similar to: R-beta: S Compatibility (again)

Using R-2.12.1 and latticeExtra-0.6-14, I would like to understand why a rootogram displaying samples from the Poisson distribution looks like I expected it, whereas a rootogram using the normal distribution does not: library(latticeExtra) rootogram(~rpois(1000, lambda = 50), dfun = function(x) dpois(x, lambda = 50)) rootogram(~rnorm(1000), dfun = function(x) dnorm(x,mean(x),sd(x))) I

Is there any R function that computes the convolution of the double exponential distribution? If not, is there a good way to integrate ((q+x)^n)*exp(-2x) over x from 0 to Inf for any value of q and for any positive integer n? I need to perform the integration within a function with q and n as arguments. The function integrate() is giving me this message: "evaluation of function gave a

Hi, Am not sure if my code itself is correct. Here's what am trying to do: Minimize integration of a function of gaussian distributed variable 'x' over the interval qnorm(0.999) to Inf by changing value of parameter 'mu'. mu is the shift in mean of 'x'. Code: # x follows gaussian distribution # fx2 to be minimized by changing values of mu # integration to be done over

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after

Full_Name: Joerg Polzehl Version: 2.3.1 OS: x86_64, linux-gnu Submission from: (NULL) (62.141.176.22) I observed an incorrect behavior of function deriv when evaluating arguments of dnorm deriv(~dnorm(z,0,s),"z") expression({ .value <- dnorm(z, 0, s) .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(z * dnorm(z))

hello, please can anyone help me out. Am a new user of R program. Am having problem with this code below, not getting the expected results. 1. Each m, the cumulative sum should be 1.000 but the 2nd and 3rd m returned 2.000 and 3.000 instead of 1.000. 2. to get the LCL(m) and UCL(m) for each m base on these instructions if out.cum > 0.025 then LCL(m)= y-1 if out.cum >0.975

Dear R-users, The numerical integration example given in Gentleman and Ihaka (2000), "Lexical Scope and Statistical Computing," JCGS, 9, 491-508, is very interesting and helpful in understanding how lexical scope is about. However, I got some questions that I just can't figure out. First all, allow me to copy the two functions given by the authors: midpoint <- function(f, a,

Dear R-people, I have to compute C - -(pnorm(B)*dnorm(B)*B + dnorm(B)^2)/pnorm(B)^2 This expression seems to be converging to -1 if B approaches to -Inf (although I am unable to prove it). R has no problems until B equals around -28 or less, where both numerator and denominator go to 0 and you get NaN. A simple workaround I did was C <- ifelse(B > -25, -(pnorm(B)*dnorm(B)*B

I believe that this may be more appropriate here in r-devel than in r-help. The normal hazard function, or reciprocal Mill's Ratio, may be obtained in R as dnorm(z)/(1 - pnorm(z)) or, better, as dnorm(z)/pnorm(-z) for small values of z. The latter formula breaks dowm numerically for me (running R 2.4.1 under Windows XP 5.1 SP 2) for values of z near 37.4 or greater. Looking at the pnorm

Hi all, I want to solve the following equation for x with rho <- 0.5 pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))==0.05 Is there a function in R to do this? Thank you very much! Hannah [[alternative HTML version deleted]]

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 everyone, I am new to R, but it's really great and helped me a lot! But now I have 2 questions. It would be great, if someone can help me: 1. I want to integrate a normal distribution, given a median and sd. The integrate function works great BUT the first argument has to be a function so I do integrate(dnorm,0,1) and it works with standard m. and sd. But I have the m and sd given.

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, I already found a conversation on the integration of a normal distribution and two suggested solutions (https://stat.ethz.ch/pipermail/r-help/2007-January/124008.html): 1) integrate(dnorm, 0,1, mean = 0, sd = 1.2) and 2) pnorm(1, mean = 0, sd = 1.2) - pnorm(0, mean = 0, sd = 1.2) where the pnorm-approach is supposed to be faster and with higher precision. I want to integrate a mixed

I'm using ks.test (mydata, dnorm) on my data. I know some of my different variable samples (mydata1, mydata2, etc) must be normally distributed but the p value is always < 2.0^-16 (the 2.0 can change but not the exponent). I want to test mydata against a normal distribution. What could I be doing wrong? I tried instead using rnorm to create a normal distribution: y = rnorm

Dear list, has anyone written a package/function in R for computing a point- biserial resp. biserial correlation? Thanks in advance Bernd

Can someone help me understand the following: > D(expression(dnorm(x, mean)), "mean") [1] 0 > sessionInfo() R version 2.2.1, 2005-12-20, i386-pc-mingw32 attached base packages: [1] "methods" "stats" "graphics" "grDevices" "utils" "datasets" [7] "base" By my computations, this should be

> version _ platform i686-pc-linux-gnu arch i686 os linux-gnu system i686, linux-gnu status major 1 minor 7.1 year 2003 month 06 day 16 language R Bug: integrate(f,lower,upper,extra_args) where f <- function(x,extra_args) { body } integrate doesn't pass the extra arguments when calling f. As a first check of this finding I integrated dnorm from

Hello all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

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