similar to: bug with dlnorm

In theory, dlnorm(x, ...) == dnorm(log(x), ...)/x, x>0. Unfortunately, when sd=0, dlnorm and plnorm return NaN, while dnorm returns (if(x != mean)0 else Inf) and pnorm returns (if(x<mean)0 else 1). [A numerical optimization, maxLik{maxLik}, reported the NaNs for me.] help('dlnorm') says, "dlnorm is calculated from the definition (in ?Details?). [pqr]lnorm are based on the

The density of a lognormal should be 0 for negative arguments, but > dlnorm(-1) [1] NaN Warning message: NaNs produced in: dlnorm(x, meanlog, sdlog, log) A simple fix is to change dlnorm's definition to: function (x, meanlog = 0, sdlog = 1, log = FALSE) .Internal(dlnorm(x*(x>0), meanlog, sdlog, log)) It might be faster to put the same sort of adjustment into the internal code, but

Dear All, Consider the following code: pdf(file="figure.pdf",family="URWPalladio") curve(dlnorm(x,0,1.5),0,10,xlim=c(0,10),ylim=c(0,0.85),axes=F,xlab="",ylab="f") segments(exp(-1.5^2),0,exp(-1.5^2),dlnorm(exp(-1.5^2),0,1.5),lty="dashed") segments(1,0,1,dlnorm(1,0,1.5),lty="dashed")

Hello, I am trying to compute the density of a variable k that is either (1) Normally distributed; (2) Log-Normally distributed; or (3) follows proportional error distribution. I tried to search R-help and the answer for normal distribution was easy to find (please see 1c). I am not sure if my formula for dlnorm is correct (please see 2c below)? I really don't know what function to use for the

Hello, I've the following setting: (1) Data from a source without truncation (x) (2) Data from an other source with left-truncation at threshold u (xu) I have to fit a model on these these two sources, thereby I assume that both are "drawn" from the same distribution (eg lognormal). In a MLE I would sum the densities and maximize. The R-Function could be:

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Hello, I tried to fit a lognormal distribution by using optim. But sadly the output seems to be incorrect. Who can tell me where the "bug" is? test = rlnorm(100,5,3) logL = function(parm, x,...) -sum(log(dlnorm(x,parm,...))) start = list(meanlog=5, sdlog=3) optim(start,logL,x=test)$par Carsten. [[alternative HTML version deleted]]

Hi Again, I think I've solved my problem, but please tell me if you think I'm wrong, or you can see a better way! A plot of the integrand showed a very sharp peak, so I was running into the integrand "feature" mentioned in the note. I resolved it by limiting the range of integration as shown here: -------------------------------------------------- function (x, meanlog = 0,

Hello Satya, > RSTP daemon runs in application and stops the kernel stp and informs > to pass the packets to application layer. > RSTP daemon waits in raw socket with stp filter. > You just need to change the a bit code for port wise that all. ok thanks, do I need the DSA (Distributed Switch Architecture) driver support to distinguish from which port a frame arrived?

Dear All, I hope the title speaks by itself. I believe that there should be a solution when I see what Mclust is able to do. However, this problem is quite particular in that d3 is not known and does not necessarily correspond to a common distribution (e.g. normal, exponential ...). However it must look like dmix and d1 to some extent. To illustrate my problem I worked out a simple example:

Hi! I have following data which is left truncated say at 10. I am trying to estimate the parameters of the Truncated Lognormal distribution to this data as given below. (I have referred to R code appearing in an earlier post - http://finzi.psych.upenn.edu/Rhelp10/2008-October/176136.html) library(MASS) x <- c(600.62,153.05,70.26,530.42,3440.29,97.45,174.51,168.47, 116.63,36.51, 219.77,

From sf.net/xen I have obtained xen2.0 and brought xen2.0 stable up on FC2. I am trying to find out a way to obtaining a tar file of xen2.0testing, without bk. I may have overlooked something here, helpful corrections solicited. Thanks in advance. -- Woody Marvel LTC Open Source Projects tel: 503-578-3833/775-3833 email: marvel@us.ibm.com

Hi again, Adding further information to my own query, this function gets to the core of the problem, which I think lies in the behaviour of 'integrate'. ------------------------------------- function (x, meanlog = 0, sdlog = 1, ...) { require(stats) integrand <- function(t, x, meanlog, sdlog) dpois(x,t)*dlnorm(t, meanlog, sdlog) mapply(function(x, meanlog, sdlog, ...) #

With reference to R News News: R version 3.0.0 (Masked Marvel) has been released on 2013-04-03. R version 2.15.3 (Security Blanket) has been released on 2013-03-01 R version 2.15.2 (Trick or Treat) .... R version 2.15.1 ("Roasted Marshmallows") ... R version 2.15.0 ("Easter Beagle") R version 2.14.0 ("Great Pumpkin") Dear R help List, How are these version types

(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

Trying to launch R with a *non-existing* encoding core dumps/crashes, e.g. R --encoding=foo -e 1. EXAMPLES: > R --encoding=foo -e 1 R version 3.0.1 (2013-05-16) -- "Good Sport" Copyright (C) 2013 The R Foundation for Statistical Computing Platform: x86_64-apple-darwin10.8.0 (64-bit) [...] > *** caught segfault *** address 0xffffffffffffffff, cause 'memory not mapped'

DeaR all, I do not have a clue with is the following NOT working like I expect to do... (and I cannot find any answer at CRAN)... # This one is for my sample > x _ seq(3,10) # This two for parameters > a _ seq(2,4) > b _ seq(2,5) # This one for the likelihood of a sample >f _ function(a,b) { + prod(dlnorm(x,meanlog=a,sdlog=b)) + } > outer(a,b,f) [,1] [,2]

Dear R-users I've got the next problem: I've got this *function*: fitcond=function(x,densfun,pcorte,start,...){ myfn <- function(parm,x,pcorte,...) -sum(log(dens(parm,x,pcorte,...))) Call <- match.call(expand.dots = TRUE) if (missing(start)) start <- NULL dots <- names(list(...)) dots <- dots[!is.element(dots, c("upper",

Hello. I am not certain even how to search the archives for this particular question, so if there is an obvious answer, please smack me with a large halibut and send me to the URLs. I have been experimenting with fitting curves by using both maximum likelihood and maximum spacing estimation techniques. Originally, I have been writing distribution-specific functions in 'R' which work

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