similar to: AR vs ARIMA question

I have just put up a new executable as a replacement for the one in rseptbeta.zip there have only been a few changes; mostly to the menu's. I am about to start on a major overhaul including getting survival to work and grabbing the 0.60 version once it's stable. Please let me know about other enhancements you want.... robert

I have just put up a new executable as a replacement for the one in rseptbeta.zip there have only been a few changes; mostly to the menu's. I am about to start on a major overhaul including getting survival to work and grabbing the 0.60 version once it's stable. Please let me know about other enhancements you want.... robert

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 everyone I've got a problem regarding the arima() and the ar() function for autoregressive series. I would simply like to combine them. To better understand my question, I first show you how I'm using these two functions individually (see file in the attachement). 1) apply(TSX,2, function(x) ar(na.omit(x),method="mle")$order # this function finds the optimal

Hi, Does anyone know a method for adding a linear/polynominal trend to a simulated arima model using the arima.sim function? Any help will be greatly appreciated. Cheers, Sam.

Dear R People: I have the output from an arima model fit in an object xxx. I want to verify that the ma1 coefficient is there, so I did the following: > xxx$coef ar1 ar2 ma1 intercept 1.3841297 -0.4985667 -0.9999996 -0.1091657 > str(xxx$coef) Named num [1:4] 1.384 -0.499 -1 -0.109 - attr(*, "names")= chr [1:4] "ar1" "ar2"

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 gives

Hello, I'm an R newbie. I've tried to search, but my search skills don't seem up to finding what I need. (Maybe I don't know the correct terms?) I need the standard errors and not the confidence intervals from an ARIMA fit. I can get fits: > coef(test) ar1 ma1 intercept time(TempVector) - 1900

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'm using the fixed argument in arima. Shouldn't ar4, ar5, and ar6 display as zero in the output? Call: arima(x = window(log(hhprice), start = c(1990, 1), end = c(2003, 3)), order = c(7, 1, 0), xreg = window(ts.union(exa1 = lag(exa, -1), exa12 = lag(exa, -12), exb1 = lag(exb, -1), exc1 = lag(exc, -1), exc12 = lag(exc, -12)), start = c(1990, 1), end = c(2003, 3)),

Full_Name: Ahmad Abu Hammour Version: rw0651 OS: windows 95 Submission from: (NULL) (63.23.128.44) Although I know that "ts package" is preliminary, I wanted to compare the results from R and SPSS. I ran ARIMA(2,1,2) in both softwares. I got NaN in standard errors of coefficients from R and real figures from SPSS. I changed "delta" in R to match that used by SPSS, I received

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

Here is my problem: Autoregressive models are very interesting in forecasting consumptions (eg water, gas etc). Generally time series of this type have a long history with relatively simple patterns and can be useful to add external regressors for calendar events (holydays, vacations etc). arima() is a very powerful function but kalman filter is very slow (and I foun difficulties of estimation)

Dear all, The standard call to ARIMA in the base package such as arima(y,c(5,0,0),include.mean=FALSE) gives a full 5th order lag polynomial model with for example coeffs Coefficients: ar1 ar2 ar3 ar4 ar5 0.4715 0.067 -0.1772 0.0256 -0.2550 s.e. 0.1421 0.158 0.1569 0.1602 0.1469 Is it possible (I doubt it but am

Hello , I have estimated the following model, a sarima: p=9 d=1 q=2 P=0 D=1 Q=1 S=12 In R 2.12.2 Call: arima(x = xdata, order = c(p, d, q), seasonal = list(order = c(P, D, Q), period = S), optim.control = list(reltol = tol)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ar8 ar9 0.3152 0.8762 -0.4413 0.0152 0.1500 0.0001 -0.0413 -0.1811

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

Hi, Can anyone explain how the standard error in arima() is calculated? Also, how can I extract it from the Arima object? I don't see it in there. > x <- rnorm(1000) > a <- arima(x, order = c(4, 0, 0)) > a Call: arima(x = x, order = c(4, 0, 0)) Coefficients: ar1 ar2 ar3 ar4 intercept -0.0451 0.0448 0.0139 -0.0688 0.0010 s.e.

Hello, I need to know how to import ARIMA coefficients. I already determined the coefficients of the model with other software, but now i need to do the forecast in R. For Example: I have a time series named x and i have fitted an ARIMA(1,0,1) (with other software) AR coef = -.172295 MA coef = .960043 (i know that this is not a good model, it's just an example) I try to

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 All, I have a model to predict time series data for example: data(LakeHuron) Lake.fit <- arima(LakeHuron,order=c(1,0,1)) then the function predict() can be used for predicting future data with the model: LakeH.pred <- predict(Lake.fit,n.ahead=5) I can see the result LakeH.pred$pred and LakeH.pred$se but I did not see residual in predict function. If I have a model: [\ Z_t =