similar to: question about xreg of arima

Hello all, I''m having fun again with the arima function. This time I read in: http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm <<It has recently been suggested (by a reliable source) that using xreg in arima() does NOT fit an ARMAX model [insert slap head icon here]. This will be investigated as soon as time permits.>> (by R.H. Shumway & D.S. Stoffer)

I am trying to understand how to fit an ARMAX model with the arima function from the stats package. I tried the simple data below, where the time series (vector x) is generated by filtering a step function (vector u, the exogenous signal) through a lowpass filter with AR coefficient equal to 0.8. The input gain is 0.3 and there is a 0.01 normal white noise added to the output: x <- u

Dear R-help-archive.. I am trying to figure out how to make arima prediction when I have a process involving multivariate time series input, and one output time series (output is to be predicted) .. (thus strictly speaking its an ARMAX process). I know that the arima function of R was not designed to handle multivariate analysis (there is dse but it doesnt handle arma multivariate analysis, only

Hello, The "arma" function in the "tseries" package allows estimation of models with specific "ar" and "ma" lags with its "lag" argument. For example: y[t] = a[0] + a[1]y[t-3] +b[1]e[t-2] + e[t] can be estimated with the following specification : arma(y, lag=list(ar=3,ma=2)). Is this possible with the "arima" function in the

Hi, I have been using this website ( http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm ) to help me to fit ARIMA models to my data. At the moment I have two possible methods to use. Method 1 If I use arima(ts.data, order=c(1,2,0), xreg=1:length(ts.data)) then the wrong value for the intercept/mean is given (checked on SPSS and Minitab) and

Hi, I want to give a summary or anova for "arima" model in R, as "summary", and "anova" for "lm". As including various intervention factors in arima(xreg = ) part, I want to assess the significancy of thse factors. I can do it using interrupted analysis of time series by linear regression, but want to see whether arima model works for the data first.

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 levels of the factor. Can/how does ARIMA handle factors in xreg? here is some example

Hello, Does anyone know how the parameter estimates are calculated for xreg variables when called as part of an arima() command, or know of any literature that provides this info? In particular, I was wondering if there is a quick way to compare different combinations of "xreg" variables in the arima() fit in the same way that you would in multiple regression (using AIC & R^2

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)(0,1,0)[12] with drift

When I run predict.Arima in my code, I get warnings like: Warning message: In cbind(intercept = rep(1, n), xreg) : number of rows of result is not a multiple of vector length (arg 1) I think this is because I'm not running predict.Arima in the same environment that I did the fit, so the data object used in the fit is no longer present. Looking at the predict.Arima source,

David Stoffer describes some challenges with R's output when fitting ARIMA models for different orders (see Issue 2 at http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm). R doesn't fit an intercept in the model if there is any differencing. David describes a workaround using the xreg parameter to force R to calculate an intercept. Assume I have a variable y and 3 explanatory variables a,

Dear R, I have find a website where they report problem with ARIMA models in R. I run the examples there and they give result as shown on the website. Does this mean that nothing has corrected in R? Maybe you not have seen the page, but the author said he contacted you. Here is the URL: http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm I like to know your opinion. Mvh. Marie [[alternative

Dear R, I have find a website where they report problem with ARIMA models in R. I run the examples there and they give result as shown on the website. Does this mean that nothing has corrected in R? Maybe you not have seen the page, but the author said he contacted you. Here is the URL: http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm I like to know your opinion. Mvh. Marie [[alternative

Dear all, When I run an arima(1,1,1) on an I(1) variable, y, I get different estimates to when I first difference the variable myself, e.g y2<-diff(y), and then run arima(1,0,1) on y2. Shouldn't these two approaches give the same output? Any help will be much appreciated. george

I am relatively new to R and object oriented programming. I have relied on SAS for most of my data analysis. I teach an introductory undergraduate forecasting course using the Diebold text and I am considering using R in addition to SAS and Eviews in the course. I work primarily with univariate or multivariate time series data. I am having a great deal of difficulty understanding and working with

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(x, order=c(1,0,0), xreg=t, include.mean=FALSE, init=init1, fixed=fixed1)

i am using sarima() function as below ___________________________________________________________________________________________ sarima=function(data,p,d,q,P=0,D=0,Q=0,S=-1,tol=.001){ n=length(data) constant=1:n xmean=matrix(1,n,1) if (d>0 & D>0) fitit=arima(data, order=c(p,d,q), seasonal=list(order=c(P,D,Q), period=S),

Hello I want to fit an AR model were two of the coefficients are fixed to zero (the second and third ar-coefficients). I used the "arima" function with the "fixed" argument but the ar3 coefficient is not set to zero: ============================================== > arima(Y, order=c(4,0,0), xreg=1:23, fixed=c(NA,0,0,NA,NA,NA)) Call: arima(x = Y, order = c(4, 0, 0), xreg =

Hi, My query is related to ARIMA function in stats package. While looking for the time series literature I found following link which highlights discrepancy in "arima" function while dealing with differenced time series. Is there a substitute function similar to "sarima" mentioned in the following website implemened in R? Any pointers would be of great help.

Hello, Does anybody know if the function arima with an external regressor (xreg) applies the auto correlation on the dependant variable or on the residuals. In comparison with SAS (proc autoreg), it seems that the auto correlation applies on the residuals but i'd like to have the confirmation. I want to estimate: Y[t] = a[1]*X[t] + a[2] + E[t] with E[t]=b[1]*E[t-1] Should I use : arima(Y,