所以，我想制作具有异常值效应的 ARIMA。然后我已经完成了异常值的检测。但是，当我想对其建模时，我的输出出现错误
#Input data
dt<-read_excel("Docs/Kurs EUR .xlsx",
sheet = "KJ KB", range = "C1:C752")
dt
dt<-data.frame(dt)
#Make data is a time series data
tsKJ<-ts(dt$`Kurs Jual`)
#Produce Return data from time series data
RtKJ<-diff(log(tsKJ))
#Modelling ARIMA using Return Data
acf(RtKJ, main="ACF Return Kurs Jual")
pacf(RtKJ, main="PACF Return Kurs Jual")
armakj1<-Arima(RtKJ, order=c(1,0,1), method="ML")
summary(armakj1)
Series: RtKJ
ARIMA(1,0,1) with non-zero mean
Coefficients:
ar1 ma1 mean
0.8610 -0.7760 1e-04
s.e. 0.0761 0.0948 3e-04
sigma^2 = 2.759e-05: log likelihood = 2874.05
AIC=-5740.11 AICc=-5740.05 BIC=-5721.63
Training set error measures:
ME RMSE MAE MPE MAPE MASE
Training set 5.310198e-06 0.005241987 0.003742169 NaN Inf 0.7384854
ACF1
Training set 0.03451706
coeftest(armakj1)
z test of coefficients:
Estimate Std. Error z value Pr(>|z|)
ar1 8.6101e-01 7.6143e-02 11.3078 < 2.2e-16 ***
ma1 -7.7604e-01 9.4776e-02 -8.1881 2.653e-16 ***
intercept 8.8872e-05 3.0949e-04 0.2872 0.774
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#Outlier Detection
mkjo<-tso(RtKJ,
cval = 4,
types = c("AO","IO", "LS", "TC"))
mkjo
Series: RtKJ
Regression with ARIMA(0,0,1) errors
Coefficients:
ma1 TC46 AO53 TC55 IO65 AO74
0.1624 0.0159 0.0265 0.0244 -0.0276 -0.0197
s.e. 0.0387 0.0038 0.0048 0.0038 0.0049 0.0048
sigma^2 = 2.351e-05: log likelihood = 2935.55
AIC=-5857.11 AICc=-5856.96 BIC=-5824.77
Outliers:
type ind time coefhat tstat
1 TC 46 47 0.01592 4.165
2 AO 53 54 0.02647 5.493
3 TC 55 56 0.02445 6.398
4 IO 65 66 -0.02757 -5.639
5 AO 74 75 -0.01967 -4.126
#Modelling from Outlier Detection
m2.o<-arimax(x=RtKJ,
order=c(1,0,1),
xreg=data.frame(AO1=1*(seq(RtKJ)==53),
AO2=1*(seq(RtKJ)==74),
io=c(65), method="ML"))
Error in svd(na.omit(xreg)) : a dimension is zero
In addition: Warning message:
In storage.mode(xreg) <- "double" : NAs introduced by coercion
那么，根据我的输出，我应该如何处理语法或数据？是什么让我的语法出错？
我期待关于我制作的模型摘要的输出。