search for: ar3

I've posted a question (http://www2.ambientdesign.com/forums/showthread.php?p=389042#post389042) at ArtRage forums regarding this issue. So, it seems that starting and closing AR3 works normally, except in one case: if the window is maximized and you close AR3 while it's still maximized. In this case, next start of AR3 will result in application running, but not showing on screen or panel. So, the temporary solution might be to make sure you "un-maximize" AR3...

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 = 1:23, fixed = c(NA, 0, 0, NA, NA, NA)) Coefficients: ar1 ar2 ar3 ar4 intercept 1...

hey can anybody help me? i have to simulate the richardson Arms race model on R.. for my simulation class...

...= 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)), include.mean = FALSE, fixed = c(NA, NA, NA, 0, 0, 0, NA, NA, NA, NA, NA, NA)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 exa1 exa12 exb1 exc1 exc12 0.0922 -0.1279 -0.2661 -0.0577 -0.0277 0.02 -0.2167 -0.3015 0.3424 0.0281 0.0519 0.1715 s.e. 0.0789 0.0801 0.0742 0.0000 0.0000 0.00 0.0853 0.0503 0.0515 0.0295...

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 0.0646 s.e. 0.0865 0.0885 0.1141 0.1181 0.1196 0.1220 0.1120 0.0908 0.0865 ma1 ma2 sma1 -0.0221 -0.9779 -0.7635 s.e. 0.0539 0.0534 0.08...

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 zero. I would like to refine the model considering ar1 and ar3 only so I fit a mode...

...a zip file with the audio files, converted to .wavs for simpler listening. https://www.dropbox.com/s/bzu4i3dmg5f91tv/20msAudioModeQuestion.zip?dl=0 <https://www.dropbox.com/s/bzu4i3dmg5f91tv/20msAudioModeQuestion.zip?dl=0> If there is one single thing to listen to, it would be ar3_20_audio.wav, loop the section "china hit" starting t=0.6s and listen for artifacts in the unvoiced speech. reference is ar3.wav. and by comparison ar2_10_audio.wav ( same segment, sounds more like the reference ar3.wav) Here is a cat of the README.txt. Thanks very mu...

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 just checking) to define a more parsimonous lag1 and lag 5 model with coeff ar1 and ar5? Or do I need one of the other TS packages? thanks G...

...ie 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 -0.44774 0.41881 0.47624 0.01406 0.06565 -0.06167 -0.01294 0.31313 ma3 ma4 ma5 intercept -0.49027 -0.55461 -0.11520 -0.10692 > modelarima<-arima(x,c(7,1,5)) > modelarima Call: arima(x = x, order = c(7, 1, 5)) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ma1...

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 0.0646 s.e. 0.0865 0.0885 0.1141 0.1181 0.1196 0.1220 0.1120 0.0908 0.0865 ma1 ma2 sma1 -0.0221 -0.9779 -0.7635 s.e. 0.0539 0.0534 0.08...

...ne 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. 0.0316 0.0316 0.0317 0.0316 0.0296 sigma^2 estimated as 0.9775: log likelihood = -1407.56, aic = 2827.12 > names(a) [1] "coef" "sigma2" "var.coef" "mask"...

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

Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e", using the function "lag"? If so, how? If not, of what use is the function "lag"? I get the same answer from y~x as y~lag(x), whether using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider the following: > set.seed(1) > x <- rep(c(rep(0, 4), 9), len=9) > y <-

Good afternoon, I would like fitting an ARIMA model without the first coefficient. For example, I want to fit an AR(3) like this : y[t]=a[1]*y[t-1]+a[2]*y[t-2]+a[3]*y[t-3], where a[1]=0. How can I specify it in the function "arima", if it is possible ? Thank you in advance. Yohann Moreau [[alternative HTML version deleted]]

...e: 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 from the arima model, the error message appears again: > reg Call: arima(x = y, order = c(7, 0, 7), xreg = isr) Coefficients: ar1 ar2 ar3 ar4 ar5 ar6 ar7 ma1 ma2 -0.3356 0.0282 -0.0124 0.0604 0.1608 0.7949 0.2286 0.4461 0.0099 s.e. NaN 0.2140 0.1711 0.2545 0.1547 0.0725 NaN NaN 0.1842 ma3 ma4 ma5 ma6 ma7 intercept isr 0.0841 0.0773 -0.054 -0...

Absolutely. I've just created a new entry (http://appdb.winehq.org/objectManager.php?sClass=version&iId=25122) considering AR3 v.3.5.4, which contains a note about this issue.

...= 9, M = 100) *** Warning during fit: unable to compute correlation matrix Coefficients: Estimate Std. Error z value Pr(>|z|) d 4.745e-01 0.000e+00 Inf <2e-16 *** ar1 8.897e-01 0.000e+00 Inf <2e-16 *** ar2 -3.386e-01 0.000e+00 -Inf <2e-16 *** ar3 3.339e-01 2.044e-17 1.634e+16 <2e-16 *** ar4 -4.406e-01 0.000e+00 -Inf <2e-16 *** ar5 3.924e-02 6.349e-18 6.182e+15 <2e-16 *** ar6 -5.184e-01 2.558e-17 -2.026e+16 <2e-16 *** ar7 8.988e-01 0.000e+00 Inf <2e-16 *** ar8 -7.568e-01 3.112e-16 -2.432e+15...

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!

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This patch set is intended for the next merge window. They are just enhancements of the already merged patches or ia64 porting from x86 paravirt techniques and that their quality is enough for merge. This patch set is for binary patch optimization for paravirt_ops. The binary patch optimization is important on native case because the paravirt_ops overhead can be reduced by converting indirect