similar to: arima function - estimated coefficients and forecasts

I've posted this before but have not been able to locate what I'm doing wrong. I cannot determine how the forecast is made using the estimated coefficients from a simple AR(2) model when there is an exogenous variable. Does anyone know what the problem is? The help file for arima doesn't show the model with any exogenous variables. I haven't been able to locate any documents

I'm trying to do the following out of sample regression with autoregressive terms and additional x variables: y(t+1)=const+B(L)*y(t)+C(1)*x_1(t)...+C(K)*x_K(t) where: B(L) = lag polynom. for AR terms C(1..K) = are the coeffs. on K exogenous variables that have only 1 lag Question 1: ----------- Suppose I use arima to fit the model:

Hello all, I use arima to fit the model with fit <- arima(y, order = c(1,0,1), xreg = list.indep, include.mean = TRUE) and would like to use predict() to forecast: chn.forecast <- rep(0,times=num.record) chn.forecast[1] <- y[1] for (j in 2:num.record){ indep <- c(aa=chn.forecast[j-1], list.indep[j,2:num.indep]) # this is the newxreg in the

Dear all, I just have a short question regarding the forecasting of ARIMA models with external regressors. I tried to program a ARX(1) model arx.mod <- arima(reihe.lern, order = c(1, 0, 0), seasonal = list(order = c(0, 0, 0), period = 52), xreg = lern.design, include.mean = TRUE) for which I need to estimate the next (105th) value. Xreg=lern.design is - at this time - 104 rows long. I

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,

Hello,guys: Recently, I am working on a seasonal ARIMA model. And I met some problem in the forecasting. Now I just want to know that How does R perform the predict procedure(the predict formula, the initial setting of errors,etc.)? I run the following commands and get the original code of the "predict" command, but I can't read it. Can anybody explain it to me? Thanks! saji from

Dear all, I am fitting an arimax (arima with some extra explanatory variables) model to a time series. Say, I have a Y (dependent variable) and an X (explanatory). Y is 100 observations (time series) and X is 100 + 20 (20 to use for the forecast horizon). I can not make xreg work with the forecast function for an arima fit. The "predict" function seems to be working but the

Hi R Users, I'm fairly new to R (about 3 months use thus far.) I wanting to use the arimax function from the TSA library to incorporate some exogenous inputs into the basic underllying arima model.Then with that newly model of type arimax, I would like to make a prediction. To avoid being bogged down with issues specific to my own work, I would like to refer to readers to the example

Hello R users, Hope everyone is doing great. I have a dataset that is in .csv format and consists of two columns: 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

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

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

Hi all there I am enjoying R since 2 weeks and I come to my first deadlock, il am trying to use predict.Arima in the ts package. I get a "Error in cbind(...) : cannot create a matrix from these types" -- Start R session ----------------------------------------------------- > fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv) Call: arima(x = data, order = c(2, 0, 3),

Hi all there I am enjoying R since 2 weeks and I come to my first deadlock, il am trying to use predict.Arima in the ts package. I get a "Error in cbind(...) : cannot create a matrix from these types" -- Start R session ----------------------------------------------------- > fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv) Call: arima(x = data, order = c(2, 0, 3),

Hello, I am calling the auto.arima method in the forecast package at it returns what seems to be valid Arima output. But when I feed this output to 'predict' I get: Error in predict.Arima(catall.fit[[.index]], n.ahead = 12) : 'xreg' and 'newxreg' have different numbers of columns Is there a way to tell what is being supplied to xreg from the Arima output? Any ideas?

to clean up some code I would like to make a list of arbitrary length to store?various objects for use in a loop sample code: ############ BEGIN SAMPLE ############## # You can see the need for a loop already linearModel1=lm(modelSource ~ .,mcReg) linearModel2=step(linearModel1) linearModel3=lm(modelSource ~ .-1,mcReg) linearModel4=step(linearModel3) #custom linearModel5=lm(modelSource ~ .

Hello everyone, Hope you all are doing great! I have been fitting arima models and performing forecasts pretty straightforwardly in R. However, I wanted to add a couple of regressors to the arima model to see if it could improve the accuracy of the forecasts but have had a hard time trying to do so. I used the following R function: arima(x, order = c(0, 0, 0), seasonal = list(order = c(0, 0,

Hello, I'm currently using auto.arima to verify the order of my arima model. I would like to use the model to conduct an intervention analysis. The problem is that, when I include a step function in auto.arima, by including a binary variable in "xreg", the arima order that auto.arima gives is different from when I don't include it. From my understanding, the

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

Hello everybody! I have an ARIMA model for a time series. This model was obtained through an auto.arima function. The resulting model is a ARIMA(2,1,4)(2,0,1)[12] with drift (my time series has monthly data). Then I perform a 12-step ahead forecast to the cited model... so far so good... but when I look the plot of my forecast I see that the result is really far from the behavior of my time

Hello all, I have been using R's time series capabilities to perform analysis for quite some time now and I am having some questions regarding its reliability. In several cases I have had substantial disagreement between R and other packages (such as gretl and the commercial EViews package). I have just encountered another problem and thought I'd post it to the list. In this case,