CHAPTER 12 TIME SERIES ANALYSIS Example 11 X ar-ar 日r-bn where ti a,ers t-er Brvt-r x EU X t-a Ev X t-A →VX2 The process 3 is said to be an ARIMA. PxIxg Sprocess if t-LS yh is a statio naI ARMA pxg processes Example 12 S.t-LSyt Xeta<t gt is an ARIMAtxx Sprocess, i.e., differencing gt once yields ARMAtx sprocess Remark 3 The slo wly decay ing po sitive sample autocorrelation suggests the appropriate ness of an ARIMA model Remark 4 We need to difference ARIMA. pxlxg S process l-times in order to obtain a stationary process Remark 5 To see whether there is a unit root or not, we perform unit root tests. That we test h da xt in the model X where u is an ARM A process. Under the null, differencing once yield ARMA processCHAPTER 12 TIME SERIES ANALYSIS 9 Example 11 yt = α1yt−1 + et + θet−1, et ∼ iid  0, σ2 √ T  αˆ1 − α1 ˆθ1 − θ1  d→ N (0, V ), where V = # (1 − α 2 1 ) −1 (1 + α1θ1) −1 (1 + α1θ1) −1  1 − θ 2 1 −1 $−1 Ut − α1Ut−1 = et , Vt − θ1Vt−1 = et ⇒ EU 2 t = σ 2 1 − α 2 1 EUtVt = σ 2 1 + α1θ1 EV 2 t = σ 2 1 − θ 2 1 ⇒ V = σ 2 # σ 2 1−α2 1 σ 2 1+α1θ1 σ 2 1+α1θ1 σ 2 1−θ 2 1 $ = # (1 − α 2 1 ) −1 (1 + α1θ1) −1 (1 + α1θ1) −1  1 − θ 2 1 −1 $ The process {yt} is said to be an ARIMA (p, d, q) process if (1 − L) d yt is a stationary ARMA (p, q) processes. Example 12 (1 − α) (1 − L) yt = et (|α| < 1) yt is an ARIMA (1, 1, 0) process, i.e., differencing yt once yields ARMA (1, 0) process. Remark 3 The slowly decaying positive sample autocorrelation suggests the appropriate￾ness of an ARIMA model. Remark 4 We need to difference ARIMA (p, d, q) process d−times in order to obtain a stationary process. Remark 5 To see whether there is a unit root or not, we perform unit root tests. That is, we test H0 : α = 1 in the model yt = αyt−1 + ut , where {ut} is an ARMA process. Under the null, differencing once yield ARMA process