Booth School of Business, University of Chicago Business 41914, Spring Quarter 2017, Mr. Ruey S. Tsay Midterm Chicago Booth Honor Code: I pledge my honor that I have not violated the Honor Code during this examination. Signature: Name: UC. ID: Notes: Open notes and books. The exam has seventeen (17) pages with questions in 8. Write your answer in the blank space provided for each question. Manage your time carefully and answer as many questions as you can. Unless stated otherwise, {a t } is a sequence of iid Gaussian random vectors with mean zero and positive-definite covariance matrix Σ a. This assumption applies to the univariate case too. For simplicity, ALL tests use the 5% significance level. In the exam, Γ i = Cov(z t, z t i ) denotes the lag-i autocovariance matrix of a k-dimensional stationary time series z t. Problem A: (51 points; 3 points per question) Answer briefly the following questions. 1. Consider the transfer function models: (a) y t = 3B 3 B 4 1 2.3B+1.7B 2 0.4B 3 x t + 1 1 0.7B a t, (b) y t = 0.6B+2B2 x 1 0.7B+0.8B 2 t + a t, where x t is an input variable independent of a t. Are the models stable? Why? Compute the steady-state gain of any stable model. 2. Compute the first five (5) impulse response weights of the transfer function y t = 2B+0.5B2 x 1 1.3B+0.4B 2 t + 1 a 1 0.7B t. 1
3. (Questions 3 to 6). Consider the 2-dimensional model z t = φ 1 z t 1 + φ 2 z t 2 + a t, where φ 1 = [ 1.5 0.6 0.3 0.2 ], φ 2 = [ 0.5 0.3 0.7 0, 2 ], Σ a = [ 2 1 1 1 ]. Let y t = (z t, z t 1). Write down a VAR model for y t. 4. Verify that z t is stationary. 5. Suppose z 100 = (1.2, 0.6) and z 99 = (0.5, 0.9). Compute the 1-step and 2-step ahead predictions at the forecast origin t = 100. 6. Compute the covariance matrix of the 2-step ahead forecast error. 7. (Questions 7 and 8) Suppose {z 1,..., z 100 } is a realization from a bivariate VAR(1) model, z t = φ 1 z t 1 + a t. The sample covariance matrices are given below: Γ 0 = [ 1.0 1.0 1.0 2.0 ], Γ 1 = [ 0.6 0.4 0.7 1.2 Using the Yule-Walker equation to obtain the estimate of φ 1. ]. 2
8. Obtain an estimate of Σ a. 9. (Questions 9 to 16). Consider the following k-dimensional linear time series model φ(b)z t = φ 0 + θ(b)a t, where z t = (z 1t,..., z kt ), φ(b) = I φ 1 B φ 2 B 2 and θ(b) = I θ 1 B are matrix polynomials with order 2 and 1. Assume that φ(b) and θ(b) are left coprime and [φ 2, θ 1 ] is of rank k. What is the invertibility condition for z t? 10. Assume that z t is stationary. What is E(z t )? 11. Assume that k = 3, what is the implied univariate ARMA model for z it? 12. Assume stationarity. What are the Yule-Walker equations for z t? 13. Assume that z t is stationary. Let z t µ = ψ(b)a t be the MA representation of z t, where µ = E(z t ). Derive the impulse response function ψ i for i = 1, 2, 3. 14. Assume k = 2. What is the necessary and sufficient condition that z 1t is an input variable and z 2t is a output variable? That is, there is a unidirectional relationship between z it. 3
15. Let w l = z 2l + z 2l 1 for l = 1, 2,.... It is easy to see that w l follows a VARMA(p, q) model. What are the maximum values of p and q? 16. Let w t = z 1t +z 2t and k = 2. It is also easy to see that w t follows a univariate ARMA(p, q) model. What are the maximum values of p and q? 17. Consider the k-dimensional MA(2) model z t = a t θ 1 a t 1 θ 2 a t 2. Show that the component z it also follows an MA(2) model. 4
Problem B. (34 points) Consider the U.S. quarterly gross domestic income (gdi) and consumption of fixed capital from 1947.I to 2016.IV for 280 observations. The data are from FRED and in billions of U.S. dollars, and log transformation was taken. R output of some analysis is attached, where yt denotes the log series. Answer the following questions. 1. (4 points) State the VAR orders selected by the four order-selection statistics (i.e., AIC, BIC, HQ and M(p)). 2. (4 points) Johansen s test is performed to test for co-integration between the two series. Draw your conclusion using the 5% significance level. 3. (4 points) An ECM is fitted. Write down the simplified ECM model. 4. (2 points) Is the fitted ECM model adequate? Why? 5. (4 points) A VAR(3) model is also entertained. Write down the simplified VAR(3) model. 6. (2 points) Compared the VAR(3) and ECM models. Which model do you prefer? Why? 5
7. (4 points) Consider the first-differenced data, dyt. Extended cross-correlation matrices are given. What are the two most parsimonious models specified by the Eccm table? 8. (4 points) A VAR(5) model is used for dyt series. Write down the simplified model. 9. (2 points) Is the VAR(5) model adequate? Why? 10. (4 points) The VAR(5) model for dyt and the ECM model for yt seem to be inconsistent? Why? Which model do you preferred? Why? 6
Problem C. (15 points) Briefly answer the following questions. 1. (4 points) Consider the transfer function model (1 B)z 1t = ɛ 1t θɛ 1,t 1, z 2t = ωz 1t + ɛ 2t, where ɛ 1t and ɛ 2t are independent Guassian white noise series with mean zero and variances σ 2 1 and σ 2 2, respectively. What is the univariate ARIMA model for z 2t? Let z t = (z 1t, z 2t ). Express the model of z t in the form (1 B)z t = (I ΘB)a t. What is Θ? Is the resulting bivariate VARMA model invertible? Why? 2. (6 points) Consider the model z t = z t 2 + a t, where {a t } is a Gaussian white noise series with mean zero and variance σ 2 a. Let {z 1,..., z T } be a realization of the process z t, where, for simplicity, assume thay z 1 = z 0 = 0. [Hint: You may use the results discussed in the lecture notes.] Obtain the limiting distribution of T 2 T t=1 z 2 t 2. 7
Obtain the limiting distribution of T 1 T t=1 z t 2 a t. 3. (5 points) Consider the bivariate time series VARMA(1,1) model z t = [ 0.4 0.8 0.45 0.4 where a t is a white noise series. ] z t 1 + a t [ 0.5 0.2 0.0 0.7 ] a t 1, (2 points) Show that each component z it is unit-root nonstationary. Write the model in an Error-correction form. 8
R output ***** Problem B ****** > gdi <- read.csv("gdi.csv") > cofc <- read.csv("cofc.csv") > gdi <- as.numeric(gdi[,2]) > cofc <- as.numeric(cofc[,2]) > zt <- cbind(gdi,cofc[-281]) > MTSplot(zt) > yt <- log(zt) > MTSplot(yt) > > VARorder(yt) selected order: aic = 13 selected order: bic = 3 selected order: hq = 3 Summary table: p AIC BIC HQ M(p) p-value [1,] 0-5.6600-5.6600-5.6600 0.0000 0.0000 [2,] 1-19.1605-19.1086-19.1397 3564.9048 0.0000 [3,] 2-21.4816-21.3778-21.4400 614.4442 0.0000 [4,] 3-21.5488-21.3930-21.4863 24.8404 0.0001 [5,] 4-21.5359-21.3282-21.4526 4.0477 0.3996 [6,] 5-21.5147-21.2551-21.4105 1.8726 0.7592 [7,] 6-21.5578-21.2463-21.4329 18.1836 0.0011 [8,] 7-21.5557-21.1922-21.4099 6.6420 0.1561 [9,] 8-21.5426-21.1271-21.3759 3.8531 0.4263 [10,] 9-21.5450-21.0776-21.3575 7.6655 0.1046 [11,] 10-21.5819-21.0626-21.3736 16.0828 0.0029 [12,] 11-21.5900-21.0188-21.3609 8.9215 0.0631 [13,] 12-21.6000-20.9769-21.3501 9.3209 0.0536 [14,] 13-21.6068-20.9317-21.3360 8.4660 0.0759 > t1 <- ca.jo(yt,k=3,spec="transitory",ecdet="const") > summary(t1) ###################### # Johansen-Procedure # ###################### Test type: maximal eigenvalue statistic (lambda max), without linear trend and constant in cointegration Eigenvalues (lambda): [1] 9.434457e-02 2.480812e-02-2.461365e-17 Values of teststatistic and critical values of test: 9
test 10pct 5pct 1pct r <= 1 6.96 7.52 9.24 12.97 r = 0 27.45 13.75 15.67 20.20 Eigenvectors, normalised to first column: (These are the cointegration relations) gdi.l1 X.l1 constant gdi.l1 1.0000000 1.0000000 1.000000 X.l1-0.9105301-0.9335164 3.286586 constant -2.6579542-2.3135957-25.970110 Weights W: (This is the loading matrix) gdi.l1 X.l1 constant gdi.d -0.03932501-0.002428561 2.657268e-16 X.d -0.00400367 0.008821827-8.327080e-17 > wt <- yt[,1]-0.911*yt[,2] > m1 <- ECMvar1(yt,3,wt) alpha: gdi [1,] -0.0297 0.00522 [1,] 0.0117 0.00373 AR coefficient matrix AR( 1 )-matrix gdi gdi 0.5537 0.382 0.0884 1.028 [1,] 0.0610 0.147 [2,] 0.0194 0.047 AR( 2 )-matrix gdi gdi 0.0422-0.0625 0.0387-0.1634 [1,] 0.0644 0.1387 [2,] 0.0205 0.0442 ----- 10
Residuals cov-mtx: gdi gdi 8.084826e-05 6.901582e-06 6.901582e-06 8.209761e-06 det(sse) = 6.16113e-10 AIC = -21.13616 BIC = -21.00635 > m1a <- refecmvar1(m1,thres=1.0) Equation: 1 npar = 3 Equation: 2 npar = 5 alpha: [1,] -0.0281 0.00522 [1,] 0.0115 0.00373 AR coefficient matrix AR( 1 )-matrix [1,] 0.5727 0.342 [2,] 0.0884 1.028 [1,] 0.0541 0.0554 [2,] 0.0194 0.0470 AR( 2 )-matrix [1,] 0.0000 0.000 [2,] 0.0387-0.163 [1,] 1.0000 1.0000 [2,] 0.0205 0.0442 ----- Residuals cov-mtx: [1,] 8.101490e-05 6.901582e-06 [2,] 6.901582e-06 8.209761e-06 det(sse) = 6.174812e-10 AIC = -21.14823 BIC = -21.04438 > MTSdiag(m1a) 11
[1] "Covariance matrix:" gdi gdi 8.04e-05 6.78e-06 6.78e-06 8.22e-06 CCM at lag: 0 [1,] 1.000 0.264 [2,] 0.264 1.000 Hit Enter to compute MQ-statistics: Ljung-Box Statistics: m Q(m) df p-value [1,] 1.00 6.38 4.00 0.17 [2,] 2.00 7.90 8.00 0.44 [3,] 3.00 12.64 12.00 0.40 [4,] 4.00 26.44 16.00 0.05 [5,] 5.00 43.56 20.00 0.00 [6,] 6.00 45.16 24.00 0.01 [7,] 7.00 48.65 28.00 0.01... [23,] 23.00 146.00 92.00 0.00 [24,] 24.00 149.02 96.00 0.00 Hit Enter to obtain residual plots: > m2 <- VAR(yt,3) Constant term: Estimates: 0.1101428-0.009768571 Std.Error: 0.03240993 0.01039896 AR coefficient matrix AR( 1 )-matrix [1,] 1.4498 0.514 [2,] 0.0763 2.128 [1,] 0.0626 0.1866 [2,] 0.0201 0.0599 AR( 2 )-matrix [1,] -0.4786-0.825 [2,] -0.0476-1.427 [1,] 0.1051 0.350 [2,] 0.0337 0.112 12
AR( 3 )-matrix [1,] -0.0130 0.349 [2,] -0.0239 0.294 [1,] 0.0639 0.1742 [2,] 0.0205 0.0559 Residuals cov-mtx: [1,] 7.513542e-05 5.679284e-06 [2,] 5.679284e-06 7.735147e-06 det(sse) = 5.489292e-10 AIC = -21.23734 BIC = -21.08156 HQ = -21.17486 > VARchi(yt,3,thres=1.0) Number of targeted parameters: 2 Chi-square test and p-value: 0.9376372 0.6257411 > m2a <- refvar(m2,thres=1) Constant term: Estimates: 0.109214 0 Std.Error: 0.03203079 0 AR coefficient matrix AR( 1 )-matrix [1,] 1.4553 0.52 [2,] 0.0722 2.14 [1,] 0.0563 0.1841 [2,] 0.0196 0.0588 AR( 2 )-matrix [1,] -0.4967-0.831 [2,] -0.0448-1.445 [1,] 0.0558 0.348 [2,] 0.0336 0.111 AR( 3 )-matrix [1,] 0.0000 0.349 13
[2,] -0.0266 0.305 [1,] 0.0000 0.1739 [2,] 0.0203 0.0546 Residuals cov-mtx: [1,] 7.514697e-05 5.676881e-06 [2,] 5.676881e-06 7.760428e-06 det(sse) = 5.509456e-10 AIC = -21.24081 BIC = -21.09802 HQ = -21.18354 > dyt <- diffm(yt) > Eccm(dyt) p-values table of Extended Cross-correlation Matrices: Column: MA order Row : AR order 0 1 2 3 4 5 6 0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1 0.0001 0.0022 0.0001 0.0000 0.0004 0.5482 0.6528 2 0.0042 0.0000 0.0000 0.0004 0.3950 0.6170 0.1572 3 0.0014 0.0027 0.0008 0.0457 0.6579 0.1214 0.2015 4 0.0859 0.0868 0.3882 0.2030 0.3629 0.2087 0.4065 5 0.3178 0.1886 0.7872 0.9856 0.6834 0.5085 0.3424 > m5 <- VAR(dyt,5) Constant term: Estimates: 0.006764087 0.0006067729 Std.Error: 0.00133671 0.0004260413 AR coefficient matrix AR( 1 )-matrix [1,] 0.4442 0.536 [2,] 0.0827 1.139 [1,] 0.0622 0.1997 [2,] 0.0198 0.0636 AR( 2 )-matrix [1,] 0.0192-0.0848 [2,] 0.0251-0.3544 14
[1,] 0.0671 0.2913 [2,] 0.0214 0.0928 AR( 3 )-matrix [1,] -0.0140-0.501 [2,] -0.0105 0.121 [1,] 0.0667 0.3014 [2,] 0.0213 0.0961 AR( 4 )-matrix [1,] -0.132 0.954 [2,] -0.011-0.114 [1,] 0.0668 0.2945 [2,] 0.0213 0.0939 AR( 5 )-matrix [1,] -0.1681-0.5130 [2,] 0.0251 0.0663 [1,] 0.0631 0.179 [2,] 0.0201 0.057 Residuals cov-mtx: [1,] 6.986697e-05 6.779035e-06 [2,] 6.779035e-06 7.097431e-06 det(sse) = 4.499207e-10 AIC = -21.37858 BIC = -21.11828 HQ = -21.27416 > m5a <- refvar(m5,thres=1.0) Constant term: Estimates: 0.006783134 0.0006197275 Std.Error: 0.001321716 0.0004206282 AR coefficient matrix AR( 1 )-matrix 15
[1,] 0.4516 0.497 [2,] 0.0902 1.147 [1,] 0.0565 0.1277 [2,] 0.0181 0.0625 AR( 2 )-matrix [1,] 0 0.000 [2,] 0-0.357 [1,] 0 0.0000 [2,] 0 0.0923 AR( 3 )-matrix [1,] 0-0.563 [2,] 0 0.114 [1,] 0 0.2259 [2,] 0 0.0957 AR( 4 )-matrix [1,] -0.136 0.978 [2,] 0.000-0.111 [1,] 0.0616 0.2847 [2,] 0.0000 0.0935 AR( 5 )-matrix [1,] -0.1684-0.5195 [2,] 0.0202 0.0655 [1,] 0.0627 0.1770 [2,] 0.0184 0.0566 Residuals cov-mtx: [1,] 6.991500e-05 6.807473e-06 [2,] 6.807473e-06 7.143451e-06 det(sse) = 4.530927e-10 AIC = -21.41457 16
BIC = -21.23235 HQ = -21.34147 > MTSdiag(m5a) [1] "Covariance matrix:" gdi gdi 7.02e-05 6.83e-06 6.83e-06 7.17e-06 CCM at lag: 0 [1,] 1.000 0.305 [2,] 0.305 1.000 Simplified matrix: CCM at lag: 1.... CCM at lag: 2.... Hit Enter to compute MQ-statistics: Ljung-Box Statistics: m Q(m) df p-value [1,] 1.00 1.13 4.00 0.89 [2,] 2.00 3.22 8.00 0.92... [9,] 9.00 37.08 36.00 0.42 [10,] 10.00 42.87 40.00 0.35 [11,] 11.00 44.30 44.00 0.46... [21,] 21.00 83.46 84.00 0.50 [22,] 22.00 85.53 88.00 0.55 [23,] 23.00 95.10 92.00 0.39 [24,] 24.00 98.28 96.00 0.42 17