CHAPTER 18 SOLUTIONS TO PROBLEMS18.1With zt1and zt2now in the model, we should use one lag each as instrumental variables, zt-1,1and zt-1,2. This gives one overidentifying restriction that can be tested. 18.3For δ≠β, yt– δzt= yt– βzt+ (β– δ)zt, which is an I(0) sequence (yt– βzt) 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. 18.5Following the hint, we have yt– yt-1= βxt– βxt-1+ βxt-1– yt-1+ utor Δyt= βΔxt– (yt-1– βxt-1) + ut. Next, we plug in Δxt= γΔxt-1+ vtto get Δyt= β(γΔxt-1+ vt) – (yt-1– βxt-1) + ut= βγΔxt-1– (yt-1– βxt-1) + ut+ βvt≡γ1Δxt-1+ δ(yt-1– βxt-1) + et, where γ1= βγ, δ= –1, and et= ut+ βvt. 18.7If unemtfollows a stable AR(1) process, then this is the null model used to test for Granger causality: under the null that gMtdoes not Granger cause unemt, we can write unemt= β0+ β1unemt-1+ utE(ut|unemt-1, gMt-1, unemt-2, gMt-2, …) = 0 and |β1| < 1. Now, it is up to us to choose how many lags of gMto add to this equation. The simplest approach is to add gMt-1and to do a ttest. But we could add a second or third lag (and probably not beyond this with annual data), and compute an Ftest for joint significance of all lags of gMt. 18.9Let be the forecast error for forecasting yn+1, and let 1ˆne+1ˆna+be the forecast error for forecasting Δyn+1. By definition, = yn+1−1ˆne+ˆnf= yn+1– (+ yn) = (yn+1– yn) −= Δyn+1−= , where the last equality follows by definition of the forecasting error for Δyn+1. ˆngˆngˆng1+ˆna110
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