应用时间序列分析习题答案解析

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AutocorrelationaCorreiaticin精品资料1.00000illillillilliIiillillillalaillilill11illalaillalaillilill1|ii|lipI|HT!TT!0.70000illillillillillillillillill11ill11illiinihiii|n|n|n|i111|11T*T1110.41212illillillillillillillillT

2、TTTT0.14848-.07879脚-.25758illillillillill.iTiTu-.37576iiiiiiiiiiiiiiiiiiiiiiia!Il2.2(1) 非平稳，时序图如下(2) -(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关图AutocarrelationsCorrelation-1887654321012345678811.00000illillillillillillillillillillillillillillillillillillillilli|ii|ii|upipi|ilUITITllUIT0.907E1illiliilul

3、lilnhiliillilliInlullilnhiliillillill1|11|ll|11|11|11|111H1I|11|ll|11|1ipiIllIH|11111110.72171illillillillill111111illillill111illill1IInil(11111(1Ijll|1|0.512E2illill111illill111111illillill|1111110.34982illillillillillillillTTTiT.0.24C90iiiiiiiiiiiiiiiTTiT.0.20809轉榊.0.21021栉榊0.2G429illiliilullillT

4、TiT.0.36433iiiiiiiiiiiiiiiiiiiii0.48472illillillillillillillillillillininin|ii|n|ii|n|i111111.0.58456iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii|ii|ii|ii|ii|ii|ii|iTii|n0.G0198illillillillillillillillillillillill1|11|ll|11|11|11|111H|11|11|ll|11|10.51841illillillillillillillillillill1JI1IlliIlliIjllIlliI1

5、JI1JI11|0.36056illillillillill1II111|10.20671O.OS1380.00135-.03248:十;-.02710出0.011240.08275艸：0.17011笊栅i.0.24320illillillillillIjll11111!1|10.25252illillillillill|ipmarkstwowtandarderrors2.3(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

6、-0.204-0.2450.0660.0062-0.139-0.0340.206-0.0100.0800.118(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AutocorrelationaIillillilliIiillil.iriiTiiTiiTiiTuTiIiillillillallillillillillillillilliiii!1!1!1111nT1T1111uillillliill11ill11illillillillillill

7、11i|iiTiipi|iiTin!11ITill1111111illallillillillilli11ill11bi|ia|iigii|ai|i|ii|ii|i|ii|11|ii|i111illillillIIiillil1-riiriiTiiTiiTiiTiiTi IiillillillilliIiillalaillallillilillilliIiillillill11 infillii|i1111111111111111iTT,T1TnI11!1T1T,T,T, Iiillillillillillill1ill1illilillillill |1l(ll|11|11|11|1111|

8、1Bill11|Bl|l111111 Ilillillillillillill11illTbiniTniTi1T11|htb1T1sum.IlillillillillillillTTiTT. Ilill11iillilli1111i1illall111il |1l|ll|11|11|11|111ll|B1|1Bill!11|. Ilillillillillillillillillillillillillillill infillii|iiT11111111111111iTT,T1T1111-1907C54321O1234EG7091WORD格式.分享2)非平稳3)非纯随机2.6(1) 平稳，非

9、纯随机序列(拟合模型参考：ARMA(1,2)(2) 差分序列平稳，非纯随机第三章习题答案(10.7)E(x)=0tE(x)=0t3.1解：E(x)二0.7-E(x)+E(s)tt1ttt(10.7B)xx=(10.7B)-is=(1+0.7B+0.72B2+)sttt1Var(x)=c2=1.9608c2t10.49ssp=e2p=0.490=0210223.2解：对于AR(2)模型：p=0p+0P=0+0P二0.5J11021121p=0p+0p=0p+0=0.321120112解得：0=7/15J0=1/1523.3解：根据该AR(2)模型的形式，易得：E(x)=0t原模型可变为：x=0.

10、8x0.15x+stt1t2tVar(x)=t102c2(1+0)(100)(1+00)21212(1+0.15)(10.15)(10.8+0.15)(1+0.8+0.15)c2=1.9823c2p=0/(10)=0.6957112Jp=0p+0p=0.406621120p=0p+0p=0.2209312210=p=0.6957111J0=0=0.152220=0333.4解：原模型可变形为：(1-B-cB2)x=tt由其平稳域判别条件知：当10i1,e+e1且e-e1时，模型平稳。22121由此可知C应满足：丨c11,c-11且c+11即当一1c1，2模型非平稳；九二1.37381九=-0.

11、873622)|e|=0.31，2e+e=0.81，21=-1.41,模型平稳。九二0.61九=0.523)|0|=0.31，20+021=0.61，0-02=-1.21，1模型可逆。九=0.45+0.2693i1九=0.450.2693i24)|0|=0.41，20+021=-0.91，21模型不可逆。九=0.25691=-1.55695)|e|=0.71，1模型平稳；九=0.71|0|=0.61，1模型可逆；九二0.616)|e|=0.51，2e+e=-0.31，2121模型非平稳。九=0.41241九=-1.2124210|=1.11，模型不可逆；尢=1.1。113.12解法1：G=1，

12、G=eG-0=0.6-0.3=0.3，01101GGk-1G=0.3X06k-1,k2k1k-111所以该模型可以等价表示为：X=+左0.3X0.6kttt-k-1k=0解法2：(1-0.6B)x=(1-0.3B)8ttx=(1-0.3B)(1+0.6B+0.62B2+)&tt=(1+0.3B+0.3*0.6B2+0.3*0.62B3+)&=s+兰0.3*0.6j-18tG二1,G=0.3*0.6j-10j3.13解：E(B)x=E3+0(B)8n(1-0.5)2E(x)=3tttE(x)二12。t113.14证明：已知0二,0二，根据ARMA(1,1)模型Green函数的递推公式得:1214

13、G=1，G=0G-0=0.50.25=02，G=0G=k-1G=协+1,k2011011k1k-1111Pk3.153.16艺GG02+艺0jj+1114=0=-艺G21+艺02(j+1)j1j=0j=1艺GG艺G(0Gjj+kj1j+k-1j=011-0202-04+051=114041-02+041+1111-0217=0.2726j=01)解：成立1)x艺G2jj=02)成立-10=0.3*(x)艺GGjj+k-1-=0j=01%j=01Pk-1j=03)成立4)不成立-10)+8，t-1txT=9.6=9.88T+1(1)=E(x)=E10+0.3*(x-10)+8t+1Tx(2)=E

14、(x)=E10+0.3*(x-10)+8=9.964Tt+2T+1T+2x(3)=E(x)=E10+0.3*(x-10)+8=9.9892Tt+3T+2T+3已知AR(1)模型的Green函数为：G=0j，j=1,2,j1WORD格式分享e(3)=Gs+Gs+Gs=w+028T0t+31t+22t+1t+31t+21t+1Vare(3)二(1+0.32+0.092)*9二9.8829Tx的95%的置信区间：9.9892-1.96*9.8829,9.9892+1.96*.98829t+3即3.8275,16.1509(2)s二x-X(1)二10.5-9.88二0.62T+1T+1TX(1)二E(

15、X)二0.3*0.62+9.964二10.15T+1t+2X(2)二E(X)二0.09*0.62+9.9892二10.045T+1t+3Vare(2)二(1+0.32)*9二9.81T+2X的95%的置信区间：10.045-1.96乂哂981,10.045+1.96911t+3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：FareceistsforYarieiblexbsForecastStdError35SConfidenceLimits6490.156322.729445.6075134.7050S583.888223.8863

16、37.1698130.60656681.908323.944034.9789128.837667SI.82923.354734.33S5128.332S881.085323.955834.1329128.03773.18 (1)平稳非白噪声序列(2) AR(1)(3) 5年预测结果如下：ForecastsforvariablexObsForecastStdError95XConfidenceLimits750.70460.27710.16151.247G760.7956D.29570.21611.3751770.82950.29810.24521.4139780.8421D.29850.257

17、11.4271790.84600.29850.28171.43193.19 (1)平稳非白噪声序列(2) MA(1)(3) 下一年95%的置信区间为(80.41,90.96)3.20 (1)平稳非白噪声序列(2)ARMA(1,3)序列第四章习题答案4.1解：八1X=(x+x+x+x)T+14TT-1T-2T31,5551(x+x+x+x)=x+x+x+x4T+1TT-1T216T16T-116T216T-35八,以，在t+2中t与XT-1前面的系数均为16。4.2解由x=ax+(1tt)xVttt-1x=ttx+(1tt)xt+1t+1t代入数据得x=5.25tt+5(1tt)Vt5.26=5

18、.5tt+(1tt)xvt解得x=5.1Vttt=0.4(舍去ttl的情况)4.3 解：(1)11x=(x+x+x+x+x)=(13+11+10+10+12=11.221520191817165精品资料WORD格式.分享x=(x+x+x+x+x)=!(11.2+13+11+10+10=11.0422521201918175x04x亠06XX=xX=X=X(2)利用兀厂04二十0.6Xt1且初始值X0一X1进行迭代计算即可。另外，22一21一20该题详见Excel。11.792773)在移动平均法下:X-X+-昱X215205ii16X-X+-X+-昱X225215205ii171116a+x在

19、指数平滑法中:x2255525xx0.4x+0.6x21202019b0.40.16。f：:.b一a0.4254.4解：根据指数平滑的定义有（1）式成立，（1）式等号两边同乘（1Q）有（2）式成立xtu+(t一10(1a)+(t一2)a(1a)2+(t2)a(1a)3+-(1)t(1a)xta(1a)+(t1)a(1a)2+(t2)a(1a)3+.(2)t-(2)得axta一a(1一a)一a(1一a)2一xt(1a)(1a)21at1aat1。a则limts4.5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

20、。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，时序图如下精品资料（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：x二T+S+1。（注：如果用乘法模型也可以）tttt首先求季节指数（没有消除趋势，并不是最精确的季节指数）0.9607220.9125751.0381691.0643021.1536271.116566

21、1.042920.9841620.9309470.9385490.9022810.955179消除季节影响，得序列y=x-Sx，使用线性模型拟合该序列趋势影响（方法不唯一）：tttT=97.70+1.79268t，t二1,2,3,t（注：该趋势模型截距无意义，主要是斜率有意义，反映了长期递增速率）得到残差序列1=xSx=yT，残差序列基本无显著趋势和周期残留。tttttX3020100102030aiJAN48aiJAM5001JAN5201JAN5401JAN5601JAN58预测1971年奶牛的月度产量序列为X=T+Sx,t=109,110,，工。ttmodG12）得到771.502183

22、9.9249739.517800.4953829.4208849.5468914.0062889.7989764.9547772.0807748.4289787.3327（3）该序列使用x11方法得到的趋势拟合为D12FinalTrendCycle-HendersonCurve13-termMovingAveragertppliedI/CRatiois1.,159YesirJANFEBMARAPR川M196260S.307608.002809.997612.172814.4226ie.62S1963613.380620.197622.556626.132630.059638.273196464

23、5.771649.468852.4866E4.405S55.241655.5521365672.427673.544673.923673.773673.316672.7211966604.132689.462894.546699.150703.20470S.735196772S.303727.237728.134729.363730.844732.615196S740.451741.853743.154744.442745.822747.51&1969751.978752.171753.571756.280759.856763.5181970771.350771.557772.252773.5

24、71776.OSS773.771脱690.900692.614634.518636.588688.761700.926DIEFinalTrendCycle-HendesonGurv亡12一t亡rmMijvingAverageMppIiedI/CRatiois1.159YeeirJULmSEPOCTmDECTotal1962618.744620.335621.173621.202620.591619.7717389.341963635.SGI636.446637.006637.7G9839.393642.1557579.731964656.079657.423659.903663.336667.

25、078670.3107887.051965672.119671.764671.959673.029S75.423679.2108053.21isee709.932713.173716.547719.787722.667724.8508484.13734.302735.554736.4&9737.289738.073733.105S795.351968743.5W751.47D752.880753.495753.296752.5818976.471969766.763763.256770.681771.206771.284771.2889157.851970784.011788.01D791.3

26、79793.GG3795.195795.97G9392.84Avs.702.980704.826706.444707.864709.222710.584Total：7574GMean：701.35S.D.：56.792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（trend=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X=x+，估计下一天的收盘价为289tt-1t5.2拟合模型不唯一，答案仅供参考。拟合ARIMA(1,1,O)模型，五年预测值为：Fcirecas

27、tsforablexbsForecastStdError35蛊ConfidenceLimits61341444.41267372.7986326993.3929355894.832362349S21.846315413.404323632.0566$76211.C3&0G3357040.971118817.529320159.2928393922.649464363454.004023594.255317210.1143409697.893865369499.988127836.917314940.G341424059.34215.3 ARIMA(l,l,O)x(1,1,0)125.4 (1)

28、AR(1)，(2)有异方差性。最终拟合的模型为x=7.472+8tt8=-0.55958+vh=11.9719+0.4127v2tt-15.5(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InxARIMA(1,3),1,0)(3)预测结果如下:bsForecastStdError85ConfidenceLimits737.52140.04057.44207.6007747.54010.06827.40647.6788757.51450.09087.33637.6926767.49490.10297.29327.6968777.48160.1

29、1017.26587.697478?.48550.11537.25957.7115797.49350.12097.25657.73055.6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(=0.06)，或者显著的趋势平稳。WORD格式.分享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)

30、谷物产量二23.5521+0.775549降雨量tt6.3(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y-0x为平稳序列。tt-2(2)被掠食者拟合乘积模型：ARIMA(0,1,0)x(1,1,0)，模型口径为：1WX=5t1+0.92874B5t拟合掠食者的序列为：y=2.9619+0.283994x+-0.47988tt-2tt-1未来一周的被掠食者预测序列为：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.