R 中的 ARIMA 模型 - 数据营练习 -
在 datacamp 练习中,在我看来根本不清楚答案
- 我们有数据: globaltemp ,由 asta 包提供,我也使用 astsa 包函数
- 该练习说使用 sarima 评估此数据并决定两个模型中的哪一个提供底部最适合:
sarima(globtemp,1,1,1) AIC 和 BIC 值:分别为 -1.716773 和 -1.630691
(此处为 sarima(globtemp,1,1,1) 的 asta 包的图形结果)
sarima(globtemp,0,1,2) AIC 和 BIC 值:分别为 -1.723268 和 -1.637185 (sarima(globtemp,0,1,2) 的 asta 包的图形结果在这里)
Datacamp 表示最适合的模型是第二个。然而,AIC 和 BIC 值较小的模型是 sarima(globtemp,1,1,1)。为什么正确答案是第二个模型?难道是那里出了错吗?
感谢您抽出时间!
我查看了 AIC 和 BIC 值,我期望第一个模型: sarima(globaltemp,1,1,1) 是与 sarima(globtemp,0,1,2) 相比的最佳模型;然而,在 datacamp 中,它说最好的是 sarima(globtemp,0,1,2)
in a datacamp exercise it seems to me not clear at all the answer
- We have the data: globaltemp , provide by asta package, also I use astsa package functions
- The exercise says that evaluate this data using sarima and decide which of the two models provide on the bottom have the best fit:
sarima(globtemp,1,1,1)
AIC and BIC values: -1.716773 and -1.630691, respectively
(graphic result with asta package for sarima(globtemp,1,1,1) is here)
sarima(globtemp,0,1,2)
AIC and BIC values: -1.723268 and -1.637185, respectively
(graphic result with asta package for sarima(globtemp,0,1,2) is here)
Datacamp says the model with the best fit is the second. However, the model with the less AIC and BIC value is sarima(globtemp,1,1,1). Why the correct answer is the second model? Is it a mistake there?
Thanks for your time!
I look at the AIC and BIC values, and I expect that the first model: sarima(globaltemp,1,1,1) is the best model in comparison of sarima(globtemp,0,1,2); however, in datacamp, it says that the best is sarima(globtemp,0,1,2)
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两种模型非常相似,但我认为越简单越好,因为 AIC 和 BIC 值的差异可以忽略不计。
在本例中,ARIMA(0,1,2) 更简单,因为它缺少公式 ARIMA(p,d,q) 中的自回归分量
p
。请参阅 fpp2 了解更多信息。
Both models are pretty similar but I guess the simpler the better due to the negligible difference in AIC and BIC values.
In this case ARIMA(0,1,2) its simpler because it lacks the autoregresive component
p
in the formulation ARIMA(p,d,q).See fpp2 for more info.