similar to: ARIMA and sig. tests

Rob J Hyndman gives great explanation here (https://robjhyndman.com/hyndsight/estimation/) for reasons why results from R's arima may differ from other softwares. @iacobus, to cite one, 'Major discrepancies between R and Stata for ARIMA' (https://stackoverflow.com/questions/22443395/major-discrepancies-between-r-and-stata-for-arima), assign the, sometimes, big diferences from R

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

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 don't think arima works exactly the way one would expect when there is differencing. What I think should happen is that by default the mean of the differenced series is estimated and if include.mean=F, then it is not. This is not what happens. Instead when there is differencing the include.mean argument is ignored. Now I guess, someone could argue that the mean of the original series

Hi *, Firstly, thank you so much for your time to read my email. I am currently interested in how to use R to predict time series from models fitted by ARIMA. The package I used is basic stats package, and the method I used is predict.Arima. What I know is that ARIMA parameters are estimated by Kalman Filter, but I have difficulty in understanding how exactly maximum likelihood (ML) estimator

Hello, I'm running a number of arima models using the "arima" function. Often, when lag length gets too high, these model don't converge and an error message appears as this: > reg <- arima(y,order=c(7,0,7),xreg=isr) Warning message: In arima(y, order = c(7, 0, 7), xreg = isr) : possible convergence problem: optim gave code=1 In this case, when you print the results

Hi, I've got a little problem using auto.arima. I run the following command auto.arima(drivers,ic="aic",d=1,D=1,max.order=10,max.p=5,max.q=5,max.P=5,max.Q=5,stepwise=FALSE,allowdrift=FALSE) and I get the following output : Series: drivers ARIMA(0,1,1)(5,1,1)[12] Coefficients: ma1 sar1 sar2 sar3 sar4 sar5 sma1 -0.6421

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

Hi I'm trying to fit arima models with the arima() function and I have two questions. ###### ##1. ## ###### I have n observations for my time series. Now, no matter what arima(p,d,q)- model I fit, I always get n residuals. How is that possible? For example: If I try this out myself on an AR(1) and calculate the fitted values from the estimated coefficients I can calculate n-1 residuals.

Hello, The "arma" function in the "tseries" package allows estimation of models with specific "ar" and "ma" lags with its "lag" argument. For example: y[t] = a[0] + a[1]y[t-3] +b[1]e[t-2] + e[t] can be estimated with the following specification : arma(y, lag=list(ar=3,ma=2)). Is this possible with the "arima" function in the

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

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 Could just anyone explain me the coefficients in the output of arima model timeseriesarima <- arima(series, order=c(1,1,2)) > timeseriesarima Series: series ARIMA(1,1,2) Coefficients: ar1 ma1 ma2 0.9744 -1.7695 0.7873 s.e. 0.0310 0.0481 0.0426 sigma^2 estimated as 337.4: log likelihood=-1096.03 AIC=2200.07 AICc=2200.23 BIC=2214.2 ****************

Dear R People: Here is some output from AR and ARIMA functions: > xb <- arima.sim(n=120,model=list(ar=0.85)) > xb.ar <- ar(xb) > xb.ar Call: ar(x = xb) Coefficients: 1 0.6642 Order selected 1 sigma^2 estimated as 1.094 > xb.arima <- arima(xb,order=c(1,0,0),include.mean=FALSE) > xb.arima Call: arima(x = xb, order = c(1, 0, 0), include.mean = FALSE)

Dear all, I stumbled upon what appears to be a troublesome issue when sampling from an ARIMA model (from Rob Hyndman's excellent 'forecast' package) that contains a seasonal AR component. Here's how to reproduce the issue. (I'm using R 2.9.2 with forecast 2.19; see sessionInfo() below). First some data: > x <- c( 0.132475, 0.143119, 0.108104, 0.247291, 0.029510,

I'm using the arima function to estimate coefficients and also using predict.Arima to forecast. This works nicely and I can see that the results are the same as using SAS's proc arima. I can also take the coefficent estimates for a simple model like ARIMA(2,1,0) and manually compute the forecast. The results agree to 5 or 6 decimal places. I can do this for models with and without

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

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

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

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.