Displaying 7 results from an estimated 7 matches for "arima1".
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arima
2011 Jul 04
1
forecast: bias in sampling from seasonal Arima model?
...0.050722, -0.109773, -0.110301,
-0.095433, 0.071133, 0.023793, 0.192476, 0.057746)
First, a CORRECT model, containing a seasonal MA component but no seasonal
AR component. After estimation, I forecast for 1 time-step, and I take the
mean of sampling 10000 times from the same model:
> my.arima1 <- Arima(x, order=c(3,0,0), seasonal=list(order=c(0,0,2),
period=7), include.mean=FALSE)
> forecast(my.arima1, 1)
Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
251 -0.03143283 -0.1882245 0.1253589 -0.271225 0.2083594
> set.seed(1827) ; mean(sapply(seq_len(10000), function(...
2007 Feb 08
2
(no subject)
Hi.
I hope you can help me...
I have fitted the following ARIMA model:
arima1<-arima(bigspring$log.volume, order=c(0,1,2))
I need to predict 30 days ahead. I used following code
predict(arima1,n.ahead=30,se=T)
However I get 30 predictions, but from predictions 2:30 I get the same
predictions. Why is this? What am I doing wrong
Thanks
Catherine
KSS Ltd
Seve...
2010 Sep 05
8
R time series analysis
...est set and training set. I would then like to
build an ARIMA model on the training set and apply this model on test set.
Below is some code:
[CODE]
data= read.table("A.txt",sep=",")
attach(data)
training = data[1:120, 6]
test = data[121:245, 6]
ts1 = ts(training)
ts2 = ts(test)
arima1 = arima(ts1)
arima2 = arima(ts2)
[/CODE]
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2010 Sep 06
2
how do I transform this to a for loop
arima1 = arima(data.ts[1:200], order = c(1,1,1))
arima2 = arima(data.ts[5:205], order = c(1,1,1))
arima3 = arima(data.ts[10:210], order = c(1,1,1))
arima4 = arima(data.ts[15:215], order = c(1,1,1))
arima5 = arima(data.ts[20:220], order = c(1,1,1))
arima6 = arima(data.ts[25:225], order = c(1,1,1))
arima7 =...
2004 Sep 30
1
Using try()
...) function. Currently I am running a loop which uses arima() for some values of p and q, which sometimes crashes. When it crashes, I want the program to just ignore it and move on to the next values to loop through. I currently have this, looping through a range of values for p and q:
lo = try( arima1 <- arima( y, order=c( p, 0, q ) )
if( !inherits( lo, "try-error" ) ){
... code here ...
}
This works perfectly for my purposes, with one exception: Everytime the program invokes try(), I have to press Esc in the R console for the program to continue. How can I avoid this...
2010 Sep 11
0
outputting arima models
sseq <- c(1, seq(5, 120, by = 5))
for(i in 1:length(sseq)){
assign(paste("arima", i, sep=""), arima0(data.ts[sseq[i]:(sseq[i]+115)],
order=c(1,1,1)))
}
pred1 = predict(arima1, n.ahead = 5, se.fit = TRUE)$pred
how do I traverse the arima models so I repeat the above prediction
procedure(bold) on all arima models. Also, how do i automatically create
one huge vector of the predictions from all the arima models?
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2007 May 07
0
Analyzing "Stacked" Time Series
...?s easy enough to do something like,
fit1 <- lm(X ~ YO + YD + ZO + ZD)
But anything involving lags is not correct, since it seems?at best?to be
treating my ?stacked? time series (i.e., 6 series of length 10) as one
series of length 60. For example, these give questionable, if any, output.
arima1 <- arima(X, order = c(1, 0, 0))
fit.gls001 <- gls(X ~ YO + YD + ZO + ZD,
correlation = corARMA(p = 2), method = "ML")
fit.gls002 <- gls(X ~ YO + YD + ZO + ZD +
lag(YO) + lag(YD) + lag(ZO) + lag(ZD),
correlation = corARMA(p = 1), method = "ML")
ar001 <- ar.ols(...