Kinh tế học - Chapter 6: Applied economitric time series 4th ed. walter enders

yt = myt + eyt zt = mzt + ezt where mit = a random walk process representing the trend in variable i eit = the stationary (irregular) component of variable i If {yt} and {zt} are cointegrated of order (1,1), there must be nonzero values of b1 and b2 for which the linear combination b1yt + b2zt is stationary. Consider the sum b1yt + b2zt = b1(myt + eyt) + b2(mzt + ezt) = (b1myt + b2mzt) + (b1eyt + b2ezt) (6.6) For b1yt + b2zt to be stationary, the term (b1myt + b2mzt) must vanish.

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Chapter 6Applied Economitric Time Series 4th ed. Walter EndersExample of Cointegration and Money DemandIn logarithms, an econometric specification for such an equation can be written as: mt = b0 + b1pt + b2yt + b3rt + et where: mt = demand for money pt = price level yt = real income rt = interest rate et = stationary disturbance term bi = parameters to be estimatedOther ExamplesConsumption function theory. Unbiased forward rate hypothesis. Commodity market arbitrage and purchasing power parity. The formal analysis begins by considering a set of economic variables in long-run equilibrium when b1x1t + b2x2t + + bnxnt = 0 Letting b and xt denote the vectors (b1, b2, , bn) and (x1t, x2t, , xnt)', the system is in long-run equilibrium when bxt = 0. The deviation from long-run equilibrium—called the equilibrium error—is et, so that et = bxtGeneralizationLetting β and xt denote the vectors (β1, β2, ..., βn) and (x1t, x2t, ..., xnt), the system is in long-run equilibrium when βxt' = 0. The deviation from long-run equilibrium--called the equilibrium error--is et, so that: et = βx'tIf the equilibrium is meaningful, it must be the case that the equilibrium error process is stationary. The scatter plot was drawn using the {y} and {z} sequences from Case 1 of Worksheet 6.1. Since both series decline over time, there appears to be a positive relationship between the two. The equilibrium regression line is shown. Three important points 1. Cointegration refers to a linear combination of non-stationary variables. If (β1, β2, ... , βn) is a cointegrating vector, then for any non-zero value of λ, (λβ1, λβ2, ... , λβn) is also a cointegrating vector. Typically, one of the variables is used to normalize the cointegrating vector by fixing its coefficient at unity. To normalize the cointegrating vector with respect to x1t, simply select λ = 1/β1. 2. The equation must be balanced in that the order of integration of the two sides must be equal3. If xt has m components, there may be as many as m-1 linearly independent cointegrating vectorsExample of Multiple Cointegrating VectorsLet the money supply rule be:mt = γ0 - γ1(yt + pt) + e1t (1.3) = γ0 - γ1yt - γ1 pt + e1twhere: {e1t} is a stationary error in the money supply feedback rule.Given the money demand function in (1.1), there are two cointegrating vectors for the money supply, price level, real income, and the interest rate. Let β be the (5 x 2) matrix:COINTEGRATION AND COMMON TRENDSyt = myt + eyt zt = mzt + ezt where mit = a random walk process representing the trend in variable i eit = the stationary (irregular) component of variable i If {yt} and {zt} are cointegrated of order (1,1), there must be nonzero values of b1 and b2 for which the linear combination b1yt + b2zt is stationary. Consider the sum b1yt + b2zt = b1(myt + eyt) + b2(mzt + ezt) = (b1myt + b2mzt) + (b1eyt + b2ezt) (6.6) For b1yt + b2zt to be stationary, the term (b1myt + b2mzt) must vanish. Granger Representation TheoremIn an error-correction model, the short-term dynamics of the variables in the system are influenced by the deviation from equilibrium. DrSt = aS(rLt–1 - b rSt–1) + eSt aS > 0 DrLt = –aL(rLt–1 - b rSt–1) + eLt aL > 0 This finding illustrates the Granger representation theorem stating that for any set of I(1) variables, error correction and cointegration are equivalent representations. The Engle-Granger MethodologyStep 1: Pretest the variables for their order of integration.Step 2: Estimate the long-run equilibrium relationship.If the results of Step 1 indicate that both {yt} and {zt} are I(1), the next step is to estimate the long-run equilibrium relationship in the form: yt = β0 + β1zt + et Consider the autoregression of the residuals: Test a1 = 0?Step 3: Estimate the error-correction modelThe Error Correction ModelInstead of a cross-equation restriction, useSpeed of adjustment coefficients The speed of adjustment coefficients ay and az are of particular interest in that they have important implications for the dynamics of the system. Direct convergence necessitates that be negative and z be positive. If we focus on (6.36) it is clear that for any given value of the deviation from long-run equilibrium, a large value of az is associated with a large value of Dzt.If one of these coefficients is (say y) is zero,the {zt} sequence does all of the correction to eliminate any deviation from long-run equilibrium. Since {yt} does not do any of the error-correcting, {yt} is said to be weakly exogenous.Problems with the EG-Method1. In practice, it is possible to find that one regression indicates the variables are cointegrated whereas reversing the order indicates no cointegration. This is a very undesirable feature of the procedure since the test for cointegration should be invariant to the choice of the variable selected for normalization. The problem is obviously compounded using three or more variables since any of the variables can be selected as the left-hand-side variable. 2. Moreover, in tests using three or more variables, we know that there may be more than one cointegrating vector. The method has no systematic procedure for the separate estimation of the multiple cointegrating vectors.3. Another serious defect of the Engle-Granger procedure is that it relies on a two-step estimator. Johansen MethodologyReconsider the n-variable first-order VAR given by (6.3): xt = A1xt-1 + εt. Subtract xt-1 from each side to obtain: Δxt = A1xt1  xt1 + εt = (A1  I)xt1 + εt = πxt1 + εtThe rank of (A1 – I) equals the number of cointegrating vectors. If (A1 – I) consists of all zeroes—so that rank(π) = 0—all of the {xit} sequences are unit root processes. If (A1 – I) is of full rank—so that rank(π) = n—each of the {xit} sequences converges to a point. The process can be modified to include a drift and seasonal dummy variables. π11x1t + π12x2t + π13x3t + ... + π1nx1n + π10 = 0π21x1t + π22x2t + π23x3t + ... + π2nx1n + π20 = 0Consider Dxt = *xt-1:If rank * = 2, Note: Adding a column of constants still means that rank(*) cannot exceed n The number of distinct cointegrating vectors can be obtained by checking the significance of the characteristic roots of π. We know that the rank of a matrix is equal to the number of its characteristic roots that differ from zero. Suppose we obtained the matrix π and ordered the n characteristic roots such that λ1 > λ2 > ... > λn. If the variables in xt are not cointegrated, the rank of π is zero and all of these characteristic roots will equal zero. Since ln(1) = 0, each of the expressions ln(1 - λi) will equal zero if the variables are not cointegrated. Similarly, if the rank of π is to unity, the first expression ln(1 - λ1) will be negative and all the other expressions are such that ln(1 - λ2) = ln(1 - λ3) = ... = ln(1 - λn) = 0. The null hypothesis that the number of distinct cointegrating vectors is less than or equal to r against a general alternative. From the previous discussion, it should be clear that λtrace equals zero when all λi = 0. The null that the number of cointegrating vectors is r against the alternative of r+1 cointegrating vectors. Again, if the estimated value of the characteristic root is close to zero, λmax will be small. In order to test other restrictions on the cointegrating vector, Johansen defines the two matrices α and β both of dimension (n x r) where r is the rank of π. The properties of α and β are such that: π = α β' In essence, we can normalize to obtain α β'Hypothesis TestingAsymptotically, the statistic has a χ2 distribution with (n - r) degrees of freedom. The value of this statistic should be zero if the restriction is not binding. Lag Length and Causality TestsEstimate the models with p and p – 1 lags. Let c denote the maximum number of regressors contained in the longest equation. The test statistic(T–c)(logr – logu) can be compared to a 2 distribution with degrees of freedom equal to the number of restrictions in the system. Alternatively, you can use the multivariate AIC or SBC to determine the lag length. If you want to test the lag lengths for a single equation, an F-test is appropriate.To difference or not to difference?DifferenceDo not differenceTests lose power if you do not difference: you estimate n2 more parameters (one extra lag of each variable in each equation).If you use first differences, you can use the standard F distribution to test for Granger causality.When the VAR has I(1) variables, the impulse responses at long forecast horizons are inconsistent estimates of the true responses. Since the impulse responses need not decay, any imprecision in the coefficient estimates will have a permanent effect on the impulse responses. If the system contains a cointegrating relationship, the system in differences is misspecified since it excludes the long-run equilibrium relationships among the variables that are contained in pxt–1.All of the coefficient estimates, t-tests, F-tests, tests of cross-equation restrictions, impulse responses and variance decompositions are not representative of the true process.Restrictions on the cointegrating vectorsTesting coefficient restrictions: As in the previous section, once you select the number of cointegrating vectors, you can test restrictions on the resulting values of β and/or α. Suppose you want to test the restriction that the intercept is zero. From the menu, you select Restrictions on subsets of β. Instead, suppose you want to test the three restrictions: β1 = β2, β1 = -β3, and β3 = 0 (so that the normalized cointegrating vector has the form yt + zt - wt = 0). In matrix form, the Linear vs Threshold CointegrationIn the simplest case, the two-step methodology entails using OLS to estimate the long-run equilibrium relationship as: x1t = β0 + β2x2t + β3x3t + ... + βnxnt + etwhere: xit are the individual I(1) components of xt, βi are the estimated parameters, and et is the disturbance term which may be serially correlated.The second-step focuses on the OLS estimate of ρ in the regression equation:Δet = ρet-1 + εt The TAR SpecificationLet the error process have the formΔet = It ρ1et-1 + (1 - It )ρ2et-1 + εt where: It is the Heaviside indicator function such that:The Momentum SpecificationTABLE 7: Estimates of the Interest Rate DifferentialFrom Enders and Siklos (JBES)Engle-GrangerThresholdMomentumMomentum-Consistentρ1a-0.068(‑2.858 )‑0.085(-2.522)-0.021(-0.628)-0.020(-0.680)ρ2aNA-0.020(-1.582)-0.117(-3.526)-0.141(-3.842)γ1a0.188(-2.782)0.190(2.787)0.183(2.730)0.186(2.790)γ2a-0.149(-2.197)-0.147(-2.153)-0.161(-2.376)-0.155(-2.312)AIC b11.7413.249.2857.022ΦcNA4.326.3637.548ρ1 = ρ2 dNA0.495(0.482)4.418(0.037)6.698(0.010)Q(4)eQ(8)Q(12)0.650.600.750.640.580.730.640.520.680.480.510.70Δxit = ρ1.iItet-1 + ρ2.i(1 - It)et-1 + ... + vit where: ρ1.i and ρ2.i are the speed of adjustment coefficients of Δxit.10. Error-Correction and ADL TestsDyt = a1(yt-1 - bzt-1) + e1t Dzt = a2(yt-1 - bzt-1) + e2twhere: e1t = e2t + vtAs such, we can always write Dyt = a(yt-1 – bzt-1) + Dzt + vt (6.67)The general problem is that Dzt will be correlated with the error term vt so that there is a simultaneity problem. However, if zt is weakly exogenous and causally prior to yet we can estimate (6.67)The ADL TestDyt = α1yt-1 - α1βzt-1 + ρDzt + vtTable F uses the work of Ericsson and MacKinnon (2002) to calculate the appropriate critical values necessary to determine whether β1 < 0. Given that the variables are cointegrated:If Dzt is unaffected by innovations in Dyt, it is appropriate to conduct inference on (6.69) using a standard t-tests and F-tests.

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