Hello,
The answer to your question is no, you don't have to do anything
special, except, of course, say how many steps ahead you want to predict.
To see this, run the second example in ?predict.Arima. It predicts 6
steps ahead, accounting for the frequency.
Hope this helps,
Rui Barradas
Em 15-03-2013 17:51, Erin Hodgess escreveu:
Dear R People:
I have the following situation. I have observations that are 128 samples
per second, which is fine. I want to fit them with ARIMA models, also fine.
My question is, please: when I do my forecasting, do I need to do anything
special to the "n.ahead" parm, please? Here is the initial setup:
xx <- ts(rnorm(128),start=0,freq=128)
str(xx)
Time-Series [1:128] from 0 to 0.992: -1.07 0.498 1.508 0.354 -0.497 ...
xx.ar <- arima(xx,order=c(1,0,0))
str(xx.ar)
List of 13
$ coef : Named num [1:2] -0.0818 0.0662
..- attr(*, "names")= chr [1:2] "ar1" "intercept"
$ sigma2 : num 1.06
$ var.coef : num [1:2, 1:2] 7.78e-03 -5.09e-05 -5.09e-05 7.07e-03
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:2] "ar1" "intercept"
.. ..$ : chr [1:2] "ar1" "intercept"
$ mask : logi [1:2] TRUE TRUE
$ loglik : num -185
$ aic : num 376
$ arma : int [1:7] 1 0 0 0 128 0 0
$ residuals: Time-Series [1:128] from 0 to 0.992: -1.133 0.338 1.477 0.406
-0.54 ...
$ call : language arima(x = xx, order = c(1, 0, 0))
$ series : chr "xx"
$ code : int 0
$ n.cond : int 0
$ model :List of 10
..$ phi : num -0.0818
..$ theta: num(0)
..$ Delta: num(0)
..$ Z : num 1
..$ a : num 0.156
..$ P : num [1, 1] 0
..$ T : num [1, 1] -0.0818
..$ V : num [1, 1] 1
..$ h : num 0
..$ Pn : num [1, 1] 1
- attr(*, "class")= chr "Arima"
predict(xx.ar,n.ahead=3)
$pred
Time Series:
Start = c(1, 1)
End = c(1, 3)
Frequency = 128
[1] 0.05346814 0.06728105 0.06615104
$se
Time Series:
Start = c(1, 1)
End = c(1, 3)
Frequency = 128
[1] 1.028302 1.031737 1.031760
Thanks for any help.
Sincerely,
Erin
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