Displaying 4 results from an estimated 4 matches for "sar2".
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2017 Jun 20
1
How to write an estimated seasonal ARIMA model from R output?
...g command.
arima (x, order = c(p,d,q), seasonal =list(order=c(P,D,Q), period=s)
How can I write an estimated seasonal ARIMA model from the outputs. To be specifically, which sign to use? I know R uses a different signs from S plus.
Is it correct that the model is:
(1-ar1*B-ar2*B^2-...)(1-sar1*B^s-sar2*B^2s-....)(1-B)^d(1-B^s)^D X_t=(1+ma1*B+ma2*B^2+...)(1+sma1*B^s+sma2*B^2s+....) a_t
For example:
> m1=arima(koeps,order=c(0,1,1),seasonal=list(order=c(0,1,1),period=4))
> m1
Call:
arima(x = koeps, order = c(0, 1, 1), seasonal = list(order = c(0, 1, 1), period = 4))
Coefficients:
ma...
2009 Jan 23
1
forecasting error?
...0, start.q = 0, start.P = 0, start.Q
= 0, stationary = FALSE, stepwise = TRUE)
Coefficients:
ar1 ar2 ma1 ma2 ma3 ma4 sar1
0.0639 -0.7820 -1.2103 1.2236 -0.9511 0.2357 1.0031
s.e. 0.0686 0.0582 0.1098 0.1558 0.1568 0.1007 0.0716
sar2 sma1 drift
-0.0711 -0.8963 -780.9456
s.e. 0.0747 0.0608 403.2112
sigma^2 estimated as 10202381: log likelihood = -1100.61
AIC = 2206.69 AICc = 2209.23 BIC = 2236.98
>for<-forecast(mods,h=12,newxreg=(1+length(x[[1]])):(length(x[[1]]+12)))
#forecast
and as I said...
2023 Jan 05
1
R 'arima' discrepancies
...pca_tsb%>%
model(ARIMA(servicos ~ 1 + pdq(2, 0, 1) + PDQ(2, 0, 2), method =
"ML", optim.method = "BFGS", optim.control = list(maxit = 1000)))
%>% report() summary(modelo) |*|Series: . ARIMA(2,0,1)(2,0,2)[12]
with non-zero mean Coefficients: ar1 ar2 ma1 sar1 sar2 sma1 sma2
mean 0.7534 0.0706 -0.5705 0.1759 0.7511 0.3533 -0.6283 0.5001 s.e.
NaN NaN 0.0011 NaN NaN NaN NaN 0.1996 sigma^2 = 0.05312: log
likelihood = 1.75 AIC=14.5 AICc=15.33 BIC=45.33 Training set error
measures: ME RMSE MAE MPE MAPE MASE ACF1 Training set -0.006082139
0.2263...
2011 Dec 17
0
auto.arima from the Forecast package
...arima.
I run the following command
auto.arima(drivers,ic="aic",d=1,D=1,max.order=10,max.p=5,max.q=5,max.P=5,max.Q=5,stepwise=FALSE,allowdrift=FALSE)
and I get the following output :
Series: drivers
ARIMA(0,1,1)(5,1,1)[12]
Coefficients:
ma1 sar1 sar2 sar3 sar4 sar5 sma1
-0.6421 -0.1341 -0.2063 -0.1076 -0.2361 -0.2205 -0.7387
s.e. 0.0718 0.1273 0.1061 0.1063 0.0938 0.1029 0.1312
sigma^2 estimated as 16916: log likelihood=-1137.18
AIC=2290.37 AICc=2291.21 BIC=2315.87
So the fitted model has an AIC...