Kinh tế học - Chapter 1: Difference equations

Chapter 1: Difference EquationsApplied Econometric Time SeriesFourth EditionTIME-SERIES MODELSSection 1The traditional use of time series models was for forecastingIf we know yt+1 = a0 + a1yt + et+1then Etyt+1 = a0 + a1ytand since yt+2 = a0 + a1yt+1 + et+2Etyt+2 = a0 + a1Etyt+1 = a0 + a1(a0 + a1yt) = a0 + a1a0 + (a1)2ytCapturing Dynamic RelationshipsWith the advent of modern dynamic economic models, the newer uses of time series models involveCapturing dynamic economic relationshipsHypothesis testingDeveloping “stylized facts” In a sense, this reverses the so-called scientific method in that modeling goes from developing models that follow from the data.The Random Walk Hypothesisyt+1 = yt + et+1orDyt+1 = et+1 where yt = the price of a share of stock on day t, and et+1 = a random disturbance term that has an expected value of zero. Now consider the more general stochastic difference equationDyt+1 = a0 + a1yt + et+1The random walk hypothesis requires the testable restriction: a0 = a1 = 0.The Unbiased Forward Rate (UFR) hypothesis Given the UFR hypothesis, the forward/spot exchange rate relationship is: st+1 = ft + et+1 (1.6)where et+1 has a mean value of zero from the perspective of time period t.Consider the regressionst+1 = a0 + a1ft + t+1The 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 long-run equilibrium when et+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 processst+2 = st+1 – a[ st+1 – ft ] + est+2 a > 0 (1.7)ft+1 = ft + b [ st+1 – ft ] + eft+1 b > 0 (1.8)where est+2 andeeft+1 both have an expected value of zero. Trend-Cycle RelationshipsWe can think of a time series as being composed of:yt = trend + “cycle” + noiseTrend: Permanent Cycle: predictable (albeit temporary) (Deviations from trend)Noise: unpredictableSeries with decidedly upward trendGDP Volatility?Stock Market VolatilityCo-MovementsCommon TrendsDIFFERENCE EQUATIONS AND THEIR SOLUTIONSSection 2Consider the function yt* = f(t*)We can then form the first differences:Dyt = f(t) – f(t–1)  yt – yt–1Dyt+1 = f(t+1) – f(t)  yt+1 – ytDyt+2 = f(t+2) – f(t+1)  yt+2 – yt+1More generally, for the forcing process xt a n-th order linear process is What is a solution?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).The key property of a solution is that it satisfies the difference equation for all permissible values of t and {xt}.SOLUTION BY ITERATIONSection 3Iteration without an Initial ConditionReconciling the Two Iterative MethodsNonconvergent SequencesSolution by IterationConsider the first-order equationyt = a0 + a1yt–1 + et (1.17)Given the value of y0, it follows that y1 will be given b y1 = a0 + a1y0 + e1In the same way, y2 must be y2 = a0 + a1y1 + e2 = a0 + a1[a0 + a1y0 + e1] + e2 = a0 + a0a1 + (a1)2y0 + a1e1 + e2Continuing the process in order to find y3, we obtain y3 = a0 + a1y2 + e3 = a0[1 + a1 + (a1)2] + (a1)3 y0 + a12e1 + a1e2 + e3Fromy3 = a0[1 + a1 + (a1)2] + (a1)3 y0 + a12e1 + a1e2 + e3you can verify that for all t > 0, repeated iteration yieldsIf | a1 | 1, the homogeneous solution is not convergent. If a1 > 1, yt approaches ∞ as t increases. If a1 01 + a1 – a2 + a3 > 01 – a1a3 + a2 – a32 > 03 + a1 + a2 – 3a3 > 0 or 3 – a1 + a2 + 3a3 > 0Given that the first three inequalities are satisfied, one of the last conditions is redundant.At least one characteristic root equals unity ifSTEP 1: form the homogeneous equation and find all n homogeneous solutions; STEP 2: find a particular solution; 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. The Solution MethodologyTHE COBWEB MODELSection 5Stability ConditionsHigher-Order SystemsSetting supply equal to demand:b + bpt–1 + et = a – gptor pt = (–b/g)pt–1 + (a – b)/g – et/gThe homogeneous equation is pt = (–b/g)pt–1. If the ratio b/g is less than unity, you can iterate (1.39) backward from pt to verify that the particular solution for the price is Stability requires | b/g | 0, d is a real number and there will be two distinct real characteristic roots. CASE 2If + 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 = 0.7yt-1 + 0.35yt2. Hence: a1 = 0.7 and a2 = 0.35 Form the homogeneous equation: yt  0.7yt-1  0.35yt2 = 0 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. THE METHOD OF UNDETERMINED COEFFICIENTSSection 8The Method of Undetermined CoefficientsConsider the simple first-order equation: yt = a0 + a1yt–1 + etPosit the challenge solution:b0 + a0et + a1et–1 + a2et–2 + = a0 + a1[b0 + a0et–1 + a1et–2 + ] + et 0 − 1 = 0a1 – a1a0 = 0a2 – a1a1 = 0b0 – a0 – a1b0 = 0The Method of Undetermined Coefficients IIConsider: yt = a0 + a1yt–1 + a2yt–2 + et (1.68)Since we have a second-order equation, we use the challenge solutionyt = b0 + b1t + b2t2 + a0et + a1et–1 + a2et–2 + where b0, b1, b2, and the ai are the undetermined coefficients. Substituting the challenge solution into (1.68) yields[b0+b1t+b2t2] + a0et + a1et–1 + a2et–2+ = a0 + a1[b0 + b1(t – 1) + b2(t – 1)2 + a0et–1 + a1et–2 + a2et–3 + ] + a2[b0 + b1(t – 2) + b2(t – 2)2 + a0et–2 + a1et–3 + a2et-4 + ] + etHence:  a0 = 1 a1 = a1a0 [so that a1 = a1] a2 = a1a1 + a2a0 [so that a2 = (a1)2 + a2] a3 = a1a2 + a2a1 [so that a3 = (a1)3 + 2a1a2] Notice that for any value of j  2, the coefficients solve the second-order difference equation aj = a1aj–1 + a2aj–2. LAG OPERATORSSection 9Lag Operators in Higher-Order SystemsLag OperatorsThe lag operator L is defined to be: Liyt = yt-i Thus, Li preceding yt simply means to lag yt by i periods.The lag of a constant is a constant: Lc = c.  The distributive law holds for lag operators. We can set: (Li + Lj)yt = Liyt + Ljyt = yt-i + yt-j  Lag Operators (cont’d)Lag operators provide a concise notation for writing difference equations. Using lag operators, the p-th order equation yt = a0 + a1yt-1 + ... + apyt-p + εt can be written as: (1 - a1L - a2L2 - ... - apLp)yt = εtor more compactly as: A(L)yt = εtAs a second example, yt = a0 + a1yt-1 + ... + apyt-p + εt + β1εt-1 + ... + βqεt-q as: A(L)yt = a0 + B(L)εtwhere: A(L) and B(L) are polynomials of orders p and q, respectively. APPENDIX 1.1: IMAGINARY ROOTS AND DE MOIVRE’S THEOREM