search for: demeaned

Dears Sirs During my computational work I encountered unexpected behavior when calling "ar" function, namely # time series x<-ts(c(-0.2052083,-0.3764986,-0.3762448,0.3740089,0.2737568,2.8235722,- 1.7783313,0.2728676,-0.3273164),start=c(1978,3),frequency=4,end=c(1980,3)) # ar function res.ar<-ar(x,aic=TRUE,demean=F) # call "ar" again and ............

Full_Name: Steven McKinney Version: 2.6.0 OS: OS X Submission from: (NULL) (142.103.207.10) Function ar() in package "stats" is showing a quirky bug. Some calls to ar() run to completion, others throw an error. The bug is reproducible by several people on different machines, however, the ar() function itself ends up throwing the error sporadically. Several calls to ar() may be

Hello, I understand that the/ demean/ argument in the *ar()* function to fit an autoregressive model selects the best AR model fitted to the mean deleted observations. What is the purpose of using this demean procedure at all? Its seems silly as the post doesn't deal with R problems.... Thanks -- View this message in context:

Dear R-helpers, I have a very simple question and I really hope that someone could help me I would like to estimate a simple fixed effect regression model with clustered standard errors by individuals. For those using Stata, the counterpart would be xtreg with the "fe" option, or areg with the "absorb" option and in both case the clustering is achieved with "vce(cluster

...89) transformation (from Baltagi's book Ch. 9). As a benchmark I use both the pmodel.response() and model.matrix() functions in package plm and the results from using dummy variables. As far as I understood the transformation (Ch.3), Q%*%y (with y being the dependent variable) should yield the demeaned series. However, ... ...I find that the results do not match, if I do so. ...that if I am looking at a balanced panel, I get the correct results when multiplying Q with the already demeaned series y_it, Q%*%y_it. ...that if I am looking at an unbalanced panel, I receive results which differ...

Hi there, we are beginners in R and we are trying to fit the following time series using ar(2): > x <- c(1.89, 2.46, 3.23, 3.95, 4.56, 5.07, 5.62, 6.16, 6.26, 6.56, 6.98, > 7.36, 7.53, 7.84, 8.09) The reason of choosing the present time series is that the we have previously calculated analitically the autoregressive coefficients using the direct inversion method as 1.1, 0.765, 0.1173.

Suppose I have longitudinal data and want to use the econometric strategy of "de-meaning" a model matrix by time. For sake of illustration 'mat' is a model matrix for 3 individuals each with 3 observations where ``1'' denotes that individual i was in group j at time t or ``0'' otherwise. mat <- matrix(c(1,1,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,0,0,1,0,0,0,1,1,0),

Dear R-helpers, First of all, sorry for those who have (eventually) already received that request. The mail has been bumped several times, so I am not sure the list has received it... and I need help (if you have time)! ;-) I have a very simple question and I really hope that someone could help me I would like to estimate a simple fixed effect regression model with clustered standard errors by

Dear R users I have two questions about estimating the spectral power of a time series: 1) I came across a funny thing with the following code: data(co2) par(mfrow=c(2,1)) co2.sp1<-spectrum(co2,detrend=T,demean=T,span=3) co2.sp2<-spectrum(co2[1:468],detrend=T,demean=T,span=3) The first plot displays the frequencies ranging from 0 to 6 whearas the second plot displays the same curve but

Hi everyone, I am doing panel data analysis using the 'plm' package. However, I have noticed that the plm() function reports a different value of R-squared from the R-squared of the lm() function with time-demeaned data. To be clear, I have tried to compute the within model both manually (run an OLS regression with time-demeaned data using lm()) and by using plm(). The two methods give me 2 different values of R-squared and I am not sure which one is the correct one for the fixed-effect estimation. I am new...

Full_Name: Ian McLeod Version: 2.3.1 OS: Windows Submission from: (NULL) (129.100.76.136) > There is a simple bug in acf as shown below: > > z <- 1 > acf(z,lag.max=1,plot=FALSE) > Error in acf(z, lag.max = 1, plot = FALSE) : > 'lag.max' must be at least 1 > This is certainly a bug. There are two problems: (i) the error message is wrong since lag.max is

Dears Sirs During my computational work I encountered unexpected behaviour when calling "ar" function. I want to select the order p of the autoregressive approximation by AIC criterion and sometimes an error occurs. Example: # time series

...o difficult for me to tell without output, references etc., although of course I trust D.B.'s opinion. For variables x,c,d,and e in object pdata_frame, I find that all sd() are reasonably similar both among the cross-sections as well as among the variables. However, I find that extracting the demeaned data from plm(), variables demXt$d and demXt$e (i.e. the demeaned variables) have sd()s that are very small compared to those of dem_yt and demXt$c (approx. by factor 1e-15). I extract the demeaned data as follows: dem_yt<-pmodel.response(res) demXt<-model.matrix(res) How is this possible?...

Hello, I propose two small changes to spec.pgram to get modest speedup when dealing with input (x) having multiple columns. With plot = FALSE, I commonly see ~10-20% speedup, for a two column input matrix and the speedup increases for more columns with a maximum close to 45%. In the function as it currently exists, only the upper right triangle of pgram is necessary and pgram is not returned by

In the R Document, the usage of the acf() is as follow: acf(x, lag.max = NULL, type = c("correlation", "covariance", "partial"), plot = TRUE, na.action = na.fail, demean = TRUE, ...) But now I want to get the result picture like: plot(x,type="l") or plot(x,type="p") How can I do this with acf function? 仭仭仭仭仭仭仭仭仭仭仭仭仭仭仭仭佒伮 伬侎仯仭

I'm tring the functions to check the cointegration of a matrix. I'm using **Phillips & Ouliaris Cointegration Test** The function in *tseries* package is **po.test** and **ca.po** in *urca* The results with **URCA** are: > ca.po(prices, demean='none') ######################################## # Phillips and Ouliaris Unit Root Test #

...______________________ #Unfortunately, I was unable to create an appropriate panel dataset with an MA process in the residuals. Maybe someone has an idea where to find such data? Nevertheless you should be able to follow my subsequent thoughts: # I should be able to get my (time- and sectionally) demeaned series as follows: res1<-plm(x~c+v,data=pdata_frame, effect="twoways", model="within", na.action=na.omit)) dem_yt<-pmodel.response(res) demXt<-model.matrix(res) # Given the demeaned series, I need to set the first observation(s) in each cross-section to NA in order t...

...the "big" dimension is some sort of nonparametric truncation. So: ** 1st (possible) solution ** In my opinion, you would actually need a panel implementation of Newey-West, which is not implemented in 'plm' yet. It might well be feasible by applying vcovHAC{sandwich} to the time-demeaned data but I'm not sure; in this case, vcovHAC should be applied this way (here: the famous Munnell data, see example(plm)) > library(plm) > fm<-log(gsp)~log(pcap)+log(pc)+log(emp)+unemp > data(Produc) > ## est. FE model > femod<-plm(fm, Produc) > ## extract time-demeane...

Hello, Thanks for the reply Rui and for pointing out that I forgot to attach my code. Please find attached in this email my code and data. Thanks in advance. Best regards, Pedro Gerhardt Gavronski. On Fri, Feb 23, 2024 at 5:50?AM Rui Barradas <ruipbarradas at sapo.pt> wrote: > > ?s 16:34 de 22/02/2024, Pedro Gavronski. escreveu: > > Hello, > > > > My name is Pedro

Hi, I'm trying to obtain confidence intervals for auto and cross correlation estimates. I've adapted code made available by Stock and Watson that uses the Bartlett Kernel and the delta method. In R it runs really, really slow because of the loops it uses and I have 9 series that I'd like to examine (81 total combinations). It was easy enough to replace one of the while loops with a