时间序列作业

#2--11
prod=c(47, 64, 23, 71, 38, 64, 55, 41, 59, 48,
71, 35, 57, 40, 58, 44, 80, 55, 37, 74,
51, 57, 50, 60, 45, 57, 50, 45, 25, 59,
50, 71, 56, 74, 50, 58, 45, 54, 36, 54,
48, 55, 45, 57, 50, 62, 44, 64, 43, 52,
38, 59, 55, 41, 53, 49, 34, 35, 54, 45,
68, 38, 50, 60, 39, 59, 40, 57, 54, 23)

mean(prod)#51.12857
var(prod)#141.8238
fivenum(prod)#23.0 44.0 51.5 58.0 80.0
plot(prod,type="b",xlab="i",ylab=expression(X[i]),col=2)#see picture1
library(tseries)#平稳性检验
adf.test(prod)#p-value = 0.01 stationary
par(mfrow=c(1,2))
acf(prod)
pacf(prod) #see picture2

ar1<-arima(prod,order=c(1,0,0))
ar2<-arima(prod,order=c(2,0,0))
ma2<-arima(prod,order=c(0,0,2))
ar1$aic#537.9579
ar2$aic#537.6573
ma2$aic#538.7055
#选择AR（2）模型,其实我们也可认为他们都不截尾，用arma去做亦可，但aic略大于ar2
ar2
#Coefficients:
#          ar1     ar2  intercept
#      -0.3407  0.1873    51.2263
#s.e.   0.1218  0.1223     1.1007
#sigma^2 estimated as 112.7:  log likelihood = -264.83,  aic = 537.66
predict(ar1, n.ahead=12)# [1] 63.11143 46.30157 53.34623 50.39396 51.63120 51.11270 51.32999 51.23893 51.27709 51.26109 51.26780 51.26499
predict(ar2, n.ahead=12)# [1] 61.36382 42.48392 56.10439 47.92625 53.26463 49.91348 52.05548 50.69778 51.56170 51.01297 51.36180 51.14013
predict(ma2, n.ahead=12)# [1] 61.59200 43.03600 51.16947 51.16947 51.16947 51.16947 51.16947 51.16947 51.16947 51.16947 51.16947 51.16947

#其实通过扩展的自相关函数来判断，ar2，ma2,arma(1,1)的效果相当
library(TSA)
eacf(prod)

#12
data(sunspots)
mean(sunspots)#51.26596
var(sunspots)#1887.813
fivenum(sunspots)#0.00  15.70  42.00  74.95 253.80
plot(sunspots,type="b",xlab="i",ylab=expression(X[i]),col=2)#see picture3
#趋势分析
decompose(sunspots)#$figure
#[1] -2.1619020 -0.1085010 -1.2754348 -0.2400181  1.1422112  0.2472504
#[7]  0.1120830  0.7560467  1.0681372  0.6220011 -0.5170302  0.3551564
#没有很强的季节性，季节性指数大多在0附近。
plot(decompose(sunspots))#see picture 4
#没有季节性，趋势性为正弦波动,没有增长趋势，这也是符合自然常识的。
adf.test(sunspots)#p-value = 0.01 stationary
#数据仍然有非常好的平稳性，可以忽略季节性吗？
par(mfrow=c(1,2))
acf(sunspots)
pacf(sunspots)#see picture 5
#这里可以看出acf没有明显的周期性，
sunspots1<-diff(sunspots,lag=1)#从acf来看，一阶差分是有必要的
adf.test(sunspots1)#stationary
par(mfrow=c(1,2))
acf(sunspots1)
pacf(sunspots1)
stl(sunspots1,"per")
plot(stl(sunspots1,"per"))
par(mfrow=c(2,1))
acf(sunspots1)
pacf(sunspots1)#得到了非常好的平稳性数据，那么前面做的平稳性检验错了吗？
#从acf来看，差分后的数据应该是一个ar（2）模型，给出模型估计
ar2<-arima(sunspots1,order=c(2,0,0))
ar2
#Coefficients:
#          ar1      ar2  intercept
#      -0.3629  -0.2064    -0.0096
#s.e.   0.0184   0.0184     0.1926
#sigma^2 estimated as 257.3:  log likelihood = -11823.37,  aic = 23654.74
#建模确定为arima（2,1,0）
sunspots.fit1<-arima(sunspots,order=c(2,1,0))
#arima(x = sunspots, order = c(2, 1, 0))
#Coefficients:
#          ar1      ar2
#      -0.3629  -0.2064
#s.e.   0.0184   0.0184
#sigma^2 estimated as 257.3:  log likelihood = -11823.37,  aic = 23652.74
library(forecast)
sunspots.forecast1 <- forecast(sunspots.fit1,12)
plot.forecast(sunspots.forecast1)
#课上我们曾确定使用ar（2）模型，这个模型好吗？
sunspots.fit<-arima(sunspots,order=c(2,0,0))
#arima(x = sunspots, order = c(2, 0, 0))
#Coefficients:
#         ar1     ar2  intercept
#      0.6704  0.2722    51.2642
#s.e.  0.0181  0.0181     5.2842
#sigma^2 estimated as 262.8:  log likelihood = -11858.25,  aic = 23724.5
#从aic判定角度来看，差别并不是十分大，这是不是说明了通过平稳性检验未必见得是真正平稳的，可能一阶差分后效果更好？
library(forecast)
sunspots.forecast <- forecast(sunspots.fit,12)
plot.forecast(sunspots.forecast)#see picture 6
#用pandit-wu法处理
aico=resi<-matrix(0,nrow=6)
for (i in 1:6) {
a<-arima(sunspots,order=c(2*i,1,2*i-1))
aico[i]<-a$aic
resi[i]=sum(a$resi^2)
}
aico
# 首先对ARMA(2,1)和ARMA(4,3)作F检验：
F<-(resi[1,1]-resi[2,1])*2809/(4*resi[2,1])  #s=4,p+q=7 N1=2820-2-2 length(sunspots)
F<qf(0.95,4,2089)#FALSE
#这与课堂上的结论不符，ARMA（4,3）是有显著改进的
F<-(resi[4,1]-resi[5,1])*2791/(4*resi[5,1])
F<qf(0.95,4,2791)#FALSE
F<-(resi[5,1]-resi[6,1])*2785/(4*resi[6,1])
F<qf(0.95,4,2785)#TRUE
arima(sunspots,order=c(10,0,9))#ar10 0.3924 se 0.0283 利用t检验知最高项系数不为0
aico1=resi1<-matrix(0,nrow=9)
for (i in 1:9) {
a<-arima(sunspots,order=c(10,0,i))
aico1[i]<-a$aic
resi1[i]=sum(a$resi^2)
}
aico1
F<-(resi1[8,1]-resi1[9,1])*2785/(resi1[8,1])
F<qf(0.95,4,2785)#FALSE,说明这里ma的阶数不可以降，所以选择模型arma（10,9）
sunspots.fit2<-arima(sunspots,order=c(10,0,9))#AIC=23473.9
#从AIC的角度来看，这个模型确实更好，但也更加的繁琐，从预测来看效果也没提高多少
library(forecast)
sunspots.forecast2 <- forecast(sunspots.fit2,12)
plot.forecast(sunspots.forecast2)#see picture 7
#结论：pandit-wu法解决问题是对数据平稳性要求较高，单纯通过平稳性检验是不够的，结合acf与pacf来看稳定性更好些
#pandit-wu运算复杂，计算量大，F检验不易通过，计算机求解尚且不易，何况手算？易导致较高阶的模型
#R对规模较大的数据处理速度较慢，用matlab，sas的效果可能好些。
#对残差进行box检验对模型判断还是有好处的，可以发现我们选择的ARIMA(2,1,0),ARMA(10,9),都是合适的模型
Box.test(sunspots.fit1$residuals)#ARIMA(2,1,0)   p-value = 0.1031
Box.test(sunspots.fit$residuals)#AR(2) p-value = 0.006665
Box.test(sunspots.fit2$residuals)#ARMA(10,9)  p-value = 0.9418
#说明了课上讲的AR（2）不是最优拟合，甚至是不好的拟合，再次说明AIC准则不太靠谱，（几个模型aic差别不大）
#综上，sunspots模型确定为ARIMA（2,1,0）最好