similar to: ARIMA prediction

Hello, I have used the arima.sim function to generate a lot of time series, but to day I got som results that I didn't quite understand. Generating two time series z0 and z1 as eps <- rnorm(n, sd=0.03) z0 <- arima.sim(list(ar=c(0.9)), n=n, innov=eps) and z1 <- arima.sim(list(ar=c(0.9)), n=n, sd=0.03), I would expect z0 and z1 to be qualitatively similar. However, with n=10 the

I am trying to simulate time series data for an ar(1) and ma(1) process. I want the error term to have either a t distribution with 1 degree of freedom or a normal distribution with mean=0 and sd=1. Here is my code: error.model=function(n){rnorm(n,mean=0, sd=1)} data<-arima.sim(model=list(ar=c(0.1)), n=1000, n.start=200, start.innov=rnorm(200,mean=0, sd=1), rand.gen=error.model ) data

Hi, I am using the arima.sim function to generate some AR time series. However, the function does not seem to produce exactly the same time series when I specify the innov parameter. For example > r <- rnorm(300) > x <- arima.sim(300, model=list(order=c(1,0,0),ar=c(.96)), innov=r, n.start=10) > y <- arima.sim(300, model=list(order=c(1,0,0),ar=c(.96)), innov=r, n.start=10) >

Hi, I have been using arima.sim from the stats package recently, and I'm wondering why I get different results when using what seem to be the same parameters. For example, I've given examples of three different ways to run arima.sim with what I believe are the same parameters. It's my understanding from the R documentation that rnorm is the default function for rand.gen if not

Hi Everyone, I'm having a problem with arima.sim. Namely specifying inital values for the series. If I generate a random walk > vs = rnorm(100,0,1) > xs = cumsum(vs) and fit an ARIMA(1,0,0) to it > xarima = arima(xs,order=c(1,0,0)) > xarima Call: arima(x = xs, order = c(1, 0, 0)) Coefficients: ar1 intercept 0.9895 8.6341 s.e. 0.0106 6.1869 I should

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,trace=TRUE,ic='aicc') The best model returned

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

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

Hi users. I'm new user in R. I'm workiing with Time series and I would like to know how can I do to undo the command DIFF(X), for exemple: If I have the model: m=arima(X, order=c(0,1,1), seasonal=list(order=c(0,0,1))) (note that have d=1 one difference), to find, in the same scale, the original numbers (like one "unDiff"), after the forecast, I need to develop some function or in

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

Apologies for thickness - I'm sure that this operates as documented and with good reason. However... My understanding of arima.sim() is obviously imperfect. In the example below I assume that x1 and x2 are similar white noise processes with a mean of 5 and a standard deviation of 1. I thought x3 should be an AR1 process but still have a mean of 5 and a sd of 1. Why does x3 have a mean of ~7?

Full_Name: Allan McRae Version: 1.6.0 OS: Win 2000 P Submission from: (NULL) (129.215.190.229) When using predict.Arima in library ts(), it appears differencing is only accounted for in the first step of prediction and so any trend is not apparent in the predictions. The example shows the difference between the predictions of an arima(1,1,1) model and the backtransformed predictions of an

Hi, I have been looking at arima.sim to simulate the output from an ARMA model fed with a normal and uncorrelated input series but I cannot find a way to pass an intercept / constant into the model. In other words, the model input in the function allows only for the AR and MA components but I need to pass a constant. Can anyone help? Thanks Paolo [[alternative HTML version deleted]]

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,

Dear colleagues, I am relatively new to R and time series and so I am experiencing difficulties in interpreting the output of "arima" in MA models (but not in AR models). I cannot make sense of the 1st innovations returned by "arima". In an AR(1) model I expect data[t]=phi1*data[t-1]+a[t] and in a MA(1) model data[t]=a[t]+theta1*a[t-1]. My interpretation from R-help is

Dear all, I'm new to R and time series analysis. I'd appreciate if you could shed light on my problem. Here is what I have been trying to do: 1. I fit the model ARIMA(1,0,0) with the training dataset xdata[1:100] fitit = arima(xdata, order=c(1,0,0) 2. I have some current observations in the buffer. Say that buf = xdata_new[1:20] 3. I'm trying to forecast the xdata_new[21] based on

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,

In some time series simulations I'm doing, I occasionally want the model to be ``white noise'', i.e. no model at all. I thought it would be nice if I could fit this into the arima.sim() context, without making an exceptional case. I.e. one ***could*** do something to the effect if(length(model)==0) x <- rnorm(n) else x <- arima.sim(model,n) but it would be more suave if one

Hi, All, This is my program ts1.sim <- arima.sim(list(order = c(1,1,0), ar = c(0.7)), n = 200) ts2.sim <- arima.sim(list(order = c(1,1,0), ar = c(0.5)), n = 200) tdata<-ts(c(ts1.sim[-1],ts2.sim[-1])) tre<-c(rep(0,200),rep(1,200)) gender<-rbinom(400,1,.5) x<-matrix(0,2,400) x[1,]<-tre x[2,]<-gender fit <- arima(tdata, c(1, 1, 0), method = "CSS",xreg=t(x))

Dear useRs, I am trying to model a time series with a transfer function. I think it can be put into the ARMA framework, and estimated with the 'arima' function (and others have made similar comments on this list). I have tried to do that, but the results have so far been disappointing. Maybe I am trying to make 'arima' do something it can't... The data are time series of