similar to: how to change one of the axis to normal probability scale

Hi, guys, I am trying to sample from a truncated normal/gamma distribution. But only the far end of the distribution (where the probability is very low) is left. e.g. mu = - 4; sigma = 0.1; The distribution is Normal(mu,sigma^2) truncated on [0,+Inf]; How can I get a sample? I tried to use inverse CDF method, but got Inf as answers. Please help me out. Also, pls help me on the similar

Dear all, I'm trying to estimate the parameters of a lognormal distribution fitted from some data. The tricky thing is that my data represent the time at which I recorded certain events. However, in many cases I don't really know when the event happened. I' only know the time at which I recorded it as already happened. Therefore I want to fit the lognormal from the cumulative

I'd like to generate some plots like you'd see on the old "normal probability graph paper", like the first plot in: <http://www.itl.nist.gov/div898/handbook/eda/section3/normprpl.htm> except the horizontal scale would have 1%, 5%, 25%, 50%, 75%, 95%, 99%, or similar quantiles, with associated tick/grid lines. [still hunting around for a good example...] something like

Dear all, Sorry for bringing up an old issue: >pexp(50, 0.5) [1] 1 In some cases, pexp() gives CDF=1. I read some discussion in 2002 saying it has been patched. However it's not working in "R2.4.1Patched". Could anyone help me out? Thanks a lot, Jeann _________________________________________________________________ Fine Dining & Fancy Food. Check Out This

Hi, I want to generate random samples from truncated normal say Normal(0,1)Indicator((0,1),(2,4)). It has more than one intervals. In the library msm, it seems to me that the 'lower' and 'upper' arguments can only be a number. I tried rtnorm(1,mean=0,sd=1, lower=c(0,2),upper=c(1,4)) and it didn't work. Can you tell me how I can do truncated normal at more than one intervals?

Does anyone know of an R-function that will generate an observation from a truncated normal (left or right) with a mu and sigma2? Any correspondence would be greatly appreciated. Best regards, Scott Summerill FAA ACB-330 SIGNAL Corp. 609-485-6377 scott.ctr.summerill at tc.faa.gov -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear list, could someone point me to the normal score transform and it's use in R? I would like to transform my data with normal score transform, do some geostatsitical predictions and would like to transform the estimated values back including the estimation variance. Any suggestions? Ulrich -- __________________________________________________ Ulrich Leopold MSc. Department of

dear all, here's a couple of questions that puzzled me in these last hours: ##### issue 1 qnorm(1-10e-100)!=qnorm(10e-100) qnorm(1-1e-10) == -qnorm(1e-10) # turns on to be FALSE. Ok I'm not a computer scientist but, # but I had a look at the R inferno so I write: all.equal(qnorm(1-1e-10) , -qnorm(1e-10)) # which turns TRUE, as one would expect, but all.equal(qnorm(1-1e-100) ,

My routine (below) works OK but misbehaves if the on-screen plot is made wider using the mouse. The problem is caused by using par("usr")[1] - 0.07 * (par("usr")[2] - par("usr")[1]) to locate two items on the y-axis. The rest of the labeling is controlled by the "line=0" parameter setting. Of course resizing changes the absolute plot width, and

Hello, I am trying to generate random survival times by inverting the function,  S(t)= exp(b*F(t)), where b is constant and F(t) is some cumulative distribution function, let say that F(t) is cdf of normal distribution or any others distributions.   as we know that S(t) has uniform distribution on  (0,1) so we can write that U= exp(b*F(t)), where U is uniform (0,1). Now to generat the time t, we

Suppose I know the value of cumulative bivariate standard normal distribution. How can I solve correlation between variables? Pekka --------------------------------- [[alternative HTML version deleted]]

Tom Cohen <tom.cohen78@yahoo.se> skrev: Thanks Prof Brian for your suggestion. I should know that for right-skewed data, one should generate the samples from a lognormal. My problem is that x and y are two instruments that were thought to be measured the same thing but somehow show a wide confidence interval of the difference between the two intruments.This may be true that these

Hello, Well, try it: p <- .Machine$double.eps^seq(0.5, 1, by = 0.05) z <- qnorm(p/2) pnorm(z) # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15 6.731134e-16 #[11] 1.110223e-16 p/2 # [1] 7.450581e-09 1.228888e-09 2.026908e-10 3.343152e-11 5.514145e-12 # [6] 9.094947e-13 1.500107e-13 2.474254e-14 4.080996e-15

You may want to look into using the log option to qnorm e.g., in round figures: > log(1e-300) [1] -690.7755 > qnorm(-691, log=TRUE) [1] -37.05315 > exp(37^2/2) [1] 1.881797e+297 > exp(-37^2/2) [1] 5.314068e-298 Notice that floating point representation cuts out at 1e+/-308 or so. If you want to go outside that range, you may need explicit manipulation of the log values. qnorm()

* fitdistr? * it seems unusual (to me) to fit directly to the data with lognormal... fitting a normal to the log of the data seems more in keeping with the assumptions associated with that distribution. -- Sent from my phone. Please excuse my brevity. On July 10, 2017 7:27:47 AM PDT, PIKAL Petr <petr.pikal at precheza.cz> wrote: >Dear all > >I am struggling to fit data which form

Dear R users, You can literally safe my life my telling me the solution of my problem. I have created matrix of a data frame with 3 columns, with each column representing data of different year. 2 1 5 3 4 4 4 3 32 3 4 4 4 32 5 3 4

Dear all I am struggling to fit data which form something like CDF by lognorm. Here are my data: proc <- c(0.9, 0.84, 0.5, 0.16, 0.1) size <- c(0.144, 0.172, 0.272, 0.481, 0.583) plot(size, proc, xlim=c(0,1), ylim=c(0,1)) fit<-nls(proc~SSfpl(size, 1, 0, xmid, scal), start=list(xmid=0.2, scal=.1)) lines(seq(0,1,.01), predict(fit, newdata=data.frame(sito=seq(0,1,.01))), col=2) I tried

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?

hello, I'm trying to improve the speed of my calculation but didn't get to a satisfying result. It's about the numerical Integration of a bivariate normal distribution. The code I'm currently using x <- qnorm(seq(.Machine$double.xmin,c(1-2*.Machine$double.eps),by=0.01), mean=0,sd=1) rho <- 0.5 integral <- function(rho,x1){

I agree with many the sentiments about the wisdom of computing very small p-values (although the example below may win some kind of a prize: I've seen people talking about p-values of the order of 10^(-2000), but never 10^(-(10^8)) !). That said, there are a several tricks for getting more reasonable sums of very small probabilities. The first is to scale the p-values by dividing the