This chapter’s objectives are to: 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,...
Applied Econometric Time Series Fourth Edition Chapter 1: Difference Equations Walter Enders, University of Alabama Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Section 1 TIMESERIES MODELS Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The traditional use of time series models was for forecasting If we know yt+1 = a0 + a1yt + εt+1 then Etyt+1 = a0 + a1yt and since yt+2 = a0 + a1yt+1 + εt+2 Etyt+2=a0+a1Etyt+1 =a0+a1(a0+a1yt) =a0+a1a0+(a1)2yt Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved CapturingDynamicRelationships ã ã Withtheadventofmoderndynamiceconomicmodels,the newerusesoftimeseriesmodelsinvolve – Capturing dynamic economic relationships – Hypothesis testing Developing “stylized facts” – In a sense, this reverses the socalled scientific method in that modeling goes from developing models that follow from the data Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The Random Walk Hypothesis yt+1 = yt + εt+1 or ∆yt+1 = εt+1 where yt = 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 ∆yt+1 = a0 + a1yt + εt+1 Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The Unbiased Forward Rate (UFR) hypothesis Given the UFR hypothesis, the forward/spot exchange rate relationship is: st+1 = ft + εt+1 (1.6) where εt+1 has a mean value of zero from the perspective of time period t Consider the regression st+1 = a0 + a1ft + t+1 The hypothesis requires a0 = 0, a1 = 1, and that the regression residuals t+1 have a mean value of zero from the perspective of time period t The spot and forward markets are said to be in longrun 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 – a[ st+1 – ft ] + εst+2 a>0 (1.7) ft+1=ft+b[st+1ft]+ft+1 b>0 (1.8) wherest+2andeft+1bothhaveanexpectedvalueofzero. Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved TrendưCycleRelationships ã We can think of a time series as being composed of: yt = trend + “cycle” + noise – – – Trend: Permanent Cycle: predictable (albeit temporary) • (Deviations from trend) Noise: unpredictable Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Series with decidedly upward trend 15000 Trillions of 2005 dollars 12500 10000 7500 5000 2500 1950 1960 GDP 1970 1980 Potential 1990 2000 Consumption Figure 3.1 Real GDP, Consumption and Investment Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved 2010 Investment GDP Volatility? 20 15 Percent per year 10 -5 -10 -15 1950 1960 1970 1980 1990 2000 Figure 3.2 Annualized Growth Rate of Real GDP Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved 2010 Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved Section6 ã StabilityConditions ã HigherưOrderSystems SOLVING HOMOGENEOUS DIFFERENCE EQUATIONS Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Consider the secondorder equation yt – a1yt–1 – a2yt–2 = 0 Aα t – a1Aα t1 – a2Aα t2 = 0 If you divide (1.46) by Aαt–2, the problem is to find the values of α that satisfy α2 – a1α – a2 = 0 There are two characteristic roots. Hence the homogeneous solution is A1(α1)t + A2(α2)t Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved THE THREE CASES Case 1 If a12 + 4a2 > 0, d is a real number and there will be two distinct real characteristic roots. CASE 2 If + 4a2 = 0, it follows that d = 0 and a1 = a2 = a1/2. A homogeneous solution is a1/2. However, when d = 0, there is a second homogeneous solution given by t(a1/2)t Case 3 If a12 + 4a2 0, the roots will be real and distinct. Substitute yt = t into the homogenous equation to obtain: t – 0.2 t 1 – 0.35 t 2 = 0 Divide by t 2 to obtain the characteristic equation: 2 – 0.2 0.35 = 0 Compute the two characteristic roots: 1 = 0.7 2 = −0.5 The homogeneous solution is: A1(0.7)t + A2( 0.5)t. Example 2: yt = 0.7yt1 + 0.35yt 2. Hence: a1 = 0.7 and a2 = 0.35 Form the homogeneous equation: yt 0.7yt1 0.35yt 2 = 0 Thus d = + 4a2 = 1.89. Given that d > 0, the roots will be real and distinct. Form the characteristic equation t – 0.7 t 1 – 0.35 t 2 = 0 Divide by t 2 to obtain the characteristic equation: 2 – 0.7 – 0.35 = 0 Compute the two characteristic roots: 1 = 1.037 2 = 0.337 The homogeneous solution is: A1(1.037)t + A2(–0.337)t. Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Section 8 THE METHOD OF UNDETERMINED COEFFICIENTS Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The Method of Undetermined Coefficients Consider the simple firstorder equation: yt = a0 + a1yt–1 + εt Posit the challenge solution: α i ε t −i y t = b0 + i=0 b0 + a0εt + a1εt–1 + a2εt–2 + … = a0 + a1[b0 + a0εt–1 + a1εt–2 + ] + εt 0 − 1 = 0 a1 – a1a0 = 0 a2 – a1a1 = 0 b0 – a0 – a1b0 = 0 y t = a0 + − a1 i =0 a1iε t −i Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The Method of Undetermined Coefficients II Consider: yt = a0 + a1yt–1 + a2yt–2 + εt (1.68) Since we have a secondorder equation, we use the challenge solution yt = b0 + b1t + b2t2 + a0εt + a1εt–1 + a2εt–2 + where b0, b1, b2, and the ai are the undetermined coefficients. Substituting the challenge solution into (1.68) yields [b0+b1t+b2t2]+a0t+a1t1+a2t2+=a0+a1[b0+ b1(t1)+b2(t1)2 +a0t1+a1t2+a2t3+]+a2[b0+b1(t2)+ b2(t2)2 Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved +a0 t2+a1t3+a2tư4+]+t Section9 ã LagOperatorsinHigherưOrderSystems LAG OPERATORS Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved Lag Operators The lag operator L is defined to be: Liyt = yti Thus, Li preceding yt simply means to lag yt by i periods The lag of a constant is a constant: Lc = c. Thedistributivelawholdsforlagoperators.Wecanset: (Li+Lj)yt=Liyt+Ljyt=ytưi+ytưj Copyrightâ2015John,Wiley&Sons,Inc.Allrightsreserved LagOperators(contd) ã Lag operators provide a concise notation for writing difference equations. Using lag operators, the pth order equation yt = a0 + a1yt1 + + apytp + εt can be written as: (1 a1L a2L2 apLp)yt = εt or more compactly as: A(L)yt = εt As a second example, yt = a0 + a1yt1 + + apytp + εt + β1εt1 + + βqεtq as: A(L)yt = a0 + B(L)εt where: A(L) and B(L) are polynomials of orders p and q, respectively. Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved APPENDIX 1.1: IMAGINARY ROOTS AND DE MOIVRE’S THEOREM Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved ...Section 1 TIME? ?SERIES? ?MODELS Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved The traditional use of? ?time? ?series? ?models was for forecasting If we know... CoMovements 16 14 percent per year 12 10 1960 1965 1970 1975 1980 1985 T -bill 1990 1995 2000 5-year Figure 3.4 Short- and Long-Term Interest Rates Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved... Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved 2010 Investment GDP Volatility? 20 15 Percent per year 10 -5 -1 0 -1 5 1950 1960 1970 1980 1990 2000 Figure 3.2 Annualized Growth Rate of Real GDP Copyright © 2015 John, Wiley & Sons, Inc. All rights reserved