similar to: How to get residuals with arima0? [fwd]

Dear all, this may be a stupid question but... The underlying model in procedure arima0 is X[t] = a[1]X[t-1] + ... + a[p]X[t-p] + e[t] + b[1]e[t-1] + ... +b[q]e[t-q] Is it possible to get an estimate of e for every point t, t-1 etc. or at least an estimate of the variance of e? Thanks a lot in advance for any hints Kai Arzheimer

Dear R-users, I would like to show you a simple example that gives an overview of one of my current issue. Although my working setting implies a different parametric model (which cannot be framed in the glm), I guess that what I'll get from the following example it would help for the next steps. Anyway here it is. Firstly I simulated from a series of exposures, a series of deaths (given a

Hi, I'm trying to go through the examples for function arima0() in ts package, i.e, >data(lh) >arima0(lh, order = c(1,0,0)) each time the call to arima0() causes a segmentation fault. I checked the earlier version (1.1.1) of R, the function arima0 works fine. Tracing the call indicates that the function "setup_starma" (in pacf.c under ts) interprets the addresses of the

Dear R People: I'm trying to use the predict command on an arima0 object. I do the following: xm.arma <- arima0(xm2,order=c(1,0,1)) predict(xm.arma,n.ahead=2) and I get the message: Error in round(x, digits) : Non-numeric argument to mathematical function Any ideas what the problem might be, please? R version 1 4 1 on Windows. Thanks in advance! Sincerely, Erin Hodgess Associate

Hi all, I'm trying to apply loess regression to my data and then use the fitted model to get the *standardized/studentized residuals. I understood that for linear regression (lm) there are functions to do that:* * * fit1 = lm(y~x) stdres.fit1 = rstandard(fit1) studres.fit1 = rstudent(fit1) I was wondering if there is an equally simple way to get the standardized/studentized residuals for a

Hi, I work on order estimation for autoregressive processes and after some inconsistencies cropped up I implemented the AIC criterion myself. Its results do not match the implementation in R and there are a few things I can not understand even after reading the source code of R. I used the function called arima0 (with empty "ma" coeficient vector), and I do not understand how some of

R 2.7 Windows XP I have two model that have been run using exactly the same data, both fit using glm(). One model is a linear regression (gaussian(link = "identity")) the other a quasipoisson(link = "log"). I have log likelihoods from each model. Is there any way I can determine which model is a better fit to the data? anova() does not appear to work as the models have the

What is the difference between arima and arima0?

Full_Name: Arto Luoma Version: 1.1.0 OS: Windows 98 Submission from: (NULL) (153.1.53.119) In arima0 it is possible to specify whether the mean of the original series is included in the model or not. However, it is not possible to specify whether the mean of the differenced series is included. It seems that it is not included. However, if differencing is used to eliminate trend, the mean of the

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

Dear R People: Suppose I want to estimate the parameters of the following AR model: (1 - phi_1 B - phi_2 B^2 - phi_9 B^9) x_t = a_t and I want to use the arima0 command from the ts library. How would I use the order subcommand, please? R Version 1.4.1 for Windows. Thanks! Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston -

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

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

Dear all, I need to understand a thing in the beheaviour of the two functions aov and anova in the following case involving an analysis of ANOVA with repeated measures: If I use the folowing command I don´t get any problem: >aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), >data=scrd) > summary(aov1) Instead if I try to fit the same model for the

Here is an example, modified from the help page to use test="Cp": -------------------------------------------------------------------------------- > fit0 <- lm(sr ~ 1, data = LifeCycleSavings) > fit1 <- update(fit0, . ~ . + pop15) > fit2 <- update(fit1, . ~ . + pop75) > anova(fit0, fit1, fit2, test="Cp") Error in `[.data.frame`(table, , "Resid.

Dear all, I am struggling with calculation of Schoenfeld residuals of my Cox Ph models. Based on the formula as attached, I calculated the Schoenfeld residuals for both non tied and tied data, respectively. And then I validated my results with R using the same data sets. However, I found that my results for non-tied data was ok but the results for tied data were different from R's. How

Here is the summary on predict when using an arima0 object: The arima0 object must be based on a time series vector. That is; x <- ts(xm1, frequency=12, start=c(1975,1)) x.ar <- arima0(x,order=c(1,1,1)) predict(x.ar,n.ahead=3) Thanks so much to Prof. Brian Ripley and David Brahm and other! Sincerely, Erin Hodgess

Full_Name: Robert King Version: 1.5.0 OS: windows Submission from: (NULL) (134.148.4.19) Also occurs in 1.6.0 on linux anova.glm(fitted.object,test="Chisq") is giving strange answers in this situation > resptime sex task time 1 m s 210 2 m s 300 3 m s 420 4 f s 250 5 f s 310 6 f s 390 7 m c 310 8 m c 400 9 m c 600

quantreg package is used. *fit1 results are* Call: rq(formula = op ~ inp1 + inp2 + inp3 + inp4 + inp5 + inp6 + inp7 + inp8 + inp9, tau = 0.15, data = wbc) Coefficients: (Intercept) inp1 inp2 inp3 inp4 inp5 -0.191528450 0.005276347 0.021414032 0.016034803 0.007510343 0.005276347 inp6 inp7 inp8 inp9 0.058708544

Dear all, I've replicated the cheese tasting example on p175 of GLM's by McCullagh and Nelder. This is a 4 treatment (rows) by 9 ordinal response (cols) table. Here's my simple code: #### cheese library(MASS) options(contrasts = c("contr.treatment", "contr.poly")) y = c(0,0, 1, 7, 8,8,19, 8,1, 6,9,12,11, 7,6, 1, 0,0, 1,1, 6, 8,23,7,