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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 =

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("$\beta_1$","$SE(\beta_1)$", "$\beta_2$",

I want to estimate the following fixed effect model: y_i,t = alpha_i + beta_1 x1_t + beta_2 x2_i,tx2_i,t = gamma_i + gamma_1 x1_t + gamma_2 Z1_i + gamma_3 Z2_i I can use ivreg from AER to do the iv regression. fm <- ivreg(y_i,t ~ x1_t + x2_i,t | x1_t + Z1_i + Z2_i, data = DataSet) But, I'm not sure how can I add the fixed effects. Thanks! [[alternative HTML

I am currently trying to find MLE of a function with four parameters. My codes run well but i don't get the results. I get the following message: BHHH maximisation Number of iterations: 0 Return code: 100 Initial value out of range. I don't know this is so because of the way i have written my loglikelihood or what. The loglikelihood LogLik<-function(param){ beta_1<-param[1]

On 16/02/18 15:28, Bert Gunter wrote: > This is really a statistical issue. What do you think the Intercept term > represents? See ?contrasts. > > Cheers, > Bert > > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along and > sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom

This is really a statistical issue. What do you think the Intercept term represents? See ?contrasts. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Thu, Feb 15, 2018 at 5:27 PM, Marc Girondot via R-help < r-help at

Dear all, I'm running a coxph model of the form: coxph(Surv(Start, End, Death.ID) ~ x1 + x2 + a1 + a2 + a3) Within this model, I would like to compare the influence of x1 and x2 on the hazard rate. Specifically I am interested in testing whether the estimated coefficient for x1 is equal (or not) to the estimated coefficient for x2. I was thinking of using a Chow-test for this but the Chow

Dear Sir/Madam, I apologize for any cross-posting. I got a simple question, which I thought the R list may help me to find an answer. Suppose we have Y_1, Y_2, ., Y_n ~ Poisson (Lambda_i) and Lambda_i ~Gamma(alpha_i, beta_i). Empirical Bayes Estimator for hyper-parameters of the gamma distr, i.e. (alpha_t, beta_t) are needed. y=c(12,5,17,14) n=4 What about a Hierarchal B ayes

Dear R-er, I try to get the standard error of fitted parameters for factors with a glm, even the reference one: a <- runif(100) b <- sample(x=c("0", "1", "2"), size=100, replace = TRUE) df <- data.frame(A=a, B=b, stringsAsFactors = FALSE) g <- glm(a ~ b, data=df) summary(g)$coefficients # I don't get SE for the reference factor, here 0:

Hello, I'm having trouble figuring out how to calculate a p-value for a 1-tailed test of beta_1 in a linear model fit using command lm. My model has only 1 continuous, predictor variable. I want to test the null hypothesis beta_1 is >= 0. I can calculate the p-value for a 2-tailed test using the code "2*pt(-abs(t-value), df=degrees.freedom)", where t-value and degrees.freedom

I would like to be able to place a normal distribution surrounding the predicted values at various places on a plot. Below is some toy code that creates a scatterplot and plots a regression line through the data. library(MASS) mu <- c(0,1) Sigma <- matrix(c(1,.8,.8,1), ncol=2) set.seed(123) x <- mvrnorm(50,mu,Sigma) plot(x) abline(lm(x[,2] ~ x[,1])) Say I want to add a normal

Hi, I'm trying to get maximum likelihood estimates of \alpha, \beta_0 and \beta_1, this can be achieved by solving the following three equations: n / \alpha + \sum\limits_{i=1}^{n} ln(\psihat(i)) - \sum\limits_{i=1}^{n} ( ln(x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} 1/(psihat(i)) - (\alpha+1) \sum\limits_{i=1}^{n} ( 1 / (x_i + \psihat(i)) ) = 0 \alpha \sum\limits_{i=1}^{n} (

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

Hi everybody! I'm looking some way to do in R a polynomial fit, say like polyfit function of Octave/MATLAB. For who don't know, c = polyfit(x,y,m) finds the coefficients of a polynomial p(x) of degree m that fits the data, p(x[i]) to y[i], in a least squares sense. The result c is a vector of length m+1 containing the polynomial coefficients in descending powers: p(x) = c[1]*x^n +

In the logistic regression model, there is no residual log (pi/(1-pi)) = beta_0 + beta_1*X_1 + ..... But glm model will return residuals What is that? How to understand this? Can we put some residual in the logistic regression model by replacing pi with pi' (the estimated pi)? log (pi'/(1-pi')) = beta_0 + beta_1*X_1 + .....+ ei Thanks! [[alternative HTML version deleted]]

Hi all I know about producing a minimal example to show my problem. But I'm having trouble producing a minimal example that displays this behaviour, so please bear with me to begin with. Observe: I create an array called model.mat. Some details on this: > str(model.mat) num [1:18, 1:4] -0.170 -0.304 -2.617 2.025 -1.610 ... - attr(*, "dimnames")=List of 2 ..$ : chr

The following is a data frame > "jjd" <- structure(list(Observations = c(6.8, 6.6, 5.3, 6.1, 7.5, 7.4, 7.2, 6.5, 7.8, 9.1, 8.8, 9.1), LevelA = structure(c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3), .Label = c("A1", "A2", "A3"), class = "factor"), LevelB = structure(c(1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2), .Label =

Hi All. I am quite a newbie to R. This is a next basic question. I have a matrix; > x <- matrix(1:10.,5) > x [,1] [,2] [1,] 1 6 [2,] 2 7 [3,] 3 8 [4,] 4 9 [5,] 5 10 I like to get a modified matrix as follows; [,1] [,2] [1,] 1 6 [2,] 2 7 [3,] 3 8 * 5 -> 40 [4,] 4 9 [5,] 5 10 The following expression does not work.

I am fitting a logistic binomial model of the form glm(y ~ a*x,family=binomial) where a is a factor (with 5 levels) and x is a continuous predictor. To assess how much ``impact'' x has, I want to compare the fitted success probability when x = its maximum value with the fitted probability when x = its mean value. (The mean and the max are to be taken by level of the factor

Hello R-Helpers, I have a slight problem with the expresion data[data==""] <- NA which works well for a data.frame. But now i must use the same for a list of data.frames. My idea is data[[]][data==""] but it don´t work. Thanks!! Dominic [[alternative HTML version deleted]]