时序第五章SAS作业

1、第五章SAS作业问题1：1867-1938年英国绵羊数量如下所示：2203 2360 2254 2165 2024 2078 2214 2292 2207 2119 2119 21372132 1955 1785 1747 1818 1909 1958 1892 1919 1853 1868 19912111 2119 1991 1859 1856 1924 1892 1916 1968 1928 1898 18501841 1824 1823 1843 1880 1968 2029 1996 1933 1805 1713 17261752 1795 1717 1648 1512 1338 1

2、383 1344 1384 1484 1597 16861707 1640 1611 1632 1775 1850 1809 1653 1648 1665 1627 17911、选择恰当模型，拟合该序列的发展；2、利用拟合模型预测1939-1945年英国绵羊的数量；3、按照书本相应例题的格式完成问题，并附上SAS程序。答：绘制该序列的时序图如下：时序图显示该序列有显著的递减趋势，为非平稳序列。考虑使用ARIMA模型拟合。使用一阶差分提取序列信息，一阶差分后作时序图和ACF图如下：一阶差分后的时序图无明显趋势和周期性，可认为是平稳序列。由自相关图可知，自相关系数衰减到零的速度较快且1阶以后大部分

3、都落在两倍标准误以内。所以可认为一阶差分后的序列为平稳序列。对一阶差分后的序列进行白噪声检验：由SAS结果可知，该序列为非白噪声序列。一阶差分后序列的PACF图如下：观察ACF图和PACF图后，可认为ACF图拖尾，PACF图3阶截尾。建立AR(3)模型。SAS输出残差序列及系数检验结果如下：结果显示AR(3)模型显著，常数项和2未通过检验，先剔除常数项，检验结果如下：2仍未通过检验，建立疏系数模型ARIMA(1,3),1,0)，检验结果如下：由上述结果可知，模型ARIMA(1,3),1,0)显著，系数显著。模型如下：即xt=0.32196xt-1-0.37616xt-3+t (1-B)xt=t

4、/(1-0.32196B+0.37616B3)预测7年绵羊数量如下：预测拟合图如下：SAS程序如下：data homework5_1;input sheep;dif1=dif(sheep); /*一阶差分*/year=intnx('year','1jan1867'd,_n_-1);format year date.;cards;2203 2360 2254 2165 2024 2078 2214 2292 2207 2119 2119 21372132 1955 1785 1747 1818 1909 1958 1892 1919 1853 1868 19912

5、111 2119 1991 1859 1856 1924 1892 1916 1968 1928 1898 18501841 1824 1823 1843 1880 1968 2029 1996 1933 1805 1713 17261752 1795 1717 1648 1512 1338 1383 1344 1384 1484 1597 16861707 1640 1611 1632 1775 1850 1809 1653 1648 1665 1627 1791;proc print data=homework5_1;proc gplot;plot sheep*year dif1*year

6、;symbol v=star c=red i=join;proc arima data=homework5_1;identify var=sheep(1); /*如果输入var=dif1，则forecast的结果为差分后的序列的预测值*/estimate p=(1 3) noint;forecast lead=7 id=year out=result;data result2;set result;year=_n_;proc gplot data=result2; /*画图数据集*/plot sheep*year=1 forecast*year=2 l95*year=3 u95*year=3/

7、overlay;symbol1 v=star c=red i=none;symbol2 v=none c=bule i=join;symbol3 v=none c=black i=join l=32;run;问题2，使用Auto-Regressive模型分析例5.9序列。（作业格式参照书“例5.6续”）答：绘制该序列时序图如下：由上图可以看出该序列既有递增趋势又有周期性。考虑建立如下结构的残差自回归模型： xt=St·(Tt+It) 对Tt和St分别构造如下四个确定性模型：(1) 趋势变量为时间t的幂函数，季节指数为给定的季节指数。Tt=0+1t ，St=StS1=1.149585

8、S2=0.932638 S3=0.924146 S4=0.993631消除季节效应后的时序图：然后构造如下确定性趋势模型：Tt=0+1t得如下结果：消除趋势后的时序图如下：对消除确定性因素后的序列t=xt/St-Tt做DW检验，结果如下：DW检验结果p值小于0.05，显示残差序列显著自相关性。尝试对残差序列t拟合AR(5)模型，确定残差自回归模型的口径：因为参数没通过检验，去除参数后再次估计模型口径：所以得到的最终拟合模型为：xt=St·0.0726t+ut , S1=1.149585 S2=0.932638 S3=0.924146 S4=0.993631ut=0.9598ut-1+

9、0.6881ut-4-0.7238ut-5+at模型解释:德国工人季度失业率有一个长期递增的趋势，每年每季度的增长速度为0.0726St.同时它还受到诸多随机因素的影响，导致随机波动序列具有短期相关性。拟合图如下:(2) 趋势变量为历史观测值，季节指数为给定的季节指数。 ，S1=1.149585 S2=0.932638 S3=0.924146 S4=0.993631消除季节效应后的时序图：然后构造如下确定性趋势模型：得如下结果：Dh检验结果p值小于0.05，显示残差序列显著自相关性。尝试对残差序列t拟合AR(2)模型：确定残差自回归模型的口径：所有参数均显著。所以得到的最终拟合模型为：xt=S

10、t·1.0029xt-1+ut , S1=1.149585 S2=0.932638 S3=0.924146 S4=0.993631ut=0.2812ut-1-0.3151ut-2+at拟合图如下: (3) 趋势变量为历史观测值，季节效应为季节自回归模型。 , ,构造如下季节自回归模型：St=0+1·xt-12得如下结果：参数p值均小于0.05，即参数显著。所以St=1.5043+0.8006 xt-12.消除季节效应后的时序图：然后构造如下确定性趋势模型：得如下结果：Dh检验结果p值小于0.05，显示残差序列显著自相关性。尝试对残差序列t拟合AR(1)模型，确定残差自回归模

11、型的口径：所有参数均显著。所以得到的最终拟合模型为：xt=(1.5043+0.8006 xt-12)·0.9824xt-1+ut , ut=0.3777ut-1+at拟合图如下:(4)趋势自变量为时间t的幂函数，季节效应为季节自回归模型。消除季节效应后的时序图：然后构造如下确定性趋势模型：Tt=0+1t得如下结果：DW检验结果p值小于0.05，显示残差序列显著自相关性。尝试对残差序列t拟合AR(5)模型，确定残差自回归模型的口径：所有参数均显著。所以得到的最终拟合模型为：xt=(1.5043+0.8006 xt-12)·0.0130t+ut , ut=1.0162ut-1+

12、0.4061ut-4-0.5190ut-5+at拟合图如下:SAS程序如下：data homework5_2;input x; t=_n_;quarter=intnx('quarter','1jan1962'd,_n_-1);format quarter date.;cards;1.10.50.40.71.60.60.50.71.30.60.50.71.20.50.40.60.90.50.51.12.92.11.722.71.30.911.60.60.50.71.10.50.50.61.20.70.711.510.91.11.5111.62.62.12.33.6

13、54.54.54.95.74.344.45.24.34.24.55.24.13.94.14.83.53.43.54.23.43.64.35.54.85.46.5877.48.510.18.98.89108.78.88.910.48.98.9910.28.68.48.49.98.58.68.79.88.68.48.28.87.67.57.68.17.16.96.66.866.26.2;proc print data=homework5_2;proc gplot; plot x*quarter;symbol v=star c=red i=join;data homework5_2a; merge

14、homework5_2;input x1; lagx1=lag(x1); cards;0.9568670250.5361134110.432832080.7044871791.3918065820.6433360930.54104010.7044871791.1308428480.6433360930.54104010.7044871791.0438549360.5361134110.432832080.6038461540.7828912020.5361134110.54104011.1070512822.5226494292.2516763251.8395363412.0128205132

15、.3486736061.3938948680.973872181.0064102561.3918065820.6433360930.54104010.7044871790.9568670250.5361134110.54104010.6038461541.0438549360.7505587750.757456141.0064102561.304818671.0722268210.973872181.1070512821.304818671.0722268211.0820802011.610256412.2616856952.2516763252.4887844613.6230769234.3

16、493955674.8250206954.8693609024.9314102564.9583109474.6105753314.3283208024.4282051284.523371394.6105753314.5447368424.5288461544.523371394.3961299674.2201127824.1262820514.1754197453.7527938743.6790726823.5224358973.6534922773.6455711923.8954887224.3275641034.7843351245.1466887425.8432330836.541666

17、6676.9590329087.5055877488.0073934848.5544871798.7857790469.5428187099.5223057649.0576923088.6987911359.3283733449.5223057648.9570512829.046742789.5428187099.6305137849.0576923088.8727669589.2211506629.0894736848.4538461548.6118032249.113927989.3058897248.7557692318.5248153129.2211506629.0894736848.

18、2525641037.6549361998.1489238418.1156015047.6487179497.0460208197.612810437.4663533836.6423076925.9151779726.4333609276.7088972436.23974359; proc print data=homework5_2a; run;proc gplot data=homework5_2a; plot x1*quarter;symbol v=star c=red i=join;run;proc autoreg data=homework5_2a; model x1=t/dwpro

19、b;output out=out1 p=xx1 residual=res;run;proc print data=out1;proc gplot data=out1; plot x1*t=1 xx1*t=2/overlay;symbol1 v=star c=red i=join;symbol2 c=green v=none i=join;run;proc autoreg data=homework5_2a; model x1=t/nlag=5 backstep method=ml noint ;output out=out2 p=xx2; run;data homework5_2b;input

20、 st; cards;1.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9

21、241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326

22、384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850

23、220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731

24、.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.9936305731.1495850220.9326384870.9241459180.993630573;data homework5_2c; merge homework5_2a homework5_2b out2;x3=xx2*st;proc gplot data=homework5_2c;plot x*t=1 x3*t=2/overlay;symbol1 v=star c=red i=none;symbol2 c=blue v=no

25、ne i=join;run;proc autoreg data=homework5_2a; model x1=lagx1/lagdep=lagx1 nlag=2 noint;output out=out3 p=xx3;run;data homework5_2d; merge homework5_2a homework5_2b out3;x4=xx3*st;proc gplot data=homework5_2d; plot x*t=1 x4*t=2/overlay;symbol1 v=star c=red i=none;symbol2 c=blue v=none i=join;run;data homework5_22;merge homework5_2;lag12x=lag12(x);proc autoreg data=homework5_22; *;model x=lag12x;output out=out1_ p=xx_ ;run;data homework5_22a; merge homework5_22 out1_;x2_=x/xx_;lag1_12x=lag(x2_); proc pri