Question

我花了很多时间，而且我认为我无处可去。我尝试在predict方法中使用预测区间。在这里，我试图根据从日志返回生成的AR模型来预测日志返回的未来值。

> model_1 <- ar(data1[,'Log Return'], aic = TRUE, order.max = NULL, method = c("mle"))
> predict(model_1, data1[,'Log Return'], n.ahead = 8, level = 0.95, interval = "prediction")

但是，R拒绝在输出中给出任何预测间隔：

$pred
Time Series:
Start = 264 
End = 271 
Frequency = 1 
[1] 0.005904645 0.006259840 0.007770717 0.007785322 0.007944444 0.007832193 0.007811966 0.007772691

$se
Time Series:
Start = 264 
End = 271 
Frequency = 1 
[1] 0.009038997 0.009569202 0.009830485 0.009831987 0.009831988 0.009834717 0.009835227 0.009835437

我到处搜索都无济于事，我开始怀疑predict方法不能给出AR模型的预测间隔。

Answer 1

由于我们没有您的数据，因此我将使用ar帮助文件中的模型进行预测。您不希望predict用于预测间隔。使用n.ahead中的ar参数，并从forecast包中获取预测间隔。

> (sunspot.ar <- ar(sunspot.year, n.ahead = 8))

Call:
ar(x = sunspot.year, n.ahead = 8)

Coefficients:
      1        2        3        4        5        6        7        8        9  
 1.1305  -0.3524  -0.1745   0.1403  -0.1358   0.0963  -0.0556   0.0076   0.1941  

Order selected 9  sigma^2 estimated as  267.5

> library(forecast)
> forecast(sunspot.ar, levels = 95)
     Point Forecast      Lo 80     Hi 80      Lo 95     Hi 95
1989      135.25933 114.299317 156.21935 103.203755 167.31491
1990      148.09051 116.455825 179.72519  99.709436 196.47158
1991      133.98476  96.875479 171.09404  77.231012 190.73851
1992      106.61344  68.200200 145.02667  47.865460 165.36141
1993       71.21921  32.673811 109.76461  12.269108 130.16932
1994       40.84057   2.193737  79.48741 -18.264662  99.94581
1995       18.70100 -20.206540  57.60853 -40.802945  78.20494
1996       11.52416 -27.675854  50.72418 -48.427088  71.47541
1997       27.24208 -12.115656  66.59982 -32.950383  87.43454
1998       56.99888  17.600443  96.39731  -3.255828 117.25359

如何在R中使用AR模型找到预测的预测间隔？

1 个答案: