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Trim Size: 6in x 9in Enders bindex.tex V1 - 08/27/2014 6:48pm Page 486 Trim Size: 6in x 9in Enders FOURTH EDITION APPLIED ECONOMETRIC TIME SERIES WALTER ENDERS University of Alabama ffirs.tex V1 - 08/21/2014 11:43am Page i Trim Size: 6in x 9in Vice President and Executive Publisher Executive Editor Sponsoring Editor Project Editor Editorial Assistant Photo Editor Cover Designer Associate Production Manager Senior Production Editor Production Management Services Cover Image Credit Enders ffirs.tex V1 - 09/02/2014 12:46pm George Hoffman Joel Hollenbeck Marian Provenzano Brian Baker Jacqueline Hughes Billy Ray Kenji Ngieng Joyce Poh Jolene Ling Laserwords Mmdi/Stone/Getty Images This book was set in 10/12 Times by Laserwords and printed and bound by Lightning Source. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2015, 2010, 2009, 1998 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the app; ropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923 (Web site: www.copyright.com). Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, or online at: www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at: www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. Library of Congress Cataloging-in-Publication Data Enders, Walter, 1948Applied econometric time series / Walter, University of Alabama. – Fourth edition. pages cm Includes index. ISBN 978-1-118-80856-6 (pbk.) 1. Econometrics. 2. Time-series analysis. I. Title. HB139.E55 2015 330.01'519233–dc23 2014013428 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 Page ii Trim Size: 6in x 9in Enders ffirs.tex To Lola V1 - 08/21/2014 11:43am Page iii Trim Size: 6in x 9in Enders CONTENTS PREFACE vii ABOUT THE AUTHOR CHAPTER 1 x DIFFERENCE EQUATIONS 1 Introduction 1 1 Time-Series Models 1 2 Difference Equations and Their Solutions 7 3 Solution by Iteration 10 4 An Alternative Solution Methodology 14 5 The Cobweb Model 18 6 Solving Homogeneous Difference Equations 22 7 Particular Solutions for Deterministic Processes 31 8 The Method of Undetermined Coefficients 34 9 Lag Operators 40 10 Summary 43 Questions and Exercises 44 CHAPTER 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 iv STATIONARY TIME-SERIES MODELS Stochastic Difference Equation Models 47 ARMA Models 50 Stationarity 51 Stationarity Restrictions for an ARMA (p, q) Model The Autocorrelation Function 60 The Partial Autocorrelation Function 64 Sample Autocorrelations of Stationary Series 67 Box–Jenkins Model Selection 76 Properties of Forecasts 79 A Model of the Interest Rate Spread 88 Seasonality 96 Parameter Instability and Structural Change 102 Combining Forecasts 109 Summary and Conclusions 112 Questions and Exercises 113 47 55 ftoc.tex V1 - 08/21/2014 11:47am Page iv Trim Size: 6in x 9in Enders ftoc.tex V1 - 08/21/2014 11:47am Page v CONTENTS CHAPTER 3 1 2 3 4 5 6 7 8 9 10 11 12 13 152 MODELS WITH TREND 181 Deterministic and Stochastic Trends 181 Removing the Trend 189 Unit Roots and Regression Residuals 195 The Monte Carlo Method 200 Dickey–Fuller Tests 206 Examples of the Dickey–Fuller Test 210 Extensions of the Dickey–Fuller Test 215 Structural Change 227 Power and the Deterministic Regressors 235 Tests with More Power 238 Panel Unit Root Tests 243 Trends and Univariate Decompositions 247 Summary and Conclusions 254 Questions and Exercises 255 CHAPTER 5 1 2 3 4 5 6 7 8 9 10 11 12 118 Economic Time Series: The Stylized Facts 118 ARCH and GARCH Processes 123 ARCH and GARCH Estimates of Inflation 130 Three Examples of GARCH Models 134 A GARCH Model of Risk 141 The ARCH-M Model 143 Additional Properties of GARCH Processes 146 Maximum Likelihood Estimation of GARCH Models Other Models of Conditional Variance 154 Estimating the NYSE U.S. 100 Index 158 Multivariate GARCH 165 Volatility Impulse Responses 172 Summary and Conclusions 174 Questions and Exercises 176 CHAPTER 4 1 2 3 4 5 6 7 8 9 10 11 12 13 MODELING VOLATILITY MULTIEQUATION TIME-SERIES MODELS 259 Intervention Analysis 261 ADLs and Transfer Functions 267 An ADL of Terrorism in Italy 277 Limits to Structural Multivariate Estimation 281 Introduction to VAR Analysis 285 Estimation and Identification 290 The Impulse Response Function 294 Testing Hypotheses 303 Example of a Simple VAR: Domestic and Transnational Terrorism Structural VARs 313 Examples of Structural Decompositions 317 Overidentified Systems 321 309 v Trim Size: 6in x 9in vi 13 14 15 Enders ftoc.tex V1 - 08/21/2014 11:47am CONTENTS The Blanchard–Quah Decomposition 325 Decomposing Real and Nominal Exchange Rates: An Example Summary and Conclusions 335 Questions and Exercises 337 CHAPTER 6 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 13 COINTEGRATION AND ERROR-CORRECTION MODELS 343 Linear Combinations of Integrated Variables 344 Cointegration and Common Trends 351 Cointegration and Error Correction 353 Testing for Cointegration: The Engle–Granger Methodology Illustrating the Engle–Granger Methodology 364 Cointegration and Purchasing Power Parity 370 Characteristic Roots, Rank, and Cointegration 373 Hypothesis Testing 380 Illustrating the Johansen Methodology 389 Error-Correction and ADL Tests 393 Comparing the Three Methods 397 Summary and Conclusions 400 Questions and Exercises 401 CHAPTER 7 331 NONLINEAR MODELS AND BREAKS 407 Linear Versus Nonlinear Adjustment 408 Simple Extensions of the ARMA Model 410 Testing for Nonlinearity 413 Threshold Autoregressive Models 420 Extensions of the TAR Model 427 Three Threshold Models 433 Smooth Transition Models 439 Other Regime Switching Models 445 Estimates of STAR Models 449 Generalized Impulse Responses and Forecasting Unit Roots and Nonlinearity 461 More on Endogenous Structural Breaks 466 Summary and Conclusions 474 Questions and Exercises 475 INDEX 479 REFERENCES (ONLINE) ENDNOTES (ONLINE) STATISTICAL TABLES (ONLINE) 453 360 Page vi Trim Size: 6in x 9in Enders fpref.tex V2 - 08/18/2014 6:29pm Page vii PREFACE When I began writing the first edition, my intent was to write a text in time-series macroeconometrics. Fortunately, a number of my colleagues convinced me to broaden the focus. Applied microeconomists have embraced time-series methods, and the political science journals have become more quantitative. As in the previous editions, examples are drawn from macroeconomics, agricultural economics, international finance, and my work with Todd Sandler on the study of domestic and transnational terrorism. You should find that the examples in the text provide a reasonable balance between macroeconomic and microeconomic applications. 1. BACKGROUND The text is intended for those with some background in multiple regression analysis. I presume the reader understands the assumptions underlying the use of ordinary least squares. All of my students are familiar with the concepts correlation and covariation; they also know how to use t-tests and F-tests in a regression framework. I use terms such as mean square error, significance level, and unbiased estimate without explaining their meaning. Two chapters of the text examine multiple time-series techniques. To work through these chapters, it is necessary to know how to solve a system of equations using matrix algebra. Chapter 1, entitled "Difference Equations," is the cornerstone of the text. In my experience, this material and a knowledge of regression analysis are sufficient to bring students to the point where they are able to read the professional journals and to embark on a serious applied study. Nevertheless, one unfortunate reader wrote, "I did everything you said in you book, and my article still got rejected." Some of the techniques illustrated in the text need to be explicitly programmed. Structural VARs need to be estimated using a package that has the capacity to manipulate matrices. Monte Carlo methods are very computer intensive. Nonlinear models need to be estimated using a package that can perform nonlinear least squares and maximum likelihood estimation. Completely menu-driven software packages are not able to estimate every form of time-series model. As I tell my students, by the time a procedure appears on the menu of an econometric software package, it is not new. To get the most from the text, you should have access to a program such as EViews, RATS, MATLAB, R, STATA, SAS, or GAUSS. I take the term applied that appears in the title earnestly. Toward this end, I believe in teaching by induction. The method is to take a simple example and build toward more general and more complicated models. Detailed examples of each procedure are provided. Each concludes with a step-by-step summary of the stages typically employed in using that procedure. The approach is one of learning by doing. vii Trim Size: 6in x 9in viii Enders fpref.tex V2 - 08/18/2014 6:29pm Page viii PREFACE A large number of solved problems are included in the body of each chapter. The "Questions and Exercises" section at the end of each chapter is especially important. You are encouraged to work through as many of the examples and exercises as possible. 2. WHAT IS NEW IN THE FOURTH EDITION? I have tried to be careful about the trade-off between being complete and being concise. In deciding on which new topics to include in the text, I relied heavily on the e-mail messages I received from instructors and from students. To keep the manuscript from becoming encyclopedic, I have included a number of new topics in the Supplementary Manual. The new material in Chapter 2 discusses the important issue of combining multiple univariate forecasts so as to reduce overall forecast error variance. Chapter 3 expands the discussion of multivariate GARCH models by illustrating volatility impulse response functions. In doing so, volatility spillovers need to be analyzed in a way that is analogous to the impulse responses from a VAR. I received a surprisingly large number of questions regarding autoregressive distributed lag (ADL) models. As such, the first few parts of Chapter 5 have been rewritten so as to show the appropriate ways to properly identify and estimate ADLs. This new material complements the material in Chapter 6 involving ADLs in a cointegrated system. Chapter 7 now discusses the so-called Davies problem involving unidentified nuisance parameters under the null hypothesis. The chapter continues to discuss the issues involved with testing for multiple endogenous breaks (i.e., potential breaks occurring at an unknown date) using the Bai–Perron procedure. Moreover, since breaks can manifest themselves slowly, the process of estimating a model with a logistic break is illustrated. Some content has been moved to the website for the Fourth Edition. This content is called out in the Table of Contents as being "online." To locate this content, go to Wiley.com/College/Enders or to time-series.net. 3. ADDITIONAL MATERIALS Since it was necessary to exclude some topics from the text, I prepared a Supplementary Manual to the text. This manual contains material that I deemed important (or interesting), but not sufficiently important for all readers, to include in the text. Often the text refers you to this Supplementary Manual to obtain additional information on a topic. To assist you in your programming, I have written a RATS Programming Manual to accompany this text. Of course, it is impossible for me to have versions of the guide for every possible platform. Most programmers should be able to transcribe a program written in RATS into the language used by their personal software package. An Instructors' Manual is available to those adopting the text for their class. The manual contains the answers to all of the mathematical questions. It also contains programs that can be used to reproduce most of the results reported in the text and all of the models indicated in the "Questions and Exercises" sections. Versions of the manual are available for EVIEWS, RATS, SAS, and STATA users. Trim Size: 6in x 9in Enders fpref.tex V2 - 08/18/2014 6:29pm Page ix PREFACE ix I have prepared a set of Powerpoint slides for each chapter. I have prepared the slides from the materials I use in my own class. As such, they emphasize on the material I deem to consider most important. Moreover, some of the slides expand in the material in the text. Wiley makes all of the manuals available to the faculty who use the text for their class. The Supplementary Manual and several versions of the Programming Manual can be downloaded (at no charge) from the Wiley website or from my personal website: www.time-series.net. The Programming Manual can also be downloaded from the ESTIMA website: www.estima.com. In spite of all my efforts, some errors have undoubtedly crept into the text. If the first three editions are any guide, the number is embarrassingly large. I will keep an updated list of typos and corrections on my website www.time-series.net. Many people made valuable suggestions for improving the organization, style, and clarity of the manuscript. I received a great number of e-mails from readers who pointed out typos and who made very useful suggestions concerning the exposition of the text. I am grateful to my students who kept me challenged and were quick to point out errors. Especially helpful were my former students Karl Boulware, Pin Chung, Selahattin Dibooglu, HyeJin Lee, Jing Li, Eric Olson, Ling Shao, and Jingan Yuan. Pierre Siklos and Mark Wohar who made a number of important suggestions concerning the revised chapters for the second edition. I learned so much about time series from Barry Falk and Junsoo Lee that they deserve a special mention. I would like to thank my loving wife, Linda, for putting up with me during my illness (especially during the time I was working on the manuscript). Just before writing the preface to the third edition, I learned that Clive Granger had died. A few months before I was to take a sabbatical at the University of Minnesota, I had the opportunity to present a seminar at UCSD. At the time, I was working with overlapping generations models and had no thoughts about being an applied econometrician. However, when I first met Clive, he stated: "It will be 100 degrees warmer here than in Minnesota next winter. Why not do the sabbatical here?" I changed my plans, thinking that I would work with the math–econ types at UCSD. Fortunately, I happened to sit through one of his classes (team-taught with Robert Engle) and fell in love with time-series econometrics. I know that it tickled Clive to tell people the story of how his class clearly changed my career. In an important way, he and Robert Engle are responsible for the approach taken in the text. Trim Size: 6in x 9in Enders flast.tex V2 - 08/06/2014 5:25pm Page x ABOUT THE AUTHOR Walter Enders holds the Bidgood Chair of Economics and Finance at the University of Alabama. He received his doctorate in economics in 1975 from Columbia University in New York. Dr. Enders has published numerous research articles in such journals as the Review of Economics and Statistics, Quarterly Journal of Economics, and the Journal of International Economics. He has also published articles in the American Economic Review (a journal of the American Economic Association), the Journal of Business and Economic Statistics (a journal of the American Statistical Association), and the American Political Science Review (a journal of the American Political Science Association). Dr. Enders (along with Todd Sandler) received the National Academy of Sciences' Estes Award for Behavioral Research Relevant to the Prevention of Nuclear War. The award recognizes "basic research in any field of cognitive or behavioral science that has employed rigorous formal or empirical methods, optimally a combination of these, to advance our understanding of problems or issues relating to the risk of nuclear war." The National Academy presented the award for their "joint work on transnational terrorism using game theory and time-series analysis to document the cyclic and shifting nature of terrorist attacks in response to defensive counteractions." x Trim Size: 6in x 9in Enders c01.tex CHAPTER V3 - 09/02/2014 12:52pm Page 1 1 DIFFERENCE EQUATIONS Learning Objectives 1. 2. 3. 4. 5. 6. 7. 8. 9. Explain how stochastic difference equations can be used for forecasting and illustrate how such equations can arise from familiar economic models. Explain what it means to solve a difference equation. Demonstrate how to find the solution to a stochastic difference equation using the iterative method. Demonstrate how to find the homogeneous solution to a difference equation. Illustrate the process of finding the homogeneous solution. Show how to find homogeneous solutions in higher order difference equations. Show how to find the particular solution to a deterministic difference equation. Explain how to use the Method of Undetermined Coefficients to find the particular solution to a stochastic difference equation. Explain how to use lag operators to find the particular solution to a stochastic difference equation. INTRODUCTION The theory of difference equations underlies all of the time-series methods employed in later chapters of this text. It is fair to say that time-series econometrics is concerned with the estimation of difference equations containing stochastic components. The traditional use of time-series analysis was to forecast the time path of a variable. Uncovering the dynamic path of a series improves forecasts since the predictable components of the series can be extrapolated into the future. The growing interest in economic dynamics has given a new emphasis to time-series econometrics. Stochastic difference equations arise quite naturally from dynamic economic models. Appropriately estimated equations can be used for the interpretation of economic data and for hypothesis testing. 1. TIME-SERIES MODELS The task facing the modern time-series econometrician is to develop reasonably simple models capable of forecasting, interpreting, and testing hypotheses concerning economic data. The challenge has grown over time; the original use of time-series 1 Trim Size: 6in x 9in 2 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 2 DIFFERENCE EQUATIONS analysis was primarily as an aid to forecasting. As such, a methodology was developed to decompose a series into a trend, a seasonal, a cyclical, and an irregular component. The trend component represented the long-term behavior of the series and the cyclical component represented the regular periodic movements. The irregular component was stochastic and the goal of the econometrician was to estimate and forecast this component. Suppose you observe the fifty data points shown in Figure 1.1 and are interested in forecasting the subsequent values. Using the time-series methods discussed in the next several chapters, it is possible to decompose this series into the trend, seasonal, and irregular components shown in the lower panel of the figure. As you can see, the trend changes the mean of the series, and the seasonal component imparts a regular cyclical pattern with peaks occurring every twelve units of time. In practice, the trend and seasonal components will not be the simplistic deterministic functions shown in this figure. The modern view maintains that a series contains stochastic elements in the trend, seasonal, and irregular components. For the time being, it is wise to sidestep these complications so that the projection of the trend and seasonal components into periods 51 and beyond is straightforward. Notice that the irregular component, while lacking a well-defined pattern, is somewhat predictable. If you examine the figure closely, you will see that the positive and negative values occur in runs; the occurrence of a large value in any period tends to 12 Observed data 10 Forecast 8 6 4 2 0 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 10 8 Trend Seasonal Irregular 6 Forecasts 4 2 0 –2 –4 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 F I G U R E 1.1 Hypothetical Time-Series Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 3 TIME-SERIES MODELS 3 be followed by another large value. Short-run forecasts will make use of this positive correlation in the irregular component. Over the entire span, however, the irregular component exhibits a tendency to revert to zero. As shown in the lower part, the projection of the irregular component past period 50 rapidly decays toward zero. The overall forecast, shown in the top part of the figure, is the sum of each forecasted component. The general methodology used to make such forecasts entails finding the equation of motion driving a stochastic process and using that equation to predict subsequent outcomes. Let yt denote the value of a data point at period t; if we use this notation, the example in Figure 1.1 assumes we observed y1 through y50 . For t = 1 to 50, the equations of motion used to construct components of the yt series are Trend: Tt = 1 + 0.1t Seasonal: St = 1.6 sin(t𝜋∕6) Irregular: It = 0.7It−1 + 𝜀t where: Tt = value of the trend component in period t St = value of the seasonal component in t It = the value of the irregular component in t 𝜀t = a pure random disturbance in t Thus, the irregular disturbance in t is 70% of the previous period's irregular disturbance plus a random disturbance term. Each of these three equations is a type of difference equation. In its most general form, a difference equation expresses the value of a variable as a function of its own lagged values, time, and other variables. The trend and seasonal terms are both functions of time and the irregular term is a function of its own lagged value and of the stochastic variable 𝜀t . The reason for introducing this set of equations is to make the point that time-series econometrics is concerned with the estimation of difference equations containing stochastic components. The time-series econometrician may estimate the properties of a single series or a vector containing many interdependent series. Both univariate and multivariate forecasting methods are presented in the text. Chapter 2 shows how to estimate the irregular part of a series. Chapter 3 considers estimating the variance when the data exhibit periods of volatility and tranquility. Estimation of the trend is considered in Chapter 4, which focuses on the issue of whether the trend is deterministic or stochastic. Chapter 5 discusses the properties of a vector of stochastic difference equations, and Chapter 6 is concerned with the estimation of trends in a multivariate model. Chapter 7 introduces the new and growing area of research involving nonlinear time-series models. Although forecasting has always been the mainstay of time-series analysis, the growing importance of economic dynamics has generated new uses for time-series analysis. Many economic theories have natural representations as stochastic difference equations. Moreover, many of these models have testable implications concerning the time path of a key economic variable. Consider the following four examples: 1. The Random Walk Hypothesis: In its simplest form, the random walk model suggests that day-to-day changes in the price of a stock should have Trim Size: 6in x 9in 4 Enders c01.tex V3 - 09/02/2014 12:52pm Page 4 CHAPTER 1 DIFFERENCE EQUATIONS a mean value of zero. After all, if it is known that a capital gain can be made by buying a share on day t and selling it for an expected profit the very next day, efficient speculation will drive up the current price. Similarly, no one will want to hold a stock if it is expected to depreciate. Formally, the model asserts that the price of a stock should evolve according to the stochastic difference equation yt+1 = yt + 𝜀t+1 or Δyt+1 = 𝜀t+1 where yt = the logarithm of the price of a share of stock on day t, and 𝜀t+1 = a random disturbance term that has an expected value of zero. Now consider the more general stochastic difference equation Δyt+1 = 𝛼0 + 𝛼1 yt + 𝜀t+1 2. The random walk hypothesis requires the testable restriction: 𝛼0 = 𝛼1 = 0. Rejecting this restriction is equivalent to rejecting the theory. Given the information available in period t, the theory also requires that the mean of 𝜀t+1 be equal to zero; evidence that 𝜀t+1 is predictable invalidates the random walk hypothesis. Again, the appropriate estimation of this type of single-equation model is considered in Chapters 2 through 4. Reduced-Forms and Structural Equations: Often it is useful to collapse a system of difference equations into separate single-equation models. To illustrate the key issues involved, consider a stochastic version of Samuelson's (1939) classic model: yt = ct + it (1.1) ct = 𝛼yt−1 + 𝜀ct 0<𝛼<1 (1.2) it = 𝛽(ct − ct−1 ) + 𝜀it 𝛽>0 (1.3) where yt , ct , and it denote real GDP, consumption, and investment in time period t, respectively. In this Keynesian model, yt , ct , and it are endogenous variables. The previous period's GDP and consumption, yt−1 and ct−1 , are called predetermined or lagged endogenous variables. The terms 𝜀ct and 𝜀it are zero mean random disturbances for consumption and investment, and the coefficients 𝛼 and 𝛽 are parameters to be estimated. The first equation equates aggregate output (GDP) with the sum of consumption and investment spending. The second equation asserts that consumption spending is proportional to the previous period's GDP plus a random disturbance term. The third equation illustrates the accelerator principle. Investment spending is proportional to the change in consumption; the idea is that growth in consumption necessitates new investment spending. The error terms 𝜀ct and 𝜀it represent the portions of consumption and investment not explained by the behavioral equations of the model. Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 5 TIME-SERIES MODELS 5 Equation (1.3) is a structural equation since it expresses the endogenous variable it as being dependent on the current realization of another endogenous variable, ct . A reduced-form equation is one expressing the value of a variable in terms of its own lags, lags of other endogenous variables, current and past values of exogenous variables, and disturbance terms. As formulated, the consumption function is already in reduced form; current consumption depends only on lagged income and the current value of the stochastic disturbance term 𝜀ct . Investment is not in reduced form because it depends on current period consumption. To derive a reduced-form equation for investment, substitute (1.2) into the investment equation to obtain it = 𝛽[𝛼yt−1 + 𝜀ct − ct−1 ] + 𝜀it = 𝛼𝛽yt−1 − 𝛽ct−1 + 𝛽𝜀ct + 𝜀it Notice that the reduced-form equation for investment is not unique. You can lag (1.2) one period to obtain: ct−1 = 𝛼yt−2 + 𝜀ct−1 . Using this expression, the reduced-form investment equation can also be written as it = 𝛼𝛽yt−1 − 𝛽(𝛼yt−2 + 𝜀ct−1 ) + 𝛽𝜀ct + 𝜀it = 𝛼𝛽(yt−1 − yt−2 ) + 𝛽(𝜀ct − 𝜀ct−1 ) + 𝜀it (1.4) Similarly, a reduced-form equation for GDP can be obtained by substituting (1.2) and (1.4) into (1.1): yt = 𝛼yt−1 + 𝜀ct + 𝛼𝛽(yt−1 − yt−2 ) + 𝛽(𝜀ct − 𝜀ct−1 ) + 𝜀it = 𝛼(1 + 𝛽)yt−1 − 𝛼𝛽yt−2 + (1 + 𝛽)𝜀ct + 𝜀it − 𝛽𝜀ct−1 so that yt can be written in the form yt = ayt−1 + byt−2 + xt (1.5) where a = 𝛼(1 + 𝛽), b = –𝛼𝛽, and xt = (1 + 𝛽)𝜀ct + 𝜀it − 𝛽𝜀ct – 1 . Equation (1.5) is a univariate reduced-form equation; yt is expressed solely as a function of its own lags and a disturbance term. A univariate model is particularly useful for forecasting since it enables you to predict a series based solely on its own current and past realizations. It is possible to estimate (1.5) using the univariate time-series techniques explained in Chapters 2 through 4. Once you have obtained estimates of a and b, it is straightforward to use the observed values of y1 through yt to predict all future values in the series (i.e., yt+1 , yt+2 , … ). Chapter 5 considers the estimation of multivariate models when all variables are treated as jointly endogenous. The chapter also discusses the restrictions needed to recover (i.e., identify) the structural model from the estimated reduced-form model. 3. Error-Correction: Forward and Spot Prices: Certain commodities and financial instruments can be bought and sold on the spot market (for immediate delivery) or for delivery at some specified future date. For example, Trim Size: 6in x 9in 6 Enders c01.tex V3 - 09/02/2014 12:52pm Page 6 CHAPTER 1 DIFFERENCE EQUATIONS suppose that the price of a particular foreign currency on the spot market is st dollars and that the price of the currency for delivery one period into the future is ft dollars. Now, consider a speculator who purchased forward currency at the price ft dollars per unit. At the beginning of period t + 1, the speculator receives the currency and pays ft dollars per unit received. Since spot foreign exchange can be sold at st+1 , the speculator can earn a profit (or loss) of st+1 − ft per unit transacted. The Unbiased Forward Rate (UFR) hypothesis asserts that expected profits from such speculative behavior should be zero. Formally, the hypothesis posits the following relationship between forward and spot exchange rates: st+1 = ft + 𝜀t+1 (1.6) where 𝜀t+1 has a mean value of zero from the perspective of time period t. In (1.6), the forward rate in t is an unbiased estimate of the spot rate in t + 1. Thus, suppose you collected data on the two rates and estimated the regression st+1 = 𝛼0 + 𝛼1 ft + 𝜀t+1 If you were able to conclude that 𝛼0 = 0, 𝛼1 = 1, and that the regression residuals 𝜀t+1 have a mean value of zero from the perspective of time period t, the UFR hypothesis could be maintained. The spot and forward markets are said to be in long-run equilibrium when 𝜀t+1 = 0. Whenever st+1 turns out to differ from ft , some sort of adjustment must occur to restore the equilibrium in the subsequent period. Consider the adjustment process st+2 = st+1 − 𝛼[st+1 − ft ] + 𝜀st+2 ft+1 = ft + 𝛽[st+1 − ft ] + 𝜀ft+1 𝛼>0 𝛽>0 (1.7) (1.8) where 𝜀st+2 and 𝜀ft+1 both have a mean value of zero. Equations (1.7) and (1.8) illustrate the type of simultaneous adjustment mechanism considered in Chapter 6. This dynamic model is called an error-correction model because the movement of the variables in any period is related to the previous period's gap from long-run equilibrium. If the spot rate st+1 turns out to equal the forward rate ft , (1.7) and (1.8) state that the spot rate and forward rates are expected to remain unchanged. If there is a positive gap between the spot and forward rates so that st+1 − ft > 0, (1.7) and (1.8) lead to the prediction that the spot rate will fall and the forward rate will rise. 4. Nonlinear Dynamics: All of the equations considered thus far are linear (in the sence that each variable is raised to the first power) with constant coefficients. Chapter 7 considers the estimation of models that allow for more complicated dynamic structures. Recall that (1.3) assumes investment is always a constant proportion of the change in consumption. It might be more realistic to assume investment responds more to positive than to negative changes in Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 7 DIFFERENCE EQUATIONS AND THEIR SOLUTIONS 7 consumption. After all, firms might want to take advantage of positive consumption growth but simply let the capital stock decay in response to declines in consumption. Such behavior can be captured by modifying (1.3) such that the coefficient on (ct − ct−1 ) is not constant. Consider the specification it = 𝛽1 (ct − ct−1 ) − 𝜆t 𝛽2 (ct − ct−1 ) + 𝜀it where 𝛽1 > 𝛽2 > 0 and 𝜆t is an indicator function such that 𝜆t = 1 if (ct − ct−1 ) < 0, otherwise 𝜆t = 0. Hence, if (ct − ct−1 ) ≥ 0, 𝜆t = 0 so that it = 𝛽1 (ct − ct−1 ) + 𝜀it and if (ct − ct−1 ) < 0, 𝜆t = 1 so that it = (𝛽1 − 𝛽2 ) (ct − ct−1 ) + 𝜀it . Since 𝛽1 − 𝛽2 > 0, investment is more responsive to positive than negative changes in consumption. 2. DIFFERENCE EQUATIONS AND THEIR SOLUTIONS Although many of the ideas in the previous section were probably familiar to you, it is necessary to formalize some of the concepts used. In this section, we will examine the type of difference equation used in econometric analysis and make explicit what it means to "solve" such equations. To begin our examination of difference equations, consider the function y = f (t). If we evaluate the function when the independent variable t takes on the specific value t∗ , we get a specific value for the dependent variable called yt∗ . Formally, yt∗ = f (t∗ ). Using this same notation, yt∗ +h represents the value of y when t takes on the specific value t∗ + h. The first difference of y is defined as the value of the function when evaluated at t = t∗ + h minus the value of the function evaluated at t∗ : Δyt∗ +h ≡ f (t∗ + h) − f (t∗ ) ≡ yt∗ +h − yt∗ (1.9) Differential calculus allows the change in the independent variable (i.e., the term h) to approach zero. Since most economic data is collected over discrete periods, however, it is more useful to allow the length of the time period to be greater than zero. Using difference equations, we normalize units so that h represents a unit change in t (i.e., h = 1) and consider the sequence of equally spaced values of the independent variable. Without any loss of generality, we can always drop the asterisk on t∗ . We can then form the first differences: Δyt = f (t) − f (t − 1) ≡ yt − yt−1 Δyt+1 = f (t + 1) − f (t) ≡ yt+1 − yt Δyt+2 = f (t + 2) − f (t + 1) ≡ yt+2 − yt+1 Often it will be convenient to express the entire sequence of values {… yt−2 , yt−1 , yt , yt+1 , yt+2 , …} as {yt }. We can then refer to any particular value in the sequence as yt . Unless specified, the index t runs from –∞ to +∞. In time-series econometric models, we use t to represent "time" and h to represent the length of a time period. Thus, yt and yt+1 might represent the realizations of the {yt } sequence in the first and second quarters of 2014, respectively. Trim Size: 6in x 9in 8 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 8 DIFFERENCE EQUATIONS In the same way we can form the second difference as the change in the first difference. Consider Δ2 yt ≡ Δ(Δyt ) = Δ(yt − yt−1 ) = (yt − yt−1 ) − (yt−1 − yt−2 ) = yt − 2yt−1 + yt−2 Δ2 yt+1 ≡ Δ(Δyt+1 ) = Δ(yt+1 − yt ) = (yt+1 − yt ) − (yt − yt−1 ) = yt+1 − 2yt + yt−1 The nth difference (Δn ) is defined analogously. At this point, we risk taking the theory of difference equations too far. As you will see, the need to use second differences rarely arises in time-series analysis. It is safe to say that third- and higher order differences are never used in applied work. Since most of this text considers linear time-series methods, it is possible to examine only the special case of an nth-order linear difference equation with constant coefficients. The form for this special type of difference equation is given by yt = a0 + n ∑ ai yt−i + xt (1.10) i=1 The order of the difference equation is given by the value of n. The equation is linear because all values of the dependent variable are raised to the first power. Economic theory may dictate instances in which the various ai are functions of variables within the economy. However, as long as they do not depend on any of the values of yt or xt , we can regard them as parameters. The term xt is called the forcing process. The form of the forcing process can be very general; xt can be any function of time, current and lagged values of other variables, and/or stochastic disturbances. From an appropriate choice of the forcing process, we can obtain a wide variety of important macroeconomic models. Re-examine equation (1.5), the reduced-form equation for real GDP. This equation is a second-order difference equation since yt depends on yt−2 . The forcing process is the expression (1 + 𝛽)𝜀ct + 𝜀it − 𝛽𝜀ct−1 . You will note that (1.5) has no intercept term corresponding to the expression a0 in (1.10). An important special case for the {xt } sequence is xt = ∞ ∑ 𝛽i 𝜀t−i i=0 where the 𝛽i are constants (some of which can equal zero) and the individual elements of the sequence {𝜀t } are not functions of the yt . At this point it is useful to allow the {𝜀t } sequence to be nothing more than a sequence of unspecified exogenous shocks. For example, let {𝜀t } be a random error term and set 𝛽0 = 1 and 𝛽1 = 𝛽2 = · · · = 0; in this case, (1.10) becomes the autoregression equation yt = a0 + a1 yt – 1 + a2 yt – 2 + · · · + an yt−n + 𝜀t Let n = 1, a0 = 0, and a1 = 1 to obtain the random walk model. Notice that equation (1.10) can be written in terms of the difference operator (Δ). Subtracting yt−1 from (1.10), we obtain yt − yt−1 = a0 + (a1 − 1) yt−1 + n ∑ i=2 ai yt−i + xt Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 9 DIFFERENCE EQUATIONS AND THEIR SOLUTIONS 9 or defining 𝛾 = (a1 − 1), we get Δyt = a0 + 𝛾yt−1 + n ∑ ai yt−i + xt (1.11) i=2 Clearly, equation (1.11) is simply a modified version of (1.10). A solution to a difference equation expresses the value of yt as a function of the elements of the {xt } sequence and t (and possibly some given values of the {yt } sequence called initial conditions). Examining (1.11) makes it clear that there is a strong analogy to integral calculus, where the problem is to find a primitive function from a given derivative. We seek to find the primitive function f (t), given an equation expressed in the form of (1.10) or (1.11). Notice that a solution is a function rather than a number. The key property of a solution is that it satisfies the difference equation for all permissible values of t and {xt }. Thus, the substitution of a solution into the difference equation must result in an identity. For example, consider the simple difference equation Δyt = 2 (or yt = yt−1 + 2). You can easily verify that a solution to this difference equation is yt = 2t + c, where c is any arbitrary constant. By definition, if 2t + c is a solution, it must hold for all permissible values of t. Thus, for period t − 1, yt−1 = 2(t − 1) + c. Now substitute the solution into the difference equation to form 2t + c ≡ 2(t − 1) + c + 2 (1.12) It is straightforward to carry out the algebra and verify that (1.12) is an identity. This simple example also illustrates that the solution to a difference equation need not be unique; there is a solution for any arbitrary value of c. Another useful example is provided by the irregular term shown in Figure 1.1; recall that the equation for this expression is: It = 0.7It−1 + 𝜀t . You can verify that the solution to this first-order equation is It = ∞ ∑ (0.7)i 𝜀t−i (1.13) i=0 Since (1.13) holds for all time periods, the value of the irregular component in t − 1 is given by ∞ ∑ (0.7)i 𝜀t−1−i (1.14) It−1 = i=0 Now substitute (1.13) and (1.14) into It = 0.7It−1 + 𝜀t to obtain 𝜀t + 0.7𝜀t−1 + (0.7)2 𝜀t−2 + (0.7)3 𝜀t−3 + · · · = 0.7[𝜀t−1 + 0.7𝜀t−2 + (0.7)2 𝜀t−3 + (0.7)3 𝜀t−4 + · · ·] + 𝜀t (1.15) The two sides of (1.15) are identical; this proves that (1.13) is a solution to the first-order stochastic difference equation It = 0.7It−1 + 𝜀t . Be aware of the distinction between reduced-form equations and solutions. Since It = 0.7It−1 + 𝜀t holds for all values of t, it follows that It−1 = 0.7It−2 + 𝜀t−1 . Combining these two equations yields It = 0.7[0.7It−2 + 𝜀t−1 ] + 𝜀t = 0.49It−2 + 0.7𝜀t−1 + 𝜀t (1.16) Trim Size: 6in x 9in 10 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 10 DIFFERENCE EQUATIONS Equation (1.16) is a reduced-form equation since it expresses It in terms of its own lags and disturbance terms. However, (1.16) does not qualify as a solution because it contains the "unknown" value of It−2 . To qualify as a solution, (1.16) must express It in terms of the elements xt , t, and any given initial conditions. 3. SOLUTION BY ITERATION The solution given by (1.15) was simply postulated. The remaining portions of this chapter develop the methods you can use to obtain such solutions. Each method has its own merits; knowing the most appropriate to use in a particular circumstance is a skill that comes only with practice. This section develops the method of iteration. Although iteration is the most cumbersome and time-intensive method, most people find it to be very intuitive. If the value of y in some specific period is known, a direct method of solution is to iterate forward from that period to obtain the subsequent time path of the entire y sequence. Refer to this known value of y as the initial condition or the value of y in time period 0 (denoted by y0 ). It is easiest to illustrate the iterative technique using the first-order difference equation yt = a0 + a1 yt−1 + 𝜀t (1.17) Given the value of y0 , it follows that y1 will be given by y1 = a0 + a1 y0 + 𝜀1 In the same way, y2 must be y2 = a0 + a1 y1 + 𝜀2 = a0 + a1 [a0 + a1 y0 + 𝜀1 ] + 𝜀2 = a0 + a0 a1 + (a1 )2 y0 + a1 𝜀1 + 𝜀2 Continuing the process in order to find y3 , we obtain y3 = a0 + a1 y2 + 𝜀3 = a0 [1 + a1 + (a1 )2 ] + (a1 )3 y0 + a1 2 𝜀1 + a1 𝜀2 + 𝜀3 You can easily verify that for all t > 0, repeated iteration yields yt = a0 t−1 ∑ i=0 aii + at1 y0 + t−1 ∑ aii 𝜀t−i (1.18) i=0 Equation (1.18) is a solution to (1.17) since it expresses yt as a function of t, the forcing process xt = Σ(a1 )i 𝜀t−i , and the known value of y0 . As an exercise, it is useful to show that iteration from yt back to y0 yields exactly the formula given by (1.18). Since yt = a0 + a1 yt−1 + 𝜀t , it follows that yt = a0 + a1 [a0 + a1 yt−2 + 𝜀t−1 ] + 𝜀t = a0 (1 + a1 ) + a1 𝜀t−1 + 𝜀t + a1 2 [a0 + a1 yt−3 + 𝜀t−2 ] Continuing the iteration back to period 0 yields equation (1.18). Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 11 SOLUTION BY ITERATION 11 Iteration without an Initial Condition Suppose you were not given the initial condition for y0 . The solution given by (1.18) would no longer be appropriate because the value of y0 is an unknown. You would not be able to select this initial value of y and iterate forward, nor would you be able to iterate backward from yt and simply choose to stop at t = t0 . Thus, suppose we continued to iterate backward by substituting a0 + a1 y−1 + 𝜀0 for y0 in (1.18): yt = a0 t−1 ∑ ai1 + at1 (a0 + a1 y−1 + 𝜀0 ) + i=0 = a0 t−1 ∑ ai1 𝜀t−i i=0 t ∑ ai1 + i=0 t ∑ ai1 𝜀t−i + at+1 y−1 1 (1.19) i=0 Continuing to iterate backward another m periods, we obtain yt = a0 t+m ∑ i=0 ai1 + t+m ∑ ai1 𝜀t−i + at+m+1 y−m−1 1 (1.20) i=0 Now examine the pattern emerging from (1.19) and (1.20). If |a1 | < 1, the term a1 t+m+1 approaches zero as m approaches infinity. Also, the infinite sum [1 + a1 + (a1 )2 + · · ·] converges to 1∕(1 − a1 ). Thus, if we temporarily assume that |a1 | < 1, after continual substitution, (1.20) can be written as yt = a0 ∕(1 − a1 ) + ∞ ∑ ai1 𝜀t−i (1.21) i=0 You should take a few minutes to convince yourself that (1.21) is a solution to the original difference equation (1.17); substitution of (1.21) into (1.17) yields an identity. However, (1.21) is not a unique solution. For any arbitrary value of A, a solution to (1.17) is given by ∞ ∑ ai1 𝜀t−i (1.22) yt = Aat1 + a0 ∕(1 − a1 ) + i=0 To verify that (1.22) is a solution for any arbitrary value of A, substitute (1.22) into (1.17) to obtain yt = Aat1 + a0 ∕(1 − a1 ) + [ = a0 + a1 ∞ ∑ ai1 𝜀t−i i=0 ] ∞ ∑ ( ) Aat−1 ai1 𝜀t−1−i + 𝜀t 1 + a0 ∕ 1 − a1 + i=0 Since the two sides are identical, (1.22) is necessarily a solution to (1.17). Reconciling the Two Iterative Methods Given the iterative solution (1.22), suppose that you are now given an initial condition concerning the value of y in the arbitrary period t0 . It is straightforward to show that we can impose the initial condition on (1.22) to yield the same solution as (1.18). Trim Size: 6in x 9in 12 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 12 DIFFERENCE EQUATIONS Since (1.22) must be valid for all periods (including t0 ), when t = 0, it must be true that ∞ ∑ ai1 𝜀−i (1.23) y0 = A + a0 ∕(1 − a1 ) + i=0 so that A = y0 − a0 ∕(1 − a1 ) − ∞ ∑ ai1 𝜀−i i=0 Since y0 is given, we can view (1.23) as the value of A that renders (1.22) as a solution to (1.17), given the initial condition. Hence, the presence of the initial condition eliminates the arbitrariness of A. Substituting this value of A into (1.22) yields ] [ ∞ ∞ ∑ ) ∑ ( i yt = y0 − a0 ∕ 1 − a1 − a1 𝜀−i at1 + a0 ∕(1 − a1 ) + ai1 𝜀t−i (1.24) i=0 i=0 Simplification of (1.24) results in yt = [y0 − a0 ∕(1 − a1 )]at1 + a0 ∕(1 − a1 ) + t−1 ∑ ai1 𝜀t−i (1.25) i=0 It is a worthwhile exercise to verify that (1.25) is identical to (1.18). Nonconvergent Sequences Given that |a1 | < 1, (1.21) is the limiting value of (1.20) as m grows infinitely large. What happens to the solution in other circumstances? If |a1 | > 1, it is not possible to move from (1.20) to (1.21) because the expression |a1 |t+m grows infinitely large as t + m approaches ∞.1 However, if there is an initial condition, there is no need to obtain the infinite summation. Simply select the initial condition y0 and iterate forward; the result will be (1.18): t−1 t−1 ∑ ∑ yt = a0 ai1 + at1 y0 + ai1 𝜀t−i i=0 i=0 Although the successive values of the {yt } sequence will become progressively larger in absolute value, all values in the series will be finite. A very interesting case arises if a1 = 1. Rewrite (1.17) as yt = a0 + yt−1 + 𝜀t or Δyt = a0 + 𝜀t As you should verify by iterating from yt back to y0 , a solution to this equation is2 yt = a0 t + t ∑ 𝜀i + y0 (1.26) i=1 After a moment's reflection, the form of the solution is quite intuitive. In every period t, the value of yt changes by a0 + 𝜀t units. After t periods, there are t such changes; hence, the total change is ta0 plus the t values of the {𝜀t } sequence. Notice that the solution contains summation of all disturbances from 𝜀1 through 𝜀t . Thus, Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 13 SOLUTION BY ITERATION 13 when a1 = 1, each disturbance has a permanent non-decaying effect on the value of yt . You should compare this result to the solution found in (1.21). For the case in which |a1 | < 1, |a1 |t is a decreasing function of t so that the effects of past disturbances become successively smaller over time. The importance of the magnitude of a1 is illustrated in Figure 1.2. Thirty random numbers with a theoretical mean equal to zero were computer-generated and denoted by 𝜀1 through 𝜀30 . Then the value of y0 was set equal to unity and the next 30 values of the {yt } sequence were constructed using the formula yt = 0.9yt−1 + 𝜀t . The result is shown by the thin line in Panel (a) of Figure 1.2. If you substitute a0 = 0 and a1 = 0.9 y t = 0.9y t–1 + εt 1.0 0.8 0.75 0.6 0.4 0.50 0.2 0.25 0.0 0.00 –0.2 –0.4 y t = 0.5y t–1 + εt 1.00 0 5 10 15 20 Panel (a) 25 30 y t = –0.5y t–1 + εt 1.00 –0.25 0.9 0.50 0.8 0.25 0.7 0.00 0.6 –0.25 0.5 –0.50 0.4 0 5 10 15 20 5 10 25 30 0.3 0 5 10 Panel (c) 200 175 150 125 100 75 50 25 0 5 10 15 20 25 30 15 20 25 30 25 30 Panel (d) y t = 1.2y t–1 + εt 0 15 20 Panel (b) y t = y t–1 + εt 1.0 0.75 –0.75 0 25 30 200 150 100 50 0 –50 –100 –150 –200 –250 y t = –1.2y t–1 + εt 0 5 10 Panel (e) F I G U R E 1.2 Convergent and Nonconvergent Sequences 15 20 Panel (f ) Trim Size: 6in x 9in 14 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 14 DIFFERENCE EQUATIONS into (1.18), you will see that the time path of {yt } consists of two parts. The first part, 0.9t , is shown by the slowly decaying thick line in the panel. This term dominates the solution for relatively small values of t. The influence of the random part is shown by the difference between the thin and the thick line; you can see that the first several values of {𝜀t } are negative. Notice that as t increases, the influence of the initial value y0 = 1 becomes less pronounced. Using the previously drawn random numbers, we again set y0 equal to unity and a second sequence was constructed using the formula yt = 0.5yt−1 + 𝜀t . This second sequence is shown by the thin line in Panel (b) of Figure 1.2. The influence of the expression 0.5t is shown by the rapidly decaying thick line. Again, as t increases, the random portion of the solution becomes more dominant in the time path of {yt }. When we compare the first two panels, it is clear that reducing the magnitude of |a1 | increases the rate of convergence. Moreover, the discrepancies between the simulated values of yt and the thick line are less pronounced in the second panel. As you can see in (1.18), each value of 𝜀t−i enters the solution for yt with a coefficient of (a1 )i . The smaller value of a1 means that the past realizations of 𝜀t−i have a smaller influence on the current value of yt . Simulating a third sequence with a1 = −0.5, yields the thin line shown in Panel (c). The oscillations are due to the negative value of a1 . The expression (−0.5)t , shown by the thick line, is positive when t is even and negative when t is odd. Since |a1 | < 1, the oscillations are dampened. The next three panels in Figure 1.2 all show nonconvergent sequences. Each uses the initial condition y0 = 1 and the same 30 values of {𝜀t } used in the other simulations. The line in Panel (d) shows the time path of yt = yt−1 + 𝜀t . Since each value of 𝜀t has an expected value of zero, Panel (d) illustrates a random walk process. Here Δyt = 𝜀t so that the change in yt is purely random. The nonconvergence is shown by the tendency of {yt } to meander. In Panel (e), the thick line representing the explosive expression (1.2)t dominates the random portion of the {yt } sequence. Also notice that the discrepancy between the simulated {yt } sequence and the thick line widens as t increases. The reason is that past values of 𝜀t−i enter the solution for yt with the coefficient (1.2)i . As i increases, the importance of these previous discrepancies becomes increasingly important. Similarly, setting a1 = −1.2 results in the exploding oscillations shown in the lower-right panel of the figure. The value (−1.2)t is positive for even values of t and negative for odd values of t. 4. AN ALTERNATIVE SOLUTION METHODOLOGY Solution by the iterative method breaks down in higher order equations. The algebraic complexity quickly overwhelms any reasonable attempt to find a solution. Fortunately, there are several alternative solution techniques that can be helpful in solving the nth-order equation given by (1.10). If we use the principle that you should learn to walk before you learn to run, it is best to step through the first-order equation given by (1.17). Although you will be covering some familiar ground, the first-order case illustrates the general methodology extremely well. To split the procedure into its component parts, Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 15 AN ALTERNATIVE SOLUTION METHODOLOGY 15 consider only the homogeneous portion of (1.17)3 yt = a1 yt−1 (1.27) The solution to this homogeneous equation is called the homogeneous solution; at times it will be useful to denote the homogeneous solution by the expression yth . Obviously, the trivial solution yt = yt−1 = · · · = 0 satisfies (1.27). However, this solution is not unique. By setting a0 and all values of {𝜀t } equal to zero, (1.18) becomes yt = at1 y0 . Hence, yt = at1 y0 must be a solution to (1.27). Yet, even this solution does not constitute the full set of solutions. It is easy to verify that the expression at1 multiplied by any arbitrary constant A satisfies (1.27). Simply substitute yt = Aat1 and yt−1 = Aa1t−1 into (1.27) to obtain Aat1 = a1 Aat−1 1 Since at1 = a1 at−1 , it follows that yt = Aat1 also solves (1.27). With the aid of the 1 thick lines in Figure 1.2, we can classify the properties of the homogeneous solution as follows: If |a1 | < 1, the expression at1 converges to zero as t approaches infinity. Convergence is direct if 0 < a1 < 1 and oscillatory if −1 < a1 < 0. 2. If |a1 | > 1, the homogeneous solution is not stable. If a1 > 1, the homogeneous solution approaches ∞ as t increases. If a1 < −1, the homogeneous solution oscillates explosively. 3. If a1 = 1, any arbitrary constant A satisfies the homogeneous equation yt = yt−1 . If a1 = −1, the system is meta-stable: at1 = 1 for even values of t and −1 for odd values of t. 1. Now consider (1.17) in its entirety. In the last section, you confirmed that (1.21) is a valid solution to (1.17). Equation (1.21) is called a particular solution to the differp ence equation; all such particular solutions will be denoted by the term yt . The term "particular" stems from the fact that a solution to a difference equation may not be unique; hence, (1.21) is just one particular solution out of the many possibilities. In moving to (1.22) you verified that the particular solution was not unique. The homogeneous solution Aat1 plus the particular solution given by (1.21) constituted the complete solution to (1.17). The general solution to a difference equation is defined to be a particular solution plus all homogeneous solutions. Once the general solution is obtained, the arbitrary constant A can be eliminated by imposing an initial condition for y0 . The Solution Methodology The results of the first-order case are directly applicable to the nth-order equation given by (1.10). In this general case, it will be more difficult to find the particular solution and there will be n distinct homogeneous solutions. Nevertheless, the solution methodology will always entail the following four steps: STEP 1: form the homogeneous equation and find all n homogeneous solutions; STEP 2: find a particular solution; Trim Size: 6in x 9in 16 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 16 DIFFERENCE EQUATIONS STEP 3: obtain the general solution as the sum of the particular solution and a linear combination of all homogeneous solutions; STEP 4: eliminate the arbitrary constant(s) by imposing the initial condition(s) on the general solution. Before we address the various techniques that can be used to obtain homogeneous and particular solutions, it is worthwhile to illustrate the methodology using the equation (1.28) yt = 0.9yt−1 − 0.2yt−2 + 3 Clearly, this second-order equation is in the form of (1.10) with a0 = 3, a1 = 0.9, a2 = −0.2, and xt = 0. Beginning with the first of the four steps, form the homogenous equation (1.29) yt − 0.9yt−1 + 0.2yt−2 = 0 In the first-order case of (1.17), the homogeneous solution was Aat1 . Section 1.6 will show you how to find the complete set of homogeneous solutions. For now, it is sufficient to assert that the two homogeneous solutions are yh1t = (0.5)t and yh2t = (0.4)t . To verify the first solution, note that yh1t−1 = (0.5)t−1 and yh1t−2 = (0.5)t−2 . Thus, yh1t is a solution if it satisfies (0.5)t − 0.9(0.5)t−1 + 0.2(0.5)t−2 = 0 If we divide by (0.5)t−2 , the issue is whether (0.5)2 − 0.9(0.5) + 0.2 = 0 Carrying out the algebra, 0.25 − 0.45 + 0.2 does equal zero so that (0.5)t is a solution to (1.29). In the same way, it is easy to verify that yh2t = (0.4)t is a solution since (0.4)t − 0.9(0.4)t−1 + 0.2(0.4)t−2 = 0 Divide by (0.4)t−2 to obtain (0.4)2 − 0.9(0.4) + 0.2 = 0.16 − 0.36 + 0.2 = 0. The second step is to obtain a particular solution; you can easily confirm that the p particular solution yt = 10 solves (1.28) as 10 = 0.9(10) − 0.2(10) + 3. The third step is to combine the particular solution and a linear combination of both homogeneous solutions to obtain yt = A1 (0.5)t + A2 (0.4)t + 10 where A1 and A2 are arbitrary constants. For the fourth step, assume you have two initial conditions for the {yt } sequence. So that we can keep our numbers reasonably round, suppose that y0 = 13 and y1 = 11.3. Thus, for periods zero and one, our solution must satisfy 13 = A1 + A2 + 10 11.3 = A1 (0.5) + A2 (0.4) + 10 Solving simultaneously for A1 and A2 , you should find A1 = 1 and A2 = 2. Hence, the solution is yt = (0.5)t + 2(0.4)t + 10 Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 17 AN ALTERNATIVE SOLUTION METHODOLOGY 17 You can substitute yt = (0.5)t + 2(0.4)t + 10 into (1.28) to verify that the solution is correct. Generalizing the Method To show that this method is applicable to higher order equations, consider the homogeneous part of (1.10): n ∑ ai yt−i (1.30) yt = i=1 As shown in Section 1.6, there are n homogeneous solutions that satisfy (1.30). For now, it is sufficient to demonstrate the following proposition: If yth is a homogeneous solution to (1.30), Ayth is also a solution for any arbitrary constant A. By assumption, yth solves the homogeneous equation so that yth = n ∑ ai yht−i (1.31) ai Ayht−i (1.32) i=1 The expression Ayth is also a solution if Ayth = n ∑ i=1 We know (1.32) is satisfied because dividing each term by A yields (1.31). Now suppose that there are two separate solutions to the homogeneous equation denoted by yh1t and yh2t . It is straightforward to show that for any two constants A1 and A2 , the linear combination A1 yh1t + A2 yh2t is also a solution to the homogeneous equation. If A1 yh1t + A2 yh2t is a solution to (1.30), it must satisfy A1 yh1t + A2 yh2t = a1 [A1 yh1t−1 + A2 yh2t−1 ] + a2 [A1 yh1t−2 + A2 yh2t−2 ] + · · · + an [A1 yh1t−n + A2 yh2t−n ] Regrouping terms, we want to know if [ ] [ ] n n ∑ ∑ h h h h A1 y1t − A1 ai y1t−i + A2 y2t − A2 ai y2t−i = 0 i=1 A1 yh1t i=1 A2 yh2t Since and are separate solutions to (1.30), each of the expressions in brackets is zero. Hence, the linear combination is necessarily a solution to the homogeneous equation. This result easily generalizes to all n homogeneous solutions of an nth-order equation. Finally, the use of Step 3 is appropriate since the sum of any particular solution and any linear combination of all homogeneous solutions is also a solution. To prove this proposition, substitute the sum of the particular and homogeneous solutions into (1.10) to obtain n ∑ p p ai (yt−i + yht−i ) + xt (1.33) yt + yth = a0 + i=1 Trim Size: 6in x 9in 18 Enders c01.tex V3 - 09/02/2014 12:52pm Page 18 CHAPTER 1 DIFFERENCE EQUATIONS Recombining the terms in (1.33), we want to know if ] [ ] [ n n ∑ ∑ p p h h yt − a0 − ai yt−i − xt + yt − ai yt−i = 0 i=1 (1.34) i=1 p Since yt solves (1.10), the expression in the first bracket of (1.34) is zero. Since yth solves the homogeneous equation, the expression in the second bracket is zero. Thus, (1.34) is an identity; the sum of the homogeneous and particular solutions solves (1.10). 5. THE COBWEB MODEL An interesting way to illustrate the methodology outlined in the previous section is to consider a stochastic version of the traditional cobweb model. Since the model was originally developed to explain the volatility in agricultural prices, let the market for a product—say, wheat—be represented by dt = a − 𝛾pt st = b + 𝛽p∗t + 𝜀t st = dt 𝛾>0 𝛽>0 (1.35) (1.36) (1.37) dt = demand for wheat in period t st = supply of wheat in t pt = market price of wheat in t p∗t = price that farmers expect to prevail at t 𝜀t = a zero mean stochastic supply shock and parameters a, b, 𝛾, and 𝛽 are all positive such that a > b.4 The nature of the model is such that consumers buy as much wheat as is desired at the market clearing price pt . At planting time, farmers do not know the price prevailing at harvest time; they base their supply decision on the expected price (p∗t ). The actual quantity produced depends on the planned quantity b + 𝛽p∗t plus a random supply shock 𝜀t . Once the product is harvested, market equilibrium requires that the quantity supplied equals the quantity demanded. Unlike the actual market for wheat, the model does not allow for the possibility of storage. The essence of the cobweb model is that farmers form their expectations in a naive fashion; let farmers use last year's price as the expected market price (1.38) p∗t = pt−1 where: Point E in Figure 1.3 represents the long-run equilibrium price and quantity combination. Note that the equilibrium concept in this stochastic model differs from that of the traditional cobweb model. If the system is stable, successive prices will tend to converge to point E. However, the nature of the stochastic equilibrium is such that the ever-present supply shocks prevent the system from remaining at E. Nevertheless, it is useful to solve for the long-run price. If we set all values of the {𝜀t } sequence equal to zero, set pt = pt−1 = · · · = p, and equate supply and demand, the long-run equilibrium Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 19 THE COBWEB MODEL 19 Price s 1 pt 2 E 3 pt+1 4 d 0 st st+1 Quantity F I G U R E 1.3 The Cobweb Model price is given by p = (a − b)∕(𝛾 + 𝛽). Similarly, the equilibrium quantity (s) is given by s = (a𝛽 + 𝛾b)∕(𝛾 + 𝛽). To understand the dynamics of the system, suppose that farmers in t plan to produce the equilibrium quantity s. However, let there be a negative supply shock such that the actual quantity produced turns out to be st . As shown by point 1 in Figure 1.3, consumers are willing to pay pt for the quantity st ; hence, market equilibrium in t occurs at point 1. Updating one period allows us to see the main result of the cobweb model. For simplicity, assume that all subsequent values of the supply shock are zero (i.e., 𝜀t+1 = 𝜀t+2 = · · · = 0). At the beginning of period t + 1, farmers expect the price at harvest time to be the price of the previous period; thus, p∗t+1 = pt . Accordingly, they produce quantity st+1 (see point 2 in the figure); consumers, however, are willing to buy quantity st+1 only if the price falls to that indicated by pt+1 (see point 3 in the figure). The next period begins with farmers expecting to be at point 4. The process continually repeats until the equilibrium point E is attained. As drawn, Figure 1.3 suggests that the market will always converge to the long-run equilibrium point. This result does not hold for all demand and supply curves. To formally derive the stability condition, combine (1.35) through (1.38) to obtain b + 𝛽pt−1 + 𝜀t = a − 𝛾pt or pt = (−𝛽∕𝛾)pt−1 + (a − b)∕𝛾 − 𝜀t ∕𝛾 (1.39) Clearly, (1.39) is a stochastic first-order linear difference equation with constant coefficients. To obtain the general solution, proceed using the four steps listed at the end of the last section: 1. Form the homogeneous equation pt = (−𝛽∕𝛾)pt−1 . In the next section you will learn how to find the solution(s) to a homogeneous equation. For now, it Trim Size: 6in x 9in 20 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 20 DIFFERENCE EQUATIONS is sufficient to verify that the homogeneous solution is pht = A(−𝛽∕𝛾)t 2. where A is an arbitrary constant. Note that (1.39) is a first-order difference equation in the form pt = a0 + a1 pt−1 + et where a0 = (a − b)∕𝛾, a1 = −(𝛽∕𝛾), and et = −𝜀t ∕𝛾. If the ratio 𝛽∕𝛾 is less than unity, you can iterate (1.39) backward from pt to verify that the particular solution for the price is p ∞ pt = 3. a − b 1∑ (−𝛽∕𝛾)i 𝜀t−i − 𝛾 + 𝛽 𝛾 i=0 (1.40) If 𝛽∕𝛾 ≥ 1, the infinite summation in (1.40) is not convergent. As discussed in the last section, it is necessary to impose an initial condition on (1.40) if 𝛽∕𝛾 ≥ 1. The general solution is the sum of the homogeneous and particular solutions; combining these two solutions, the general solution is ∞ pt = 4. a − b 1∑ (−𝛽∕𝛾)i 𝜀t−i + A(−𝛽∕𝛾)t − 𝛾 + 𝛽 𝛾 i=0 (1.41) In (1.41), A is an arbitrary constant that can be eliminated if we know the price in some initial period. For convenience, let this initial period have a time subscript of zero. Since the solution must hold for every period, including period zero, it must be the case that ∞ a − b 1∑ (−𝛽∕𝛾)i 𝜀−i + A(−𝛽∕𝛾)0 − p0 = 𝛾 + 𝛽 𝛾 i=0 Since (−𝛽∕𝛾)0 = 1, the value of A is given by ∞ A = p0 − a − b 1∑ (−𝛽∕𝛾)i 𝜀−i + 𝛾 + 𝛽 𝛾 i=0 Substituting this solution for A back into (1.41) yields ] [ ]t [ ∞ ∞ 𝛽 a − b 1∑ a − b 1∑ i i p0 − (−𝛽∕𝛾) 𝜀t−i + − − + pt = (−𝛽∕𝛾) 𝜀−i 𝛾 + 𝛽 𝛾 i=0 𝛾 𝛾 + 𝛽 𝛾 i=0 and, after simplifying the two summations, pt = [ ]t [ ] t−1 −𝛽 a − b 1∑ a−b (−𝛽∕𝛾)i 𝜀t−i + p0 − − 𝛾 + 𝛽 𝛾 i=0 𝛾 𝛾 +𝛽 (1.42) We can interpret (1.42) in terms of Figure 1.3. In order to focus on the stability of the system, temporarily assume that all values of the {𝜀t } sequence are zero. Subsequently, we will return to a consideration of the effects of supply shocks. If the system Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 21 THE COBWEB MODEL 21 begins in long-run equilibrium, the initial condition is such that p0 = (a − b)∕(𝛾 + 𝛽). In this case, inspection of equation (1.42) indicates that pt = (a − b)∕(𝛾 + 𝛽). Thus, if we begin the process at point E, the system remains in long-run equilibrium. Instead, suppose that the process begins at a price below long-run equilibrium: p0 < (a − b)∕(𝛾 + 𝛽). Equation (1.42) tells us that p1 is p1 = (a − b)∕(𝛾 + 𝛽) + [p0 − (a − b)∕(𝛾 + 𝛽)] (−𝛽∕𝛾)1 (1.43) Since p0 < (a − b)∕(𝛾 + 𝛽) and −𝛽∕𝛾 < 0, it follows that p1 will be above the long-run equilibrium price (a − b)∕(𝛾 + 𝛽). In period 2, p2 = (a − b)∕(𝛾 + 𝛽) + [p0 − (a − b)∕(𝛾 + 𝛽)] (−𝛽∕𝛾)2 Although p0 < (a − b)∕(𝛾 + 𝛽), (−𝛽∕𝛾)2 is positive; hence, p2 is below the long-run equilibrium. For the subsequent periods, note that (−𝛽∕𝛾)t will be positive for even values of t and negative for odd values of t. Just as we found graphically, the successive values of the {pt } sequence will oscillate above and below the long-run equilibrium price. Since (𝛽∕𝛾)t goes to zero if 𝛽 < 𝛾 and explodes if 𝛽 > 𝛾, the magnitude of 𝛽∕𝛾 determines whether the price actually converges toward the long-run equilibrium. If 𝛽∕𝛾 < 1, the oscillations will diminish in magnitude, and if 𝛽∕𝛾 > 1, the oscillations will be explosive. The economic interpretation of this stability condition is straightforward. The slope of the supply curve (i.e., 𝜕pt ∕𝜕st ) is 1∕𝛽 and the absolute value of the slope of the demand curve [i.e., −𝜕pt ∕𝜕(dt )] is 1∕𝛾. If the supply curve is steeper than the demand curve, it must be the case that 1∕𝛽 > 1∕𝛾 or 𝛽∕𝛾 < 1 so that the system is stable. This is precisely the case illustrated in Figure 1.3. As an exercise, you should draw a diagram with the demand curve steeper than the supply curve and show that the price oscillates and diverges from the long-run equilibrium. Now consider the effects of the supply shocks. The contemporaneous effect of a supply shock on the price of wheat is the partial derivative of pt with respect to 𝜀t ; from (1.42) 𝜕pt 1 =− (1.44) 𝜕𝜀t 𝛾 Equation (1.44) is called the impact multiplier since it shows the impact effect of a change in 𝜀t on the price in t. In terms of Figure 1.3, a negative value of 𝜀t implies a price above the long-run price p; the price in t rises by 1∕𝛾 units for each unit decline in current period supply. Of course, this terminology is not specific to the cobweb model; in terms of the nth-order model given by (1.10), the impact multiplier is the partial derivative of yt with respect to the partial change in the forcing process.5 The effects of the supply shock in t persist into future periods. Updating (1.42) by one period yields the one-period multiplier: 𝜕pt+1 1 = − (−𝛽∕𝛾) 𝜕𝜀t 𝛾 = 𝛽∕𝛾 2 Trim Size: 6in x 9in 22 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 22 DIFFERENCE EQUATIONS Point 3 in Figure 1.3 illustrates how the price in t + 1 is affected by the negative supply shock in t. It is straightforward to derive the result that the effects of the supply shock decay over time. Since 𝛽∕𝛾 < 1, the absolute value of 𝜕pt ∕𝜕𝜀t exceeds 𝜕pt+1 ∕𝜕𝜀t . All of the multipliers can be derived analogously; updating (1.42) by two periods 𝜕pt+2 ∕𝜕𝜀t = −(1∕𝛾)(−𝛽∕𝛾)2 and after n periods: 𝜕pt+n ∕𝜕𝜀t = −(1∕𝛾)(−𝛽∕𝛾)n The time path of all such multipliers is called the impulse response function. This function has many important applications in time-series analysis because it shows how the entire time path of a variable is affected by a stochastic shock. Here, the impulse function traces the effects of a supply shock in the wheat market. In other economic applications, you may be interested in the time path of a money supply shock or a productivity shock on real GDP. In actuality, the function can be derived without updating (1.42) because it is always the case that 𝜕pt+j 𝜕pt = 𝜕𝜀t 𝜕𝜀t−j To find the impulse response function, simply find the partial derivative of (1.42) with respect to the various 𝜀t−j . These partial derivatives are nothing more than the coefficients of the {𝜀t−j } sequence in (1.42). Each of the three components in (1.42) has a direct economic interpretation. The deterministic portion of the particular solution (a − b)∕(𝛾 + 𝛽) is the long-run equilibrium price; if the stability condition is met, the {pt } sequence tends to converge to this long-run value. The stochastic component of the particular solution captures the short-run price adjustments due to the supply shocks. The ultimate decay of the coefficients of the impulse response function guarantees that the effects of changes in the various 𝜀t are of a short-run duration. The third component is the expression (−𝛽∕𝛾)t A = (−𝛽∕𝛾)t [p0 − (a − b)∕(𝛾 + 𝛽)]. The value of A is the initial period's deviation of the price from its long-run equilibrium level. Given that 𝛽∕𝛾 < 1, the importance of this initial deviation diminishes over time. 6. SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS Higher order difference equations arise quite naturally in economic analysis. Equation (1.5)—the reduced-form GDP equation resulting from Samuelson's (1939) model—is an example of a second-order difference equation. Moreover, in time-series econometrics it is quite typical to estimate second- and higher order equations. To begin our examination of homogeneous solutions, consider the second-order equation yt − a1 yt−1 − a2 yt−2 = 0 (1.45) Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 23 SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS 23 Given the findings in the first-order case, you should suspect that the homogeneous solution has the form yth = A𝛼 t . Substitution of this trial solution into (1.45) yields A𝛼 t − a1 A𝛼 t−1 − a2 A𝛼 t−2 = 0 (1.46) Clearly, any arbitrary value of A is satisfactory. If you divide (1.46) by A𝛼 t−2 , the problem is to find the values of 𝛼 that satisfy 𝛼 2 − a1 𝛼 − a2 = 0 (1.47) Solving this quadratic equation—called the characteristic equation—yields two values of 𝛼, called the characteristic roots. Using the quadratic formula, we find that the two characteristic roots are √ a1 ± a21 + 4a2 𝛼1 , 𝛼2 = √2 = (a1 ± d)∕2 (1.48) where d is the discriminant [a21 + 4a2 ]. Each of these two characteristic roots yields a valid solution for (1.45). Again, these solutions are not unique. In fact, for any two arbitrary constants A1 and A2 , the linear combination A1 (𝛼1 )t + A2 (𝛼2 )t also solves (1.45). As proof, simply substitute yt = A1 (𝛼1 )t + A2 (𝛼2 )t into (1.45) to obtain A1 (𝛼1 )t + A2 (𝛼2 )t = a1 [A1 (𝛼1 )t−1 + A2 (𝛼2 )t−1 ] + a2 [A1 (𝛼1 )t−2 + A2 (𝛼2 )t−2 ] Now, regroup terms as follows: A1 [(𝛼1 )t − a1 (𝛼1 )t−1 − a2 (𝛼1 )t−2 ] + A2 [(𝛼2 )t − a1 (𝛼2 )t−1 − a2 (𝛼2 )t−2 ] = 0 Since 𝛼1 and 𝛼2 each solve (1.45), both terms in brackets must equal zero. As such, the complete homogeneous solution in the second-order case is yth = A1 (𝛼1 )t + A2 (𝛼2 )t Without knowing the specific values of a1 and a2 , we cannot find the two characteristic roots 𝛼1 and 𝛼2 . Nevertheless, it is possible to characterize the nature of the solution; three possible cases are dependent on the value of the discriminant d. CASE 1 If a1 2 + 4a2 > 0, d is a real number and there will be two distinct real characteristic roots. Hence, there are two separate solutions to the homogeneous equation denoted by (𝛼1 )t and (𝛼2 )t . We already know that any linear combination of the two is also a solution. Hence, yth = A1 (𝛼1 )t + A2 (𝛼2 )t Trim Size: 6in x 9in 24 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 24 DIFFERENCE EQUATIONS WORKSHEET 1.1 SECOND-ORDER EQUATIONS Example 1: yt = 0.2yt−1 + 0.35yt−2 . Hence: a1 = 0.2 and a2 = 0.35 Form the homogeneous equation: yt − 0.2yt−1 − 0.35yt−2 = 0 A check of the discriminant reveals: d = a21 + 4a2 so that d = 1.44. Given that d > 0, the roots will be real and distinct. Let the trial solution have the form: yt = 𝛼 t . Substitute the trial solution into the homogenous equation to obtain: 𝛼 t − 0.2𝛼 t−1 − 0.35 𝛼 t−2 = 0 Divide by 𝛼 t−2 to obtain the characteristic equation: 𝛼 2 − 0.2𝛼 − 0.35 = 0 Compute the two characteristic roots: 𝛼1 = 0.5(a1 + d1∕2 ) 𝛼2 = 0.5(a1 − d1∕2 ) 𝛼1 = 0.7 𝛼2 = −0.5 The homogeneous solution is: A1 (0.7)t + A2 (−0.5)t . The first graph shows the time path of this solution for the case in which the arbitrary constants equal unity and t runs from 1 to 20. Example 2: yt = 0.7yt−1 + 0.35yt−2 . Hence: a1 = 0.7 and a2 = 0.35 Form the homogeneous equation: yt − 0.7yt−1 − 0.35yt−2 = 0 A check of the discriminant reveals: d = a21 + 4a2 so that d = 1.89. Given that d > 0, the roots will be real and distinct. Form the characteristic equation 𝛼 t − 0.7𝛼 t−1 − 0.35𝛼 t−2 = 0 Divide by 𝛼 t−2 to obtain the characteristic equation: 𝛼 2 − 0.7𝛼 − 0.35 = 0 Compute the two characteristic roots: 𝛼1 = 0.5(a1 + d1∕2 ) 𝛼2 = 0.5(a1 − d1∕2 ). 𝛼1 = 1.037 𝛼2 = −0.337 The homogeneous solution is: A1 (1.037)t + A2 (−0.337)t . The second graph shows the time path of this solution for the case in which the arbitrary constants equal unity and t runs from 1 to 20. Example 1 1 1.5 0.5 0 Example 2 2.5 0 10 20 0.5 0 10 20 Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 25 SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS 25 It should be clear that if the absolute value of either 𝛼1 or 𝛼2 exceeds unity, the homogeneous solution will explode. Worksheet 1.1 examines two second-order equations showing real and distinct characteristic roots. In the first example, yt = 0.2yt−1 + 0.35yt−2 , the characteristic roots are shown to be 𝛼1 = 0.7 and 𝛼2 = −0.5. Hence, the full homogeneous solution is yt h = A1 (0.7)t + A2 (−0.5)t . Since both roots are less than unity in absolute value, the homogeneous solution is convergent. As you can see in the graph on the bottom left-hand side of Worksheet 1.1, convergence is not monotonic because of the influence of the expression (−0.5)t . In the second example, yt = 0.7yt−1 + 0.35yt−2 . The worksheet indicates how to obtain the solution for the two characteristic roots. Given that one characteristic root is 1.037, the {yt } sequence explodes. The negative root (𝛼2 = −0.337) is responsible for the nonmonotonicity of the time path. Since (−0.337)t quickly approaches zero, the dominant root is the explosive value 1.037. CASE 2 If a21 + 4a2 = 0, it follows that d = 0 and 𝛼1 = 𝛼2 = a1 ∕2. Hence, a homogeneous solution is a1 ∕2. However, when d = 0, there is a second homogeneous solution given by t(a1 ∕2)t . To demonstrate that yth = t(a1 ∕2)t is a homogeneous solution, substitute it into (1.45) to determine whether t(a1 ∕2)t − a1 [(t − 1)(a1 ∕2)t−1 ] − a2 [(t − 2)(a1 ∕2)t−2 ] = 0 Divide by (a1 ∕2)t−2 and form −[(a21 ∕4) + a2 ]t + [(a21 ∕2) + 2a2 ] = 0 Since we are operating in the circumstance where a21 + 4a2 = 0, each bracketed expression is zero; hence, t(a1 ∕2)t solves (1.45). Again, for arbitrary constants A1 and A2 , the complete homogeneous solution is yth = A1 (a1 ∕2)t + A2 t(a1 ∕2)t Clearly, the system is explosive if |a1 | > 2. If |a1 | < 2, the term A1 (a1 ∕2)t converges, but you might think that the effect of the term t(a1 ∕2)t is ambiguous [since the diminishing (a1 ∕2)t is multiplied by t]. This ambiguity is correct in the limited sense that the behavior of the homogeneous solution is not monotonic. As illustrated in Figure 1.4 for a1 ∕2 = 0.95, 0.9, and –0.9, as long as |a1 | < 2, lim[t(a1 ∕2)t ] is necessarily zero as t → ∞; thus, there is always convergence. For 0 < a1 < 2, the homogeneous solution appears to explode before ultimately converging to zero. For −2 < a1 < 0, the behavior is wildly erratic; the homogeneous solution appears to oscillate explosively before the oscillations dampen and finally converge to zero. Trim Size: 6in x 9in 26 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 26 DIFFERENCE EQUATIONS 8 6 t·(0.95t ) t·(0.9t ) 4 2 0 10 20 40 60 80 60 80 t 4 2 t·(–0.9)t 0 –2 –4 0 20 40 100 t F I G U R E 1.4 The homogeneous solution t(a1 )t CASE 3 If a1 2 + 4a2 < 0, it follows that d is negative so that the characteristic roots are imaginary. Since a1 2 ≥ 0, imaginary roots can occur only if a2 < 0. Although this might be hard to interpret directly, if we switch to polar coordinates it is possible to transform the roots into more easily understood trigonometric functions. The technical details are presented in Appendix 1.1 of the Supplementary Manual. For now, write the two characteristic roots as √ √ 𝛼2 = (a1 − i −d)∕2 𝛼1 = (a1 + i −d)∕2 √ where i = −1. Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 27 SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS 27 As shown in Appendix 1.1, de Moivre's theorem allows us to write the homogeneous solution as (1.49) yth = 𝛽1 rt cos(𝜃t + 𝛽2 ) where 𝛽1 and 𝛽2 are arbitrary constants, r = (−a2 )1∕2 , and the value of 𝜃 is chosen so as to satisfy (1.50) cos(𝜃) = a1 ∕[2(−a2 )1∕2 ] The trigonometric functions impart a wavelike pattern to the time path of the homogeneous solution; note that the frequency of the oscillations is determined by 𝜃. Since cos(𝜃t) = cos(2𝜋 + 𝜃t), the stability condition is determined solely by the magnitude of r = (−a2 )1∕2 . If |a2 | = 1, the oscillations are of unchanging amplitude; the homogeneous solution is periodic. The oscillations will dampen if |a2 | < 1 and explode if |a2 | > 1. Example: It is worthwhile to work through an exercise using an equation with imaginary roots. The left-hand side of Worksheet 1.2 examines the behavior of the equation yt = 1.6yt−1 − 0.9yt−2 . A quick check shows that the discriminant d is negative so that the characteristic roots are imaginary. If we transform to polar coordinates, the value of r is given by (0.9)1∕2 = 0.949. From (1.50), cos(𝜃) = 1.6∕(2 ∗ 0.949) = 0.843. You can use a trig table or a calculator to show that 𝜃 = 0.567 (i.e., if cos(𝜃) = 0.843, 𝜃 = 0.567). Thus, the homogeneous solution is yth = 𝛽1 (0.949)t cos(0.567t + 𝛽2 ) (1.51) The graph on the left-hand side of Worksheet 1.2 sets 𝛽1 = 1 and 𝛽2 = 0 and plots the homogeneous solution for t = 1, … , 30. Case 2 uses the same value of a2 (hence, r = 0.949) but sets a1 = −0.6. Again, the value of d is negative; however, for this set of calculations, cos(𝜃) = −0.316 so that 𝜃 is 1.89. Comparing the two graphs, you can see that increasing the value of 𝜃 acts to increase the frequency of the oscillations. Stability Conditions The general stability conditions can be summarized using triangle ABC in Figure 1.5. Arc A0B is the boundary between Cases 1 and 3; it is the locus of points where d = a1 2 + 4a2 = 0. The region above A0B corresponds to Case 1 (since d > 0), and the region below A0B corresponds to Case 3 (since d < 0). In Case 1 (in which the roots are real and distinct), stability requires that the largest root be less than unity √ and the smallest root be greater than –1. The largest characteristic root, 𝛼1 = (a1 + d)∕2, will be less than unity if a1 + (a1 2 + 4a2 )1∕2 < 2 or (a1 2 + 4a2 )1∕2 < 2 − a1 Square each side to obtain the condition: a1 2 + 4a2 < 4 − 4a1 + a1 2 Trim Size: 6in x 9in 28 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 28 DIFFERENCE EQUATIONS WORKSHEET IMAGINARY ROOTS 1.2 Example 1 Example 2 yt − 1.6yt−1 + 0.9yt−2 yt + 0.6yt−1 + 0.9yt−2 (a) Check the discriminant d = (a𝟏 )𝟐 + 𝟒a𝟐 d = (1.6)2 + 4(−0.9) = −1.04 d = (−0.6)2 + 4(−0.9) = −3.24 Hence, the roots are imaginary. The homogeneous solution has the form yth = 𝛽1 rt cos(𝜃t + 𝛽2 ) b) Obtain the value of r = ( − a𝟐 )𝟏∕𝟐 where 𝛽1 and 𝛽2 are arbitrary constants. r = (0.9)1∕2 r = (0.9)1∕2 = 0.949 = 0.949 c) Obtain 𝜃 from cos(𝜃) = a𝟏 ∕ [𝟐( − a𝟐 )𝟏∕𝟐 ] cos(𝜃) = 1.6∕[2(0.9)1∕2 ] = 0.843 cos(𝜃) = −0.6∕[2(0.9)1∕2 ] = −0.316 Given cos(𝜃), use a calculator or a trig-table to find 𝜃: 𝜃 = 0.567 𝜃 = 1.89 d) Form the homogeneous solution: yht = 𝛽1 rt cos(𝜃t + 𝛽2 ) yth = 𝛽1 (0.949)t cos(0.567t + 𝛽2 ) yth = 𝛽1 (0.949)t cos(1.89t + 𝛽2 ) For 𝛽1 = 1 and 𝛽2 = 0, the time paths of the homogeneous solutions are: 1 1 0.5 0.5 0 1 11 21 0 –0.5 –0.5 –1 –1 1 11 21 Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 29 SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS 29 a2 a2 = 1 + a1 a2 = 1 – a1 C 0 a1 A B a1 = –2 a2 = –1 a1 = 2 F I G U R E 1.5 Characterizing the Stability Conditions or a1 + a2 < 1 √ The smallest root, 𝛼2 = (a1 − d)∕2, will be greater than minus one if a1 − (a1 2 + 4a2 )1∕2 > −2 or (1.52) 2 + a1 > (a1 2 + 4a2 )1∕2 Square each side to obtain the condition: 4 + 4a1 + a1 2 > a1 2 + 4a2 or a2 < 1 + a1 (1.53) Thus, the region of stability in Case 1 consists of all points in the region bounded by A0BC. For any point in A0BC, conditions (1.52) and (1.53) hold and d > 0. In Case 2 (repeated roots), a1 2 + 4a2 = 0. The stability condition is |a1 | < 2. Thus, the region of stability in Case 2 consists of all points on arc A0B. In Case 3 (d < 0), the stability condition is r = (−a2 )1∕2 < 1. Hence, −a2 < 1 (where a2 < 0) (1.54) Thus, the region of stability in Case 3 consists of all points in region A0B. For any point in A0B, (1.54) is satisfied and d < 0. A succinct way to characterize the stability conditions is to state that the characteristic roots must lie within the unit circle. Consider the semicircle drawn in Figure 1.6. Real numbers are measured on the horizontal axis and imaginary numbers are measured on the vertical axis. If the characteristic roots 𝛼1 and 𝛼2 are both real, they can be plotted on the horizontal axis. Stability requires that they lie within a circle of radius one. Complex roots will lie somewhere in the complex plane. If a1 > 0, the √ √ roots 𝛼1 = (a1 + i d)∕2 and 𝛼2 = (a1 − i d)∕2 can be represented by the two points Trim Size: 6in x 9in 30 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 30 DIFFERENCE EQUATIONS Imaginary α1 d1/2/2 r 0 a1/2 Real r α2 F I G U R E 1.6 Characteristic Roots and the Unit Circle shown √ in Figure 1.6. For example, 𝛼1 is drawn by moving a1 ∕2 units along the real axis and d∕2 units along the imaginary axis. Using the distance formula, the length of the radius r is given by √ r= (a1 ∕2)2 + (d1∕2 i∕2)2 and, using the fact that i2 = −1, we obtain r = (−a2 )1∕2 The stability condition requires that r < 1. Therefore, when plotted on the complex plane, the two roots 𝛼1 and 𝛼2 must lie within a circle of radius equal to unity. In the time-series literature it is simply stated that stability requires that all characteristic roots lie within the unit circle. Higher Order Systems The same method can be used to find the homogeneous solution to higher order difference equations. The homogeneous equation for (1.10) is yt − n ∑ ai yt−i = 0 (1.55) i=1 Given the results in Section 1.4, you should suspect each homogeneous solution to have the form yth = A𝛼 t where A is an arbitrary constant. Thus, to find the value(s) of 𝛼, we seek the solution for n ∑ ai A𝛼 t−i = 0 (1.56) A𝛼 t − i=1 or, dividing through by 𝛼 t−n , we seek the values of 𝛼 that solve 𝛼 − a1 𝛼 n−1 − a2 𝛼 n−2 − · · · − an = 0 n (1.57) Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 31 PARTICULAR SOLUTIONS FOR DETERMINISTIC PROCESSES 31 This nth-order polynomial will yield n solutions for 𝛼. Denote these n characteristic roots by 𝛼1 , 𝛼2 , … , 𝛼n . As in Section 1.4, the linear combination A1 𝛼1t + A2 𝛼2t + · · · + An 𝛼nt is also a solution. The arbitrary constants A1 through An can be eliminated by imposing n initial conditions on the general solution. The 𝛼i may be real or complex numbers. Stability requires that all real valued 𝛼i be less than unity in absolute value. Complex roots will necessarily come in pairs. Stability requires that all roots lie within the unit circle shown in Figure 1.6. In most circumstances there is little need to directly calculate the characteristic roots of higher order systems. Many of the technical details are included in Section 1.2 of the Supplementary Manual (Appendix 1.2 of this chapter). However, there are some useful rules for checking the stability conditions in higher order systems. 1. In an nth-order equation, a necessary condition for all characteristic roots to lie inside the unit circle is n ∑ ai < 1 i=1 2. Since the values of the ai can be positive or negative, a sufficient condition for all characteristic roots to lie inside the unit circle is n ∑ |ai | < 1 i=1 3. At least one characteristic root equals unity if n ∑ ai = 1 i=1 4. Any sequence that contains one or more characteristic roots that equal unity is called a unit root process. For a third-order equation, the stability conditions can be written as 1 − a1 − a2 − a3 > 0 1 + a1 − a2 + a3 > 0 1 − a1 a3 + a2 − a3 2 > 0 3 + a1 + a2 − 3a3 > 0 or 3 − a1 + a2 + 3a3 > 0 Given that the first three inequalities are satisfied, either of the last two can be checked. One of the last conditions is redundant, given that the other three hold. 7. PARTICULAR SOLUTIONS FOR DETERMINISTIC PROCESSES Finding the particular solution to a difference equation is often a matter of ingenuity and perseverance. The appropriate technique depends heavily on the form of the {xt } process. We begin by considering those processes that contain only deterministic Trim Size: 6in x 9in 32 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 32 DIFFERENCE EQUATIONS components. Of course, in econometric analysis, the forcing process will contain both deterministic and stochastic components. CASE 1 xt = 𝟎. When all elements of the {xt } process are zero, the difference equation becomes (1.58) yt = a0 + a1 yt−1 + a2 yt−2 + · · · + an yt−n Intuition suggests that an unchanging value of y (i.e., yt = yt−1 = · · · = c) should solve the equation. Substitute the trial solution yt = c into (1.58) to obtain c = a0 + a1 c + a2 c + · · · + an c so that c = a0 ∕(1 − a1 − a2 − · · · − an ) (1.59) As long as (1 − a1 − a2 − · · · − an ) does not equal zero, the value of c given by (1.59) is a solution to (1.58). Hence, the particular solution to (1.58) is given by p yt = a0 ∕(1 − a1 − a2 − · · · − an ). If 1 − a1 − a2 − · · · − an = 0, the value of c in (1.59) is undefined; it is necessary to try some other form for the solution. The key insight is that {yt } is a unit root process if Σai = 1. Since {yt } is not convergent, it stands to reason that the constant solution does not work. Instead, recall equations (1.12) and (1.26); these solutions suggest that a linear time trend can appear in the solution of a unit p root process. As such, try the solution yt = ct. For ct to be a solution it must be the case that ct = a0 + a1 c(t − 1) + a2 c(t − 2) + · · · + an c(t − n) or, combining like terms, (1 − a1 − a2 − · · · − an )ct = a0 − c(a1 + 2a2 + 3a3 + · · · + nan ) Since 1 − a1 − a2 − · · · − an = 0, select the value of c such that c = a0 ∕(a1 + 2a2 + 3a3 + · · · + nan ) For example, let yt = 2 + 0.75yt−1 + 0.25yt−2 Here, a1 = 0.75 and a2 = 0.25; {yt } is a unit root process because a1 + a2 = 1. The particular solution has the form ct, where c = 2∕[0.75 + 2(0.25)] = p 1.6. In the event that the solution ct fails, sequentially try the solutions yt = 2 3 n ct , ct , … , ct . For an nth-order equation, one of these solutions will always be the particular solution. Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 33 PARTICULAR SOLUTIONS FOR DETERMINISTIC PROCESSES 33 CASE 2 The Exponential Case. Let xt have the exponential form b(d)rt , where b, d, and r are constants. Since r has the natural interpretation as a growth rate, we would expect to encounter this type of forcing process case in a growth context. We illustrate the solution procedure using the first-order equation yt = a0 + a1 yt−1 + bdrt (1.60) To try to gain an intuitive feel for the form of the solution, notice that if b = 0, (1.60) is a special case of (1.58). Hence, you should expect a constant to appear in the particular solution. Moreover, the expression drt grows at the constant rate r. p Thus, you might expect the particular solution to have the form yt = c0 + c1 drt , where c0 and c1 are constants. If this equation is actually a solution, you should be able to substitute it back into (1.60) and obtain an identity. Making the appropriate substitutions, we get c0 + c1 drt = a0 + a1 [c0 + c1 dr(t−1) ] + bdrt (1.61) For this solution to work, it is necessary to select c0 and c1 such that c0 = a0 ∕(1 − a1 ) and c1 = [bdr ]∕(dr − a1 ) Thus, a particular solution is p yt = a0 bdr rt + r d 1 − a1 d − a1 p The nature of the solution is that yt equals the constant a0 ∕(1 − a1 ) plus an expression that grows at the rate r. Note that for |dr | < 1, the particular solution converges to a0 ∕(1 − a1 ). If either a1 = 1 or a1 = dr , use the trick suggested in Case 1. If a1 = 1, try the solution c0 = ct, and if a1 = dr , try the solution c1 = tb. Use precisely the same methodology in higher order systems. CASE 3 Deterministic Time Trend. In this case, let the {xt } sequence be represented by the relationship xt = btd where b is a constant and d is a positive integer. Hence, yt = a0 + n ∑ ai yt−i + dtd (1.62) i=1 Since yt depends on td , it follows that yt−1 depends on (t − 1)d , yt−2 depends on (t − 2)d , and so on. As such, the particular solution has the form p yt = c0 + c1 t + c2 t2 + · · · + cd td . To find the value of each ci , substitute the particular solution into (1.62). Then select the value of each ci that results in an identity. Although various values of d are possible, in economic applications it is Trim Size: 6in x 9in 34 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 34 DIFFERENCE EQUATIONS common to see models incorporating a linear time trend (d = 1). For illustrative purposes, consider the second-order equation yt = a0 + a1 yt−1 + a2 yt−2 + bt. p Posit the solution yt = c0 + c1 t where c0 and c1 are undetermined coefficients. Substituting this "challenge solution" into the second-order difference equation yields c0 + c1 t = a0 + a1 [c0 + c1 (t − 1)] + a2 [c0 + c1 (t − 2)] + bt (1.63) Now select values of c0 and c1 so as to force equation (1.63) to be an identity for all possible values of t. If we combine all constant terms and all terms involving t, the required values of c0 and c1 are c1 = b∕(1 − a1 − a2 ) c0 = [a0 − (2a2 + a1 )c1 ]∕(1 − a1 − a2 ) so that c0 = [a0 ∕(1 − a1 − a2 )] − [b∕(1 − a1 − a2 )2 ](2a2 + a1 ) Thus, the particular solution will also contain a linear time trend. You should have no difficulty foreseeing the solution technique if a1 + a2 = 1. In this circumstance—which is applicable to higher order cases, as well—try multiplying the original challenge solution by t. 8. THE METHOD OF UNDETERMINED COEFFICIENTS At this point, it is appropriate to introduce the first of two useful methods for finding particular solutions when there are stochastic components in the {yt } process. The key insight of the method of undetermined coefficients is that linear equations have linear solutions. Hence, the particular solution to a linear difference equation is necessarily linear. Moreover, the solution can depend only on time, a constant, and the elements of the forcing process {xt }. Thus, it is often possible to know the exact form of the solution even though the coefficients of the solution are unknown. The technique involves positing a solution—called a challenge solution—that is a linear function of all terms thought to appear in the actual solution. The problem becomes one of finding the set of values for those undetermined coefficients that solve the difference equation. The actual technique for finding the coefficients is straightforward. Substitute the challenge solution into the original difference equation and solve for the values of the undetermined coefficients that yield an identity for all possible values of the included variables. If it is not possible to obtain an identity, the form of the challenge solution is incorrect. Try a new trial solution and repeat the process. In fact, we used the method p of undetermined coefficients when positing the challenge solutions yt = c0 + c1 drt and p yt = c0 + c1 t for Cases 2 and 3 in Section 1.7. To begin, reconsider the simple first-order equation yt = a0 + a1 yt−1 + 𝜀t . Since you have solved this equation using the iterative method, the equation is useful for illustrating the method of undetermined coefficients. The nature of the {yt } process Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 35 THE METHOD OF UNDETERMINED COEFFICIENTS 35 is such that the particular solution can depend only on a constant term, time, and the individual elements of the {𝜀t } sequence. Given that t does not explicitly appear in the forcing process, t can be in the particular solution only if the characteristic root is unity. Since the goal is to illustrate the method, posit the challenge solution: yt = b0 + b1 t + ∞ ∑ 𝛼i 𝜀t−i (1.64) i=0 where b0 , b1 , and all the 𝛼i are the coefficients to be determined. Substitute (1.64) into the original difference equation to form b0 + b1 t + 𝛼0 𝜀t + 𝛼1 𝜀t−1 + 𝛼2 𝜀t−2 + · · · = a0 + a1 [b0 + b1 (t − 1) + 𝛼0 𝜀t−1 + 𝛼1 𝜀t−2 + · · ·] + 𝜀t Collecting like terms, we obtain (b0 − a0 − a1 b0 + a1 b1 ) + b1 (1 − a1 )t + (𝛼0 − 1)𝜀t + (𝛼1 − a1 𝛼0 )𝜀t−1 + (𝛼2 − a1 𝛼1 )𝜀t−2 + (𝛼3 − a1 𝛼2 )𝜀t−3 + · · · = 0 (1.65) Equation (1.65) must hold for all values of t and all possible values of the {𝜀t } sequence. Thus, each of the following conditions must hold: 𝛼0 − 1 = 0 𝛼1 − a1 𝛼0 = 0 𝛼2 − a1 𝛼1 = 0 ⋮ b0 − a0 − a1 b0 + a1 b1 = 0 b1 − a1 b1 = 0 Notice that the first set of conditions can be solved for the 𝛼i recursively. The solution of the first condition entails setting 𝛼0 = 1. Given this solution for 𝛼0 , the next equation requires 𝛼1 = a1 . Moving down the list, 𝛼2 = a1 𝛼1 or 𝛼2 = a1 2 . Continuing the recursive process, we find 𝛼i = a1 i . Now consider the last two equations. There are two possible cases depending on the value of a1 . If a1 ≠ 1, it immediately follows that b1 = 0 and b0 = a0 ∕(1 − a1 ). For this case, the particular solution is ∞ yt = ∑ a0 + ai 𝜀 1 − a1 i=0 1 t−i Compare this result to (1.21); you will see that it is precisely the same solution found using the iterative method. The general solution is the sum of this particular solution plus the homogeneous solution Aa1 t . Hence, the general solution is ∞ yt = ∑ a0 + ai 𝜀 + Aat1 1 − a1 i=0 1 t−i Now, if there is an initial condition for y0 , it follows that ∞ y0 = ∑ a0 + ai 𝜀 + A 1 − a1 i=0 1 −i Trim Size: 6in x 9in 36 CHAPTER 1 Enders c01.tex V3 - 09/02/2014 12:52pm Page 36 DIFFERENCE EQUATIONS Combining these two equations so as to eliminate the arbitrary constant A, we obtain [ ] ∞ ∞ ∑ ( ) ∑ a0 i t i yt = + a 𝜀 + a1 y0 − a0 ∕ 1 − a1 − a1 𝜀−i 1 − a1 i=0 1 t−i i=0 so that ∑ a0 + ai 𝜀 + at1 [y0 − a0 ∕(1 − a1 )] 1 − a1 i=0 1 t−i t−1 yt = (1.66) It can be easily verified that (1.66) is identical to (1.25). Instead, if a1 = 1, b0 can be any arbitrary constant and b1 = a0 . The improper form of the solution is yt = b0 + a0 t + ∞ ∑ 𝜀t−i i=0 The form of the solution is "improper" because the sum of the {𝜀t } sequence may not be finite. Therefore, it is necessary to impose an initial condition. If the value y0 is given, it follows that ∞ ∑ y0 = b0 + 𝜀−i i=0 Imposing the initial condition on the improper form of the solution yields (1.26) yt = y0 + a0 t + t ∑ 𝜀i i=1 To take a second example, consider the equation yt = a0 + a1 yt−1 + 𝜀t + 𝛽1 𝜀t−1 (1.67) Again, the solution can depend only on a constant, the elements of the {𝜀t } sequence, and t raised to the first power. As in the previous example, t does not need to be included in the challenge solution if the characteristic root differs from unity. To reinforce this point, use the challenge solution given by (1.64). Substitute this tentative solution into (1.67) to obtain [ ] ∞ ∞ ∑ ∑ b0 + b1 t + 𝛼i 𝜀t−i = a0 + a1 b0 + b1 (t − 1) + 𝛼i 𝜀t−1−i + 𝜀t + 𝛽1 𝜀t−1 i=0 i=0 Matching coefficients on all terms containing 𝜀t , 𝜀t−1 , 𝜀t−2 , … , yields 𝛼0 = 1 𝛼1 = a1 𝛼0 + 𝛽1 [so that 𝛼1 = a1 + 𝛽1 ] 𝛼2 = a1 𝛼1 𝛼3 = a1 𝛼2 [so that 𝛼2 = a1 (a1 + 𝛽1 )] [so that 𝛼3 = (a1 )2 (a1 + 𝛽1 )] ⋮ 𝛼i = a1 𝛼i−1 [so that 𝛼i = (a1 )i−1 (a1 + 𝛽1 )] Trim Size: 6in x 9in Enders c01.tex V3 - 09/02/2014 12:52pm Page 37 THE METHOD OF UNDETERMINED COEFFICIENTS 37 Matching coefficients of intercept terms and coefficients of terms containing t, we get b0 = a0 + a1 b0 − a1 b1 b1 = a1 b1 Again, there are two cases. If a1 ≠ 1, then b1 = 0 and b0 = a0 ∕(1 − a1 ). The particular solution is ∞ ∑ a0 + 𝜀t + (a1 + 𝛽1 ) ai−1 yt = 1 𝜀t−i 1 − a1 i=1 The general solution augments the particular solution with the term Aat1 . You are left with the exercise of imposing the initial condition for y0 on the general solution. Now consider the case in which a1 = 1. The undetermined coefficients are such that b1 = a0 and b0 is an arbitrary constant. The improper form of the solution is yt = b0 + a0 t + 𝜀t + (1 + 𝛽1 ) ∞ ∑ 𝜀t−i i=1 If y0 is given, it follows that y0 = b0 + 𝜀0 + (1 + 𝛽1 ) ∞ ∑ 𝜀−i i=1 Hence, imposing the initial condition, we obtain yt = y0 + a0 t + 𝜀t + (1 + 𝛽1 ) t−1 ∑ 𝜀t−i i=1 Higher Order Systems The identical procedure is used for higher order systems. As an example, let us find the particular solution to the second-order equation yt = a0 + a1 yt−1 + a2 yt−2 + 𝜀t (1.68) Since we have a second-order equation, we use the challenge solution yt = b0 + b1 t + b2 t2 + 𝛼0 𝜀t + 𝛼1 𝜀t−1 + 𝛼2 𝜀t−2 + · · · where b0 , b1 , b2 , and the 𝛼i are the undetermined coefficients. Substituting the challenge solution into (1.68) yields [b0 + b1 t + b2 t2 ] + 𝛼0 𝜀t + 𝛼1 𝜀t−1 + 𝛼2 𝜀t−2 + · · · = a0 + a1 [b0 + b1 (t − 1) + b2 (t − 1)2 + 𝛼0 𝜀t−1 + 𝛼1 𝜀t−2 + 𝛼2 𝜀t−3 + · · ·] + a2 [b0 + b1 (t − 2) + b2 (t − 2)2 + 𝛼0 𝜀t−2 + 𝛼1 𝜀t−3 + 𝛼2 𝜀t−4 + · · ·] + 𝜀t Trim Size: 6in x 9in 38 Enders c01.tex CHAPTER 1 V3 - 09/02/2014 12:52pm Page 38 DIFFERENCE EQUATIONS There are several necessary and sufficient conditions for the values of the 𝛼i 's to render the equation above an identity for all possible realizations of the {𝜀t } sequence: 𝛼0 = 1 𝛼1 = a1 𝛼0 [so that 𝛼1 = a1 ] 𝛼2 = a1 𝛼1 + a2 𝛼0 [so that 𝛼2 = (a1 )2 + a2 ] 𝛼3 = a1 𝛼2 + a2 𝛼1 [so that 𝛼3 = (a1 )3 + 2a1 a2 ] ⋮ Notice that for any value of j ≥ 2, the coefficients solve the second-order difference equation 𝛼j = a1 𝛼j−1 + a2 𝛼j−2 . Since we know 𝛼0 and 𝛼1 , we can solve for all the 𝛼j iteratively. The properties of the coefficients will be precisely those discussed when considering homogeneous solutions: 1. 2. Convergence necessitates that |a2 | < 1, a1 + a2 < 1, and that a2 − a1 < 1. Notice that convergence implies that past values of the {𝜀t } sequence ultimately have a successively smaller influence on the current value of yt . If the coefficients converge, convergence will be direct or oscillatory if (a21 + 4a2 ) > 0, will follow a sine/cosine pattern if (a21 + 4a2 ) < 0, and will "explode" and then converge if (a21 + 4a2 ) = 0. Appropriately setting the 𝛼i , we are left with the remaining expression: b2 (1 − a1 − a2 )t2 + [b1 (1 − a1 − a2 ) + 2b2 (a1 + 2a2 )]t + [b0 (1 − a1 − a2 ) − a0 + a1 (b1 − b2 ) + 2a2 (b1 − 2b2 )] = 0 (1.69) Equation (1.69) must equal zero for all values of t. First, consider the case in which a1 + a2 ≠ 1. Since (1 − a1 − a2 ) does not vanish, it is necessary to set the value of b2 equal to zero. Given that b2 = 0 and that the coefficient of t must equal zero, it follows that b1 must also be set equal to zero. Finally, given that b1 = b2 = 0, we must set b0 = a0 ∕(1 − a1 − a2 ). Instead, if a1 + a2 = 1, the solutions for the bi depend on the specific values of a0 , a1 , and a2 . The key point is that the stability condition for the homogeneous equation is precisely the co

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