search for: arima

HI, there, When i use R, i tried to use function arima(), it complains: Error: couldn't find function "arima" But when I type "help.search("arima") ", I got arima() poped up.. arima(ts) ARIMA Modelling of Time Series arima.sim(ts) Simulate from an ARIMA Model arima0(ts) ARIMA Mode...

Dear R-helpers, I would appreicate if someone can help me on the transfer parameter in ARIMAX and also see what I am doing is correct. I am using ARIMAX with 2 Exogeneous Variables and 10 years data are as follows: DepVar Period, depVar, IndepVar1 Period, indepVar1, IndepVar2 Period, indepVar2 Jan 1998,708,Jan 1998,495,Jan 1998,245.490 Feb 1998,670,Feb 1998,421.25,Feb 1998,288.170 Mar 19...

Hello everybody! I have a problem when I try to perform a forecast of an ARIMA model produced by an auto.arima function. Here is what I'm doing: c<-auto.arima(fil[[1]],start.p=0,start.q=0,start.P=0,start.Q=0,stepwise=TRUE,stationary=FALSE,trace=TRUE) # fil[[1]] is time series of monthly data ARIMA(0,0,0)(0,1,0)[12] with drift : 1725.272 ARIMA(0,0,0)(0,1...

Hi, I have been using this site ( http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm) to help me with some ARIMA modelling in R. Unfortunately the methods mentioned do not appear to work with second order differencing; arima(*, 2, *). I have used some dummy data to illustrate my point. When I use the xreg=... method, the estimate of intercept is *way* off. This can be seen by the high s.e but I *have* chec...

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...

Dear R-helpers. Would appreciate if someone can explain the warning messages below, after auto.arima. I couldn't find any clue in the archived help. Also, how do I retrieve the AICs of each tried model in auto.arima? The purposes are (1) to output to a text file, and (2) to find the 2nd best model by finding 2nd lowest AIC instead of eyeballing thru the value at the console scree...

Another pathological test. arima does not crash for that series that crashes arma: arima(c(2.01, 2.22, 2.09, 2.17, 2.42), order=c(1,0,0)) However, arima crashes for this: arima(c(1.71, 1.78, 1.95, 1.59, 2.13), order=c(1,0,0)) arima seems pretty consistent in its crashing behaviour, since crashing for one series means crashing...

I am using arima to develop a time series regression model, I am using arima b/c I have autocorrelated errors. Several of my independent variables are categorical and I have coded them as factors . When I run ARIMA I don't get any warning or error message, but I do not seem to get estimates for all the level...

Hello, I have a question about function arima.sim I tried to somulate a AR(1) process, with no innovation, no error term. I used this code: library(forecast) e=rnorm(100,mean=0,sd=0) series=arima.sim(model=list(ar=0.75),n=100,innov=e)+20 Then I tried to applicate ti this series auto.arima function: mod1<-auto.arima(series,stepwise=FALSE,...

Hello All, I trying to use the function auto.arima(....) from package forecast but I have a problem. My steps after I used the function auto.arima(...) I create the time series like this: >bbrass = scan("C:/Program Files/R/data PTIN/my_file.dat") >regts.start = ISOdatetime(2006, 7, 1, hour=0, min=0, sec=0, tz="GMT") #2...

I have been trying to simulate from a time series with trend but I don't see how to include the trend in the arima.sim() call. The following code illustrates the problem: # Begin demonstration program x <- c(0.168766559, 0.186874000, 0.156710548, 0.151809531, 0.144638812, 0.142106888, 0.140961714, 0.134054659, 0.138722419, 0.134037018, 0.122829846, 0.120188714, 0.122060497, 0.13742...

Dear all R users, I am really struggling to determine the most appropriate lag order of ARIMA model. My understanding is that, as for MA [q] model the auto correlation coeff vanishes after q lag, it says the MA order of a ARIMA model, and for a AR[p] model partial autocorrelation vanishes after p lags it helps to determine the AR lag. And most appropriate model choosed by this argument give...

Hi, I have the following code from Splus that I'd like to migrate to R. So far, the only problem is the arima.filt function. This function allows me to filter an existing time-series through a previously estimated arima model, and obtain the residuals for further use. Here's the Splus code: # x is the estimation time series, new.infl is a timeseries that contains new information # a.mle is estimated r...

Dear R People: Here is some output from AR and ARIMA functions: > xb <- arima.sim(n=120,model=list(ar=0.85)) > xb.ar <- ar(xb) > xb.ar Call: ar(x = xb) Coefficients: 1 0.6642 Order selected 1 sigma^2 estimated as 1.094 > xb.arima <- arima(xb,order=c(1,0,0),include.mean=FALSE) > xb.arima Call: arima(x = xb, order =...

...olumns: one named Period (which contains dates in the format yyyy_mm) and goes from 1995_10 to 2007_09 and the second column named pcumsdry which is a volumetric measure and has been formatted as numeric without any commas or decimals. I imported the dataset as pauldataset and made use of the auto.arima function which is great when you want R to suggest an appropriate arima model to fit to your data. I successfully got the arima fit by coding: > paulfit<-auto.arima(pauldataset$pcumsdry) > paulfit which produced the following results: Series: pauldataset$pcumsdry ARIMA(2,1,1) with drif...

Dear List, A posting to R-Help exposed this problem with the print method for objects of class Arima: > set.seed(1) > x <- arima.sim(n = 100, list(ar = 0.8897, ma = -0.2279)) > mod <- arima(x, order = c(1,0,1)) > coefs <- coef(mod) > mod2 <- arima(x, order = c(1,0,1), fixed = coefs) > mod2 Call: arima(x = x, order = c(1, 0, 1), fixed = coefs) Coefficients: Error in...

Dear forum, I would like to forecast e.g. with the arima-model. To figure out which model works best I am going to predict with this models. my first code: for(ar.ord in 1:3){ for(ma.ord in 1:3){ print(predict(arima(para_qtr[1:(n-8),1],order=c(ar.ord,1,ma.ord)), n.ahead=8)$pred) } } this one works. but I want to "save" my results in a matrix...

As I am reading ?arima, only NA entries in the argument fixed= imports. The following seems to indicate otherwise: x <- arima.sim(model=list(ar=0.8), n=100) + (1:100)/50 > t <- 1:100 > mod1 <- lm(x ~ t) > > init1 <- c(0, coef(mod1)[2]) > fixed1 <- c(as.numeric(NA), 0) > > arima...

Hello, I have a several time series, which I would like to check for their best fitted Arima model (I am checking for the lowest aic value). Which lets me raise two questions: 1) is there are more efficient way, than using 6 for-loops? 2) sometimes the system cannot calculate with given parameters - is there a more efficient solution than I found? I hope, you can help me to make this ca...

Hello R People: When using the arima function with the seasonal option, are the seasonal options only good for monthly and quarterly data, please? Also, I believe that weekly and daily data are not appropriate for seasonal parm estimation via arima. Is that correct, please? Thanks, Sincerely, Laura Holt mailto: lauraholt_983 at h...