similar to: Calculate the natural log of cdf between 2 intervals

In the R source, nmath/pnorm.c contains the code for a rational function approximation for the normal cdf. These constants are listed: const double a[5] = { 2.2352520354606839287, 161.02823106855587881, 1067.6894854603709582, 18154.981253343561249, 0.065682337918207449113 }; The source file cites a paper by Cody (1969) and states that these

Hi, I'm trying to find a way to take evaluate the Normal CDF over an interval and return the result on the log scale. This works, but I think it isn't numerically stable: log(pnorm(a, mean = x, sd = y) - pnorm(b, mean = x, sd = y)) Does anyone know of a single function that does the above? Or knows of a way to make it more stable? I'd really appreciate any suggestions!

Dear R-users, I have been using the code below in order to verify how the CDF of a skew-normal distribution was calculated: library(sn) s=seq(-30,30,by=0.1) a<-matrix(nrow=length(s),ncol=5) lambda=1 for(i in 1:length(s)){ a[i,1]=pnorm(s[i],mean=0,sd=1); a[i,2]=T.Owen(s[i],lambda); a[i,3]=a[i,5]-2*a[i,6]; a[i,4]=pnorm(s[i])-2*T.Owen(s[i],lambda);

Dear Experts, I need to calculate the cdf of the standard normal distribution, i.e. H(x) = 1/sqrt(2*pi) integral(exp(-z^2/2) dz), where z is b/w -infi to infi. I know there should be a way to do it in R, but did not know to do it. I'd appreciate any help you could offer. Charlie Liu Graduate student intern at EPA/ECO

Full_Name: Morten Welinder Version: 2 OS: Solaris/space/gcc2.95.2 Submission from: (NULL) (65.213.85.217) I changed src/nmath/standalone/test.c to read: --------------------------------------------------------------------------------- #define MATHLIB_STANDALONE 1 #include <Rmath.h> #include <stdio.h> int main() { double x; for (x = 99990; x <= 100009; x++) printf

I am trying to make a loglikelihood function using copulas. I am trying to use mvdc to find the density function. When I run this I got the error that the pdf and cdf of my function tobit doesn't exist. Can somebody guide me where my mistake is? dtobit <- function(beta,sigma, x, y) {ifelse(y>0, dnorm(y,x%*%beta, sigma),(1-pnorm((x%*%beta)/sigma)))} ptobit <- function(beta,sigma, x,

This practical problem in maximum likelihood estimation must be encountered quite a bit. What do you do when a data point has a probability that comes out in numerical evaluation to zero? In calculating the log likelihood you then have a log(0) problem. Here is a simple example (probit) which illustrates the problem: x<-c(1,2,3,4,100) ntrials<-100 yes<-round(ntrials*pnorm((x-3)/1))

Hi, I have a data set which is assumed to follow weibull distr'. How can I find of cdf for this data. For example, for normal data I used (package - lmomco) >cdfnor(15,parnor(lmom.ub(c(df$V1)))) Also, lmomco package does not have functions for finding cdf for some of the distributions like lognormal. Is there any other package, which can handle these distributions?

Under platform x86_64-pc-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 13.1 year 2011 month 07

Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y

I found the following strange behavior using qnorm() and pnorm(): > x<-8.21;x-qnorm(pnorm(x)) [1] 0.0004638484 > x<-8.22;x-qnorm(pnorm(x)) [1] 0.01046385 > x<-8.23;x-qnorm(pnorm(x)) [1] 0.02046385 > x<-8.24;x-qnorm(pnorm(x)) [1] 0.03046385 > x<-8.25;x-qnorm(pnorm(x)) [1] 0.04046385 > x<-8.26;x-qnorm(pnorm(x)) [1] 0.05046385 > x<-8.27;x-qnorm(pnorm(x))

Hi all, I have the following two function f1 and f2. f1 <- function(lambda,z,p1){ lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8} f2 <- function(p1,cl, cu){ 0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-pnorm(cu-4))))}-0.05 First fix p1 to be 0.15. (i) choose a lambda value, say lamda=0.6, (ii)

Hi folks, I'm trying to estimate bias-corrected percentile (BCP) confidence intervals on a vector from a simple for loop used for resampling. I am attempting to follow steps in Manly, B. 1998. Randomization, bootstrap and monte carlo methods in biology. 2nd edition., p. 48. PDF of the approach/steps should be available here: https://wyocoopunit.box.com/s/9vm4vgmbx5h7um809bvg6u7wr392v6i9 If

On a recent FreeBSD 8.0-CURRENT (i386) building R (any version) breaks with the following messages: ---------------------------------------------------------------------- [...snip...] gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include -DHAVE_CONFIG_H -g -O2 -c wilcox.c -o wilcox.o gcc -std=gnu99 -I. -I../../src/include -I../../src/include -I/usr/local/include

Hi, I'm trying to evaluate a character vector within pnorm. I have a vector with values and names x = c(2,3) names(x) = c("mean", "sd") so that i tried the following temp = paste(names(x), x, sep = "=") #gives #> temp #[1] "mean=2" "sd=3" #Problem is that both values 2 and 3 are taken as values for the mean argument in pnorm pnorm(0,

Hi all. I may be wrong, (and often am), but in trying to determine how to calculate the erf function, the documentation for 'pnorm' states: ## if you want the so-called 'error function' erf <- function(x) 2 * pnorm(x * sqrt(2)) - 1 ## and the so-called 'complementary error function' erfc <- function(x) 2 * pnorm(x * sqrt(2), lower=FALSE) Should, instead, it read:

Dear all, Can anyone take a look at my program below? There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu). I fixed p1=0.15 for both functions. For any fixed value of lambda (between 0.01 and 0.99), I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu values. Then I plug the calculated cl and cu back into the function f2. Eventually, I want to find the lambda value

I stumbled across this and I am wondering if this is unexpected behavior or if I am missing something. > pnorm(-1.0e+307, log.p=TRUE) [1] -Inf > pnorm(-1.0e+308, log.p=TRUE) [1] NaN Warning message: In pnorm(q, mean, sd, lower.tail, log.p) : NaNs produced > pnorm(-1.0e+309, log.p=TRUE) [1] -Inf I don't know C and am not that skilled with R, so it would be hard for me to look into

Dear R user I follow the steps defined in Modern applied statistics page(453) to use optim. However, when I run the following code the parameters seems way off and the third parameter(p3) stayed as the initial value. below is the code: ## data da=c(418,401,416,360,411,425,537,379,484,388,486,380,394,363,405,383,392,363,398,526) ### initial values pars=c(392.25, 507.25, 0.80)

Hi, I would have thought that these two constructions would produce the same result but they do not. Resp <- rbinom(10, 1, 0.5) Stim <- rep(0:1, 5) mm <- model.matrix(~ Stim) Xb <- mm %*% c(0, 1) ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) pnorm(as.vector(Xb), lower.tail = Resp, log.p = TRUE) > ifelse(Resp, log(pnorm(Xb)), log(1 - pnorm(Xb))) [1] -0.6931472 -1.8410216