similar to: Bug in rt() ? (PR#9422)

Full_Name: Long Qu Version: 2.3.1 OS: Windows XP Submission from: (NULL) (64.113.93.235) The QQ-plot of two versions of simulating noncentral F-distributed random numbers has quite different scales: > qqplot(rf(1000,2,15,3),qf(runif(1000),2,15,3)) The rf() function reads: > rf function (n, df1, df2, ncp = 0) { if (ncp == 0) .Internal(rf(n, df1, df2)) else rchisq(n, df1,

help(qt) states that: "ncp non-centrality parameter delta; currently except for rt(), only for abs(ncp) <= 37.62" so I would expect that calling qt with non-centrality parameter exceeding 37.62 should fail, instead e.g. calling > mapply(function(x) qt(p = 0.9, df = 55, ncp = x),35:45) gives: [1] 40.21448 41.35293 42.49164 43.68862 44.82945 45.97048 47.11170 48.25310 [9]

Hi there: I'd thought these two versions of noncentral t-distribution are essentially the same: > qqplot(rt(1000,df=20,ncp=3),qt(runif(1000),df=20,ncp=3)) But, the scales of the x-axis and the y-axis are quite different according to the QQ-plot. Did I make any mistakes somewhere? Thanks, Long ---------------------------------

Dear R useRs, i wanted to ask if the folded normal destribution (Y = abs(X) with X normal distributed) with density and random number generator is implemented in R or in any R-related package so far? Maybe i can use the non-central chi-square distribution and rchisq(n, df=1, ncp>0) here? Thanks and best regards Andreas

Dear all Just for fun, I have just downloaded the paper mentioned below and checked it with R-1.6.1. Everything is ok with exception of Table 2b, where I get always 1 instead of 0.5: > pbinom(1e15,2e15,0.5) [1] 1 Which value should be correct? Best regards Christian Stratowa ============================================== Christian Stratowa, PhD Boehringer Ingelheim Austria Dept NCE Lead

Hi Folks, I note that, while the "chisq" functions dchisq(x, df, ncp=0, log = FALSE) pchisq(q, df, ncp=0, lower.tail = TRUE, log.p = FALSE) qchisq(p, df, ncp=0, lower.tail = TRUE, log.p = FALSE) rchisq(n, df, ncp=0) all have a slot for the non-centrality parameter "ncp", of the functions for the t and F distributions: dt(x, df, log = FALSE)

Hi! This is my first time posting. I've read the general rules and guidelines, but please bear with me if I make some fatal error in posting. Anyway, I have a continuous response and 29 predictors made up of continuous variables and nominal and ordinal categorical variables. I'd like to do lasso on these, but I get an error. The way I am using "lars" doesn't allow for the

Not real data. It was gererated randomly. The original codes are the following: par(mfrow=c(2,1)) n <- 500 ######################### #DATA GENERATING PROCESS# ######################### x1 <- rnorm(n,0,1) x2 <- rchisq(n,df=3,ncp=0)-3 sigma <- 1 u1 <- rnorm(n,0,sigma) ylatent1 <-x1+x2+u1 y1 <- (ylatent1 >=0) # create the binary indicator ####################### #THE

The function 'pchisq' from the 'stats' library gives a wrong result if the argument equals exactly zero: # Upper tail of central 1-df chi^2 distribution > pchisq(1 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1] 0.3173105 > pchisq(0.5 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1] 0.4795001 > pchisq(0.01 , 1, ncp=0, lower.tail = F, log.p = FALSE) [1]

Dear List: The square of the noncentral t-statistic with noncentrality parameter \delta is a noncentral F with noncentrality parameter \lambda=\delta^2. So, t^2_{\nu,\delta} = F_{1,\nu,\lambda=\delta^2}. Consequently, it should follow that t^2_{1-\alpha/2,\nu,\delta} = f_{1-alpha,1,\vu,\lambda=\delta^2}. However, this is not what is happening with the following code. The central

Bug Description: Problem with the function pf() when the non-centrality parameter is large. Here is a sample command. You should see a smooth line from 0 to about 55, and then the values of pf() go crazy from 55 to 100. ############################ ncp <- seq(0,100,length=200) plot(ncp,pf(5,7,2,ncp=ncp)) ############################ Version: platform = i686-pc-linux-gnu arch = i686 os =

Hi, I've written some C code for density evaluation of the non-central t distribution. It works with the R-1.7.1 source code if placed in the src/nmath directory and after appropriate changes are made to Makefiles, to the dt function in src/library/base/R/distn.R etc. I haven't read a lot of R source code so it may need some R-ification, but I've tried to use the dt.c file as a

Full_Name: Jerry W. Lewis Version: 2.6.1 OS: Windows XP Professional Submission from: (NULL) (24.147.191.250) pchisq(0,0,ncp=lambda) returns 0 instead of exp(-lambda/2) pchisq(x,0,ncp=lambda) returns NaN instead of exp(-lambda/2)*(1 + SUM_{r=0}^infty ((lambda/2)^r / r!) pchisq(x, df + 2r)) qchisq(.7,0,ncp=1) returns 1.712252 instead of 0.701297103 qchisq(exp(-1/2),0,ncp=1) returns 1.238938

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that

Full_Name: David Clayton Version: 1.8.1 OS: Linux Submission from: (NULL) (131.111.126.242) qchisq behaves very strangely when ncp is passed as zero (forcing internal qnchisq to be called) when first argument is small. Eg > qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) [1] 1024 while, if ncp is unspecified, > qchisq(1-1e-6, 1) qchisq(1-1e-6, 1)

Hi all, I compiled 64-bit R on an Apple Mac G5 running OS X, but it failed make check. Simon Urbanek suggested I post results to R-devel. > On Nov 6, 2007, at 10:23 PM, Steven McKinney wrote: > > > Hi Simon, > > > > Would you be able to give more guidance on how to compile 64-bit > > libiconv for Tiger, > > You can get the sources from Apple and compile

Dear all, I want to see the source codes for "dchisq(x, df, ncp=0, log = FALSE)", but cannot find it. I input "dchisq" in the R interface, and then enter, the following message return: > dchisq /*****************************************************/ function (x, df, ncp = 0, log = FALSE) { if (missing(ncp)) .Internal(dchisq(x, df, log)) else

Hi Folks, I'm looking for a neat procedure for the following: Given t0 such that pt(t0,df,ncp=0) = alpha (given) find ncp0 such that for given beta pt(t0,df,ncp=ncp0) = (1 - beta) (In other words, what's the ncp such that you get power (1-beta) to detect it, using a 1-sided test with size alpha when ncp = 0?) In the past I've done the groping by hand, but this time it needs

Hello all, Here's my query: Running R 1.6.2 on FreeBSD 5.0, and on WinXP, and I find that the following hangs the process: dchisq(alpha=0.01, df=1, ncp=295) but it does work for ncp < 294.92. Is this general? Best wishes to all, Andrew Andrew Robinson Ph: 208 885 7115 Department of Forest Resources Fa: 208 885 6226 University of Idaho E : andrewr at uidaho.edu PO

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) qchisq(1,10) works well but qchisq(1,10,ncp=0) does not work whereas ncp=0 is the default value of the function qchisq(1,10). (of course, 10 will be replaced by any integer value). Let us notice that this bug occurs only when applying probability one. (qchisq(seq(0,.9,.1),10,ncp=0) works very well).