similar to: Regression coefficients

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

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

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

Hello, I have a question on unit root test with urca toolbox. First, to run a unit root test with lags selected by BIC, I type: > CPILD4UR<-ur.df(x1$CPILD4[5:nr1], type ="drift", lags=12, selectlags ="BIC") > summary(CPILD4UR) The results indicate that the optimal lags selected by BIC is 4. Then I run the same unit root test with drift and 4 lags:

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 &lt;- as.Date(data.df[,1]) selection&lt;-which(dates&gt;=as.Date("1986-1-1") &amp; dates&lt;=as.Date("2007-12-31")) dep &lt;- ts(data.df[selection,c("dep")]) indep.ret1

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,

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

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

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

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 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

Hi all, In my data analysis, I have created a random matrix M ( of order 500 X 7). I want to use the same matrix when I start a new session, or suppose I want to send this matrix to one of my friends (because this matrix is randomly generated, and I dont want to use any other 500X7 matrix randomly generated by R). How can I save and call this matrix in the later sessions as well? Appreciate

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 all, I have 4 matrices, each having 5 columns and 4 rows .....denoted by B1,B2,B3,B4. I have generated a vector of 7 indices, say (1,2,4,3,2,3,1} which refers to the index of the matrices to be chosen and then appended one on the top of the next: like, in this case, I wish to have the following mega matrix: B1over B2 over B4 over B3 over B2 over B3 over B1. 1> How can I achieve this?

Hi all, I have a data set containing variables LOSS, GDP, HPI and UE. (I have attached it in case it is required). Having renamed the variables as l,g,h and u, I wish to run a Lasso Regression with l as the dependent variable and all the other 3 as the independent variables. data=read.table("data.txt", header=T) l=data$LOSS h=data$HPI u=data$UE g=data$GDP matrix=data.frame(l,g,h,u)

Hi, I want to fit the following model to my data: Y_t= a+bY_(t-1)+cY_(t-2) + Z_t +Z_(t-1) + Z_(t-2) + X_t + M_t i.e. it is an ARMA(2,2) with some additional regressors X and M. [Z_t's are the white noise variables] How do I find the estimates of the coefficients in R? And also I would like to know what technique R employs to find the estimates? Any help is appreciated. Thanks,

Hi all, 1>i have 3 vectors a,b and c, each of length 25....... i want to define a new data frame z such that z[1] = (a[1] b[1] c[1]), z[2] = (a[2] b[2] c[2]) and so on...how do i do it in R 2> Then i want to draw bootstrap samples from z. Kindly suggest how i can do this in R. Thanks, Preetam -- Preetam Pal (+91)-9432212774 M-Stat 2nd Year,

Hi guys, I am running glm(y~., data = history,family=binomial)-essentially, logistic regression for credit scoring (y = 0 or 1). The dataset 'history' has 14 variables, a few examples: history <- read.csv("history.csv". header = TRUE) 1> 'income = 100,200,300 (these are numbers in my dataset; however interpretation is that these are just tags or labels,for every

Hi all, I want to fit the following model to my data: Y_t= a+bY_(t-1)+cY_(t-2) + Z_t +Z_(t-1) + Z_(t-2) + X_t + M_t i.e. it is an ARMA(2,2) with some additional regressors X and M. [Z_t's are the white noise variables] So, I run the following code: for (i in 1:rep) { index=sample(4,15,replace=T) final<-do.call(rbind,lapply(index,function(i)

Hi all, Suppose I am fitting an arma(p,q) model to a time series y_t. So, my model should contain (q+1) white noise variables. As far as I know, each of them should have the same variance. How do I get the estimate of this variance by running the arma(y) function (or is there any other way)? Appreciate your help. Thanks, Preetam -- Preetam Pal (+91)-9432212774 M-Stat 2nd Year,