UNIVERSITI SAINS MALASIA Second Semester Examination 009/00 Academic Session April/May 00 MSG 367 Time Series Analysis [Analisis Siri Masa] Duration : 3 hours [Masa : 3 jam] Please check that this examination paper consists of THIRTEEN pages of printed material before you begin the examination. [Sila pastikan bahawa kertas peperiksaan ini mengandungi TIGA BELAS muka surat yang bercetak sebelum anda memulakan peperiksaan ini.] Instructions: Answer all four [4] questions. [Arahan: Jawab semua empat [4] soalan.] In the event of any discrepancies, the English version shall be used. [Sekiranya terdapat sebarang percanggahan pada soalan peperiksaan, versi Bahasa Inggeris hendaklah diguna pakai]. /-
. (a) Discuss the following with example: [MSG 367] (i) (ii) (iii) Why diagnostic checking procedure is important in the process of building a time series model? What is meant by over-fitting and its use in the diagnostic checking process? How model selection criteria such as AIC and BIC can be used in choosing the best among competing models? [45 marks] (c) Consider a process defined as: t 0 0t Xt where X t is a process defined as below: X t X t t t with X and that Var X. Show that: (i) E X t 0 (ii) t is not stationary. Find the mean, variance and autocovariance function of t t t. Is Z t a stationary process? Briefly explain your reason. [35 marks] Z (d) Rewrite each of the models below using the backward operator B and state the form of ARIMA(p,d,q) or SARIMA(p,d,q)(P,D,Q). [p, d, q, P, D, and Q are positive finite numbers]. (i) t t t t 3 t t t (ii) t t t t (iii) t t t t t t t 3 (iv) t t 0.7 t 0.49 t 4 0.343 t 6 0. 8 t [0 marks]...3/-
3 [MSG 367]. (a) Bincangkan yang berikut dengan contoh: (i) Mengapa prosedur pemeriksaan dianostik adalah penting dalam proses pembentukkan suatu model siri masa? (ii) Apakah yang dimaksudkan dengan terlebih-suai dan kegunaannya dalam proses pemeriksaan dianostik? (iii) Bagaimanakah criteria pemilihan model seperti AIC dan BIC boleh digunakan dalam memilih yang terbaik dikalangan model yang berkemungkinan? [45 markah] (b) Pertimbangkan suatu proses yang dinyatakan sebagai: t 0 0t Xt yang mana X t adalah suatu proses yang diberikan seperti di bawah: X t X t t t dengan X dan juga Var X Tunjukkan bahawa: (i) E X t 0 (ii) t adalah tidak pegun. Cari min, varians dan fungsi autokovarians bagi Z t t t. Adakah Z t merupakan suatu proses yang pegun? Terangkan secara ringkas alasan kamu? [35 markah] (c) Tulis semula setiap model di bawah menggunakan pengoperasi anjak kebelakang B dan nyatakan bentuk ARKPB(p,d,q) atau bermusim ARKPBR(p,d,q)(P,D,Q). [p, d, q, P, D dan Q adalah nombor-nombor positif terhingga] (i) t t t t 3 t t t (ii) t t t t (iii) t t t t t t t 3 (iv) t t 0.7 t 0.49 t 4 0.343 t 6 0. 8 t [0 markah]...4/-
4 [MSG 367]. (a) Given two processes of autoregressive of order one, AR(): A: t t t B: t t t t t with Show that process A is stationary with constant mean, that. E A t such Show that process B is not only non-stationary but having a deterministic mean that increases over time. [30 marks] (b) Given an ARMA(,) process: B B t B t Show that: 0 3 k k k for k A series of 5 observations was collected and an ARMA(,) model has been fitted with the following estimates: ˆ 0.55, ˆ 0. 5 and ˆ 0.0. Calculate the values of autocorrelation, acf for lag k =,, 3, 4, 5, and partial autocorrelation, pacf for lag k = and. What can you say about the calculated values of acf and pacf and its underlying process. [Given the values of acf at lag 6 through to 0 are 0.56, -0.485, 0.440, - 0.95 and 0.350 respectively, and pacf at lag 3 through to 8 are -0.030, - 0.05, -0.08, 0.40, 0.7, 0.06 respectively]. [30 marks] (c ) A newly employed trainee at Company ZZ has been given a time series of length 300. She has been asked to fit a suitable time series model to the data. Appendix A shows the procedures and steps that she has conducted in her analysis. Explain with reason each of the output in Appendix A. [40 marks]...5/-
5 [MSG 367]. (a) Diberi dua proses autoregressive peringkat pertama, AR(): A: t t t B: t t t t t dengan Tunjukkan bahawa proses A adalah pegun dengan min konstan, E t seperti mana. Tunjukkan bahawa proses B bukan sahaja tidak pegun malah mempunyai min tertentu yang mana ianya meningkat mengikut masa. [30 markah] (b) Diberi suatu proses ARPB(,): B B t B t Tunjukkan bahawa: 0 3 k k k untuk k Suatu siri denagn 5 cerapan telah dikumpul dan suatu model ARPB(,) telah disuai dengan anggaran-anggaran berikut: ˆ 0. 55, ˆ 0. 5 dan ˆ 0. 0. Hitung nilai autokorelasi, fak untuk susulan k =,, 3, 4,5 dan 6, dannilai autokorelai separa, faks untuk susulan k = dan. Apa yang boleh kamu katakana tentang nilai terhitung bagi fak dan faks dan juga proses yang diwakilkan. [Diberi bahawa nilai-nilai fak pada susulan 6 hingga 0 masing-masing adalah??,??,??,??,?? dan nilai faks pada susulan 3 hingga 8 masingmasing adalah??,??,??,??,?? dan??] [30 markah] (c) Seorang pelatih yang baru diambil bekerja di Syarikat ZZ telah diberi suatu siri masa dengan panjang 300. Beliau telah disuruh untuk menyuaikan suatu model siri masa terhadap data tersebut. Lampiran A menunjukkan prosedur dan juga langkah yang telah dijalankan oleh pelatih tersebut. Terangkan dengan alasan bagi setiap output di Lampiran A. [40 markah]...6/-
6 [MSG 367] 3. (a) Consider a seasonal AR() process, SAR() as given by: B B4 4 t t Using method of moments, show that the estimates of and 4 are given by: ˆ ˆ ˆ 4 and ˆ ˆ4 4 [5 marks] (b) A non-stationary seasonal time series S t has 50 observations and is believed to follows an invertible SARIMA 0,0,,,0 model given by: S t St St 4 t t (i) Show that t St St has the variance and autocorrelation function given by: 0 k k for k,, k k k for k 0,,, (ii) Table and Table in the Appendix B show the sample acf and sample pacf of S t and St. The mean and standard deviation for the original and differenced series are also given. Discuss the appropriateness of the SARIMA 0,0,,,0 model for the series based on the given sample acf and sample pacf. Calculate the estimate for, and. [50 marks] (c) Consider the following SARIMA model: t t t t Show that the forecast error variance is given by: Var m k n for k r m, k = 0,,, and 0 r. What can you say about the model and its forecast error variance? [5 marks]...7/-
7 [MSG 367] 3. (a) Pertimbangkan suatu proses bermusim AR(), SAR() yang diwakili oleh: B B4 4 t t Menggunakan kaedah momen, tunjukkan bahawa anggaran bagi dan 4 adalah diberikan oleh: ˆ ˆ ˆ 4 and ˆ ˆ4 4 [5 markah] (b) Suatu siri masa bermusim tak pegun S t mempunyai 50 cerapan dan dipercayai mengikuti model bolehsongsang bermusim ARKPB 0,0,,,0 yang diberikan oleh: S t St St 4 t t (i) Tunjukkan bahawa t St St mempunyai varians dan fungsi autokorelasi yang diberikan oleh: k 0 k untuk k,, k k k untuk k 0,,, (ii) Jadual dan Jadual 4 di Lampiran B menunjukkan sampel fak dan sampel faks bagi S t dan St. Min serta sisihan piawai bagi siri asal dan siri yang telah dibezakan juga diberikan. 0,0,,,0 Bincang kesesuaian bagi model bermusim ARKPB untuk siri tersebut berdasarkan sampel fak dan sampel faks yang diberi. Hitung anggaran bagi, dan. [50 markah] (c) Pertimbangkan model bermusim ARKPB yang berikut: t t t t Tunjukkan bahawa varians bagi ralat telahan adalah diberikan oleh: untuk k r Var n m k m, k = 0,,, dan 0 r. Apakah yang boleh anda katakana mengenai model serta varians bagi ralat telahan yang sepadan? [5 markah]...8/-
8 [MSG 367] 4. Consider an ARMA(,) model for a series with non-zero mean: B t B B t (a) Consider a special case with d 0. (i) Show that the MA coefficient is given by: k k. Show that the m-step ahead forecasts made at time t = n is given by: ˆ ˆ N m N m for m and that it can be rewritten as: ˆ for m m m m N m N N (ii) Show that the corresponding variance of forecast error is given by: Var N m m (iii) Finally, show that as m : ˆ N m and Var N m [40 marks] (b) Show that the one-step and two-step ahead forecasts made at t = n are respectively given by: ˆ n n n n, n n n and also show that the m-step-ahead forecast is given by: ˆ n m ˆn m for m 3 ˆ ˆ...9/-
9 [MSG 367] 4. Pertimbangkan suatu model ARPB ARMA(,)bagi suatu siri dengan min bukan kosong: B t B B t (a) Pertimbangkan kes khas dengan d 0. (i) Tunjukkan bahawa koefisien PB adalah diberikan oleh: k k. Tunjukkan bahawan telahan m-langkah kehadapan yang dibuat pada waktu t = n adalah diberikan oleh: ˆ ˆ N m N m untuk m Dan bahawa ia boleh ditulis semula sebagai: ˆ untuk m m m m N m N N (ii) Tunjukkan bahawa varians bagi ralat telahan yang sepadan diberikan oleh: Var N m m (iii) Akhir sekali, tunjukkan bahawa apabila m : ˆ N m dan Var N m [40 markah] (b) Tunjukkan bahawa telahan satu-langkah dan dua-langkah kehadapan yang dibuat pada t = n masing-masing diberikan oleh: ˆ n n n n, n n n Dan juga tunjukkan bahawa telahan m-langkah kehadapan adalah diberikan oleh: ˆ n m ˆn m untuk m 3 ˆ ˆ...0/-
(i) (ii) (iii) 0 [MSG 367] Consider n = 50. If estimated values for the coefficients are ˆ 0.7, ˆ 0. 3, ˆ 0. 73, ˆ 0. 35, ˆ 00, s with 50 6, 50 4 and 49, obtain multi-step (dynamic) forecasts for ˆ 50 m for m =,,, 6. Construct a 95% forecast interval for 5, 5, 53 and 54. Comment on the six forecast values obtained above. At time t = 5 the observed value is found to be 88. Calculate the updated forecast of 5, 56. Compare these new forecasts with those calculated in (i) above and discuss. At time t = 5 and t = 53 the observed value is noted as 97 and 94 respectively. Together with the information in (ii) above, obtain the -step-ahead forecast of ˆ5, ˆ 5 and ˆ53 and its corresponding 95% confidence interval. Compare and what can you say about the multi-step-ahead forecast and -step-ahead forecast for 5, 5, 53 and 54. [60 marks].../-
[MSG 367] (i) (ii) (iii) Pertimbangkan n = 50. Sekiranya nilai teranggar bagi koefisienkoefisien adalah ˆ 0. 7, ˆ 0. 3, ˆ 0. 73, ˆ 0. 35, ˆ 00, s dengan 6, 4 50 50 dan 49, dapatkan nilai telahan banyak-langkah (dinamik) bagi ˆ 50 m untuk m =,,, 6. Bina selang telahan 95% bagi 5, 5, 53 dan 54. Komen terhadap enam nilai telahan yang diperoleh di atas. Pada waktu t = 5 nilai dicerap dijumpai sebagai 88. Hitung nilai telahan kemaskini bagi 5, 56. Bandingkan nilai-nilai telahan ini dengan nilai-nilai yang diperoleh dalam (i) di atas dan bincangkan,. Pada waktu t = 5 dan t = 53 nilai yang dicerap masing-masing dicatatkan sebagai 97 dan 94. Bersama maklumat yang terdapat dalam (ii) di atas, dapatkan nilai telahan -langkah kehadapan bagi ˆ5, ˆ 5 dan ˆ53 dan selang keyakinan 95% yang sepadan. Bandingkan dan apakah yang boleh dikatakan mengenai telahan banyak-langkah kehadapan dan telahan -langkah kehadapan bagi 5, 5, 53 dan 54. [60 markah].../-
ACF Observed value [MSG 367] STEP 700 APPENDIX A Time Series Plot of Variable "X" 600 500 400 300 00 00 0 4 8 6 0 4 8 Time (Daily).0 ACF of Variable "X".5 0.0 -.5 -.0 6 6 6 3 36 Lag Number...3/-
Partial ACF ACF Observed value 3 [MSG 367] 0 TS Plot of First Difference of "X" 0 0-0 -0-30 4 8 6 0 4 8 Time (Daily) Transforms: difference ().0 ACF of st Diff. of "X".5 0.0 -.5 -.0 6 6 6 3 36 Lag Number Transforms: difference ().0 PACF of st Diff. of "X".5 0.0 -.5 -.0 6 6 6 3 36 Lag Number Transforms: difference ()...4/-
STEP Dependent Variable: st Diff. of X Method: Least Squares Included observations: 98 after adjustments 4 [MSG 367] Variable Coefficient Std. Error t-statistic Prob. AR() 0.9644 0.0554 6.86407 0.0000 MA() 0.8407 0.03394 3.997 0.0000 R-squared 0.977806 Mean dependent var -0.64904 Adjusted R-squared 0.97773 S.D. dependent var 7.69079 S.E. of regression.3846 Akaike info criterion 3.039 Log likelihood -460.4844 Schwarz criterion 3.8735 Residuals Analysis of ARMA(,) Residuals Residuals-squared Lag ACF PACF Q-Stat Prob ACF PACF Q-Stat Prob 0.88 0.88 5.04 0.0 0.0 4.60 0.96 0.4 36.70 0.55 0.7 34.3 3-0.070-0.7 38.0 0.000 0.60 0.075 4.97 0.000 4 0.037 0.084 38.63 0.000 0.30 0.040 47.09 0.000 5-0.077-0.073 40.46 0.000 0.05 0.030 50.4 0.000 6 0.06 0.036 40.67 0.000 0.00-0.07 50.4 0.000 9-0.03 0.030 4.76 0.000 0.00 0.08 50.58 0.000-0.078-0.075 45.3 0.000 0.00-0.005 50.80 0.000 8-0.08 0.08 48.49 0.000-0.043-0.08 5.99 0.000 4-0.086-0.078 5.93 0.000 0.3 0.3 87.38 0.000 36 0.036 0.005 56.8 0.008-0.09-0.07 97.36 0.000 ARCH-LM Test: Lag Obs*R-squared 4.40693 Probability 0.00047 ARCH-LM Test: Lag Obs*R-squared 30.765 Probability 0.0039...5/-
Partial ACF ACF 5 [MSG 367].3 ACF of Residuals from ARMA(,).. -.0 -. -. -.3 6 6 6 3 36 Lag Number.3 PACF of Residuals from ARMA(,).. -.0 -. -. -.3 6 6 6 3 36 Lag Number STEP 3a Dependent Variable: st Diff of X Method: Least Squares Included observations: 97 after adjustments Variable Coefficient Std. Error t-statistic Prob. AR().399340 0.06336.8446 0.0000 AR() -0.447636 0.06078-7.38954 0.0000 MA() 0.653330 0.0505.5605 0.0000 Schwarz criterion.9939 Akaike info criterion.95408 STEP 3b Dependent Variable: st Diff of X Method: Least Squares Included observations: 98 after adjustments Variable Coefficient Std. Error t-statistic Prob. AR() 0.939440 0.00006 46.95776 0.0000 MA().0365 0.053847 0.49604 0.0000 MA() 0.375905 0.05363 7.00900 0.0000 Schwarz criterion.99740 Akaike info criterion.9599...6/-
6 [MSG 367] STEP 3c Residuals Analysis of ARMA(,) Residuals Residuals-squared Lag ACF PACF Q-Stat Prob ACF PACF Q-Stat Prob 3-0.039-0.038 0.68 0.4 0.4 59.70 4 0.00 0.0 0.8 0.369 0.7 0.73 8.5 0.000 5-0.003-0.005 0.8 0.668 0.03-0.064 85.34 0.000 6 0.056 0.055.75 0.66 0.03-0.089 85.39 0.000 9 0.045 0.048.8 0.83-0.037-0.036 86.57 0.000-0.047-0.048 3.66 0.93-0.03-0.06 87.6 0.000 4-0.09-0.099.37 0.955 0.094 0.004 5.58 0.000 ARCH-LM Test: Lag Obs*R-squared 30.850 Probability 0.000000 STEP 4a Dependent Variable: st Diff of X Method: ML - ARCH (Marquardt) - Normal distribution Included observations: 97 after adjustments Coefficient Std. Error z-statistic Prob. AR().398897 0.0770 9.35645 0.0000 AR() -0.44557 0.070304-6.337780 0.0000 MA() 0.673700 0.053530.58550 0.0000 Variance Equation C 0.0566 0.0889.33746 0.00 RESID(-)^ 0.367633 0.44 3.0835 0.003 GARCH(-) 0.463 0.35647 3.40685 0.0007 Schwarz criterion.860635 Akaike info criterion.78604 Inverted AR Roots.9.49 Inverted MA Roots -.67 Residuals Analysis of ARMA(,)-GARCH(,) Residuals Residuals-squared Lag ACF PACF Q-Stat Prob ACF PACF Q-Stat Prob 3-0.0-0.00 0.4 0.087 0.083 3.86 4-0.078-0.078.07 0.50 0.046 0.047 4.49 0.034 5-0.09-0.08.7 0.337 0.098 0.3 7.39 0.05 6 0.077 0.073 3.97 0.65-0.04-0.09 7.56 0.056 9 0.07 0.070 5.75 0.45-0.06-0.04 9.37 0.54-0.057-0.064 7.40 0.596-0.037-0.0 0.05 0.346 4-0.007-0.0 3.4 0.893-0.036-0.044 9.54 0.550...7/-
7 [MSG 367] ARCH-LM Test: Lag Obs*R-squared 0.5553 Probability 0.630 STEP 4b Dependent Variable: DIFFX Method: ML - ARCH (Marquardt) - Normal distribution Included observations: 98 after adjustments Coefficient Std. Error z-statistic Prob. AR() 0.950597 0.0538 6.0707 0.0000 MA() 0.80457 0.03886 0.7 0.0000 Variance Equation C 0.3037 0.083.0496 0.075 RESID(-)^ 0.363746 0.09405 3.865304 0.000 GARCH(-) 0.475396 0.373 3.596763 0.0003 Schwarz criterion.979936 Schwarz criterion.979936 Residuals Analysis of ARMA(,)-GARCH(,) Residuals Residuals-squared Lag AC PAC Q-Stat Prob AC PAC Q-Stat Prob 0.34 0.07 5.9-0.065-0.066.4 3-0.065-0.5 7.0 0.000 0.057 0.06.4 0. 4-0.044-0.00 7.79 0.000 0.06 0.055 3.57 0.68 5-0.070-0.036 9.8 0.000 0.056 0.06 4.5 0. 6 0.050 0.08 30.05 0.000-0.007-0.006 4.53 0.339-0.086-0.093 33.05 0.000-0.04 0.003 0.58 0.39 4 0.05 0.000 39.75 0.0 0.033 0.03 8.97 0.647 STEP 5 Dependent Variable: DIFFX Included observations: 97 after adjustments LOG(GARCH) = C(4) + C(5)*ABS(RESID(-)/@SQRT(GARCH(-))) + C(6)*RESID(-)/@SQRT(GARCH(-)) + C(7)*LOG(GARCH(-)) Coefficient Std. Error z-statistic Prob. AR().38488 0.0708 9.49768 0.0000 AR() -0.4386 0.067947-6.37068 0.0000 MA() 0.660503 0.0566.76079 0.0000 Variance Equation C(4) -0.4957 0.33786-3.70563 0.000...8/-
8 [MSG 367] C(5) 0.59069 0.58697 3.7458 0.000 C(6) -0.0590 0.0875-0.7637 0.783 C(7) 0.8045 0.0978 8.8958 0.0000 Schwarz criterion.86954 Mean dependent var -0.685 APPENDIX/LAMPIRAN B Table : Series S t, mean = 7.497, std deviation = 9.447 lag 3 4 5 6 7 8 9 0 ACF 0.436-0.047 0.005-0.044 0.58 0.384 0.3-0.074-0.09-0.067 PACF 0.436-0.93 0.6-0.7 0.457 0.007-0.08-0.048 0.044-0. lag 3 4 5 6 7 8 9 0 ACF 0.398 0.96 0.375-0.085-0.04-0.06 0.45 0.358 0.09-0. PACF 0.85 0.58-0.576 0.066-0.30-0.00-0.034 0.059 0.033-0.06 lag 3 4 5 6 8 30 3 34 ACF -0.05-0.085 0.353 0.839 0.33-0.9-0.07 0.340-0.37-0.09 PACF -0.009 0.084-0.53 0.047 0.096-0.07 0.088-0.0 0.04-0.006 lag 35 36 37 38 40 4 44 46 47 48 ACF 0.308 0.75 0.6-0.43-0.067 0.35-0.5-0.09 0.66 0.670 PACF -0.074 0.07 0.065-0.003 0.03-0.0-0.009-0.005-0.03 0.04 Table : Series S t, mean = 0.758, std deviation =.575 lag 3 4 5 6 7 8 9 0 ACF 0.450-0.09-0.08-0.080-0.05-0.087-0.04 0.0-0.044-0.050 PACF 0.450-0.9 0.78-0.7 0.077-0.37 0.0-0.086-0.06-0.034 lag 3 4 5 6 7 8 9 0 ACF 0.97 0.5 0.8-0.005 0.060-0.07-0.8-0.05 0.007-0.05 PACF 0.49 0.08-0.95 0.9-0.07 0.03-0.067 0.096-0.04-0.065 lag 3 4 5 6 8 30 3 34 ACF -0.60-0.060 0. 0.95-0.078-0.035-0. -0.040-0.044-0.03 PACF -0.084 0.097-0.049-0.9-0.048-0.03-0.047 0.030 0.03 0.038 lag 35 36 37 38 40 4 44 46 47 48 ACF 0. -0.03-0.74-0.04-0.098-0.05-0. 0.05 0.36-0.05 PACF -0.048-0.053-0.058 0.069 0.09-0.0 0.005 0.057-0.00 0.066 - ooo O ooo -...9/-