similar to: bug in arma.sim (PR#322)

Dear R users, I need to fit an ARMA model. As far as I've seen, EACF (extended ACF) is not available in R. 1. Let's say I fit a series of ARMA models in a loop. Given the code/output included below, how do I pull 'Model' and 'Fit' (AIC) from each summary() so that I can combine them into an array/data frame to be sorted by AIC? 2. Apart from EACF, are you aware perhaps

In the summary of the output of arma, there's a number Pr(>|t|), however, I don't know what is its meaning - at least, it doesn't _seem_ to be a Student's t distribution. Reproducible test case: x <- c(0.5, sin(1:9)) reg <- arma(x, c(1,0)) summary(reg) <output> Call: arma(x = x, order = c(1, 0)) Model: ARMA(1,0) Residuals: Min 1Q Median 3Q

Hi dear sirs, I am wondering why the fit of the time serie x with an arima and the fit of diff(x) with an arma (same coeff p & d) differ one from another here are the output of R: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% > modelarma<-arma(diff(x),c(7,5)) > modelarma Call: arma(x = diff(x), order = c(7, 5)) Coefficient(s): ar1 ar2 ar3 ar4 ar5 ar6 ar7 ma1 ma2 0.06078

Hi all. I have a periodiv arma model and I want to simulate it. In S-plus, the following works for me: phi <- 0.9 theta <- 0 p <- 1 # period model <- list(ar=phi, ma=theta, period=p) Yt <- arima.sim(model, n=250) How do I do something like this "period=12" in R? I read help(arima.sim) but it doesn't tell. Thanks! Philip.

> It might clarify your thinking to note that a seasonal ARIMA model > is just an ``ordinary'' ARIMA model with some coefficients > constrained to be 0 in an efficient way. E.g. a seasonal AR(1) s = > 4 model is the same as an ordinary (nonseasonal) AR(4) model with > coefficients theta_1, theta_2, and theta_3 constrained to be 0. You > can get the same answer as from

When I do ARMA(2,2) using one lag of LCPIH data This is eview result > > *Dependent Variable: DLCPIH > **Method: Least Squares > **Date: 08/12/11 Time: 12:44 > **Sample (adjusted): 1970Q2 2010Q2 > **Included observations: 161 after adjustments > **Convergence achieved after 14 iterations > **MA Backcast: 1969Q4 1970Q1 > ** > **Variable Coefficient Std.

I am having problems with the package dse. I just installed R 2.2.1 and reinstalled all packages. I am running Windows XP Pro with all updates. Below there are two examples of error messages generated when trying to execute some simple programs. The code was taken directly from the package documentation. Any help on this will be greatly appreciated. Merry Christmas Fernando

Hi all, I want to fit the following model to my data: Y_t= a+bY_(t-1)+cY_(t-2) + Z_t +Z_(t-1) + Z_(t-2) + X_t + M_t i.e. it is an ARMA(2,2) with some additional regressors X and M. [Z_t's are the white noise variables] So, I run the following code: for (i in 1:rep) { index=sample(4,15,replace=T) final<-do.call(rbind,lapply(index,function(i)

Dear all, I am new to R. I need to implement an ARMA filter, some thing like: y(n) = a0*x(n) + a1*x(n-1) + b1*y(n-1) + b2*y(n-2) I checked out the filter manual page. It doesn't seem that the filter function can do this job for me. Can any one help me out? Thanks a lot! Best regards, Jingzhao

Hi given an ARMA process and the AR and MA coefficients I need the residuals. arima() calculates the residuals together with the best AR and MA coefficients, but I need the coefficients to take known values. In S-PLUS there is a function arima.filt(). Is there something similar in R? Thanks for any help, Matthias Budinger

Dear users, I tried to fit an AR(2) model to data. This the result: > arima(vw,c(3,0,0)) Call: arima(x = vw, order = c(3, 0, 0)) Coefficients: ar1 ar2 ar3 intercept 0.1052 -0.0102 -0.1203 0.0099 s.e. 0.0337 0.0339 0.0338 0.0018 sigma^2 estimated as 0.002934: log likelihood = 1293.16, aic = -2576.33 Now, ar2 is not significantly different from

(My apologies if this is a repeated posting. I couldn't find any trace of my previous attempt in the archive.) I'm having trouble with forecast() in the dse2 package. It works fine for me on a model without a trend, but gives me NaN output for the forecast values when using a model with a trend. An example: # Set inputs and outputs for the ARMA model fit and test periods

Dear R-users, I am trying to use the following code to reproduce results from Prof. Gelman's book, but have the listed error for R2WinBUGS version (the openbugs version is good). I am using R-2.6.1 on windows XP, and all the R packages are most current ones. schools.bug can be found at http://www.stat.columbia.edu/~gelman/bugsR/runningbugs.html . Can anyone help me to figure out what's

Hello, I am running an ARMA model to run forecast for changes in S&P 500 prices. My ARMA calculations look as follows armacal <- arma( spdata, order = c(0,4), lag = list(ma = c(1,2,4)) ) Output: Call: arma(x = spdata, order = c(0, 4), lag = list(ma = c(1, 2, 4)) ) Coefficient(s): ma1 ma2 ma4 intercept -0.073868 0.058020 -0.081292 0.007082 All's

I'd like to fit an ARMA(1,1) model to some data (Federal Reserve Bank interest rates) that looks like: ... 30JUN2006, 5.05 03JUL2006, 5.25 04JUL2006, N &lt;---- here! 05JUL2006, 5.25 ... One problem is that holidays have that "N" for their data. As a test, I tried fitting ARMA(1,1) with and without the holidays deleted. In other words, I fit the above data

Hi all, I am fitting an ARMA(1,(1,4)) model. y(t) = a*y(t-1) + e(t) + b1*e(t-1) + b4*e(t-4) > arma1.14 <- arma(series, lag=list(ar=1, ma=c(1,4)), + include.intercept = F, qr.tol = 1e-07) works fine: Coefficient(s): ar1 ma1 ma4 0.872 -0.445 0.331 I want to forecast 50 periods. I could not find a 'predict' function for ARMA models. I

Hello Following some standard textbooks on ARMA(1,1)-GARCH(1,1) (e.g. Ruey Tsay's Analysis of Financial Time Series), I try to write an R program to estimate the key parameters of an ARMA(1,1)-GARCH(1,1) model for Intel's stock returns. For some random reason, I cannot decipher what is wrong with my R program. The R package fGarch already gives me the answer, but my customized function

Dear useRs, I want to simulate a time series (stationary; the distribution of values is skewed to the right; quite a few ARMA absolute standardized residuals above 2 - about 8% of them). Is this the right way to do it? #-------------------------------- load("rdtb") #the time series > summary(rdtb) Min. 1st Qu. Median Mean 3rd Qu. Max. -1.11800 -0.65010 -0.09091

Dear R experts, I am trying to estimate an ARMA 2,2 model with garch errors. I used the following code on R 2.9. #library library(fGarch) #data data1<-ts(read.table("C:/Users/falcon/Desktop/Time Series/exports/goods1.csv"), start=c(1992,1), frequency=12) head(data1) #garch garchFit(formula.mean= ~arma(2,2),formula.var=~garch(1,1), data=data1) but get this error: >

Dear R users, According Brockwell & Davis (1991, Section 9.3, p.304), the penalty term for computing the AIC criteria is "p+q+1" in the context of a zero-mean ARMA(p,q) time series model. They arrived at this criterion (with this particular penalty term) estimating the Kullback-Leibler discrepancy index. In practice, the user usually chooses the model whose estimated index is