search for: sar1

hallo to all I've to calculate an arima model and I need only the first and 365 th parameter and also the sar1 and the intercept, so I'm traing with: arima(X,order=c(365,0,0),seasonal=list(order=c(1,0,0),..),fixed=c(NA,rep(0,363),NA,NA,NA),transform.pars=F) but the error answer is: Error in polyroot(z) : polynomial degree too high (49 max) also there are problems in allocating memory (I've 512 mb...

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)) ----------...

...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 X_t=(1+ma1*B+ma2*B^2+...)(1+sma1*B^s+sma2*B^2s+....) a_t For example: > m1=arima(koeps,order=c(0,1,1),seasonal=list(order=c(0,1,1),period=4)) > m1 Call: arima(x = koeps, order = c(0, 1, 1), seasonal = list(order = c(0, 1, 1), period = 4)) Coefficients:...

...start.Q=0,stepwise=TRUE,stationary=FALSE) >ARIMA(2,1,4)(2,0,1)[12] with drift # the output Call: auto.arima(x = x[[k]], start.p = 0, start.q = 0, start.P = 0, start.Q = 0, stationary = FALSE, stepwise = TRUE) Coefficients: ar1 ar2 ma1 ma2 ma3 ma4 sar1 0.0639 -0.7820 -1.2103 1.2236 -0.9511 0.2357 1.0031 s.e. 0.0686 0.0582 0.1098 0.1558 0.1568 0.1007 0.0716 sar2 sma1 drift -0.0711 -0.8963 -780.9456 s.e. 0.0747 0.0608 403.2112 sigma^2 estimated as 10202381: log likelihood = -1100.61 AIC = 2206...

...ts: ipca_tsb%>% model(ARIMA(servicos ~ 1 + pdq(2, 0, 1) + PDQ(2, 0, 2), method = "ML", optim.method = "BFGS", optim.control = list(maxit = 1000))) %>% report() summary(modelo) |*|Series: . ARIMA(2,0,1)(2,0,2)[12] with non-zero mean Coefficients: ar1 ar2 ma1 sar1 sar2 sma1 sma2 mean 0.7534 0.0706 -0.5705 0.1759 0.7511 0.3533 -0.6283 0.5001 s.e. NaN NaN 0.0011 NaN NaN NaN NaN 0.1996 sigma^2 = 0.05312: log likelihood = 1.75 AIC=14.5 AICc=15.33 BIC=45.33 Training set error measures: ME RMSE MAE MPE MAPE MASE ACF1 Training set -0.006082139 0...

...phi.1 and Phi.1, by phi.21 = phi.1 + Phi.1 phi.22 = -phi.1*Phi.1 (just multiplying out (1 - phi.1*x)(1 - Phi.1*x) and comparing with 1 - phi.21*x - phi.22*x^2). But I get > fx2$coef ar1 ar2 intercept 0.06465065 0.05896674 -0.10499616 > fx1$coef ar1 sar1 intercept 0.03339341 0.03339341 -0.10562032 so phi.1 + Phi.1 = 0.06678682 --- close but no cigar, and -phi.1*Phi.1 = -0.001115120 What am I missing/misunderstanding? Thanks for any insight. cheers, Rolf Turner rolf at math.unb.ca -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-...

..." option which does put the labels horizontally, however, it also draws rectangles around the labels which I don't want. How can I create a dendrogram with horizontal labels, without those rectangles? Some example data for the script: code;name;sex;date1;exp_type;exp1;category1;comment1;sar1;distance1;order1;n_trials;run1_name_cond1_rt;run1_name_cond2_rt;run1_name_cond3_rt;run1_name_cond1_er;run1_name_cond2_er;run1_name_cond3_er;run1_col_cond1_rt;run1_col_cond2_rt;run1_col_cond4_rt;run1_col_cond1_er;run1_col_cond2_er;run1_col_cond4_er;run2_name_cond1_rt;run2_name_cond2_rt;run2_name_con...

...ing 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 -0.1341 -0.2063 -0.1076 -0.2361 -0.2205 -0.7387 s.e. 0.0718 0.1273 0.1061 0.1063 0.0938 0.1029 0.1312 sigma^2 estimated as 16916: log likelihood=-1137.18 AIC=2290.37 AICc=2291.21 BIC=2315.87 So the fitted model ha...

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!

Hi , I have a problem with NT 4.0 workstation with service pack 3 cant able to login to hp system. From FAQ what i understood is to modify the registry but even after modifying i couldn't able to login from NT 4.0 workstation. Can anybody can able to help me in this matter. Thx Ramesh -- ___________________________ Ramesh Ragineni Ilog (s) Pte Ltd Systems Engineer