Kinh tế học - Chapter 7: Applied econometric time series 4rd ed. walter enders

Chapter 7Applied Econometric Time Series 4rd ed.Walter EndersAETS 3rd. edition1LINEAR VERSUS NONLINEAR ADJUSTMENTOn a long automobile trip to a new location, you might take along a road atlas. For most trips, such a linear approximation is extremely useful. Try to envision the nuisance of a nonlinear road atlas. For other types of trips, the linearity assumption is clearly inappropriate. It would be disastrous for NASA to use a flat map of the earth to plan the trajectory of a rocket launch. Similarly, the assumption that economic processes are linear can provide useful approximations to the actual time-paths of economic variables. Nevertheless, policy makers could make a serious error if they ignore the empirical evidence that unemployment increases more sharply than it decreases.The Use of Nonlinear ModelsIt is now generally agreed that linear econometric models do not capture the dynamic relationships present in many economic time-series. The observation that firms are more apt to raise than to lower prices is a key feature of many macroeconomic models. Neftci (1984), Falk(1986), DeLong and Summers (1988), Granger and Lee (1989), and Teräsvirta and Anderson (1992) establish the result that many real variables display non-linear adjustment over the course of the business cycle. In several papers, Enders and Sandler model many terrorist incident series as nonlinear.However, adopting an incorrect non-linear specification may be more problematic than simply ignoring the non-linear structure in the data. It is not surprising, therefore, that non-linear model selection is an important area of current research.The Interest Rate SpreadThere is evidence that interest rate spreads (st) display a nonlinear adjustment pattern.As long as | a2 | > | a1 |, periods when st–1 0 is the smoothness parameter.The ACF Can be Misleading in a Nonlinear ModelCorrelation is only a measure of linear association. Consider: yt = byt-1xt-1 + μtwhere: yt is observable but μt and xt are both white noise.Here, all rk (for k > 0) are zero. E[yt yt-k] = b2E[(yt-1xt-1 + μt)(yt-kxt-k + μt-k)] = b2*0Also, all cross correlations are zero. Consider: E[yt xt-k] = bE[(yt-1xt-1 + μt)xt-k] = b*0 for k ¹ 1 = bVar(x)Eyt-1 = 0However, the optimal non-linear one-step ahead forecast is: βytxt.Also, data generated by an explosive process AR(1) process will have an ACF like that from a stationary AR(1) process. Some Tests for NonlinearityMcLeod–Li (1983) test: Since we are interested in nonlinear relationships in the data, a useful diagnostic tool is to examine the ACF of the squares or cubed values of a series. Let i denote the sample correlation coefficient between squared residuals and use the Ljung–Box statistic to determine whether the squared residuals exhibit serial correlation. Regression Error Specification Test (RESET)STEP 1: Estimate the best-fitting linear model. Let {et} be the residuals from the modelSTEP 2: Select a value of H (usually 3 or 4) and estimate the regression equation:where zt is the vector that contains the variables included in the model estimated in Step 1.Hence, you can reject linearity if the sample value of the F-statistic for the null hypothesis 2 = = H = 0 exceeds the critical value from a standard F-table. The idea is that this regression should have little explanatory power if the model is truly linear.Specific Testing for NonlinearityLagrange Multiplier TestsYou need not estimate the nonlinear modelThey have a specific alternative hypothesisUnfortunately, they detect many types of nonlinearityMethodology—H0: The model has a particular linear form against a specific alternative.Step 1. Estimate the linear portion of the model to get the residuals et (i.e., estimate the model under H0)Step 2. Regress et on f( )/b evaluated at the constrained values of b. Step 3. From the regression in Step 2, it can be shown that: TR2 ~ χ2 with degrees of freedom equal to the number of restrictions. Thus, if the calculated value of TR2 exceeds that in a χ2 table, reject H0.With a small sample, it is standard to use an F-test. Example 1yt = a0 + a1yt-1 + a2yt-1yt-2 + et H0: a2 = 0In this case you can estimate the nonlinear model and perform a t-test on a2. However to illustrate the procedure:Step 1: Estimate the model under H0 to get the estimated residuals; i.e., estimate yt = a0 + a1yt-1 + et Step 2: The partial derivatives of yt w.r.t. parameters are 1, yt-1 and yt-1yt-2. Hence, regress the residuals on a constant, yt-1 and yt-1yt-2 Step 3: Find TR2. This is χ2 with 1 degree of freedom Example 2: Bilinear Modelyt = a0 + a1yt-1 + a2yt-2+ yt-2εt-1 + εtHo: γ = 0Regress yt on a constant yt-1 and yt-2 to obtain etRegress et on a constant, yt-1 , yt-2, and yt-2εt-1THRESHOLD AUTOREGRESSIVE MODELSAs in the equation for the spread, if we include a disturbance term, the basic TAR model isIf we assume that the variances of the two error terms are equal [i.e., var(1t) = var(2t)]where It = 1 if yt–1 > 0 and It = 0 if yt–1  0.The indicator can also be set using DytThe Standard TAR ModelConsiderwhere It = 1 if yt–1 >  and It = 0 if yt–1  .The M-TAR ModelThe momentum threshold autoregressive (M-TAR) model used by Enders and Granger (1998) allows the regime to change according to the first-difference of {yt-1}. Hence, equation is replaced with:It is argued that the M-TAR model is useful for capturing situations in which the degree of autoregressive decay depends on the direction of change in {yt}. Enders and Granger (1998) and Enders and Siklos (2001) show that interest rate adjustments to the term-structure relationship display M-TAR behavior. It is important to note that for the TAR and M-TAR models, if all 1i = 2i the TAR and M-TAR models are equivalent to an AR(p) model. See TAR_figure.prgExtensionsSelecting the Delay ParameterMultiple Regimesband-TAR st = +a1(st–1  m ) + t when st–1 > + c st = st–1 + t when – c  and It = 0 otherwise. Pretesting for a TAR Model However, the F-statistic needs to be bootstrapped. (see Hansen above)TAR Models and Endogenous BreaksThe threshold model is equivalent to a model with a structural break. The only difference is that in a model with structural breaks, time is the threshold variable.Carrasco (2002) shows that the usual tests for structural breaks (i.e., those using dummy variables) have little power if the data are actually generated by a threshold processHowever, a test for a threshold process using yt-d as the threshold variable has power to detect both threshold behavior and structural change. Even if there is a single structural break at time period t, using yt-d as the threshold variable will mimic this type of behavior. As such, she recommends using the threshold model as a general test for parameter instability.Asymmetric Monetary PolicyConsider the linear Taylor Rule it = 0.269 + 0.464t + 0.345yt + 0.810it-1 (1.47) (6.05) (5.16) (21.83) AIC = –27.72 and SBC = –16.85The TAR Taylor RuleSince we do not know the delay factor, we can estimate four threshold regressions with πt-1, πt-2, yt-1 and yt-2 as the threshold variables it = 1.421 + 1.051t + 0.469yt + 0.374it-1 when πt-2  3.527 (3.15) (10.55) (6.22) (5.74)  and   it = –0.456 + 0.232t + 0.302yt + 0.959it-1 when πt-2 τhigh = 1.1185 and I2t = 1 if pHt-1 τhigh, I1t = 1 the intercept is 10929 and when pHt-1 0 is a scale parameter The LSTAR ‘AR’ Coefficient Pretesting for an LSTAR ModelAETS 3rd. edition43For the LSTAR model:Use a third-order Taylor series approximation of  with respect to ht–d evaluated ht–d = 0. Of course, this is identical to evaluating the expansion at  = 0.  / ht–d exp(–ht–d)/[1 + exp(–ht–d)]21/42nd deriv.exp(–ht–d)[1 – exp(–ht–d)]/[1 + exp(–ht–d)]303rd derivexp(–ht–d)[1+exp(–2ht–d) – 4exp(–ht–d)]/[1 + exp(–ht–d)]4–1/8 yt = 0 + 1yt–1 + + pyt–p + (0 + 1yt–1 + + pyt–p )(1ht–d + 3 (ht-d)3) + tBecause ht–d depends only on the value of yt–d, we can write the model in the more compact form:can test for the presence of LSTAR behavior by estimating an auxiliary regression: et = a0 + a1yt–1 + + apyt–p + a11yt–1yt–d + + a1pyt–pyt–d + a21yt–1 + + a2pyt–1 + a31yt–1 + + a3pyt–p + t. (7.21)The ESTAR ModelThe exponential form of the model (ESTAR) uses (7.19), but replaces (7.20) with  = 1 – exp [ (yt–1 – c)2]  > 0.Note that for an ESTAR model Timo’s test has a quadratic term but not a cubic termAETS 3rd. edition45Pretesting for an ESTAR ModelAETS 3rd. edition46 Let  be:  = 1 – exp(–h2t-d) so that ht-d = 1/2(yt–d – c))Teräsvirta’s (1994) PretestThe key insight in Teräsvirta (1994) is that the auxiliary equation for the ESTAR model is nested within that for an LSTAR model. If the ESTAR is appropriate, it should be possible to exclude all of the terms multiplied by the cubic expression from the Taylor series expansin. Hence, the testing procedure follows these steps:STEP 1: Estimate the linear portion of the AR(p) model to determine the order p and to obtain the residuals {et}. STEP 2: Estimate the auxiliary equation (7.21). Test the significance of the entire regression by comparing TR2 to the critical value of 2. If the calculated value of TR2 exceeds the critical value from a 2 table, reject the null hypothesis of linearity and accept the alternative hypothesis of a smooth transition model. (Alternatively, you can perform an F-test). STEP 3: If you accept the alternative hypothesis (i.e., if the model is nonlinear), test the restriction a31 = a32 = = a3n = 0 using an F-test. If you reject the hypothesis a31 = a32 = = a3n = 0, the model has the LSTAR form. If you accept the restriction, conclude that the model has the ESTAR form. Estimation IssuesMany of the numerical methods used to estimate the parameter values have difficulty in simultaneously finding  and c. It is crucial to provide the numerical routine with very good initial guesses. Estimate  using a grid search. Fix  at its smallest possible value and estimate all of the remaining parameters using NLLS. Slightly increase the value of  and reestimate the model. Continue this process until the plausible values of  are exhausted. Use the value of  yielding the best fit. If  is large and convergence to a solution is a problem, it could be easier to estimate a TAR model instead of the LSTAR model. Terasverta (1994) notes that rescaling the expressions in  can aid in finding a numerical solution. For example, with an LSTAR model, standardize by dividing exp[(yt-d  c)] by the standard deviation of the {yt} series. For an ESTAR model, standardize by dividing exp[(yt-d  c)2] by the variance of the {yt} series. In this way, the threshold value c is measured in standardized units so that a reasonable value for the initial guess (e.g., c = 1 standard deviation) can be readily made. Michael, Nobay, and Peel (1997)yt = 0.40yt–1 + [1 – exp(–532.4(yt–1 – 0.038)2] (–yt–1 + 0.59yt–2 + 0.57yt–4 – 0.017)The point estimates imply that when the real rate is near 0.038, there is no tendency for mean reversion since a1 = 0. However, when (yt–1 – 0.038)2 is very large, the speed of adjustment coefficient is quite rapid. Hence, the adjustment of the real exchange rate is consistent with the presence of transaction costs. 11. UNIT ROOTS AND NONLINEARITYyt = It1(yt–1 – ) + (1 – It)2(yt–1 – ) + t (7.30) STEP 1: If you know the value of  (for example  = 0), estimate (7.30). Otherwise, use Chan’s method: select the value of  from the regression containing the smallest value for the sum of squared residuals. STEP 2: If you are unsure as to the nature of the adjustment process, repeat Step 1 using the M-TAR model. STEP 3: Calculate the F-statistic for the null hypothesis 1 = 2 = 0. For the TAR model, compare this sample statistic with the appropriate critical value in Table G. STEP 4: If the alternative hypothesis is accepted (i.e., if there is an attractor), it is possible to test for symmetric versus asymmetric adjustment since the asymptotic joint distribution of 1 and 2 converges to a multivariate normal. Old School versus New School‘Old School’ forecasting techniques, such as exponential smoothing and the Box-Jenkins methodology, do not attempt to explicitly model or to estimate the breaks in the series. Exponential smoothing: place relatively large weights on the most recent values of the series. The Box-Jenkins: first-difference or second difference the series in order to control for the lack of mean reversion. Differencing can be chosen by the autocorrelation function or by some type of Dickey-Fuller test. ‘New School’ forecasters attempt to estimate the number and magnitudes of the breaks. Given that the breaks are well-estimated, it is possible to control for the regime shifts when forecasting. Exponential forecasts place a relatively large weight on the most recent values of the series and quickly captures the mean shift.Panel d estimates the series in first-differences to remove the effects of the level-shifts from the series. Endogenous Structural BreaksEquation (7.34) is a partial break model where the break is assumed to affect only the intercept whereas (7.35) is a pure break model in that all parameters are allowed to change. You can use the Andrews and Ploeberger test (1994) Recall that an endogeneous break model is a threshold model with time as the threshold variable. As such, you can estimate (7.34) or (7.35) by performing a grid search for the best-fitting break date. The test is feasible since the selection of the best fitting regression amounts to a supremum test.With the sample sizes typically used in applied work, it is standard to use Hansen’s (1997) bootstrapping test for a threshold model.Bai and Peron: Multiple BreaksBai and Perron develop a supremum test for the null hypothesis of no structural change (m = 0) versus the alternative hypothesis of m = k breaks.The second method of selecting the number of breaks is to use a sequential test.SupremumEstimate models for every possible combination of breaks (given the trimming and minimum break size) and select the best fitting combination of break dates. The appropriate F-statistic, called the F(k; q) statistic, is nonstandard; the critical values depend on the number of breaks, k, and the number of breaking parameters, q. If the null hypothesis of no breaks is rejected, they select the actual number of breaks using the SBC. For q = 1, 2, and 3, the 95% critical for 1, 2, and 5 breaks are: Sequential MethodBegin with the null hypothesis of no-breaks versus the alternative of a single break. If the null hypothesis of no breaks is rejected, proceed to test the null of a single break versus two breaks, and so forth. The method is sequential in that the test for break l + 1 takes the first l breaks as given. At each stage, the so-called sup F(l+1| l) statistic is calculated as the maximum F-statistic for the null hypothesis of no additional against the alternative of one additional break. For q = 1, 2, and 3, the 95% critical for = 0, 1, 2, and 5 are: Fourier Breaks (see www.time-series.net) A simple modification of the standard autoregressive model is to allow the intercept to be a time-dependent function Although (1) is linear in {yt}, the specification is reasonably general in that d(t) can be a deterministic polynomial expression in time, a p-th order difference equation, a threshold function, or a switching function. UDmaxit also seems reasonable to test the null of no breaks against the alternative of some breaks. If the largest of the F(k; q) statistics for k = 1, 2, 3 exceeds the UDmax statistic reported above, you can conclude that there are some breaks and then go on to select the number using the SBC.The Fourier Approximation Under very weak conditions, the behaviour of almost any function can be exactly represented by a sufficiently long Fourier series: Note that the linear specification emerges as the special case in which all values of dsi and dci are set equal to zero. Thus, instead of positing a specific nonlinear model, the specification problem becomes one of selection the most appropriate frequencies to include. Logistic Breaksyt = 0 + 1yt–1 + + pyt–p +  [0 + 1yt–1 + + pyt–p] + t = [1 + exp(-(t - t*))]-1