similar to: BHHH algorithm on duration time models for stock prices

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

Dear R users, Could you tell me where a I find some references of BHHH algorithm ? I need write it in R. Thank you.

Dear R users, I am new to R. I would like to find *maximum likelihood estimators for psi and alpha* based on the following *log likelihood function*, c is consumption data comprising 148 entries: fn<-function(c,psi,alpha) { s1<-sum(for(i in 1:n){(c[i]-(psi^(-1/alpha)*(lag(c[i],-1))))^2* (lag(c[i],-1)^((-2)*(alpha+1)) )}); s2<- sum(for(m in 1:n){log(lag(c[m],-1)^(((2)*alpha)+2))});

Hola, Esta es una forma de hacerlo... Mira que lo primero que he modificado es el fichero "x.csv" para sustituir los espacios en los nombres por "_". Y también he quitado los acentos y las eñes... He utilizado el paquete RNNS y la función "mlp()" para ajustar la red. #------------------------------------------- > x <- read.csv("x.csv",

Dear R helpers Suppose I have a data frame as given below mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8, 8, 8, 7, 7, 2, 2, 4)) mydat         x     y 1      1     10 2      1     10 3      1     10 4      2       8 5      2       8 6      2       8 7      2       7 8      2       7 9      5       2 10    5       2 11    6       4 unique(mydat$x) will give me 1,

A data set(dat),has 2 variables: x and a, and 100 rows. I wanna add 2 variables,and call the new data set dat1: var1:f = a/median(a) var2:x_new = x*f My solution: dat1<-transform(dat,f = a/median(a),x_new = x*f) But gets error reply which says that "f" is not exits since dat has no variables called "f". So I have to do through 2 steps:

Dear all, I have several character strings with a high number of different levels. unique(x) gives me values in the range of 100-200. This creates problems as I would like to use them as predictors in a coxph model. I therefore would like to convert each of these strings to a new string (x_new). x_new should be equal to x for the top n categories (i.e. the top n levels with the highest

Dear all, I am using modified LARS algorithm (ref: The Adaptive Lasso and Its Oracle Properties, Zou 2006) for adaptive lasso penalized linear regression. 1. w(j) <- |beta_ols(j)|^(-gamma) gamma>0 and j = 1,...,p 2. define x_new(j) <- x(j)*w(j) 3. apply LARS to solve modified lasso problem out.adalasso <- lars(X_new,y,type="lasso") or enet(X_new,

Hi, folks, Happy work in weekends >_< My question is how to plot the dependent variable against one of the predictors with other predictors as constant. Not for the original data, but after prediction. It means y is the predicted value of the dependent variables. The constane value of the other predictors may be the average or some fixed value. ####### y=1:10 x=10:1 z=2:11

Hi all, Are there functions in R that could help me do the following? We have a special type of regression which is called Geometric Mean Regression. We have done some search and found the following: https://stat.ethz.ch/pipermail/r-help/2011-July/285022.html The question is: how to do the statistical inference on GMR results? More specifically, we are looking for the prediction interval:

I have some data fit to a smooth.spline object as follows: (x=vector of data for the predictor variable, y=vector of data for the response variable) fit <- smooth.spline(x,y) Now, given a spline fit point y_new, I want to be able to find out what value of x_new yielded this fit value. How to do so? (This problem is the inverse of the predict.smooth.spline function, which takes x_new as input

Dear R Helpers, I'm struggling with a data preparation problem. I feel that it is a quite easy task but I don't get it done. I hope you can help me with that. I have a data frame looking like this: ID <- c(1,1,1,2,2,3,3,3,3) T <- c(1,2,3,1,4,3,5,6,8) x <- rep(1,9) df <- data.frame(ID,T,x) >df ID T x 1 1 1 1 2 1 1 3 1 2 1 1 2 4 1 3 3 1 3 5 1 3 6 1 3 8 1 I want to

Hi, This is a practical question and I am sure there are many statisticians can give me a hand. I have 500 time series data (500 rows), each row contains 100 intervals, i.e., on each row, I have X1, X2, ..... X100. I am trying to reduce the dimension of this input because the data at the end of each row does not have significant meaning to the project I am doing. I used cubic splines on ea.

Hi, I'm trying to fit the following model to data using 'nls': y = alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x) and the call I've been using is: nls(y ~ alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x), start=list(alpha_1=4, alpha_2=2, beta_1=3.5, beta_2=2.5), trace=TRUE, control=nls.control(maxiter =

Estimados Les consulto por redes neuronales, hay diversos artículos como los siguientes (el último tienen un error actualmente). Pero mi pregunta va un poco por otro lado. http://www.r-bloggers.com/build-your-own-neural-network-classifier-in-r/ http://www.r-bloggers.com/classification-using-neural-net-in-r/ Básicamente se puede calcular un valor, por ejemplo doblar 2,4 grados a la derecha, luego 1

Hello all, I have a problem estimating Random Effects model using censReg function. small part of code: UpC <- censReg(Power ~ Windspeed, left = -Inf, right = 2000,data=PData_In,method="BHHH",nGHQ = 4) Error in maxNRCompute(fn = logLikAttr, fnOrig = fn, gradOrig = grad, hessOrig = hess, : NA in the initial gradient ...then I tried to set starting values myself and here is

Hello, I would like to create a matrix with one of the columns named $\delta$. I have also created columns $\beta_1$ , $\beta_2$, etc. However, it seems like \d is an escape sequence which gets automatically removed. (Using these names such that they work right in xtable -> latex) colnames(simpleReg.mat) <- c("$\beta_1$","$SE(\beta_1)$", "$\beta_2$",

Hello R community, I have been using R's inbuilt maximum likelihood functions, for the different methods (NR, BFGS, etc). I have figured out how to use all of them except the maxBHHH function. This one is different from the others as it requires an observation level gradient. I am using the following syntax: maxBHHH(logLik,grad=nuGradient,finalHessian="BHHH",start=prm,iterlim=2)

Have a look at ?optim. I don't think it has the BHHH algorithm as an option, though. =========================================== David Barron Jesus College University of Oxford -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Harold Doran Sent: 10 July 2003 15:43 To: Fohr, Marc [AM]; R-help at stat.math.ethz.ch

Dear list, this is not a particular R question but perhaps someone can help. I am running a maximum likelihood estimation (competing risk duration model with unobserved heterogeneity) on 30 different datasets. The problem is that on 2 datasets the model does not converge. I am interested if there are any methods, based on the gradients or (an approximation of) the hessian which helps to