Econometrics for Policy Analysis A Train The Trainer Workshop Oct 22-28, 2016 Organized by African Heritage Institution

Transcription

1 Econometrics for Policy Analysis A Train The Trainer Workshop Oct 22-28, 2016 Organized by African Heritage Institution Delivered by Dr. Nathaniel E. Urama Department of Economics, University of Nigeria, Nsukka

2 Loading Time Series data in E-views: Review For the purpose of this training, we will make use of E-views. The first thing therefore is to load the time series data in E-views. Steps: Save the data in Excel (xlsx or csv comma delimited). Open the E-view software; For Version 8 and above, click on Open Foreign files, select the file, open, next, next, finish. Double click any of the series to cross check. If you see NA in place of data values, your loading was wrong, if you see the data values, you are good to go!! Delivered by Dr. Nathaniel E. Urama Oct 22-28,

3 Loading Time Series data in E-views: Review cont. For Versions below 8, Click on file, new, workfile On the Workfile create Dialogue box below, select the frequency, Enter start date e.g.(1970) oor 1970Q1 and the end date, type the workfile name e.g. (TTT1), ok. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

4 Loading Time Series data in E-views: Review cont. On the workfile TTT1 user interface, click; Proc, Import, Read text lotus excel. Select the file containing the data saved in csv comma delimited, Enter the number of series, ok. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

5 Class Activity 1 1. Load the data set of Nigerian Macro Variables provided. 2. Follow the above steps to load the data set in EViews depending your version. 3. Revision on how to: i. manipulate the data like; logging, differencing and lagging a variable; generating another variable that is a mathematical combination of other variables etc. ii. Plotting graph/ Chats of the Series to visualise there statistical properties. iii. Test the variables for Structural breaks iv. Test the variables for Stationarity or level of Integration. v. Test for stationarity Under structural Breaks. Delivered by Dr. Nathaniel E. Urama vi. Cointegration tests Oct 22-28,

6 Univariate Time Series models AR, MA, ARMA, ARIMA, ARMAX, ARIMAX or ARFIMA? What are these Models? Compare to structural Models a-theoretical Why use them? Serial correlation: presence of correlation between residuals and their lagged values OLS is no longer efficient among linear estimators. Standard errors computed using the textbook OLS formula are not correct, and are generally understated. If there are lagged dependent variables on the right-hand side of the equation specification, OLS estimates are biased and inconsistent. Lack of data date for structural models ( absence, incorrect frequency, immeasurable) Better for Forecasting Oct 22-28, 2016 Delivered by Dr. Nathaniel E. Urama 6

7 AR(p) Model: A review An autoregressive model is one where the current value of a variable depends upon only its previous values and a white noise error term. AR p Y t = α 1 Y t 1 + α 2 Y t α p Y t p + ε t p = j=1 α j Y t j + ε t (1) Using a lag operator L such that L k Y t = Y t k p The AR(p) is given as: Y t = j=1 α j L j Y t + ε t p Y t j=1 p The term: 1 j=1 the AR model. α j L j p Y t = ε t or 1 j=1 α j L j Y t = ε t (2) α j L j is the characteristics polynomial of AR 1 is given as 1 αl Y t = ε t (3) Oct 22-28, 2016 Delivered by Dr. Nathaniel E. Urama 7

8 MA(p) Model : A review An MA(q) model a linear combination of white noise processes, so that yt depends on the current and previous values of a white noise disturbance term. MA q Y t = ε t + θ 1 ε t 1 + θ 2 ε t θ p ε t q q = ε t + j=1 = θ(l)ε t θ j ε t j (4) q For θ L = 1 + j=1 θ j L j (5) q The term: 1 + j=1 θ j L j is the characteristics polynomial of the MA model. 8 where ε t are the independent and identically distributed innovations for the process Delivered by Dr. Nathaniel E. Urama Oct 22-28, 2016

9 MA(p) Model : A review The distinguishing properties of the moving average process of order q given above are: 1. E y t = μ 2. var y t = γ 0 = 1 + θ θ θ q 2 σ 2 3. cov y t = γ s = 1 + θ θ θ 2 q σ 2 for s = 1,2,, q 0 for s > q Oct 22-28, Delivered by Dr. Nathaniel E. Urama

10 ARMA model: A review ARMA p, q is a combination of Ar(p) amd MA(q) as follows: p ARMA p, q Y t = j=1 q α j Y t j + ε t + j=1 θ j ε t j (6) ARMA(1,1) is given as: 1 αl Y t = 1 + θl ε t (7) Seasonal AR and MA Terms: Due to seasonal patterns in most monthly and quarterly data, Box and Jenkins (1976) recommend the use of seasonal autoregressive (SAR) and seasonal moving average (SMA) terms in the ARMA process. SAR p is a seasonal AR term with lag p and it adds to an existing AR, a polynomial with lag p given as 1 p L p : A second order AR process for quarterly data can be written as; 1 α 1 L 1 α 2 L L 4 Delivered by Dr. Nathaniel E. Urama Y t = ε t (8) Oct 22-28,

11 AR, MA and ARMA, : A review (8) on expansion will give: Y t = α 1 Y t 1 + α 2 Y t 2 4 Y t 4 α 1 4 Y t 5 α 2 4 Y t 6 + ε t (9) For seasonal moving average with lag q, the resulting MA lag structure is obtained from the product of the lag polynomial specified by the MA terms and the one specified by any SMA terms. For a second order MA without seasonality, the process is written as : Y t = ε t + θ 1 ε t 1 + θ 2 ε t 2 2 = ε t + j=1 θ j ε t j (10) This in the lag form is given as: Y t = 1 + θ 1 L 1 θ 2 L 2 ε t (11) Delivered by Dr. Nathaniel E. Urama Oct 22-28,

12 AR, MA and ARMA, : A review If the data for (11) is quarterly for example, we introduce the SMA(4) given as 1 + φ 4 L 4 in the MA term. This will give: Y t = 1 + θ 1 L 1 θ 2 L φ 4 L 4 ε t (12) (12) on Expansion will give: Y t = ε t + θ 1 ε t 1 + θ 2 ε t 2 + φ 4 ε t 4 + θ 1 φ 4 ε t 5 + θ 2 φ 4 ε t 6 (13) The parameter φ is associated with the seasonal part of the MA process. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

13 ARIMA and ARIMAX models The AR, MA and ARMA models discussed before assumes that the series in question is at least weakly stationary. Why? (see Gujarati, 2004, pp. 840) Since most time series are not stationary, there is need to account for this in our ARMA model. Hence, the need for ARIMA model. In our previous class, a series that must be differenced d times for it to become stationary is said to integrated of order d i.e. I(d) ARIMA (p,d,q) is an ARMA(p,q) model of non-stationary series Oct 22-28, 2016 differenced d times to make it stationary. Delivered by Dr. Nathaniel E. Urama 13

14 Estimating ARIMA models:the BJ [Box Jenkins] Methods Revisited How do we identify the value of P, d and q for an ARIMA(p, d, q) models? The BJ methodology has an answer and consists of the following steps: Differencing to achieve Stationarity Identification Estimation Diagnostic Checking Forecasting Delivered by Dr. Nathaniel E. Urama Oct 22-28,

15 The BJ [Box Jenkins] Methods Differencing the Series to achieve stationarity Identify the model to be tentatively estimated Estimate the parameters of the tentative model Diagnostic Checking of the adequacy of the model Is the model adequate? No Yes Use the model for Forecasting and control Oct 22-28, 2016 Delivered by Dr. Nathaniel E. Urama 15

16 The BJ [Box Jenkins] Methods cont. 1. Differencing to achieve stationarity: If a series is integrated of order d, it has to be differenced d times to make it stationary. 1. Identification: This involves determining the order of the model required to capture the dynamic features of the data. 2. The chief tools are autocorrelations (AC), the and partial autocorrelations (PAC), and the correlogram of the series. Covariance at lag K AC at lag k is given as ρ k = With γ k = t=1 n k Y t Y n Y t+k Y, variance = γ k γ 0 γ 0 = Y t Y 2 n A plot of ρk against k is known as the sample correlogram. Note if n is small, use n-1 as the divisor. Asymptotically however, both give same result 16 Delivered by Dr. Nathaniel E. Urama Oct 22-28, 2016

17 Finding the value of d in ARIMA (p, d, q) Model The autocorrelation is the correlation coefficient of the current value of the series with the series lagged a certain number of periods. For the purely white noise(stationary) process the autocorrelations bars/values are not significantly different from zero at various lags. For pure RWM, the autocorrelation coefficients at various lags are statistically different from zero. Generating AC and PAC in Eviews: Click Quick, series statistics, correlogram, on the dialogue box that will appear, select level, ok Or Double click the series, View, correlogram, Level, Enter the lag length OK. Delivered by Dr. Nathaniel E. Urama 17 Oct 22-28, 2016

18 Correlogram of a white noise random process Delivered by Dr. Nathaniel E. Urama Oct 22-28,

19 Correlogram of a random walk time series Delivered by Dr. Nathaniel E. Urama Oct 22-28,

20 Finding the value of d in ARIMA (p, d, q) Model cont. If a time series exhibits white noise, the sample autocorrelation coefficients ρ k are approximately ρ k ~N 0, 1 n Hence, following the properties of the standard normal distribution, the 95% confidence interval for any (population) ρ k is: ρ k ± n If the preceding interval includes the value of zero, we do not reject the hypothesis that the true ρk is zero. If this interval does not include 0, we reject the hypothesis that the true ρk is zero. Alternatively, we can use the asymptotic confidence limit for the ρ k which is ± Delivered by Dr. Nathaniel E. Urama Oct 22-28, 2016

21 Finding the value of d in ARIMA (p, d, q) Model cont. If the autocorrelation coefficients ρ k are statistically zero, the value of d=0. Otherwise, it may not be zero. Run the correlogram for the 1 st, 2 nd, until the dth higher order difference of the series where autocorrelation coefficients ρ k becomes statistically zero. Steps in Eviews: Double click the series, View, correlogram, select the order and enter the lag length, ok For series needing more than 2 nd differencing, Quick, series statistics, correlogram, type D(Y, n) as the series name where y is the series name and d is the number of differencing, select level, enter the lag length, ok. Oct 22-28, Delivered by Dr. Nathaniel E. Urama

22 Finding the value of d in ARIMA (p, d, q) Model cont. For series having seasonal terms, click Quick, series statistics, correlogram type d(y, n, s) as the series name where y is the series, with seasonal difference at lag s, and d the number of differencing, level, enter the lag length, ok. d y, n, s = d 1 L 2 (1 L s )y More accurately, conduct the structural break and the stationarity test to determine the order of the integration of the series which is the d. Why Structural Break? The presence of structural breaks in a series may lead to wrong conclusion that the series is not stationary. See the example below for the annual series of OER in our data set. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

23 The BJ [Box Jenkins] Methods cont. Correlogram of OER below show that it is not stationary Delivered by Dr. Nathaniel E. Urama Oct 22-28,

24 The BJ [Box Jenkins] Methods cont. Correlogram of OER below show that it is stationary after the first difference i.e., the D may be 1 Delivered by Dr. Nathaniel E. Urama Oct 22-28,

25 The BJ [Box Jenkins] Methods cont. The ADF unit root test confirm that ORE is I(1) Delivered by Dr. Nathaniel E. Urama Oct 22-28,

26 Finding the value of d in ARIMA (p, d, q) Model cont. The ADF results are confirmed by the KPSS results below. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

27 Finding the value of d in ARIMA (p, d, q) Model cont. A test of the series OER for structural breaks shows the rejection of null of no structural break as shown in the next 2 Slides below. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

28 Finding the value of d in ARIMA (p, d, q) Model cont. A test of the series OER for structural Using the multiple breakpoint test. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

29 Finding the value of d in ARIMA (p, d, q) Model cont. A Look at the series graph reveals the breaks. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

30 Finding the value of d in ARIMA (p, d, q) Model cont. Zivo-Andrews unit root test shows a rejection of the null hypothesis that OER has a unit root with a structural break. This shows that OER may be I(0). Delivered by Dr. Nathaniel E. Urama Oct 22-28,

31 Finding the value of d in ARIMA (p, d, q) Model cont. In E-views 9 and above, one can conduct unit root with structural break test directly. The result as shown below also show that considering the structural breaks, OER may be I(0). Delivered by Dr. Nathaniel E. Urama Oct 22-28,

32 Finding the values of p and q in ARIMA (p, d, q) Model The PAC measures the correlation of Y values that are K periods apart after removing the correlation from the intermediate lags. This means that it measures the added predictive power of the kth lag in a model containing Y t-1 to Y t-k terms as the regressors. The last two columns reported in the correlogram are the Ljung-Box Q-statistics and their p-values. 2 m ρ Ljung-Box (LB) = n(n + 2) k k=1 n k ~χ2 m The Q-statistics test the joint hypothesis that all the ρk up to certain lags are simultaneously equal to zero. if the computed Q exceeds the critical Q value, in the Chi Square distribution, reject the Oct 22-28, null 2016 hypothesis that all the (true) ρks are zero; at least some of them must be nonzero. 32 Delivered by Dr. Nathaniel E. Urama

33 The correlogram of the ARIMA Model The correlogram view compares the autocorrelation pattern of the structural residuals and that of the estimated model The structural residuals are the residuals after removing the effect of the fitted exogenous regressors but not the ARIMA terms For a properly specified model, the residual and theoretical (estimated) autocorrelations and partial autocorrelations should be close. Steps in E-view: From the ARIMA Diagnostic Views dialog box, select correlogram, enter number of lags, graph, OK, to display the graph or, table, OK, to displays the table Delivered by Dr. Nathaniel E. Urama Oct 22-28,

34 Autocorrelation Function (ACF) and Correlogram How does a sample correlogram enable us to find out if a particular series is AR(p) MA(p) or ARMA(p,q)? Delivered by Dr. Nathaniel E. Urama Oct 22-28,

35 BJ Identification Methods: An e.g. AC and PAC of AR(1), MA(1) and ARMA(1, 1) and ARIMA(1, 1,1) for the Annual OER Septs for AR(1) in Eviews: Quick, estimate equation, enter the equation as oer c ar 1, ok. Or d(oer) c ar 1, Ok if the series is I(1). On the equation dialogue box (DB), click on View, ARMA structure. On the ARMA Diagnostic views DB, click Correlogram, graph, ok. 1 Oct 22-28, Delivered by Dr. Nathaniel E. Urama

36 AC and PAC of AR(1) process of Annual OER From E-views 9 Delivered by Dr. Nathaniel E. Urama Oct 22-28,

37 AC and PAC of MA(1) process of OER Delivered by Dr. Nathaniel E. Urama Oct 22-28,

38 AC and PAC of ARMA(1, 1) process of OER From E-views 9 Delivered by Dr. Nathaniel E. Urama Oct 22-28,

39 AC and PAC of ARMA(1, 1, 1) process of Annual OER Delivered by Dr. Nathaniel E. Urama Oct 22-28,

40 AC and PAC of AR(1) process of Quarterly OER From E-views 8 From E-views 9 Delivered by Dr. Nathaniel E. Urama Oct 22-28,

41 AC and PAC of MA(1) process of Quarterly OER Delivered by Dr. Nathaniel E. Urama Oct 22-28,

42 AC and PAC of ARMA(1,1,1) process of Quarterly OER From E-views 8 Delivered by Dr. Nathaniel E. Urama Oct 22-28,

43 ARIMA (p, d, q) Model Selection: Information Criteria Since the actual data ACF and PACF rarely exhibits the theoretical pattern shown, identification will be difficult using them An alternative is to use the information criteria namely: Akaike (AIC), Schwarz (SBIC) and Hannan-Queen (HQIC). AIC = ln σ 2 SBIC = ln σ 2 HQIC = ln σ 2 + 2k T + k ln T T + 2k T ln(ln( T)) Where σ 2 is the residual variance = RSS T The objective is the select the model with the least IC Oct 22-28, Delivered by Dr. Nathaniel E. Urama

44 ARIMA (p, d, q) Selection: Information Criteria Cont. Which criterion should be preferred if they suggest different model order? SBIC embodies much stiffer penalty term than the AIC. SBIC is strongly consistent but inefficient. AIC is not consistent but generally efficient. Hence, SBIC asymptotically gives the correct model order than the AIC. However, the average variation in selected model orders from different samples within a given population is greater with SBIC than AIC Oct 22-28, 2016 In conclusion then, no criterion is absolutely superior to others. 44 Delivered by Dr. Nathaniel E. Urama

45 BJ Identification Methods: An e.g. Cont. Since the ACFs and PACFs didn t show the exact resemblance of of any of the theoretical AR(p), MA(q), nor ARMA(p, d, q), though closer to ARIMA(1, 1, 1) we resort to the Information criteria method. The Information criteria involves estimating different models and comparing their IC values. Tedious!!! testing up to ARMA(p, q) means running (P + 1)X(Q + 1) Models for non seasonal series and(p + 1)X(Q + 1)X4 models for Quarterly series with seasonal pattern. Thanks to E-views for simplifying the case. Steps in Eview 8 and above: Double click the series, Proc. Add-ins, Automatic ARIMA selection, Select the max AR,and MA terms and the IC, OK. For this particular series, the result is as shown below Oct 22-28, This shows ARIMA( 1, 1, 1) with seasonal pattern in the 4 th lag of the moving average Delivered by Dr. Nathaniel E. Urama

46 SBIC selection for ARMA(P,Q) for p and q up to 3 each Model Selection Criterion Model Selection Criterion Model Selection Criterion Model Selection Criterion Model Selection Criterion 1 (0,0)(0,0) (0,3)(0,4) (1,2)(0,0) (2,1)(4,0) (3,0)(4,4) (0,0)(0,0) (0,3)(0,0) (1,2)(0,4) (2,1)(4,0) (3,1)(0,0) (0,0)(0,4) (0,3)(0,0) (1,2)(4,0) (2,1)(4,4) (3,1)(0,0) (0,0)(0,0) (0,3)(0,4) (1,2)(4,0) (2,2)(0,0) (3,1)(0,4) (0,0)(0,0) (0,3)(4,0) (1,2)(4,4) (2,2)(0,0) (3,1)(0,0) (0,0)(0,4) (0,3)(4,0) (1,3)(0,0) (2,2)(0,4) (3,1)(0,0) (0,0)(4,0) (0,3)(4,4) (1,3)(0,0) (2,2)(0,0) (3,1)(0,4) (0,0)(4,0) (1,0)(0,0) (1,3)(0,4) (2,2)(0,0) (3,1)(4,0) (0,0)(4,4) (1,0)(0,0) (1,3)(0,0) (2,2)(0,4) (3,1)(4,0) (0,1)(0,0) (1,0)(0,4) (1,3)(0,0) (2,2)(4,0) (3,1)(4,4) (0,1)(0,0) (1,0)(0,0) (1,3)(0,4) (2,2)(4,0) (3,2)(0,0) (0,1)(0,4) (1,0)(0,0) (1,3)(4,0) (2,2)(4,4) (3,2)(0,0) (0,1)(0,0) (1,0)(0,4) (1,3)(4,0) (2,3)(0,0) (3,2)(0,4) (0,1)(0,0) (1,0)(4,0) (1,3)(4,4) (2,3)(0,0) (3,2)(0,0) (0,1)(0,4) (1,0)(4,0) (2,0)(0,0) (2,3)(0,4) (3,2)(0,0) (0,1)(4,0) (1,0)(4,4) (2,0)(0,0) (2,3)(0,0) (3,2)(0,4) (0,1)(4,0) (1,1)(0,0) (2,0)(0,4) (2,3)(0,0) (3,2)(4,0) (0,1)(4,4) (1,1)(0,0) (2,0)(0,0) (2,3)(0,4) (3,2)(4,0) (0,2)(0,0) (1,1)(0,4) (2,0)(0,0) (2,3)(4,0) (3,2)(4,4) (0,2)(0,0) (1,1)(0,0) (2,0)(0,4) (2,3)(4,0) (3,3)(0,0) (0,2)(0,4) (1,1)(0,0) (2,0)(4,0) (2,3)(4,4) (3,3)(0,0) (0,2)(0,0) (1,1)(0,4) (2,0)(4,0) (3,0)(0,0) (3,3)(0,4) (0,2)(0,0) (1,1)(4,0) (2,0)(4,4) (3,0)(0,0) (3,3)(0,0) (0,2)(0,4) (1,1)(4,0) (2,1)(0,0) (3,0)(0,4) (3,3)(0,0) (0,2)(4,0) (1,1)(4,4) (2,1)(0,0) (3,0)(0,0) (3,3)(0,4) (0,2)(4,0) (1,2)(0,0) (2,1)(0,4) (3,0)(0,0) (3,3)(4,0) (0,2)(4,4) (1,2)(0,0) (2,1)(0,0) (3,0)(0,4) (3,3)(4,0) Delivered by Dr. Nathaniel E. Urama Oct 22-28, (0,3)(0,0) (1,2)(0,4) (2,1)(0,0) (3,0)(4,0) (3,3)(4,4) (0,3)(0,0) (1,2)(0,0) (2,1)(0,4) (3,0)(4,0)

47 Estimating the ARIMA Model Quick Estimate equation Select LS- Least Square(NLS and ARMA) Type D(OER) ar(1) ma(1) Sma(4) Delivered by Dr. Nathaniel E. Urama Oct 22-28,

48 Result of ARIMA(1, 1, 1) SMA(4) Model Delivered by Dr. Nathaniel E. Urama Oct 22-28,

49 AC and PAC for ARIMA(1, 1, 1) SMA(4) Model Delivered by Dr. Nathaniel E. Urama Oct 22-28,

50 Class Activity Determine the values of p, d and q ARIMA (p, d, q) model of the following Nigerian series: CPI, INF, RIR., Ycul, using, 1. Correlagram 2. Information Criteria 7.2 Estimate the ARIMA (p, d, q) model of the above Nigerian series and use the AC and PAC to check the correctness f the Model Order. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

51 ARMA Diagnostic Checking. Over fitting : Deliberately fitting a larger model than that identified to see if all other extra terms would be insignificant. Need for Parsimonious Model Inclusion of irrelevant variables will increase the coefficient standard error and decrease the parameter significance. Whether this happens of not depends on how much the RSS falls and the size of T(number of observations) relative to K(number of Parameters). If T is large, relative to K, the decrease in RSS will outweigh the decrease in Degrees of freedom (T-K) and hence, coefficient standard error will increase. In line with Brooks, Models that are profligate might be inclined to fit the Oct 22-28, data specific features, which would not be replicated out-of-sample. Delivered by Dr. Nathaniel E. Urama

52 ARIMA Diagnostic Check: Overfitting The added ar(2) and ar(3) are all insignificant. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

53 ARMA Diagnostic Checking Cont. Residual diagnostics: Checks the residual for evidence of linear dependence which, if present, suggest that the model is inadequate. The correlogram ( ACF and PACF) If the acf and pacf confirms stationarity of the residual, the model is correctly specified, Steps in E-view: View/Residual Diagnostics/Correlogram-Qstatistic Oct 22-28, 2016 Delivered by Dr. Nathaniel E. Urama 53

54 ARIMA Diagnostic Check: Correlogram Delivered by Dr. Nathaniel E. Urama Oct 22-28,

55 ARMA Diagnostic Checking Cont. Residual diagnostics through the serial Correlation LM test If the LM test fail to reject the null hypothesis of no serial Correlation, the model is correctly specified. Steps in E-view: Ensure that ML is bot the Used ARMA method. In Estimate DB, click option and ensure that ML is not the selected method View/Residual Diagnostics/Serial Correlation LM Test Oct 22-28, Delivered by Dr. Nathaniel E. Urama

56 ARIMA Diagnostic Check: LM Test Delivered by Dr. Nathaniel E. Urama Oct 22-28,

57 Notes on BJ diagnostic tests It essentially involves only autocorrelation tests I can only find under and not over parameterized model(brooks. 2008, P.231) 57 Delivered by Dr. Nathaniel E. Urama Oct 22-28, 2016

58 Further ARIMA Equation Diagnostics The Stationarity and the Invertibility condition The Stationarity Condition Needed to avoid the previous values of the error term having nondeclining effect on the current values of Y t as the time progresses. Given a general ARMA model in the lag polynomial form α L and θ L as α L Y t = θ L ε t The reported roots are the inverse roots of the characteristics polynomials: α x 1 = 0, and θ x 1 = 0 58 Which may be imaginary, but should have modulus not greater than 1 Delivered by Dr. Nathaniel E. Urama Oct 22-28, 2016

59 Further ARIMA Diagnostic Tests Cont. The above is same as the roots of the characteristics equation, 1 α 1 L 1 α 2 L 2 α p L p = 0 all being outside the unit circle. (see Brooks, 2008, p.216) The roots view displays the inverse roots of the AR and/or MA characteristic polynomial, either as a graph or as a table. The graph view plots the inverse roots in the complex plane the horizontal axis is the real part and the vertical axis is the imaginary part of each root. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

60 Further ARIMA Diagnostic Tests Cont. If the estimated ARMA process is (covariance) stationary, all the AR inverse roots should be inside the unit circle. If the estimated ARMA process is invertible, all MA inverse roots should be inside the unit circle. If the AR has a real Inverse roots with r > 1, a pair of complex reciprocal roots with modulus greater than one, it means that the autoregressive process is explosive. If the MA has reciprocal roots outside the unit circle, we say that the MA process is noninvertible, which makes interpreting and using the MA results difficult. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

61 Further ARIMA Diagnostic Tests Cont. If the MA has roots with Modulus close to 1, It is a sign that we have over differenced the series. It makes estimation and forecasting difficult. According to Hamilton (1994a, p. 65) however, noninvertibility is not a serious problem since there is always an equivalent representation of the MA model with the inverse roots inside the unit circle. The Iteration should continue with different staring values until invertibility is achieved. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

62 Further ARIMA Diagnostic Tests Cont. Steps in E-view: On the ARIMA Equation DB, click view, ARMA structure and the ARMA diagnostic views below will appear Select Roots, Graph, ok and the graph in the next slide will appear. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

63 The Inverse Roots Diagnostic test for ARIMA Delivered by Dr. Nathaniel E. Urama 63 Oct 22-28, 2016 All the Inverse roots of both the AR and MA lies inside the unit circle. The AR is stationary and the MA is ivertible

64 The Inverse Roots of AR/MA in a Tabular form Both the AR and the MA has no root outside the unit circle. Hence, AR is stationary and MA is invertible. However, the MA has roots close to 1, a sign that we may have over differenced the Delivered by Dr. Nathaniel E. Urama Oct 22-28, series.

65 The Inverse Roots Diagnostic test for ARMA Reducing the number of differencing by 1 gives us: an ARIMA(1, 0, 1) model or an ARMA(1, 1) model with the result as below: Delivered by Dr. Nathaniel E. Urama Oct 22-28,

66 Inverse roots of the AR/MA Polynomial for the ARMA(1,1) All the Inverse roots of both the AR and MA lies inside the unit circle. Delivered by Dr. Nathaniel E. Urama Oct 22-28,

67 Inverse roots of the AR/MA Polynomial for the ARMA(1,1) All the Inverse roots of both the AR and MA still lies inside the unit circle but the roots of the MA are now far from 1. Oct 22-28, How ever, since the value of the IC criteria are all smaller for the differenced series, we will adopt it. Delivered by Dr. Nathaniel E. Urama

68 Impulse Response of an ARIMA model Traces the response of the ARIMA part of the estimated equation to one time shock in the innovation. The accumulated response is the response to step impulse where the same shock occurs in every period from the first. If the estimated ARMA model is stationary, the impulse responses will asymptote to zero, while the accumulated responses will asymptote to its longrun value. Steps in E-view: From the ARIMA Diagnostic Views dialog box, select Impulse Response, enter number of periods, graph, OK, to display the graph or, table, Oct 22-28, OK, to display the table. Delivered by Dr. Nathaniel E. Urama

69 Impulse Response of our ARIMA model Delivered by Dr. Nathaniel E. Urama Oct 22-28,

70 Class Activity 8 Carry out a diagnostic check on the ARIMA models you estimated during Class activity 7 and interpret them. Conduct an impulse response on all of them. End of day 6 Assignment: Read Brooks section 5.9 and adapt the model there to test if Uncovered Interest rate Parity held in Nigeria between 1980 and Oct 22-28, Delivered by Dr. Nathaniel E. Urama

71 Day 7: Course Details Forecasting with ARIMA and ARIMAX models In-sample vs Out-of Sample Dynamic vs static Forecast Accuracy Decomposition of the Forecast Errors Class Activity 9: Multivariate models VAR, Parsimonious VAR (PVAR), VEC, ARDL Series stationarity and the VAR model Stability and Stationarity in VAR VAR lag length Selection, Impulse Response and Variance Decomposition Forecasting with VAR In-sample vs Out-ofSample Dynamic vs static VAR diagnostic Checking Class activity 10: Delivered by Dr. Nathaniel E. Urama Oct 22-28,