应用时间序列分析课后习题答案

1、第二章习题答案2.1(1)非平稳(2)0.01730.7000.4120.148-0.079-0.258-0.376(3)典型的具有单调趋势的时间序列样本自相关图2.2(1)非平稳，时序图如下AulocorreIatlionsCorrelation-19S7654321u1J515679911.OOODOEG巾,IHii!qi亦造厘巾TfelTim甲断大0J00PV*i_Ii;Lii|itl.1liil.aliiltlul.-XbiX-iJiiX-EFE用rp-Ffir事即即ffJ0J12120.14348w.-.07878.-.25758,强小Hi不强-.37576部踹*:*;1“marks

2、ttfostandarderrors(2)关图-(3)样本自相关系数及自相关图如下：典型的同时具有周期和趋势序列的样本自相AutocorrelationsCorrelation-13f76513?101284567831-0.139-0.0340.206-0.0100.0800.118LOOOQO0.30751illill1n1illFbill1III111Fl1111J11ill3IIillall1111IIIBill吊巾亦面新巾小巾巾巾巾巾面ThTiTuriTiiii-iliniili|liilaHiiTailiiliiliilraliiliilidilialiili,TilyisisiT

3、isTT1T-l11f*TiTltiijirjup,.yiayi.T*0,72171HiHifcill111111allillalliiipiitii|hiTiniriniQiEiTiiTE0,5125211由*底*版）:0,34982郴1Mi梯*0.24690*第中*,0.20303就*.0.21021辛柳出,0.264290.3B433不出格出雅曲本0.484720.58456甯出格膈用市0.B019604518411dikljJiJfdi也rp!1111ilrrrp0.2067110.031381o.ooias-*03240.郭-.0271D*0.01124口.08”5:0,17011*

4、MS.0.24320奉帆曲|e.0,25352率帆*.J.markstosteindarderrors2.3(1)自相关系数为：0.20230.0130.042-0.043-0.179-0.251-0.0940.0248-0.068-0.0720.0140.1090.2170.3160.0070-0.0250.075-0.141-0.204-0.2450.0660.0062(2)平稳序列(3)白噪声序列2.4LB=4.83,LB统计量对应的分位点为0.9634,P值为0.0363。显著性水平=0.05,序列不能视为纯随机序列。2.5(1)时序图与样本自相关图如下1234667491M1fJj山

5、Lliill.sig.jL.hsT,i|r-111f,iprii-ifpiipTi,nT，,iliiiiKairiii1JrlL*IIciillpf.T,BT,I1*1Is!|：iT!lB.吊忖布删(2)非平稳(3)非纯随机平稳，非纯随机序列(拟合模型参考:差分序列平稳，非纯随机第三章习题答案3.2解:对于AR(2)模型:22000.490.50.3解得:7/151/153.3解:根据该AR(2)模型的形式，易得:E(xt)0原模型可变为：xt0.8xt10.15xt2Var(xt)122(125(1125(11253.4(10.15)(1i/(1(10.15)0.80.15)(10.80.1

6、5)2=1.982320.69570.40660.22091122330.69570.150解：原模型可变形为:_2(1BcB)xtt由其平稳域判别条件知：当|2I模型平稳。由此可知c应满足：|c|1,c13.5即当1c0时，该AR(2)模型平稳。证明：已知原模型可变形为：(1BcB2cB3)xt3.6其特征方程为：32不论c取何值，都会有一特征根等于1,1)(c)因此模型非平稳解：(1)错,Var(xt)/(1(2)错,E(%)(xt1)/(1(3)错,?t(I)l1xT0(4)错,eT(l)GiTlG2Gl1(5)错,limVarxTXT(l)limVareT(l)lim3.7解:1_11

7、2MA(1)模型的表达式为:Xt3.8解法1:由xt=+t1xt0.5xt1=0.5+t(10.5)t1(20.51)t2+0.52t3,与xt=10+0.5xt1+t0.8t2+Ct3对照系数得0.510,20,0.500.510.8故0.5,0.55,00.50.275解法2:将xt100.5xt10.8t2Ct3等价表达为xt2010.8B2CB310.5B10.8B2CB3(10.5B0.52B20.53B3L)t展开等号右边的多项式,整理为0.5B2_20.5B0.8B2330.5B0.8CB30.5B3440.5B240.80.52B40.5CB4合并同类项，原模型等价表达为Xt2

8、010.5B0.55B20.5k(0.530.4C)B3kt0当0.530.4C0时,该模型为MA(2)模型，解出C0.275。3.9解：E(xt)Var(xt)(1222)1.650.981.650.59390.41.650.2424k0,k3。3.10解法1:(DxtC(xt1C(t2xtxtxt1(C1)t1即(1B)xt1(C1)Bt显然模型的AR部分的特征根是1,模型非平稳(2)ytxtxt1t(C1)t1为MA模型，平稳1C1122112C22C2解法2:(1)因为Var(xt)lim(1kC2)2,所以该序列为非平稳序列。k(2)ytxtxt1t(C1)t1,该序列均值、方差为常

9、数,22E(yt)0,Var(yt)1(C1)自相关系数只与时间间隔长度有关，与起始时间无关C121(C1)k0,k2所以该差分序列为平稳序列。3.11解：(1)|2|1.21,模型非平稳;1.37382-0.8736(2)|2|0.31,210.81,211.41,模型平稳1.21,模型可逆10.620.5(3) |2|0.31,10.45+0.2693i20.45-0.2693i(4)|2|0.41,0.91,211.71,模型不可逆。10.25692-1.5569(5) |J0.71,模型平稳；10.7|1|0.61,模型可逆；10.6(6) |2|0.51,210.31,211.31,

10、模型非平稳0.4124-1.2124|111.11,模型不可逆;3.12解法1：G01,G11G010.60.30.3,GkG11k1Gl0.30.6k1,k2所以该模型可以等价表示为:Xttk0.30.6k0解法2:(10.6B)xt(10.3B)t3.133.14Goxt(10.3B)(1_2_20.6B0.6B(1Go_2_0.3B0.3*0.6B20.3*0.3*0.6j1tjj1.i11,Gj0.3*0.6j解：E(B)xtE3E(xt)12。证明：已知11,G1G0Gjj0Gjj0g2j0(B)t(10.62B3)t0.5)2E(xt)3ARMA(1,1)模型Green函数的递推公

11、式得:0.52j10.251Gk1k1GT,k22(j1)11j151二412141215141260.27GjGjk1j0,kGj2j0g2j03.15(1)成立(2)成立(3)成立(4)不成立xT9.63.16解：(1)xt100.3*(xt110)/(1)E(xt1)E100.3*(xt10)t19.88XT(2)E(xt2)E100.3*(xt110)t29.964xT(3)E(xt3)E100.3*(xt210)t39.9892已知AR(1)模型的Green函数为：Gj/,j1,2,2eT(3)G0t3G1t2G2t1t31t2111VareT(3)(10.320.092)*99.8

12、829xt3的95%的置信区间：9.9892-1.96y9.8829,9.9892+1.96*79.8829即3.8275,16.1509(2)T1Xt1xT(1)10.59.880.62於1(1)E(xt2)0.3*0.629.96410.15XT1(2)E(xt3)0.09*0.629.989210.045Var02(2)(10.32)*99.81xt3的95%的置信区间：10.045-1.96XJ丽，10.045+1.96*7911即3.9061,16.1839。3.17 (1)平稳非白噪声序列(2) AR(1)(3) 5年预测结果如下：forvarieibIex的ForecastStd

13、Error95XConfidenseLimitsS422.723415-6075134.7050第蛭,既吃23.336837,1898190.60655631.90BS23.944034.37S91?e.637B5731.282925.S54734,5325128.2332SIJI05323J55834.1329123.03773.18 (1)平稳非白噪声序列(2) AR(1)(3) 5年预测结果如下:Forecistsforvariablen.CbsForecastStdErrorSEXCtonfidenceLinits750.7046I1.27710.1G16L2476760即56O.28

14、G70.2161L3751770.82950.29910.24521.4139730.34210.25711.4271790.94GeD.29GE0.961?L4S193.19(1)(2)平稳非白噪声序列MA(1)下一年95%勺置信区间为80.41,90.963.20(1)(2)(3)平稳非白噪声序列ARMA(1,3)序歹U拟合及5年期预测图如下:第四章习题答案4.1解:xT1XT2XT3)xT1xTxT1Xt2)5Xt165Xt1165Xt2161-Xt316所以,xT2中xT匕xT1前面的系数均为16。4.2xt(1)%1%1xt1(1)%代入数据得%5.255(1)5.265.5(1)%

15、解得%5.10.4(舍去1的情况)4.3解：(1)c1,、1卷一(x20x19x18x17+x16)(13+11+10+10+10=11.255八1,八、1一一一&(X21+x20x19x18x17)(11.2+13+11+10+10)=11.04(2)利用0.4xt550.6%1且初始值X进行迭代计算即可。另外，？22尺1%该题详见Excel。11.79277(3)在移动平均法下：X21920195i16Xi1X2019Xii171116a-55525在指数平滑法中：&%0.4x200.6%9b0.4(1)式等号两边同乘(1)有(2)式成立)2(t2)(1)3L(1)2(t2)(1)3L(2

16、)ba0.40.16。254.4 解：根据指数平滑的定义有(1)式成立,%t(t1)(1)(t2)(1(1)磔t(1)(t1)(1(1) -(2)得%t(1)(1)2L%t(1)(1)2L则lim版limttt4.5 该序列为显著的线性递增序列，利用本章的知识点，可以使用线性方程或者holt两参数指数平滑法进行趋势拟合和预测，答案不唯一，具体结果略。4.6 该序列为显著的非线性递增序列，可以拟合二次型曲线、指数型曲线或其他曲线，也能使用holt两参数指数平滑法进行趋势拟合和预测，答案不唯一，具体结果略。4.7 本例在混合模型结构，季节指数求法，趋势拟合方法等处均有多种可选方案，如下做法仅是可选

17、方法之一，结果仅供参考（1）该序列有显著趋势和周期效应，时序图如下（2）该序列周期振幅几乎不随着趋势递增而变化，所以尝试使用加法模型拟合该序列：XtTtStIt。（注：如果用乘法模型也可以）首先求季节指数（没有消除趋势，并不是最精确的季节指数）0.9607220.9125751.0381691.0643021.1536271.1165661.042920.9841620.9309470.9385490.9022810.955179消除季节影响，得序列ytxtStX,使用线性模型拟合该序列趋势影响（方法不唯一）Tt97.701.79268t,t1,2,3,L（注：该趋势模型截距无意义，主要是斜率

18、有意义，反映了长期递增速率）得到残差序列ItXtStxytTt,残差序列基本无显著趋势和周期残留。3020100-203001MN48O1MNM01JAN位DIJ部1540UAN560UAN56t预测1971年奶牛的月度产量序列为XtTtSm0dti12X,t109,110,L,120得到771.5021739.517829.4208849.5468914.0062889.7989839.9249800.4953764.9547772.0807748.4289787.3327D12Fira17rendCycle-H&rwi&rsonCurve18temMovingAverageAppIiedI

19、/GRatioIsL158YeeirJ限FEB岫RAPRM好JUN1QG2EU队307eoe.oos609.99?612.172Cl4,422G1G.C2B1963619.38062C.197622.556625J32&30.059C33.2731964C45J71C49.4BS652.48GGE4.40E%5.241呻用219B5672.427673,544873.923673.773S73.3I6B72.721004.192CSS-462694.546SS9.160703.20470ft.735197726.303727.237728.1i4729.388780.844782.61519的

20、740,451741.063743,154744.442746.622747.G16761.97K752J7I758.6717SS.28O7BS.85S78S.51S1970771,360771.5的172.232773.571776.089775,77190.900602.614634*518636m588588.7617m.昵&012FinalTrendCycle-HendersonOjrve13-terwMovin匿AverageAppliedl/CRtiois1.159YeeirJULmSEPOJTMO?DECTotal19B2BW.74462C,335BZ1.173521.20262

21、0.591619.7717309.3419党C95.9G1G3C-44C637.000837.7C9S99.993C42.1557579.751964C56.073657-423659.903663.338S67.O7567a.31D78S7.0518附Q72J19671JM671,959&79.029-5.428C7J.2196089.21703.982713.173716.647718.787722.S67724.85984S4.1S1967794J02735.5547辨,4部737.289733.079739,105/限展196S743,51076L47C7E2.GS07E3.4857E

22、8.28&752,5818978.471863766,76376A.2GB770.601771.206771,264771,26881574651870704,01178B.CKI791.379黑3/E3795.19S79E,9;G9892.94702,3807IM.82E孙.444707,064709.2S!710,561Tcrtal!7574店Nea,n:7CL36S.D.：66.792x11方法得到的趋势拟合为(3)该序列使用趋势拟合图为4.8这是一个有着曲线趋势，但是有没有固定周期效应的序列，所以可以在快速预测程序中用曲线拟合(stepar)或曲线指数平滑(expo)进行预测(tren

23、d=3)。具体预测值略。第五章习题5.1 拟合差分平稳序列，即随机游走模型xt=xt-1+t,估计下一天的收盘价为2895.2 拟合模型不唯一，答案仅供参考。拟合ARIMA(1,1,0)模型，五年预测值为：ForecsistsforvirlablexForeesistSidError95,Conf123456GG-G341444.4126349921.8463357040.37115G3464.004036S499.9SS17S72.739613413.40415S17.52823594.2552783.91Z326393.盹2g323632.0566320158.2828317210,114

24、33U940.8341iI.二仁。Lifiiit:355盟4,B323376211.E3E033392?.6434409G07.9S30424059.34215.3 ARIMA(1,1,0)(1,1,0)125.4(1)AR(1),(2)有异方差性。最终拟合的模型为xt=7.472+tt=-0.5595t-1+vtvt=hen=11.9719+0.41275.5(1)非平稳(2)取对数消除方差非齐，对数序列一节差分后，拟合疏系数模型AR(1,3)所以拟合模型为lnxARIMA(1,3),1,0)(3)预测结果如下:5.6口ForecastStdErrorS&KConfidenceLimits7

25、37.52140,0405?.44207.6007747.54010/。般7,40647.673S?.&1450.0909LC92S76L羽430.10237.29327.6966777.4S1S0.11017.2S5S7.C97+7F.48550.11537.25957.71157S?.49350.12037.25657.7305原序列方差非齐，差分序列方差非齐，对数变换后，差分序列方差齐性。第六章习题6.1 单位根检验原理略。例2.1原序列不平稳，一阶差分后平稳例2.2原序列不平稳，一阶与12步差分后平稳例2.3原序列带漂移项平稳例2.4原序列不带漂移项平稳例2.5原序列带漂移项平稳(=0

26、.06),或者显著的趋势平稳。6.2 (1)两序列均为带漂移项平稳(2)谷物产量为带常数均值的纯随机序列，降雨量可以拟合AR(2)疏系数模型。(3)两者之间具有协整关系(4)谷物产量t23.55210.775549降雨量t6.3 (1)掠食者和被掠食者数量都呈现出显著的周期特征，两个序列均为非平稳序列。但是掠食者和被掠食者延迟2阶序列具有协整关系。即yt-殁_2为平稳序列。(2)被掠食者拟合乘积模型：ARIMA(0,1,0)(1,1,0)5,模型口径为：15%=5t1+0.92874B5拟合掠食者的序列为：yt=2.9619+0.28399442+t-0.47988t.1未来一周的被掠食者预测

27、序列为：ForecastsforvariablexObsForecastStdError95%ConfidenceLimits4970.792449.4194-26.0678167.652650123.835869.8895-13.1452260.816751195.098485.596827.3317362.865152291.637698.838797.9173485.357953150.0496110.5050-66.5363366.63555463.5621122.5322-176.5965303.72085580.3352133.4800-181.2807341.95115655.5269143.5955-225.9151336.96905773.8673153.0439-226.0932373.82795875.2471161.9420-242.1534392.64755970.005318