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 am trying to carry out time series modelling and would like to copy the data input we get from running a data model onto a dataframe. For eg, when I run the following code:
a<-auto.arima(tt1[1:33,1])
summary(a)
I get the following output, which I want to copy to a dataframe to analyse in future-to see how many instances across different models have I got similar values.
Series: tt1[1:33, 1]
ARIMA(0,0,1) with non-zero mean
Coefficients:
ma1 mean
-0.4421 219.4943
s.e. 0.2079 27.2563
sigma^2 estimated as 79580: log likelihood=-232.1
AIC=470.19 AICc=471.02 BIC=474.68
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set 3.103363 273.4178 233.6786 NaN Inf 0.6599926 0.07472074
You can use "broom" package to solve this issue. Check the package vignette for more information. Here is a reproducible example:
library(broom)
library(forecast)
library(fpp2) # To get a data set to work on
# Fit a linear regression with AR errors
fit <- Arima(uschange[,"Consumption"], xreg = uschange[,"Income"], order = c(1,0,0))
# Use broom to convert summary output into a data.frame
tidy(fit) # coefficients and standard error
# Get main statistics
glance(fit)
I have a return time series in daily frequency, which is stationary(proofed by ADF test) , has no autocorrelation up to lag10(proofed by lbq test with lag10) and has ARCH effect(proofed by LM test). My initial though is just directly applying GARCH model. Rather than the usual procedure: first using ARMA(p,q) to get the residuals, and then fit GARCH to this ARMA residuals.
However, out of curiosity, I still use ARMA(p,q) model loop through (p,q) lags with range [0,1,..,10] to see whether ARMA(0,0) has the smallest AIC among all. After looping though those 121 (p,q) combinations, I find the smallest AIC does NOT belong to ARMA(0,0) model, but ARMA(2,7). Then I check the coef of this ARMA(2,7) model and find the may lags included are significant. The two AR lags are both significant at 1% level.
Now, I am quite confusing. Based on the result of lbq(10) test, I should use ARMA(0,0). Based on the results of smallest AIC of ARMA models, I should use ARMA(2,7). May I please ask, in this case, should I use ARMA(0,0) or ARMA(2,7)? My preference is to use ARMA(2,7), but how can I explain to others when they ask: why still use ARMA model when lbq test shows no autocorrelation?
Any of your kind thoughts is greatly appreciated!
Please see the code and results below
lbqtest(returns,'Lags',1:10)
I could also use the following code to only get autocorrelation up to lag10
lbqtest(returns,'Lags',10)
The p results of lbq(1) till lbq(10) are:
p =
0.3425 0.5612 0.4180 0.5356 0.6637 0.7696 0.7770 0.8448 0.8995 0.9198
The AIC results of ARMA(2,7) and ARMA(0,0) are
AIC AR MA
-1498.252431 2 7
-1494.028 0 0
The estimation result of ARMA(2,7) using R is
arima(x = returns, order = c(2, 0, 7))
Coefficients:
ar1 ar2 ma1 ma2 ma3 ma4 ma5 ma6 ma7 intercept
-1.6786 -0.8756 1.6808 0.8128 -0.1691 -0.1736 -0.1065 0.0419 0.0411 -0.0006
s.e. 0.0381 0.0308 0.0660 0.1044 0.1078 0.1097 0.1082 0.1017 0.0642 0.0015
I am trying to use the R-package forecast to fit arima models (with the function Arima) and automatically select an appropriate model (with the function auto.arima). I first estimated two possible models with the function Arima:
tt.1 <- Arima(x, order=c(1,0,1), seasonal=list(order=c(0,1,1)),
include.drift=F)
tt.2 <- Arima(x, order=c(1,0,1), seasonal=list(order=c(0,1,0)),
include.drift=F)
Then, I used the function auto.arima to automatically select an appropriate model for the same data. I fixed d=0 and D=1 just as in the two models above. Furthermore, I set the maximum to 1 for all other parameters, did not use approximation of the selection criterion and did not use stepwise selection (note that the settings I use here are only for demonstration of the strange behavior, not what I really intend to use). I used BIC as criterion for selection the model. Here is the function call:
tt.auto <- auto.arima(x, ic="bic", approximation=F, seasonal=T, stepwise=F,
max.p=1, max.q=1, max.P=1, max.Q=1, d=0, D=1, start.p=1,
start.q=1, start.P=1, start.Q=1, trace=T,
allowdrift=F)
Now, I would have expected that auto.arima selects the model with the lower BIC from the two models above or a model not estimated above by Arima. Furthermore, I would have expected that the output generated by auto.arima when trace=T is exactly the same as the BIC calculated by Arima for the two models above. This is indeed true for the second model but not for the first one. For the first model, the BIC calculated by Arima is 10405.81 but the screen output of auto.arima for the model (1,0,1)(0,1,1) is Inf. Consequently, the second model is selected by auto.arima although the first model has a lower BIC when comparing the two models estimated by Arima. Does anyone have an idea why the BIC calculated by Arima does not correspond to the BIC calculated by auto.arima in case of the first model?
Here is the screen output of auto.arima:
ARIMA(0,0,0)(0,1,0)[96] : 11744.63
ARIMA(0,0,0)(0,1,1)[96] : Inf
ARIMA(0,0,0)(1,1,0)[96] : Inf
ARIMA(0,0,0)(1,1,1)[96] : Inf
ARIMA(0,0,1)(0,1,0)[96] : 11404.67
ARIMA(0,0,1)(0,1,1)[96] : Inf
ARIMA(0,0,1)(1,1,0)[96] : Inf
ARIMA(0,0,1)(1,1,1)[96] : Inf
ARIMA(1,0,0)(0,1,0)[96] : 11120.72
ARIMA(1,0,0)(0,1,1)[96] : Inf
ARIMA(1,0,0)(1,1,0)[96] : Inf
ARIMA(1,0,0)(1,1,1)[96] : Inf
ARIMA(1,0,1)(0,1,0)[96] : 10984.75
ARIMA(1,0,1)(0,1,1)[96] : Inf
ARIMA(1,0,1)(1,1,0)[96] : Inf
ARIMA(1,0,1)(1,1,1)[96] : Inf
And here are summaries of the models calculated by Arima:
> summary(tt.1)
Series: x
ARIMA(1,0,1)(0,1,1)[96]
Coefficients:
ar1 ma1 sma1
0.9273 -0.5620 -1.0000
s.e. 0.0146 0.0309 0.0349
sigma^2 estimated as 867.7: log likelihood=-5188.98
AIC=10385.96 AICc=10386 BIC=10405.81
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set 0.205128 28.16286 11.14871 -7.171098 18.42883 0.3612059 -0.03466711
> summary(tt.2)
Series: x
ARIMA(1,0,1)(0,1,0)[96]
Coefficients:
ar1 ma1
0.9148 -0.4967
s.e. 0.0155 0.0320
sigma^2 estimated as 1892: log likelihood=-5481.93
AIC=10969.86 AICc=10969.89 BIC=10984.75
Training set error measures:
ME RMSE MAE MPE MAPE MASE ACF1
Training set 0.1942746 41.61086 15.38138 -8.836059 24.55919 0.49834 -0.02253845
Note: I am not allowed to make the data available. But I would be happy to provide more output or run modified calls of the functions if necessary.
EDIT: I now looked at the source code of auto.arima and found out that the behavior is caused by a check on the roots which sets the information criterion used for selecting a model to Inf if the model fails the check. The paper cited in the help for auto.arima confirms that (Hyndman, R.J. and Khandakar, Y. (2008) "Automatic time series forecasting: The forecast package for R", Journal of Statistical Software, 26(3), page 11). Sorry for the question, I should have read the paper before asking a question here!
auto.arima tries to find the best model subject to some constraints, avoiding models with parameters that are close to the non-stationarity and non-invertibility boundaries.
Your tt.1 model has a seasonal MA(1) parameter of -1 which lies on the non-invertibility boundary. So you don't want to use that model as it will lead to numerical instabilities. The seasonal difference operator is confounded with the seasonal MA operator.
Internally, auto.arima gives an AIC/AICc/BIC value of Inf to any model that doesn't satisfy the constraints to avoid it being selected.
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?