I am using the forecast package in R and I want to know what the list of models are that the auto.arima function is going through in order to decide which ARIMA model fits best. Is there a way I can extract a list of all the models being tested to ensure it wasn't missing anything or so that it isn't so much of a blackbox?
Here is an example:
library(forecast)
fit <- auto.arima(WWWusage)
fit
Series: WWWusage
ARIMA(1,1,1)
Coefficients:
ar1 ma1
0.6504 0.5256
s.e. 0.0842 0.0896
sigma^2 estimated as 9.995: log likelihood=-254.15
AIC=514.3 AICc=514.55 BIC=522.08
plot(forecast(fit,h=20))
Thanks!
See here and here for documentation/explanation on the auto.arima() algorithm.
Related
First time using doing any forecasting and was looking into using auto.arima but I'm not really sure what the results mean, basically from 'Coefficients' onwards.
Can someone please be kind enough to give me an explanation
Series: total[, "total"]
ARIMA(2,0,0) with non-zero mean
Coefficients:
ar1 ar2 mean
1.1055 -0.4207 138805.107
s.e. 0.2020 0.2002 4664.756
sigma^2 = 53468931: log likelihood = -205.36
AIC=418.72 AICc=421.39 BIC=422.7
It means that the main model fitted by autoarima() converges to:
Y_t = 138805.107 - 0.4207·Y_(t-1) + 1.1055·Y_(t-2)
For the rest of the output, the standard deviation (s.e.) for each of the coefficients is also shown (necessary to see if the coefficients are significant). Finally, sigma^2 is the variance of the residual values, the log-likelihood is a "quality measure" of the model (the closer to zero the better) and necessary for comparing this fit against others) and the AIC, AICc and BIC are other "quality measures" based of the log-likelihood, sample size and amount of coefficients estimated.
I used auto.arima to forecast in my time series data set, and let the auto.arima function be arima.model. Then I typed in the console arima.model and it showed these values:
Series: train.data$M4
ARIMA(2,1,1)
Coefficients:
ar1 ar2 ma1
0.2138 0.0284 -0.9424
s.e. 0.0493 0.0482 0.0260
sigma^2 estimated as 6.9: log likelihood=-1361.02
Id like to know what does 2,1,1 mean? Is that P,D,Q? same with coefficients. Im not that good in both R and ARIMA so Id like to know. Thank you so much.
Id
AIC=2730.03 AICc=2730.1 BIC=2747.42
yes,You are correct.(2,1,1) is p,d,q found by auto.arima process using given Information criterion.which means you have 2 AR terms,1 difference and 1 Moving average term in your series.
I used an R code with an auto.arima function on a time series data set to forecast. From here, Id like to know how to find the p,d,q values for the arima. Is there a quick way to determine that, thank you.
The forecast::auto.arima() function was written to pick the optimal p, d, and q with respect to some optimization criterion (e.g. AIC). If you want to see which model was picked, use the summary() function.
For example:
fit <- auto.arima(lynx)
summary(fit)
Series: lynx
ARIMA(2,0,2) with non-zero mean
Coefficients:
ar1 ar2 ma1 ma2 mean
1.3421 -0.6738 -0.2027 -0.2564 1544.4039
s.e. 0.0984 0.0801 0.1261 0.1097 131.9242
sigma^2 estimated as 761965: log likelihood=-932.08
AIC=1876.17 AICc=1876.95 BIC=1892.58
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set -1.608903 853.5488 610.1112 -63.90926 140.7693 0.7343143 -0.01267127
Where you can see the particular specification in the second row of the output. In this example, auto.arima picks an ARIMA(2,0,2).
Note that I did this naively here for demonstration purposes. I didn't check whether this is an accurate representation of the dependency structure in the lynx data set.
Other than summary(), you could also use arimaorder(fit) to get the vector c(p,d,q) or as.character(fit) to get "ARIMA(p,d,q)".
I am performing a time series analysis on my data and I have run the auto arima function to determine the best coefficients to use in my ARIMA model.
model1 <- auto.arima(log(mydata_ts))
model1
Series: log(mydata_ts)
ARIMA(2,1,1)(1,0,0)[12]
Coefficients:
ar1 ar2 ma1 sar1
-1.1413 -0.3872 0.9453 0.7572
s.e. 0.1432 0.1362 0.0593 0.0830
sigma^2 estimated as 0.006575: log likelihood=48.35
AIC=-86.69 AICc=-85.23 BIC=-77.44
I understand that (2,1,1) in the result above refer to the values of p, d and q that will be used in the ARIMA model.
But what about (1,0,0)?
ARIMA(2,1,1)(1,0,0)[12] is seasonal ARIMA. [12] stands for number of periods in season, i.e. months in year in this case. (1,0,0) stands for seasonal part of model. Take a look at this.
This is out of my curiosity trying to compare time series input to an ARMA model and reconstructed series after an ARMA estimate is obtained. These are the steps I am thinking:
construct simulation time series
arma.sim <- arima.sim(model=list(ar=c(0.9),ma=c(0.2)),n = 100)
estimate the model from arma.sim, assuming we know it is a (1,0,1) model
arma.est1 <- arima(arma.sim, order=c(1,0,1))
also say we get arma.est1 in this form, which is close to the original (0.9,0,0.2):
Coefficients:
ar1 ma1 intercept
0.9115 0.0104 -0.4486
s.e. 0.0456 0.1270 1.1396
sigma^2 estimated as 1.15: log likelihood = -149.79, aic = 307.57
If I try to reconstruct another time series from arma.est1, how do I incorporate intercept or s.e. in arima.sim? Something like this doesn't seem to work well because arma.sim and arma.rec are far off:
arma.rec <- arima.sim(n=100, list(ar=c(0.9115),ma=c(0.0104)))
Normally we use predict() to check the estimate. But is this a legit way to look at the estimate?