search for: beta0

...\alpha \sum\limits_{i=1}^{n} ( i / \psihat(i) ) - (\alpha + 1) \sum\limits_{i=1}^{n} ( i / (x_i + \psihat(i)) ) = 0 where \psihat=\beta_0 + \beta_1 * i. Now i want to get iterated values for \alpha, \beta_0 and \beta_1, so i used the following implementation # first equation l1 <- function(beta0,beta1,alpha,x) { n<-length(x) s2<-length(x) for(i in 1:n) { s2[i]<-log(beta0+beta1*i)-log(x[i]+beta0+beta1*i) } s2<-sum(s2) return((n/alpha)+s2) } # second equation l2 <- function(beta0,beta1,alpha,x) { n<-length(x) s1<-length(x) s2<-length(x)...

Hello, The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0 and beta1 via GLM) 2) with lambda, estimate alpha via ML estimation 3) with updataed alpha, replicate 1) and 2) until alpha is converged to a value I coded 1) and 2) (it works),...

Dear All I'm attempting to speed up my model building procedure, but need some help with the loops I've created...please bear with me through the explanation! My basic model call is something like: m0sulf.nlme<-nlme(conc~beta0*exp(-beta1*day)+beta2*exp(-beta3*day), data=m0sulf, fixed=(beta0+beta1+beta2+beta3~1), random=pdDiag(beta0+beta1+beta2+beta3~1), start=c(beta0=-300,beta1=15, beta2=300,beta3=0.5),...

Hi all, I am trying to create simulated data for exploring reclassfication measures in a logistic setting with two continuous predictors and I would like to set the average population probability of outcome rather than the logistic beta0. Is there a way to find a beta0 that will generate the desired average population probability of outcome given x,y and their odds ratios? #Here is an outline of what I would like to do: pop.d=0.1 xvar=rnorm(5000,0,0.5) yvar=rnorm(5000,0,0.5) orx=16 ory=2 #find beta0 beta0=f(x...

...zero_trt -0.200134813 Three of the four look right (ie, converging to population values), but the count_trt is stuck at zero, regardless of sample size (when it should be ~ -0.20). Does anyone see what's wrong? Thanks for any input. cheers, Dave mysim <- function(n, beta0, beta1, alpha0, alpha1, theta){ trt <- c(rep(0,n), rep(1,n)) ### mean function logit model p0 <- exp(alpha0 + alpha1*trt)/(1 + exp(alpha0 + alpha1*trt)) ### 0 / 1 based on p0 y1 <- as.numeric(runif(n)>p0) ### mean function count portion mu <- exp(beta0 + beta1*trt) ### estima...

HI >i have a little problem please help me to solve it >this is the code in R: >> beta0 [1] 64.90614 > beta1 [1] 17.7025 > beta [1] 17 64 >her beta<- c(beta0, beta1) thank you in advance hafida -- View this message in context: http://r.789695.n4.nabble.com/Decimal-number-tp4639428.html Sent from the R help mailing list archive at Nabble.com.

...am new to both of things, so do not blame me too much... I am busy with semiparametric regression and use WinBUGS to sample posteriors. The code to call Winbugs is as follows: data <- list("y","X","n","m") #My variables inits.beta <- rep(0,K) inits.beta0 <- 0 inits <- function(){list(beta=inits.beta,beta0=inits.beta0,taueps=1.0E-3)} parameters <- list("sigma","beta","beta0","y.star") fitm <- bugs(data,inits,parameters,model.file="model.bug", n.chains=3, n.iter=n.iter, n.burnin=n....

I use "while" loop but it produces an errro. I have no idea about this. Error in "[<-"(`*tmp*`, i, value = numeric(0)) : nothing to replace with The problem description is The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0 and beta1 via GLM) 2) with lambda, estimate alpha via ML estimation 3) with updated alpha, replicate 1) and 2) until alpha is converged to a value My source code is alpha=rep...

...l.b1 = current.b1 = ratio1 = double(Nsim) # blank vector index <- 1:Nsim # creating index a0 <- 10 # initial value (assumed) for tau b0 <- (.01) # initial value (assumed) for tau fit <- glm((y/n)~x,family=binomial, weight = n, data=dta) # initial value for beta parameters <- c("Beta0", "Beta1", "Tau") parameter.matrix <- array(NA,c(Nsim,3)) # blank array parameter.matrix <- as.data.frame(parameter.matrix) # creating data frame parameter.matrix[1,] <- c(fit$coef[1],fit$coef[2],rgamma(1, shape=a0, scale = b0)) # putting initial values for (i in...

...in 1 : length(sim[,1])) { if (tmp[k,1]==sim[n,1] && v[k]<=sim[n,2]){tmp[k,2]=1} if(tmp[k,1]==sim[n,1] && v[k]>sim[n,2]) {tmp[k,2]=0} } } return(tmp) } # The model to be fitted Model_1=function(beta0_mod,mu_mod,k_mod,I_mod) { parms = c(beta0=beta0_mod, mu=mu_mod, k=k_mod) my.atol = 1e-6 times1=c(0,0.002884615,0.003846154,0.005769231,0.009615385,0.01,0.019230769) times2=c(0.028846154,0.038461538,0.057692308, 0.076923077,0.083333333,0.096153846) times3=c(0.115384615,0...

...ction. However, I can't seem to duplicate these predictions in a separate program using the model coefficients. From duplicating simpler models, I understand the input variables are adjusted by the intercept term. (e.g. for an arima(1,0,0) the prediction equation is y(t) = beta1 * ( y(t-1) - beta0 ) + beta0 ....where beta0 is the intercept) Currently, I've expected the prediction to follow the equation below: y(t) = beta0 + [beta1*( y(t-1) - beta0 )] + [beta2 * epsilon(t-1)] + [beta3 * (y(t-12) - beta0)] + [beta4 * epsilon(t-12)] This has proved unsuccessful. What equatio...

...39 resources_used.cpupercent=98 resources_used.cput=00:25:02 resources_used.mem=233536kb resources_used.ncpus=1 resources_used.vmem=250252kb resources_used.walltime=00:25:13 ===========End of original message text=========== The second source file contains: rmnlRwMetrop1=function(Data,Prior,Mcmc,beta0) { # # purpose: # draw from posterior for MNL using Independence Metropolis # # Arguments: # Data - list of m,X,y # m is number of alternatives # X is nobs*m x nvar matrix # y is nobs vector of values from 1 to m # Prior - list of A, betabar # A is nvar x nvar prior preci matr...

Hello R-Experts, I am facing the problem that I have to estimate several parameters for a lot of different dependent variables. One single regression looks something like this: y = beta0 + beta1 * x1 + beta2 * x2 + beta3 * x1 * x2 + beta4 * x4 + beta5 * lag(x4,-1) where y is the dependent variable and xi are the independent ones. Important to me are the different estimates of betai and their respective p-values only. Now I have aprx. 50 different data sets of y and x1 to x4. So...

Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best Regards,Bo------------------------model {for(i...

Hi, I am trying to covert a Winbugs code into R code. Here is the winbugs code model{# model’s likelihoodfor (i in 1:n){time[i] ~ dnorm( mu[i], tau ) # stochastic componenent# link and linear predictormu[i] <- beta0 + beta1 * cases[i] + beta2 * distance[i]}# prior distributionstau ~ dgamma( 0.01, 0.01 )beta0 ~ dnorm( 0.0, 1.0E-4)beta1 ~ dnorm( 0.0, 1.0E-4)beta2 ~ dnorm( 0.0, 1.0E-4)# definition of sigmas2<-1/taus <-sqrt(s2)# calculation of the sample variancefor (i in 1:n){ c.time[i]<-time[i]-mean(tim...

...trend and season), the top line s determines the number of seasons in a year (3 and 4 work fine, 2 gives the above error), and at the end tries to fit the model. I am using R version 2.11.1, Windows XP. # ----- Basic structural model ----- s <- 2 # frequency or period mu0 <- 6 beta0 <- 1 it0 <- 0 it.states <- sample(1:6,s) mu <- numeric() # mean beta <- numeric() # slope it <- numeric() # seasonal state mu[1] <- mu0 + beta0 + rnorm(1,0,2) beta[1] <- beta0 + rnorm(1,0,0.5) it[1] <- - sum(it.states[2:s]) + rnorm...

...por png funciona, aparentemente hay un inconveniente para guardar en jpg. Un código reproducible de internet con una modificación en el tipo de archivo es: names = LETTERS[1:26] ## Gives a sequence of the letters of the alphabet beta1 = rnorm(26, 5, 2) ## A vector of slopes (one for each letter) beta0 = 10 ## A common intercept for(i in 1:26){ x = rnorm(500, 105, 10) y = beta0 + beta1[i]*x + 15*rnorm(500) mypath <- file.path("C:","R","SAVEHERE",paste("myplot_", names[i], ".png", sep = "")) png(file=mypath) mytitle =...

...al = sample(setdiff(1:ncol(gt),which(freq==0)),45) beta.sign = rep(1,45) c.value = 0.402 beta.abs = c.value*abs(log10(freq[causal])) beta.val = beta.sign*beta.abs x.bar = apply(gt[,causal],2,mean) x.bar = as.matrix(x.bar) beta.val = t(as.matrix(beta.val)) #disease prvalance = 1% #beta0 = -log(99)-beta.val %*% x.bar #disease prvalance = 1.5% beta0 = 0-beta.val %*% x.bar eta = beta.val %*% t(gt[,causal]) eta = as.vector(eta) + rep(beta0,nrow(gt)) prob = exp(eta)/(1+exp(eta)) genocase = rep(NA, sample.size) set.seed(seed.num) for(i in 1:sample.size){ genocase[...

Dear all I used R2WinBUGS package's bugs() function to generate MCMC results. Then I tried to save the simulation draws in R, using read.bugs() function. Here is a simple test: ###################### library(coda) library(R2WinBUGS) #fake some data to test beta0=1 beta1=1.5 beta2=-1 beta3=2 N=200 x1=rnorm(N, mean=0,sd=1) x2=rnorm(N, mean=0,sd=1) x3=rnorm(N, mean=0,sd=1) lambda2= exp(beta0+ beta1*x1+ beta2 * x2 + beta3 * x3) y=rep(NA,N) for (i in 1: N) { y[i]=rpois(1,lambda2[i]) } #generate inputs required by winbugs data <- list(Y20=y,x1=x1, x2=x2, x3=x...

...e change is occurring, in other words, the break point Use of segmented(): Segmented does a very good job of capturing the breakpoints and fitting three distinct slopes, i.e. linear models. It does not, however give me Pr(>|t|) for the break point coefficients. I need to answer the question H:0 Beta0=Beta with a certainty metric, i.e. probability statistic. This is especially important for my, flat looking data class. davies.test() question: davies.test() only excepts lm() or glm() objects as input. If I run segmented to find 1 breakpoint instead of 2, I get a totally bogus answer. Without kno...