时序答案第三章习题12

1、习题 17（1）xiti17-scan(3.17.txt)Read 63 items ts.plot(xiti17)acf(xiti17)pacf(xiti17,xlab=jieshu,xlim=c(0,20),lag.max=20)时序没有显示明显的趋势性或者周期性，自相关系数二阶截尾，偏自相关系数拖尾是个平稳序列Box.test(xiti17,type=Ljung,lag=1)Box-Ljung testdata:xiti17X-squared = 6.1906, df = 1, p-value = 0.01284Box.test(xiti17,type=Ljung,lag=2)Box-L

2、jung testdata:xiti17X-squared = 12.064, df = 2, p-value = 0.002401Box.test(xiti17,type=Ljung,lag=3)Box-Ljung testdata:xiti17X-squared = 12.1592, df = 3, p-value = 0.006857Box.test(xiti17,type=Ljung,lag=6)Box-Ljung testdata:xiti17X-squared = 13.284, df = 6, p-value = 0.03874原假设，应该是非随机的（2）利用 AR（2）模型来拟

3、合xiti17.m1 pt(0.2541/0.1262,df=61,lower.tail=F)*2 1 0.04848292 pt(0.2374/0.1262,df=61,lower.tail=F)*2 1 0.06472759 pt(81.5418/5.2980,df=61,lower.tail=F)*2 1 1.120962e-22如果用AR（1）模型？Box.test(xiti17.m1$residuals,type=Ljung,lag=3,fitdf=2)调用 help(Box.test)Box-Ljung testdata:xiti17.m1$residualsX-squared =

4、 0.9072, df = 1, p-value = 0.3409Box.test(xiti17.m1$residuals,type=Ljung,lag=4,fitdf=2)Box-Ljung testdata:xiti17.m1$residualsX-squared = 1.3824, df = 2, p-value = 0.501Box.test(xiti17.m1$residuals,type=Ljung,lag=6,fitdf=2)Box-Ljung testdata:xiti17.m1$residualsX-squared = 1.4037, df = 4, p-value = 0.

5、8435P 值都大于显著性水平，所以模型残差是白噪声序列(3)a-predict(xiti17.m1,n.ahead=5)a$predTime Series:Start = 64End = 68Frequency = 11 92.44339 91.06750 86.55037 85.07581 83.62883该城市未来五年的降雪量$seAR（1）模型结果： Coefficients:ar1ercept0.330280.8793s.e.0.12364.1723pt(0.3302/0.1236,df=62,lower.tail=F)*2 1 0.009633125pt(80.8793/4.172

6、3,df=62,lower.tail=F)*2 1 5.266425e-28各参数均显著Time Series:Start = 64End = 68Frequency = 11 21.67058 22.35943 23.29727 23.48584 23.59899ts.plot(xiti17,xlim=c(0,70)lines(a$pred,col=red)在序列图上作出值U-a$pred+1.96*a$seL-a$pred-1.96*a$selines(U,col=blue)lines(L,col=blue)习题 18（1）xiti18-scan(3.18.txt) Read 74 ite

7、msts.plot(xiti18)acf(xiti18)pacf(xiti18)Box.test(xiti18,type=Ljung,lag=1)Box-Ljung testdata:xiti18X-squared = 10.1752, df = 1, p-value = 0.001423Box.test(xiti18,type=Ljung,lag=2)Box-Ljung testdata:xiti18X-squared = 15.71, df = 2, p-value = 0.0003878Box.test(xiti18,type=Ljung,lag=3)Box-Ljung testdata

8、:xiti18X-squared = 19.7793, df = 3, p-value = 0.0001886Box.test(xiti18,type=Ljung,lag=6)Box-Ljung testdata:xiti18X-squared = 29.8725, df = 6, p-value = 4.156e-05平稳序列，P 值较小，原假设，是非白噪声序列（2）利用 ARMA(1,2)模拟xiti18.m1-arima(xiti18,order=c(1,0,2),method=ML)xiti18.m1Call:arima(x = xiti18, order = c(1, 0, 2),

9、method = ML)Coefficients:ar1 0.9462s.e.0.0736ma1ma2ercept0.83520.1187-0.6959-0.05930.14260.1233sigma2 estimated as 0.0688:log likelihood = -6.25,aic = 22.51ercept=（1-0.9462）*0.8352=0.04493376得出的模型为： = 0.04493376 + 0.94621 + 0.69591 0.05932？pt(0.9462/0.0736,df=71,lower.tail=F)*2 1 3.259342e-20pt(0.69

10、59/0.1426,df=71,lower.tail=F)*2 1 6.295711e-06pt(0.0593/0.1233,df=71,lower.tail=F)*21 0.6320374pt(0.8352/0.1187,df=71,lower.tail=F)*2 1 1.007717e-09ARMA（1,2）模型中，MA2 参数不显如果采用 ARMA（1,1）模型，各参数显著Box.test(xiti18.m2$residuals,type=Ljung,lag=3,fitdf=2)Box-Ljung testdata:xiti18.m2$residualsX-squared = 0.196

11、1, df = 1, p-value = 0.6579Box.test(xiti18.m2$residuals,type=Ljung,lag=4,fitdf=2)Box-Ljung test如果用 ARMA（1,1）模型xiti18.m2-arima(xiti18,order=c(1,0,1),method=ML)xiti18.m2著Call:arima(x = xiti18, order = c(1, 0, 1), method = ML)Coefficients:ar1ma1ercept 0.9354-0.72650.8321s.e.0.08380.14760.1134sigma2 est

12、imated as 0.06903:log likelihood = -6.37,aic = 20.74pt(0.9354/0.0838,df=72,lower.tail=F)*2 1 2.30512e-17pt(0.7265/0.1476,df=72,lower.tail=F)*2 1 5.254795e-06pt(0.8321/0.1134,df=72,lower.tail=F)*2 1 2.622869e-10data:xiti18.m2$residualsX-squared = 0.1966, df = 2, p-value = 0.9064Box.test(xiti18.m2$res

13、iduals,type=Ljung,lag=5,fitdf=2)Box-Ljung testdata:xiti18.m2$residualsX-squared = 0.1986, df = 3, p-value = 0.9778Box.test(xiti18.m2$residuals,type=Ljung,lag=6,fitdf=2)Box-Ljung testdata:xiti18.m2$residualsX-squared = 0.4817, df = 4, p-value = 0.9753拟合的 arma（1,1）模型残差为白噪声序列 ercept=（1-0.9354）*0.8321 =

14、 0.05375366 + 0.93541 0.72651xiti18forecast-predict(xiti18.m2,n.ahead=5)xiti18forecast$predTime Series:Start = 75End = 79Frequency = 11 0.5588057 0.5764527 0.5929604 0.6084024 0.6228474+ $seTime Series:Start = 75End = 79Frequency = 11 0.2627327 0.2684068 0.2732753 0.2774654 0.2810806如果用AR（1）模型？习题 19

15、（1）xiti19-scan(3.19.txt) Read 201 itemsts.plot(xiti19)acf(xiti19)pacf(xiti19)ACF 拖尾、PACF 截尾，结合时序图认为其平稳Box.test(xiti19,type=Ljung,lag=1)Box-Ljung testdata:xiti19X-squared = 18.1473, df = 1, p-value = 2.045e-05Box.test(xiti19,type=Ljung,lag=2)Box-Ljung testdata:xiti19X-squared = 21.1223, df = 2, p-val

16、ue = 2.59e-05Box.test(xiti19,type=Ljung,lag=6)Box-Ljung testdata:xiti19X-squared = 31.0829, df = 6, p-value = 2.444e-05P 值较小，原假设，非白噪声序列（2）拟用AR（3）模型xiti19.m1-arima(xiti19,order=c(3,0,0),method=ML)xiti19.m1Call:arima(x = xiti19, order = c(3, 0, 0), method = ML)Coefficients:ar1-0.41180.0695ar2-0.29630.

17、0722ar3-0.18870.0701ercept84.12730.0987s.e.sigma2 estimated as 6.989:log likelihood = -480.77,aic = 971.53ercept=（1+0.4118+0.2963+0.1887）*84.1273=159.5727所以拟合的模型为 = 159.5727 + 0.41181 0.29632 0.18873各参数显著性检验：pt(0.4118/0.0695,df=198,lower.tail = F)*2 1 1.362645e-08pt(0.2963/0.0722,df=198,lower.tail = F)*2 1 5.935127e-05pt(0.1887/0.0701,df=198,lower.tail = F)*2 1 0.007713479P 值均较小，各参数皆是显著有效的（3）xiti19forecast-predict(xiti19.m1,n.ahead=1)xiti19forecast$predTime Series: Start = 202End = 202Frequency