Dickey-Fuller Test - an Overview - Assignment Example

Question 1: Test for stationarity: Dickey fuller test: This test involves specifying the null and alternative hypothesis; the null hypothesis in this case is that the variable contains a unit root whereas the alternative hypothesis is that the variable is stationary. In this case only one time lag is considered given that the yield spread model considered is stated as follows: St (nm) = rtn - rtm (i) Number of lags 4: Dickey fuller test results: (appendix 1) dfuller r03, trend regress lags (4) T statistics= -3.315 T critical 5% = -3.427 Information criteria: AIC = 574.1521 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.315, the 5% critical value is -3.427, given that the critical value is less than the t statistics value then the null hypothesis that the variable contains a unit root is rejected, therefore given 4 lags of R3 is stationary. Number of lags 3: Dickey fuller test results: (appendix 2) dfuller r03, trend regress lags(3) T statistics= -3.564 T critical 5% = -3.427 Information criteria: AIC = 573.8277 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.564, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 3 lags of R3 is non stationary. Number of lags 2: Dickey fuller test results: (appendix 3) dfuller r03, trend regress lags(2) T statistics= -4.075 T critical 5% = -3.427 Information criteria: AIC = 576.2418 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -4.075, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 2 lags of R3 is non stationary. Number of lags 1: Dickey fuller test results: (appendix 4) dfuller r03, trend regress lags(1) T statistics= -3.811 T critical 5% = -3.427 Information criteria: AIC = 581.509 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.315, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 1 lag of R3 is non stationary. Best model: Information criteria: Number of lags AIC 4 lags- 574.1521 3lags- 573.8277 2 lags- 576.2418 1 lag - 581.509 According to the AIC criteria by Akaike (1974) the model with the lowest AIC value is the best model; from the above table the model with that contains 3 lags of R3 is therefore the best model. (ii) Number of lags 4: Dickey fuller test results: (appendix 5) dfuller r12, trend regress lags(4) T statistics= -3.347 T critical 5% = -3.427 Information criteria: AIC = 539.232 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.347, the 5% critical value is -3.427, given that the critical value is less than the t statistics value then the null hypothesis that the variable contains a unit root is rejected, therefore given 4 lags of R12 is stationary. Number of lags 3: Dickey fuller test results: (appendix 6) dfuller r12, trend regress lags(3) T statistics= -3.445 T critical 5% = -3.427 Information criteria: AIC = 540.192 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.445, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 3 lags of R12 is non stationary. Number of lags 2: Dickey fuller test results: (appendix 7) dfuller r12, trend regress lags(2) T statistics= -3.774 T critical 5% = -3.427 Information criteria: AIC = 540.2859 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.774, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 2 lags of R12 is non stationary. Number of lags 1: Dickey fuller test results: (appendix 8) dfuller r12, trend regress lags(1) T statistics= -3.797 T critical 5% = -3.427 Information criteria: AIC = 539.0715 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -3.797, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 1 lag of R12 is non stationary. Best model: Information criteria: Number of lags AIC 4 lags- 539.232 3lags- 540.192 2 lags- 540.2859 1 lag - 539.0715 According to the AIC criteria by Akaike (1974) the model with the lowest AIC value is the best model; from the above table the model with that contains 1 lags of R12 is therefore the best model. (iii) In order to test for stationarity in this variable the first step is to generate the St (nm) variable as follows: Where n = 12 and m = 3 St (nm) = rtn - rtm St (12,3) = rt12 - rt3 After generating the St (12,3) variable the Dickey Fuller test is undertaken and the following are the results: Number of lags 4: Dickey fuller test results: (appendix 5) dfuller S123, trend regress lags(4) T statistics= -4.352 T critical 5% = -3.427 Information criteria: AIC = 102.1329 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -4.352, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 4 lags of St (12,3) is non stationary. Number of lags 3: Dickey fuller test results: (appendix 6) dfuller S123, trend regress lags(3) T statistics= -4.259 T critical 5% = -3.427 Information criteria: AIC = 100.389 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -4.259, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 3 lags of St (12,3) is non stationary. Number of lags 2: Dickey fuller test results: (appendix 7) dfuller S123, trend regress lags(2) T statistics= -4.793 T critical 5% = -3.427 Information criteria: AIC = 99.8413 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -4.793, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 2 lags of St (12,3) is non stationary. Number of lags 1: Dickey fuller test results: (appendix 8) Dfuller S123, trend regress lags(1) T statistics= -5.125 T critical 5% = -3.427 Information criteria: AIC = 539.0715 From the above results testing the null hypothesis that the variable has a unit root the T statistics value is -5.125, the 5% critical value is -3.427, given that the critical value is greater than the t statistics value then the null hypothesis that the variable contains a unit root is accepted, therefore given 1 lag of St (12,3) is non stationary. Best model: Information criteria: Number of lags AIC 4 lags- 102.1329 3lags- 100.389 2 lags- 99.8413 1 lag - 100.8465 According to the AIC criteria by Akaike (1974) the model with the lowest AIC value is the best model; from the above table the model with that contains 2 lags of St (12,3) is therefore the best model. Question 2: VAR models i. (,) for (n, m) = (6, 3) The first step is to generate the St (nm) variable as follows: Where n = 6 and m = 3 St (nm) = rtn - rtm St (6, 3) = rt6 - rt3 According to Shiller and Campbell (1987) lagged variables of St (nm) will granger cause values of ∆rt3 The following are the estimated coefficients and standard deviation and Akaike Information Criterion, 4 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (6, 3) + b2 St-2 (6, 3) + b3 St -3(6, 3) + b4 St-4 (6, 3) The following are the results: rc03 Coef. Std. Err. t P>t L1. .8610917 .1672577 5.15 0.000 L2. -.262706 .1994481 -1.32 0.189 L3. .1193578 .1984507 0.60 0.548 L4. -.1619977 .1651313 -0.98 0.327 _cons -.0037966 .0304774 -0.12 0.901 AIC = 555.9051 3 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (6, 3) + b2 St-2 (6, 3) + b3 St -3(6, 3) rc03 Coef. Std. Err. t P>t s63 L1. .8479222 .1663897 5.10 0.000 L2. -.2236259 .1992993 -1.12 0.263 L3. -.0102674 .1643259 -0.06 0.950 _cons -.0002271 .0305536 -0.01 0.994 AIC = 559.4457 2 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (6, 3) + b2 St-2 (6, 3) rc03 Coef. Std. Err. t P>t L1. .8293439 .16369 5.07 0.000 L2. -.2233154 .1632623 -1.37 0.172 _cons -.001601 .0304343 -0.05 0.958 AIC = 558.6906 1 lag of St (nm) Estimated model ∆rt3= c + b1 St-1 (6, 3) rc03 Coef. Std. Err. t P>t L1. .6630354 .1063788 6.23 0.000 _cons -.0006129 .030389 -0.02 0.984 AIC = 559.446 ii. (, ) for (n,m)= (12,3) the St (nm) variable is as follows: Where n = 12 and m = 3 St (nm) = rtn - rtm St (12,3) = rt6 - rt3 According to Shiller and Campbell (1987) lagged variables of St (nm) will granger cause values of ∆rt3 The following are the estimated coefficients and standard deviation and Akaike Information Criterion, 4 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (12, 3) + b2 St-2 (12, 3) + b3 St -3(12, 3) + b4 St-4 (12, 3) The following are the results: rc03 Coef. Std. Err. t P>t L1. .4407564 .1109741 3.97 0.000 L2. -.2434574 .1483392 -1.64 0.102 L3. .0135227 .148309 0.09 0.927 L4. .0095088 .1100596 0.09 0.931 _cons -.0087208 .0310009 -0.28 0.779 AIC = 567.6822 3 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (12, 3) + b2 St-2 (12, 3) + b3 St -3(12, 3) rc03 Coef. Std. Err. t P>t L1. .4536051 .1107301 4.10 0.000 L2. -.2367571 .1483585 -1.60 0.111 L3. .0145325 .1099526 0.13 0.895 _cons -.0052312 .0310381 -0.17 0.866 AIC = 570.2997 2 lags of St (nm) Estimated model ∆rt3= c + b1 St-1 (12, 3) + b2 St-2 (12, 3) rc03 Coef. Std. Err. t P>t L1. .4495303 .1100156 4.09 0.000 L2. -.2219743 .1096318 -2.02 0.044 _cons -.0059039 .0309037 -0.19 0.849 AIC = 569.1742 1 lag of St (nm) Estimated model ∆rt3= c + b1 St-1 (12, 3) rc03 Coef. Std. Err. t P>t L1. .2540567 .0508684 4.99 0.000 _cons -.0056542 .030964 -0.18 0.855 AIC = 572.2121 iii. (, ) for (n,m)= (12,6) 4 lags: ∆rt6= c + b1 St-1 (12, 6) + b2 St-2 (12, 6) + b3 St -3(12, 6) + b4 St-4 (12, 6) rc06 Coef. Std. Err. t P>t L1. .6131173 .2076933 2.95 0.003 L2. -.4213587 .2705973 -1.56 0.120 L3. -.3956654 .2699411 -1.47 0.144 L4. .4886279 .2058101 2.37 0.018 _cons -.0147604 .0305285 -0.48 0.629 AIC= 557.4276 3 lags: ∆rt6= c + b1 St-1 (12, 6) + b2 St-2 (12, 6) + b3 St -3(12, 6) rc06 Coef. Std. Err. t P>t L1. .6476953 .2088784 3.10 0.002 L2. -.4172124 .2724892 -1.53 0.127 L3. .0323436 .2081147 0.16 0.877 _cons -.0101589 .0308471 -0.33 0.742 AIC= 566.078 2 lags: ∆rt6= c + b1 St-1 (12, 6) + b2 St-2 (12, 6) rc06 Coef. Std. Err. t P>t s126 L1. .650026 .2077572 3.13 0.002 L2. -.391159 .2069921 -1.89 0.060 _cons -.0103669 .0307099 -0.34 0.736 AIC= 564.8065 1 lag: ∆rt6= c + b1 St-1 (12, 6) rc06 Coef. Std. Err. t P>t L1. .3035827 .0875686 3.47 0.001 _cons -.0091344 .0308122 -0.30 0.767 AIC= 568.7308 Best models: According to Akaike (1974) the model with the lowest AIC value is the best model; from the above results the best models are: ∆rt3= c + b1 St-1 (6, 3) + b2 St-2 (6, 3) + b3 St -3(6, 3) + b4 St-4 (6, 3) and ∆rt6= c + b1 St-1 (12, 6) + b2 St-2 (12, 6) + b3 St -3(12, 6) + b4 St-4 (12, 6) Hypothesis: 1. ∆rt3= c + b1 St-1 (6, 3) + b2 St-2 (6, 3) + b3 St -3(6, 3) + b4 St-4 (6, 3) Lagged values of do not Granger cause current values of . Granger causality Wald tests +------------------------------------------------------------------+ Equation Excluded chi2 df Prob > chi2 --------------------------------------+--------------------------- _ ALL 452.95 4 0.000 +------------------------------------------------------------------+ From the above results it is evident that the p value of the test is less than 0.01 and this means that that lagged values of St (n, m) granger cause current value of ∆rtm 2. ∆rt6= c + b1 St-1 (12, 6) + b2 St-2 (12, 6) + b3 St -3(12, 6) + b4 St-4 (12, 6) Lagged values of do not Granger cause current values of . Granger causality Wald tests +------------------------------------------------------------------+ Equation Excluded chi2 df Prob > chi2 --------------------------------------+--------------------------- _ ALL 1563.9 4 0.000 +------------------------------------------------------------------+ From the above results it is evident that the p value of the test is less than 0.01 and this means that that lagged values of St (n, m) granger cause current value of ∆rtm Question 3: For (n,m)=(6,3) and (12,6), estimate the coefficients and standard errors of the following regression: OLS (n,m)=(6,3) F3.r6_3 Coef. Std. Err. t P>t s63 .5369155 .045004 11.93 0.000 _cons -.0087129 .0128947 -0.68 0.500 null hypothesis H0:=0 constant P value =0. 5, this value is greater than 0.01 and therefore the coefficient is not statistically significant null hypothesis H0:=1 P value =0.000, this value is less than 0.01 and therefore the coefficient is statistically significant (n,m)= (12,6) F3.r12_6 Coef. Std. Err. t P>t s126 .7517247 .0351218 21.40 0.000 _cons -.0032273 .0123946 -0.26 0.795 null hypothesis H0:=0 constant P value =0.795, this value is greater than 0.01 and therefore the coefficient is not statistically significant null hypothesis H0:=1 P value =0.000, this value is less than 0.01 and therefore the coefficient is statistically significant (ii) the Newey-West estimator (n,m)=(6,3) Newey-West F3.r6_3 Coef. Std. Err. t P>t s63 .5369155 .0666543 8.06 0.000 _cons -.0087129 .0177988 -0.49 0.625 null hypothesis H0:=0 constant P value =0.625, this value is greater than 0.01 and therefore the coefficient is not statistically significant null hypothesis H0:=1 P value =0.000, this value is less than 0.01 and therefore the coefficient is statistically significant (n,m)= (12,6) Newey-West F6.r12_6 Coef. Std. Err. t P>t s126 .5603886 .1058352 5.29 0.000 _cons -.0040385 .0285614 -0.14 0.888 Null hypothesis H0:=0 Constant P value =0.888, this value is greater than o.01 and therefore the coefficient is not statistically significant Null hypothesis H0:=1 P value =0.000, this value is less than 0.01 and therefore the coefficient is statistically significant References: Shiller, R.J. and Campbell, J.Y. (1987). Cointegration and tests of present value models. Journal of Political Economy, volume 95, pages 1062 to 1088. Read More