similar to: Arms Race

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

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 =

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

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.

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

Hello R People: When using the arima function with the seasonal option, are the seasonal options only good for monthly and quarterly data, please? Also, I believe that weekly and daily data are not appropriate for seasonal parm estimation via arima. Is that correct, please? Thanks, Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com download!

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

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 Kevin, Are you saying that the quality is good at 20 ms and bad at 10 ms, or the reverse? Also, is this speech or music? What tool, what options? In general, it helps a lot if you post the sample (input and output). Cheers, Jean-Marc On 06/03/2016 12:48 PM, Kevin Connor wrote: > Hi Opus list, > > I'm noticing a discontinuity in the quality between use of 10ms and > 20ms

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

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 R-users, I am currently modifying a previously developed predator prey model and was curious if there was a way to add in a disturbance to the model (let's say at time t=100). The disturbance can be the introduction of 40 prey (N=40) and 10 predators (Pred = 10). I would like to see my model go from a state of equilibrium (up to t = 99), show this disturbance (at t = 100) and then

Evening, I was writing some code that tried to insert calls to the llvm.annotation intrinsic function, which has a signature of (i32, i8*, i8*, i32). The code is below. void addAnnotation( BasicBlock *block, Function *F) { string foo = "foo"; string bar = "barr"; Type *charTy = Type::getInt8Ty(block->getContext()); ArrayType *s1Ty =

To whomever may be of help, I am a student in a graduate modeling class at the University of North Carolina at Wilmington. I am trying to get a STELLA model converted into R. I am in the process of trying to 're-write' the script into R, but I seem to be missing pieces (i.e. parm values) that are keeping me from being able to replicate this model into R. Does anyone have an idea of a

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

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

Dear list The deSolve package allows you to specify the model code in C or Fortran. Thanks to the excellent vignette this works fine. However I have not yet managed to use forcing functions in C code. In pure R code this works very well with approxfun() specified outside the model: ############################################### #Model lvml <- function(t, x, parms) {

Dear List, Below is a simple, standard loss model that takes into account the terms of an insurance policy: deductible <- 15 coverage.limit <- 75 insurance.threshold <- deductible + coverage.limit tmpf <- function() { loss <- rlnorm(rpois(1, 3), 2, 5) sum(ifelse(loss > insurance.threshold, loss - coverage.limit, pmin(loss, deductible))) } net <- replicate(1000000, tmpf())

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