similar to: A question on Unit Root Test using "urca" toolbox

Full_Name: Ben Cooper Version: 1.5.0 OS: linux Submission from: (NULL) (134.174.187.90) The help page for glm.nb (in MASS package) says that it takes "Any other arguments for the glm() function except family" One such argument is start "starting values for the parameters in the linear predictor." However, when called with starting values glm.nb returns: Error in

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates <- as.Date(data.df[,1]) selection<-which(dates>=as.Date("1986-1-1") & dates<=as.Date("2007-12-31")) dep <- ts(data.df[selection,c("dep")]) indep.ret1

Dear Members, Greetings! I would like to know how to create the lag variable for my data. # Load data and create time series object ---- oil <- read_xlsx("crudefinal.xlsx") pricet=ts(oil$price, start = c(2020, 22), frequency = 365) roilt=ts(diff(log(oil$price))*100,start=c(2020,22),freq=365) # Fit MSW model ---- roilt.lag0 =

Hello guys .. I would like to have some help about as.table . I made a table with the autocorrelations of the returns whit 10 lags and i get this : autocorrelazione2 <- as.table(c((cor(r2[-1151,],lag(r2))),(cor(r2[- c(1151,1150),],lag(r2, k=2))),(cor(r2[- c(1151,1150,1149),],lag(r2, k=3))),(cor(r2[- c(1151,1150,1149,1148),],lag(r2, k=4))),(cor(r2[- c(1151,1150,1149,1148,1147),],lag(r2,

Hi, is there any function in R that shifts elements of a vector to the opposite direction of what Lag() of the Hmisc package does? (something like, Lag(x, shift = -1) ) Thanks Zava -------------------------------------------------------- This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Hello all, I need to figure out a way to lag a variable in by a number of days without using the zoo package. I need to use a remote R connection that doesn't have the zoo package installed and is unwilling to do so. So that is, I want a function where I can specify the number of days to lag a variable against a Date formatted column. That is relatively easy to do. The problem arises when I

I am using R 63.0. Now let's try this simple image plot. Here is the data file: ============================ lag1 lag2 cif2d 1 1 11 1 2 12 1 3 13 2 1 21 2 2 22 2 3 23 3 1 31 3 2 32 3 3 33 ==================== data<-read.table("~/r/rt/data/unif/junk.out",header=TRUE) x<-unique(data$lag1) y<-unique(data$lag2) z<-matrix(data$cif2d,length(y),length(x)) At this point, see

Hi all, I have run a ridge regression on a data set 'final' as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it

I am using R on unix version 63.0 I am doing an image plot of the following data file: ================================ lag1 lag2 cif2d 0.000 0.000 NaN 0.000 1.000 0.500000 0.000 2.000 0.489831 0.000 3.000 0.492986 0.000 4.000 0.493409 0.000 5.000 0.492727 0.000 6.000 0.494485 1.000 0.000 0.500000 1.000 1.000 NaN 1.000 2.000 0.495098 1.000 3.000 0.489831 1.000 4.000 0.492986 1.000 5.000

Hello I'm only user of R and have many little knowledge in programming but I permit to send you some whishes/suggestions for R. which.min like which(), which.min() should also include an argument arr.ind. Note that one can have it with which(a==min(a), arr.ind=TRUE) but if there is a reason to build a special function which.min, why not add also this nice argument? lag() If one wants to

Hi all R users, I'm trying to replicate the same results that are given in a published article after been granted the same data that the authors use. I'm having problems to determine the cointegration rank of my data set using the Johnasen's trace test. This trace test is already programmed in the package ur.ca and can be found in the function ca.jo(). After I run the ca.jo()

I think my problems are coused by a fundamental R incompatibility in how matrices are stored and the usual way of specifying Cartesian coordinates. When I do data<-read.table("~/r/rt/data/unif/6cbcif2d.out",header=TRUE) x<-unique(data$lag1) y<-unique(data$lag2) z<-matrix(data$cif2d,length(y),length(x)) This z matrix is printed apparently correctly from a Cartesian point of

Hi All, I'm new to R and am trying to run a unit root test on the vector "y" (a time series of inflation (i.e. changes in the Consumer Price Index quarter on quarter)). I've run the Augmented-Dickey-Fuller Test below (R's URCA package). It gives me an error that it cannot find the function ur.df unless I comment out the third last line of code (see below). I try to call

Hi, I am trying to test whether a series is return series stationary, but before proceeding I wanted to make sure I understand correctly how to use the adf.test function and interpret its output... Could you please let me know whether I am correct in my interpretations? ex: I take x such as I know it doesn't have a unit root, and is therefore stationary 1/ > x <- rnorm(1000) >

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 #

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

I'm currently working with some time series data with the xts package, and would like to generate a forecast 12 periods into the future. There are limited observations, so I am unable to use an ARIMA model for the forecast. Here's the regression setup, after converting everything from zoo objects to vectors. hire.total.lag1 <- lag(hire.total, lag=-1, na.pad=TRUE) lm.model <-

Hi, While doing the ADF test in R using the following command I am getting the error and the result.."> x.ct=ur.df(rev$REVENUE,start=1,end=length(rev$REVENUE),frequency=1) Error in ur.df(rev$REVENUE, start = 1, end = length(rev$REVENUE), frequency = 1) : unused argument(s) (start = 1, end = 4, frequency = 1) >