31 Page 366

Hence the specification of the AR 1� � under the deterministic trend alternativehypothesis includes both the intercept and the trend in the regression. The randomwalk with drift is a null hypothesis that frequently arises in economic applications;stationary deviations from linear trend are a natural alternative. The distribution ofthe DF statistic ττ, which allows for linear trend under the alternative hypothesis, hasbeen tabulated under the unit root hypothesis by Fuller.

Example:We test the hypothesis of a random walk with drift for the weekly S&P500 stock index during March 2009 to May 2013 against the alternative of adeterministic trend at a confidence level of 95 percent.

We run a regression of the form Δyt � α � δt � πyt�1 � εt.

We find Δyt � �0:009yt�1 � 0:022t � 9:3498, ττ � π

SE π� � � �3:91.The Dickey-Fuller critical value at 5 percent is�3.42.We reject the hypothesis of a

unit-root: π � 0. The S&P 500 stock index is driven by a deterministic trend. Shockshave only a transitory effect.

In sum, the basic objective of the unit root test is to test the null hypothesis thatβ � 1 in

yt � βyt�1 � εt (16.37)

against the one-sided alternative β < 1, so we haveH0 the time series contains a unit-root versus H1 the series is stationary. We run a regression of the form

Δyt � πyt�1 � εt (16.38)

so that a test of β � 1 is equivalent to a test of π � 0, π � β � 1. We may perform theDickey-Fuller tests in three possible setups. The nullH0 and the alternativeH1 modelsin each case are:

i. H0: yt � yt�1 � εt; H1 : yt � βyt�1 � εt, β < 1.This is a test for a randomwalk against a stationary autoregressive process of

order one (AR(1)).ii. H0: yt � yt�1 � εt; H1 : yt � α � βyt�1 � εt, β < 1.

This is a test for a random walk against a stationary AR(1) with drift.iii. H0: yt � yt�1 � εt; H1 : yt � α � δt � βyt�1 � εt, β < 1.

This is a test for a random walk against a stationary AR(1) with drift and atime trend.

We can write the null Δyt � εt where Δyt � yt � yt�1 and the alternatives may beexpressed as

yt � α � δt � πyt�1 � εt (16.39)

π � 1 � β; we have in case (i), α � δ � 0; in case (ii), α ≠ 0; δ � 0; and in case(iii), α ≠ 0; δ ≠ 0. In each case, the tests are based on the t-ratio on the yt�1 termin the estimated regression of Δyt�1 on yt�1, plus a constant in case (ii) and a constantand trend in case (iii). The test statistic does not follow the usual t distribution underthe null, since the null is one of nonstationarity, but rather follows a nonstandarddistribution. Critical values are derived from Monte Carlo experiments in, for

366 STATISTICS

3GC16 05/15/2014 12:35:32 Page 367

example, Fuller (1976). Software statistical packages run unit-tests with a need forconsulting Fuller’s tables.

THE AUGMENTED DICKEY-FULLER TEST (ADF)

The preceding tests were for a unit root process against an AR 1� �. In fact, an AR 1� �may not be an adequate representation of the series andmay result in serial correlationof the error terms εt. To remedy the serial correlation we increase the number of lags.The general autoregressive AR p� � model is

yt � α � δt � βyt�1 �Xp�1

j�1 θjΔyt�j � εt (16.40)

The null H0 : β � 1 is tested against H1: β < 1. We may consider three possiblealternative models: (i) stationary AR p� �, α � δ � 0; (ii) stationary AR p� � with driftδ � 0; and (iii) stationary process stationary AR p� � with drift and trend. These tests,called the augmented Dickey-Fuller (ADF) tests, involve estimating the equation:

Δyt � α � δt � πyt�1 � θ1Δyt�1 � θ2Δyt�2� ∙ ∙ ∙�θp�1Δyt�p�1 � εt (16.41)

The null is H0 : π � 0 and the alternative is: H1: π < 0, where π � 1 � β. The

statistics of the test are τ, τμ, and ττ and are given in each model byπ

SE π� �.Example: Using an AR 3� �, we test for unit-root random walk for the weekly S&P

500 stock index duringMarch 2009 toMay 2013 at a confidence level of 95 percent intwo models. We test the hypothesis of a random walk against (i) a stationary processwith drift and (ii) a stationary process with drift and trend.

i. We run a regression of the form

yt � α � βyt�1 � ρ1yt�2 � ρ2yt�3 � εt

The model may be written as

Δyt � α � πyt�1 � θ1Δyt�1 � θ2Δyt�2 � εt

The estimated model using OLS is

Δyt � 3:164 � 0:0022yt�1 � 0:082Δyt�1 � 0:043Δyt�2The t-statistic is τμ � �2:41. The critical ADF value is�2.87. Hence, we fail to

reject the random walk hypothesis despite the use of an AR 3� �.ii. We run a regression of the form

yt � α � δt � βyt�1 � ρ1yt�2 � ρ2yt�3 � εt

The model may be written as

Δyt � α � δt � πyt�1 � θ1Δyt�1 � θ2Δyt�2 � εt


3GC16 05/15/2014 12:35:32 Page 368

The estimated model using OLS is

Δyt � 10:126 � 0:023t � 0:0099yt�1 � 0:046Δyt�1 � 0:079Δyt�2

The t-statistic is ττ � �4:05. The critical ADF value is –3.43. Hence, we rejectthe random walk hypothesis.

SUMMARY

Islamic finance deals with nonstationary time series. In fact, most of financial data isnonstationary. Portfolio managers need to determine the driving forces of marketdata. Islamic asset management companies use nonstationary time series analysis tofollow market trends and trade stocks based on predictable trends.

This chapter covered the notions of random walk, the decomposition of anonstationary time series, the forecasting of a random walk, the meaning andimplications of nonstationary processes, the Dickey-Fuller unit-root tests, and theAugmented Dickey-Fuller test (ADF).

QUESTIONS

1. Simulate 200 observations for a random walk with drift δ = 0.27. Choose y0 � 0.Plot the graph of the random walk. You want to forecast a random walk 10periods ahead. Compute the forecast value and the variance of the forecast error.

2. Download data for the weekly S&P 500 stock index during January 2009 toDecember 2013. Test the hypothesis of a random walk against the alternatives ofa stationary process, a stationary process with drift, and a stationary process witha drift and time trend.

3. Using anAR 3� �, test for unit root walk for the weekly S&P 500 stock index duringJanuary 2009 to December 2013 at a confidence level of 95 percent. Consider thehypothesis of a random walk against (i) a stationary process with drift and (ii) astationary process with drift and trend.

368 STATISTICS

3GC17 05/15/2014 12:39:56 Page 369

CHAPTER 17Vector Autoregressive

Analysis (VAR)

V ector Autoregressive Model (VAR) is a statistical model used to capture the linearinterdependencies among multiple time series. VAR models generalize the uni-

variate autoregression (AR) models by allowing for more than one evolving variable.For instance, we may model the stock price index as a univariate ARMA and omit theinfluence of any other variable that may affect the stock index. Financial analysts,however, want to test relationships between the stock price index and other relevantvariables such as the interest rate and assess causality between two or more variables.It may happen that variable x causes y, but y does not cause x. In this instance, x isexogenous and y is endogenous. If there is two-way causality between x and y, the twovariables are endogenous to each other.

To motivate analyzing one time series in relation to another pertinent variable weplot the S&P 500 stock index and the interest rate, measured by the U.S. federal fundsrate during 1970 to 2013 in Figure 17.1. Close scrutiny indicates that stock prices maybe influenced by the interest rate. Low interest rates increase the leverage of specula-tors and leads to a stock market boom. Moreover, asset prices are inversely related tointerest rates. In contrast, high interest rates reduce the leverage of speculators.Moreover, asset prices are lowered by high interest rates. It may happen that stockprices influence interest rates. For instance, the central bank puts interest rates at near-zero levels to boost stock prices following a speculative crash. This practice is knownas unorthodox money policy and is used in many industrial countries with developedstockmarkets. It may also happen that the central bank lowers interest rates to preventany downward adjustment of stock prices.

Figure 17.2 and Figure 17.3 illustrate the behavior of monthly gold prices andcrude oil prices in relation to monthly interest rates during 1970 to 2012, respectively.An inspection of the two figures indicates that low interest rates may lead to acommodity boom, and inversely.

FORMULATION OF THE VAR

A VAR model provides the tool to study the dynamics shown in Figures 17.1, 17.2,and 17.3 and analyze interaction between two or more variables. It describes theevolution of a set of n variables (called endogenous variables) over the same sampleperiod t � 1; . . . ;T� � as a linear function of only their past values. The variables are

369

3GC17 05/15/2014 12:39:56 Page 373

collected in a n � 1 vector yt, which has as the ith element, yit, the time t observation ofthe ith variable. For example, if the ith variable is the S&P 500 stock index, then yit isthe value of S&P 500 index at time t.

A pth order VAR, denoted VAR p� �, is

yt � μ � A1yt�1 � A2yt�2 � ∙ ∙ ∙ � Apyt�p � εt (17.1)

where μ is a n � 1 vector of constants (intercepts), Ai is a time-invariant n � n matrix,and εt is a n � 1 vector of error terms satisfying

i. E εt� � � 0, every error term has mean zero;ii. E εtεt

� � � Ω, the contemporaneous covariance matrix of error terms Ω is a n � npositive-semi-definite matrix; and

iii. E εtεt�j� �

� 0, for any nonzero j there is no correlation across time; in particular,

no serial correlation in individual error terms.

A pth order VAR is also called a VAR with p lags. The process of choosing themaximum lag p in the VAR model requires special attention because inference isdependent on correctness of the selected lag order.

Note that all variables have to be of the same order of integration. The followingcases are distinct:

■ All the variables are I(0) (stationary): we have the standard case, that is, a VAR inlevel.

■ All the variables are I(d) (nonstationary) with d > 0. If the variables are co-integrated, the error correction term has to be included in the VAR. The modelbecomes a vector error correction model (VECM), which can be seen as arestricted VAR. If the variables are not co-integrated, the variables have firstto be differenced d times and one has a VAR in difference.

The key point is that, in contrast to the univariate case, vector autoregressionsallow for cross-variable dynamics. Each variable is related not only to its ownpast, but also to the past of all other variables in the system. In a two-variableVAR(1), for example, we have two equations, one for each variable: y1 and y2.We write

y1t � α11y1t�1 � α12y2t�1 � ε1t (17.2)

y2t � α21y1t�1 � α22y2t�1 � ε2t (17.3)

Each variable depends on one lag of the other variable in addition to one lag ofitself; that is, one obvious source of multivariate interaction captured by the VAR thatmay be useful for forecasting. In addition, the disturbances may be correlated, so thatwhen one equation is shocked, the other will typically be shocked as well, which isanother type of multivariate interaction that univariate models lack. We summarize

Vector Autoregressive Analysis (VAR) 373

3GC17 05/15/2014 12:39:57 Page 374

the disturbance variance-covariance structure as

ε1t∼iid 0; σ21� �

; ε2t∼iid 0; σ22� �

; cov ε1t; ε2t� � � σ12 (17.4)

The innovations could be uncorrelated, which occurs when σ12 � 0, but they neednot be. VARs are very easy to estimate because we need to run only regressions. That isone reason VARs are so popular.

Example: Using EViews, we estimate a VAR(1) for the monthly returns of theS&P 500 stock price index �y1t� and the interest rate y2t

� �, measured by the U.S.

federal funds rate, during January 2000 to September 2012:

y1t � 0:12y1t�1 � 0:20y2t�1 � 0:54

y2t � 0:009y1t�1 � 0:99y2t�1 � 0:02

We want to determine the lag of a VAR. A VAR with a long lag causes overfittingand may not be appropriate for analysis or projection. A VAR with short-lag may notcapture the interdependent dynamics of the variables. The information criteria areoften used as a guide in model selection. The notion of an information criterion is toprovide a measure of information that strikes a balance between this measure ofgoodness of fit and parsimonious specification of the VAR. The various informationcriteria differ in how to strike this balance. The basic information criteria are given by

Akaike info criterion�AIC� : �2 ℓT

� �� 2 k=T� � (17.5)

Schwarz criterion�SC� : �2 ℓT

� �� k log T� �=T� � (17.6)

Where ℓ is the log of the likelihood function, k is the number of the parametersestimated using T observations. The various information criteria are all based on �2times the average log likelihood function, adjusted by a penalty function. For systemsof equations, where applicable, the information criteria are computed using the fullsystem log likelihood. The log likelihood value is computed assuming a multivariatenormal (Gaussian) distribution as

ℓ � �T � n2

1 � log 2π� � � T2

Ω�� (17.7)

where Ω�� is the determinant of the residuals covariance matrix, and n is the number of

equations.When we use the information criteria as a model selection guide, we select themodel with the smallest information criterion. The information criterion has been widelyused in time series analysis to determine the appropriate length of the lag in VARmodels.

Example: Using EViews, we estimate VAR(1), VAR(2), and VAR(3) for themonthly returns of the S&P 500 stock index and interest rates during January 2000 toSeptember 2012. We report the respective information criteria in Table 17.1. Weobserve that a VAR(2) has smaller Akaike and Schwartz criterion and is preferable toVAR(1) or VAR(3).

374 STATISTICS

3GC17 05/15/2014 12:40:0 Page 375

FORECASTING WITH VAR

We construct VAR forecasts in a way that precisely parallels the univariate case. Wecan construct one-step ahead point forecasts immediately, because all variables arelagged by one period. Armed with one-step ahead forecasts, we can construct the two-step-ahead forecasts, from which we can construct the three-step-ahead forecast, andso on in the usual way, following Wold’s chain rule.

Example: We assume y1T � 2:42, y2T � 0:13, and the following VAR:

y1t � 0:80y1t�1 � 0:14y2t�1y2t � 0:017y1t�1 � 0:65y2t�1

In Table 17.2, we make a five period-ahead forecast:

THE IMPULSE RESPONSE FUNCTION

The impulse-response function is a device that helps us to learn about the dynamicproperties of VARs that are to be used by forecasters. We introduce it in the univariatecontext, and then in VARs. The question is simple: How does a unit innovation to aseries affect it, now and in the future? To answer this question, we simply read off thecoefficients in the moving average (MA) representation of the process. We are used tonormalize the coefficient on εt to unity inMA representations, but we don’t have to doso; more generally, we can write

yt � εt � b1εt�1 � b2εt�2 � b3εt�3 � ∙ ∙ ∙ (17.8)

εt∼ 0; σ2� �

We observe that if there is a shock εt at time t, the contemporaneous effect of thisshock on yt is εt; the effect on yt�1 is b1εt; the effect on yt�2 is b2εt; the effect on yt�3 isb3εt. The coefficients 1; b1; b2; b3; . . .f g are called the impulse response of the future

TABLE 17.1 Comparing Information Criteria

VAR(1) VAR(2) VAR(3)

Akaike information criterion 5.47 4.77 4.81Schwarz criterion 5.59 4.97 5.08

TABLE 17.2 Forecast Using VAR

T T � 1 T � 2 T � 3 T � 4 T � 5

y1 2.42 2.42 1.92 1.52 1.20 0.94y2 0.13 0.13 0.13 0.11 0.10 0.09


3GC17 05/15/2014 12:40:1 Page 376

values of yt to a shock εt at time t. They describe the complete dynamic response of y toa shock ε. Since yt is stationary, the coefficients bi tend to zero as i ! ∞.

By multiplying and dividing each term by the constant σ we may write the MArepresentation as

yt � σεtσ

� �� σb1

εt�1σ

� �� σb2

εt�2σ

� �� σb3

εt�3σ

� �� ∙ ∙ ∙

� σεt � σb1εt�1 � σb2εt�2 � σb3εt�3 � ∙ ∙ ∙(17.9)

where εt � εtσ and Var εt

� � � Var εt� �σ2 � σ2

σ2 � 1; hence, εt∼iid 0; 1� �.We have converted shocks from εt to εt. Since εt � σεt, a unit shock to εt

corresponds to one standard deviation shock σ to εt.Example: We run an AR(2) for the monthly returns of the S&P 500 stock index

during January 2000 to September 2012. We show in Figure 17.4 the impulseresponse to one standard deviation shock. We observe that weekly returns respondpositively by amultiple of 4.5 standard deviations; the effect dies off completely withinfive months.

Now we can consider the multivariate case. The idea is the same, but there aremore shocks to track. The key question is: How does a unit shock to εt affect yt nowand in the future taking into account the VAR relationships? Consider for example thebivariate VAR(1),

y1t � α11y1t�1 � α12y2t�1 � ε1t (17.10)

y2t � α21y1t�1 � α22y2t�1 � ε2t (17.11)

ε1t∼iid 0; σ21� �

; ε2t∼iid 0; σ22� �

; cov ε1t; ε2t� � � σ12 (17.12)

written as

yt � Ayt�1 � εt;E εt� � � 0; and E εtεt� � � Ω (17.13)

yt � A2yt�2 � Aεt�1 � εt � A3yt�3 � A2εt�2 � Aεt�1 � εt (17.14)

–2

–1

0

1

2

3

4

5

6

1 2 3 4 5 6 7 8 9

FIGURE 17.4 Response of Stock Returns to Standardized Own ShockInvention

376 STATISTICS

3GC17 05/15/2014 12:40:1 Page 377

We may continue the back substitution to obtain anMA representation assumingthe process is invertible:

yt � εt � C1εt�1 � C2εt�2 � C3εt�3 � ∙ ∙ ∙ : (17.15)

EachCj is a square matrix 2 � 2� �. In the case of two variables: y1t and y2t, theMArepresentation may be written as

y1t � ε1t � c11ε1t�1 � c12ε2t�1 � ∙ ∙ ∙ (17.16)

y2t � ε2t � c21ε1t�1 � c22ε2t�1 � ∙ ∙ ∙ (17.17)

Each yit variable is expressed in terms of its error terms and the error terms of theother VAR variables. Contrary to the univariate case where we have one singlerandom shock, in the bivariate or multivariate case we have many shocks that arecorrelated. Hence, if we shock the VAR by ε1t, the other shock ε2t changes simulta-neously and the VAR is subject to the simultaneous effect of two shocks. Hence, wecannot isolate the effect of one single shock because of the covariance of the shocks isnot zero. We need therefore to operate a transformation of the shocks to obtainnoncorrelated shocks.

We exploit the properties of the covariance matrix Ω; the latter is a symmetricmatrix that can be subjected to a Cholesky decomposition and be written as

Ω � PP´ or P�1ΩP´�1 � I (17.18)

where P is a lower triangular matrix and I is the identity matrix, all of the samedimensions as Ω. If we make the transformation

ηt � P�1εt (17.19)

we notice immediately that

E ηt� � � 0 and E ηtηt

� � � P�1εtεtP�1´ � P�1ΩP�1´ � P�1ΩP´�1 � I (17.20)

Obviously the shocks η1t, and η2t have a unit standard deviation and are notcorrelated since cov η1t; η2t

� � � 0.The MA representation can be rewritten as

yt � PP�1εt � C1PP�1εt�1 � C2PP�1εt�2 � C3PP�1εt�3 � ∙ ∙ ∙

� Pηt � C1Pηt�1 � C2Pηt�2 � C3Pηt�3 � ∙ ∙ ∙(17.21)

After normalizing the system for a given ordering, say y1 is first, we compute foursets of impulse-response functions for the bivariate model:

1. Response of y1 to a unit normalized innovation to y1, b011; b111; b

211; . . .

n o;

2. Response of y1 to a unit normalized innovation to y2, b112; b212; . . .

n o;


222; . . .

n o; and


221; . . .

n o.


3GC17 05/15/2014 12:40:2 Page 378

Typically, we examine the set of impulse-response functions graphically.Example: We run a VAR(2) for the monthly returns of the S&P 500 stock index

and interest rates during January 2000 to September 2012. We display in Figure 17.5the impulse response of the stock returns to own shock and to interest rate shock.

VARIANCE DECOMPOSITION

Another way of characterizing the dynamics associated with VARs, closely related toimpulse-response functions, is the variance decomposition. Variance decompositionhas an immediate link to forecasting. It answers the question: Howmuch of the h-step-ahead forecast error variance of variable i is explained by innovation to variable j, forh � 1; 2; . . . ? As with the impulse-response functions, we make a separate graph forevery i; j� � pair. Impulse response functions and the variance decomposition providethe same information, although they do so in different ways. For that reason it is notstrictly necessary to present both, and impulse response analysis has gained greaterpopularity.

The forecast error of one period-ahead forecast is

eT�1 � yT�1 � E yT�1 jyT� � �PηT�1 � C1PηT � C2PηT�1 � ∙ ∙ ∙ :� � � C1PηT � C2PηT�1 � ∙ ∙ ∙

� � � PηT�1(17.22)

Var eT�1� � � PE�ηT�1 � ηT�1�P´ � PP´ � Ω (17.23)

The forecast error of two period-ahead forecasts is

eT�2 � PηT�2 � C1PηT�1 (17.24)

The variance of eT�2 is

Var eT�2� � � PP´ � C1PP´C1 (17.25)

–2

0

2

4

6

1 2 3 4 5 6 7 8 9 10

Response of stock returnsto own shock

–2

0

2

4

6

1 2 3 4 5 6 7 8 9 10

Response of stock returns tointerest rate shock

FIGURE 17.5 Impulse Response of the Monthly Stock Returns

378 STATISTICS

3GC17 05/15/2014 12:40:2 Page 379

The forecast error of h period-ahead forecast is

eT�h � PηT�h � C1PηT�h�1 � ∙ ∙ ∙ � Ch�1PηT�1 (17.26)

Var eT�2� � � PP´ � C1PP´C1 � ∙ ∙ ∙ � Ch�1PP´Ch�1 (17.27)

Example:WerunaVAR(2) for themonthly returns of the S&P500 stock index andinterest rates during January 2000 to September 2012. We display in Figure 17.6 thevariance decomposition of the stock returns to own shock and to interest rate shock.

SUMMARY

Islamic finance operates in an environment of interdependent financial variables;change in one variable affects many other variables. For instance, the U.S. FederalReserve decision to lower interest rates in 2002 to 2005 led to the commodity andhousing boom followed by a financial crash. Variables such as exchange rates affectstock and commodity prices. Islamic finance uses VAR analysis to assess the interac-tion between variables and to make portfolio decisions based on the knowledge of thisinteraction. The chapter covers the formulation of the VAR, forecasting with VAR,the impulse response functions, and variance decompositions of forecasting errors.

QUESTIONS

1. TheMicrosoft Excel file “Data” contains data onmonthly S&P 500 stock indexesand U.S. federal rates during January 2000 to June 2013; using EViews estimate aVAR(2) comprising these two variables. Based on the information criteria whatshould be the lag of the VAR?

2. Run a VAR(4) for the monthly returns of the S&P 500 stock index and the U.S.federal rate during January 2000 to June 2013. Display the impulse response ofthe stock returns to own shock and to interest rate shock. Display the variancedecomposition for the monthly stock returns in response to own shock and to aninterest rate shock.

0

20

40

60

80

100

5 10 15 20 25 30 35 40 45 50

Percent stock returns variancedue to own shock

0

20

40

60

80

100

5 10 15 20 25 30 35 40 45 50

Percent stock returns variance due to interest rate shock

FIGURE 17.6 Variance Decomposition


3GC17 05/15/2014 12:40:2 Page 380

3. TheMicrosoft Excel file “Data” contains data onmonthly S&P 500 stock indexesand crude oil prices during January 2000 to June 2013. Using EViews estimate aVAR with appropriate lag determined by the information criteria. Display theimpulse response function and the variance decomposition.Make a forecast of theS&P 500 index and crude oil prices for July and August 2013. Compare yourforecast with actual data.

4. We assume y1T � 2:40, y2T � 0:15, and the following VAR:

y1t � 0:80y1t�1 � 0:16y2t�1y2t � 0:017y1t�1 � 0:70y2t�1

Using Microsoft Excel, make a five-period-ahead forecast for y1 and y2.

380 STATISTICS

3GC18 05/15/2014 12:44:17 Page 381

CHAPTER 18Co-Integration: Theory

and Applications

V ariables such as gross domestic product, investment, and consumption have closerelations in the long-run; however, they may deviate from each other in the short-

run. Likewise, stock prices and dividends of the issuing enterprise are closely related inthe long-run; however, in the short-run, they may wander from each other. We maymultiply the examples of long-run relationships between economic variables such asthe level of education and the level of development in an economy, or the degree ofmachinery use and agriculture production. When we deal with two integratedprocesses, for example, of order I, and, it is useful to distinguish short-run relation-ships from long-run relationships. The former relates to links that do not persist. Thelong-run relationships are closely associated with concepts of equilibrium in economictheory and persistence of co-movements of economic time series.

An examination of economic or financial relationships between time series willlead to discuss aspects of time series analysis, co-integration, and error-correction.The first step is to clarify the statistical notion of stationarity and its links to theconcept of equilibrium. We say that an equilibrium relationship holds between twovariables, and if the error by which actual observations deviate from this equili-brium is a mean-stationary process. That is the error or discrepancy betweenoutcome and postulated equilibrium has a fixed distribution centered on zerothat does not change over time.

We introduce the concept of co-integration, its relation to the definition of thelong-run equilibrium between series given above, and its use as part of a statisticaldescription of the behavior of time series that satisfy some equilibrium relationship.We discuss tests for co-integration that show whether two or more variables are co-integrated, that is, have long-run relationships. An important issue in econometricsis the need to integrate short-run dynamics with long-run equilibrium. The analysisof short-run dynamics is often done by first eliminating trends in the variables,usually by differencing. This procedure, however, throws away potential valuableinformation about long-term relationships about which economic theory has a lot tosay. The error-correction model of co-integration exploits both the levels and thefirst differences of nonstationary variables and shows the short-run dynamics of agiven variable in relation to the long-run relations with other variables. Co-integrated variables often share common shocks and trends, meaning that a singlerandom shock affects more than one variable at a time. For instance, a technologicalinnovation increases at the same time as production, investment, and consumption.

381

3GC18 05/15/2014 12:44:18 Page 382

Similarly, a monetary shock, such as injecting more money in the economy, affectsmany variables at the same time such as prices, consumption, exports, and imports. If asystem of n variables has r co-integration relations, then it has n � r common trends.Each system of n variables admits decomposition into common trends stochasticcomponent and a transitory component.

SPURIOUS REGRESSION

Some examples of what can emerge when standard regression techniques are usedwith nonstationary data can be outlined:

yt � yt�1 � ut ut ∼ iid 0; σ2u� �

(18.1)

xt � xt�1 � vt vt ∼ iid 0; σ2v� �

(18.2)

E utvs� � � 0∀ t; s; E utut�j� � � E vtvt�j

� � � 0∀j ≠ 0 (18.3)

We have assumed that xt and yt are uncorrelated randomwalks.We formulate thefollowing regression

yt � β0 � β1xt � εt (18.4)

Since xt neither affects nor is affected by yt, one would hope that the coefficientβ1 in the regression model would converge to zero, reflecting the lack of a relationbetween the series, and that the coefficient of determination R2 from this regres-sion would tend to zero. However, this is not the case. Regression methods detectthat correlations may persist in large samples despite the absence of any connec-tion between underlying series. If two time series are each growing, for example,they may be correlated even though they are increasing for entirely differentreasons and by increments that are uncorrelated. Hence, a correlation betweenintegrated series cannot be interpreted in the way that it could be if it arose amongstationary series.

In the regression equation both the null hypothesis β1 � 0 (implying yt � β0 � εt)and the alternative β1 ≠ 0 lead to false models, because the true data-generatingprocess is not nested within the regression model. From this perspective it is notsurprising that the null hypothesis, implying that yt

� �is awhite noise, is rejected; the

autocorrelation in the random walk yt� �

tends to project onto xtf g, also a randomwalk, and therefore also strongly autocorrelated. Tests based on badly specifiedmodels can often be misleading. Co-integration techniques provide tools to solvespurious regression.

Example: Using Microsoft Excel, we generate 100 observations of ut ∼ iid 0; σ2u� �

and vt ∼ iid 0; σ2v� �

; we generate yt � 0:45 � yt�1 � ut.

xt � 0:4 � xt�1 � vt

382 STATISTICS

3GC18 05/15/2014 12:44:18 Page 383

We regress yt on xt; we obtain the following regression:

yt � 1:74 � 1:22xt;R2 � 0:98t � 4:88� � t � 76:4� �:

Although we have highly significant regression coefficients and very high coef-ficients of determination, R2, this regression is spurious.

Example: A regression of the U.S. consumption USC� � on the United Kingdomdisposable income UKY� � during 1960 to 1998 yields the following relationship:

log USC� � � �5:57 � 1:2log UKY� �;R2 � 0:976t � �32:8� � t � 79:0� �

Economic theory stipulates little or no relation between U.S. consumption andU.K. disposable income; that is, we expect the latter to have a negligible effect on theformer. Yet, the regression between the two variables is highly significant.Unfortunately, it is spurious and has no theoretical or policy implications.

STATIONARITY AND LONG-RUN EQUILIBRIUM

We consider two nonstationary processes of order I 1� � denoted by yt� �

and xtf g. Weshow that integrated processes can be reduced to a stationary process by suitabletransformations that take advantage of co-integrated (equilibrium) relationships.

When we deal with two integrated processes, for example, of order I 1� �, yt� �

andxtf g, it is useful to distinguish short-run relationships from long-run relationships. Wewant to analyze long-run equilibrium relationships. An equilibrium state is defined asone in which there is no inherent tendency to change. A dis-equilibrium is any statethat tends to self-correct and return to equilibrium. We write the equation represent-ing co-movements as

yt � βxt (18.5)

todenote a linear long-run relationbetween yt� �

and xtf g. Even if shocks toa systemareconstantly occurring so that the economic system is never in equilibrium, the concept oflong-run equilibriummay nonetheless be useful. A long-run equilibriumwill often holdon average over time. Methods for investigating such long-run relationships are ourconcern. An examination of these methods leads to a discussion of the aspects of time-series analysis, co-integration, and error correction. The first step is to clarify thestatistical notion of stationarity and its links to the concept of equilibrium.

We say that an equilibrium relationship yt � βxt holds between two variables yt� �

and xtf g if the error

εt � yt � βxt (18.6)

by which actual observations deviate from this equilibrium is a mean-stationaryprocess. That is the error or discrepancy between outcome and postulated equilibrium

Co-Integration 383

3GC18 05/15/2014 12:44:19 Page 384

has a fixed distribution centered on zero that does not change over time. Thisdefinition of an equilibrium relationship holds automatically when applied to seriesthat are themselves stationary. However, this concept of statistical equilibrium isnecessary in examining equilibrium relationships between variables that tend to growover time. In such cases, if the actual relationship is yt � βxt, the discrepancy εt �yt � βxt will be nonstationary for any b ≠ β, because the discrepancy deviates from thetrue relationship by the constant proportion b � β� � of the growing variable xt; onlythe true relationship can yield a stationary discrepancy.

Suppose the equilibrium relationship is y � βx, then the discrepancy, or error,yt � βxt

� �should be a useful explanatory variable for the next direction of the

movement of yt. In particular when yt � βxt is positive, yt is too high relative to xt; onaverage, we might expect a fall in y in future periods relative to its trend growth.Inversely, when yt � βxt is negative, yt is too low relative to xt; on average, we mightexpect a rise in y in future periods relative to its trend growth. The termyt�1 � βxt�1� �

, called an error correctionmechanism, is therefore included in dynamicregressions.

The practice of exploiting information contained in the current deviationfrom an equilibrium relationship, in explaining the path of a variable, hasbenefited from the formalization of the concept of co-integration. The informaldefinition of equilibrium is based on a special case of the definition of co-integration. Further, the practice of modeling co-integrated series is closely relatedto error correction mechanisms: error correcting behavior on the part of economicagents will induce co-integrating relationships among the corresponding variablesand vice versa.

CO-INTEGRATION

We discussed the notion of equilibrium. The idea that variables linked by sometheoretical economic relationship should not diverge from each other in the long runis a fundamental one. We may cite the example of the quantity theory of money,permanent income hypothesis of consumption, and purchasing power parity. Forinstance, the quantity of money may increase rapidly in the short run; however,prices will catch up with some delay. Such variables may drift apart in the short run,but if they were to diverge without bound, an equilibrium relationship among suchvariables could not be said to exist. The divergence from a stable equilibrium statemust be stochastically bounded and, at some point, diminishing over time. Co-integration may be viewed as the statistical expression of the nature of suchequilibrium relationships.

We can now introduce the concept of co-integration, its relation to the definitionof the long-run equilibrium between series given earlier, and its use as part of astatistical description of the behavior of time series that satisfy some equilibriumrelationship. A simple example concerns two series, each of which is integrated oforder 1. Assume a long-run equilibrium relationship holds between them, and that it islinear: y � βx. Then y � βxmust be equal to zero in equilibrium and the series yt � βxthas a constant unconditional mean of zero. This need not imply that yt � βxt isstationary; the variance of yt � βxt might be nonconstant, for example. The definition

384 STATISTICS

3GC18 05/15/2014 12:44:20 Page 385

of co-integration does require stationary deviations: yt � βxt. When stationarity doeshold, we say that y and x are co-integrated.*

The concept of co-integration is a powerful one because it describes the existenceof an equilibrium, or stationary, relationship among two or more series, each of whichis individually nonstationary. That is while the component time series may havemoments such as means, variances, and covariances varying with time, some linearcombination of these series, which defines the equilibrium relationship, has time-invariant linear properties. Thus, for example, if xtf g and yt

� �are integrated of order

1 and are also co-integrated, then Δxtf g, Δyt� �

, and xt � αyt� �

, for some α, are allstationary series.

In order to illustrate the preceding discussion, consider a simple example. Twoseries xtf g and yt

� �are each integrated of order 1 and evolve according to the

following data-generating process:

xt � βyt � ut (18.7)

xt � αyt � et (18.8)

ut � ut�1 � ε1t (18.9)

et � ρet�1 � ε2t with ρj j < 1 (18.10)

ε1t and ε2t are distributed identically and independently as white noises with

E ε1t� � � E ε2t� � � 0;Var ε1t� � � σ21;Var ε2t� � � σ22;Cov ε1t; ε2t� � � σ12 (18.11)

Solving for xt and yt from the above system with α ≠ β gives

xt � α α � β� ��1ut � β α � β� ��1et (18.12)

yt � � α � β� ��1ut � β α � β� ��1et (18.13)

Since utf g is a random walk and xtf g and yt� �

depend linearly on utf g these maytherefore be classified as I 1� � variables. Nonetheless, xt � αyt

� �is I 0� � because et is

stationary. In this example the vector 1; α� � is the co-integrating vector and xt � αyt isthe equilibrium relationship. In the long run, the variables move toward the equili-brium: xt � αyt � 0, recognizing that this relationship need not be realized exactlyeven as t ! ∞ .

Example: We assume in the above model α � 1 and β � 2, we obtain

xt � �ut � 2et (18.14)

yt � ut � 2et (18.15)

*A humoristic example of a drunk lady and her dog was provided by M. Murray (1994) toillustrate a co-integration relationship between two random walks. The drunk lady walksrandomly and so does her puppy. However, neither loses distance with the other; either thedrunk lady has to catch upwith the puppy or the inverse, so they keep the distance between themstationary.

Co-Integration 385

3GC18 05/15/2014 12:44:20 Page 386

We generate a random walk and a white noise in Microsoft Excel with 100 datapoints. We compute xt and yt, which we display in Figure 18.1. The two variables areco-integrated.

Although this is a simple example, much of the method and reasoning can begeneralized to more complex cases. What is crucial is that, while xtf g and yt

� �are

integrated processes, not tied to any fixed means, a linear combination of the twovariables makes the resulting series a stationary process and the variables x and ymaybe said to be linked by the corresponding equilibrium relationship.

The concept of co-integration is central to econometric modeling with integratedvariables as well as to the examination of long-run relationships among thosevariables. The identification of co-integration with equilibrium yields a meaningfulregression instead of a spurious regression. Regressions involving levels of time seriesof nonstationary variables make sense if and only if these variables are co-integrated.A test for co-integration yields a useful method of distinguishing meaningful regres-sions from those called nonsense or spurious. A set of co-integrated variables is knownto have, among other representations, an error-correction representation; that is, therelationship may be expressed so that a term representing the deviation of observedvalues from the long-run equilibrium enters the model.

In Islamic finance, co-integration is an essential technique for detecting long-runequilibrium relationships between financial series. Financial analysts have to studyfinancial data such as stock prices, commodity prices, rates of returns, exchange rates,output, exports, and imports. They may have to invest in commodities or hedgepositions in commodities. In Figure 18.2, we plot monthly gold and crude oil pricesfrom January 2000 to September 2012. We observe that the two series display thesame trend; price inflation was proceeding at a two-digit rate per year during theperiod under consideration. Money was depreciating at a fast rate during this period.The dollar has depreciated too much with regard to every commodity, including oil

–15

–10

–5

0

5

10

15

1 5 7 9 11

13

15

17

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

51

53

55

57

59

61

63

65

67

69

71

73

75

77

79

81

83

85

87

89

91

93

95

97

99 3

FIGURE 18.1 Simulation of Co-Integrated Variables

386 STATISTICS

3GC18 05/15/2014 12:44:20 Page 388

and gold. However, in the long run gold and crude oil seemed to keep their real valuein respect to each other. Their relative price may change in the short run due tochanges in demand and supply and speculation; however, in the long run, the relativeprice is determined by real cost of production and normal profit margins. We are notinterested in whether the gold price is pushing the oil price upward or whether the oilprice is pushing the gold price upward. More specifically, we are not interested inwhich variable is endogenous and which is exogenous. We want only to determine along-run relationship between gold and oil prices. The search for a long run betweentwo or more variables is called co-integration analysis.

TEST FOR CO-INTEGRATION

An important issue in econometrics is the need to integrate short-run dynamics withlong-run equilibrium. The analysis of short-run dynamics is often done by firsteliminating trends in the variables, usually by differencing. This procedure, however,throws away potential valuable information about long-term relationships aboutwhich economic theory has a lot to say. The theory of co-integration addresses theissue of integrating short-run dynamics with long-run equilibrium. For instance, if wespecify the regression model,

yt � βxt � εt (18.16)

we have to make sure that yt and xt are integrated to the same order. For instance, ifyt is I 1� � and xt is I 0� � there will not be any β that will satisfy the relationship yt �βxt � εt . Suppose yt is I 1� � and xt is I 1� �; if there is a nonzero β such that yt � β xt isI 0� �, then yt and xt are said to be co-integrated.What this means is that the regressionyt � βxt � εt makes sense because yt and xt do not drift too far apart each other overtime. There is a long-run equilibrium relationship between them. If yt and xt are notco-integrated, that is, yt � βxt � εt is also I 1� �, they can drift apart from each othermore and more as time goes on. Thus there is no long-run equilibrium between them.In this case the relationship between yt and xt that we obtain by regressing yt on xt isspurious.

Suppose that yt and xt are both random walks, so that they are both I 1� �. Ifyt � βxt� �

is stationary, then an equation in first differences, as shown below, is a validequation since Δyt;Δxt; yt � βxt

� �and εt are all I 0� �:

Δyt � αΔxt � λ yt�1 � βxt�1� � � εt (18.17)

In this case, the equation yt�1 � βxt�1� �

is considered as a long-run relationshipbetween yt and xt and the equation in first differences describes short-run dynamics.Thus if xtf g and yt

� �are integrated of order 1 and are also co-integrated, then Δxtf g,

Δyt� �

, and the error correction term yt � βxt� �

, for some β, are all stationary series.We may estimate yt � βxt � εt by OLS, obtaining the estimator β of β and

substituting it in an equation error-correction model to estimate the parameters αand λ. This two-step estimation procedure, however, rests on the assumption that ytand xt are co-integrated. It is therefore important to test for co-integration. We may

388 STATISTICS

3GC18 05/15/2014 12:44:21 Page 389

estimate yt � βxt � εt by OLS, getting the residuals εt, and then apply the Dickey-Fuller test based on εt. What this test amounts to is testing the hypothesis ρ � 1 inεt � ρεt�1 � et.

Since the unit root test will be applied to εt the null hypothesis and the alternate inthe co-integration tests are H0: εt has a unit root εt is I 1� � or xt and yt are not co-integrated.

H1: xt and yt are co-integrated, εt is I 0� �:In essence we are testing the null hypothesis that yt and xt are not co-integrated.

Note that yt is I 1� � and xt is I 1� �, so we are trying to see that εt is not I 1� �. Theadditional problem is that εt is not observed. Hence, we use the estimated residuals εtfrom the co-integration regression.

The spurious regression problem is a serious one, the practice of differencingintegrated series to achieve stationarity, and of treating the resulting time series as theproper objects of econometric analysis, is not without costs; it entails loss ofinformation incorporated in the level of variables. Error-correction mechanisms(ECMs) are intended to provide a way of combining the advantages of modelingboth levels and differences. In an error-correction model the dynamics of both short-run (changes) and long-run (levels) adjustment processes are modeled simultaneously.

Example: We consider data on monthly S&P 500 stock index xtf g and the goldprices yt

� �during January 2000 to September 2012. To investigate the existence of a

long-run stationary relationship we explore a simple relationship of the form

log yt � β log xt � εt

We find log yt � 0:90 log xt ,

t � 130:1� �The unit-root test on the residuals εtf g of the equation yields a Dickey-Fuller

statistic equal to �0.64 with probability value of 0.43; we cannot reject the nullhypothesis of unit root of the residuals εtf g. We conclude that monthly S&P 500 stockindex xtf g and the gold prices yt

� �during January 2000 to September 2012 were not

co-integrated.Example: We consider data on monthly crude oil prices yt

� �and the gold prices

xtf g during January 2000 to September 2012. To investigate the existence of a long-run stationary relationship we explore a simple relationship of the form

log yt � β log xt � εt

We find

log yt � 0:62 log xt ; t � 201:1� �

The unit-root test on the residuals εtf g of the equation yields a Dickey-Fullerstatistic equal to �2.83 with probability value of 0.004; we reject the null hypothesisof unit root of the residuals εtf g. We conclude that monthly crude oil prices and goldprices during January 2000 to September 2012 were co-integrated.

Co-Integration 389

3GC18 05/15/2014 12:44:22 Page 390

We are permitted to estimate an error-correction (ECM) relationship of the form

Δyt � αΔxt � λ yt�1 � βxt�1� � � εt

We find

Δyt � 0:425Δxt � 5:547 yt�1 � 0:62xt�1� � � 0:717

t � 2:47� � t � 2:00� � t � 1:01� �The ECM describes the short dynamics of crude oil prices; they are impacted by

changes in gold prices (0.425) and they adjust to the long-run relation by a coefficientof 5.547.

CO-INTEGRATION AND COMMON TRENDS

Co-integrated variables share common stochastic trends. This property provides auseful way to understand co-integration relationships. Consider two co-integratedI 1� � variables yt and xt. We can write each variable as

yt � zt � ut (18.18)

xt � wt � vt (18.19)

where zt and wt are random walk processes representing the (stochastic) trends of thevariables and ut and vt are stationary processes. Because yt and xt are co-integratedthere is a constant a ≠ 0 such that yt � axt is stationary. Assume for simplicity thata � 1. Then

yt � xt � zt �wt� � � ut � vt� � (18.20)

Because this is stationary the random walk component must be zero, that is, zt �wt � 0 or zt � wt.

Thus co-integration of yt and xt implies that they share the same commonstochastic random walk component: zt � wt. In general, if yt

� �and xtf g are co-

integrated, there must be nonzero values of β1 and β2 for which the linear combinationis stationary,

β1yt � β2xt � β1 zt � ut� � � β2 wt � vt� � � β1zt � β2wt� � � β1ut � β2vt

� �(18.21)

For �β1yt � β2xt� to be stationary the term β1zt � β2wt� �

must vanish. After all, ifeither of the two trends appears in the linear combination β1yt � β2xt will also have atrend. Since the second term β1zt � β2wt

� �is stationary, the necessary and sufficient

condition for yt� �

and xtf g to be co-integrated is

β1zt � β2wt � 0 (18.22)

Clearly, zt andwt are variables whose realized values will be continually changingover time. Since we preclude both β1 and β2 from being equal to zero, it follows that

390 STATISTICS

3GC18 05/15/2014 12:44:22 Page 391

β1zt � β2wt � 0 holds for all t if and only if

zt � �β2wt=β1 (18.23)

For nonzero values of β1 and β2 the only way to ensure equality is for the stochastictrends to be identical up to a scalar. Thus, up to a scalar β2=β1, two I 1� � stochasticprocesses yt

� �and xtf g must have the same common trend if they are co-integrated.

Example:We assume: yt � zt � ut, xt � wt � vt, zt andwt are random walks withzt � �β2wt=β1. We assume β1 � 0:8 and β2 � �2:4. We simulate 100 observations forxt and yt, which we display in Figure 18.3. The latter shows that xt and yt share acommon trend.

CO-INTEGRATED VARs

In this section we extend the notion of co-integration, error correction, and commontrends to a general case of n integrated I 1� � variables. We discussed the notion of co-integration and its role to analyze long-run equilibrium; in particular, the notion thatvariables linked by some theoretical economic relationship should not diverge fromeach other in the long run is a fundamental one. Consider three I(1) variables, xtf g,wtf g, and ztf g. These variables may be pair-wise non–co-integrated; however,considered together they may be co-integrated, which means there may a co-integrat-ing vector β1; β2; β3

� �such that

εt � β1xt � β2wt � β3zt is I 0� � (18.24)

A vector of I 1� � variables yt, for example, xt;wt; ztf g, is said to be co-integrated ifthere exists a nonzero vector βi such that

βiyt is trend stationary; that is; βiyt ∼ I 0� � (18.25)

–15

–10

–5

0

5

10

1 3 5 7 9 11

13

15

17

19

21

23

25

27

29

31

33

35

37

39

41

43

45

47

49

51

53

55

57

59

61

63

65

67

69

71

73

75

77

79

81

83

85

87

89

91

93

95

97

99

FIGURE 18.3 Common Trend in Co-Integrated Variables

Co-Integration 391

3GC18 05/15/2014 12:44:23 Page 392

The vector β is called the co-integrating vector. It is possible for several equili-brium relationships to govern the joint evolution of these variables. If there existexactly r linearly independent co-integrating vectors with r � n � 1, then these can begathered into an n � rmatrix β. The rank of β will be r and is called the co-integratingrank. The vector yt is said to be co-integrated with co-integrating rank r. If yt has n � 2components, then there may be more than one co-integrating vector β.

Note that βiyt is an r-dimensional vector of stationary variables. Also note thatthis definition is symmetric in the variables, that is, there is no designated left-hand sidevariable. This means that variables move together and we are not interested in whichvariable or variables are driving the others. For instance, consumption, income, andinvestment move together and we are not interested in the causation among thesevariables. This is usually an advantage for the statistical testing but it makes it harderfor the economic intuition. Note the βi vectors are individually identified only up toscale because βiyt stationary implies that cβiyt is stationary for a constant c ≠ 0.Finally note that the treatment of constants and drift terms are suppressed here. Onehas to consider those for any practical applications of co-integration methods.

REPRESENTATION OF A CO-INTEGRATED VAR

A co-integrated system of variables can be represented in error correction or movingaverage forms.

Vector Error-Correction (VEC) Representation

Let yt be an I 1� � vector of n components, each with a possibly deterministic trend inmean. Suppose that the system can be written as a finite order vector autoregressionAR p� �,

yt � μ � A1yt�1 � A2yt�2 � ∙ ∙ ∙ � Apyt�p � εt; t � 1; 2; . . . ;T (18.26)

where εt is stationary. The model can be written as

Δyt � μ � Γ1Δyt�1 � Γ2Δyt�2 � ∙ ∙ ∙ � ΓpΔxt�p�1 � Πyt�p � εt (18.27)

Clearly the long-run properties of the system are described by the properties of then; n� � matrix Π. There are three cases of interest:

1. Rank Π� � � 0. The system is nonstationary with no co-integration between thevariables considered. This is the only case in which nonstationarity is correctlyremoved by simply taking the first differences of the variables.

2. Rank Π� � � n, full rank; then all yt must be stationary since the left-hand side andthe other right-hand side variables are stationary. The system yt is stationary.

3. Rank Π� � � r < n. Themost interesting case is whenΠ has less than full rank, but isnot equal to zero. This is the case of co-integration. In this case Π can be written as

Π � αβ´ (18.28)

392 STATISTICS

3GC18 05/15/2014 12:44:24 Page 393

This β corresponds to the matrix of the co-integrating relationships; the matrices αand β are n � r matrices. The matrix α can be interpreted as a speed of adjustmenttoward equilibrium. The system yt is nonstationary but there are r stationaryintegrating relationships among the considered variables. Under the assumption ofco-integration of order r the model can be written in a vector error correction (VEC)form as

Δyt � μ � Γ1Δyt�1 � ∙ ∙ ∙ � Γk�1Δyt�p�1 � αβ´yt�p � εt (18.29)

Therefore, the rankΠ is crucial in determining the numberof co-integrationvectors.The Johansen procedure is based on the fact that the rank of a square matrix equals thenumber of its characteristic roots that differ from zero. Here is the intuition on how thetests can be constructed. Having obtained estimates for the parameters in Πmatrix, weassociatewith themestimates for then characteristic roots andweorder themas follows:λ1 > λ2 > . . . > λn. If the variables are not co-integrated, then the rank ofΠ is zero andall the characteristic roots are equal to zero. In this case each of the expression ln 1 � λi� �equals zero, too. If, instead, the rank of Π is one, and 0 < λ1 < 1, then ln 1 � λ1� � isnegative, and ln 1 � λ2� � � ln 1 � λ3� � � ∙ ∙ ∙ � ln 1 � λn� � � 0. Johansen derived a test onthe number of the characteristic roots that are different from zero by considering thefollowing statistics,

λtrace r� � � �T Xni�r�1

ln 1 � λi� �

(18.30)

λmax r; r � 1� � � �T ln 1 � λr�1� �

(18.31)

whereT is the number of observations used to estimate the VAR. The first statistic teststhe null of at most r co-integration vectors against a generic alternative. The test shouldbe run in sequence starting from the null of at most zero co-integrating vector up to thecase of atmostn co-integrating vectors. The second statistic tests the null of atmost r co-integrating vectors against the alternative of at most r � 1 co-integrating vectors. Bothstatistics are small under the null hypothesis. Critical values are tabulated by Johansenand they depend on the number of nonstationary components under the null hypothe-sis and on the specification of the deterministic component of the VAR.

Assume yt has four I 1� � variables. Often we do not really want to test whetherthere are (say) three co-integrating vectors against no co-integrating vectors; rather wewant to make a decision on to what is the number of co-integrating vectors. In thesituation where we directly want to test r � 1 co-integrating vectors against r co-integrating vectors we should use the λmax test, but this test will not give us a consistentway of deciding the co-integration rank. A consistent way to do this, using the tracetest, is to start by testing for zero co-integrating vectors; that is, if our system is four-dimensional, we compare the test statistic �TP4

i�1 ln 1 � λi� �

. If we reject zero co-integrating vectors, we then test for (at most) one co-integrating vector. In the four-dimensional case, we compare the test statistic

λtrace r� � � �TX4i�2

ln 1 � λi� �

(18.32)

Co-Integration 393

3GC18 05/15/2014 12:44:25 Page 394

If this is not rejected we stop and decide that r � 1. If we reject this we move onuntil we can no longer reject and stop there. Johansen has shown in the past somepreference for the trace test, based on the argument that the maximum eigenvalue testdoes not give rise to a coherent testing strategy.

Example: We consider data on monthly crude oil and gold data for the period2000M01–2012M09. Using EViews we run a Johansen trace test; we find one co-integration vector between crude oil and gold prices during January 2000 toSeptember 2012.

Sample: 2000M01 2012M09Included observations: 153

Trend assumption: No deterministic trend (restricted constant)Series: LOIL LGOLD

Lags interval (in first differences): 1 to 4Unrestricted Co-integration Rank Test (Trace)

Hypothesized Trace 0.05No. of CE(s) Eigenvalue Statistic Critical Value Probability Value

None * 0.088092 20.59245 20.26184 0.0450At most 1 0.041490 6.483371 9.164546 0.1566

Trace test indicates 1 co-integrating equation at the 0.05 level*Denotes rejection of the hypothesis at the 0.05 level

We estimate a co-integration relationship, in logarithm, between oil and goldprices. We find the following relationship:

log oil price� � � 0:658 � log gold price� � � 0:240

This relationship shows that gold has appreciated considerably in relation to oilduring the period under study, that is, January 2000 to September 2012. In otherwords, if gold prices increase by 10 percent a year, oil prices increase by 6.58 percentper year. However, both commodities have appreciated considerably in relation topaper money, that is, the dollar. Gold has appreciated by 16.3 percent a year andcrude oil by 12 percent a year. Otherwise, if you hold your savings in dollars, you losein real terms 16.3 percent per year in relation to gold and 12 percent in relation tocrude oil. This is called inflation tax collected by the government or the debtors orboth. Every year you buy much less oil with your dollar.

Co-Integration and Moving Average Representation

We assume that the system yt is I 1� � and that the n � n matrix Π has reduced rankr < n and is therefore expressible as the product of two n � rmatrices α and β, whereα and β have rank r. Thus Π � αβ´. We obtain the following six results:

i. Δyt is stationary;ii. β´yt is stationary;

394 STATISTICS

3GC18 05/15/2014 12:44:26 Page 395

iii. Δyt has a Wold moving-average representation:*

Δyt � μ � C L� �εt (18.33)

iv. The vector yt has a stochastic trend representation as

yt � y0 � C 1� �Xt

i�1εi � μt � C* L� �εt (18.34)

where C L� � � C 1� � � 1 � L� �C* L� �, C 1� � is n � n matrix;v. βĆ 1� � � 0r�n; andvi. C 1� �α � 0n�r.

Without loss of generality, we assume y0 � 0 and μ � 0 and let zt � Pti�1 εi a n � 1

random walk vector. We may write yt as

yt � C 1� �zt � C* L� �εt (18.35)

Here, we have a representation of yt in terms of stochastic trends C 1� �zt and acyclical or transitory components C* L� �εt. Now βýt is stationary in the case of co-integration, so that

βýt � βĆ 1� �zt � βĆ* L� �εt (18.36)

is stationary, which implies that βĆ 1� �zt � 0. Hence, βĆ 1� � � 0r�n. This gives anothercharacterization of co-integration that may be useful for testing. If we multiply βĆ 1� �by α we obtain βĆ 1� �α � 0n�r. Since β is a nonzero vector, we have, therefore,

C 1� �α � 0n�r (18.37)

SUMMARY

Islamic finance applies co-integration analysis to estimate equilibrium long-runrelationships between financial variables. Co-integration analysis avoids difficultiesof nonstationarity and spurious regressions. This chapter covers the notions ofspurious regression, stationarity, and long-run equilibrium, co-integration, test forco-integration, common trends, and co-integrated VAR; the latter admits a vectorerror-correction (VEC) representation as well as a moving average representation.With co-integration techniques a financial analyst is able to determine long-runequilibrium relationships between variables of interest and how short-run deviations

*The lag operator L is used to represent lagged variables; for instance, εt�1 is written as Lεt, εt�2is written as L2εt, and εt�j is written as Ljεt. The notation C L� � designates the lag polynomial1 � γ1L � γ2L

2 � ∙ ∙ ∙ γjLj � ∙ ∙ ∙ where γ1; γ2; . . . :; γj; . . . : are the coefficients of the polynomial.

Hence, C 1� � � 1 � γ1 � γ2 � ∙ ∙ ∙ � γj � ∙ ∙ ∙ .

Co-Integration 395

3GC18 05/15/2014 12:44:26 Page 396

from this equilibrium become self-correcting. For instance, gold and crude oil pricesmay deviate from each other temporarily; however, they may return to a long-runequilibrium determined by the production cost of each commodity.

QUESTIONS

1. Using Microsoft Excel, we generate 100 observations of ut ∼ iid 0; σ2u� �

andvt ∼ iid 0; σ2v

� �. Generate yt � 0:645 � yt�1 � ut and xt � 0:574 � xt�1 � vt and

regress yt on xt ; comment on your findings.

2. We consider two nonstationary processes of order I 1� � denoted by yt� �

and xtf g.Show how they can be reduced to a stationary process by a suitable transforma-tion that takes advantage of co-integrated (equilibrium) relationship. If yt

� �and

xtf g are not co-integrated, is this transformation possible?

3. Read the article “A Drunk and Her Dog” by M. Murray (1994). Provide a briefdescription of the co-integration and error-correction model.

4. We assume the following data-generating process:

xt � yt � utxt � 3yt � etut � ut�1 � ε1t

et � 0:3et�1 � ε2t

ε1t and ε2t are distributed identically and independently as white noises withE ε1t� � � E ε2t� � � 0, Var ε1t� � � σ11, Var ε2t� � � σ22, and Cov ε1t; ε2t� � � σ12. UseMicrosoft Excel to simulate 200 observations of xt and yt. Plot xt and yt in onegraph. Comment on the graph.

5. We assume yt � zt � ut, xt � wt � vt, zt and wt are random walks withzt � �β2wt=β1. We assume β1 � 1:8 and β2 � �4:4. Simulate 200 observationsfor xt and yt. Plot the simulations in a graph. Do you observe a common trend inxt and yt?

6. Microsoft Excel file “Data” contains monthly data on the S&P 500 stock indexand gold prices. Consider data onmonthly S&P 500 stock index xtf g and the goldprices yt

� �during January 2000 to August 2013. Investigate the existence of a co-

integration relation between the logarithms of these two variables. If such arelationship exists, estimate an ECMof the formΔyt � αΔxt � λ yt�1 � βxt�1

� � � εt.7. Microsoft Excel file “Data” contains on monthly crude oil prices yt

� �and the

gold prices xtf g for January 2000 to August 2013. Investigate the existence of aco-integration relation between the logarithms of these two variables. If such arelationshipexists, estimateanECMof the formΔyt � αΔxt � λ yt�1 � βxt�1

� � � εt.8. Microsoft Excel file “Data” contains monthly data on the S&P 500 stock index,

gold prices, and crude oil prices from January 2000 to August 2013. Run aco-integration to determine the existence of co-integration vectors between thesetime series.

396 STATISTICS

WEBC19 05/20/2014 10:17:35 Page 397

CHAPTER 19Modeling Volatility:ARCH-GARCH Models

M odeling volatility is key for pricing assets and asset derivatives. Time-varyingvolatility of returns was initially studied in the context of an autoregressive

conditional heteroskedasticity (ARCH) model. The ARCHmodel concentrates on thevolatility dynamic. Financial markets display high volatility. Stock prices seem to gothrough a period of high volatility with significant changes in returns. Hence, volatilityof returns may be a time-varying variable. If the variance of returns is time-varyingthen the risk variable has a time-varying variance. Many economic time seriesexhibit volatility clustering; that is, periods of unusually large volatility are followedby periods of relative tranquillity. Large changes in prices tend to be followed bylarge changes, of either sign, and small changes of prices tend to be followedby small changes. The assumption of linear effect of past shocks on returns is notrealistic; likewise, the assumption of constant variance of returns is not supportedby data.

We describe the ARCHmodel in this chapter. The term heteroskedasticity refersto changing volatility (i.e., variance) of returns. But it is not the variance itself thatchanges with time according to an ARCH model; rather, it is the conditionalvariance of returns that changes, in a specific way, depending on the available data.We analyze the properties of the ARCH model. The key insight offered by theARCH model lies in the distinction between the conditional and the unconditionalvariances. While the unconditional variance for the variable of interest may be timeinvariant, the conditional variance often depends on past shocks. Understanding theexact nature of this temporal dependence is crucially important for many issues infinance. The ARCH model has some shortcomings; it assumes that positive andnegative shocks have the same effects on volatility because it depends on the squareof the previous shocks. In practice, it is well known that the price of a financial assetresponds differently to positive and negative shocks. Further, the ARCH modelis not parsimonious and requires estimation of a large number of parameters.The generalized ARCH (GARCH) model remedies these shortcomings of theARCH model.

We describe the ARCH-GARCH inmean. In fact, many theories in finance call foran explicit trade-off between the expected returns and the variance, or the covarianceamong the returns. In the case of financial assets we expect higher risk to becompensated by a higher return. For example, according to the traditional capitalasset pricing model (CAPM) the excess returns on all risky assets are proportional to

397

WEBC19 05/20/2014 10:17:35 Page 398

the nondiversifiable risk as measured by the covariances with the market portfolio.This implies that the expected excess return on the market portfolio is simplyproportional to its own conditional variance. As such, it would be useful to havea model in which the return is partly determined by its risk.

We address the testing for the ARCH effects. The intuition behind the test forARCH effects is simple. If the data are homoskedastic, then the variance cannot bepredicted and variations in ε2t will be purely random. However, if ARCH effects arepresent, large values of the past squared residuals will predict large values ofcontemporaneous residuals. If there is conditional heteroskedasticity, the correlogramof the residuals should be suggestive of such process. We discuss the Lagrangianmultiplier statistic as well as the Ljung-Box statistic in testing for ARCH effects.

MOTIVATION FOR ARCH MODELS

In conventional econometrics, the variance of the disturbance term is assumed to beconstant. The assumption that the variance of the errors is constant is known ashomoskedasticity. If the variance of the errors is not constant we have heteroske-dasticity, and estimates of the standard errors of the coefficients could be wrong.Financial time series hardly fulfill constant variance of errors. Many economic timeseries exhibit volatility clustering. That is, periods of unusually large volatility arefollowed by periods of relative tranquillity; large changes of prices tend to be followedby large changes of either sign, and small changes of prices tend to be followed bysmall changes. Volatility clustering phenomenon is immediately apparent when assetreturns are plotted through time. Common examples of such series include stockprices, foreign exchange rates, and other prices determined in financial markets (i.e.,where volatility seems to vary over time). In such circumstances, the assumption ofconstant variance (homoskedasticity) is inappropriate. Autoregressive conditionalheteroskedasticity (ARCH) models are methods of modeling such volatility. Engle(1982) showed that it is possible to simultaneously model the mean and variance of aseries. ARCH models are designed to model and forecast conditional variances, withthe variance of the dependent variable modeled as a function of past values of thedependent variable and independent, exogenous variables. Bollerslev (1986) intro-duced generalized ARCH (GARCH).

Let Yt, t � 1; . . . ;T be a time series of prices of a financial asset, for example, dailyquotes on a share, stock index, currency exchange rate, or a commodity price. Insteadof analyzing Yt, which often displays unit-root behavior and thus cannot be modeledas stationary, we often analyze log-returns on Yt, that is, the series

yt � log Yt � log Yt�1 (19.1)

The mean return is μy �XT

t�1 yt=T (19.2)

Excess return rt is defined as

rt � yt � μy (19.3)

398 STATISTICS

WEBC19 05/20/2014 10:17:35 Page 399

The mean excess return μr is zero. The variance of excess return σ2t is equal to r2t .*

As an illustration, consider the time series of monthly closing values of the S&P500 stock index from January 2000 to May 2013. Figure 19.1 shows the actual seriesYt. Figure 19.2 shows actual series yt, which displays many of the typical stylized facts

600

800

1,000

1,200

1,400

1,600

1,800

2000 2002 2004 2006 2008 2010 2012

FIGURE 19.1 Monthly S&P 500 Stock Price Index, January 2000to May 2013Source: Yahoo.finance.

–25

–20

–15

–10

–5

0

5

10

15

2000 2002 2004 2006 2008 2010 2012

FIGURE 19.2 Monthly Returns on the S&P 500 Index, January2000 to May 2013Source: Yahoo.finance.

*The sample variance is defined as σ2 �Pn

i�1 xi�x� �2n . We may think of rt as one observation

sample; that is, n � 1. Since the mean is μr � 0, by replacing in the formula the variance of rt istherefore r2t .

Modeling Volatility 399

WEBC19 05/20/2014 10:17:36 Page 400

present in financial log-return series. Figure 19.3 shows actual series y2t . The series yt isuncorrelated; however, the correlogram in Figure 19.4 for squared series y2t displaysauto-correlation even for large lags. It is also typical of financial log-return series to beheavy-tailed. Finally, the series yt displays the so-called leverage effect: the series ytresponds differently to its own positive and negative movements, or in other words theconditional distribution of yt

�� j yt�1 > 0� ��

is different from that yt�� j yt�1 < 0

� �� .

The explanation is that the market responds differently to good and bad news, whichis only too natural.

Stationarity enables us to estimate parameters globally, using the entire availabledata set. However, to propose a stationary model for yt that captures the abovestylized facts is not easy, as the series does not look stationary: the local variance

0

100

200

300

400

500

2000 2002 2004 2006 2008 2010 2012

FIGURE 19.3 Squared Returns of S&P 500, January 2000 toMay 2013Source: Yahoo.finance.

–0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6 7 8 9 10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

FIGURE 19.4 Correlogram of Squared Returns

400 STATISTICS

WEBC19 05/20/2014 10:17:36 Page 401

(volatility) is clearly clustered in bunches of low/high values. If we were to fit a lineartime-series model (such as ARMA) to yt, the estimated parameters would be affectedby the presence of serial correlation in squared returns.

FORMALIZATION OF THE ARCH MODEL

Recent developments in financial econometrics suggest the use of nonlinear time-seriesstructures to model the attitude of investors toward risk and expected return. Whendealing with nonlinearities, we make the distinction between:

i. Linear time series: Shocks are assumed to be uncorrelated but not necessarilyidentically independent distributed iid� �; and

ii. Nonlinear time series: Shocks are assumed to be iid, but there is a nonlinearfunction relating the observed time series yt

� � ∞t�0 and the underlying shocks

εtf g ∞t�0.A drawback of linear stationary models is their failure to account for changing

volatility. They imply that past shocks have a linear effect and that the width of theforecast intervals remains constant even as new data become available, unless theparameters of the model are changed. We consider a linear model AR 1� �,

yt � βyt�1 � εt; with β < 1 (19.4)

εt ∼white noise 0; σ2ε� �

(19.5)

The model admits a Wold representation as

yt �X ∞

i�0 αiεt�i and α0 � 1 (19.6)

We note the effect of the shock εt�i is linear and equal to αi; moreover, the width ofthe forecast interval is proportional to the square root of the one-step forecast errorvariance:

Var yt � E yt jyt�1� �� Var εt� � � σ2ε � constant (19.7)

The conditional variance, Var yt jyt�1� � � Var εt� � � σ2ε , remains constant regard-less of the given data. However, actual financial time series often show sudden burstsof high volatility and contradict the linearity of past shocks and constant conditionalvariance. For example, if a recent innovationwas strongly negative (indicating a crash,etc.), a period of high volatility will often follow. In the ARCH model, the forecastintervals are able to widen immediately to account for sudden changes in volatility,without changing the parameters of the model. Because of this feature, ARCH (andother related) models have become a very important element in the analysis offinancial time series.

The novelty of the ARCH model is that it allows the conditional variance todepend on the data. But it is not the variance itself, which changes with time accordingto an ARCH model; rather, it is the conditional variance that changes, in a specific


WEBC19 05/20/2014 10:17:36 Page 402

way, depending on the available data. The conditional variance quantifies ouruncertainty about the future observation, given everything we have seen so far.This is of more practical interest to the forecaster than the volatility of the seriesconsidered as a whole.

Modeling volatility is key for pricing asset derivatives. Time-varying volatilitywas initially studied in the context of an autoregressive conditional heteroskedasticity(ARCH) model. ARCH model concentrates on the volatility dynamic. A volatilitymodel is defined by its first and second moment, which can be referred to as the meanand variance equation. A simple example of nonlinear dynamic model, introduced tocapture the variability of risk, measured by the “volatility” is ARCH of order 1, orARCH(1) is presented as

Mean equation: yt � σtεt; εt ∼N 0; 1� � (19.8)

Variance equation: σ2t � ω � ϕy2t�1 (19.9)

yt = excess returns on asset prices, and σ2t = volatility of the asset excess returns.Volatility is a measure of the risk on returns. Each observed data point yt has astandard deviation σt. In the ARCHmodel, volatility is a deterministic function of thesquares of past returns and therefore can be estimated from these returns. Thevariance equation specifies the way in which the conditional variance is determinedby the available information. Note that it is defined as a linear function of squares ofpast innovations. The conditional variance of yt is dependent on the realized value ofy2t�1. If the realized value of y2t�1 is large, the conditional variance in t will be large aswell. The conditional variance follows a first-order autoregressive process denoted byARCH(1). As opposed to a usual autoregression the coefficients ω and ϕ have to berestricted. In order to ensure that the conditional variance is never negative, it isnecessary that both ω > 0 and ϕ � 0. Moreover, to ensure stability of the autore-gressive process, ϕ is restricted such that 0 < ϕ < 1.

The ARCH model for the series yt� �

is defined by specifying the conditionaldistribution of yt via its moments; here the conditional mean using the mean equationand conditional variance using the variance equation, given the information available upto time t � 1. LetΨt�1 denote this information; it consists of the knowledge of all availablevalues of the series, and anything that can be computed from these values, for example,innovations and squared observations. In principle, it may even include the knowledge ofthevaluesofother related time series, andanythingelse thatmightbeuseful for forecastingand is available by time t � 1. The conditional distribution of yt is written as

yt jΨt�1 ∼ E yt jΨt�1� �

;Var yt jΨt�1� ��

(19.10)

We note that

E yt jΨt�1� � � E σtεt� � � σtE εt� � � 0 (19.11)

and

Var yt jΨt�1� � � E yt � E yt jΨt�1

� �� 2 � Ey2t � σ2t Var εt� � � σ2t (19.12)

402 STATISTICS

WEBC19 05/20/2014 10:17:37 Page 403

We write the conditional distribution of yt as

yt jΨt�1 ∼ 0; σ2t � ω � ϕy2t�1� �

(19.13)

We note that y2t is a random variable and its conditional mean is σ2t . We maydefine its dispersion as

ηt � y2t � σ2t (19.14)

The random variable ηt is called a martingale difference because it has aconditional mean equal to zero. The variance equation may be rewritten as anautoregressive process:

y2t � ω � ϕy2t�1 � ηt (19.15)

For this process to be stationary the coefficient is restricted to ϕ < 1.Wemay thusestimate the coefficients ω, and ϕ by estimating an AR process in y2t .

Example: We consider excess returns on the weekly S&P 500 stock index duringJanuary 2000 to May 2013. Using EVIEWS, we estimate an ARCH model:

Mean equation: yt � σtεt; εt ∼N 0; 1� �Variance equation: σ2t � ω � ϕε2t�1

σ2t � 4:0699 � 0:4057ε2t�1

We find ε2T � 1:7948; we forecast a one-period ahead excess return, yT�1 �E yT�1 jyT� � � 0; the conditional variance of the forecast is σ2T�1 � 4:0699 � 0:4057�1:7948 � 4:5484.

ARCH model has been specified in alternative forms. Assuming an AR p� � for ytARCH may be written as

Mean equation: yt � μ � β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � εt; εt ∼N 0; σ2t� �

(19.16)

Variance equation: σ2t � ω � ϕε2t�1 (19.17)

It is possible to write an ARCH model in the following form:

Mean equation: yt � μ � β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � εt; εt � νtσt; νt ∼N 0; 1� �(19.18)

Variance equation: σ2t � ω � ϕε2t�1 (19.19)

An ARCH(1) has one lag in the variance equation. An ARCH q� � relates the errorvariance to q lags of squared errors:

σ2t � ω � α1ε2t�1 � α2ε2t�2 � ∙ ∙ ∙ � αqε2t�q (19.20)


WEBC19 05/20/2014 10:17:38 Page 404

Often, the ARCH model uses ht for the variance σ2t , that is, ht � σ2t ; the ARCHmodel is written

Mean equation: yt � εtffiffiffiffiffiht

p; εt ∼white noise 0; 1� � (19.21)

Variance equation: ht � ω � ϕε2t�1 (19.22)

The ARCH model may also be written as the conditional distribution of theprocess {εt}, given the available information ψt�1,

Mean equation: εt jψt�1 ∼N 0; ht� � (19.23)

Variance equation: ht � ω �Xq

i�1 αiε2t�i (19.24)

where ω > 0, αi � 0 for all i andPq

i�1 αi < 1. The conditional distribution of {εt} givenψt�1 is normal with conditional mean E εt jψt�1

� � � 0 and conditional varianceVar εt jψt�1

� � � ht.

PROPERTIES OF THE ARCH MODEL

Parallel to the success of standard linear time series models, arising from the use of theconditional versus the unconditional mean, the key insight offered by the ARCHmodel lies in the distinction between the conditional and the unconditional variances.While the unconditional variance for the variable of interest may be time invariant, theconditional variance often depends on past shocks. Understanding the exact nature ofthis temporal dependence is crucially important for many issues in finance, such asirreversible investments, option pricing, the term structure of interest rates, andgeneral dynamic asset pricing relationships. Also, from the perspective of econometricinference, the loss in asymptotic efficiency from neglected heteroskedasticity may bearbitrarily large and, when evaluating economic forecasts, a much more accurateestimate of the forecast error uncertainty is generally available by conditioning on thecurrent information set.

To illustrate the importance of conditional versus unconditional expectation (i.e.,moments) we consider an ARCH(1) for the excess return as


Variance equation: σ2t � ω � ϕy2t�1 (19.26)

The conditional mean is

E yt jψt�1� � � 0 (19.27)

and conditional variance is

Var yt jψt�1� � � σ2t � ht (19.28)

404 STATISTICS

WEBC19 05/20/2014 10:17:38 Page 405

The unconditional mean is

E yt� � � 0 (19.29)

The unconditional variance is

Var yt� � � E yt � E yt

� �� 2 � Ey2t (19.30)

Since the variance equation may be written as

y2t � ω � ϕy2t�1 � ηt (19.31)

The unconditional mean of y2t is

Ey2t � ω1 � ϕ

(19.32)

We note that random values of the process yt� �

are uncorrelated, that is,

E ytyt�j �

� 0 for any t ≠ j (19.33)

However, the random values of the process y2t� �

are correlated because

E y2t y2t�j

�� E ω � ϕy2t�1 � ηt

� �y2t�j

�� E ωy2t�j

�� E ϕy2t�1y2t�j

�≠ 0: (19.34)

Hence, the values of the process yt� �

are uncorrelated, but they are notindependent. In ARCH, the εtf g sequence is serially uncorrelated since for allj ≠ 0, Eεtεt�j � 0. The key point is that the errors are not independent becausethey are related through their second moment. The conditional variance itself is anautoregressive process resulting in conditionally heteroskedastic errors. In accordancewith the stylized facts for asset returns, there is a tendency for large (small) absolutevalues of the process to be followed by other large (small) values of unpredictable sign.When the realized value of εt�1 is far from zero, so that α1ε2t�1 is large, the variance ofεt will tend to be large. The conditional heteroskedasticity in εtf g will result in yt

� �being an ARCH process. Thus, the ARCH model is able to capture periods oftranquillity and volatility in the yt

� �series.

We undertake the computation of conditional and unconditional moments for anARCH q� � of the form

Mean equation: yt � μ � β1yt�1 � εt; εt � νtσt; νt ∼N 0; 1� � (19.35)

Variance equation: σ2t � ω � α1ε2t�1 � α2ε2t�2 � ∙ ∙ ∙ � αqε2t�q (19.36)

The conditional mean is

E yt jyt�1� � � μ � β1yt�1 (19.37)


WEBC19 05/20/2014 10:17:39 Page 406

The conditional variance is

Var yt jyt�1� � � E yt � E yt jyt�1� �� 2 � E μ � β1yt�1 � εt � μ � β1yt�1� �2 � E εt� �2 � σ2t

(19.38)

For β < 1, the solution of yt in terms of εt is

yt � μ 1 � β1 � β21 � β31 � ∙ ∙ ∙� � � εt � β1εt�1 � β21εt�2 � β31εt�3 � ∙ ∙ ∙ (19.39)

The unconditional mean is

E yt� � � μ

1 � β1(19.40)


Var yt� �� E yt �E yt

� �� 2� E μ 1�β1�β21�β31� ∙ ∙ ∙

� �� εt �β1εt�1�β21εt�2�β31εt�3� ∙ ∙ ∙ � μ1�β1

� 2

� Eε2t �β21Eε2t�1�β41Eε2t�2�β61Eε2t�3� ∙ ∙ ∙

From the variance equation, we obtain

Eε2t � ω � α1Eε2t�1 � α2Eε2t�2 � ∙ ∙ ∙ � αqEε2t�q

The unconditional variances of εt�i’s are equal; namely, Eε2t � Eε2t�1� Eε2t�2 � Eε2t�3 � ∙ ∙ ∙ .We have, therefore, ε2t � ω

1�Pq

i�1 αi. By substituting into the formula forVar yt

� �we

find

Var yt� � � ω

1 �Pqi�1 αi

� �1

1 � β21

" #(19.41)

Example: We consider excess returns on the weekly S&P 500 stock index duringJanuary 2000 to May 2013. EViews estimates ARCH(2) of the form

Mean equation: yt � μ � β1yt�1 � εt; εt � νtσt; νt ∼N 0; 1� �Variance equation: σ2t � ω � α1ε2t�1 � α2ε2t�2

We obtain

yt � 0:21 � 0:06yt�1 and σ2t � 3:53 � 0:38ε2t�1 � 0:105ε2t�2The unconditional mean is

E yt� � � μ

1 � β1� 0:211 � 0:06

� 0:22

406 STATISTICS

WEBC19 05/20/2014 10:17:40 Page 407


Var yt� � � ω

1 �Pqi�1 αi

� �1

1 � β21

" #� 3:53

1 � 0:38 � 0:105

� �1

1 � 0:062

� �� 6:88

THE GENERALIZED ARCH (GARCH) MODEL

The ARCHmodel assumes that positive and negative shocks have the same effects onvolatility because it depends on the square of the previous shocks. In practice, it is wellknown that price of a financial asset responds differently to positive and negativeshocks. Further, it is not a parsimonious representation of the model and requiresestimation of a large number of parameters. Due to large persistence in volatility theARCHmodel often requires a large number of lags q to fit the data. The ARCHmodelwas extended by Bollerslev (1986) to become a generalized autoregressive conditionalheteroskedasticity (GARCH) model. GARCH is probably the most commonly usedfinancial time series model. GARCH is more parsimonious than ARCH and avoidsoverfitting. A GARCH(1,1) has the following form


Variance equation: σ2t � ω � ϕy2t�1 � βσ2t�1 (19.43)

In this version of the model, the volatility, the conditional variance, is a determi-nistic function of the squares of past returns as well as of past volatility. The GARCHformulation introduces terms analogous to the moving average terms in an ARMAmodel, thereby making forecast of volatility a function of a distributed lag of pastsquared volatility. Parameters are restricted as follows: ω > 0, ϕ � 0, β � 0, andβ � ϕ < 1.

Example: We consider excess returns on the weekly S&P 500 stock index duringJanuary 2000 to May 2013. We estimate GARCH(1,1). We obtain

σ2t � 0:38 � 0:21ε2t�1 � 0:75σ2t�1

We display the conditional variance of the returns in Figure 19.5.A GARCH(p; q� model is defined by

Mean equation: yt � σtεt; εt ∼ iid 0; 1� � (19.44)

Variance equation: σ2t � w �Xpi�1

αiy2t�i �Xqj�1

βiσ2t�j (19.45)

where w > 0, αi � 0, and βi � 0,Pp

i�1 αi �Pqj�1 βi < 1. The main idea is that σ2t , the

conditional variance of yt given information available up to time t � 1, has anautoregressive structure and is positively correlated to its own recent past and torecent values of the squared returns y2. This captures the idea of volatility (i.e.,


WEBC19 05/20/2014 10:17:40 Page 408

conditional variance) being persistent: large (small) values of y2t are likely to befollowed by large (small) values.

Now let the error process be such that

εt � etffiffiffiffiffiht

p; et ∼white noise 0; σ2e � 1

� �(19.46)

ht � α0 �Xqi�1

αiε2t�i �Xpi�1

δiht�i (19.47)

Since etf g is a white-noise process that is independent of past realizations of εt�i,the conditional and unconditional means of εt are equal to zero. By taking theexpected value of εt, it is easy to verify that

Eεt � Eetffiffiffiffiffiht

p � 0 (19.48)

The important point is that the conditional variance of εt is given by

E ε2t jΨt�1� � � ht (19.49)

The key feature of GARCH models is that the conditional variance of thedisturbances of the yt

� �sequence constitutes an ARMA process. Hence, it is to be

expected that the residuals from a fitted ARMA model should display this character-istic pattern. To explain, suppose you estimate yt

� �as an ARMA process. If this

model is adequate, the autocorrelation function (ACF) and the partial ACF (PACF)should be indicative of a white-noise process. The ACF of the squared residuals canhelp identify the order of the GARCH process. Since E ε2t jΨt�1

� � � ht, we can write

ht � E ε2t jΨt�1� � � α0 �

Xqi�1

αiε2t�i �Xpi�1

δiht�i (19.50)

0

20

40

60

80

100

120

2000 2002 2004 2006 2008 2010 2012

FIGURE 19.5 Conditional Variance of the Returns on the S&P500 Stock Index, January 2000 to May 2013Source: Yahoo.finance.

408 STATISTICS

WEBC19 05/20/2014 10:17:41 Page 409

The previous equation looks very much like an ARMA q; p� � process in the ε2t� �

sequence.

ARCH-GARCH IN MEAN

Many theories in finance call for an explicit trade-off between the expected returnsand the variance. In the case of financial assets we expect higher risk to be compen-sated by a higher return. For example, according to the traditional capital asset pricingmodel (CAPM) the excess returns on all risky assets are proportional to the non-diversifiable risk as measured by the covariances with the market portfolio. Thisimplies that the expected excess return on the market portfolio is simply proportionalto its own conditional variance. As such, it would be useful to have a model in whichthe return is partly determined by its risk. The ARCH-M model is often used infinancial applications where the expected return on an asset is related to the expectedasset risk. If we introduce the conditional variance into the mean equation, we get theARCH-in-Mean (ARCH-M) model. A GARCH-M may be represented as follows:

Mean equation: yt � μ � δσt�1 � εt εt ∼N 0; σ2t� �

(19.51)

Variance equation: σ2t � α0 � α1ε2t�1 � β1σ2t�1 (19.52)

δ can be interpreted as a sort of risk premium. If δ > 0 and it is significant (t-test),there is a trade-off between the mean (return) and the conditional variance (timevarying risk).

Example:We consider returns on the weekly S&P 500 stock index during January2000 to May 2013. We estimate a GARCH-M of the form yt � μ � δσt�1 � εt andσ2t � α0 � α1ε2t�1 � β1σ2t�1. We find

yt � 0:0255 � 0:097σt�1

σ2t � 0:395 � 0:229ε2t�1 � 0:726σ2t�1

Example:We consider returns on the weekly S&P 500 stock index during January2000 to May 2013. We estimate a GARCH-M of the form yt � μ � δσ2t�1 � εt, andσ2t � α0 � α1ε2t�1 � β1σ2t�1. We find

yt � 0:146 � 0:0164σ2t�1

σ2t � 0:395 � 0:229ε2t�1 � 0:726σ2t�1

TESTING FOR THE ARCH EFFECTS

The intuition behind the test for ARCH effects is very clear. If the data arehomoskedastic, then the variance cannot be predicted and variations in ε2t will bepurely random. If ARCH effects are present, large values of the past squared residuals


WEBC19 05/20/2014 10:17:41 Page 410

will predict large values of ε2t . If there is conditional heteroskedasticity, the correlo-gram of ε2t should be suggestive of such process. The technique to construct thecorrelogram of the squared residuals is:

i. Estimate yt� �

sequence use the best-fitting ARMA model and obtain the squaresof fitted error ε2t . Calculate the sample variance of the residuals σ2 defined as

σ2 � XTi�1

ε2t =T (19.53)

where T � number of residuals.ii. Calculate and plot the sample autocorrelations of the squared residuals as

ρ i� � �PT

t�i�1 ε2t � σ2� �

ε2t�i � σ2� �

PTt�1 ε2t � σ2

� �2 (19.54)

iii. In large samples, the standard deviation of ρ i� � can be approximated by 1=ffiffiffiffiT

p.

Individual values of ρ i� � with a value that is significantly different from zero areindicative of GARCH errors. Ljung-Box Q-statistics can be used to test for groupsof significant coefficients; the statistic is

Q � T T � 2� �Xhi�1

ρ i� �= T � i� � (19.55)

Q has an asymptotic χ2 distribution with h degrees of freedom if theε2t are uncorrelated. Rejecting the null hypothesis that the ε2t are uncor-related is equivalent to rejecting the null hypothesis of no ARCH or GARCHerrors.

The more formal Lagrangian multiplier LM� � test for ARCH disturbances hasbeen proposed by Engle. The two steps of the tests are:

i. Use OLS to estimate the most appropriate AR p� � model:

yt � β0 � β1yt�1 � β2yt�2 � ∙ ∙ ∙ � βpyt�p � εt (19.56)

ii. Obtain the squares of the fitted errors ε2t . Regress these squared residuals on aconstant and on the q lagged values, ε2t�1, ε2t�2, . . . , ε2t�q; that is, estimate

ε2t � α0 � α1ε2t�1 � α2ε2t�2 � ∙ ∙ ∙ αqε2t�q (19.57)

If there are no ARCHorGARCH effects, the estimated values of α1; . . . ; αq shouldbe zero. Hence, this regression will have little explanation power so that the coefficientof determination R2 will be quite low. With a sample of T residuals, under the null

410 STATISTICS

WEBC19 05/20/2014 10:17:42 Page 411

hypothesis of no ARCH errors, the test statistic T � R2 converges to χ2q distribution. IfT � R2 is large, rejection of the null hypothesis that α1; . . . ; αq are jointly equal to zerois equivalent to rejecting the null hypothesis of no ARCH errors. On the other hand, ifT � R2 is sufficiently low, it is possible to conclude that there are no ARCH effects.

Example: We consider excess returns on the weekly S&P 500 stock index duringJanuary 2000 to May 2013. We estimate an GARCH(1,1) of the form

Mean equation: yt � μ � β1yt�1 � β2yt�2 � εt; εt � νtσt; νt ∼N 0; 1� �Variance equation: σ2t � ω � α1ε2t�1 � β1σ

2t�1

We obtain yt � 0:235 � 0:081yt�1 � 0:00726yt�2 and σ2t � 0:398 � 0:232ε2t�1�0:721σ2t�1.

We perform an LM test for the GARCH effects; we find T � R2 � 6:2, and theprobability value of χ2 1� � is 0.0128. We reject the null hypothesis of no GARCHeffects.

SUMMARY

Islamic finance applies ARCH-GARCH models to estimate volatility and assess risksof assets. This chapter provides motivation for ARCH models, namely returns andshocks have a nonlinear relationship that has to be accounted in predicting returns.The chapter provides a formalization of the ARCH model; it studies the properties ofthe ARCH model; it describes the generalized ARCH (GARCH) model, the ARCH-GARCH in mean, and the testing for the ARCH effects.

QUESTIONS

1. Assume the returns of a stock price index are shown by an MA 2� �yt � μ � εt � α1εt�1 � α2εt�2

The model predicts that if the market is hit by a negative shock of the order of 2percent, returns will be affected by �α1 � 2 percent at t � 1 and �α2 � 2 percent att � 2; likewise, if the market is hit by a negative shock of the order of 20 percent,returns will be affected by �α1 � 20 percent at t � 1 and �α2 � 20 percent at t � 2.The model predicts also that if the market is hit by a positive shock of 25 percent,returns will be affected by α1 � 25 percent at t � 1 and α2 � 25 percent at t � 2.Are these linear predictions confirmed by the weekly S&P 500 stock index duringJanuary 2009 to September 2013?

2. Consider the excess returns on the weekly S&P 500 stock index duringJanuary 2009 to September 2013; using EViews, estimate an ARCH model ofthe form

Mean equation: yt � σtεt; εt ∼N 0; 1� �Variance equation: σ2t � ω � ϕε2t�1


WEBC19 05/20/2014 10:17:42 Page 412

Plot the graph of the conditional variance. Make a forecast of excess return three-period ahead and a forecast of conditional variance three-period ahead.

3. Consider returns on the weekly S&P 500 stock index during January 2009 toSeptember 2013; estimate ARCH(2) of the form

Mean equation: yt � μ � β1yt�1 � εt; εt � νtσt; νt ∼N 0; 1� �Variance equation: σ2t � ω � α1ε2t�1 � α2ε2t�2

Compute the unconditional mean and the unconditional variance.

4. Consider excess returns on the weekly S&P 500 stock index during January 2009to September 2013. Estimate GARCH(1,1). Plot the conditional variance; make aforecast of the conditional variance three-period ahead.

5. Consider returns on the weekly S&P 500 stock index during January 2009 toSeptember 2013. Estimate a GARCH(1,1) of the form

Mean equation: yt � μ � β1yt�1 � β2yt�2 � εt; εt � νtσt; νt ∼N 0; 1� �Variance equation: σ2t � ω � α1ε2t�1 � β1σ

2t�1

Test for the GARCH effects using LM test.

6. Consider returns on the weekly S&P 500 stock index during January 2009 toSeptember 2013. Estimate a GARCH-M of the form yt � μ � δσ2t�1 � εt andσ2t � α0 � α1ε2t�1 � β1σ2t�1. Test for the GARCH effects using the LM test.

412 STATISTICS

3GC20 05/15/2014 13:13:26 Page 413

CHAPTER 20Asset Pricing under Uncertainty

A sset pricing under uncertainty uses the theory of stochastic processes. Asset pricingunder certainty is fairly simple: the price of an asset is the present value of its

certain future payoffs discounted by a risk-free return. However, with uncertainty,stocks’ and sukuks’ payoffs are uncertain; pricing of these assets is no longer as simpleas under certainty. Moreover, with uncertainty, many assets that do not exist in acertainty environment appear; they are called derivatives and have also to be priced.Theories of asset pricing under uncertainty cover stocks, sukuks, and derivatives.

Uncertainty is described in terms of a statistical probability distribution withexpected mean and standard deviation. Covariance plays a role in measuring riskamong assets. It is used to determine the systemic risk of an asset in relation to marketportfolio. Uncertainty is also described using two random processes that dominatedebate in capital market efficiency theory; these are the random walk and themartingale processes.

Even though market participants have subjective probabilities regarding thefuture payoff of a specific asset and may be risk averse or risk lover, they stillhave to agree on one common set of probabilities called risk-neutral probabilities toestablish equilibrium price. The probability distribution of a financial asset cannot beapplied to pricing the asset or derivatives on the asset without transforming it into arisk-neutral probability distribution. Moreover, assets or derivatives have to be pricedat equilibrium price that admits no arbitrage; the asset should not be undervalued orovervalued. For instance, if a stock is priced differently from the capital asset pricingmodel (CAPM) price, traders will make a costless profit by selling the stock if it isovervalued or buying it if it is undervalued.Moreover, if an asset is priced at a differentprice than its replicating portfolio, arbitrageurs will make a costless profit byexploiting this price difference. If an asset is overvalued, they will sell it and buyits replicating portfolio. If it is undervalued, they will buy it and sell its replicatingportfolio. Asset pricing theory under uncertainty relies on three equivalentapproaches:

1. Arbitrage-free pricing2. Replication of an asset3. Risk-neutral pricing

The theory of arbitrage-free pricing shows that each asset can be replicated by ahedging portfolio. Efficiency of capital markets requires that asset prices be free ofarbitrage. The price of an asset is equal to that of its replicating portfolio to preclude

413

3GC20 05/15/2014 13:13:26 Page 414

arbitrage opportunities, defined as enjoying profit with zero cost. Risk-neutral pricing isequivalent to portfolio replication.However, it ismuch simpler, because it computes theasset pricewithout requiring the knowledge of the arbitrage-free replicating portfolio. Itcomputes a risk-neutral distribution by transforming the asset price process into amartingale.The fundamental pricingprinciple asserts that thepriceof anasset is equal tothe expected value of its payoffs under the risk-neutral distribution using the risklessdiscount rate.We compute state prices from risk-neutral probabilities and show that thepriceof anyasset is its discounted expectedvalueunder the risk-neutral distribution.Thediscount rate in risk-neutral pricing is necessarily the risk-free rate. Risk-neutral pricingasserts that asset pricing should be fair game; namely, once assets are adjusted for theirrisk, they should yield the same return, which is the risk-free rate.

MODELING RISK AND RETURN

Financial markets may exhibit high uncertainty; therefore, investorsmay face high risk.The price of an asset at time t= 0 is knownand is denoted by S0. The price of an asset at afuture time T is not known today; it is denoted by ST. It is a random variable and theremay be great uncertainty regarding the value it will take. Yet studying uncertainty of STis essential to pricing assets andderivatives.Uncertaintyhasbeen studied in two settings:in a discrete-time setting using a binomial tree approach and in continuous-time settingusing a Brownian motion, known also asWiener process, because Norbert Wiener wasthe mathematician who formulated the properties of the Brownian motion. The twoapproaches are closely linked because a Brownian motion can be constructed from abinomial tree as the time interval becomes infinitesimally small. We describe thebinomial model for modeling uncertainty. We discuss popular models used for stockprice behavior based on the market efficiency hypotheses; these are the random walkand the martingale stochastic processes.

Consider a binomial probability distribution for the random share price STwhereST assumes two possible outcomes (Figure 20.1, Tree 1). It may go up from S0 to ST =u.S0 with probability θ > 0 or down from S0 to ST = d.S0 with probability 1 � θ� �. Thetwo possible outcomes u.S0 and d. S0 are called state of the world at time T. Only one

Time 0 Time 0 Time 0Time T

ST = u.S0

ST = d.S0

ST = u.S0 = $110

ST = d.S0 = $90

ST = d.S0 = $70

S0 S0 = $100S0 = $100

ST = u.S0 = $130

Time T Time T

Tree 1 Tree 3Tree 2

FIGURE 20.1 Uncertainty Described by a Binomial Tree

414 STATISTICS

3GC20 05/15/2014 13:13:27 Page 415

Time T states of the worldST (ω1)

Transitionprobabilities

ST (ω2)

S0

ST (ωN)

θ2

θ1

θN

FIGURE 20.2 Uncertainty Described by a Multinomial Tree

state of the world will occur at time T. If you toss a coin, only one outcome will occur,either heads or tails.

Consider a binomial tree (Figure 20.1, Tree 2) with initial share price ST = $100;the share price may go up by u � 1:1 to ST = $110 with a probability θ � 0:65 or godown by d � 0:9 to ST = $90 with probability 1 � θ� � � 0:35. Consider anotherbinomial tree (Figure 20.1, Tree 3) with S0 � $100. The share price may go up byu � 1:3 to ST = $130 with a probability θ � 0:65 or go down by d � 0:7 to ST = $70with probability 1 � θ� � � 0:35. Comparing Tree 2 and Tree 3, there is moreuncertainty regarding ST in Tree 3 than in Tree 2. The wider the jaw of the binomialtree, the higher is the uncertainty. The ratio u=d is larger in Tree 3 than in Tree 2; morespecifically (1.3/0.7) = 1.86 > (1.1/0.9) = 1.22. Hence, uncertainty is measured by thevariance of a random variable and not by its probability distribution.

In general, the states of the world at future time T could be more than two states(Figure 20.2). Assume there areN possible states of the world at timeT. Denote state 1by ω1, state 2 by ω2, . . . , and state N by ωN. Then the possible values of ST can bedescribed by the random variables ST ω1� �, ST ω2� �, . . . , ST ωN� �. Let the probability oftransition from S0 to state ω1 be θ1 > 0, to state ω2 be θ2 > 0; . . . , and to state ωN beθN > 0, with

θ1 � θ2 � ∙ ∙ ∙ � θN � XNj�1

θj � 1; j � 1; 2; . . . ;N (20.1)

The random variable ST has a probability distribution described by its states of theworld at time T and the transition probabilities from S0 to each possible state of theworld. A probability distribution is described by its moments. First moment is calledthe expectedmean, secondmoment is called variance, third moment is called skewness,and fourth moment is called kurtosis. The expected mean of ST is denoted by Eθ

0 ST� �; itindicates that expectations are computed at time t= 0 under the probability distributionθ1; θ2; . . . ; θN. It may be simply written as E ST� �. It is defined as

Eθ0 ST� � � θ1ST ω1� � � θ2ST ω2� � � ∙ ∙ ∙ � θjST ωj

� � � ∙ ∙ ∙ � θNST ωN� �� XN

j�1θjST ωj

� �j � 1; 2 . . . ;N

(20.2)

Asset Pricing under Uncertainty 415

3GC20 05/15/2014 13:13:29 Page 416

The uncertainty, or the risk, associated with ST is measured by the variance andthe standard deviation of ST. More specifically, the variance is

Var ST� � � Eθ0 ST � Eθ

0 ST� �� 2(20.3)

The expectation operator Eθ0 indicates the variance is computed at time t � 0

under the probability distribution θ1; θ2; . . . ; θN: Notation may be simplified toVar ST� � � E ST � E ST� �� 2. The variance is written in summation form as

Var ST� � � XNj�1

θj ST ωj� � � E ST� �� 2

; j � 1; 2 . . . ;N (20.4)

The variance is expressed in squared unit of the random variable. To be able to usethe same unit of measurement as the random variable, we compute the standarddeviation of ST ; it is denoted by σT

σT � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiVar ST� �p

(20.5)

The mean and standard deviation are expressed in the same unit as the randomvariable; the standard deviation analyzes the spread of the statistical distributionaround the mean.

Example: Measuring uncertainty, expected mean, and standard deviation.Consider a stock with S0 � $120. It has a payoff (i.e., dividends + capital gain or

loss) for each of the 10 scenarios that may occur next year, as shown in Table 20.1.We assume a probability distribution associated with the payoff of each state. Theexpected mean payoff is $10.85, standard deviation is $15.29, and expected returnis 9.04 percent.

UNCERTAINTY AND EFFICIENT CAPITAL MARKETS: RANDOMWALK AND MARTINGALE

The analysis of the random process of a financial variable is a fundamental topic inpricing of financial assets and derivatives. Louis Bachelier (1900) analyzed the randomprocess of financial series. He modeled the random process of a financial variable incontinuous time in terms of Brownian motion. His work provides the basic model foranalyzing market efficiency and asset pricing. Theories of capital market efficiencyhave defined market efficiency in terms of two random processes: the random walkand the martingale process. Namely, markets are efficient if asset prices follow either

TABLE 20.1 Stock Payoff Scenarios and Associated Probability Distribution One Year (t = 1)from Today (t = 0)

State 1 2 3 4 5 6 7 8 9 10Stock payoff in $ 30 �10 �5 8 34 23 �4 �3 0 26Probabilities 0.08 0.06 0.07 0.1 0.09 0.12 0.17 0.06 0.11 0.14

416 STATISTICS

3GC20 05/15/2014 13:13:30 Page 417

process. Both processes continue to dominate the debate on capital market efficiencyand have proponents and opponents. Both processes are formulated in discrete timeand in continuous time.*

The Random Walk

The basic hypothesis is that markets are efficient; namely, the current price of an assetSt reflects all available information and expectations of traders and follows a randomwalk. The random walk theory asserts that price movements will not follow anypattern or trend and that past price movements cannot be used to predict future pricemovements. The theory states that stock price changes have the same distribution andare independent of each other, so the past movement or trend of a stock price ormarket cannot be used to predict its future movement. The simplest version of therandom walk hypothesis is the independently and identically distributed (iid) incre-ments case in which

St�1 � St � εt�1 εt�1 ∼ iid 0; σ2� �

(20.6)

The expected change in stock prices can be written as

E St�1 � St� � � E εt�1� � � 0 (20.7)

Equation (20.7) satisfies the definition of a fair game—that is, a game that isneither in your favor nor in your opponent’s. Independence of εt�1f g implies that arandom walk is a fair game but in a much stronger sense than the martingale;independence implies not only that increments εt�1 are uncorrelated but that anynonlinear function of increments, such as ε2t�1 or ε3t�1, are also uncorrelated. Prices ofthe asset will move under unexpected news and purely random factors that cannot bediscounted by the market in the present time t. If company XYZ has projected higherdividends next semester, then St moves up now to reflect this information. When thecompany actually distributes higher dividends than planned, this event will have noadditional effect on stock prices because the effect had already been incorporated onesemester earlier.

The random walk theory states that securities’ prices are random and notinfluenced by past events. The idea is also referred to as the weak-form efficient-market hypothesis. Randomness of stock prices renders attempts to find price patternsor take advantage of new information futile. In short, random walk is the idea thatstocks take a random and unpredictable path. A follower of the random walk theorybelieves it is impossible to outperform the market without assuming additional risk.For instance, strategies of active trading do not beat the buy-and-hold strategy. Criticsof the theory, however, contend that stocks do maintain price trends over time—inother words, it is possible to outperform the market by carefully selecting entry andexit points for equity investments.

Example: Testing the random walk hypothesis.We test the random walk hypothesis for the weekly S&P 500 stock index during

March 2009 to May 2013. We compute the excess returns. Using EViews, we

*In continuous time, a random walk is approximated by a Brownian motion.


3GC20 05/15/2014 13:13:32 Page 418

compute the sample autocorrelation at lags 1, 2, 3, 4, 5, and 6, along the Ljung-BoxQ-statistic and the probability values as shown in Table 20.2.

The test does not reject the random walk hypothesis; weekly returns of the stockindex were highly uncorrelated. For instance, at lag 1, the Ljung-Box Q-statistic is1.3507 with probability value of 0.245. The test confirms randomness of stock pricesand market efficiency.

The Martingale

The martingale process has become a fundamental model of uncertainty, pricingassets, and studying capital market efficiency. Knowledge of martingale theory isimportant to understand the literature on market capital efficiency and asset pricing.In probability theory, a martingale is a model of a fair game where no knowledge ofpast events can help to predict future winnings. A fair game is the essence of amartingale. In particular, a martingale is a sequence of random variables (i.e., astochastic process) for which, at a particular time in the realized sequence, theexpectation of the next value in the sequence is equal to the present observed value,even given knowledge of all prior observed values at a current time. A basic definitionof a discrete-time martingale is a discrete-time stochastic process Stf g (i.e., a sequenceof random variables) that satisfies for any time t the following condition:

E St�1 jSt; St�1; . . . :� � � St (20.8)

Due to the linearity of expectation this requirement is equivalent to

E St�1 � St jSt; St�1; . . . :� � � 0 (20.9)

Themartingale is a process that has no drift. Equation (20.9) defines a fair game; itstates that the average winnings from observation t to observation t + 1 are 0. If Strepresents one’s cumulative winnings or wealth at date t from playing some game ofchance each period, then a fair game is one for which the expected wealth next periodis simply equal to this period’s wealth conditioned on the history game. If St is taken tobe an asset’s price at date t, then the martingale hypothesis states that tomorrow’sprice is expected to be equal to today’s price, given the asset’s entire price history.Alternatively, the asset’s expected price change is zero when conditioned on the asset’sprice history; hence, its price is just as likely to rise as it is to fall. From a forecastingperspective, the martingale hypothesis implies that the best forecast of tomorrow’sprice is simply today’s price, where best means minimal mean-squared error.

Another aspect of the martingale is that nonoverlapping price changes areuncorrelated at all leads and lags, which implies the ineffectiveness of all linear

TABLE 20.2 Testing Market Efficiency

Lag 1 2 3 4 5 6Sample autocorrelation �0.078 �0.024 0.008 �0.073 �0.028 �0.004Q-statistic 1.3507 1.4826 1.4958 2.6956 2.8775 2.8811Probability value 0.245 0.476 0.683 0.61 0.719 0.824

418 STATISTICS

3GC20 05/15/2014 13:13:33 Page 419

forecasting rules for future price changes based on historical price alone. In fact, themartingale was long considered to be a necessary condition for an efficient assetmarket, one in which information contained in past prices is instantly, fully, andperpetually reflected in asset’s current price. The martingale hypothesis is calledweak-formmarket efficiency. If the market is efficient, then it should not be possible to profitby trading on the information contained in the asset’s price history. Asset-pricingmodels based on risk-neutral pricing have established that once asset returns areproperly adjusted for risk, the martingale property does hold. In particular, marginal-utility-weighted prices do followmartingales under quite general conditions. The risk-adjusted martingale property of asset prices has led to a great simplification in thepricing of complex financial options such as options, swaps, and other derivativesecurities. More specifically, the pricing based on a martingale is simpler than pricingbased on portfolio replication.

Examples: Illustrating Martingale.

■ Random walk without drift:A random walk without drift is an example of a martingale.

■ Tossing a fair coin:Suppose Sn is a gambler’s fortune after n tosses of a fair coin, where the

gambler wins $1 if the coin comes up heads and loses $1 if the coin comes up tails.The gambler’s conditional expected fortune after the next trial, given the history,is equal to the gambler’s present fortune, so this sequence is a martingale.

■ Computing martingale probabilities:A stock price is at S0 � $100 today. For next year, investors predict the stock

price may go up to $130 with probability θ � 0:7 or down to $83with probability1 � θ � 0:3. We can verify that E S1 jS0� � � 0:7 � $130 � 0:3 � $83 � $115:9.Hence E S1 jS0� �

is not equal to S0 � $100. The stock price under the initialprobabilities is not a martingale. To get a martingale, we have to compute pseudo-probabilities that may be different from the true probabilities; however, theysatisfy the martingale condition. If we denote the martingale probabilities by qand 1 � q� �, these probabilities satisfy the martingale property,

E S1 jS0� � � q � $130 � 1 � q� � � $83 � $100:

We find q � 0:3617 and 1 � q� � � 0:6383. We have operated a change ofprobabilities, also called a changeofmeasureor adjustment for risk.Wehave changedtrueprobabilities θ’s intomartingale probabilitiesq’s. The effect ofmaking the gameafair game is to reduce theexpectedvalue from$115.9 to$100,which is the initialprice.Anypricingof aderivativebasedon this stock suchas futuresoranoptioncontracthasto use this martingale probability and not the true or subjective probability.

■ A martingale process:A stock price is today at S0 � $107. Next month, it may be in any of the

following states: $130, $120, $112, $90, and $78, with respective probabilities0.24, 0.15, 0.23, 0.20, and 0.18.We can verify that the stock price process satisfiesthe martingale property; namely, E S1 jS0� � � S0. Taking expectation we find

0:24 � $130 � 0:15 � $120 � 0:23 � $112 � 0:20 � $90 � 0:18 � $78 � $107

■ Simulating a martingale process:


3GC20 05/15/2014 13:13:33 Page 420

We may simulate a martingale using Microsoft Excel. We enter 0.0 in the A1(top left) cell, and in the cell below it (A2) enter =A1+NORMINV(RAND(),0,1).We copy that cell by dragging down to create 300 or so copies. This will create amartingale series with a mean of 0 and standard deviation of 1. With the cells stillhighlighted, we go to the chart creation tool and create a chart of these values.Now, every time a recalculation happens (in Microsoft Excel, the F9 key doesthis), the chart will display another martingale series.

Relationship between the Random Walk and Martingale Models

In the random walk model, the notion of independence implies that the current returndoes not depend on past returns. Consequently, it is impossible to predict the futurereturn using past returns. Clearly, the random walk hypothesis implies the martingalehypothesis but the inverse is not always true. In other words, if {St} follows a randomwalk, then the market is efficient. However, {St} may not follow a random walk whenthe market is efficient. A random walk is more restrictive than a martingale. Themartingale only rules out serial dependence in conditional mean, whereas the randomwalk rules out not only this but also serial dependence involving the higher orderconditional moments of εt�1. More specifically, Var εt�1� � is a constant and indepen-dent of time in a randomwalk. Empirically, however, this is not the case formost high-frequency financial time series that display strong volatility clustering. That is, a highvolatility tends to be followed by another high volatility, and a low volatility tends tobe followed by a low volatility. Statistically speaking, there exists positive auto-correlation in the squares of returns or the absolute returns. Hence, serial correlationof higher-order moments is allowed in martingale, but it is ruled out in the randomwalk process.

MARKET EFFICIENCY AND ARBITRAGE-FREE PRICING

Capital market efficiency is a basic hypothesis of asset pricing. The pricing theory offinancial assets assumes perfect markets, which means there are no transaction costsand short selling is unlimited. This means that traders can sell a commodity that theydo not possess by borrowing that commodity from a broker or counterparty. Itassumes market efficiency, which means absence of arbitrage. It is free-of-arbitragepricing. Assets can be priced by replication. If an asset is replicated by a portfolio ofassets and has the same payoff as the replicating portfolio at every moment in timeprior to the expiration of the maturity, then the price of the replicated asset must equalthe price of the replicating portfolio. Arbitrage is defined as a gain obtained now or atmaturity with no cost through exploiting distortions in prevailing prices of assets.Availability of arbitrage opportunities means capital markets are inefficient.

In economics and finance, arbitrage is the practice of taking advantage of a pricedifferential between two or more markets: striking a combination of matching dealsthat capitalize on the imbalance, the profit being the difference between the marketprices.When used by academics, an arbitrage is a transaction that involves no negativecash flow at any probabilistic or temporal state and a positive cash flow in at least onestate; in simple terms, a risk-free profit is called a free lunch. If the market prices do notallow for profitable arbitrage, the prices are said to constitute arbitrage equilibrium or

420 STATISTICS

3GC20 05/15/2014 13:13:33 Page 421

arbitrage-free prices. Arbitrage equilibrium is a precondition for a general economicequilibrium. The assumption that there is no arbitrage is used in quantitative financeto calculate unique risk-neutral prices for derivatives.

Arbitrage is possible when one of the following conditions is met:

■ The same asset does not trade at the same price on all markets (“the law of oneprice”).

■ Two assets with identical cash flows do not trade at the same price.■ An asset with a known price in the future does not today trade at its future price

discounted at the risk-free interest rate (or, the asset does not have negligible costsof storage; as such, for example, this condition holds for grain but not forsecurities).

■ Two portfolios can be created that have identical payoffs in every state but havedifferent costs.

■ Two portfolios can be created with equal costs, but where the first portfolio has atleast the same payoff as the second in all states but has higher payoff in at least onestate.

■ A portfolio can be created that has zero cost but has a non-negative payoff in allstates and positive payoff in at least one state.

Arbitrage happens when we are able to construct at time 0 some portfolio that hasnet value zero (thus a nontrivial portfolio will have a mixture of positive and negativeholdings that cost zero in total, for example, we borrow cash and buy stocks, or short-sell one type of stock and buy another type of stock) and at some fixed time T in thefuture this portfolio will give us a sure profit, a free lunch. Suppose we can invest in nassets. Asset i has price Pi t� � at time t per unit with no dividends or coupons payable.Suppose we have xi units of asset i; the portfolio of these assets has a value at time t:

V t� � � Xni�1

xiPi t� � (20.10)

The definition of arbitrage then implies that arbitrageurs incur no cost today,

V 0� � � Xni�1

xiPi 0� � � 0 (20.11)

Moreover, they think today with certainty that their arbitrage strategy incurs noloss and probably may yield profit:

Probability V T� � � 0� � � 1 and Probability V T� � > 0� � > 0 (20.12)

Besides the definition already given, the principle of no arbitrage has the followingequivalent forms: We cannot construct a risk-free portfolio that returns more than therisk-free rate of return; if two portfolios A and B give rise to identical (but possiblyrandom) future cash flowwith certainty, then A and Bmust have the same value at thepresent time (the law of one price). Arbitrage is a condition resulting from the fact thattwo identical combinations of assets are selling for different prices. An investor whospots such an opportunity will buy the lower-price combination and sell the higher-price combination.


3GC20 05/15/2014 13:13:34 Page 422

Example: Detecting arbitrage opportunities.Share 1 has an expected return of 8.25 percent and risk ratio defined as expected

return divided by standard deviation of returns equal to 0.5. Its current market price is$120. Share 2 has an expected return of 6.60 percent and a risk ratio equal to 0.41. Itscurrent price is $97. Arbitrageurs notice that share 1 is underpriced and share 2 isoverpriced. Consequently, they short-sell shares 2 and buy shares 1. Their initialinvestment is zero; they make profits by repurchasing shares 2 at a lower price andselling shares 1 at a higher price.

Pricing of Assets by Arbitrage

Here we show how to construct an arbitrage portfolio to exploit mispricing of assets.We also illustrate how arbitrage is applied to price option and forward contracts.

Construction of Arbitrage Portfolio Arbitrageplaysa role in asset pricing.Arbitrageurstry to detect arbitrage opportunities by constructing arbitrage portfolios, especiallywhen arbitrage does not involve one single asset but many assets. In fact, if anarbitrageur compares the price of Company XYZ shares on two stock exchangesand finds difference in prices, then he can easily exploit these differences by selling in ahigher-price market and buying in a lower-price market. However, arbitrage opportu-nities may not be as apparent as in the case of one share. If there are many sharesinvolved, arbitrageurs have to construct and compare portfolios. Let us consider howanarbitrageur can produce an arbitrage opportunity involving three shares: A, B, and C.These shares can be purchased today at prices shown in Table 20.3, and each canproduce only one of two payoffs, referred to as state 1 and state 2, a year from now.

Although it is not obvious from the data constructed in Table 20.3, an investorcan construct a portfolio of assets A and B that will have the identical return as share Cin both state 1 and state 2. Let wA and wB be the proportion of shares A and B,respectively, in the portfolio. Then the payoff (that is, the terminal value of theportfolio) under the two states can be expressed as:

■ If state 1 occurs: $50wA � $30wB � $38■ If state 2 occurs: $100wA � $120wB � $112

We have created a portfolio consisting of A and B that will reproduce the payoff ofC regardless of the state that occurs one year from now. Solving, we findwA � 0:4 andwB � 0:60. The cost of the portfolio today is 0:4� � $70� � � 0:6� � $60� � � $64.

Ourportfolio (i.e., packageof assets) comprising sharesAandBhas the samepayoffin state 1 and state 2 as the payoff of asset C. The cost of share C is $80, while the cost ofthe portfolio is only $64. This is a market inefficiency that can be exploited by buying

TABLE 20.3 Pricing by Arbitrage

Asset Price today Future payoff state 1 Future payoff state 2

A $70 $50 $100B 60 30 120C 80 38 112

422 STATISTICS

3GC20 05/15/2014 13:13:36 Page 423

assets A and B in the proportions given above and shorting (selling) share C. Forexample, suppose that $1million is invested to create the portfolio with shares A and B.The $1million is obtained by selling short shares C. The proceeds from the short sale ofshares C provide the funds to purchase shares A and B. Thus, there would be no cashoutlay by the investor. The payoffs for states 1 and 2 are shown in Table 20.4.

In either state 1 or 2, the investor profits without risk. The arbitrage-free modelassumes that the marketplace would quickly eliminate such an opportunity. Thismeans that short-selling asset C will drive its price to arbitrage-free price equal to $64.

Pricing an Option by Arbitrage Using the replication portfolio of an asset, we price acall option on stock XYZ. The replication portfolio of the option consists of theunderlying stock XYZ and a risk-free sukuk. The prices of stock XYZ and the risk-freesukuk, as well as the future prices, are as indicated in Table 20.5.

Let wA and wB be the proportion of stock XYZ and risk-free sukuk, respectively,in the portfolio. Then the payoff (i.e., the terminal value of the portfolio) under the twostates can be expressed as:

■ If state 1 occurs: $115wA � $100wB � $25■ If state 2 occurs: $80wA � $100wB � $0

We have created a portfolio consisting of stock XYZ and risk-free sukuk that willreproduce the payoff of the call option regardless of the state that occurs one year fromnow. Solving, we find wA � 0:714286 and wB � �0:57143. The cost of the portfoliotoday is 0.714286 × $87 � 0.57143 × $90 = $10.71.

This cost is the arbitrage free-price V0 of the call option. Arbitrageurs short-sell0.714286 shares XYZ for 0.714286 × $87= $62.14; with the proceeds they buy a calloption for $10.71 and invest 0.57143 × $90 = $51.43 in risk-free sukuks. Any othervalue of the call option will create market inefficiency and induces arbitrage betweenthe option and its replicating portfolio. If V0 > $10:71, the arbitrageurwill sell the option and buy the replicating portfolio. If V0 < $10:71, the arbitrageurwill buy the option and sell the replicating portfolio.

TABLE 20.4 Payoffs of an Arbitrage Strategy

Asset Investment today Payoff state 1 Payoff state 2

A $400,000 $285,715 $571,429B 600,000 300,000 1,200,000C �1000,000 �475,000 �1,400,000Total 0 110,715 371,429

TABLE 20.5 Pricing an Option Using Arbitrage


Asset A = stock XYZ $87 $115 $80Asset B = risk-free sukuk 90 100 100Call option V0? 25 0


3GC20 05/15/2014 13:13:37 Page 424

Pricing the Forward Exchange Rate by Arbitrage Denote the local currency by $, andthe foreign currency by £. The exchange rate is defined as the number of local currencyper one unit of foreign currency and is written as $/£. The spot exchange rate isdenoted by S0 (e.g., $1.7/£1). The forward exchange rate of the local currency in termsof a foreign currency FT is the rate agreed on today for delivery at time T. This rate isdetermined as a free-of-arbitrage price that satisfies the covered yield parity andprecludes costless profits.

Let the sukuk yield rate in local currency be denoted by r$. One unit of localcurrency invested in the home country becomes $1 � 1 � r$

� �one year later. If

investors convert today the unit of local currency into foreign currency at the spotexchange rate S0, they acquire 1=S0 units of foreign currency £. If they invest today£ � 1=S0 in the foreign country at a foreign sukuk yield rate r£ they will receive£ � 1=S0� � � 1 � r£� � at the end of the year. To compare foreign investment to localinvestment, investors have to transform their foreign investment into local currency.Applying the forward exchange rate FT , the foreign investment is transformed intolocal currency $ as follows: £ � 1=S0� � � 1 � r£� � � FT . To prevent arbitrage, theremust be equality of yields in local and foreign investments:

$ 1 � r$� � � $ 1=S0� � � 1 � r£� � � FT (20.13)

More specifically, the forward exchange rate must satisfy the covered yield parity:

FT � S0 � 1 � r$� �

= 1 � r£� � (20.14)

If the forward rate exchange rate is mispriced, then arbitrage opportunity exitsand arbitrageurs will earn costless profits. Let the mispriced market forward rate bedenoted by Fmisp

T . If FmispT > FT , then $ 1=S0� � � 1 � r£� � � Fmisp

T > $ 1 � r$� �

. Thearbitrageurs borrow now in local currency, invest in foreign currency, and sellnow forward their foreign currency proceeds. They pay back their local debt atthe end of the year and pocket a costless net gain equal to proceeds of the foreigninvestment converted in local currency minus the local cost of the investment, that is,$ 1=S0� � � 1 � r£� � � Fmisp

T � $ 1 � r$� �

.

If FmispT < FT , then $ 1=S0� � � 1 � r£� � � Fmisp

T < $ 1 � r$� �

. The arbitrageurs bor-row now in foreign currency, invest in local currency, and purchase now forwardforeign currency to repay their foreign debt. Their proceeds at the end of the year willbe $S0 � 1 � r$

� �, the cost of their investment will be $ 1 � r£� � � Fmisp

T , and their net

costless gain will be $S0 � 1 � r$� � � $ 1 � r£� � � Fmisp

T .Example. Arbitraging a mispriced forward exchange rate.Let S0 � $2=£1, r$ � 4 percent, r£ � 6 percent, then FT � 1:962264. If the for-

ward rate is mispriced and is equal to FmispT � 2:1, then arbitrageurs will borrow in

local currency $1 and invest in foreign currency £0.5. At the end of the year, theyreceive £0.53; they convert it to $1.113 = (0.53 × 2.1); they pay back their local loan$1.04 and achieve a net gain of $0.073. However, if Fmisp

T � 1:9, arbitrageurs willborrow in foreign currency £1 and invest $2 in local currency. At the end of year, theyreceive $2.08; they redeem their foreign loan at the cost $1.9 × 1.06 = $2.014 andachieve a costless net gain equal to $0.066.

424 STATISTICS

3GC20 05/15/2014 13:13:37 Page 425

The relationship FT � S01�ri$1�ri£ applies in simple compounding. In continuous

compounding, the formula is written

FT � S0e r$�r£� �T (20.15)

The covered yield parity is an important relationship between the spot andforward exchange rates and the yield rates in two countries. The yield parity theoremasserts that yield rates and exchange rates form one system. According to the yield rateparity theorem, foreign exchange rates will adjust to ensure that a trader earns thesame return by investing in risk-free instruments of any currency, assuming that theproceeds from investment are repatriated into home currency by a forward contract atthe outset of the holding period. In other words, the forward exchange rate premiumon the two currencies is equal to the yield rate differential on the currencies (assumingno transaction costs).

Pricing of a Forward Contract by Arbitrage The forward price of an asset agreed ontoday for delivery at time T is expressed as

FT � er�TS0 (20.16)

where S0 is the spot price of the asset and r is the riskless yield rate. If the forward priceis mispriced, then arbitrage opportunity becomes available. Let Fmisp

T > FT , thenarbitrageurs will borrow S0 at a yield rate r, buy the asset, and sell a future contractat Fmisp

T . At maturity date T, they deliver the asset, cash FmispT , redeem their debt er�TS0,

and make a costless arbitrage profit equal to FmispT � er�TS0. If Fmisp

T < FT , thearbitrageur will short-sell the asset, lend S0 at a riskless rate r, and buy a futurecontract at a mispriced price Fmisp

T . At maturity, arbitrageurs realize an income from

their lending equal to er�TS0, buy the asset at FmispT , and deliver the asset to the person it

was initially borrowed from. Their net cash flow is er�TS0 � FmispT :

The no-arbitrage pricing is characteristic of market efficiency and applies inpricing of all assets and derivatives. The arbitrage takes place between the asset and itsreplicating portfolio. The replicating portfolio often includes the underlying asset andcash or a riskless sukuk. Asset and derivative pricing satisfies no-arbitrage conditionbetween the derivative and its replicating portfolio. Moreover, no-arbitrage pricingsatisfies the martingale condition.

BASIC PRINCIPLES OF DERIVATIVES PRICING

In this section we describe a unified fundamental principle for pricing assets. Morespecifically all assets are priced according to the same principle of computing expectedpayoffs under risk-neutral distribution.

Principles of Derivatives Pricing Theory

The pricing of a derivativemay use replication, no-arbitrage, or risk-neutral probabilitydistribution methods. Each method necessarily satisfies the other two methods. If an


3GC20 05/15/2014 13:13:38 Page 426

asset (e.g., a derivative) is replicated by a hedging portfolio of assets and has the samepayoff as the replicating portfolio at every moment in time prior to expiration atmaturity, then thepriceof the replicatedassetmust be equal to thepriceof the replicatingportfolio. This is referred to as free-of-arbitrage pricing, since if the price of the asset andthe hedging portfolio differ, arbitrageurs will have riskless profits. The existence of areplicating portfolio implies no-arbitrage pricing; it implies also the existence of a risk-neutral probability distribution,whichmakes an adjustment for themarket price of riskand under which the expected return of any derivative or asset is equal to the risk-freeinterest rate. A risk-neutral probability distribution is also called a martingale distribu-tion. The replicating hedging portfolio is a riskless portfolio and, therefore, must earnthe same return as a riskless sukuk under the no-arbitrage condition (risk-free interestrate). The equivalence of three pricing methods can be stated as:

Replicating portfolio$No-arbitrage pricing$Risk-neutral �martingale�probability distribution(20.17)

Fundamental Principle for Pricing Derivatives

The fundamental principle for pricing derivatives is that the price of any derivativetoday,V0, is equal to the discounted value of its expected payoff at the maturity dateTof the derivative. The discounting has to use necessarily the risk-free rate r. Theexpectation has to be computed today under a risk-neutral (martingale) probabilitydistribution:

V0 � e�rTEQ0 ∏T

� �(20.18)

The initiation date of a derivative contract is today t = 0. The maturity date of thederivative contract is T; and the life of the derivative contract is the time interval0;T� �:* V0 is the price that is agreed on today for the derivative and is written in thederivative contract. ∏T is a random variable denoting the payoff of the derivative at

time T, it takes only one value according to the state of the world at time T. EQ0 is the

expectation operator, the subscript indicates t = 0 meaning that expectation iscomputed today, and the superscript Q indicates that expectation is computed usinga risk-neutral probability distribution Q and not under the true probability of ∏T .

Consider a derivative that is written today at time t = 0. Assume that there areNpossible states of the world at thematurity dateT. Only one state will prevail at timeT.Denote these states by ω1;ω2; . . .ωj; . . . :ωN. Let the probability associated with eachstate be θ1; θ2; . . . θj; . . . :θN. Note that

θ1 > 0; θ2 > 0; . . . ; θj > 0; . . . ; θN > 0; (20.19)

θ1 � θ2 � ∙ ∙ ∙ θj � ∙ ∙ ∙ � θN � XNj�1

θj � 1; j � 1; . . . ;N (20.20)

The payoff of the derivative associated with each state is a random variable thatdepends on the state that prevails at time T. Hence ∏ ω1� � is the payoff of the

*For an American option, the life of the option is determined by the time of its exercise, whichcan be any time prior to expiry date, that is, t ∈ 0;T� �.

426 STATISTICS

3GC20 05/15/2014 13:13:38 Page 427

derivative in state ω1, ∏ ω2� � is the payoff of the derivative in state ω2, ∏ ωj� �

is thepayoff of the derivative in state ωj, and ∏ ωN� � is the payoff of the derivative in stateωN. Simplifying the notation, we denote these payoffs as ∏1, ∏2; . . .∏j; . . . ;∏N ,j � 1; . . . ;N. Let the risk-neutral probabilities associated with each state of the worldbe denoted by q1; q2; . . . qj; . . . :qN. Note that

q1 > 0; q2 > 0; . . . qj > 0; . . . : qN > 0 (20.21)

q1 � q2 � ∙ ∙ ∙ qj � ∙ ∙ ∙ � qN � XNj�1

qj � 1; j � 1; . . . ;N (20.22)

By virtue of risk-neutral pricing, the value of the derivative today is its discountedexpected payoff under the risk-neutral distribution and can be written as


� � � e�rT q1∏1 � q2∏2 � ∙ ∙ ∙ � qj∏j � ∙ ∙ ∙ � qN∏N

h i(20.23)

V0 � e�rTXNj�1

qj∏j

" #; j � 1; . . . ;N (20.24)

Why do we discount? A dollar at time T is not the same dollar as at time 0. To beable to compare a dollar at time Twith a dollar at time 0, we have to compute presentvalue of time T dollar. The annualized rate of return of the derivative computed from

erT � PNj�1 qj∏j

h i=V0 is the risk-free rate r.

Why do we use a risk-neutral distribution? If we compute expected payoff usingtrue probabilities θ1; θ2; . . . ; θj; . . . ; θN we violate the fair game rule and there will bearbitrage. An arbitrageur will be able to make riskless profit by selling the derivativeand buying its replicating portfolio if the derivative is overpriced, or buying thederivative and selling the replicating portfolio if the derivative is underpriced. Hence,as arbitrageurs rush to exploit arbitrage opportunity, they will bid down the price ofthe derivative and bid up the price of the replicating portfolio when the derivative isoverpriced, or bid up the price of the derivative and bid down the price of thereplicating portfolio when the derivative is underpriced. In doing so, arbitrage profit iseliminated and the price of the derivative is reestablished to its no-arbitrage level.

Examples: Pricing a derivative.

1. Pricing a derivative using risk-neutral probabilities.Consider derivative A that matures in one year. Assume that the states of the

world in one year from today could have three possible outcomes: high, medium,and low. Let the derivative payoff be ∏1 � $25 in the high scenario, ∏2 � $5 inthe medium scenario, and ∏3 � �$10 in the low scenario. Let the probability of∏1 be θ1 � 0:25, the probability of∏2 be θ2 � 0:45, and the probability of∏3 beθ3 � 0:30. Let the risk-free discount rate be r = 5 percent. Let the risk-neutralprobabilities for ∏1, ∏2, and ∏3 be: q1 � 0:16, q2 � 0:35, and q3 � 0:49,respectively. Applying risk-neutral theory, the no-arbitrage price of derivativeA today is

V0 � e�0:05 � 0:16 � $25 � 0:35 � $5 � 0:49 � �$10� � � $0:808


3GC20 05/15/2014 13:13:39 Page 428

If we use simple compounding,

V0 � 0:16 � $25 � 0:35 � $5 � 0:49 � �$10� �1 � 0:05

� $0:809

Assume that, by mistake, the price of derivative A was computed using trueprobabilities: θ1 � 0:25, θ2 � 0:45, and θ3 � 0:30. Then,

Vmispriced0 � e�0:05 � 0:25 � $25 � 0:45 � $5 � 0:30 � �$10� � � $5:23

Arbitrageurs will rush to make riskless profit at zero cost. They will sell theoverpriced derivative and buy its replicating portfolio. Notice the effect of therisk-neutral probabilities in reducing the price of derivative from $5.23 to $0.809.They adjusted the price of the derivative so that it reflects a fair game and can bebought by risk-neutral traders.

2. Illustration of the effect of risk-neutral probabilities.Consider derivative B that matures in one year. Assume the states of the world

in one year from today are the same as for derivative A in the preceding example:high, moderate, and low. Let the derivative payoff be ∏1 � $20 in the highscenario,∏2 � $8:7 in the moderate scenario, and∏3 � �$6 in the low scenario.Let the true probability for ∏1 be θ1 � 0:15, ∏2 be θ2 � 0:29, and ∏3 beθ3 � 0:56. Let the discount rate be r = 5 percent. Since risk-neutral probabilitiesdepend on the states and not on the securities, they are the same for derivative B asfor derivative A; namely, q1 � 0:16, q2 � 0:35, and q3 � 0:49.

The no-arbitrage price of derivative B today is

V0 � e�0:05 � 0:16 � 20 � 0:35 � 8:7 � 0:49 � �6� � � $3:14

Assume that by mistake, the price of derivative B was computed using the trueprobabilities: θ1 � 0:15, θ2 � 0:29, and θ3 � 0:56. Then,

Vmispriced0 � e�0:05 � 0:15 � 20 � 0:29 � 8:7 � 0:56 � �6� � � $2:06

Arbitrageurs will rush to make riskless profit at zero cost. They will buy theunderpriced derivative and sell its replicating portfolio. Note the effect of risk-neutral probabilities was to adjust for risk and make the derivative moreexpensive so as to correspond to valuation of risk-neutral traders.

STATE PRICES

Risk-neutral probabilities are the same for every derivative. They depend on the stateof the world and not on the derivative. Let the state of the world at the maturity data Tbe described by N possible states denoted by ω1;ω2; . . .ωj; . . . : ωN . Assume that thereare traded today in the economy N state-contingent securities called Arrow-Debreuad� � securities and denoted by

ad�ω1�; ad ω2� �; . . . ad�ωj�; . . . ad ωN� �

428 STATISTICS

3GC20 05/15/2014 13:13:43 Page 429

TABLE 20.6 Arrow-Debreu Primitive Securities

States attime T

ω1 ω2 . . . ωj . . . ωN

Arrow-Debreusecurities

ad ω1� � ad ω2� � . . . ad ωj� �

. . . ad ωN� �

State pricestoday indollars

π1 π2 . . . πj . . . πN

Payofftime T

$1 in state ω1,$0 in allother states

$1 in state ω2,$0 in allother states

. . . $1 in state ωj,$0 in allother states

. . . $1 in state ωN ,$0 in allother states

The state-contingent security ad ω1� � pays one dollar if state ω1 occurs and zero inany other state. The state-contingent security ad ωj

� �pays one dollar if state ωj occurs

and zero in any other state. The contingent claim ad ωN� � pays one dollar if state ωN

occurs and zero in any other state. Let the price today of ad ω1� � be denoted by π1dollars, the price of ad ωj

� �be denoted by πj dollars, and the price of ad ωN� � be denoted

by πN dollars. These prices are called state prices because they price state-contingentsecurities. The prices and payoffs of the state-contingent securities are described inTable 20.6.

If the payoffs of a derivative are ∏1, ∏2, . . . ∏j, . . . , ∏N at time T in statesω1;ω2; . . .ωj; . . . ;ωN then these payoffs can be replicated by a portfolio of state-contingent securities. For instance, payoff $∏1 can be replicated by purchasing anumber of ad securities equal to ∏1 � ad ω1� � � $1. Likewise, payoff $∏2 can bereplicated by purchasing a number of ad securities equal to ∏2 � ad ω2� � � $1. Payoff$∏N can be replicated by ∏N � ad ωM� � � $1. Hence, the portfolio of primitive adsecurities that replicate the derivative’s payoffs in N states is

AD� ∏1�ad ω1� ��$1;∏2�ad ω2� ��$1; . . . ;∏j�ad ωj� ��$1; . . . ;∏N �ad ωN� ��$1

h i(20.25)

The price of the replicating portfolio today is

V0 � π1∏1 � π2∏2 � ∙ ∙ ∙ � πj∏j � ∙ ∙ ∙ � πN∏N (20.26)

Comparing the price of AD portfolio with the price of the derivative under therisk-neutral distribution,


� � � e�rT q1∏1 � q2∏2 � ∙ ∙ ∙ � qj∏j � ∙ ∙ ∙ � qN∏N

h i(20.27)

we find that state prices in dollars are related to risk-neutral probabilities as

π1 � e�rTq1; π2 � e�rTq2; . . . ; πj � e�rTqj; . . . ; πN � e�rTqN (20.28)


3GC20 05/15/2014 13:13:43 Page 430

The state prices π1; π1; π2; . . . ; πj; . . . ; πN are called the pricing kernel because theyprice any derivative in the economy. They are also called stochastic discount factors.

MARTINGALE DISTRIBUTION AND RISK-NEUTRAL PROBABILITIES

A martingale distribution is an essential condition of the theory for pricing assets andderivatives. A martingale distribution is also known as risk-neutral distribution, orstate price distribution. The definition of a martingale was given in Equations (20.8)and (20.9). However, this definition has to be modified when applied to future payoffsand future asset prices. Namely, dollars earned from an investment or a game at afuture time T and dollars earned now are not the same dollars from the consumer’spoint of view. They have to be discounted and expressed in present value. Accord-ingly, the martingale definition is modified as

E St�1 jSt� � � erSt (20.29)

This says that a fair game should allow investment to grow at the risk-free rate r. Ifit grows more or less than the risk-free rate it will benefit one party at the expense ofthe other. If the time internal is (0, T), the martingale definition can be rewritten as

e�rTE ST jS0� � � S0 (20.30)

The underlying asset price process is transformed into a martingale probabilitydistribution Q if

S0 � e�rTEQ0 ST ωj

� �jS0� �(20.31)

An example of a martingale is illustrated in Figure 20.3 in a two-year period,three-state setting. The discounted conditional expectation of the next year asset price

Time = 0 Time = 1Time = 2

S2,1∣1

S1,1 = e–rE(S2(ωj)∣S1,1)

S1,2 = e–rE(S2(ωj)∣S1,2)

S1,3 = e–rE(S2(ωj)∣S1,3)

S0 = e–rE(S1(ωj)∣S0)

S0 = e–2rE(S2(ωj)∣S1j∣S0)

S2,2∣1

Node 0

Node 1

Node 3

Node 2

Probabilities qi,j satisfy the martingale condition

S2,3∣1S2,1∣2

S2,2∣2

S2,3∣2

S2,1∣3

q3,3

S2,2∣3

S2.3∣3

q2,3

q1,3

q3,2

q2,2

q1,2

q3,1

q2,1

q1,1

q1,0

q2,0

q3,0

FIGURE 20.3 The Martingale Process

430 STATISTICS

3GC20 05/15/2014 13:13:44 Page 431

at each node is equal to the asset price at that node. Because S0 is the only observationknown today, we have

S0 � e�rE S1 jS0� � � e�rE e�rE S2jS1� �jS0� � � e�2rE S2jS0� �(20.32)

If we are forecasting the asset price for T years, the law of iterated expectationsimplies that

S0 � e�rE S1 jS0� � � e�rE e�rE S2 jS1� �jS0� � � ∙ ∙ ∙ � e�rTE ST jS0� �(20.33)

The risk-neutral probabilities at each node add to 1. Hence,

■ q1;0 � q2;0 � q3;0 � 1■ q1;1 � q2;1 � q3;1 � 1■ q1;2 � q2;2 � q3;2 � 1■ q1;3 � q2;3 � q3;3 � 1

The risk-neutral probabilities for each state at maturity are obtained as a productof the risk-neutral probabilities along the branches that are traveled to reach that state.For instance, in Figure 20.3 there are nine states at time T � 2. The risk-neutralprobability for state ω1 is q ω1� � � q1 � q1;0q1;1; for state ω2; q ω2� � � q2 � q1;0q2;1;for state ω3, q ω3� � � q3 � q1;0q3;1; for state ω4, q ω4� � � q4 � q2;0q1;2; for stateω5, q ω5� � � q5 � q2;0q2;2; for state ω6, q ω6� � � q6 � q2;0q3;2; for state ω7,q ω7� � � q7 � q3;0q1;3; for state ω8, q ω8� � � q8 � q3;0q2;3; and for state ω9,q ω9� � � q9 � q3;0q3;3, with

q1 � q2 � q3 � q4 � q5 � q6 � q7 � q8 � q9 � 1

State prices at time t � 0 for the states of the world at time T � 2 are computed bydiscounting the risk-neutral probabilities. The state price for state ω1 is π1 � e�2rq1;for state ω2; π2 � e�2rq2; for state ω3, π3 � e�2rq3; for state ω4, π4 � e�2rq4; for stateω5, π5 � e�2rq5; for state ω6, π6 � e�2rq6; for state ω7, π7 � e�2rq7; for state ω8,π8 � e�2rq8; and for state ω9, π9 � e�2rq9.

The value of any contingent claim at time 0 based on the martingale process ofFigure 20.3 is

V0 � π1∏1 � π2∏2 � π3∏3 � π4∏4 � π5∏5 � π6∏6 � π7∏7 � π8∏8�π9∏9

(20.34)

Example: Pricing derivatives using martingale probabilities and state prices.S0 � $100; the asset price may go up by u � 1:18 to ST � $118 or go down by

d � 0:87 to ST � $87 (Figure 20.4a). Let r � 0:05. The asset price process is turnedinto a martingale by applying the definition S0 � e�rTE ST jS0� �

. The conditionalexpectation is

$100 � e�0:05 q � $118 � 1 � q� � � $87� �


3GC20 05/15/2014 13:13:45 Page 432

By solving this equation, the risk-neutral probabilities are q � 0:585 and1 � q � 0:415.

The state prices are π1 � e�0:05q � $0:556 and π2 � e�0:05 1 � q� � � $0:395.Let us price today a call option with a strikeK = $104. The payoff to the option in

the up state is $14 and $0 in the down state (Figure 20.4b). The call price isV0 � πu∏u � πd∏d. Using the state prices we find V0 � π1∏u � π2∏d � $0:556�14 � $0:395 � 0 � $7:787.

Let us price today a put option with a strike K = $104. The payoff to the option inthe up state is $0 and in the down state, $17 (Figure 20.4c). The put priceis V0 � πu∏u � πd∏d. Using the state prices we find V0 � πu∏u � πd∏d �$0:556� 0 � $0:395 � 17 � $6:715.

Let us price a straddle that involves buying a call and a put with the same strikeK = $104 and the same expiration date (Figure 20.4d). The straddle price isV0 � πu∏u � πd∏d. Using the state prices we find V0 � πu∏u � πd∏d � $0:556�14 � $0:395 � 17 � $14:502, which is the sum of the price of a call ($7.787) and a put($6.715).

Based on the samemartingale, let us price a butterfly spread that involves buying acall option with a relatively low strike price (K1), buying a call option with a relativelyhigh strike price (K3), and selling two call options with a strike price of K2. LetK1 = $94, K2 = $104, and K3 = $106. The payoff in the up state is∏u � $118 � $95� � � $118 � $106� � � 2 � $118 � $104� � � $7. The payoff in thedown state is zero. The price of the butterfly is V0 � πu∏u � πd∏d. Using the stateprices we find V0 � πu∏u � πd∏d � $0:556 � 7 � $0:395 � 0 � $3:894.

Example: Two-period martingale.Let S0 � $100; at time T � 1, the asset price may go up to ST � $118 or go down

to ST � $87. At time T � 2, the asset price may go up to $140 or down to $107 giventhe upstate at T � 1, or it may go up $101 or down to $75 given the down state atT � 1 (Figure 20.5). Let r = 0.05 per year. The stochastic process is transformed into amartingale. The risk-neutral probabilities for time T � 1 are q � 0:585 and1 � q� � � 0:415. The risk-neutral probabilities for time T � 2 are computed by

a. Martingale process b. Call option pricing

c. Put option pricing d. Straddle option pricing

t = 0 t = 0

t = 0 t = 0t = 1

V0 = ? V0 = ?

V0 = ?

t = 1

t = 1

t = 1S1,u = $118

S1,d = $87

S0 = $100

∏1,u = $0

∏1,d = $17

∏1,u = $14

∏1,d = $0

∏1,u = $14

∏1,d = $17

FIGURE 20.4 Martingale Pricing

432 STATISTICS

3GC20 05/15/2014 13:13:46 Page 433

applying the martingale property at each node. In the up-node, the martingalecondition is

$118 � e�0:05 q � $140 � 1 � q� � � $107� �The solutions are q � 0:16 and 1 � q� � � 0:84. In the down node, the martingale

condition is

$87 � e�0:05 q � $101 � 1 � q� � � $75� �The solutions are q � 0:30 and 1 � q� � � 0:70.There are four states at T � 2. The risk-neutral probability for each state at T�2 is

qu;u�0:585�0:16�0:093, qd;u�0:585�0:84�0:492, qu;d�0:415�0:30�0:124,and qd;d �0:415�0:70�0:291. The state prices are πu;u� e�0:05�2qu;u�$0:084,πd;u�e�0:05�2qd;u�$0:445,πd;u�e�0:05�2qd;u�$0:112, andπd;d�e�0:05�2qd;d�$0:264.

Let us price a call option expiring at T � 2 with a strike K � $98. The payoffsof the option are ∏u;u � $42, ∏d;u � $9, ∏u;d � $3, and ∏d;d � $0. The price ofthe call today is

V0 � πu;u∏u;u � πd;u∏d;u � πu;d∏u;d � πd;d∏d;d

Applying the state prices we have computed we find

V0 � $0:084 � 42 � $0:445 � 9 � $0:112 � 3 � $0:264 � $0 � $7:873

Let us price a put option with a strike K = $108. The payoffs of the option are∏u;u � $0, ∏d;u � $1, ∏u;d � $7, and ∏d;d � $33. The price of the put today is

V0 � πu;u∏u;u � πd;u∏d;u � πu;d∏u;d � πd;d∏d;d

Applying the state prices we have computed we find

V0 � $0:084 � 0 � $0:445 � 1 � $0:112 � 7 � $0:264 � 33 � $9:93

Time 0 Time 1 Time 2

Su,u = $140 qu,u = 0.093qu = 0.16

Su = $118qu = 0.585

qd = 0.84

qd = 0.415Sd = $100

S0 = $100 qd = 0.30

Sd,u = $107

Su,d = $101 qu,d = 0.124

qd,u = 0.492

qd,d = 0.291Sd,d = $100qd = 0.70

FIGURE 20.5 Example of a Two-Period Martingale


3GC20 05/15/2014 13:13:49 Page 434

MARTINGALE AND COMPLETE MARKETS

Let stock A have S0 � $100; assume at time T = 1, there are three states of the worldω1, ω2, and ω3, with S1;1 � $122, S1;2 � $106, and S1;3 � $92. The risk-neutralprobabilities are q1, q2, and q3 with q3 � 1 � q1 � q2. Let r � 5 percent per year.The martingale condition is

$100 � e�0:05 q1 � $122 � q2 � $106 � q3 � $92� �

(20.35)

Noting that q3 � 1 � q1 � q2, we have therefore two equations and threeunknowns, q1, q2, and q3. The martingale probability distribution is, therefore,not unique. The markets are called incomplete when the number of the states ofthe world exceeds the number of securities. We have two securities, which are a risk-free sukuk and stock A, and three states, ω1, ω2, and ω3. There is no perfect hedge inthis case. To make the market complete, we need another security.

Let’s introduce stock B, whose price is Z. Let Z0 � $97:25; assume thatZ1;1 � $128, Z1;2 � $110, and Z1;3 � $75. The martingale condition applied tothe Z process is

$97:25 � e�0:05 q1 � $128 � q2 � $110 � q3 � $75� �

(20.36)

Along with themartingale condition for stock A, we now have three equations andthree unknowns. The solutions are q1 � 0:25, q2 � 0:40, and q3 � 1 � q1 � q2 � 0:35.

Example: Computing martingale probabilities in a complete markets model.LetthestochasticprocessesforsecurityA,denotedbyS,andforsecurityB,denotedby

Z, be described by Table 20.7 in a two-period, three-states setting.Wewant to computethe common martingale probabilities implied by these processes. We assume r = 0.05.

The common risk-neutral probabilities for T = 1 are generated by the followingequations,

$100 � e�0:05 q1 � $122 � q2 � $106 � q3 � $92� �

and$97:5 � e�0:05 q1 � $128 � q2 � $110 � q3 � $75

� �

TABLE 20.7 Complete Markets

Time 0 Time 1 Time 2 Time 0 Time 1 Time 2

142 155122 128.1 128 127

101 108.9

120 123S0 � 100 106 112 Z0 � 97:5 110 118

100.4 100.74

105 75 9292 103 76.15

79.4 64Security A Security B

434 STATISTICS

3GC20 05/15/2014 13:13:50 Page 435

Noting that q3 � 1 � q1 � q2, we find, as above, q1 � 0:25, q2 � 0:40, andq3 � 0:35.

The common risk-neutral probabilities for T � 2 states conditional on states inT � 1 are given by the following equations for node 1:

$122 � e�0:05 q1 � $142 � q2 � $128:1 � q3 � $101� �

$128 � e�0:05 q1 � $155 � q2 � $127 � q3 � $108:9� �

Noting that q3 � 1 � q1 � q2, we find q1 � 0:40, q2 � 0:40, and q3 � 0:20.The common risk-neutral probabilities for T = 2 states are given by the following

equations for node 2:

$106 � e�0:05 q1 � $120 � q2 � $112 � q3 � $100:4� �

$110 � e�0:05 q1 � $123 � q2 � $118 � q3 � $100:74� �

Noting that q3 � 1 � q1 � q2, we find q1 � 0:22, q2 � 0:58, and q3 � 0:20.The common risk-neutral probabilities for T = 2 states are given by the following

equations for node 3:

$92 � e�0:05 q1 � $105 � q2 � $103 � q3 � $79:4� �

$75 � e�0:05 q1 � $92 � q2 � $76:15 � q3 � $64� �

Noting that q3 � 1 � q1 � q2, we find q1 � 0:40, q2 � 0:30, and q3 � 0:30.The common risk-neutral probabilities for the states at T = 2 are q1 � 0:1,

q2 � 0:1, q3 � 0:05, q4 � 0:088, q5 � 0:232, q6 � 0:08, q7 � 0:14, q8 � 0:105, andq9 � 0:105.

By discounting risk-neutral probabilities by e�0:05�2, the common state prices areπ1 � $0:09, π2 � $0:09, π3 � $0:045, π4 � $0:079, π5 � $0:209, π6 � $0:072,π7 � $0:127, π8 � $0:095, and π9 � $0:095.

Let’s price a portfolio composed of a put on the security Awith a strike K = $110and a call on security B with a strike K = $102. The payoffs of this portfolio are∏1 � $53, ∏2 � $25, ∏3 � $15:9, ∏4 � $21, ∏5 � $16, ∏6 � $9:6, ∏7 � $5,∏8 � $7, and ∏9 � $30:6. The price of the portfolio is

V0 � $0:09 � 53 � $0:09 � 25 � $0:045 � 15:9 � $0:079 � 21 � $0:209 � 16

� $0:072 � 9:6 � $0:127 � 5 � $0:095 � 7 � $0:095 � 30:6 � $17:3

SUMMARY

Islamic finance applies the theory of asset pricing under uncertainty. Asset prices haveto be equilibrium, arbitrage free prices. This chapter covers the modeling of risk andreturn, the efficient market hypothesis, including the random walk and martingaleprinciples, the arbitrage-free pricing, the basic principles of asset pricing, the stateprices, the martingale distribution and risk-neutral probabilities, and the completemarkets condition. The chapter stresses the equivalence of replication, arbitrage-free,and martingale theory. Each principle implies the others.


3GC20 05/15/2014 13:13:51 Page 436

Islamic financial institutions operate in integrated world capital markets and tradein a wide range of assets including foreign securities, foreign currencies, and sukuks;they enter into Murabaha transactions, and deal in structured products. They have todetect arbitrage opportunities and ensure that prices at which financial transactionsare concluded are efficient prices. If their prices are lower than equilibrium prices, theymay hurt their profitability; if they are higher, they may lose business.

QUESTIONS

1. Stocks A and B are at $100. Stock A may go up to $135 or down to $87 withprobabilities of 0.62 and 0.38, respectively. Stock Bmay go up to $118 or down to$85 with the same probabilities as for stock A. Are stocks A and B equally risky?

2. Simulate a martingale processes with 252 observations using Microsoft Excel.Present a graph for the process.

3. Download the weekly S&P 500 stock index for January 2009 toNovember 2013.Present a graph for the data. Compute the returns in percent, their mean, andstandard deviation. How uncertain is the stock index? Using EViews, compute thesample auto-correlation at lags 1, 2, 3, and 4. Test the efficient capital markethypothesis.

4. The spot price of crude oil is $85/barrel. Futures contract for one-year delivery isquoted at $95/barrel. The yield rate on a one-year Treasury note is 2 percent.Explain the arbitrage strategy and compute the arbitrage profit. What should theequilibrium forward price for crude oil be?

5. The following table shows today’s prices for securities A and B and future payoffsof securities A, B, and C for each state at maturity time. Compute the price todayof security C. If C is priced at $68, what kind of arbitrage takes place? If C ispriced at $60, what kind of arbitrage takes place?


A $70 $50 $100B 60 30 120C ? 39 111

6. The spot exchange rate betweenMalaysia ringgit (RM) and the euro is RM4.328/euro. Let the yield rate be 6 percent in Malaysia and 3 percent in Euro-Eurozone.Compute the forward exchange rate for a futures contract at six-month maturity.

7. Today, the yield rate is 2 percent in the United States. It is 7 percent in Malaysia.The spot rate is RM3.72/$1. Compute the free-of-arbitrage forward exchangerate RM/$ for a one-year futures contract. If the forward exchange rate happensto be quoted at $3.1/$1, what kind of arbitrage takes place? Compute thearbitrage profits. What will happen to the exchange rate as greater amountsof a currency are sold forward?

8. A stock price is now at $100. It can go up next year to $120 or fall to $80. Therisk-free sukuk rate is 8 percent. Compute the martingale (risk-neutral) proba-bilities and today’s state prices of Arrow-Debreu securities.

436 STATISTICS

3GC20 05/15/2014 13:13:53 Page 437

9. A stock price is now at $100. It can go up after 77 days to $120 or fall to $80. Therisk-free sukuk rate is 8 percent. Compute the martingale probabilities (risk-neutral) andtoday’s statepricesofArrow-Debreusecurities.Acalloptionmaturingin 77 days with a strike at $106 is written on the stock. Compute its price today.

10. a. The following table describes a stock’s price today and in two possible statesnext year. A put option on the stock is to be bought today. The option has aone-year maturity. The strike price is $90. What is the put option price today?Explain the composition of the replicating portfolio, more specifically, whichasset to buy, sell, borrow, or lend?


Stock $87 $115 $80Sukuk 90 100 100Put option ? 0 10

b. Compute martingale probabilities and state prices and apply them to computethe put option price. Compare your answer in a and b.

11. A stock price is described by the binomial tree that follows:

T = 0 T = 1 T = 2

$135$118

$107$100

$101$87

$79

a. The sukuk risk-free rate is 7.8 percent. Compute the martingale probabilities andtoday’s state prices ofArrow-Debreu securities for each of the four states in year 2.

b. A call option with a strike at $97 and maturing at end-year 2 is written.Compute the payoffs of the call. Compute the price of the call.

c. A put option with a strike of $102 and maturing at end-year 2 is written.Compute the payoffs of the put. Compute the price of the put.

12. The prices of stocks S and Z may be in three different states next year, asillustrated by the following table. The risk-free sukuk yield rate is 7.5 percent.

Security Today’s price

Next year

State 1 State 2 State 3

Security S $97.6 $128 $110 $75Security Z 100 125 106 92Sukuk 93 100 100 100Call option ? 23 4 0


3GC20 05/15/2014 13:13:56 Page 438

a. Compute martingale probabilities.b. A call option is written on stock Z with a strike K = $102. Compute its price

today.c. Replicate the call option. Find its price using replication method. Which

method do you prefer to use for pricing assets: martingale or replicationpricing? Why?

13. Securities S and Z can move to three different states at each node. At end-year 2,each security will be in only one of nine different states as described by thefollowing table. The risk-free yield is 6 percent. A portfolio composed of a putoption on the security S with a strike K = $110 and a call option on security Zwith a strike K = $102 and maturing at end-year 2 has been put in place.

Security Time 0 Year 1 Year 2 Security Time 0 Year 1 Year 2

$160.0 $150.0$128.0 125.0 $125.0 128.0

91.2 92.0123.0 120.0

S $97.6 110.0 118.0 Z $100.0 106.0 112.0107.5 103.998.0 110.0

75.0 79.0 92.0 95.055.3 83.4

a. Compute the martingale probability and state price for each of the nine states.b. Compute the payoff of the portfolio at end-year 2.c. Compute the price of the portfolio today.

14. Stocks S andZ are traded today at prices as indicated in the following table. In oneyear from now, the prices of these stocks may move to three states. The risk-freerate is 5 percent per year.

Security Time 0 Year 1 Security Time 0 Year 1

$128.0 $125.0S $99.4 110.0 Z $101.9 106.0

75.0 92.0

a. Compute the risk-neutral probabilities and state prices.b. A portfolio manager bought a call on stock S at a strike price of $105 and sold

a put on stock Z at a strike price of $107. What is his net cash flow today?Compute the probability with which the call will be exercised. Compute theprobability with which the put will be exercised.

438 STATISTICS

3GC21 05/15/2014 13:18:43 Page 439

CHAPTER 21The Consumption-Based Pricing Model

T he purpose of this chapter is to analyze capital asset pricing using the inter-temporal-consumption model. This approach aims to determine the price of a

capital asset in terms of an investor’s intertemporal choice of his future consumptionas in the case of retirement planning and social security contributions. Consump-tion-based pricing model is called equilibrium pricing model because it applies thestandard consumer utility maximization model under budget constraint and derivesfirst-order conditions for equilibrium. From the first-order optimization conditions,the model shows the price of an asset to be equal to its expected discounted futurepayoffs; the stochastic discount factor is the marginal rate of substitution betweenpresent and future stochastic consumption. The consumption-based model showsthe equivalence of pricing payoffs and returns, and enables us to derive the capitalasset pricing model (CAPM), portfolio theory, mean-variance efficiency frontier,and risk-neutral pricing. The chapter stresses equivalence of asset pricing methodsunder uncertainty; each method implies the others. For instance, risk-neutral pricingshould be free of arbitrage and should satisfy the optimality conditions of investors’choices.

The merit of the consumption-based asset pricing is to relate asset pricing toeconomic growth and capital theory. Consumption is the ultimate objective ofinvestment. Growth theory has been concerned with achieving higher per capitaconsumption through capital accumulation. Themore an economy invests, the more itproduces, and the more consumers enjoy higher consumption in the form of food,energy, clothing, cars, housing, medical services, and so on. In a similar fashion,investors who invest in stocks want to earn future returns so they can enjoy higherfuture consumption. This is the case for pension funds and social security. People haverelatively high labor income during youth, but low or zero labor income during oldage; they would like to save now part of their labor income so that they can maintainthe same living standard during retirement.

INTERTEMPORAL OPTIMIZATION AND IMPLICATIONTO ASSET PRICING

Investors are facing a choice of increasing or reducing present consumption. In thesecond alternative, investors save part of their present labor income, invest now insecurities, and earn future returns from these securities that will enable moreconsumption in the future.

439

3GC21 05/15/2014 13:18:44 Page 440

The investor is facing an intertemporal optimization problem between presentand future consumption. To study the investor’s choice, we use standard consumeroptimization framework, namely, maximization of a utility function under a budgetconstraint. We model an investor by a utility function defined over current and futurevalues of consumption:

maxct ;ct�1

U ct; ct�1� � � u ct� � � 11 � δ� �Et u ct�1� �� (21.1)

This utility function is time-separable for convenience. It simply states thatinvestors’ intertemporal is a function of their present consumption, ct, and their futureconsumption, ct�1. Consumption in t � 1 is stochastic, that is, random. It depends, inpart, on the performance of stocks inwhich investors have invested their savings, whichare random. Investors do not knowwhat theirwealthwill be tomorrow, and hence howmuch they will decide to consume tomorrow. We use therefore the expectationoperator Et to compute average utility over possible scenarios of consumption intime t + 1. The subscript in the expectation operator is crucial; it indicates thatexpectation is made today, conditioned by the information available to investors today.The parameter δ describes the rate of intertemporal preference of the investor, or theimpatience of the investor. A high value of δ indicates high preference for presentconsumption. The investor’s intertemporal optimization often uses a convenient powerutility form,

u ct� � � 11 � γ

c1�γt (21.2)

The parameter γ describes the investor’s risk aversion. The limit as γ ! 1 isu�c� � ln�c�. This formalism captures investors’ impatience and their aversion to risk,so we can quantitatively correct for the risk and delay of cash flows.

After formulating the objective function, we need to formulate the intertemporalbudget constraint. We assume that the investor has a fixed labor income streamyt; yt�1� �

, whichmay describe labor income during youth and labor income during oldage. If investors consume all their labor income yt, then they have no pension forretirement; their future consumption may fall when their labor income yt�1 becomesvery low or even negligible. However, often people rearrange their consumptionprofile and plan for retirement when income yt�1 becomes low and does not allowthem to maintain the same living standard as during youth. Let investors invest theirsavings in present time t in equity shares; the price of each share today is pt. This priceis a main theme of capital asset pricing theory and we want to show how it isdetermined in intertemporal consumption optimization model. The payoff for inves-tors (e.g., retirees) in time t + 1 is the sum of the price at which they liquidate the sharept�1 and the dividends from the share dt�1, both are random variables. If we denote thepayoff by xt�1, then we have

xt�1 � pt�1 � dt�1 (21.3)

If we designate the number of shares investors wish to buy today by I, that is,investment, then their investment (i.e., saving) today is ptI and their random payoff

440 STATISTICS

3GC21 05/15/2014 13:18:44 Page 441

tomorrow is xt�1I. Their consumption today and tomorrow can be formulated,respectively, as

ct � yt � ptI and ct�1 � yt�1 � xt�1I (21.4)

Investors decide how many shares I to purchase by maximizing their utilitysubject to their budget constraint,

maxIf g U ct; ct�1� � � u ct� � � 1

1 � δ� �Et u ct�1� �� (21.5)

subject to

ct � yt � ptI; ct�1 � yt�1 � xt�1I (21.6)

Substituting the constraints into the objective function we find

maxIf g U ct; ct�1� � � u ct� � � 1

1 � δ� �Et u ct�1� ��

� u yt � ptI� � � 1

1 � δ� �Et u yt�1 � xt�1I� ��

(21.7)

The first-order condition for a maximum is @U@I � 0. Hence, by taking the deriva-

tive of U with respect to I we find

ptu´ ct� � � 1

1 � δ� �Et xt�1u´ ct�1� �� (21.8)

This equation is known as the Euler equation. It is a condition for an optimalconsumption and portfolio choice; it simply states that if an investor decides to buy anadditional share, his consumption today will be reduced by pt dollars and his loss ofutility is ptu

´ ct� �. That is, the marginal utility per dollar of consumptionu´ ct� � multi-plied by the number of dollars pt expended on the share. If the marginal utility perdollar of consumption is 10 utils, and the price per share is $100, then the investor’sloss in utility is 1,000 utils for each additional share purchased. Investors expect theirfuture consumption to increase by xt�1 dollars. Their utility will increase by anexpected value 1

1�δ� �Et xt�1u´ ct�1� �� . The optimality condition requires that this dis-counted gain in utility be equal to the loss in utility today. If the loss in utility todayexceeds the discounted expected gain, then the investors will buy less of stocks andincrease their present consumption. In contrast, if present loss in utility is less thandiscounted expected utility gain, investors will reduce their present consumption untilequality between utility loss and discounted expected utility gain is established. Themost important result that derives from the optimality condition is its asset pricingimplication, namely,

pt � Et1

1 � δ� �u´ ct�1� �u´ ct� � xt�1

� �(21.9)

The Consumption-Based Pricing Model 441

3GC21 05/15/2014 13:18:45 Page 442

Given the payoff xt�1 and given the investor’s consumption choice ct and ct�1, thiscondition tells what market price pt to expect. The price of an asset today isdetermined by random payoff xt�1 and by the intertemporal marginal rate ofsubstitution between future and present consumptions, 1

1�δ� �u´ ct�1� �u´ ct� � . In asset pricing,

the variable 11�δ� �

u´ ct�1� �u´ ct� � is called stochastic discount factor because it discounts the

random payoff and itself is random due to randomness of ct�1. We denote thestochastic discount factor by mt�1; we have

mt�1 � 11 � δ� �

u´ ct�1� �u´ ct� � (21.10)

The asset pricing formula is rewritten as

pt � Et mt�1xt�1� � (21.11)

Equation (21.11) is the central asset pricing formula. All pricing models amountto alternative ways of connecting the stochastic discount factor to investor’s riskpreferences and impatience. In particular, Equation (21.11) can cover stocks, sukuks,and derivatives, and makes clear that there is one theory for all asset pricing. Manyimportant results can be derived from this equation. Let us define gross return on assetby R, namely, R = 1 + r, where r is simple return. The one-period gross rate of returnon a share is

Rt�1 � pt�1 � dt�1pt

� xt�1pt

(21.12)

Accordingly, by dividing by pt Equation (21.11) can be rewritten as

1 � Et mt�1xt�1pt

� �� Et mt�1Rt�1� � (21.13)

The price of a return is always equal to $1. It is very important to emphasize theequivalence between Equation (21.11) and (21.13). Often in asset pricing, especiallyin the CAPM, we deal with return, instead directly with payoffs. Nonetheless, it has tobe stressed that when we find a price for expected return, Rt�1 such as in the CAPM,this price implies a price pt for the asset’s payoff xt�1. Returns and payoffs are nomorethan a different language for the same phenomenon.

ASSET-SPECIFIC PRICING AND CORRECTION FOR RISK

Risk varies across assets. Some assets are riskier than others, with implications on theirrespective prices and returns. They fetch lower prices and higher returns. Since thereare many assets in the economy, we denote gross return specific to each asset i by

Rit�1 � pit�1� di

t�1pit

(21.14)

442 STATISTICS

3GC21 05/15/2014 13:18:45 Page 443

Accordingly, the gross return on any asset Rit�1 is priced as

1 � Et mt�1Rit�1

� �(21.15)

If there is a risk-free asset with gross returnRft�1 defined as a fixed constant known

with certainty in the present time t, its price is given by

1 � Et mt�1Rft�1

h i� Rf

t�1Et mt�1� � (21.16)

We obtain as a result:*

Rft�1 � 1

Et mt�1� � (21.17)

The price of the risk-free asset is

pft � Et mt�1xft�1h i

� xft�1Et mt�1� � � xft�1Rf

t�1(21.18)

Since a risk-free asset has constant risk-free cash flow xft�1, its price today pft isobtained by discounting by the risk-free rate Rf

t�1.The price of a risky asset i is obtained by applying equation pt � Et mt�1xt�1� � to

that asset, namely,

pit � Et mt�1xit�1� �

(21.19)

Equation (21.19) is obviously a generalization. There is a common stochasticdiscount factor mt�1 for all assets in the economy.y This common discount factor isalso called pricing kernel or state price density. Although mt�1 is common for allassets, the correlation between the random components of the common discountfactor mt�1 and the asset-specific payoff xit�1 generates asset-specific risk corrections.High-risk assets command lower price today pit compared to low-risk assets; theirpayoffs are discounted by a higher discount rate compared to low-risk assets.

Asset prices are adjusted in relation to the risk-free price by a risk premium. Thehigher the risk, the greater the risk premium investors require for holding the riskierassets. To understand the correction for risk, we use the covariance formula, whichcan be written asz

cov m; x� � � E mx� � � E m� �E x� � (21.20)

*The relation Rf � 1Et mt�1� � implies that risk-free rate is influenced by Bohm-Bawerk’s time

preference and capital productivity.More specifically, high impatience parameter δ implies highrisk-free rate; high consumption growth emanating from high economic growth also implieshigh risk-free rate.yThe application of discount factors to all assets is the same as the application of martingaleprobabilities to all assets. More specifically, martingale probabilities depend on the states andnot on the securities.zSubscripts are dropped from variables and expectation operators to simplify presentation. Weoften drop subscripts to make formulas easier to write.


3GC21 05/15/2014 13:18:46 Page 444

Using this formula and the equation for the risk-free rate, Rft�1 � 1

Et mt�1� �, the priceof any asset can be rewritten as

p � E mx� � � E m� �E x� � � cov m; x� � � E x� �Rf

� cov m; x� � (21.21)

The term E x� �=Rf in equation (21) is the standard discounted present-valueformula. The term cov m; x� � is a risk adjustment. An asset whose payoff co-variespositively with the discount factor has its price raised, and vice versa. To understandthe risk adjustment substitute back for m in terms of consumption, we obtain:

p � E x� �Rf

�cov

11 � δ� � u

´ ct�1� �; xt�1� �

u´ ct� � (21.22)

Marginal utility u´ ct�1� � declines as ct�1 rises. Thus, an asset’s price is lowered if itspayoff covaries positively with consumption. Conversely, an asset’s price is raised if itcovaries negatively with consumption. For instance, an asset that pays low payoffduring retirement age commands a lower price than an asset that has high payoffduring retirement.

We use returns so often that it is worth restating the same correction of risk for thespecial case that the price is 1 and the payoff is a return. We start with the basic pricingequation for returns, 1 � E mRi� �

. The asset pricing model says that, althoughexpected returns can vary across time and assets, expected discounted returns shouldalways be the same, 1. Applying the covariance decomposition,

1 � Et mt�1� �Et Rit�1

� � � cov mt�1;Rit�1

� �(21.23)

and, using Rft�1 � 1=Et mt�1� �, we obtain

Et Rit�1

� � � Rft�1 � Rf

t�1cov mt�1;Rit�1

� �(21.24)

All assets have an expected return equal to the risk-free rate, plus a risk correctionterm. Assets whose returns covary positively with consumption make consumptionmore volatile, and must promise higher expected return to induce investors to holdthem. Conversely, assets that covary negatively with consumption, such as insurance,can offer expected rates of return that are lower than the risk-free rate, or evennegative (net) expected returns.

RELATIONSHIP BETWEEN EXPECTED RETURN AND BETA

The consumption-based asset-pricing model provides an expected return–beta repre-sentation of asset prices and returns. This representation can be obtained simply byrearranging Equation (21.24) using the definition the risk-free rate Rf � 1=E m� � andintroducing the variance of the discount factor m denoted by Var m� �, we obtain

E Ri� � � Rf � cov Ri;m� �

Var m� � !

�Var m� �E m� �

� (21.25)

444 STATISTICS

3GC21 05/15/2014 13:18:46 Page 445

Defining,

βi;m � cov Ri;m� �

Var m� � and λm � �Var m� �E m� � (21.26)

The coefficient βi;m is a ratio; it measures contribution of asset’s i risk to total riskVar m� �. The pricing equation (25) becomes

E Ri� � � Rf � βi;mλm (21.27)

Consequently, we can write the price of an asset alternatively as

pi � E mxi� �

(21.28)

or as

E Ri� � � Rf � βi;mλm (21.29)

where βi;m is the regression coefficient of the return Ri on λm. These two representa-tions are equivalent and imply each other. Equation (21.27) illustrates a beta pricingmodel. It says that expected returns on asset i should be proportional to its beta βi;m ina regression of returns on the discount factor λm. Notice that the coefficient λm is thesame for all assets i, while the coefficient βi;m varies from asset to asset. The coefficientλm is often interpreted as the price of risk and the βi;m as the quantity of risk in eachasset i. Expected returns should increase linearly with their betas on consumptiongrowth. In addition, though it is treated as a free parameter in many applications, thefactor risk premium λm is determined by risk aversion and the volatility of consump-tion. The more risk averse people are, or the riskier their environment, the larger anexpected return premium one must pay to get investors to hold risky (high beta) assets.

THE MEAN VARIANCE (mv) FRONTIER

The consumption-based model enables us to derive an important concept in portfoliotheory, which is the mean-variance (mv) frontier, that is, the frontier within whichassets defined by their risk and return lie (Figure 21.1). The derivation of the mean-variance frontier is done through manipulation of Equation (21.25). Note thatcov Ri;m� � � ρi;mσiσm, where ρi;m is the coefficient of correlation between Ri and

m, σi is the standard deviation of Ri, that is, σ Ri� �, and σm is standard deviation

of m, that is, σ m� �. Using the covariance definition, we can rewrite the pricingEquation (21.25) as

E Ri� � � Rf � � cov Ri;m� �E m� �

!� � ρi;mσiσm

E m� � (21.30)

By dividing through by σi,

E Ri� � � Rf

σi� � ρi;mσm

E m� � (21.31)


3GC21 05/15/2014 13:18:47 Page 446

Since �1 � ρi;m � 1, that is, ρi;m � 1, it follows that

E Ri� � � Rf

σi

�

ρi;mσmE m� �

� σmE m� � (21.32)

Inequality (21.32) asserts that means and variances of asset returns must lie in aregion formed by the intersection at Rf of two lines with slope � σm=E m� �� whenρi;m � 1 and σm=E m� �� when ρi;m � �1. The boundary of the mean-variance region inwhich assets can lie is called the mean-variance frontier. It provides the highestexpected return for a given level of risk, or minimum risk for a targeted expectedreturn. All returns on the frontier are perfectly correlated with the discount factor—that is, the frontier is generated by ρi;m

� 1. Returns on the upper part of the frontierare perfectly negatively correlated with the discount factor and hence positivelycorrelated with consumption. They are “maximally risky” and thus get the highestexpected returns. Returns on the lower part of the frontier are perfectly positivelycorrelated with the discount factor and hence negatively correlated with consumption.They provide the best insurance against consumption fluctuations. All frontier returnsare also perfectly correlated with each other, since they are all perfectly correlated withthe discount factor. This fact implies that any mean-variance frontier return can beconstructed from two such returns. For example, if any single frontier return Rm ispicked up, then all frontier returns Rmv must be expressible as

Rmv � Rf � μ Rm � Rf� �

(21.33)

where μ is a parameter. In CAPM, Rm is considered as the return to the stock marketportfolio to which individual stocks are assessed. Thus, any mean-variance efficientreturn carries all pricing information.

Given a mean-variance efficient return and the risk-free rate, there is a discountfactor that prices all assets. The CAPM assumes that the typical investor’s consump-tion stream is perfectly correlated with the return to the stock market. Subsequently,the risk of a financial security is measured by its covariance with the return to the stock

Expected return

Idiosyncraticrisk

Mean-variance region

Systemicrisk

Security Si

0

Rmv

Rm

Rf

E(Ri) – Rf

Smv

Sm

ρi,m = –1

σi

ρi,m = 1

βi,m = cov(Ri,m)Var (m)

Risk = σi = σ(Ri)

FIGURE 21.1 Mean Variance Frontier in Consumption-BasedModel

446 STATISTICS

3GC21 05/15/2014 13:18:47 Page 447

market. Given a discount factor, a beta representation can be constructed for expectedreturn of an asset i using any mean-variance efficient return:

E Ri� � � Rf � βi;m E Rm� � � Rfh i

(21.34)

Since the beta model applies to every return includingRm itself, andRm has a betaof one on itself, the factor risk premium can be identified as λm � E Rm� � � Rf ; it is theprice of risk, and the risk premium is βi;mλm.

Clearly, Equation (21.34) states that E Ri� �is not directly related to σ Ri� �

; it is

influenced by βi;m � cov Ri;m� �Var m� � . Accordingly, an asset risk can be decomposed into a

priced or systematic risk and a residual or idiosyncratic risk as shown in Figure 21.1.The priced part is perfectly correlated with the discount factor, and hence perfectlycorrelated with any frontier asset. The residual or idiosyncratic part generates noexpected return, so it lies flat as shown in Figure 21.1, and it is uncorrelated with thediscount factor or any frontier asset.

These derivations suggest an intimate relationship between discount factor, betamodel, and mean-variance frontier. The equilibrium consumption-based pricingmodel can be exploited further to establish a relationship between the Sharpe ratioand the volatility of the discount factor. The Sharpe ratio is limited by the volatility ofthe discount factor σm. For any asset i, we have

Sharpe ratio � E Ri� � � Rf

σi

�

σmE m� � � Rfσm (21.35)

However, for a return of a portfolio on the frontier, the slope of the frontier is

E Ri� � � Rf

σi

�

σmE m� � � Rfσm (21.36)

Thus, the slope of the frontier is governed by the volatility of the discount factor.

RISK-NEUTRAL PRICING IMPLIED BY THE GENERAL PRICINGFORMULA pt = Et(mt+1xt+1)

The consumption-based pricing formula, pt � Et mt�1xt�1� �, encompasses also risk-neutral pricing and pricing using Arrow-Debreu contingent securities. Let us pricetoday time t an asset that promises to have a payoff xt�1 ωj

� �contingent on the state of

the world ωj in time t + 1. Let us assume that there are N possible states in time t + 1with j � 1; 2; . . . ;N. The true probability of each state ωj occurring is

θ j� � > 0; with θ 1� � � θ 2� � � ∙ ∙ ∙ � θ N� � � 1

We assume that there are traded todayNArrow-Debreu contingent securities thatpromise each to pay one dollar in state ωj and zero dollars in all other states. The price


3GC21 05/15/2014 13:18:48 Page 448

today of each Arrow-Debreu security is π j� �, j � 1; 2; . . . ;N. Consequently, the pricetoday of the asset is

pt � π 1� �xt�1 1� � � π 2� �xt�1 2� � � ∙ ∙ ∙ � π N� �xt�1 N� � �XNj�1

π j� �xt�1 j� � (21.37)

We would like to show that there exist stochastic discount factors m j� � such thatEquation (21.37) becomes

pt � Et mt�1xt�1� �The risk-neutral probability q j� �, j � 1; 2; . . . ;N, is related to Arrow-Debreu

security price by the relation

π j� � � q j� �=Rft�1 (21.38)

Using the risk-neutral probabilities q j� �´s, Equation (21.38) can be rewritten as

pt � 1

Rft�1

XNj�1

q j� �xt�1 j� � � 1

Rft�1

XNj�1

θ j� �θ j� � q j� �xt�1 j� � (21.39)

The multiplication of each term j in the summation by θ j� �θ j� � � 1 does not affect that

term. We define the stochastic discount factor,

mt�1 j� � � 1

Rft�1

q j� �θ j� � (21.40)

The asset price Equation (21.39) can thus be expressed as

pt �XNj�1

θ j� �mt�1 j� �xt�1 j� � (21.41)

Because: θ j� �, j � 1; 2; . . . ;N, are probabilities, pt is equal to expected value ofmt�1 j� �xt�1 j� �. Consequently,

pt � Et mt�1xt�1� �This establishes equivalence between consumption-based model and risk-neutral

asset pricing. The expected value of mt�1, E mt�1� �, under the probabilities θ j� �, j �1; 2; . . . ;N is

E mt�1� � �XNj�1

θ j� �mt�1 j� � �XNj�1

θ j� � 1

Rft�1

q j� �θ j� � �

1

Rf

XNj�1

q j� � � 1

Rft�1

(21.42)

Because q j� �´s) are risk-neutral probabilities, the last equality in Equation (21.42)follows from

PNj�1 q j� � � 1. Therefore, we have E mt�1� � � 1

Rft�1, or simply E m� � � 1

Rf .

448 STATISTICS

3GC21 05/15/2014 13:18:48 Page 449

CONSUMPTION-BASED CONTINGENT DISCOUNT FACTORS

We have shown the existence of discount factors that establish equivalence betweenpricing consumption-based pricing and the Arrow-Debreu contingent claims model.We want to show how the contingent discount factors can be derived from theconsumption-based model. The investor’s optimization problem can be formulated as

maxct ;ct�1

U ct; ct�1� � � u ct� � � 11 � δ� �Et u ct�1� �� u ct� � � 1

1 � δ� �XNj�1

θ j� �u ct�1 j� �� (21.43)

subject to

ct � yt � ptI; ct�1 j� � � yt�1 � I xt�1 j� �; j � 1; 2; . . .N (21.44)

The first-order condition for maximization is

ptu´ ct� � � 1

1 � δ� �Et xt�1u´ ct�1� �� 11 � δ� �

XNj�1

θ j� �u´ ct�1 j� �� xt�1 j� � (21.45)

It may be rewritten as

pt � 11 � δ� �

XNj�1

θ j� � u´ ct�1 j� �� u0 ct� � xt�1 j� � (21.46)

The stochastic discount factor is therefore

mt�1 j� � � 11 � δ� �

u´ ct�1 j� �� u´ ct� � (21.47)

It is exactly the intertemporal marginal rate of substitution between future andpresent consumptions. The price of the asset can therefore be stated as

pt �XNj�1

θ j� �mt�1 j� �xt�1 j� � � Et mt�1xt�1� � (21.48)

where mt�1 j� � are obtained from the consumption-based model. We note also thatmt�1 j� � � 1

Rft�1

q j� �θ j� �. This establishes the theoretical equivalence between the consump-

tion-based model and the Arrow-Debreu contingent pricing model.The consumption-based model, called equilibrium model, offers a setup for asset

pricing. An asset price derives from first-order optimization condition. Pricing anasset’s payoff is equivalent to pricing its return; one implies the other. The consump-tion-based model has been shown to encompass CAPM, risk-neutral pricing, andoffers a unified approach to all pricing theories. All pricing theories are equivalent andimply each other; namely, a price computed under one approach cannot be differentfrom price computed under another approach. An equilibrium asset price has to be


3GC21 05/15/2014 13:18:49 Page 450

free of arbitrage, satisfy the fair game principle, and be coherent with an investor’soptimization condition. Namely, equilibrium prices in general equilibrium theoryhave to satisfy consumer and producers optimization conditions. Prices that violateoptimization conditions cannot be considered as equilibrium prices and cause marketinefficiencies.

SUMMARY

The intertemporal-consumption model explains investors’ optimal choices betweenpresent and future consumption and provides the pricing of securities consistent withoptimization conditions. The chapter covers intertemporal optimization and implica-tion to asset pricing, asset-specific pricing and correction for risk, the relationshipbetween expected return and beta, the mean variance frontier, the risk-neutral pricingimplied by the consumption model, and the consumption-based stochastic discountfactors. The chapter stresses equivalence of asset-pricing methods under uncertainty;each method implies the others. For instance, risk-neutral pricing should be free ofarbitrage and should satisfy the optimality conditions of investors’ choices.

QUESTIONS

1. Formulate the consumption-based model in terms of objective function andbudget constraints. Derive the equilibrium asset-pricing equation.

2. Derive an expression for the stochastic discount factor.

3. Show the equivalence between pricing payoffs and returns.

4. State the general formula for pricing a risky asset i.

5. Derive an expression relating the risk-free rate to the stochastic discount factor.How does the stochastic discount factor influence the risk-free rate?

6. Derive the expected return-beta representation from the equilibrium consump-tion model.

7. Provide a definition for the market price of risk.

8. Derive the mean-variance efficiency frontier from the equilibrium consumptionmodel.

9. Show how the stochastic discount factor influences the Sharpe ratio.

10. A trader has priced his asset according to the following expected value formula:

pt � 1

Rft�1

XNj�1

θ j� �xt�1 j� �

Is his pricing correct?Do the stochastic discount factors depend on the security being priced or do

they depend only on the state of the world?

11. Show the equivalence between the state prices method and the consumptionmodel in pricing assets.

450 STATISTICS

3GC22 05/15/2014 13:23:40 Page 451

CHAPTER 22Brownian Motion, Risk-Neutral

Processes, and the Black-Scholes Model

I n this chapter we cover some basic elements and results of continuous time finance.In fact, considerable advances in finance theory have been made in continuous time.

Many asset pricing, risk analysis, and rate of return models have been developed incontinuous time. There is a close relationship between discrete time and continuoustime analysis. Time series are stochastic processes defined on discrete time intervals.More specifically, the observations were made at fixed points in time such as at theclose of the market, or end of the month, or every hour. However, modern finance hasalso used continuous-time stochastic processes with infinitesimal time intervals. Therandom variable is assumed to be continuous in time. Continuous-time stochasticprocesses are widely applied in finance theory.Many pricingmodels such as the Black-Scholes option pricing formula were developed in continuous time. In this chapter, weintroduce some basic concepts of continuous-time stochastic models and show theirapplications in asset pricing theory.

We study a continuous stochastic process in the same way as a time series. We tryto characterize the probability law of the random variable, that is, the data generatingprocess; then, we determine the mean and variance of the process. We use the processeither to predict the future values of the variable, or to study risk, and price an asset.We examine how to transform a process into a risk-neutral, or equivalently amartingale, process in order to be able to use it for pricing.

BROWNIAN MOTION

Brownian motion is a main element of continuous-time finance. We need to under-stand this concept in order to appreciate the many advances in finance theory.Brownian motion was originally described by the botanist R. Brown (1828) whodescribed the irregular and random motion of a pollen particle suspended in a fluid.Hence, this motion is called Brownian motion. The theory is far from completewithout a mathematical model developed later by Wiener (1931). The stochasticmodel used for Brownian motion is also called the Wiener process. We see that themathematical model has many desirable properties. Since 1931, Brownian motion hasbeen used in mathematical theory for stock prices. It is nowadays a fashion to useWiener process to study financial markets.

451

3GC22 05/15/2014 13:23:40 Page 452

AWiener process, also known as Brownianmotion, may be thought of as the limitof a discrete-time random walk as the time interval between realizations goes to zero.AWiener process has a normal distribution with mean zero and variance equal to thetime interval during which the process moves. We denote a Wiener process by zt. Theprocess is disturbed during a small interval of timeΔt by an independent normal shockεt ∼N 0;1� � and has the following random move:

Δzt � εtffiffiffiffiffiffiΔt

p;Δt ! 0; εt ∼N 0; 1� �;Δzt ∼N 0;Δt� � (22.1)

To understand the analogy between a random walk and a Wiener process weconsider a random walk yt generated by independent increments εtf g with y0 � 0:

yt � yt�1 � εt (22.2)

We assume εtf g to be an independent, normally distributed, zero-mean, unitvariance, stationary process so that εt ∼N 0; 1� �, E εtεt�j

� � � 0 for any j ≠ t. Let

yT �XTt�1

εt (22.3)

For 0 � κ < τ < T, we state the following properties:

E yT � yκ� � � 0

E yT � yτ� �

yτ�1 � yκ� �� 0

E yT � yκ� �2h i

� EPT

κ�1 εt� �2

� T � κ

yT is a normal (Gaussian) I 1� � process with independent increments, yT ∼N 0;T� �AWiener process is like a continuous random walk defined on the interval 0;T� �

but has unbounded variation despite being continuous, and so can be imagined asmoving extremely erratically in the vertical direction. The Wiener process can bederived from the simple random walk, replacing the time sequence εtf g by time series

εtffiffiffiffiffiffiΔt

pn owhen the time interval becomes smaller and smaller and approaches zero.

We let z0 � 0 and

zt�Δt � zt � εtffiffiffiffiffiffiΔt

p(22.4)

or equivalently,

Δzt � εtffiffiffiffiffiffiΔt

p(22.5)

We have

E Δzt� � � E εtffiffiffiffiffiffiΔt

p� �� ffiffiffiffiffiffi

Δtp

E εt� � � 0 (22.6)

andVar Δzt� � � Var εtffiffiffiffiffiffiΔt

p� �� ΔtVar εt� � � Δt (22.7)

452 STATISTICS

3GC22 05/15/2014 13:23:42 Page 453

We state the following four properties:

1. E zT � zκ� � � 0;

2. E zT � zτ� � zτ�Δt � zκ� �� 0;

3. E zT � zκ� �2� � � EPT

κ�1 εtffiffiffiffiffiffiΔt

p� �2 � T � κ; and

4. zT is Gaussian I 1� � process with independent increments, zT ∼N 0;T� �.The process starts at z0 � 0. Condition (1) says that zT � zκ has mean zero and

reflects the fact that the process is as likely to go up as to go down with no drift.Condition (2) reflects a lack of memory. The displacement zτ�Δt � zκ the processundergoes during κ; τ � Δt� � in no way influences the displacement zT � zτ it under-goes during τ;T� �. Condition (3) states that the variance is equal to the length of thetime interval T � κ and increases with the time interval. Condition (4) states that theprocess has a normal distribution with mean zero and variance T. In fact, if wepartition the interval 0;T� � into equal length intervalsΔt, the number of these intervalsis n � T=Δt. The variance of zT is the sum of n independent incrementsΔzt. Therefore,

Var zt� � � n � Δt � T=Δt� � � Δt � T (22.8)

In general, a continuous process z t� �, t � 0, is a Wiener process if E z t� �� 0 forall t � 0.

For all fixed t � 0, z t� �∼N 0; t� �; z t� � has independent increments; and z 0� � � 0.

DYNAMICS OF THE STOCK PRICE: THE DIFFUSION PROCESS

We may start with the daily S&P 500 stock index during January 2012 to June 2013as reported in Figure 22.1.

We observe that the daily S&P 500 stock index resembles a random walk with apositive drift. More specifically, there is a strong drift that is pushing the indexupward, and there are unpredictable random movements around the drift. Based onobserved behavior, we are interested in modeling the process of a financial variable.We let the share price be St. The change of St during a unit-time interval (e.g., one year)is ΔSt � St � St�1� � and may be written in the form of a stochastic differential equation(SDE), called a diffusion process, as

ΔSt � E ΔSt� � � σΔSνt (22.9)

where νt ∼N 0; 1� �, E ΔSt� � is the expected value of ΔSt, and σ2ΔS is the variance of thedifference ΔSt per unit-time interval. We assume E ΔSt� � � ϕ, where ϕ is a constantindicating an increase (or a decrease) of the share price. Accordingly, we have thefollowing SDE:

ΔSt � ϕ � σΔSνt (22.10)

The interpretation of the equation is simple; we expect St to change by a constantϕ, that is, E ΔSt� � � ϕ; however, there are random disturbances represented by σΔSνt.The total change of the share price per unit-time interval is the sum of a deterministiccomponent ϕ and a stochastic component, called diffusion, σΔSνt.

Brownian Motion, Risk-Neutral Processes, and the Black-Scholes Model 453

3GC22 05/15/2014 13:23:43 Page 455

Example: We consider the change of the daily S&P 500 index during January2012 to June 2013. We find ϕ � 0.96 and σΔS � 11.085. We expect the index tochange according the following scheme:

ΔSt � 0.96 � 11.085νt

The index is expected to increase by 0.96; however, there is a random shockνt ∼N 0; 1� �, which will cause significant fluctuation around the trend of the order of�11.085νt.

We consider change ΔSt�Δt � St�Δt � St per Δt time-interval; we have

ΔSt�Δt � ϕΔt � σΔSνtffiffiffiffiffiffiΔt

p(22.11)

If we replace νtffiffiffiffiffiffiΔt

pby Δzt we obtain

ΔSt�Δt � ϕΔt � σΔSΔzt (22.12)

The process is called arithmetic Brownian motion. However, financial marketsrarely use changes in value. They often use relative changes, that is, percentage change.If we define the rate of price change per unit-time interval as St�1�St

St� ΔSt�1

St, then we

write the total rate of return per unit-time interval as

ΔSt�1St

� EΔSt�1St

� �� σεt (22.13)

εt ∼N 0; 1� �. We assume E ΔSt�1St

� �� μ � constant, and the variance of returns

Var ΔSt�1St

� �� σ2 � constant. Total return is therefore expected return μ plus the effect

of a random shock εt:

ΔSt�1St

� μ � σεt (22.14)

If we consider the rate of return over a small interval of time Δt, then the rate ofreturn is expressed as

ΔSt�ΔtSt

� μΔt � σεtffiffiffiffiffiffiΔt

p(22.15)

We note that εtffiffiffiffiffiffiΔt

pis a Brownian motion Δzt; total return can be expressed as

ΔSt�ΔtSt

� μΔt � σΔzt (22.16)

Example:We consider the daily returns of S&P 500 index during January 2012 toJune 2013. We find μ � 0.07 percent and σ � 0.77 percent. We expect the daily indexto change according the following diffusion scheme:

ΔSt�ΔtSt

� 0.07%Δt � 0.77Δzt


3GC22 05/15/2014 13:23:44 Page 456

We assume Δt � 1; the daily index is expected to increase by 0.07 percent.However, there is a normal random shock εt, which will cause the daily index tofluctuate by 0.77Δzt around the trend.

The process ΔSt�ΔtSt

is called geometric Brownian motion because we are dealingwith the growth rate of St; often, it is written as

ΔSt�Δt � μStΔt � σStΔzt

Or in continuous-time form as

dSt � μStdt � σStdzt (22.17)

We observe that, in general, the stock price will not change by μSΔt; there arerandom shocks that cause the asset return to fluctuate around the mean. The extent offluctuation is determined by volatility of the returns σ. We note that μ and σ may beconsidered as smooth functions of St and t; we express them as μ St; t� � and σ St; t� �. Thedynamics of the return process are rewritten as

ΔSt � μ St; t� �Δt � σ St; t� �Δzt (22.18)

Or in continuous form,

dSt � μ St; t� �dt � σ St; t� �dzt (22.19)

Example:Weassumeμ � 12 percent per year,σ � 30 percent per year, εt ∼ iid 0; 1� �,and Δt � 1 month; we want to compute ΔSt�Δt

St. Since εt is a random shock, which may

assume any positive or negative value, we compute a forecast interval for ΔSt�ΔtSt

defined as

�μ � σffiffiffiffiffiffiΔt

p �. Converting the values of the parameters μ andσ intomonthly data,we find aforecast interval given by �1% � 30%

ffiffiffiffiffiffiffiffiffiffiffi1=12

p � � 9.66%;�7.66%� �.

APPROXIMATION OF A GEOMETRIC BROWNIAN MOTION BY ABINOMIAL TREE

Assume the stock price starts at S. In the interval of time Δt, assume that stock pricehas a binomial move; it can either go up with probability θ to u � S � u:S or go downwith probability 1 � θ� � to d � S � d:S. In successive time intervals Δt the stock pricefollows the same binomial process as shown in Figure 22.2.

For this tree to approximate the model ΔS � μSΔt � σSΔz, the parameters u; d,and θmust be chosen so that the expected change in the stock price and the variance ofthe change on the tree over the period Δt are equal to those of the Brownian motion.

The expected change in the stock implied by the Brownian motion is μSΔt; henceSt�Δt � SteμΔt.

The expected change in the stock price implied by the binomial tree isθ:u:S � 1 � θ� �:d:S. We equate the expected changes; we obtain

θ:u:S � 1 � θ� �:d:S � SeμΔt (22.20)

456 STATISTICS

3GC22 05/15/2014 13:23:45 Page 457

The variance of the change in the stock price implied by the Brownian motion isσ2S2Δt. On the tree, the variance of the price change over Δt is

θ u:S � S:eμΔt� �2 � 1 � θ� � d:S � S:eμΔt

� �2(22.21)

We equate the two variances to obtain

θ u:S � S:eμΔt� �2 � 1 � θ� � d:S � S:eμΔt

� �2 � σ2S2Δt (22.22)

which simplifies, after eliminating S2, to

θ u � eμΔt� �2 � 1 � θ� � d � eμΔt

� �2 � σ2Δt (22.23)

We have two equations in three unknowns. We add a condition u:d � 1, whichrequires that the tree be centered at St as shown in Figure 22.2. The solution is

u � eσffiffiffiffiΔt

p; d � e�σ

ffiffiffiffiΔt

p; θ � eμΔt � d

u � d(22.24)

Example: Suppose the stock price starts at $100; it has an expected return μ �18 percent per year and a volatility σ � 14 percent per year. We want to construct atree that approximates the weekly movements of the stock price. We assume 50 weeksper business year, that is, Δt � 0.02 year. Consequently, we have

u � eσffiffiffiffiΔt

p� e0.14

ffiffiffiffiffiffiffi0.02

p� 1.02; d � e�σ

ffiffiffiffiΔt

p� e�0.14

ffiffiffiffiffiffiffi0.02

p� 0.98;

θ � eμΔt � du � d

� e0.18�0.02 � 0.981.02 � 0.98

� 0.59

For instance, the stock may move in three weeks to u3:S � 1.023 � 100 � $106.1with a probability θ3 � 0.583 � 0.20. It may move to d2:u:S � 0.982 � 1.02 � 100 �$98 with a probability 1 � θ� �2θ � 0.412 � 0.59 � 0.10. In fact, we may compute the

Time

Now: t

u3. S

u2. S

u2. d . S

u. d . S d2. u. S

d3. Sd2. S

d.S

t + Δt t + 2Δt t + 3Δt

u.S

S

FIGURE 22.2 Approximation of a Brownian Motion by aBinomial Tree


3GC22 05/15/2014 13:23:46 Page 458

move in 10 weeks to d7u3S and the corresponding probability by using the parametersof the tree.

ITO’S LEMMA

Ito’s lemma is a fundamental tool of modern finance. It is basic in derivatives pricingtheory. Often, the price of a derivative on asset, f St� �, is a function of the price St of theunderlying asset. For instance, the price of an option on Apple stock is a function ofthe Apple’s stock price. Consequently, we need to determine the change in the functionf St� � of the asset price where St follows


This result is known as Ito’s lemma and is fundamental for asset pricing:

df St; t� � � @f St; t� �@t

� @f St; t� �@St

μ St; t� � � 12@2f St; t� �

@S2tσ2 St; t� �

( )dt � @f St; t� �

@Stσ St; t� �

�dzt

(22.26)

Ito’s lemma has many applications in finance; here we illustrate two applications:the log-normal model and the futures contract model.

The Log-Normal Model

As an application of Ito lemma, we consider the log-normal model defined asf St; t� � � Ln St� �; we suppose that

dSt � μStdt � σStdzt

In this case, μ St; t� � � μSt and σ St; t� � � σSt; we have then

@f@t

� 0;@f@St

� 1St; and

@2f

@S2t� � 1

S2t

Applying Ito’s lemma we find

df St; t� � � @f St; t� �@t

� @f St; t� �@St

μ St; t� � � 12@2f St; t� �


( )dt

� @f St; t� �@St

σ St; t� � �

dzt � 1StμSt � 1

2� 1

S2t

!σ2S2t

( )dt

� 1StσSt

�dzt � μ � 1

2σ2

�dt � σdzt

(22.27)

458 STATISTICS

3GC22 05/15/2014 13:23:47 Page 459

Since μ and σ are constant, it follows that

dLn St� � � μ � 12σ2

�dt � σdzt ∼N μ � 1

2σ2

�dt; σ2dt

� �(22.28)

Ln St� � is an arithmetic Brownian motion. The solution to this stochastic differ-ential equation (SDE) at time t is

St � S0exp μ � 12σ2

� �t � σzt

(22.29)

Example: A Sharia-compliant stock has a diffusion process,dSt � μStdt � σStdzt � 0.12 � Stdt � 0.16 � Stdzt.We assume S0 � $100. The solution St is

St � S0 exp μ � 12σ2

� �t � σzt

� $100 exp 0.12 � 120.162

� �t � 0.16zt

� $100 exp 0.107� �t � 0.16zt� �The model dSt � μStdt � σStdzt is called the log-normal because its log, dLn St� �,

has a normal distribution. We note also that

Ln ST� � � Ln St� �f g∼N μ � 12σ2

�T � t� �; σ2 T � t� �

� �(22.30)

Now Ln ST� � � Ln St� �f g � Ln STSt

� �is the log of asset rate of return between t

and T.By applying Ito’s lemma, it is easy to show that the solution to dSt � σStdzt is

St � S0exp � 12σ2t � σzt

(22.31)

The Futures Contract Model

We apply the Ito lemma to the stock index futures. Suppose the index price follows

dSt � μStdt � σStdzt

The value at t of an index futures contract specifying delivery of one share of theindex at the fixed date T is

Ft � Ste r�δ� � T�t� � � f St; t� � (22.32)


3GC22 05/15/2014 13:23:48 Page 460

where r is the riskless interest rate and δ the dividend yield of the index (both assumedconstant). Then,

@f St; t� �@t

� � r � δ� �Ste r�δ� � T�t� � � � r � δ� �Ft

@f St; t� �@St

� e r�δ� � T�t� � � Ft

St;@2f St; t� �

@S2t� 0

Applying Ito’s lemma we find:

dFt � df St; t� �� @f St; t� �

@t� @f St; t� �

@Stμ St; t� � � 1

2@2f St; t� �


( )dt

� @f St; t� �@St

σ�St; t� �

dzt

� � r � δ� �Ft � Ft

StμSt

�dt � Ft

StσSt

�dzt

dFt � μ � r � δ� �Ftdt � σFtdzt (22.33)

So like St, Ft follows a geometric Brownian motion. Its expected rate of growth isμ � r � δ� � instead of μ but its volatility σ is the same as that of St.

DISCRETE APPROXIMATIONS

The Wiener process is a continuous process; however, when applied to financialmarkets, the data is observed at given time intervals such as minute, hour, day, week,and month. We need to approximate the model at a discrete-time interval Δt. Thecontinuous-time model is


The approximation of the dynamics of St at an interval Δt is

St�Δt � St � μ St; t� �Δt � σ St; t� �εtffiffiffiffiffiffiΔt

pwhere εt ∼N 0; 1� � (22.35)

This is called a Euler approximation or Euler discretization scheme. The sizeof the time interval is Δt; starting from t0 � 0, the time ti is ti � iΔt. When the step sizeis Δt ! 0, the Euler scheme will converge to the continuous solution,

limΔt!0E ST � SEulerT

�� h i! 0 for any date T.

The exact solution of the log-normal stochastic differential equation (SDE) atti � iΔt is

Sti � St0 exp μ � 12σ2

� �ti � σ

Xij�0

Δztj

( )(22.36)

Starting from St0 , we can simulate recursively the values St at future dates.

460 STATISTICS

3GC22 05/15/2014 13:23:49 Page 461

Example: A Sharia-compliant stock has the process dSt � μStdt � σStdzt withμ � 1.5 percent, σ � 1.7 percent, Δt � 0.1, and St0 � 1.0. Using Microsoft Excel, wedraw random normal increments:

Δztj �ffiffiffiffiffiffiffi0.1

p �N 0; 1� � for j � 1; 2; ∙ ∙ ∙ ; n � 250

The solution is

Sti � St0 exp 0.014856ti � 0.017Xij�0

Δztj

( )(22.37)

We show the simulated process in Figure 22.3.

ARBITRAGE PRICING: BLACK-SCHOLES MODEL

Arbitrage pricing was an approach used to price assets; the principle is that the price ofan asset has to be the same as the price of the replicating portfolio at all times prior tothe expiration date. If this principle does not hold, then there is costless profit fromselling the asset if it is overvalued and buying the replication portfolio and inverselybuying the asset if it is undervalued and selling the replication portfolio. The Black-Scholes formula for option pricing is based on the arbitrage principle between anoption and its hedging or replication portfolio.

Many assumptions underlie the Black-Scholes model. Stock price dynamics areassumed to be log-normal:

dSt � μStdt � σStdzt (22.38)

The stock pays no dividends between t and T; the markets are frictionless,implying no taxes, no transactions costs, no restrictions on short sales, all assets areperfectly divisible, and trading takes place continuously. The interest rate r is constantbetween t and T, and is the same for borrowing or lending.

The change in the price of the derivative is f St; t� � is related to changes in St via theIto’s Lemma:

df St ; t� � � @f St; t� �@t

� @f St; t� �@St

μ St; t� � � 12@2f St ; t� �


( )dt � @f St; t� �

@Stσ St; t� �

�dzt

(22.39)

The seller of a derivative has to hedge his position; the hedging uses the notion ofdelta of an option. The latter is defined as

Δ � @f St; t� �@St

(22.40)

The delta indicates the change in the price of a derivative per one unit of change inthe price of an asset.

The sellers of the call option establish a hedge portfolio as illustrated inTable 22.1; they sell one call for C and buy ΔS of the stock. The payoff of their


3GC22 05/15/2014 13:23:51 Page 463

portfolio in the up state is ΔSu � Cu; in the down state, it is ΔSd � Cd. To make theportfolio riskless, the payoff has to be the same in each state, that is,ΔSu � Cu � ΔSd � Cd. Accordingly, the option writer chooses Δ to satisfy thiscondition. We find

Δ � Cu � Cd

Su � Sd� @C

@S(22.41)

Example: Let Su � 120, Sd � 80, K � 100, C � max 0; S � K� �, Cu � 20, Cd � 0.We compute Δ as Δ � Cu�Cd

Su�Sd � 20�0120�80 � 0.5. The seller has to hedge the written call by

buying 0.5 shares.We have shown that a hedge portfolio consisting of buying Δ units of the stock

and selling one unit of the derivative is a riskless portfolio. The value of this portfolioat t (i.e., now) is

Π � Δ � S � f S; t� � (22.42)

The change of value of the portfolio is

dΠ S; t� � � ΔdS � df S; t� �� ΔμStdt � ΔσStdzt � @f

@t� @f@St

μSt � 12@2f

@S2tσ2S2t

( )dt � @f

@StσSt

�dzt

� � @f@t

� 12@2f

@S2tσ2S2t

( )dt

(22.43)

We observe that the random shock, dzt, which is the source of uncertainty, hasdisappeared. The portfolioΠ is riskless and therefore must earn the risk-free rate ofinterest r by no-arbitrage; thus

dΠΠ

� rdt (22.44)

or equivalently,

dΠ � rΠdt � @f@t

� 12@2f

@S2tσ2S2t

( )dt � r

@f@St

S � f �

dt (22.45)

TABLE 22.1 Hedge Portfolio, the Delta of an Option

Maturity

Portfolio Today Up State Down State

Asset ΔS ΔSu ΔSdCall option (call) C Cu Cd

Payoff ΔSu � Cu ΔSd � Cd


3GC22 05/15/2014 13:23:52 Page 464

Hence,

@f@t

� @f@St

rS � 12@2f

@S2tσ2S2t � rf � 0 (22.46)

This partial differential equation (PDE) must be satisfied by every derivativesecurity whose underlying price is S.What distinguishes derivative securities is the typeof final and boundary conditions. For a call option with strike K, the final conditionat T is f S;T� � � max 0; S � K� �. For a put option, the final condition at T isf S;T� � � max 0;K � S� �. For a futures contract with delivery price K, f S;T� � � S � K.

For a European call option, the solution is the Black-Scholes formula:*

C S;K; t;T� � � SN d1� � � Ke�r T�t� �N d2� � (22.47)

N di� � is the cumulative normal distribution:

d1 � Ln St=Kt� � � r � σ2=2� �

T � t� �σffiffiffiffiffiffiffiffiffiffiffiT � t

p (22.48)

d2 � d1 � σffiffiffiffiffiffiffiffiffiffiffiT � t

p(22.49)

If t � 0, the formula becomes

C S;K;T� � � SN d1� � � Ke�rTN d2� � (22.50)

d1 � Ln S0=K� � � r � σ2=2� �

T

σffiffiffiffiT

p (22.51)

d2 � d1 � σffiffiffiffiT

p(22.52)

Example: Applying the Black-Scholes formula, we price a call option on a stock inwhich S0 � $110, K � $102.5, σ � 14% per year, r � 6.3 percent per year, andT � 8months. We compute

d1 � Ln S0=K� � � r � σ2=2� �

T

σffiffiffiffiT

p

� Ln 110=102.5� � � 0.063 � 0.142=2� � � 8=12� �0.14

ffiffiffiffiffiffiffiffiffiffiffi8=12

p � 1.042


p � 1.042 � 0.14


r� 0.928

N d1� � � NORM:S:DIST 1.042;TRUE� � � 0.851376

N d2� � � NORM:S:DIST 0.928;TRUE� � � 0.823308

*There are many online calculators of option prices using the Black-Scholes formula.

464 STATISTICS

3GC22 05/15/2014 13:23:53 Page 465

C S;K;T� � � SN d1� � � Ke�rTN d2� �� $110 � 0.851376 � $102.5 � exp 0.063 � 8

12

� �

� 0.823308 � $12.73

If the asset pays dividends continuously between t and t � dt the asset holderreceives as income δSdt per share held. Then the riskless portfolio earns in total rΠdtduring dt, including dividends:

dΠ � ΔδSdt � rΠdt (22.53)

Total return is equal to capital gains + dividends. Note that the hedge portfoliocontains Δ shares of the asset and �1 share of the derivative, and income is earned onthe Δ shares of the underlying. The pricing partial differential equation is

@f@t

� @f@St

r � δ� �S � 12@2f

@S2tσ2S2t � rf � 0 (22.54)

THE MARKET PRICE OF RISK

We can extend the arbitrage concept to price derivatives that depend on a singleunderlying asset xt. We assume a geometric Brownian motion for xt:

dxt � μxxtdt � σxxtdzt (22.55)

The parameters μx and σx are the expected growth in xt and the volatility of xt,respectively.

Suppose that f 1 and f 2 are the prices of two derivatives dependent on xt and t.These can be instruments that provide a payoff equal to some function of xt at somefuture time.We assume that during the time period until maturity f 1 and f 2 provide noincome. Suppose the processes followed by f 1 and f 2 are

df 1f 1

� μ1dt � σ1dz (22.56)

df 2f 2

� μ2dt � σ2dz (22.57)

where μ1, μ2, σ1 and σ2 are functions of xt and t. The term dz is the same Wienerprocess as in dxt because it is the only source of the uncertainty in the prices of f 1 andf 2. The discrete versions of the processes are

Δf 1 � μ1f 1Δt � σ1f 1Δz (22.58)

Δf 2 � μ2f 2Δt � σ2f 2Δz (22.59)


3GC22 05/15/2014 13:23:53 Page 466

We can eliminateΔz by forming an instantaneously riskless portfolio consistent ofσ2f 2 of the first derivative and �σ1f 1 of the second derivative. If Π is the value of theportfolio, then

Π � σ2f 2� �

f 1 � σ1f 1� �

f 2 (22.60)

and

ΔΠ � σ2f 2� �

Δf 1 � σ1f 1� �

Δf 2 (22.61)

Substituting for Δf 1 and Δf 2 we obtain

ΔΠ � μ1σ2f 1f 2 � μ2σ1f 1f 2� �

Δt (22.62)

Because the portfolio is instantaneously riskless, it must earn the risk-free rate.Hence,

ΔΠ � rΠΔt (22.63)

Substituting into this equation for Π and ΔΠ we obtain

μ1σ2 � μ2σ1 � rσ2 � rσ1 (22.64)

or

μ1σ2 � rσ2 � μ2σ1 � rσ1 (22.65)

This yields

μ1 � rσ1

� μ2 � rσ2

(22.66)

Define λ as

λ � μ1 � rσ1

� μ2 � rσ2

(22.67)

The parameter λ is known as the market price of risk of xt. It can be dependent onxt and t, but it does not depend on the nature of the derivatives f 1, f 2, and so on.At any given time, λmust be the same for all derivatives that are dependent on xt and t.The market price of risk of xt measures the trade-off between risk and return that aremade for securities dependent on xt . Dropping the subscripts, the equation for λ can bewritten as

μ � r � λσ (22.68)

We may assume that σ measure the quantity of xt-risk present in f . On the rightside of the equation, we are multiplying the quantity of x risk by the price of x risk. Theleft-hand side is the expected return in excess of the risk-free rate that is required tocompensate for risk. Equation (22.68) is similar to the capital asset pricing model,which relates expected excess return on a stock to its risk.

466 STATISTICS

3GC22 05/15/2014 13:23:54 Page 467

Example: Suppose two Sharia-compliant stocks A and B are influenced by onesingle source of uncertainty dzt. This implies

μA�rσA � μB�r

σB � λ. Assume r � 7 percent,

μA � 8 percent, σA � 12 percent, and σB � 15 percent. Equilibrium pricing imposes

μB�r � σB μA�rσA

� ��7 percent�15 percent 8%�7%

12%

� � � 8.25 percent. If μB>8.25 percent,

stock B is undervalued. It will be purchased at the expense of stock A; ifμB < 8.25 percent, stock B is overvalued. It will be sold in favor of stock A.

RISK-NEUTRAL PRICING

Instead of forming an arbitrage portfolio, asset pricing has relied on an equivalentprinciple, which is risk-neutral pricing. If we transform the stochastic process into arisk-neutral process, then the price of an asset is the discounted payoff under the risk-neutral distribution using the riskless rate of return as a discount factor; no replicatingportfolio is needed. Let the returns of the asset dSt=St be described as

dSt=St � μdt � σdzt (22.69)

To obtain a risk-neutral process, we need to perform two economically importantoperations. First, we need to align the return of the asset with the riskless return, r.Second, we need to factor in the market price of risk attached to the asset, given byλ � μ�r

σ , meaning that the risk of the asset has to be adjusted for its risk in relation to themarket. These two economic operations are implemented by a simple rearrangementof the return equation as

dStSt

� μdt � σdzt � rdt � μ � r� �dt � σdzt � rdt � σd~zt (22.70)

The adjusted risk is described by

d~zt � μ � rσ

dt � dzt � λdt � dzt (22.71)

where λ � μ�rσ is the market price of risk. The new process ~zt is a standard Wiener

process with d~zt ∼N 0; dt� � under a risk neutral distribution. The transformed processis a Wiener process:

dStSt

� rdt � σd~zt (22.72)

The distribution law of transformed process, under the risk-neutral measure, is anormal distribution with expected mean equal to r and variance equal to σ2dt.

We define a discounted process yt � e�rtSt. We show that dyt is martingale;namely E dyt

� � � 0 under the risk-neutral process. We compute dyt as

dyt � �re�rtStdt � e�rtdSt � �re�rtStdt � e�rtrStdt � e�rtσStd~zt � e�rtσStd~zt (22.73)


3GC22 05/15/2014 13:23:55 Page 468

Hence the process yt satisfies the stochastic differential equation:

dyt � ytσd~zt (22.74)

Since ~zt is a standard Wiener process under the risk-neutral measure, we have

E�dyt=yt� � σE d~zt� � � 0 (22.75)

The solution yt is

yt � y0 exp �12σ2t � σ~zt

(22.76)

This establishes that yt � e�rtSt is martingale under the risk-neutral measure andthat all assets have an expected payoff under the risk-neutral measure equal to theriskless rate of return r.

Example: A Sharia-compliant stock has the process dStSt

� μdt � σdzt � 0.09dt �0.14dzt and r � 6 percent per year. We let

d~zt � μ � rσ

dt � dzt � λdt � dzt � 0.09 � 0.060.14

dt � dzt

� 0.214dt � dzt

The risk neutral process is dStSt

� rdt � σd~zt � 0.06dt � 0.14d~zt.Under the no arbitrage condition, the price of a derivative must satisfy a partial

differential equation:

@f@t

� @f@St

rS � 12@2f

@S2tσ2S2t � rf � 0 (22.77)

Let the boundary condition, or payoff, be given by f ST;T� � � g ST� �, then the priceof a derivative is the solution to this partial differential equation and is given asexpected mean given by the formula

f ST ;T� � � e�r T�t� �EQ g ST� �jSt� �(22.78)

The conditional expectation is computed with respect to the risk-neutral transi-tion probability density.

We apply risk-neutral distribution to value the call option. To value the calloption using risk-neutral pricing, we must take the expectation under the risk-neutralprobability of the discounted payoffs from the call. We may recall that yt � e�rtSt; thediscounted payoff from the call option is defined as

e�rTmax ST � K; 0� � � max yT � e�rTK; 0� �

(22.79)

where yT � e�rTST . To price the call option we need to compute the expectation of thepayoff under the risk-neutral probability. To do this, we recall that:

468 STATISTICS

3GC22 05/15/2014 13:23:55 Page 469

■ ~zT is a standard Brownian motion under the risk-neutral probability:~zT ∼N 0;T� �;

■ yT is a random variable computed as: yT � y0exp � 12 σ

2T � σ~zT� �

; and■ Substituting yT in the payoff function, we obtain

max yT � e�rTK; 0� � � max y0 exp � 1

2σ2T � σ~zT

� e�rTK; 0

(22.80)

We show that the expectation

E max y0 exp � 12σ2T � σ~zT

� e�rTK; 0

�(22.81)

under the risk neutral distribution is precisely equal to the Black-Scholes formula. Weproceed in several steps. We let f ~zT� � denote the probability density of ~zT . Since~zT ∼N 0;T� �, we have

f ~zT� � � 1ffiffiffiffiffiffiffiffiffi2πT

p exp � ~z2T2T

(22.82)

Thus the expectation is computed as

∫∞

� ∞max y0 exp �1

2σ2T � σ~zT

� e�rTK; 0

� 1ffiffiffiffiffiffiffiffiffi

2πTp exp � ~z2T

2T

d~zT (22.83)

For the call option to be exercised we need ST � K, or equivalently,e�rTST � e�rTK, that is, yT � e�rTK.

This condition is written as

y0 exp � 12σ2T � σ~zT

� e�rTK (22.84)

Taking log on both sides we find

Lny0 � 12σ2T � σ~zT � �rT � LnK (22.85)

We note that y0 � S0, we obtain

~zT � 1σ

�LnS0K

� 12σ2 � r

� �T

(22.86)

We let

h � 1σ

�LnS0K

� 12σ2 � r

� �T

(22.87)


3GC22 05/15/2014 13:23:56 Page 470

The value of the option, given by the discounted expectation, is

∫∞

hy0 exp �1

2σ2T � σ~zT

� 1ffiffiffiffiffiffiffiffiffi

2πTp exp � ~z2T

2T

d~zT � ∫

∞

he�rTK � 1ffiffiffiffiffiffiffiffiffi

2πTp exp � ~z2T

2T

d~zT

(22.88)

Consider the first term of the expectation. After rearrangement, we obtain

y0∫∞

h

1ffiffiffiffiffiffiffiffiffi2πT

p exp �12σ2T � σ~zT � ~z2T

2T

d~zT � y0∫

∞

h

1ffiffiffiffiffiffiffiffiffi2πT

p exp �12

~zT � σTffiffiffiffiT

p !2

24

35d~zT

(22.89)

The term under the integral is the density of a normal distribution with mean σTand varianceT. Therefore, the integral itself is the area under this distribution betweenh and ∞ . A standard transformation shows that this area may be represented using astandard normal distribution as N �d*

1

� �where d*1 � h�σTffiffiffi

Tp . Thus the first term of the

expectation reduces to y0N �d*1

� �. We substitute the value of h in the formula for d*

1;we find

1σ

�LnS0K

� 12σ2 � r

� �T

� σT

ffiffiffiffiT

p ��Ln S0

K� �1

2σ2 � r

� �T

σffiffiffiffiT

p � �Ln

S0K

� 12σ2 � r

� �T

σffiffiffiffiT

p(22.90)

We let

d1 �Ln

S0K

� 12σ2 � r

� �T

σffiffiffiffiT

p (22.91)

and using y0 � S0, the first term of the expectation is S0N d1� � as established by Black-Scholes formula.

A similar, but much simpler argument as in the previous step, shows that thesecond term

∫∞

he�rTK � 1ffiffiffiffiffiffiffiffiffi

2πTp exp � ~z2T

2T

d~zT (22.92)

is expectation under normal distribution with mean zero and variance T. Thisexpectation is equal to e�rTKN �d*

2

� �where d*

2 � hffiffiffiT

p . Substituting the value of h inthe formula for d*

2 we have

d*2 �

1σ

�Ln S0K

� 12σ2 � r

� �T

ffiffiffiffiT

p � �Ln

S0K

� r � 12σ2

� �T

σffiffiffiffiT

p (22.93)

470 STATISTICS

3GC22 05/15/2014 13:23:56 Page 471

We let

d2 �Ln

S0K

� r � 12σ2

� �T

σffiffiffiffiT

p (22.94)

We have

e�rTKN �d*2

� � � e�rTKN d2� � (22.95)

which is the second term in the Black-Scholes formula. We observe that


p(22.96)

Example: Traders versant with martingale theory were given the following processfor the stock price of Malay Palm Oil Corp: dSt

St� μdt � σdzt � 0.095dt � 0.14dzt

where the drift and diffusion parameters μ � 9.5 percent and σ � 14 percent are yearlyparameters. They wrote a call option maturing in seven months; S0 � $110,K � $102.5, and the riskless rate of return was r � 5.5 percent per year. Usingmartingale theory, he computed the call option according to Black-Sholes formula;he found C � $12.11.

SUMMARY

Islamic finance applies continuous-time models in areas of asset pricing, hedging, riskmanagement, and rates of return modeling. In fact, capital markets around the worlduse continuous-timemodels such as the Black-Scholes model or rates of return models.This chapter covers basic elements of continuous-time finance. It introduces theBrownian motion, the diffusion process, the approximation of a geometric Brownianmotion by a binomial tree, Ito’s Lemma, discrete approximations of a diffusionprocess, the Black-Scholes model, the market price of risk, and risk-neutral pricing.The material of the chapter is basic for practicing Islamic finance and understandingtools of the finance industry built on continuous-time models.

QUESTIONS

1. Define the Brownian motion; show the importance of time in the definition.

2. Show the relationship between the random walk and the Brownian motion.

3. What is the distribution of the Brownian motion? Compute the expectation andvariance of the Brownian motion Δzt.

4. Using Microsoft Excel, simulate 200 steps of a Brownian motion with Δt � 0.1.

5. Show a graph for the daily movement of the S&P 500 stock index during January2010 to September 2013. What are the main features of the graph?

6. Approximate the daily returns of the S&P 500 stock index during January 2010to September 2013 by a geometric Brownian motion.

7. Assume μ � 17 percent per year, σ � 30 percent per year, εt ∼ iid 0; 1� �, andΔt � 1 day; provide an expression for the daily geometric Brownian motionof ΔSt�ΔtSt

. Compute a daily forecast interval for ΔSt�ΔtSt

.


3GC22 05/15/2014 13:23:57 Page 472

8. Assume the price of a Sharia-compliant stock is described by a geometricBrownian motion with an expected return μ � 15 percent per year and a volatilityσ � 14 percent per year. The stock price is $100 today. Find a binomial approxi-mation for the weekly movements of the stock price. Compute u6d4S and u3d7S,and their respective probabilities.

9. Let dSt � μStdt � σStdzt � 0.17 � Stdt � 0.11 � Stdzt. Assume S0 � $100. Findthe solution St.

10. Let dSt � σStdzt � 0.11 � Stdzt. Assume S0 � $100. Find the solution St.

11. Consider the process dSt � μStdt � σStdzt with μ � 1.5 percent, σ � 2.1 percent,Δt � 0.02, and St0 � 1.0. Find the solution St. Simulate 300 steps of the processand plot the graph of the simulation.

12. A call option is written on a stock; the payoff of the option is C � max 0; S � K� �.The stock price may go up to Su � $115 or down to Sd � $85; the strike price isK � $105. Compute the delta Δ of the call option.

13. Applying the Black-Scholes formula, find the price of a call option on a stock forwhich S0 � $108, K � $99, σ � 15 percent per year, r � 6.3 percent per year, andT � 9months. Recompute the price of the call with σ � 10 percent. What do younotice?

14. Let dStSt

� μdt � σdzt � 0.095dt � 0.124dzt and r � 6.5 percent. Compute themarket price of risk and find the risk neutral process of dSt

St.

15. Let dSt � μStdt � σStdzt; provide an expression for the distribution ofln ST� � � ln St� �f g.

16. a. A call option is written on stock A, whose stochastic process isdStSt

� μdt � σdzt � 0.095dt � 0.124dzt. The parameters μ and σ are yearlyparameters. Stock A is priced today at $120, the strike price is K � $106,the riskless rate is r � 6.5 percent, and the maturity of the option is 10 months.Find the price of the call option on stock A.

b. A call option is written on stock B, whose stochastic process isdStSt

� μdt � σdzt � 0.125dt � 0.124dzt. Stock B is priced today at $120, thestrike price is K � $106, the riskless rate is r � 6.5 percent, and the maturity ofthe option is 10 months. Find the price of the call option on stock B. Compareprices in (a) and (b).

17. A trader not versant of risk neutral pricing wrote a call option on stock whosedynamics are

dStSt

� μdt � σdzt � 0.095dt � 0.124dzt

The stock is priced today at $100, the strike is at K � $100, the riskless isr � 6.5 percent, and the maturity of the option is 10 months. The trader did notmake the transformation into a risk neutral process. Provide the normal distri-bution, which the trader wrongly used in the computation of the option price.Provide an expression of the expected payoff. Compare with the normal distri-bution and the expected payoff when the risk-neutral process is used.

472 STATISTICS

3GBREF 05/15/2014 13:28:56 Page 473

References

PART ONE: MATHEMATICS

Chiang, A. C. 1984. Fundamental Methods of Mathematical Economics. 3rd ed. New York:McGraw-Hill.

Dorfman, R., P. A. Samuelson, and R. M. Solow. 1958. Linear Programming and EconomicAnalysis. New York: McGraw-Hill.

Gale, D. 1960. The Theory of Linear Economic Models. New York: McGraw-Hill.Griva, I., S. G. Nash, and S. Ariela. 2009. Linear and Nonlinear Optimization. 2nd ed.

Philadelphia: Society for Industrial and Applied Mathematics (SIAM).Schneider, H., andG. P. Barker. 1968.Matrices and Linear Algebra. NewYork: Holt, Rinehart,

and Winston.Strang, G. 1986. Introduction to Applied Mathematics. Wellesley, MA: Wellesley-Cambridge

Press.Taylor,A.E., andW.R.Mann. 1983.AdvancedCalculus. 3rd ed.NewYork: JohnWiley&Sons.

PART TWO: STATISTICS

Bachelier, Louis. 1900. The Theory of Speculation. Paris: Gauthier-Villars.Bollerslev, T. 1986. “Generalized Autoregressive Conditional Heteroskedasticity.” Journal of

Econometrics 31: 307–327.Enders, W. 2009. Applied Econometric Time Series. 3rd ed. Hoboken, NJ: JohnWiley & Sons.Engle, R. F. 1982. “Autoregressive Conditional Heteroskedasticity with Estimates of the

Variance of U.K. Inflation.” Econometrica 50: 987–1008.Fuller, W. A. 1976. Introduction to Statistical Time Series. New York: John Wiley & Sons.Granger, C. W. J., and P. Newbold. 1986. Forecasting Economic Time Series. 2nd ed. Orlando,

FL: Academic Press.Hamilton, J. D. 1994. Time Series Analysis. Princeton, NJ: Princeton University Press.Hull, J. C. 2003. Options, Futures, and Other Derivatives. 5th ed. Upper Saddle River,

NJ: Prentice Hall.Krichene, N. 2012. Islamic Capital Markets. Singapore: John Wiley & Sons.Maddala. G. S., and K. Lahiri. 2010. Introduction to Econometrics. 4th ed. Hoboken, NJ: John

Wiley & Sons.Miller, M. B. 2012. Mathematics and Statistics for Financial Risk Management. Hoboken,

NJ: John Wiley & Sons.Murray, M. 1994. “A Drunk and Her Dog: An Illustration of Co-integration and Error-

Correction.” American Statistician 48 (1): 37–39.Pindyck, R. S., and D. L. Rubinfeld. 1998. Econometric Models and Econometric Forecasts.

4th ed. Boston: McGraw-Hill.Taylor, S. 1996. Modeling Financial Time Series. 2nd ed. New York: John Wiley & Sons.Wonnacott, T. H., and Wonnacott, R. J. 1990. Introductory Statistics. 5th ed. New York:

John Wiley & Sons.

473

3GBREF 05/15/2014 13:28:57 Page 474

3GBINDEX 05/15/2014 13:34:29 Page 475

Index

Page numbers in italic type refer to figures

Absolute value, 5Addition, of vectors, 93–94ADF (augmented Dicker-Fuller) test, 367–368Allocation of resources, 195–196, 212–213Alternate hypothesis, 283Analysis of variance (ANOVA), 317Angle, 6Anti-derivative, 51–53, 88Approximation curve, 302Arbitrage-free pricing, 413, 420–425, 426Arbitrage pricing, 461–465ARCH (autoregressive conditional

heteroskedasticity) model, 397–412applications, 397–398formulation of, 401–404generalized (GARCH), 397, 398, 407–409in mean (ARCH-M), 409motivation for, 398–401properties of, 404–407shortcomings, 397testing, 409–411

Arithmetic Brownian motion, 455ARMA (autoregressive-moving average) models,

336, 344–346, 351–353Array, 6Arrow-Debreu (ad) security, 428–430, 447–448AR time series models. See Autoregressive (AR)

time series modelsAsset pricing under uncertainty, 413–438

applications, 436arbitrage-free pricing, 413, 420–425, 426basic principles, 425–428complete markets, 434–435efficient market hypothesis, 416–420future payoffs and asset prices, 430–433risk and return modeling, 414–416state prices, 428–430theories, 413–414See also Consumption-based pricing model

Asset-specific pricing, 442–444Augmented Dicker-Fuller (ADF) test, 367–368Autocorrelation function, 333–336, 346–348Autoregression, 347Autoregressive-moving average (ARMA) models,

336, 344–346, 351–353Autoregressive representation, 340Autoregressive (AR) time series models

features, 336, 340–343forecasting with, 349–350invertible process, 343

See also Vector autoregressive analysis (VAR)Auxiliary equation, 126, 149Axioms of probability, 230–231

Bachelier, Louis, 416Banker’s Equation, 117–118Base of logarithm, 78Base of vector space, 97–98Base year, 7Bayes’ Theorem, 236–237Bernoulli distribution, 252–253, 265–266Best estimator, 278Best-fitting curve, 304Biased estimate, 276Bi-infinite sequence, 16Binomial, 8Binomial distribution, 253, 266Binomial tree, geometric Brownian motion

approximation, 456–457Black-Scholes model, 461–465Bounding face solution, 200Box-PierceQ-statistic, 335Brownian motion, 414, 451–458Budget constraint, 30–31

Canonical form linear programming, 198–199,204–205

Capital asset pricing model (CAPM)consumption-based model and, 439description of, 34–35, 301risk and return, 25, 397–398, 409significance testing, 318–319

Cash flow sequence, 18Cauchy sequence, 17Central limit theorem, 249Certain event, 228Ceteris paribus condition, 61Chain rule, 44, 63–64Characteristic equation, 126, 149Chebyshev’s inequality, 247–248Chi-square distribution, 257–258, 282Cholesky decomposition, 112CI (confidence interval), 278–282, 290–296,

348–349Classical programming, 169, 183–186Closed interval, 43Cobweb model, 147–149Coefficient of correlation, 312Coefficient of determination, 311–312Co-integrating vector, 392

475

3GBINDEX 05/15/2014 13:34:30 Page 476

Co-integration, 381–396applications, 381, 395–396common trends for variables, 390–391definition of, 384–388long-run equilibrium and, 383–388spurious regression problem and, 382–383stationarity and, 383–384test for, 388–390vector autoregressive analysis, 391–394

Combination, 233–234Combinatorial analysis, 232–234Commodity, 203Complementary slackness theorem, 216–217Complete markets, 434–435“Completing the square,” 11–12Complex number, 4–5Component movements of time series, 327–330Components, 91Composite function, 27Conditional probability, 234–237Confidence coefficient, 278, 290Confidence interval (CI), 278–282, 290–296,

348–349Constant payment, 21Constant rule, 44Constrained optimization, 168, 178–183, 184–186Constraint, external problems with, 72–75Constraint constant, 169Constraint function, 169Consumer price index (CPI), 8Consumption-based pricing model, 439–450advantages of, 439asset-specific pricing and risk correction,

442–444contingent discount factors, 449–450expected return-beta representation, 444–445features, 439mean-variance frontier, 445–447optimization and, 439–442risk-neutral pricing implied by, 447–448

Consumption function, 33Continuous random variable, 239–240Continuous-time pricing models, 451–472Black-Scholes model, 461–465Brownian motion, 414, 451–458diffusion process, 453–456discrete approximations, 460–461introduction, 451Ito’s lemma, 458–460market price of risk, 465–467risk-neutral pricing, 467–471time series analysis vs., 451

Contours, 38Convergence, 17, 19Convexity of sukuk’s price, 57–58Correlogram, 336Cosine, 89Cost function, 33Cotangent, 89Counting techniques, 232–234CPI (consumer price index), 8Critical point, 176Critical region, 285Critical value, 278, 290Cumulative distribution function, 239

Curve fitting, 301–304Cyclical component, 355Cyclical movements, 329

Decomposition of time series, 328Definite integral, 51Degrees, 88–89Delta, 62Demand, 29–30Dependent variable, 202Derivatives (functions)computation of, 41–43definition of, 117directional, 65–66integration, 41, 51–55left-hand, 43right-hand, 43See also Differentiation; Partial derivatives

Derivatives pricingIto’s lemma, 458–460principles, 425–428volatility modeling, 402See also Asset pricing under uncertainty

Determinant of matrix, 105Deterministic trend, 359Diagnostic test of regression results, 319–321Diagonable matrix, 111Diagonal matrix, 100, 102Dickey-Fuller (DF) distribution, 364Dickey-Fuller (DF) test, 355, 364–368Differenceconfidence intervals for, 281–282sampling distribution of, 274–276

Difference equations, 141–166applications, 141cobweb model, 147–149definition of, 141–143equilibrium and stability, 158–165first-order linear, 142, 144–149, 159–161impulse response function, 146–147linear systems, 154–158, 161–162second-order linear, 149–154

Difference stationary, 363Differential equations, 117–140applications, 117definition of, 117examples of, 117–120first-order linear, 123–125linear, 120linear systems, 128–132nonlinear, 120phase diagrams, 133–139second-order linear, 125–128, 129–130solution methods, 120–122

Differentiation, 41–50derivative computation, 41–43maximum and minimum of function, 44–47mean value theorem, 48rules, 44sukuk duration and convexity application,

55–58Taylor expansion, 49–50, 57–58

Diffusion process, 453–456Direction, 6Directional derivative, 65–66

476 INDEX

3GBINDEX 05/15/2014 13:34:31 Page 477

Discontinuous function, 28Discounted value, 18Discriminant, 12–13, 71Distribution function, 232Dot product, 94–95Double integral, 54–55Double root, 12Duality

economic theory and, 219–222interpretation of variables, 217–219problem of, 195–196, 212–213theorem, 216

Duration of sukuk, 55–57Durbin-Watson statistic, 320–321Dynamic multiplier, 146

Echelon matrix, 104–105ECM (error correction mechanism), 384,

389–390Econometrics, 113–114Economic data, reporting of, 7–8Efficient estimate, 278Efficient estimator of mean, 278Efficient market hypothesis, 416–420Eigenvalue, 109–110, 130–132Eigenvector, 109–110, 130–132Elasticity, 8Elementary event, 228Elements, 15Elimination method, 73Empirical distributions, 260–262Endogenous variable, 369Entries, 98Equations

auxiliary, 126, 149Banker’s, 117–118curve approximation, 302difference (See Difference equations)differential (See Differential equations)Euler, 441of higher order, 14–15homogeneous systems of linear equations,107–108

mean and variance, 402–404quadratic, 11–14

Equilibrium, 160–161, 383–388Equilibrium pricing model, 439, 449–450. See also

Consumption-based pricing modelError, 302Error correction mechanism (ECM), 384, 389–390Estimated error term, 304, 305, 310–311Estimation of parameters, 276–278Euler approximation or discretization scheme, 460Euler equation, 441Events

definition of, 228–230independence of, 237–238

Existence theorem, 215Expected mean, 241–242, 415Experiment, 228Explained part of yi, 311Exponential function

applications, 86–87definition of, 82–83general, 85–86

power series, 84Exponential number, 5Exponential trend model, 329Extrema of functions, 44–47, 70–75Extremum value theorem, 45–47

Factorial, 10Factorial notation, 232Factorization of polynomial, 10–11Farmer’s linear programming problem, 200–201Fat-tailed distributions, 244, 321F distribution, 259–260, 296–297Fibonacci numbers, 16Financial data, reporting of, 7–8Finite sequence, 16First fundamental theorem of calculus, 53First moment of random variable, 241–242First-order linear difference equations, 142,

144–149, 159–161First-order linear differential equations, 123–125Fitted value, 305Forecast confidence interval, 348–349Forward contracts, pricing by arbitrage, 425Forward exchange rate, 37, 424–425Fourth moment of random variable, 244–245Frequency, 327Functions

autocorrelation, 333–336, 346–348budget constraint, 30–31composite, 27constraint, 169consumption, 33cost, 33cumulative distribution, 239definition of, 25–27demand and supply, 29–30differentiation, 41–50discontinuous, 28distribution, 232exponential, 82–87forward exchange rate, 37graphs of, 28implicit, 64impulse response, 146–147, 375–378integration, 51–55investment, 33likelihood (l), 309linear, 28logarithm, 77–87maximum and minimum of, 44–47, 70–75mean value theorem, 48moment generating, 263–268money demand, 33multi valued, 28multivariate, 37–38option price, 36parabolic, 28parametric form, 27–29payoff of futures contract, 35payoff of option contract, 35–36payoff to swap, 36polynomial, 26–27polynomial approximations, 49–50present value, 34production, 32–33

Index 477

3GBINDEX 05/15/2014 13:34:32 Page 478

Functions (Continued )production possibility frontier, 31profit, 33quantity theory of money, 33sales’, 33trigonometric, 88–90utility, 32, 37–38, 61vector-valued, 28–29See also Models

Fundamental theorems of calculus, 53–54Futures contracts, 35, 459–460

Gaussian distribution. SeeNormal (Gaussian)distribution

Generalized ARCH (GARCH) model, 397, 398,407–411

Generalized inverse matrix, 108Geometric Brownian motion, 456–458Geometry of linear programming, 199–201Geometry of optimization problem, 171–173Gradient, 66–67, 184Gradient theorem, 67Graphics, 6–7Gross domestic product (GDP), 5, 7

Heteroskedasticity, 397, 398, 401High-frequency variable, 327Histogram of probability distribution, 238–239Homogeneous difference equations, 149–150Homogeneous differential equations, 123, 125–126Homogeneous systems of equations, 107–108Homoskedasticity, 398Hypothesis testing, 282–289alternate hypothesis, 283applications, 298level of significance, 284–286MPLE differences, 288–289null hypothesis, 283probability value, 286–287regression coefficients, 315–319special cases, 287–288statistical hypothesis, 283type I or type II errors, 283–284

Identities, 10, 33–34, 78Identity matrix, 101Implicit function, 64Impossible event, 228Impulse response function, 146–147, 375–378Income, sequence for, 18Incomplete markets, 434Indefinite integralchange in variables in, 54definition of, 51, 52differential equation solution method, 120–121

Independence of events, 237–238Independent random variable, 247Independent variable, 302Indexing notation, 16Indicators, 7–8Indices, 7–8Indifference curve, 32Inefficient estimator, 278Inequality constraints, 186–192Infinite sequence, 16

Infinite series, 19Inner product, 94–95Innovations, 337Instruments, 168–169Integral, 51–55Integrand, 51Integration, 41, 51–55. See also Co-integrationIntegration by parts theorem, 88Intercept, 303Interestcontinuous compounding, 86–87simple compounding, 86–87

Internal rate of return (IRR), 15Intertemporal optimization, 439–442Interval estimate, 278Inverse matrix, 108Investment function, 33Irrational number, 3Irregular movements, 329Ito’s lemma, 458–460

Jarque-Bera statistic, 262, 321Johansen procedure, 393, 394Joint probability, 245–247

Kth moment, 261Kuhn-Tucker (K-T) conditions, 188–192, 214–215Kurtosis, 244–245, 321, 415

Lagrange multipliers, 73–75, 179–186, 214–217Large numbers, law of, 248Laurent series, 20Law of large numbers, 248Leading entry, 103Least-square curve, 304Least squares estimators of ß, 114Least squares method, 304Left-hand derivative, 43Leptokurtosis, 321Level curves, 38Level of significance, 284–286Likelihood function (l), 309Linear algebra, 91–116applications, 113–114homogeneous systems of equations, 107–108matrices, 98–112stability of linear system, 112–113vectors, 91–98

Linear combinations of vectors, 96Linear difference equationsfirst-order, 142, 144–149, 159–161second-order, 149–154systems, 154–158, 161–162

Linear differential equationsfirst order, 123–125second order, 125–128, 129–130systems, 128–132

Linear function, 28Linearly dependent or independent vector, 97Linear programming (LP), 195–224applications, 222complementary slackness theorem, 216–217duality theorem, 216dual problem of, 195–196, 212–213economic theory and duality, 219–222

478 INDEX

3GBINDEX 05/15/2014 13:34:33 Page 479

existence theorem, 215features of, 195–197formation, 170–171, 197–199geometry of, 199–201importance of, 195interpretation of dual variables, 217–219Lagrangian approach, 214–217simplex method, 201–212

Linear regression analysis, 304–312Linear regression line, 303Linear system, stability of, 112–113Linear time series models, 336–338, 401Linear trend model, 329Ljung-BoxQ-statistic, 335Location parameter, 241Logarithm function

applications, 86–87change of base, 78definition of, 77general, 85–86identities, 78natural, 78–82power series, 84

Logarithm number, 5Log linear trend model, 329Long-run equilibrium, 383–388Low-frequency variable, 327LP. See Linear programming (LP)

Maclaurin series, 50, 84Magnitude, 91, 92–93Main diagonal, 100Marginal distribution, 245Marginal productivity, 61Marginal propensity of consumption, 153Marginal utility, 61Market model, 29–30Market price of risk, 465–467Market stability analysis, 133–139, 158–165Market value of portfolio, 6–7Martingale difference, 403Martingale pricing

aspects of, 418–419complete markets and, 434–435examples of, 419–420future payoffs and asset prices, 430–433principles, 426–428random walk vs., 420

MA time series models. SeeMoving average (MA)time series models

Matrices, 98–112Cholesky decomposition, 112definition of, 6, 98–99echelon, 104–105generalized inverse, 109inverse, 108multiplication, 100rank of, 103–105transposes, 99See also Square matrices

Maxima point, 176Maximum likelihood method, 306, 309–310Maximum of function, 44–47, 70–75Mean

confidence intervals for, 279–280

hypothesis testing, 287–288sampling distribution of, 272–273

Mean and variance equation, 402–404Mean of squared errors (MSE), 310Mean value theorem, 48Mean-variance frontier, 445–447Method of moments, 306, 308–309MGF (moment generating function), 263–268Minima point, 176Minimum mean-squared-error forecasts, 348Minimum of function, 44–47, 70–75Models

capital asset pricing, 34–35definition of, 25, 29market, 29–30See also Functions

Modulus, 92–93Moment generating function (MGF), 263–268Moments, method of, 306, 308–309Money demand function, 33Monomial, 9Monotone sequence, 16Monotonically increasing or decreasing sequence, 16Monte Carol methods, 251Most efficient estimator, 278Moving average (MA) time series models

co-integration and, 394–395features, 336, 338–340forecasting with, 350–351

MSE (mean of squared errors), 310Multiple correlation, 323–324Multiplication

matrices, 100vectors, 94–96

Multiplier-accelerator model, 153–154Multi valued function, 28Multivariate function, 37–38

National income identity, 33–34Natural logarithmic function, 78–82Neoclassical marginalism, 167NLP (nonlinear programming), 169–170, 186–192Nonhomogeneous difference equations, 150–153Nonhomogeneous differential equations, 124–125,

127–128Nonlinear programming (NLP), 169–170, 186–192Nonlinear relationship, 302Nonlinear time series, 401Nonsense (spurious) regression, 382–383, 386, 389Nonsingular matrix, 108Nonstationary time series analysis, 355–368applications, 355, 368decomposition of nonstationary time series,359–360

random walk, 355–362unit-root test, 355, 362–368

Norm, 92–93Normal (Gaussian) distribution

definition of, 255–256diagnostic test of regression results, 321MGF of, 267–268probability density function, 87

Normal equations method, 306, 308–309Normality test, 321Normal line, 70

Index 479

3GBINDEX 05/15/2014 13:34:34 Page 480

Null hypothesis, 283, 315Number line, 3

OLS (ordinary least-squares method), 306–308One-sided test, 286One-tailed test, 286Opportunity cost, 202, 206Optimization, 167–194application, 167–168classical programming, 169, 183–186constrained, 168, 178–183, 184–186definition of, 167–168geometry of, 171–173intertemporal, 439–442nonlinear programming, 169–170, 186–192programming problem formulation, 168–173unconstrained, 168, 173–178See also Linear programming (LP)

Option contracts, payoff of, 35–36Option pricingby arbitrage, 423Black-Scholes model, 461–465delta, 61–62function, 36

Ordered pair, 91Ordered sample, 233Ordinary least-squares method (OLS), 306–308Orthogonal matrix, 103Orthogonal vector, 95

Parabolic function, 28Parameter estimation, 276–278Parametric representation, 38Partial autocorrelation function, 346–348Partial derivatives, 61–76chain rule, 63–64computation of, 62–64definition of, 62directional derivatives, 65–66extremal problems with constraints, 72–75extrema of functions of several variables, 70–72gradients, 66–67of implicit functions, 64importance of, 61–62normal line, 70tangent planes, 68–69total differential of function with many

variables, 64Percent change, 8Permanent component, 355Permutations, 232–233Phase diagram, 133–139, 163–165Point estimate, 278Poisson distribution, 254–255, 266–267Polynomial, 9–11Polynomial approximations of function, 49–50Polynomial function, 26–27Polynomial lags, 9–10Portfolio risk, 13–14Positive definite matrix, 102Power series, 19PPF (production possibility frontier), 31Prediction, 321–323Present value, 20–21, 87Present value function, 34Price index, 7–8

Pricing kernel, 430, 443Primal linear programming, 195–196, 212–213Prime number, 15–16Probability, 227–250applications, 227axioms of, 230–231central limit theorem, 249Chebyshev’s inequality, 247–248combinatorial analysis, 232–234conditional, 234–237definition of, 227, 230events, 228–230independence of events, 237–238law of large numbers, 248random variables, 231–232sample space, 228See also Probability distribution

Probability distributionapplications, 268Bernoulli, 252–253, 265–266binomial, 253, 266chi-square, 257–258, 290–296continuous random variables, 239–240cumulative distribution function, 239empirical, 260–262F, 259–260, 296–297histogram for, 238–239joint distribution of random variables, 245–247moment generating function and, 263–268moments of, 241–245normal (Gaussian), 87, 255–256, 267–268, 321Poisson, 254–255, 266–267of regression coefficients, 312–315t, 258–259, 289–292uniform, 251–252, 264–265

Probability value (p-value), 286–287Production function, 32–33Production possibility frontier (PPF), 31Product rule, 44Profit function, 33Programming problem formulation, 168–173Proportionsconfidence intervals for, 280–281hypothesis testing, 288sampling distribution of, 273–274

Pth order VAR, 373p-value, 286–287

Quadratic equation, 11–14Quadratic matrix form, 102Quadratic trend model, 329Quantity theory of money, 33Quotient rule, 44

Radians, 88–89Random error vector, 305Random event, 230Random movements, 329Random sampling, 271Random variablesdefinition of, 231–232probability distribution, 238–247

Random walk, 355–362asset pricing under uncertainty, 417–418, 420decomposition of nonstationary time series,

359–360

480 INDEX

3GBINDEX 05/15/2014 13:34:35 Page 481

definition of, 355–358forecasting, 360–362

Rank of matrix, 103–105Rate of return, 15, 86–87, 238–239Rational number, 3Ratios, 7Real gross domestic product (GDP), 5, 7Realization, 330Real line, 3Real number, 3–4, 91Rectangular distribution, 251–252Region of acceptance, 285Region of insignificance, 285Region of rejection, 285Region of significance, 285Regression analysis, 301–326

applications, 301curve fitting, 301–304diagnostic test of results, 319–321hypothesis testing of regression coefficients,315–319

linear regression analysis, 304–312multiple correlation, 323–324prediction, 321–323probability distribution of regression coefficients,312–315

software packages, 301nRelative frequency, 230Replication of asset, 413–414, 426Residuals, 304, 305, 310–311Resource allocation and valuation, 195–196,

212–213Return, modeling for asset pricing, 414–416Right-hand derivative, 43Risk

forms of, 227market price of, 465–467modeling for asset pricing, 414–416

Risk-neutral probability or pricingdefinition of, 413principles, 414, 426, 467–471See also Martingale pricing

Root, 149

Saddle point, 139, 176Sales’ function, 33Sample autocorrelation function, 333–334Sample mean, 261Sample moments, 261Sample point, 228Sample space, 228Sample statistics, 271Sample variance, 261–262Sampling distribution, 271–276Sampling theory, 271–282

applications, 298confidence intervals, 278–282, 290–296parameter estimation, 276–278sampling distribution, 271–276small samples, 289–298

Sampling with or without replacement, 233Scalar, 91Scalar product, 94–95Scatter diagram, 302Seasonal movements, 329Second fundamental theorem of calculus, 53–54

Second moment of random variable, 242–243Second-order linear difference equations, 149–154Second-order linear differential equations,

125–128, 129–130Separable variable, differential equation solution

method, 121–122Sequence, 15–18SER (standard error of regression), 319–320Serial correlation, 320–321Series, 18–21Sharpe ratio, 447Significance level, 284–286Significance testing, 315–319Similarity of square matrices, 110–111Simplex method, 201–212Simulation analysis, 251–252Sine, 89Singular matrix, 108Skewness, 243–244, 415Small sampling theory, 289–297Spurious regression, 382–383, 386, 389Square matrices

definition of, 100–101determinant, 105–106diagonable, 111diagonal, 100eigenvalues and eigenvectors, 109–110generalized inverse, 109identity, 101inverse, 108nonsingular, 108orthogonal, 103positive definite, 102quadratic form, 102similarity, 110–111singular, 108symmetric, 101–102upper triangular, 101

Stability analysis, 133–139, 158–165Stability of linear system, 112–113Stable arm, 139Standard deviation, 242–243, 282Standard error of regression (SER), 319–320Standard form linear programming, 197–198Standard & Poor’s 500 Stock Index (S&P 500)

crude oil prices and, 372gold prices and, 371interest rates and, 370monthly returns, 399as time series, 327, 328weekly returns, 262

State of world, 414–415, 428–430State price density, 443State prices, 428–433Stationarity, 383–384, 400–401Stationary point, 176Stationary time series, 330–333Stationary value, 160–161Statistical decisions, 282–283Statistical hypothesis, 283Statistical significance, 284–286, 315–319Statistics, definition of, 227Stochastic discount factor, 430, 442Stochastic process, 330, 413Stochastic trend, 356, 359Stock prices

Index 481

3GBINDEX 05/15/2014 13:34:41 Page 482

Stock prices (Continued )Brownian motion, 451–458diffusion process, 453–456equilibrium value, 21random walk hypothesis, 417–418sample space, 228vector autoregressive analysis, 369volatility, 398See also Standard & Poor’s 500 Stock Index

(S&P 500)Strings, 15Student’s t-distribution, 289–292Subtraction, of vectors, 93–94Success, 273Sukuk, 55–58Sum rule, 44Supply, 29–30Sure event, 228Swaps, 36Symmetric matrix, 101–102

Tangent, 89Tangent plane, 68–69, 176Taylor expansion, 49–50, 57–58, 84Taylor series, 19, 84T-distribution, 258–259, 289–292Technical equivalence, 202–203Temporary shock, 359Terms of sequence, 15, 16Tests of significance, 315–319Test statistic, 286Third moment of random variable, 243–244Time series analysis, 327–354applications, 353autocorrelation function, 333–336autoregressive linear models, 336, 340–343,

349–350autoregressive-moving average linear models,

336, 344–346, 351–353component movements, 327–330continuous time finance vs., 451definition of, 327forecasting with, 348–353moving average linear models, 336, 338–340,

350–351nonstationary, 355–368partial autocorrelation function, 346–348stationary time series, 330–333Wold decomposition of stationary process,

336–338Trajectories, 134Transitory component, 355Transposes of matrices, 99Tree diagram, 234, 235Trend component, 355Trend movements, 328Trend stationary, 363Trigonometric function, 88–90Trigonometric number, 5Trivial solution, 107Two-sided test, 286Two-tailed test, 286Type I error, 283–284Type II error, 283–284

Unbiased estimate, 276–278Uncertainty, 227, 413. See also Asset pricing under

uncertaintyUnconstrained optimization, 168, 173–178Unexplained part of yi, 311Uniform distribution, 251–252, 264–265Unit-root test, 355, 362–368Unit vector, 93Unorthodox money policy, 369Unstable arm, 139Upper triangular matrix, 101Utility function, 32, 37–38, 61

Valuation of resources, 195–196, 212–213Value-at-Risk (VaR), 256Value theory, 167VAR. See Vector autoregressive analysis (VAR)Variableschange in indefinite integrals, 54definition of, 8dependent, 202dual, 217–219endogenous, 369high-frequency, 327independent, 302of integration, 51low-frequency, 327random, 231–232, 238–247separable, 121–122

Variance, 242–243, 415–416Variance decomposition, 378–379Vector autoregressive analysis (VAR), 369–380applications, 369, 379co-integration, 391–394forecasting with, 375formulation of, 369–375impulse response function, 375–378variance decomposition, 378–379

Vector error-correction (VEC), 392–394Vectorsaddition or subtraction, 93–94definition of, 5–6, 91–93linear combinations, 96linear dependence or independence, 97magnitude, 91, 92–93multiplication, 94–96unit, 93

Vector space, 96, 97–98Vector-valued function, 28–29Vertex solution, 200Volatility models, 401–402. See also ARCH

(autoregressive conditional heteroskedasticity)model

Weak-form efficient market hypothesis, 417Weak-form market efficiency, 419White noise, 333, 355Wiener process, 414, 451–453Wold decomposition, 336–338, 359Wold representation, 337Words (finite sequence), 15

Zero solution, 107Zero vector, 92Z-score, 285, 287

482 INDEX

introductory mathematics and statistics for islamic finance...as an introductory text, no...

Documents