chapter-2时间序列.doc

CHAPTER 2STATIONARY TIME-SERIES MODELS Answers to Questions 1. In the coin-tossing example of Section 1, your winnings on the last four tosses (wt) can be denoted by wt = 1/4et + 1/4et-1 + 1/4et-2 + 1/4et-3A. Find the expected value of your winnings. Find the expected value given that et-3 = et-2 = 1. Answers: Throughout the text, the term et denotes a white-noise disturbance. The properties of the et sequence are such that:i. Eet = Eet-1 = Eet-2 = . = 0, ii. Eetet-i = 0 for i 0, and iii. E(et)2 = E(et-i)2 = . = s2. Hence:Ewt = E(1/4et + 1/4et-1 + 1/4et-2 + 1/4et-3)Since the expectation of a sum is the sum of the expectations, it follows thatEwt = (1/4)(Eet + Eet-1 + Eet-2 + Eet-3) = 0.Given the information et-3 = et-2 = 1, the conditional expectation of wt is Et-2wt = E(wt et-3 = et-2 = 1) = (1/4)(Et-2et + Et-2et-1 + Et-2et-2 + Et-2et-3) so thatEt-2wt = 0.25(0 + 0 + 1 + 1) = 0.5B. Find var(wt). Find var(wt) conditional on et-3 = et-2 = 1.Answers: var(wt) = E(wt)2 - E(wt)2 so thatvar(wt) = E(1/4et + 1/4et-1 + 1/4et-2 + 1/4et-3)2 = (1/16)E(et)2 + 2etet-1 + 2 etet-2 + 2etet-3 + (et-1)2 + 2et-1et-2 + 2et-1et-3+ (et-2)2 + 2et-2et-3 + (et-3)2Page 31: Stationary ModelsSince the expected values of all cross-products are zero, and it follows that:var(wt) = (1/16)4s2 = 0.25s2Given the information et-3 = et-2 = 1, the conditional variance isvar(wtet-3 = et-2 = 1) = Et-2(1/4et + 1/4et-1 + 1/4et-2 + 1/4et-3)2 - (Et-2wt)2 = (1/16)Et-2+ 2etet-1 + 2etet-2 + 2etet-3 + (et-1)2 + 2et-1et-2 + 2et-1et-3 + (et-2)2 + 2et-2et-3 + (et-3)2 - (0.5)2Since Et-2et-2 = Et-2et-3 = 1, it follows that var(wt et-3 = et-2 = 1) = (1/16)(s2 + s2 + 1 + 1 + 2) - 0.25, so thatvar(wt et-3 = et-2 = 1) = (1/8)s2C. Find: i. Cov(wt, wt-1)ii. Cov(wt, wt-2) iii. Cov(wt, wt-5)Answers: Using the same techniques as in Part B:i. Cov(wt, wt-1) = Ewtwt-1-E(wt)E(wt-1) = (1/16)E(et + et-1 + et-2 + et-3)(et-1 + et-2 + et-3 + et-4)= (1/16)E(et-1)2 + (et-2)2 + (et-3)2 + cross-product termsSince the expected values of the cross-product terms are all zeroCov(wt, wt-1) = (1/16)3s2ii. Cov(wt, wt-2) = (1/16)E(et + et-1 + et-2 + et-3)(et-2 + et-3 + et-4 + et-5) = (1/16)E(et-2)2 + (et-3)2 + cross-product termsSince the expected values of the cross-product terms are all zeroCov(wt, wt-2) = (1/16)2s2iii. Cov(wt, wt-5) = (1/16)E(et + et-1 + et-2 + et-3)(et-5 + et-6 + et-7 + et-8) = (1/16)Ecross-product terms. Hence:Cov(wt, wt-5) = 02. Substitute (2.10) into yt = a0 + a1yt-1 + et. Show that the resulting equation is an identity. Answer: For (2.10) to be a solution, it must satisfy:a01 + a1 + a12 + . + a1t-1 + a1ty0 + et + a1et-1 + a12et-2 + . + a1t-1e1 = a0 + a1a01 + a1 + a12 + . + a1t-2 + a1t-1y0 + et-1 + a1et-2 + a12et-3 + .+ a1t-2e1 + etNotice that all terms cancel. Specifically:a01 + a1 + a12 + . + a1t-1 a0 + a1a01 + a1 + a12 + . + a1t-2a1ty0 a1a1t-1y0 and:et + a1et-1 + a12et-2 + . + a1t-1e1 = a1 et-1 + a1et-2 + a12et-3 + . + a1t-2e1 + etA. Find the homogeneous solution to: yt = a0 + a1yt-1 + et.Answer: Attempt a challenge solution of the form yt = Aat. For this solution to solve the homogeneous equation it follows that a = a1 and A can be any arbitrary constant. B. Find the particular solution given that a1 1.Answer: Using lag operators, write the equation as (1 - a1L)yt = a0 + et. Since a0/(1-a1L) = a0/(1-a1) and et/(1-a1L) = et + a1et-1 + a12et-2 + . + a1t-1e1 + a1te0 + a1t+1e-1 + ., it follows that the particular solution isC. Show how to obtain (2.10) by combining the homogeneous and particular solutions.Answer: Combining the homogeneous and particular solutions yields the general solution:so that when t = 0Solve for A and substitute the answer into the general solution to obtain (2.10). 3. Consider the second-order autoregressive process yt = a0 + a2yt-2 + et ,where a2 1. A. Find: i. Et-2ytii. Et-1ytiii. Etyt+2 iv. Cov(yt, yt-1) v. Cov(yt, yt-2)vi. the partial autocorrelations f11 and f22Answers: i) Et-2yt = Et-2(a0 + a2yt-2 + et) = a0 + a2yt-2ii) Et-1yt = Et-1(a0 + a2yt-2 + et) = a0 + a2yt-2Note the Et-1yt = Et-2yt since information obtained in period (t-1) does not help to predict the value of yt.iii) Etyt+2 can be obtained directly from the answer to Part i. Simply update the time index by two periods to obtain: Etyt+2 = a0 + a2ytThe simplest way to answer Parts iv. and v. is to obtain the particular solution for yt. Students should be able to show:yt = a0/(1-a2) + et + a2et-2 + (a2)2et-4 + (a2)3et-6 + (a2)4et-8 + .iv) Cov(yt, yt-1) = E(yt - Eyt)(yt-1 - Eyt-1) = Eet + a2et-2 + (a2)2et-4 + (a2)3et-6 + (a2)4et-8 + .et-1 + a2et-3 + (a2)2et-5 + (a2)3et-7 + .so thatCov(yt, yt-1) = 0v) Cov(yt, yt-2) = Eet + a2et-2 + (a2)2et-4 + (a2)3et-6 + .et-2 + a2et-4 + (a2)2et-6 + (a2)3et-8 + .= a2E(et-2)2 + (a2)2(et-4)2 + (a2)4(et-6)2 + (a2)6(et-8)2 + . Given a2 1, forecasts conditioned on the information in t+1 and t can be made using Et+1yt+s = as-1yt+1 + asyt.E. Find: Eyt, Eyt+1, Var(yt), Var(yt+1), and Cov(yt+1, yt). Answers: The point is to illustrate that the yt sequence is not stationary. Consider:i. Eyt = at-1y1 + aty0. Since at-1 and at are functions of time, the mean is not constant.ii. Eyt+1 = aty1 + at+1y0. Note that Eyt+1 Eyt.iii. var(yt) = 1 + (a1)2 + (a2)2 + . + (at-2)2s2iv. var(yt+1) = 1 + (a1)2 + (a2)2 + . + (at-1)2s2 so that var(yt+1) var(yt ).v. cov(yt+1, yt) = a0a1 + a1a2 + . + at-3at-2s2.7. The file entitled SIM_2.XLS contains the simulated data sets used in this chapter. The first column contains the 100 values of the simulated AR(1) process used in Section 7. This first series is entitled Y1. The following programs will perform the tasks indicated in the text. Due to differences in data handling and rounding, your answers need only approximate those presented in the text.Sample Program for RATS Users all 100;* The first 3 lines read in the data setopen data a:sim_2.xls;* Modify this if your data is not on drive a:data(format=xls,org=obs) cor(partial=pacf,qstats,number=24,span=8) y1;* calculates the ACF, PACF, and Q-statistics graph 1;* plots the simulated series# y1boxjenk(ar=1) y1 / resids;* estimates an AR(1) model and saves the ;*residuals in the series called resids* The next 3 lines compute and display the AIC and SBCcompute aic = %nobs*log(%rss) + 2*%nregcompute sbc = %nobs*log(%rss) + %nreg*log(%nobs)display aic = AIC sbc = sbc* Obtain the ACF, PACF, and Q-statistics of the residuals cor(partial=pacf,qstats,number=24,span=8,dfc=%nreg) resids* Now estimate the model with a MA term at lag 12boxjenk(ar=1,ma=|12|) y1 / residscor(partial=pacf,qstats,number=24,span=8,dfc=%nreg) residscompute aic = %nobs*log(%rss) + 2*%nregcompute sbc = %nobs*log(%rss) + %nreg*log(%nobs)display aic = AIC sbc = sbcboxjenk(ar=2) y1 / resids;* Estimates the AR(2) modelboxjenk(ar=1,ma=1) y1 / resids;* Estimates an ARMA(1, 1) model8. The second column in file entitled SIM_2. XLS contains the 100 values of the simulated ARMA(1, 1) process used in Section 7. This series is entitled Y2. The following programs will perform the tasks indicated in the text. Due to differences in data handling and rounding, your answers need only approximate those reported in the text. Sample Program for RATS Users all 100;* The first 3 lines read in the data setopen data a:sim_2.xls;* Modify this if your data is not on drive a:data(format=xls,org=obs) cor(partial=pacf,qstats,number=24,span=8) y2 ;* calculates the ACF, PACF, and Q-statistics graph 1 ;* plots the simula