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1、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 =
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 s
3、um 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
4、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)2Since the expected values of all cross-produc
5、ts 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-2
6、et-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) =
7、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-
8、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 (2.1
9、0) 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 + etNotic
10、e 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
11、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) = e
12、t + 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
13、 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-
14、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 t
15、o 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 + a
