similar to: using exists with coef from an arima fit

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'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

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

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, I want to model different timeseries with ARMAX models in R because I think that ARMAX models will map best to these data. Besides I don't want to use the order of the AR or MA part but the lag e.g. AR Part =ar1, ar2, ar7; MA Part =ma1, ma3 and I want to use exogenous variables as well. I coudn't find any solutions in the R help and therefore I want to ask all of you. Does anyone

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))

Hello R users, I want to export to an xls or .csv some predictions I produced with the auto.arima and forecast functions. A detail of all my work is presented below. I loaded a package called dataframes2xls and tried to use the function write.xls without any success. Can anybody help me figure this out? How could I get R to export the output to an xls file? Any help will be greatly

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 all, I have a problem that may be someone of you can help. I am a newbie and do not find how to do it in manuals. I have two arrays, for example: ar1 <- array(data=c(1:16),dim=c(4,4)) ar2 <- array(data=c(1:16),dim=c(4,4)) colnames(ar1)<-c("A","B","D","E") colnames(ar2)<-c("C","A","E","B") > ar1

Hi, A quick question: I have to vectors, say ar1 and ar2 > ar1 [1] "a" "c" "c" "a" attr(,"levels") [1] "a" "b" "c" > ar2 [1] TRUE TRUE FALSE TRUE > table(ar1, ar2) ar2 ar1 FALSE TRUE a 0 2 c 1 1 I would like to obtain: T F a 2 0 b 0 0 c 1 1

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

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 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)

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

Hello, Guys: I'm from China, my English is poor and I'm new to R. The first message I sent to R help meets some problems, so I send again. Hope that I can get useful suggestions from you warm-hearted guys. Thanks. I builded a multiplicative seasonal ARMA model to a series named "cDownRange". And the order is (1,1)*(0,1)45 The regular AR=1; regular MA=1; seasonal AR=0; seasonal

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'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)),

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

Estimated people, I have two matrices: ar1 <- array(data=c(1:16),dim=c(4,4)) ar2 <- array(data=c(1,2,3,3,5:16),dim=c(4,4)) They only differ in the fourth row. I would like to compare them in order to know which columns are equal. The following works, but I would like to have a better solution, and not to use what someone called "prehistorical loops": for(i in

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