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chapter18_s - CHAPTER 18 SOLUTIONS TO PROBLEMS 18.1 With...

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CHAPTER 18 SOLUTIONS TO PROBLEMS 18.1 With z t 1 and z t 2 now in the model, we should use one lag each as instrumental variables, z t- 1,1 and z t- 1,2 . This gives one overidentifying restriction that can be tested. 18.3 For δ β , y t δ z t = y t β z t + ( β δ ) z t , which is an I(0) sequence ( y t β z t ) plus an I(1) sequence. Since an I(1) sequence has a growing variance, it dominates the I(0) part, and the resulting sum is an I(1) sequence. 18.5 Following the hint, we have y t y t -1 = β x t β x t -1 + β x t -1 y t -1 + u t or Δ y t = β Δ x t – ( y t -1 β x t -1 ) + u t . Next, we plug in Δ x t = γ Δ x t -1 + v t to get Δ y t = β ( γ Δ x t -1 + v t ) – ( y t -1 β x t -1 ) + u t = βγ Δ x t -1 – ( y t -1 β x t -1 ) + u t + β v t γ 1 Δ x t -1 + δ ( y t -1 β x t -1 ) + e t , where γ 1 = βγ , δ = –1, and e t = u t + β v t . 18.7 If unem t follows a stable AR(1) process, then this is the null model used to test for Granger causality: under the null that gM t does not Granger cause unem t , we can write unem t = β 0 + β 1 unem t- 1 + u t E( u t | unem t -1 , gM t -1 , unem t -2 , gM t -2 , ) = 0 and | β 1 | < 1. Now, it is up to us to choose how many lags of gM to add to this equation. The simplest approach is to add gM t -1 and to do a t test. But we could add a second or third lag (and probably not beyond this with annual data), and compute an F test for joint significance of all lags of gM t . 18.9 Let be the forecast error for forecasting y n +1 , and let 1 ˆ n e + 1 ˆ n a + be the forecast error for forecasting Δ y n +1 . By definition, = y n +1 1 ˆ n e + ˆ n f = y n +1 – ( + y n ) = ( y n +1 y n ) = Δ y n +1 = , where the last equality follows by definition of the forecasting error for Δ y n +1 . ˆ n g ˆ n g ˆ n g 1 + ˆ n a 110
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