similar to: arima0() (PR#314)

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

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.

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)

I did a regression with ARMA errors using arima0 with ari<-arima0(y,order=c(2,0,2),xreg=reg1,delta=-1) or ari<-arima0(y,order=c(2,0,2),xreg=reg1) where reg1 is the matrix of the regressors and when I see diag(ari$var.coef) I get negative terms. Do you know what this mean ? I try to change transform.pars to 0 or 1 but this crash R on Windows. Is it possible to test the significativity

I'm trying to use the following command. arima (x, order = c(p,d,q), seasonal =list(order=c(P,D,Q), period=s) How can I write an estimated seasonal ARIMA model from the outputs. To be specifically, which sign to use? I know R uses a different signs from S plus. Is it correct that the model is: (1-ar1*B-ar2*B^2-...)(1-sar1*B^s-sar2*B^2s-....)(1-B)^d(1-B^s)^D

Dear R People: I'm trying to use the predict command on an arima0 object. I do the following: xm.arma <- arima0(xm2,order=c(1,0,1)) predict(xm.arma,n.ahead=2) and I get the message: Error in round(x, digits) : Non-numeric argument to mathematical function Any ideas what the problem might be, please? R version 1 4 1 on Windows. Thanks in advance! Sincerely, Erin Hodgess Associate

Hi everyone, ----------------------------- Coefficients: ar1 ar2 ma1 ma2 sar1 intercept drift 1.5283 -0.7189 -1.9971 0.9999 0.3982 0.0288 -9e-04 s.e. 0.0869 0.0835 0.0627 0.0627 0.1305 NaN NaN sigma^2 estimated as 0.04383: log likelihood = 4.34, aic = 7.32 Warning message: NaNs produced in: sqrt(diag(object$var.coef))

Hi !            I am new to R. Can somebody help me in reformatting huge output files ,i.e, rearranging sets of columns in specific order. For example: I have data for three compunds 1, 2 and 3 file1: ID CA1 CA3 CA2 MA2 MA1 MA3 1 14 15 13 7 12 3 2 19 7 12 10 14 5 3 21 12 19 6 8 9   to File 2:   ID CA1 CA2 CA3 MA1 MA2 MA3 1 14 13 15 12 7 3 2 19 12 7 14 10 5 3 21 19 12 8 6 9   or File3: ID

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 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

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"

Hi I am trying to define a function fu() in the following way but when I try to run I get the error that ma1 is not found. I am not sure where I am going wrong? Does the scope of ma1 not extend to an expr.frame object? expr.frame() is under library tradesys. function (y,ma1,ma2) { x <- y[, c("Open","Close")] d <- expr.frame(x, list(MAf=quote(SMA(Close, ma1)),

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),

I can get an array of strings for the data that I want using 'paste()' as follows: paste('ma', 1:am$arma[2], '=', coef(am)[1:am$arma[2] + am$arma[1]], sep='') This results in a vector of strings like: [1] "ma1=1.17760133668255" "ma2=0.649795570407939" "ma3=0.329456750858276" What I would like is fixed.pars <-

Hi, I am using R interactively for estimation and even data manipulation. I want to use lag() function within lm() function in a way looks like: mymodel<- lm(y~x+I(lag(y,-1)-1, data=anydata) What I get is always perfect fit; (that is, coefficient=1) which is not true. If the model looks like mymodel<- lm(y~z+x+I(lag(x,-1)-1, data=anydata) summary returns only one coefficient for x and

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)),

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

Hi, I used RODBC to import data from an Excel spreadsheet "*.xls", but some columns were returned as zeros. When I looked at these columns in Excel, I found that thery are results of formula calculations and not entry. My question is: Is there any parameter or command I need to use in order to overcome this problem? Thank you Ahmad Abu Hammour

Dear all, this may be a stupid question but... The underlying model in procedure arima0 is X[t] = a[1]X[t-1] + ... + a[p]X[t-p] + e[t] + b[1]e[t-1] + ... +b[q]e[t-q] Is it possible to get an estimate of e for every point t, t-1 etc. or at least an estimate of the variance of e? Thanks a lot in advance for any hints Kai Arzheimer

What is the difference between arima and arima0?