我一直在使用R编程语言使用ARIMA模型进行将来的预测,但是当我从预测包中运行ARIMA模型时,我的标准误差未正确计算,并出现了NA
错误。您能否建议我以解决此问题?
arima(5,1,5)
arima515 <- Arima(GSPC$SP500,
order = c(5, 1, 5),
include.constant = TRUE,
optim.control = list(maxit = 500),)
coeftest(arima515)
输出:
z test of coefficients:
Estimate Std. Error z value Pr(>|z|)
ar1 0.074793 NA NA NA
ar2 0.142322 NA NA NA
ar3 0.754132 NA NA NA
ar4 0.179091 NA NA NA
ar5 -0.370530 NA NA NA
ma1 -0.122067 NA NA NA
ma2 -0.180075 NA NA NA
ma3 -0.751949 NA NA NA
ma4 -0.147119 NA NA NA
ma5 0.381992 NA NA NA
drift 0.387899 0.132093 2.9366 0.003319 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Warning message:
In sqrt(diag(se)) : NaNs produced
正如您在上面看到的那样,错误,z
值和pr
为NA
。
我不知道我是否正确理解了这个问题,但是,如果只需要参数标准错误,则只能像这样调用arima515
或summary(arima515)
arima515 <- Arima(cop[,1],
order = c(5, 1, 5),
include.constant = TRUE,
optim.control = list(maxit = 500))
summary(arima515)
Series: cop[, 1]
ARIMA(5,1,5) with drift
Coefficients:
ar1 ar2 ar3 ar4 ar5 ma1 ma2 ma3
-0.7413 -1.2671 -0.5753 -0.3700 0.0221 -0.2412 0.5399 -0.7129
s.e. 1.4269 0.6982 1.1093 0.5289 0.0852 1.4266 1.1724 0.8139
ma4 ma5 drift
-0.1744 -0.4114 -1e-04
s.e. 1.0093 0.5184 1e-04
sigma^2 estimated as 0.5005: log likelihood=-801.31
AIC=1626.63 AICc=1627.05 BIC=1682.05
Training set error measures:
ME RMSE MAE MPE MAPE MASE
Training set -0.005256166 0.7017616 0.513783 -Inf Inf 0.6721109
ACF1
Training set -0.0009624799
您将获得参数估计值和请参阅coeftest
的文档,并查看可用于此功能的对象。相反,如果您想为您的Arima模型测试其他任何东西,请告诉我们您想做什么。