Suppose we have the following adl11 model yt 1 2 yt

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Unformatted text preview: ab the OLS residuals perform an Augmented Dickey Fuller test on them stationarity means that the null is rejected Note that for cointegration to even be possible, the variables X and Y must be of the same order. 40 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration F Cointegration Yt = β1 + β2 Xt + εt Two variables are cointegrated if they move together. If this is true, then the error term in the regression should be stationary. That is the idea behind the test for cointegration. run the regression with OLS grab the OLS residuals perform an Augmented Dickey Fuller test on them stationarity means that the null is rejected Note that for cointegration to even be possible, the variables X and Y must be of the same order. 41 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Regressions Testing for Nonstationarity Cointegration F Cointegration Yt = β1 + β2 Xt + εt Two variables are cointegrated if they move together. If this is true, then the error term in the regression should be stationary. That is the idea behind the test for cointegration. run the regression with OLS grab the OLS residuals perform an Augmented Dickey Fuller test on them stationarity means that the null is rejected Note that for cointegration to even be possible, the variables X and Y must be of the same order. 42 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Stationary Difference for Nonstationarity Cointegration F Trend Regressions Testing Stationary Trend Stationary For trend stationary, detrending does the job: ˜ Xt ˆ Xt ˆ = Xt − Xt = b1 + b2 t This does not work if X is difference stationary (e.g. a random walk). For the random walk, b2 is likely to be insignificant according to simulation results. 43 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Stationary Difference for Nonstationarity Cointegration F Trend Regressions Testing Stationary Difference Stationary One solution to solve nonstationarity problems is just to take first differences: Yt ∆Y t = β1 + β2 Xt + εt = β2 ∆Xt + ∆εt The problem is that this method only reveals short run dynamics 44 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Stationary Difference for Nonstationarity Cointegration F Trend Regressions Testing Stationary Difference Stationary One solution to solve nonstationarity problems is just to take first differences: Yt ∆Y t = β1 + β2 Xt + εt = β2 ∆Xt + ∆εt The problem is that this method only reveals short run dynamics 45 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Stationary Difference for Nonstationarity Cointegration F Trend Regressions Testing Stationary Difference Stationary We would like to uncover long run relationships as well. Suppose we have the following ADL(1,1) model: Yt = β1 + β2 Yt −1 + β3 Xt + β4 Xt −1 + εt in equilibrium, we have: ¯ Y ¯ ¯ ¯ = β1 + β2 Y + β3 X + β4 X Rearranging we get the following long run relationship: ¯ Y = β3 + β4 ¯ β1 + X 1 − β2 1 − β2 46 / 52 Introduction Stationary Processes Nonstationary Processes Spurious Stationary Difference for Nonstationarity Cointegration F Trend Regressions Testing Stationary Difference Stationary We would like to uncover long run relationships as well. Suppose we have the following ADL(1,1) model: Yt = β1 + β2 Yt −1 + β3 Xt + β4 Xt −1 + εt in equ...
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This document was uploaded on 03/12/2014 for the course ECON 202 at University of London University of London International Programmes (Distance Learning).

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