解释 ARIMA 模型的预测 [英] Explaining the forecasts from an ARIMA model
问题描述
我试图向自己解释将 ARIMA 模型应用于时间序列数据集的预测结果.数据来自M1-Competition，系列为MNB65.我正在尝试将数据拟合到 ARIMA(1,0,0) 模型并获得预测.我正在使用 R.以下是一些输出片段:
I am trying to explain to myself the forecasting result from applying an ARIMA model to a time-series dataset. The data is from the M1-Competition, the series is MNB65. I am trying to fit the data to an ARIMA(1,0,0) model and get the forecasts. I am using R. Here are some output snippets:
> arima(x, order = c(1,0,0))
Series: x
ARIMA(1,0,0) with non-zero mean
Call: arima(x = x, order = c(1, 0, 0))
Coefficients:
ar1 intercept
0.9421 12260.298
s.e. 0.0474 202.717
> predict(arima(x, order = c(1,0,0)), n.ahead=12)
$pred
Time Series:
Start = 53
End = 64
Frequency = 1
[1] 11757.39 11786.50 11813.92 11839.75 11864.09 11887.02 11908.62 11928.97 11948.15 11966.21 11983.23 11999.27
我有几个问题:
(1) 我如何解释虽然数据集显示出明显的下降趋势，但该模型的预测却呈上升趋势?这也发生在 ARIMA(2,0,0) 上，这是最适合使用 auto.arima(预测包)的数据和 ARIMA(1,0,1) 模型的 ARIMA.
(1) How do I explain that although the dataset shows a clear downward trend, the forecast from this model trends upward? This also happens for ARIMA(2,0,0), which is the best ARIMA fit for the data using auto.arima (forecast package) and for an ARIMA(1,0,1) model.
(2) ARIMA(1,0,0) 模型的截距值为 12260.298.截距不应该满足方程:C = mean * (1 - sum(AR coeffs))，在这种情况下，值应该是715.52.我一定在这里遗漏了一些基本的东西.
(2) The intercept value for the ARIMA(1,0,0) model is 12260.298. Shouldn't the intercept satisfy the equation: C = mean * (1 - sum(AR coeffs)), in which case, the value should be 715.52. I must be missing something basic here.
(3) 这显然是一个具有非平稳均值的序列.为什么 AR(2) 模型仍然被 auto.arima 选为最佳模型?有没有直观的解释?
(3) This is clearly a series with non-stationary mean. Why is an AR(2) model still selected as the best model by auto.arima? Could there be an intuitive explanation?
谢谢.
推荐答案
没有 ARIMA(p,0,q) 模型允许趋势，因为模型是固定的.如果您确实想包含趋势，请使用带有漂移项的 ARIMA(p,1,q) 或 ARIMA(p,2,q).auto.arima() 暗示 0 差异的事实通常表明没有明显的趋势.
No ARIMA(p,0,q) model will allow for a trend because the model is stationary. If you really want to include a trend, use ARIMA(p,1,q) with a drift term, or ARIMA(p,2,q). The fact that auto.arima() is suggesting 0 differences would usually indicate there is no clear trend.
arima() 的帮助文件显示截距实际上是均值.也就是说，AR(1) 模型是 (Y_t-c) = ϕ(Y_{t-1} - c) + e_t 而不是 Y_t = c + ϕY_{t-1} + e_t 正如你所料.
The help file for arima() shows that the intercept is actually the mean. That is, the AR(1) model is (Y_t-c) = ϕ(Y_{t-1} - c) + e_t rather than Y_t = c + ϕY_{t-1} + e_t as you might expect.
auto.arima() 使用单位根检验来确定所需的差异数.所以检查单位根测试的结果，看看发生了什么.如果您认为单位根测试没有导致合理的模型，您始终可以在 auto.arima() 中指定所需的差异数量.
auto.arima() uses a unit root test to determine the number of differences required. So check the results from the unit root test to see what's going on. You can always specify the required number of differences in auto.arima() if you think the unit root tests are not leading to a sensible model.
以下是对您的数据进行两次测试的结果:
Here are the results from two tests for your data:
R> adf.test(x)
Augmented Dickey-Fuller Test
data: x
Dickey-Fuller = -1.031, Lag order = 3, p-value = 0.9249
alternative hypothesis: stationary
R> kpss.test(x)
KPSS Test for Level Stationarity
data: x
KPSS Level = 0.3491, Truncation lag parameter = 1, p-value = 0.09909
因此，ADF 表示强烈的非平稳性(在这种情况下为原假设)，而 KPSS 并不完全拒绝平稳性(该测试的原假设).auto.arima() 默认使用后者.如果您想要第一次测试，可以使用 auto.arima(x,test="adf").在这种情况下，它建议模型 ARIMA(0,2,1) 确实有趋势.
So the ADF says strongly non-stationary (the null hypothesis in that case) while the KPSS doesn't quite reject stationarity (the null hypothesis for that test). auto.arima() uses the latter by default. You could use auto.arima(x,test="adf") if you wanted the first test. In that case, it suggests the model ARIMA(0,2,1) which does have a trend.
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