similar to: decimal accuracy in pnorm( )

Dear R people The function below should be decreasing, convex, and tend to zero when x tends to infinity. curve((1-pnorm(x))/dnorm(x),from=0, to=9) >From the plot we see that for x between 8.0 and 8.3 the function is fluctuating. As far as I understand, this is due to the function pnorm() not being sufficiently accurate in the tails. I am using pnorm() in a way that has probably not been

Hi to all, maybe the last question was not clear enough. I did not found any hints how to decide whether it should use lower.tail or not. As it is an extra R-feature ( written in http://finzi.psych.upenn.edu/R/Rhelp02a/archive/66250.html ) I do not find anything about it in any statistical books of me. Regards Carmen

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

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))

Dear R list, I calculated a two-sided p values according to 2*(1-pnorm(8.104474)), which gives 4.440892e-16. However, it appears to be 5.30E-16 by a colleague and 5.2974E-16 from SAS. I tried to get around with mvtnorm package but it turns out to be using pnorm for univariate case. I should have missed some earlier discussions, but for the moment is there any short answer for a higher

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,

As to the function"pnorm",the default degree of freedom(df) is infinite. I wanna know how to set the df as I want. Help on pnorm doesn't have df setting.The only choice are:"mean, sd, lower.tail, log.p",but no df. For instance: sample size=6 df=6-1=5 t value=9.143 I wanna to the corresponding p value by using function "pnorm". How can I do it? Thanks a lot

Hi, I've read in a csv file (test.csv) which gives me the following table: Hin1 Hin2 Hin3 Hin4 Hin5 Hin6 HAI1 9534.83 4001.74 157.16 3736.93 484.60 59.25 HAI2 13272.48 1519.88 36.35 33.64 46.68 82.11 HAI3 12587.71 5686.94 656.62 572.29 351.60 136.91 HAI4 15240.81 10031.57 426.73 275.29 561.30 302.38 HAI5 15878.32 10517.14 18.93 22.00 16.91

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

Full_Name: James Michael Rath Version: all (I think) OS: doesn't matter Submission from: (NULL) (129.116.226.162) The code for pnorm in R was adapted from a Fortran library published in the ACM TOMS journal. The published version had a typographical error, though, which was pointed out in a second article published three years after the original. The error was that a macro/variable named

Full_Name: Julian Faraway Version: R-Release (June 15) OS: Linux (redhat) Submission from: (NULL) (141.211.66.172) R fails to compile on the current released version. Some constants are undefined in pnorm.c. It appears that adding #include "nmath.h" to pnorm.c solves this problem. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list

This is hopefully a quick question on decimal accuracy. Is any decimal accuracy lost when casting a numeric vector as a matrix? And then again casting the result back to a numeric? I'm finding that my calculation values are different when I run for loops that manually calculate matrix multiplication as compared to when I cast the vectors as matrices and multiply them using "%*%".

--=-YFjXKq8/D/t1qWmIzQ9D Content-Type: text/plain Content-Transfer-Encoding: quoted-printable I was going over the source in src/nmath/pnorm.c and noticed a little bug in pnorm_both (in R 1.7.0). The else-if on line 205 covers the entire real line. Seems you want an &&, not an ||. Doesn't make a big difference (you still get a 0 or 1 from extreme starting values) but your log

Full_Name: Marco Vezzoli Version: 2.2.0 2.1.0 2.0.0 OS: Solaris, Windows, Linux Submission from: (NULL) (57.78.11.38) The dnorm functions yield a wrong value when the standard deviation is near to 1e-1 e.g. > dnorm(0,mean=0.04,sd=0.3) [1] 1.318039 this error is consistent in various version and os.

Hello, Rlisters I have to compute p-values that are on the tail of the distribution, P-values < 10^-20. However, my current implementations enable one to estimate P-values up to 10^-12, or so. A typical example is found below, where t is my critical value. ########### example - code adapted from Rassoc ####################### rho01 = 0.5 rho105 = 0.5 rho005 = 0.5 t = 8 z = 2

Trenkler, Dietrich said: > > I found the following strange behavior using qnorm() and pnorm(): > > > x<-8.21;x-qnorm(pnorm(x)) > [1] 0.0004638484 > > x<-8.28;x-qnorm(pnorm(x)) > [1] 0.07046385 > > x<-8.29;x-qnorm(pnorm(x)) > [1] 0.08046385 > > x<-8.30;x-qnorm(pnorm(x)) > [1] -Inf > qnorm(1-.Machine$double.eps) [1] 8.12589

Hi, I've got an ugly but fairly simple function: mdevstdev <- function(a){ l <- dnorm(a)/(1-pnorm(a)) integrand <- function(z)(abs(z-l)*dnorm(z)) inted <- integrate(integrand, a, Inf) inted[[1]]/((1- pnorm(a))*sqrt((1 + a*l - l^2))) } I wanted to quickly produce a graph of this over the range [-3,3] so I used: plotit <-function(x=seq(-3,3,0.01),...){

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)

In checking over the solutions to some homework that I had assigned I observed the fact that in R (version 2.4.0) pnorm(-1.46) gives 0.07214504. The tables in the text book that I am using for the course give the probability as 0.0722. Fascinated, I scanned through 5 or 6 other text books (amongst the dozens of freebies from publishers that lurk on my shelf) and found that some agree with R

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: