Kinh tế học - Chapter 2: Stationary time - Series models

Chapter 2: STATIONARY TIME-SERIESMODELS Applied Econometric Time Series4th EditionCopyright © 2015 John Wiley & Sons, Inc.STOCHASTIC DIFFERENCE EQUATION MODELSSection 1Example of a time-series modelAlthough the money supply is a continuous variable, (2.2) is a discrete difference equation. Since the forcing process {et} is stochastic, the money supply is stochastic; we can call (2.2) a linear stochastic difference equation. If we knew the distribution of {et}, we could calculate the distribution for each element in the {mt} sequence. Since (2.2) shows how the realizations of the {mt} sequence are linked across time, we would be able to calculate the various joint probabilities. Notice that the distribution of the money supply sequence is completely determined by the parameters of the difference equation (2.2) and the distribution of the {et} sequence. Having observed the first t observations in the {mt} sequence, we can make forecasts of mt+1, mt+2, . (2.2)White NoiseE(et) = E(et–1) = = 0E(et)2 = E(et–1) 2 = = s2 [or var(et) = var(et–1) = = s2]E(et et-s) = E(et-j et-j-s) = 0 for all j and s[or cov(et, et-s) = cov(et-j, et-j-s) = 0]A sequence formed in this manner is called a moving average of order q and is denoted by MA(q)2. ARMA MODELSIn the ARMA(p, q) modelyt = a0 + a1yt–1 + + apyt-p + et + b1et–1 + + bqet-qwhere et series are serially uncorrelated “shocks”The particular solution is:Note that all roots must lie outside of the unit circle. If this is the case, we have the MA RepresentationSTATIONARITYSection 3Stationarity Restrictions for an AR(1) ProcessCovariance Stationary SeriesMean is time-invariantVariance is constantAll covariances are constantall autocorrelations are constantExample of a series that are not covariance stationaryyt = a + b timeyt = yt-1 + et (Random Walk)Formal DefinitionA stochastic process having a finite mean and variance is covariance stationary if for all t and t  s,1. E(yt) = E(yt-s) = m 2. E[(yt – m)2] = E[(yt-s – m)2]or [var(yt) = var(yts) = ] 3. E[(yt – m)(yt-s – m)] = E[(yt-j – m)(yt-j-s – m)] = gs or cov(yt, yt-s) = cov(yt-j, yt-j-s) = s  where , and s are all constants.4. STATIONARITY RESTRICTIONSFOR AN ARMA(p, q) MODELStationarity Restrictions for the Autoregressive Coefficientsyt = a0 + a1yt–1 + et with an initial conditionOnly if t is large is this stationary: E[(yt – m)(yt-s – m)] = E{[et + a1et–1 + (a1)2et–2 + ] [et-s + a1et-s–1 + (a1)2et-s–2 + ]} = s2(a1)s[1 + (a1)2 + (a1)4 + ] = s2(a1)s/[1 – (a1)2]Stationarity of an AR(1) ProcessRestrictions for the AR Coefficients We know that the sequence {ci} will eventually solve the difference equation  ci – a1ci–1 – a2ci–2 – – apci−p = 0 (2.21)If the characteristic roots of (2.21) are all inside the unit circle, the {ci} sequence will be convergent. The stability conditions can be stated succinctly:1. The homogeneous solution must be zero. Either the sequence must have started infinitely far in the past or the process must always be in equilibrium (so that the arbitrary constant is zero). 2. The characteristic root a1 must be less than unity in absolute value.A Pure MA Process1. Take the expected value of xt   E(xt) = E(et + b1et–1 + b2et–2 + ) = Eet + b1Eet–1 + b2Eet–2 + = 0 E(xt-s) = E(et-s + b1et-s–1 + b2et-s–2 + ) = 0 Hence, all elements in the {xt} sequence have the same finite mean (m = 0). 2. Form var(xt) as  var(xt) = E[(et + b1et–1 + b2et–2 + )2]   = s2[1 + (b1)2 + (b2)2 + ]  As long as S(bi)2 is finite, it follows that var(xt) is finite. var(xt-s) = E[(et-s + b1et-s–1 + b2et-s–2 + )2] = s2[1 + (b1)2 + (b2)2 + ] Thus, var(xt) = var(xt-s) for all t and t−s.Are all autocovariances finite and time independent?  E[xtxt−s] = E[(et + b1et–1 + b2et–2 + )(et-s + b1et−s–1 + b2et−s–2 + )] = s2(bs + b1bs+1 + b2bs+2 + )Restricting the sum bs + b1bs+1 + b2bs+2 + to be finite means that E(xtxt−s) is finite.5. THE AUTOCORRELATION FUNCTION The Autocorrelation Function of an AR(2) ProcessThe Autocorrelation Function of an MA(1) ProcessThe Autocorrelation Function of an ARMA(1, 1) ProcessThe Autocorrelation Function of an MA(1) Process Consider yt = et + bet–1. Again, multiply yt by each yt-s and take expectations g0 = var(yt) = Eytyt = E[(et + bet–1)(et + bet–1)] = (1 + b2)s2g1 = cov(ytyt–1) = Eytyt–1 = E[(et + bet–1)(et–1 + bet–2)] = bs2 andgs = Eytyt−s = E[(et + bet–1)(et−s + bet−s–1)] = 0 for all s > 1 The ACF of an ARMA(1, 1) Process: Let yt = a1yt–1 + et + b1et–1.  Eytyt = a1Eyt–1yt + Eetyt + b1Eet–1yt  g0 = a1g1 + s2 + b1(a1+b1)s2Eytyt–1 = a1Eyt–1yt–1 + Eetyt–1 + b1Eet–1yt–1  g1 = a1g0 + b1s2Eytyt–2 = a1Eyt–1yt–2 + Eetyt–2 + b1Eet–1yt–2  g2 = a1g1 . Eytyt−s = a1Eyt–1yt−s + Eetyt−s + b1Eet–1yt−s  gs = a1gs–1 Eytyt = a1Eyt–1yt + a2Eyt–2yt + EetytEytyt–1 = a1Eyt–1yt–1 + a2Eyt–2yt–1 + Eetyt–1Eytyt–2 = a1Eyt–1yt–2 + a2Eyt–2yt–2 + Eetyt–2 . .Eytyt−s = a1Eyt–1yt−s + a2Eyt–2yt−s + Eetyt−sACF of an AR(2) ProcessSo that g0 = a1g1 + a2g2 + s2 g1 = a1g0 + a2g1  r1= a1/(1 - a2) gs = a1gs–1 + a2gs–2  ri = a1ri-1 + a2ri-26. THE PARTIAL AUTOCORRELATION FUNCTION PACF of an AR Processyt = a + a1yt-1 + et yt = a + a1yt-1 + a2yt-2+ et yt = a + a1yt-1 + a2yt-2 + a3yt-3 + etThe successive estimates of the ai are the partial autocorrelationsPACF of a MA(1)yt = t + 1t-1 but t-1 = yt-1 - 1t-2 yt = t + 1[ yt-1 - 1t-2 ]  = t + 1yt-1 – (1)2 t-2 yt = t + 1yt-1 – (1)2[yt-2 - 1t-3 ] It looks like an MA()or Using Lag Operatorsyt = t + 1t1 = (1 + 1L)tyt /(1 + 1L) = tRecall yt/(1  a1L) = yt + a1yt1 + a12yt2 + a13yt3 + so that 1 plays the role of a1 yt /(1 + 1L)t = yt /[1  (1)L]t = yt  1yt1 + 12yt2  a13yt3 + = toryt =1yt1  12yt2 +  13yt3 + = tSummary: Autocorrelations and Partial AutocorrelationsACFAR(1)geometric decayMA(q)cuts off at lag qPACFAR(p)Cuts off at lag pMA(1)Geometric DecayFor stationary processes, the key points to note are the following:1. The ACF of an ARMA(p,q) process will begin to decay after lag q. After lag q, the coefficients of the ACF (i.e., the ri) will satisfy the difference equation (ri = a1ri–1 + a2ri–2 + + apri-p). 2. The PACF of an ARMA(p,q) process will begin to decay after lag p. After lag p, the coefficients of the PACF (i.e., the fss) will mimic the ACF coefficients from the model yt /(1 + b1L + b2L2 + + bqLq). TABLE 2.1: Properties of the ACF and PACFTesting the significance of iUnder the null i = 0, the sample distribution of is:approximately normal (but bounded at -1.0 and +1.0) when T is large distributed as a students-t when T is small.  The standard formula for computing the appropriate t value to test significance of a correlation coefficient is: with df = T - 2SD(r) = [ ( 1 – r2) / (T – 2) ]1/2In reasonably large samples, the test for the null that i = 0 is simplified to T1/2. Alternatively, the standard deviation of the correlation coefficient is (1/T)0.5. Significance LevelsA single autocorrelationst.dev(r) = [ ( 1 – r2) / (T – 2) ] ½For small r and large T, st.dev( r ) is approx. (1/T)1/2If the autocorrelation exceeds | 2/T1/2 | we can reject the null that r = 0.A group of k autocorrelations:Is a Chi-square with degrees of freedom = k7. SAMPLE AUTOCORRELATIONSOF STATIONARY SERIESModel Selection CriteriaEstimation of an AR(1) ModelEstimation of an ARMA(1, 1) ModelEstimation of an AR(2) ModelSample AutocorrelationsForm the sample autocorrelationsTest groups of correlationsIf the sample value of Q exceeds the critical value of c2 with s derees of freedom, then at least one value of rk is statistically different from zero at the specified significance level. The Box–Pierce and Ljung–Box Q-statistics also serve as a check to see if the residuals from an estimated ARMA(p,q) model behave as a white-noise process. However, the degrees of freedom are reduced by the number of estimated parametersModel Selection AIC = T ln(sum of squared residuals) + 2nSBC = T ln(sum of squared residuals) + n ln(T) where n = number of parameters estimated (p + q + possible constant term) T = number of usable observations. ALTERNATIVEAIC* = –2ln(L)/T + 2n/TSBC* = –2ln(L)/T + n ln(T)/T where n and T are as defined above, and L =maximized value of the log of the likelihood function.For a normal distribution, –2ln(L) = Tln(2π) +Tln(σ2) + (1/σ2) (sum of squared residuals)Model 1yt = a1yt-1 + etModel 2yt = a1yt-1 + et + b12et-12Degrees of Freedom9897Sum of Squared Residuals 85.10 85.07Estimated a1(standard error)0.7904(0.0624) 0.7938(0.0643) Estimated b(standard error)  ‑0.0250 (0.1141) AIC / SBCAIC = 441.9 ; SBC = 444.5AIC = 443.9 ; SBC = 449.1Ljung‑Box Q-statistics for the residuals(significance level in parentheses)Q(8) = 6.43 (0.490) Q(16) = 15.86 (0.391) Q(24) = 21.74 (0.536) Q(8) = 6.48 (0.485) Q(16) = 15.75 (0.400) Q(24) = 21.56. (0.547) Table 2.2: Estimates of an AR(1) ModelTable 2.3: Estimates of an ARMA(1,1) ModelEstimatesQ-StatisticsAIC / SBCModel 1a1: -0.835 (.053)Q(8) = 26.19 (.000)Q(24) = 41.10 (.001)AIC = 496.5SBC = 499.0Model 2a1: -0.679 (.076)b1: -0.676 (.081)Q(8) = 3.86 (.695)Q(24) = 14.23 (.892) AIC = 471.0SBC = 476.2Model 3a1: -1.16 (.093)a2: -0.378 (.092) Q(8) = 11.44 (.057)Q(24) = 22.59 (.424)AIC = 482.8SBC = 487.9ACF of Nonstationary Series8. BOX–JENKINS MODEL SELECTIONParsimonyStationarity and InvertibilityGoodness of FitPost-Estimation EvaluationBox Jenkins Model SelectionParsimonyExtra AR coefficients reduce degrees of freedom by 2Similar processes can be approximated by very different modelsCommon Factor Problemyt = t and yt = 0.5 yt-1 + t - 0.5t-1Hence: All t-stats should exceed 2.0Model should have a good fit as measured by AIC or BIC (SBC)Box-Jenkins IIStationarity and Invertibilityt-stats, ACF, Q-stats, all assume that the process is stationaryBe suspicious of implied roots near the unit circleInvertibility implies the model has a finite AR representation.No unit root in MA part of the modelDiagnostic CheckingPlot residuals—look for outliers, periods of poor fitResiduals should be serially uncorrrelatedExamine ACF and PACF of residuals Overfit the modelDivide sample into subperiodsF = (ssr – ssr1 – ssr2)/(p+q+1) / (ssr1 + ssr2)/(T-2p-2q-2)Residuals PlotWhat can we learn by plotting the residuals?What if there is a systematic pattern in the residuals?Requirements for Box-JenkinsSuccessful in practice, especially short term forecasts Good forecasts generally require at least 50 observationsmore with seasonalityMost useful for short-term forecastsYou need to ‘detrend’ the data.DisadvantagesNeed to rely on individual judgmentHowever, very different models can provide nearly identical forecasts9. PROPERTIES OF FORECASTSHigher-Order ModelsForecast EvaluationThe Granger–Newbold TestThe Diebold–Mariano TestForecasting with ARMA ModelsThe MA(1) Model yt = 0 + 1t-1 + t Updating 1 period: yt+1 = 0 + 1t + t+1 Hence, the optimal 1-step ahead forecast is: Etyt+1 = 0 + 1tNote: Etyt+j is a short-hand way to write the conditional expectation of yt+j The 2-step ahead forecast is:Etyt+2 = Et[ 0 + 1t+1 + t+2 ] = 0 Similarly, the n-step ahead forecasts are all 0Forecast errors The 1-step ahead forecast error is: yt+1 - Etyt+1 = 0 + 1t + t+1 - 0 - 1t = t+1 Hence, the 1-step ahead forecast error is the "unforecastable" portion of yt+1 The 2-step ahead forecast error is: yt+2 - Etyt+2 = 0 + 1t+1 + t+2 - 0 = 1t+1 + t+2Forecast error variance The variance of the 1-step ahead forecast error is: var(t+1) = 2 The variance of the 2-step ahead forecast error is: var(1t+1 + t+2) = (1 + 12)2Confidence intervalsThe 95% confidence interval for the 1-step ahead forecast is: 0 + 1t  1.96The 95% confidence interval for the 2-step ahead forecast is: 0  1.96(1 + 12)1/2In the general case of an MA(q), the confidence intervals increase up to lag q.The AR(1) Model: yt = a0 + a1yt-1 + εt. Updating 1 period, yt+1 = a0 + a1yt + εt+1, so that Etyt+1 = a0 + a1yt [ * ] The 2-step ahead forecast is: Etyt+2 = a0 + a1Etyt+1 and using [ * ] Etyt+2 = a0 + a0a1 + a12ytIt should not take too much effort to convince yourself that: Etyt+3 = a0 + a0a1 + a0a12 + a13ytand in general: Etyt+j = a0[ 1 + a1 + a12 + ... + a1j-1 ] + a1jyt If we take the limit of Etyt+j we find that Etyt+j = a0/(1 - a1). This result is really quite general; for any stationary ARMA model, the conditional forecast of yt+j converges to the unconditional mean.Forecast errorsThe 1-step ahead forecast error is: yt+1 – Etyt+1 = a0 + a1yt + εt+1 - a0 - a1yt = t+1The 2-step ahead forecast error is: yt+2 - Etyt+2. Since yt+2 = a0 + a1a0 + a12yt + εt+2 + a1εt+1 and Etyt+2 = a0 + a1a0 + a12yt , it follows that: yt+2 - Etyt+2 = εt+2 + a1εt+1 Continuing is this fashion, the j-step ahead forecast error is : yt+j - Etyt+j = εt+j + a1εt+j-1 + a12εt+j-2 + a13εt+j-3 + ... + a1j-1εt+1 Forecast error variance: The j-step ahead forecast error variance is: σ2[ 1 + a12 + a14 + a16 + ... + a12(j-1) ]The variance of the forecast error is an increasing function of j. As such, you can have more confidence in short-term forecasts than in long-term forecasts. In the limit the forecast error variance converges to σ2/(1-a12); hence, the forecast error variance converges to the unconditional variance of the {yt} sequence. Confidence intervalsThe 95% confidence interval for the 1-step ahead forecast is: a0 + a1yt  1.96σ Thus, the 95% confidence interval for the 2-step ahead forecast is: a0(1+a1) + a12yt  1.96σ(1+a12)1/2.Forecast EvaluationOut-of-sample Forecasts:1. Hold back a portion of the observations from the estimation process and estimate the alternative models over the shortened span of data.2. Use these estimates to forecast the observations of the holdback period. 3. Compare the properties of the forecast errors from the two models. Example: 1. If {yt} contains a total of 150 observations, use the first 100 observations to estimate an AR(1) and an MA(1) and use each to forecast the value of y101. Construct the forecast error obtained from the AR(1) and from the MA(1). 2. Reestimate an AR(1) and an MA(1) model using the first 101 observations and construct two more forecast errors. 3. Continue this process so as to obtain two series of one-step ahead forecast errors, each containing 50 observations. A regression based method to assess the forecasts is to use the 50 forecasts from the AR(1) to estimate an equation of the form y100+t = a0 + a1f1t + v1t If the forecasts are unbiased, an F-test should allow you to impose the restriction a0 = 0 and a1 = 1. Repeat the process with the forecasts from the MA(1). In particular, use the 50 forecasts from the MA(1) to estimate y100+t = b0 + b1f2t + v2t t = 1, , 50If the significance levels from the two F-tests are similar, you might select the model with the smallest residual variance; that is, select the AR(1) if var(v1t) < var(v2t).  Instead of using a regression-based approach, many researchers would select the model with the smallest mean square prediction error (MSPE). If there are H observations in the holdback periods, the MSPE for the AR(1) can be calculated asThe Diebold–Mariano TestLet the loss from a forecast error in period i be denoted by g(ei). In the typical case of mean-squared errors, the loss is et2We can write the differential loss in period i from using model 1 versus model 2 as di = g(e1i) – g(e2i). The mean loss can be obtained as  10. A MODEL OF THE INTEREST RATE SPREADOut-of-Sample Forecasts AR(7) AR(6) AR(2) p = 1, 2, and 7ARMA(1, 1)ARMA(2, 1)p = 2ma = (1, 7)µy1.201.201.191.191.191.191.20 (6.57)(7.55)(6.02)(6.80)(6.16)(5.56)(5.74)a11.111.091.051.040.760.430.36 (15.76)(15.54)(15.25)(14.83)(14.69)(2.78)(3.15)a2-0.45-0.43-0.22-0.20 0.310.38 (-4.33)(-4.11)(-3.18)(-2.80) (2.19)(3.52)a30.400.36      (3.68)(3.39)     a4-0.30-0.25      (-2.70)(-2.30)     a50.220.16      (2.02)(1.53)     a6-0.30-0.15      (-2.86)(-2.11)     a70.14  -0.03    (1.93)  (-0.77)   1    0.380.690.77     (5.23)(5.65)(9.62)7      -0.14       (-3.27)        SSR43.8644.6848.0247.8746.9345.7643.72AIC791.10792.92799.67801.06794.96791.81784.46SBC817.68816.18809.63814.35804.93805.10801.07        Q(4)0.180.298.998.566.631.180.76Q(8)5.6910.9321.7422.3918.4812.272.60Q(12)13.6716.7529.3729.1624.3819.1411.13 Table 2.4: Estimates of the Interest Rate Spread11. SEASONALITYModels of Seasonal DataSeasonal DifferencingSeasonality in the Box-Jenkins frameworkSeasonal AR coefficientsyt = a1yt-1+a12yt-12 + a13yt-13yt = a1yt-1+a12yt-12 + a1a12yt-13(1 – a1L)(1 – a12L12)ytSeasonal MA CoefficientsSeasonal differencing: Dyt = yt – yt-1 versus D12yt = yt – yt-12 NOTE: You do not difference 12 timesIn RATS you can use: dif(sdiffs=1) y / sdyModel 1: AR(1) with Seasonal MA mt = a0 + a1mt–1 + t + 4t–4 Model 2: Multiplicative Autoregressive mt = a0 + (1 + a1L)(1 + a4L4)mt–1 + et Model 3: Multiplicative Moving Average mt = a0 + (1 + b1L)(1 + b4L4)et Three Models of Money growthModel 1Model 2Model 3a10.541(8.59)0.496(7.66) a4 0.476(7.28) 1  0.453(6.84)40.759(15.11) 0.751(14.87)SSR0.01770.02140.0193AICSBC735.9726.2701.3691.7720.1710.4Q(4)Q(8)Q(12) 1.39 (0.845) 6.34 (0.609)14.34 (0.279) 3.97 (0.410)24.21 (0.002)32.75 (0.001)22.19 (0.000)30.41 (0.000)42.55 (0.000)Table 2.5 Three Models of Money Growth12. PARAMETER INSTABILITY AND STRUCTURAL CHANGETesting for Structural ChangeEndogenous BreaksParameter InstabilityAn Example of a BreakParameter Instability and the CUSUMsBrown, Durbin and Evans (1975) calculate whether the cumulated sum of the forecast errors is statistically different from zero. Define: N = n, , T  1n = date of the first forecast error you constructed, e is the estimated standard deviation of the forecast errors. Example: With 150 total observations (T = 150), if you start the procedure using the first 10 observations (n = 10), 140 forecast errors (T  n) can be created. Note thate is created using all T – n forecast errors. To create CUSUM10, use the first ten observations to create e10(1)/e. Now let N = 11 and create CUSUM11 as [e10(1)+e11(1)]/e. Similarly, CUSUMT-1 = [e10(1)++eT-1(1)]/e. If you use the 5% significance level, the plot value of each value of CUSUMN should be within a band of approximately  0.948 [ (T  n)0.5 + 2(N – n) (T  n)-0.5 ]. COMBINING FORECASTSSection 1313 Combining Forecasts A Simple ExampleTo keep the notation simple, let n = 2. Subtract yt from each side of (2.71) to obtain fct  yt = w1(f1t  yt) + (1  w1)(f2t  yt) Now let e1t and e2t denote the series containing the one-step-ahead forecast errors from models 1 and 2 (i.e., eit = yt  fit) and let ect be the composite forecast error. As such, we can writeect = w1e1t + (1  w1)e2t The variance of the composite forecast error is var(ect) = w12var(e1t) + (1  w1)2var(e2t) + 2w1(1  w1)cov(e1te2t) (2.72)Suppose that the forecast error variances are the same size and that cov(e1te2t) =0. If you take a simple average by setting w1 = 0.5, (2.72) indicates that the variance of the composite forecast is 25% of the variances of either forecast: var(ect) = 0.25var(e1t) = 0.25var(e2t). Optimal Weightsvar(ect) = (w1)2var(e1t) + (1  w1)2var(e2t) + 2w1(1  w1)cov(e1te2t)Select the weight w1 so as to minimize var(ect):In the n-variable case:Bates and Granger (1969), recommend constructing the weights excluding the covariance terms. Alternative methodsConsider the regression equation yt = 0 + 1f1t + 2f2t + + nfnt + vt (2.75)It is also possible to force 0 = 0 and 1 + 2 + + n = 1. Under these conditions, the i’s would have the direct interpretation of optimal weights. Here, an estimated weight may be negative. Some researchers would reestimate the regression without the forecast associated with the most negative coefficient. Granger and Ramanathan recommend the inclusion of an intercept to account for any bias and to leave the i’s unconstrained. As surveyed in Clemen (1989), not all researchers agree with the Granger–Ramanathan recommendation and a substantial amount of work has been conducted so as to obtain optimal weights. The SBCLet SBCi be the SBC from model i and let SBC* be the SBC from the best fitting model. Form i = exp[(SBC*  SBCi)/2] and then construct the weightsSince exp(0) = 1, the model with the best fit has the weight 1/i. Since i is decreasing in the value of SBCi, models with a poor fit with have smaller weights than models with large values of the SBC. Example of the Spread AR(7)AR(6)AR(2)AR(||1,2,7||)ARMA(1,1)ARMA(2,1)ARMA(2,||1,7||)fi2013:10.7750.7750.7090.6870.7290.7250.799I estimated seven different ARMA models of the interest rate spread. The data ends in April 2012 and if I use each of the seven models to make a one-step-ahead forecast for January 2013:Simple averaging of the individual forecasts results in a combined forecast of 0.743. Construct 50 1-step-ahead out-of-sample forecasts for each model so as to obtain  AR(7)AR(6)AR(2)AR(||1,2,7||)ARMA(1,1)ARMA(2,1)ARMA(2,||1,7||)var(eit)0.6350.6180.5830.5870.5820.6000.606wi0.1350.1390.1470.1460.1480.1430.141Next, use the spread (st) to estimate a regression in the form of (5). If you omit the intercept and constrain the weights to unity, you should obtain: st = 0.55f1t – 0.25f2t −2.37f3t + 2.44f4t + 0.84f5t – 0.28f6t + 1.17f7t (6)Although some researchers would include the negative weights in (6), most would eliminate those that are negative. If you successively reestimate the model by eliminating the forecast with the most negative coefficient, you should obtain:st = 0.326f4t + 0.170f5t + 0.504f7tThe composite forecast using the regression method is: 0.326(0.687) + 0.170(0.729) + 0.504(0.799) = 0.751.If you use the values of the SBC as weights, you should obtain: AR(7)AR(6)AR(2)AR(||1,2,7||)ARMA(1,1)ARMA(2,1)ARMA(2,||1,7||)wi0.0000.0000.0110.0010.1120.1030.773The composite forecast using SBC weights is 0.782. In actuality, the spread in 2013:1 turned out to be 0.74 (the actual data contains only two decimal places). Of the four methods, simple averaging and weighting by the forecast error variances did quite well. In this instance, the regression method and constructing the weights using the SBC provided the worst composite forecasts. APPENDIX 2.1: ML ESTIMATION OF A REGRESSIONLet et = yt – bxtML ESTIMATION OF AN MA(1)Now let yt = t–1 + t. The problem is to construct the {t} sequence from the observed values of {yt}. If we knew the true value of  and knew that 0 = 0, we could construct 1, , T recursively. Given that 0 = 0, it follows that: 1 = y1 2 = y2 – 1 = y2 – y1 3 = y3 – 2 = y3 –  (y2 – y1 ) 4 = y4 – 3 = y4 –  [y3 –  (y2 – y1 ) ]In general, t = yt – t–1 so that if L is the lag operator