chapter2时间序列

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 ¹ 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

3、 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(wt½ 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)

4、= 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)2Since the expected values of all

5、 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(wt½et-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-

6、3 + (et-2)2 + 2et-2et-3 + (et-3)2 - (0.5)2Since Et-2et-2 = Et-2et-3 = 1, it follows that var(wt½ et-3 = et-2 = 1) = (1/16)(s2 + s2 + 1 + 1 + 2) - 0.25, so thatvar(wt½ 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 a

7、s 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 +

8、 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(

9、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 +

10、 a12et-3 + .+ a1t-2e1 + etNotice that all terms cancel. Specifically:a01 + a1 + a12 + . + a1t-1 º a0 + a1a01 + a1 + a12 + . + a1t-2a1ty0 º 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

11、+ 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 ½ a1 ½ < 1.Answer: Using lag operators, write the equation as (1 - a

12、1L)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 sol

13、utions 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 ½a2½ < 1. A. Find: i. Et-2ytii. Et-1ytiii. Etyt+2 iv. Cov(yt, yt-1) v.

14、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

15、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

16、) 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(e

17、t-6)2 + (a2)6(et-8)2 + . Given ½a2½ < 1, the infinite summation S(a2)2i = 1/1 - (a2)2 so that:cov(yt, yt-2) = a2s2/1 - (a2)2vi). As shown in Part iv, the covariance between yt and yt-1 is zero. Hence, from (2.35), r1 = f11 = 0. Given r1 = 0, (2.36) indicates that f22 = r2. Given the ans

18、wer to v and that var(yt) = var (yt-i) = . = s2/1 - (a2)2, it follows that f22 = cov(yt, yt-2)/var(yt) = a2s2/1 - (a2)2/s2/1 - (a2)2 = a2B. Find the impulse response function. Given yt-2, trace out the effects on an et shock on the yt sequence. Answer: One way to answer the question is to use the pa

19、rticular solution for yt:yt = a0/(1-a2) + et + a2et-2 + (a2)2et-4 + (a2)3et-6 + (a2)4et-8 + .Hence: ¶yt/¶et = 1. By the simple change of subscripts: ¶yt+1/¶et = ¶yt/¶et-1 = 0; ¶yt+2/¶et = ¶yt/¶et-2 = a2;¶yt+3/¶et = ¶yt/¶et-3 = 0;

20、¶yt+4/¶et = ¶yt/¶et-4 = (a2)2 .C. Determine the forecast function: Etyt+s. The forecast error is the difference between yt+s and Etyt+s. Derive the correlogram of the sequence. Hint: Find Et, Var, and for j = 0 to s. Answer: To find the forecast function, first find the general s

21、olution for yt in terms of y0. For a given value of y0, students should be able to show that if t is even, then:yt = a01 + a2 + (a2)2 + . + (a2)t/2-1 + et + a2et-2 + (a2)2et-4 + (a2)3et-6 + . + (a2)t/2y0Hence: E0yt = a01 + a2 + (a2)2 + . + (a2)t/2-1 + (a2)t/2y0. To find Etyt+2s, update the time subs

22、cripts such that Etyt+2s = a01 + a2 + (a2)2 + . + (a2)s-1 + (a2)sytGeneralizing the result from Part A above, note that for odd-period forecasts, Etyt+2s-1 = Etyt+2s-2. To find the forecast error, subtract Etyt+s from yt+s. For even values of s, we can let s = j/2 and form= et+s + a2et+s-2 + (a2)2et

23、+s-4 + . + (a2)s/2-1et+2The forecast error has a mean of zero since: Etet+s + a2et+s-2 + (a2)2et+s-4 + . + (a2)s/2-1et+2 = 0Similarly, the variance is 1 + (a2)2 + (a2)4 + . + (a2)s-2s2. The correlations between the forecast error for any period t+s and the forecast error for any odd period is zero.

24、For even periods, we can begin by forming the covariance between et(2) and et(4) as:E(et+4 + a2et+2)(et+2) = a2s2 so that the correlation coefficient is a2/(1 + (a2)2)1/2.Similarly, the covariance between et(2) and et(6) as:E(et+6 + a2et+4 + (a2)2et+2)(et+2) = (a2)2s2 so that the correlation coeffic

25、ient is(a2)2/1 + (a2)2 + (a2)41/24. Two different balls are drawn from a jar containing three balls numbered 1, 2, and 4. Let x = number on the first ball drawn and y = sum of the two balls drawn.  A. Find the joint probability distribution for x and y; that is, find prob(x = 1, y = 3), prob(x

26、= 1, y = 5), . , and prob(x = 4, y = 6). Answer: Let x = number on the first ball; z = number on the second ball; and y = x + z. Consider the contingency tableSummationsx = 1x = 2x = 4z = 1y = 3Y = 5z = 2y = 3y = 6z = 4y = 5y = 6Notice that the same outcome for x and z is not possible since two diff

27、erent balls are drawn from the jar. Thus, prob(x=1, y=2), prob(x=2, y=4), and prob(x=3, y=6) are all equal to zero. The remaining six outcomes are equally likely. The probabilities (x=1, y=3), (x=1, y=5), . (x=4, y=6) all equal 1/6.B. Find each of the following: E(x), E(y), E(y½x = 1), E(x½

28、;y = 5), Var(x½y = 5), and E(y2)Answers: Each outcome for x has a probability of 1/3. Thus: i) E(x) = (1/3)(1 + 2 + 4) = 7/3ii) The expected value of y is the summation of each possible outcome for y multiplied by the probability of that outcome. Since each cell has a probability of 1/6, readin

29、g across the rows of the table yields:E(y) = (1/6)(3 + 5 + 3 + 6 + 5 + 6) = 14/3iii) When x = 1, y can take on values 3 or 5. Hence:E(y½x = 1) = 0.5(3) + 0.5(5) = 4.0iv) When y = 5, x can take on values 1 or 4. Hence: E(x½y =5) = 0.5(1 + 4) = 2.5v) Var(x½y = 5) = E(x2½y = 5) - E(

30、x½y =5)2 = 0.5(12 + 42) - (2.5)2 = 8.5 - 6.25 = 2.25Var(x½y = 5) = 2.25vi) The expected value of y2 is the summation of each possible squared outcome for y multiplied by the probability of that outcome. Hence: E(y2) = (1/3)(32 + 52 + 62) = 70/3C. Consider the two functions: w1 = 3x2 and w2

31、 = x-1. Find: E(w1 + w2) and E(w1 + w2½y = 3).Answers: The expected value of a function of x (i.e., F(x) is equal to the value of the function evaluated at each possible realization of x multiplied by the probability of the associated realizations. Moreover, since the expectation of a sum is th

32、e sum of the expectations, E(w1 + w2) = E(3x2) + Ex-1 = 3(1/3)(12 + 22 + 42) + (1/3)(1 + 1/2 + 1/4) = 21 + 1.75/3. Hence:E(w1 + w2) = 21.5833When y = 3, x can only take on the values 1 or 2. Hence, E(w1 + w2½y = 3) = E(w1½y = 3) + E(w2½y = 3) = 3(1/2)(12 + 22) + (1/2)(1 + 1/2) =15/2 +

33、 3/4 = 33/4. Hence:E(w1+ w2½y = 3) = 33/4D. How would your answers change if the balls were drawn with replacement? Answer: The contingency table changes since it is possible for x and z to take on the same values. The new contingency table becomes:Summations x = 1x = 2x = 4z = 1y = 2y = 3y = 5

34、z = 2y = 3y = 4y = 6z = 4y = 5y = 6y = 8Each entry from (x=1, y=2) through (x=4, y=8) has a probability of 1/9. Part B becomes:i) E(x) = 1/3(1 + 2 + 4) = 7/3.ii) E(y) = (1/9)(2 + 3 + 5 + 3 + 4 + 6 + 5 + 6 + 8) = 42/9 = 14/3iii) E(y½x=1) = (1/3)(2 + 3 + 5) = 10/3.iv) E(x½y=5) = (1/2)(1 + 4)

35、 = 2.5.v) Var(x½y=5) = E(x2½y=5) - E(x½y=5)2 = (1/2)(1 + 42) - (2.5)2 = 17/2 - 6.25 = 2.25.vi). E(y2) = (1/9)22 + 2(3)2 + 42 + 2(5)2 + 2(6)2 + 82 = 224/9.Note that the answer to Part C is unchanged. 5. The general solution to an n-th order difference equation requires n arbitrary cons

36、tants. Consider the second-order equation: yt = a0 + 0.75yt-1 - 0.125yt-2 + et. A. Find the homogeneous and particular solutions. Discuss the shape of the impulse response function.Answer: The homogeneous equation is yt = 0.75yt-1 - 0.125yt-2. Try the challenge solution yt = Aat and obtainAat - 0.75

37、Aat-1 + 0.125Aat-2 = 0so that the characteristic equation is a2 - 0.75a + 0.125 = 0.The two characteristic roots are 0.5 and 0.25 and A can be any arbitrary constant. Since a linear combination of the two homogeneous solutions is also a solution, the complete form of the homogeneous solution is yt =

38、 A1(0.5)t + A2(0.25)t where A1 and A2 are arbitrary constants. The particular solution has the form yt = b0 + Saiet-i. The constant b0 can easily be found as b0 = a0(1 - 0.75 + 0.125) = 8a0/3. To find the ai substitute yt = Saiet-i into yt = 0.75yt-1 - 0.125yt-2 + et to obtain:a0et + a1et-1 + a2et-2

39、 + a3et-3 + . = 0.75(a0et-1 + a1et-2 + a2et-3 + .) - 0.125(a0et-2 + a1et-3 + a2et-4 + .) + etMatching coefficients on like terms, it follows that:a0 = 1; a1 = 0.75; and all subsequent ai are such that ai = 0.75ai-1 - 0.125ai-2For example, a2 = 0.4375, a3 = 0.2348, a4 = 0.1211, and a5 = 0.0615.The im

40、pulse responses are given by the coefficients of the particular solution. For example, ¶yt/¶et = 1; ¶yt+1/¶et = ¶yt/¶et-1 = 0.75; ¶yt+2/¶et = ¶yt/¶et-2 = 0.4375. Since both characteristic roots are positive and less than unity, the impulse responses

41、converge directly toward the long-run value yt = 8a0/3. B. Find the values of the initial conditions (i.e., y0 and y1) that ensure yt sequence is stationary (Note: A1 and A2 are the arbitrary constants in the homogeneous solution). Answer: The general solution is the sum of the homogeneous and parti

42、cular solutions: yt = A1(0.5)t + A2(0.25)t + 8a0/3 + et + 0.75et-1 + 0.4375et-2 + . so that for periods 0 and 1: y0 = A1+ A2 + 8a0/3 + e0 + 0.75e-1 + 0.4375e-2 + . y1 = A1(0.5) + A2(0.25) + 8a0/3 + e1 + 0.75e0 + 0.4375e-1 + . The issue is to select y0 and y1 such that A1 = A2 = 0. If y0 = 8a0/3 + e0

43、 + 0.75e-1 + 0.4375e-2 + . and y1 = 8a0/3 + e1 + 0.75e0 + 0.4375e-1 + . , then A1 = A2 = 0, so that the general solution is: yt = 8a0/3 + et + 0.75et-1 + 0.4375et-2 + .C. Given your answer to part B, derive the correlogram for the yt sequence. Answer: Given the initial conditions found in Part A, th

44、e Yule-Walker equations can be used to derive the correlogram. For simplicity, abstract from the constant a0 since it does not affect the autocorrelations. Hence, set a1 = 0.75 and a2 = -0.125, and use (2.28) and (2.29) to find the autocorrelations. Given that r0 = 1, it follows that r1 = 0.75 - 0.1

45、25r1 andr1 = 2/3 and rs = 0.75rs-1 -0.125rs-2For example, r2 = 0.375; r3 = 0.1979; r4 = 0.1012; r5 = 0.0512.6. Consider the second-order stochastic difference equation: yt = 1.5yt-1 - 0.5yt-2 + et.A. Find the characteristic roots of the homogeneous equation. Answer: The homogeneous equation is yt -

46、1.5yt-1 + 0.5yt-2 = 0. If you try the challenge solution yt = Aat, A and a must satisfy: Aat - 1.5Aat-1 + 0.5Aat-2 = 0. Dividing by Aat-2, the characteristic equation is a2 - 1.5a + 0.5 = 0. Thus, A can be any arbitrary constant and the characteristic roots, (i.e., a) can be 1 or 0.5. The homogeneou

47、s solution is:yt = A1 + A2(0.5)tB. Demonstrate that the roots of 1 - 1.5L + 0.5L2 are the reciprocals of your answer in Part A.Answer: To solve the inverse characteristic equation for L, form (1 - L)(1 - 0.5L) = 0. The solutions are L = 1 and L = 1/0.5 = 2. Hence, the two inverse characteristic root

48、s are the reciprocals of characteristic roots found in Part A. Thus, the stability condition is for the characteristic roots to lie inside of the unit circle or for the roots of the inverse characteristic equation to lie outside of the unit circle. C. Given initial conditions for y0 and y1, find the

49、 solution for yt in terms of the current and past values of the et sequence. Explain why it is not possible to obtain the backward looking solution for yt unless such initial conditions are given. Answer: One way to solve the problem is to use the initial conditions y0 and y1 and iterate forward. Si

50、nce, y2 = 1.5y1 - 0.5y0 + e2, and y3 = 1.5y2 - 0.5y1 + e3, it follows that:y3 = e3 + 1.5e2 + 1.75y1 - 0.75y0. Similarly, y4 = 1.5y3 - 0.5y2 + e4. Substituting for y3 and y2 yields: y4= e4 + 1.5e3 + 1.75e2 + 1.875y1 - 0.875y0Continuing in this fashion yields:y5 = e5 + 1.5e4 + 1.75e3 + 1.875e2 + 1.937

51、5y1 - 0.9375y0y6 = e6 + 1.5e5 + 1.75e4 + 1.875e3 + 1.9375e2 + 1.96875y1 - 0.96875y0The solution has the form:where: a0 = 1, a1 = 1.5, at = 1- at-1, and the remaining coefficients ai solve the difference equation ai = 1.5ai-1 - 0.5ai-2. Since the coefficients will grow progressively larger, the backw

52、ard particular solution will not be convergent. The initial conditions are necessary for yt to be finite for finite values of t.D. Find the forecast function for yt+s. Answer: First write the answer to Part C in terms of s rather than in terms of t:Then update by t periods to obtain:For s > 1, fo

53、recasts 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

54、 time, the mean is not constant.ii. Eyt+1 = aty1 + at+1y0. Note that Eyt+1 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) ¹ var(yt ).v. cov(yt+1, yt) = a0a1 + a1a2 + . + at-3at-2s2.7. The file entitled SIM_2.XLS con

55、tains 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

56、 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

57、, 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)