similar to: variance of a linear model

When I saw the subject of the original message on R-help, I was 95% confident that I knew the answer (before I had seen the question). This made me think that perhaps for some functions there should be a 'Troubleshooting' section in the help file. The current help file for 'integrate' does say, as Sundar points out, what the requirements are. However, I think more people would

Dear all, I am stuck on the following problem with integrate(). I have been out of luck using RSiteSearch().. My function is g2<-function(b,theta,xi,yi,sigma2){ xi<-cbind(1,xi) eta<-drop(xi%*%theta) num<-exp((eta + rep(b,length(eta)))*yi) den<- 1 + exp(eta + rep(b,length(eta))) result=(num/den)*exp((-b^2)/sigma2)/sqrt(2*pi*sigma2)

Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it

Hi, folks, x=1:10 y=rep(2:6,2) lin=lm(y~x) x=3:12 new=predict(lin,se.fit=T) #se.fit: the standard error of the predicted means, namely, the square root of Var( E[y|x] | x) # How can I generate the variances of the new observations? Namely the square root of var(y|x), ## Which I think should be much larger than the values from se.fit=T. The reason why I need to know the estimations of

I need to construct confidence intervals for the binomial variance. This is the usual estimate v = x*(n-x)/n or its unbiased counterpart v' = x*(n-x)/(n-1) where x = binomial number of successes observed in n Bernoulli trials from proportion p. The usual X^2 method for variance confidence intervals will not work, because of the strong non-normal character of the sampling

Hi fellow R-users, I am interested in fitting the following model yi=log(a+xi*b*e), e~N(0,sigma2) and where x is a known covariate and a and b are parameters to be estimated as is sigma2. I am not sure how to fit such a model using lm or nls. Is there another function I can use to fit this? Thanks for any help. Stacey --------------------------------- [[alternative HTML version

Hi all, I have probably a basic question, but I can't seem to find the answer in the literature or in the R-archives. I would like to do a robust ANCOVA (using either rlm or lmRob of the MASS and robust packages) - my response variable deviates slightly from normal and I have some "outliers". The data consist of 2 factor variables and 3-5 covariates (fdepending on the model).

Hi everyone, I am looking to do some manual maximum likelihood estimation in R. I have done a lot of work in Stata and so I have been using output comparisons to get a handle on what is happening. I estimated a simple linear model in R with lm() and also my own maximum likelihood program. I then compared the output with Stata. Two things jumped out at me. Firstly, in Stata my coefficient

Dear friends, After fitting a generalized linear models ,i hope to get the variance of alpha,variance of beta and their covariance, that is , the variance-covariance matrix/information of alpha and beta , suppose *B* is the object of GLMs, i use attributes(B) to look for the options ,but can't find it, anybody knows how to get it? > attributes(B) $names [1] "coefficients"

I am trying to write a loop to forecast realized volatility over successive days for the purpose of VaR prediction using the HAR-RV-CJ model which is as follows: log(RV_t+1) = ?_0 + ?_CD log(CV_t) + ?_CW log(CV_t-5) + ?_CM log(CV_t-22) + ?_JD log(J_t + 1) + ?_JW log(J_t-5 + 1) + ?_JM log(J_t-22 + 1) + e_t where RV is realized volatility, CV is continuous volatility and J is the jump which is

Hi. I am trying to analyze with BRugs the Box-Tiao variance components example in WinBUGS. The output from BRugs, mean sd MC_error val2.5pc median val97.5pc start sample sigma2.btw 681.9 1161 10.89 0.7016 253.8 4232 25001 100000 sigma2.with 4266.0 1246 4.92 2480.0000 4057.0 7262 25001 100000 doesn't match the output from WinBUGS, node mean

Dear R friends, When fitting GLMs in R, we may need to specify the variance function to do our analysis. I had thought it's the mean value, but it seems not. Could anybody expain the correct meaning of *mu* in the variance function of GLMs? The following content is from the R-hlep. variance for all families other than quasi, the variance function is determined by the family. The quasi

Dear R users, I have a linear model of the kind outcome ~ treatment + covariate where 'treatment' is a factor with three levels ("0", "1", and "2"), and the covariate is continuous. Treatments "1" and "2" both have regression coefficients significantly different from 0 when using treatment contrasts with treatment "0" as the

Dear WizaRds, dear Thomas, First of all, I want to tell you how grateful I am for all your support. I wish I will be able to help others along one day the same way you do. Thank you so much. I am struggling with a multistage sampling design: library(survey) multi3 <- data.frame(cluster=c(1,1,1,1 ,2,2,2, 3,3), id=c(1,2,3,4, 1,2,3, 1,2), nl=c(4,4,4,4, 3,3,3, 2,2), Nl=c(100,100,100,100,

I wrote a simple log likelihood (for the ordinary least squares (OLS) model), in two ways. The first works out the likelihood. The second merely calls the first, but after transforming the variance parameter, so as to allow an unconstrained maximisation. So the second suffers a slight cost for one exp() and then it pays the cost of calling the first. I did performance measurement. One would

Dear Kailas Gokhale, The negative individual variance is not a problem with your code or plm. It a property of your data. Please check the posts of Giovanni Millo on this topic: [R] R: plm random effect: the estimated variance of the individual effect is negative Millo Giovanni Giovanni_Millo at Generali.com Sat Jan 5 10:10:01 CET 2013 You can find the posts in the archive by rseek.org.

Hi, folks, As I understand, Least-squares Estimate (second-moment assumption) and the Method of Maximum Likelihood (full distribtuion assumption) are used for linear regression. I do >?lm, but the help file does not tell me the model employed in R. But in the book 'Introductory Statistics with R', it indicates R estimate the parameters using the method of Least-squares. However it

I wonder whether the functions for skewness and kurtosis in the e1071 package are based on correct formulas. The functions in the package e1071 are: # -------------------------------------------- skewness <- function (x, na.rm = FALSE) { if (na.rm) x <- x[!is.na(x)] sum((x - mean(x))^3)/(length(x) * sd(x)^3) } # -------------------------------------------- and #

Dear All, have a general question about coefficients estimation of the mixed model. I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni); b follows N(0,\psi) #i.e. bivariate normal where b is the latent variable, Z and X are ni*2 design matrices, sigma is the error variance, Y are longitudinal data, i.e. there are ni

Mark Leeds <mleeds at mlp.com> wrote (to S-News): > does anyone know the R equivalent of the SPlus rowVars function ? Andy Liaw <andy_liaw at merck.com> replied: > More seriously, I seem to recall David Brahms at one time had created an R > package with these dimensional summary statistics, using C code. (And I > pointed him to the `two-pass' algorithm for variance.)