当时间序列不平稳时建议ARIMA(2,0,0)的auto.arima()函数

时间:2019-12-01 00:20:23

标签: r time-series

我目前正在处理VALE3.SA https://finance.yahoo.com/quote/VALE3.SA/history?p=VALE3.SA的时间序列。

在R上,程序包预测的auto.arima()函数返回ARIMA(2,0,0)。但是当我使用adf.test()和pp.test()时,它们表示不稳定。我还尝试了diff(),这使我的序列变成平稳的,但是当我尝试强制auto.arima使用1个微分时,它表明ARIMA(0,1,0)。我应该使用什么型号?没有平稳性,ARIMA(2,0,0)甚至有效吗?

auto <- auto.arima(tsvale[,"Close"])

ARIMA(2,0,0) with non-zero mean 

Coefficients:
         ar1      ar2     mean
      1.1156  -0.1518  35.9059
s.e.  0.0739   0.0742   6.0635

sigma^2 estimated as 10.89:  log likelihood=-464.95
AIC=937.91   AICc=938.14   BIC=950.63 ```


> adf.test(tsvale[,"Close"])

    Augmented Dickey-Fuller Test

data:  tsvale[, "Close"]
Dickey-Fuller = -2.2874, Lag order = 5, p-value = 0.4561
alternative hypothesis: stationary

> pp.test(tsvale[,"Close"])

    Phillips-Perron Unit Root Test

data:  tsvale[, "Close"]
Dickey-Fuller Z(alpha) = -8.1416, Truncation lag parameter = 4, p-value = 0.649
alternative hypothesis: stationary```



auto.arima(tsvale[,"Close"], d = 1)
Series: tsvale[, "Close"] 
ARIMA(0,1,0) 

sigma^2 estimated as 11.13:  log likelihood=-464.39
AIC=930.78   AICc=930.8   BIC=933.96 ```

0 个答案:

没有答案